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<div>SOLVED PROBLEMS 67<br /> PROBLEM SETS & INSIGHTS<br /> Solved Problems<br /> S1. A diploid cell has eight chromosomes, four per set. For the<br /> following diagram, in what phase of mitosis, meiosis I or<br /> meiosis II, is this cell?<br /> Answer: The cell is in metaphase of meiosis II. You can tell because the<br /> chromosomes are lined up in a single row along the metaphase plate,<br /> and the cell has only four pairs of sister chromatids. If it were mitosis,<br /> the cell would have eight pairs of sister chromatids.<br /> S2. An unaffected woman (i.e., without disease symptoms) who is<br /> heterozygous for the X-linked allele causing Duchenne muscular<br /> dystrophy has children with a man with a normal allele. What are<br /> the probabilities of the following combinations of offspring?<br /> A. An unaffected son<br /> B. An unaffected son or daughter<br /> C. A family of three children, all of whom are affected<br /> Answer: The first thing we must do is construct a Punnett square to<br /> determine the outcome of the cross. N represents the normal allele, n<br /> the recessive allele causing Duchenne muscular dystrophy. The mother<br /> is heterozygous, and the father has the normal allele.<br /> Female gametes<br /> X N<br /> X n<br /> Male gametes<br /> X N<br /> X N X N<br /> X N X n<br /> Y<br /> X N Y<br /> X n Y<br /> Phenotype ratio is<br /> 2 normal daughters :<br /> 1 normal son :<br /> 1 affected son<br /> A. There are four possible children, one of whom is an unaffected<br /> son. Therefore, the probability of an unaffected son is 1/4.<br /> B. Use the sum rule: 1/4 + 1/2 = 3/4.<br /> C. You could use the product rule because there would be three<br /> offspring in a row with the disorder: (1/4)(1/4)(1/4) = 1/64 =<br /> 0.016 = 1.6%.<br /> S3. What are the major differences between prophase, metaphase, and<br /> anaphase when comparing mitosis, meiosis I, and meiosis II?<br /> Answer: The table summarizes key differences.<br /> A Comparison of Mitosis, Meiosis I, and Meiosis II<br /> Phase Event Mitosis Meiosis I Meiosis II<br /> Prophase Synapsis: No Yes No<br /> Prophase Crossing Rarely Commonly Rarely<br /> over:<br /> Metaphase Alignment Sister Bivalents Sister<br /> along the chro- chrometaphase<br /> matids matids<br /> plate:<br /> Anaphase Separation Sister Bivalents Sister<br /> of: chro- chromatids<br /> matids<br /> S4. Among different plant species, both male and female gametophytes<br /> can be produced by single individuals or by separate sexes. In some<br /> species, such as the garden pea, a single individual can produce<br /> both male and female gametophytes. Fertilization takes place via<br /> self-fertilization or cross-fertilization. A plant species that has a<br /> single type of flower producing both pollen and eggs is termed a<br /> monoclinous plant. In other plant species, two different types of<br /> flowers produce either pollen or eggs. When both flower types are<br /> on a single individual, such a species is termed monoecious. It is<br /> most common for the “male flowers” to be produced near the top<br /> of the plant and the “female flowers” toward the bottom. Though<br /> less common, some species of plants are dioecious. For dioecious<br /> species, one individual makes either male flowers or female<br /> flowers, but not both.<br /> Based on your personal observations of plants, try to give<br /> examples of monoclinous, monoecious, and dioecious plants.<br /> What would be the advantages and disadvantages of each?<br /> Answer: Monoclinous plants—pea plant, tulip, and roses. The same<br /> flower produces pollen on the anthers and egg cells within the ovary.<br /> Monoecious plants—corn and pine trees. In corn, the tassels are<br /> the male flowers and the ears result from fertilization within the female<br /> flowers. In pine trees, pollen is produced in cones near the top of the<br /> tree, and eggs cells are found in larger cones nearer the bottom.<br /> Dioecious plants—holly and ginkgo trees. Certain individuals<br /> produce only pollen, while others produce only eggs.<br /> An advantage of being monoclinous or monoecious is that<br /> fertilization is relatively easy because the pollen and egg cells are<br /> produced on the same individual. This is particularly true for<br /> monoclinous plants. The proximity of the pollen to the egg cells makes<br /> it more likely for self-fertilization to occur. This is advantageous if the<br /> plant population is relatively sparse. On the other hand, a dioecious<br /> species can reproduce only via cross-fertilization. The advantage of<br /> cross-fertilization is that it enhances genetic variation. Over the long<br /> run, this can be an advantage because cross-fertilization is more likely<br /> to produce a varied population of individuals, some of which may<br /> possess combinations of traits that promote survival.<br /> S5. To test the chromosome theory of inheritance, Calvin Bridges<br /> made crosses involving the inheritance of X-linked traits. One of<br /> his experiments concerned two different X-linked genes affecting<br /> eye color and wing size. For the eye color gene, the red-eye allele</div> <div>358 CHAPTER 13 :: TRANSLATION OF mRNA<br /> C26. How does a eukaryotic ribosome select its start codon? Describe<br /> the sequences in eukaryotic mRNA that provide an optimal<br /> context for a start codon.<br /> C27. For each of the following sequences, rank them in order (from best<br /> to worst) as sequences that could be used to initiate translation<br /> according to Kozak’s rules.<br /> GACGCCAUGG<br /> GCCUCCAUGC<br /> GCCAUCAAGG<br /> GCCACCAUGG<br /> C28. Explain the functional roles of the A, P, and E sites during<br /> translation.<br /> C29. An mRNA has the following sequence: 5'–AUG UAC UAU<br /> GGG GCG UAA–3'. Describe the amino acid sequence of the<br /> polypeptide that would be encoded by this mRNA. Be specific<br /> about the amino and carboxyl terminal ends.<br /> C30. For eukaryotic proteins, what is the function of a sorting signal?<br /> Describe an example.<br /> C31. What is the difference between posttranslational and<br /> cotranslational protein sorting? What cellular components are<br /> required for cotranslational sorting?<br /> C32. Which steps during the translation of bacterial mRNA involve an<br /> interaction between complementary strands of RNA?<br /> C33. What is the function of the nucleolus?<br /> C34. In which of the ribosomal sites, the A site, P site, and/or E site,<br /> could the following be found?<br /> A. A tRNA without an amino acid attached<br /> B. A tRNA with a polypeptide attached<br /> C. A tRNA with a single amino acid attached<br /> C35. What is a polysome?<br /> C36. According to Figure 13.20, explain why the ribosome translocates<br /> along the mRNA in a 5' to 3' direction rather than a 3' to 5'<br /> direction.<br /> C37. The lactose permease of E. coli is a protein composed of a single<br /> polypeptide that is 417 amino acids in length. By convention, the<br /> amino acids within a polypeptide are numbered from the amino<br /> terminal end to the carboxyl terminal end. Are the following<br /> questions about the lactose permease true or false?<br /> A. Because the 64th amino acid is glycine and the 68th amino acid<br /> is aspartic acid, the codon for glycine-64 is closer to the 3' end<br /> of the mRNA compared to the codon for aspartic acid-68.<br /> B. The mRNA that encodes the lactose permease must be greater<br /> than 1,241 nucleotides in length.<br /> C38. An mRNA encodes a polypeptide that is 312 amino acids in<br /> length. The 53rd codon in this polypeptide is a tryptophan codon.<br /> A mutation in the gene that encodes this polypeptide changes<br /> this tryptophan codon into a stop codon. How many amino acids<br /> would be in the resulting polypeptide: 52, 53, 259, or 260?<br /> C39. Explain what is meant by the coupling of transcription and<br /> translation in bacteria. Does coupling occur in bacterial and/or<br /> eukaryotic cells? Explain.<br /> Experimental Questions<br /> E1. In the experiment of Figure 13.3, what would be the predicted<br /> amounts of amino acids incorporated into polypeptides if the<br /> RNA was a random polymer containing 50% C and 50% G?<br /> E2. With regard to the experiment described in Figure 13.11, answer<br /> the following questions:<br /> A. Why was a polyUG mRNA template used?<br /> B. Would you radiolabel the cysteine with the isotope 14 C or 35 S?<br /> Explain your choice.<br /> C. What would be the expected results if the experiment was<br /> followed in the same way except that a polyGC template was<br /> used? Note: A polyGC template could contain two different<br /> alanine codons (GCC and GCG), but it could not contain any<br /> cysteine codons.<br /> E3. An experimenter has a chemical reagent that modifies threonine<br /> to another amino acid. Following the protocol described in Figure<br /> 13.11, an mRNA is made composed of 50% C and 50% A. The<br /> amino acid composition of the resultant polypeptides is 12.5%<br /> lysine, 12.5% asparagine, 25% serine, 12.5% glutamine, 12.5%<br /> histidine, and 25% proline. One of the amino acids present in this<br /> polypeptide is due to the modification of threonine. Which amino<br /> acid is it? Based on the structure of the amino acid side chains,<br /> explain how the structure of threonine has been modified.<br /> E4. Polypeptides can be translated in vitro. Would a bacterial mRNA<br /> be translated in vitro by eukaryotic ribosomes? Would a eukaryotic<br /> mRNA be translated in vitro by bacterial ribosomes? Why or why<br /> not?<br /> E5. Discuss how the elucidation of the structure of the ribosome can<br /> help us to understand its function.<br /> E6. Figure 13.22 shows an electron micrograph of a bacterial gene as it<br /> is being transcribed and translated. In this figure, label the 5' and<br /> 3' ends of the DNA and RNA strands. Place an arrow where you<br /> think the start codons are found in the mRNA transcripts.<br /> E7. Chapter 18 describes a blotting method known as Western blotting<br /> that can be used to detect the production of a polypeptide that is<br /> translated from a particular mRNA. In this method, a protein is<br /> detected with an antibody that specifically recognizes and binds<br /> to its amino acid sequence. The antibody acts as a probe to detect<br /> the presence of the protein. In a Western blotting experiment, a<br /> mixture of cellular proteins is separated using gel electrophoresis<br /> according to their molecular masses. After the antibody has bound<br /> to the protein of interest within a blot of a gel, the protein is<br /> visualized as a dark band. For example, an antibody that recognizes<br /> the b-globin polypeptide could be used to specifically detect the b-<br /> globin polypeptide in a blot. As shown here, the method of Western<br /> blotting can be used to determine the amount and relative size of a<br /> particular protein that is produced in a given cell type.<br /> Western blot<br /> 1 2 3</div> <div>840 INDEX<br /> Prokaryotic gene transfer, 685<br /> Proline, 335<br /> Prometaphase<br /> in meiosis, 56, 57<br /> in mitosis, 52, 53<br /> Promoter, 362, 363<br /> –35 and –10 sequences, 301<br /> computer analysis of, 587<br /> conventional numbering sequence of, 301<br /> defined, 298, 298<br /> eukaryotic, 306<br /> in transcription initiation, 301<br /> Proofreading function, 283, 283<br /> Prophage, 143, 378<br /> Prophase<br /> in meiosis, 57<br /> in mitosis, 51, 52, 53<br /> Protease, 224, 381<br /> Protein dimerization, 391<br /> Protein microarrays, 583–84<br /> Protein(s)<br /> accessory, 4<br /> activator, 361, 362<br /> basic amino acids in, 335<br /> in cell composition, 4–5<br /> cellular, 335–37, 336, 354<br /> chaperone, 334<br /> checkpoint, 620<br /> DNA-binding, 238–39<br /> in DNA replication, 276–80, 277<br /> in eukaryotic chromosome compaction, 264<br /> evolution of, 737<br /> gene expression and, 6<br /> genetic sequences in predicting structure of, 593–94<br /> in homologous recombination, 461–62<br /> in interphase, 49<br /> mass spectrometry in identification of, 581–83<br /> methyl-CpG-binding, 404<br /> in mismatch repair system, 446<br /> noncapsid, 247<br /> in overdominance, 78<br /> primary structure of, 336<br /> radioisotopes in distinguishing from DNA,<br /> 225–26<br /> recessive alleles in suppression of, 73<br /> regulatory, 361<br /> release factors, 351<br /> repressor, 361, 362<br /> in ribosome assembly, 344–46<br /> SMC, 264, 265<br /> structural levels in, 337<br /> traits and, 7<br /> in transcription, 307<br /> transgenic animals in production of, 525–26<br /> vs. polypeptides, 326<br /> Western blotting in detection of, 501–2<br /> Protein sorting, 352, 353, 354<br /> Protein synthesis, DNA in, 5–6<br /> Proteome, 571<br /> Proteomics, 543, 571, 580–84<br /> mass spectrometry in identification of proteins,<br /> 581–83<br /> protein microarray technology, 583–84<br /> proteome size compared to genome, 581<br /> RNA and protein structure prediction in,<br /> 594, 593<br /> two-dimensional gel electrophoresis of cellular<br /> proteins, 581<br /> Protists, 564<br /> Proto-oncogenes, 614<br /> genetic changes in conversion to oncogenes,<br /> 616–18<br /> mutation into oncogenes, 616<br /> Protoplast, 213<br /> Prototroph, 135<br /> Proximo-distal axis, 637<br /> Prusiner, Stanley, 608<br /> Pseudoautosomal inheritance, 83<br /> Pseudo-dominance, 719<br /> Pseudohypoaldosteronism, 2<br /> Pseudomonas syringae, 519<br /> Pugh, J.E., 405<br /> Pulse/chase experiment, 280<br /> Pulsed-field gel electrophoresis (PFGE), 559–60<br /> Punctuated equilibrium, 733<br /> Punnett, Reginald, 23, 87, 101<br /> Punnett square<br /> allele frequencies compared with union of<br /> alleles, 669<br /> for coat color in rats, 89<br /> for comb morphology, 87<br /> for cream-eyed fruit flies, 90<br /> in cross outcome prediction, 23–25<br /> for dihybrid cross, 28<br /> for hemophilia, 604<br /> quantitative traits and, 703<br /> in solving independent assortment<br /> problems, 28<br /> for X-linked muscular dystrophy, 82<br /> Purines, 228<br /> P values, 34<br /> Pyrimidines, 228<br /> Q<br /> Qβ virus, 228<br /> QTL mapping, 708, 709<br /> Quagga, 744<br /> Quantitative genetics, 697–719<br /> heritability, 708–19<br /> polygenic inheritance in, 702–8<br /> quantitative traits, 697–702<br /> Quantitative trait locus, 702<br /> DDT resistance in Drosophila, 705–7<br /> genetic markers in mapping, 705<br /> mapped by linkage to molecular markers,<br /> 708, 709<br /> Quantitative traits, 697–702<br /> continuum of phenotypic variation in, 698<br /> fingerprints as, 713<br /> heritability of, 715<br /> normal distribution of, 698, 699, 700<br /> selective breeding in altering, 716–18<br /> statistical methods in evaluation of, 698–700<br /> types of, 698<br /> Quinolones, 250<br /> R<br /> Rabbit coat color, 79<br /> Radding, Charles, 461<br /> Radial loop domains, 259–61, 260<br /> RAG1, 465<br /> RAG2, 465<br /> Random genetic drift<br /> in allele frequency alteration, 672–74<br /> hypothetical simulation of, 672<br /> in microevolution, 670<br /> Random mutation theory, 434<br /> Random sampling error, 31<br /> Raney nickel, 339<br /> Ras protein, 616<br /> Rb protein, 618<br /> Reactive end, chromosomal, 197<br /> Reading frame, of mRNA, 328<br /> Realized heritability, 718<br /> RecA protein, 381, 461, 462<br /> RecBCD protein complex, 461, 462<br /> Recessive alleles, 72–74<br /> Recessive epistasis, 87<br /> Recessive traits, 22<br /> RecG protein, 461, 462<br /> Reciprocal crosses<br /> extranuclear inheritance in non-Mendelian<br /> results in, 176–77<br /> inheritance pattern of X-linked genes revealed<br /> by, 81–82<br /> in snail shell coiling, 162<br /> Reciprocal translocation, chromosomal, 190, 190,<br /> 197, 199–200<br /> Recognition helix, 391<br /> Recombinant cells, 101<br /> Recombinant DNA technology. See also<br /> Biotechnology<br /> analysis and alteration of DNA sequences, 503–7<br /> detection of cloned genes, 497–503<br /> gene cloning, 483–97<br /> Recombinant microorganisms<br /> biological control agent production, 518–19<br /> bioremediation by, 520–21<br /> Ice � strain, 520<br /> medicine production by, 518<br /> patents on, 520<br /> release into environment, 519–20<br /> somatostatin production by, 517–18<br /> Recombinant offspring, 112<br /> Recombinant vector, 486<br /> Recombination<br /> between direct repeats, 478<br /> homologous, 146, 455–63<br /> defined, 455<br /> double-strand break model, 461<br /> in gene conversion, 462, 463<br /> harlequin chromosome staining in revealing,<br /> 456–59<br /> Holliday model of, 459, 460<br /> molecular steps of, 459–61<br /> proteins needed for, 461–62<br /> resolution in, 459<br /> between inverted repeats, 478<br /> site-specific, 464–66<br /> Recombination signal sequence, 465<br /> Regression analysis, 702<br /> Regression coefficient, 721<br /> Regulatory elements, 306, 390<br /> Regulatory proteins, 361, 362<br /> Regulatory sequences, 298<br /> Regulatory transcription factors, 390–97<br /> acting through TFIID and mediator, 393<br /> affect on transcription, 394<br /> binding to DNA, 391<br /> dimerization of, 391<br /> domains of, 391<br /> heterodimer, 391, 393<br /> homodimer, 391, 393<br /> modulation of, 393, 395<br /> by covalent modification, 395, 396–97<br /> recognition of response elements, 292–93<br /> steroid hormone binding, 393–96<br /> SWI5P, 403<br /> Relaxosome, 137<br /> Release factors, 351<br /> Renaturation<br /> Cot curve and, 255<br /> in evaluation of DNA sequence complexity,<br /> 254–55<br /> Renin inhibitor, 518<br /> Repetitive sequences, 249<br /> genetic variation via changes in, 686–87<br /> Replica plating, 433<br /> Replication forks, 274<br /> DNA synthesis at, 279<br /> protein requirements at, 276–80<br /> termination sequences and, 281<br /> Replicative transposition, 469, 470, 471–72, 473<br /> Replicative transposons, 471<br /> Replisome, 281, 282</div> <div>QUESTIONS FOR STUDENT DISCUSSION/COLLABORATION 757<br /> Questions for Student Discussion/Collaboration<br /> 1. The raw material for evolution is random mutation. Discuss<br /> whether or not you view evolution as a random process.<br /> 2. Compare the forms of speciation that are slow with those that<br /> occur more rapidly. Make a list of the slow and fast forms. With<br /> regard to mechanisms of genetic change, what features do slow<br /> and rapid speciation have in common? What features are different?<br /> 3. Do you think that Darwin would object to the neutral theory of<br /> evolution?<br /> Note: All answers appear at the website for this textbook; the answers to<br /> even-numbered questions are in the back of the textbook.<br /> www.mhhe.com/brookergenetics3e<br /> Visit the Online Learning Center for practice tests, answer keys, and other learning aids for this chapter. Enhance your understanding of genetics with<br /> our interactive exercises, quizzes, animations, and much more.</div> <div>308 CHAPTER 12 :: GENE TRANSCRIPTION AND RNA MODIFICATION<br /> Next, TFIIB, TFIIE, and TFIIH dissociate, and RNA polymerase<br /> II is free to proceed to the elongation stage of transcription.<br /> In vitro, when researchers mix together TFIID, TFIIB,<br /> TFIIF, TFIIE, TFIIH, RNA polymerase II, and a DNA sequence<br /> containing a TATA box and transcriptional start site, the DNA is<br /> transcribed into RNA. Therefore, these components are referred<br /> to as the basal transcription apparatus. In a living cell, however,<br /> additional components regulate transcription and allow it to proceed<br /> at a reasonable rate.<br /> In addition to GTFs and RNA polymerase II, another<br /> component required for transcription is a large protein complex<br /> termed mediator. This complex was discovered by Roger Kornberg<br /> and colleagues in 1990. In 2006, Kornberg was awarded<br /> the Nobel Prize for his studies regarding the molecular basis of<br /> eukaryotic transcription. Mediator derives its name from the<br /> observation that it mediates interactions between RNA polymerase<br /> II and regulatory transcription factors that bind to<br /> enhancers or silencers. It serves as an interface between RNA<br /> polymerase II and many, diverse regulatory signals. The subunit<br /> composition of mediator is quite complex and variable. The core<br /> subunits form an elliptical-shaped complex that partially wraps<br /> around RNA polymerase II. Mediator itself may phosphorylate<br /> the CTD of RNA polymerase II and it may regulate the ability of<br /> TFIIH to phosphorylate the CTD. Therefore, it can play a pivotal<br /> role in the switch between transcriptional initiation and elongation.<br /> The function of mediator during eukaryotic gene regulation<br /> is explored in greater detail in Chapter 15.<br /> Chromatin Structure Plays<br /> a Key Role in Gene Transcription<br /> As we have learned, eukaryotic transcription involves the binding<br /> of general transcription factors and RNA polymerase to the promoter<br /> region and the subsequent movement of RNA polymerase<br /> along the DNA double helix, allowing one strand to function as<br /> a template for transcription. The compaction of DNA to form<br /> chromatin can be an obstacle to the transcription process. During<br /> interphase, when most transcription occurs, the chromatin<br /> of eukaryotes is found in 30 nm fibers organized into radial loop<br /> domains. Within the 30 nm fiber, the DNA is wound around<br /> histone octamers to form nucleosomes (refer back to Chapter<br /> 10, Figure 10.16). The size of a histone octamer is roughly<br /> five times smaller than the complex of RNA polymerase II and<br /> GTFs. Therefore, because RNA polymerase is a very large enzyme<br /> compared to a nucleosome, the tight wrapping of DNA within a<br /> nucleosome is expected to inhibit the ability of RNA polymerase<br /> to transcribe the DNA.<br /> How is chromatin structure loosened so that transcription<br /> can occur? Two common mechanisms alter chromatin structure.<br /> First, the amino terminal tails of the core histone proteins are<br /> covalently modified in a variety of ways, including the acetylation<br /> of lysines, methylation of lysines, and phosphorylation of serines<br /> (refer back to Chapter 10, Figure 10.20). These covalent modifications<br /> play a key role in the level of chromatin compaction.<br /> For example, positively charged lysine residues within the amino<br /> terminal tails of the core histone proteins can be acetylated by<br /> Core histone<br /> protein<br /> (a) Histone acetylation<br /> Change in the relative positions<br /> of a few nucleosomes<br /> Histone<br /> acetyltransferase<br /> Histone<br /> deacetylase<br /> (b) ATP-dependent chromatin remodeling<br /> COCH 3<br /> ATP-dependent<br /> chromatin remodeling<br /> complex<br /> COCH 3<br /> Acetyl<br /> group<br /> Change in the spacing<br /> of nucleosomes over a<br /> long distance<br /> COCH 3<br /> FIGURE 12.15 Mechanisms that disrupt the tight binding<br /> of DNA and histones within nucleosomes. (a) The covalent<br /> modification of the amino terminal tails of histones by acetylation,<br /> for example, alters their ability to bind to the DNA backbone. Histone<br /> acetyltransferase catalyzes the attachment of acetyl groups to lysine<br /> residues, thereby loosening the interactions between histones and<br /> DNA. Histone deacetylase reverses this process. (b) In ATP-dependent<br /> chromatin remodeling, proteins of the SWI/SNF family catalyze an<br /> ATP-dependent change in the locations of nucleosomes. This may<br /> involve a shift in nucleosomes to a new location (left) or a change in<br /> the spacing of nucleosomes over a long stretch of DNA (right). These<br /> effects may significantly alter gene transcription.<br /> enzymes called histone acetyltransferases (Figure 12.15a). The<br /> attachment of the acetyl group (–COCH 3 ) eliminates the positive<br /> charge on the lysine side chain and thereby disrupts the favorable<br /> interaction between the histone protein and the negatively<br /> charged DNA backbone. This may loosen the structure of nucleosomes<br /> and the 30 nm fiber and thereby facilitate the ability of<br /> RNA polymerase to transcribe a gene. Some studies suggest that<br /> histones are completely displaced, whereas others suggest they<br /> are loosened but remain attached to the DNA. In addition, covalently<br /> modified histones can be recognized by proteins that alter<br /> the compaction level of the chromatin. Other enzymes, called</div> <div>24.2 FACTORS THAT CHANGE ALLELE AND GENOTYPE FREQUENCIES IN POPULATIONS 675<br /> encoded protein, which, in turn, may alter the function of<br /> the protein.<br /> 2. Some alleles may encode proteins that enhance an<br /> individual’s survival or reproductive capability compared<br /> to that of other members of the population. For example,<br /> an allele may produce a protein that is more efficient at a<br /> higher temperature, conferring on the individual a greater<br /> probability of survival in a hot climate.<br /> 3. Individuals with beneficial alleles are more likely to survive<br /> and contribute to the gene pool of the next generation.<br /> 4. Over the course of many generations, allele frequencies<br /> of many different genes may change through this process,<br /> thereby significantly altering the characteristics of a<br /> species. The net result of natural selection is a population<br /> that is better adapted to its environment and more<br /> successful at reproduction.<br /> As mentioned at the beginning of the chapter, Fisher,<br /> Wright, and Haldane developed mathematical relationships to<br /> explain the theory of natural selection. As our knowledge of the<br /> process of natural selection has increased, it has become apparent<br /> that it operates in many different ways. In this chapter, we<br /> will consider a few examples of natural selection involving a<br /> single trait or a single gene that exists in two alleles. In reality,<br /> however, natural selection acts on populations of individuals in<br /> which many genes are polymorphic and each individual contains<br /> thousands or tens of thousands of different genes.<br /> To begin our quantitative discussion of natural selection,<br /> we must examine the concept of Darwinian fitness—the relative<br /> likelihood that a genotype will contribute to the gene pool of the<br /> next generation as compared to other genotypes. Natural selection<br /> acts on phenotypes that are derived from individuals’ genotypes.<br /> Although Darwinian fitness often correlates with physical<br /> fitness, the two ideas should not be confused. Darwinian fitness<br /> is a measure of reproductive superiority. An extremely fertile<br /> genotype may have a higher Darwinian fitness than a less fertile<br /> genotype that appears more physically fit.<br /> To consider Darwinian fitness, let’s use our example of a<br /> gene existing in the A and a alleles. If the three genotypes have<br /> the same level of mating success and fertility, we can assign fitness<br /> values to each of the three genotype classes based on their<br /> likelihood of surviving to reproductive age. For example, let’s<br /> suppose that the relative survival to adulthood of each of the<br /> three genotype classes is as follows: For every five AA individuals<br /> that survive, four Aa individuals survive, and one aa individual<br /> survives. By convention, the genotype with the highest reproductive<br /> ability is given a fitness value of 1.0. Relative fitness values<br /> are denoted by the variable W. The fitness values of the other<br /> genotypes are assigned values relative to this 1.0 value:<br /> Fitness of AA: W AA � 1.0<br /> Fitness of Aa: W Aa � 4/5 � 0.8<br /> Fitness of aa: W aa � 1/5 � 0.2<br /> Keep in mind that differences in reproductive achievement<br /> among genotypes may stem from various reasons. In this case,<br /> the fittest genotype is more likely to survive to reproductive age.<br /> In other situations, the most fit genotype is more likely to mate.<br /> For example, a bird with brightly colored feathers may have an<br /> easier time attracting a mate than a bird with duller plumage.<br /> Finally, a third possibility is that the fittest genotype may be more<br /> fertile. It may produce a higher number of gametes or gametes<br /> that are more successful at fertilization.<br /> By studying species in their native environments, population<br /> geneticists have discovered that natural selection can occur<br /> in several ways. The patterns of natural selection depend on<br /> the relative fitness values of the different genotypes and on the<br /> variation of environmental effects. The four patterns of natural<br /> selection that we will consider are called directional, stabilizing,<br /> disruptive, and balancing selection. In most of the examples<br /> described next, natural selection leads to adaptation so that a<br /> species is better able to survive to reproductive age.<br /> Directional Selection Favors the Extreme Phenotype<br /> Directional selection favors individuals at one extreme of a phenotypic<br /> distribution that are more likely to survive and reproduce<br /> in a particular environment. Different phenomena may<br /> initiate the process of directional selection. One way that directional<br /> selection may arise is that a new allele may be introduced<br /> into a population by mutation, and the new allele may promote<br /> a higher fitness in individuals that carry it (Figure 24.8). If the<br /> homozygote carrying the favored allele has the highest fitness<br /> value, directional selection may cause this favored allele to eventually<br /> become the predominant allele in the population, perhaps<br /> even becoming a monomorphic allele.<br /> Another possibility is that a population may be exposed<br /> to a prolonged change in its living environment. Under the new<br /> environmental conditions, the relative fitness values may change<br /> to favor one genotype, and this will promote the elimination of<br /> other genotypes. As an example, let’s suppose a population of<br /> finches on the mainland already has genetic variation in beak<br /> size. A small number of birds migrate to an island where the<br /> seeds are generally larger than they are on the mainland. In this<br /> new environment, birds with larger beaks would have a higher<br /> fitness because they would be better able to crack open the larger<br /> seeds and thereby survive to reproductive age. Over the course of<br /> many generations, directional selection would produce a population<br /> of birds carrying alleles that promote larger beak size.<br /> In the case of directional selection, allele frequencies may<br /> change in a step-by-step, generation-per-generation way. To<br /> appreciate how this occurs, let’s take a look at how fitness affects<br /> the Hardy-Weinberg equilibrium and allele frequencies. Again,<br /> let’s suppose a gene exists in two alleles, A and a. The three fitness<br /> values, which are based on relative survival levels, are<br /> W AA � 1.0<br /> W Aa � 0.8<br /> W aa � 0.2<br /> In the next generation, we expect that the Hardy-Weinberg equilibrium<br /> will be modified in the following way due to directional<br /> selection:</div> <div>CONCEPTUAL SUMMARY 383<br /> The O R region contains three operator sites, designated<br /> O R1 , O R2 , and O R3 . These operator sites control two promoters<br /> called P RM and P R that transcribe in opposite directions. The λ<br /> repressor protein or the cro protein can bind to any or all of the<br /> three operator sites. The binding of these two proteins at these<br /> sites governs the switch between the lysogenic and lytic cycles.<br /> Two critical issues influence this binding event. The first is the<br /> relative affinities that the regulatory proteins have for these operator<br /> sites. The second is the concentrations of the λ repressor<br /> protein and the cro protein within the cell.<br /> Let’s consider how an increasing concentration of the λ<br /> repressor protein can switch on the lysogenic cycle and switch<br /> off the lytic cycle (Figure 14.20, left side). This protein was first<br /> isolated by Mark Ptashne and his colleagues in 1967. Their studies<br /> showed that λ repressor binds with highest affinity to O R1 ,<br /> followed by O R2 and O R3 . As the concentration of the λ repressor<br /> builds within the cell, a dimer of the λ repressor protein first<br /> binds to O R1 because it has the highest affinity for this site. Next,<br /> a second λ repressor dimer binds to O R2 . This occurs very rapidly,<br /> because the binding of the first dimer to O R1 favors the binding<br /> of a second dimer to O R2 . This is called a cooperative interaction.<br /> The binding of the λ repressor to O R1 and O R2 inhibits transcription<br /> from P R and thereby switches off the lytic cycle.<br /> Early in the lysogenic cycle, the λ repressor protein concentration<br /> may become so high that it occupies O R3 . Eventually,<br /> however, the λ repressor concentration begins to drop, because<br /> the inhibition of P R decreases the synthesis of cII, which activates<br /> the λ repressor gene from P RE . As the λ repressor concentration<br /> gradually falls, it is first removed from O R3 . This allows transcription<br /> from P RM . As mentioned earlier, the term λ repressor<br /> is somewhat misleading because the binding of the λ repressor<br /> at only O R1 and O R2 acts as an activator of P RM . The ability of the<br /> λ repressor to activate its own transcription allows the switch to<br /> the lysogenic cycle to be maintained.<br /> In the lytic cycle (Figure 14.20, right side), the binding of<br /> the cro protein controls the switch. The cro protein has its highest<br /> affinity for O R3 and has a similar affinity for O R2 and O R1 .<br /> Under conditions that favor the lytic cycle, the cro protein accumulates,<br /> and a cro dimer first binds to O R3 . This blocks transcription<br /> from P RM and thereby switches off the lysogenic cycle.<br /> Later in the lytic cycle, the cro protein concentration continues<br /> to rise, so eventually it binds to O R2 and O R1 . This turns down<br /> expression from P R , which is not needed in the later stages of<br /> the lytic cycle.<br /> Genetic switches, like the one just described for phage λ,<br /> represent an important form of genetic regulation. As we have<br /> just seen, a genetic switch can be used to control two alternative<br /> reproductive cycles of a bacteriophage. In addition, genetic<br /> switches are also important in the developmental pathways for<br /> bacteria and eukaryotes. For example, certain species of bacteria<br /> can grow in a vegetative state when nutrients are abundant but<br /> will produce spores when conditions are unfavorable for growth.<br /> The choice between vegetative growth and sporulation involves<br /> genetic switches. Likewise, genetic switches operate in the developmental<br /> pathways in eukaryotes. As we will discover in Chapter<br /> 23, they are key events in the initiation of cell differentiation during<br /> development. Studies of the phage λ life cycle have provided<br /> fundamental information with which to understand how these<br /> other switches can operate at the molecular level.<br /> CONCEPTUAL SUMMARY<br /> Genes can be regulated at any step in the pathway of gene<br /> expression. In bacterial cells, it is common for structural genes<br /> to be organized into a multigene unit called an operon. An<br /> operon typically consists of a promoter, an operator site, two or<br /> more structural genes, and a terminator. The organization of an<br /> operon allows two or more genes to be regulated as a single unit.<br /> In most cases, transcriptional regulation involves the actions of<br /> regulatory proteins that bind to the operon and can greatly influence<br /> the rate of transcription. A repressor is a protein that inhibits<br /> transcription, while an activator is a protein that increases the<br /> rate of transcription. Furthermore, the binding of small effector<br /> molecules to regulatory proteins can influence whether or not<br /> the regulatory proteins can bind to the DNA. Some regulatory<br /> proteins that exert negative control, such as the lac repressor and<br /> trp repressor, inhibit transcription. Others, such as the catabolite<br /> activator protein (CAP), exert positive control by enhancing<br /> transcription. The AraC protein is interesting because it can act<br /> as both a repressor and an activator, depending on the presence<br /> or absence of arabinose. Experiments involving the use of F' factors<br /> showed that regulatory proteins are synthesized and can diffuse<br /> within the cell to ultimately bind to a distant operator site<br /> and cause repression; this form of genetic regulation is called a<br /> trans-effect. In contrast, a cis-effect on gene expression is due to<br /> DNA sequences adjacent to the gene.<br /> Our understanding of transcriptional regulation has been<br /> greatly aided by studying the reproductive cycles of viruses. In<br /> this chapter, we considered the ability of phage λ to proceed<br /> along one of two alternative cycles: the lytic and lysogenic cycles.<br /> This choice is determined by the actions of several genetic regulatory<br /> proteins that can bind to the λ DNA and influence the<br /> transcription of nearby genes. The O R region provides a genetic<br /> switch for the lytic or lysogenic cycle. If the λ repressor controls<br /> the switch, the lysogenic cycle is favored, whereas binding of the<br /> cro protein to this switch favors the lytic cycle. Cellular proteases,<br /> which are influenced by environmental conditions, play a key<br /> role in the choice between the lytic and lysogenic cycles.<br /> In addition to direct transcriptional regulation, gene<br /> expression in bacteria can be affected at later stages in the expression<br /> process. During attenuation of the trp operon, for example,<br /> transcription actually begins, but it is terminated before the entire<br /> mRNA is made; the length of the transcript is determined by the<br /> translation of the trpL gene. This form of regulation is unique to<br /> bacteria because they couple the processes of transcription and<br /> translation. The translation of a complete mRNA transcript can<br /> also be regulated. One form of translational regulation involves<br /> the binding of translational repressor proteins that prevent translational<br /> initiation. A second form involves the binding of antisense<br /> RNA to an mRNA to block translation. Finally, the function of</div> <div>::<br /> 16<br /> GENE MUTATION<br /> AND DNA REPAIR<br /> CHAPTER OUTLINE<br /> 16.1 Consequences of Mutation<br /> 16.2 Occurrence and Causes of Mutation<br /> 16.3 DNA Repair<br /> The effects of a mutation. A mutation during<br /> embryonic development has caused this sheep to<br /> have a black spot on its side.<br /> The primary function of DNA is to store information for the synthesis<br /> of cellular proteins. A key aspect of the gene expression process is that the DNA<br /> itself does not normally change. This allows DNA to function as a permanent<br /> storage unit. However, on relatively rare occasions, a mutation can occur. The<br /> term mutation refers to a heritable change in the genetic material. This means<br /> that the structure of DNA has been changed permanently, and this alteration<br /> can be passed from mother to daughter cells during cell division. If a mutation<br /> occurs in reproductive cells, it may also be passed from parent to offspring.<br /> The topic of mutation is centrally important in all fields of genetics,<br /> including molecular genetics, Mendelian inheritance, and population genetics.<br /> Mutations provide the allelic variation that we have discussed throughout<br /> this textbook. For example, phenotypic differences, such as tall versus dwarf<br /> pea plants, are due to mutations that alter the expression of particular genes.<br /> With regard to their phenotypic effects, mutations can be beneficial, neutral,<br /> or detrimental. On the positive side, mutations are essential to the continuity<br /> of life. They provide the variation that enables species to change and adapt to<br /> their environment. Mutations are the foundation for evolutionary change. On<br /> the negative side, however, new mutations are much more likely to be harmful<br /> rather than beneficial to the individual. The genes within each species have<br /> evolved to work properly. They have functional promoters, coding sequences,<br /> terminators, and so on, that allow the genes to be expressed. Random mutations<br /> are more likely to disrupt these sequences rather than improve their function.<br /> For example, many inherited human diseases result from mutated genes. In<br /> addition, diseases such as skin and lung cancer can be caused by environmental<br /> agents that are known to cause DNA mutations. For these and many other reasons,<br /> understanding the molecular nature of mutations is a deeply compelling<br /> area of research. In this chapter, we will consider the nature of mutations and<br /> their consequences on gene expression at the molecular level.<br /> Because mutations can be quite harmful, organisms have developed several<br /> ways to repair damaged DNA. DNA repair systems reverse DNA damage<br /> before it results in a mutation that could potentially have negative consequences.<br /> DNA repair systems have been studied extensively in many organisms, particularly<br /> Escherichia coli, yeast, mammals, and plants. A variety of systems repair<br /> different types of DNA lesions. In this chapter, we will examine the ways that<br /> several of these DNA repair systems operate.</div> <div>INDEX 835<br /> Induced fit, 283<br /> Induced mutations, 432, 438–40<br /> Inducer molecule, 361, 362<br /> Inducible genes, 361, 362<br /> Inducible operon, 375–76<br /> Induction, 634<br /> Influenza virus, 228<br /> Inheritance<br /> autosomal recessive, 601–2<br /> blending, 17, 702<br /> of blood type, 79–81<br /> chromosome theory of, 60–66<br /> of comb morphology in chickens, 87<br /> concordance in twins, 600<br /> of defective enzyme, in trait inheritance, 325<br /> of disease-causing alleles RFLP analysis in<br /> following, 551–53<br /> of DNA methylation, 404–6, 406<br /> maternal, 177<br /> Mendel’s laws of, 9–10<br /> particulate theory of, 22–23<br /> patterns involving single genes, 72<br /> pedigree analysis, 29, 29–30<br /> polygenic, 702–8<br /> probability calculations on, 30–35<br /> pseudoautosomal, 83<br /> sex-influenced, 83–84<br /> sex-limited, 72, 84, 85<br /> of traits, 9–10<br /> X-linked, 61<br /> Inheritance, non-Mendelian<br /> epigenetic inheritance, 164–74<br /> DNA methylation in, 172–74<br /> dosage compensation in, 164–66<br /> genomic imprinting in, 170–72<br /> X inactivation in, 168–69<br /> extranuclear inheritance, 174–81<br /> in Chlamydomonas, 177–79, 179<br /> genetic material in mitochondria and<br /> chloroplasts, 174–76, 175<br /> human diseases caused by mitochondrial<br /> mutations, 179–80, 180<br /> non-Mendelian results in reciprocal crosses,<br /> 176–77<br /> in yeast, 177–79<br /> maternal effect genes in, 161–64<br /> in snail shell coiling morphology, 162–64, 163<br /> Inheritance patterns<br /> ABO blood type, 80<br /> in corn, 107–9<br /> in Manx cats, 85<br /> in pattern baldness, 84<br /> in sickle-cell disease, 77<br /> single genes, 71–86<br /> of snail shell coiling, 162<br /> two genes involved in, 86<br /> of X-linked genes, 81–82<br /> Inhibitor molecule, 361, 362<br /> Initiation phase<br /> in transcription, 299–300<br /> in X inactivation, 169–70, 170<br /> Initiator tRNA, 348<br /> Insects, sex determination in, 63<br /> Insertion sequence, 470<br /> In situ hybridization, 544–45, 607<br /> Insulin<br /> functions of, 336, 337<br /> human, E.coli in synthesizing, 1<br /> synthetic, 518, 518<br /> bacteria in production of, 519<br /> Insulin-like growth factor, 526<br /> Insulin-like growth factor II, 172<br /> Insulin receptor, 336<br /> Integrins, 336<br /> Intelligence quotient testing, 715, 716<br /> Interference<br /> in number of double crossovers, 117<br /> positive, 117<br /> Intergenic regions, 248<br /> Intergenic suppressor mutation, 427, 428<br /> Intergenic suppressors, 86, 92, 93<br /> Internal loops, in RNA, 239, 240<br /> Internal nuclear matrix, 259<br /> Interphase, 49<br /> chromatin structure during, 261<br /> condensin localization, 265<br /> in mitosis, 52<br /> Interrupted mating, 139<br /> Interspecies crosses, 685<br /> Interstitial deletion, chromosomal, 190<br /> Intervening sequences. See Introns<br /> Intracellular signaling pathway, 614<br /> Intracellular signaling proteins, 616<br /> Intragenic mapping, in bacteriophages, 148–55<br /> complementation tests, 150–51<br /> deletion mapping, 153–54<br /> insight into relationship between traits and<br /> molecular genetics, 154–55<br /> map construction, 151–55<br /> mutants with altered plaque morphology,<br /> 149–50<br /> in the rII region, 152<br /> Intragenic recombination, 151, 153<br /> Intragenic suppressor mutation, 427, 428<br /> Intrinsic termination, 304, 305<br /> Introns, 253, 309<br /> discovery of, 311–14<br /> occurrence of, 315<br /> Invasive cells, 610<br /> Inversion, chromosomal, 189–90, 190, 195,<br /> 195–96, 196<br /> Inversion heterozygote, 196<br /> Inversion loop, 196<br /> Invertebrate development<br /> body pattern depends on positional information<br /> each cell receives, 633–34<br /> early embryonic development determines body<br /> pattern in adults, 635–37<br /> fate of cells in Caenorhabditis, 643–44<br /> heterochronic mutations in Caenorhabditis,<br /> 644–47<br /> homeotic genes control phenotypic<br /> characteristics of segments, 641–43<br /> maternal effect genes in, 637–38<br /> mutations in revealing genes controlling<br /> development, 634–35<br /> segments and parasegments in Drosophila, 637,<br /> 638–41, 639<br /> Inverted repeats, 470<br /> Iron assimilation, 415–17<br /> Iron regulatory protein, 416<br /> Iron response element, 416<br /> Isoacceptor rules, 344<br /> Isoelectric focusing, 581<br /> Isogamous organism, 54<br /> Isoleucine, 335<br /> Itakura, Keiichi, 515<br /> J<br /> Jackson, David, 482<br /> Jacob, François, 139, 300, 361, 366<br /> Jacobs syndrome, 203<br /> Jaenisch, Rudolf, 527<br /> Janssens, F.A., 103<br /> Jellyfish, fluorescent gene of, in mice, 3, 3–4<br /> Johannsen, Wilhelm, 23<br /> “Jumping genes.” See Transposable elements<br /> K<br /> Kaiser, A. Dale, 482<br /> Kan, Yuet, 551<br /> kanamycin R gene, 488<br /> Kappa particles, 181<br /> Karpechenko, Georgi, 210–11<br /> Karyotype, 45–47, 46, 607, 608<br /> of hybrid animal, 211<br /> numbering system for, 188<br /> Khorana, H. Gobind, 329, 332, 488<br /> Killer trait, 181<br /> Kimura, Motoo, 736, 737<br /> Kinesin, 336<br /> Kinetochore, 49, 50, 253, 253<br /> Kinetochore microtubules, 51, 53, 58<br /> Klebsiella pneumoniae, 589<br /> Klinefelter syndrome, 203<br /> Knockout Mouse Project, 523<br /> Knowledge-based modeling, 594<br /> Knudson, Alfred, 618<br /> Kohne, David, 254<br /> Kölreuter, Joseph, 17<br /> Kornberg, Arthur, 285, 286<br /> Kornberg, Roger, 256, 308<br /> Kozak, Marilyn, 349<br /> Kraatz, Ernst Gustav, 641<br /> Krüppel gene, 640<br /> Kuru, 609<br /> L<br /> lac genes<br /> lacA, 364, 365<br /> lacI, 366–69, 369<br /> lacI � mutation, 366<br /> lacO c mutants, 370<br /> lacO � mutants, 370<br /> lacY, 364, 365, 589<br /> lacZ, 364, 365<br /> Lack, David, 678<br /> lac operon<br /> induction and repression cycle of, 365<br /> mechanism of induction of, 364<br /> operator sites for lac repressor, 370–71<br /> protein encoding in lactose metabolism,<br /> 363–64<br /> regulation of<br /> by activator protein, 369–70<br /> by repressor protein, 364–69<br /> lac repressor, 365<br /> in lac operon regulation, 364, 370<br /> operator sites for, 370–71, 371<br /> β-lactamase, 484<br /> Lactic acidosis, 180<br /> Lactoferin, 526<br /> Lactose metabolism, 363<br /> Lactose permease, 336<br /> Lagging strand, 277, 279, 280, 282<br /> Lamarck, Jean Baptiste, 433<br /> Lampbrush chromosome, 398<br /> Lander, Eric, 708<br /> Lathyrus odoratus, 87<br /> Law of independent assortment<br /> chromosome behavior and, 62, 62<br /> Mendel’s law of, 27–28<br /> Punnett square in solving, 28<br /> two traits not assorting independently,<br /> 101–2<br /> vs. linkage, Chi square analysis in distinguishing,<br /> 103–5<br /> Law of segregation, 23, 24, 61, 61–62, 105<br /> Laws of inheritance, 9–10<br /> Leading strand, 277, 282<br /> Leber’s hereditary optic neuropathy, 179–80, 180</div> <div>108 CHAPTER 5 :: LINKAGE AND GENETIC MAPPING IN EUKARYOTES<br /> ■<br /> THE DATA<br /> Phenotype of<br /> F 1 Kernel<br /> Number of<br /> Kernels<br /> Analyzed<br /> Cytological Appearance of a Chromosome in F 1 Offspring*<br /> Did a Crossover<br /> Occur During<br /> Gamete Formation<br /> in Parent A?<br /> Colored/waxy 3 Knobbed/translocation Normal<br /> No<br /> C wx<br /> c wx<br /> Colorless/starchy 11 Knobless/normal<br /> Normal<br /> No<br /> c<br /> Wx<br /> c<br /> or<br /> Wx<br /> Colorless/starchy<br /> c wx<br /> 4 Knobless/translocation Normal<br /> Yes<br /> c wx<br /> c Wx<br /> Colorless/waxy 2 Knobless/translocation Normal<br /> Yes<br /> c wx<br /> c wx<br /> Normal<br /> Colored/starchy 5 Knobbed/normal<br /> c or Wx<br /> C Wx<br /> Total<br /> 25<br /> c wx<br /> *In this table, the chromosome on the left was inherited from parent A, and the chromosome on the right was inherited from parent B.<br /> Yes<br /> ■<br /> INTERPRETING THE DATA<br /> By combining the gametes in a Punnett square, the following<br /> types of offspring can be produced:<br /> Parent A<br /> C wx<br /> c Wx<br /> C Wx<br /> c wx<br /> c Wx<br /> Cc Wxwx<br /> Colored,<br /> starchy<br /> cc WxWx<br /> Colorless,<br /> starchy<br /> Cc WxWx<br /> Colored,<br /> starchy<br /> cc Wxwx<br /> Colorless,<br /> starchy<br /> Parent B<br /> c wx<br /> Cc wxwx<br /> Colored,<br /> waxy<br /> cc Wxwx<br /> Colorless,<br /> starchy<br /> Cc Wxwx<br /> Colored,<br /> starchy<br /> cc wxwx<br /> Colorless,<br /> waxy<br /> Nonrecombinant<br /> Nonrecombinant<br /> Recombinant<br /> Recombinant<br /> In this experiment, the researchers were interested in whether or<br /> not crossing over had occurred in parent A, which was heterozygous<br /> for both genes. This parent could produce four types of<br /> gametes, while parent B could produce only two types.<br /> Parent A<br /> C wx (nonrecombinant)<br /> c Wx (nonrecombinant)<br /> C Wx (recombinant)<br /> c wx (recombinant)<br /> Parent B<br /> c Wx<br /> c wx<br /> As seen in the Punnett square, two of the phenotypic categories,<br /> colored, starchy (Cc Wxwx or Cc WxWx) and colorless, starchy<br /> (cc WxWx or cc Wxwx), were ambiguous because they could arise<br /> from a nonrecombinant and from a recombinant gamete. In other<br /> words, these phenotypes could be produced whether or not recombination<br /> occurred in parent A. Therefore, let’s focus on the two<br /> unambiguous phenotypic categories: colored, waxy (Cc wxwx) and<br /> colorless, waxy (cc wxwx). The colored, waxy phenotype could happen<br /> only if recombination did not occur in parent A and if parent<br /> A passed the knobbed, translocated chromosome to its offspring.<br /> As shown in the data, three kernels were obtained with this phenotype,<br /> and all of them had the knobbed, translocated chromosome.<br /> By comparison, the colorless, waxy phenotype could be obtained<br /> only if genetic recombination occurred in parent A and this parent<br /> passed a chromosome 9 that had a translocation but was knobless.<br /> Two kernels were obtained with this phenotype, and both of<br /> them had the expected chromosome that had a translocation but<br /> was knobless. Taken together, these results showed a perfect correlation<br /> between genetic recombination of alleles and the cytological<br /> presence of a chromosome displaying a genetic exchange of chromosomal<br /> pieces from parent A.</div> <div>24.1 GENES IN POPULATIONS, AND THE HARDY-WEINBERG EQUATION 667<br /> FIGURE 24.2 Polymorphism in the Hawaiian happy-face spider.<br /> GenesgTraits These three spiders are members of the same species and carry the same genes. However, several genes that affect pigmentation patterns are polymorphic,<br /> meaning there is more than one allele for each gene within the population. This polymorphism within the Hawaiian happy-face spider population produces members that look<br /> quite different from each other.<br /> Region of the human β-globin gene<br /> These<br /> are three<br /> different<br /> alleles of<br /> the human<br /> β-globin<br /> gene. Many<br /> more have<br /> been<br /> identified.<br /> A<br /> T<br /> C T<br /> G A<br /> A<br /> C T<br /> T<br /> G A<br /> A<br /> C T<br /> T<br /> G A<br /> C<br /> G<br /> C<br /> G<br /> C<br /> G<br /> C<br /> G<br /> C<br /> G<br /> C<br /> G<br /> T<br /> G A G<br /> A<br /> C T C<br /> T<br /> G<br /> A<br /> C<br /> A<br /> T<br /> A<br /> T<br /> T<br /> A<br /> Site of a 5 bp deletion<br /> G<br /> C<br /> G<br /> C<br /> G<br /> C<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> Loss-of-function allele<br /> Hb A allele<br /> These two alleles are an<br /> example of a single<br /> nucleotide polymorphism<br /> in the human population.<br /> Hb S allele<br /> FIGURE 24.3 The relationship between alleles and various<br /> types of mutations. The DNA sequence shown here is a small portion<br /> of the β-globin gene in humans. Mutations have altered the gene<br /> to create the three different alleles in this figure. The top two alleles<br /> differ by a single base pair and are referred to as a single nucleotide<br /> polymorphism (SNP). The bottom allele has a 5 bp deletion that results<br /> in a nonfunctional polypeptide. It is a loss-of-function allele.<br /> by a single nucleotide, so they are an example of a single nucleotide<br /> polymorphism. As discussed in Chapter 4, the Hb S allele causes<br /> sickle-cell disease in a homozygote. The bottom sequence contains a<br /> short, 5 bp deletion compared to the other two alleles. This deletion<br /> results in a nonfunctional β-globin polypeptide. Therefore, the bottom<br /> sequence is an example of a loss-of-function allele.<br /> Population Genetics Is Concerned<br /> with Allele and Genotype Frequencies<br /> As we have seen, population geneticists want to understand the<br /> prevalence of polymorphic genes within populations. Much of<br /> their work evaluates the frequency of alleles in a quantitative way.<br /> Two fundamental calculations are central to population genetics:<br /> allele frequencies and genotype frequencies. The allele and<br /> genotype frequencies are defined as<br /> Allele frequency �<br /> Genotype frequency �<br /> Number of copies of an allele<br /> in a population<br /> Total number of all alleles for<br /> that gene in a population<br /> Number of individuals with a particular<br /> genotype in a population<br /> Total number of individuals<br /> in a population</div> <div>19.2 GENETICALLY MODIFIED ANIMALS 527<br /> How was Dolly created? As shown in Figure 19.10, the<br /> researchers removed mammary cells from an adult female sheep<br /> and grew them in the laboratory. The researchers then extracted<br /> the nucleus from an egg cell of a different sheep and used electrical<br /> pulses to fuse the diploid mammary cell with the enucleated<br /> egg cell. After fusion, the zygote began embryonic development<br /> and the resulting embryo was implanted into the uterus of a surrogate<br /> mother sheep. One hundred and forty-eight days later,<br /> Dolly was born.<br /> While Dolly was clearly a clone of the initial adult female<br /> sheep, tests conducted when she was three years old suggested<br /> that she was “genetically older” than her actual age indicated.<br /> As mammals age, chromosomes in somatic cells tend to shorten<br /> from the telomeres—the ends of eukaryotic chromosomes.<br /> Therefore, older individuals have shorter chromosomes in their<br /> somatic cells compared to younger ones. This shortening does<br /> not seem to occur in the cells of the germ line, however. When<br /> researchers analyzed the chromosomes in Dolly’s somatic cells<br /> when she was about three years old, the lengths of her chromosomes<br /> were consistent with a sheep that was significantly older,<br /> say, nine or 10 years old. The sheep that donated the somatic<br /> cell that produced Dolly was six years old, and her mammary<br /> cells had been grown in culture for several cell doublings before<br /> a mammary cell was fused with an oocyte. This led researchers<br /> to suspect that Dolly’s shorter telomeres were a result of<br /> chromosome shortening in the somatic cells of the sheep that<br /> donated the nucleus. In 2003, the Roslin Institute announced<br /> the decision to euthanize six-year-old Dolly after an examination<br /> showed progressive lung disease. Her death has raised concerns<br /> among experts that the techniques used to produce Dolly<br /> could have caused premature aging.<br /> With regard to telomere length, research in mice and cattle<br /> has shown different results; the telomeres of these cloned animals<br /> appear to be the correct length. For example, cloning was<br /> conducted on mice via the method described in Figure 19.10 for<br /> six consecutive generations. The cloned mice of the sixth generation<br /> had normal telomeres. Further research will be necessary to<br /> determine if cloning via somatic cells has an effect on the length<br /> of telomeres in subsequent generations. However, other studies<br /> in mice point to various types of genetic flaws in cloned animals.<br /> For example, Rudolf Jaenisch and his colleagues used DNA<br /> microarray technology (described in Chapter 21) and analyzed<br /> the transcription patterns of over 10,000 genes in cloned mice.<br /> Up to 4% of those genes were not expressed normally. Furthermore,<br /> research has shown that cloned mice die at a younger age<br /> than their naturally bred counterparts.<br /> Mammalian cloning is still at an early stage of development.<br /> Nevertheless, the breakthrough of creating Dolly has<br /> shown that it is technically possible. In recent years, cloning from<br /> somatic cells has been achieved in several mammalian species,<br /> including sheep, cattle, mice, goats, and pigs. In 2002, the first<br /> pet was cloned, which was named Carbon Copy, also called Copy<br /> Cat (Figure 19.11). Mammalian cloning has the potential to provide<br /> many practical applications. With regard to livestock, cloning<br /> would enable farmers to use the somatic cells from their best<br /> individuals to create genetically homogeneous herds. This could<br /> Mammary<br /> cell<br /> Donor sheep<br /> Mammary cell<br /> Donor sheep's mammary cell<br /> is extracted and grown in a<br /> tissue culture flask. Another<br /> sheep's unfertilized egg is<br /> extracted, and the nucleus is<br /> removed.<br /> Nucleus<br /> The donor nucleus from<br /> the mammary cell and<br /> the maternal proteins<br /> within the enucleated egg<br /> initiate development of<br /> the egg into an embryo.<br /> The embryo is<br /> transferred into a<br /> surrogate ewe.<br /> Surrogate<br /> ewe<br /> Allow pregnancy<br /> to proceed.<br /> A lamb genetically<br /> identical to the donor<br /> sheep is then born.<br /> Unfertilized egg<br /> The cells are fused<br /> together with electrical<br /> pulses.<br /> Egg with nucleus<br /> removed<br /> FIGURE 19.10 Protocol for the successful cloning of sheep.<br /> Genes g Traits Dolly was (almost) genetically identical to the sheep that donated<br /> a mammary cell to create her. Dolly and the donor sheep were (almost) genetically<br /> identical in the same way that identical twins are; they carried the same set of<br /> genes and looked remarkably similar. However, they may have had minor genetic<br /> differences due to possible variation in their mitochondrial DNA and may have<br /> exhibited some phenotypic differences due to maternal effect or imprinted genes.</div> <div>614 CHAPTER 22 :: MEDICAL GENETICS AND CANCER<br /> ■<br /> THE DATA<br /> Number of<br /> Malignant Foci<br /> Source of DNA Recipient Cells Found on 12 Plates<br /> Malignant Cell Lines<br /> MC5-5-0 NIH3T3 48*<br /> (normal fibroblasts)<br /> MCA16 “ 5<br /> MB66 MCA ad 36 “ 8<br /> MB66 MCA ACL 6 “ 0<br /> MB66 MCA ACL 13 “ 0<br /> Normal Cell Lines<br /> NIH3T3 “ 1<br /> C3H10T1/2 “ 0<br /> *In this experiment, two of the plates were contaminated, so this<br /> is 48 foci on 10 plates.<br /> ■ INTERPRETING THE DATA<br /> As shown in the data, the DNA isolated from some (but not all)<br /> malignant cell lines could transform normal mouse cells, which<br /> then proliferated and produced malignant foci. These results are<br /> consistent with the hypothesis that oncogenes had been taken up<br /> and expressed in the normal mouse cells, converting them into<br /> malignant cells. By comparison, the DNA isolated from normal<br /> cells did not cause a significant amount of transformation. From<br /> these experiments, it is not clear why the DNA from two of the<br /> malignant cell lines, MB66 MCA ACL 6 and MB66 MCA ACL 13,<br /> could not transform the normal mouse cells. One possibility is<br /> that some malignancies are caused by dominant oncogenes, while<br /> others involve genes that act recessively. Recessive genes would<br /> be unable to transform normal mouse cells that already contain<br /> the normal (nonmalignant) dominant allele. Later, we will learn<br /> how another category of genes involved in cancer, called tumorsuppressor<br /> genes, act recessively.<br /> Just two years later, in 1981, the laboratories of Weinberg<br /> and Geoffrey Cooper identified the first cellular oncogene in<br /> humans. They isolated chromosomal DNA from human bladder<br /> carcinoma cells and used it to transform mouse cells in vitro.<br /> This chromosomal DNA contained a human oncogene that was<br /> the result of a mutation. The identity of the gene was eventually<br /> determined by transposon tagging, as described in solved problem<br /> S2 at the end of this chapter. This observation paved the way<br /> for the isolation of many human cellular oncogenes.<br /> A self-help quiz involving this experiment can be found at<br /> the Online Learning Center.<br /> Many Oncogenes Have Abnormalities That Affect<br /> Proteins Involved in Cell Division Pathways<br /> As described in Chapter 3, eukaryotic cells destined to divide<br /> progress through a series of stages known as the cell cycle (Figure<br /> 22.11a). The phases consist of G 1 (first gap), S (synthesis of DNA,<br /> the genetic material), G 2 (second gap), and M phase (mitosis and<br /> cytokinesis). The G 1 phase is a period in a cell’s life when it may<br /> become committed to divide. Depending on the conditions, a cell<br /> in the G 1 phase may accumulate molecular changes that cause it<br /> to progress through the rest of the cell cycle. When this occurs,<br /> cell biologists say that a cell has reached a special control point<br /> called the restriction point. The commitment to divide is based<br /> on a variety of factors. For example, environmental conditions,<br /> such as the presence of sufficient nutrients, are important for cell<br /> division. In addition, multicellular organisms rely on signaling<br /> molecules to coordinate cell division throughout the body. These<br /> signaling molecules are often called growth factors because they<br /> promote cell division.<br /> As researchers began to identify oncogenes, they wanted to<br /> understand how these mutant genes promote abnormal cell division.<br /> In parallel with cancer research, cell biologists have studied<br /> the roles that normal cellular proteins play in cell division. As<br /> mentioned, the cell cycle is regulated in part by growth factors<br /> that bind to cell surface receptors and initiate a cascade of cellular<br /> events that lead eventually to cell division. Figure 22.11b<br /> considers a protein called epidermal growth factor (EGF) that<br /> is secreted from endocrine cells and stimulates epidermal cells,<br /> such as skin cells, to divide. As seen here, EGF binds to its receptor,<br /> leading to the activation of an intracellular signaling pathway.<br /> This pathway, also known as a signal cascade, leads to a<br /> change in gene transcription. In other words, the transcription<br /> of specific genes is activated in response to the growth hormone.<br /> Once these genes are transcribed and translated, the gene products<br /> function to promote the progression through the cell cycle.<br /> Figure 22.11b is just one example of a pathway between a growth<br /> factor and gene activation. Eukaryotic species produce many different<br /> growth factors, and the signaling pathways are often more<br /> complex than the one shown here.<br /> What is the relationship between normal genes and oncogenes?<br /> A normal, nonmutated gene that has the potential to<br /> become an oncogene is termed a proto-oncogene. To become an<br /> oncogene, a proto-oncogene must incur a mutation that causes<br /> its expression to be abnormally active. The mutation typically<br /> has one of three possible effects:<br /> 1. The amount of the encoded protein is greatly increased.<br /> 2. A change occurs in the structure of the encoded protein<br /> that causes it to be overly active.<br /> 3. The encoded protein is expressed in a cell type where it is<br /> not normally expressed.<br /> The mutations that convert proto-oncogenes into oncogenes<br /> have been analyzed in many types of cancers. Oncogenes<br /> commonly encode proteins that function in cell growth signaling<br /> pathways (Table 22.7). These include growth factors, growth</div> <div>12.4 RNA MODIFICATION 313<br /> ■ TESTING THE HYPOTHESIS — FIGURE 12.19 RNA hybridization to the β-globin gene reveals an intron.<br /> Starting material: A cloned fragment of chromosomal DNA that contains the mouse β-globin gene.<br /> Experimental level<br /> Conceptual level<br /> 1. Isolate mature mRNA for the mouse<br /> β-globin gene. Note: Globin mRNA is<br /> abundant in reticulocytes, which are<br /> immature red blood cells.<br /> Solution of<br /> β-globin<br /> mRNA<br /> mRNA<br /> Cloned<br /> β-globin<br /> DNA<br /> 2. Mix together the β-globin mRNA and<br /> cloned DNA of the β-globin gene.<br /> 70%<br /> formamide<br /> Solution<br /> of cloned<br /> β-globin<br /> DNA<br /> β-globin DNA<br /> 3. Separate the double-stranded DNA and<br /> allow the mRNA to hybridize. This is<br /> done using 70% formamide, at 52°C, for<br /> 16 hours.<br /> Incubator<br /> 4. Dilute the sample to decrease the<br /> formamide concentration. This allows<br /> the DNA to re-form a double-stranded<br /> structure. Note: The DNA cannot form<br /> a double-stranded structure in regions<br /> where the mRNA has already hybridized.<br /> 5. Spread the sample onto a microscopy<br /> grid.<br /> 6. Stain with uranyl acetate and shadow<br /> with heavy metal. Note: The technique<br /> of electron microscopy is described in<br /> the Appendix.<br /> 7. View the sample under the electron<br /> microscope.<br /> Vacuum evaporator<br /> Platinum<br /> electrode<br /> Specimen<br /> Platinum<br /> atoms</div> <div>CONCEPTUAL QUESTIONS 97<br /> C15. A human disease known as vitamin D−resistant rickets is inherited<br /> as an X-linked dominant trait. If a male with the disease produces<br /> children with a female who does not have the disease, what is the<br /> expected ratio of affected and unaffected offspring?<br /> C16. Hemophilia is a X-linked recessive trait in humans. If a<br /> heterozygous woman has children with a an unaffected man, what<br /> are the odds of the following combinations of children?<br /> A. An affected son<br /> B. Four unaffected offspring in a row<br /> C. An unaffected daughter or son<br /> D. Two out of five offspring that are affected<br /> C17. Explain whether the following events could or could not happen.<br /> A. A male has the same X chromosome as his paternal grandfather.<br /> B. A male has the same X chromosome as his maternal grandfather.<br /> C. A female has the same X chromosome as her maternal<br /> grandmother.<br /> D. A female has the same X chromosome as her maternal<br /> great-great-grandmother.<br /> C18. Incontinentia pigmenti is a rare, X-linked dominant disorder<br /> in humans characterized by swirls of pigment in the skin. If an<br /> affected female, who had an unaffected father, has children with<br /> an unaffected male, what would be the predicted ratios of affected<br /> and unaffected sons and daughters?<br /> C19. With regard to pattern baldness in humans (a sex-influenced<br /> trait), a woman who is not bald and whose mother is bald has<br /> children with a bald man whose father is not bald. What are their<br /> probabilities of having the following types of families?<br /> A. Their first child will not become bald.<br /> B. Their first child will be a male who will not become bald.<br /> C. Their first three children will be females who are not bald.<br /> C20. In rabbits, the color of body fat is controlled by a single gene with<br /> two alleles, designated Y and y. The outcome of this trait is affected<br /> by the diet of the rabbit. When raised on a standard vegetarian diet,<br /> the dominant Y allele confers white body fat, and the y allele confers<br /> yellow body fat. However, when raised on a xanthophyll-free diet,<br /> the homozygote yy animal has white body fat. If a hetero zygous<br /> animal is crossed to a rabbit with yellow body fat, what are the<br /> proportions of offspring with white and yellow body fat when raised<br /> on a standard vegetarian diet? How do the proportions change if the<br /> offspring are raised on a xanthophyll-free diet?<br /> C21. A Siamese cat that spends most of its time outside was accidentally<br /> injured in a trap and required several stitches in its right front paw.<br /> The veterinarian had to shave the fur from the paw and leg, which<br /> originally had rather dark fur. Later, when the fur grew back, it was<br /> much lighter than the fur on the other three legs. Do you think<br /> this injury occurred in the hot summer or cold winter? Explain<br /> your answer.<br /> C22. A true-breeding male fly with eosin eyes is crossed to a white-eyed<br /> female that is heterozygous for the wild-type (C) and cream alleles<br /> (c a ). What are the expected proportions of their offspring?<br /> C23. In Chapter 4, we considered the trait of hen- versus cock-feathering.<br /> Starting with two heterozygous fowl that are hen-feathered, explain<br /> how you would obtain a true-breeding line that always produced<br /> cock-feathered males.<br /> C24. In the pedigree shown here for a trait determined by a single gene<br /> (affected individuals are shown in black), state whether it would be<br /> possible for the trait to be inherited in each of the following ways:<br /> A. Recessive<br /> B. X-linked recessive<br /> C. Dominant, complete penetrance<br /> D. Sex influenced, dominant in males<br /> E. Sex limited<br /> F. Dominant, incomplete penetrance<br /> III-1<br /> I-1 I-2<br /> II-1 II-2 II-3 II-4<br /> III-2<br /> IV-1 IV-2 IV-3<br /> II-5<br /> III-3 III-4 III-5 III-6 III-7<br /> C25. The pedigree shown here also concerns a trait determined by a<br /> single gene (affected individuals are shown in black). Which of the<br /> following patterns of inheritance are possible?<br /> A. Recessive<br /> B. X-linked recessive<br /> C. Dominant<br /> D. Sex influenced, recessive in males<br /> E. Sex limited<br /> I-1 I-2<br /> II-1 II-2 II-3 II-4<br /> III-1<br /> IV-1 IV-2 IV-3<br /> II-5<br /> III-2 III-3 III-4 III-5<br /> C26. Let’s suppose you have pedigree data from thousands of different<br /> families involving a particular genetic disease. How would you<br /> decide whether the disease is inherited as a recessive trait as<br /> opposed to one that is dominant with incomplete penetrance?</div> <div>CONCEPTUAL QUESTIONS 661<br /> C3. If you observed fruit flies with the following developmental<br /> abnormalities, would you guess that a mutation has occurred in a<br /> segmentation gene or a homeotic gene? Explain your guess.<br /> A. Three abdominal segments were missing.<br /> B. One abdominal segment had legs.<br /> C. A fly with the correct number of segments had two additional<br /> thoracic segments and two fewer abdominal segments.<br /> C4. Which of the following statements are true with regard to<br /> positional information in Drosophila?<br /> A. Morphogens are a type of molecule that conveys positional<br /> information.<br /> B. Morphogenetic gradients are established only in the oocyte,<br /> prior to fertilization.<br /> C. Cell adhesion molecules also provide a way for a cell to obtain<br /> positional information.<br /> C5. Discuss the morphological differences between the parasegments<br /> and segments of Drosophila. Discuss the evidence, providing<br /> specific examples, that suggests the parasegments of the embryo<br /> are the subdivisions for the organization of gene expression.<br /> C6. Here are schematic diagrams of mutant larvae.<br /> (a)<br /> (c)<br /> The left side of each pair shows a wild-type larva, with gray boxes<br /> showing the sections that are missing in the mutant larva. Which<br /> type of gene is defective in each larva: a gap gene, a pair-rule gene,<br /> or a segment-polarity gene?<br /> C7. Describe what a morphogen is and how it exerts its effects. What<br /> do you expect will happen when a morphogen is expressed in the<br /> wrong place in an embryo? List five examples of morphogens that<br /> function in Drosophila.<br /> C8. What is the meaning of positional information? Discuss three<br /> different ways that cells can obtain positional information. Which<br /> of these three ways do you think is the most important for the<br /> formation of a segmented body pattern in Drosophila?<br /> (b)<br /> C9. Gradients of morphogens can be preestablished in the oocyte.<br /> Also, later in development, morphogens can be secreted from cells.<br /> How are these two processes similar and different?<br /> C10. Discuss how the anterior portion of the antero-posterior axis is<br /> established. What aspects of oogenesis are critical in establishing<br /> this axis? What do you think would happen if the bicoid mRNA<br /> was not trapped at the anterior end but instead diffused freely<br /> throughout the oocyte?<br /> C11. Describe the function of the Bicoid protein. Explain how its ability<br /> to exert its effects in a concentration-dependent manner is a<br /> critical feature of its function.<br /> C12. With regard to development, what are the roles of the maternal<br /> effect genes versus the zygotic genes? Which types of genes are<br /> needed earlier in the development process?<br /> C13. Discuss the role of homeotic genes in development. Explain<br /> what happens to the phenotype of a fly when a gain-of-function<br /> homeotic gene mutation occurs in an abnormal region of the<br /> embryo. What are the consequences of a loss-of-function mutation<br /> in such a gene?<br /> C14. Describe the molecular features of the homeobox and<br /> homeodomain. Explain how these features are important in the<br /> function of homeotic genes.<br /> C15. What would you predict to be the phenotype of a Drosophila larva<br /> whose mother was homozygous for a loss-of-function allele in the<br /> nanos gene?<br /> C16. Based on the photographs in Figure 23.12, in which segments<br /> would the Antp gene normally be expressed?<br /> C17. If a mutation in a homeotic gene produced the following<br /> phenotypes, would you expect the mutation to be a loss-offunction<br /> or a gain-of-function allele? Explain your answer.<br /> A. An abdominal segment has antennae attached to it.<br /> B. The most anterior abdominal segment resembles the most<br /> posterior thoracic segment.<br /> C. The most anterior thoracic segment resembles the most<br /> posterior abdominal segment.<br /> C18. Explain how loss-of-function mutations in the following categories<br /> of genes would affect the morphologies of Drosophila larvae:<br /> A. Gap genes<br /> B. Pair-rule genes<br /> C. Segment-polarity genes<br /> C19. What is the difference between a maternal effect gene and a<br /> zygotic gene? Of the following genes that play a role in Drosophila<br /> development, which would be maternal effect genes and which<br /> would be zygotic? Explain your answer.<br /> A. nanos<br /> B. Antp<br /> C. bicoid<br /> D. lab<br /> C20. Cloning of mammals (such as Dolly) is described in Chapter 19.<br /> Based on your understanding of animal development, explain why<br /> an enucleated egg is needed to clone mammals. In other words,<br /> what features of the oocyte are essential for animal development?</div> <div>5.3 GENETIC MAPPING IN HAPLOID EUKARYOTES 121<br /> alleles A (orange pigmentation) and a (albino, which results in a<br /> white phenotype). In Figure 5.14a, a crossover has not occurred,<br /> so the octad contains a linear arrangement of four haploid cells<br /> carrying the A allele, which are adjacent to four haploid cells that<br /> contain the a allele. This 4:4 arrangement of spores within the<br /> ascus is called first-division segregation (FDS) or an M1 pattern.<br /> It is called a first-division segregation pattern because the A<br /> and a alleles have segregated from each other after the first meiotic<br /> division.<br /> In contrast, as shown in Figure 5.14b, if a crossover occurs<br /> between the centromere and the gene of interest, the ordered<br /> octad will deviate from the 4:4 pattern. Depending on the relative<br /> locations of the two chromatids that participated in the<br /> crossover, the ascus will contain a 2:2:2:2 or 2:4:2 pattern. These<br /> patterns are called second-division segregation (SDS) or M2<br /> patterns. In this case, the A and a alleles do not segregate until<br /> the second meiotic division is completed.<br /> Because a pattern of second-division segregation is a result<br /> of crossing over, the percentage of SDS asci can be used to calculate<br /> the map distance between the centromere and the gene of<br /> interest. To understand why this is possible, let’s consider the relationship<br /> between a crossover site and the centromere. As shown<br /> in Figure 5.15, a crossover will separate a gene from its original<br /> centromere only if it begins in the region between the centromere<br /> and that gene. Therefore, the chances of getting a 2:2:2:2 or<br /> 2:4:2 pattern depend on the distance between the gene of interest<br /> and the centromere.<br /> To determine the map distance between the centromere<br /> and a gene, the experimenter must count the number of SDS asci<br /> and the total number of asci. In SDS asci, only half of the spores<br /> are actually the product of a crossover. Therefore, the map distance<br /> is calculated as<br /> Map distance 5<br /> (1>2) (Number of SDS asci)<br /> Total number of asci<br /> 3 100<br /> Unordered Tetrad Analysis Can Be Used<br /> to Map Genes in Dihybrid Crosses<br /> Unordered tetrads contain a group of spores that are the product of<br /> meiosis and randomly arranged in an ascus. An experimenter can<br /> conduct a dihybrid cross, remove the spores from each ascus, and<br /> determine the phenotypes of the spores. This analysis can determine<br /> if two genes are linked or assort independently. If two genes are<br /> linked, a tetrad analysis can also be used to compute map distance.<br /> Figure 5.16 illustrates the possible outcomes starting with two<br /> haploid yeast strains. One strain carries the wild-type alleles ura +<br /> and arg + , which are required for uracil and arginine biosynthesis,<br /> respectively. The other strain has defective alleles ura-2 and arg-3;<br /> these result in yeast strains that require uracil and arginine in the<br /> growth medium. A diploid zygote with the genotype ura + ura-2 arg +<br /> arg-3 was produced from the fusion of haploid cells from these two<br /> strains. The diploid cell then proceeds through meiosis to produce<br /> four haploid cells. After the completion of meiosis, three distinct<br /> types of tetrads could be produced. One possibility is that the tetrad<br /> will contain four spores with the parental combinations of alleles.<br /> This ascus is said to have the parental ditype (PD). Alternatively, an<br /> ascus may have two parental cells and two nonparental cells, which<br /> is called a tetratype (T). Finally, an ascus with a nonparental ditype<br /> (NPD) contains four cells with nonparental genotypes.<br /> Centromere<br /> A<br /> A<br /> Centromere<br /> A<br /> A<br /> A<br /> a<br /> A<br /> A<br /> a<br /> A<br /> a<br /> a<br /> a<br /> a<br /> Result: The gene is<br /> separated from its<br /> original centromere.<br /> a<br /> a<br /> Result: The gene is<br /> not separated from its<br /> original centromere.<br /> (a) Crossover begins between centromere<br /> and gene of interest.<br /> (b) Crossover does not begin between<br /> centromere and gene of interest.<br /> FIGURE 5.15 The relationship between a crossover site and the separation of an allele from its original centromere. (a) If a crossover<br /> initially forms between the centromere and the gene of interest, the gene will be separated from its original centromere. (b) If a crossover initiates<br /> outside this region, the gene remains attached to its original centromere.</div> <div>25.3 HERITABILITY 713<br /> EXPERIMENT 25B<br /> Heritability of Dermal Ridge Count<br /> in Human Fingerprints Is Very High<br /> Fingerprints are inherited as a quantitative trait. It has long been<br /> known that identical twins have fingerprints that are very similar,<br /> whereas fraternal twins show considerably less agreement. Galton<br /> was the first researcher to study fingerprint patterns, but this<br /> trait became more amenable to genetic studies in the 1920s, when<br /> Kristine Bonnevie, a Norwegian geneticist, developed a method<br /> for counting the number of ridges within a human fingerprint.<br /> As shown in Figure 25.9, human fingerprints can be categorized<br /> as having an arch, loop, or whorl, or a combination of<br /> these patterns. The primary difference among these patterns is<br /> the number of triple junctions, each known as a triradius (Figure<br /> 25.9b and c). At a triradius, a ridge emanates in three different<br /> Frequency in population<br /> 160<br /> Males<br /> 140<br /> 120<br /> 100<br /> 80<br /> 60<br /> 40<br /> 20<br /> 0<br /> 0 20 40 60 80 100 120 140 160180 200 220 240 260 280 300<br /> Total ridge count<br /> Based on these results, Holt decided to conduct a more<br /> detailed analysis of ridge counts by examining the fingerprint<br /> patterns of a large group of people and their close relatives. In<br /> the experiment of Figure 25.10, the ridge counts for pairs of<br /> related individuals were determined by the method described<br /> in Figure 25.9. The correlation coefficients for ridge counts<br /> were then calculated among the pairs of related or unrelated<br /> individuals. To estimate the narrow-sense heritability, the<br /> directions. An arch has zero triradii, a loop has one, and a whorl<br /> has two. In Bonnevie’s method of counting, a line is drawn from<br /> a triradius to the center of the fingerprint. The ridges that touch<br /> this line are then counted. (Note: The triradius ridge itself is not<br /> counted, and the last ridge is not counted if it forms the center of<br /> the fingerprint.) With this method, one can obtain a ridge count<br /> for all 10 fingers. Bonnevie conducted a study on a small population<br /> and found that ridge count correlations were relatively high<br /> in genetically related individuals.<br /> Sarah Holt, who was also interested in the inheritance of<br /> this quantitative trait, carried out a more exhaustive study of ridge<br /> counts in a British population. As shown in the two graphs below,<br /> in groups of 825 males and 825 females, the ridge count on all 10<br /> fingers varied from 0 to 300, with mean values of approximately<br /> 145 for males (SD � 51.1) and 127 for females (SD � 52.5).<br /> 160<br /> 140<br /> 120<br /> 100<br /> 80<br /> 60<br /> 40<br /> 20<br /> Females<br /> 0<br /> 0 20 40 60 80 100 120 140 160180 200 220 240 260 280 300<br /> Total ridge count<br /> observed correlations were then divided by the expected correlations<br /> based on the known genetic relationships.<br /> ■ THE HYPOTHESIS<br /> Dermal ridge count has a genetic component. The goal of this<br /> experiment is to determine the contribution of genetics in the<br /> variation of dermal ridge counts.<br /> Triradius<br /> Triradius<br /> (a) Arch (no triradius) (b) Loop (1 triradius) (c) Whorl (2 triradii)<br /> Triradius<br /> FIGURE 25.9 Human fingerprints and the ridge count method of Bonnevie. (a) This print has an arch rather than a triradius. The ridge<br /> count is zero. (b) This print has one triradius. A straight line is drawn from the triradius to the center of the print. The number of ridges dissecting<br /> this straight line is 13. (c) This print has two triradii. Straight lines are drawn from both triradii to the center. There are 16 ridges touching the left line<br /> and 7 touching the right line, giving a total ridge count of 23.</div> <div>12.2 TRANSCRIPTION IN BACTERIA 301<br /> A Promoter Is a Short Sequence of DNA<br /> That Is Necessary to Initiate Transcription<br /> The type of DNA sequence known as the promoter gets its name<br /> from the idea that it “promotes” gene expression. More precisely,<br /> this sequence of bases directs the exact location for the initiation<br /> of RNA transcription. Most of the promoter region is located<br /> just ahead of or upstream from the site where transcription of<br /> a gene actually begins. By convention, the bases in a promoter<br /> sequence are numbered in relation to the transcriptional start<br /> site (Figure 12.4). This site is the first base used as a template for<br /> RNA transcription and is denoted +1. The bases preceding this<br /> site are numbered in a negative direction. No base is numbered<br /> zero. Therefore, most of the promoter region is labeled with negative<br /> numbers that describe the number of bases preceding the<br /> beginning of transcription.<br /> Although the promoter may encompass a region several<br /> dozen nucleotides in length, short sequence elements are particularly<br /> critical for promoter recognition. By comparing the<br /> sequence of DNA bases within many promoters, researchers<br /> have learned that certain sequences of bases are necessary to create<br /> a functional promoter. In many promoters found in E. coli<br /> and similar species, two sequence elements are important. These<br /> are located at approximately the –35 and –10 sites in the promoter<br /> region (Figure 12.4). The sequence at the –35 region is<br /> 5'–TTGACA–3', and the one at the –10 region is 5'–TATAAT–3'.<br /> The TATAAT sequence is sometimes called the Pribnow box after<br /> David Pribnow, who initially discovered it in 1975.<br /> The sequences at the –35 and –10 sites can vary among different<br /> genes. For example, Figure 12.5 illustrates the sequences<br /> found in several different E. coli promoters. The most commonly<br /> occurring bases within a sequence element form the consensus<br /> sequence. This sequence is efficiently recognized by proteins that<br /> Coding strand<br /> Template strand<br /> Promoter region<br /> Transcriptional<br /> start site<br /> –35 sequence 16 –18 bp –10 sequence<br /> 5′<br /> +1<br /> 3′<br /> T T<br /> G A<br /> C A<br /> T<br /> A<br /> T<br /> A A<br /> T<br /> A<br /> A A<br /> C T<br /> G T<br /> A<br /> T<br /> A<br /> T T<br /> A<br /> T<br /> 3′ 5′<br /> 5′ 3′<br /> RNA<br /> A<br /> Transcription<br /> FIGURE 12.4 The conventional numbering system of<br /> promoters. The first nucleotide that acts as a template for transcription<br /> is designated +1. The numbering of nucleotides to the left of this spot is<br /> in a negative direction, while the numbering to the right is in a positive<br /> direction. For example, the nucleotide that is immediately to the left of<br /> the +1 nucleotide is numbered –1, and the nucleotide to the right of<br /> the +1 nucleotide is numbered +2. There is no zero nucleotide in this<br /> numbering system. In many bacterial promoters, sequence elements at<br /> the –35 and –10 regions play a key role in promoting transcription.<br /> lac operon<br /> lacI<br /> trp operon<br /> rrnX<br /> recA<br /> lexA<br /> tRNA tyr<br /> Consensus<br /> –35 region –10 region<br /> Transcribed<br /> initiate transcription. For many bacterial genes, a strong correlation<br /> is found between the maximal rate of RNA transcription<br /> and the degree to which the –35 and –10 regions agree with their<br /> consensus sequences.<br /> Bacterial Transcription Is Initiated<br /> When RNA Polymerase Holoenzyme<br /> Binds at a Promoter Sequence<br /> Thus far, we have considered the DNA sequences that constitute<br /> a functional promoter. Let’s now turn our attention to the proteins<br /> that recognize those sequences and carry out the transcription<br /> process. The enzyme that catalyzes the synthesis of RNA is<br /> RNA polymerase. In E. coli, the core enzyme is composed of five<br /> subunits, a 2 bb'v. The association of a sixth subunit, sigma (S)<br /> factor, with the core enzyme is referred to as RNA polymerase<br /> holoenzyme. The different subunits within the holoenzyme play<br /> distinct functional roles. The two a subunits are important in the<br /> proper assembly of the holoenzyme and in the process of binding<br /> to DNA. The b and b' subunits are also needed for binding<br /> to the DNA and carry out the catalytic synthesis of RNA. The v<br /> (omega) subunit is important for the proper assembly of the core<br /> enzyme. The holoenzyme is required to initiate transcription; the<br /> primary role of s factor is to recognize the promoter. Proteins,<br /> such as s factor, that influence the function of RNA polymerase<br /> are types of transcription factors.<br /> +1<br /> TTTACA N 17 TATGTT N 6 A<br /> GCGCAA N 17 CATGAT N 7 A<br /> TTGACA N 17 TTAACT N 7 A<br /> TTGTCT N 16 TAATAT N 7 A<br /> TTGATA N 16 TATAAT N 7 A<br /> TTCCAA N 17 TATACT N 6 A<br /> TTTACA<br /> TTGACA<br /> N 16 TATGAT N 7<br /> TATAAT<br /> FIGURE 12.5 Examples of –35 and –10 sequences within a<br /> variety of bacterial promoters. This figure shows the –35 and –10<br /> sequences for seven different bacterial and bacteriophage promoters.<br /> The consensus sequence is shown at the bottom. The spacer regions<br /> contain the designated number of nucleotides between the –35 and –10<br /> region or between the –10 region and the transcriptional start site. For<br /> example, N 17 means there are 17 nucleotides between the end of the –35<br /> region and the beginning of the –10 region.<br /> A</div> <div>120 CHAPTER 5 :: LINKAGE AND GENETIC MAPPING IN EUKARYOTES<br /> A<br /> A<br /> A<br /> A<br /> A<br /> A<br /> 4<br /> A<br /> A<br /> A<br /> A<br /> a<br /> Meiosis I<br /> a<br /> Meiosis II<br /> a<br /> Mitosis<br /> a<br /> a<br /> (a) First-division segregation (FDS)<br /> No crossing over produces a 4:4 arrangement.<br /> a<br /> a<br /> a<br /> a<br /> 4<br /> a<br /> A<br /> A<br /> A<br /> A<br /> A<br /> 2<br /> A<br /> a<br /> a<br /> a<br /> Meiosis I Meiosis II Mitosis<br /> A<br /> A<br /> a<br /> a<br /> A<br /> 2<br /> 2<br /> a<br /> a<br /> a<br /> A<br /> a<br /> 2<br /> a<br /> A<br /> A<br /> A<br /> A<br /> A<br /> 2<br /> A<br /> a<br /> a<br /> a<br /> Meiosis I Meiosis II Mitosis<br /> a<br /> a<br /> a<br /> a<br /> a<br /> 4<br /> a<br /> (b) Second-division segregation (SDS)<br /> A single crossover can produce a 2:2:2:2 or 2:4:2 arrangement.<br /> A<br /> A<br /> a<br /> A<br /> A<br /> 2<br /> FIGURE 5.14 A comparison of the arrangement of cells within an ordered octad, depending on whether or not crossing over has<br /> occurred. (a) If no crossing over has occurred, the octad will have a 4:4 arrangement of spores known as an FDS or M1 pattern. (b) If a crossover<br /> has occurred between the centromere and the gene of interest, a 2:2:2:2 or 2:4:2 pattern, known as an SDS or M2 pattern, is observed.</div> <div>5.1 LINKAGE AND CROSSING OVER 105<br /> Of course, a statistical analysis cannot prove that a hypothesis<br /> is true. If the chi square value is high, we accept the linkage<br /> hypothesis because we are assuming that only two explanations<br /> for a genetic outcome are possible: the genes are either linked<br /> or not linked. However, if other factors affect the outcome of<br /> the cross, such as a decreased viability of particular phenotypes,<br /> these may result in large deviations between the observed and<br /> expected values and cause us to reject the independent assortment<br /> hypothesis even though it may be correct.<br /> Let’s consider Morgan’s data concerning body color and<br /> eye color (refer back to page 103). This cross produced the following<br /> offspring: 1,159 gray body, red eyes; 1,017 yellow body,<br /> white eyes; 17 gray body, white eyes; and 12 yellow body, red eyes.<br /> However, when a heterozygous female (X y+w+ X yw ) is crossed to a<br /> hemizygous male (X yw Y), the laws of segregation and independent<br /> assortment predict the following outcome:<br /> F 1 female gametes<br /> F 1 male gametes<br /> X yw<br /> X y+ w + X yw X y+ w + Y<br /> X y+ w +<br /> Gray body, Gray body,<br /> red eyes red eyes<br /> X y+w X yw X y+w Y<br /> X y+ w<br /> Gray body, Gray body,<br /> white eyes white eyes<br /> X yw+ X yw X yw+ Y<br /> X yw+<br /> Yellow body, Yellow body,<br /> red eyes red eyes<br /> X yw X yw X yw Y<br /> X yw Yellow body, Yellow body,<br /> white eyes white eyes<br /> Mendel’s laws predict a 1:1:1:1 ratio among the four phenotypes.<br /> The observed data obviously seem to conflict with this expected<br /> outcome. Nevertheless, we stick to the strategy just discussed. We<br /> begin with the hypothesis that the two genes are not linked, and<br /> then we conduct a chi square analysis to see if the data fit this<br /> hypothesis. If the data do not fit, we will reject the idea that the<br /> genes assort independently and conclude the genes are linked.<br /> A step-by-step outline for applying the chi square test to<br /> distinguish between linkage and independent assortment is<br /> described next.<br /> Step 1. Propose a hypothesis. Even though the observed data<br /> appear inconsistent with this hypothesis, we propose that<br /> the two genes for eye color and body color obey Mendel’s<br /> law of indepen dent assortment. This hypothesis allows<br /> us to calculate expected values. Because the data seem<br /> Y<br /> to conflict with this hypothesis, we actually anticipate<br /> that the chi square analysis will allow us to reject the<br /> independent assortment hypothesis in favor of a linkage<br /> hypothesis. We are also assuming the alleles follow the<br /> law of segregation, and the four phenotypes are equally<br /> viable.<br /> Step 2. Based on the hypothesis, calculate the expected values of<br /> each of the four phenotypes. Each phenotype has an equal<br /> probability of occurring (see the Punnett square given<br /> previously). Therefore, the probability of each phenotype<br /> is 1/4. The observed F 2 generation had a total of 2,205<br /> individuals. Our next step is to calculate the expected<br /> numbers of offspring with each phenotype when the total<br /> equals 2,205; 1/4 of the offspring should be each of the<br /> four phenotypes:<br /> 1/4 × 2,205 = 551 (expected number of each phenotype,<br /> rounded to the nearest whole number)<br /> Step 3. Apply the chi square formula, using the data for the<br /> observed values (O) and the expected values (E) that have<br /> been calculated in step 2. In this case, the data consist of<br /> four phenotypes.<br /> x 2 5 (O 1 2 E 1 ) 2<br /> E 1<br /> 1 (O 2 2 E 2 ) 2<br /> E 2<br /> 1 (O 3 2 E 3 ) 2<br /> E 3<br /> 1 (O 4 2 E 4 ) 2<br /> x 2 5<br /> 1<br /> (1,159 2 551)2<br /> 551<br /> (12 2 551)2<br /> 551<br /> 1<br /> 1<br /> (17 2 551)2<br /> 551<br /> (1,017 2 551)2<br /> 551<br /> x 2 5 670.9 1 517.5 1 527.3 1 394.1 5 2,109.8<br /> Step 4. Interpret the calculated chi square value. This is done with<br /> a chi square table, as discussed in Chapter 2. The four phenotypes<br /> are based on the law of segregation and the law of<br /> independent assortment. By itself, the law of independent<br /> assortment predicts only two categories, recombinant and<br /> nonrecombinant. Therefore, based on a hypothesis of<br /> independent assortment, the degrees of freedom equals<br /> n�1, which is 2�1, or 1.<br /> The calculated chi square value is enormous! This means<br /> that the deviation between observed and expected values is very<br /> large. With one degree of freedom, such a large deviation is<br /> expected to occur by chance alone less than 1% of the time (see<br /> Table 2.1). Therefore, we reject the hypothesis that the two genes<br /> assort independently. As an alternative, we accept the hypothesis<br /> that the genes are linked.<br /> E 4</div> <div>SOLVED PROBLEMS 95<br /> Answer: The inheritance pattern for this trait is sex-influenced<br /> inheritance. The M allele is dominant in males but recessive in females,<br /> while the m allele is dominant in females but recessive in males.<br /> S3. For the following pedigree involving a single gene causing an<br /> inherited disease, indicate which modes of inheritance are not<br /> possible. (Affected individuals are shown as filled symbols.)<br /> I-1 I-2<br /> II-1 II-2 II-3 II-4<br /> III-1 III-2 III-3 III-4<br /> IV-1 IV-2 IV-3<br /> A. Recessive<br /> B. Dominant<br /> C. X linked, recessive<br /> D. Sex influenced, dominant in females<br /> E. Sex limited, recessive in females<br /> Answer:<br /> A. It could be recessive.<br /> B. It is probably not dominant unless it is incompletely penetrant.<br /> C. It could not be X-linked recessive because individual IV-2 does<br /> not have an affected father.<br /> D. It could not be sex influenced, dominant in females because<br /> individual II-3 (who would have to be homozygous) has an<br /> unaffected mother (who would have to be heterozygous and<br /> affected).<br /> E. It is not sex limited because individual II-3 is an affected male<br /> and IV-2 is an affected female.<br /> S4. Red-green color blindness is inherited as a recessive X-linked trait.<br /> What are the following probabilities?<br /> A. A woman with phenotypically normal parents and a color-blind<br /> brother will have a color-blind son. Assume that she has no<br /> previous children.<br /> B. The next child of a phenotypically normal woman, who has<br /> already had one color-blind son, will be a color-blind son.<br /> C. The next child of a phenotypically normal woman, who has<br /> already had one color-blind son, and who is married to a<br /> color-blind man, will have a color-blind daughter.<br /> Answer:<br /> A. The woman’s mother must have been a heterozygote. So there is<br /> a 50% chance that the woman is a carrier. If she has children, 1/4<br /> (i.e., 25%) will be affected sons if she is a carrier. However, there<br /> is only a 50% chance that she is a carrier. We multiply 50% times<br /> 25%, which equals 0.5 � 0.25 = 0.125, or a 12.5% chance.<br /> B. If she already had a color-blind son, then we know she must be<br /> a carrier, so the chance is 25%.<br /> C. The woman is heterozygous and her husband is hemizygous<br /> for the color-blind allele. This couple will produce 1/4<br /> offspring that are color-blind daughters. The rest are 1/4<br /> carrier daughters, 1/4 normal sons, and 1/4 color-blind sons.<br /> Answer is 25%.<br /> S5. Pattern baldness is an example of a sex-influenced trait that is<br /> dominant in males and recessive in females. A couple, neither of<br /> whom is bald, produced a bald son. What are the genotypes of the<br /> parents?<br /> Answer: Because the father is not bald, we know he must be<br /> homozygous, bb. Otherwise, he would be bald. A female who is not bald<br /> can be either Bb or bb. Because she has produced a bald son, we know<br /> that she must be Bb in order to pass the B allele to her son.<br /> S6. Two pink-flowered four-o’clocks were crossed to each other. What<br /> are the following probabilities for the offspring?<br /> A. A red-flowered plant.<br /> B. The first three plants examined will be white.<br /> C. A plant will be either white or pink.<br /> D. A group of six plants contain one pink, two whites, and three<br /> reds.<br /> Answer: The first thing we need to do is construct a Punnett square to<br /> determine the individual probabilities for each type of offspring.<br /> Because flower color is incompletely dominant, the cross is Rr × Rr.<br /> R<br /> r<br /> R<br /> RR<br /> Red<br /> Rr<br /> Pink<br /> r<br /> Rr<br /> Pink<br /> rr<br /> White<br /> The phenotypic ratio is 1 red : 2 pink : 1 white. In other words, 1/4 are<br /> expected to be red, 1/2 pink, and 1/4 white.<br /> A. The probability of a red-flowered plant is 1/4, which equals<br /> 25%.<br /> B. Use the product rule.<br /> 1/4 × 1/4 × 1/4 = 1/64 = 1.6%<br /> C. Use the sum rule because these are mutually exclusive events. A<br /> given plant cannot be both white and pink.<br /> 1/4 + 1/2 = 3/4 = 75%<br /> D. Use the multinomial expansion equation. See solved problem S7<br /> in Chapter 2 for an explanation of the multinomial expansion<br /> equation. In this case, three phenotypes are possible.</div> <div>686 CHAPTER 24 :: POPULATION GENETICS<br /> Gene 1<br /> Exon 1 Intron 1 Exon 2 Intron 2 Exon 3 Intron 3<br /> Protein 1<br /> Domain 1<br /> 3<br /> 2<br /> Gene 2<br /> Exon 1 Intron 1 Exon 2 Intron 2 Exon 3 Intron 3<br /> Protein 2<br /> 1<br /> 2<br /> 3<br /> Gene 1<br /> A segment of gene 1, including<br /> exon 2 and parts of the flanking<br /> introns, is inserted into gene 2.<br /> Exon 1 Exon 3<br /> Protein 1<br /> 1<br /> 3<br /> Domain 2 is missing. Natural<br /> selection may eliminate this<br /> gene from the population if it<br /> is no longer functional.<br /> Gene 2<br /> Exon 1 Exon 2<br /> Gene 1 is missing exon 2.<br /> Exon 2 Exon 3<br /> Gene 2 has exon 2 from gene 1.<br /> Protein 2<br /> 1<br /> 2<br /> 2<br /> Domain 2 from gene 1 has<br /> been inserted into gene 2. If<br /> this protein provides a new,<br /> beneficial trait, natural<br /> selection may increase its<br /> prevalence in a poulation.<br /> 3<br /> FIGURE 24.19 The process of exon shuffling. In this example, a segment of one gene containing an exon and its flanking introns has been<br /> inserted into another gene. A rare, abnormal crossing over event called nonhomologous recombination may cause this to happen. This results in<br /> proteins that have new combinations of domains and possibly new functions.<br /> Eukaryotic<br /> cell<br /> Bacterial<br /> cell<br /> Bacterial<br /> chromosome<br /> Bacterial gene<br /> Endocytic<br /> vesicle<br /> Gene<br /> transfer<br /> FIGURE 24.20 Horizontal gene transfer from a<br /> bacterium to a eukaryote. In this example, a bacterium<br /> is engulfed by a eukaryotic cell, and a bacterial gene is<br /> transferred to one of the eukaryotic chromosomes.<br /> E. coli and Salmonella typhimurium, roughly 17% of their genes<br /> have been acquired via horizontal gene transfer during the past<br /> 100 million years. The roles of these acquired genes are quite<br /> varied, though they commonly involve functions that are readily<br /> acted upon by natural selection. These include genes that confer<br /> antibiotic resistance, the ability to degrade toxic compounds, and<br /> pathogenicity (the ability to cause disease).<br /> Genetic Variation Is Produced Via Changes<br /> in Repetitive Sequences<br /> Another source of genetic variation comes from changes in<br /> repetitive sequences—short sequences typically a few base pairs<br /> to a few thousand base pairs that are repeated many times within a<br /> species’ genome. Repetitive sequences usually come from two types<br /> of sources. First, transposable elements are genetic sequences that<br /> can move from place to place in a species’ genome (see Chapter<br /> 17). The prevalence and movement of transposable elements provides<br /> a great deal of genetic variation between species and within a<br /> single species. In certain eukaryotic species, transposable elements<br /> have become fairly abundant (see Table 17.3).<br /> A second type of repetitive sequence is nonmobile and<br /> involves short sequences that are tandemly repeated. In a<br /> mi crosatellite (also called short tandem repeats, STRs), the<br /> repeat unit is usually 1 to 6 bp, and the whole tandem repeat is<br /> less than a couple hundred bp in length. For example, the most<br /> common microsatellite encountered in humans is a sequence<br /> (CA) N , where N may range from 5 to more than 50. In other<br /> words, this dinucleotide sequence can be tandemly repeated 5 to<br /> 50 or more times. The (CA) N microsatellite is found, on average,<br /> about every 10,000 bases in the human genome. In a minisatellite,<br /> the repeat unit is typically 6 to 80 bp in length, and the size<br /> of the minisatellite ranges from 1 kbp to 20 kbp. An example of<br /> a minisatellite in humans is telomeric DNA. In a human sperm<br /> cell, for example, the repeat unit is 6 bp and the size of a telomere<br /> is about 15 kbp. (Note: Tandem repeat sequences are called<br /> satellites because they sediment away from the rest of the chromosomal<br /> DNA during equilibrium gradient centrifugation.)</div> <div>SOLUTIONS TO EVEN-NUMBERED<br /> PROBLEMS<br /> ::<br /> B<br /> CHAPTER 1<br /> Conceptual Questions<br /> C2. A chromosome is a very long polymer of DNA. A gene is a specific<br /> sequence of DNA within that polymer; the sequence of bases creates<br /> a gene and distinguishes it from other genes. Genes are located in<br /> chromosomes, which are found within living cells.<br /> C4. At the molecular level, a gene (a sequence of DNA) is first transcribed<br /> into RNA. The genetic code within the RNA is used to synthesize a<br /> protein with a particular amino acid sequence. This second process is<br /> called translation.<br /> C6. Genetic variation involves the occurrence of genetic differences<br /> within members of the same species or different species. Within any<br /> population, variation may occur in the genetic material. Variation may<br /> occur in particular genes so that some individuals carry one allele<br /> and other individuals carry a different allele. An example would be<br /> differences in coat color among mammals. There also may be variation<br /> in chromosome structure and number. In plants, differences in<br /> chromosome number can affect disease resistance.<br /> C8. You could pick almost any trait. For example, flower color in petunias<br /> would be an interesting choice. Some petunias are red and others are<br /> purple. There must be different alleles in a flower color gene that affect<br /> this trait in petunias. In addition, the amount of sunlight, fertilizer,<br /> and water also affects the intensity of flower color.<br /> C10. A DNA sequence is a sequence of nucleotides. Each nucleotide may<br /> have one of four different bases (i.e., A, T, G, or C). When we speak of<br /> a DNA sequence, we focus on the sequence of bases.<br /> C12. A. A gene is a segment of DNA. For most genes, the expression of<br /> the gene results in the production of a functional protein. The<br /> functioning of proteins within living cells affects the traits of an<br /> organism.<br /> B. A gene is a segment of DNA that usually encodes the information<br /> for the production of a specific protein. Genes are found within<br /> chromosomes. Many genes are found within a single chromosome.<br /> C. An allele is an alternative version of a particular gene. For example,<br /> suppose a plant has a flower color gene. One allele could produce<br /> a white flower, while a different allele could produce an orange<br /> flower. The white allele and orange allele are alleles of the flower<br /> color gene.<br /> D. A DNA sequence is a sequence of nucleotides. The information<br /> within a DNA sequence (which is transcribed into an RNA<br /> sequence) specifies the amino acid sequence within a protein.<br /> C14. A. How genes and traits are transmitted from parents to offspring<br /> B. How the genetic material functions at the molecular and cellular<br /> levels<br /> C. Why genetic variation exists in populations, and how it changes<br /> over the course of many generations<br /> 766<br /> Experimental Questions<br /> E2. This would be used primarily by molecular geneticists. The sequence<br /> of DNA is a molecular characteristic of DNA. In addition, as we will<br /> learn throughout this textbook, the sequence of DNA is interesting to<br /> transmission and population geneticists as well.<br /> E4. A. Transmission geneticists. Dog breeders are interested in how<br /> genetic crosses affect the traits of dogs.<br /> B. Molecular geneticists. This is a good model organism to study<br /> genetics at the molecular level.<br /> C. Both transmission geneticists and molecular geneticists. Fruit flies<br /> are easy to cross and study the transmission of genes and traits<br /> from parents to offspring. Molecular geneticists have also studied<br /> many genes in fruit flies to see how they function at the molecular<br /> level.<br /> D. Population geneticists. Most wild animals and plants would be<br /> the subject of population geneticists. In the wild, you cannot<br /> make controlled crosses. But you can study genetic variation<br /> within populations and try to understand its relationship to the<br /> environment.<br /> E. Transmission geneticists. Agricultural breeders are interested in<br /> how genetic crosses affect the outcome of traits.<br /> CHAPTER 2<br /> Conceptual Questions<br /> C2. In the case of plants, cross-fertilization occurs when the pollen and<br /> eggs come from different plants, whereas in self-fertilization, they<br /> come from the same plant.<br /> C4. A homozygote that has two copies of the same allele<br /> C6. Diploid organisms contain two copies of each type of gene. When they<br /> make gametes, only one copy of each gene is found in a gamete. Two<br /> alleles cannot stay together within the same gamete.<br /> C8. Genotypes: 1:1 Tt and tt<br /> Phenotypes: 1:1 Tall and dwarf<br /> C10. c is the recessive allele for constricted pods; Y is the dominant allele<br /> for yellow color. The cross is ccYy � CcYy. Follow the directions for<br /> setting up a Punnett square, as described in Chapter 2. The genotypic<br /> ratio is 2 CcYY : 4 CcYy : 2 Ccyy : 2 ccYY : 4 ccYy : 2 ccyy. This<br /> 2:4:2:2:4:2 ratio could be reduced to a 1:2:1:1:2:1 ratio.<br /> The phenotypic ratio is 6 smooth pods, yellow seeds : 2 smooth pods,<br /> green seeds : 6 constricted pods, yellow seeds : 2 constricted pods,<br /> green seeds. This 6:2:6:2 ratio could be reduced to a 3:1:3:1 ratio.<br /> C12. Offspring with a nonparental phenotype are consistent with<br /> the idea of independent assortment. If two different traits were<br /> always transmitted together as unit, it would not be possible to get<br /> nonparental phenotypic combinations. For example, if a true-breeding<br /> parent had two dominant traits and was crossed to a true-breeding<br /> parent having the two recessive traits, the F 2 generation could not have<br /> offspring with one recessive and one dominant phenotype. However,<br /> because independent assortment can occur, it is possible for F 2<br /> offspring to have one dominant and one recessive trait.</div> <div>476 CHAPTER 17 :: RECOMBINATION AND TRANSPOSITION AT THE MOLECULAR LEVEL<br /> Chromosomal<br /> DNA<br /> Red-eyed flies carrying<br /> the copia element<br /> x<br /> Rare white-eyed fly<br /> Isolate chromosomal DNA.<br /> Eye color<br /> gene<br /> On rare occasions, the copia element may<br /> transpose into the X-linked eye color gene,<br /> thereby inactivating its function.<br /> copia<br /> transposon<br /> Eye color<br /> gene<br /> Chromosomal DNA is isolated from such<br /> white-eyed flies, digested into fragments,<br /> and then inserted into viral vectors to create<br /> a DNA library. Only 4 clones in the library<br /> are shown here.<br /> The researchers reasoned that the white-eye phenotype could be<br /> due to the insertion of the copia element into the wild-type gene,<br /> thereby inactivating it.<br /> To clone the eye color gene, chromosomal DNA from this<br /> white-eyed strain was isolated, digested with restriction enzymes,<br /> and cloned into vectors that were derived from a naturally occurring<br /> virus (viral vectors). This procedure creates a DNA library, a<br /> collection of vectors that contain different pieces of chromosomal<br /> DNA (described in Chapter 18). If a transposon had “jumped”<br /> into the eye color gene, vectors that contain this gene will also<br /> contain the transposon sequence. In other words, the presence of<br /> the transposon tags the eye color gene. Each vector is introduced<br /> into a bacterial cell, which produces a virus and ultimately a viral<br /> plaque, a clear area where the bacteria have been lysed. A radiolabeled<br /> fragment of DNA that is complementary to the transposon<br /> sequence can be used as a probe to identify plaques that also contain<br /> the eye color gene. The method of using a probe to screen a<br /> DNA library is also described in Chapter 18. In the example of<br /> Figure 17.18, the method of transposon tagging was successful at<br /> cloning an eye color gene in Drosophila.<br /> A<br /> B<br /> Vector<br /> Vector<br /> C<br /> D<br /> Vector<br /> Vector<br /> Each vector is introduced into a bacterial<br /> cell, which produces virus and ultimately<br /> a viral plaque.<br /> A<br /> C<br /> B<br /> D<br /> Viral plaque<br /> X-ray film<br /> The plaques are lifted onto a filter, probed<br /> with a radiolabeled DNA fragment that is<br /> complementary to the copia element, and<br /> then exposed to X-ray film.<br /> Viral plaque with<br /> the copia element<br /> FIGURE 17.18 The procedure of transposon tagging.<br /> GenesgTraits A white-eyed fly may occur due to the insertion of a transposable<br /> element into a gene that confers red eye color. As discussed in Chapter 13, the wildtype<br /> eye color gene encodes a protein that is necessary for red pigment production.<br /> When a TE inserts into this gene, it disrupts the coding sequence and thereby causes<br /> the gene to produce a nonfunctional protein. Therefore, no red pigment can be made,<br /> and a white-eye phenotype results. In many cases, transposons affect the phenotypes<br /> of organisms by inactivating individual genes.</div> <div>GENETIC TRANSFER<br /> AND MAPPING IN<br /> BACTERIA AND<br /> BACTERIOPHAGES<br /> 6<br /> ::<br /> One reason researchers are so interested in bacteria and viruses is<br /> related to their impact on health. Infectious diseases caused by these agents are<br /> a leading cause of human death, accounting for a quarter to a third of deaths<br /> worldwide. The spread of infectious diseases results from human behavior, and<br /> in recent times has been accelerated by increased trade and travel, and the inappropriate<br /> use of antibiotic drugs. Although the incidence of fatal infectious diseases<br /> in the United States is relatively low compared to the worldwide average,<br /> an alarming increase in more deadly strains of bacteria and viruses has occurred<br /> over the past few decades. Since 1980, the number of deaths in the United States<br /> due to infectious diseases has approximately doubled.<br /> Thus far, our attention in Part II of this textbook has focused on genetic<br /> analyses of eukaryotic species such as fungi, plants, and animals. As we have seen,<br /> these organisms are amenable to genetic studies for two reasons. First, allelic differences,<br /> such as white versus red eyes in Drosophila and tall versus dwarf pea<br /> plants, provide readily discernible traits among different individuals. Second,<br /> because most eukaryotic species reproduce sexually, crosses can be made, and<br /> the pattern of transmission of traits from parent to offspring can be analyzed.<br /> The ability to follow allelic differences in a genetic cross is a basic tool in the<br /> genetic examination of eukaryotic species.<br /> In Chapter 6, we turn our attention to the genetic analysis of bacteria.<br /> Like their eukaryotic counterparts, bacteria often possess allelic differences<br /> that affect their cellular traits. Common allelic variations among bacteria that<br /> are readily discernible involve traits such as sensitivity to antibiotics and differences<br /> in their nutrient requirements for growth. In these cases, the allelic differences<br /> are between different strains of bacteria, because any given bacterium<br /> is usually haploid for a particular gene. In fact, the haploid nature of bacteria<br /> is one advantage that makes it easier to identify mutations that produce phenotypes<br /> such as altered nutritional requirements. Loss-of-function mutations,<br /> which are often recessive in diploid eukaryotes, are not masked by dominant<br /> alleles in haploid species. Throughout this chapter, we will consider interesting<br /> experiments that examine bacterial strains with allelic differences.<br /> Compared to eukaryotes, another striking difference in prokaryotic species<br /> is their mode of reproduction. Because bacteria reproduce asexually, researchers<br /> do not use crosses in the genetic analysis of bacterial species. Instead, they rely on<br /> a similar mechanism, called genetic transfer, in which a segment of bacterial DNA<br /> is transferred from one bacterium to another. In the first part of this chapter,<br /> CHAPTER OUTLINE<br /> 6.1 Genetic Transfer and Mapping<br /> in Bacteria<br /> 6.2 Intragenic Mapping in Bacteriophages<br /> Conjugating bacteria. The bacteria shown here<br /> are transferring genetic material by a process<br /> called conjugation.</div> <div>24.3 SOURCES OF NEW GENETIC VARIATION 687<br /> Tandem repetitive sequences such as microsatellites and<br /> minisatellites tend to undergo mutation in which the number<br /> of tandem repeats changes. For example, a microsatellite with a<br /> 4 bp repeat unit and a length of 64 bp may undergo a mutation<br /> that adds three more repeat units and become 76 bp long. The<br /> mechanism whereby micro- and minisatellites change in length<br /> is not well understood, but it may involve errors in DNA replication<br /> and homologous recombination.<br /> Because repetitive sequences tend to vary within a population,<br /> they have become a common tool that geneticists use in a<br /> variety of ways. For example, as described in Chapters 20 and 22,<br /> microsatellites can be used as molecular markers to map the locations<br /> of genes (see Figure 22.5). Likewise, population geneticists<br /> analyze microsatellites or minisatellites to study variation at the<br /> population level and to determine the relationships among individuals<br /> and neighboring populations. The sizes of microsatellites<br /> and minisatellites found in closely related individuals tend to be<br /> more similar compared to unrelated individuals. As described<br /> next, this phenomenon is the basis for DNA fingerprinting.<br /> FIGURE 24.21 A comparison of two DNA fingerprints. The<br /> chromosomal DNA from two different individuals (Suspect 1—S1, and<br /> Suspect 2—S2) was subjected to DNA fingerprinting. The DNA evidence<br /> at a crime scene, E(vs), was also subjected to DNA fingerprinting.<br /> Following the hybridization of a radiolabeled probe, the DNA appears as<br /> a series of bands on a gel. The dissimilarity in the pattern of these bands<br /> distinguishes different individuals, much as the differences in physical<br /> fingerprint patterns can be used for identification. As seen here, S2<br /> matches the DNA found at the crime scene.<br /> DNA Fingerprinting Is Used for Identification<br /> and Relationship Testing<br /> The technique of DNA fingerprinting, also known as DNA profiling,<br /> analyzes individuals based on the occurrence of repetitive<br /> sequences in their genome. When subjected to traditional DNA fingerprinting,<br /> the chromosomal DNA gives rise to a series of bands<br /> on a gel (Figure 24.21). The sizes and order of bands is an individual’s<br /> DNA fingerprint. Like the human fingerprint, the DNA of<br /> each individual has a distinctive pattern. It is the unique pattern of<br /> these bands that makes it possible to distinguish individuals.<br /> A comparison of the DNA fingerprints among different<br /> individuals has found two applications. First, DNA fingerprinting<br /> can be used as a method of identification. In forensics, DNA<br /> fingerprinting can identify a crime suspect. In medicine, the<br /> technique can identify the type of bacterium that is causing an<br /> infection in a particular patient. A second use of DNA fingerprinting<br /> is relationship testing. Closely related individuals have<br /> more similar fingerprints compared to distantly related ones (see<br /> solved problem S6). In humans, this can be used in paternity<br /> testing. In population genetics, DNA fingerprinting can provide<br /> evidence regarding the degree of relatedness among members of<br /> a population. Such information may help geneticists determine if<br /> a population is likely to be suffering from inbreeding depression.<br /> The development of DNA fingerprinting has relied on the<br /> identification of DNA fragments that are quite variable among<br /> members of a population. This naturally occurring variation<br /> causes each individual to have a unique DNA fingerprint. In the<br /> 1980s, Alec Jeffries and his colleagues found that certain locations<br /> within human chromosomes are particularly variable in<br /> their lengths. These minisatellites tend to vary within the human<br /> population due to changes in the number of tandem repeats at<br /> each minisatellite. For this reason, minisatellites are also referred<br /> to as loci with a variable number of tandem repeats (VNTRs).<br /> The occurrence of minisatellites is shown schematically in<br /> Figure 24.22, where the chromosomal DNA from two individuals<br /> is compared. The diagram emphasizes two features of the<br /> chromosomal DNA. First, it shows the sites recognized by a particular<br /> restriction enzyme (designated with an arrow). As a matter<br /> of chance, these sites are interspersed throughout the genome<br /> of both individuals. Second, minisatellites (designated by their<br /> orange color with a series of repeat units labeled as R’s) are also<br /> found. For the minisatellite shown on the left, the two individuals<br /> have the same number of repeats. For the other two minisatellites,<br /> the number of repeats differs substantially. The variation in<br /> the sizes of minisatellites affects the sizes of fragments produced<br /> when the chromosomal DNAs are digested with the restriction</div> <div>702 CHAPTER 25 :: QUANTITATIVE GENETICS<br /> To illustrate the use of Table 25.2, let’s consider the correlation<br /> we have just calculated for 5-year weights of cows and their<br /> female offspring. In this case, we obtained a value of 0.237 for<br /> r, and the value of N was 10. Under these conditions, df equals<br /> 8. To be valid at a 5% confidence interval, the value of r would<br /> have to be 0.632 or higher. Because the value that we obtained is<br /> much less than this, it is fairly likely that this value could have<br /> occurred as a matter of random sampling error. In this case, we<br /> cannot reject the null hypothesis, and, therefore, we cannot conclude<br /> the positive correlation is due to a true association between<br /> the weights of mothers and offspring.<br /> In an actual experiment, however, a researcher would examine<br /> many more pairs of cows and offspring, perhaps 500 to 1,000. If a<br /> correlation of 0.237 was observed for N � 1,000, the value would be<br /> significant at the 1% level. We would therefore reject the null hypothesis<br /> that weights are not associated with each other. Instead, we<br /> would conclude that a real association occurs between the weights of<br /> mothers and their offspring. In fact, these kinds of experiments have<br /> been done for cattle weights, and the correlations between mothers<br /> and offspring have often been found to be significant.<br /> If a statistically significant correlation is obtained, how do<br /> we interpret its meaning? An r value that is statistically significant<br /> need not imply a cause-and-effect relationship. When parents<br /> and offspring display a significant correlation for a trait, we<br /> should not jump to the conclusion that genetics is the underlying<br /> cause of the positive association. In many cases, parents and<br /> offspring share similar environments, so the positive association<br /> might be rooted in environmental factors. In general, correlations<br /> are quite useful in identifying positive or negative associations<br /> between two variables. We should use caution, however, because<br /> this statistic, by itself, cannot prove that the association is due to<br /> cause and effect. When it has been established that the variables<br /> are related due to cause and effect—that one variable (the independent<br /> variable) affects the outcome of another (the dependent<br /> variable)—researchers may use a regression analysis to predict<br /> how much the dependent variable will change in response to the<br /> independent variable. This approach is described in solved problem<br /> S4 at the end of the chapter.<br /> 25.2 POLYGENIC INHERITANCE<br /> In Section 25.1, we saw that quantitative traits tend to show a<br /> continuum of variation and can be analyzed with various statistical<br /> tools. At the beginning of the 1900s, a great debate focused<br /> on the inheritance of quantitative traits. The biometric school,<br /> founded by Francis Galton and Karl Pearson, argued that these<br /> types of traits are not controlled by discrete genes that affect<br /> phenotypes in a predictable way. To some extent, the biometric<br /> school favored the idea of blending inheritance, which had been<br /> proposed many years earlier (see Chapter 2).<br /> Alternatively, the followers of Mendel, led by William Bateson<br /> in England and William Castle in the United States, held<br /> firmly to the idea that traits are governed by genes, which are<br /> inherited as discrete units. As we know now, Bateson and Castle<br /> were correct. However, as we will see in this section, the difficulty<br /> of studying quantitative traits lies in the fact that these traits<br /> are controlled by multiple genes and substantially influenced by<br /> environmental factors.<br /> Most quantitative traits are polygenic and exhibit a continuum<br /> of phenotypic variation. The term polygenic inheritance<br /> refers to the transmission of a trait governed by two or more different<br /> genes. The location on a chromosome that harbors one<br /> or more genes that affect the outcome of a quantitative trait is<br /> called a quantitative trait locus (QTL). As discussed later, QTLs<br /> are chromosomal regions that are identified by genetic mapping.<br /> Because such mapping usually locates the QTL to a relatively<br /> large chromosomal region, a QTL may contain a single gene or<br /> two or more closely linked genes that affect a quantitative trait.<br /> Just a few years ago, it was extremely difficult for geneticists<br /> to determine the inheritance patterns for genes underlying polygenic<br /> traits, particularly those determined by three or more genes<br /> having multiple alleles for each gene. Recently, however, molecular<br /> genetic tools (described in Chapters 19 and 20) have greatly<br /> enhanced our ability to find regions in the genome where QTLs<br /> are likely to reside. This has been a particularly exciting advance<br /> in the field of quantitative genetics. In some cases, the identification<br /> of QTLs may allow the improvement of quantitative traits in<br /> agriculturally important species.<br /> Polygenic Inheritance and Environmental Factors<br /> Create Overlaps Between Genotypes<br /> and Phenotypes<br /> The first experiment demonstrating that continuous variation is<br /> related to polygenic inheritance was conducted by the Swedish<br /> geneticist Herman Nilsson-Ehle in 1909. He studied the inheritance<br /> of red pigment in the hull of bread wheat, Triticum aestivum<br /> (Figure 25.3a). When true-breeding plants with white hulls<br /> were crossed to a variety with red hulls, the F 1 generation had an<br /> intermediate color. When the F 1 generation was allowed to selffertilize,<br /> great variation in redness was observed in the F 2 generation,<br /> ranging from white, light red, intermediate red, medium<br /> red, and dark red. An unsuspecting observer might conclude that<br /> this F 2 generation displayed a continuous variation in hull color.<br /> However, as shown in Figure 25.3b, Nilsson-Ehle carefully categorized<br /> the colors of the hulls and discovered that they followed<br /> a 1:4:6:4:1 ratio. He concluded that this species is diploid for two<br /> different genes that control hull color, each gene existing in a red<br /> or white allelic form. He hypothesized that these two loci must<br /> contribute additively to the color of the hull; the contribution of<br /> each red allele to the color of the hull is additive.<br /> Later, researchers discovered a third gene that also affects<br /> hull color. The strains that Nilsson-Ehle had used in his original<br /> experiments must have been homozygous for the white allele of<br /> this third gene. It makes sense that wheat would be diploid for<br /> three genes that affect hull color because we now know that T.<br /> aestivum is a hexaploid derived from three closely related diploid<br /> species, as discussed in Chapter 8. Therefore, T. aestivum has six<br /> copies of many genes.<br /> As we have just seen, Nilsson-Ehle categorized wheat hull<br /> colors into several discrete genotypic categories. However, for<br /> many polygenic traits, this is difficult or impossible. In general,<br /> as the number of genes controlling a trait increases and the influ-</div> <div>CHAPTER 7 775<br /> E8. A. If we extrapolate these lines back to the x-axis, the hisE intersects<br /> at about 3 minutes, and the pheA intersects at about 24 minutes.<br /> These are the values for the times of entry. Therefore, the distance<br /> between these two genes is 21 minutes (i.e., 24 minus 3).<br /> B. _____________________________________________<br /> h 4 h 17 h<br /> hisE pabB pheA<br /> E10. One possibility is that you could treat the P1 lysate with DNase I, an<br /> enzyme that digests DNA. (Note: If DNA were digested with DNase<br /> I, the function of any genes within the DNA would be destroyed.) If<br /> the DNA were within a P1 phage, it would be protected from DNase I<br /> digestion. This would allow you to distinguish between transformation<br /> (which would be inhibited by DNase I) versus transduction (which<br /> would not be inhibited by DNase I). Another possibility is that you<br /> could try to fractionate the P1 lysate. Naked DNA would be smaller<br /> than a P1 phage carrying DNA. You could try to filter the lysate to<br /> remove naked DNA, or you could subject the lysate to centrifugation<br /> and remove the lighter fractions that contain naked DNA.<br /> E12. Cotransduction frequency � (1 � d/L) 3<br /> For the normal strain,<br /> Cotransduction frequency � (1 � 0.7 / 2 ) 3 � 0.275, or 27.5%<br /> For the new strain,<br /> Cotransduction frequency � (1 � 0.7 / 5 ) 3 � 0.64, or 64%<br /> The experimental advantage is that you could map genes that are<br /> farther than 2 minutes apart. You could map genes that are up to<br /> 2 minutes apart.<br /> E14. Cotransduction frequency � (1 � d / L ) 3<br /> 0.53 � (1 � d / 2 minutes) 3<br /> 3<br /> (1 � d / 2 minutes) � "0.53<br /> (1 � d / 2 minutes) � 0.81<br /> d � 0.38 minutes<br /> E16. A. We first need to calculate the cotransformation frequency, which<br /> equals 2/70, or 0.029.<br /> Contransformation frequency � (1 � d / L ) 3<br /> 0.029 � (1 � d / 2 minutes) 3<br /> d � 1.4 minutes<br /> B. Contransformation frequency � (1 � d / L ) 3<br /> � (1 � 1.4 / 4 ) 3<br /> � 0.27<br /> As you may have expected, the cotransformation frequency is<br /> much higher when the transformation involves larger pieces of<br /> DNA.<br /> E18. Benzer could use this observation as a way to evaluate if intragenic<br /> recombination had occurred. If two rII mutations recombined to<br /> make a wild-type gene, the phage would produce plaques in this E. coli<br /> K12(l) strain.<br /> E20. rIIA: L47 and L92<br /> rIIB: L33, L40, L51, L62, L65, and L91<br /> E22. Benzer first determined the individual nature of each gene by showing<br /> that mutations within the same gene did not complement each other.<br /> He then could map the distance between two mutations within the<br /> same gene. The map distances defined each gene as a linear, divisible<br /> unit. In this regard, the gene is divisible due to crossing over.<br /> Questions for Student Discussion/Collaboration<br /> 2. Flower color in sweet peas is another example. When two different<br /> genes affect flower color, the white allele of one gene may be epistatic<br /> to the purple allele of another gene. A heterozygote for both genes<br /> would have purple flowers. In other words, the two purple alleles<br /> are complementing the two white alleles. Other similar examples are<br /> discussed in Chapter 4 in the section on Gene Interactions.<br /> CHAPTER 7<br /> Conceptual Questions<br /> C2. A maternal effect gene is one in which the genotype of the mother<br /> determines the phenotype of the offspring. At the cellular level,<br /> this happens because maternal effect genes are expressed in diploid<br /> nurse cells and then the gene products are transported into the egg.<br /> These gene products play key roles in the early steps of embryonic<br /> development.<br /> C4. The genotype of the mother must be bic � bic � . That is why it<br /> produces abnormal offspring. Because the mother is alive and able to<br /> produce offspring, its mother (the maternal grandmother) must have<br /> been bic � bic � and passed the bic � allele to its daughter (the mother<br /> in this problem). The maternal grandfather also must have passed the<br /> bic � allele to its daughter. The maternal grandfather could be either<br /> bic � bic � or bic � bic � .<br /> C6. The mother must be heterozygous. She is phenotypically abnormal<br /> because her mother must have been homozygous for the abnormal<br /> recessive allele. However, because she produces all normal offspring,<br /> she must have inherited the normal dominant allele from her<br /> father. She produces all normal offspring because this is a maternal<br /> effect gene, and the gene product of the normal dominant allele is<br /> transferred to the egg.<br /> C8. Maternal effect genes exert their effects because the gene products<br /> are transferred from nurse cells to eggs. The gene products, mRNA<br /> and proteins, do not last a very long time before they are eventually<br /> degraded. Therefore, they can exert their effects only during early<br /> stages of embryonic development.<br /> C10. Dosage compensation refers to the phenomenon that the level of<br /> expression of genes on the sex chromosomes is the same in males<br /> and females, even though they have different numbers of sex<br /> chromosomes. In many species it seems necessary so that the balance<br /> of gene expression between the autosomes and sex chromosomes is<br /> similar between the two sexes.<br /> C12. In mammals, one of the X chromosomes is inactivated in females; in<br /> Drosophila, the level of transcription on the X chromosome in males is<br /> doubled; in C. elegans, the level of transcription of the X chromosome<br /> in hermaphrodites is decreased by 50% of that of males.<br /> C14. X inactivation begins with the counting of Xics. If there are two<br /> X chromosomes, in the process of initiation, one is targeted for<br /> inactivation. During embryogenesis, this inactivation begins at the<br /> Xic locus and spreads to both ends of the X chromosome until it<br /> becomes a highly condensed Barr body. The Tsix gene may play a role<br /> in the choice of the X chromosome that remains active. The Xist gene,<br /> which is located in the Xic region, remains transcriptionally active<br /> on the inactivated X chromosome. It is thought to play an important<br /> role in X inactivation by coating the inactive X chromosome.<br /> After X inactivation is established, it is maintained in the same X<br /> chromosome in somatic cells during subsequent cell divisions. In<br /> germ cells, however, the X chromosomes are not inactivated, so<br /> an egg can transmit either copy of an active (noncondensed) X<br /> chromosome.<br /> C16. A. One<br /> B. Zero<br /> C. Two<br /> D. Zero<br /> C18. The offspring inherited X B from its mother and X O and Y from its<br /> father. It is an XXY animal, which is male (but somewhat feminized).<br /> C20. Erasure and reestablishment of the imprint occurs during<br /> gametogenesis. It is necessary to erase the imprint because each sex<br /> will transmit either inactive or active alleles of a gene. In somatic cells,<br /> the two alleles for a gene are imprinted according to the sex of the<br /> parent from which the allele was inherited.<br /> C22. A person born with paternal uniparental disomy 15 would have<br /> Angelman syndrome, because this individual would not have an active<br /> copy of the AS gene; the paternally inherited copies of the AS gene are</div> <div>588 CHAPTER 21 :: GENOMICS II: FUNCTIONAL GENOMICS, PROTEOMICS, AND BIOINFORMATICS<br /> By comparing the amino acid sequences and known functions<br /> of proteins in thousands of cases, researchers have also<br /> found amino acid sequence motifs that carry out specialized functions<br /> within proteins. For example, researchers have determined<br /> that the amino acid sequence motif asparagine–X–serine (where<br /> X is any amino acid except proline) within eukaryotic proteins is<br /> a glycosylation site (i.e., it may have a carbohydrate attached to<br /> it). The Prosite database (refer back to Table 21.3) contains a collection<br /> of all amino acid sequence motifs known to be functionally<br /> important. Researchers can use computer programs to determine<br /> whether an amino acid sequence contains any of the motifs<br /> found in the Prosite database. This may help them to understand<br /> the role of a newly found protein of unknown function.<br /> Several Computer-Based Approaches Can Identify<br /> Structural Genes Within a Nucleotide Sequence<br /> A structural gene is composed of nucleotide sequences organized<br /> in a particular way. A typical gene contains a promoter, followed<br /> by a start codon, a coding sequence, a stop codon, and a transcriptional<br /> termination site. In addition, most genes contain regulatory<br /> sequences (e.g., eukaryotic response elements or prokaryotic<br /> operator sites), and eukaryotic genes are likely to contain introns.<br /> After researchers have sequenced a long segment of chromosomal<br /> DNA, they frequently want to know if the sequence contains any<br /> genes. In an attempt to answer this question, geneticists can use<br /> computer programs that are aimed at identifying genes in long<br /> genomic DNA sequences.<br /> How do computer programs identify a gene within a long<br /> genetic sequence? These programs employ different strategies.<br /> A search by signal approach relies on known sequences such as<br /> promoters, start and stop codons, and splice sites to help predict<br /> whether or not a DNA sequence contains a structural gene.<br /> The program tries to locate an organization of known sequence<br /> elements that normally are found within a gene. It would try to<br /> locate a region that contains a promoter sequence, followed by<br /> a start codon, a coding sequence, a stop codon, and a transcriptional<br /> terminator.<br /> A second strategy is a search by content approach. The<br /> goal here is to identify sequences with a nucleotide content that<br /> differs significantly from a random distribution. Within structural<br /> genes, this occurs primarily due to codon usage. Although<br /> there are 64 codons, most organisms display a codon bias within<br /> structural genes. This means that certain codons are used much<br /> more frequently than others. For example, UUA, UUG, CUU,<br /> CUC, CUA, and CUG all specify leucine. In yeast, however, 80%<br /> of the leucine codons are UUG. Codon bias allows organisms to<br /> more efficiently rely on a smaller population of tRNA molecules.<br /> A search by content strategy, therefore, attempts to locate coding<br /> regions by identifying regions where the nucleotide content displays<br /> a known codon bias.<br /> A third way to locate coding regions within a DNA<br /> sequence is to examine translational reading frames. Recall that<br /> the reading frame is a sequence of codons determined by reading<br /> bases in groups of three. In a new DNA sequence, researchers<br /> must consider that the reading of codons (in groups of three<br /> nucleotides) could begin with the first nucleotide (reading frame<br /> 1), the second nucleotide (reading frame 2), or the third nucleotide<br /> (reading frame 3). An open reading frame (ORF) is a region<br /> of a nucleotide sequence that does not contain any stop codons.<br /> Because most proteins are several hundred amino acids in length,<br /> a relatively long reading frame is required to encode them. In<br /> prokaryotic species, long ORFs are contained within the chromosomal<br /> gene sequences. In eukaryotic genes, however, the coding<br /> sequence may be interrupted by introns. As described earlier,<br /> one way to determine eukaryotic ORFs is to clone and sequence<br /> cDNA, which is complementary to mRNA. Alternatively, a computer<br /> program can translate a genomic DNA sequence in all three<br /> reading frames, seeking to identify a long ORF. In Figure 21.8, a<br /> DNA sequence has been translated in all three reading frames.<br /> Only one of the three reading frames (3) contains a very long<br /> open reading frame without any stop codons, suggesting that this<br /> DNA sequence encodes a protein. In Figure 21.8, it was assumed<br /> that the reading frame was from left to right. In an uncharacterized<br /> genetic sequence, however, a reading frame could proceed<br /> from right to left. Therefore, in a newly discovered genetic<br /> sequence, six reading frames are possible—three in the forward<br /> direction and three in the reverse direction.<br /> Even though computer programs are a valuable tool, they are<br /> not always accurate in their prediction of gene sequences. In particular,<br /> it is often difficult for programs to predict the correct start<br /> codon and the precise intron/exon boundaries. In some cases, computer<br /> programs may even suggest that a region encodes a gene when<br /> it does not. Therefore, while a bioinformatic approach is a relatively<br /> easy tool to identify potential genes, it should not be viewed as a<br /> definitive method. The confirmation that a DNA region encodes an<br /> actual gene requires laboratory experimentation to show that it is<br /> truly transcribed into RNA.<br /> DNA sequence<br /> Reading frame<br /> 1<br /> 2<br /> SS SSS S S S<br /> S S S S S S S S S<br /> 3 S<br /> FIGURE 21.8 Translation of a DNA sequence in all three<br /> reading frames. The three lines represent the translation of a gene<br /> sequence in each of three forward reading frames; the reading frames<br /> proceed from left to right. The letter S indicates the location of a stop<br /> codon. Reading frame 3 has a very long open reading frame, suggesting<br /> that it may be the reading frame for a structural gene. Reading frames<br /> 1 and 2 are not likely to be the reading frames for a structural gene,<br /> because they contain many stop codons. During the cloning of DNA,<br /> the orientation of a gene may become flipped so that the coding<br /> sequence is inverted. Therefore, when analyzing many cloned DNA<br /> fragments, six reading frames (i.e., three forward and three reverse) are<br /> evaluated. Only the three forward frames are shown here.</div> <div>154 CHAPTER 6 :: GENETIC TRANSFER AND MAPPING IN BACTERIA AND BACTERIOPHAGES<br /> Deletion<br /> strain<br /> 1272<br /> 1241<br /> J3<br /> PT1<br /> Deleted region<br /> Deleted region<br /> Wild-type recombinants<br /> when coinfected with r103?<br /> Deleted region<br /> Deleted region<br /> No<br /> No<br /> No<br /> No<br /> example, 24 mutations were deletion-mapped to a region called<br /> A5d. Pairwise coinfection experiments were conducted among<br /> this group of 24 mutants to precisely map their locations relative<br /> to each other in the A5d region. Similarly, all mutants in each<br /> of the 46 other groups were mapped by pairwise coinfections.<br /> In this way, a fine structure map was constructed depicting the<br /> locations of hundreds of different rII mutations (Figure 6.20).<br /> As seen in this figure, certain locations contained a relatively<br /> high number of mutations compared to other sites. These were<br /> termed hot spots for mutation.<br /> PB242<br /> A105<br /> 638<br /> Deleted region<br /> Deleted region<br /> Del. reg.<br /> A1 A2 A3 A4 A5 A6 B1–B10<br /> gene rIIA<br /> gene rIIB<br /> FIGURE 6.19 The use of deletion strains to localize rII<br /> mutants to short regions within the rIIA or rIIB gene. The deleted<br /> regions are shown in gray.<br /> Yes<br /> Yes<br /> Yes<br /> Intragenic Mapping Experiments Provided<br /> Insight into the Relationship Between Traits<br /> and Molecular Genetics<br /> Intragenic mapping studies were a pivotal achievement in our<br /> early understanding of gene structure. Since the time of Mendel,<br /> geneticists had considered a gene to be the smallest unit of heredity,<br /> which provided an organism with its inherited traits. In the late<br /> 1950s, however, the molecular nature of the gene was not understood.<br /> Because it is a unit of heredity, some scientists envisioned a<br /> gene as being a particle-like entity that could not be further subdivided<br /> into additional parts. However, intragenic mapping studies<br /> revealed, convincingly, that this is not the case. These studies<br /> Start of<br /> rIIA gene<br /> A4e<br /> A4f<br /> Hot spot<br /> A1a A1b1 A1b2 A2a A2c A2e A2f A2g<br /> A2b A2d<br /> A4d A4c A4b A4a A3i A3h A3g A3f A3e A3a–d<br /> A2h3<br /> Hot spot<br /> A2h1<br /> A2h2<br /> A4g<br /> A5a A5b A5c1 A5c2 A5d A6a1 A6a2 A6b<br /> A6c<br /> B6<br /> B5<br /> B4<br /> B3<br /> B2<br /> B1<br /> A6d<br /> B7<br /> Hot spots<br /> Hot spots<br /> End of<br /> rIIB gene<br /> Start of<br /> rIIB gene<br /> End of<br /> rIIA gene<br /> Hot spot<br /> B8 B9a B9b B10<br /> FIGURE 6.20 The outcome of intragenic mapping of many rII mutations. The blue line represents the linear sequence of the rIIA and<br /> rIIB genes, which are found within the T4 phage’s genetic material. Each small purple box attached to the blue line symbolizes a mutation that was<br /> mapped by intragenic mapping. Among hundreds of independent mutant phages, several mutations sometimes mapped to the same site. In this figure,<br /> mutations at the same site form columns of boxes. Hot spots contain a large number of mutations and are represented as a group of boxes attached to<br /> a column of boxes. A hot spot contains many mutations at the same site within the rIIA or rIIB gene.</div> <div>5.2 GENETIC MAPPING IN PLANTS AND ANIMALS 117<br /> With regard to eye color and wing shape, the recombinant<br /> offspring have red eyes and vestigial wings (61 + 1) or<br /> purple eyes and long wings (2 + 60). The total number<br /> is 124. The map distance between the eye color and wing<br /> shape genes is<br /> Map distance 5<br /> 124<br /> 881 1 124<br /> 3 100 5 12.3 mu<br /> With regard to body color and wing shape, the<br /> recombinant offspring have gray bodies and vestigial wings<br /> (61 + 30) or black bodies and long wings (28 + 60). The<br /> total number is 179. The map distance between the body<br /> color and wing shape genes is<br /> Map distance 5<br /> 179<br /> 826 1 179<br /> 3 100 5 17.8 mu<br /> Step 5. Construct the map. Based on the map unit calculation,<br /> the body color (b) and wing shape (vg) genes are farthest<br /> apart. The eye color gene (pr) must lie in the middle. As<br /> mentioned earlier, this order of genes is also confirmed<br /> by the pattern of traits found in the double crossovers. To<br /> construct the map, we use the distances between the genes<br /> that are closest together.<br /> 12.3<br /> 6.1<br /> b pr vg<br /> In our example, we have placed the body color gene first<br /> and the wing shape gene last. The data also are consistent<br /> with a map in which the wing shape gene comes first<br /> and the body color gene comes last. In detailed genetic<br /> maps, the locations of genes are mapped relative to the<br /> centromere.<br /> You may have noticed that our calculations underestimate<br /> the distance between the body color and wing shape genes. We<br /> obtained a value of 17.8 map units even though the distance<br /> seems to be 18.4 map units when we add together the distance<br /> between body color and eye color genes (6.1 mu) and the distance<br /> between eye color and wing shape genes (12.3 mu). What<br /> accounts for this discrepancy? The answer is double crossovers. If<br /> you look at the data in Table 5.1, the offspring with gray bodies,<br /> purple eyes, and long wings or with black bodies, red eyes, and<br /> vestigial wings are due to a double crossover. From a phenotypic<br /> perspective, these offspring are not recombinant with regard to<br /> the body color and wing shape alleles. Even so, we know that<br /> they arose from a double crossover between these two genes.<br /> Therefore, we should consider these crossovers when calculating<br /> the distance between the body color and wing shape genes. In<br /> this case, three offspring (2 + 1) were due to double crossovers.<br /> Because they are double crossovers, we multiply two times the<br /> number of double crossovers (2 + 1) and add this number to our<br /> previous value of recombinant offspring:<br /> Map distance 5<br /> 179 1 2(2 1 1)<br /> 826 1 179<br /> 3 100 5 18.4 mu<br /> Interference Can Influence the Number<br /> of Double Crossovers That Occur in a Short Region<br /> In Chapter 2, we considered the product rule to determine the<br /> probability that two independent events will both occur. The product<br /> rule allows us to predict the expected likelihood of a double<br /> crossover provided we know the individual probabilities of each<br /> single crossover. Let’s reconsider the data of the trihybrid testcross<br /> just described to see if the frequency of double cross overs is what<br /> we would expect based on the product rule. If each crossover is<br /> an independent event, we can multiply the likelihood of a single<br /> crossover between b and pr (0.061) times the likelihood of a single<br /> crossover between pr and vg (0.123). The product rule predicts<br /> Expected likelihood of a double crossover =<br /> 0.061 × 0.123 = 0.0075 = 0.75%<br /> Based on a total of 1,005 offspring produced:<br /> Expected number of offspring due to a double crossover =<br /> 1,005 × 0.0075 = 7.5<br /> In other words, we would expect about seven or eight offspring<br /> to be produced as a result of a double crossover. The observed<br /> number of offspring was only three (namely, two with gray bodies,<br /> purple eyes, and long wings, and one with a black body, red eyes,<br /> and vestigial wings). What accounts for the lower number? This<br /> lower-than-expected value is probably not due to random sampling<br /> error. Instead, the likely cause is a common genetic phenomenon<br /> known as positive interference, in which the occurrence of a<br /> crossover in one region of a chromosome decreases the probability<br /> that a second crossover will occur nearby. In other words, the first<br /> crossover interferes with the ability to form a second crossover in<br /> the immediate vicinity. To provide interference with a quantitative<br /> value, we first calculate the coefficient of coincidence (C), which<br /> is the ratio of the observed number of double crossovers to the<br /> expected number.<br /> C 5<br /> Interference (I) is expressed as<br /> Observed number of double crossovers<br /> Expected number of double crossovers<br /> I � 1 � C<br /> For the data of the trihybrid testcross, the observed number<br /> of crossovers is 3 and the expected number is 7.5, so the coefficient<br /> of coincidence equals 3/7.5 = 0.40. In other words, only<br /> 40% of the expected number of double crossovers were actually<br /> observed. The value for interference equals 1 � 0.4 = 0.60, or<br /> 60%. This means that 60% of the expected number of crossovers<br /> did not occur. Because I has a positive value, this is called positive<br /> interference. Rarely, the outcome of a testcross yields a negative<br /> value for interference. A negative interference value suggests<br /> that a first crossover enhanced the rate of a second crossover in<br /> a nearby region. Although the molecular mechanisms that cause<br /> interference are not entirely understood, in most organisms the<br /> number of crossovers is regulated so that very few occur per<br /> chromosome. The reasons for positive and negative interference<br /> will require further research.</div> <div>7.2 EPIGENETIC INHERITANCE 171<br /> lgf2 - lgf2 -<br /> (mother’s<br /> genotype)<br /> x<br /> lgf2 lgf2<br /> (father’s<br /> genotype)<br /> lgf2 lgf2<br /> (mother’s<br /> genotype)<br /> x<br /> lgf2 - lgf2 -<br /> (father’s<br /> genotype)<br /> Establishment of the imprint<br /> In this example, imprinting occurs during gametogenesis in the<br /> lgf2 gene, which exists in the lgf2 allele from the male and the<br /> lgf2 - allele from the female. This imprinting occurs so that only<br /> the paternal allele is expressed.<br /> Sperm<br /> Egg<br /> Sperm<br /> Egg<br /> lgf2 lgf2 -<br /> Normal offspring<br /> lgf2 lgf2 -<br /> Dwarf offspring<br /> lgf2<br /> lgf2 -<br /> lgf2<br /> lgf2 -<br /> (Only the lgf2 allele is<br /> expressed in somatic cells<br /> of this heterozygous offspring.)<br /> (Only the lgf2 - allele is<br /> expressed in somatic cells<br /> of this heterozygous offspring.)<br /> Denotes an allele that is silent in the offspring<br /> Denotes an allele that is expressed in the offspring<br /> Maintenance of the imprint<br /> After fertilization, the imprint pattern is<br /> maintained throughout development. In<br /> this example, the maternal lgf2 - allele<br /> will not be expressed in the somatic cells.<br /> Note that the offspring on the left is a<br /> female and the one on the right is a male;<br /> both are normal in size.<br /> lgf2<br /> lgf2 -<br /> Somatic cell<br /> lgf2<br /> lgf2 -<br /> Somatic cell<br /> Erasure and reestablishment<br /> During gametogenesis, the imprint is<br /> erased. The female mouse produces<br /> eggs in which the gene is silenced.<br /> The male produces sperm in which the<br /> gene can be transcribed into mRNA.<br /> FIGURE 7.9 An example of genomic<br /> imprinting in the mouse. In the cross on the left, a<br /> homozygous male with the normal Igf 2 allele is crossed<br /> to a homozygous female carrying a defective allele, designated Igf2 � . An<br /> offspring is heterozygous and normal because the paternal allele is active.<br /> In the reciprocal cross on the right, a homozygous male carrying the<br /> defective allele is crossed to a homozygous normal female. In this case, the<br /> offspring is heterozygous and dwarf. This is because the paternal allele is<br /> defective due to mutation and the maternal allele is not expressed. The<br /> photograph shows normal-size (left) and dwarf littermates (right) derived<br /> from a cross between a wild-type female and a heterozygous male carrying<br /> a loss-of-function Igf2 allele (courtesy of A. Efstratiadis). The loss-offunction<br /> allele was created using gene knockout methods described in<br /> Chapter 19 (see Figure 19.7).<br /> lgf2 lgf2 -<br /> lgf2 lgf2 -<br /> Eggs carry<br /> silenced alleles<br /> Silenced allele<br /> Transcribed allele<br /> lgf2 lgf2 -<br /> lgf2 lgf2 -<br /> Sperm carry<br /> expressed alleles<br /> alleles to her offspring. The male mouse on the right will transmit<br /> transcriptionally active alleles. However, because this male is<br /> a heterozygote, it will transmit either a functionally active Igf 2<br /> allele or a functionally defective mutant allele (Igf 2 � ). An Igf 2 �<br /> allele, which is inherited from a male mouse, can be expressed<br /> into mRNA (i.e., it is transcriptionally active), but it will not produce<br /> a functional Igf 2 protein due to the deleterious mutation<br /> that created the Igf 2 � allele; a dwarf phenotype will result.<br /> As seen in Figure 7.10, genomic imprinting is permanent in<br /> the somatic cells of an animal, but the marking of alleles can be<br /> altered from generation to generation. For example, the female<br /> mouse on the left possesses an active copy of the Igf 2 allele, but<br /> any allele this female transmits to its offspring will be transcriptionally<br /> inactive.<br /> Genomic imprinting occurs in several species, including<br /> numerous insects, plants, and mammals. Imprinting may involve<br /> FIGURE 7.10 Genomic imprinting during gametogenesis.<br /> This example involves a mouse gene Igf 2, which is found in two<br /> alleles designated Igf 2 and Igf2 � . The left side shows a female mouse<br /> that was produced from a sperm carrying the Igf 2 allele and an egg<br /> carrying the Igf2 � allele. In the somatic cells of this female animal,<br /> the Igf 2 allele is active. However, when this female produces eggs,<br /> both alleles are transcriptionally inactive when they are transmitted<br /> to offspring. The right side of this figure shows a male mouse that<br /> was also produced from a sperm carrying the Igf 2 allele and an egg<br /> carrying the Igf2 � allele. In the somatic cells of this male animal, the<br /> Igf 2 allele is active. However, the sperm from this male will contain<br /> either a functionally active Igf 2 allele or a functionally defective<br /> Igf2 � allele. Note: The Igf2 � allele transmitted by the sperm can<br /> be expressed into mRNA in the somatic cells of offspring, but the<br /> expression will not produce a functional Igf 2 protein due to the<br /> mutation that inhibits the protein’s function.</div> <div>472 CHAPTER 17 :: RECOMBINATION AND TRANSPOSITION AT THE MOLECULAR LEVEL<br /> Transposition<br /> IR<br /> TE— A few hundred to several<br /> thousand base pairs in length<br /> 5′ 3′<br /> 3′ 5′<br /> IR<br /> Transposase recognizes the inverted<br /> repeats and cleaves at both ends of<br /> the transposable element, releasing it<br /> from its original site.<br /> 5′ 3′<br /> DNA replication proceeds past the<br /> point where the TE has been<br /> inserted. The top copy of the TE<br /> then transposes ahead of the fork,<br /> where it is copied again.<br /> TE<br /> TE<br /> TE<br /> TE<br /> TE<br /> 3′ 5′<br /> The bottom copy of DNA has 2 TEs.<br /> TE<br /> IR<br /> IR<br /> IR<br /> IR<br /> Transposase<br /> Transposase carries the TE to a new site and<br /> cleaves the target DNA at staggered sites.<br /> 5′ 3′<br /> T<br /> Target<br /> T<br /> A G C T<br /> A<br /> C G A<br /> DNA<br /> 3′ 5′<br /> The transposable element is<br /> inserted into the target site.<br /> 5′ 3′<br /> T<br /> A C G<br /> A<br /> T G C T<br /> A<br /> 3′ 5′<br /> Transposable<br /> element<br /> DNA gap<br /> repair synthesis<br /> 5′ 3′<br /> T<br /> A G C T<br /> A<br /> T<br /> T C G A<br /> A G C T<br /> A<br /> T C G A<br /> 3′ 5′<br /> Transposable<br /> element<br /> Direct<br /> repeats<br /> FIGURE 17.14<br /> IR<br /> IR<br /> Simple transposition.<br /> FIGURE 17.15 Increase in TE copy number via simple<br /> transposition. In this example, a TE that has already been replicated<br /> transposes to a new site that has not yet replicated. Following the<br /> completion of DNA replication, the TE has increased in number.<br /> transposon. The net result of replicative transposition is that a<br /> transposable element occurs at a new site, and a TE also remains<br /> in its original location. Figure 17.16 describes a model for replicative<br /> transposition between two circular DNA molecules. One<br /> DNA molecule already has a TE, whereas the other does not. In<br /> this mechanism, transposase initially makes one cut at each end<br /> of the TE and two cuts in the target DNA. Note that this differs<br /> from simple transposition, in which the transposase makes four<br /> cuts and completely removes the TE from its original site (refer<br /> back to Figure 17.14). In replicative transposition, the TE is left<br /> at its original location.<br /> Following ligation, both the target DNA and the transposable<br /> element have a long gap. DNA gap repair synthesis copies<br /> the target DNA gap as well as the TE. This creates two copies of<br /> the TE within a large circular molecule known as a cointegrant.<br /> The enzyme resolvase catalyzes homologous recombination<br /> within the TEs so that the cointegrant can be resolved into two<br /> separate DNA molecules. One of these molecules contains the TE<br /> in its original location, and the other has a TE at a new location.<br /> Retroelements Use Reverse Transcriptase<br /> and Integrase for Retrotransposition<br /> Thus far, we have considered how DNA elements—transposons—can<br /> move throughout the genome. By comparison, retroelements<br /> use an RNA intermediate in their transposition mechanism.<br /> As shown in Figure 17.17, the movement of retroelements<br /> also requires two key enzymes: reverse transcriptase and integrase.<br /> In this example, the cell already contains a retroelement known<br /> as the Alu sequence within its genome. This retroelement, which</div> <div>110 CHAPTER 5 :: LINKAGE AND GENETIC MAPPING IN EUKARYOTES<br /> X chromosome<br /> composition of<br /> fertilized egg<br /> Normal mitotic divisions<br /> to produce embryo<br /> y +<br /> sn<br /> FIGURE 5.7 Mitotic recombination in Drosophila that<br /> produces twin spotting.<br /> y<br /> sn +<br /> Genes gTraits In this illustration, the genotype of the fertilized egg was y + y sn + sn.<br /> During development, a mitotic crossover occurred in a single embryonic cell. After<br /> mitotic crossing over, the separation of the chromatids occurred, so one embryonic<br /> cell became y + y + snsn and the adjacent sister cell became yy sn + sn + . The embryonic<br /> cells then continued to divide by normal mitosis to produce an adult fly. The cells<br /> in the adult that were derived from the y + y + snsn embryonic cell produced a spot<br /> on the adult body that was gray and had singed bristles. The cells derived from the<br /> yy sn + sn + embryonic cell produced an adjacent spot on the body that was yellow<br /> and had long bristles. In this case, mitotic recombination produced an unusual<br /> trait known as a twin spot. The characteristics of this twin spot differ from the<br /> surrounding tissue, which is gray with long bristles.<br /> Rare mitotic recombination<br /> in one embryonic cell<br /> Sister<br /> chromatids<br /> y +<br /> sn<br /> y +<br /> sn<br /> y y<br /> y +<br /> sn + sn + sn<br /> y<br /> sn +<br /> y +<br /> sn<br /> y<br /> sn +<br /> Embryonic cell<br /> Mitotic<br /> crossover<br /> Subsequent separation<br /> of sister chromatids<br /> y +<br /> sn<br /> y<br /> sn +<br /> y +<br /> sn<br /> y<br /> sn +<br /> Cytokinesis produces 2<br /> adjacent cells<br /> y +<br /> sn<br /> y +<br /> sn<br /> y<br /> sn +<br /> y<br /> sn +<br /> Continued normal mitotic<br /> divisions to produce adult fly<br /> with a twin spot</div> <div>22<br /> MEDICAL GENETICS<br /> AND CANCER<br /> ::<br /> Genetic information is highly personal and unique. Our genes underlie<br /> every aspect of human health, both in function and dysfunction. Obtaining<br /> a detailed understanding of how genes work together and interact with environmental<br /> factors ultimately will allow us to appreciate the differences between the<br /> events in normal cellular processes and those that occur in disease pathogenesis.<br /> Such knowledge will have a profound impact on the way many diseases are<br /> defined, diagnosed, treated, and prevented. Genetic insight is expected to bring<br /> about revolutionary changes in medical practices. In fact, changes are already<br /> beginning. Currently, several hundred genetic tests are in clinical use, with many<br /> more under development. Most of these tests detect mutations associated with<br /> rare genetic disorders that follow Mendelian inheritance patterns. These include<br /> Duchenne muscular dystrophy, cystic fibrosis, sickle-cell disease, and Huntington<br /> disease. In addition, genetic tests are available to detect the predisposition to<br /> develop certain forms of cancer.<br /> Approximately 5,000 genetic diseases are known to afflict people, but this<br /> is almost certainly an underestimate. Given enough time and effort, scientists<br /> can learn how to prevent or treat a great many of them. Most of the genetic disorders<br /> discussed in the first part of this chapter are the direct result of a mutation<br /> in one gene. However, many diseases have a complex pattern of inheritance<br /> involving several genes. These include common medical disorders such as diabetes,<br /> asthma, and mental illness. In these cases, a single mutant gene does not<br /> determine whether a person has a disease. Instead, a number of genes may each<br /> make a subtle contribution to a person’s susceptibility to a disease. Unraveling<br /> these complexities will be a challenge for some time to come. The availability of<br /> the human genome sequence, discussed in Chapter 20, will be of great help.<br /> In this chapter, we will focus our attention on ways that mutant genes contribute<br /> to human disease. In the first part of the chapter, we will explore the<br /> molecular basis of several genetic disorders and their patterns of inheritance. We<br /> will also examine how genetic testing can determine if an individual carries a<br /> defective allele. The second part of the chapter will concern cancer, a disease that<br /> involves the uncontrolled growth of somatic cells. We will examine the underlying<br /> genetic basis for cancer and discuss the roles that many different genes may<br /> play in the development of this disease.<br /> CHAPTER OUTLINE<br /> 22.1 Genetic Analysis of Human Diseases<br /> 22.2 Genetic Basis of Cancer<br /> Cigarette smoking and lung cancer. Cigarette<br /> smoke contains chemicals that are known to<br /> mutate genes in the cells of a person’s lungs,<br /> thereby leading to lung cancer. Lung cancer<br /> remains the top cause of cancer death in the<br /> United States, with 87% of those deaths linked<br /> to smoking.<br /> PART VI<br /> Genetic Analysis of Individuals and Populations</div> <div>606 CHAPTER 22 :: MEDICAL GENETICS AND CANCER<br /> I<br /> II<br /> III<br /> IV<br /> AB<br /> AA<br /> V<br /> AA<br /> AB AB AB<br /> BC AB AB AB AB BC AB AB BC BB BC AC AA BC CD BB BC<br /> VI<br /> AC<br /> AB AC AC AC AC AA BC AA BC AA<br /> BC BC CC<br /> VII<br /> AC<br /> BC<br /> BC<br /> FIGURE 22.5 The transmission pattern of a molecular marker for Huntington disease. Affected members are shown with black symbols. The<br /> letters under each person indicate the forms of the G8 marker (A, B, C, or D) the individual carries. Affected individuals always carry the C version<br /> of this marker. Symbols with slashes indicate deceased individuals. Data from Gusella et al. (1983) “A polymorphic DNA marker genetically linked to<br /> Huntington’s disease.” Nature 306, 234–239.<br /> different genes, though the number can vary greatly. Therefore,<br /> mapping does not definitively tell researchers which gene may<br /> play a role in human disease, but it usually narrows down the list<br /> to a few candidate genes.<br /> How does a researcher determine which of the candidate<br /> genes is the correct one? To further narrow down the list, researchers<br /> may also consider biological function. As the scientific community<br /> explores the functions of genes experimentally, the data are<br /> published in the research literature and placed into databases. In<br /> some cases, this information may help to narrow down the list of<br /> candidate genes. For example, if the disease of interest is neurological,<br /> researchers may discover that only certain genes in the mapped<br /> region are expressed in nerve cells. Also, researchers may compare<br /> data from other organisms. If a mutant gene in a mouse causes<br /> neurological problems and a human homolog of the mouse gene<br /> is found in the mapped region, this human gene would be a good<br /> candidate for being responsible for the human disease symptoms.<br /> After researchers have narrowed down the number of candidate<br /> genes to as short a list as possible, the next phase is to<br /> sequence the candidate gene(s) from many affected and unaffected<br /> individuals, using the dideoxy sequencing method described in<br /> Chapter 18. The goal is to identify a gene in which affected individuals<br /> always carry a mutation. Once a gene has been identified that<br /> has a genetic change found in affected individuals, this is strong<br /> evidence that the candidate gene is causing the disease symptoms.<br /> Why is it useful to identify disease-causing mutant genes?<br /> In many cases, the identification of these genes will help us to<br /> understand how genes contribute to pathogenesis and may even<br /> aid in developing strategies aimed at the treatment of the disease.<br /> As described next, the identification of such genes may also<br /> result in the development of genetic tests that can enable people<br /> to determine if they are carrying disease-causing alleles.<br /> Genetic Testing Can Identify Many<br /> Inherited Human Diseases<br /> Because genetic abnormalities occur in the human population at<br /> a significant level, people have sought ways to determine whether<br /> individuals possess disease-causing alleles. The term genetic testing<br /> refers to the use of testing methods to determine if an individual<br /> carries a genetic abnormality. Table 22.4 describes several<br /> different testing strategies. The term genetic screening refers to<br /> population-wide genetic testing.<br /> In many cases, single-gene mutations that affect the function<br /> of cellular proteins can be examined at the protein level. If<br /> a gene encodes an enzyme, biochemical assays to measure that<br /> enzyme’s activity may be available. As mentioned earlier, Tay-<br /> Sachs disease involves a defect in the enzyme hexosaminidase A<br /> (hexA). Enzymatic assays for this enzyme involve the use of an<br /> artificial substrate in which 4-methylumbelliferone (MU) is covalently<br /> linked to N-acetylglucosamine (GlcNAc). HexA cleaves this<br /> covalent bond and releases MU, which is fluorescent.<br /> hexA<br /> MU–GlcNAc MU + GlcNAc<br /> (nonfluorescent)<br /> (fluorescent)</div> <div>25.3 HERITABILITY 717<br /> Wild mustard plant<br /> FIGURE 25.12 Crop plants developed by selective breeding of<br /> the wild mustard plant (Brassica oleracea).<br /> GenesgTraits The wild mustard plant carries a large amount of genetic (i.e., allelic)<br /> variation, which was used by plant breeders to produce modern strains that are<br /> agriculturally desirable and economically important. For example, by selecting for<br /> alleles that promote the formation of large lateral leaf buds, the strain of Brussels<br /> sprouts was created. By selecting for alleles that alter the leaf morphology, kale<br /> was developed. Although these six agricultural plants look quite different from each<br /> other, they carry many of the same alleles as the wild mustard. However, they differ<br /> in alleles affecting the formation of stems, leaves, flower buds, and leaf buds.<br /> Strain<br /> Modified trait<br /> Kohlrabi<br /> Kale<br /> Broccoli<br /> Brussels sprouts<br /> Cabbage<br /> Cauliflower<br /> Stem<br /> Leaves<br /> Flower buds and stem<br /> Lateral leaf buds<br /> Terminal leaf bud<br /> Flower buds<br /> were divided into two separate groups. In one group, members<br /> with the highest oil content were chosen as parents of the next<br /> generation. In the other group, members with the lowest oil content<br /> were chosen. After 80 generations, the oil content in the first<br /> group rose to over 18%; in the other group, it dropped to less<br /> than 1%. These results show that selective breeding can modify<br /> quantitative traits in a very directed manner.<br /> Similar results have been obtained for many other quantitative<br /> traits. Figure 25.13b shows an experiment by Kenneth<br /> Mather conducted in the 1940s, in which flies were selected on<br /> the basis of their bristle number. The starting group had an average<br /> of 40 bristles for females and 35 bristles for males. After eight<br /> generations, the group selected for high bristle number had an<br /> average of 46 bristles for females and 40 for males, while the<br /> group selected for low bristle number had an average of 36 bristles<br /> for females and 30 for males.<br /> When comparing the curves in Figure 25.13, keep in mind<br /> that quantitative traits are often at an intermediate value in<br /> unselected populations. Therefore, artificial selection can increase<br /> or decrease the magnitude of the trait. Oil content can go up or<br /> down, and bristle number can increase or decrease. Artificial<br /> selection tends to be the most rapid and effective in changing the<br /> frequency of alleles that are at intermediate range in a starting<br /> population, such as 0.2 to 0.8.<br /> Figure 25.13 also shows the phenomenon known as a<br /> selection limit—after several generations a plateau is reached<br /> where artificial selection is no longer effective. A selection limit<br /> may occur for two reasons. Presumably, the starting population<br /> possesses a large amount of genetic variation, which contributes<br /> to the diversity in phenotypes. By carefully choosing the parents,<br /> each succeeding generation has a higher proportion of the desirable<br /> alleles. However, after many generations, the population may<br /> be nearly monomorphic for all or most of the desirable alleles that<br /> affect the trait of interest. At this point, additional selective breeding<br /> will have no effect. When this occurs, the heritability for the<br /> trait is near zero, because nearly all genetic variation for the trait<br /> of interest has been eliminated from the population. Without the<br /> introduction of new mutations into the population, further selection<br /> is not possible. A second reason for a selection limit is related<br /> to fitness. Some alleles that accumulate in a population due to artificial<br /> selection have a negative impact on the population’s overall<br /> fitness. A selection limit is reached in which the desired effects of<br /> artificial selection are balanced by the negative effects on fitness.</div> <div>132 CHAPTER 5 :: LINKAGE AND GENETIC MAPPING IN EUKARYOTES<br /> E26. On chromosome 4 in Neurospora, the allele pyr-1 results in a<br /> pyrimidine requirement for growth. A cross was made between a<br /> pyr-1 and a pyr + (wild-type) strain, and the following results were<br /> obtained:<br /> pyr-1<br /> pyr-1<br /> pyr +<br /> pyr +<br /> pyr-1<br /> pyr-1<br /> pyr-1<br /> pyr-1<br /> pyr +<br /> pyr +<br /> pyr +<br /> pyr +<br /> E27. On chromosome 3 in Neurospora, the pro-1 gene is located<br /> approximately 9.8 mu from the centromere. Let’s suppose a cross<br /> was made between a pro-1 and a pro + strain and 1,000 asci were<br /> analyzed.<br /> A. What are the six types of asci that can be produced?<br /> B. What are the expected numbers of each type of ascus?<br /> pyr +<br /> pyr-1<br /> pyr +<br /> pyr-1<br /> pyr-1<br /> pyr +<br /> pyr +<br /> pyr-1<br /> pyr +<br /> pyr-1<br /> pyr-1<br /> pyr +<br /> pyr +<br /> pyr-1<br /> pyr-1<br /> pyr +<br /> pyr +<br /> pyr-1<br /> pyr +<br /> pyr-1<br /> pyr-1<br /> pyr +<br /> pyr +<br /> pyr-1<br /> pyr-1<br /> pyr +<br /> pyr +<br /> pyr +<br /> pyr-1<br /> pyr-1<br /> pyr-1<br /> pyr +<br /> pyr +<br /> pyr +<br /> pyr-1<br /> pyr-1<br /> Total: 22<br /> 21<br /> 21<br /> 451<br /> 23<br /> 455<br /> What is the distance between the pyr-1 gene and the centromere?<br /> Questions for Student Discussion/Collaboration<br /> 1. In mice, a dominant gene that causes a short tail is located on<br /> chromosome 2. On chromosome 3, a recessive gene causing<br /> droopy ears is 6 mu away from another recessive gene that causes<br /> a flaky tail. A recessive gene that causes a jerker (uncoordinated)<br /> phenotype is located on chromosome 4. A jerker mouse with<br /> droopy ears and a short, flaky tail was crossed to a normal mouse.<br /> All F 1 generation mice were phenotypically normal, except they<br /> had short tails. These F 1 mice were then testcrossed to jerker<br /> mice with droopy ears and long, flaky tails. If this cross produced<br /> 400 offspring, what would be the proportions of the 16 possible<br /> phenotypic categories?<br /> 2. In Chapter 3, we discussed the idea that the X and Y chromosomes<br /> have a few genes in common. These genes are inherited in a<br /> pseudoautosomal pattern. With this phenomenon in mind, discuss<br /> whether or not the X and Y chromosomes are really distinct<br /> linkage groups.<br /> 3. Mendel studied seven traits in pea plants, and the garden pea<br /> happens to have seven different chromosomes. It has been pointed<br /> out that Mendel was very lucky not to have conducted crosses<br /> involving two traits governed by genes that are closely linked on<br /> the same chromosome because the results would have confounded<br /> his theory of independent assortment. It has even been suggested<br /> that Mendel may not have published data involving traits that were<br /> linked! An article by Stig Blixt (“Why Didn’t Gregor Mendel Find<br /> Linkage?” Nature 256:206, 1975) considers this issue. Look up this<br /> article and discuss why Mendel did not find linkage.<br /> Note: All answers appear at the website for this textbook; the answers to<br /> even-numbered questions are in the back of the textbook.<br /> www.mhhe.com/brookergenetics3e<br /> Visit the Online Learning Center for practice tests, answer keys, and other learning aids for this chapter. Enhance your understanding of genetics with<br /> our interactive exercises, quizzes, animations, and much more.</div> <div>288 CHAPTER 11 :: DNA REPLICATION<br /> TABLE 11.3<br /> Examples of ts Mutants Involved in DNA Replication in E. coli<br /> Gene Name<br /> Rapid-Stop Mutants<br /> dnaE<br /> dnaX<br /> dnaN<br /> dnaZ<br /> dnaG<br /> dnaB<br /> Slow-Stop Mutants<br /> dnaA<br /> dnaC<br /> Protein Function<br /> a subunit of DNA polymerase III, synthesizes DNA<br /> t subunit of DNA polymerase III, promotes the<br /> dimerization of two DNA polymerase III proteins<br /> together at the replication fork and stimulates<br /> DNA helicase<br /> b subunit of DNA polymerase III, functions as<br /> a clamp protein that makes DNA polymerase a<br /> processive enzyme<br /> g subunit of DNA polymerase III, helps the b<br /> subunit bind to the DNA<br /> Primase, needed to make RNA primers<br /> Helicase, needed to unwind the DNA strands<br /> during replication<br /> DnaA protein that recognizes the DnaA boxes at<br /> the origin<br /> DnaC protein that recruits DNA helicase to the<br /> origin<br /> genes were purified, and their functions were studied in vitro.<br /> This work contributed greatly to our modern understanding of<br /> DNA replication at the molecular level.<br /> 11.3 EUKARYOTIC DNA REPLICATION<br /> Eukaryotic DNA replication is not as well understood as bacterial<br /> replication. Much research has been carried out on a variety<br /> of experimental organisms, particularly yeast and mammalian<br /> cells. Many of these studies have found extensive similarities<br /> between the general features of DNA replication in prokaryotes<br /> and eukaryotes. For example, DNA helicases, topoisomerases,<br /> single-strand binding proteins, primases, DNA polymerases,<br /> and DNA ligases—the types of bacterial enzymes described in<br /> Table 11.1—have also been identified in eukaryotes. Nevertheless,<br /> at the molecular level, eukaryotic DNA replication appears<br /> to be substantially more complex. These additional intricacies<br /> of eukaryotic DNA replication are related to several features of<br /> eukaryotic cells. In particular, eukaryotic cells have larger, linear<br /> chromosomes, the chromatin is tightly packed within nucleosomes,<br /> and cell cycle regulation is much more complicated. This<br /> section will emphasize some of the unique features of eukaryotic<br /> DNA replication.<br /> Initiation Occurs at Multiple Origins of Replication<br /> on Linear Eukaryotic Chromosomes<br /> Because eukaryotes have long, linear chromosomes, the chromosomes<br /> require multiple origins of replication so that the DNA can<br /> be replicated in a reasonable length of time. In 1968, Joel Huberman<br /> and Arthur Riggs provided evidence for multiple origins of<br /> replication by adding a radiolabeled nucleoside ( 3 H-thymidine)<br /> to a culture of actively dividing cells, followed by a chase with<br /> nonlabeled thymidine. The radiolabeled thymidine was taken up<br /> by the cells and incorporated into their newly made DNA strands<br /> for a brief period. The chromosomes were then isolated from the<br /> cells and subjected to autoradiography. As seen in Figure 11.20,<br /> radiolabeled segments were interspersed among nonlabeled segments.<br /> This result is consistent with the hypothesis that eukaryotic<br /> chromosomes contain multiple origins of replication.<br /> As shown schematically in Figure 11.21a, DNA replication<br /> proceeds bidirectionally from many origins of replication during<br /> S phase of the cell cycle. The multiple replication forks eventually<br /> make contact with each other to complete the replication process.<br /> Figure 11.21b shows a region of eukaryotic DNA in the process<br /> of replication. As DNA replication radiates bidirectionally from<br /> each origin, regions are formed that contain two double helices.<br /> These regions are punctuated by other regions that have not yet<br /> replicated and consist of one double helix.<br /> The molecular features of eukaryotic origins of replication<br /> may have some similarities to the origins found in bacteria. At<br /> the molecular level, eukaryotic origins of replication have been<br /> extensively studied in the yeast Saccharomyces cerevisiae. In this<br /> organism, several replication origins have been identified and<br /> sequenced. They have been named ARS elements (for Autonomously<br /> Replicating Sequence). ARS elements, which are about<br /> 50 bp in length, are necessary to initiate chromosome replication.<br /> ARS elements have unique features of their DNA sequences.<br /> First, they contain a higher percentage of A and T bases than the<br /> rest of the chromosomal DNA. In addition, they contain a copy<br /> of the ARS consensus sequence (ACS), ATTTAT(A or G)TTTA,<br /> along with additional elements that enhance origin function. This<br /> arrangement is similar to bacterial origins. In S. cerevisiae, origins<br /> of replication are determined primarily by their DNA sequences.<br /> 100 μm<br /> Radiolabeled segments<br /> FIGURE 11.20 Evidence for multiple origins of replication in<br /> eukaryotic chromosomes. In this experiment, cells were given a pulse/<br /> chase of 3 H-thymidine and unlabeled thymidine. The chromosomes<br /> were isolated and subjected to autoradiography. In this micrograph,<br /> radiolabeled segments were interspersed among nonlabeled segments,<br /> indicating that eukaryotic chromosomes contain multiple origins of<br /> replication.</div> <div>25.3 HERITABILITY 711<br /> shorter because they cannot take up a sufficient supply of certain<br /> minerals to support maximal growth, while the TT plants may<br /> not be limited by mineral uptake. According to this hypothetical<br /> scenario, adding minerals to the soil would enhance the growth<br /> rate of tt plants by a large amount and the Tt plants by a smaller<br /> amount (Figure 25.8). The height of TT plants would not be<br /> affected by mineral supplementation. When the environmental<br /> effects on phenotype differ according to genotype, this phenomenon<br /> is called a genotype-environment interaction. Variation<br /> due to interactions between genetic and environmental factors is<br /> termed V G�E as noted earlier.<br /> Interactions of genetic and environmental factors are common.<br /> As an actual example, Table 25.3 shows results from a study<br /> conducted in 2000 by Cristina Vieira, Trudy Mackay, and colleagues<br /> in which they investigated genotype-environment interaction<br /> for quantitative trait loci affecting life span in Drosophila<br /> TABLE 25.3<br /> Longevity of Two Strains of Drosophila melanogaster *<br /> Strain A<br /> Strain B<br /> Temperature Male Female Male Female<br /> Standard 33.6 39.5 37.5 28.9<br /> High 36.3 33.9 23.2 28.6<br /> Low 77.5 48.3 45.8 77.0<br /> *Longevity was measured in the mean number of days of survival. Strains A and B were inbred<br /> strains of D. melanogaster called Oregon and 2b, respectively. The standard, high, and low<br /> temperature conditions were 25˚C, 29˚C, and 14˚C, respectively.<br /> Genotype<br /> TT<br /> Tt<br /> tt<br /> Standard<br /> soil<br /> Soil supplemented<br /> with minerals<br /> No<br /> environmental<br /> effect<br /> Small<br /> environmental<br /> effect<br /> Large<br /> environmental<br /> effect<br /> FIGURE 25.8 A schematic example of genotype-environment<br /> interaction. When grown in standard soil, the three genotypes TT,<br /> Tt, and tt show large, medium, and small heights, respectively. When<br /> the soil is supplemented with minerals, a great effect is seen on the tt<br /> genotype and a smaller effect on the Tt genotype. The TT genotype is<br /> unaffected by the environmental change.<br /> melanogaster. The data seen here compare the life span in days of<br /> male and female flies from two different strains of D. melanogaster<br /> raised at different temperatures. Because males and females<br /> differ in their sex chromosomes and gene expression patterns,<br /> they can be viewed as having different genotypes. The effects of<br /> environmental changes depended greatly on the strain and the<br /> sex of the flies. Under standard culture conditions, the females of<br /> strain A had the longest life span, while females of strain B had<br /> the shortest. In strain A, high temperature increased the longevity<br /> of males and decreased the longevity of females. In contrast,<br /> under hotter conditions, the longevity of males of strain B was<br /> dramatically reduced, while females of this same strain were not<br /> significantly affected. Lower growth temperature also had different<br /> effects in these two strains. While low temperature increased<br /> the longevity of both strains, the effects were most dramatic in<br /> the males of strain A and the females of strain B. Taken together,<br /> these results illustrate the potential complexity of genotypeenvironmental<br /> interaction when measuring a quantitative trait<br /> such as life span.<br /> Another issue confronting geneticists is that genotypes<br /> may not be randomly distributed in all possible environments.<br /> When certain genotypes are preferentially found in<br /> particular environments, this phenomenon is called a genotype-environment<br /> association (V Gfg E ). When such an association<br /> occurs, the effects of genotype and environment are<br /> not independent of each other, and the association needs to<br /> be considered when determining the effects of genetic and<br /> environmental variation on the total phenotypic variation.<br /> Gen otype-environment associations are very common in the<br /> study of human genetics, in which large families tend to have<br /> more similar environments compared to the population as a<br /> whole. One way to evaluate this effect is to compare different<br /> genetic relationships, such as identical versus fraternal twins. We<br /> will examine this approach later in the chapter. Another strategy<br /> that geneticists might follow is to analyze siblings that have been<br /> adopted by different parents at birth. Their environment conditions<br /> tend to be more disparate, and this may help to minimize<br /> the effects of genotype-environment association.</div> <div>6 CHAPTER 1 :: OVERVIEW OF GENETICS<br /> DNA<br /> Gene<br /> Transcription<br /> RNA (messenger RNA)<br /> Translation<br /> Protein<br /> (sequence of<br /> amino acids)<br /> Functioning of proteins within living<br /> cells influences an organism’s traits.<br /> FIGURE 1.5 A micrograph of the 46 chromosomes found in a<br /> cell from a human male.<br /> million nucleotides. Along the immense length of a chromosome,<br /> the genetic information is parceled into functional units known<br /> as genes. An average-sized human chromosome is expected to<br /> contain about 1,000 different genes.<br /> The Information in DNA Is Accessed<br /> During the Process of Gene Expression<br /> To synthesize its proteins, a cell must be able to access the information<br /> that is stored within its DNA. The process of using a gene<br /> sequence to affect the characteristics of cells and organisms is<br /> referred to as gene expression. At the molecular level, the information<br /> within genes is accessed in a stepwise process. In the first<br /> step, known as transcription, the DNA sequence within a gene<br /> is copied into a nucleotide sequence of ribonucleic acid (RNA).<br /> Most RNAs contain the information for the synthesis of a particular<br /> polypeptide. This type of RNA is called messenger RNA<br /> (mRNA). For polypeptide synthesis to occur, the sequence of<br /> nucleotides transcribed in an mRNA must be translated (using<br /> the genetic code) into the amino acid sequence of a polypeptide<br /> (Figure 1.6). After a polypeptide is made, it folds into a threedimensional<br /> structure. As mentioned, a protein is a functional<br /> unit. Some proteins are composed of a single polypeptide, and<br /> other proteins consist of two or more polypeptides.<br /> Gene expression results in the production of proteins<br /> with specific structures and functions. The unique relationship<br /> between gene sequences and protein structures is of paramount<br /> FIGURE 1.6 Gene expression at the molecular<br /> level. The expression of a gene is a multistep process.<br /> During transcription, one of the DNA strands is used as a<br /> template to make an RNA strand. During translation, the RNA strand is<br /> used to specify the sequence of amino acids within a polypeptide. One<br /> or more polypeptides produce a protein that functions within the cell,<br /> thereby influencing an organism’s traits.<br /> importance because the distinctive structure of each protein<br /> determines its function within a living cell or organism. Mediated<br /> by the process of gene expression, therefore, the sequence of<br /> nucleotides in DNA stores the information required for synthesizing<br /> proteins with specific structures and functions.<br /> The Molecular Expression of Genes<br /> Within Cells Leads to an Organism’s Traits<br /> A trait is any characteristic that an organism displays. In genetics,<br /> we often focus our attention on morphological traits that<br /> affect the appearance of an organism. The color of a flower and<br /> the height of a pea plant are morphological traits. Geneticists frequently<br /> study these types of traits because they are easy to evaluate.<br /> For example, an experimenter can simply look at a plant and<br /> tell if it has red or white flowers. However, not all traits are morphological.<br /> Physiological traits affect the ability of an organism<br /> to function. For example, the rate at which a bacterium metabolizes<br /> a sugar such as lactose is a physiological trait. Like morphological<br /> traits, physiological traits are controlled, in part, by the<br /> expression of genes. Behavioral traits also affect the ways that an<br /> organism responds to its environment. An example would be the<br /> mating calls of bird species. In animals, the nervous system plays<br /> a key role in governing such traits.</div> <div>468 CHAPTER 17 :: RECOMBINATION AND TRANSPOSITION AT THE MOLECULAR LEVEL<br /> location within the C gene and the rejoining of the two ends, not<br /> simply due to chromosome breakage. According to this hypothesis,<br /> the red phenotype should be associated with two observations.<br /> First, Ds should have moved to a new location. Second,<br /> the restored C allele should no longer be mutable because the<br /> mutable locus had been removed.<br /> To further study this phenomenon, McClintock carried out<br /> the experiment shown in Figure 17.11. As seen here, a cross produced<br /> kernels that were not entirely colorless. In most cases, red<br /> sectoring occurred because the Ds element left the C gene in a<br /> few cells during the growth of the kernel. However, in the ear of<br /> corn shown in Figure 17.11, one kernel was completely red. This<br /> suggests that the Ds element transposed out of the C gene prior<br /> to kernel growth. In this case, the Ds element transposed out of<br /> the C gene during the formation of the haploid male gametophyte<br /> that produced the pollen nuclei that fertilized this particular<br /> red kernel. Therefore, all of the cells in this kernel would have<br /> a red phenotype. (Note: Every kernel in an ear of corn is equivalent<br /> to a distinct offspring, because each is produced from the<br /> union of haploid cells from the male and female gametophytes.)<br /> ■ THE HYPOTHESIS<br /> The transposition of the Ds element into the normal C gene prevents<br /> kernel pigmentation. When the Ds element transposes back<br /> out of the C gene, the normal C allele is restored, which gives a<br /> red phenotype.<br /> ■ TESTING THE HYPOTHESIS — FIGURE 17.11 Evidence for transposable elements in corn.<br /> Starting material: The male pollen of the corn plant was homozygous for a chromosome in which the Ds element had moved into the C<br /> gene: CDsC Sh wx. The male plant also contained the Ac locus. The plant contributing the female gamete was homozygous for a chromosome<br /> carrying c sh Wx.<br /> 1. Cross a plant that is homozygous for c,<br /> sh, and Wx to a plant that is<br /> homozygous for CDsC, Sh, wx.<br /> Experimental level<br /> x<br /> Conceptual level<br /> c sh Wx x CDsC Sh wx<br /> Expected endosperm genotype:<br /> 2. In this cross, most kernels will have a<br /> colorless background with red sectoring.<br /> On rare occasions, Ds may have<br /> transposed during male gametophyte<br /> formation, producing pollen that gives<br /> rise to a completely red kernel.<br /> 3. Identify those occasional kernels that<br /> are completely red.<br /> (Contributes<br /> eggs)<br /> (Contributes<br /> pollen)<br /> CDsC<br /> Sh wx<br /> c sh Wx<br /> c sh Wx<br /> Endosperm genotype of<br /> solid red kernels:<br /> Ds has transposed out of<br /> C gene to a new site.<br /> C Sh wx<br /> 4. Germinate the solid red kernels.<br /> Germinate<br /> c sh Wx<br /> c sh Wx</div> <div>604 CHAPTER 22 :: MEDICAL GENETICS AND CANCER<br /> TABLE 22.3<br /> Examples of Human Disorders Inherited in an X-linked Recessive<br /> Manner<br /> Disorder Gene Product Effects of Disease-Causing<br /> Allele<br /> Duchenne muscular Dystrophin Progressive degeneration of<br /> dystrophy<br /> muscles that begins in early<br /> childhood<br /> Hemophilia A Clotting factor VIII Defect in blood clotting<br /> Hemophilia B Clotting factor IX Defect in blood clotting<br /> Androgen insensitivity Androgen receptor Missing male steroid hormone<br /> syndrome<br /> receptor; XY individuals have<br /> external features that are<br /> feminine but internally have<br /> undescended testes and no<br /> uterus<br /> chromosome. Males are hemizygous—have a single copy—for<br /> these genes. Therefore, a female heterozygous for an X-linked<br /> recessive gene will pass this trait to 50% of her sons, as shown<br /> in the following Punnett square for hemophilia. In this example,<br /> X h-A is the chromosome that carries a mutant allele causing<br /> hemophilia.<br /> X H<br /> X h-A<br /> X H<br /> X H X H<br /> Normal<br /> female<br /> X H X h-A<br /> Carrier<br /> female<br /> Y<br /> X H Y<br /> Normal<br /> male<br /> X h-A Y<br /> Male with<br /> hemophilia<br /> As mentioned previously, hemophilia is a disorder in<br /> which the blood cannot clot properly when a wound occurs. For<br /> individuals with this trait, a minor cut may bleed for a very long<br /> time, and small injuries can lead to large bruises, because internal<br /> broken capillaries may leak blood profusely before they are<br /> repaired. For hemophiliacs, common injuries pose a threat of<br /> severe internal or external bleeding. Hemophilia A, also called<br /> classical hemophilia, is caused by a defect in an X-linked gene<br /> that encodes the protein clotting Factor VIII. This disease has also<br /> been called the “Royal disease,” because it affected many members<br /> of European royal families. The pedigree shown in Figure 22.4<br /> illustrates the prevalence of hemophilia A among the descendants<br /> of Queen Victoria of England. The pattern of X-linked recessive<br /> inheritance is revealed by the following observations:<br /> 1. Males are much more likely to exhibit the trait.<br /> 2. The mothers of affected males often have brothers or fathers<br /> who are affected with the same trait.<br /> 3. The daughters of affected males will produce, on average,<br /> 50% affected sons.<br /> Many Genetic Disorders Exhibit Locus Heterogeneity<br /> Hemophilia can be used to illustrate another concept in genetics<br /> called locus heterogeneity. This term refers to the phenomenon in<br /> which a particular type of disease may be caused by mutations in<br /> two or more different genes. For example, blood clotting involves the<br /> participation of several different proteins that take part in a cellular<br /> cascade that leads to the formation of a clot. Hemophilia is usually<br /> caused by a defect in one of three different clotting factors. In hemophilia<br /> A, also called classic hemophilia, a protein called Factor VIII<br /> is missing. Hemophilia B is a deficiency in a different clotting factor,<br /> called Factor IX. Both Factor VIII and IX are encoded by different<br /> genes on the X chromosome. These two types of hemophilia show<br /> an X-linked recessive pattern of inheritance. By comparison, hemophilia<br /> C is due to a Factor XI deficiency. The gene encoding Factor<br /> XI is found on chromosome 4, and this form of hemophilia follows<br /> an autosomal recessive pattern of inheritance.<br /> In hemophilia, locus heterogeneity arises from the participation<br /> of several proteins in a common cellular process. Another<br /> mechanism that may lead to locus heterogeneity occurs when<br /> proteins are composed of two or more different subunits, with<br /> each subunit being encoded by a different gene. The disease thalassemia<br /> is an example of locus heterogeneity caused by a mutation<br /> in a protein composed of multiple subunits. This potentially<br /> life-threatening disease involves defects in the ability of the<br /> red blood cells to transport oxygen. The underlying cause is an<br /> alteration in hemoglobin. In adults, hemoglobin is a tetrameric<br /> protein composed of two α-globin and two β-globin subunits;<br /> α globin and β globin are encoded by separate genes (namely,<br /> the α-globin and β-globin genes). Two main types of thalassemia<br /> have been discovered in human populations: α thalassemia,<br /> in which the α-globin subunit of hemoglobin is defective; and β<br /> thalassemia, in which the β-globin subunit is defective.<br /> Unfortunately, locus heterogeneity may greatly confound<br /> pedigree analysis. For example, a human pedigree might contain<br /> individuals with X-linked hemophilia and other individuals<br /> with hemophilia C. A geneticist who assumed all affected individuals<br /> had defects in the same gene would be unable to explain<br /> the resulting pattern of inheritance. For disorders such as hemophilia<br /> and thalassemia, pedigree analysis is not a major problem<br /> because the biochemical basis for these diseases is well understood.<br /> However, for rare diseases that are poorly understood at<br /> the molecular level, locus heterogeneity may profoundly obscure<br /> the pattern of inheritance.<br /> Disease-Causing Mutant Genes Are Identified<br /> by Mapping and DNA Sequencing<br /> How do geneticists identify genes that cause disease when they<br /> are mutant? While a variety of approaches may be followed, the<br /> hunt often begins with family pedigrees. As described in Chapter<br /> 20, the human genome has been extensively mapped with many<br /> molecular markers such as microsatellites (refer back to Figure<br /> 20.12). By comparing the transmission patterns of many molecular<br /> markers with the occurrence of an inherited disease, researchers<br /> can pinpoint particular markers that are closely linked to the<br /> disease-causing mutant gene.</div> <div>7.3 EXTRANUCLEAR INHERITANCE 177<br /> in Mirabilis jalapa (the four-o’clock plant). Leaves can be green,<br /> white, or variegated with both green and white sectors. Correns<br /> demonstrated that the pigmentation of the offspring depended<br /> solely on the maternal parent (Figure 7.16). If the female parent<br /> had white pigmentation, all offspring had white leaves. Similarly,<br /> if the female was green, all offspring were green. When the female<br /> was variegated, the offspring could be green, white, or variegated.<br /> The pattern of inheritance observed by Correns is a type of<br /> extrachromosomal inheritance called maternal inheritance (not<br /> to be confused with maternal effect). Chloroplasts are a type of<br /> plastid that makes chlorophyll, a green photosynthetic pigment.<br /> Maternal inheritance occurs because the chloroplasts are inherited<br /> only through the cytoplasm of the egg. The pollen grains of<br /> M. jalapa do not transmit chloroplasts to the offspring.<br /> The phenotypes of leaves can be explained by the types of<br /> chloroplasts within the leaf cells. The green phenotype, which is<br /> the wild-type condition, is due to the presence of normal chloroplasts<br /> that make green pigment. By comparison, the white<br /> Cross 1 Cross 2<br /> x<br /> x<br /> phenotype is due to a mutation in a gene within the chloroplast<br /> DNA that diminishes the synthesis of green pigment. A cell may<br /> contain both types of chloroplasts, a condition known as heteroplasmy.<br /> A leaf cell containing both types of chloroplasts is green<br /> because the normal chloroplasts produce green pigment.<br /> How does a variegated phenotype occur? Figure 7.17 considers<br /> the leaf of a plant that began from a fertilized egg that contained<br /> both types of chloroplasts (i.e., a heteroplasmic cell). As a<br /> plant grows, the two types of chloroplasts are irregularly distributed<br /> to daughter cells. On occasion, a cell may receive only the<br /> chloroplasts that have a defect in making green pigment. Such a<br /> cell continues to divide and produce a sector of the plant that is<br /> entirely white. In this way, the variegated phenotype is created.<br /> Similarly, if we consider the results of Figure 7.16, a female parent<br /> that is variegated may transmit green, white, or a mixture of<br /> these types of chloroplasts to the egg cell and thereby produce<br /> green, white, or variegated offspring, respectively.<br /> Studies in Yeast and Chlamydomonas Provided<br /> Genetic Evidence for Extranuclear Inheritance<br /> of Mitochondria and Chloroplasts<br /> The research of Correns and others indicated that some traits,<br /> such as leaf pigmentation, are inherited in a non-Mendelian<br /> manner. However, such studies did not definitively determine that<br /> All white offspring<br /> Green, white, or<br /> variegated offspring<br /> Reciprocal cross of cross 1 Reciprocal cross of cross 2<br /> x<br /> x<br /> Variegated<br /> leaf<br /> All white<br /> chloroplasts<br /> All green offspring<br /> All green offspring<br /> FIGURE 7.16 Maternal inheritance in the<br /> four-o’clock plant, Mirabilis jalapa. The reciprocal<br /> crosses of four-o’clock plants by Carl Correns consisted<br /> of a pair of crosses between white-leaved and green-leaved plants, and a<br /> pair of crosses between variegated-leaved and green-leaved plants.<br /> Genes g Traits In this example, the white phenotype is due to chloroplasts that<br /> carry a mutant allele that diminishes green pigmentation. The variegated phenotype<br /> is due to a mixture of chloroplasts, some of which carry the normal (green) allele<br /> and some of which carry the white allele. In the crosses shown here, the parent<br /> providing the eggs determines the phenotypes of the offspring. This is due to<br /> maternal inheritance. The egg contains the chloroplasts that are inherited by the<br /> offspring. (Note: The defective chloroplasts that give rise to white sectors are not<br /> completely defective in chlorophyll synthesis. Therefore, entirely white plants can<br /> survive, though they are smaller than green or variegated plants.)<br /> Green chloroplast<br /> White chloroplast<br /> Nucleus<br /> or<br /> All green<br /> chloroplasts<br /> FIGURE 7.17 A cellular explanation of the variegated<br /> phenotype in Mirabilis jalapa. This plant inherited two types of<br /> chloroplasts—those that can produce green pigment and those that<br /> are defective. As the plant grows, the two types of chloroplasts are<br /> irregularly distributed to daughter cells. On occasion, a cell may receive<br /> only the chloroplasts that are defective at making green pigment. Such<br /> a cell continues to divide and produces a sector of the plant that is<br /> entirely white. Cells that contain both types of chloroplasts or cells<br /> that contain only green chloroplasts would produce green tissue, which<br /> may be adjacent to a sector of white tissue. This is the basis for the<br /> variegated phenotype of the leaves.</div> <div>718 CHAPTER 25 :: QUANTITATIVE GENETICS<br /> Percentage of oil content<br /> 19<br /> 18<br /> 17<br /> 16<br /> 15<br /> 14<br /> 13<br /> 12<br /> 11<br /> 10<br /> 9<br /> 8<br /> 7<br /> 6<br /> 5<br /> 4<br /> 3<br /> 2<br /> 1<br /> 0<br /> 0 10 20 30 40 50 60 70 80<br /> Bristle number<br /> 46<br /> 44<br /> 42<br /> 40<br /> 38<br /> 36<br /> 34<br /> 32<br /> 30<br /> Females<br /> Males<br /> 28<br /> 0 1 2 3 4 5 6 7 8<br /> Generations<br /> Generations<br /> (a) Results of selective breeding for high and low oil content<br /> in corn<br /> (b) Results of selective breeding for high and low bristle number<br /> in flies<br /> FIGURE 25.13 Common results of selective breeding for a quantitative trait. (a) Selection for high and low oil content in corn. (b) Selection<br /> for high and low bristle number in flies.<br /> Using artificial selection experiments, the response to selection<br /> is a common way to estimate the narrow-sense heritability<br /> in a starting population. The narrow-sense heritability measured<br /> in this way is also called the realized heritability. It is calculated<br /> as<br /> h 2 N 5 R S<br /> where<br /> R is the response in the offspring to selection, or the<br /> difference between the mean of the offspring and the<br /> mean of the population of the parents’ generation.<br /> S is the selection differential in the parents, or the<br /> difference between the mean of the parents and the<br /> mean of the population.<br /> Here,<br /> where<br /> So,<br /> R 5 X O 2 X<br /> S 5 X P 2 X<br /> X is the mean of the starting population<br /> X O is the mean of the offspring<br /> X P is the mean of the parents<br /> h N 2 5 X O 2 X<br /> X P 2 X<br /> The narrow-sense heritability is the proportion of the variance in<br /> phenotype that can be used to predict changes in the population<br /> mean when selection is practiced.<br /> As an example, let’s suppose we began with a population of<br /> fruit flies in which the average bristle number for both sexes was<br /> 37.5. The parents chosen from this population had an average<br /> bristle number of 40. The offspring of the next generation had<br /> an average bristle number of 38.7. With these values, the realized<br /> heritability is<br /> h N 2 5<br /> 38.7 2 37.5<br /> 40 2 37.5<br /> h 2 N 5 1.2<br /> 2.5<br /> h<br /> 2<br /> N � 0.48<br /> This result tells us that about 48% of the phenotypic variation is<br /> due to the additive effects of alleles.<br /> An important aspect of narrow-sense heritabilities is their<br /> ability to predict the outcome of selective breeding. Solved problem<br /> S1 at the end of the chapter illustrates this idea.<br /> Heterosis May Be Explained by Dominance<br /> or Overdominance<br /> As we have just seen, selective breeding can alter the phenotypes<br /> of domesticated species in a highly directed way. An unfortunate<br /> consequence of inbreeding, however, is that it may inadvertently</div> <div>16.3 DNA REPAIR 447<br /> resynthesizing the correct sequence using the parental DNA as a<br /> template.<br /> Eukaryotic species also have homologs to MutS and MutL,<br /> along with many other proteins that are needed for mismatch<br /> repair. However, a MutH homolog has not been identified in<br /> eukaryotes, and the mechanism by which the eukaryotic mismatch<br /> repair system distinguishes between the parental and daughter<br /> strands is not well understood. As with defects in nucleotide excision<br /> repair systems, mutations in the human mismatch repair system<br /> are associated with particular types of cancer. For example,<br /> mutations in two human mismatch repair genes, hMSH2 and<br /> hMLH1, play a role in the development of a type of colon cancer<br /> known as hereditary nonpolyposis colorectal cancer.<br /> An identical<br /> region<br /> between<br /> sister<br /> chromatids<br /> Double-strand break<br /> End processing<br /> Double-Strand Breaks Can Be Repaired<br /> by Homologous Recombination Repair<br /> and by Nonhomologous End Joining<br /> Of the many types of DNA damage that can occur within living<br /> cells, the breakage of chromosomes—called a DNA doublestrand<br /> break (DSB)—is perhaps the most dangerous. DSBs can<br /> be caused by ionizing radiation (X-rays or gamma rays), chemical<br /> mutagens, and certain drugs used for chemotherapy. In addition,<br /> free radicals that are the byproducts of cellular metabolism<br /> can cause double-strand breaks. Surprisingly, researchers estimate<br /> that naturally occurring double-strand breaks in a typical<br /> human cell occur at a rate of between 10 and 100 breaks per cell<br /> per day! Such breaks can be harmful in a variety of ways. First,<br /> DSBs can result in chromosomal rearrangements such as inversions<br /> and translocations (see Figure 8.2). In addition, DSBs can<br /> lead to terminal or interstitial deficiencies (see Figure 8.3). Such<br /> genetic changes have the potential to result in detrimental phenotypic<br /> effects.<br /> How are DSBs repaired? The two main mechanisms are<br /> homologous recombination repair (HRR) and nonhomologous<br /> end joining (NHEJ). Homologous recombination repair, also<br /> called homology-directed repair, occurs when homologous DNA<br /> strands, usually from a sister chromatid, are used to repair a<br /> DSB in the other sister chromatid (Figure 16.22). First, the DSB<br /> is processed by the short digestion of DNA strands at the break<br /> site. This processing event is followed by the exchange of DNA<br /> strands between the broken and unbroken sister chromatids.<br /> The unbroken strands are then used as templates to synthesize<br /> DNA in the region where the break occurred. Finally, the crisscrossed<br /> strands are resolved, which means that they are broken<br /> and then rejoined in a way that produces separate chromatids.<br /> Because sister chromatids are genetically identical, an advantage<br /> is that homologous recombination repair can be an error-free<br /> mechanism to repair a DSB. A disadvantage is that sister chromatids<br /> are available only during the S and G 2 phases of the cell<br /> cycle in eukaryotes or following DNA replication in bacteria.<br /> Although sister chromatids are strongly preferred, HRR may also<br /> occur between homologous regions that are not identical. Therefore,<br /> HRR may occasionally happen when sister chromatids are<br /> unavailable. The proteins involved in homologous recombination<br /> repair are described in Chapter 17.<br /> Strand exchange<br /> DNA synthesis<br /> Resolution and ligation<br /> FIGURE 16.22 DNA repair of double-strand breaks (DSBs)<br /> via homologous recombination repair.<br /> During nonhomologous end joining, the two broken ends<br /> of DNA are simply pieced back together (Figure 16.23). This<br /> mechanism requires the participation of several proteins that<br /> play key roles in the process. First, the DSB is recognized by<br /> end-binding proteins. These proteins then recognize additional</div> <div>496 CHAPTER 18 :: RECOMBINANT DNA TECHNOLOGY<br /> Template DNA<br /> Region that will be amplified<br /> Mix together template DNA,<br /> present in low amounts,<br /> with dNTPs, Taq<br /> polymerase, and 2 primers<br /> present in high amounts.<br /> Cycle 1<br /> During each cycle, the DNA<br /> strands are separated via heating.<br /> The temperature is then lowered<br /> to allow the primers to bind, and<br /> a complementary strand is made.<br /> Cycle 2<br /> +<br /> Primerannealing<br /> site<br /> 5′ 3′<br /> T<br /> A<br /> G C<br /> C G<br /> A<br /> T<br /> C C<br /> G G<br /> A G<br /> T C<br /> C<br /> G<br /> A<br /> T<br /> T<br /> A<br /> C C<br /> G G<br /> G<br /> C<br /> A<br /> T<br /> T C<br /> A G<br /> 3′ 5′<br /> Primer<br /> +<br /> +<br /> +<br /> Cycle 3<br /> +<br /> +<br /> +<br /> +<br /> +<br /> With each successive<br /> cycle, the relative amount<br /> of this type of DNA fragment<br /> increases. Therefore, after<br /> many cycles, the vast majority<br /> of DNA fragments only contain<br /> the region that is flanked by the<br /> 2 primers.<br /> +<br /> +<br /> FIGURE 18.6 The technique of polymerase chain reaction (PCR). During each cycle, oligonucleotides that are complementary<br /> to the ends of the targeted DNA sequence bind to the DNA and act as primers for the synthesis of this DNA region. The primers used<br /> in actual PCR experiments are usually about 20 nucleotides in length. The region between the two primers is typically hundreds of<br /> nucleotides in length, not just several nucleotides as shown here. The net result of PCR is the synthesis of many copies of DNA in the region flanked by<br /> the two primers. Note: The DNA template strands are purple, and those strands made during the first, second, and third cycles are blue, red, and gray,<br /> respectively.</div> <div>4.1 INHERITANCE PATTERNS OF SINGLE GENES 83<br /> few genes are located in these homologous regions. One example<br /> is a human gene called Mic2, which encodes a cell surface antigen.<br /> The Mic2 gene is found on both the X and Y chromosomes.<br /> It follows a pattern of inheritance called pseudoautosomal<br /> inheritance. The term pseudoautosomal refers to the idea that<br /> the inheritance pattern of the Mic2 gene is the same as the inheritance<br /> pattern of a gene located on an autosome even though<br /> the Mic2 gene is actually located on the sex chromosomes. As in<br /> autosomal inheritance, males have two copies of pseudoautosomally<br /> inherited genes, and they can transmit the genes to both<br /> daughters and sons.<br /> Some Traits Are Influenced<br /> by the Sex of the Individual<br /> As we have just seen, the transmission pattern of sex-linked genes<br /> depends on the sex of the parents and offspring. Sex can influence<br /> traits in other ways as well. The term sex-influenced inheritance<br /> refers to the phenomenon in which an allele is dominant in one<br /> sex but recessive in the opposite sex. Therefore, sex influence is a<br /> phenomenon of heterozygotes. Sex-influenced inheritance should<br /> not be confused with sex-linked inheritance. The genes that govern<br /> sex-influenced traits are usually autosomal, not on the X or<br /> Y chromosome.<br /> In humans, the common form of pattern baldness provides<br /> an example of sex-influenced inheritance. As shown in<br /> Figure 4.15, the balding pattern is characterized by hair loss on<br /> the front and top of the head but not on the sides. This type of<br /> pattern baldness is inherited as an autosomal trait. (A common<br /> misconception is that this gene is X linked.) When a male is heterozygous<br /> for the baldness allele, he will become bald.<br /> Genotype<br /> Males<br /> Phenotype<br /> Females<br /> BB Bald Bald<br /> Bb Bald Nonbald<br /> bb Nonbald Nonbald<br /> In contrast, a heterozygous female will not be bald. Women who<br /> are homozygous for the baldness allele will develop the trait, but<br /> it is usually characterized by a significant thinning of the hair<br /> that occurs relatively late in life.<br /> The sex-influenced nature of pattern baldness is related<br /> to the production of the male sex hormone testosterone. The<br /> gene that affects pattern baldness encodes an enzyme called 5-<br /> α-reductase, which converts testosterone to 5-α-dihydrotestosterone<br /> (DHT). DHT binds to cellular receptors and affects the<br /> expression of many genes, including those in the cells of the scalp.<br /> The allele that causes pattern baldness results in an overexpression<br /> of this enzyme. Because mature males normally make more<br /> testosterone than females, this allele has a greater phenotypic<br /> impact in males. However, a rare tumor of the adrenal gland can<br /> cause the secretion of abnormally large amounts of testosterone<br /> in females. If this occurs in a woman who is heterozygous Bb, she<br /> will become bald. If the tumor is removed surgically, her hair will<br /> return to its normal condition.<br /> The autosomal nature of pattern baldness has been revealed<br /> by the analysis of many human pedigrees. An example is shown<br /> in Figure 4.16a. A bald male may inherit the bald allele from<br /> either parent, and thus a striking observation is that bald fathers<br /> can pass this trait to their sons. This could not occur if the trait<br /> was X linked, because fathers do not transmit an X chromosome<br /> to their sons. The analyses of many human pedigrees have<br /> shown that bald fathers, on average, have at least 50% bald sons.<br /> They are expected to produce an even higher percentage of bald<br /> male offspring if they are homozygous for the bald allele and/or<br /> the mother also carries one or two copies of the bald allele. For<br /> example, a heterozygous bald male and heterozygous (nonbald)<br /> female will produce 75% bald sons, while a homozygous bald<br /> male or homozygous bald female will produce all bald sons.<br /> Figure 4.16b shows the predicted offspring if two heterozygotes<br /> produce offspring. In this Punnett square, the phenotypes<br /> are designated for both sons and daughters. BB offspring<br /> are bald, while bb offspring are nonbald. Bb offspring are bald if<br /> they are sons and nonbald if they are daughters. The predicted<br /> (a) John Adams (father) (b) John Quincy Adams (son) (c) Charles Francis Adams<br /> (grandson)<br /> FIGURE 4.15 Pattern baldness in the Adams family line.<br /> (d) Henry Adams<br /> (great-grandson)</div> <div>CONCEPTUAL QUESTIONS 385<br /> a bacterium to coordinately regulate a group of genes whose encoded<br /> proteins have a common function. For example, an operon may contain<br /> a group of genes involved in lactose breakdown or a group of genes<br /> involved in tryptophan synthesis. The genes within an operon usually<br /> encode proteins within a common metabolic pathway or cellular function.<br /> S5. The sequential use of two sugars by a bacterium is known as<br /> diauxic growth. It is a common phenomenon among many<br /> bacterial species. When glucose is one of the two sugars available,<br /> it is typical that the bacterium metabolizes glucose first, and then a<br /> second sugar after the glucose has been used up. Among E. coli and<br /> related species, diauxic growth is regulated by intracellular cAMP<br /> levels and the catabolite activator protein (CAP). Summarize the<br /> effects of glucose and lactose on the ability of the lac repressor and<br /> the cAMP-CAP complex to regulate the lac operon.<br /> Answer: In the absence of lactose, the lac repressor has the dominating<br /> effect of shutting off the lac operon. Even when glucose is also absent<br /> and the cAMP-CAP complex is formed, the presence of the bound lac<br /> repressor prevents the expression of the lac operon. The effects of the<br /> cAMP-CAP complex are exerted only in the presence of lactose. When<br /> lactose is present and glucose is absent, the cAMP-CAP complex acts to<br /> enhance the rate of transcription. However, when both lactose and glucose<br /> are present, the inability of CAP to bind to the lac operon decreases the<br /> rate of transcription. The table shown here summarizes these effects.<br /> lac Operon Regulation<br /> Sugar Present<br /> None<br /> Lactose<br /> Glucose<br /> Lactose and glucose<br /> Conceptual Questions<br /> Transcription of the lac Operon<br /> The operon is turned off due to the<br /> dominating effect of the lac repressor protein.<br /> The operon is maximally turned on. The<br /> repressor protein is removed from the<br /> operator site, and the cAMP-CAP complex is<br /> bound to the CAP site.<br /> The operon is turned off due to the<br /> dominating effect of the lac repressor protein.<br /> The expression of the lac operon is greatly<br /> decreased. The lac repressor is removed<br /> from the operator site, and the catabolite<br /> activator protein is not bound to the CAP<br /> site. The absence of CAP at the CAP site<br /> makes it difficult for RNA polymerase to<br /> begin transcription. However, a little more<br /> transcription occurs under these conditions<br /> than in the absence of lactose, when the<br /> repressor is bound.<br /> C1. What is the difference between a constitutive gene and a regulated<br /> gene?<br /> C2. In general, why is it important to regulate genes? Discuss examples<br /> of situations in which it would be advantageous for a bacterial cell<br /> to regulate genes.<br /> C3. If a gene is repressible and under positive control, describe what<br /> kind of effector molecule and regulatory protein are involved.<br /> Explain how the binding of the effector molecule affects the<br /> regulatory protein.<br /> S6. The ability of DNA-binding proteins to promote a loop in DNA<br /> structure is an interesting phenomenon that is important in the<br /> structure and function of DNA. Besides the regulation of genes<br /> and operons, DNA looping is required in the compaction of DNA<br /> within the nucleoid of a bacterium and the nucleus of a eukaryotic<br /> cell (see Chapter 10). In addition, DNA looping is frequently<br /> involved in the expression of eukaryotic genes (see Chapter 15).<br /> In this solved problem, we will examine an experimental approach<br /> that made it possible for Robert Schleif and his colleagues to<br /> determine that the AraC protein causes a loop to form in the<br /> DNA. This work relied on the mobility of DNA in an acrylamide<br /> gel. A segment of DNA that contains a loop is more compact than<br /> the same DNA segment without a loop. Therefore, when these two<br /> alternative structures (looped versus unlooped) are run through<br /> a gel, the looped structure migrates more quickly to the bottom<br /> of the gel because it can more easily penetrate the gel matrix. As<br /> a starting material, Schleif had a sample of DNA that contained a<br /> portion of the ara operon including both the araI and araO 2 sites.<br /> In the gel show here, this segment of DNA was exposed to the<br /> following conditions before it was run on the gel:<br /> Unlooped<br /> Looped<br /> 1 2 3 4<br /> Lane 1. No further additions<br /> Lane 2. Add AraC protein<br /> Lane 3. Add arabinose<br /> Lane 4. Add AraC protein<br /> and arabinose<br /> Explain these results.<br /> Answer: In lane 2, AraC protein was added and arabinose was not. As<br /> expected from the DNA looping hypothesis, a DNA loop was formed<br /> as evidenced from the faster mobility on the acrylamide gel. In lane 1,<br /> no AraC protein was added, so no DNA loop was able to form. These<br /> results confirm the idea that the AraC protein causes a loop to form<br /> in the sample loaded into lane 2. In lane 3, only the sugar arabinose is<br /> present but not the AraC protein. Because the DNA remains unlooped,<br /> the sugar by itself has no effect. In the sample loaded into lane 4, AraC<br /> protein was added, and arabinose was added as well. In this case, no<br /> DNA loop is observed. These results are consistent with the idea that<br /> arabinose binds to the AraC protein and breaks the DNA loop that is<br /> promoted by the AraC protein.<br /> C4. Transcriptional regulation often involves a regulatory protein that<br /> binds to a segment of DNA and a small effector molecule that<br /> binds to the regulatory protein. Do the following terms apply to a<br /> regulatory protein, a segment of DNA, or a small effector molecule?<br /> A. Repressor<br /> E. Activator<br /> B. Inducer<br /> F. Attenuator<br /> C. Operator site G. Inhibitor<br /> D. Corepressor</div> <div>38 CHAPTER 2 :: MENDELIAN INHERITANCE<br /> A second approach that is analogous to the multiplication<br /> method is the forked-line method. In this case, the genetic proportions<br /> are determined by multiplying together the probabilities of each<br /> phenotype.<br /> Tall or<br /> dwarf<br /> Round or<br /> wrinkled<br /> Yellow or<br /> green Observed product Phenotype<br /> 3 / 4 tall<br /> 3 / 4 round<br /> 1 / 4 wrinkled<br /> 3 / 4 yellow<br /> 1 / 4 green<br /> 3 / 4 yellow<br /> 1 / 4 green<br /> ( 3 / 4 )( 3 / 4 )( 3 / 4 ) =<br /> ( 3 / 4 )( 3 / 4 )( 1 / 4 ) =<br /> ( 3 / 4 )( 1 / 4 )( 3 / 4 ) =<br /> ( 3 / 4 )( 1 / 4 )( 1 / 4 ) =<br /> 27 / 64 tall, round, yellow<br /> 9 / 64 tall, round, green<br /> 9 / 64 tall, wrinkled, yellow<br /> 3 / 64 tall, wrinkled, green<br /> 1 / 4 dwarf<br /> 3 / 4 round<br /> 1 / 4 wrinkled<br /> 3 / 4 yellow<br /> 1 / 4 green<br /> 3 / 4 yellow<br /> 1 / 4 green<br /> ( 1 / 4 )( 3 / 4 )( 3 / 4 ) = 9 / 64 dwarf, round, yellow<br /> ( 1 / 4 )( 3 / 4 )( 1 / 4 ) = 3 / 64 dwarf, round, green<br /> ( 1 / 4 )( 1 / 4 )( 3 / 4 ) = 3 / 64 dwarf, wrinkled, yellow<br /> ( 1 / 4 )( 1 / 4 )( 1 / 4 ) = 1 / 64 dwarf, wrinkled, green<br /> S5. For an individual expressing a dominant trait, how can you tell if<br /> it is a heterozygote or a homozygote?<br /> Answer: One way is to conduct a testcross with an individual that<br /> expresses the recessive version of the same trait. If the individual<br /> is heterozygous, half of the offspring will show the recessive trait,<br /> whereas if the individual is homozygous, none of the offspring will<br /> express the recessive trait.<br /> Dd � dd or DD � dd<br /> i<br /> i<br /> 1Dd (dominant trait)<br /> All Dd (dominant trait)<br /> 1 dd (recessive trait)<br /> Another way to determine heterozygosity involves a more careful<br /> examination of the individual at the cellular or molecular level.<br /> At the cellular level, the heterozygote may not look exactly like the<br /> homozygote. This phenomenon is described in Chapter 4. Also, gene<br /> cloning methods described in Chapter 18 can be used to distinguish<br /> between heterozygotes and homozygotes.<br /> S6. In dogs, black fur color is dominant to white. Two heterozygous<br /> black dogs are mated. What would be the probability of the<br /> following combinations of offspring?<br /> A. A litter of six pups, four with black fur and two with white fur.<br /> B. A litter of six pups, the firstborn with white fur, and among the<br /> remaining five pups, two with white fur and three with black<br /> fur.<br /> C. A first litter of six pups, four with black fur and two with white<br /> fur, and then a second litter of seven pups, five with black fur<br /> and two with white fur.<br /> D. A first litter of five pups, four with black fur and one with<br /> white fur, and then a second litter of seven pups in which the<br /> firstborn is homozygous, the second born is black, and the<br /> remaining five pups are three black and two white.<br /> Answer:<br /> A. Because this is an unordered combination of events, we use the<br /> binomial expansion equation, where n � 6, x � 4, p � 0.75<br /> (probability of black), and q � 0.25 (probability of white).<br /> The answer is 0.297, or 29.7%, of the time.<br /> B. We use the product rule because the order is specified. The first<br /> pup is white and then the remaining five are born later. We also<br /> need to use the binomial expansion equation to determine the<br /> probability of the remaining five pups.<br /> (probability of a white pup)(binomial expansion for the<br /> remaining five pups)<br /> The probability of the white pup is 0.25. In the binomial<br /> expansion equation, n � 5, x � 2, p � 0.25, and q � 0.75.<br /> The answer is 0.066, or 6.6%, of the time.<br /> C. The order of the two litters is specified, so we need to use the<br /> product rule. We multiply the probability of the first litter times<br /> the probability of the second litter. We need to use the binomial<br /> expansion equation for each litter.<br /> (binomial expansion of the first litter)(binomial expansion of<br /> the second litter)<br /> For the first litter, n � 6, x � 4, p � 0.75, q � 0.25. For the<br /> second litter, n � 7, x � 5, p � 0.75, q � 0.25.<br /> The answer is 0.092, or 9.2%, of the time.<br /> D. The order of the litters is specified, so we need to use the<br /> product rule to multiply the probability of the first litter<br /> times the probability of the second litter. We use the binomial<br /> expansion equation to determine the probability of the first<br /> litter. The probability of the second litter is a little more<br /> complicated. The firstborn is homozygous. There are two<br /> mutually exclusive ways to be homozygous, BB and bb. We use<br /> the sum rule to determine the probability of the first pup, which<br /> equals 0.25 + 0.25 � 0.5. The probability of the second pup is<br /> 0.75, and we use the binomial expansion equation to determine<br /> the probability of the remaining pups.<br /> (binomial expansion of first litter)([0.5][0.75][binomial<br /> expansion of second litter])<br /> For the first litter, n � 5, x � 4, p � 0.75, q � 0.25. For the last<br /> five pups in the second litter, n � 5, x � 3, p � 0.75, q � 0.25.<br /> The answer is 0.039, or 3.9%, of the time.<br /> S7. In Chapter 2, the binomial expansion equation was used in<br /> situations where only two phenotypic outcomes are possible.<br /> When more than two outcomes are possible, we use a multinomial</div> <div>18.1 GENE CLONING 485<br /> bind to a specific base sequence and then cleave the DNA backbone<br /> at two defined locations, one in each strand. Proposed by<br /> Werner Arber in the 1960s and discovered by Hamilton Smith<br /> and Daniel Nathans in the 1970s, restriction enzymes are made<br /> naturally by many different species of bacteria and protect bacterial<br /> cells from invasion by foreign DNA, particularly that of<br /> bacteriophages. Researchers have isolated and purified restriction<br /> enzymes from many bacterial species and now use them in their<br /> cloning experiments.<br /> Figure 18.1 shows the role of a restriction enzyme, called<br /> EcoRI, in producing a recombinant DNA molecule. Certain<br /> types of restriction enzymes are useful in cloning because they<br /> digest DNA into fragments with “sticky ends.” As shown in Figure<br /> 18.1, the sticky ends are single-stranded regions of DNA that<br /> can hydrogen bond to a complementary sequence of DNA from<br /> a different source. The ends of two different DNA pieces will<br /> hydrogen bond to each other because of their complementary<br /> sticky ends.<br /> The hydrogen bonding between the sticky ends of DNA<br /> fragments promotes a temporary interaction between the two<br /> fragments. However, this interaction is not stable because it<br /> involves only a few hydrogen bonds between complementary<br /> bases. How can this interaction be made more permanent? The<br /> answer is that the sugar–phosphate backbones within the DNA<br /> DNA from 2 different sources<br /> 5′ 3′<br /> 5′ 3′<br /> G A A<br /> T T C<br /> G A A<br /> T T C<br /> G A A<br /> T T C<br /> G A A<br /> T T C<br /> C T T<br /> A A G<br /> C T T<br /> A A G<br /> C T T<br /> A A G<br /> C T T<br /> A A G<br /> 3′ 5′<br /> 3′ 5′<br /> EcoRI recognition<br /> sequence<br /> 5′ 3′<br /> A “sticky<br /> end”<br /> A A<br /> T T C<br /> G<br /> G<br /> C T T<br /> A A<br /> 3′ 5′<br /> Incubate both DNAs with EcoRI,<br /> which cuts the DNA backbone<br /> between G and A.<br /> 5′ 3′<br /> A A<br /> T T C<br /> G<br /> G<br /> C T T<br /> A A<br /> A “sticky<br /> end” 3′ 5′<br /> Incubate the DNAs<br /> together, allowing sticky<br /> ends to hydrogen bond.<br /> 5′<br /> 3′<br /> A<br /> A<br /> T<br /> T<br /> 3′<br /> C<br /> G<br /> G<br /> C<br /> A<br /> T<br /> A<br /> T<br /> T<br /> A<br /> T<br /> A<br /> C<br /> G<br /> G<br /> C<br /> T<br /> T<br /> A<br /> A<br /> 5′<br /> Add DNA ligase, which<br /> covalently links the<br /> DNA backbones.<br /> Covalent bond<br /> 5′<br /> 3′<br /> A<br /> A<br /> T<br /> T<br /> 3′<br /> C<br /> G<br /> G<br /> C<br /> A<br /> T<br /> A<br /> T<br /> T<br /> A<br /> T<br /> A<br /> C<br /> G<br /> G<br /> C<br /> T<br /> T<br /> A<br /> A<br /> 5′<br /> Covalent bond<br /> A recombinant DNA molecule<br /> FIGURE 18.1 The action of a restriction enzyme and the production of recombinant DNA. The restriction enzyme EcoRI binds to a specific<br /> sequence, in this case 5'–GAATTC–3'. It then cleaves the DNA backbone between G and A, producing DNA fragments. The single-stranded ends of<br /> the DNA fragments can hydrogen bond with each other, because they have complementary sequences. The enzyme DNA ligase can then catalyze the<br /> formation of covalent bonds in the DNA backbones of the fragments.</div> <div>438 CHAPTER 16 :: GENE MUTATION AND DNA REPAIR<br /> EXPERIMENT 16A<br /> X-Rays Were the First Environmental Agent<br /> Shown to Cause Induced Mutations<br /> As shown in Table 16.5, changes in DNA structure can also be<br /> caused by environmental agents, either chemical or physical agents.<br /> These agents are called mutagens, and the mutations they cause<br /> are referred to as induced mutations. In 1927, Hermann Müller<br /> devised an approach to show that X-rays can cause induced mutations<br /> in Drosophila melanogaster. Müller reasoned that a mutagenic<br /> agent might cause some genes to become defective. His experimental<br /> approach focused on the ability of a mutagen to cause defects<br /> in X-linked genes that result in a recessive lethal phenotype.<br /> To determine if X-rays increase the rate of recessive, X-<br /> linked lethal mutations, Müller sought an easy way to detect<br /> the occurrence of such mutations. He cleverly realized that he<br /> had a laboratory strain of fruit flies that could make this possible.<br /> In particular, he conducted his crosses in such a way that<br /> a female fly that inherited a new mutation causing a recessive X-<br /> linked lethal allele would not be able to produce any male offspring.<br /> This made it very easy for him to detect lethal mutations;<br /> he had to count only the number of female flies that could not<br /> produce sons. To understand Müller’s crosses, we need to take a<br /> closer look at a peculiar version of one of the X chromosomes in<br /> a strain of flies that he used in his crosses. This X chromosome,<br /> designated ClB, had three important genetic alterations.<br /> C: Contained a large inversion that prevents it from<br /> Crossing over with the other X chromosome in female<br /> flies. The letter C is a reminder that this region of the<br /> chromosome cannot cross over.<br /> l:<br /> Carried a lethal recessive X-linked gene. If males<br /> (XY) inherit this chromosome, they will die.<br /> heterozygous for recessive lethal mutations in two different genes.<br /> However, because each X chromosome would have a lethal mutation,<br /> this female would not be able to produce any living sons.<br /> The steps in Müller’s protocol are shown in Figure 16.12.<br /> He began with wild-type males and exposed them to X-rays.<br /> These X-rays may mutate the X chromosome in sperm cells,<br /> resulting in a recessive lethal allele. These males, and a control<br /> group of males that were not exposed to X-rays, were then<br /> mated to females carrying the ClB chromosome. Daughters with<br /> bar eyes were saved from this cross and mated to nonirradiated<br /> males. You should look carefully at this cross and realize that if<br /> these daughters also contained a lethal allele on the X chromosome<br /> they inherited from their father (e.g., an irradiated male in<br /> step 1), they would not be able to produce living sons.<br /> ■ THE HYPOTHESIS<br /> The exposure of flies to X-rays will increase the rate of<br /> mutation.<br /> C<br /> +<br /> I<br /> B<br /> X<br /> I<br /> +<br /> B +<br /> X<br /> Lethal allele caused<br /> by X rays<br /> This is a female<br /> that can survive,<br /> since each lethal<br /> gene has a normal<br /> allele on the other<br /> X chromosome.<br /> x<br /> +<br /> +<br /> X<br /> Y<br /> B: Carried a dominant mutation that causes the eyes<br /> of the fly to have a Bar shape.<br /> +<br /> +<br /> I<br /> +<br /> +<br /> B + X Y<br /> I<br /> (Note: C and B are uppercase because they are inherited in a dominant<br /> manner, and l is lowercase because it is a recessive allele.)<br /> A female fly that has one copy of this X chromosome<br /> would have bar-shaped eyes, because bar is a dominant allele.<br /> Even though this X chromosome has a recessive lethal allele, a<br /> female fly can survive if the corresponding gene on the other X<br /> chromosome is a normal allele. In Müller’s experiments, the goal<br /> was to determine if exposure to X-rays caused a mutation on the<br /> normal X chromosome (not the ClB chromosome) that created<br /> a recessive lethal allele in any essential X-linked gene except for<br /> the gene that already had a lethal allele on the ClB chromosome<br /> (Figure 16.11). If a recessive lethal mutation occurred on the normal<br /> X chromosome, this female could survive because it would be<br /> C<br /> I<br /> +<br /> B B +<br /> X X<br /> Living female<br /> +<br /> +<br /> B + B +<br /> X X<br /> Living female<br /> B<br /> X Y<br /> Dead male<br /> No living male offspring are produced from this cross.<br /> FIGURE 16.11 A strategy to detect the presence of lethal<br /> X-linked mutations. X-rays may cause a recessive lethal mutation to<br /> occur in the normal X chromosome. This female also contains another<br /> lethal allele in the ClB chromosome. Nevertheless, this female could<br /> survive because it would be heterozygous for recessive lethal mutations<br /> in two different genes. Because each X chromosome would have a lethal<br /> mutation, this female would not be able to produce any living sons.<br /> C<br /> I<br /> +<br /> B +<br /> Dead male</div> <div>576 CHAPTER 21 :: GENOMICS II: FUNCTIONAL GENOMICS, PROTEOMICS, AND BIOINFORMATICS<br /> 2. Beginning at 9 hours after inoculation,<br /> take out samples of cells and isolate<br /> mRNA. This involves breaking open<br /> cells and running the cell contents over<br /> a poly-dT column under high-salt<br /> conditions. Since mRNA has a polyA<br /> tail, it will bind to this column while<br /> other cell components flow through.<br /> The purified mRNA can be then<br /> eluted by adding a buffer to the top<br /> of the column that contains a low<br /> concentration of salt. This breaks the<br /> interaction between the poly-dT and<br /> polyA tails. In this experiment, mRNA<br /> samples were isolated at approximately<br /> 2-hour intervals, beginning at 9 hours<br /> and ending at 21 hours.<br /> Break open cells.<br /> Load onto<br /> column.<br /> Collection<br /> tubes<br /> Purified mRNA<br /> TTTTT<br /> AAAAA<br /> AAAAA<br /> TTTTT<br /> Column<br /> bead<br /> AAAAA<br /> TTTTT<br /> AAAAA<br /> TTTTT<br /> High salt<br /> mRNA<br /> binds, while<br /> other<br /> cellular<br /> components<br /> do not bind.<br /> Low salt<br /> mRNA is<br /> released.<br /> High<br /> salt<br /> Low<br /> salt<br /> mRNA from:<br /> Cells grown<br /> 9 hours<br /> Later time<br /> points<br /> mRNA from:<br /> Cells grown<br /> 9 hours<br /> Later time<br /> points<br /> 3. Add reverse transcriptase, poly-dT<br /> primers, and fluorescently labeled<br /> nucleotides to make complementary<br /> strands of fluorescently labeled cDNA.<br /> Note: The sample at 9 hours was used to<br /> make green cDNA, while mRNA<br /> samples collected at later time points<br /> were used to make red cDNA.<br /> Incubate with reverse<br /> transcriptase and<br /> deoxyribonucleotides,<br /> one of which is<br /> fluorescently labeled.<br /> Add reverse transcriptase<br /> with a fluorescently<br /> labeled nucleotide.<br /> (continued)</div> <div>660 CHAPTER 23 :: DEVELOPMENTAL GENETICS<br /> 3 and 4 hours after fertilization, but they do survive if subjected to<br /> the nonpermissive temperature at other times of development. These<br /> results indicate that this protein plays a crucial role at the 3–4-hour<br /> stage of development.<br /> S4. An intriguing question in developmental genetics is, How can a<br /> particular gene, such as even-skipped, be expressed in a multiple<br /> banding pattern as seen in Figure 23.10? Another way of asking<br /> this question is, How is the positional information within the<br /> broad bands of the gap genes able to be deciphered in a way<br /> that causes the pair-rule genes to be expressed in this alternating<br /> banding pattern? The answer lies in a complex mechanism of<br /> genetic regulation. Certain pair-rule genes have several stripespecific<br /> enhancers that are controlled by multiple transcription<br /> factors. A stripe-specific enhancer is typically a short segment of<br /> DNA, 300 to 500 bp in length, that contains binding sequences<br /> recognized by several different transcription factors. This term is<br /> a bit misleading because a stripe-specific enhancer is a regulatory<br /> region that contains both enhancer and silencer elements.<br /> In 1992, Michael Levine and his colleagues investigated<br /> stripe-specific enhancers that are located near the promoter of<br /> the even-skipped gene. A segment of DNA, termed the stripe 2<br /> enhancer, controls the expression of the even-skipped gene; this<br /> enhancer is responsible for the expression of the even-skipped gene<br /> in stripe 2, which corresponds to parasegment 3 of the embryo.<br /> The stripe 2 enhancer is a segment of DNA that contains binding<br /> sites for four transcription factors that are the products of the<br /> Krüppel, bicoid, hunchback, and giant genes. The Hunchback and<br /> Bicoid transcription factors bind to this enhancer and activate<br /> the transcription of the even-skipped gene. In contrast, the<br /> transcription factors encoded by the Krüppel and giant genes<br /> bind to the stripe 2 enhancer and repress transcription. The<br /> figure shown below describes the concentrations of these four<br /> transcription factor proteins in the region of parasegment 3 (i.e.,<br /> stripe 2) in the Drosophila embryo.<br /> To study stripe-specific enhancers, researchers have constructed<br /> artificial genes in which the enhancer is linked to a reporter gene,<br /> the expression of which is easy to detect. The next figure shows the<br /> results of an experiment in which an artificial gene was made by<br /> putting the stripe 2 enhancer next to the β-galactosidase gene. This<br /> artificial gene was introduced into Drosophila, and then embryos<br /> containing this gene were analyzed for β-galactosidase activity. If a<br /> region of the embryo is expressing β galactosidase, the region will<br /> stain darkly because β galactosidase converts a colorless compound<br /> into a dark blue compound.<br /> Explain these results.<br /> Answer: As shown in the first figure to this problem, the concentrations<br /> of the Hunchback and Bicoid transcription factors are relatively<br /> high in the region of the embryo corresponding to stripe 2 (which<br /> is parasegment 3). The levels of Krüppel and Giant are very low in<br /> this region. Therefore, the high levels of activators and low levels<br /> of repressors cause the even-skipped gene to be transcribed. In this<br /> experiment, β galactosidase was made only in stripe 2 (i.e., parasegment<br /> 3). These results show that the stripe-2-specific enhancer controls<br /> gene expression only in parasegment 3. Because we know that the<br /> even-skipped gene is expressed as several alternating stripes (as seen in<br /> Figure 23.10), the even-skipped gene must contain other stripe-specific<br /> enhancers that allow it to be expressed in these other alternating<br /> parasegments.<br /> Concentration of transcription<br /> factor protein<br /> Giant<br /> Hunchback<br /> Bicoid<br /> Krüppel<br /> Parasegment 2 Parasegment 3 Parasegment 4<br /> Region of Drosophila embryo<br /> Conceptual Questions<br /> C1. In the case of multicellular animals, what are the four types of<br /> cellular processes that occur so a fertilized egg can develop into an<br /> adult organism? Briefly discuss the role of each process.<br /> C2. The arrangement of body axes of the fruit fly are shown in Figure<br /> 23.4g. Are the following statements true or false with regard to<br /> body axes in the mouse?<br /> A. Along the antero-posterior axis, the head is posterior to the tail.<br /> B. Along the dorso-ventral axis, the vertebrae of the back are<br /> dorsal to the stomach.<br /> C. Along the dorso-ventral axis, the feet are dorsal to the hips.<br /> D. Along the proximo-distal axis, the toes on the hind legs are<br /> distal to the hips of the hind legs.</div> <div>12.3 TRANSCRIPTION IN EUKARYOTES 307<br /> have discovered that three categories of proteins are needed for<br /> basal transcription at the core promoter: RNA polymerase II,<br /> general transcription factors, and mediator (Table 12.2).<br /> Five different proteins called general transcription factors<br /> (GTFs) are always needed for RNA polymerase II to initiate<br /> transcription of structural genes. Figure 12.14 describes the<br /> assembly of GTFs and RNA polymerase II at the TATA box. As<br /> shown here, a series of interactions leads to the formation of the<br /> open complex. Transcription factor IID (TFIID) first binds to<br /> the TATA box and thereby plays a critical role in the recognition<br /> of the core promoter. TFIID is composed of several subunits,<br /> including TATA-binding protein (TBP), which directly binds to<br /> the TATA box, and several other proteins called TBP-associated<br /> factors (TAFs). After TFIID binds to the TATA box, it associates<br /> with TFIIB. TFIIB promotes the binding of RNA polymerase II<br /> and TFIIF to the core promoter. Lastly, TFIIE and TFIIH bind to<br /> the complex. This completes the assembly of proteins to form a<br /> closed complex, also known as a preinitiation complex.<br /> TFIIH plays a major role in the formation of the open<br /> complex. TFIIH has several subunits that perform different functions.<br /> Certain subunits act as helicases, which break the hydrogen<br /> bonding between the double-stranded DNA and thereby promote<br /> the formation of the open complex. Another subunit hydrolyzes<br /> ATP and phosphorylates a domain in RNA polymerase II known<br /> as the carboxyl terminal domain (CTD). Phosphorylation of the<br /> CTD releases the contact between RNA polymerase II and TFIIB.<br /> TFIID binds to the TATA box. TFIID is<br /> a complex of proteins that includes the<br /> TATA-binding protein (TBP) and several<br /> TBP-associated factors (TAFs).<br /> TFIID<br /> TATA box<br /> TFIIB binds to TFIID.<br /> TFIID<br /> TFIIB<br /> TABLE<br /> TABLE<br /> 12.1<br /> 12.2<br /> Proteins Needed for Transcription via the Core Promoter of<br /> Eukaryotic Structural Genes<br /> TABLE 12.2<br /> RNA polymerase II: The enzyme that catalyzes the linkage of ribonucleotides<br /> in the 5' to 3' direction, using DNA as a template. Essentially all eukaryotic RNA<br /> polymerase II proteins are composed of 12 subunits. The two largest subunits are<br /> structurally similar to the b and b' subunits found in E. coli RNA polymerase.<br /> General transcription factors:<br /> TFIID: Composed of TATA-binding protein (TBP) and other TBP-associated<br /> factors (TAFs). Recognizes the TATA box of eukaryotic structural<br /> gene promoters.<br /> TFIIB: Binds to TFIID and then enables RNA polymerase II to bind to the<br /> core promoter. Also promotes TFIIF binding.<br /> TFIIF: Binds to RNA polymerase II and plays a role in its ability to bind to<br /> TFIIB and the core promoter. Also plays a role in the ability of TFIIE<br /> and TFIIH to bind to RNA polymerase II.<br /> TFIIE: Plays a role in the formation and/or the maintenance of the open<br /> complex. It may exert its effects by facilitating the binding of TFIIH to<br /> RNA polymerase II and regulating the activity of TFIIH.<br /> TFIIH: A multisubunit protein that has multiple roles. First, certain subunits<br /> act as helicases and promote the formation of the open complex.<br /> Other subunits phosphorylate the carboxyl terminal domain (CTD)<br /> of RNA polymerase II, which releases its interaction with TFIIB and<br /> thereby allows RNA polymerase II to proceed to the elongation phase.<br /> Mediator: A multisubunit complex that mediates the effects of regulatory<br /> transcription factors on the function of RNA polymerase II. Though mediator<br /> typically has certain core subunits, many of its subunits vary, depending on the<br /> cell type and environmental conditions. The ability of mediator to affect RNA<br /> polymerase II function is thought to occur via the CTD of RNA polymerase II.<br /> Mediator can influence the ability of TFIIH to phosphorylate CTD, and subunits<br /> within mediator itself have the ability to phosphorylate CTD. Because CTD<br /> phosphorylation is needed to release RNA polymerase II from TFIIB, mediator<br /> plays a key role in the ability of RNA polymerase II to switch from the initiation<br /> to the elongation stage of transcription.<br /> TFIIB<br /> TFIIE<br /> FIGURE 12.14<br /> complex.<br /> TFIIH<br /> TFIIB acts as a bridge to bind<br /> RNA polymerase II and TFIIF.<br /> TFIID<br /> TFIID<br /> TFIIB<br /> TFIIB<br /> TFIIF<br /> RNA polymerase II<br /> TFIIE and TFIIH bind to RNA<br /> polymerase II to form a preinitiation<br /> or closed complex.<br /> Preinitiation complex<br /> TFIIH<br /> TFIIE<br /> TFIIF<br /> TFIIH acts as a helicase to form an<br /> open complex. TFIIH also phosphorylates<br /> the CTD domain of RNA polymerase II.<br /> CTD phosphorylation breaks the contact<br /> between TFIIB and RNA polymerase II.<br /> TFIIB, TFIIE, and TFIIH are released.<br /> TFIID<br /> TFIIF<br /> Open complex<br /> PO 4<br /> PO 4<br /> CTD domain of<br /> RNA polymerase II<br /> Steps leading to the formation of the open</div> <div>24.2 FACTORS THAT CHANGE ALLELE AND GENOTYPE FREQUENCIES IN POPULATIONS 683<br /> Nonrandom Mating May Occur in Populations<br /> I-1 I-2 I-3<br /> II-1 II-2 II-3 II-4<br /> III-1 III-2 III-3 III-4<br /> IV-1<br /> FIGURE 24.18 A human pedigree containing inbreeding.<br /> Individual IV-1 is the result of inbreeding because her parents are<br /> related.<br /> As mentioned earlier, one of the conditions required to establish<br /> the Hardy-Weinberg equilibrium is random mating. This means<br /> that individuals choose their mates irrespective of their genotypes<br /> and phenotypes. In many cases, particularly in human populations,<br /> this condition is violated frequently.<br /> When mating is nonrandom in a population, the process<br /> is called assortative mating. Positive assortative mating occurs<br /> when individuals with similar phenotypes choose each other as<br /> mates. The opposite situation, where dissimilar phenotypes mate<br /> preferentially, is called negative assortative mating. In addition,<br /> individuals may choose a mate that is part of the same genetic<br /> lineage. The mating of two genetically related individuals, such<br /> as cousins, is called inbreeding. This is also termed consanguinity.<br /> Inbreeding sometimes occurs in human societies and is more<br /> likely to take place in nature when population size becomes very<br /> limited. In Chapter 25, we will examine how inbreeding is a useful<br /> strategy for developing agricultural breeds or strains with<br /> desirable characteristics. Conversely, outbreeding, which involves<br /> mating between unrelated individuals, can create hybrids that are<br /> heterozygous for many genes.<br /> In the absence of other evolutionary processes, inbreeding<br /> and outbreeding do not affect allele frequencies in a population.<br /> However, these patterns of mating do disrupt the balance<br /> of genotypes that is predicted by the Hardy-Weinberg equation.<br /> Let’s first consider inbreeding in a family pedigree. Figure 24.18<br /> illustrates a human pedigree involving a mating between cousins.<br /> Individuals III-2 and III-3 are cousins and have produced the<br /> daughter labeled IV-1. She is said to be inbred, because her parents<br /> are genetically related to each other.<br /> During inbreeding, the gene pool is smaller, because the<br /> parents are related genetically. In the 1940s, Gustave Malécot<br /> developed methods to quantify the degree of inbreeding. The<br /> inbreeding coefficient is the probability that two alleles in a particular<br /> individual will be identical for a given gene because both<br /> copies are due to descent from a common ancestor. An inbreeding<br /> coefficient (F) can be computed by analyzing the degree of<br /> relatedness within a pedigree.<br /> As an example, let’s determine the inbreeding coefficient<br /> for individual IV-1. To begin this problem, we must first identify<br /> all of this individual’s common ancestors. A common ancestor<br /> is anyone who is an ancestor to both of an individual’s parents.<br /> In Figure 24.18, IV-1 has one common ancestor, I-2, her greatgrandfather.<br /> I-2 is the grandfather of III-2 and III-3.<br /> Our next step is to determine the inbreeding paths. An<br /> inbreeding path for an individual is the shortest path through<br /> the pedigree that includes both parents and the common ancestor.<br /> In a pedigree, there is an inbreeding path for each common<br /> ancestor. The length of each inbreeding path is calculated<br /> by adding together all of the individuals in the path except the<br /> individual of interest. In this case, there is only one path because<br /> IV-1 has only one common ancestor. To add the members of the<br /> path, we begin with individual IV-1, but we do not count her. We<br /> then move to her father (III-2); to her grandfather (II-2); to I-2,<br /> her great-grandfather (the common ancestor); back down to her<br /> other grandmother (II-3); and finally to her mother (III-3). This<br /> path has five members. Finally, to calculate the inbreeding coefficient,<br /> we use the following formula:<br /> where<br /> F � �(1/2) n (1 � F A )<br /> F is the inbreeding coefficient of the individual of<br /> interest<br /> n is the number of individuals in the inbreeding path,<br /> excluding the inbred offspring<br /> F A is the inbreeding coefficient of the common<br /> ancestor<br /> � indicates that we add together (1/2) n (1 � F A ) for<br /> each inbreeding path<br /> In this case, there is only one common ancestor and, therefore,<br /> only one inbreeding path. Also, we do not know anything about<br /> the heritage of the common ancestor, so we assume that F A is<br /> zero. Thus, in our example of Figure 24.18,<br /> F � �(1/2) n (1 � 0)<br /> � (1/2) 5 � 1/32 � 3.125%<br /> What does this value mean? Our inbreeding coefficient,<br /> 3.125%, tells us the probability that a gene in the inbred individual<br /> (IV-1) is homozygous due to its inheritance from a common<br /> ancestor (I-2). In this case, therefore, each gene in individual<br /> IV-1 has a 3.125% chance of being homozygous because she has<br /> inherited the same allele twice from her great-grandfather (I-2),<br /> once through each parent.<br /> As an example, let’s suppose that the common ancestor<br /> (I-2) is heterozygous for the gene involved with cystic fibrosis. His<br /> genotype would be Cc, where c is the recessive allele that causes<br /> cystic fibrosis. There is a 3.125% probability that the inbred individual<br /> (IV-1) is homozygous (CC or cc) for this gene because</div> <div>17.2 SITE-SPECIFIC RECOMBINATION 465<br /> sites within those genes and catalyze the breakage and reunion of<br /> DNA segments. Before we discuss the details of this mechanism,<br /> let’s first consider the biology of antibodies.<br /> Antibodies, or immunoglobulins (Igs), are proteins produced<br /> by the B cells of the immune system. Their function is to<br /> recognize foreign material, such as viruses and bacteria, and target<br /> them for destruction. A foreign material that elicits an immune<br /> response is called an antigen. Antibodies recognize sites within<br /> antigens. The recognition between an antibody and antigen is<br /> very specific. Within the immune system, each B cell produces a<br /> single type of antibody. However, our bodies have millions of B<br /> cells. Site-specific recombination allows each B cell to produce an<br /> antibody with a different amino acid sequence. These differences<br /> in the amino acid sequences of antibody proteins enable them to<br /> recognize different antigens. In this way, site-specific recombination<br /> plays an important role in the ability of the immune system<br /> to identify an impressive variety of foreign materials as being<br /> antigens and thereby target them for destruction.<br /> From a genetic viewpoint, the production of millions of<br /> different antibodies poses an interesting problem. If a distinct<br /> gene was needed to produce each different antibody polypeptide,<br /> the genome would need to contain millions of different antibody<br /> genes. By comparison, consider that the entire human genome<br /> contains only about 20,000 to 25,000 different genes. How then<br /> is it possible for the genome to produce millions of different<br /> antibody proteins, which are encoded by genes? The answer to<br /> this question baffled geneticists for several decades until research<br /> revealed that antibodies with different polypeptide sequences<br /> are generated by an unusual mechanism in which the DNA is<br /> cut and reconnected by site-specific recombination. We might<br /> call this “DNA splicing,” although the term splicing is usually<br /> reserved for the cutting and rejoining of RNA molecules. With<br /> this mechanism, only a few large antibody precursor genes are<br /> needed to produce millions of different antibodies. These precursor<br /> genes are spliced in many different ways to produce a vast<br /> array of polypeptides with differing amino acid sequences.<br /> Figure 17.9 shows the site-specific recombination of an antibody<br /> precursor gene. Antibodies are tetrameric proteins composed<br /> of two heavy polypeptide chains and two light chains. One type of<br /> light chain is the κ (kappa) light chain, which is a component of a<br /> class of antibodies known as immunoglobulin G (IgG). The organization<br /> of the precursor gene for the κ light chain is shown at the<br /> top of Figure 17.9. At the left side, the gene has approximately 300<br /> regions known as variable (V) sequences or domains. In addition,<br /> the gene contains four different joining (J) sequences and a single<br /> constant (C) sequence. Each variable domain or joining domain<br /> encodes a different amino acid sequence.<br /> During the maturation of B cells, the κ light-chain precursor<br /> gene is cut and rejoined so that one variable domain becomes<br /> adjacent to a joining domain. At the end of every V domain and<br /> the beginning of every J domain is located a recombination signal<br /> sequence that functions as a recognition site for site-specific<br /> recombination between the V and J regions. The recombination<br /> event is initiated by two proteins called RAG1 and RAG2. RAG<br /> is an acronym for recombination-activating gene. These proteins<br /> recognize recombination signal sequences and generate two<br /> Organization of domains in the precursor gene for the K light-chain<br /> 5′<br /> NH 2<br /> NH 2<br /> NH 2<br /> V 1 V 2 V 3 V . . . . J1 J J 2 3<br /> J 4 C<br /> 4 V 300<br /> . . . .<br /> V 1 V 2 V 3 V 4 . . . . V 78<br /> V 1 V 2 V 3 V 4 . . . . V 78 J 2<br /> NH 2<br /> 5′<br /> NH 2<br /> V 78<br /> V 78<br /> V 78<br /> RAG1 and RAG2 recognize recombination<br /> signal sequences and catalyze the<br /> breakage at the end of a variable domain<br /> and beginning of a joining domain. In this<br /> example, it occurs at the end of V 78 and<br /> beginning of J 2 .<br /> V 300 J 1<br /> J J 2 3<br /> J 4 C<br /> The intervening DNA is lost. NHEJ<br /> proteins catalyze the joining of the<br /> last V domain and first J domain in<br /> the remaining DNA.<br /> J 3<br /> J 4<br /> The gene is transcribed into a<br /> pre-mRNA starting at the last<br /> variable domain.<br /> J<br /> J J 3 4 C<br /> 2<br /> V 78<br /> V 78<br /> J 2 C<br /> J 2 C<br /> J 2<br /> J 2<br /> The region between the first joining<br /> domain and constant domain in the<br /> pre-mRNA is spliced out.<br /> C<br /> The mRNA is translated into a<br /> polypeptide containing 1 variable,<br /> 1 joining, and 1 constant domain.<br /> C<br /> 3′<br /> Two light-chain polypeptides and 2<br /> heavy-chain polypeptides assemble to<br /> form a functional antibody protein.<br /> COOH<br /> COOH<br /> COOH<br /> C<br /> Heavy-chain<br /> polypeptide<br /> COOH<br /> COOH<br /> 3′<br /> A functional<br /> antibody<br /> made in<br /> 1 B cell<br /> FIGURE 17.9 Site-specific recombination within the<br /> precursor gene that encodes the κ light chain for immunoglobulin G<br /> (IgG) proteins.<br /> double-strand breaks, one at the end of a V domain and one at<br /> the beginning of a J domain. For example, in the recombination<br /> event shown in Figure 17.9, RAG1 and RAG2 have made cuts</div> <div>796 SOLUTIONS TO EVEN-NUMBERED PROBLEMS<br /> E26. PFGE is a method of electrophoresis that is used to separate small<br /> chromosomes and large DNA fragments. The electrophoresis devices<br /> used in PFGE have two sets of electrodes. The two sets of electrodes<br /> produce alternating pulses of current, and this facilitates the<br /> separation of large DNA fragments.<br /> It is important to handle the sample gently to prevent the breakage of<br /> the DNA due to mechanical forces. Cells are first embedded in agarose<br /> blocks, and then the blocks are loaded into the wells of the gel. The<br /> agarose keeps the sample very stable and prevents shear forces that<br /> might mechanically break the DNA. After the blocks are in the gel, the<br /> cells within the blocks are lysed, and, if desired, restriction enzymes<br /> can be added to digest the DNA. For PFGE, a restriction enzyme that<br /> cuts very infrequently might be used.<br /> PFGE can be used as a preparative technique to isolate and purify<br /> individual chromosomes or large DNA fragments. PFGE can also be<br /> used, in conjunction with Southern blotting, as a mapping technique.<br /> E28. Note that the insert of cosmid B is contained completely within the<br /> insert of cosmid C.<br /> 7,000 bp D<br /> 2,200 bp B<br /> 11,500 bp C<br /> 6,000 bp A<br /> E30. A. The general strategy is shown in Figure 20.18. The researcher<br /> begins at a certain location and then walks toward the gene of interest.<br /> You begin with a clone that has a marker that is known to map<br /> relatively close to the gene of interest. A piece of DNA at the end of<br /> the insert is subcloned and then used in a Southern blot to identify<br /> an adjacent clone in a cosmid DNA library. This is the first “step.”<br /> The end of this clone is subcloned to make the next step. And so on.<br /> Eventually, after many steps, you will arrive at your gene of interest.<br /> B. In this example, you would begin at STS-3. If you walked a few<br /> steps and happened upon STS-2, you would know that you were<br /> walking in the wrong direction.<br /> C. This is a difficult aspect of chromosome walking. Basically, you<br /> would walk toward gene X using DNA from a normal individual<br /> and DNA from an individual with a mutant gene X. When you<br /> have found a site where the sequences are different between<br /> the normal and mutant individual, you may have found gene<br /> X. You would eventually have to confirm this by analyzing the<br /> DNA sequence of this region and determining that it encodes a<br /> functional gene.<br /> E32. We can calculate the probability that a base will not be sequenced<br /> using this approach with the following equation:<br /> P � e �m<br /> where<br /> P is the probability that a base will be left unsequenced<br /> e is the base of the natural logarithm; e � 2.72<br /> m is the number of bases sequenced divided by the total genome<br /> size<br /> In this case, m � 19 divided by 4.4, which equals 4.3<br /> P � e �m � e �4.3 � 0.0136 � 1.36%<br /> This means that if we randomly sequence 19.0 Mb, we are likely to<br /> miss only 1.36% of the genome. With a genome size of 4.4 Mb, we<br /> would miss about 57,840 base pairs out of approximately 4,400,000.<br /> Questions for Student Discussion/Collaboration<br /> 2. A molecular marker is a segment of DNA, not usually encoding a<br /> gene, that has a known location within a particular chromosome. It<br /> marks the location of a site along a chromosome. RFLPs and STSs are<br /> examples. It is easier to use these types of markers because they can be<br /> readily identified by molecular techniques such as restriction digestion<br /> analysis and PCR. The locations of functional genes is usually<br /> more difficult because this relies on conventional genetic mapping<br /> approaches whereby allelic differences in the gene are mapped by<br /> making crosses or following a pedigree. For monomorphic genes, this<br /> approach doesn’t work.<br /> CHAPTER 21<br /> Conceptual Questions<br /> C2. There are two main reasons why the proteome is larger than the<br /> genome. The first reason involves the processing of pre-mRNA, a<br /> phenomenon that occurs primarily in eukaryotic species. RNA splicing<br /> and editing can alter the codon sequence of mRNA and thereby<br /> produce alternative forms of proteins that have different amino acid<br /> sequences. The second reason for protein diversity is posttranslational<br /> modifications. There are many ways that a given protein’s structure<br /> can be covalently modified by cellular enzymes. These include<br /> proteolytic processing, disulfide bond formation, glycosylation,<br /> attachment of lipids, phosphorylation, methylation, and acetylation, to<br /> name a few.<br /> C4. Centromeric sequences, origins of replication, telomeric sequences,<br /> repetitive sequences, and enhancers. Other examples are possible.<br /> C6. There are a few interesting trends. Sequences 1 and 2 are similar to<br /> each other, as are sequences 3 and 4. There are a few places where<br /> amino acid residues are conserved among all five sequences. These<br /> amino acids may be particularly important with regard to function.<br /> C8. A. Correct<br /> B. Correct<br /> C. This is not correct. These are short genetic sequences that happen<br /> to be similar to each other. The lac operon and trp operon are not<br /> derived from the same primordial operon.<br /> D. This is not correct. The two genes are homologous to each other. It<br /> is correct to say that their sequences are 60% identical.<br /> Experimental Questions<br /> E2. As described in solved problem S1, one reason for making a<br /> subtractive DNA library is to determine which RNAs are produced<br /> when environmental conditions change. You want to load a small<br /> amount of cDNA on the column that was derived from cells that had<br /> been exposed to mercury. Remember that the cDNA, which is derived<br /> from mRNA that is made in the absence of mercury, is already bound<br /> to the column. You want the cDNA that is made in the presence of<br /> mercury to bind to this cDNA if it is complementary. If too much of<br /> this cDNA is loaded, all of the cDNAs will have complementary cDNA<br /> bound to them, and some of them will not bind to the column, even<br /> though they may be complementary to the cDNAs. In other words, if<br /> you load too much cDNA (derived from the mercury-exposed cells),<br /> you will have saturated the binding sites for cDNAs that are made<br /> in the absence of mercury. You do not want this to happen, because<br /> you want only the cDNAs that are derived from mRNAs that are<br /> specifically expressed in the presence of mercury to flow through the<br /> column. These cDNAs are not complementary to any cDNAs attached<br /> to the column.<br /> E4. The cDNA labeled with a green dye is derived from mRNA obtained<br /> from cells at an early time point, when glucose levels were high. The<br /> other samples of cDNA were derived from cells collected at later time<br /> points, when glucose levels were falling and when the diauxic shift was</div> <div>:: ROBERT J. BROOKER<br /> UNIVERSITY OF MINNESOTA–MINNEAPOLIS<br /> third edition<br /> GENETICS<br /> ANALYSIS & PRINCIPLES</div> <div>SOLVED PROBLEMS 155<br /> showed that mutations can occur at many different sites within a<br /> single gene. Furthermore, intragenic crossing over could recombine<br /> these mutations, resulting in wild-type genes. Therefore,<br /> rather than being an indivisible particle, a gene must be composed<br /> of a large structure that can be subdivided during crossing over.<br /> Benzer’s results were published in the late 1950s and early<br /> 1960s, not long after the physical structure of DNA had been elucidated<br /> by Watson and Crick. We now know that a gene is a segment<br /> of DNA that is composed of smaller building blocks called<br /> nucleotides. A typical gene is a linear sequence of several hundred<br /> to many thousand base pairs. As the genetic map of Figure<br /> 6.20 indicates, mutations can occur at many sites along the linear<br /> structure of a gene; intragenic crossing over can recombine mutations<br /> that are located at different sites within the same gene.<br /> CONCEPTUAL SUMMARY<br /> This chapter has been concerned with genetic transfer and mapping<br /> in bacteria. Three mechanisms are known for the transfer<br /> of genetic material from one bacterial cell to another: conjugation,<br /> transduction, and transformation. During conjugation, bacteria<br /> make direct physical contact with each other. Donor cells<br /> can be either F + , F', or Hfr cells. An F + cell contains a circular<br /> piece of genetic material, a plasmid called an F factor that is<br /> transferred to the recipient (F – ) cell during mating. An F' factor<br /> is an F factor that also carries genes derived from the bacterial<br /> chromosome. An Hfr cell (for high frequency of recombination)<br /> has an integrated F factor and can transfer chromosomal DNA<br /> to a recipient cell. A second route of genetic transfer is transduction.<br /> In this case, a bacteriophage accidentally packages a portion<br /> of the bacterial chromosome, which is then transferred to<br /> another bacterium upon infection. Finally, a third mechanism of<br /> genetic transfer is transformation. For this to occur, a dead bacterial<br /> cell must release its genetic material into the environment.<br /> A living bacterial cell in a competent state subsequently imports<br /> the DNA, and then recombination causes the imported DNA to<br /> replace segments of genetic material along the bacterial chromosome.<br /> Overall, these three mechanisms promote gene transfer<br /> within bacterial species and also allow horizontal gene transfer<br /> among different bacterial species. From a medical perspective,<br /> this phenomenon has led to a dramatic rise in the prevalence of<br /> strains that are resistant to antibiotics.<br /> Chapter 6 concluded with a description of intragenic mapping<br /> studies in T4 bacteriophages. Benzer constructed a fine<br /> structure genetic map of two bacteriophage genes, rIIA and rIIB.<br /> In the early 1960s, his fine structure map provided important<br /> insights into the molecular nature of the gene. It revealed that a<br /> gene is not an indivisible unit. Rather, his results indicated that<br /> mutations can occur at many sites along the linear structure of a<br /> gene and that intragenic crossing over can recombine mutations<br /> located at different sites within the same gene. These results are<br /> consistent with our current knowledge of the molecular structure<br /> of genes.<br /> EXPERIMENTAL SUMMARY<br /> Research studies of genetic transfer in bacteria have provided a<br /> unique experimental strategy for mapping the linear order and<br /> relative locations of bacterial genes. Conjugation has been used<br /> to map the locations of many genes along the bacterial chromosome.<br /> In this approach, map distances are determined by the<br /> number of minutes it takes for a gene to enter a recipient cell<br /> during conjugation. In addition, cotransduction and cotransformation<br /> experiments have been commonly used to accurately<br /> map bacterial genes that are relatively close to each other on the<br /> chromosome.<br /> As mentioned in the Conceptual Summary, intragenic mapping<br /> studies of T4 bacteriophages have provided great insight into<br /> the molecular nature of genes. Benzer constructed a fine structure<br /> genetic map of two bacteriophage genes, rIIA and rIIB. This was<br /> accomplished using a coinfection strategy, in which the very low<br /> percentage of intragenic recombinants yielding wild-type phages<br /> could be identified by their unique ability to infect a strain of bacteria<br /> known as E. coli K12(λ). This approach illustrates the power<br /> of phage genetics compared to the analysis of offspring in eukaryotic<br /> species. The selective inability of rII mutants to lyse E. coli<br /> K12(λ) allowed detection of the very few intragenic recombinants<br /> that produced wild-type phages that could lyse E. coli K12(λ). In<br /> this way, the very short distance that occurs between mutations<br /> within a single gene could be determined.<br /> PROBLEM SETS & INSIGHTS<br /> Solved Problems<br /> S1. In E. coli, the gene bioD + encodes an enzyme involved in biotin<br /> synthesis, and galK + encodes an enzyme involved in galactose<br /> utilization. An E. coli strain that contained wild-type versions of both<br /> genes was infected with P1, and then a P1 lysate was obtained. This<br /> lysate was used to transduce (infect) a strain that was bioD – and<br /> galK – . The cells were plated on media containing galactose as the sole<br /> carbon source for growth to select for transduction of the galK + gene.<br /> These media also were supplemented with biotin. The colonies were<br /> then restreaked on media that lacked biotin to see if the bioD + gene<br /> had been cotransduced. The following results were obtained:</div> <div>756 CHAPTER 26 :: EVOLUTIONARY GENETICS<br /> Experimental Questions<br /> E1. Two populations of snakes are separated by a river. The snakes<br /> cross the river only on rare occasions. The snakes in the two<br /> populations look very similar to each other except that the<br /> members of the population on the eastern bank of the river have<br /> a yellow spot on the top of their head, while the members of<br /> the western population have an orange spot on the top of their<br /> head. Discuss two experimental methods that you might follow to<br /> determine whether the two populations are members of the same<br /> species or members of different species.<br /> E2. Sympatric speciation by allotetraploidy has been proposed as<br /> a common mechanism for speciation. Let’s suppose you were<br /> interested in the origin of certain grass species in Southern<br /> California. Experimentally, how would you go about determining<br /> if some of the grass species are the result of allotetraploidy?<br /> E3. Two diploid species of closely related frogs, which we will call<br /> species A and species B, were analyzed with regard to genes that<br /> encode an enzyme called hexokinase. Species A has two distinct<br /> copies of this gene: A1 and A2. In other words, this diploid species<br /> is A1A1 A2A2. The other species has three copies of the hexokinase<br /> gene, which we will call B1, B2, and B3. A diploid individual of<br /> species B would be B1B1 B2B2 B3B3. These hexokinase genes<br /> from the two species were subjected to DNA sequencing, and the<br /> percentage of sequence identity was compared among these genes.<br /> The results are shown here:<br /> Percentage of DNA Sequence Identity<br /> A1 A2 B1 B2 B3<br /> A1 100 62 54 94 53<br /> A2 62 100 91 49 92<br /> B1 54 91 100 67 90<br /> B2 94 49 67 100 64<br /> B3 53 92 90 64 100<br /> If we assume that hexokinase genes were never lost in the<br /> evolution of these frog species, how many distinct hexokinase<br /> genes do you think there were in the most recent ancestor that<br /> preceded the divergence of these two species? Explain your answer.<br /> Also explain why species B has three distinct copies of this gene,<br /> whereas species A has only two.<br /> E4. A researcher sequenced a portion of a bacterial gene and obtained<br /> the following sequence, beginning with the start codon, which is<br /> underlined:<br /> ATG CCG GAT TAC CCG GTC CCA AAC AAA ATG<br /> ATC GGC CGC CGA ATC TAT CCC<br /> The bacterial strain that contained this gene has been maintained<br /> in the laboratory and grown serially for many generations.<br /> Recently, another person working in the laboratory isolated DNA<br /> from the bacterial strain and sequenced the same region. The<br /> following results were obtained.<br /> ATG CCG GAT TAT CCG GTC CCA AAT AAA ATG<br /> ATC GGC CGC CGA ATC TAC CCC<br /> Explain why these sequencing differences may have occurred.<br /> E5. F 1 hybrids between two species of cotton, Gossypium barbadense<br /> and G. hirsutum, are very vigorous plants. However, F 1 crosses<br /> produce many seeds that do not germinate and a low percentage of<br /> very weak F 2 offspring. Suggest two reasons for these observations.<br /> E6. A species of antelope contains 20 chromosomes per set. The<br /> species is divided by a mountain range into two separate<br /> populations, which we will call the eastern and western population.<br /> When comparing the karyotypes between these two populations,<br /> it was discovered that the members of the eastern population are<br /> homozygous for a large inversion within chromosome 14. How<br /> would this inversion affect the interbreeding between the two<br /> populations? Could such an inversion play an important role in<br /> speciation?<br /> E7. Explain why molecular techniques were needed as a way to provide<br /> evidence for the neutral theory of evolution.<br /> E8. Prehistoric specimens often contain minute amounts of ancient<br /> DNA. What technique can be used to increase the amount of DNA<br /> in an older sample? Explain how this technique is performed and<br /> how it increases the amount of a specific region of DNA.<br /> E9. In the experiment of Figure 26.15, explain how we know that the<br /> kiwis are more closely related to the emu and cassowary than to<br /> the moas. Cite particular regions in the sequences that support<br /> your answer.<br /> E10. In Chapter 20, we learned about a technique called fluorescence in<br /> situ hybridization (FISH), during which a labeled piece of DNA is<br /> hybridized to a set of chromosomes. Let’s suppose that we cloned<br /> a piece of DNA from G. pubescens (see Figure 26.3) and used it as<br /> a labeled probe for in situ hybridization. What would you expect<br /> to happen if we hybridized it to the G. speciosa, the natural G.<br /> tetrahit, or the artificial G. tetrahit strains? Describe your expected<br /> results.<br /> E11. A team of researchers has obtained a dinosaur bone<br /> (Tyrannosaurus rex) and has attempted to extract ancient DNA<br /> from it. Using primers to the 12S rRNA gene, they have used PCR<br /> and obtained a DNA segment that yields a sequence homologous<br /> to crocodile DNA. Other scientists are skeptical that this sequence<br /> is really from the dinosaur. Instead, they believe that it may be due<br /> to contamination from more recent DNA, such as the remains of a<br /> reptile that lived much more recently. What criteria might you use<br /> to establish the credibility of the dinosaur sequence?<br /> E12. Discuss how the principle of parsimony can be used in a cladistics<br /> approach of constructing a phylogenetic tree.<br /> E13. As discussed in this chapter and Chapter 24, genes are sometimes<br /> transferred between different species via horizontal gene transfer.<br /> Discuss how horizontal gene transfer might give misleading results<br /> when constructing a phylogenetic tree. How could you overcome<br /> this problem?<br /> E14. If a researcher used genetic engineering techniques to express<br /> the Drosophila eyeless gene in the embryo at the region that will<br /> become the tip of the mouse’s tail, what results would you expect<br /> in the resulting offspring?</div> <div>3.3 SEXUAL REPRODUCTION 59<br /> In female animals, oogenesis, the production of egg cells,<br /> occurs within specialized diploid cells of the ovary known as<br /> oogonia. Quite early in the development of the ovary, the oogonia<br /> initiate meiosis to produce primary oocytes. For example,<br /> in humans, approximately 1 million primary oocytes per ovary<br /> are produced before birth. These primary oocytes are arrested—<br /> enter a dormant phase—at prophase of meiosis I, remaining at<br /> this stage until the female becomes sexually mature. Beginning at<br /> this stage, primary oocytes are periodically activated to progress<br /> through the remaining stages of oocyte development.<br /> During oocyte maturation, meiosis produces only one cell<br /> that is destined to become an egg, as opposed to the four gametes<br /> produced from each primary spermatocyte during spermatogenesis.<br /> How does this occur? As shown in Figure 3.14b, the first<br /> meiotic division is asymmetric and produces a secondary oocyte<br /> and a much smaller cell, known as a polar body. Most of the<br /> cytoplasm is retained by the secondary oocyte and very little by<br /> the polar body, allowing the oocyte to become a larger cell with<br /> more stored nutrients. The secondary oocyte then begins meiosis<br /> II. In mammals, the secondary oocyte is released from the<br /> ovary—an event called ovulation—and travels down the oviduct<br /> toward the uterus. During this journey, if a sperm cell penetrates<br /> the secondary oocyte, it is stimulated to complete meiosis II; the<br /> secondary oocyte produces a haploid egg and a second polar<br /> body. The haploid egg and sperm nuclei then unite to create the<br /> diploid nucleus of a new individual.<br /> Anther<br /> Microsporocyte<br /> Meiosis<br /> Microspores (4)<br /> Megasporocyte<br /> Megaspore<br /> Ovary<br /> Meiosis<br /> Plant Species Alternate Between Haploid<br /> (Gametophyte) and Diploid (Sporophyte)<br /> Generations<br /> Most species of animals are diploid, and their haploid gametes<br /> are considered to be a specialized type of cell. By comparison, the<br /> life cycles of plant species alternate between haploid and diploid<br /> generations. The haploid generation is called the gametophyte,<br /> whereas the diploid generation is called the sporophyte. Meiosis<br /> produces haploid cells called spores, which divide by mitosis to<br /> produce the gametophyte. In simpler plants, such as mosses, a<br /> haploid spore can produce a large multicellular gametophyte by<br /> repeated mitoses and cellular divisions. In flowering plants, however,<br /> spores develop into gametophytes that contain only a few<br /> cells. In this case, the organism that we think of as a “plant” is the<br /> sporophyte, while the gametophyte is very inconspicuous. In fact,<br /> the gametophytes of many plant species are small structures produced<br /> within the much larger sporophyte. Certain cells within<br /> the haploid gametophytes then become specialized as haploid<br /> gametes.<br /> Figure 3.15 provides an overview of gametophyte development<br /> and gametogenesis in flowering plants. Meiosis occurs<br /> within two different structures of the sporophyte: the anthers<br /> and the ovaries, which produce male and female gametophytes,<br /> respectively. This diagram depicts a flower from an angiosperm,<br /> which is a plant that produces seeds within an ovary.<br /> In the anther, diploid cells called microsporocytes undergo<br /> meiosis to produce four haploid microspores. These separate into<br /> Mitosis<br /> Each of 4<br /> microspores<br /> Pollen grain<br /> (the male<br /> gametophyte)<br /> Nucleus of the<br /> tube cell<br /> Generative cell<br /> (will divide to form<br /> 2 sperm cells)<br /> Ovule<br /> Cell<br /> division<br /> Three<br /> antipodal<br /> cells<br /> One central<br /> cell with 2<br /> polar nuclei<br /> Egg cell<br /> Two synergid<br /> cells<br /> Embryo sac<br /> (the female<br /> gametophyte)<br /> FIGURE 3.15 The formation of male and female gametes by<br /> the gametophytes of angiosperms.<br /> individual microspores. In many angiosperms, each microspore<br /> undergoes mitosis to produce a two-celled structure containing one<br /> tube cell and one generative cell, both of which are haploid. This<br /> structure differentiates into a pollen grain, which is the male gametophyte<br /> with a thick cell wall. Later, the generative cell undergoes</div> <div>522 CHAPTER 19 :: BIOTECHNOLOGY<br /> + Cloned gene<br /> + Cloned gene<br /> Normal<br /> gene<br /> Normal<br /> gene<br /> FIGURE 19.5 The introduction<br /> of a cloned gene into a cell can lead<br /> to gene replacement or gene addition.<br /> In these two examples, the cloned gene<br /> is similar to a normal gene already<br /> present in the genome of a cell. (a) If<br /> the cloned gene undergoes homologous<br /> recombination and replaces the<br /> normal gene, this event is called gene<br /> replacement. (b) Alternatively, the cloned<br /> gene may recombine nonhomologously<br /> at some other chromosomal location,<br /> leading to gene addition.<br /> (a) Gene replacement<br /> Homologous<br /> recombination<br /> Cloned<br /> gene<br /> (b) Gene addition<br /> Normal<br /> gene<br /> Nonhomologous<br /> recombination<br /> Cloned<br /> gene<br /> FIGURE 19.6 The use of gene<br /> addition to produce fish that glow. The<br /> aquarium fish shown here, which are named<br /> GloFish®, are transgenic organisms that have<br /> received a gene from jellyfish or sea corals<br /> that encodes a fluorescent protein, causing<br /> them to glow green, red, or yellow.<br /> Molecular Biologists Can Produce Mice<br /> That Contain Gene Replacements<br /> In bacteria and yeast, which have relatively small genomes,<br /> homologous recombination between cloned genes and the host<br /> cell chromosome occurs at a relatively high rate, so gene replacement<br /> is commonly achieved. Gene replacement is useful when a<br /> researcher or biotechnologist wants to make a gene mutation in<br /> the laboratory and then introduce the mutant gene into a living<br /> organism to compare the effects of the normal and mutant genes<br /> on the phenotype. However, in more complex eukaryotes with very<br /> large genomes, the introduction of cloned genes into cells is much<br /> more likely to result in gene addition rather than gene replacement.<br /> For example, when a cloned gene is introduced into a mouse cell,</div> <div>15.3 REGULATION OF RNA PROCESSING, RNA STABILITY, AND TRANSLATION 409<br /> repressor causes exon 2 to be spliced out of the mature mRNA,<br /> an event called exon skipping. Alternatively, other splicing factors<br /> enhance the ability of the spliceosome to recognize particular<br /> splice sites. In Figure 15.17b, splicing enhancers bind to the 3'<br /> and 5' splice sites that flank exon 3, which results in the inclusion<br /> of exon 3 in the mature mRNA.<br /> Alternative splicing in different tissues is thought to occur<br /> because each cell type has its own characteristic concentration of<br /> many kinds of splicing factors. Furthermore, much like transcription<br /> factors, splicing factors may be regulated by the binding of<br /> small effector molecules, protein–protein interactions, and covalent<br /> modifications. Overall, the differences in the composition of<br /> splicing factors, and the regulation of their activities, form the<br /> basis for alternative splicing decisions.<br /> TABLE 15.3<br /> Examples of RNA Editing<br /> Organism Type of Editing Found in<br /> Trypanosomes Primarily additions but Many mitochondrial mRNAs<br /> (protozoa) occasionally deletions<br /> of uracil nucleotides<br /> Land plants C-to-U conversion Many mitochondrial and<br /> chloroplast mRNAs, tRNAs,<br /> and rRNAs<br /> Slime mold C additions Many mitochondrial mRNAs<br /> Mammals C-to-U conversion Apolipoprotein B mRNA, and<br /> NFI mRNA, which encodes a<br /> tumor-suppressor protein<br /> A-to-I conversion<br /> Glutamate receptor mRNA,<br /> many tRNAs<br /> Drosophila A-to-I conversion mRNA for calcium and<br /> sodium channels<br /> The Nucleotide Sequence of RNA<br /> Can Be Modified by RNA Editing<br /> The term RNA editing refers to a change in the nucleotide<br /> sequence of an RNA molecule that involves additions or deletions<br /> of particular bases, or a conversion of one type of base to<br /> another, such as a cytosine to a uracil. In the case of mRNAs,<br /> editing can have various effects, such as generating start codons,<br /> generating stop codons, and changing the coding sequence for a<br /> polypeptide.<br /> The phenomenon of RNA editing was first identified in<br /> trypanosomes, the protists that cause sleeping sickness. As with<br /> the discovery of RNA splicing, the initial finding of RNA editing<br /> was met with great skepticism. Since that time, however, RNA<br /> editing has been shown to occur in various organisms and in<br /> a variety of ways, although its functional significance is slowly<br /> emerging (Table 15.3).<br /> The molecular mechanisms of RNA editing have been<br /> the subject of many investigations. In the specific case of trypanosomes,<br /> which are single-celled protists, the editing process<br /> involves the participation of a guide RNA. This guide RNA<br /> directs the addition of one or more uracil nucleotides into an<br /> RNA or the deletion of one or more uracils. As shown in Figure<br /> 15.18, the guide RNA has two important characteristics. First,<br /> its 5' anchor is complementary to the RNA that is to be edited.<br /> Second, it has uracil nucleotides at its 3' end. During the editing<br /> process, the 5' anchor binds to the target RNA. The RNA is<br /> cleaved at a defined location by an endonuclease, and the 3' end<br /> of the guide RNA becomes displaced from the target RNA. For<br /> additions, an enzyme called terminal U-transferase then attaches<br /> uracil-containing nucleotides. For deletions, an enzyme called 3'-<br /> U-exonuclease removes one or more uracils. Finally, RNA ligase<br /> rejoins the two pieces of RNA.<br /> A more widespread mechanism for RNA editing involves<br /> changes of one type of base to another. In this form of editing,<br /> a base in the RNA is deaminated—an amino group is removed<br /> from the base. When cytosine is deaminated, uracil is formed,<br /> and when adenine is deaminated, inosine is formed (Figure<br /> 15.19). Inosine is recognized as guanine during translation.<br /> An example of RNA editing occurs in mammals involving<br /> an mRNA that encodes a protein called apolipoprotein B. In the<br /> liver, the RNA editing process produces apolipoprotein B-100,<br /> a protein essential for the transport of cholesterol in the blood.<br /> In intestinal cells, the mRNA may be edited so that a single C is<br /> changed to a U. What is the significance of this base substitution?<br /> This change converts a glutamine codon (CAA) to a stop codon<br /> (UAA) and thereby results in a shorter apolipoprotein. In this<br /> case, RNA editing produces an apolipoprotein B with an altered<br /> structure. Therefore, RNA editing can produce two proteins from<br /> the same gene, much like the phenomenon of alternative splicing<br /> described earlier in this chapter.<br /> The Stability of mRNA Influences<br /> mRNA Concentration<br /> In eukaryotes, the stability of mRNAs can vary considerably.<br /> Certain mRNAs have very short half-lives, such as several minutes,<br /> whereas others can persist for many days or even months.<br /> In some cases, the stability of an mRNA can be regulated so that<br /> its half-life is shortened or lengthened. A change in the stability<br /> of mRNA can greatly influence the cellular concentration of that<br /> mRNA molecule. In this way, mechanisms that control mRNA<br /> stability can dramatically affect gene expression.<br /> Various factors can play a role in mRNA stability. One<br /> important structural feature is the length of the polyA tail. As you<br /> may recall from Chapter 12, most newly made mRNAs contain a<br /> polyA tail that averages 200 nucleotides in length. The polyA tail<br /> is recognized by the polyA-binding protein. The binding of this<br /> protein enhances RNA stability. However, as an mRNA ages, its<br /> polyA tail tends to be shortened by the action of cellular exonucleases.<br /> Once it becomes less than 10 to 30 adenosines in length,<br /> the polyA-binding protein can no longer bind, and the mRNA is<br /> rapidly degraded by exo- and endonucleases.<br /> Certain mRNAs, particularly those with short half-lives,<br /> contain sequences that act as destabilizing elements. While these<br /> destabilizing elements can be located anywhere within the mRNA,</div> <div>CHAPTER 26 805<br /> each gene that controls weight. For example, let’s suppose there<br /> are 20 genes that affect weight, with each gene existing in a light<br /> and heavy allele. During the early stages of selective breeding,<br /> when Mary and Hector picked their 10 plants as seed producers<br /> for the next generation, as a matter of random chance, some of<br /> these plants may have been homozygous for the light alleles at<br /> a few of the 20 genes that control weight. Therefore, just as a<br /> matter of chance, they probably “lost” a few of the heavy alleles<br /> that affect weight. So, after 12 generations of breeding, they have<br /> predominantly heavy alleles but also have light alleles for some<br /> of the genes. If we represent heavy alleles with a capital letter and<br /> light alleles with a lowercase letter, Mary’s and Hector’s strains<br /> could be the following:<br /> Mary’s strain:<br /> AA BB cc DD EE FF gg hh II JJ KK LL mm<br /> NN OO PP QQ RR ss TT<br /> Hector’s strain: AA bb CC DD EE ff GG HH II jj kk LL MM<br /> NN oo PP QQ RR SS TT<br /> As we see here, Mary’s strain is homozygous for the heavy allele at<br /> 15 of the genes but carries the light allele at the other 5. Similarly,<br /> Hector’s strain is homozygous for the heavy allele at 15 genes and<br /> carries the light allele at the other 5. It is important to note, however,<br /> that the light alleles in Mary’s and Hector’s strains are not in the same<br /> genes. Therefore, when Martin crosses them together, he will initially<br /> get the following:<br /> Martin’s F 1 offspring: AA Bb Cc DD EE Ff Gg Hh II Jj Kk LL<br /> Mm NN Oo PP QQ RR Ss TT<br /> If the alleles are additive and contribute equally to the trait, we<br /> would expect about the same weight (1.5 lb), because this hybrid<br /> has a total of 10 light alleles. However, if heterosis is occurring,<br /> genes (which were homozygous recessive in Mary’s and Hector’s<br /> strains) will become heterozygous in the F 1 offspring, and this<br /> may make the plants healthier and contribute to a higher weight.<br /> If Martin’s F 1 strain is subjected to selective breeding, the 10 genes<br /> that are heterozygous in the F 1 offspring may eventually become<br /> homozygous for the heavy allele. This would explain why Martin’s<br /> tomatoes achieved a weight of 2.0 pounds after five generations of<br /> selective breeding.<br /> E18. A.<br /> hN<br /> 2<br /> � X O � X<br /> X P � X<br /> h N<br /> 2<br /> � 269 � 254 � 0.56<br /> 281 � 254<br /> B. 0.56 =<br /> 275 � 254<br /> X P � 254<br /> X P = 291.5 lb<br /> Questions for Student Discussion/Collaboration<br /> 2. Most traits depend on the influence of many genes. Also, genetic<br /> variation is a common phenomenon in most populations. Therefore,<br /> most individuals have a variety of alleles that contribute to a given<br /> trait. For quantitative traits, some alleles may make the trait bigger,<br /> and other alleles may make the trait turn out smaller. If a population<br /> contains many different genes and alleles that govern a quantitative<br /> trait, most individuals will have an intermediate phenotype<br /> because they will have inherited some large and some small alleles.<br /> Fewer individuals will inherit a predominance of large alleles or a<br /> predominance of small alleles. An example of a quantitative trait that<br /> does not fit a normal distribution is snail pigmentation. The dark<br /> snails and light snails are favored rather than the intermediate colors<br /> because they are less susceptible to predation.<br /> CHAPTER 26<br /> Conceptual Questions<br /> C2. Evolution is unifying because all living organisms on this planet<br /> evolved from the same primordial organism. At the molecular level,<br /> all organisms have a great deal in common. With the exception of<br /> a few viruses, they all use DNA as their genetic material. This DNA<br /> is found within chromosomes, and the sequence of the DNA is<br /> organized into units called genes. Most genes are structural genes<br /> that encode the amino acid sequence of polypeptides. Polypeptides<br /> fold to form functional units called proteins. At the cellular level, all<br /> living organisms also share many similarities. For example, living cells<br /> share many of the same basic features including a plasma membrane,<br /> ribosomes, enzymatic pathways, and so on. In addition, as discussed<br /> in Chapter 7, the mitochondria and chloroplasts of eukaryotic cells are<br /> evolutionarily derived from bacterial cells.<br /> C4. 1. Phylogenetic—Each species has distinct morphological, cellular,<br /> and/or molecular traits.<br /> 2. Biological—Each species is considered distinct if it is<br /> reproductively isolated from other species.<br /> 3. Evolutionary—Each species is derived from a single lineage that<br /> is distinct from other lineages and has its own evolutionary<br /> tendencies and historical fate.<br /> 4. Ecological—Each species occupies its own ecological niche.<br /> C6. Anagenesis is the evolution of one species into another, whereas<br /> cladogenesis is the divergence of one species into two or more<br /> species. Of the two, cladogenesis is more prevalent. There may be<br /> many reasons why. It is common for an abrupt genetic change such<br /> as alloploidy to produce a new species from a preexisting one. Also,<br /> migrations of a few members of species into a new region may<br /> lead to the formation of a new species in the region (i.e., allopatric<br /> speciation).<br /> C8. A. Allopatric<br /> B. Sympatric<br /> C. At first, it may involve parapatric speciation with a low level of<br /> intermixing. Eventually, when smaller lakes are formed, allopatric<br /> speciation will occur.<br /> C10. The main evidence in favor of punctuated equilibrium is the fossil<br /> record. Paleontologists rarely find a gradual transition of fossil<br /> forms. The transition period in which environment pressure and<br /> genetic changes cause a previous species to evolve into a new<br /> species is thought to be so short that few, if any, of the transitional<br /> members would be preserved as fossils. Therefore, the fossil record<br /> primarily contains representatives from the long equilibrium periods.<br /> Also, rapid evolutionary change is consistent with known genetic<br /> phenomena, including single-gene mutations that have dramatic<br /> effects on phenotypic characteristics, the founder effect, and genetic<br /> events such as changes in chromosome structure (e.g., inversions<br /> and translocations) or chromosome number, which may abruptly<br /> create individuals with new phenotypic traits. In some cases, however,<br /> gradual changes are observed in certain species over long periods of<br /> time. In addition, the gradual accumulation of mutations is known to<br /> occur from the molecular analyses of DNA.<br /> C12. Allopatric speciation involves a physical separation of a species<br /> into two or more separate populations. Over time, each population<br /> accumulates mutations that alter the characteristics of each<br /> population. Because the populations are separated, each will evolve<br /> different characteristics and eventually become distinct species. In<br /> parapatric speciation, there is some physical separation of two or more<br /> populations, but the separation is not absolute. On occasion, members<br /> of different populations can interbreed. Even so, the (somewhat)<br /> separated populations will tend to accumulate different genetic</div> <div>12.4 RNA MODIFICATION 317<br /> As an example, let’s suppose a mammalian pre-mRNA<br /> contained seven exons. In one cell type (e.g., muscle cells), it<br /> may be spliced to produce a mature mRNA with the following<br /> pattern of exons: 1-2-4-5-6-7. In a different cell type (e.g., nerve<br /> cells), it might be spliced in an alternative pattern: 1-2-3-5-6-7.<br /> In this example, the mRNA from muscle cells contains exon 4,<br /> while the mRNA from nerve cells contains exon 3. When alternative<br /> splicing occurs, proteins with significant differences in their<br /> amino acid sequences are produced. In most cases, the alternative<br /> versions of a protein will have similar functions, because much<br /> of their amino acid sequences will be identical to each other.<br /> Nevertheless, alternative splicing produces differences in amino<br /> acid sequences that provide each protein with its own unique<br /> characteristics. The biological advantage of alternative splicing is<br /> that two or more different proteins can be derived from a single<br /> gene. This allows an organism to carry fewer genes in its genome.<br /> The molecular mechanism of alternative splicing is examined in<br /> greater detail in Chapter 15. It involves the actions of proteins<br /> (not shown in Figure 12.22) that influence whether or not U1<br /> and U2 can begin the splicing process.<br /> The Ends of Eukaryotic Pre-mRNAs<br /> Have a 5' Cap and a 3' Tail<br /> In addition to splicing, pre-mRNAs in eukaryotes are also subjected<br /> to modifications at their 5' and 3' ends. At their 5' end,<br /> most mature mRNAs have a 7-methylguanosine covalently<br /> attached, an event known as capping. Capping occurs while the<br /> pre-mRNA is being made by RNA polymerase II, usually when<br /> the transcript is only 20 to 25 nucleotides in length. As shown<br /> in Figure 12.23, it is a three-step process. The nucleotide at the<br /> 5' end of the transcript has three phosphate groups. First, an<br /> enzyme called RNA 5'-triphosphatase removes one of the phosphates,<br /> and then a second enzyme, guanylyltransferase, uses GTP<br /> to attach a guanosine monophosphate (GMP) to the 5' end.<br /> Finally, a methyltransferase attaches a methyl group to the guanine<br /> base.<br /> What are the functions of the 7-methylguanosine cap?<br /> The cap structure is recognized by cap-binding proteins, which<br /> perform various roles. For example, cap-binding proteins are<br /> required for the proper exit of most mRNAs from the nucleus.<br /> Also, the cap structure is recognized by initiation factors that<br /> are needed during the early stages of translation. Finally, the cap<br /> structure may be important in the efficient splicing of introns,<br /> particularly the first intron located nearest the 5' end.<br /> Let’s now turn our attention to the 3' end of the RNA molecule.<br /> Most mature mRNAs have a string of adenine nucleotides,<br /> referred to as a polyA tail, which is important for mRNA stability<br /> and in the synthesis of polypeptides. The polyA tail is not<br /> encoded in the gene sequence. Instead, it is added enzymatically<br /> after the pre-mRNA has been completely transcribed. This process<br /> is termed polyadenylation.<br /> The steps required to synthesize a polyA tail are shown in<br /> Figure 12.24. To acquire a polyA tail, the pre-mRNA contains a<br /> polyadenylation sequence near its 3' end. In higher eukaryotes,<br /> the consensus sequence is AAUAAA. This sequence is downstream<br /> NH 2<br /> 5′<br /> O<br /> O – P<br /> O<br /> O –<br /> O<br /> O O<br /> Base<br /> O– – P<br /> O P<br /> O P O<br /> CH 2 O<br /> O –<br /> O –<br /> O –<br /> H<br /> H<br /> H<br /> H<br /> O OH<br /> O<br /> P O– –<br /> O CH 2<br /> 3′<br /> P i<br /> H<br /> O<br /> N<br /> N<br /> N<br /> PP i<br /> N H<br /> OH<br /> O<br /> O P<br /> O<br /> O –<br /> O<br /> P<br /> O<br /> O –<br /> O<br /> P<br /> O<br /> O –<br /> O<br /> Base<br /> P O CH 2 O<br /> O –<br /> H<br /> H<br /> H<br /> H<br /> O<br /> OH<br /> O P O –<br /> O CH 2<br /> Rest of mRNA<br /> RNA 5′-triphosphatase<br /> removes a phosphate.<br /> O<br /> Base<br /> P O CH 2 O<br /> O –<br /> H<br /> H<br /> H<br /> H<br /> O<br /> OH<br /> O P O –<br /> O CH 2<br /> Rest of mRNA<br /> Guanylyltransferase<br /> hydrolyzes GTP. The GMP is<br /> attached to the 5′ end, and<br /> PP i is released.<br /> O<br /> O<br /> Base<br /> H HO CH 2 P O<br /> P O CH 2<br /> O –<br /> O<br /> H<br /> O<br /> O –<br /> O –<br /> O H<br /> H<br /> H<br /> H<br /> H<br /> O<br /> OH<br /> O P O –<br /> O CH<br /> H<br /> 2<br /> O<br /> N<br /> Rest of mRNA<br /> NH 2<br /> CH 3 Methyltransferase attaches<br /> N<br /> +<br /> a methyl group.<br /> N<br /> N H<br /> OH<br /> O<br /> O<br /> O<br /> Base<br /> H HO CH 2 O P<br /> O P<br /> O<br /> P O CH 2<br /> O –<br /> O<br /> H<br /> O<br /> O –<br /> O –<br /> O H<br /> H<br /> H<br /> H<br /> H<br /> 7-methylguanosine cap<br /> O<br /> OH<br /> O P O –<br /> O CH 2<br /> Rest of mRNA<br /> FIGURE 12.23 Attachment of a 7-methylguanosine cap to the<br /> 5' end of mRNA. When the transcript is about 20 to 25 nucleotides in<br /> length, RNA 5'-triphosphatase removes one of the three phosphates, and<br /> then a second enzyme, guanylyltransferase, attaches GMP to the 5' end.<br /> Finally, a methyltransferase attaches a methyl group to the guanine base.</div> <div>688 CHAPTER 24 :: POPULATION GENETICS<br /> Minisatellite<br /> Chromosomal DNA<br /> (individual #1)<br /> RRR RRR RRRRR<br /> 1,500 bp<br /> 2,500 bp<br /> 3,000 bp<br /> FIGURE 24.22 A comparison of minisatellites between<br /> two individuals. The restriction sites found in both individuals are<br /> represented by arrows. Each individual carries three minisatellites,<br /> shown in orange. The repeat units within each minisatellite are depicted<br /> with the letter R. The variation in the number of repeats affects the<br /> sizes of the DNA fragments produced when the DNA is digested<br /> with the restriction enzyme that cleaves at the designated sites. (Note:<br /> Minisatellites actually have more repeat units than shown here.)<br /> Chromosomal DNA<br /> (individual #2)<br /> 4,500<br /> 4,000<br /> 3,500<br /> 3,000<br /> 2,500<br /> 2,000<br /> 1,500<br /> RRR<br /> 1,500 bp<br /> RRRRRRRRRRR<br /> 4,000 bp<br /> #1 #2<br /> Cut the DNA with a restriction<br /> enzyme that cleaves at the<br /> recognition sites labeled with<br /> arrows. Separate DNA fragments<br /> by gel electrophoresis.<br /> R<br /> 2,000 bp<br /> enzyme; these differences yield a distinct pattern of DNA fragments<br /> when analyzed via gel electrophoresis.<br /> Figure 24.22 depicts only a short segment of DNA. When<br /> the chromosomal DNA from a sample of actual cells is digested<br /> with a restriction enzyme, this would yield too many DNA fragments<br /> to analyze. In traditional DNA fingerprinting, DNA probes<br /> were used that hybridized specifically to the repeat sequence<br /> located within selected minisatellites. DNA fingerprinting probes<br /> were made that recognized a selected minisatellite sequence found<br /> at a relatively small number of sites (say, 5 to 30 sites) in the<br /> human genome. Using such probes, a DNA fingerprinting analysis<br /> examined the length of the DNA fragments at the sites of the<br /> chosen minisatellite sequence. This involved the analysis of 5 to 30<br /> bands that correspond to minisatellite sites, as in Figure 24.21.<br /> Figure 24.23a outlines the steps in a traditional DNA fingerprinting<br /> experiment. This procedure is a Southern blot using<br /> a radiolabeled probe complementary to a selected minisatellite<br /> sequence. The chromosomal DNA is isolated from a sample and<br /> digested with a restriction enzyme. The resulting DNA fragments<br /> are then separated by gel electrophoresis. The fragments in the<br /> gel are blotted onto a nylon membrane, the DNA is denatured,<br /> and the membrane is exposed to the radiolabeled probe. Because<br /> the probe is complementary to a selected minisatellite sequence,<br /> it hybridizes to approximately 5 to 30 fragments of DNA that<br /> contain this sequence and thereby labels 5 to 30 bands. In Figure<br /> 24.23a, the results of a DNA fingerprinting experiment on two<br /> samples are compared. Because the pattern of band sizes does not<br /> match between the two samples, it is concluded that the samples<br /> came from different individuals.<br /> In the past decade, the technique of DNA fingerprinting<br /> has become automated, much like the automation that changed<br /> the procedure of DNA sequencing described in Chapter 18.<br /> DNA fingerprinting is now done using the technique of polymerase<br /> chain reaction (PCR), which amplifies microsatellites.<br /> Like minisatellites, microsatellites are found in multiple sites in<br /> the genome of humans and other species and are variable among<br /> different individuals. In this procedure, the microsatellites from<br /> a sample of DNA are amplified by PCR using primers that flank<br /> the repetitive region and then separated by gel electrophoresis<br /> according to their molecular masses. As in automated DNA<br /> sequencing, the amplified microsatellite fragments are fluorescently<br /> labeled. A laser excites the fluorescent molecule within a<br /> microsatellite, and a detector records the amount of fluorescence<br /> emission for each microsatellite. As shown in Figure 24.23b, this<br /> type of DNA fingerprint yields a series of peaks, each peak having<br /> a characteristic molecular mass. In this automated approach,<br /> the pattern of peaks rather than bands constitutes an individual’s<br /> DNA fingerprint.</div> <div>520 CHAPTER 19 :: BIOTECHNOLOGY<br /> species. These Ice + species synthesize cellular proteins that promote<br /> ice nucleation—the initiation of ice crystals. Using recombinant<br /> DNA technology, Lindow constructed an Ice – strain of<br /> P. syringae in the early 1980s, which lacks the gene responsible<br /> for the production of ice-nucleation proteins. When applied to<br /> the surface of plants, an Ice – strain can compete with and thereby<br /> reduce the proliferation of Ice + bacteria, thus inhibiting the formation<br /> of frost.<br /> Lindow sought approval for field tests of an Ice – recombinant<br /> strain in Tulelake, California. For several years, these tests<br /> were delayed because of a lawsuit from the Foundation on Economic<br /> Trends. During that time, Lindow made great efforts to<br /> ensure the safety of this project by studying the local environment.<br /> He also consulted with local townspeople where the field<br /> test was to take place. Initially, the idea was well received by the<br /> local residents. However, another company tested similar bacteria<br /> on the roof of an Oakland facility without Environmental<br /> Protection Agency (EPA) approval. The media reported this incident,<br /> and it caused many Tulelake townspeople to become apprehensive<br /> about the release of recombinant bacteria. Nevertheless,<br /> in 1987, approval was finally granted for the field testing of the<br /> recombinant P. syringae (Figure 19.3).<br /> During the first test on several thousand strawberry plants,<br /> the plants were ripped out by vandals. In a second field test, the<br /> ability of Ice – bacteria to protect potato plants was tested. While<br /> some of the plants were destroyed by vandals, the results of this<br /> field experiment showed that the Ice – bacteria did protect potato<br /> plants from frost damage. In addition, soil sampling showed that<br /> the recombinant bacteria were contained at the field site and did<br /> not proliferate into surrounding areas. Even so, the release of<br /> recombinant microorganisms into the environment remains controversial.<br /> Since this first test, relatively few recombinant strains<br /> have been released. In the case of Ice – bacteria, further research<br /> was discouraged by a variety of factors, including governmental<br /> regulation and the expense of doing additional experiments. The<br /> recombinant Ice – bacteria were never commercialized.<br /> FIGURE 19.3<br /> test.<br /> The release of recombinant P. syringae in a field<br /> Microorganisms Can Reduce<br /> Environmental Pollutants<br /> The term bioremediation refers to the use of living organisms<br /> or their products to decrease pollutants in the environment. As<br /> its name suggests, this is a biological remedy for pollution. During<br /> bioremediation via microorganisms, enzymes produced by<br /> a microorganism modify a toxic pollutant by altering or transforming<br /> its structure. This event is called biotransformation.<br /> In many cases, biotransformation results in biodegradation,<br /> in which the toxic pollutant is degraded, yielding less complex,<br /> nontoxic metabolites. Alternatively, biotransformation without<br /> biodegradation can also occur. For example, toxic heavy metals<br /> can often be rendered less toxic by oxidation or reduction reactions<br /> carried out by microorganisms. Another way to alter the<br /> toxicity of organic pollutants is by promoting polymerization. In<br /> many cases, polymerized toxic compounds are less likely to leach<br /> from the soil and, therefore, are less environmentally toxic than<br /> their parent compounds.<br /> Since the early 1900s, microorganisms have been used in<br /> the treatment and degradation of sewage. More recently, the field<br /> of bioremediation has expanded into the treatment of hazardous<br /> and refractory chemical wastes—chemicals that are difficult<br /> to degrade and usually associated with industrial activity. These<br /> pollutants include petroleum hydrocarbons, halogenated organic<br /> compounds, pesticides, herbicides, and organic solvents. Many<br /> new applications that use microorganisms to degrade these pollutants<br /> are being tested. The field of bioremediation has been<br /> fostered, to a large extent, by better knowledge of how pollutants<br /> are degraded by microorganisms, the identification of new and<br /> useful strains of microbes, and the ability to enhance bioremediation<br /> through genetic engineering.<br /> Molecular genetic technology is key in identifying genes<br /> that encode enzymes involved in bioremediation. The characterization<br /> of the relevant genes greatly enhances our understanding<br /> of how microbes can modify toxic pollutants. In addition, recombinant<br /> strains created in the laboratory can be more efficient at<br /> degrading certain types of pollutants.<br /> In 1980, in a landmark case (Diamond v. Chakrabarty), the<br /> U.S. Supreme Court ruled that a live, recombinant microorganism<br /> is patentable as a “manufacture or composition of matter.”<br /> The first recombinant microorganism to be patented was an “oileating”<br /> bacterium that contained a laboratory-constructed plasmid.<br /> This strain can oxidize the hydrocarbons commonly found<br /> in petroleum. It grew faster on crude oil than did any of the natural<br /> isolates tested. However, it has not been a commercial success<br /> because this recombinant strain metabolizes only a limited<br /> number of toxic compounds; the number of compounds actually<br /> present in crude oil is over 3,000. Unfortunately, the recombinant<br /> strain did not degrade many higher-molecular-weight compounds,<br /> which tend to persist in the environment.<br /> Currently, bioremediation should be considered a developing<br /> industry. This field will need well-trained molecular geneticists<br /> to conduct research aimed at elucidating the mechanisms<br /> whereby microorganisms degrade toxic pollutants. In the future,<br /> recombinant microorganisms may provide an effective way to</div> <div>396 CHAPTER 15 :: GENE REGULATION IN EUKARYOTES<br /> How does the glucocorticoid receptor regulate the expression<br /> of particular genes? In the nucleus, the glucocorticoid<br /> receptor homodimer binds to glucocorticoid response elements<br /> (GREs) with the following consensus sequence:<br /> 5'–AGRACA–3'<br /> 3'–TCYTGT–5'<br /> where R is a purine and Y is pyrimidine. This sequence is found<br /> next to many genes and functions as an enhancer. The binding<br /> of the glucocorticoid receptor homodimer to adjacent GREs activates<br /> the transcription of the nearby gene, eventually leading to<br /> the synthesis of the encoded protein.<br /> Mammalian cells usually have a large number of glucocorticoid<br /> receptors within the cytoplasm. Because GREs are<br /> located near dozens of different genes, the uptake of many hormone<br /> molecules can activate many glucocorticoid receptors and<br /> thereby stimulate the transcription of many different genes. For<br /> this reason, a cell can respond to the presence of the hormone<br /> in a very complex way. Glucocorticoid hormones stimulate many<br /> genes that encode proteins involved in several different cellular<br /> processes, including the synthesis of glucose, the breakdown of<br /> proteins, and the mobilization of fats. Although the genes are<br /> not physically connected to each other, the regulation of multiple<br /> genes via glucocorticoid hormones is much like the ability<br /> of bacterial operons to simultaneously control the expression of<br /> several genes.<br /> The CREB Protein Is an Example<br /> of a Regulatory Transcription Factor<br /> Modulated by Covalent Modification<br /> As we have just seen, steroid hormones function as signaling<br /> molecules that bind directly to regulatory transcription factors to<br /> alter their function. This enables a cell to respond to a hormone<br /> by up regulating a particular set of genes. Most extracellular signaling<br /> molecules, however, do not enter the cell or bind directly<br /> to transcription factors. Instead, most signaling molecules must<br /> bind to receptors in the plasma membrane. This binding activates<br /> the receptor and may lead to the synthesis of an intracellular<br /> signal that causes a cellular response. One type of cellular<br /> response is to affect the transcription of particular genes within<br /> the cell.<br /> As our second example of regulatory transcription factor<br /> function within living cells, we will examine the cAMP response<br /> element–binding (CREB) protein. The CREB protein is a regulatory<br /> transcription factor that becomes activated in response to<br /> cell-signaling molecules that cause an increase in the cytoplasmic<br /> concentration of the molecule cyclic adenosine monophosphate<br /> (cAMP). This transcription factor recognizes a response element<br /> with the following consensus sequence.<br /> 5'–TGACGTCA–3'<br /> 3'–ACTGCAGT–5'<br /> This response element, which is found near many different genes,<br /> has been termed a cAMP response element (CRE).<br /> Figure 15.7 shows the steps leading to the activation of the<br /> CREB protein. A wide variety of hormones, growth factors, neurotransmitters,<br /> and other signaling molecules can bind to plasma<br /> membrane receptors to initiate an intracellular response. In this<br /> case, the response involves the production of a second messenger,<br /> cAMP. The extracellular signaling molecule itself is considered<br /> the primary messenger. When the signaling molecule binds to the<br /> receptor, it activates a G protein that subsequently activates the<br /> enzyme adenylyl cyclase. The activated adenylyl cyclase catalyzes<br /> the synthesis of cAMP. The cAMP molecule then binds to a second<br /> enzyme, protein kinase A, and activates it. This enzyme travels<br /> into the nucleus and phosphorylates several different cellular<br /> proteins, including the CREB protein. When phosphorylated,<br /> Extracellular<br /> signaling<br /> molecule<br /> Protein<br /> kinase A<br /> Phosphorylates<br /> Plasma<br /> membrane<br /> receptor<br /> Activates<br /> ATP<br /> CREB<br /> protein<br /> dimer<br /> G protein<br /> Activates<br /> 2–<br /> PO 4<br /> CRE<br /> Activates<br /> 2–<br /> PO 4<br /> cAMP<br /> Target gene<br /> Adenylyl<br /> cyclase<br /> ATP<br /> Core<br /> promoter<br /> FIGURE 15.7 The activity of the CREB protein. When an<br /> extracellular signaling molecule binds to a receptor in the plasma<br /> membrane, this activates a G protein, which activates adenylyl cyclase,<br /> leading to the synthesis of cAMP. Next, cAMP binds to protein kinase<br /> A, which activates it. Protein kinase A then travels into the nucleus and<br /> phosphorylates the CREB protein. Once phosphorylated, the CREB<br /> protein acts as a transcriptional activator.</div> <div>722 CHAPTER 25 :: QUANTITATIVE GENETICS<br /> Now here is the question. Using the data found in this chapter<br /> regarding weight in cattle, what is the predicted weight of an<br /> offspring if its mother weighed 660 lb?<br /> Answer: We first need to calculate a and b.<br /> b 5 CoV (X, Y)<br /> V X<br /> We need to use the data on page 701 to calculate V X , which is the<br /> variance for the mothers’ weights. The variance equals 445.1. The<br /> covariance is already calculated on page 701; it equals 152.6.<br /> b 5 152.6<br /> 445.1<br /> b � 0.34<br /> a 5 Y 2 bX<br /> a � 598 � (0.34)(596) � 395.4<br /> Now we are ready to calculate the predicted weight of the offspring<br /> using the equation<br /> Y � bX � a<br /> In this problem, X � 660 pounds<br /> Y � 0.34(660) � 395.4<br /> Y � 619.8 lb<br /> The average weight of the offspring is predicted to be 619.8 lb.<br /> S5. Genetic variance can be used to estimate the number of genes<br /> affecting a quantitative trait by using the following equation:<br /> n 5 D2<br /> 8V G<br /> where<br /> n is the number of genes affecting the trait<br /> D is the difference between the mean values of the trait in<br /> two strains that have allelic differences at every gene that<br /> influences the trait<br /> V G is the genetic variance for the trait; it is calculated using<br /> data from both strains<br /> For this method to be valid, several assumptions must be met.<br /> In particular, the alleles of each gene must be additive, each<br /> gene must contribute equally to the trait, all the genes must<br /> assort independently, and the two strains must be homozygous<br /> for alternative alleles of each gene. For example, if three genes<br /> affecting a quantitative trait exist in two alleles each, one strain<br /> could be AA bb CC and the other would be aa BB cc. In addition,<br /> the strains must be raised under the same environmental<br /> conditions. Unfortunately, these assumptions are not typically<br /> met with regard to most quantitative traits. Even so, when one or<br /> more assumptions are invalid, the calculated value of n is smaller<br /> than the actual number. Therefore, this calculation can be used to<br /> estimate the minimum number of genes that affect a quantitative<br /> trait.<br /> Now here is the question. The average bristle number in<br /> two strains of flies was 35 and 42. The genetic variance for bristle<br /> number calculated for both strains was 0.8. What is the minimum<br /> number of genes that affect bristle number?<br /> Answer: We apply the equation described previously.<br /> n 5 D2<br /> 8V G<br /> (35 2 42)2<br /> n 5<br /> 8(0.8)<br /> n � 7.7 genes<br /> Because genes must come in whole numbers and because this<br /> calculation is a minimum estimate, one would conclude there must be<br /> at least eight genes that affect bristle number.<br /> S6. Are the following statements regarding heritability true or false?<br /> A. Heritability applies to a specific population raised in a<br /> particular environment.<br /> B. Heritability in the narrow sense takes into account all types of<br /> genetic variance.<br /> C. Heritability is a measure of the amount that genetics contributes<br /> to the outcome of a trait.<br /> Answer:<br /> A. True<br /> B. False. Narrow-sense heritability considers only the effects of<br /> additive alleles.<br /> C. False. Heritability is a measure of the amount of phenotypic<br /> variation that is due to genetic variation; it applies to the<br /> variation of a specific population raised in a particular<br /> environment.<br /> Conceptual Questions<br /> C1. Give several examples of quantitative traits. How are these<br /> quantitative traits described within groups of individuals?<br /> C2. At the molecular level, explain why quantitative traits often exhibit<br /> a continuum of phenotypes within a population. How does the<br /> environment help produce this continuum?<br /> C3. What is a normal distribution? Discuss this curve with regard to<br /> quantitative traits within a population. What is the relationship<br /> between the standard deviation and the normal distribution?<br /> C4. Explain the difference between a continuous trait and a<br /> discontinuous trait. Give two examples of each. Are quantitative<br /> traits likely to be continuous or discontinuous? Explain why.<br /> C5. What is a frequency distribution? Explain how the graph is made<br /> for a quantitative trait that is continuous.<br /> C6. The variance for weight in a particular herd of cattle is 484 lb 2 .<br /> The mean weight is 562 lb. How heavy would an animal have to be<br /> if it was in the top 2.5% of the herd? The bottom 0.13%?<br /> C7. Two different varieties of potatoes both have the same mean<br /> weight of 1.5 lb. One group has a very low variance, and the other<br /> has a much higher variance.</div> <div>618 CHAPTER 22 :: MEDICAL GENETICS AND CANCER<br /> molecule fits into the active site of the ABL protein, preventing<br /> ATP from binding there. Without ATP, the ABL protein cannot<br /> phosphorylate its target proteins. This prevents the ABL protein<br /> from stimulating cell division. In a clinical trial, almost 90% of<br /> the CML patients treated with the drug showed no further progression<br /> of their disease!<br /> Other forms of cancer also involve chromosomal translocations<br /> that cause an overexpression of an oncogene. In Burkitt’s<br /> lymphoma, for example, a region of chromosome 8 is translocated<br /> to either chromosome 2, 14, or 22. The breakpoint in chromosome<br /> 8 is near the c-myc gene, and the sites on chromosomes<br /> 2, 14, and 22 correspond to locations of different immunoglobulin<br /> genes that are normally expressed in lymphocytes. The translocation<br /> of the c-myc gene near the immunoglobulin genes leads<br /> to the overexpression of the c-myc gene and thereby promotes<br /> malignancy in lymphocytes.<br /> A fourth way that oncogenes may occur is via viral integration.<br /> As part of their reproductive cycle, certain viruses integrate<br /> their genomes into the chromosomal DNA of their host cell. If<br /> the integration occurs next to a proto-oncogene, a viral promoter<br /> or enhancer sequence may cause the proto-oncogene to be<br /> overexpressed. For example, in certain lymphomas that occur in<br /> birds, the genome of the avian leukosis virus has been found to<br /> be integrated next to the c-myc gene and enhances its level of<br /> transcription.<br /> Tumor-Suppressor Genes Play a Role<br /> in Preventing the Proliferation of Cancer Cells<br /> Thus far, we have considered how oncogenes promote cancer.<br /> An oncogene is an abnormally activated gene that leads to<br /> uncontrolled cell growth. We will now turn our attention to a<br /> second category of genes called tumor-suppressor genes. As the<br /> name suggests, the role of a tumor-suppressor gene is to prevent<br /> cancerous growth. Therefore, when a tumor-suppressor gene<br /> becomes inactivated by mutation, it becomes more likely that<br /> cancer will occur.<br /> The first identification of a human tumor-suppressor gene<br /> involved studies of retinoblastoma, a tumor that occurs in the<br /> retina of the eye. Some people have inherited a predisposition to<br /> develop this disease within the first few years of life. By comparison,<br /> the noninherited form of retinoblastoma, which is caused by<br /> environmental agents, tends to occur later in life but only rarely.<br /> Based on these differences, in 1971, Alfred Knudson proposed<br /> a “two-hit” model for retinoblastoma. According to this<br /> model, retinoblastoma requires two mutations to occur. People<br /> with the hereditary form already have received one mutant gene<br /> from one of their parents. They need only one additional mutation<br /> in the other copy of this tumor-suppressor gene to develop<br /> the disease. Because the retina has more than 1 million cells, it is<br /> relatively likely that a mutation may occur in one of these cells<br /> at an early age, leading to the disease. However, people with the<br /> noninherited form of the disease must have two mutations in the<br /> same retinal cell to cause the disease. Because two rare events are<br /> much less likely to occur than a single such event, the noninherited<br /> form of this disease is expected to occur much later in life<br /> and only rarely. Therefore, this hypothesis explains the different<br /> populations typically affected by the inherited and noninherited<br /> forms of retinoblastoma.<br /> Since Knudson’s original hypothesis, molecular studies<br /> have confirmed the two-hit hypothesis for retinoblastoma. In<br /> this case, the gene in which mutations occur is designated rb (for<br /> retinoblastoma). This tumor suppressor gene is found on the<br /> long arm of chromosome 13. Most people have two normal copies<br /> of the rb gene. Persons with hereditary retinoblastoma have<br /> inherited one normal and one defective copy. In nontumorous<br /> cells throughout the body, they have one functional copy and one<br /> defective copy of rb. However, in retinal tumor cells, the normal<br /> rb gene has also suffered the second hit (i.e., a mutation), which<br /> renders it defective. Without the tumor-suppressor ability, cells<br /> are allowed to grow and divide in an unregulated manner, which<br /> ultimately leads to cancer. (In contrast, as discussed later, most<br /> other forms of cancer involve mutations in several genes.)<br /> More recent studies have revealed how the Rb protein suppresses<br /> the proliferation of cancer cells (Figure 22.14). The Rb<br /> protein regulates a transcription factor called E2F, which activates<br /> genes required for cell cycle progression. (The eukaryotic cell<br /> cycle is described earlier in Figure 22.11a.) The binding of the Rb<br /> protein to E2F inhibits its activity and prevents the cell from progressing<br /> through the cell cycle. As discussed later in this chapter,<br /> when a normal cell is supposed to divide, cellular proteins called<br /> cyclins bind to cyclin-dependent protein kinases. This activates<br /> Inactive<br /> complex<br /> PO 4<br /> Active<br /> transcription<br /> factor<br /> E2F<br /> E2F<br /> E2F<br /> Rb<br /> Phosphorylation of Rb by cyclindependent<br /> protein kinases causes<br /> it to dissociate from E2F.<br /> Target gene transcription occurs<br /> when E2F binds to DNA.<br /> Target gene<br /> FIGURE 22.14 Interactions between the Rb and E2F proteins.<br /> The binding of the Rb protein to the transcription factor E2F inhibits<br /> the ability of E2F to function. This prevents cell division. For cell<br /> division to occur, cyclins bind to cyclin-dependent protein kinases,<br /> which then phosphorylate the Rb protein. The phosphorylated Rb<br /> protein is released from E2F. The free form of E2F can activate target<br /> genes needed to progress through the cell cycle.</div> <div>9.2 NUCLEIC ACID STRUCTURE 239<br /> as histone proteins, the double helix becomes greatly twisted and<br /> folded. Figure 9.21 depicts the relationship between the DNA<br /> double helix and the compaction that occurs within a eukaryotic<br /> chromosome. Chapter 10 is devoted to the topic of chromosome<br /> organization and the molecular mechanisms responsible for the<br /> packaging of genetic material in cells.<br /> RNA Molecules Are Composed of Strands That Fold<br /> into Specific Structures<br /> Let’s now turn our attention to RNA structure, which bears<br /> many similarities to DNA structure. The structure of an RNA<br /> strand is much like a DNA strand (Figure 9.22). Strands of RNA<br /> are typically several hundred or several thousand nucleotides in<br /> length, which is much shorter than chromosomal DNA. When<br /> RNA is made during transcription, the DNA is used as a template<br /> to make a copy of single-stranded RNA. In most cases,<br /> only one of the two DNA strands is used as a template for RNA<br /> synthesis. Therefore, only one complementary strand of RNA is<br /> usually made. Nevertheless, relatively short sequences within one<br /> RNA molecule or between two separate RNA molecules can form<br /> double-stranded regions.<br /> The helical structure of RNA molecules is due to the ability<br /> of complementary regions to form base pairs between A and U<br /> and between G and C. This base pairing allows short segments<br /> to form a double-stranded region. As shown in Figure 9.23, different<br /> types of structural patterns are possible. These include<br /> bulge loops, internal loops, multibranched junctions, and stemloops<br /> (also called hairpins). These structures contain regions<br /> of complementarity punctuated by regions of noncomplementarity.<br /> As shown in Figure 9.23, the complementary regions are<br /> held together by connecting hydrogen bonds, while the noncomplementary<br /> regions have their bases projecting away from the<br /> double-stranded region.<br /> Many factors contribute to the structure of RNA molecules.<br /> These include the base-paired double-stranded helices,<br /> stacking between bases, and hydrogen bonding between bases<br /> and backbone regions. In addition, interactions with ions, small<br /> molecules, and large proteins may influence RNA structure.<br /> Figure 9.24 depicts the structure of a transfer RNA molecule<br /> known as tRNA phe , which is a tRNA molecule that carries the<br /> amino acid phenylalanine. It was the first naturally occurring<br /> RNA to have its structure elucidated. This RNA molecule has<br /> several double-stranded and single-stranded regions. RNA double<br /> helices are antiparallel and right-handed with 11 to 12 base<br /> pairs per turn. In a living cell, the various regions of an RNA<br /> molecule fold and interact with each other to produce the threedimensional<br /> structure.<br /> Radial loops<br /> (300 nm in diameter)<br /> Metaphase<br /> chromosome<br /> 30 nm fiber<br /> Nucleosomes<br /> (11 nm in diameter)<br /> DNA<br /> (2 nm in diameter)<br /> Histone<br /> protein<br /> Each chromatid<br /> (700 nm in diameter)<br /> FIGURE 9.21 The steps in eukaryotic chromosomal compaction leading to the metaphase chromosome. The DNA double helix is wound<br /> around histone proteins and then is further compacted to form a highly condensed metaphase chromosome. The levels of DNA compaction will be<br /> described in greater detail in Chapter 10.</div> <div>164 CHAPTER 7 :: NON-MENDELIAN INHERITANCE<br /> early steps of embryogenesis. The accumulation of maternal gene<br /> products in the egg allows embryogenesis to proceed quickly<br /> after fertilization. Maternal effect genes often play a role in cell<br /> division, cleavage pattern, and body axis orientation. Therefore,<br /> defective alleles in maternal effect genes tend to have a dramatic<br /> effect on the phenotype of the individual, altering major features<br /> of morphology, often with dire consequences.<br /> Our understanding of maternal effect genes has been<br /> greatly aided by their identification in experimental organisms<br /> such as Drosophila melanogaster. In such organisms with a short<br /> generation time, geneticists have successfully searched for mutant<br /> alleles that prevent the normal process of embryonic development.<br /> In Drosophila, geneticists have identified several maternal<br /> effect genes with profound effects on the early stages of development.<br /> The pattern of development of a Drosophila embryo<br /> occurs along axes, such as the anteroposterior axis and the dorsoventral<br /> axis. The proper development of each axis requires a<br /> distinct set of maternal gene products. For example, the maternal<br /> effect gene called bicoid produces a gene product that accumulates<br /> in a region of the egg that will eventually become anterior<br /> structures in the developing embryo. Mutant alleles of maternal<br /> effect genes often lead to abnormalities in the anteroposterior or<br /> the dorsoventral pattern of development. More recently, several<br /> maternal effect genes have been identified in mice and humans<br /> that are required for proper embryonic development. Chapter 23<br /> examines the relationships among the actions of several maternal<br /> effect genes during embryonic development.<br /> 7.2 EPIGENETIC INHERITANCE<br /> As we have just seen, events during oogenesis can cause the<br /> inheritance pattern of traits to deviate from a Mendelian pattern.<br /> Likewise, epigenetic inheritance is a pattern in which a modification<br /> occurs to a nuclear gene or chromosome that alters gene<br /> expression, but is not permanent over the course of many generations.<br /> As we will see, epigenetic inheritance patterns are the<br /> result of DNA and chromosomal modifications that occur during<br /> oogenesis, spermatogenesis, or early stages of embryogenesis.<br /> Once they are initiated during these early stages, epigenetic<br /> changes alter the expression of particular genes in a way that may<br /> be fixed during an individual’s lifetime. Therefore, epigenetic<br /> changes can permanently affect the phenotype of the individual.<br /> However, epigenetic modifications are not permanent over the<br /> course of many generations, and they do not change the actual<br /> DNA sequence. For example, a gene may undergo an epigenetic<br /> change that inactivates it for the lifetime of an individual. However,<br /> when this individual makes gametes, the gene may become<br /> activated and remain operative during the lifetime of an offspring<br /> who inherits the active gene.<br /> In this section, we will examine two examples of epigenetic<br /> inheritance called dosage compensation and genomic imprinting.<br /> The effect of dosage compensation is to offset differences in<br /> the number of sex chromosomes. One of the sex chromosomes is<br /> altered, with the result that males and females have similar levels<br /> of gene expression, even though they do not possess the same<br /> complement of sex chromosomes. In mammals, dosage compensation<br /> is initiated during the early stages of embryonic development.<br /> By comparison, genomic imprinting happens prior to fertilization;<br /> it involves a change in a single gene or chromosome<br /> during gamete formation. Depending on whether the modification<br /> occurs during spermatogenesis or oogenesis, imprinting<br /> governs whether an offspring expresses a gene that has been<br /> inherited from its mother or father.<br /> Dosage Compensation Is Necessary to Ensure<br /> Genetic Equality Between the Sexes<br /> Dosage compensation refers to the phenomenon that the level<br /> of expression of many genes on the sex chromosomes (such as<br /> the X chromosome) is similar in both sexes even though males<br /> and females have a different complement of sex chromosomes.<br /> This term was coined in 1932 by Hermann Muller to explain<br /> the effects of eye color mutations in Drosophila. Muller observed<br /> that female flies homozygous for certain X-linked eye color<br /> alleles had a similar phenotype to hemizygous males. He noted<br /> that an X-linked gene conferring an apricot eye color produces<br /> a very similar phenotype in homozygous females and hemizygous<br /> males. In contrast, a female that has one copy of the apricot<br /> allele and a deletion of the apricot gene on the other X chromosome<br /> has eyes of paler color. Therefore, one copy of the allele<br /> in the female is not equivalent to one copy of the allele in the<br /> male. Instead, two copies of the allele in the female produce a<br /> phenotype that is similar to that produced by one copy in the<br /> male. In other words, the difference in gene dosage—two copies<br /> in females versus one copy in males—is being compensated at<br /> the level of gene expression.<br /> Since these initial studies, dosage compensation has been<br /> studied extensively in mammals, Drosophila, and Caenorhabditis<br /> elegans (a nematode). Depending on the species, dosage compensation<br /> occurs via different mechanisms (Table 7.1). Female<br /> mammals equalize the expression of X-linked genes by turning<br /> off one of their two X chromosomes. This process is known as X<br /> inactivation. In Drosophila, the male accomplishes dosage compensation<br /> by doubling the expression of most X-linked genes. In<br /> C. elegans, the XX animal is a hermaphrodite that produces both<br /> sperm and egg cells, while an animal carrying a single X chromosome<br /> is a male that produces only sperm. The XX hermaphrodite<br /> diminishes the expression of X-linked genes on both X chromosomes<br /> to approximately 50% of that in the male.<br /> In birds, the Z chromosome is a large chromosome, usually<br /> the fourth or fifth largest, which contains almost all of the<br /> known sex-linked genes. The W chromosome is generally a<br /> much smaller microchromosome containing a high proportion<br /> of repeat sequence DNA that does not encode genes. Males are<br /> ZZ, and females are ZW. Several years ago, researchers studied<br /> the level of expression of a Z-linked gene that encodes an enzyme<br /> called aconitase. They discovered that males express twice as<br /> much aconitase as females do. These results suggested that dosage<br /> compensation does not occur in birds. However, more recently<br /> the expression of nine Z-linked genes was examined in chickens,<br /> and at least six of the genes showed expression levels that were</div> <div>22.1 GENETIC ANALYSIS OF HUMAN DISEASES 609<br /> TABLE 22.5<br /> Neurodegenerative Diseases Caused by Prions*<br /> Disease<br /> Infectious Diseases<br /> Kuru<br /> Description<br /> A human disease that was once common<br /> in New Guinea. It begins with a loss of<br /> coordination, usually followed by dementia.<br /> Infection was spread by cannibalism, a practice<br /> that ended in 1958.<br /> Nucleus<br /> Cytoplasm<br /> Neuron<br /> Scrapie<br /> Mad cow disease<br /> Human Inherited Diseases<br /> A disease of sheep and pigs characterized by<br /> intense itching in which the animals tend to<br /> scrape themselves against trees, followed by<br /> neurodegeneration<br /> Begins with changes in posture and<br /> temperament, followed by loss of coordination<br /> and neurodegeneration<br /> PrP Sc<br /> PrP C<br /> Creutzfeldt-Jakob disease<br /> Gerstmann-Straussler-<br /> Scheinker disease<br /> Familial fatal insomnia<br /> *All of these diseases are eventually fatal.<br /> Characterized by loss of coordination and<br /> dementia<br /> Characterized by loss of coordination and<br /> dementia<br /> Begins with sleeping and autonomic nervous<br /> system disturbances followed by insomnia and<br /> dementia<br /> Prion-related diseases arise from the ability of the prion<br /> protein to exist in two conformational states: a normal form<br /> PrP C , which does not cause disease, and an abnormal form, PrP Sc ,<br /> which does. (Note: The superscript C refers to the normal conformation,<br /> while the superscript Sc refers to the abnormal conformation,<br /> such as the one found in the disease called scrapie.)<br /> The gene encoding the prion protein (PrP) is found in humans<br /> and other mammals, and the protein is expressed at low levels<br /> in certain types of cells such as nerve cells. The abnormal conformation<br /> of the prion protein can come from two sources. An<br /> individual can be infected with the abnormal protein by taking<br /> the abnormal protein into their bodies. For example, someone<br /> may eat products from an animal that had the disease. Alternatively,<br /> some people carry alleles of the PrP gene that cause their<br /> prion protein to convert spontaneously to the abnormal conformation<br /> at a very low rate. These individuals have an inherited<br /> predisposition to develop a prion-related disease. An example of<br /> an inherited prion disease is familial fatal insomnia (Table 22.5).<br /> What is the molecular mechanism through which prions<br /> cause disease? As noted, the prion protein can exist in two conformations,<br /> PrP C and PrP Sc . As shown in Figure 22.7, the abnormal<br /> conformation, PrP Sc , acts as a catalyst to convert normal<br /> prion proteins within the cell to the misfolded conformation. As<br /> a prion disease progresses, the PrP Sc protein forms dense aggregates<br /> in the cells of the brain and peripheral nervous tissues.<br /> This deposition is correlated with the disease symptoms affecting<br /> the nervous system. Some of the abnormal prion protein is also<br /> excreted from infected cells and can travel through the nervous<br /> system to infect other cells.<br /> A prion protein in the PrP Sc<br /> conformation comes in contact<br /> with a PrP C protein, converting it<br /> into a PrP Sc protein.<br /> Original PrP Sc<br /> molecule<br /> PrP Sc<br /> converted<br /> from PrP C<br /> The 2 PrP Sc proteins bind<br /> to 2 PrP C proteins, and<br /> convert them to PrP Sc .<br /> Over time, all of the PrP C<br /> proteins will be converted<br /> to PrP Sc proteins.<br /> FIGURE 22.7 A proposed molecular mechanism<br /> of prion diseases. The PrP Sc protein catalyzes the<br /> conversion of PrP C to PrP Sc . Over time, the PrP Sc<br /> conformation will accumulate to high levels, leading to symptoms of<br /> the prion disease.</div> <div>752 CHAPTER 26 :: EVOLUTIONARY GENETICS<br /> Normal eye<br /> Eye where<br /> an antenna<br /> is normally<br /> found<br /> (a) Abnormal expression<br /> of Drosophila eyeless gene<br /> Eye on<br /> the side<br /> of a leg<br /> (b) Abnormal expression of<br /> mouse Pax6 gene in a<br /> fruit fly leg<br /> FIGURE 26.22 Formation of additional eyes in Drosophila<br /> due to the abnormal expression of a master control gene for eye<br /> morphogenesis. (a) When the Drosophila eyeless gene is expressed in<br /> the antenna region, eyes are formed where antennae should be located.<br /> (b) When the mouse Pax6 gene is expressed in the leg region of<br /> Drosophila, a small eye is formed there.<br /> In 1995, Walter Gehring and his colleagues were able to<br /> show experimentally that the abnormal expression of the eyeless<br /> gene in other parts of the fruit fly body could promote the formation<br /> of additional eyes. For example, using genetic engineering<br /> techniques, they were able to express the eyeless gene in the<br /> region where antennae should form. As seen in Figure 26.22a,<br /> this resulted in the formation of fruit fly eyes where antennae<br /> are normally found! Remarkably, the expression of the mouse<br /> Pax6 gene in Drosophila can also cause the formation of eyes in<br /> unusual places. Figure 26.22b shows the formation of an eye on<br /> the leg of a fruit fly. This eye was caused by the expression of the<br /> mouse Pax6 gene in this region.<br /> The mouse Pax6 gene switches on eye formation in Drosophila,<br /> but the eye produced is a Drosophila eye, not a mouse<br /> eye. This happens because the genes activated by the Pax6 gene<br /> are all from the Drosophila genome. In Drosophila, the eyeless<br /> gene switches on a cascade involving 2,500 different genes<br /> required for eye morphogenesis. The Pax6 gene and its Drosophila<br /> homolog, eyeless, are master control genes that promote the<br /> formation of an eye. In addition, Gehring has suggested that the<br /> eyes of Drosophila and mammals are evolutionarily derived from<br /> the modification of an eye that arose once during evolution. If<br /> Drosophila and mammalian eyes had arisen independently, the<br /> Pax6 gene from mice would not be expected to induce the formation<br /> of eyes in Drosophila.<br /> Since the initial discovery of the Pax6 and eyeless genes,<br /> homologs of this gene have been discovered in many different<br /> species. In all cases where it has been tested, this gene directs eye<br /> development. The Pax6 gene and its homologs encode a transcription<br /> factor protein that controls the expression of many<br /> different genes. Gehring and colleagues have hypothesized that<br /> the eyes from many different species all evolved from a common<br /> ancestral form consisting of, as proposed by Darwin, one photoreceptor<br /> cell and one pigment cell (Figure 26.23). As mentioned<br /> previously, such a very simple eye can accomplish some rudimentary<br /> form of vision by detecting light and its direction. Eyes<br /> such as these are still found in modern species such as the larvae<br /> of certain types of mollusks. Over the course of evolution, simple<br /> eyes were transformed into more complex types of eyes by modifications<br /> that resulted in the addition of more types of cells such<br /> as lens and muscle cells.<br /> The ancestral Pax6<br /> gene controlled<br /> other genes that<br /> produced a<br /> primitive 2-celled<br /> eye.<br /> During evolution, species<br /> diverged from each other,<br /> but each species retained<br /> a Pax6 homolog.<br /> Drosophila eyeless gene<br /> Drosophila eye<br /> Photoreceptor<br /> cell<br /> Pigment<br /> cell<br /> Ancestral Pax6 gene<br /> Over time, gene duplications and other<br /> genetic changes produced many more<br /> genes that added to eye complexity.<br /> These additional genes remained under<br /> the control of the Pax6 gene and its<br /> homologs.<br /> Ommatidia<br /> Retina<br /> Optic nerve<br /> Simple eye<br /> Mammal Pax6 gene<br /> Lens<br /> Mammal eye<br /> Cornea<br /> FIGURE 26.23 Eye evolution. In this diagram, genetic changes,<br /> under control of the ancestral Pax6 gene, led to the evolution of<br /> different types of eyes. Ommatidia are units of the insect compound<br /> eye.<br /> Iris<br /> CONCEPTUAL SUMMARY<br /> Biological evolution involves heritable changes in one or more<br /> characteristics in a population or species over the course of many<br /> generations. Charles Darwin’s theory of evolution is based on two<br /> fundamental principles: genetic variation and natural selection. In<br /> the first part of this chapter, we were concerned with how these<br /> features of evolution result in the formation of new species. Biologists<br /> use four different species concepts—the phylogenetic, biological,<br /> evolutionary, and ecological species concepts—to define and<br /> identify species. Speciation is often a branching process (cladogenesis),<br /> although anagenesis (transformation of a single species)<br /> occasionally occurs. Speciation may be allopatric, parapatric, or<br /> sympatric, depending on whether or not geographic barriers are</div> <div>25.3 HERITABILITY 709<br /> Strain producing large fruit<br /> Strain producing small fruit<br /> 1A<br /> 1A<br /> 1B<br /> 1B<br /> X<br /> 2A<br /> 3A<br /> 4A<br /> 5A<br /> X<br /> 2A<br /> 3A<br /> 4A<br /> 5A<br /> 6A<br /> 7A<br /> 8A<br /> 9A<br /> 10A<br /> 6A<br /> 7A<br /> 8A<br /> 9A<br /> 10A<br /> X<br /> 11A<br /> 12A<br /> 13A<br /> 14A<br /> 15A<br /> X<br /> 11A<br /> 12A<br /> 13A<br /> 14A<br /> 15A<br /> 16A<br /> 17A<br /> 18A<br /> 19A<br /> X<br /> 20A<br /> X<br /> 20A<br /> X<br /> X<br /> X – a QTL promoting large-size fruit<br /> Note: The locations of QTLs are not known at the start of this experiment.<br /> 16A<br /> 17A<br /> 18A<br /> 19A<br /> X<br /> 2B<br /> 3B<br /> 4B<br /> 5B<br /> X<br /> X<br /> 2B<br /> 3B<br /> 4B<br /> 5B<br /> X<br /> Cross the 2 strains to<br /> produce F 1 offspring.<br /> 6B<br /> 7B<br /> 8B<br /> 9B<br /> 10B<br /> 6B<br /> 7B<br /> 8B<br /> 9B<br /> 10B<br /> X<br /> 11B<br /> 12B<br /> 13B<br /> 14B<br /> 15B<br /> X<br /> 11B<br /> 12B<br /> 13B<br /> 14B<br /> 15B<br /> X – a QTL promoting small-size fruit<br /> 16B<br /> 17B<br /> 18B<br /> 19B<br /> X<br /> 20B<br /> 16B<br /> 17B<br /> 18B<br /> 19B<br /> X<br /> 20B<br /> 1A<br /> 1B<br /> X<br /> 2A<br /> X<br /> 2B<br /> 6A<br /> 7A<br /> 6B<br /> 7B<br /> X<br /> 11A<br /> 12A<br /> X<br /> 11B<br /> 12B<br /> 16A<br /> 16B<br /> 3A<br /> 3B<br /> 8A<br /> 8B<br /> 17A<br /> 17B<br /> 9A<br /> 9B<br /> 13A<br /> 13B<br /> 18A<br /> 18B<br /> 4A<br /> 4B<br /> 14A<br /> 14B<br /> 19A<br /> 19B<br /> 5A<br /> 5B<br /> 10A<br /> 10B<br /> 15A<br /> 15B<br /> X<br /> 20A<br /> X<br /> 20B<br /> X<br /> X<br /> Backcross the F 1 offspring<br /> to both parental strains.<br /> F 2 offspring contain<br /> different combinations<br /> of parental chromosomes.<br /> For many F 2 offspring,<br /> determine fruit size<br /> and molecular marker<br /> composition.<br /> 2A, 5A, 11A, and 19A are strongly associated with large fruit<br /> size. This suggests that 4 QTLs lie close to these markers.<br /> FIGURE 25.7 The general strategy for QTL mapping via molecular markers. Two different inbred strains have four chromosomes per set. The<br /> strain on the left produces large fruit, and the strain on the right produces small fruit. The goal of this mapping strategy is to locate the unknown genes<br /> affecting this trait, which are designated with the letter X. A black X indicates a site promoting large fruit, and a blue X is a site promoting small fruit.<br /> The two strains differ with regard to many molecular markers designated 1A and 1B, 2A and 2B, and so forth. The two strains are mated, and then the<br /> F 1 offspring are backcrossed to the parental strains. Many F 2 offspring are then examined for their fruit size and to determine which molecular markers<br /> are found in their chromosomes. The data are analyzed by computer programs that can statistically associate the phenotype (e.g., fruit size) with<br /> particular markers. Markers found throughout the genome of this species provide a way to locate many different genes that may affect the outcome<br /> of a single quantitative trait. In this case, the analysis would predict four QTLs promoting heavier fruit weight that would be linked to regions of the<br /> chromosomes containing the following markers: 2A, 5A, 11A, and 19A.</div> <div>638 CHAPTER 23 :: DEVELOPMENTAL GENETICS<br /> for the formation of the terminal ends of the embryo. The dorsoventral<br /> axis is governed by the actions of several proteins, including<br /> a receptor protein known as Toll. This receptor is found in<br /> plasma membranes throughout the embryo but is activated by<br /> ligand binding only along the ventral midline of the embryo.<br /> Let’s now take a closer look at the molecular mechanism of<br /> one morphogen, Bicoid. The bicoid gene got its name because a<br /> larva defective in this gene develops with two posterior ends (Figure<br /> 23.6). This allele exhibits a maternal effect pattern of inheritance,<br /> in which the genotype of the mother determines the phenotypic<br /> traits of the offspring (see Chapter 7). A female fly that<br /> is phenotypically normal (because its mother was heterozygous<br /> for the normal bicoid allele), but genotypically homozygous for an<br /> inactive bicoid allele (because it inherited an inactive allele from<br /> both its mother and father), will produce 100% abnormal offspring<br /> even when mated to a male that is homozygous for the<br /> normal bicoid allele. In other words, the genotype of the mother<br /> determines the phenotype of the offspring. This occurs because the<br /> bicoid gene product is provided to the oocyte via the nurse cells.<br /> In the ovaries of female flies, the nurse cells are localized<br /> asymmetrically toward the anterior end of the oocyte. During<br /> oogenesis, gene products are transferred from nurse cells into<br /> the oocyte via cell-to-cell connections called cytoplasmic bridges.<br /> Maternally encoded gene products enter one side of the oocyte,<br /> which will eventually become the anterior end of the embryo<br /> (Figure 23.7a). The bicoid gene is actively transcribed in the<br /> nurse cells, and bicoid mRNA is transported into the anterior end<br /> of the oocyte. The 3' end of bicoid mRNA contains a signal that<br /> is recognized by RNA-binding proteins thought necessary for<br /> the transport of this mRNA into the oocyte. After it enters the<br /> oocyte, the bicoid mRNA is trapped at the anterior end.<br /> How can researchers determine the location of bicoid mRNA<br /> in the oocyte and resulting zygote? Figure 23.7b shows an in situ<br /> hybridization experiment in which a Drosophila egg was examined<br /> via a probe complementary to the bicoid mRNA. (The technique<br /> of in situ hybridization is described in Chapter 20.) As seen here,<br /> the bicoid mRNA is highly concentrated near the anterior end of<br /> the egg cell. When the bicoid mRNA subsequently is translated, a<br /> gradient of Bicoid protein is established as shown in Figure 23.7c.<br /> After fertilization occurs, the Bicoid protein functions as a<br /> transcription factor. A remarkable feature of this protein is that<br /> its ability to influence gene expression is tuned exquisitely to its<br /> concentration. Depending on the distribution of the Bicoid protein,<br /> this transcription factor will activate genes only in certain<br /> regions of the embryo. For example, Bicoid stimulates the transcription<br /> of a gene called hunchback in the anterior half of the<br /> embryo, but its concentration is too low in the posterior half to<br /> activate the hunchback gene there.<br /> Gap, Pair-Rule, and Segment-Polarity Genes Act<br /> Sequentially to Divide the Drosophila Embryo<br /> into Segments<br /> After the antero-posterior and dorso-ventral regions of the embryo<br /> have been established by maternal effect genes, the next develop-<br /> Nurse cell<br /> Follicle cell<br /> Anterior end<br /> of oocyte<br /> Oocyte<br /> (a) Transport of maternal effect gene products<br /> into the oocyte<br /> (b) In situ hybridization of bicoid mRNA<br /> Spiracle<br /> (a) bicoid + embryo<br /> Spiracle<br /> (b) bicoid – embryo<br /> Spiracle<br /> FIGURE 23.6 The bicoid mutation in Drosophila. (a) A<br /> normal bicoid � embryo. (b) A bicoid � embryo in which both ends of<br /> the larva develop posterior structures. For example, both ends develop<br /> a spiracle, which normally is found only at the posterior end. This is a<br /> lethal condition.<br /> (c) Immunostaining of Bicoid protein<br /> FIGURE 23.7 Asymmetrical localization of gene products<br /> during oogenesis in Drosophila. (a) The nurse cells transport gene<br /> products into the anterior (left) end of the developing oocyte. (b) An in<br /> situ hybridization experiment showing that the bicoid mRNA is trapped<br /> near the anterior end. (c) The bicoid mRNA is translated into protein<br /> soon after fertilization. The location of the Bicoid protein is revealed by<br /> immunostaining using an antibody that specifically recognizes this protein.</div> <div>710 CHAPTER 25 :: QUANTITATIVE GENETICS<br /> these two extremes. For example, both genes and diet affect the<br /> size that an individual will attain. Some individuals will inherit<br /> alleles that tend to make them large, and a proper diet will also<br /> promote larger size. Other individuals will inherit alleles that<br /> make them small, and an inadequate diet may contribute to small<br /> size. Taken together, both genetics and the environment influence<br /> the phenotypic results.<br /> In the study of quantitative traits, a primary goal is to determine<br /> how much of the phenotypic variation arises from genetic<br /> variation and how much comes from environmental variation. In<br /> this section, we will examine how geneticists analyze the genetic<br /> and environmental components that affect quantitative traits. As<br /> we will see, this approach has been applied with great success in<br /> breeding strategies to produce domesticated species with desirable<br /> and commercially valuable characteristics.<br /> Genetic Variance and Environmental Variance<br /> Both May Contribute to Phenotypic Variance<br /> Earlier, we examined the amount of phenotypic variation within<br /> a group by calculating the variance. In studying quantitative trait<br /> variation, geneticists partition this variation into components<br /> that are attributable to different causes. These include<br /> Genetic variation (V G )<br /> Environmental variation (V E )<br /> Variation due to interactions between genetic and<br /> environmental factors (V G�E )<br /> Variation due to associations between genetic and<br /> environmental factors (V Gfg E )<br /> Let’s begin by considering a simple situation in which V G and V E<br /> are the only factors that determine phenotypic variance and they<br /> are independent of each other. If so, then the total variance for a<br /> trait in a group of individuals is<br /> V T � V G � V E<br /> where<br /> V T is the total variance. It reflects the amount of<br /> variation that is measured at the phenotypic<br /> level.<br /> V G is the relative amount of variance due to<br /> genetic variation.<br /> V E is the relative amount of variance due to<br /> environmental variation.<br /> Why is this equation useful? The partitioning of variance<br /> into genetic and environmental components allows us to estimate<br /> their relative importance in influencing the variation within<br /> a group. If V G is very high and V E is very low, genetics plays a<br /> greater role in promoting variation within a group. Alternatively,<br /> if V G is low and V E is high, environmental factors underlie much<br /> of the phenotypic variation. As described later in this chapter, a<br /> livestock breeder might want to apply selective breeding if V G for<br /> an important (quantitative) trait is high. In this way, the characteristics<br /> of the herd may be improved. Alternatively, if V G is negligible,<br /> it would make more sense to investigate (and manipulate)<br /> the environmental causes of phenotypic variation.<br /> With experimental and domesticated species, one possible<br /> way to determine V G and V E is by comparing the variation<br /> in traits between genetically identical and genetically disparate<br /> groups. For example, researchers have followed the practice of<br /> inbreeding to develop genetically homogeneous strains of mice.<br /> Inbreeding in mice involves many generations of brother-sister<br /> matings, which eventually produces strains that are monomorphic<br /> for all of their genes. The term monomorphic means that all<br /> the members of a population are homozygous for the same allele<br /> of a given gene. Within such an inbred strain of mice, V G equals<br /> zero. Therefore, all phenotypic variation is due to V E . When<br /> studying quantitative traits such as weight, an experimenter<br /> might want to know the genetic and environmental variance<br /> for a different, genetically heterogeneous group of mice. To do<br /> so, the genetically homogeneous and heterogeneous mice could<br /> be raised under the same environmental conditions and their<br /> weights measured. The phenotypic variance for weight could<br /> then be calculated as described earlier. Let’s suppose we obtained<br /> the following results:<br /> V T � 0.30 sq oz for the group of genetically homogeneous mice<br /> V T � 0.52 sq oz for the group of genetically heterogeneous mice<br /> In the case of the homogeneous mice, V T � V E , because V G<br /> equals zero. Therefore, V E equals 0.30 sq oz. To estimate V G for<br /> the heterogeneous group of mice, we could assume that V E (i.e.,<br /> the environmentally produced variance) is the same for them<br /> as it is for the homogeneous mice, because the two groups were<br /> raised in identical environments. This assumption allows us to<br /> calculate the genetic variance for the heterogeneous mice.<br /> V T � V G � V E<br /> 0.52 � V G � 0.30<br /> V G � 0.22 sq oz<br /> This result tells us that some of the phenotypic variance in<br /> the genetically heterogeneous group is due to the environment<br /> (namely, 0.30 sq oz) and some (0.22 sq oz) is due to genetic variation<br /> in alleles that affect weight.<br /> Phenotypic Variation May Also Be Influenced<br /> by Interactions and Associations Between<br /> Genotype and the Environment<br /> Thus far, we have considered the simple situation in which<br /> genetic variation and environmental variation are independent<br /> of each other and affect the phenotypic variation in an additive<br /> way. As another example, let’s suppose that three genotypes,<br /> TT, Tt, and tt, affect height, producing tall, medium, and dwarf<br /> plants, respectively. Greater sunlight makes the plants grow taller<br /> regardless of their genotypes. In this case, our assumption that<br /> V T � V G � V E would be reasonably valid.<br /> However, let’s consider a different environmental factor<br /> such as minerals in the soil. Perhaps the Tt and tt plants are</div> <div>168 CHAPTER 7 :: NON-MENDELIAN INHERITANCE<br /> Clone of cells<br /> S<br /> S<br /> 3. Take nine isolated clones and grow in<br /> liquid cultures. (Only three are shown<br /> here.)<br /> Grow each<br /> clone in a<br /> separate<br /> flask.<br /> S S S S<br /> Many cell divisions<br /> Produces a clone of cells<br /> 4. Take cells from the liquid cultures, lyse<br /> cells to obtain proteins, and subject to<br /> gel electrophoresis. (This technique is<br /> described in the Appendix.)<br /> Note: As a control, lyse cells from step 1,<br /> and subject the proteins to gel<br /> electrophoresis. This control sample is<br /> not from a clone. It is a mixture of cells<br /> derived from a woman’s skin sample.<br /> –<br /> Slow<br /> Fast<br /> +<br /> Control<br /> From a clone with<br /> the fast allele active<br /> Sample<br /> Origin<br /> From a clone with<br /> the slow allele active<br /> ■<br /> All cells<br /> THE DATA<br /> Clones<br /> 1 2 3 4 5 6 7 8 9 10<br /> Slow G-6-PD<br /> Fast G-6-PD<br /> ■ INTERPRETING THE DATA<br /> In the data shown in Figure 7.6, the control (lane 1) was a protein<br /> sample obtained from a mixture of epithelial cells from a heterozygous<br /> woman who produced both types of G-6-PD enzymes.<br /> Bands corresponding to the fast and slow enzymes were observed<br /> in this lane. As described in steps 2 to 4, this mixture of epithelial<br /> cells was also used to generate nine clones. The proteins obtained<br /> from these clones are shown in lanes 2 to 10. Each clone was a<br /> population of cells independently derived from a single epithelial<br /> cell. Because the epithelial cells were obtained from an adult<br /> female, the Lyon hypothesis predicts that each epithelial cell<br /> would already have one of its X chromosomes permanently inactivated<br /> and would pass this trait to its progeny cells. For example,<br /> suppose that an epithelial cell had inactivated the X chromosome<br /> that encoded the fast G-6-PD. If this cell was allowed to form a<br /> clone of cells on a plate, all cells in this clonal population would<br /> be expected to have the same X chromosome inactivated—the X<br /> chromosome encoding the fast G-6-PD. Therefore, this clone of<br /> cells should express only the slow G-6-PD. As shown in the data,<br /> all nine clones expressed either the fast or slow G-6-PD protein,<br /> but not both. These results are consistent with the hypothesis<br /> that X inactivation has already occurred in any given epithelial<br /> cell and that this pattern of inactivation is passed to all of its<br /> progeny cells.<br /> A self-help quiz involving this experiment can be found at<br /> the Online Learning Center.<br /> X Inactivation in Mammals Depends<br /> on the X-Inactivation Center and the Xist Gene<br /> Since the Lyon hypothesis was confirmed, the genetic control of<br /> X inactivation has been investigated further by several laboratories.<br /> Research has shown that mammalian cells possess the ability<br /> to count their X chromosomes and allow only one of them to<br /> remain active. How was this determined? A key observation came<br /> from comparisons of the chromosome composition of people<br /> who have been born with normal or abnormal numbers of sex<br /> chromosomes.</div> <div>406 CHAPTER 15 :: GENE REGULATION IN EUKARYOTES<br /> 5′ 3′<br /> CH 3 C<br /> G<br /> G<br /> C<br /> 3′ 5′<br /> 5′ 3′<br /> C<br /> G<br /> G<br /> C<br /> 3′ 5′<br /> de novo methylation<br /> 5′ 3′<br /> CH 3 C<br /> G<br /> G<br /> C<br /> 3′ 5′<br /> CH 3<br /> CH 3<br /> DNA replication<br /> Hemimethylated<br /> DNA<br /> 5′ 3′<br /> C<br /> G<br /> G<br /> C<br /> 3′ 5′<br /> is made (Table 15.2). One mechanism is pre-mRNA processing.<br /> Following transcription, a pre-mRNA transcript is processed<br /> before it becomes a functional mRNA. These processing events,<br /> which include splicing, capping, polyA tailing, and RNA editing,<br /> were described in Chapter 12. We will begin this section by<br /> examining how the processes of alternative splicing and RNA<br /> editing are regulated at the RNA level.<br /> Another strategy for regulating gene expression is to influence<br /> the concentration of mRNA. This can be accomplished by<br /> regulating the rate of transcription. When the transcription of a<br /> gene is increased, a higher concentration of the corresponding<br /> RNA results. In addition, RNA concentration is greatly affected<br /> by the stability or half-life of a particular RNA. Factors that<br /> increase RNA stability are expected to raise the concentration<br /> of that RNA molecule. We will explore how sequences within<br /> mRNA molecules greatly affect their stability.<br /> We will also consider a newly discovered mechanism of<br /> mRNA silencing, known as RNA interference, which involves<br /> double-stranded RNAs that may direct the breakdown of specific<br /> TABLE<br /> TABLE<br /> 15.1<br /> 15.2<br /> Gene Regulation via RNA Processing and Translation<br /> Maintenance<br /> methylation<br /> Fully methylated<br /> DNA<br /> 5′ 3′ 5′ 3′<br /> CH 3 C<br /> CH<br /> G<br /> 3 C<br /> G<br /> G<br /> C CH<br /> G<br /> 3<br /> C CH 3<br /> 3′ 5′ 3′ 5′<br /> FIGURE 15.15 A molecular model for the inheritance of DNA<br /> methylation. The DNA initially undergoes de novo methylation, which<br /> is a rare, highly regulated event. Once this occurs, DNA replication<br /> produces hemimethylated DNA molecules, which are then fully<br /> methylated by DNA methyltransferase. This process, called maintenance<br /> methylation, is a routine event that is expected to occur for all<br /> hemimethylated DNA.<br /> will remain unmethylated. Along these lines, geneticists are also<br /> eager to determine how variations in DNA methylation patterns<br /> may be important for cell differentiation. It may be a key way to<br /> silence genes in different cell types. However, additional research<br /> will be necessary to understand how specific genes may be targeted<br /> for de novo methylation or demethylation during different<br /> developmental stages or in specific cell types.<br /> Effect<br /> Alternative splicing<br /> RNA editing<br /> RNA stability<br /> RNA interference<br /> Description<br /> Certain pre-mRNAs can be spliced in more than one<br /> way, leading to polypeptides that have different amino<br /> acid sequences. Alternative splicing is often cell specific<br /> so that a protein can be fine-tuned to function in a<br /> particular cell type. It is an important form of gene<br /> regulation in multicellular eukaryotic species.<br /> The sequence of RNAs can be altered after the RNA is<br /> made. This usually involves the addition or deletion of<br /> one or a few bases, or a change of C to U, or A to I.<br /> This does not appear to be a widespread phenomenon,<br /> though the study of RNA editing is still relatively recent.<br /> The amount of RNA is greatly influenced by the halflife<br /> of RNA transcripts. A long polyA tail on mRNAs<br /> promotes their stability due to the binding of polyAbinding<br /> protein. Some RNAs with a relatively short<br /> half-life contain sequences that target them for rapid<br /> destruction. Some RNAs are stabilized by specific RNAbinding<br /> proteins that usually bind near the 3’ end.<br /> Double-stranded RNA can mediate the degradation of<br /> homologous mRNAs in the cell. This is a mechanism of<br /> gene regulation. Also, it probably provides eukaryotic<br /> cells with protection from invasion by certain types of<br /> viruses and may prevent the movement of transposable<br /> elements.<br /> 15.3 REGULATION OF RNA<br /> PROCESSING, RNA STABILITY,<br /> AND TRANSLATION<br /> Thus far, we have considered a variety of mechanisms that regulate<br /> the level of gene transcription. These mechanisms control<br /> the amount of RNA transcribed from a given gene. In addition,<br /> eukaryotic gene expression is commonly regulated after the RNA<br /> General regulation of<br /> translation<br /> The function of translational initiation factors may be<br /> regulated to permit or inhibit translation. This<br /> regulation affects the translation of all cellular mRNAs.<br /> Inhibition of translation is desirable if a cell has been<br /> exposed to a virus or to toxic materials.<br /> Translational regulation Some mRNAs are regulated via binding proteins that<br /> of specific mRNAs inhibit the ability of the ribosomes to initiate<br /> translation. These proteins usually bind at the 5’ end<br /> of the mRNA and thereby prevent the ribosome from<br /> binding.</div> <div>622 CHAPTER 22 :: MEDICAL GENETICS AND CANCER<br /> of malignancy. However, further research will be needed to determine<br /> why tumor-suppressor genes are aberrantly methylated<br /> in cancer cells. Third, many types of cancer are associated with<br /> aneuploidy. As discussed in Chapter 8, aneuploidy involves the<br /> loss or addition of one or more chromosomes, so the total number<br /> of chromosomes is not an even multiple of a set. In some<br /> cases, chromosome loss may contribute to the progression of<br /> cancer because the lost chromosome carries one or more tumorsuppressor<br /> genes.<br /> Most Forms of Cancer Involve Multiple Genetic<br /> Changes Leading to Malignancy<br /> The discovery of oncogenes and tumor-suppressor genes, along<br /> with molecular techniques that can detect genetic alterations, has<br /> enabled researchers to study the progression of certain forms of<br /> cancer at the molecular level. Many cancers begin with a benign<br /> genetic alteration that, over time and with additional mutations,<br /> progresses eventually to malignancy. Furthermore, a malignancy<br /> can continue to accumulate genetic changes that make it even more<br /> difficult to treat. For example, some tumors may acquire mutations<br /> that cause them to be resistant to chemotherapeutic agents.<br /> In 1990, Eric Fearon and Bert Vogelstein proposed a series<br /> of genetic changes that lead to colorectal cancer, the second most<br /> common cancer in the United States. As shown in Figure 22.17,<br /> colorectal cancer is derived from cells in the mucosa of the colon.<br /> The loss of function of APC, a tumor-suppressor gene on chromosome<br /> 5, leads to an increased proliferation of mucosal cells<br /> and the development of a benign polyp, a noncancerous growth.<br /> Additional genetic changes involving the loss of other tumor-suppressor<br /> genes and the activation of an oncogene (namely, ras) lead<br /> eventually to the development of a carcinoma. In Figure 22.17, the<br /> genetic changes that lead to colon cancer are portrayed as occurring<br /> in an orderly sequence. While the growth of a tumor often<br /> begins with mutations in APC, it is the total number of genetic<br /> changes, not their exact order, that is important. A key issue is<br /> that multiple mutations are needed to disable checkpoints.<br /> Among different types of tumors, researchers have identified<br /> a large number of genes that are mutated in cancer cells.<br /> Though not all of these mutant genes have been directly shown<br /> to affect the growth rate of cells, such mutations are likely to<br /> be found in tumors because they provide some type of growth<br /> advantage for the cell population from which the cancer developed.<br /> For example, certain mutations may enable cells to metastasize<br /> to neighboring locations. These mutations may not affect<br /> growth rate, but they provide the growth advantage that cancer<br /> cells are not limited to growing in a particular location, but can<br /> migrate to new locations.<br /> Researchers have estimated that about 300 different genes<br /> may play a role in the development of human cancer. With an<br /> approximate genome size of 20,000 to 25,000 genes, this observation<br /> indicates that over 1% of our genes have the potential to<br /> promote cancer if their function is altered by a mutation.<br /> In addition to mutations within specific genes, another<br /> common genetic change associated with cancer are abnormalities<br /> in chromosome structure and number. Figure 22.18 compares the<br /> FIGURE 22.17<br /> cancer.<br /> Other mutations<br /> Colon<br /> Normal mucosa<br /> cells of colon<br /> Loss of APC tumor-suppressor<br /> gene, chromosome 5<br /> Cell division<br /> continues<br /> Activation of<br /> ras oncogene,<br /> chromosome 12<br /> Small polyp<br /> (benign)<br /> Class I adenoma<br /> (benign)<br /> Class II adenoma<br /> (benign)<br /> Loss of DCC tumor-suppressor<br /> gene, chromosome 18<br /> Class III<br /> adenoma<br /> (benign)<br /> Loss of p53 tumor-suppressor<br /> gene, chromosome 17<br /> Class IV<br /> carcinoma<br /> (malignant)<br /> Metastasis<br /> Multiple genetic changes leading to colorectal</div> <div>294 CHAPTER 11 :: DNA REPLICATION<br /> newly made strand? Indicate the 5' and 3' ends of the newly made<br /> strand.<br /> C9. List and briefly describe the three types of sequences within<br /> bacterial origins of replication that are functionally important.<br /> C10. As shown in Figure 11.5, five DnaA boxes are found within the<br /> origin of replication in E. coli. Take a look at these five sequences<br /> carefully.<br /> A. Are the sequences of the five DnaA boxes very similar to each<br /> other? (Hint: Remember that DNA is double stranded; think<br /> about these sequences in the forward and reverse direction.)<br /> B. What is the most common sequence for the DnaA box? In other<br /> words, what is the most common base in the first position,<br /> second position, and so on until the ninth position? The most<br /> common sequence is called the consensus sequence.<br /> C. The E. coli chromosome is about 4.6 million bp long. Based on<br /> random chance, is it likely that the consensus sequence for a<br /> DnaA box occurs elsewhere in the E. coli chromosome? If so,<br /> why aren’t there multiple origins of replication in E. coli?<br /> C11. Obtain two strings of different colors (e.g., black and white) that<br /> are the same length. A length of 20 inches is sufficient. Tie a knot<br /> at one end of the black string, and tie a knot at one end of the<br /> white string. Each knot designates the 5' end of your strings. Make<br /> a double helix with your two strings. Now tape one end of the<br /> double helix to a table so that the tape is covering the knot on the<br /> black string.<br /> A. Pretend your hand is DNA helicase and use your hand to<br /> unravel the double helix, beginning at the end that is not taped<br /> to the table. Should your hand be sliding along the white string<br /> or the black string?<br /> B. As in Figure 11.14, imagine that your two hands together form<br /> a dimeric replicative DNA polymerase. Unravel your two strings<br /> halfway to create a replication fork. Grasp the black string with<br /> your left hand and the white string with your right hand. Your<br /> thumbs should point toward the 5' end of each string. You need<br /> to loop one of the strings so that one of the DNA polymerases<br /> can synthesize the lagging strand. With such a loop, the dimeric<br /> replicative DNA polymerase can move toward the replication<br /> fork and synthesize both DNA strands in the 5' to 3' direction.<br /> In other words, with such a loop, your two hands can touch<br /> each other with both of your thumbs pointing toward the fork.<br /> Should the black string be looped, or should the white string be<br /> looped?<br /> C12. Sometimes DNA polymerase makes a mistake, and the wrong<br /> nucleotide is added to the growing DNA strand. With regard to<br /> pyrimidines and purines, two general types of mistakes are<br /> possible. The addition of an incorrect pyrimidine instead of the<br /> correct pyrimidine (e.g., adding cytosine where thymine should be<br /> added) is called a transition. If a pyrimidine is incorrectly added to<br /> the growing strand instead of purine (e.g., adding cytosine where<br /> an adenine should be added), this type of mistake is called<br /> a transversion. If a transition or transversion is not detected by<br /> DNA polymerase, this will create a mutation that permanently<br /> changes the DNA sequence. Though both types of mutations are<br /> rare, transition mutations are more frequent than transversion<br /> mutations. Based on your understanding of DNA replication and<br /> DNA polymerase, offer three explanations why transition<br /> mutations are more common.<br /> C13. A short genetic sequence, which may be recognized by DNA<br /> primase, is repeated many times throughout the E. coli<br /> chromosome. Researchers have hypothesized that DNA primase<br /> may recognize this sequence as a site to begin the synthesis of<br /> an RNA primer for DNA replication. The E. coli chromosome is<br /> roughly 4.6 million bp in length. How many copies of the DNA<br /> primase recognition sequence would be necessary to replicate the<br /> entire E. coli chromosome?<br /> C14. Single-strand binding proteins keep the two parental strands of<br /> DNA separated from each other until DNA polymerase has an<br /> opportunity to replicate the strands. Suggest how single-strand<br /> binding proteins keep the strands separated and yet do not impede<br /> the ability of DNA polymerase to replicate the strands.<br /> C15. The ability of DNA polymerase to digest a DNA strand from one<br /> end is called its exonuclease activity. Exonuclease activity is used<br /> to digest RNA primers and also to proofread a newly made DNA<br /> strand. Note: DNA polymerase I does not change direction while<br /> it is removing an RNA primer and synthesizing new DNA. It does<br /> change direction during proofreading.<br /> A. In which direction, 5' to 3' or 3' to 5' is the exonuclease activity<br /> occurring during the removal of RNA primers and during the<br /> proofreading and removal of mistakes following DNA<br /> replication?<br /> B. Figure 11.15 shows a drawing of the 3' exonuclease site. Do you<br /> think this site would be used by DNA polymerase I to remove<br /> RNA primers? Why or why not?<br /> C16. In the following drawing, the top strand is the template DNA, and<br /> the bottom strand shows the lagging strand prior to the action<br /> of DNA polymerase I. The lagging strand contains three Okazaki<br /> fragments. The RNA primers have not yet been removed.<br /> The top strand is the template DNA<br /> 3'———————————————————————————5'<br /> 5'*************———***************———*************————3'<br /> RNA primer h RNA primer h RNA primer<br /> |—————————||—————————||—————————|<br /> Left Okazaki Middle Okazaki Right Okazaki<br /> fragment fragment fragment<br /> A. Which Okazaki fragment was made first, the one on the left or<br /> the one on the right?<br /> B. Which RNA primer would be the first one to be removed by<br /> DNA polymerase I, the primer on the left or the primer on the<br /> right? For this primer to be removed by DNA polymerase I and<br /> for the gap to be filled in, is it necessary for the Okazaki<br /> fragment in the middle to have already been synthesized?<br /> Explain why.<br /> C. Let’s consider how DNA ligase connects the left Okazaki<br /> fragment with the middle Okazaki fragment. After DNA<br /> polymerase I removes the middle RNA primer and fills in the<br /> gap with DNA, where does DNA ligase function? See the arrows<br /> on either side of the middle RNA primer. Is ligase needed at the<br /> left arrow, at the right arrow, or both?<br /> D. When connecting two Okazaki fragments, DNA ligase needs to<br /> use NAD + or ATP as a source of energy to catalyze this reaction.<br /> Explain why DNA ligase needs another source of energy to<br /> connect two nucleotides, but DNA polymerase needs</div> <div>248 CHAPTER 10 :: CHROMOSOME ORGANIZATION AND MOLECULAR STRUCTURE<br /> capsid proteins. This cleavage produces a capsid protein that<br /> is somewhat smaller and able to assemble correctly. For many<br /> viruses, the cleavage of capsid proteins into smaller units is an<br /> important event that precedes viral assembly.<br /> 8 μm<br /> 10.2 BACTERIAL CHROMOSOMES<br /> Let’s now turn our attention to the organization of chromosomes<br /> found in bacterial species. Inside a bacterial cell, the chromosome<br /> is highly compacted and found within a region of the cell known<br /> as the nucleoid. Although bacteria usually contain a single type<br /> of chromosome, more than one copy of that chromosome may<br /> be found within one bacterial cell. Depending on the growth<br /> conditions and phase of the cell cycle, bacteria may have one to<br /> four identical chromosomes per cell. In addition, the number<br /> of copies varies depending on the bacterial species. As shown in<br /> Figure 10.3, each chromosome occupies its own distinct nucleoid<br /> region within the cell. Unlike the eukaryotic nucleus, the bacterial<br /> nucleoid is not a separate cellular compartment bounded by<br /> a membrane. Rather, the DNA in a nucleoid is in direct contact<br /> with the cytoplasm of the cell.<br /> In this section, we will explore two important features of<br /> bacterial chromosomes. First, the organization of DNA sequences<br /> along the chromosome will be examined. Second, we will consider<br /> the mechanisms that cause the chromosome to become a<br /> compacted structure within a nucleoid of the bacterium.<br /> FIGURE 10.3 The localization of nucleoids within Bacillus<br /> subtilis bacteria. The nucleoids are fluorescently labeled and seen as<br /> bright, oval-shaped regions within the bacterial cytoplasm. Note that<br /> two or more nucleoids are found within each cell. Some of the cells<br /> seen here are in the process of dividing.<br /> Bacterial Chromosomes Contain a Few Thousand<br /> Gene Sequences Interspersed with Other<br /> Functionally Important Sequences<br /> Bacterial chromosomal DNA is usually a circular molecule,<br /> though some bacteria have linear chromosomes. A typical chromosome<br /> is a few million base pairs (bp) in length. For example,<br /> the chromosome of one strain of Escherichia coli has approximately<br /> 4.6 million bp, and the Haemophilus influenzae chromosome<br /> has roughly 1.8 million bp. A bacterial chromosome<br /> commonly has a few thousand different genes. These genes are<br /> interspersed throughout the entire chromosome (Figure 10.4).<br /> Structural genes—nucleotide sequences that encode proteins—<br /> account for the majority of bacterial DNA. The nontranscribed<br /> regions of DNA located between adjacent genes are termed intergenic<br /> regions.<br /> Other sequences in chromosomal DNA influence DNA<br /> replication, gene transcription, and chromosome structure. For<br /> Key features:<br /> Origin of<br /> replication<br /> • Most, but not all, bacterial species<br /> contain circular chromosomal DNA.<br /> • A typical chromosome is a few<br /> million base pairs in length.<br /> • Most bacterial species contain a<br /> single type of chromosome, but it<br /> may be present in multiple copies.<br /> • Several thousand different genes are<br /> interspersed throughout the chromosome.<br /> The short regions between adjacent genes<br /> are called intergenic regions.<br /> Genes<br /> Intergenic regions<br /> Repetitive sequences<br /> • One origin of replication is required to<br /> initiate DNA replication.<br /> • Repetitive sequences may be interspersed<br /> throughout the chromosome.<br /> FIGURE 10.4<br /> Organization of sequences in bacterial chromosomal DNA.</div> <div>17.3 TRANSPOSITION 471<br /> 5'–CTGACTCTT–3' and 5'–AAGAGTCAG–3'<br /> 3'–GACTGAGAA–5'<br /> 3'–TTCTCAGTC–5'<br /> Depending on the particular element, the lengths of inverted<br /> repeats range from 9 to 40 bp in length. In addition, insertion<br /> sequences may contain a central region that encodes the enzyme<br /> transposase, which catalyzes the transposition event.<br /> Composite transposons contain additional genes that are<br /> not necessary for transposition per se. They commonly contain<br /> genes that confer a selective advantage to the organism. Composite<br /> transposons are prevalent in bacteria, where they often<br /> contain genes that provide resistance to antibiotics or toxic<br /> heavy metals. For example, the composite transposon shown in<br /> Figure 17.13a contains two insertion sequences flanking a gene<br /> that confers antibiotic resistance. During transposition of a composite<br /> transposon, only the inverted repeats at the ends of the<br /> transposon are involved in the transpositional event. Whenever<br /> insertion sequences are found at both ends of a gene, they create<br /> a composite transposon. Both insertion sequences and composite<br /> transposons are elements that move via simple transposition,<br /> also called cut-and-paste transposition.<br /> Replicative transposons, elements that move by replicative<br /> transposition, have a sequence organization that is similar<br /> to insertion sequences except that replicative transposons have a<br /> resolvase gene that is found between the inverted repeats (Figure<br /> 17.13b). As discussed later in this chapter, both transposase and<br /> resolvase are needed to catalyze the transposition of replicative<br /> transposons.<br /> The organization of retroelements can be quite variable,<br /> and they are categorized based on their evolutionary relationship<br /> to retroviral sequences. Retroviruses are RNA viruses that make a<br /> DNA copy that integrates into the host’s genome. The viral-like<br /> retroelements are evolutionarily related to known retro viruses.<br /> These transposable elements have retained the ability to move<br /> around the genome, though, in most cases, they do not produce<br /> mature viral particles. Viral-like retroelements contain long terminal<br /> repeats (LTRs) at both ends of the element (Figure 17.13c).<br /> The LTRs are typically a few hundred nucleotides in length. Like<br /> their viral counterparts, viral-like retroelements encode virally<br /> related proteins such as reverse transcriptase and integrase that<br /> are needed for the transposition process.<br /> By comparison, nonviral-like retroelements appear less<br /> like retroviruses in their sequence, although some similarity, such<br /> as the occurrence of a reverse transcriptase gene, may be present<br /> (Figure 17.13c). Many nonviral-like retroelements, however, do<br /> not share any sequence similarity with known viruses. Instead,<br /> some nonviral-like retroelements are evolutionarily derived from<br /> normal eukaryotic genes. For example, the Alu family of repetitive<br /> sequences found in humans is derived from a single ancestral<br /> gene known as the 7SL RNA gene (a component of the complex<br /> called signal recognition particle, which is described in Chapter<br /> 13). This gene sequence has been copied by retrotransposition to<br /> achieve the current number of approximately 1,000,000 copies.<br /> Transposable elements are considered to be complete or<br /> autonomous elements when they contain all the information<br /> necessary for transposition or retrotransposition to take place.<br /> However, TEs are often incomplete or nonautonomous. A nonautonomous<br /> element typically lacks a gene such as transposase<br /> or reverse transcriptase that is necessary for transposition. The Ds<br /> locus described in the experiment of Figure 17.11 is a nonautonomous<br /> element, because it lacks a transposase gene. An element<br /> that is similar to Ds but contains a functional transposase gene<br /> is called the Ac locus or Ac element, which stands for Activator<br /> element. As mentioned earlier, an Ac locus provides a transposase<br /> gene that enables Ds to transpose. Therefore, nonautonomous<br /> TEs such as Ds can transpose only when the Ac locus is present at<br /> another region in the genome.<br /> Transposase Catalyzes the Excision<br /> and Insertion of Transposable Elements<br /> Now that we have an understanding of the typical organization<br /> of transposable elements, let’s examine the steps of the transposition<br /> process. The enzyme transposase catalyzes the removal<br /> of a TE from its original site in the chromosome and its subsequent<br /> insertion at another location. A general scheme for simple<br /> transposition is shown in Figure 17.14. Transposase binds to the<br /> inverted repeat sequences at the ends of the TE and brings them<br /> close together. The DNA is cleaved at the ends of the TE, excising<br /> it from its original site within the chromosome. The transposase<br /> carries the TE to a new site and cleaves the target DNA sequence<br /> at staggered recognition sites. The TE is then inserted and ligated<br /> to the target DNA.<br /> As noted in Figure 17.14, the ligation of the transposable<br /> element into its new site initially leaves short gaps in the target<br /> DNA. Notice that the DNA sequences in these gaps are complementary<br /> to each other (in this case, ATGCT and TACGA).<br /> Therefore, when they are filled in by DNA gap repair synthesis,<br /> the DNA base pair sequences that flank both ends of the TE are<br /> identical. These direct repeats are common features found adjacent<br /> to all TEs (see Figure 17.13).<br /> Although the transposition process depicted in Figure<br /> 17.14 does not directly alter the number of transposable elements,<br /> simple transposition is known to increase the number of<br /> TEs in genomes, in some cases to fairly high levels. How can this<br /> happen? The answer is that transposition often occurs around<br /> the time of DNA replication (Figure 17.15). After a replication<br /> fork has passed a region containing a TE, two TEs will be found<br /> behind the fork—one in each of the replicated regions. One of<br /> these TEs could then transpose from its original location into a<br /> region ahead of the replication fork. After the replication fork<br /> has passed this second region and DNA replication is completed,<br /> two TEs will be found in one of the chromosomes and one TE in<br /> the other chromosome. In this way, simple transposition can lead<br /> to an increase in TEs. We will discuss the biological significance<br /> of transposon proliferation later in this chapter.<br /> Replicative Transposition Requires<br /> Both Transposase and Resolvase<br /> Replicative transposition has been studied in several bacterial<br /> transposons and in bacteriophage μ (mu), which behaves like a</div> <div>628 CHAPTER 22 :: MEDICAL GENETICS AND CANCER<br /> C8. Gaucher disease (type I) is due to a defect in a gene that encodes<br /> a protein called acid β glucosidase. This enzyme plays a role in<br /> carbohydrate metabolism within the lysosome. The gene is located<br /> on the long arm of chromosome 1. Persons who inherit two<br /> defective copies of this gene exhibit Gaucher disease, the major<br /> symptoms of which include an enlarged spleen, bone lesions,<br /> and changes in skin pigmentation. Let’s suppose a phenotypically<br /> unaffected woman, whose father had Gaucher disease, has a child<br /> with a phenotypically unaffected man, whose mother had Gaucher<br /> disease.<br /> A. What is the probability that this child will have the disease?<br /> B. What is the probability that this child will have two normal<br /> copies of this gene?<br /> C. If this couple has five children, what is the probability that<br /> one of them will have Gaucher disease and four will be<br /> phenotypically unaffected?<br /> C9. Ehler-Danlos syndrome is a relatively rare disorder due to a<br /> mutation in a gene that encodes a protein called collagen (type<br /> 3 A1). Collagen is a protein found in the extracellular matrix<br /> that plays an important role in the formation of skin, joints,<br /> and other connective tissues. Persons with this syndrome have<br /> extraordinarily flexible skin and very loose joints. The pedigree<br /> shown here contains several members affected with Ehler-Danlos<br /> syndrome, shown with black symbols. Based on this pedigree, does<br /> this syndrome appear to be an autosomal recessive, autosomal<br /> dominant, X-linked recessive, or X-linked dominant trait? Explain<br /> your reasoning.<br /> II-1<br /> III-1<br /> III-2<br /> II-2<br /> III-3<br /> I-1 I-2<br /> II-3<br /> III-4<br /> II-4<br /> III-5<br /> II-5<br /> II-6<br /> III-6<br /> IV-1 IV-2 IV-3 IV-4 IV-5<br /> III-7<br /> II-7<br /> III-8<br /> C10. Hurler syndrome is due to a mutation in a gene that encodes a<br /> protein called α-L-iduronidase. This protein functions within the<br /> lysosome as an enzyme that breaks down mucopolysaccharides<br /> (a type of polysaccharide that has many acidic groups attached).<br /> When this enzyme is defective, excessive amounts of the<br /> mucopolysaccharides dermatan sulfate and heparin sulfate<br /> accumulate within the lysosomes, especially in liver cells and<br /> connective tissue cells. This leads to symptoms such as an enlarged<br /> liver and spleen, bone abnormalities, corneal clouding, heart<br /> problems, and severe neurological problems. The pedigree shown<br /> here contains three members affected with Hurler syndrome,<br /> indicated with black symbols. Based on this pedigree, does<br /> this syndrome appear to be an autosomal recessive, autosomal<br /> II-1<br /> III-1<br /> III-2<br /> II-2<br /> III-3<br /> I-1 I-2<br /> II-3<br /> III-4<br /> II-4<br /> III-5<br /> II-5<br /> II-6<br /> III-6 III-7 III-8<br /> IV-1 IV-2 IV-3<br /> IV-4 IV-5 IV-6<br /> C11. Like Hurler syndrome, Fabry disease involves an abnormal<br /> accumulation of substances within lysosomes. However, the lysosomes<br /> of individuals with Fabry disease show an abnormal accumulation<br /> of lipids. The defective enzyme is α-galactosidase A, which is a<br /> lysosomal enzyme that functions in lipid metabolism. This defect<br /> causes cell damage, especially to the kidneys, heart, and eyes. The gene<br /> that encodes α-galactosidase A is found on the X chromosome. Let’s<br /> suppose a phenotypically unaffected couple produces two sons with<br /> Fabry disease and one phenotypically unaffected daughter. What is<br /> the probability that the daughter will have an affected son?<br /> C12. Achondroplasia is a rare form of dwarfism caused by an autosomal<br /> dominant mutation that affects the gene that encodes a fibroblast<br /> growth factor receptor. Among 1,422,000 live births, the number of<br /> babies born with achondroplasia was 31. Among those 31 babies,<br /> 18 of them had one parent with achondroplasia. The remaining<br /> babies had two unaffected parents. How do you explain these 13<br /> babies, assuming that the mutant allele has 100% penetrance?<br /> What are the odds that these 13 individuals will pass this mutant<br /> gene to their offspring?<br /> C13. Lesch-Nyhan syndrome is due to a mutation in a gene<br /> that encodes a protein called hypoxanthine-guanine<br /> phosphoribosyltransferase (HPRT). HPRT is an enzyme that<br /> functions in purine metabolism. Persons afflicted with this<br /> syndrome have severe neurodegeneration and loss of motor<br /> control. The pedigree shown here contains several members with<br /> Lesch-Nyhan syndrome. Affected members are shown with black<br /> symbols. Based on this pedigree, does this syndrome appear to be<br /> an autosomal recessive, autosomal dominant, X-linked recessive, or<br /> X-linked dominant trait? Explain your reasoning.<br /> II-1<br /> III-1<br /> dominant, X-linked recessive, or X-linked dominant trait? Explain<br /> your reasoning.<br /> III-2<br /> II-2<br /> III-3<br /> I-1 I-2<br /> II-3<br /> III-4<br /> II-4<br /> III-5<br /> II-5<br /> II-6<br /> III-6 III-7 III-8<br /> IV-1 IV-2 IV-3 IV-4 IV-5 IV-6 IV-7 IV-8</div> <div>450 CHAPTER 16 :: GENE MUTATION AND DNA REPAIR<br /> environmental agents, known as chemical or physical mutagens,<br /> are known to induce mutations. Mutagens can alter DNA structure<br /> by modifying or removing bases, causing errors in DNA replication,<br /> forming thymine dimers, and producing breaks in DNA<br /> strands. Testing methods such as the Ames test can determine<br /> whether an agent is mutagenic.<br /> Due to the prevalence of mutagens in the environment,<br /> organisms have evolved DNA repair systems that can detect and<br /> repair different types of DNA lesions. Some DNA repair systems<br /> can recognize altered bases and repair them directly. For example,<br /> photolyase is a critical DNA repair enzyme for many plant<br /> species that removes thymine dimers, and alkyltransferase is a<br /> protein that can remove methyl or ethyl groups from guanine<br /> bases that have been mutagenized by alkylating agents. Base<br /> excision repair and nucleotide excision repair systems recognize<br /> alterations in DNA structure and excise the abnormal base or<br /> region, respectively. Certain proteins, such as the transcriptionrepair<br /> coupling factor (TRCF) of E. coli, can target the nucleotide<br /> excision repair system to actively transcribing genes that<br /> have damaged DNA. Mismatch repair systems detect and repair<br /> base mismatches between nucleotides in parental and newly made<br /> strands. The breakage of chromosomes—called a DNA doublestrand<br /> break (DSB)—is a dangerous type of DNA damage.<br /> The two main mechanisms for repair of DSBs are homologous<br /> recombination repair (HRR) and nonhomologous end joining<br /> (NHEJ). Finally, in translesion synthesis (TLS), specialized DNA<br /> polymerases are able to replicate over damaged DNA.<br /> EXPERIMENTAL SUMMARY<br /> Experimentally, the occurrence, causes, and consequences of<br /> mutation have been studied in many ways. Luria and Delbrück<br /> conducted a fluctuation test to explore whether mutations occur<br /> randomly in a population or whether they are a physiological<br /> adaptation. Similarly, the Lederbergs used replica plating to<br /> distinguish between the random mutation hypothesis and the<br /> physiological adaptation hypothesis. Results from both studies<br /> supported what is now known as the random mutation theory.<br /> According to this theory, mutations are a random process that<br /> can occur in any gene and do not involve exposure of an organism<br /> to particular conditions that select for specific types of mutations.<br /> However, certain regions of genes, called hot spots, occur<br /> where mutations are more frequent.<br /> Müller was the first scientist to establish that environmental<br /> agents such as X-rays can cause induced mutations. His work<br /> showed that X-rays dramatically increased the likelihood of X-<br /> linked, recessive lethal mutations. Since these studies, biochemists,<br /> microbiologists, and geneticists have discovered an enormous<br /> array of agents that can act as mutagens that permanently<br /> alter the structure of DNA. Many kinds of testing methods, such<br /> as the commonly used Ames test, can ascertain whether or not<br /> an agent is a mutagen.<br /> PROBLEM SETS & INSIGHTS<br /> Solved Problems<br /> S1. Mutant tRNAs may act as nonsense and missense suppressors. At<br /> the molecular level, explain how you think these suppressors work.<br /> Answer: A suppressor is a second-site mutation that suppresses<br /> the phenotypic effects of a first mutation. Intergenic suppressor<br /> mutations in tRNA genes can act as nonsense or missense suppressors.<br /> For example, let’s suppose a first mutation puts a stop codon into<br /> a structural gene. A second mutation in a tRNA gene can alter the<br /> anticodon region of a tRNA so that the anticodon recognizes a stop<br /> codon but inserts an amino acid at this site. A missense suppressor is<br /> a mutation in a tRNA gene that changes the anticodon so that it puts<br /> in the wrong amino acid at a normal codon that is not a stop codon.<br /> These mutant tRNAs are termed missense tRNAs. For example, a<br /> tRNA that normally recognizes glutamic acid may incur a mutation<br /> that changes its anticodon sequence so that it recognizes a glycine<br /> codon instead. Like nonsense suppressors, missense suppressors can be<br /> produced by mutations in the anticodon region of tRNAs so that the<br /> tRNA recognizes an incorrect codon. Alternatively, missense suppressors<br /> can also be produced by mutations in aminoacyl-tRNA synthetases that<br /> cause them to attach the incorrect amino acid to a tRNA.<br /> S2. If the rate of mutation is 10 –5 per gene, how many new mutations<br /> per gene would you expect in a population of 1 million bacteria?<br /> Answer: If we multiply the mutation rate times the number of bacteria<br /> (10 –5 × 10 6 ), we obtain a value of 10 new mutations per gene in this<br /> population. This answer is correct, but it is an oversimplification of<br /> mutation rate. For any given gene, the mutation rate is based on a<br /> probability that an event will occur. Therefore, when we consider a<br /> particular population of bacteria, we should be aware that the actual<br /> rate of new mutation would vary. Even though the rate may be 10 –5 , we<br /> would not be surprised if a population of 1 million bacteria had 9 or<br /> 11 new mutations per gene instead of the expected number of 10. We<br /> would be surprised if it had 5,000 new mutations per gene, because this<br /> value would deviate much too far from our expected number.<br /> S3. In the Ames test, several Salmonella strains are used that contain<br /> different types of mutations within the gene that encodes an<br /> enzyme necessary for histidine biosynthesis. These mutations<br /> include transversions, transitions, and frameshift mutations. Why<br /> do you think it would be informative to test a mutagen with these<br /> different strains?<br /> Answer: Different types of mutagens have different effects on<br /> DNA structure. For example, if a mutagen caused transversions,<br /> an experimenter would want to use a Salmonella strain in which a<br /> transversion would convert a his – strain into a his + strain. This type of<br /> strain would make it possible to detect the effects of the mutagen.<br /> S4. In Chapter 14, we discussed how bacterial genes can be arranged<br /> in an operon structure in which a polycistronic mRNA contains<br /> the coding sequences for two or more genes. For genes in an<br /> operon, a relatively short distance occurs between the stop codon</div> <div>10.3 EUKARYOTIC CHROMOSOMES 253<br /> Telomere<br /> Origin of<br /> replication<br /> Origin of<br /> replication<br /> Kinetochore<br /> proteins<br /> Centromere<br /> Origin of<br /> replication<br /> Origin of<br /> replication<br /> Telomere<br /> FIGURE 10.11<br /> Key features:<br /> • Eukaryotic chromosomes are usually linear.<br /> • A typical chromosome is tens of millions to<br /> hundreds of millions of base pairs in length.<br /> • Eukaryotic chromosomes occur in sets.<br /> Many species are diploid, which means that<br /> somatic cells contain 2 sets of chromosomes.<br /> • Genes are interspersed throughout the<br /> chromosome. A typical chromosome<br /> contains between a few hundred and several<br /> thousand different genes.<br /> • Each chromosome contains many origins of<br /> replication that are interspersed about every<br /> 100,000 base pairs.<br /> • Each chromosome contains a centromere<br /> that forms a recognition site for the<br /> kinetochore proteins.<br /> • Telomeres contain specialized sequences<br /> located at both ends of the linear<br /> chromosome.<br /> • Repetitive sequences are commonly found<br /> near centromeric and telomeric regions, but<br /> they may also be interspersed throughout<br /> the chromosome.<br /> Genes<br /> Repetitive sequences<br /> Organization of eukaryotic chromosomes.<br /> mere. The centromere serves as an attachment site for the kinetochore,<br /> a group of cellular proteins that link the centromere to<br /> the spindle apparatus during mitosis and meiosis, ensuring the<br /> proper segregation of the chromosomes to each daughter cell.<br /> Finally, at the ends of linear chromosomes are found specialized<br /> regions known as telomeres. Telomeres serve several important<br /> functions in the replication and stability of the chromosome. As<br /> discussed in Chapter 8, telomeres prevent chromosomes from<br /> having “sticky ends” and thereby inhibit chromosomal rearrangements<br /> such as translocations. In addition, they prevent<br /> chromosome shortening in two ways. First, the telomeres protect<br /> chromosomes from digestion via enzymes called exonucleases<br /> that recognize the ends of DNA. Secondly, as discussed in Chapter<br /> 11, an unusual form of DNA replication occurs at the telomere<br /> to ensure that eukaryotic chromosomes do not become<br /> shortened with each round of DNA replication.<br /> Genes are located between the centromeric and telomeric<br /> regions along the entire eukaryotic chromosome. A single chromosome<br /> usually has a few hundred to several thousand different<br /> genes. The sequence of a typical eukaryotic gene is a few<br /> thousand to tens of thousands of base pairs in length. In less<br /> complex eukaryotes such as yeast, genes are relatively small and<br /> primarily contain nucleotide sequences that encode the amino<br /> acid sequences within proteins. In more complex eukaryotes<br /> such as mammals and higher plants, structural genes tend to be<br /> much longer due to the presence of introns—noncoding intervening<br /> sequences. Introns range in size from less than 100 bp to<br /> more than 10,000 bp. Therefore, the presence of large introns can<br /> greatly increase the lengths of eukaryotic genes.<br /> The Genomes of Eukaryotes Contain Sequences<br /> That Are Unique, Moderately Repetitive,<br /> or Highly Repetitive<br /> The term sequence complexity refers to the number of times<br /> a particular base sequence appears throughout the genome.<br /> Unique or nonrepetitive sequences are those found once or a few<br /> times within the genome. Structural genes are typically unique<br /> sequences of DNA. The vast majority of proteins in eukaryotic<br /> cells are encoded by genes present in one or a few copies. In the<br /> case of humans, unique sequences make up roughly 40% of the<br /> entire genome.<br /> Moderately repetitive sequences are found a few hundred<br /> to several thousand times in the genome. In a few cases, moderately<br /> repetitive sequences are multiple copies of the same gene.<br /> For example, the genes that encode ribosomal RNA (rRNA)<br /> are found in many copies. Ribosomal RNA is necessary for the<br /> functioning of ribosomes. Cells need a large amount of rRNA<br /> for making ribosomes, and this is accomplished by having multiple<br /> copies of the genes that encode rRNA. Likewise, the histone<br /> genes are also found in multiple copies because a large number<br /> of histone proteins are needed for the structure of chromatin.<br /> In addition, other types of functionally important sequences<br /> can be moderately repetitive. For example, moderately repetitive<br /> sequences may play a role in the regulation of gene transcription<br /> and translation. By comparison, some moderately repetitive<br /> sequences do not play a functional role and are derived from<br /> transposable elements—segments of DNA that have the ability<br /> to move within a genome. This category of repetitive sequences<br /> is discussed in greater detail in Chapter 17.<br /> Highly repetitive sequences are found tens of thousands<br /> or even millions of times throughout the genome. Each copy of a<br /> highly repetitive sequence is relatively short, ranging from a few<br /> nucleotides to several hundred in length. A widely studied example<br /> is the Alu family of sequences found in humans and other<br /> primates. The Alu sequence is approximately 300 bp long. This<br /> sequence derives its name from the observation that it contains<br /> a site for cleavage by a restriction enzyme known as AluI. (The<br /> function of restriction enzymes is described in Chapter 18.) The<br /> Alu sequence is present in about 1,000,000 copies in the human</div> <div>676 CHAPTER 24 :: POPULATION GENETICS<br /> Dark brown coloration arises<br /> by a new mutation that<br /> creates a dark allele. Dark<br /> brown wings make the<br /> butterflies less susceptible to<br /> predation. The dark brown<br /> butterflies have a higher<br /> Darwinian fitness than do the<br /> light butterflies.<br /> Number of<br /> individuals<br /> Light wings<br /> Starting<br /> population<br /> Dark wings<br /> Many generations<br /> This population has a higher<br /> mean fitness than the starting<br /> population because the darker<br /> butterflies are less susceptible<br /> to predation and therefore are<br /> more likely to survive and<br /> reproduce.<br /> Number of<br /> individuals<br /> Population<br /> after directional<br /> selection<br /> Light wings<br /> Dark wings<br /> (a) An example of directional selection<br /> (b) Graphical representation of directional selection<br /> FIGURE 24.8 Directional selection. (a) A new mutation arises in a population that confers higher Darwinian fitness. In this example,<br /> butterflies with dark wings are more likely to survive and reproduce. Over many generations, directional selection will favor the prevalence of darker<br /> individuals. (b) A graphical representation of directional selection.<br /> Frequency of AA: p 2 W AA<br /> Frequency of Aa:<br /> Frequency of aa:<br /> 2pqW Aa<br /> q 2 W aa<br /> In a population that is changing due to natural selection, these<br /> three terms may not add up to 1.0, as they would in the Hardy-<br /> Weinberg equilibrium. Instead, the three terms sum to a value<br /> known as the mean fitness of the population (W ):<br /> p 2 W AA 1 2pqW Aa 1 q 2 W aa 5 W<br /> Dividing both sides of the equation by the mean fitness of the<br /> population,<br /> p 2 W AA<br /> W<br /> 1 2pqW Aa<br /> W<br /> 1 q2 W aa<br /> W 5 1<br /> Using this equation, we can calculate the expected genotype and<br /> allele frequencies after one generation of directional selection:<br /> Frequency of AA genotype:<br /> p 2 W AA<br /> W<br /> Frequency of Aa genotype:<br /> Frequency of aa genotype:<br /> 2pqW Aa<br /> W<br /> q 2 W aa<br /> W<br /> Allele frequency of A: p A 5 p2 W AA<br /> W<br /> Allele frequency of a: q a 5 q2 W aa<br /> W<br /> 1 pqW Aa<br /> W<br /> 1 pqW Aa<br /> W<br /> As an example, let’s suppose that the starting allele frequencies<br /> are A � 0.5 and a � 0.5, and use fitness values of 1.0,<br /> 0.8, and 0.2 for the three genotypes, AA, Aa, and aa, respectively.<br /> We begin by calculating the mean fitness of the population:<br /> p 2 W AA 1 2pqW Aa 1 q 2 W aa 5 W<br /> W 5 (0.5) 2 (1) 1 2(0.5)(0.5)(0.8) 1 (0.5) 2 (0.2)<br /> W 5 0.25 1 0.4 1 0.05 5 0.7<br /> After one generation of directional selection,</div> <div>506 CHAPTER 18 :: RECOMBINANT DNA TECHNOLOGY<br /> An important innovation in the method of dideoxy<br /> sequencing is automated sequencing. Instead of having four separate<br /> tubes, with a single type of dideoxyribonucleotide in each<br /> tube, automated sequencing uses one tube containing all four<br /> types of dideoxyribonucleotides. However, each type of dideoxyribonucleotide<br /> (ddGTP, ddATP, ddTTP, and ddCTP) has a different-colored<br /> fluorescent label attached. After incubating the<br /> target DNA with deoxyribonucleotides, the four types of fluorescent<br /> dideoxyribonucleotides, and DNA polymerase, the sample is<br /> then loaded into a single lane of a gel. A schematic example of a<br /> lane of automated sequencing is shown in Figure 18.16a.<br /> In automated sequencing, a sample is loaded at the top of a<br /> gel, and then the fragments are separated by electrophoresis. Theoretically,<br /> it would be possible to read this sequence directly from<br /> the gel. From a practical perspective, however, it is more efficient<br /> to automate the procedure using a laser and fluorescence detector.<br /> Electrophoresis is continued until each band emerges from<br /> the bottom of the gel. As each band comes off, a laser excites the<br /> fluorescent dye, and a fluorescence detector records the amount<br /> of fluorescence emission. The detector reads the level of fluorescence<br /> at four wavelengths, corresponding to the four dyes. An<br /> example of the printout from the fluorescence detector is shown<br /> in Figure 18.16b. As seen here, the peaks of fluorescence correspond<br /> to the DNA sequence that is complementary to the target<br /> DNA.<br /> G<br /> T<br /> C<br /> A<br /> G<br /> G<br /> A<br /> A<br /> T<br /> G<br /> C<br /> C<br /> A<br /> C<br /> (a) Automated sequencing<br /> gel<br /> C A CCG T AAGGA C T G<br /> (b) Output from automated sequencing<br /> FIGURE 18.16 Automated DNA sequencing. (a) This<br /> diagram schematically depicts a series of bands on a gel; the four<br /> colors of the bands occur because each type of dideoxynucleotide is<br /> labeled with a different colored fluorescent molecule. (b) The mixture<br /> of DNA fragments is electrophoresed off the end of the gel. As each<br /> band is released from the bottom of the gel, the fluorescent dye is<br /> excited by a laser, and the fluorescence emission is recorded by a<br /> fluorescence detector. The detector reads the level of fluorescence at<br /> four wavelengths, corresponding to the four dyes. As shown in the<br /> printout, the peaks of fluorescence correspond to the DNA sequence<br /> that is complementary to the target DNA described in Figure 18.15.<br /> Site-Directed Mutagenesis Is a Technique<br /> to Alter DNA Sequences<br /> As we have seen, dideoxy sequencing provides a way to determine<br /> the base sequence of DNA. To understand how the genetic material<br /> functions, researchers often analyze mutations that alter the<br /> normal DNA sequence and thereby affect the expression of genes<br /> and the outcome of traits. For example, geneticists have discovered<br /> that many inherited human diseases, such as sickle-cell disease<br /> and hemophilia, involve mutations within specific genes.<br /> These mutations provide insight into the function of the genes in<br /> unaffected individuals. Hemophilia, for example, involves deleterious<br /> mutations in genes that encode blood clotting factors.<br /> Because the analysis of mutations can provide important<br /> information about normal genetic processes, researchers often<br /> wish to obtain mutant organisms. As we discussed in Chapter 16,<br /> mutations can arise spontaneously or can be induced by environmental<br /> agents. Mendel’s pea plants are a classic example of allelic<br /> strains with different phenotypes that arose from spontaneous<br /> mutations. X-rays and UV light are physical agents that can cause<br /> induced mutations. In addition, experimental organisms can be<br /> treated with mutagens that increase the rate of mutations.<br /> More recently, researchers have developed molecular techniques<br /> to make mutations within cloned genes or other DNA<br /> segments. One widely used method, known as site-directed<br /> mutagenesis, allows a researcher to produce a mutation at a specific<br /> site within a cloned DNA segment. For example, if a DNA<br /> sequence is 5'–AAATTTCTTTAAA–3', a researcher can use sitedirected<br /> mutagenesis to change it to 5'–AAATTTGTTTAAA–3'.<br /> In this case, the researcher deliberately changed the seventh base<br /> from a C to a G. Why is this method useful? The site-directed<br /> mutant can then be introduced into a living organism to see how<br /> the mutation affects the expression of a gene, the function of a<br /> protein, and the phenotype of an organism.<br /> The first successful attempts at site-directed mutagenesis<br /> involved changes in the sequences of viral genomes. These<br /> studies were conducted in the 1970s. Mark Zoller and Michael<br /> Smith also developed a protocol for the site-directed mutagenesis<br /> of DNA that has been cloned into a viral vector. Since these<br /> early studies, many approaches have been used to achieve sitedirected<br /> mutagenesis. Figure 18.17 describes the general steps in<br /> the procedure. Prior to this experiment, the DNA was denatured<br /> into single strands; only the single strand needed for site-directed<br /> mutagenesis is shown. As in PCR, this single-stranded DNA is<br /> referred to as the template DNA, because it is used as a template<br /> to synthesize a complementary strand.<br /> As shown in Figure 18.17, an oligonucleotide primer is<br /> allowed to hybridize or anneal to the template DNA. The primer,<br /> typically 20 or so nucleotides in length, is synthesized chemically.<br /> (A shorter version of the primer is shown in Figure 18.17 for<br /> simplicity.) The scientist designs the base sequence of the primer.<br /> The primer has two important characteristics. First, most of the<br /> sequence of the primer is complementary to the site in the DNA<br /> where the mutation is to be made. However, a second feature is<br /> that the primer contains a region of mismatch where the primer<br /> and template DNA are not complementary. The mutation will</div> <div>706 CHAPTER 25 :: QUANTITATIVE GENETICS<br /> rough eyes (R) and minute bristles (M) inherited one copy of<br /> chromosome 3 from the DDT-resistant strain and one copy from<br /> the DDT-sensitive strain. The transmission of the other Drosophila<br /> chromosomes can also be followed in a similar way. Therefore,<br /> the phenotypes of the offspring from the backcross provide<br /> a way to discern whether particular chromosomes were inherited<br /> from the DDT-resistant or DDT-sensitive strain.<br /> Figure 25.6 shows the protocol followed by James Crow.<br /> He began with a DDT-resistant strain that had been produced by<br /> exposing flies to DDT for many generations. This DDT-resistant<br /> strain was crossed to a sensitive strain. As described previously in<br /> Figure 25.5, the two strains had allelic markers that made it possible<br /> to determine the origins of the different Drosophila chromosomes.<br /> Recall that Drosophila has four chromosomes. In this<br /> study, only chromosomes X, 2, and 3 were marked with alleles.<br /> Chromosome 4 was neglected due to its very small size. The F 1<br /> flies were backcrossed to both parental strains, and then the F 2<br /> female progeny were examined in two ways. First, their phenotypes<br /> were examined to determine whether particular chromosomes<br /> were inherited from the DDT-resistant or DDT-sensitive<br /> strain. Next, the female flies were exposed to filter paper impregnated<br /> with DDT. It was then determined if the flies survived this<br /> exposure for 18 to 24 hours.<br /> ■ THE HYPOTHESIS<br /> DDT resistance is a polygenic trait.<br /> ■ TESTING THE HYPOTHESIS — FIGURE 25.6 Polygenic inheritance of DDT-resistance alleles in Drosophila<br /> melanogaster.<br /> Starting material: DDT-resistant and DDT-sensitive strains of fruit flies.<br /> 1. Cross the DDT-resistant strain to the<br /> sensitive strain. In each strain,<br /> chromosomes X, 2, and 3 were marked<br /> with alleles that provided easily<br /> discernible phenotypes.<br /> DDT-resistant<br /> flies<br /> Experimental level<br /> x<br /> DDT-sensitive<br /> flies<br /> Conceptual level<br /> Resistance to DDT may be<br /> due to multiple genes on<br /> different chromosomes.<br /> 2. Take the F 1 flies and backcross to both<br /> parental strains.<br /> 3. Identify the origin of the chromosomes<br /> in the F 2 flies according to their<br /> phenotypes.<br /> 4. Expose the F 2 female flies to DDT on a<br /> filter paper for 18–24 hours.<br /> F 1 offspring<br /> Backcross the F 1<br /> offspring to both<br /> parental strains.<br /> Record phenotypes of F 2 offspring<br /> Expose F 2 flies to DDT.<br /> F 2 offspring will have different<br /> combinations of chromosomes<br /> from the DDT-resistant and<br /> DDT-sensitive strains. The<br /> phenotypes reveal whether<br /> particular chromosomes were<br /> inherited from the DDT-resistant<br /> or DDT-sensitive strain.</div> <div>Test Yourself<br /> Take a quiz at the Genetics Online<br /> Learning Center to gauge your mastery<br /> of chapter content. Each chapter quiz<br /> is specifically constructed to test your<br /> comprehension of key concepts.<br /> Immediate feedback on your responses<br /> explains why an answer is correct or<br /> incorrect. You can even e-mail your<br /> quiz results to your professor!<br /> Tweaking the Experiment<br /> For the in-depth experiments that are<br /> found in each chapter, a self-help quiz<br /> is found. These questions ask you to<br /> predict how changes in the experimental<br /> parameters may affect the outcome of<br /> the experiment.<br /> Interactive Activities<br /> Fun and exciting learning experiences await you at the Genetics Online<br /> Learning Center! Each chapter offers a series of interactive crossword<br /> puzzles, vocabulary flashcards, and other engaging activities designed<br /> to reinforce learning.</div> <div>23.2 VERTEBRATE DEVELOPMENT 647<br /> 23.2 VERTEBRATE DEVELOPMENT<br /> Biologists have studied the morphological features of development<br /> in many vertebrate species. Historically, amphibians and birds have<br /> been studied extensively, because their eggs are rather large and<br /> easy to manipulate. For example, certain developmental stages of<br /> the frog and chicken have been described in great detail. In more<br /> recent times, the successes obtained in Drosophila have shown<br /> the great power of genetic analyses in elucidating the underlying<br /> molecular mechanisms that govern biological development. With<br /> this knowledge, many researchers are attempting to understand<br /> the genetic pathways that govern the development of the more<br /> complex body structure found in vertebrate organisms.<br /> Several vertebrate species have been the subject of developmental<br /> studies. These include the mouse (Mus musculus), the<br /> frog (Xenopus laevis), and the small aquarium zebrafish (Danio<br /> rerio). In this section, we will primarily discuss the genes that are<br /> important in mammalian development, particularly those that<br /> have been characterized in the mouse, one of the best-studied<br /> mammals. As we will see, several genes affecting its developmental<br /> pathways have been cloned and characterized. In this section,<br /> we will examine how these genes affect the course of vertebrate<br /> development.<br /> Researchers Have Identified Homeotic Genes<br /> in Vertebrates<br /> Vertebrates typically have long generation times and produce<br /> relatively few offspring. Therefore, it is usually not practical<br /> to screen large numbers of embryos or offspring in search of<br /> mutant phenotypes with developmental defects. As an alternative,<br /> a successful way of identifying genes that affect vertebrate<br /> development has been the use of molecular techniques to identify<br /> vertebrate genes similar to those that control development in<br /> simpler organisms such as Drosophila.<br /> As discussed in Chapters 21 and 26, species that are evolutionarily<br /> related to each other often contain genes with similar<br /> DNA sequences. When two or more genes have similar sequences<br /> because they are derived from the same ancestral gene, they are<br /> called homologous genes. Homologous genes found in different<br /> species are termed orthologs. Experimentally, a DNA strand<br /> from one gene will hybridize to a complementary strand of a<br /> homologue, because they have similar sequences.<br /> Researchers initially followed a strategy of using cloned Drosophila<br /> genes as probes to identify homologous vertebrate genes,<br /> which has been quite successful. With this method, researchers<br /> have found complexes of homeotic genes in many vertebrate species<br /> that are homologous to those in the fruit fly. These groups<br /> of adjacent homeotic genes are called Hox complexes. As shown<br /> in Figure 23.16, the mouse has four Hox complexes, designated<br /> HoxA (on chromosome 6), HoxB (on chromosome 11), HoxC<br /> (on chromosome 15), and HoxD (on chromosome 2). A total<br /> of 38 genes are found in the four complexes. Thirteen different<br /> types of homeotic genes occur within the four Hox complexes,<br /> although none of the four complexes contains representatives of<br /> all 13 types of genes. The addition of Hox genes into the genomes<br /> Fly<br /> Mouse<br /> Anterior<br /> Posterior<br /> lab pb Dfd Scr Antp Ubx abd-A AbdB<br /> HoxA<br /> HoxB<br /> HoxC<br /> HoxD<br /> Antennapedia<br /> complex<br /> 1 2 3 4 5 6 7 9 10 11 13<br /> 1 2 3 4 5 6 7 8 9<br /> 1<br /> bithorax<br /> complex<br /> 4 5 6 8 9 10 1112 13<br /> 3 4 8 9 10 11 12 13<br /> FIGURE 23.16 A comparison of homeotic genes in<br /> Drosophila and the mouse. The mouse contains four gene clusters,<br /> HoxA–D, that correspond to certain homeotic genes found in<br /> Drosophila. Thirteen different types of homeotic genes are found in<br /> the mouse, although each Hox gene cluster does not contain all 13<br /> genes. In this drawing, orthologous genes are aligned in columns. For<br /> example, lab is the ortholog to HoxA-1, HoxB-1, and HoxD-1; Ubx is<br /> orthologous to HoxA-7 and HoxB-7.<br /> of animals has allowed animals to develop more complex body<br /> plans. We will consider the evolutionary origin of the Hox genes<br /> and their importance in the evolution of animal species in<br /> Chapter 26.<br /> Remarkably, several of the homeotic genes in fruit flies<br /> and mammals are strikingly similar. Among the first six types of<br /> genes, five of them are homologous to genes found in the Antennapedia<br /> complex of Drosophila. Among the last seven, three are<br /> homologous to the genes of the bithorax complex. These results<br /> indicate there are fundamental similarities in the ways that animals<br /> as different as fruit flies and mammals undergo embryonic<br /> development. This suggests that a “universal body plan” underlies<br /> animal development.<br /> Like the Antennapedia and bithorax complexes in Drosophila,<br /> the arrangement of Hox genes along the mouse chromosomes<br /> reflects their pattern of expression from the anterior<br /> to the posterior end (Figure 23.17a). This phenomenon is seen<br /> in more detail in Figure 23.17b, which shows the expression<br /> pattern for a group of HoxB genes in a mouse embryo. Overall,<br /> these results are consistent with the idea that the Hox genes play<br /> a role in determining the fates of segments along the anteroposterior<br /> axis.<br /> Currently, researchers are trying to understand the functional<br /> roles of the genes within the Hox complexes in vertebrate<br /> development. In Drosophila, great advances in developmental<br /> genetics have been made by studying mutant alleles in genes that<br /> control development. In mice, however, few natural mutations<br /> have been identified that affect development. This has made it</div> <div>750 CHAPTER 26 :: EVOLUTIONARY GENETICS<br /> and ducks. The BMP4 protein, which is stained in blue, causes<br /> cells to undergo apoptosis and die. The gremlin protein, which is<br /> stained brown in Figure 26.20b, inhibits the function of BMP4<br /> and thereby allows cells to survive. In the developing chicken<br /> limb, the gremlin gene is expressed throughout the limb, except<br /> in the regions between each digit. Therefore, these cells die, and a<br /> chicken develops a nonwebbed foot (Figure 26.20c). In contrast,<br /> gremlin expression in the duck occurs throughout the entire limb,<br /> including the interdigit regions, resulting in a webbed foot. Interestingly,<br /> researchers have been able to introduce gremlin protein<br /> into the interdigit regions of developing chicken limbs. This produces<br /> a chicken with webbed feet!<br /> How do these observations relate to bird evolution? As we<br /> have seen, variation in the expression of these genes determines<br /> whether or not a bird’s feet are webbed. At some point in the<br /> evolution of birds, mutations occurred that provided variation in<br /> the expression of the BMP4 and gremlin genes, which resulted<br /> in nonwebbed or webbed feet. In terrestrial settings, having nonwebbed<br /> feet is an advantage because these are more effective at<br /> holding on to perches, running along the ground, and snatching<br /> prey. Therefore, natural selection would maintain nonwebbed<br /> feet in terrestrial environments. This process explains the occurrence<br /> of nonwebbed feet in chickens, hawks, crows, and many<br /> other terrestrial birds. In aquatic environments, webbed feet<br /> are an advantage because they act as paddles for swimming, so<br /> genetic variation that produced webbed feet would have been<br /> promoted by natural selection. Over the course of many generations,<br /> this gave rise to webbed feet that are now found in ducks,<br /> geese, penguins, and other aquatic birds.<br /> How does having webbed or nonwebbed feet influence<br /> speciation? This trait may not directly affect the ability of two<br /> individuals to mate. However, due to natural selection, birds with<br /> webbed feet would become more prevalent in aquatic environments,<br /> while birds with nonwebbed feet would be found in terrestrial<br /> locations. Therefore, reproductive isolation would occur<br /> because the populations would occupy different environments.<br /> The Evolution of Animal Body Plans Is Related<br /> to Changes in Hox Gene Number and Expression<br /> Hox genes, which are discussed in Chapter 23, are found in all<br /> animals. Developmental biologists have speculated that genetic<br /> variation in the Hox genes may have spawned the formation<br /> of many new body types, yielding many different animal species.<br /> As shown in Figure 26.21, the number and arrangement of<br /> Hox genes varies considerably among different types of animals.<br /> Sponges<br /> Sponges are the simplest animals, with bodies that are not organized along a<br /> body axis.<br /> Anemones<br /> Anemones have a primitive body axis, showing radial symmetry.<br /> Flatworms<br /> The other animals shown in this figure have a more complex form of symmetry<br /> called bilateral symmetry, meaning that their bodies are organized along a welldefined<br /> anteroposterior axis, with right and left sides that show a mirror symmetry.<br /> Such organisms are called bilaterians. Flatworms are very simple bilaterians.<br /> Insects<br /> Simple chordates<br /> Mammals<br /> Bilaterians<br /> Chordates<br /> Vertebrates<br /> Invertebrates such as insects are structurally more complex than flatworms, but<br /> less complex than organisms with a spinal cord.<br /> Animals with spinal cords are known as chordates. The simple chordates lack<br /> bony vertebrae that enclose the spinal cord.<br /> The vertebrates, such as mammals, have vertebrae and possess a very complex<br /> body structure.<br /> Anterior Group 3 Central Posterior<br /> FIGURE 26.21 Hox gene composition in different types of animals. Researchers speculate that the duplication of Hox genes and Hox gene<br /> clusters played a key role in the evolution of more complex body plans in animals. The Hox genes are divided into four groups, called anterior, group 3,<br /> central, and posterior, based on their relative similarities. Each group is represented by a different color in this figure.</div> <div>58 CHAPTER 3 :: REPRODUCTION AND CHROMOSOME TRANSMISSION<br /> Kinetochore<br /> FIGURE 3.13 Attachment of the kinetochore microtubules<br /> to replicated chromosomes during meiosis. The kinetochore<br /> microtubules from a given pole are attached to one pair of chromatids<br /> in a bivalent, but not both. Therefore, each pair of sister chromatids is<br /> attached to only one pole.<br /> MEIOSIS I<br /> In Animals, Spermatogenesis Produces Four<br /> Haploid Sperm Cells and Oogenesis Produces<br /> a Single Haploid Egg Cell<br /> In male animals, spermatogenesis, the production of sperm,<br /> occurs within glands known as the testes. The testes contain<br /> spermatogonial cells that divide by mitosis to produce two cells.<br /> One of these remains a spermatogonial cell, and the other cell<br /> becomes a primary spermatocyte. As shown in Figure 3.14a, the<br /> spermatocyte progresses through meiosis I and meiosis II to produce<br /> four haploid cells, which are known as spermatids. These<br /> cells then mature into sperm cells. The structure of a sperm cell<br /> includes a long flagellum and a head. The head of the sperm<br /> contains little more than a haploid nucleus and an organelle at<br /> its tip, known as an acrosome. The acrosome contains digestive<br /> enzymes that are released when a sperm meets an egg cell. These<br /> enzymes enable the sperm to penetrate the outer protective layers<br /> of the egg and gain entry into the egg cell’s cytosol. In animal<br /> species without a mating season, sperm production is a continuous<br /> process in mature males. A mature human male, for example,<br /> produces several hundred million sperm each day.<br /> MEIOSIS II<br /> Primary<br /> spermatocyte<br /> (diploid)<br /> (a) Spermatogenesis<br /> Spermatids<br /> Sperm cells<br /> (haploid)<br /> Secondary oocyte<br /> Primary<br /> oocyte<br /> (diploid)<br /> Polar<br /> bodies<br /> Egg cell<br /> (haploid)<br /> (b) Oogenesis<br /> Polar body<br /> FIGURE 3.14 Gametogenesis in animals. (a) Spermatogenesis. A diploid spermatocyte undergoes meiosis to produce four haploid (n)<br /> spermatids. These differentiate during spermatogenesis to become mature sperm. (b) Oogenesis. A diploid oocyte undergoes meiosis to produce one<br /> haploid egg cell and two or three polar bodies. For some species, the first polar body divides, while in other species, it does not. Because of asymmetric<br /> cytokinesis, the amount of cytoplasm the egg receives is maximized. The polar bodies degenerate.</div> <div>580 CHAPTER 21 :: GENOMICS II: FUNCTIONAL GENOMICS, PROTEOMICS, AND BIOINFORMATICS<br /> If they get a PCR product, this means that the protein of interest<br /> must have been bound (either directly or through other proteins) to<br /> this DNA site in living cells.<br /> Alternatively, a researcher may want to determine where<br /> the protein of interest binds across the whole genome. In this<br /> case, a DNA microarray can be used (see bottom right side of<br /> Figure 21.4). Because a DNA microarray is found on a chip,<br /> this is called a ChIP-on-chip assay. The ends of the precipitated<br /> DNA are first ligated to short DNA pieces called linkers.<br /> PCR primers are then added that are complementary to<br /> the linkers and, therefore, amplify the DNA regions between<br /> the linkers. During PCR, the DNA is fluorescently labeled. The<br /> labeled DNA is then hybridized to a DNA microarray. Because<br /> the DNA was isolated using an antibody against the protein of<br /> interest, the fluorescent spots on the microarray identify sites<br /> in the genome where the protein binds. In this way, researchers<br /> may be able to determine where a protein binds to locations<br /> in the genome, even if those site(s) had not been previously<br /> determined by other methods.<br /> Protein of interest<br /> Add formaldehyde to crosslink<br /> protein to DNA. Lyse the cells.<br /> Sonicate DNA into small pieces.<br /> Add antibodies against protein<br /> of interest. The antibodies cause<br /> the DNA–protein complexes to<br /> precipitate. The sample is<br /> subjected to centrifugation.<br /> 21.2 PROTEOMICS<br /> Thus far, we have considered ways to characterize the genome of<br /> a given species and study its function. Because most genes encode<br /> proteins, a logical next step is to examine the functional roles of<br /> the proteins that a species can make. As mentioned, this field is<br /> called proteomics, and the entire collection of a species’ proteins<br /> is its proteome.<br /> Genomics represents only the first step in our comprehensive<br /> understanding of protein structure and function. Researchers<br /> often use genomic information to initiate proteomic studies,<br /> but such information must be followed up with research that<br /> involves the direct analysis of proteins. For example, as discussed<br /> in the preceding section, a DNA microarray may provide insights<br /> into the transcription of particular genes under a given set of<br /> conditions. However, mRNA levels may not provide an accurate<br /> measure of the abundance of a protein that is encoded by<br /> a given gene. Protein levels are greatly affected, not only by the<br /> level of mRNA, but also by the rate of mRNA translation and by<br /> the turnover rate of a given protein. Therefore, DNA microarray<br /> data must be corroborated using other methods, such as Western<br /> blotting (discussed in Chapter 18), which directly determine the<br /> abundance of a protein in a given cell type.<br /> A second way that genomic data can elucidate the workings<br /> of the proteome is via homology. As discussed later in this chapter,<br /> homology between the genes of different species can be used<br /> to predict protein structure and/or function, when the structure<br /> or function of the protein in one of the species is already<br /> known. However, homology may not provide direct information<br /> regarding the regulation of protein structure and function. Also,<br /> it may not reveal potential types of protein–protein interactions<br /> in which a given protein may participate. Therefore, even though<br /> Conduct PCR using primers<br /> to a specific DNA region.<br /> If PCR amplifies the DNA,<br /> the protein was bound to<br /> the DNA region that is<br /> recognized by the primers.<br /> or<br /> Antibody against<br /> protein of interest<br /> Pellet<br /> Collect complexes in pellet.<br /> Add chemical that breaks the<br /> crosslinks to remove the protein.<br /> Covalently attach DNA linkers to<br /> the ends of the DNA.<br /> Linker<br /> Conduct PCR using primers<br /> that are complementary to<br /> the linkers. Incorporate<br /> fluorescently labeled<br /> nucleotides during PCR.<br /> Denature DNA and<br /> hybridize to a microarray.<br /> FIGURE 21.4 Chromatin immunoprecipitation (ChIP). This<br /> method can determine whether proteins bind to a particular region of DNA<br /> found within the chromatin of living cells.<br /> See Figure 21.2</div> <div>9.2 NUCLEIC ACID STRUCTURE 231<br /> H<br /> C<br /> C N<br /> H<br /> O C<br /> H C<br /> N C<br /> C N<br /> O<br /> C<br /> O<br /> H<br /> C<br /> H<br /> O N C<br /> C N<br /> H<br /> O C<br /> H C<br /> N C N<br /> C O<br /> C<br /> HO<br /> C<br /> H<br /> O N C<br /> C N<br /> H<br /> O C<br /> H C<br /> N<br /> C<br /> C N<br /> O<br /> O<br /> (a) An � helix in a protein<br /> Carbonyl<br /> oxygen<br /> Amide<br /> hydrogen<br /> Hydrogen<br /> bond<br /> (b) Linus Pauling<br /> FIGURE 9.12 Linus Pauling and the A-helix protein structure.<br /> (a) An a helix is a secondary structure found in proteins. This structure<br /> emphasizes the polypeptide backbone (shown as a tan ribbon), which is<br /> composed of amino acids linked together in a linear fashion. Hydrogen<br /> bonding between hydrogen and oxygen atoms stabilizes the helical<br /> conformation. (b) Linus Pauling with a ball-and-stick model.<br /> (a) Rosalind Franklin<br /> X rays diffracted<br /> by DNA<br /> Wet DNA fibers<br /> X-ray beam<br /> The pattern represents the<br /> atomic array in wet fibers.<br /> A second important development that led to the elucidation<br /> of the double helix was X-ray diffraction data. When a purified<br /> substance, such as DNA, is subjected to X-rays, it will produce<br /> a well-defined diffraction pattern if the molecule is organized<br /> into a regular structural pattern. An interpretation of the diffraction<br /> pattern (using mathematical theory) can ultimately provide<br /> information concerning the structure of the molecule. Rosalind<br /> Franklin (Figure 9.13a), working in the same laboratory as Maurice<br /> Wilkins, used X-ray diffraction to study wet DNA fibers.<br /> Franklin made marked advances in X-ray diffraction techniques<br /> while working with DNA. She adjusted her equipment to produce<br /> an extremely fine beam of X-rays. She extracted finer DNA fibers<br /> than ever before and arranged them in parallel bundles. Franklin<br /> also studied the fibers’ reactions to humid conditions. The<br /> (b) X-ray diffraction of wet DNA fibers<br /> FIGURE 9.13<br /> X-ray diffraction of DNA.<br /> diffraction pattern of Franklin’s DNA fibers is shown in Figure<br /> 9.13b. This pattern suggested several structural features of DNA.<br /> First, it was consistent with a helical structure. Second, the diameter<br /> of the helical structure was too wide to be only a single-stranded<br /> helix. Finally, the diffraction pattern indicated that the helix contains<br /> about 10 base pairs per complete turn. These observations<br /> were instrumental in solving the structure of DNA.<br /> EXPERIMENT 9B<br /> Chargaff Found That DNA Has a Biochemical<br /> Composition in Which the Amount of A Equals T<br /> and the Amount of G Equals C<br /> Another piece of information that led to the discovery of the<br /> double-helix structure came from the studies of Erwin Chargaff.<br /> In the 1940s and 50s, he pioneered many of the biochemical<br /> techniques for the isolation, purification, and measurement of<br /> nucleic acids from living cells. This was not a trivial undertaking,<br /> because the biochemical composition of living cells is complex.<br /> At the time of Chargaff’s work, researchers already knew that<br /> the building blocks of DNA are nucleotides containing the bases<br /> adenine, thymine, guanine, or cytosine. Chargaff analyzed the<br /> base composition of DNA, which was isolated from many different<br /> species. He expected that the results might provide important<br /> clues concerning the structure of DNA.</div> <div>19.1 THE USES OF MICROORGANISMS IN BIOTECHNOLOGY 517<br /> 4. Transform the plasmids into E. coli by<br /> amp<br /> Mix DNA with<br /> S<br /> cells<br /> ampicillin-sensitive<br /> (amp S ) bacterial cells.<br /> treatment with CaCl 2 . The transformed<br /> cells are spread on plates containing<br /> ampicillin. Grow overnight to obtain<br /> bacterial colonies that are ampicillin<br /> resistant because they contain the<br /> plasmid.<br /> CaCl 2 solution<br /> with ampicillinresistant<br /> plasmids<br /> Transform<br /> plasmid<br /> into cells.<br /> 5. Pick cells from a bacterial colony and<br /> grow these recombinant bacteria in<br /> liquid media; activate the lac promoter<br /> with isopropyl thiodigalactoside (IPTG).<br /> Ampicillinresistant<br /> colony<br /> lacZ<br /> P O som<br /> lacZ<br /> P O som<br /> Note: The lac promoter in this plasmid<br /> is controlled by the lac repressor<br /> (see Chapter 15). IPTG is an inducer<br /> that activates transcription of the lacZ<br /> gene by removing the lac repressor.<br /> Add cells<br /> to tube.<br /> Add<br /> IPTG.<br /> Suspension of<br /> bacterial cells<br /> lacZ<br /> P O som<br /> +IPTG<br /> lacZ<br /> P O som<br /> Fusion<br /> protein<br /> β-galactosidase<br /> 6. Place in a tube and centrifuge to obtain<br /> a bacterial cell pellet.<br /> Fusion<br /> proteins<br /> No fusion<br /> proteins<br /> Pellet<br /> containing<br /> bacterial cells<br /> β-galactosidase<br /> Methionine<br /> 7. Resuspend the pellet in 70% formic acid<br /> and cyanogen bromide (5 mg/ml).<br /> Note: This breaks open the cells and<br /> cleaves polypeptides at methionine<br /> residues.<br /> 70%<br /> formic acid<br /> + CNBr<br /> Somatostatin<br /> P O lacZ<br /> som<br /> P O lacZ<br /> som<br /> 8. Determine the amount of somatostatin<br /> using a radioimmunoassay. (See the<br /> Appendix for a description of this<br /> procedure.)<br /> See Appendix.<br /> Broken cell<br /> See Appendix.<br /> Broken cell</div> <div>9<br /> MOLECULAR<br /> STRUCTURE OF DNA<br /> AND RNA<br /> ::<br /> In Chapters 2 through 8, we focused on the relationship between the inheritance<br /> of genes and chromosomes, and the outcome of an organism’s traits. In<br /> Chapter 9, we will shift our attention to molecular genetics—the study of DNA<br /> structure and function at the molecular level. An exciting goal of molecular<br /> genetics is to use our knowledge of DNA structure to understand how DNA<br /> functions as the genetic material. Using molecular techniques, researchers have<br /> determined the organization of many genes. This information, in turn, has<br /> helped us understand how the expression of such genes governs the outcome of<br /> an individual’s inherited traits.<br /> The past several decades have seen dramatic advances in techniques and<br /> approaches to investigate and even to alter the genetic material. These advances<br /> have greatly expanded our understanding of molecular genetics and also have<br /> provided key insights into the mechanisms underlying transmission and population<br /> genetics. Molecular genetic technology is also widely used in supporting<br /> disciplines such as biochemistry, cell biology, and microbiology.<br /> To a large extent, our understanding of genetics comes from our knowledge<br /> of the molecular structure of DNA (deoxyribonucleic acid) and RNA<br /> (ribonucleic acid). In this chapter, we will begin by considering classic experiments<br /> that showed DNA is the genetic material. We will then survey the molecular<br /> features of DNA and RNA that underlie their function.<br /> CHAPTER OUTLINE<br /> 9.1 Identification of DNA<br /> as the Genetic Material<br /> 9.2 Nucleic Acid Structure<br /> 9.1 IDENTIFICATION OF DNA<br /> AS THE GENETIC MATERIAL<br /> In his pioneering experiments, Gregor Mendel studied several different traits<br /> in pea plants. By conducting the appropriate crosses, he showed that traits are<br /> inherited as discrete units as they pass from parent to offspring. This implies<br /> that living organisms contain a genetic material that governs an individual’s<br /> traits, a substance that is transferred during the process of reproduction.<br /> To fulfill its role, the genetic material must meet several criteria.<br /> 1. Information: The genetic material must contain the information<br /> necessary to construct an entire organism. In other words, it must<br /> provide the blueprint to determine the inherited traits of an organism.<br /> 2. Transmission: During reproduction, the genetic material must be passed<br /> from parents to offspring.<br /> A molecular model showing the structure of the<br /> DNA double helix.<br /> PART III<br /> Molecular Structure and Replication of the Genetic Material</div> <div>156 CHAPTER 6 :: GENETIC TRANSFER AND MAPPING IN BACTERIA AND BACTERIOPHAGES<br /> Number of Colonies<br /> Non- That Grew On: Cotrans-<br /> Selected selected Galactose + Galactose � duction<br /> Gene Gene Biotin Biotin Frequency<br /> galK + bioD + 80 10 0.125<br /> How far apart are these two genes?<br /> Answer: We can use the cotransduction frequency to calculate the<br /> distance between the two genes (in minutes) using the equation<br /> Cotransduction frequency = (1 – d/2) 3<br /> 0.125 = (1 – d/2) 3<br /> 3<br /> 1 – d/2 = "0.125<br /> 1 – d/2 = 0.5<br /> d/2 = 1 – 0.5<br /> d = 1.0 minute<br /> The two genes are approximately 1 minute apart on the E. coli<br /> chromosome.<br /> S2. By conducting mating experiments between a single Hfr strain<br /> and a recipient strain, Wollman and Jacob mapped the order of<br /> many bacterial genes. Throughout the course of their studies, they<br /> identified several different Hfr strains in which the F factor DNA had<br /> been integrated at different places along the bacterial chromosome. A<br /> sample of their experimental results is shown in the following table:<br /> Order of Transfer of Several Different<br /> Hfr<br /> Bacterial Genes<br /> strain Origin First Last<br /> H O thr leu azi ton pro lac gal str met<br /> 1 O leu thr met str gal lac pro ton azi<br /> 2 O pro ton azi leu thr met str gal lac<br /> 3 O lac pro ton azi leu thr met str gal<br /> 4 O met str gal lac pro ton azi leu thr<br /> 5 O met thr leu azi ton pro lac gal str<br /> 6 O met thr leu azi ton pro lac gal str<br /> 7 O ton azi leu thr met str gal lac pro<br /> A. Explain how these results are consistent with the idea that the<br /> bacterial chromosome is circular.<br /> B. Draw a map that shows the order of genes and the locations of<br /> the origins of transfer among these different Hfr strains.<br /> Answer:<br /> A. In comparing the data among different Hfr strains, the order<br /> of the nine genes was always the same or the reverse of the same order.<br /> For example, HfrH and Hfr1 transfer the same genes but their orders are<br /> reversed relative to each other. In addition, the Hfr strains showed an<br /> overlapping pattern of transfer with regard to the origin. For example,<br /> Hfr1 and Hfr2 had the same order of genes, but Hfr1 began with leu and<br /> ended with azi, while Hfr2 began with pro and ended with lac. From these<br /> findings, Wollman and Jacob concluded that the segment of DNA that<br /> was the origin of transfer had been inserted at different points within a<br /> circular E. coli chromosome in different Hfr strains. They also concluded<br /> that the origin can be inserted in either orientation, so the direction of<br /> gene transfer can be clockwise or counterclockwise around the circular<br /> bacterial chromosome.<br /> B. A genetic map consistent with these results is shown here.<br /> ton<br /> 7<br /> pro<br /> 2<br /> 1<br /> azi<br /> lac<br /> 3<br /> leu<br /> E. coli<br /> genetic<br /> map<br /> gal<br /> thr<br /> str<br /> H<br /> met<br /> S3. An Hfr strain that is leuA + and thiL + was mated to a strain that<br /> is leuA – and thiL – . In the data points shown here, the mating was<br /> interrupted and the percentage of recombinants for each gene was<br /> determined by streaking on media that lacked either leucine or<br /> thiamine. The results are shown.<br /> % of F – recipient cells<br /> that have received the<br /> gene during conjugation<br /> 30<br /> 20<br /> 10<br /> leuA + thiL +<br /> 0<br /> 0 10 20 30 40 50<br /> Duration of mating (minutes)<br /> What is the map distance (in minutes) between these two genes?<br /> Answer: This problem is solved by extrapolating the data points to the<br /> x-axis to determine the time of entry. For leuA + , they extrapolate back to<br /> 10 minutes. For thiL + , they extrapolate back to 20 minutes. Therefore, the<br /> distance between the two genes is approximately 10 minutes.<br /> S4. Genetic transfer via transformation can also be used to map genes<br /> along the bacterial chromosome. In this approach, fragments of<br /> chromosomal DNA are isolated from one bacterial strain and<br /> used to transform another strain. The experimenter examines the<br /> transformed bacteria to see if they have incorporated two or more<br /> different genes. For example, the DNA may be isolated from a donor<br /> E. coli bacterium that has functional copies of the araB and leuD<br /> genes. Let’s call these genes araB + and leuD + to indicate the genes are<br /> functional. These two genes are required for arabinose metabolism<br /> and leucine synthesis, respectively. To map the distance between these<br /> two genes via transformation, a recipient bacterium would be used<br /> that is araB – and leuD – . Following transformation, the recipient<br /> bacterium may become araB + and leuD + . This phenomenon is called<br /> cotransformation because two genes from the donor bacterium<br /> have been transferred to the recipient via transformation. In<br /> 6<br /> 4<br /> 5</div> <div>recombination<br /> 524 CHAPTER 19 :: BIOTECHNOLOGY<br /> The gene of interest has been cloned. A<br /> neomycin resistance gene is inserted into<br /> the center of this gene, and a thymidine<br /> kinase gene is inserted next to the gene.<br /> TK<br /> Neo R<br /> Gene<br /> of<br /> interest<br /> – Same gene as normal,<br /> chromosomal gene except<br /> it has Neo R inserted into it.<br /> TK<br /> Normal<br /> gene<br /> Neo R<br /> Chromosomal<br /> DNA<br /> This cloned DNA is then introduced into<br /> embryonic stem cells. In this case, the<br /> cells were derived from a mouse with dark<br /> fur color. The cells are grown in the<br /> presence of neomycin and gancyclovir.<br /> Only those cells that contain the Neo R<br /> gene but are lacking the TK gene will<br /> survive.<br /> Dies<br /> Nonhomologous<br /> Embryonic stem cells<br /> Homologous<br /> recombination<br /> Lives<br /> Neo R<br /> Blastocyst<br /> Surviving cells are injected into embryonic<br /> blastocysts derived from a mouse with<br /> white coat color. The injected blastocysts<br /> are reimplanted into the uterus of a<br /> female mouse.<br /> Following birth, chimeric mice are<br /> identified as those that contain a<br /> coat with both dark and white fur.<br /> The appropriate crosses are made<br /> in order to produce mice that have<br /> two copies of the target gene.<br /> Chimeric offspring<br /> FIGURE 19.7 Producing a gene<br /> replacement in mice. The bottom of this figure<br /> shows a photograph of a chimeric mouse. Note the<br /> patches of black and white fur.</div> <div>536 CHAPTER 19 :: BIOTECHNOLOGY<br /> FIGURE 19.20 Methods of gene<br /> transfer used in gene therapy. (a) In this<br /> example, the DNA containing the gene<br /> of interest is complexed with cationic<br /> liposomes. These complexes are taken into<br /> cells by endocytosis, in which a portion<br /> of the plasma membrane invaginates and<br /> creates an intracellular vesicle known as<br /> an endosome. After it is released from the<br /> endosome, the DNA may then integrate into<br /> the chromosomal DNA via recombination.<br /> (b) In this example, the gene of interest is<br /> cloned into a retrovirus. When the retrovirus<br /> infects a cell, the RNA genome is reverse<br /> transcribed into double-stranded DNA, which<br /> then integrates into the chromosome. Viruses<br /> used in gene therapy have been genetically<br /> altered so they cannot proliferate after entry<br /> into the target cell.<br /> Target cell<br /> Liposome<br /> DNA carrying the<br /> gene of interest<br /> DNA-liposome complex is<br /> taken into the target cell by<br /> endocytosis.<br /> The liposome is degraded<br /> within the endosome and the<br /> DNA is released into the<br /> cytosol.<br /> The DNA is imported into the<br /> cell nucleus.<br /> RNA<br /> genome<br /> Target cell<br /> Retrovirus-<br /> RNA genome contains<br /> gene of interest.<br /> Retrovirus is taken into the<br /> target cell via endocytosis.<br /> The viral coat is disassembled<br /> in the endosome, and two<br /> copies of the RNA genome are<br /> released into the cytosol.<br /> The RNA is reverse<br /> transcribed into DNA, which<br /> travels into the nucleus.<br /> Endosome<br /> Endosome<br /> By recombination, the DNA<br /> carrying the gene of interest is<br /> integrated into a chromosome of<br /> the target cell.<br /> Reverse<br /> transcriptase<br /> By recombination, the viral<br /> DNA, carrying the gene of<br /> interest, is integrated into a<br /> chromosome of the target cell.<br /> Integrated<br /> gene<br /> Integrated<br /> gene<br /> (a) Nonviral approach<br /> (b) Viral approach</div> <div>780 SOLUTIONS TO EVEN-NUMBERED PROBLEMS<br /> E6. This is really a matter of opinion. The Avery, MacLeod, and McCarty<br /> experiment seems to indicate directly that DNA is the genetic material,<br /> because DNase prevented transformation and RNase and protease did<br /> not. However, one could argue that the DNA is required for the rough<br /> bacteria to take up some other contaminant in the DNA preparation.<br /> It would seem that the other contaminant would not be RNA or<br /> protein. The Hershey and Chase experiments indicate that DNA is<br /> being injected into bacteria, although quantitatively the results are not<br /> entirely convincing. Some 35 S-labeled protein was not sheared off, so<br /> the results do not definitely rule out the possibility that protein could<br /> be the genetic material. But the results do indicate that DNA is the<br /> more likely candidate.<br /> E8. A. The purpose of chromatography was to separate the different types<br /> of bases.<br /> B. It was necessary to separate the bases and determine the total<br /> amount of each type of base. In a DNA strand, all the bases are<br /> found within a single molecule, so it is difficult to measure the<br /> total amount of each type of base. When the bases are removed<br /> from the strand, each type can be purified, and then the total<br /> amount of each type of base can be measured by spectroscopy.<br /> C. Chargaff’s results would probably not be very convincing if done<br /> on a single species. The strength of his data was that all species<br /> appeared to conform to the AT/GC rule, suggesting that this is<br /> a consistent feature of DNA structure. In a single species, the<br /> observation is that A � T and G � C could occur as a matter of<br /> chance.<br /> Questions for Student Discussion/Collaboration<br /> 2. There are many possibilities. You could use a DNA-specific chemical<br /> and show that it causes heritable mutations. Perhaps you could inject<br /> an oocyte with a piece of DNA and produce a mouse with a new trait.<br /> CHAPTER 10<br /> Conceptual Questions<br /> C2. Viruses also need sequences that enable them to be replicated. These<br /> sequences are equivalent to the origins of replication found in<br /> bacterial and eukaryotic chromosomes.<br /> C4. A bacterium with two nucleoids is similar to a diploid eukaryotic<br /> cell because it would have two copies of each gene. The bacterium<br /> is different, however, with regard to alleles. A eukaryotic cell can<br /> have two different alleles for the same gene. For example, a cell<br /> from a pea plant could be heterozygous, Tt, for the gene that affects<br /> height. By comparison, a bacterium with two nucleoids has two<br /> identical chromosomes. Therefore, a bacterium with two nucleoids is<br /> homozygous for its chromosomal genes. Note: As discussed in Chapter<br /> 6, a bacterium can contain another piece of DNA, called an F' factor,<br /> that can carry a few genes. The alleles on an F' factor can be different<br /> from the alleles on the bacterial chromosome.<br /> C6. A. One loop is 40,000 bp. One base pair is 0.34 nm, which equals<br /> 0.34 � 10 �3 μm. If we multiply the two together:<br /> (40,000)(0.34 � 10 �3 ) � 13.6 μm<br /> B. Circumference � pD<br /> 13.6 μm � pD<br /> D � 4.3 μm<br /> C. No, it is too big to fit inside of E. coli. Supercoiling is needed to<br /> make the loops more compact.<br /> C8. These drugs would diminish the amount of negative supercoiling in<br /> DNA. Negative supercoiling is needed to compact the chromosomal<br /> DNA, and it also aids in strand separation. Bacteria might not be able<br /> to survive and/or transmit their chromosomes to daughter cells if their<br /> DNA was not compacted properly. Also, because negative supercoiling<br /> aids in strand separation, these drugs would make it more difficult<br /> for the DNA strands to separate. Therefore, the bacteria would have<br /> C10.<br /> a difficult time transcribing their genes and replicating their DNA,<br /> because both processes require strand separation. As discussed in<br /> Chapter 11, DNA replication is needed to make new copies of the<br /> genetic material to transmit from mother to daughter cells. If DNA<br /> replication was inhibited, the bacteria could not grow and divide into<br /> new daughter cells. As discussed in Chapters 12–14, gene transcription<br /> is necessary for bacterial cells to make proteins. If gene transcription<br /> was inhibited, the bacteria could not make many proteins that are<br /> necessary for survival.<br /> Gyrase<br /> Negative<br /> supercoil<br /> C12. The centromere is the attachment site for the kinetochore, which<br /> attaches to the spindle. If a chromosome is not attached to the<br /> spindle, it is free to “float around” within the cell, and it may not be<br /> near a pole when the nuclear membrane re-forms during telophase.<br /> If a chromosome is left outside of the nucleus, it is degraded during<br /> interphase. That is why the chromosome without a centromere may<br /> not be found in daughter cells.<br /> C14. Highly repetitive DNA, as its name suggests, is a DNA sequence that<br /> is repeated many times, from tens of thousands to millions of times<br /> throughout the genome. It can be interspersed in the genome or found<br /> clustered in a tandem array, in which a short nucleotide sequence is<br /> repeated many times in a row. In DNA renaturation studies, highly<br /> repetitive DNA renatures at a much faster rate because there are many<br /> copies of the complementary sequences.<br /> C16. During interphase (i.e., G 1 , S, and G 2 ), the euchromatin is found<br /> primarily as a 30 nm fiber in a radial loop configuration. Most<br /> interphase chromosomes also have some heterochromatic regions<br /> where the radial loops are more highly compacted. During M phase,<br /> each chromosome becomes entirely heterochromatic, which is needed<br /> for the proper sorting of the chromosomes during nuclear division.<br /> C18.<br /> Scaffold<br /> Magnified<br /> DNA loop<br /> AT-rich<br /> sequences<br /> (MAR)<br /> C20. During interphase, the chromosomes are found within the cell nucleus.<br /> They are less tightly packed and are transcriptionally active. Segments<br /> of chromosomes are anchored to the nuclear matrix. During M<br /> phase, the chromosomes become highly condensed, and the nuclear<br /> membrane is fragmented into vesicles. The chromosomal DNA<br /> remains anchored to a scaffold, formed from the nuclear matrix. The<br /> chromosomes eventually become attached to the spindle apparatus via<br /> microtubules that attach to the kinetochore, which is attached to the<br /> centromere.<br /> C22. There are 146 bp around the core histones. If the linker region is<br /> 54 bp, we expect 200 bp of DNA (i.e., 146 � 54) for each nucleosome<br /> and linker region. If we divide 46,000 bp by 200 bp, we get 230.<br /> Because there are two molecules of H2A for each nucleosome, there<br /> would be 460 molecules of H2A in a 46,000-bp sample of DNA.<br /> C24. The role of the core histones is to form the nucleosomes. In a<br /> nucleosome, the DNA is wrapped 1.65 times around the core histones.<br /> Histone H1 binds to the linker region. It may play a role in compacting<br /> the DNA into a 30 nm fiber.<br /> C26. The answer is B and E. A Barr body is composed of a type of highly<br /> compacted chromatin called heterochromatin. Euchromatin is not so<br /> compacted. A Barr body is not composed of euchromatin. A Barr body<br /> is one chromosome, the X chromosome. The term genome refers to all<br /> the types of chromosomes that make up the genetic composition of an<br /> individual.</div> <div>5.2 GENETIC MAPPING IN PLANTS AND ANIMALS 113<br /> two genes. Those with long bristles and ebony bodies or short<br /> bristles and gray bodies have inherited a chromosome that is the<br /> product of a crossover during meiosis in the heterozygous parent.<br /> As noted in Figure 5.9, the recombinant offspring are fewer<br /> in number than are the nonrecombinant offspring.<br /> The frequency of recombination can be used as an estimate<br /> of the physical distance between two genes on the same chromosome.<br /> The map distance is defined as the number of recombinant<br /> offspring divided by the total number of offspring, multiplied<br /> by 100. We can calculate the map distance between these<br /> two genes using this formula:<br /> Map distance 5<br /> 5<br /> Number of recombinant offspring<br /> Total number of offspring<br /> 76 1 75<br /> 537 1 542 1 76 1 75 3 100<br /> 5 12.3 map units<br /> 3 100<br /> The units of distance are called map units (mu) or sometimes<br /> centiMorgans (cM) in honor of Thomas Hunt Morgan. One<br /> map unit is equivalent to a 1% frequency of recombination. In this<br /> example, we would conclude that the s and e alleles are 12.3 map<br /> units apart from each other along the same chromosome.<br /> EXPERIMENT 5B<br /> Alfred Sturtevant Used the Frequency<br /> of Crossing Over in Dihybrid Crosses<br /> to Produce the First Genetic Map<br /> In 1911, the first individual to construct a (very small) genetic<br /> map was Alfred Sturtevant, an undergraduate who spent time in<br /> the laboratory of Thomas Hunt Morgan. Sturtevant wrote: “In<br /> conversation with Morgan . . . I suddenly realized that the variations<br /> in the strength of linkage, already attributed by Morgan<br /> to differences in the spatial separation of the genes, offered the<br /> possibility of determining sequences [of different genes] in the<br /> linear dimension of a chromosome. I went home and spent most<br /> of the night (to the neglect of my undergraduate homework) in<br /> producing the first chromosome map, which included the sexlinked<br /> genes, y, w, v, m, and r, in the order and approximately the<br /> relative spacing that they still appear on the standard maps.”<br /> In the experiment of Figure 5.10, Sturtevant considered<br /> the outcome of crosses involving six different mutant alleles that<br /> altered the phenotype of flies. All of these alleles were known to<br /> be recessive and X linked. They are y (yellow body color), w<br /> (white eye color), w-e (eosin eye color), v (vermilion eye color),<br /> m (miniature wings), and r (rudimentary wings). The w and<br /> w-e alleles are alleles of the same gene. In contrast, the v allele<br /> (vermilion eye color) is an allele of a different gene that also<br /> affects eye color. The two alleles that affect wing length, m and<br /> r, are also in different genes. Therefore, Sturtevant studied the<br /> inheritance of six recessive alleles, but since w and w-e are alleles<br /> of the same gene, his genetic map contained only five genes.<br /> The corresponding wild-type alleles are y + (gray body), w + (red<br /> eyes), v + (red eyes), m + (long wings), and r + (long wings).<br /> ■ THE HYPOTHESIS<br /> When genes are located on the same chromosome, the distance<br /> between the genes can be estimated from the proportion of<br /> recombinant offspring. This provides a way to map the order of<br /> genes along a chromosome.<br /> ■ TESTING THE HYPOTHESIS — FIGURE 5.10 The first genetic mapping experiment.<br /> Starting materials: Sturtevant began with several different strains of Drosophila that contained the six alleles already described.<br /> Experimental level<br /> Conceptual level<br /> 1. Cross a female that is heterozygous for<br /> two different genes to a male that is<br /> hemizygous recessive for the same two<br /> genes. In this example, cross a female<br /> that is X y+ w + X yw to a male that is X yw Y.<br /> x<br /> y +<br /> w +<br /> y<br /> w<br /> y<br /> w<br /> This strategy was employed for many<br /> dihybrid combinations of the six alleles<br /> already described.<br /> Y chromosome<br /> Offspring</div> <div>562 CHAPTER 20 :: GENOMICS I: ANALYSIS OF DNA<br /> Isolate<br /> chromosomal<br /> DNA<br /> Isolate<br /> chromosomal<br /> DNA<br /> BAC contig<br /> Clone large chromosomal<br /> DNA fragments into BACs and<br /> create a contig for each<br /> chromosome.<br /> BAC<br /> vector<br /> Chromosomal<br /> DNA<br /> For each BAC, shear into<br /> smaller pieces and clone DNA<br /> pieces into vectors.<br /> Vector<br /> Chromosomal<br /> DNA<br /> Shear DNA into small and<br /> large pieces. Clone<br /> chromosomal DNA pieces<br /> into vectors.<br /> Clones from<br /> one BAC insert:<br /> Vector<br /> Chromosomal<br /> DNA<br /> From the clones of each BAC,<br /> determine the chromosomal<br /> DNA sequence, usually at<br /> one end, by shotgun<br /> sequencing. The results<br /> below show the sequences<br /> from three chromosomal DNA<br /> clones.<br /> CCGACCTTACCGACCA<br /> GACCACCCGATTAAT<br /> TTAATCGCGAATTG<br /> Based on overlapping<br /> regions, create one<br /> contiguous sequence.<br /> CCGACCTT ACCGACCACCCGAT T AATCGCGAATTG<br /> Determine the chromosomal<br /> DNA sequence usually, at<br /> both ends, by shotgun<br /> sequencing. The results<br /> below show sequences of<br /> three chromosomal DNA<br /> clones.<br /> TTACCGGTAGGCACCT<br /> CACCTGTTACGGGTC<br /> GGGT CAA ACCT AGG<br /> Based on overlapping<br /> regions, create one<br /> contiguous sequence.<br /> TTACCGGT AGGCACCTGTTACGG GTCAAACCTAGG<br /> (a) Hierarchical genome shotgun sequencing<br /> (b) Whole genome shotgun sequencing<br /> FIGURE 20.20 Two approaches of shotgun sequencing of genomes. Note: the BAC vector and chromosomal DNA are not drawn to scale. The<br /> chromosomal DNA would be much longer than the BAC vector DNA.</div> <div>GLOSSARY 821<br /> sigma factor a transcription factor that recognizes<br /> bacterial promoter sequences and facilitates the<br /> binding of RNA polymerase to the promoter.<br /> silencer a DNA sequence that functions as a regulatory<br /> element. The binding of a regulatory transcription<br /> factor to the silencer decreases the level of<br /> transcription.<br /> silent mutation a mutation that does not alter the amino<br /> acid sequence of the encoded polypeptide even<br /> though the nucleotide sequence has changed.<br /> similarity with regard to DNA, refers to a comparison<br /> of DNA sequences that have regions where the bases<br /> match up. Similarity may be due to homology.<br /> simple Mendelian inheritance an inheritance pattern<br /> involving a simple, dominant/recessive relationship<br /> that produces observed ratios in the offspring that<br /> readily obey Mendel’s laws.<br /> simple translocation when one piece of a chromosome<br /> becomes attached to a different chromosome.<br /> simple transposition a cut-and-paste mechanism for<br /> transposition in which a transposable element is<br /> removed from one site and then inserted into another<br /> site.<br /> SINEs in mammals, short interspersed elements that are<br /> less than 500 base pairs in length.<br /> single-factor cross see monohybrid cross.<br /> single-nucleotide polymorphism a genetic<br /> polymorphism within a population in which two<br /> alleles of the gene differ by a single nucleotide.<br /> single-strand binding protein a protein that binds to<br /> both of the single strands of DNA during DNA<br /> replication and prevents them from re-forming a<br /> double helix.<br /> sister chromatid exchange (SCE) the phenomenon in<br /> which crossing over occurs between sister chromatids,<br /> thereby exchanging identical genetic material.<br /> sister chromatids pairs of replicated chromosomes that<br /> are attached to each other at the centromere. Sister<br /> chromatids are genetically identical to each other.<br /> site-directed mutagenesis a technique that enables<br /> scientists to change the sequence of cloned DNA<br /> segments.<br /> site-specific recombination when two different DNA<br /> segments break and rejoin with each other at a<br /> specific site. This occurs during the integration of<br /> certain viruses into the host chromosome and during<br /> the rearrangement of immunoglobulin genes.<br /> SMC proteins proteins that use energy from ATP to<br /> catalyze changes in chromosome structure.<br /> snRNP refers to a complex containing small nuclear RNA<br /> and a set of proteins, which are components of the<br /> spliceosome.<br /> somatic cell refers to any cell of the body except for<br /> germ-line cells that give rise to gametes.<br /> somatic mutation a mutation in a somatic cell.<br /> sorting signal an amino acid sequence or<br /> posttranslational modification that directs a protein<br /> to the correct region of the cell.<br /> SOS response a response to extreme environmental stress<br /> in which bacteria replicate their DNA using DNA<br /> polymerases that are likely to make mistakes.<br /> Southern blotting a technique used to detect the presence<br /> of a particular genetic sequence within a mixture of<br /> many chromosomal DNA fragments.<br /> speciation the process by which new species are formed<br /> via evolution.<br /> species a group of organisms that maintains a distinctive<br /> set of attributes in nature.<br /> species concepts different approaches for distinguishing<br /> species.<br /> spectrophotometer a device used by re searchers to<br /> determine how much radiation at various wavelengths<br /> a sample can absorb.<br /> spermatids immature sperm cells produced from<br /> spermatogenesis.<br /> spermatogenesis the production of sperm cells.<br /> sperm cell a male gamete. Sperm are small and usually<br /> travel relatively far distances to reach the female<br /> gamete.<br /> spindle see mitotic spindle apparatus.<br /> spliceosome a multisubunit complex that functions in the<br /> splicing of eukaryotic pre-mRNA.<br /> splicing see RNA splicing.<br /> splicing factor a protein that regulates the process of<br /> RNA splicing.<br /> spontaneous mutation a change in DNA structure that<br /> results from random abnormalities in biological<br /> processes.<br /> spores haploid cells that are produced by certain species<br /> such as fungi (i.e., yeast and molds).<br /> sporophyte the diploid generation of plants.<br /> SR protein a type of splicing factor.<br /> stabilizing selection natural selection that favors<br /> individuals with an intermediate phenotype.<br /> stamen the structure found in the flower of higher plants<br /> that produces the male gametophyte (i.e., pollen).<br /> standard deviation a statistic that is computed as the<br /> square root of the variance.<br /> start codon a three-base sequence in mRNA that initiates<br /> translation. It is usually 5'-AUG-3' and encodes<br /> methionine.<br /> stem cell a cell that has the capacity to divide and to<br /> differentiate into one or more specific cell types.<br /> steroid receptor a category of transcription factors that<br /> respond to steroid hormones. An example is the<br /> glucocorticoid receptor.<br /> stigma the structure in flowering plants on which the<br /> pollen land and the pollen tube starts to grows so that<br /> sperm cells can reach the egg cells.<br /> stop codon a three-base sequence in mRNA that signals<br /> the end of translation of a polypeptide. The three stop<br /> codons are 5'–UAA–3', 5'–UAG–3', and<br /> 5'–UGA–3'.<br /> strain a variety that continues to produce the same<br /> characteristic after several generations.<br /> strand in DNA or RNA, nucleotides covalently linked<br /> together to form a long, linear polymer.<br /> stripe-specific enhancer in Drosophila, a regulatory<br /> region that controls the expression of a gene so that it<br /> occurs only in a particular parasegment during early<br /> embryonic development.<br /> STRs see short tandem repeat sequences.<br /> structural gene a gene that encodes the amino acid<br /> sequence within a particular polypeptide or protein.<br /> STS see sequence-tagged site.<br /> subcloning the procedure of making smaller DNA clones<br /> from a larger one.<br /> submetacentric describes a chromosome in which the<br /> centromere is slightly off center.<br /> subtractive cDNA library a cDNA library that contains<br /> cDNA inserts derived from mRNA that is expressed<br /> only under certain conditions.<br /> subtractive hybridization a method used to create a<br /> subtractive cDNA library.<br /> subunit this term may have multiple meanings. In a<br /> protein, each subunit is a single polypeptide.<br /> sum rule the probability that one of two or more<br /> mutually exclusive events will occur is equal to the<br /> sum of their individual probabilities.<br /> supercoiling see DNA supercoiling.<br /> supergroups a relatively recent way for evolutionary<br /> biologists to subdivide the eukaryotic domain.<br /> supernatant following centrifugation, the fluid that is<br /> found above the pellet.<br /> suppressor (or suppressor mutation) a mutation at a<br /> second site that suppresses the phenotypic effects of<br /> another mutation.<br /> SWI/SNF family a group of related proteins that catalyze<br /> chromatin remodeling.<br /> sympatric speciation (Greek, sym, “together”; Latin,<br /> patria, “homeland”) a form of speciation that occurs<br /> when members of a species diverge while occupying<br /> the same habitat within the same range.<br /> synapomorphy see shared derived character.<br /> synapsis the event in which homologous chromosomes<br /> recognize each other and then align themselves along<br /> their entire lengths.<br /> synaptonemal complex a complex of proteins that<br /> promote the interconnection between homologous<br /> chromosomes during meiosis.<br /> synonymous codons two different codons that specify<br /> the same amino acid.<br /> synteny group a group of genes that are found in the<br /> same order on the chromosomes of different species.<br /> systematists biologists who study the evolutionary<br /> relationships among different species.<br /> TT an abbreviation for thymine.<br /> tandem array (or tandem repeat) a short nucleotide<br /> sequence that is repeated many times in a row.<br /> tandem mass spectrometry the sequential use of two<br /> mass spectrometers. It can be used to determine the<br /> sequence of amino acids in a polypeptide.<br /> Taq polymerase a thermostable form of DNA polymerase<br /> used in PCR experiments.<br /> TATA box a sequence found within eukaryotic core<br /> promoters that determines the starting site for<br /> transcription. The TATA box is recognized by a TATAbinding<br /> protein, which is a component of TFIID.<br /> tautomer refers to the forms of certain small molecules,<br /> such as bases, which can spontaneously interconvert<br /> between chemically similar forms.<br /> tautomeric shift a change in chemical structure such as<br /> an alternation between the keto- and enol-forms of<br /> the bases that are found in DNA.<br /> T DNA a segment of DNA found within a Ti plasmid that<br /> is transferred from a bacterium to infected plant cells.<br /> The T DNA from the Ti plasmid becomes integrated<br /> into the chromosomal DNA of the plant cell by<br /> recombination.<br /> TE see transposable element.<br /> telocentric describes a chromosome with its centromere<br /> at one end.<br /> telomerase the enzyme that recognizes telomeric<br /> sequences at the ends of eukaryotic chromosomes and<br /> synthesizes additional numbers of telomeric repeat<br /> sequences.<br /> telomeres specialized DNA sequences found at the ends<br /> of linear eukaryotic chromosomes.<br /> telophase the fifth stage of M phase. The chromosomes<br /> have reached their respective poles and decondense.<br /> temperate phage a bacteriophage that usually exists in the<br /> lysogenic cycle.<br /> temperature-sensitive allele an allele in which the<br /> resulting phenotype depends on the environmental<br /> temperature.<br /> temperature-sensitive (ts) lethal allele an allele that is<br /> lethal at a certain environmental temperature.<br /> template DNA a strand of DNA that is used to synthesize<br /> a complementary strand of DNA or RNA.<br /> template strand see template DNA.<br /> terminal deficiency see terminal deletion.<br /> terminal deletion when a segment is lost from the end of<br /> a linear chromosome.<br /> termination (1) in transcription, the release of the<br /> newly made RNA transcript and RNA polymerase<br /> from the DNA; (2) in translation, the release of the<br /> polypeptide and the last tRNA and the disassembly of<br /> the ribosomal subunits and mRNA.<br /> termination codon see stop codon.<br /> termination sequences (ter sequences) in E. coli, a pair<br /> of sequences in the chromosome that bind a protein<br /> known as the termination utilization substance (Tus),<br /> which stops the movement of the replication forks.<br /> terminator a sequence within a gene that signals the end<br /> of transcription.<br /> tertiary structure the three-dimensional structure of a<br /> macromolecule, such as the tertiary structure of a<br /> polypeptide.<br /> testcross an experimental cross between a recessive<br /> individual and an individual whose genotype the<br /> experimenter wishes to determine.<br /> tetrad (1) the association among four sister chromatids<br /> during meiosis; (2) a group of four fungal spores<br /> contained within an ascus.<br /> tetraploid having four sets of chromosomes (i.e., 4n).<br /> tetratype (T) an ascus that has two parental cells and two<br /> nonparental cells.<br /> TFIID a type of general transcription factor in eukaryotes<br /> that is needed for RNA polymerase II function. It<br /> binds to the TATA box and recruits RNA polymerase<br /> II to the core promoter.<br /> thermocycler a device that automates the timing<br /> of temperature changes in each cycle of a PCR<br /> experiment.<br /> 30 nm fiber the association of nucleosomes to form a<br /> more compact structure that is 30 nm in diameter.</div> <div>21.2 PROTEOMICS 583<br /> How does this information lead to the identification of a<br /> specific protein? After a researcher has obtained a few short peptide<br /> sequences from a given protein, genomic information can<br /> readily predict the entire amino acid sequence of the protein.<br /> For example, if a peptide had the sequence serine–histidine–<br /> leucine–asparagine–serine–asparagine, one could determine the<br /> possible codon sequences that could encode such a peptide.<br /> More than one sequence is possible due to the degeneracy of<br /> the genetic code. Using computer software described later, the<br /> codon sequences would be used as query sequences to search an<br /> entire genomic sequence. This program would locate a match<br /> between the predicted codon sequence and a specific gene within<br /> N<br /> N<br /> Digest protein into<br /> small fragments<br /> using a protease.<br /> Purified protein<br /> C<br /> C<br /> Determine the mass<br /> of these fragments with<br /> a first spectrometer.<br /> the genome. In this way, mass spectrometry makes it possible<br /> to identify the gene that encodes the entire protein. The gene<br /> sequence, in turn, can be used to predict the remaining amino<br /> acid sequence of the entire protein.<br /> Mass spectrometry can also identify protein covalent modifications.<br /> For example, if an amino acid within a peptide was<br /> phosphorylated, the mass of the peptide would be increased by<br /> the mass of a phosphate group. This increase in mass can be<br /> determined via mass spectrometry.<br /> Protein Microarrays Can Be Used to Study<br /> Protein Expression and Function<br /> Earlier in this chapter, we learned about DNA microarrays,<br /> which have gained widespread use to study gene expression at<br /> the RNA level. The technology to make DNA microarrays is also<br /> being applied to make protein microarrays. In this type of technology,<br /> proteins, rather than DNA molecules, are spotted onto a<br /> glass or silica slide. The development of protein microarrays is<br /> more challenging because proteins are much more easily damaged<br /> by the manipulations that occur during microarray formation.<br /> For example, the three-dimensional structure of a protein<br /> may be severely damaged by drying, which usually occurs during<br /> the formation of a microarray. This has created additional<br /> challenges for researchers who are developing the technology of<br /> protein microarrays. In addition, the synthesis and purification<br /> of proteins tend to be more time-consuming compared to the<br /> production of DNA, which can be amplified by PCR or directly<br /> synthesized on the microarray itself. In spite of these technical<br /> difficulties, the last few years have seen progress in the production<br /> and uses of protein microarrays (Table 21.2).<br /> Abundance<br /> 1,652 daltons<br /> 0 4,000<br /> Mass/charge<br /> Analyze this fragment with<br /> a second spectrometer.<br /> The peptide is fragmented<br /> from one end.<br /> TABLE 21.2<br /> Some Applications of Protein Microarrays<br /> Application<br /> Protein expression<br /> Description<br /> An antibody microarray can measure protein<br /> expression, because each antibody in a given spot<br /> recognizes a specific amino acid sequence. This can<br /> be used to study the expression of proteins in a<br /> cell-specific manner. It can also be used to determine<br /> how environmental conditions affect the levels of<br /> particular proteins.<br /> Abundance<br /> 900<br /> 1,201 1,428 1,652<br /> 1,008 1,315 1,565<br /> 1,114<br /> –Asn<br /> Mass/charge<br /> Ser Asn Leu His Ser–<br /> 1,800<br /> FIGURE 21.6 The use of tandem mass spectrometry to<br /> determine the amino acid sequence of a peptide.<br /> Protein function<br /> Protein–protein<br /> interactions<br /> Pharmacology<br /> The substrate specificity and enzymatic activities of<br /> groups of proteins can be analyzed by exposing a<br /> functional protein microarray to a variety of substrates.<br /> The ability of two proteins to interact with each<br /> other can be determined by exposing a functional<br /> protein microarray to fluorescently labeled proteins.<br /> The ability of drugs to bind to cellular proteins can<br /> be determined by exposing a functional protein<br /> microarray to different kinds of labeled drugs. This<br /> can help to identify the proteins within a cell to<br /> which a given drug may bind.</div> <div>402 CHAPTER 15 :: GENE REGULATION IN EUKARYOTES<br /> ■<br /> THE DATA<br /> Source of Nuclei % Hybridization of DNA Probe<br /> Reticulocytes 25%<br /> Brain cells >94%<br /> Fibroblasts >94%<br /> ■ INTERPRETING THE DATA<br /> As shown in the data of Figure 15.10, a much smaller percentage<br /> of the radiolabeled DNA probe hybridized to the chromosomal<br /> DNA in reticulocytes compared to brain cells and fibroblasts.<br /> What do these results mean? The data are consistent with<br /> the idea that DNase I had digested the globin gene in the chromatin<br /> of reticulocytes into small fragments that were too small<br /> to hybridize to the radiolabeled DNA probe. In other words, the<br /> globin genes in reticulocytes are more sensitive to DNase I digestion.<br /> By comparison, the globin genes in brain cells and fibroblasts<br /> are relatively resistant. Because the globin genes are expressed in<br /> reticulocytes but not in brain cells and fibroblasts, these results are<br /> consistent with the hypothesis that the globin gene is less tightly<br /> packed when it is being expressed. In cells where the globin genes<br /> should not be expressed, such as brain cells and fibroblasts, the<br /> chromatin containing the β-globin gene is tightly compacted.<br /> Changes in chromatin compaction provide one way to regulate<br /> globin gene expression among different cell types.<br /> A self-help quiz involving this experiment can be found at<br /> the Online Learning Center.<br /> Transcriptional Activators Recruit Chromatin-<br /> Remodeling Enzymes to the Promoter Region<br /> In recent years, geneticists have been trying to identify the steps<br /> that promote the interconversion between the closed and open<br /> conformations of chromatin. As we saw for globin genes, changes<br /> in chromatin structure are associated with the level of gene transcription.<br /> Based on the analysis of many genes, researchers have<br /> discovered that a key role of some transcriptional activators is to<br /> orchestrate changes in chromatin compaction from the closed to<br /> the open conformation.<br /> As discussed in Chapter 12, chromatin structure is altered<br /> in eukaryotes in two common ways: ATP-dependent chromatin<br /> remodeling and covalent modification of histones. The SWI/SNF<br /> family is a major category of proteins found in all eukaryotic species<br /> that catalyze ATP-dependent chromatin remodeling. The net<br /> result of ATP-dependent chromatin remodeling is a change in<br /> the locations of nucleosomes. This may involve a shift in nucleosomes<br /> to a new location or a change in the spacing of nucleosomes<br /> over a long stretch of DNA, both of which may loosen the<br /> level of chromatin compaction.<br /> Nucleosomes have been shown to change position in cells<br /> that normally express a particular gene but not in cells where the<br /> gene is inactive. For example, in fibroblasts that do not express the<br /> β-globin gene, nucleosomes are positioned at regular intervals from<br /> nucleotides –3,000 to +1,500 (Figure 15.11a). However, in reticulocytes<br /> that express the β-globin gene, a disruption in nucleosome<br /> positioning occurs in the region from nucleotide –500 to +200<br /> (Figure 15.11b). This disruption is a key step in gene activation.<br /> The position of a nucleosome may greatly influence whether or<br /> not a gene can be transcribed. For example, if the TATA box is<br /> tightly bound to a core histone protein, it may be inaccessible to<br /> general transcription factors and RNA polymerase.<br /> A second mechanism that leads to an alteration in chromatin<br /> structure involves the covalent modification of histones. A<br /> histone code appears to play a key role in controlling the level of<br /> chromatin compaction. The amino terminal ends of histone proteins<br /> are covalently modified in several ways, including the acetylation<br /> of lysines, methylation of lysines, and phosphorylation of<br /> –3,000<br /> Nucleosomes<br /> Inactive promoter<br /> (a) Nucleosomes in fibroblasts, which do not express<br /> the �-globin gene<br /> Active promoter<br /> –3,000 –500 +200<br /> (b) Nucleosomes in reticulocytes, which express<br /> the �-globin gene<br /> +1,500<br /> +1,500<br /> FIGURE 15.11 Changes in nucleosome position during<br /> the activation of the �-globin gene. (a) In fibroblasts, which do not<br /> express β globin, nucleosome positioning is uninterrupted. (b) In<br /> reticulocytes, which express this gene, disruption occurs in the positions<br /> of nucleosomes from the –500 to +200 region. (Position +1 is the<br /> beginning of transcription.) It is not yet clear whether the histones are<br /> removed, the histones are partially displaced, and/or other proteins are<br /> bound to this region.<br /> serines (see Chapter 10, Figure 10.20). Because these covalent<br /> modifications influence the compaction of chromatin, they play<br /> a key role in transcriptional regulation. For example, positively<br /> charged lysines within the core histone proteins can be acetylated<br /> by enzymes called histone acetyltransferases. The attachment<br /> of the acetyl group (–COCH 3 ) may have two effects. First, it<br /> eliminates the positive charge on the lysine side chain and thereby<br /> disrupts the electrostatic attraction between the histone protein<br /> and the negatively charged DNA backbone. Secondly, covalently<br /> modified histones are recognized by DNA-binding proteins that<br /> promote changes in the compaction level of the chromatin. In the<br /> case of acetylation, the net effect is to diminish the interactions</div> <div>52 CHAPTER 3 :: REPRODUCTION AND CHROMOSOME TRANSMISSION<br /> Two centrosomes,<br /> each with centriole pairs<br /> Microtubules<br /> forming mitotic spindle<br /> Sister chromatids<br /> Nuclear membrane<br /> fragmenting into vesicles<br /> Nuclear<br /> membrane<br /> Mitotic<br /> spindle<br /> Chromosomes<br /> Nucleolus<br /> Spindle<br /> pole<br /> (a) INTERPHASE (b) PROPHASE (c) PROMETAPHASE<br /> Astral microtubule<br /> Metaphase<br /> plate<br /> Chromosomes<br /> 25 μm<br /> Nuclear<br /> membrane<br /> re-forming<br /> Kinetochore<br /> proteins atached<br /> to centromere<br /> Kinetochore<br /> microtubule<br /> Polar<br /> microtubule<br /> Cleavage<br /> furrow<br /> Chromosomes<br /> decondensing<br /> (d) METAPHASE (e) ANAPHASE (f) TELOPHASE AND CYTOKINESIS<br /> FIGURE 3.8 The process of mitosis in an animal cell. The top panels illustrate cells of a fish embryo progressing through<br /> mitosis. The bottom panels are schematic drawings that emphasize the sorting and separation of the chromosomes. In this case, the<br /> original diploid cell had six chromosomes (three in each set). At the start of mitosis, these have already replicated into 12 chromatids. The<br /> final result is two daughter cells each containing six chromosomes.</div> <div>278 CHAPTER 11 :: DNA REPLICATION<br /> DNA polymerase III is responsible for most of the DNA<br /> replication. It is a large enzyme consisting of 10 different subunits<br /> that play various roles in the DNA replication process<br /> (Table 11.2). The a subunit actually catalyzes the bond formation<br /> between adjacent nucleotides, while the remaining nine<br /> subunits fulfill other functions. The complex of all 10 subunits<br /> together is called DNA polymerase III holoenzyme. By comparison,<br /> DNA polymerase I is composed of a single subunit. Its role<br /> during DNA replication is to remove the RNA primers and fill in<br /> the vacant regions with DNA.<br /> Though the various DNA polymerases in E. coli and other<br /> bacterial species vary in their subunit composition, several common<br /> structural features have emerged. The catalytic subunit of all<br /> DNA polymerases has a structure that resembles a human hand.<br /> As shown in Figure 11.8, the template DNA is threaded through<br /> the palm of the hand; the thumb and fingers are wrapped around<br /> the DNA. The incoming dNTPs enter the catalytic site, bind to<br /> the template strand according to the AT/GC rule, and then are<br /> covalently attached to the 3' end of the growing strand. DNA<br /> polymerase also contains a 3' exonuclease site that removes mismatched<br /> bases, as described later.<br /> As researchers began to unravel the function of DNA polymerase,<br /> two features seemed unusual (Figure 11.9). DNA polymerase<br /> cannot begin DNA synthesis by linking together the first<br /> two individual nucleotides. Rather, this type of enzyme can elongate<br /> only a preexisting strand starting with an RNA primer or<br /> existing DNA strand (Figure 11.9a). A second unusual feature is<br /> the directionality of strand synthesis. DNA polymerase can attach<br /> nucleotides only in the 5' to 3' direction, not in the 3' to 5' direction<br /> (Figure 11.9b).<br /> Due to these two unusual features, the synthesis of the leading<br /> and lagging strands shows distinctive differences (Figure 11.10).<br /> The synthesis of RNA primers by DNA primase allows DNA<br /> polymerase III to begin the synthesis of complementary daughter<br /> strands of DNA. DNA polymerase III catalyzes the attachment of<br /> nucleotides to the 3' end of each primer, in a 5' to 3' direction. In<br /> the leading strand, one RNA primer is made at the origin, and then<br /> DNA polymerase III can attach nucleotides in a 5' to 3' direction as<br /> it slides toward the opening of the replication fork. The synthesis of<br /> the leading strand is therefore continuous.<br /> TABLE 11.2<br /> Subunit Composition of DNA Polymerase III Holoenzyme<br /> from E. coli<br /> Subunit(s)<br /> a<br /> e<br /> u<br /> b<br /> Function<br /> Synthesizes DNA<br /> 3' to 5' proofreading (removes mismatched nucleotides)<br /> Accessory protein that stimulates the proofreading<br /> function<br /> Clamp protein, which allows DNA polymerase to slide<br /> along the DNA without falling off<br /> t, g, d, d', c, and x Clamp loader complex, involved with helping the clamp<br /> protein bind to the DNA<br /> 3′<br /> Thumb<br /> DNA polymerase<br /> catalytic site<br /> 5′ 5′<br /> 3′ Palm<br /> Fingers<br /> Template<br /> strand<br /> (a) Schematic side view of DNA polymerase III<br /> 3′ exonuclease<br /> site<br /> Incoming<br /> deoxyribonucleoside triphosphates<br /> (dNTPs)<br /> (b) Molecular model for DNA polymerase bound to DNA<br /> (Reprinted by permission from Macmillan Publishers Ltd. The Embo Journal. Crystal structures<br /> of open and closed forms of binary and ternary complexes of the large fragment of<br /> Thermus aquaticus DNA polymerase I: structural basis for nucleotide incorporation. Ying Li<br /> et al. 17:24, 7514–7525, 1998.)<br /> FIGURE 11.8 The action of DNA polymerase. (a) DNA<br /> polymerase slides along the template strand as it synthesizes a new<br /> strand by connecting deoxyribonucleoside triphosphates (dNTPs) in<br /> a 5' to 3' direction. The catalytic subunit of DNA polymerase resembles<br /> a hand that is wrapped around the template strand. In this regard, the<br /> movement of DNA polymerase along the template strand is similar<br /> to a hand that is sliding along a rope. (b) The molecular structure of<br /> DNA polymerase I from the bacterium Thermus aquaticus. This model<br /> shows a portion of DNA polymerase I that is bound to DNA. This<br /> molecular structure depicts a front view of DNA polymerase; part<br /> (a) is a schematic side view.<br /> In the lagging strand, the synthesis of DNA also elongates<br /> in a 5' to 3' manner, but it does so in the direction away from the<br /> replication fork. In the lagging strand, RNA primers must repeatedly<br /> initiate the synthesis of short segments of DNA; thus, the<br /> synthesis has to be discontinuous. The length of these fragments<br /> in bacteria is typically 1,000 to 2,000 nucleotides. In eukaryotes,<br /> the fragments are shorter—100 to 200 nucleotides. Each fragment<br /> contains a short RNA primer at the 5' end, which is made<br /> by primase. The remainder of the fragment is a strand of DNA</div> <div>3.2 CELL DIVISION 49<br /> Interphase<br /> S<br /> G 1<br /> G 2<br /> M<br /> G 0<br /> Cytokinesis<br /> Anaphase<br /> Telophase<br /> Mitosis<br /> Metaphase<br /> Prometaphase<br /> Prophase<br /> Two<br /> daughter<br /> cells<br /> FIGURE 3.5 The eukaryotic cell cycle. Dividing cells progress through a series of phases, denoted G 1 , S, G 2 , and M phases (mitosis). This<br /> diagram shows the progression of a cell through mitosis to produce two daughter cells. The original diploid cell had three pairs of chromosomes, for a<br /> total of six individual chromosomes. During S phase, these have replicated to yield 12 chromatids found in six pairs of sister chromatids. After mitosis<br /> is complete, each of the two daughter cells contains six individual chromosomes, just like the mother cell. Note: The chromosomes in G 0 , G 1 , S, and G 2<br /> phases are not condensed. In this drawing, they are shown partially condensed so they can be easily counted.<br /> tion and division process that is more complicated than simple<br /> binary fission. Eukaryotic cells that are destined to divide progress<br /> through a series of phases known as the cell cycle (Figure<br /> 3.5). These phases are G for gap, S for synthesis (of the genetic<br /> material), and M for mitosis. There are two G phases, G 1 and G 2 .<br /> The term “gap” originally described the gaps between S phase<br /> and mitosis in which it was not microscopically apparent that<br /> significant changes were occurring in the cell. However, we now<br /> know that both gap phases are critical periods in the cell cycle<br /> that involve many molecular changes. In actively dividing cells,<br /> the G 1 , S, and G 2 phases are collectively known as interphase. In<br /> addition, cells may remain permanently, or for long periods of<br /> time, in a phase of the cell cycle called G 0 . A cell in the G 0 phase<br /> is either temporarily not progressing through the cell cycle or, in<br /> the case of terminally differentiated cells, such as most nerve cells<br /> in an adult mammal, will never divide again.<br /> During the G 1 phase, a cell may prepare to divide. Depending<br /> on the cell type and the conditions that it encounters, a cell<br /> in the G 1 phase may accumulate molecular changes (e.g., synthesis<br /> of proteins) that cause it to progress through the rest of<br /> the cell cycle. When this occurs, cell biologists say that a cell<br /> has reached a restriction point and is committed on a pathway<br /> that leads to cell division. Once past the restriction point,<br /> the cell will then advance to the S phase, during which the chromosomes<br /> are replicated. After replication, the two copies are<br /> called chromatids. They are joined to each other at a region of<br /> DNA called the centromere to form a unit known as a pair of<br /> sister chromatids (Figure 3.6). The kinetochore is a group of<br /> proteins that are bound to the centromere. These proteins help<br /> to hold the sister chromatids together and also play a role in<br /> chromosome sorting, as discussed later. When S phase is completed,<br /> a cell actually has twice as many chromatids compared<br /> to the number of chromosomes in the G 1 phase. For example, a<br /> human cell in the G 1 phase has 46 distinct chromosomes, whereas<br /> in G 2 , it would have 46 pairs of sister chromatids, for a total of<br /> 92 chromatids. The term chromosome—meaning colored body—<br /> can be a bit confusing because it originally meant a distinct structure<br /> that is observable with the microscope. Therefore, the term</div> <div>320 CHAPTER 12 :: GENE TRANSCRIPTION AND RNA MODIFICATION<br /> Conceptual Questions<br /> C1. Genes may be structural genes that encode polypeptides, or they<br /> may be nonstructural genes.<br /> A. Describe three examples of genes that are not structural genes.<br /> B. For structural genes, one DNA strand is called the template<br /> strand, and the complementary strand is called the coding<br /> strand. Are these two terms appropriate for nonstructural genes?<br /> Explain.<br /> C. Do nonstructural genes have a promoter and terminator?<br /> C2. In bacteria, what event marks the end of the initiation stage of<br /> transcription?<br /> C3. What is the meaning of the term consensus sequence? Give an<br /> example. Describe the locations of consensus sequences within<br /> bacterial promoters. What are their functions?<br /> C4. What is the consensus sequence of the following six DNA<br /> molecules?<br /> GGCATTGACT<br /> GCCATTGTCA<br /> CGCATAGTCA<br /> GGAAATGGGA<br /> GGCTTTGTCA<br /> GGCATAGTCA<br /> C5. Mutations in bacterial promoters may increase or decrease the<br /> level of gene transcription. Promoter mutations that increase<br /> transcription are termed up promoter mutations, and those that<br /> decrease transcription are termed down promoter mutations.<br /> As shown in Figure 12.5, the sequence of the –10 region of<br /> the promoter for the lac operon is TATGTT. Would you expect<br /> the following mutations to be up promoter or down promoter<br /> mutations?<br /> A. TATGTT to TATATT<br /> B. TATGTT to TTTGTT<br /> C. TATGTT to TATGAT<br /> C6. According to the examples shown in Figure 12.5, which positions<br /> of the –35 sequence (i.e., first, second, third, fourth, fifth, and/or<br /> sixth) are more tolerant of changes? Do you think that these<br /> positions play a more or less important role in the binding of s<br /> factor? Explain why.<br /> C7. In Chapter 9, we considered the dimensions of the double helix (see<br /> Figure 9.17). In an a helix of a protein, there are 3.6 amino acids<br /> per complete turn. Each amino acid advances the a helix by<br /> 0.15 nm; a complete turn of an a helix is 0.54 nm in length. As<br /> shown in Figure 12.6, two a helices of a transcription factor occupy<br /> the major groove of the DNA. According to Figure 12.6, estimate<br /> the number of amino acids that bind to this region. How many<br /> complete turns of the a helices occupy the major groove of DNA?<br /> C8. A mutation within a gene sequence changes the start codon to a<br /> stop codon. How will this mutation affect the transcription of this<br /> gene?<br /> C9. What is the subunit composition of bacterial RNA polymerase<br /> holoenzyme? What are the functional roles of the different<br /> subunits?<br /> C10. At the molecular level, describe how s factor recognizes bacterial<br /> promoters. Be specific about the structure of s factor and the type<br /> of chemical bonding.<br /> C11. Let’s suppose a DNA mutation changes the consensus sequence at<br /> the –35 location so that s factor is no longer able to bind there.<br /> Explain how a mutation would prevent s factor from binding<br /> to the DNA. Look at Figure 12.5 and describe two specific base<br /> substitutions you think would inhibit the binding of s factor.<br /> Explain why you think your base substitutions would have this<br /> effect.<br /> C12. What is the complementarity rule that governs the synthesis of an<br /> RNA molecule during transcription? An RNA transcript has the<br /> following sequence:<br /> 5'–GGCAUGCAUUACGGCAUCACACUAGGGAUC–3'<br /> What is the sequence of the template and coding strands of the<br /> DNA that encodes this RNA? On which side (5' or 3') of the<br /> template strand is the promoter located?<br /> C13. Describe the movement of the open complex along the DNA.<br /> C14. Describe what happens to the chemical bonding interactions when<br /> transcriptional termination occurs. Be specific about the type of<br /> chemical bonding.<br /> C15. Discuss the differences between r-dependent and r-independent<br /> termination.<br /> C16. In Chapter 11, we discussed the function of DNA helicase, which is<br /> involved in DNA replication. The structure and function of DNA<br /> helicase and r protein are rather similar to each other. Explain how<br /> the function of these two proteins is similar and how it is different.<br /> C17. Discuss the similarities and differences between RNA polymerase<br /> (described in this chapter) and DNA polymerase (described in<br /> Chapter 11).<br /> C18. Mutations that occur at the end of a gene may alter the sequence<br /> of the gene and prevent transcriptional termination.<br /> A. What types of mutations would prevent r-independent<br /> termination?<br /> B. What types of mutations would prevent r-dependent termination?<br /> C. If a mutation prevented transcriptional termination at the end of<br /> a gene, where would gene transcription end? Or would it end?<br /> C19. If the following RNA polymerases were missing from a eukaryotic<br /> cell, what types of genes would not be transcribed?<br /> A. RNA polymerase I<br /> B. RNA polymerase II<br /> C. RNA polymerase III<br /> C20. What sequence elements are found within the core promoter<br /> of structural genes in eukaryotes? Describe their locations and<br /> specific functions.<br /> C21. For each of the following transcription factors, how would<br /> eukaryotic transcriptional initiation be affected if it were missing?<br /> A. TFIIB<br /> B. TFIID<br /> C. TFIIH</div> <div>832 INDEX<br /> Fingerprints, heritability of dermal ridge count in,<br /> 713–15<br /> Fire, Andrew, 412<br /> First-division segregation, 121<br /> Fisher, Ronald, 665, 675<br /> Flagellum, 45<br /> Flavr Savr tomato, 532–33, 534<br /> Flemming, Walter, 50, 60<br /> Fletcher factor deficiency, 2<br /> Flowers. See also Plants<br /> homeotic genes in development of,<br /> 653–54<br /> incomplete dominance in, 74<br /> Fluctuation test, 433, 434<br /> Fluorescence in situ hybridization (FISH), 544–45,<br /> 545, 546<br /> Focus, in assay, 612<br /> Forked-line method, 28, 38<br /> Founder, of disease-causing allele, 551<br /> Founder effect, 673, 731<br /> Four o’clock flower<br /> color in, 74<br /> variegated phenotype in, 176–77, 177<br /> Fragile X syndrome, 429<br /> Frameshift mutation, 425, 426<br /> Franklin, Rosalind, 230, 231, 234<br /> Freiberger, Mary, 744<br /> Frequency distribution of traits, 698–700<br /> Frog, euploidy in, 204<br /> Fruit fly. See Drosophila melanogaster<br /> FtsZ protein, 48<br /> Functional genomics, 571–80<br /> identification of expressed genes in cDNA<br /> library, 572–73<br /> microarray in identification of transcribed genes,<br /> 573–74<br /> Functional protein microarray, 584<br /> Fungi<br /> genetic mapping of, 118<br /> sequenced genomes, 564<br /> f2 virus, 228<br /> G<br /> Gain-of-function mutations, 603, 642<br /> Galactose, 81<br /> β-galactosidase, 336, 364<br /> Galápagos finches, 681–82, 728<br /> Galas, David, 502<br /> Gallo, Robert, 616, 617<br /> Galton, Francis, 698, 702<br /> Gametes, 10, 20, 47<br /> gene assortment, 25<br /> plant, 59<br /> Gametogenesis, 54, 58<br /> genomic imprinting during, 171<br /> Gametophyte, 59<br /> Gap genes, 638, 641<br /> Garrod, Archibald, 325<br /> Garter snakes, 8<br /> Gastrulation, 635, 636<br /> G banding, 188, 189<br /> Gearhart, John, 529<br /> Gehring, Walter, 752<br /> Gel electrophoresis, 492, 506<br /> in DNA fingerprinting, 689<br /> pulse-field, in genome comparisons,<br /> 559–60<br /> two-dimensional, in separation of cellular<br /> proteins, 581, 582<br /> Gel retardation assay, 502, 511<br /> Gene addition, 521, 522<br /> Gene amplification, 397, 397, 617<br /> Gene assortment, 25<br /> Gene chip, 573<br /> Gene cloning, 483–97. See also Cloning<br /> cDNA production from mRNA via reverse<br /> transcriptase, 493<br /> chromosomal DNA in, 483–84<br /> cutting DNA into pieces and joining pieces<br /> together, 484–86<br /> defined, 483<br /> detection of cloned genes, 497–503<br /> first cloning experiments, 488–92<br /> gene addition in, 521<br /> gene knockout in, 521<br /> gene replacement in, 521<br /> insertion of DNA fragments into vectors, 486–88<br /> restriction mapping, 493–95, 494<br /> steps in, 487<br /> uses of, 483<br /> using PCR to make many copies of DNA,<br /> 495–97, 496<br /> vector DNA in, 483–84<br /> Gene conversion, 459<br /> by DNA mismatch repair, 463<br /> by gap repair synthesis, 463<br /> homologous recombination in, 462<br /> Gene detection, 497–503<br /> by colony hybridization, 497, 498<br /> DNA library, 497, 498<br /> Northern blotting to detect RNA, 501<br /> Southern blotting to detect DNA sequences,<br /> 499–501, 500<br /> using protein-DNA interactions, 502–3<br /> Western blotting to detect proteins, 501–2<br /> Gene duplication, 191, 737<br /> Gene expression, 6<br /> aneuploidy in imbalance in, 201–2<br /> base sequences in, 298–99<br /> cell-specific, DNA microarray in identification<br /> of, 573<br /> changes in chromosome structure affecting, 430<br /> chromatin structural changes effect on, 397–406<br /> cis-effect on, 369<br /> of cloned genes, 483<br /> cluster analysis of, 579<br /> colinearity of, 309<br /> in developmental genes, 749–50<br /> electron microscopy observation of, 351<br /> euploid vs. aneuploid individuals, 202<br /> identification in cDNA library, 572–73<br /> methylation at DMR in inhibiting, 172<br /> in milk of domestic animals, 526<br /> monoallelic, 170<br /> polarity, 425<br /> position effect, 430<br /> process of, 6<br /> regulation of, 390<br /> sex of parent from which gene was inherited in,<br /> 170–72<br /> traits and, 6–8, 7<br /> transposition effect on, 475<br /> Gene family, 192, 735<br /> formation of, via chromosomal duplication,<br /> 191–93<br /> homologous genes in, 589–90<br /> Southern blotting in identification of, 499<br /> Gene flow, 674<br /> Gene interactions, 86–93<br /> Gene knockin, 523–25, 525<br /> Gene knockout, 91, 521, 523–25, 648<br /> Gene modifier effect, 89<br /> Gene mutations, 8, 424–54. See also Mutations<br /> Genentech Inc., 515, 518<br /> Gene pool, 665, 666<br /> Gene probes, 483<br /> Generalized transduction, 144<br /> General regulation of translation, 406<br /> General transcription factors, 306–8, 307<br /> Generation-skipping traits, 75<br /> Gene rearrangement, 397, 397<br /> Gene redundancy, 86, 91, 91–92, 523–25<br /> Gene regulation<br /> in bacteria, 360–77<br /> in bacteriophage reproductive cycle, 377–83<br /> benefits of, 360, 390<br /> chromatin structural changes, 397–406<br /> in eukaryotes, 390–417<br /> genetic switches in, 383<br /> iron assimilation and, 415–17<br /> posttranslational regulation, 376–77<br /> regulatory transcription factors, 390–97<br /> RNA interference in, 414<br /> transcriptional, 361–76<br /> translational, 376–77<br /> via RNA processing, 406<br /> Gene replacement, 521, 522<br /> production of mice containing, 522–23, 524<br /> Gene(s)<br /> computer-based approaches to identification<br /> of, 588<br /> constitutive, 360<br /> coordinate regulation of, 575–79<br /> defined, 4, 23<br /> in encoding of enzymes, 325<br /> essential, 84–85<br /> evolution at molecular level, 737–38<br /> holandric, 82<br /> homeotic, 635<br /> homologous, 192, 588, 647<br /> housekeeping, 404<br /> inducible, 361, 362<br /> inheritance of, 9–10<br /> loci of, 47, 111<br /> maternal effect, 161–64<br /> modifying, 86, 89<br /> mutant, in disease, 2<br /> nonessential, 85<br /> nuclear, 161<br /> oncogene, 611<br /> one gene-one enzyme hypothesis, 325–26<br /> orthologs, 588, 735<br /> paralogous, 588, 735<br /> repressible, 361, 362<br /> selectable marker, 484<br /> sex-linked, 82<br /> single<br /> inheritance patterns of, 71–86, 72<br /> pleiotropic effects of, 86<br /> structural, 248, 297, 324<br /> tissue-specific, 404<br /> traits and, 4–10, 7<br /> transgene, 514<br /> X-linked, 65<br /> Gene sequences, 483<br /> of bacterial chromosomes, 248–49<br /> repetitive, 249<br /> Gene therapy, 535–38<br /> for adenosine deaminase deficiency, 535–38<br /> for cystic fibrosis, 538<br /> ex vivo approach, 537<br /> future prospects in, 535<br /> gene transfer methods in, 536<br /> introduction of cloned genes into human<br /> cells, 535<br /> Genetically modified organisms (GMOs), 514<br /> Genetic approach, 13<br /> Genetic code, 5, 327<br /> deciphering<br /> RNA copolymers in, 332–33<br /> synthetic RNA in, 329–32, 330–31<br /> triplet binding assay in, 332–33</div> <div>::<br /> 5<br /> LINKAGE<br /> AND<br /> GENETIC MAPPING<br /> IN EUKARYOTES<br /> CHAPTER OUTLINE<br /> 5.1 Linkage and Crossing Over<br /> 5.2 Genetic Mapping in Plants and Animals<br /> 5.3 Genetic Mapping in Haploid Eukaryotes<br /> In Chapter 2, we were introduced to Mendel’s laws of inheritance. According<br /> to these principles, we expect that two different genes will segregate and<br /> independently assort themselves during the process that creates gametes. After<br /> Mendel’s work was rediscovered at the turn of the twentieth century, chromosomes<br /> were identified as the cellular structures that carry genes. The chromosome<br /> theory of inheritance explained how the transmission of chromosomes is<br /> responsible for the passage of genes from parents to offspring.<br /> When geneticists first realized that chromosomes contain the genetic material,<br /> they began to suspect that a conflict might sometimes occur between the<br /> law of independent assortment of genes and the behavior of chromosomes during<br /> meiosis. In particular, geneticists assumed that each species of organism must<br /> contain thousands of different genes, yet cytological studies revealed that most<br /> species have at most a few dozen chromosomes. Therefore, it seemed likely, and<br /> turned out to be true, that each chromosome would carry many hundreds or<br /> even thousands of different genes. The transmission of genes located close to<br /> each other on the same chromosome violates the law of independent assortment.<br /> In this chapter, we will consider the pattern of inheritance that occurs<br /> when different genes are situated on the same chromosome. In addition, we<br /> will briefly explore how the data from genetic crosses are used to construct a<br /> genetic map—a diagram that describes the order of genes along a chromosome.<br /> Newer strategies for gene mapping are described in Chapter 20. However, an<br /> understanding of traditional mapping studies, as described in this chapter, will<br /> strengthen our appreciation for these newer molecular approaches. More importantly,<br /> traditional mapping studies further illustrate how the location of two or<br /> more genes on the same chromosome can affect the transmission patterns from<br /> parents to offspring.<br /> Crossing over during meiosis. This event provides<br /> a way to reassort the alleles of genes that are<br /> located on the same chromosome.<br /> 5.1 LINKAGE AND CROSSING OVER<br /> In eukaryotic species, each linear chromosome contains a very long segment<br /> of DNA. A chromosome contains many individual functional units—called<br /> genes—that influence an organism’s traits. A typical chromosome is expected<br /> to contain many hundreds or perhaps a few thousand different genes. The term<br /> linkage has two related meanings. First, linkage refers to the phenomenon that<br /> two or more genes may be located on the same chromosome. The genes are<br /> physically linked to each other, because each eukaryotic chromosome contains<br /> a single, continuous, linear molecule of DNA. Second, genes that are close</div> <div>::<br /> 15<br /> GENE REGULATION<br /> IN EUKARYOTES<br /> Gene regulation provides many benefits to eukaryotic organisms, a<br /> category that includes protists, fungi, plants, and animals. Like their prokaryotic<br /> counterparts, eukaryotic cells need to adapt to changes in their environment.<br /> For example, eukaryotic cells can respond to changes in nutrient availability by<br /> enzyme adaptation, much as prokaryotic cells do. Eukaryotic cells also respond<br /> to environmental stresses such as ultraviolet (UV) radiation by inducing genes<br /> that provide protection against harmful environmental agents. An example is<br /> the ability of humans to develop a tan. The tanning response helps to protect a<br /> person’s cells against the damaging effects of UV rays.<br /> Among plants and animals, multicellularity and a more complex cell<br /> structure also demand a much greater level of gene regulation. The life cycle of<br /> complex eukaryotic organisms involves the progression through several developmental<br /> stages to achieve a mature organism. Some genes are expressed only<br /> during early stages of development, such as the embryonic stage, whereas others<br /> are expressed in the adult. In addition, complex eukaryotic species are composed<br /> of many different tissues that contain a variety of cell types. Gene regulation is<br /> necessary to ensure the differences in structure and function among distinct cell<br /> types. It is amazing that the various cells within a multicellular organism usually<br /> contain the same genetic material, yet phenotypically may look quite different.<br /> For example, the appearance of a human nerve cell seems about as similar to a<br /> muscle cell as an amoeba is to a paramecium. In spite of these phenotypic differences,<br /> a human nerve cell and muscle cell actually contain the same complement<br /> of human chromosomes. Nerve and muscle cells are strikingly different<br /> because of gene regulation rather than differences in DNA content. Many genes<br /> are expressed in the nerve cell and not the muscle cell, and vice versa.<br /> The molecular mechanisms that underlie gene regulation in eukaryotes<br /> bear many similarities to the ways that bacteria regulate their genes. As in prokaryotes,<br /> regulation in eukaryotes can occur at any step in the pathway of gene<br /> expression (Figure 15.1). We will begin this chapter with an exploration of gene<br /> regulation at the level of transcription, an important form of control. In addition,<br /> research in the past few decades has revealed that eukaryotic organisms<br /> frequently regulate gene expression at points other than transcription. We will<br /> discuss some well-studied examples in which genes are regulated at many of<br /> these other control points.<br /> CHAPTER OUTLINE<br /> 15.1 Regulatory Transcription Factors<br /> 15.2 Changes in Chromatin Structure<br /> 15.3 Regulation of RNA Processing, RNA<br /> Stability, and Translation<br /> Certain proteins, known as regulatory<br /> transcription factors, have the ability to bind<br /> into the major groove of DNA and regulate gene<br /> transcription.</div> <div>10.3 EUKARYOTIC CHROMOSOMES 259<br /> 30 nm<br /> 30 nm<br /> Core<br /> histone<br /> proteins<br /> (a) Micrograph of<br /> a 30 nm fiber<br /> (b) Solenoid model<br /> (c) Zigzag model<br /> FIGURE 10.16 The 30 nm fiber. (a) A photomicrograph of the 30 nm fiber. (b) In the solenoid model, the nucleosomes are packed in a spiral<br /> configuration. (c) In the zigzag model, the linker DNA forms a more irregular structure, and less contact occurs between adjacent nucleosomes. The<br /> zigzag model is consistent with more recent data regarding chromatin conformation.<br /> connected by deformable linker regions. In 2005, Timothy Richmond<br /> and colleagues were the first to solve the crystal structure<br /> of a segment of DNA containing multiple nucleosomes, in this<br /> case four. The structure with four nucleosomes revealed that the<br /> linker DNA zigzags back and forth between each nucleosome, a<br /> feature consistent with the zigzag model.<br /> Chromosomes Are Further Compacted<br /> by Anchoring the 30 nm Fiber into Radial<br /> Loop Domains Along the Nuclear Matrix<br /> Thus far, we have examined two mechanisms that compact<br /> eukaryotic DNA. These involve the wrapping of DNA within<br /> nucleosomes and the arrangement of nucleosomes to form a<br /> 30 nm fiber. Taken together, these two events shorten the DNA<br /> nearly 50-fold. A third level of compaction involves interactions<br /> between the 30 nm fibers and a filamentous network of proteins<br /> in the nucleus called the nuclear matrix. As shown in Figure<br /> 10.17a, the nuclear matrix consists of two parts. The nuclear lamina<br /> is a collection of fibers that line the inner nuclear membrane.<br /> These fibers are composed of intermediate filament proteins. The<br /> second part is an internal nuclear matrix, which is connected to<br /> the nuclear lamina and fills the interior of the nucleus. The internal<br /> nuclear matrix, whose structure and functional role remain<br /> controversial, is hypothesized to be an intricate fine network<br /> of irregular protein fibers plus many other proteins that bind<br /> to these fibers. Even when the chromatin is extracted from the<br /> nucleus, the internal nuclear matrix may remain intact (Figure<br /> 10.17b and c). However, the matrix should not be considered<br /> a static structure. Research indicates that the protein composition<br /> of the internal nuclear matrix is very dynamic and complex,<br /> consisting of dozens or perhaps hundreds of different proteins.<br /> The protein composition varies depending on species, cell type,<br /> and environmental conditions. This complexity has made it difficult<br /> to propose models regarding its overall organization. Further<br /> research will be necessary to understand the structure and<br /> dynamic nature of the internal nuclear matrix.<br /> The proteins of the nuclear matrix are involved in compacting<br /> the DNA into radial loop domains, similar to those described<br /> for the bacterial chromosome. During interphase, chromatin is<br /> organized into loops, often 25,000 to 200,000 bp in size, which<br /> are anchored to the nuclear matrix. The chromosomal DNA of<br /> eukaryotic species contains sequences called matrix-attachment<br /> regions (MARs) or scaffold-attachment regions (SARs), which<br /> are interspersed at regular intervals throughout the genome. The<br /> MARs bind to specific proteins in the nuclear matrix, thus forming<br /> chromosomal loops (Figure 10.17d).<br /> Why is the attachment of radial loops to the nuclear matrix<br /> important? In addition to compaction, the nuclear matrix serves<br /> to organize the chromosomes within the nucleus. Each chromosome<br /> in the cell nucleus is located in a discrete chromosome<br /> territory. As shown in studies by Thomas Cremer, Christoph Cremer,<br /> and others, these territories can be viewed when interphase<br /> cells are exposed to multiple fluorescent molecules that recognize</div> <div>13.1 THE GENETIC BASIS FOR PROTEIN SYNTHESIS 325<br /> In this section, we will begin by considering early experiments<br /> that showed the role of genes is to encode proteins. We<br /> will examine the general features of the genetic code—the<br /> sequence of bases in a codon that specifies an amino acid—and<br /> explore the experiments through which the code was deciphered<br /> or “cracked.” Finally, we will look at the biochemistry of polypeptide<br /> synthesis to see how this determines the structure and function<br /> of proteins, which are ultimately responsible for the characteristics<br /> of living cells and an organism’s traits.<br /> Archibald Garrod Proposed That Some Genes Code<br /> for the Production of a Single Enzyme<br /> The idea that a relationship exists between genes and the production<br /> of proteins was first suggested at the beginning of the<br /> twentieth century by Archibald Garrod, a British physician. Prior<br /> to Garrod’s studies, biochemists had studied many metabolic<br /> pathways within living cells. These pathways consist of a series of<br /> metabolic conversions of one molecule to another, each step catalyzed<br /> by a specific enzyme. Each enzyme is a distinctly different<br /> protein that catalyzes a particular chemical reaction. Figure 13.1<br /> illustrates part of the metabolic pathway for the degradation of<br /> phenylalanine, an amino acid commonly found in human diets.<br /> The enzyme phenylalanine hydroxylase catalyzes the conversion<br /> of phenylalanine to tyrosine, and a different enzyme, tyrosine<br /> aminotransferase, converts tyrosine into p-hydroxyphenylpyruvic<br /> acid, and so on. In all of the steps shown in Figure 13.1, a specific<br /> enzyme catalyzes a single type of chemical reaction.<br /> Garrod studied patients who had defects in their ability to<br /> metabolize certain compounds. He was particularly interested in<br /> the inherited disease known as alkaptonuria. In this disorder,<br /> the patient’s body accumulates abnormal levels of homogentisic<br /> acid (also called alkapton), which is excreted in the urine, causing<br /> it to appear black on exposure to air. In addition, the disease<br /> is characterized by bluish black discoloration of cartilage and<br /> skin (ochronosis). Garrod proposed that the accumulation of<br /> homogentisic acid in these patients is due to a missing enzyme,<br /> namely, homogentisic acid oxidase (see Figure 13.1).<br /> How did Garrod realize that certain genes encode enzymes?<br /> He already knew that alkaptonuria is an inherited trait that follows<br /> a recessive pattern of inheritance. Therefore, an individual<br /> with alkaptonuria must have inherited the mutant gene that<br /> causes this disorder from both parents. From these observations,<br /> Garrod proposed that a relationship exists between the<br /> inheritance of the trait and the inheritance of a defective enzyme.<br /> Namely, if an individual inherited the mutant gene (which causes<br /> a loss of enzyme function), he or she would not produce any<br /> normal enzyme and would be unable to metabolize homogentisic<br /> acid. Garrod described alkaptonuria as an inborn error of<br /> metabolism. This hypothesis was the first suggestion that a connection<br /> exists between the function of genes and the production<br /> of enzymes. At the turn of the century, this idea was particularly<br /> insightful, because the structure and function of the genetic<br /> material were completely unknown.<br /> HO<br /> Phenylketonuria<br /> Dietary<br /> protein<br /> CH 2 C COOH<br /> NH 2<br /> Phenylalanine<br /> CH 2<br /> Phenylalanine<br /> hydroxylase<br /> NH 2<br /> Tyrosine<br /> Tyrosine<br /> aminotransferase<br /> p-hydroxyphenylpyruvic<br /> acid<br /> Tyrosinosis<br /> Alkaptonuria<br /> H<br /> Homogentisic<br /> acid<br /> COOH<br /> Hydroxyphenylpyruvate<br /> oxidase<br /> Homogentisic<br /> acid oxidase<br /> Maleylacetoacetic<br /> acid<br /> FIGURE 13.1 The metabolic pathway of phenylalanine<br /> breakdown. This diagram shows part of the pathway of phenylalanine<br /> metabolism, which consists of enzymes that successively convert<br /> one molecule to another. Certain human genetic diseases (shown<br /> in red boxes) are caused when enzymes in this pathway are missing<br /> or defective.<br /> GenesgTraits When a person inherits two defective copies of the gene that<br /> encodes homogentisic acid oxidase, he or she cannot convert homogentisic acid into<br /> maleylacetoacetic acid. Such a person accumulates large amounts of homogentisic<br /> acid in the urine and has other symptoms of the disease known as alkaptonuria.<br /> Similarly, if a person has two mutant alleles of the gene encoding phenylalanine<br /> hydroxylase, he or she is unable to synthesize the enzyme phenylalanine hydroxylase<br /> and has the disease called phenylketonuria (PKU).<br /> Beadle and Tatum’s Experiments<br /> with Neurospora Led Them to Propose<br /> the One Gene–One Enzyme Hypothesis<br /> In the early 1940s, George Beadle and Edward Tatum were also<br /> interested in the relationship among genes, enzymes, and traits.<br /> They developed an experimental system for investigating the<br /> relationship between genes and the production of particular<br /> enzymes. Consistent with the ideas of Garrod, the underlying<br /> H<br /> C</div> <div>CONCEPTUAL QUESTIONS 693<br /> Conceptual Questions<br /> C1. What is the gene pool? How is a gene pool described in a<br /> quantitative way?<br /> C2. In genetics, what does the term population mean? Pick any species<br /> you like and describe how its population might change over the<br /> course of many generations.<br /> C3. What is a genetic polymorphism? What is the source of genetic<br /> variation?<br /> C4. For each of the following, state whether it is an example of an<br /> allele, genotype, and/or phenotype frequency:<br /> A. Approximately 1 in 2,500 Caucasians is born with cystic fibrosis.<br /> B. The percentage of carriers of the sickle-cell allele in West Africa<br /> is approximately 13%.<br /> C. The number of new mutations for achondroplasia, a genetic<br /> disorder, is approximately 5 � 10 �5 .<br /> C5. The term polymorphism can refer to both genes and traits. Explain<br /> the meaning of a polymorphic gene and a polymorphic trait. If a<br /> gene is polymorphic, does the trait that the gene affects also have<br /> to be polymorphic? Explain why or why not.<br /> C6. Cystic fibrosis is a recessive autosomal trait. In certain Caucasian<br /> populations, the number of people born with this disorder is<br /> about 1 in 2,500. Assuming Hardy-Weinberg equilibrium for this<br /> trait,<br /> A. What are the frequencies for the normal and CF alleles?<br /> B. What are the genotype frequencies of homozygous normal,<br /> heterozygous, and homozygous affected individuals?<br /> C. Assuming random mating, what is the probability that two<br /> phenotypically unaffected heterozygous carriers will choose<br /> each other as mates?<br /> C7. Does inbreeding affect allele frequencies? Why or why not? How<br /> does it affect genotype frequencies? With regard to rare recessive<br /> diseases, what are the consequences of inbreeding in human<br /> populations?<br /> C8. For a gene existing in two alleles, what are the allele frequencies<br /> when the heterozygote frequency is at its maximum value? What if<br /> there are three alleles?<br /> C9. In a population, the frequencies of two alleles are B � 0.67 and<br /> b � 0.33. The genotype frequencies are BB � 0.50, Bb � 0.37, and<br /> bb � 0.13. Do these numbers suggest inbreeding? Explain why or<br /> why not.<br /> C10. The ability to roll your tongue is inherited as a recessive trait.<br /> The frequency of the rolling allele is approximately 0.6, and the<br /> dominant (nonrolling) allele is 0.4. What is the frequency of<br /> individuals who can roll their tongues?<br /> C11. In the pedigree shown here, answer the following questions for<br /> individual VI-1:<br /> II-1<br /> III-1<br /> IV-1<br /> I-1 I-2<br /> II-2 II-3 III-4 II-5<br /> III-2<br /> IV-2<br /> III-3<br /> III-4<br /> II-6<br /> III-6<br /> V-1 V-2 V-3 V-4<br /> VI -1<br /> III-5<br /> IV-3<br /> A. Is this individual inbred?<br /> B. If so, who are her common ancestor(s)?<br /> C. Calculate the inbreeding coefficient for VI-1.<br /> D. Are the parents of VI-1 inbred?<br /> C12. A family pedigree is shown here.<br /> III-1<br /> II-1<br /> I-1<br /> III-2<br /> I-2<br /> I-3<br /> II-2 II-3 II-4 II-5 II-6<br /> III-3<br /> IV-1<br /> III-4<br /> IV-2<br /> IV-3<br /> IV-4<br /> IV-4<br /> V-5<br /> III-5 III-6 III-7 III-8<br /> IV-5<br /> A. What is the inbreeding coefficient for individual IV-3?<br /> B. Based on the data shown in this pedigree, is individual IV-4<br /> inbred?</div> <div>262 CHAPTER 10 :: CHROMOSOME ORGANIZATION AND MOLECULAR STRUCTURE<br /> p<br /> Ser<br /> Amino terminal tail<br /> ac<br /> Lys<br /> 5<br /> ac<br /> Lys<br /> 10<br /> Globular domain<br /> 15<br /> 20<br /> ac<br /> Lys<br /> 5<br /> ac<br /> p<br /> ac Lys<br /> Ser 15<br /> Lys<br /> 10<br /> H2A<br /> ac<br /> 20 Lys<br /> ac<br /> p ac Lys 10<br /> m<br /> Lys<br /> ac<br /> Ser<br /> m<br /> Arg Lys 15 ac m<br /> Arg<br /> 5 Lys<br /> m<br /> m<br /> Lys<br /> Lys ac ac<br /> Arg<br /> 20<br /> 5 m Lys Lys<br /> acp 15<br /> 20<br /> LysSer<br /> H3<br /> 10<br /> H2B<br /> FIGURE 10.20 Examples of covalent modifications that may<br /> occur in the amino terminal tails of core histone proteins. The amino<br /> acids are numbered from the amino terminus. The modifications<br /> shown here are m for methylation, p for phosphorylation, and ac for<br /> acetylation. Many more modifications can occur to the amino terminal<br /> tails; the ones shown here represent common examples.<br /> amino acid, a lysine, in H2B and methylation of the third amino<br /> acid in H4, which is an arginine.<br /> The pattern of covalent modifications to the amino terminal<br /> tails provides binding sites for proteins that subsequently<br /> affect the degree of chromatin compaction. One pattern of histone<br /> modification may attract proteins that cause the chromatin<br /> to become even more compact, which would silence the<br /> transcription of genes in the region. A different combination of<br /> histone modifications may attract proteins, such as the chromatin<br /> remodeling enzymes discussed in Chapter 15, which serve<br /> to loosen the chromatin and thereby promote gene transcription.<br /> In this way, the histone code plays a key role in making the<br /> information within the genomes of eukaryotic species accessible.<br /> Researchers are trying to unravel which patterns of histone modifications<br /> promote compaction and which promote a loosening<br /> of chromatin structure. In other words, they are trying to decipher<br /> the effects of the covalent modifications that comprise the<br /> histone code.<br /> Condensin and Cohesin Promote the Formation<br /> of Metaphase Chromosomes<br /> As we have seen, several mechanisms can alter the level of chromosomal<br /> compaction. Furthermore, the degree of compaction<br /> H4<br /> can vary along a single eukaryotic chromosome. When cells prepare<br /> to divide, the chromosomes become even more compacted<br /> or condensed. This aids in their proper sorting during metaphase.<br /> Figure 10.21 illustrates the levels of compaction that lead to a<br /> metaphase chromosome. During interphase, most of the chromosomal<br /> DNA is found in euchromatin, in which the 30 nm fibers<br /> form radial loop domains that are attached to a protein scaffold.<br /> The average distance that loops radiate from the protein scaffold<br /> is approximately 300 nm. This structure can be further compacted<br /> via additional folding of the radial loop domains and protein<br /> scaffold. This additional level of compaction greatly shortens<br /> the overall length of a chromosome and produces a diameter of<br /> approximately 700 nm, which is the compaction level found in<br /> heterochromatin. During interphase, most chromosomal regions<br /> are euchromatic, and some localized regions, such as those near<br /> centromeres, are heterochromatic.<br /> As cells enter M phase, the level of compaction changes dramatically.<br /> By the end of prophase, sister chromatids are entirely<br /> heterochromatic. Two parallel chromatids have a larger diameter<br /> of approximately 1,400 nm but a much shorter length compared<br /> to interphase chromosomes. These highly condensed metaphase<br /> chromosomes undergo little gene transcription because it is difficult<br /> for transcription proteins to gain access to the compacted<br /> DNA. Therefore, most transcriptional activity ceases during M<br /> phase, although a few specific genes may be transcribed. M phase<br /> is usually a short period of the cell cycle.<br /> In highly condensed chromosomes, such as those found<br /> in metaphase, the radial loops are highly compacted and remain<br /> anchored to a scaffold, which is formed from nonhistone proteins<br /> of the nuclear matrix. Experimentally, researchers can<br /> delineate the nonhistone proteins of the scaffold that hold the<br /> loops in place. Figure 10.22a shows a human metaphase chromosome.<br /> In this condition, the radial loops of DNA are in a<br /> very compact configuration. If this chromosome is treated with<br /> a high concentration of salt to remove both the core and linker<br /> histones, the highly compact configuration is lost, but the bottoms<br /> of the elongated loops remain attached to the scaffold composed<br /> of nonhistone proteins. In Figure 10.22b, an arrow points<br /> to an elongated DNA strand emanating from the darkly staining<br /> scaffold. Remarkably, the scaffold retains the shape of the original<br /> metaphase chromosome even though the DNA strands have<br /> become greatly elongated. These results illustrate that the structure<br /> of metaphase chromosomes is determined by the nuclear<br /> matrix proteins, which form the scaffold, and by the histones,<br /> which are needed to compact the radial loops.<br /> Researchers are trying to understand the steps that lead<br /> to the formation and organization of metaphase chromosomes.<br /> During the past several years, studies in yeast and frog oocytes<br /> have been aimed at the identification of proteins that promote the<br /> conversion of interphase chromosomes into metaphase chromosomes.<br /> In yeast, mutants have been characterized that have alterations<br /> in the condensation or the segregation of chromosomes.<br /> Similarly, biochemical studies using frog oocytes resulted in the<br /> purification of protein complexes that promote chromosomal<br /> condensation or sister chromatid alignment. These two lines of<br /> independent research produced the same results. Researchers</div> <div>292 CHAPTER 11 :: DNA REPLICATION<br /> by which telomerase works. The telomerase enzyme contains<br /> both protein subunits and RNA. The RNA part of telomerase<br /> contains a sequence complementary to the DNA sequence found<br /> in the telomeric repeat. This allows telomerase to bind to the 3'<br /> overhang region of the telomere. Following binding, the RNA<br /> sequence beyond the binding site functions as a template allowing<br /> the synthesis of a six-nucleotide sequence at the end of the<br /> DNA strand. This is called polymerization, because it is analogous<br /> to the function of DNA polymerase. It is catalyzed by two<br /> identical protein subunits called telomerase reverse transcriptase<br /> (TERT). TERT’s name indicates that is uses an RNA template to<br /> synthesize DNA. Following polymerization, the telomerase can<br /> then move—a process called translocation—to the new end of<br /> this DNA strand and attach another six nucleotides to the end.<br /> This binding-polymerization-translocation cycle occurs many<br /> times in a row and thereby greatly lengthens the 3' end of the<br /> DNA strand in the telomeric region. The complementary strand<br /> is then synthesized by DNA polymerase.<br /> CONCEPTUAL SUMMARY<br /> The structural basis for DNA replication is the double-stranded<br /> helix of DNA, in which the AT/GC rule is obeyed. This complementarity<br /> between strands allows DNA to be replicated in a<br /> semiconservative fashion. DNA replication begins at a specific<br /> site within the DNA, known as the origin of replication. Circular<br /> bacterial chromosomes have a single origin of replication. Specific<br /> proteins, such as the DnaA protein found in E. coli, recognize<br /> the origin and initiate the DNA replication process. The synthesis<br /> of DNA strands proceeds bidirectionally from the origin.<br /> This synthesis requires various proteins. DNA helicase breaks the<br /> hydrogen bonding between the parental strands, topoisomerase<br /> removes positive supercoils, and single-strand binding proteins<br /> hold the parental strands in a single-stranded state. Primase synthesizes<br /> RNA primers, which are necessary for DNA polymerase<br /> to elongate new daughter strands in a 5' to 3' direction. In the<br /> leading strand, synthesis of the daughter strand is continuous. In<br /> the lagging strand, the synthesis of DNA occurs in short Okazaki<br /> fragments. To complete the synthesis of Okazaki fragments<br /> within the lagging strand, the RNA primers are removed, DNA<br /> polymerase fills in the gaps, and DNA ligase covalently attaches<br /> the fragments together. DNA replication is terminated when<br /> both replication forks reach the ter sequences. If replication has<br /> resulted in two intertwined DNA molecules known as catenanes,<br /> the catenanes are separated into individual circular molecules.<br /> Coordination of cell division and DNA replication is<br /> accomplished via the cellular regulation of replication. In bacteria,<br /> the control of DNA replication occurs at the origin. Several<br /> mechanisms can prevent premature DNA replication. Two examples<br /> are the availability of the DnaA protein and the methylation<br /> of GATC sites.<br /> DNA replication in eukaryotes has several unique features.<br /> Eukaryotic chromosomes contain multiple origins of replication.<br /> Eukaryotes have multiple types of DNA polymerases involved in<br /> synthesizing the leading and lagging strand, DNA repair, and replicating<br /> of DNA lesions. Also, because eukaryotic chromosomes<br /> are linear, a specialized mechanism exists for the replication at<br /> the telomeres. An enzyme known as telomerase recognizes the<br /> sequences at the ends of eukaryotic chromosomes and synthesizes<br /> additional repeats of telomeric sequences. This prevents<br /> chromosome shortening with each round of DNA replication.<br /> EXPERIMENTAL SUMMARY<br /> The semiconservative mechanism of DNA replication was initially<br /> proposed based on the double helix structure determined by Watson<br /> and Crick. Nevertheless, researchers needed to experimentally demonstrate<br /> that this mechanism was correct. The results of the isotope<br /> labeling experiments of Meselson and Stahl were consistent with a<br /> semiconservative mode of replication in which DNA contains one<br /> parental strand and one newly synthesized or daughter strand.<br /> Once the semiconservative model was established, researchers<br /> focused their attention on the details of DNA replication within<br /> living cells. Arthur Kornberg and his colleagues developed methods<br /> to synthesize bacterial DNA in vitro. In conjunction with the isolation<br /> of ts mutants, this allowed researchers to identify the role of<br /> individual proteins in DNA replication. Many of these components<br /> were found to be enzymes, such as DNA polymerase, primase, and<br /> DNA helicase. In eukaryotes, similar findings have been obtained,<br /> but more enzymes are involved and the sequences of the origins<br /> of replication are more complex. These added complexities have<br /> made it more difficult to analyze DNA replication in eukaryotes,<br /> although much progress has been made. As we will see in Chapter<br /> 22, the regulation of DNA replication is an important topic and an<br /> area of intense research because it may underlie the proliferation<br /> of cancer cells.<br /> PROBLEM SETS & INSIGHTS<br /> Solved Problems<br /> S1. Describe three ways to account for the high fidelity of DNA replication.<br /> Discuss the quantitative contributions of each of the three ways.<br /> Answer:<br /> First: AT and GC pairs are preferred in the double helix structure.<br /> This provides fidelity to around one mistake per 1,000.<br /> Second: Induced fit by DNA polymerase prevents covalent<br /> bond formation unless the proper nucleotides are in place. This<br /> increases fidelity another 100- to 1,000-fold, to about one error<br /> in 100,000 to 1 million.<br /> Third: Exonuclease proofreading increases fidelity another 100-<br /> to 1,000-fold, to about one error per 100 million nucleotides<br /> added.</div> <div>6.1 GENETIC TRANSFER AND MAPPING IN BACTERIA 137<br /> Donor<br /> cell (F + )<br /> Coupling<br /> factor<br /> Exporter<br /> Conjugation<br /> bridge<br /> Recipient<br /> cell (F – )<br /> Relaxosome<br /> F factor<br /> Bacterial chromosome<br /> Origin of<br /> transfer<br /> Inner membrane<br /> Outer membrane<br /> Relaxosome makes a<br /> cut at the origin of transfer<br /> and begins to separate the<br /> T-DNA strand.<br /> T DNA<br /> Pilus<br /> (b) Conjugating E. coli<br /> Relaxase<br /> Coupling<br /> factor<br /> Exporter<br /> F + cell<br /> F + cell<br /> (a) Transfer of an F factor via conjugation<br /> Most proteins of the<br /> relaxosome are released.<br /> The T-DNA/relaxase<br /> complex is recognized<br /> by the coupling factor and<br /> transferred to the exporter.<br /> The exporter pumps the<br /> T-DNA/relaxase complex<br /> into the recipient cell.<br /> In the donor cell, the<br /> F-factor DNA is replicated<br /> to become double stranded.<br /> In the recipient cell, relaxase<br /> joins the ends of the T-DNA<br /> strand. It is then replicated<br /> to become double stranded.<br /> Once contact is made, the pili shorten and thereby draw the<br /> donor and recipient cells closer together. A conjugation bridge is<br /> then formed between the two cells, which provides a passageway<br /> for DNA transfer.<br /> FIGURE 6.4 The transfer of an F factor during<br /> bacterial conjugation. (a) The mechanism of transfer.<br /> The end result is that both cells have an F factor. (b) Two<br /> E. coli cells in the act of conjugation. The cell on the left is F + , and the<br /> one on the right is F – . The two cells make contact with each other via<br /> sex pili that are made by the F + cell.<br /> The successful contact between a donor and recipient cell<br /> stimulates the donor cell to begin the transfer process. Genes within<br /> the F factor encode a protein complex called the relaxosome. This<br /> complex first recognizes a DNA sequence in the F factor known<br /> as the origin of transfer. Upon recognition, the site is cut in one<br /> DNA strand. This cut strand of DNA is called T DNA because it is<br /> the strand that will be transferred to the recipient cell. The relaxosome<br /> also catalyzes the separation of the T-DNA strand from its<br /> complementary strand. As the DNA strands separate, most of the<br /> proteins within the relaxosome are released, but one protein, called<br /> relaxase, remains bound to the end of the T DNA. The complex<br /> between the T DNA and relaxase is called a nucleoprotein because<br /> it contains both nucleic acid (DNA) and protein (relaxase).<br /> The next phase of conjugation involves the export of the<br /> nucleoprotein complex from the donor cell to the recipient cell.<br /> To begin this process, the T-DNA/relaxase complex is recognized<br /> by a coupling factor that promotes the entry of the nucleoprotein<br /> into the exporter, a complex of proteins that spans both inner<br /> and outer membranes of the donor cell. In bacterial species,<br /> this complex is formed from 10 to 15 different proteins that are<br /> encoded by genes within the F factor.<br /> Once the T-DNA/relaxase complex is pumped out of the<br /> donor cell, it travels through the conjugation bridge and then<br /> into the recipient cell. As shown in Figure 6.4a, the other strand<br /> of the F factor DNA remains in the donor cell, where DNA replication<br /> restores the F factor DNA to its original double-stranded<br /> condition. After the recipient cell receives a single strand of the F<br /> factor DNA, relaxase catalyzes the joining of the ends of the linear<br /> DNA molecule to form a circular molecule. This single-stranded<br /> DNA is replicated in the recipient cell to become double stranded.<br /> The result of conjugation is that the recipient cell has<br /> acquired an F factor, converting it from an F – to an F + cell. The</div> <div>336 CHAPTER 13 :: TRANSLATION OF mRNA<br /> 1<br /> Lys<br /> NH 3<br /> +<br /> 50<br /> Ser<br /> Met<br /> 129<br /> Gly<br /> Val<br /> Asn<br /> Thr<br /> Gly<br /> Asp<br /> Tyr<br /> Cys<br /> Asn<br /> Leu<br /> Ile<br /> Trp<br /> Asp<br /> Phe Gly Arg Cys Glu Leu Ala<br /> Ala<br /> Ser<br /> Thr<br /> Leu<br /> Val<br /> Gly<br /> Tyr<br /> Asn<br /> Asn<br /> 30<br /> Cys<br /> Ile<br /> 100<br /> Gly<br /> Arg<br /> Arg<br /> 110<br /> Ala<br /> Leu<br /> Arg<br /> Asn<br /> Ser<br /> Ala<br /> Gln<br /> 20<br /> Tyr<br /> Thr<br /> Gly<br /> Lys<br /> 70<br /> Asn<br /> Ala<br /> Phe<br /> Asp<br /> Gln<br /> 60<br /> Glu<br /> 40<br /> 120<br /> 10<br /> Leu<br /> Thr<br /> Ala<br /> Ser<br /> Met<br /> Gly<br /> Asn<br /> Phe<br /> Asn<br /> Lys<br /> His<br /> Ile Asn Ser Arg Trp Trp<br /> Pro<br /> Thr<br /> Arg<br /> Asp<br /> Pro 80<br /> Cys<br /> Ser Ala Leu Leu Ser Ser<br /> Gly Asp Gly Asp Ser Val Ile Lys Lys Ala<br /> Asn<br /> Ala<br /> Leu<br /> Arg<br /> COO –<br /> Trp<br /> Cys<br /> Val<br /> Gly<br /> Ala<br /> Arg<br /> Trp<br /> Ile<br /> Gly<br /> Arg Asn Arg Cys<br /> Lys<br /> Trp<br /> Ala<br /> Gln<br /> Val<br /> Asp<br /> Arg<br /> Cys<br /> Asn<br /> Ile<br /> Thr<br /> 90<br /> Ala<br /> Cys<br /> Asn<br /> Val<br /> Asp<br /> FIGURE 13.7 An example of a protein’s primary structure.<br /> This is the amino acid sequence of the enzyme lysozyme, which<br /> contains 129 amino acids in its primary structure. As you may have<br /> noticed, the first amino acid is not methionine; instead, it is lysine. The<br /> first methionine residue in this polypeptide sequence is removed after<br /> or during translation. The removal of the first methionine occurs in<br /> many (but not all) proteins.<br /> Gly<br /> Thr<br /> Ser<br /> traits of multicellular organisms are determined by the properties<br /> of their cells. Proteins perform a variety of functions critical<br /> to the life of cells and to the morphology and function of organisms<br /> (Table 13.6). Some proteins are important in determining<br /> the shape and structure of a given cell. For example, the protein<br /> tubulin assembles into large cytoskeletal structures known as<br /> microtubules, which provide eukaryotic cells with internal structure<br /> and organization. Some proteins are inserted into the cell<br /> membrane and aid in the transport of ions and small molecules<br /> across the membrane. An example is a sodium channel that transports<br /> sodium ions into nerve cells. Another interesting category<br /> of proteins are those that function as biological motors, such as<br /> myosin, which is involved in the contractile properties of muscle<br /> cells. Within multicellular organisms, certain proteins function in<br /> cell signaling and cell surface recognition. For example, proteins,<br /> TABLE 13.6<br /> Functions of Selected Cellular Proteins<br /> Function<br /> Cell shape and<br /> organization<br /> Transport<br /> Movement<br /> Cell signaling<br /> Cell surface<br /> recognition<br /> Enzymes<br /> Examples<br /> Tubulin: Forms cytoskeletal structures known as<br /> microtubules<br /> Ankyrin: Anchors cytoskeletal proteins to the plasma<br /> membrane<br /> Sodium channels: Transport sodium ions across the nerve<br /> cell membrane<br /> Lactose permease: Transports lactose across the bacterial<br /> cell membrane<br /> Hemoglobin: Transports oxygen in red blood cells<br /> Myosin: Involved in muscle cell contraction<br /> Kinesin: Involved in the movement of chromosomes during<br /> cell division<br /> Insulin: A hormone that influences target cell metabolism<br /> and growth<br /> Epidermal growth factor: A growth factor that promotes cell<br /> division<br /> Insulin receptor: Recognizes insulin and initiates a cell<br /> response<br /> Integrins: Bind to large extracellular proteins<br /> Hexokinase: Phosphorylates glucose during the first step in<br /> glycolysis<br /> b-galactosidase: Cleaves lactose into glucose and galactose<br /> Glycogen synthetase: Uses glucose molecules as building<br /> blocks to synthesize a large carbohydrate known as<br /> glycogen<br /> Acyl transferase: Links together fatty acids and glycerol<br /> phosphate during the synthesis of phospholipids<br /> RNA polymerase: Uses ribonucleotides as building blocks<br /> to synthesize RNA<br /> DNA polymerase: Uses deoxyribonucleotides as building<br /> blocks to synthesize DNA</div> <div>24.3 SOURCES OF NEW GENETIC VARIATION 689<br /> Isolate DNA from cells<br /> (blood, skin, hair roots,<br /> or semen, etc.)<br /> Fluorescence units<br /> Digestion of<br /> chromosomal<br /> DNA yields<br /> thousands of<br /> fragments with<br /> varying sizes.<br /> Sample 1<br /> –<br /> +<br /> Sample 2<br /> Cut DNA with a restriction<br /> enzyme and separate the<br /> fragments by gel<br /> electrophoresis.<br /> Blot the gel to a nylon<br /> membrane. Denature the<br /> DNA, and add a radiolabeled<br /> probe that is complementary<br /> to a selected minisatellite<br /> sequence. Allow the probe to<br /> hybridize, and then wash<br /> away the excess probe.<br /> Increasing size<br /> (b) Automated DNA fingerprinting<br /> Add probe.<br /> X-ray film<br /> Nylon membrane<br /> Sample 1 Sample 2<br /> (a) Traditional DNA fingerprinting<br /> Expose the filter to X-ray<br /> film, and compare the<br /> results. The outcome of this<br /> experiment indicates that<br /> the 2 samples came from<br /> different individuals.<br /> FIGURE 24.23 Protocol for DNA fingerprinting.<br /> (a) In traditional DNA fingerprinting, chromosomal<br /> DNA is isolated from a sample of cells (often a blood<br /> sample, hair roots, or semen) and cut with a restriction enzyme, and<br /> then the DNA fragments are separated by gel electrophoresis. The DNA<br /> fragments within the gel are then transferred to a nylon membrane.<br /> The DNA is denatured, and a radiolabeled probe is added that is<br /> complementary to a selected minisatellite sequence. The probe is given<br /> sufficient time to hybridize to any complementary DNA fragments,<br /> and any unbound probe is washed away. Finally, the membrane is<br /> exposed to X-ray film. The outcome of this experiment indicates that<br /> the two samples came from different individuals. (b) In automated<br /> DNA fingerprinting, a sample of DNA is amplified, using primers that<br /> recognize the ends of microsatellites. The microsatellite fragments are<br /> fluorescently labeled and then separated by gel electrophoresis. The<br /> fluorescent molecules within each microsatellite are excited with a laser,<br /> and the amount of fluorescence is measured via a fluorescence detector.<br /> A printout from the detector is shown here.</div> <div>16.3 DNA REPAIR 443<br /> different kinds of mutations within the coding sequence. This<br /> makes it possible to determine if a mutagen causes transitions,<br /> transversions, or frameshift mutations.<br /> 16.3 DNA REPAIR<br /> Because most mutations are deleterious, DNA repair systems are<br /> vital to the survival of all organisms. If DNA repair systems did<br /> not exist, spontaneous and environmentally induced mutations<br /> would be so prevalent that few species, if any, would survive. The<br /> necessity of DNA repair systems becomes evident when they are<br /> missing. Bacteria contain several different DNA repair systems.<br /> Yet, when even a single system is absent, the bacteria have a much<br /> higher rate of mutation. In fact, the rate of mutation is so high<br /> that these bacterial strains are sometimes called mutator strains.<br /> Likewise, in humans, an individual who is defective in only a single<br /> DNA repair system may manifest various disease symptoms,<br /> including a higher risk of skin cancer. This increased risk is due<br /> to the inability to repair UV-induced mutations.<br /> Living cells contain several DNA repair systems that can<br /> fix different types of DNA alterations (Table 16.7). Each repair<br /> system is composed of one or more proteins that play specific<br /> roles in the repair mechanism. In most cases, DNA repair is a<br /> multistep process. First, one or more proteins in the DNA repair<br /> system detect an irregularity in DNA structure. Next, the abnormality<br /> is removed by the action of DNA repair enzymes. Finally,<br /> normal DNA is synthesized via DNA replication enzymes. In this<br /> section, we will examine several different repair systems that have<br /> TABLE<br /> TABLE 16.7<br /> Common Types of DNA Repair Systems<br /> System<br /> Direct repair<br /> Base excision repair<br /> and nucleotide excision<br /> repair<br /> Mismatch repair<br /> Description<br /> An enzyme recognizes an incorrect alteration in<br /> DNA structure and directly converts it back to a<br /> correct structure.<br /> An abnormal base or nucleotide is first<br /> recognized and removed from the DNA. A<br /> segment of DNA in this region is excised, and<br /> then the complementary DNA strand is used<br /> as a template to synthesize a normal DNA<br /> strand.<br /> Similar to excision repair except that the DNA<br /> defect is a base pair mismatch in the DNA,<br /> not an abnormal nucleotide. The mismatch<br /> is recognized, and a segment of DNA in this<br /> region is removed. The parental strand is used<br /> to synthesize a normal daughter strand of DNA.<br /> been characterized in bacteria, yeast, mammals, and plants. Their<br /> diverse ways of repairing DNA underscore the extreme necessity<br /> for the structure of DNA to be maintained properly.<br /> Damaged Bases Can Be Directly Repaired<br /> In a few cases, the covalent modification of nucleotides by mutagens<br /> can be reversed by specific cellular enzymes. As discussed<br /> earlier in this chapter, UV light causes the formation of thymine<br /> dimers. Yeast, such as Saccharomyces cerevisiae, and most plants<br /> produce an enzyme called photolyase that can repair thymine<br /> dimers by splitting the dimers, which returns the DNA to its<br /> original condition (Figure 16.17a). This process directly restores<br /> the structure of DNA. Because plants are exposed to sunlight<br /> throughout the day, photolyase is a critical DNA repair enzyme<br /> for many plant species.<br /> A protein known as alkyltransferase can remove methyl or<br /> ethyl groups from guanine bases that have been mutagenized by<br /> alkylating agents such as nitrogen mustard and ethyl methanesulfonate.<br /> This protein is called alkyltransferase because it transfers<br /> the methyl or ethyl group from the base to a cysteine side chain<br /> within the alkyltransferase protein (Figure 16.17b). Surprisingly,<br /> this permanently inactivates alkyltransferase, which means it can<br /> be used only once!<br /> Base Excision Repair Removes a Damaged Base<br /> A second type of repair system, called base excision repair (BER),<br /> involves the function of a category of enzymes known as DNA<br /> N-glycosylases. This type of enzyme can recognize an abnormal<br /> base and cleave the bond between it and the sugar in the<br /> DNA backbone, creating an apurinic or apyrimidinic site (Figure<br /> 16.18). Depending on the species, this repair system can eliminate<br /> abnormal bases such as uracil, 3-methyladenine, 7-methylguanine,<br /> and pyrimidine dimers.<br /> Figure 16.18 illustrates the general steps involved in DNA<br /> repair via N-glycosylase. In this example, the DNA contains a<br /> uracil in its sequence. This could have happened spontaneously<br /> or by the action of a chemical mutagen. N-glycosylase recognizes<br /> a uracil within the DNA and cleaves (nicks) the bond between the<br /> sugar and base. This releases the uracil base and leaves behind an<br /> apyrimidinic nucleotide. This abnormal nucleotide is recognized<br /> by a second enzyme, AP endonuclease, which makes a cut on<br /> the 5' side. DNA polymerase, which has a 5' to 3' exonuclease<br /> activity, removes the abnormal region and, at the same time,<br /> replaces it with normal nucleotides. This process is called nick<br /> translation (although DNA replication, not mRNA translation,<br /> actually occurs). Finally, DNA ligase closes the nick.<br /> Homologous recombination<br /> repair<br /> Nonhomologous end joining<br /> Occurs at double-strand breaks or when DNA<br /> damage causes a gap in synthesis during DNA<br /> replication. The strands of a normal sister<br /> chromatid are used to repair a damaged sister<br /> chromatid.<br /> Occurs at double-strand breaks. The broken<br /> ends are recognized by proteins that keep the<br /> ends together; the broken ends are eventually<br /> rejoined.<br /> Nucleotide Excision Repair Systems<br /> Remove Segments of Damaged DNA<br /> An important general process for DNA repair is the nucleotide<br /> excision repair (NER) system. This type of system can repair<br /> many different types of DNA damage, including thymine dimers,<br /> chemically modified bases, missing bases, and certain types of<br /> cross-links. In NER, several nucleotides in the damaged strand<br /> are removed from the DNA, and the intact strand is used as a</div> <div>QUESTIONS FOR STUDENT DISCUSSION/COLLABORATION 99<br /> that either an rr homozygote or a cc homozygote will produce<br /> white flowers. In other words, rr is epistatic to C, and cc is epistatic<br /> to R. To test your hypothesis, you allowed heterozygous plants<br /> (RrCc) to self-fertilize and counted the offspring. You obtained<br /> the following data: 201 plants with red flowers and 144 with white<br /> flowers. Conduct a chi square analysis to see if your observed data<br /> are consistent with your hypothesis.<br /> E12. In Drosophila, red eyes is the wild-type phenotype. There are<br /> several different genes (with each gene existing in two or more<br /> alleles) that affect eye color. One allele causes purple eyes, and a<br /> different allele causes sepia eyes. Both of these alleles are recessive<br /> compared to red eye color. When flies with purple eyes were<br /> crossed to flies with sepia eyes, all of the F 1 offspring had red eyes.<br /> When the F 1 offspring were allowed to mate with each other, the<br /> following data were obtained:<br /> 146 purple eyes<br /> 151 sepia eyes<br /> 50 purplish sepia eyes<br /> 444 red eyes<br /> Explain this pattern of inheritance. Conduct a chi square analysis<br /> to see if the experimental data fit with your hypothesis.<br /> E13. As mentioned in Experimental question E12, red eyes is the wildtype<br /> phenotype in Drosophila, and there are several different genes<br /> (with each gene existing in two or more alleles) that affect eye<br /> color. One allele causes purple eyes, and a different allele causes<br /> vermilion eyes. The purple and vermilion alleles are recessive<br /> compared to red eye color. The following crosses were made, and<br /> the following data were obtained:<br /> Cross 1: Males with vermillion eyes x females with purple eyes<br /> 354 offspring, all with red eyes<br /> Cross 2: Males with purple eyes x females with vermillion eyes<br /> 212 male offspring with vermillion eyes<br /> 221 female offspring with red eyes<br /> Explain the pattern of inheritance based on these results. What<br /> additional crosses might you make to confirm your hypothesis?<br /> E14. Let’s suppose that you were looking through a vial of fruit flies in<br /> your laboratory and noticed a male fly that has pink eyes. What<br /> crosses would you make to determine if the pink allele is an X-<br /> linked gene? What crosses would you make to determine if the<br /> pink allele is an allele of the same X-linked gene that has white<br /> and eosin alleles? Note: The white and eosin alleles are discussed in<br /> Figure 4.22.<br /> E15. When examining a human pedigree, what features do you look<br /> for to distinguish between X-linked recessive inheritance versus<br /> autosomal recessive inheritance? How would you distinguish<br /> X-linked dominant inheritance from autosomal dominant<br /> inheritance in a human pedigree?<br /> Questions for Student Discussion/Collaboration<br /> 1. Let’s suppose a gene exists as a functional wild-type allele and a<br /> nonfunctional mutant allele. At the organism level, the wild-type<br /> allele is dominant. In a heterozygote, discuss whether dominance<br /> occurs at the cellular or molecular level. Discuss examples in which<br /> the issue of dominance depends on the level of examination.<br /> 2. A true-breeding rooster with a rose comb, feathered shanks, and<br /> cock-feathering was crossed to a hen that is true-breeding for pea<br /> comb and unfeathered shanks but is heterozygous for hen-feathering.<br /> If you assume these genes can assort independently, what is the<br /> expected outcome of the F 1 generation?<br /> 3. In oats, the color of the chaff is determined by a two-gene<br /> inter action. When a true-breeding black plant was crossed to a<br /> true-breeding white plant, the F 1 generation was composed of all<br /> black plants. When the F 1 offspring were crossed to each other, the<br /> ratio produced was 12 black : 3 gray : 1 white. First, construct a<br /> Punnett square that accounts for this pattern of inheritance. Which<br /> genotypes produce the gray phenotype? Second, at the level of<br /> protein function, how would you explain this type of inheritance?<br /> Note: All answers appear at the website for this textbook; the answers to<br /> even-numbered questions are in the back of the textbook.<br /> www.mhhe.com/brookergenetics3e<br /> Visit the Online Learning Center for practice tests, answer keys, and other learning aids for this chapter. Enhance your understanding of genetics with<br /> our interactive exercises, quizzes, animations, and much more.</div> <div>502 CHAPTER 18 :: RECOMBINANT DNA TECHNOLOGY<br /> Lane 1: Red blood cells<br /> Lane 2: Brain cells<br /> Lane 3: Intestinal cells<br /> 1 2 3<br /> –<br /> FIGURE 18.11 The results of Western blotting. The black<br /> band indicates where the primary antibody has recognized the protein<br /> of interest. In lane 1, proteins were isolated from mouse red blood<br /> cells. As seen here, the β-globin polypeptide is made in these cells. By<br /> comparison, lanes 2 and 3 were samples of protein from brain cells and<br /> intestinal cells, which do not synthesize β globin.<br /> the protein of interest in a gel blot. For example, it is common<br /> for the secondary antibody to be linked to the enzyme alkaline<br /> phosphatase. When the colorless dye, XP (5-bromo-4-chloro-<br /> 3-indolyl-phosphate), is added to the blotting solution, alkaline<br /> phosphatase converts the dye to a black compound. Because the<br /> secondary antibody binds to the primary antibody, a protein<br /> band that is recognized by the primary antibody will become<br /> black. In the example shown in Figure 18.11, the primary antibody<br /> recognizes β globin. In lane 1, proteins were isolated from<br /> mouse red blood cells. The black band indicates that the β-globin<br /> polypeptide is made in these cells. By comparison, the proteins<br /> loaded into lanes 2 and 3 were from brain cells and intestinal<br /> cells, respectively. The absence of any bands indicates that these<br /> cell types do not synthesize β globin.<br /> Techniques Can Be Used to Detect the Binding<br /> of Proteins to DNA Sequences<br /> In addition to detecting the presence of genes and gene products<br /> using blotting techniques, researchers often want to study<br /> the binding of proteins to specific sites on a DNA molecule. For<br /> example, the molecular investigation of transcription factors<br /> requires methods that can identify interactions between transcription<br /> factor proteins and specific DNA sequences. A technically<br /> simple, widely used method for identifying this type of<br /> interaction is the gel retardation assay, also known as the band<br /> shift assay. This technique was used originally to study rRNA–<br /> protein interactions and quickly became popular after its success<br /> in studying protein–DNA interactions in the lac operon. Now it<br /> is commonly used as a technique to detect interactions between<br /> eukaryotic transcription factors and DNA regulatory elements.<br /> The technical basis for a gel retardation assay is that the<br /> binding of a protein to a DNA fragment will retard the fragment’s<br /> ability to move within a polyacrylamide or agarose gel.<br /> During electrophoresis, DNA fragments are pulled through the<br /> gel matrix toward the bottom of the gel by a voltage gradient.<br /> Smaller fragments of DNA migrate more quickly through a gel<br /> FIGURE 18.12 The results of a gel retardation assay. The<br /> binding of protein to a labeled fragment of DNA retards its rate of<br /> movement through a gel. For the results shown in the lane on the<br /> right, if the concentration of the DNA fragment was higher than the<br /> concentration of the protein, there would be two bands: one band with<br /> protein bound (at a higher molecular mass) and one band without<br /> protein bound (corresponding to the band found in the left lane).<br /> matrix than do larger fragments. As you might expect, therefore,<br /> the binding of a protein to a DNA fragment will retard<br /> the DNA’s rate of movement through the gel matrix, because a<br /> protein–DNA complex has a higher mass. When comparing a<br /> DNA fragment and a protein–DNA complex after electrophoresis,<br /> the complex is shifted to a higher band than the DNA alone,<br /> because the complex migrates more slowly to the bottom of the<br /> gel (Figure 18.12).<br /> A gel retardation assay must be carried out under nondenaturing<br /> conditions. This means that the buffers and gel cannot<br /> cause the unfolding of proteins or the separation of the DNA<br /> double helix. This is necessary so that the proteins and DNA<br /> retain their proper structure and thereby can bind to each other.<br /> The nondenaturing conditions of a gel retardation assay differ<br /> from the more common SDS–gel electrophoresis, in which the<br /> proteins are denatured by the detergent SDS.<br /> A second method for studying protein–DNA interactions<br /> is DNase I footprinting, a technique described originally by<br /> David Galas and Albert Schmitz in 1978. A DNase I footprinting<br /> experiment attempts to identify one or more regions of DNA<br /> that interact with a DNA-binding protein. In their original study,<br /> Galas and Schmitz identified a site in the lac operon, known as<br /> the lac operator site, that is bound by a DNA-binding protein<br /> called the lac repressor.<br /> To understand the basis of a DNase I footprinting experiment,<br /> we need to consider the molecular interactions among<br /> three things: a fragment of DNA, DNA-binding proteins, and<br /> agents that can alter DNA structure. As an example, let’s examine<br /> the binding of RNA polymerase to a bacterial promoter, a topic<br /> discussed in Chapter 12. When RNA polymerase holoenzyme<br /> binds to the promoter to form a closed complex, it binds tightly<br /> to the –35 and –10 promoter regions, but the protein covers up<br /> an even larger region of the DNA. Therefore, holoenzyme bound<br /> at the promoter prevents other molecules from gaining access to<br /> this region of the DNA. The enzyme DNase I, which can cleave<br /> +</div> <div>798 SOLUTIONS TO EVEN-NUMBERED PROBLEMS<br /> CHAPTER 22<br /> Conceptual Questions<br /> C2. When a disease-causing allele affects a trait, it is causing a deviation<br /> from normality, but the gene involved is not usually the only gene<br /> that governs the trait. For example, an allele causing hemophilia<br /> prevents the normal blood clotting pathway from operating correctly.<br /> It follows a simple Mendelian pattern because a single gene affects the<br /> phenotype. Even so, it is known that normal blood clotting is due to<br /> the actions of many genes.<br /> C4. Changes in chromosome number and unbalanced changes in<br /> chromosome structure tend to affect phenotype because they create an<br /> imbalance of gene expression. For example, in Down syndrome, there<br /> are three copies of chromosome 21 and, therefore, three copies of all<br /> the genes on chromosome 21. This leads to a relative overexpression<br /> of genes that are located on chromosome 21 compared to the<br /> other chromosomes. Balanced translocations and inversions often<br /> are without phenotypic consequences because the total amount of<br /> genetic material is not altered, and the level of gene expression is not<br /> significantly changed.<br /> C6. There are lots of possible answers; here are a few. Dwarfism occurs<br /> in people and dogs. Breeds like the dachshund and basset hound are<br /> types of dwarfism in dogs. There are diabetic people and mice. There<br /> are forms of inherited obesity in people and mice. Hip dysplasia is<br /> found in people and dogs.<br /> C8. A. Because a person must inherit two defective copies of this gene<br /> and it is known to be on chromosome 1, the mode of transmission<br /> is autosomal recessive. Both members of this couple must be<br /> heterozygous, because they have one affected parent (who had<br /> to transmit the mutant allele to them) and their phenotypes are<br /> unaffected (so they must have received the normal allele from<br /> their other parent). Because both parents are heterozygotes, there<br /> is a 1 / 4 chance of producing an affected child (a homozygote) with<br /> Gaucher disease. If we let G represent the nonmutant allele and g<br /> the mutant allele:<br /> G<br /> g<br /> G<br /> GG<br /> Normal<br /> Gg<br /> Normal<br /> g<br /> Gg<br /> Normal<br /> gg<br /> Gaucher<br /> B. From this Punnett square, we can also see that there is a 1 / 4 chance<br /> of producing a homozygote with both normal copies of the gene.<br /> C. We need to apply the binomial expansion equation to solve this<br /> problem (see Chapter 2 for a description of this equation). In this<br /> problem, n � 5, x � 1, p � 0.25, q � 0.75. The answer is 0.396, or<br /> 39.6%.<br /> C10. The mode of transmission is autosomal recessive. All of the affected<br /> individuals do not have affected parents. Also, the disorder is found in<br /> both males and females. If it were X-linked recessive, individual III-1<br /> would have to have an affected father, which she does not.<br /> C12. The 13 babies have acquired a new mutation. In other words, during<br /> spermatogenesis or oogenesis, or after the egg was fertilized, a new<br /> mutation occurred in the fibroblast growth factor gene. These 13<br /> individuals have the same chances of passing the mutant allele to their<br /> offspring as the 18 individuals who inherited the mutant allele from a<br /> parent. The chance is 50%.<br /> C14. Because this is a dominant trait, the mother must have two normal<br /> copies of the gene, and the father (who is affected) is most likely to<br /> be a heterozygote. (Note: The father could be a homozygote, but this<br /> is extremely unlikely because the dominant allele is very rare.) If we<br /> let M represent the mutant Marfan allele and m the normal allele, the<br /> following Punnett square can be constructed:<br /> m<br /> m<br /> M<br /> Mm<br /> Marfan<br /> Mm<br /> Marfan<br /> m<br /> mm<br /> Normal<br /> mm<br /> Normal<br /> A. There is a 50% chance that this couple will have an affected child.<br /> B. We use the product rule. The odds of having an unaffected child<br /> are 50%. So if we multiply 0.5 � 0.5 � 0.5, this equals 0.125, or a<br /> 12.5% chance of having three unaffected offspring.<br /> C16. A prion is a protein that behaves like an infectious agent. The<br /> infectious form of the prion protein has an abnormal conformation.<br /> This abnormal conformation is termed PrP Sc , and the normal<br /> conformation of the protein is termed PrP C . An individual can be<br /> “infected” with the abnormal conformation of the protein by eating<br /> products from another animal that had the disease, or the prion<br /> protein may convert spontaneously to the normal conformation. A<br /> prion protein in the PrP Sc conformation can bind to a prion protein<br /> in the PrP C conformation and convert it to the PrP Sc form. An<br /> accumulation of prions in the PrP Sc form is what causes the disease<br /> symptoms.<br /> C18. A. Keep in mind that a conformational change is a stepwise process.<br /> It begins in one region of a protein and involves a series of small<br /> changes in protein structure. The entire conformational change<br /> from PrP C to PrP Sc probably involves many small changes in<br /> protein structure that occur in a stepwise manner. Perhaps the<br /> conformational change (from PrP C to PrP Sc ) begins in the vicinity<br /> of position 178 and then proceeds throughout the rest of the<br /> protein. If there is a methionine at position 129, the complete<br /> conformational change (from PrP C to PrP Sc ) can take place.<br /> However, perhaps the valine at position 129 somehow blocks one<br /> of the steps needed to complete the conformational change.<br /> B. Once the PrP Sc conformational change is completed, a<br /> PrP Sc protein can bind to another prion protein in the PrP C<br /> conformation. Perhaps it begins to convert it (to the PrP Sc<br /> conformation) by initiating a small change in protein structure in<br /> the vicinity of position 178. The conversion would then proceed in<br /> a stepwise manner until the PrP Sc conformation has been achieved.<br /> If a valine is at position 129, this could somehow inhibit one of the<br /> steps that are needed to complete the conformational change. If an<br /> individual had Val-129 in the polypeptide encoded by the second<br /> PrP gene, half of their prion proteins would be less sensitive to<br /> conversion by PrP Sc , compared to individuals who had Met-129.<br /> This would explain why individuals with Val-129 in half of the<br /> prion proteins would have disease symptoms that would progress<br /> more slowly.<br /> C20. A proto-oncogene is a normal cellular gene that typically plays a<br /> role in cell division. It can be altered by mutation to become an<br /> oncogene and thereby cause cancer. At the level of protein function, a<br /> proto-oncogene can become an oncogene by synthesizing too much<br /> of a protein or synthesizing the same amount of a protein that is<br /> abnormally active.</div> <div>SOLVED PROBLEMS 127<br /> nonparental offspring. The male flies could contribute the car and<br /> B + alleles (on a cytologically normal X chromosome) or contribute<br /> a Y chromosome. In the absence of crossing over, the female flies<br /> could contribute a short X chromosome with the car and B alleles or<br /> a long X chromosome with the car + and B + alleles. If crossing over<br /> occurred in the region between these two genes, the female flies would<br /> contribute recombinant X chromosomes. One possible recombinant X<br /> chromosome would be normal-sized and carry the car and B + alleles,<br /> while the other recombinant X chromosome would be deleted at one<br /> end with a piece of the Y chromosome at the other end and carry the<br /> car + and B alleles. When combined with an X or Y chromosome from<br /> the males, the parental offspring would have carnation, bar eyes or<br /> wild-type eyes; the nonparental offspring would have carnation, round<br /> eyes or red, bar eyes.<br /> Male gametes<br /> carB +<br /> Y<br /> Phenotype<br /> X chromosome<br /> from female<br /> Answer: A double crossover between the two genes could involve two<br /> chromatids, three chromatids, or four chromatids. The possibilities for<br /> all types of double crossovers are shown here:<br /> A<br /> A<br /> a<br /> a<br /> Double crossover (involving 4 chromatids)<br /> A<br /> B<br /> B<br /> b<br /> b<br /> B<br /> A<br /> A<br /> a<br /> a<br /> A<br /> b<br /> b<br /> B<br /> B<br /> B<br /> 100%<br /> recombinants<br /> carB<br /> carB<br /> carB +<br /> carB<br /> Y<br /> Carnation,<br /> bar eyes<br /> Short X<br /> chromosome<br /> A<br /> a<br /> B<br /> b<br /> A<br /> a<br /> b<br /> B<br /> 50%<br /> recombinants<br /> Female gametes<br /> car + B +<br /> carB +<br /> car + B + car + B +<br /> carB + Y<br /> carB + carB +<br /> carB + Y<br /> Red, round<br /> eyes<br /> Carnation,<br /> round eyes<br /> Long X<br /> chromosome<br /> with a piece of Y<br /> Normal-sized X<br /> chromosome<br /> a b<br /> Double crossover (involving 3 chromatids)<br /> A B<br /> a<br /> A<br /> b<br /> B<br /> car + B<br /> car + B<br /> carB +<br /> car + B<br /> Y<br /> Red, bar<br /> eyes<br /> Short X<br /> chromosome<br /> with a piece of Y<br /> A<br /> a<br /> B<br /> b<br /> A<br /> a<br /> b<br /> B<br /> 50%<br /> recombinants<br /> The results shown in the Punnett square are the actual results<br /> that Stern observed. His interpretation was that crossing over between<br /> homologous chromosomes—in this case, the X chromosome—accounts<br /> for the formation of offspring with recombinant phenotypes.<br /> S5. Researchers have discovered a limit to the relationship between<br /> map distance and the percentage of recombinant offspring. Even<br /> though two genes on the same chromosome may be much more<br /> than 50 mu apart, we do not expect to obtain greater than 50%<br /> recombinant offspring in a testcross. You may be wondering why<br /> this is so. The answer lies in the pattern of multiple crossovers.<br /> At the pachytene stage of meiosis, a single crossover in the<br /> region between two genes will produce only 50% recombinant<br /> chromosomes (see Figure 5.1b). Therefore, to exceed a 50%<br /> recombinant level, it would seem necessary to have multiple<br /> crossovers within the tetrad.<br /> Let’s suppose that two genes are far apart on the same<br /> chromosome. A testcross is made between a heterozygous<br /> individual, AaBb, and a homozygous individual, aabb. In the<br /> heterozygous individual, the dominant alleles (A and B) are linked<br /> on the same chromosome, and the recessive alleles (a and b) are<br /> linked on the same chromosome. Draw out all of the possible<br /> double crossovers (between two, three, or four chromatids) and<br /> determine the average number of recombinant offspring, assuming<br /> an equal probability of all of the double crossover possibilities.<br /> a<br /> Double crossover (involving 3 chromatids)<br /> A<br /> A<br /> a<br /> a<br /> Double crossover (involving 2 chromatids)<br /> b<br /> B<br /> B<br /> b<br /> b<br /> a<br /> A<br /> A<br /> a<br /> a<br /> b<br /> B<br /> B<br /> b<br /> b<br /> 0%<br /> recombinants<br /> Overall average<br /> is 50% for all<br /> 4 possibilities.<br /> This drawing considers the situation where two crossovers<br /> are expected to occur in the region between the two genes. Because<br /> the tetrad is composed of two pairs of homologues, there are several<br /> possible ways that a double crossover could occur between homologues.<br /> In this illustration, the crossover on the right has occurred first. Because<br /> all of these double crossing over events are equally probable, we take the<br /> average of them to determine the maximum recombination frequency.<br /> This average equals 50%.</div> <div>13.1 THE GENETIC BASIS FOR PROTEIN SYNTHESIS 335<br /> CH 3<br /> H<br /> +<br /> H 3 N C COO –<br /> CH 3 CH 3<br /> CH 3 CH<br /> +<br /> H 3 N C COO –<br /> +<br /> H 3 N C COO –<br /> CH 3 CH 3<br /> CH<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> CH 3<br /> CH 2<br /> CH 3 CH<br /> +<br /> H 3 N C COO –<br /> CH 2<br /> CH 2 CH 2<br /> +<br /> H 2 N C COO –<br /> S<br /> SH<br /> CH 2<br /> CH 2<br /> CH 2<br /> +<br /> H 3 N C COO – +<br /> H 3 N C COO –<br /> H<br /> Glycine (Gly) G<br /> H<br /> Alanine (Ala) A<br /> H<br /> Valine (Val) V<br /> H<br /> Leucine (Leu) L<br /> H<br /> Isoleucine (Ile) I<br /> H<br /> Proline (Pro) P<br /> H<br /> H<br /> Cysteine (Cys) C Methionine (Met) M<br /> (a) Nonpolar, aliphatic amino acids<br /> OH<br /> H<br /> N<br /> OH<br /> O NH 2<br /> C<br /> O NH 2<br /> C<br /> CH 2<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> CH 3<br /> HCOH<br /> +<br /> H 3 N C COO –<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> H<br /> H<br /> Phenylalanine (Phe) F Tyrosine (Tyr) Y<br /> H<br /> Tryptophan (Trp) W<br /> H<br /> Serine (Ser) S<br /> H<br /> Threonine (Thr) T<br /> H<br /> Asparagine (Asn) N<br /> H<br /> Glutamine (Gln) Q<br /> (b) Nonpolar, aromatic amino acids<br /> (c) Polar, neutral amino acids<br /> NH 2<br /> + +<br /> NH 3<br /> C NH 2<br /> O O –<br /> C<br /> O O –<br /> C<br /> CH 2<br /> HN<br /> +<br /> NH<br /> CH 2<br /> CH 2<br /> CH 2<br /> NH<br /> CH 2<br /> CH 2<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> CH 2<br /> +<br /> H 3 N C COO –<br /> H<br /> H<br /> Aspartic acid (Asp)D Glutamic acid (Glu) E<br /> H<br /> Histidine (His) H<br /> H<br /> Lysine (Lys) K<br /> H<br /> Arginine (Arg) R<br /> (d) Polar, acidic amino acids<br /> (e) Polar, basic amino acids<br /> FIGURE 13.6<br /> The 20 amino acids found within proteins.<br /> are more likely to be found in a b-sheet conformation. Secondary<br /> structures within polypeptides are primarily stabilized by the<br /> formation of hydrogen bonds.<br /> The short regions of secondary structure within a polypeptide<br /> are folded relative to each other to make the tertiary<br /> structure of a polypeptide. As shown in Figure 13.8c, a-helical<br /> regions and b-sheet regions are connected by irregularly shaped<br /> segments to determine the tertiary structure of the polypeptide.<br /> The folding of a polypeptide into its secondary and then tertiary<br /> conformation can usually occur spontaneously because it<br /> is a thermodynamically favorable process. The structure is determined<br /> by various interactions, including the tendency of hydrophobic<br /> amino acids to avoid water, ionic interactions among<br /> charged amino acids, hydrogen bonding among amino acids in<br /> the folded polypeptide, and weak bonding known as van der<br /> Waals interactions.<br /> A protein is a functional unit that can be composed of one<br /> or more polypeptides. Some proteins are composed of a single<br /> polypeptide. Many proteins, however, are composed of two or<br /> more polypeptides that associate with each other to make a functional<br /> protein with a quaternary structure (Figure 13.8d). The<br /> individual polypeptides are called subunits of the protein, each<br /> of which has its own tertiary structure. The association of multiple<br /> subunits is the quaternary structure of a protein.<br /> Cellular Proteins Are Primarily Responsible<br /> for the Characteristics of Living Cells<br /> and an Organism’s Traits<br /> Why is the genetic material largely devoted to storing the information<br /> to make proteins? To a great extent, the characteristics of<br /> a cell depend on the types of proteins that it makes. In turn, the</div> <div>218 CHAPTER 8 :: VARIATION IN CHROMOSOME STRUCTURE AND NUMBER<br /> C10. An individual has the following reciprocal translocation: C17. A diploid fruit fly has eight chromosomes. How many total<br /> chromosomes would be found in the following flies?<br /> A<br /> A<br /> H H<br /> A. Tetraploid<br /> B B I I<br /> B. Trisomy 2<br /> C. Monosomy 3<br /> C C J J<br /> D. 3n<br /> D D K K<br /> L L<br /> E E<br /> E. 4n + 1<br /> M M<br /> What would be the outcome of alternate and adjacent-1<br /> C18. A person is born with one X chromosome, zero Y chromosomes,<br /> segregation?<br /> trisomy 21, and two copies of the other chromosomes. How many<br /> chromosomes does this person have altogether? Explain whether<br /> this person is euploid or aneuploid.<br /> C11. A phenotypically normal individual has the following<br /> combinations of abnormal chromosomes:<br /> 11<br /> 15 18<br /> The normal chromosomes are shown on the left of each pair.<br /> Suggest a series of events (breaks, translocations, crossovers, etc.)<br /> that may have produced this combination of chromosomes.<br /> C12. Two phenotypically normal parents produce a phenotypically<br /> abnormal child in which chromosome 5 is missing part of its long<br /> arm but has a piece of chromosome 7 attached to it. The child also<br /> has one normal copy of chromosome 5 and two normal copies of<br /> chromosome 7. With regard to chromosomes 5 and 7, what do you<br /> think are the chromosomal compositions of the parents?<br /> C13. In the segregation of centromeres, why is adjacent-2 segregation<br /> less frequent than alternate or adjacent-1 segregation?<br /> C14. Which of the following types of chromosomal changes would you<br /> expect to have phenotypic consequences? Explain your choices.<br /> A. Pericentric inversion<br /> B. Reciprocal translocation<br /> C. Deletion<br /> D. Unbalanced translocation<br /> C15. Explain why a translocation cross occurs during metaphase of<br /> meiosis I when a cell contains a reciprocal translocation.<br /> C16. A phenotypically abnormal individual has a phenotypically normal<br /> father with an inversion on one copy of chromosome 7 and a<br /> normal mother without any changes in chromosome structure.<br /> The order of genes along chromosome 7 in the father is as follows:<br /> R T D M centromere P U X Z C (normal chromosome 7)<br /> R T D U P centromere M X Z C (inverted chromosome 7)<br /> The phenotypically abnormal offspring has a chromosome 7 with<br /> the following order of genes:<br /> R T D M centromere P U D T R<br /> With a sketch, explain how this chromosome was formed. In your<br /> answer, explain where the crossover occurred (i.e., between which<br /> two genes).<br /> 11<br /> 18 15<br /> C19. Two phenotypically unaffected parents produce two children with<br /> familial Down syndrome. With regard to chromosomes 14 and 21,<br /> what are the chromosomal compositions of the parents?<br /> C20. Aneuploidy is typically detrimental, whereas polyploidy is<br /> sometimes beneficial, particularly in plants. Discuss why you think<br /> this is the case.<br /> C21. Explain how aneuploidy, deletions, and duplications cause genetic<br /> imbalances. Why do you think that deletions and monosomies are<br /> more detrimental than duplications and trisomies?<br /> C22. Female fruit flies homozygous for the X-linked white-eye allele<br /> are crossed to males with red eyes. On very rare occasions, an<br /> offspring is a male with red eyes. Assuming these rare offspring are<br /> not due to a new mutation in one of the mother’s X chromosomes<br /> that converted the white-eye allele into a red-eye allele, explain<br /> how this red-eyed male arose.<br /> C23. A cytogeneticist has collected tissue samples from members of<br /> the same butterfly species. Some of the butterflies were located<br /> in Canada, and others were found in Mexico. Upon karyotyping,<br /> the cytogeneticist discovered that chromosome 5 of the Canadian<br /> butterflies had a large inversion compared to the Mexican<br /> butterflies. The Canadian butterflies were inversion homozygotes,<br /> whereas the Mexican butterflies had two normal copies of<br /> chromosome 5.<br /> A. Explain whether a mating between the Canadian and Mexican<br /> butterflies would produce phenotypically normal offspring.<br /> B. Explain whether the offspring of a cross between Canadian and<br /> Mexican butterflies would be fertile.<br /> C24. Why do you think that human trisomies 13, 18, and 21 can survive<br /> but the other trisomies are lethal? Even though X chromosomes<br /> are large, aneuploidies of this chromosome are also tolerated.<br /> Explain why.<br /> C25. A zookeeper has collected a male and female lizard that look like<br /> they belong to the same species. They mate with each other and<br /> produce phenotypically normal offspring. However, the offspring<br /> are sterile. Suggest one or more explanations for their sterility.<br /> C26. What is endopolyploidy? What is its biological significance?<br /> C27. What is mosaicism? How is it produced?<br /> C28. Explain how polytene chromosomes of Drosophila melanogaster<br /> are produced and how they form a six-armed structure.<br /> C29. Describe some of the advantages of polyploid plants. What are the<br /> consequences of having an odd number of chromosome sets?<br /> C30. While conducting field studies on a chain of islands, you decide<br /> to karyotype two phenotypically identical groups of turtles,</div> <div>17.1 HOMOLOGOUS RECOMBINATION 457<br /> DNA replication<br /> +<br /> BrdU<br /> DNA replication<br /> +<br /> BrdU<br /> Division<br /> One strand<br /> with BrdU<br /> Both strands<br /> with BrdU<br /> DNA replication<br /> +<br /> BrdU<br /> Stains<br /> darker<br /> Stains<br /> lightly<br /> FIGURE 17.2<br /> Harlequin chromosomes.<br /> Harlequin chromosomes<br /> appears very dark. In this way, the two sister chromatids can be<br /> distinguished microscopically. Chromosomes stained in this way<br /> are referred to as harlequin chromosomes, because they are reminiscent<br /> of a harlequin character’s costume with its variegated<br /> pattern of light and dark patches. In these chromosomes, SCEs<br /> can be clearly identified as exchanges between light and dark<br /> chromatids.<br /> The steps in Perry and Wolff’s protocol are shown in Figure<br /> 17.3. They began with Chinese hamster ovary cells, a commonly<br /> used mammalian cell line, and exposed the cells to BrdU<br /> for two rounds of DNA replication. Near the end of the second<br /> round, colcemid was added to prevent the completion of mitosis.<br /> The cells were treated with KCl to spread out the chromosomes,<br /> which were subsequently fixed and then stained with Hoechst<br /> 33258 and Giemsa.<br /> ■ THE HYPOTHESIS<br /> Crossing over may occur between sister chromatids.<br /> ■ TESTING THE HYPOTHESIS — FIGURE 17.3 The staining of harlequin chromosomes reveals sister chromatid<br /> exchange.<br /> Starting material: A laboratory cell line of Chinese hamster ovary (CHO) cells.<br /> 1. Expose CHO cells to BrdU for two cell<br /> generations (approximately 24 hours).<br /> Note: CHO cells are Chinese hamster<br /> ovary cells, which is a commonly used<br /> mammalian cell line.<br /> CHO<br /> cells<br /> Experimental level<br /> BrdU<br /> Conceptual level<br /> 24 hours First<br /> replication</div> <div>5.1 LINKAGE AND CROSSING OVER 101<br /> together on the same chromosome tend to be transmitted as a<br /> unit. This second meaning indicates that linkage has an influence<br /> on inheritance patterns.<br /> Chromosomes are sometimes called linkage groups,<br /> because a chromosome contains a group of genes that are physically<br /> linked together. In each species, the number of linkage<br /> groups equals the number of chromosome types. For example,<br /> human somatic cells have 46 chromosomes, which are composed<br /> of 22 types of autosomes that come in pairs plus one pair of sex<br /> chromosomes, the X and Y. Therefore, humans have 22 autosomal<br /> linkage groups, an X chromosome linkage group, and a Y<br /> chromosome linkage group.<br /> Geneticists are often interested in the transmission of two<br /> or more traits in a genetic cross. When a geneticist follows the<br /> variants of two different traits in a cross, this is called a dihybrid<br /> cross; when three traits are followed, it is a trihybrid cross; and<br /> so on. The outcome of a dihybrid or trihybrid cross depends on<br /> whether or not the genes are linked to each other along the same<br /> chromosome. In this section, we will examine how linkage affects<br /> the transmission patterns of two or more traits.<br /> Crossing Over May Produce<br /> Recombinant Genotypes<br /> Even though the alleles for different genes may be linked along<br /> the same chromosome, the linkage can be altered during meiosis.<br /> In diploid eukaryotic species, homologous chromosomes can<br /> exchange pieces with each other, a phenomenon called crossing<br /> over. This event occurs during prophase of meiosis I. As discussed<br /> in Chapter 3, the replicated chromosomes, known as sister<br /> chromatids, associate with the homologous sister chromatids<br /> to form a structure known as a bivalent. A bivalent is composed<br /> of two pairs of sister chromatids. In prophase of meiosis I, a sister<br /> chromatid of one pair commonly crosses over with a sister<br /> chromatid from the homologous pair.<br /> Figure 5.1 considers meiosis when two genes are linked on<br /> the same chromosome. One of the parental chromosomes carries<br /> the A and B alleles, while the homologue carries the a and<br /> b alleles. In Figure 5.1a, no crossing over has occurred. Therefore,<br /> the resulting haploid cells contain the same combination<br /> of alleles as the original chromosomes. In this case, two haploid<br /> cells carry the dominant A and B alleles, and the other two carry<br /> the recessive a and b alleles. The arrangement of linked alleles<br /> has not been altered.<br /> In contrast, Figure 5.1b illustrates what can happen when<br /> crossing over occurs. Two of the haploid cells contain combinations<br /> of alleles, namely A and b or a and B, which differ from<br /> those in the original chromosomes. In these two cells, the grouping<br /> of linked alleles has been changed. An event such as this, leading<br /> to a new combination of alleles, is known as genetic recombination.<br /> The haploid cells carrying the A and b, or the a and B,<br /> alleles are called nonparental or recombinant cells. Likewise, if<br /> such haploid cells were gametes that participated in fertilization,<br /> the resulting offspring are called nonparental or recombinant<br /> offspring. These offspring can display combinations of traits that<br /> are different from those of either parent. In contrast, offspring<br /> B<br /> A<br /> b<br /> a<br /> b<br /> a<br /> Diploid cell after<br /> chromosome replication<br /> B<br /> A<br /> b<br /> a<br /> B<br /> A<br /> Meiosis<br /> Possible haploid cells<br /> B<br /> A<br /> b<br /> a<br /> (a) Without crossing over, linked<br /> alleles segregate together.<br /> FIGURE 5.1 Consequences of crossing over<br /> during meiosis. (a) In the absence of crossing over, the<br /> A and B alleles and the a and b alleles are maintained in<br /> the same arrangement found in the parental chromosomes. (b) Crossing<br /> over has occurred in the region between the two genes, creating two<br /> nonparental haploid cells with a new combination of alleles.<br /> that have inherited the same combination of alleles that are<br /> found in the chromosomes of their parents are known as parental<br /> or nonrecombinant offspring.<br /> In this section, we will consider how crossing over affects<br /> the pattern of inheritance for genes linked on the same chromosome.<br /> In Chapter 17, we will consider the molecular events that<br /> cause crossing over to occur.<br /> Bateson and Punnett Discovered Two Traits<br /> That Did Not Assort Independently<br /> An early study indicating that some traits may not assort independently<br /> was carried out by William Bateson and Reginald Punnett<br /> in 1905. According to Mendel’s law of indepen dent assortment,<br /> a dihybrid cross between two individuals, heterozygous for<br /> two genes, should yield a 9:3:3:1 phenotypic ratio among the offspring.<br /> However, a surprising result occurred when Bateson and<br /> Punnett conducted a cross in the sweet pea involving two different<br /> traits—flower color and pollen shape.<br /> As seen in Figure 5.2, they began by crossing a truebreeding<br /> strain with purple flowers (PP) and long pollen (LL) to<br /> a<br /> Diploid cell after<br /> chromosome replication<br /> B<br /> A<br /> b<br /> a<br /> B<br /> A<br /> b<br /> a<br /> Meiosis<br /> Possible haploid cells<br /> B<br /> a<br /> b<br /> A<br /> (b) Crossing over can reassort<br /> linked alleles.</div> <div>370 CHAPTER 14 :: GENE REGULATION IN BACTERIA AND BACTERIOPHAGES<br /> Allolactose<br /> Repressor<br /> (inactive)<br /> Allolactose<br /> CAP site Promoter Operator<br /> cAMP<br /> (b) No lactose or glucose (high cAMP)<br /> Repressor<br /> (inactive)<br /> (Inactive)<br /> Transcription<br /> is blocked<br /> by repressor.<br /> CAP site Promoter Operator<br /> (Inactive)<br /> (c) Lactose and glucose (low cAMP)<br /> CAP<br /> CAP site Promoter Operator<br /> cAMP<br /> CAP<br /> (a) Lactose, no glucose (high CAMP)<br /> CAP<br /> CAP<br /> CAP site Promoter Operator<br /> (d) Glucose, no lactose (low cAMP)<br /> Transcription occurs<br /> Binding of RNA polymerase<br /> to promoter is enhanced<br /> by CAP binding.<br /> Transcription<br /> is inhibited by<br /> lack of CAP binding.<br /> Repressor<br /> Transcription is inhibited<br /> by lack of CAP binding and<br /> presence of repressor.<br /> FIGURE 14.8 The roles of the lac repressor and catabolite<br /> activator protein (CAP) in the regulation of the lac operon. This<br /> figure illustrates how the lac operon will be regulated depending on its<br /> exposure to lactose or glucose.<br /> GenesgTraits The mechanism of catabolite repression provides the bacterium with<br /> the trait of being able to choose between two sugars. When exposed to both glucose<br /> and lactose, the bacterium chooses glucose first. After the glucose is used up, it then<br /> expresses the genes necessary for lactose metabolism. This trait allows the bacterium<br /> to more efficiently use sugars from its environment.<br /> The effect of glucose, called catabolite repression, may seem<br /> like a puzzling way to describe this process because this regulation<br /> involves the action of an inducer (cAMP) and an activator<br /> protein (CAP), not a repressor. The term was coined before the<br /> action of the cAMP-CAP complex was understood. At that time,<br /> the primary observation was that glucose (a catabolite) inhibited<br /> (repressed) lactose metabolism.<br /> Many bacterial promoters that transcribe genes involved in<br /> the breakdown of other sugars, such as maltose, arabinose, and<br /> melibiose, also have binding sites for CAP. Therefore, when glucose<br /> levels are high, these operons are inhibited, thereby promoting<br /> diauxic growth.<br /> Further Studies Have Revealed That the lac Operon<br /> Has Three Operator Sites for the lac Repressor<br /> Our traditional view of the regulation of the lac operon has been<br /> modified as we have gained a greater molecular understanding<br /> of the process. In particular, detailed genetic and crystallography<br /> studies have shown that the binding of the lac repressor is<br /> more complex than originally realized. The site in the lac operon<br /> that is commonly called the operator site was first identified by<br /> mutations that prevented lac repressor binding. These mutations,<br /> called lacO � or lacO C mutants, resulted in the constitutive<br /> expression of the lac operon even in strains that make a normal<br /> lac repressor protein. LacO C mutations were localized in the lac<br /> operator site, which is now known as O 1 . This led to the view<br /> that a single operator site was bound by the lac repressor to<br /> inhibit transcription, as in Figure 14.4.<br /> In the late 1970s and 1980s, two additional operator sites<br /> were identified. As shown at the top of Figure 14.9, these sites<br /> are called O 2 and O 3 . O 1 is the operator site slightly downstream<br /> from the promoter. O 2 is located farther downstream in the<br /> lacZ coding sequence, and O 3 is located slightly upstream from<br /> the promoter. The O 2 and O 3 operator sites were initially called<br /> pseudo-operators, because substantial repression occurred in the<br /> absence of either one of them. However, studies by Benno Müller-<br /> Hill and his colleagues revealed a surprising result. As shown in<br /> Figure 14.9 (fourth example down), if both O 2 and O 3 are missing,<br /> repression is dramatically reduced even when O 1 is present.<br /> When O 1 is missing, even in the presence of one of the other<br /> operator sites, repression is nearly abolished.<br /> How were these results interpreted? The data of Figure 14.9<br /> supported a hypothesis that the lac repressor must bind to O 1 and<br /> either O 2 or O 3 to cause full repression. According to this view,<br /> the lac repressor can readily bind to O 1 and O 2 , or to O 1 and O 3 ,<br /> but not to O 2 and O 3 . If either O 2 or O 3 were missing, maximal<br /> repression is not achieved because it is less likely for the repressor<br /> to bind when only two operator sites are present. If you take<br /> a look at Figure 14.9, you will notice that the operator sites are<br /> a fair distance away from each other. For this reason, it was proposed<br /> from these studies that the binding of the lac repressor to<br /> two operator sites requires the DNA to form a loop. A loop in the<br /> DNA would bring the operator sites closer together and thereby<br /> facilitate the binding of the repressor protein (Figure 14.10a).</div> <div>PREFACE<br /> ::<br /> In the third edition of Genetics: Analysis & Principles, the<br /> content has been updated to reflect current trends in the field. In<br /> addition, the presentation of the content has been improved in<br /> a way that fosters active learning. As an author, researcher, and<br /> teacher, I want a textbook that gets students actively involved<br /> in learning genetics. To achieve this goal, I have worked with a<br /> talented team of editors, illustrators, and media specialists who<br /> have helped me to make the third edition of Genetics: Analysis &<br /> Principles a fun learning tool. The features that we feel are most<br /> appealing to students, and which have been added to or improved<br /> upon in the third edition, are the following.<br /> • Interactive exercises Education specialists have crafted<br /> interactive exercises in which the students can make their<br /> own choices in problem-solving activities, and predict<br /> what the outcomes will be. These exercises often focus<br /> on inheritance patterns and human genetic diseases. (For<br /> example, see Chapters 4 and 22.)<br /> • Animations Our media specialists have created over<br /> 50 animations for a variety of genetic processes. These<br /> animations were made specifically for this textbook, and use<br /> the art from the textbook. The animations literally make<br /> many of the figures in the textbook “come to life.”<br /> • Experiments As in the previous editions, each chapter<br /> (beginning with Chapter 2) incorporates one or two experiments<br /> that are presented according to the scientific method.<br /> These experiments are not “boxed off” from the rest of<br /> the chapter. Rather, they are integrated within the chapters<br /> and flow with the rest of the text. As you are reading the<br /> experiments, you will simultaneously explore the scientific<br /> method and the genetic principles that have been discovered<br /> using this approach. For students, I hope this textbook will<br /> help you to see the fundamental connection between scientific<br /> analysis and principles. For both students and instructors,<br /> I expect that this strategy will make genetics much<br /> more fun to explore.<br /> • Art The art has been further refined for clarity and completeness.<br /> This makes it easier and more fun for students to<br /> study the illustrations without having to go back and forth<br /> between the art and the text.<br /> • Engaging text A strong effort has been made in the third<br /> edition to pepper the text with questions. Sometimes these<br /> are questions that scientists considered when they were conducting<br /> their research. Sometimes they are questions that<br /> the students might ask themselves when they are learning<br /> about genetics.<br /> Overall, an effective textbook needs to accomplish three<br /> goals. First, it needs to provide comprehensive, accurate, and upto-date<br /> content in its field. Second, it needs to provide students<br /> with an exposure to the techniques and skills that are needed for<br /> them to become successful in that field. And finally, it should<br /> inspire students so they want to pursue that field as a career. The<br /> hard work that has gone into the third edition of Genetics: Analysis<br /> & Principles has been aimed at achieving all three of these goals.<br /> HOW WE EVALUATED YOUR NEEDS<br /> ORGANIZATION<br /> In surveying many genetics instructors, it became apparent that<br /> most people fall into two camps: Mendel first versus Molecular<br /> first. I have taught genetics both ways. As a teaching tool, this<br /> textbook has been written with these different teaching strategies<br /> in mind. The organization and content lend themselves to various<br /> teaching formats.<br /> Chapters 2 through 8 are largely inheritance chapters, while<br /> Chapters 24 through 26 examine population and quantitative<br /> genetics. The bulk of the molecular genetics is found in Chapters 9<br /> through 23, although I have tried to weave a fair amount of molecular<br /> genetics into Chapters 2 through 8 as well. The information in<br /> ix</div> <div>ndhK<br /> 176 CHAPTER 7 :: NON-MENDELIAN INHERITANCE<br /> genes encode proteins that function within the mitochondrion.<br /> In addition, mtDNA carries genes that encode ribosomal RNA<br /> and transfer RNA. These rRNAs and tRNAs are necessary for the<br /> synthesis of the 13 proteins that are encoded by the mtDNA. The<br /> primary role of mitochondria is to provide cells with the bulk of<br /> their adenosine triphosphate (ATP), which is used as an energy<br /> source to drive cellular reactions. These 13 proteins function in<br /> a process known as oxidative phosphorylation, which enables the<br /> mitochondria to synthesize ATP. However, mitochondria require<br /> many additional proteins to carry out oxidative phosphorylation<br /> and other mitochondrial functions. Most mitochondrial proteins<br /> are encoded by genes within the cell nucleus. When these nuclear<br /> genes are expressed, the mitochondrial proteins are first synthesized<br /> outside the mitochondria in the cytosol of the cell and then<br /> transported into the mitochondria.<br /> Chloroplast genomes tend to be larger than mitochondrial<br /> genomes, and they have a correspondingly greater number<br /> of genes. A typical chloroplast genome is approximately 100,000<br /> to 200,000 bp in length, which is about 10 times larger than the<br /> mitochondrial genome of animal cells. Figure 7.15 shows the<br /> chloroplast DNA (cpDNA) of the tobacco plant, which is a circular<br /> DNA molecule that contains 156,000 bp of DNA and carries<br /> between 110 and 120 different genes. These genes encode<br /> ribosomal RNAs, transfer RNAs, and many proteins required for<br /> photosynthesis. As with mitochondria, many chloroplast proteins<br /> are encoded by genes found in the plant cell nucleus. These proteins<br /> contain chloroplast-targeting signals that direct them into<br /> the chloroplasts.<br /> Extranuclear Inheritance Produces<br /> Non-Mendelian Results in Reciprocal Crosses<br /> In diploid eukaryotic species, most genes within the nucleus obey<br /> a Mendelian pattern of inheritance because the homologous pairs<br /> of chromosomes segregate during gamete formation. Except for<br /> sex-linked traits, offspring inherit one copy of each gene from<br /> both the maternal and paternal parents. The sorting of chromosomes<br /> during meiosis explains the inheritance patterns of nuclear<br /> genes. By comparison, the inheritance of extranuclear genetic<br /> material does not display a Mendelian pattern. Mitochondria and<br /> chloroplasts are not sorted during meiosis and therefore do not<br /> segregate into gametes in the same way as nuclear chromosomes.<br /> In 1909, Carl Correns discovered a trait that showed a<br /> non-Mendelian pattern of inheritance involving pigmentation<br /> ORF2280<br /> psbH<br /> petB<br /> petD<br /> psbB<br /> ORF34<br /> rp133<br /> rps18<br /> rps8<br /> rp12<br /> rp123<br /> psaJ<br /> rp114<br /> rp116<br /> rps3<br /> rp122<br /> rps19<br /> trnI<br /> trnL<br /> petG<br /> rpoA<br /> rps11<br /> rp136<br /> ψinfA<br /> ndhB<br /> rps7<br /> ORF31<br /> clpP<br /> psbN<br /> ORF229<br /> petA<br /> 3′–rps12<br /> ORF184<br /> trnP<br /> rp120<br /> trnW<br /> 5′–rps12<br /> psaI<br /> psbF<br /> psbE<br /> accD<br /> psbL<br /> psbJ<br /> rbcL<br /> atpB<br /> trnM<br /> atpE<br /> trnV<br /> ndhC<br /> trnF<br /> ndhJ<br /> trnL<br /> trnT<br /> trnS<br /> rps4<br /> ORF168<br /> psaA<br /> psaB<br /> rps14<br /> trnfM<br /> trnS<br /> trnG<br /> trnE<br /> ORF62<br /> trnY<br /> psbC<br /> trnD<br /> psbD<br /> psbM<br /> ORF2280<br /> trnT<br /> rpoB<br /> rpoC1<br /> ORF29<br /> rpoC2<br /> trnC<br /> atpA<br /> trnS<br /> trnQ<br /> rps16<br /> trnK<br /> matK<br /> psbA<br /> trnH<br /> rps2<br /> atpI<br /> atpH<br /> atpF<br /> rpl2<br /> trnR<br /> trnG<br /> psbI<br /> psbK<br /> trnV<br /> 16S<br /> trnI<br /> trnA<br /> 23S<br /> 5S<br /> 4.5S<br /> trnR<br /> rpl32<br /> trnN<br /> trnL<br /> ndhF<br /> ORF313<br /> ndhD<br /> psaC<br /> ndhE<br /> ndhG<br /> ndhI<br /> ndhA<br /> ndhH<br /> rps15<br /> ORF1901<br /> trnR<br /> 5S<br /> trnN<br /> 4.5S<br /> 23S<br /> trnA<br /> trnI<br /> 16S<br /> trnV<br /> rps7<br /> ndhB<br /> trnL<br /> rpl23<br /> trnI<br /> FIGURE 7.15 A genetic map of tobacco chloroplast DNA (cpDNA). This diagram illustrates the locations of many genes along the circular<br /> chloroplast chromosome. The gene names shown in blue encode transfer RNAs. The genes that encode ribosomal RNA are shown in red. The<br /> remaining genes shown in black encode polypeptides that function within the chloroplast. The genes designated ORF (open reading frame) encode<br /> polypeptides with unknown functions.</div> <div>372 CHAPTER 14 :: GENE REGULATION IN BACTERIA AND BACTERIOPHAGES<br /> The ara Operon Can Be Regulated Positively<br /> or Negatively by the Same Regulatory Protein<br /> Now that we have considered the regulation of the lac operon,<br /> let’s compare its regulation to that of other genes in the bacterial<br /> chromosome. Another operon in E. coli involved in sugar metabolism<br /> is the ara (arabinose) operon. The sugar arabinose is a constituent<br /> of the cell walls of a few types of plants. As shown in Figure<br /> 14.11, the ara operon contains three structural genes, araB,<br /> araA, and araD, encoding a polycistronic mRNA for the three<br /> enzymes involved in arabinose metabolism. The actions of the<br /> three enzymes metabolize arabinose into D-xylulose-5-phosphate.<br /> Like the lac operon, the ara operon contains a single promoter,<br /> designated P BAD . The operon also contains a CAP site for<br /> the binding of the catabolite activator protein. The araC gene,<br /> which has its own promoter (P C ), is adjacent to the ara operon.<br /> AraC encodes a regulatory protein, called the AraC protein, that<br /> can bind to operator sites designated araI, araO 1 , and araO 2 .<br /> As we have seen with the lac operon, some regulatory proteins<br /> such as the lac repressor inhibit transcription, whereas others<br /> such as CAP turn on transcription. AraC is a rather interesting<br /> and unusual protein, because it can act as either a negative<br /> or positive regulator of transcription, depending on whether or<br /> not arabinose is present. How can a protein function in both<br /> ways? In the absence of arabinose, the AraC protein binds to the<br /> araI, araO 1 , and araO 2 operator sites (Figure 14.12a). An AraC<br /> protein dimer is bound to araO 1 , while monomers are bound at<br /> araO 2 and araI. The binding of the AraC proteins to the araO 1<br /> site inhibits the transcription of the araC gene. In other words,<br /> the AraC protein is a negative regulator of the araC gene. This<br /> keeps AraC protein levels fairly low. The AraC proteins bound at<br /> araO 2 and araI repress the ara operon, but only in the absence of<br /> arabinose. As shown in Figure 14.12a, the AraC proteins at araO 2<br /> and araI can bind to each other by causing a loop in the DNA, as<br /> originally proposed by Robert Schleif and his colleagues in 1990.<br /> This DNA loop prevents RNA polymerase from binding to the<br /> DNA and transcribing the ara operon via P BAD . Therefore, in the<br /> absence of arabinose, the ara operon is turned off.<br /> Figure 14.12b illustrates the activation of the ara operon in<br /> the presence of arabinose. When arabinose is bound to the AraC<br /> protein, the interaction between the AraC proteins at the araO 2<br /> and araI sites is broken. This opens the DNA loop. In addition,<br /> a second AraC protein binds at the araI site. This AraC dimer at<br /> the araI operator site activates transcription by directly interacting<br /> with RNA polymerase. This activation can occur in conjunction<br /> with the activation of the ara operon by CAP and cAMP<br /> if glucose levels are low. When the ara operon is activated, the<br /> bacterial cell can efficiently metabolize arabinose.<br /> The trp Operon Is Regulated by a Repressor<br /> Protein and Also by Attenuation<br /> The trp operon (pronounced “trip”) encodes enzymes that are<br /> needed for the biosynthesis of the amino acid tryptophan. The<br /> trpE, trpD, trpC, trpB, and trpA genes encode enzymes involved<br /> in tryptophan biosynthesis. The trpR and trpL genes are involved<br /> in regulating the trp operon in two different ways. The trpR gene<br /> encodes the trp repressor protein. When tryptophan levels within<br /> the cell are very low, the trp repressor cannot bind to the operator<br /> site. Under these conditions, RNA polymerase transcribes the<br /> trp operon (Figure 14.13a). In this way, the cell expresses the<br /> genes required for the synthesis of tryptophan. When the tryptophan<br /> levels within the cell become high, tryptophan acts as a<br /> corepressor that binds to the trp repressor protein. This causes a<br /> conformational change in the trp repressor that allows it to bind<br /> to the trp operator site (Figure 14.13b). This inhibits the ability<br /> of RNA polymerase to transcribe the operon. Therefore, when<br /> a high level of tryptophan is present within the cell—when the<br /> cell does not need to make more tryptophan—the trp operon is<br /> turned off.<br /> Transcription<br /> araC<br /> cAMP<br /> CAP<br /> ara operon<br /> Transcription<br /> araO 2 PC araO araB araA<br /> 1 CAP site araI araD<br /> P BAD<br /> Polycistronic mRNA<br /> araC<br /> araB<br /> araA<br /> araD<br /> AraA AraB AraD<br /> Arabinose L-Ribulose L-Ribulose-<br /> 5-P<br /> D-Xylulose-<br /> 5-P<br /> FIGURE 14.11 Organization of the ara operon and other genes involved in arabinose metabolism. AraC encodes a genetic regulatory<br /> protein called the AraC protein. This protein can bind to three different regulatory sites, called araI, araO 1 , and araO 2 . P C is the promoter for the araC<br /> gene; P BAD is the promoter for the ara operon, which contains three genes (araB, araA, and araD) encoding proteins involved in arabinose metabolism.</div> <div>21.3 BIOINFORMATICS 593<br /> itself (100%). The next nine sequences are in order of similarity.<br /> The next most similar sequence is from the orangutan (99%), a<br /> close relative of humans. This is followed by two mammals, the<br /> mouse and rat, and then five vertebrates that are not mammals.<br /> The tenth best match is from Drosophila, an invertebrate.<br /> You can see two trends in Table 21.5. First, the order of<br /> the matches follows the evolutionary relatedness of the various<br /> species to humans. The similarity between any two sequences<br /> is related to the time that has passed since they diverged from<br /> a common ancestor. Among the species listed in this table, the<br /> human sequence is most similar to the orangutan, a closely<br /> related primate. The next most similar sequences are found in<br /> other mammals, followed by other vertebrates, and finally invertebrates.<br /> A second trend you may have noticed is that several of<br /> the matches involve species that are important from a research,<br /> medical, or agricultural perspective. Currently, our genetic databases<br /> are biased toward organisms that are of interest to humans,<br /> particularly model organisms such as mice and Drosophila. Over<br /> the next several decades, the sequencing of genomes from many<br /> different species will tend to lessen this bias.<br /> The results shown in Table 21.5 illustrate the remarkable<br /> computational abilities of current computer technology. In<br /> minutes, the human phenylalanine hydroxylase sequence can be<br /> compared to hundreds of thousands of different sequences.<br /> Genetic Sequences Can Be Used to Predict<br /> the Structure of RNA and Proteins<br /> Another topic in which bioinformatics has impacted functional<br /> genomics and proteomics is the area of structure prediction.<br /> The function of macromolecules such as DNA, RNA, and proteins<br /> relies on their three-dimensional structure, which, in turn,<br /> depends on the linear sequences of their building blocks. In the<br /> case of DNA and RNA, this means a linear sequence of nucleotides;<br /> proteins are composed of a linear sequence of amino acids.<br /> Currently, the three-dimensional structure of macromolecules is<br /> determined primarily through the use of biophysical techniques<br /> such as X-ray crystallography and nuclear magnetic resonance<br /> (NMR). These methods are technically difficult and very timeconsuming.<br /> DNA sequencing, by comparison, requires much<br /> less effort. Therefore, because the three-dimensional structure<br /> of macromolecules depends ultimately on the linear sequence of<br /> their building blocks, it would be far easier if we could predict<br /> the structure (and function) of DNAs, RNAs, and proteins from<br /> their sequence of building blocks.<br /> RNA molecules typically are folded into a secondary structure,<br /> which commonly contains double-stranded regions. This<br /> secondary structure is further folded and twisted to adopt a tertiary<br /> conformation. Such structural features of RNA molecules<br /> are functionally important. For example, the folding of RNA into<br /> secondary structures, such as stem-loops, affects transcriptional<br /> termination and other regulatory events. Therefore, geneticists<br /> are interested in the secondary and tertiary structures that RNA<br /> molecules can adopt.<br /> Many approaches are available for investigating RNA structure.<br /> In addition to biophysical and biochemical techniques,<br /> computer modeling of RNA structure has become an important<br /> tool. Modeling programs can consider different types of information.<br /> For example, the known characteristics of RNA secondary<br /> structure, such as the ability to form double-stranded regions,<br /> can provide parameters for use in a modeling program.<br /> A comparative approach can also be used in RNA structure<br /> prediction. This method assumes that RNAs of similar function<br /> and sequence have a similar structure. For example, the genes that<br /> encode certain types of RNAs, such as the 16S rRNA that makes<br /> up most of the small ribosomal subunit, have been sequenced<br /> from many different species. Among different species, the 16S<br /> rRNAs have similar but not identical sequences. Computer programs<br /> can compare many different 16S rRNA sequences to aid<br /> in the prediction of secondary structure. Figure 21.12 illustrates<br /> a secondary structural model for 16S rRNA from E. coli based<br /> on a comparative sequence analysis of many bacterial 16S rRNA<br /> sequences. This large RNA contains 45 stem-loop regions. As<br /> you can imagine, it would be rather difficult to deduce such a<br /> model without the aid of a computer! In addition, RNA secondary<br /> structure prediction may be aimed at predicting the lowest<br /> energy state of a folded molecule. This approach, called free<br /> energy minimization, is also related to the base sequence of an<br /> RNA molecule.<br /> Structure prediction is also used in the area of proteomics.<br /> As described in Chapter 12, proteins contain repeating secondary<br /> structural patterns known as α helices and β sheets. Several<br /> FIGURE 21.12<br /> rRNA.<br /> 5′ end<br /> 3′ end<br /> A secondary structural model for E. coli 16S</div> <div>CHAPTER 26 807<br /> E8. The technique of PCR is used to amplify the amount of DNA in a<br /> sample. To accomplish this, one must use oligonucleotide primers that<br /> are complementary to the region that is to be amplified. For example,<br /> as described in the experiment of Figure 26.15, PCR primers that<br /> were complementary to and flank the 12S rRNA gene can be used to<br /> amplify the 12S rRNA gene. The technique of PCR is described in<br /> Chapter 18.<br /> E10. We would expect the probe to hybridize to the natural G. tetrahit and<br /> also the artificial G. tetrahit, because both of these strains contain two<br /> sets of chromosomes from G. pubescens. We would expect two bright<br /> spots in the in situ experiment. Depending on how closely related G.<br /> pubescens and G. speciosa are, the probe may also hybridize to two sites<br /> in the G. speciosa genome, but this is difficult to predict a priori. If so,<br /> the G. tetrahit species would show four spots.<br /> E12. The principle of parsimony chooses a phylogenetic tree that requires<br /> the fewest number of evolutionary changes. When using molecular<br /> data, researchers can use computer programs that compare DNA<br /> sequences from homologous genes of different species and construct a<br /> tree that requires the fewest numbers of mutations. Such a tree is the<br /> most likely pathway for the evolution of such species.<br /> E14. Possibly, the mouse would have an eye at the tip of its tail!<br /> Questions for Student Discussion/Collaboration<br /> 2. The founder effect and allotetraploidy are examples of rapid forms<br /> of evolution. In addition, some single gene mutations may have a<br /> great impact on phenotype and lead to the rapid evolution of new<br /> species by cladogenesis. Geological processes may promote the slower<br /> accumulation of alleles and alter a species’ characteristics more<br /> gradually. In this case, it is the accumulation of many phenotypically<br /> minor genetic changes that ultimately leads to reproductive isolation.<br /> Slow and fast mechanisms of evolution have the common theme<br /> that they result in reproductive isolation. This is a prerequisite for<br /> the evolution of new species. Fast mechanisms tend to involve small<br /> populations and a few number of genetic changes. Slower mechanisms<br /> may involve larger populations and involve the accumulation of a large<br /> number of genetic changes that each contributes in a small way.</div> <div>276 CHAPTER 11 :: DNA REPLICATION<br /> E. coli<br /> chromosome<br /> oriC<br /> AT-rich region<br /> 5′– GGATCCTGGGTATTAAAAAGAAGATCTATTTATTTAGAGATCTGTTCTAT<br /> CCTAGGACCCAT<br /> AATTTTTCTTCTAGATAAATAAATCTCTAGACAAGATA<br /> 1<br /> DnaA box<br /> 50<br /> TGTGATCTCTTATTAGGATCGCACTGCCCTGTGGATAACAAGGATCGG<br /> CT<br /> ACACTAGAGAAT<br /> AATCCTAGCGTGACGGGACACCT<br /> ATTGTTCCTAGCC<br /> GA<br /> 51 DnaA box<br /> 100<br /> TTTAAGATCAACAACCTGGAAAGGATCATTAACTGTGAATGATCGGTGAT<br /> AAATTCT<br /> AGT<br /> TGTTGGACCTTTCCTAGTAATTGACACTTACT<br /> AGCCAC<br /> TA<br /> 101 DnaA box<br /> 150<br /> CCTGGACCGTATAAGCTGGGA<br /> TCAGAATGAGGGTTATACACAGCTCAAAA<br /> GGACCT<br /> GGCATATTCGACCCTAGTCTTACTCCCAAT<br /> ATGTGTCGAGTT<br /> TT<br /> 151 DnaA box<br /> 200<br /> ACTGAACAACGGTTGTTCTTTGGATAACTACCGGT<br /> TGATCCAAGCTTCCT<br /> TGACTTGTTGCC<br /> AACAAGAAACCT<br /> ATTGATGGCCAACTAGGT<br /> TCGAAGGA<br /> 201 DnaA box<br /> 250<br /> GACAGAGTTATCCACAGTAGATCGC–3′<br /> CTGTCTCAATAGGTGTCATCTAGCG<br /> 251 275<br /> 5′ 3′<br /> 3′ 5′<br /> AT-rich region<br /> DnaA boxes<br /> DnaA proteins bind to DnaA boxes and to<br /> each other. Additional proteins that cause<br /> the DNA to bend also bind (not shown).<br /> This causes the region to wrap around<br /> the DnaA proteins and separates the<br /> AT-rich region.<br /> FIGURE 11.5 The sequence<br /> of oriC in E. coli. The AT-rich region is<br /> composed of three tandem repeats that are<br /> 13 base pairs long and highlighted in blue.<br /> The five DnaA boxes are highlighted in<br /> orange. The GATC methylation sites<br /> are underlined.<br /> it breaks the hydrogen bonds between the two strands, thereby<br /> generating two single strands. Two DNA helicases begin strand<br /> separation within the oriC region and continue to separate the<br /> DNA strands beyond the origin. These enzymes use the energy<br /> from ATP hydrolysis to catalyze the separation of the doublestranded<br /> parental DNA. In E. coli, DNA helicases bind to singlestranded<br /> DNA and travel along the DNA in a 5' to 3' direction to<br /> keep the replication fork moving. As shown in Figure 11.6, the<br /> action of DNA helicases promotes the movement of two replication<br /> forks outward from oriC in opposite directions. This initiates<br /> the replication of the bacterial chromosome in both directions,<br /> an event termed bidirectional replication.<br /> 5′<br /> 3′<br /> Helicase<br /> 5′<br /> 3′<br /> ATrich<br /> region<br /> DnaA protein<br /> 3′<br /> 5′<br /> DNA helicase (DnaB protein) binds to the<br /> origin. DnaC protein (not shown) assists<br /> this process.<br /> 3′<br /> 5′<br /> Several Proteins Are Required for DNA Replication<br /> at the Replication Fork<br /> Figure 11.7 provides an overview of the molecular events that<br /> occur as one of the two replication forks moves around the bacterial<br /> chromosome, and Table 11.1 summarizes the functions<br /> of the major proteins involved in E. coli DNA replication. Let’s<br /> begin with strand separation. To act as a template for DNA replication,<br /> the strands of a double helix must separate. As mentioned<br /> previously, the function of DNA helicase is to break the hydrogen<br /> bonds between base pairs and thereby unwind the strands; this<br /> action generates positive supercoiling ahead of each replication<br /> 5′<br /> 3′<br /> Fork<br /> DNA helicase separates the DNA in both<br /> directions, creating 2 replication forks.<br /> Fork<br /> 3′<br /> 5′<br /> FIGURE 11.6 The events that occur at oriC to initiate<br /> the DNA replication process. To initiate DNA replication,<br /> DnaA proteins bind to the five DnaA boxes, which causes the<br /> DNA strands to separate at the AT-rich region. DnaA and DnaC proteins then<br /> recruit DNA helicase (DnaB) into this region. Each DNA helicase is composed<br /> of six subunits, which form a ring around one DNA strand and migrates in<br /> the 5' to 3' direction. As shown here, the movement of two DNA helicase<br /> proteins serves to separate the DNA strands beyond the oriC region.</div> <div>124 CHAPTER 5 :: LINKAGE AND GENETIC MAPPING IN EUKARYOTES<br /> can also yield a tetratype. Therefore, the total number of tetratypes<br /> overestimates the true number of single crossovers. Fortunately, we<br /> can compensate for this overestimation. Because two types of tetratypes<br /> are due to a double crossover, the actual number of tetratypes<br /> arising from a double crossover should equal 2NPD. Therefore, the<br /> true number of single crossovers is calculated as T�2NPD.<br /> Now we have accurate measures of both single and double<br /> crossovers. The number of single crossovers equals T�2NPD,<br /> and the number of double crossovers equals 4NPD. We can substitute<br /> these values into our previous equation.<br /> Map distance 5<br /> (T 2 2NPD) 1 (2)(4NPD)<br /> Total number of asci<br /> 3 0.5 3 100<br /> T 1 6NPD<br /> 5<br /> 3 0.5 3 100<br /> Total number of asci<br /> This equation provides a more accurate measure of map distance<br /> because it considers both single and double crossovers.<br /> CONCEPTUAL SUMMARY<br /> Linkage refers to the phenomenon that many different genes<br /> may be located on the same chromosome. Chromosomes are<br /> sometimes called linkage groups because they contain a group of<br /> linked genes. Linkage affects the pattern of inheritance, because<br /> closely linked genes do not assort independently during meiosis.<br /> This produces a greater percentage of offspring that display<br /> parental phenotypes. Nevertheless, nonparental offspring can be<br /> produced as a result of crossing over.<br /> The likelihood of crossing over depends on the distance<br /> between two genes. If two genes are far apart from each other on<br /> the same chromosome, it is more likely that crossing over will<br /> occur between them. Therefore, when two genes are widely separated,<br /> a substantial percentage of recombinant offspring will be<br /> obtained from a testcross. However, the percentage of recombinant<br /> offspring cannot exceed a value of 50%, even when two genes are<br /> more than 50 map units apart on the same chromosome. The relationship<br /> between the percentage of recombinant offspring and the<br /> linear distance between genes is the basis for genetic mapping.<br /> EXPERIMENTAL SUMMARY<br /> Experimentally, the phenomenon of linkage was deduced from<br /> genetic crosses. Bateson and Punnett were the first scientists to<br /> notice that certain genes do not assort independently. Morgan<br /> conducted crosses involving X-linked traits in fruit flies and correctly<br /> proposed that linkage is due to the location of particular<br /> genes on the same chromosome. He also hypothesized that<br /> recombinant phenotypes occur because of crossing over during<br /> meiosis. Morgan realized that the likelihood of crossing over<br /> depends on the distance between two genes. The hypothesis that<br /> crossing over can result in genetic recombination was confirmed<br /> cytologically by the studies of Creighton and McClintock, which<br /> showed that the production of recombinant offspring correlates<br /> with the physical exchange of material between chromosomes.<br /> Genetic mapping is the determination of gene order and<br /> distance along chromosomes. In this chapter, we have considered<br /> how testcrosses are conducted as a method to map genes.<br /> Sturtevant was the first person to understand that the percentage<br /> of recombinant offspring in a testcross could be used as a<br /> measure of the relative distance between two genes. Map distance<br /> is computed as the number of recombinant offspring divided by<br /> the total number of offspring multiplied by 100. This approach<br /> can be readily applied to map genes using dihybrid and trihybrid<br /> testcrosses. Genetic mapping is most accurate when map<br /> distances are calculated between closely linked genes. As the map<br /> distance approaches 50 map units (mu) and above, the percentage<br /> of observed recombinant offspring is not a reliable measure<br /> of map distance. In Chapter 20, several molecular methods of<br /> genetic mapping are described.<br /> The chapter ended with a discussion of gene mapping<br /> methods in fungi. A group of fungi known as the ascomycetes<br /> have been extensively used in genetic studies, because all products<br /> of a single meiosis are enclosed within an ascus. For fungi such as<br /> Neurospora that make an ordered octad, the spores are arranged<br /> in a manner that reflects their relationship to each other during<br /> meiosis and mitosis. Ordered octads can be analyzed to map the<br /> location of a single gene relative to the centromere. Fungal species<br /> and unicellular algae that produce unordered asci have also<br /> been used in mapping studies. In yeast, for example, dihybrid<br /> crosses are made, and the distance between the two genes can be<br /> computed by determining the proportions of parental ditypes,<br /> tetratypes, and nonparental ditypes.<br /> In Chapter 6, we will consider the linkage of genes within<br /> bacterial chromosomes and bacteriophages. Although bacteria<br /> normally reproduce asexually, they still can transfer genetic material<br /> by various mechanisms. As we will see, these mechanisms also<br /> provide a way to map genes along the bacterial chromosome.</div> <div>192 CHAPTER 8 :: VARIATION IN CHROMOSOME STRUCTURE AND NUMBER<br /> because they provide raw material for the addition of more genes<br /> into a species’ chromosomes. Over the course of many generations,<br /> this can lead to the formation of a gene family consisting<br /> of two or more genes that are similar to each other. As shown in<br /> Figure 8.6, the members of a gene family are derived from the<br /> same ancestral gene. Over time, two copies of an ancestral gene<br /> Gene<br /> Gene<br /> Gene<br /> Paralogs (homologous genes)<br /> Abnormal genetic event that<br /> causes a gene duplication<br /> Gene<br /> Over the course of many generations,<br /> the 2 genes may differ due to the<br /> gradual accumulation of DNA<br /> mutations.<br /> Gene<br /> Mutation<br /> FIGURE 8.6 Gene duplication and the evolution<br /> of paralogs. An abnormal crossover event like the one<br /> described in Figure 8.5 leads to a gene duplication. Over<br /> time, each gene accumulates different mutations.<br /> can accumulate different mutations. Therefore, after many generations,<br /> the two genes will be similar but not identical. During<br /> evolution, this type of event can occur several times, creating a<br /> family of many similar genes.<br /> When two or more genes are derived from a single ancestral<br /> gene, the genes are said to be homologous. Homologous<br /> genes within a single species are called paralogs and constitute<br /> a gene family. A well-studied example of a gene family is shown<br /> in Figure 8.7, which illustrates the evolution of the globin gene<br /> family found in humans. The globin genes encode polypeptides<br /> that are subunits of proteins that function in oxygen binding. For<br /> example, hemoglobin is a protein found in red blood cells; its<br /> function is to carry oxygen throughout the body. The globin gene<br /> family is composed of 14 paralogs that were originally derived<br /> from a single ancestral globin gene. According to an evolutionary<br /> analysis, the ancestral globin gene first duplicated about 500 million<br /> years ago and became separate genes encoding myoglobin<br /> and the hemoglobin group of genes. The primordial hemoglobin<br /> gene duplicated into an a-chain gene and a b-chain gene, which<br /> subsequently duplicated to produce several genes located on<br /> chromosomes 16 and 11, respectively. Currently, 14 globin genes<br /> are found on three different human chromosomes.<br /> Why is it advantageous to have a family of globin genes?<br /> Although all globin polypeptides are subunits of proteins that<br /> play a role in oxygen binding, the accumulation of different<br /> mutations in the various family members has produced globins<br /> that are more specialized in their function. For example, myoglobin<br /> is better at binding and storing oxygen in muscle cells, and<br /> the hemoglobins are better at binding and transporting oxygen<br /> Mb ζ ψ ζ ψ α2<br /> ψ α1 α 2 φ<br /> ε γ G γ A ψ β δ β<br /> α 1<br /> 0<br /> 200<br /> α chains<br /> β chains<br /> Millions of years ago<br /> 400<br /> 600<br /> Myoglobins<br /> Hemoglobins<br /> 800<br /> Ancestral globin<br /> 1,000<br /> FIGURE 8.7 The evolution of the globin gene family in humans. The globin gene family evolved from a single ancestral globin gene. The first<br /> gene duplication produced two genes that accumulated mutations and became the genes encoding myoglobin (on chromosome 22) and the group of<br /> hemoglobins. The primordial hemoglobin gene then duplicated to produce several a-chain and b-chain genes, which are found on chromosomes 16<br /> and 11, respectively. The four genes shown in gray are nonfunctional pseudogenes.</div> <div>78 CHAPTER 4 :: EXTENSIONS OF MENDELIAN INHERITANCE<br /> (unaffected, malaria resistant) and 1/4 are Hb S Hb S (sickle-cell<br /> disease). This 1:2:1 ratio deviates from a simple Mendelian 3:1<br /> phenotypic ratio.<br /> Overdominance is usually due to two alleles that produce<br /> proteins with slightly different amino acid sequences. How can<br /> we explain the observation that two protein variants in the heterozygote<br /> produce a more favorable phenotype? There are three<br /> common explanations. In the case of sickle-cell disease, the<br /> phenotype is related to the infectivity of Plasmodium (Figure<br /> A1A1<br /> Normal homozygote<br /> (sensitive to infection)<br /> (a) Disease resistance<br /> 27°–32°C<br /> (optimum<br /> temperature<br /> range)<br /> Pathogen can<br /> successfully<br /> propagate.<br /> A1 A1 A2 A2 A1<br /> (b) Homodimer formation<br /> E1<br /> E2<br /> 30°–37°C<br /> (optimum<br /> temperature<br /> range)<br /> (c) Variation in functional activity<br /> A1A2<br /> Heterozygote<br /> (resistant to infection)<br /> A2<br /> Pathogen<br /> cannot<br /> successfully<br /> propagate.<br /> FIGURE 4.8 Three possible explanations for overdominance<br /> at the molecular level. (a) The successful infection of cells by certain<br /> microorganisms depends on the function of particular cellular proteins.<br /> In this example, functional differences between A1A1 and A1A2 proteins<br /> affect the ability of a pathogen to propagate in the cells. (b) Some proteins<br /> function as homodimers. In this example, a gene exists in two alleles<br /> designated A1 and A2, which encode polypeptides also designated A1<br /> and A2. The homozygotes that are A1A1 or A2A2 will make homodimers<br /> that are A1A1 and A2A2, respectively. The A1A2 heterozygote can make<br /> A1A1 and A2A2 and can also make A1A2 homodimers, which may have<br /> better functional activity. (c) In this example, a gene exists in two alleles<br /> designated E1 and E2. The E1 allele encodes an enzyme that functions<br /> well in the temperature range of 27° to 32°C. E2 encodes an enzyme<br /> that functions in the range of 30° to 37°C. A heterozygote, E1E2, would<br /> produce both enzymes and have a broader temperature range (i.e., 27° to<br /> 37°C) in which the enzyme would function.<br /> 4.8a). In the heterozygote, the infectious agent is less likely to<br /> propagate within red blood cells. Interestingly, researchers have<br /> speculated that other alleles in humans may confer disease resistance<br /> in the heterozygous condition but are detrimental in the<br /> homozygous state. These include PKU, in which the heterozygous<br /> fetus may be resistant to miscarriage caused by a fungal<br /> toxin, and Tay-Sachs disease, in which the heterozygote may be<br /> resistant to tuberculosis.<br /> A second way to explain overdominance is related to the<br /> subunit composition of proteins. In some cases, proteins function<br /> as a complex of multiple subunits; each subunit is composed<br /> of one polypeptide. A protein composed of two subunits<br /> is called a dimer. When both subunits are encoded by the same<br /> gene, the protein is a homodimer. The prefix homo- means that<br /> the subunits come from the same type of gene although the gene<br /> may exist in different alleles. Figure 4.8b considers a situation in<br /> which a gene exists in two alleles that encode polypeptides designated<br /> A1 and A2. Homozygous individuals can produce only<br /> A1A1 or A2A2 homodimers, whereas a heterozygote can also<br /> produce an A1A2 homodimer. Thus, heterozygotes can produce<br /> three forms of the homodimer, homozygotes only one. For some<br /> proteins, A1A2 homodimers may have better functional activity<br /> because they are more stable or able to function under a wider<br /> range of conditions. The greater activity of the homodimer protein<br /> may be the underlying reason why a heterozygote has characteristics<br /> superior to either homozygote.<br /> A third molecular explanation of overdominance is that<br /> the proteins encoded by each allele exhibit differences in their<br /> functional activity. For example, suppose that a gene encodes<br /> a metabolic enzyme that can be found in two forms (corresponding<br /> to the two alleles), one that functions better at a<br /> lower temperature and the other that functions optimally at<br /> a higher temperature (Figure 4.8c). The heterozygote, which<br /> makes a mixture of both enzymes, may be at an advantage<br /> under a wider temperature range than either of the corresponding<br /> homozygotes.<br /> Before ending this topic, let’s also compare overdominance<br /> with a related phenomenon. Among plant and animal breeders,<br /> a common mating strategy is to begin with two different highly<br /> inbred strains and cross them together to produce hybrids.<br /> When the hybrids display traits that are superior to both corresponding<br /> parental strains, the outcome is known as heterosis,<br /> or hybrid vigor. Within the field of agriculture, heterosis has<br /> been particularly valuable in improving quantitative traits such<br /> as size, weight, growth rate, and disease resistance. It was first<br /> described by George Shull in crosses involving different strains<br /> of corn. Heterosis differs from overdominance, because the<br /> hybrid may be heterozygous for many genes, not just a single<br /> gene. Some of the beneficial effects of heterosis may be caused<br /> by the occurrence of overdominance in one or more heterozygous<br /> genes. However, as we will see in Chapter 25, heterosis can<br /> also result from the masking of deleterious recessive alleles that<br /> tend to accumulate in highly inbred domesticated strains.</div> <div>13.4 STAGES OF TRANSLATION 349<br /> 3′ AUUC CCAC U A G<br /> C U<br /> 16S rRNA<br /> mRNA<br /> 5′ AUCUAGUAAGGAGGUUGUAUGGUUCAGCGC A CG C AG 3′<br /> Shine-Dalgarno<br /> sequence<br /> Start<br /> codon<br /> FIGURE 13.19 The locations of the<br /> Shine-Dalgarno sequence and the start codon in<br /> bacterial mRNA. The Shine-Dalgarno sequence<br /> is complementary to a sequence in the 16S rRNA.<br /> It hydrogen bonds with the 16S rRNA to promote<br /> initiation. The start codon is typically a few nucleotides<br /> downstream from the Shine-Dalgarno sequence.<br /> In eukaryotes, the assembly of the initiation complex bears<br /> similarities to that in bacteria. However, as described in Table<br /> 13.7, additional factors are required for the initiation process.<br /> Note that the initiation factors are designated eIF (eukaryotic Initiation<br /> Factor) to distinguish them from bacterial initiation factors.<br /> The initiator tRNA in eukaryotes carries methionine rather<br /> than formylmethionine, as in bacteria. A eukaryotic initiation<br /> factor, eIF2, binds directly to tRNA met to recruit it to the 40S subunit.<br /> Eukaryotic mRNAs do not have a Shine-Dalgarno sequence.<br /> How then are eukaryotic mRNAs recognized by the ribosome?<br /> Several initiation factors (CBPI, eIF4A, eIF4B, eIF4F, and others)<br /> bind to the mRNA. Cap-binding protein I (CBPI) recognizes the<br /> 7-methylguanosine cap structure and aids in the recruitment of<br /> the other initiation factors. Collectively, these initiation factors<br /> unwind any secondary structure in the mRNA and promote its<br /> binding to the ribosome.<br /> The identification of the correct AUG start codon in<br /> eukaryotes differs greatly from that in bacteria. After the initial<br /> binding of mRNA to the ribosome, the next step is to locate an<br /> AUG start codon that is somewhere downstream from the 5' cap<br /> structure. In 1986, Marilyn Kozak proposed that the ribosome<br /> begins at the 5' end and then scans along the mRNA in the 3'<br /> direction in search of an AUG start codon. In many, but not all,<br /> cases, the ribosome uses the first AUG codon that it encounters<br /> as a start codon. When a start codon is identified, the 60S subunit<br /> assembles with the aid of eIF5.<br /> By analyzing the sequences of many eukaryotic mRNAs,<br /> researchers have found that not all AUG codons near the 5' end<br /> of mRNA can function as start codons. In some cases, the scanning<br /> ribosome passes over the first AUG codon and chooses an<br /> AUG farther down the mRNA. The sequence of bases around the<br /> AUG codon plays an important role in determining whether or<br /> not it will be selected as the start codon by a scanning ribosome.<br /> The consensus sequence for optimal start codon recognition is<br /> shown here.<br /> Start<br /> Codon<br /> G C C (A/G) C C A U G G<br /> –6 –5 –4 –3 –2 –1 +1 +2 +3 +4<br /> Aside from an AUG codon itself, a guanine at the +4 position<br /> and a purine, preferably an adenine, at the –3 position are the<br /> most important sites for start codon selection. These rules for<br /> optimal translation initiation are called Kozak’s rules.<br /> TABLE 13.7<br /> A Comparison of Translational Protein Factors in Bacteria<br /> and Eukaryotes<br /> Bacterial Factors Eukaryotic Factors Function<br /> Initiation Factors<br /> IF1 eIF2 Involved in forming the<br /> initiation complex<br /> IF2 eIF3 Involved in forming the<br /> initiation complex<br /> IF3 eIF4C Involved in forming the<br /> initiation complex<br /> CBPI Binds to the 7-<br /> methylguanosine cap<br /> eIF4A, eIF4B, eIF4F<br /> eIF5<br /> eIF6<br /> Involved in the search for a<br /> start codon<br /> Involved in forming the<br /> initiation complex; helps<br /> dissociate eIF2, eIF3, and<br /> eIF4C<br /> Helps dissociate 60S subunit<br /> from inactive ribosomes<br /> Elongation Factors<br /> EF-Tu eEF1a Binding of tRNAs to the A<br /> site<br /> EF-Ts eEF1bg Recycling factor for EF-Tu<br /> and eEF1a, respectively.<br /> EF-G eEF2 Required for translocation<br /> Release Factors<br /> RF1 eRF Recognizes stop codon and<br /> promotes termination<br /> RF2<br /> RF3<br /> Polypeptide Synthesis Occurs<br /> During the Elongation Stage<br /> During the elongation stage of translation, amino acids are<br /> added, one at a time, to the polypeptide chain (Figure 13.20).<br /> Even though this process is rather complex, it occurs at a remarkable<br /> rate. Under normal cellular conditions, a polypeptide chain</div> <div>19.4 HUMAN GENE THERAPY 535<br /> are producing transgenic strains that carry more than one toxin<br /> gene, which makes it more difficult for insect resistance to arise.<br /> Despite these and other concerns, many farmers are embracing<br /> transgenic crops, and their use continues to rise.<br /> 19.4 HUMAN GENE THERAPY<br /> Throughout this textbook, we have considered examples in<br /> which mutant genes cause human diseases. As discussed in<br /> Chapter 22, these include rare inherited disorders and more<br /> common diseases such as cancer. Because mutant genes cause<br /> disease, geneticists are actively pursuing the goal of using normal,<br /> cloned genes to compensate for defects in mutant genes.<br /> Gene therapy is the introduction of cloned genes into living cells<br /> in the treatment of disease. It is a potential method of treating a<br /> wide variety of illnesses.<br /> Many current research efforts in gene therapy are aimed<br /> at alleviating inherited human diseases. Over 4,000 human<br /> genetic diseases are known to involve a single gene abnormality.<br /> Familiar examples include cystic fibrosis, sickle-cell disease, and<br /> hemophilia. In addition, gene therapies have also been aimed<br /> at treating diseases such as cancer and cardiovascular disease,<br /> which may occur later in life. Some scientists are even pursuing<br /> research that will use gene therapy to combat infectious diseases<br /> such as AIDS. Even though gene therapy is still at an early stage<br /> of development, a large amount of research has already been<br /> conducted. Unfortunately, success has been limited, and relatively<br /> few patients have been treated with gene therapy. Nevertheless,<br /> a few results have been somewhat promising. Table 19.6<br /> describes several types of diseases that are being investigated as<br /> potential targets for gene therapy. In this section, we will examine<br /> the approaches to gene therapy and how it may be used to<br /> treat human disease.<br /> TABLE 19.2<br /> TABLE 19.1 16.6<br /> TABLE 19.6 18.1<br /> Future Prospects in Gene Therapy<br /> Type of Disease<br /> Blood<br /> Metabolic<br /> Treatment of<br /> Sickle-cell disease, hemophilia, severe combined<br /> immunodeficiency disease (SCID)<br /> Glycogen storage diseases, lysosomal storage diseases,<br /> and phenylketonuria<br /> Gene Therapy Involves the Introduction<br /> of Cloned Genes into Human Cells<br /> A key step in gene therapy is the introduction of a cloned gene<br /> into the cells of people. The techniques to transfer a cloned gene<br /> into human cells can be categorized as nonviral and viral gene<br /> transfer methods. The most common nonviral technique involves<br /> the use of liposomes, which are lipid vesicles (Figure 19.20a).<br /> The DNA containing the gene of interest is complexed with liposomes<br /> that carry a positive charge (i.e., cationic liposomes). The<br /> DNA-liposome complexes are taken into cells via endocytosis, in<br /> which a portion of the plasma membrane invaginates and creates<br /> an intracellular vesicle known as an endosome; the liposome is<br /> degraded within the endosome. The DNA is then released into<br /> the cytosol, imported into the nucleus, and then integrated into<br /> a chromosome of the target cell. An advantage of gene transfer<br /> via liposomes is that the liposomes do not elicit an immune<br /> response. A disadvantage is that the efficiency of gene transfer<br /> may be very low.<br /> A second way to transfer genes into human cells is via<br /> viruses. Commonly used viruses for gene therapy include retroviruses,<br /> adenoviruses, and parvoviruses. The genetic modification<br /> of these viral genomes has led to the development of gene therapy<br /> vectors with a capacity to infect cells or tissues, much like the<br /> ability of wild-type viruses to infect cells. However, in contrast<br /> to wild-type viruses, gene therapy viral vectors have been genetically<br /> engineered so they have lost their capacity for replication<br /> in target cells. Nevertheless, the genetically engineered viruses are<br /> naturally taken up by cells via endocytosis (Figure 19.20b). The<br /> viral coat disassembles, and the viral genome is released into the<br /> cytosol. In the case of retroviruses, the genome is RNA, which<br /> is reverse transcribed into DNA. The viral DNA, which carries<br /> a gene of interest, travels into the nucleus and is then integrated<br /> into a chromosome of the target cell.<br /> A key advantage of viral vectors is their ability to efficiently<br /> transfer cloned genes to a variety of human cell types. However, a<br /> major disadvantage of viral-mediated gene therapy is the potential<br /> to evoke an undesirable immune response when injected<br /> into a patient. The inflammatory responses induced by adenovirus<br /> particles, for example, can be very strong and even fatal.<br /> Therefore, much effort has been aimed at preventing the inflammatory<br /> responses mediated by virus particles. These include the<br /> application of immunosuppressive drugs and the generation of<br /> less immunogenic viral vectors by further genetic modification<br /> within the viral genome.<br /> Muscular<br /> Lung<br /> Cancer<br /> Cardiovascular<br /> Infectious<br /> Duchenne muscular dystrophy, myotonic muscular<br /> dystrophy<br /> Cystic fibrosis<br /> Brain tumors, breast cancer, colorectal cancer,<br /> malignant melanoma, ovarian cancer, and several<br /> other types of malignancies<br /> Atherosclerosis, essential hypertension<br /> AIDS, possibly other viral diseases that involve latent<br /> infections<br /> Adenosine Deaminase Deficiency Was the First<br /> Inherited Disease Treated with Gene Therapy<br /> Adenosine deaminase (ADA) is an enzyme involved in purine<br /> metabolism. If both copies of the ADA gene are defective, deoxyadenosine<br /> will accumulate within the cells of the individual. At<br /> high concentrations, deoxyadenosine is particularly toxic to lymphocytes<br /> in the immune system, namely, T cells and B cells. In<br /> affected individuals, the destruction of T and B cells leads to a<br /> form of severe combined immunodeficiency disease (SCID). If</div> <div>224 CHAPTER 9 :: MOLECULAR STRUCTURE OF DNA AND RNA<br /> 1 2 3 4 5<br /> Type IIR<br /> cells<br /> Type IIR<br /> cells<br /> Type IIIS<br /> DNA<br /> extract<br /> Type IIR<br /> cells<br /> Type IIIS<br /> DNA<br /> extract<br /> +<br /> DNase<br /> Type IIR<br /> cells<br /> Type IIIS<br /> DNA<br /> extract<br /> +<br /> RNase<br /> Type IIR<br /> cells<br /> Type IIIS<br /> DNA<br /> extract<br /> +<br /> protease<br /> Mix Mix Mix Mix<br /> Allow sufficient time for the DNA to be taken up by the type IIR bacteria. Only a small percentage of the type IIR bacteria will be transformed to type IIIS.<br /> Add an antibody that aggregates type IIR bacteria (that have not been transformed). The aggregated bacteria are removed by gentle centrifugation.<br /> Plate the remaining bacteria on petri plates. Incubate overnight.<br /> Transformed Transformed Transformed<br /> FIGURE 9.3 Experimental protocol used by Avery, MacLeod, and McCarty to identify the transforming principle. Samples of S. pneumonia<br /> cells were either not exposed to a type IIIS DNA extract (tube 1) or exposed to a type IIIS DNA extract (tubes 2–5). Tubes 3, 4, and 5 also contained<br /> DNase, RNase, or protease, respectively. After incubation, the cells were exposed to antibodies, which are molecules that can specifically recognize the<br /> molecular structure of macromolecules. In this experiment, the antibodies recognized the cell surface of type IIR bacteria and caused them to clump<br /> together. The clumped bacteria were removed by a gentle centrifugation step. Only the bacteria that were not recognized by the antibody (namely, the<br /> type IIIS bacteria) remained in the supernatant. The cells in the supernatant were plated on solid growth media. After overnight incubation, visible<br /> colonies may be observed.<br /> the DNA extract with enzymes that digest DNA (called DNase),<br /> RNA (RNase), or protein (protease)(see Figure 9.3). When the<br /> DNA extracts were treated with RNase or protease, they still converted<br /> type IIR bacteria into type IIIS. These results indicated<br /> that any remaining RNA or protein in the extract was not acting<br /> as the genetic material. However, when the extract was treated<br /> with DNase, it lost its ability to convert type IIR into type IIIS<br /> bacteria. These results indicated that the degradation of the DNA<br /> in the extract by DNase prevented conversion of type IIR to type<br /> IIIS. This interpretation is consistent with the hypothesis that<br /> DNA is the genetic material. A more elegant way of saying this is<br /> that the transforming principle is DNA.<br /> EXPERIMENT 9A<br /> Hershey and Chase Provided Evidence That<br /> the Genetic Material Injected into the Bacterial<br /> Cytoplasm Is T2 Phage DNA<br /> A second experimental approach indicating that DNA is the<br /> genetic material came from the studies of Alfred Hershey and<br /> Martha Chase in 1952. Their research centered on the study of<br /> a virus known as T2. This virus infects Escherichia coli bacterial<br /> cells and is therefore known as a bacteriophage or simply<br /> a phage. As shown in Figure 9.4, the external structure of the<br /> T2 phage, known as the capsid or phage coat, consists of a head,<br /> sheath, tail fibers, and base plate. Biochemically, the phage coat<br /> is composed entirely of protein, which includes several different<br /> polypeptides. DNA is found inside the head of the T2 capsid.<br /> From a molecular point of view, this virus is rather simple,<br /> because it is composed of only two types of macromolecules:<br /> DNA and proteins.<br /> Although the viral genetic material contains the blueprint<br /> to make new viruses, a virus itself cannot synthesize new viruses.<br /> Instead, a virus must introduce its genetic material into the</div> <div>13.2 STRUCTURE AND FUNCTION OF tRNA 339<br /> Phenylalanine<br /> tRNA phe<br /> Proline<br /> tRNA pro<br /> Phenylalanine<br /> anticodon<br /> A A G<br /> Phenylalanine<br /> codon<br /> GGC<br /> U U C C C<br /> G<br /> Proline<br /> codon<br /> Proline<br /> anticodon<br /> 5′ 3′ mRNA FIGURE 13.10 Recognition between tRNAs and mRNA. The anticodon<br /> in the tRNA binds to a complementary sequence in the mRNA. At its other end,<br /> the tRNA carries the amino acid that corresponds to the codon in the mRNA via<br /> the genetic code.<br /> EXPERIMENT 13B<br /> tRNA Functions as the Adaptor Molecule Involved<br /> in Codon Recognition<br /> In 1962, François Chapeville and his colleagues conducted experiments<br /> aimed at testing the adaptor hypothesis. Their technical<br /> strategy was similar to that of the Nirenberg experiments that<br /> helped to decipher the genetic code (see Experiment 13A). In<br /> this approach, a cell-free translation system was made from cell<br /> extracts that contained the components necessary for translation.<br /> These components include ribosomes, tRNAs, and other translation<br /> factors. A cell-free translation system can synthesize polypeptides<br /> in vitro if mRNA and amino acids are added. Such a<br /> translation system can be used to investigate the role of specific<br /> factors by adding a particular mRNA template and varying individual<br /> components required for translation.<br /> According to the adaptor hypothesis, the amino acid<br /> attached to a tRNA is not directly involved in codon recognition.<br /> Chapeville reasoned that if this were true, the alteration of an<br /> amino acid already attached to a tRNA should cause that altered<br /> amino acid to be incorporated into the polypeptide instead of the<br /> normal amino acid. For example, consider a tRNA cys that carries<br /> the amino acid cysteine. If the attached cysteine were changed<br /> to an alanine, this tRNA cys should insert an alanine into a polypeptide<br /> where it would normally put a cysteine. Fortunately,<br /> Chapeville could carry out this strategy because he had a reagent,<br /> known as Raney nickel, that can chemically convert cysteine<br /> to alanine.<br /> An elegant aspect of the experimental design was the choice<br /> of the mRNA template. Chapeville and his colleagues synthesized<br /> an mRNA template that contained only U and G. Therefore, this<br /> template contained only the following codons (refer back to the<br /> genetic code in Table 13.2):<br /> UUU = phenylalanine<br /> UUG = leucine<br /> UGG = tryptophan<br /> GGG = glycine<br /> GGU = glycine<br /> GUU = valine<br /> GUG = valine<br /> UGU = cysteine<br /> Among the eight possible codons, one cysteine codon occurs, but<br /> no alanine codons can be formed from a polyUG template.<br /> As shown in the experiment of Figure 13.11, Chapeville<br /> began with a cell-free translation system that contained tRNA<br /> molecules. Amino acids, which would become attached to tRNAs,<br /> were added to this mixture. Of the 20 amino acids, only cysteine<br /> was radiolabeled. After allowing sufficient time for the amino<br /> acids to become attached to the correct tRNAs, the sample was<br /> divided into two tubes. One tube was treated with Raney nickel,<br /> while the control tube was not. As mentioned, Raney nickel converts<br /> cysteine into alanine by removing the –SH (sulfhydryl)<br /> group. However, it would not remove the radiolabel, which was<br /> a 14 C-label within the cysteine amino acid. Next, the polyUG<br /> mRNA was added as a template, and the samples were incubated<br /> to allow the translation of the mRNA into a polypeptide.<br /> In the control tube, we would expect the polypeptide to contain<br /> phenylalanine, leucine, tryptophan, glycine, valine, and cysteine,<br /> because these are the codons that contain only U and G. However,<br /> in the Raney nickel-treated sample, if the tRNA cys was using<br /> its anticodon region to recognize the mRNA, we would expect to<br /> see alanine instead of cysteine.<br /> Following translation, the polypeptides were isolated and<br /> hydrolyzed via a strong acid treatment, and then the individual<br /> amino acids were separated by column chromatography. The<br /> column separated cysteine from alanine; alanine eluted in a later<br /> fraction. The amount of radioactivity in each fraction was determined<br /> by liquid scintillation counting.<br /> ■ THE HYPOTHESIS<br /> Codon recognition is dictated only by the tRNA anticodon; the<br /> chemical structure of the amino acid attached to the tRNA does<br /> not play a role.</div> <div>394 CHAPTER 15 :: GENE REGULATION IN EUKARYOTES<br /> Activator<br /> protein<br /> ON<br /> Repressor<br /> protein<br /> TFIID<br /> OFF<br /> TFIID<br /> Coding sequence<br /> Coding sequence<br /> Silencer<br /> Enhancer<br /> Core<br /> promoter<br /> Silencer<br /> Enhancer<br /> Core<br /> promoter<br /> The activator protein recruits TFIID to the core promoter<br /> and/or activates its function. Transcription will be activated.<br /> The repressor protein inhibits the binding of TFIID<br /> or inhibits its function. Transcription is repressed.<br /> (a) Regulatory transcription factors and TFIID<br /> Core<br /> promoter<br /> Mediator<br /> ON<br /> Core<br /> promoter<br /> Mediator<br /> OFF<br /> TFIID<br /> RNA polymerase<br /> and general transcription<br /> factors<br /> TFIID<br /> RNA polymerase<br /> and general transcription<br /> factors<br /> Coding sequence<br /> Coding sequence<br /> Activator protein<br /> Enhancer<br /> Enhancer<br /> Repressor protein<br /> Silencer<br /> Silencer<br /> The activator protein interacts with mediator. This enables<br /> RNA polymerase to form a preinitiation complex that can<br /> proceed to the elongation phase of transcription.<br /> The repressor protein interacts with mediator so<br /> that transcription is repressed.<br /> (b) Regulatory transcription factors and mediator<br /> FIGURE 15.4 Ability of regulatory transcription factors to affect transcription. (a) Some regulatory transcription factors exert their effects<br /> through TFIID. Transcriptional activators may recruit TFIID to the core promoter and/or activate its function (left), while repressors inhibit TFIID<br /> binding or its activity (right). (b) Regulatory transcription factors may also exert their effects via mediator. By interacting with mediator in different<br /> ways, activators stimulate transcription (left), while repressors inhibit transcription (right). In this example, the enhancer and silencer are relatively far<br /> away from the core promoter. Therefore, a loop must form in the DNA so that the regulatory transcription factor can interact with mediator.<br /> is a category that responds to steroid hormones. This type of<br /> regulatory transcription factor is known as a steroid receptor,<br /> because the steroid hormone binds directly to the protein.<br /> The ultimate action of a steroid hormone is to affect gene<br /> transcription. Steroid hormones act as signaling molecules that<br /> are synthesized by endocrine glands of animals and secreted into<br /> the bloodstream. The hormones are then taken up by cells that<br /> can respond to the hormones in different ways. For example,<br /> glucocorticoid hormones influence nutrient metabolism in most<br /> body cells. Other steroid hormones, such as estrogen and testosterone,<br /> are called gonadocorticoids because they influence the<br /> growth and function of the gonads.<br /> Figure 15.6 shows the stepwise action of glucocorticoid<br /> hormones, which are produced in mammals. In this example,<br /> the hormone enters the cytosol of a cell by diffusing through the<br /> plasma membrane. Once inside, the hormone specifically binds<br /> to glucocorticoid receptors. Prior to hormone binding, the glucocorticoid<br /> receptor is complexed with proteins known as heat<br /> shock proteins (HSP), one example being HSP90. After the hormone<br /> binds to the glucocorticoid receptor, HSP90 is released.<br /> This exposes a nuclear localization signal (NLS)—a signal that<br /> directs a protein into the nucleus. Two glucocorticoid receptors<br /> form a homodimer and then travel through a nuclear pore into<br /> the nucleus.</div> <div>12 CHAPTER 1 :: OVERVIEW OF GENETICS<br /> 0.3 μm<br /> 7 μm<br /> (a) Escherichia coli (b) Saccharomyces cerevisiae (c) Drosophila melanogaster<br /> 133 μm<br /> (d) Caenorhabditis elegans (e) Danio rerio (f) Mus musculus<br /> (g) Arabidopsis thaliana<br /> FIGURE 1.13 Examples of model organisms studied by geneticists. (a) Escherichia coli (a bacterium),<br /> (b) Saccharomyces cerevisiae (a yeast), (c) Drosophila melanogaster (fruit fly), (d) Caenorhabditis elegans (a<br /> nematode worm), (e) Danio rerio (zebrafish), (f) Mus musculus (mouse), and (g) Arabidopsis thaliana (a flowering<br /> plant).<br /> how can two brown-eyed parents produce a blue-eyed child? Or<br /> why do tall parents tend to produce tall children, but not always?<br /> Our modern understanding of transmission genetics began with<br /> the studies of Gregor Mendel. His work provided the conceptual<br /> framework for transmission genetics. In particular, he originated<br /> the idea that genetic determinants, which we now call genes, are<br /> passed as discrete units from parents to offspring via sperm and<br /> egg cells. Since these pioneering studies of the 1860s, our knowledge<br /> of genetic transmission has greatly increased. Many patterns<br /> of genetic transmission are more complex than the simple Mendelian<br /> patterns that are described in Chapter 2. The additional<br /> complexities of transmission genetics are examined in Chapters 3<br /> through 8.<br /> Experimentally, the fundamental approach of a transmission<br /> geneticist is the genetic cross. A genetic cross involves breedding<br /> two selected individuals and the subsequent analysis of their<br /> offspring in an attempt to understand how traits are passed from<br /> parents to offspring. In the case of experimental organisms, the<br /> researcher chooses two parents with particular traits and then<br /> categorizes the offspring according to the traits they possess. In<br /> many cases, this analysis is quantitative in nature. For example,<br /> an experimenter may cross two tall pea plants and obtain 100<br /> offspring that fall into two categories: 75 tall and 25 dwarf. As we<br /> will see in Chapter 2, the ratio of tall and dwarf offspring provides<br /> important information concerning the inheritance pattern<br /> of this trait.<br /> Throughout Chapters 2 to 8, we will learn how researchers<br /> seek to answer many fundamental questions concerning the passage<br /> of traits from parents to offspring. Some of these questions<br /> are as follows:<br /> What are the common patterns of inheritance for genes?<br /> Chapters 2–4</div> <div>22 CHAPTER 2 :: MENDELIAN INHERITANCE<br /> 2. Collect many seeds. The following<br /> spring, plant the seeds and allow the<br /> plants to grow. These are the plants of<br /> the F 1 generation.<br /> Note: The P<br /> cross produces<br /> seeds that are<br /> part of the F 1<br /> generation.<br /> F 1 seeds<br /> F 1 plants<br /> All Tt<br /> Tt<br /> All<br /> tall<br /> Selffertilization<br /> 3. Allow the F 1 generation plants to selffertilize.<br /> This produces seeds that are<br /> part of the F 2 generation.<br /> Selffertilization<br /> F 2 seeds<br /> TT + 2 Tt + tt<br /> 4. Collect the seeds and plant them the<br /> following spring to obtain the F 2<br /> generation plants.<br /> F 2 plants<br /> 5. Analyze the characteristics found<br /> in each generation. Tall Tall Dwarf Tall<br /> ■<br /> THE DATA<br /> P Cross F 1 Generation F 2 Generation Ratio<br /> Tall × All tall 787 tall,<br /> dwarf stem 277 dwarf 2.84:1<br /> Purple × All purple 705 purple,<br /> white flowers 224 white 3.15:1<br /> Axial × All axial 651 axial,<br /> terminal flowers 207 terminal 3.14:1<br /> Yellow × All yellow 6,022 yellow,<br /> green seeds 2,001 green 3.01:1<br /> Round × All round 5,474 round,<br /> wrinkled seeds 1,850 wrinkled 2.96:1<br /> Green × All green 428 green,<br /> yellow pods 152 yellow 2.82:1<br /> Smooth × All smooth 882 smooth,<br /> constricted pods 299 constricted 2.95:1<br /> Total All dominant 14,949 dominant,<br /> 5,010 recessive 2.98:1<br /> ■ INTERPRETING THE DATA<br /> The data shown in Figure 2.5 are the results of producing an F 1<br /> generation via cross-fertilization, and an F 2 generation via selffertilization<br /> of the F 1 monohybrids. A quantitative analysis of<br /> these data allowed Mendel to propose three important ideas:<br /> 1. Mendel’s data argued strongly against a blending mechanism<br /> of heredity. In all seven cases, the F 1 generation displayed<br /> characteristics that were distinctly like one of the two parents<br /> rather than traits intermediate in character. Using genetic<br /> terms that Mendel originated and are still used today, his<br /> first proposal was that the variant for one trait is dominant<br /> over another variant. For example, the variant of green pods<br /> is dominant to that of yellow pods. The term recessive is<br /> used to describe a variant that is masked by the presence of<br /> a dominant trait but reappears in subsequent generations.<br /> Yellow pods and dwarf stems are examples of recessive<br /> variants. They can also be referred to as recessive traits.<br /> 2. When a true-breeding plant with a dominant trait<br /> was crossed to a true-breeding plant with a recessive<br /> trait, the dominant trait was always observed in the F 1<br /> generation. In the F 2 generation, some offspring displayed<br /> the dominant characteristic, while a smaller proportion<br /> showed the recessive trait. However, none of the offspring<br /> exhibited intermediate traits. How did Mendel explain this<br /> observation? Because the recessive trait appeared in the<br /> F 2 generation, he formed a second proposal—the genetic<br /> determinants of traits are passed along as “unit factors” from<br /> generation to generation. His data were consistent with a<br /> particulate theory of inheritance, in which the genes that<br /> govern traits are inherited as discrete units that remain</div> <div>240 CHAPTER 9 :: MOLECULAR STRUCTURE OF DNA AND RNA<br /> 5′<br /> Backbone<br /> Bases<br /> H<br /> N<br /> O –<br /> Uracil (U)<br /> H<br /> O<br /> P O<br /> CH<br /> N O<br /> 2<br /> 5′<br /> O<br /> O – 4′ 1′<br /> H<br /> H<br /> H<br /> H<br /> 3′<br /> 2′<br /> NH OH<br /> 2<br /> N<br /> N<br /> Phosphodiester<br /> H<br /> Adenine (A)<br /> linkage<br /> O<br /> N<br /> N<br /> O<br /> P O<br /> CH 2<br /> 5′ O<br /> O – 4′ 1′<br /> H<br /> H<br /> H<br /> H<br /> 3′<br /> 2′<br /> NH 2<br /> OH<br /> H<br /> N<br /> Cytosine (C)<br /> RNA<br /> nucleotide<br /> O<br /> O<br /> O<br /> H<br /> N<br /> H<br /> N<br /> P O CH 2<br /> 5′<br /> O<br /> O – 4′ 1′<br /> H<br /> H<br /> Guanine<br /> H<br /> H<br /> (G)<br /> 3′<br /> 2′<br /> O<br /> OH<br /> N<br /> H<br /> N<br /> H<br /> N<br /> NH<br /> O<br /> N<br /> 2<br /> O<br /> P O<br /> CH 2<br /> 5′ O<br /> O – 4′ 1′<br /> H<br /> H<br /> H<br /> H<br /> Phosphate<br /> 3′<br /> 2′<br /> OH<br /> OH<br /> Sugar (ribose)<br /> 3′<br /> FIGURE 9.22 A strand of RNA. This structure is very similar to a DNA strand (see Figure 9.11), except that the sugar is ribose instead of<br /> deoxyribose, and uracil is substituted for thymine.<br /> H<br /> O<br /> A U<br /> U A<br /> G C<br /> C G<br /> C G<br /> A U<br /> A<br /> A<br /> C<br /> U<br /> G<br /> C<br /> U<br /> G<br /> C<br /> A<br /> U<br /> C<br /> U<br /> A U<br /> U A<br /> C G<br /> C G<br /> U A<br /> C G<br /> C G<br /> C G<br /> G<br /> C<br /> C<br /> G<br /> G<br /> C<br /> C<br /> G<br /> A<br /> U<br /> A<br /> U<br /> G<br /> C<br /> G C<br /> U A<br /> A U<br /> C G<br /> C G<br /> G C<br /> A U<br /> A U<br /> C<br /> G<br /> C<br /> G<br /> G<br /> C<br /> A<br /> U<br /> A<br /> U<br /> A<br /> U<br /> C<br /> G<br /> G<br /> U<br /> G<br /> C<br /> U<br /> C<br /> U<br /> A<br /> U<br /> A<br /> C G<br /> G C<br /> A U<br /> A U<br /> A<br /> G<br /> C<br /> G<br /> U<br /> C<br /> A<br /> G<br /> C<br /> C<br /> G A U<br /> (a) Bulge loop (b) Internal loop (c) Multibranched junction (d) Stem-loop<br /> FIGURE 9.23 Possible structures of RNA molecules. The double-stranded regions are depicted by connecting hydrogen bonds. Loops are<br /> noncomplementary regions that are not hydrogen bonded with complementary bases. Double-stranded RNA structures can form within a single RNA<br /> molecule or between two separate RNA molecules.</div> <div>322 CHAPTER 12 :: GENE TRANSCRIPTION AND RNA MODIFICATION<br /> Lane 1 is a sample of RNA isolated from nerve cells.<br /> Lane 2 is a sample of RNA isolated from kidney cells. Nerve cells<br /> produce twice as much of this RNA compared to kidney<br /> cells.<br /> Lane 3 is a sample of RNA isolated from spleen cells. Spleen cells<br /> produce an alternatively spliced version of this RNA that<br /> is about 200 nucleotides longer than the RNA produced<br /> in nerve and kidney cells.<br /> Let’s suppose a researcher was interested in the effects of mutations<br /> on the expression of a particular structural gene in eukaryotes.<br /> The gene has one intron that is 450 nucleotides long. After this<br /> intron is removed from the pre-mRNA, the mRNA transcript is<br /> 1,100 nucleotides in length. Diploid somatic cells have two copies<br /> of this gene. Make a drawing that shows the expected results of<br /> a Northern blot using mRNA from the cytosol of somatic cells,<br /> which were obtained from the following individuals:<br /> Lane 1: A normal individual<br /> Lane 2: An individual homozygous for a deletion that removes the<br /> –50 to –100 region of the gene that encodes this mRNA<br /> Lane 3: An individual heterozygous in which one gene is normal<br /> and the other gene had a deletion that removes the –50 to<br /> –100 region<br /> Lane 4: An individual homozygous for a mutation that introduces<br /> an early stop codon into the middle of the coding<br /> sequence of the gene<br /> Lane 5: An individual homozygous for a three-nucleotide deletion<br /> that removes the AG sequence at the 3' splice site<br /> E5. A gel retardation assay can be used to study the binding of<br /> proteins to a segment of DNA. This method is described in<br /> Chapter 18. When a protein binds to a segment of DNA, it retards<br /> the movement of the DNA through a gel, so the DNA appears at a<br /> higher point in the gel (see the following).<br /> 1 2<br /> E6. As described in Chapter 18 and in experimental question E5, a gel<br /> retardation assay can be used to determine if a protein binds to<br /> DNA. This method can also determine if a protein binds to RNA. In<br /> the combinations described here, would you expect the migration of<br /> the RNA to be retarded due to the binding of a protein?<br /> A. mRNA from a gene that is terminated in a r-independent<br /> manner plus r-protein<br /> B. mRNA from a gene that is terminated in a r-dependent manner<br /> plus r-protein<br /> C. pre-mRNA from a structural gene that contains two introns<br /> plus the snRNP called U1<br /> D. Mature mRNA from a structural gene that contains two introns<br /> plus the snRNP called U1<br /> E7. The technique of DNA footprinting is described in Chapter 18. If<br /> a protein binds over a region of DNA, it will protect chromatin<br /> in that region from digestion by DNase I. To carry out a DNA<br /> footprinting experiment, a researcher has a sample of a cloned<br /> DNA fragment. The fragments are exposed to DNase I in the<br /> presence and absence of a DNA-binding protein. Regions of the<br /> DNA fragment not covered by the DNA-binding protein will<br /> be digested by DNase I, and this will produce a series of bands<br /> on a gel. Regions of the DNA fragment not digested by DNase<br /> I (because a DNA-binding protein is preventing DNase I from<br /> gaining access to the DNA) will be revealed, because a region of<br /> the gel will not contain any bands.<br /> In the DNA footprinting experiment shown here, a researcher<br /> began with a sample of cloned DNA 300 bp in length. This DNA<br /> contained a eukaryotic promoter for RNA polymerase II. For the<br /> sample loaded in lane 1, no proteins were added. For the sample<br /> loaded in lane 2, the 300 bp fragment was mixed with RNA<br /> polymerase II plus TFIID and TFIIB.<br /> 300<br /> 250<br /> 1 2<br /> 900 bp<br /> Lane 1: 900 bp fragment alone<br /> Lane 2: 900 bp fragment plus a protein that binds to the 900 bp<br /> fragment<br /> In this example, the segment of DNA is 900 bp in length, and the<br /> binding of a protein causes the DNA to appear at a higher point<br /> in the gel. If this 900 bp fragment of DNA contains a eukaryotic<br /> promoter for a structural gene, draw a gel that shows the relative<br /> locations of the 900 bp fragment under the following conditions:<br /> Lane 1: 900 bp plus TFIID<br /> Lane 2: 900 bp plus TFIIB<br /> Lane 3: 900 bp plus TFIID and TFIIB<br /> Lane 4: 900 bp plus TFIIB and RNA polymerase II<br /> Lane 5: 900 bp plus TFIID, TFIIB, and RNA polymerase II/TFIIF<br /> 150<br /> 75<br /> 1<br /> A. How long of a region of DNA is “covered up” by the binding of<br /> RNA polymerase II and the transcription factors?</div> <div>CHAPTER 23 799<br /> C22. The predisposition to develop cancer is inherited in a dominant<br /> fashion because the heterozygote has the higher predisposition. The<br /> mutant allele is actually recessive at the cellular level. But because<br /> we have so many cells in our bodies, it becomes relatively likely that<br /> a defective mutation will occur in the single normal gene and lead<br /> to a cancerous cell. Some heterozygous individuals may not develop<br /> the disease as a matter of chance. They may be lucky and not get a<br /> defective mutation in the normal gene, or perhaps their immune<br /> system is better at destroying cancerous cells once they arise.<br /> C24. If an oncogene was inherited, it may cause uncontrollable cell growth<br /> at an early stage of development and thereby cause an embryo to<br /> develop improperly. This could lead to an early spontaneous abortion<br /> and thereby explain why we do not observe individuals with inherited<br /> oncogenes. Another possibility is that inherited oncogenes may<br /> adversely affect gamete survival, which would make it difficult for<br /> them to be passed from parent to offspring. A third possibility would<br /> be that oncogenes could affect the fertilized zygote in a way that would<br /> prevent the correct implantation in the uterus.<br /> C26. The role of p53 is to sense DNA damage and prevent damaged cells<br /> from proliferating. Perhaps, prior to birth, the fetus is in a protected<br /> environment, so DNA damage may be minimal. In other words, the<br /> fetus may not really need p53. After birth, agents such as UV light may<br /> cause DNA damage. At this point, p53 is important. A p53 knockout is<br /> more sensitive to UV light because it cannot repair its DNA properly<br /> in response to this DNA-damaging agent, and it cannot kill cells that<br /> have become irreversibly damaged.<br /> C28. The p53 protein is a regulatory transcription factor; it binds to<br /> DNA and influences the transcription rate of nearby genes. This<br /> transcription factor (1) activates genes that promote DNA repair; (2)<br /> activates a protein that inhibits cyclin/CDK protein complexes that<br /> are required for cell division; and (3) activates genes that promote<br /> apoptosis. If a cell is exposed to DNA damage, it has a greater<br /> potential to become malignant. Therefore, an organism wants to<br /> avoid the proliferation of such a cell. When exposed to an agent that<br /> causes DNA damage, a cell will try to repair the damage. However, if<br /> the damage is too extensive, the p53 protein will stop the cell from<br /> dividing and program it to die. This helps to prevent the proliferation<br /> of cancer cells in the body.<br /> Experimental Questions<br /> E2. Perhaps the least convincing is the higher incidence of the disease<br /> in particular populations. Because populations living in specific<br /> geographic locations are exposed to their own unique environment,<br /> it is difficult to distinguish genetic versus environmental causes for<br /> a particular disease. The most convincing evidence might be the<br /> higher incidence of a disease in related individuals and/or the ability<br /> to correlate a disease with the presence of a mutant gene. Overall,<br /> however, the reliability that a disease has a genetic component should<br /> be based on as many observations as possible.<br /> E4. You would probably conclude that it is less likely to have a genetic<br /> component. If it were rooted primarily in genetics, it would be likely<br /> to be found in the Central American population. Of course, there is a<br /> chance that very few or none of the people who migrated to Central<br /> America were carriers of the mutant gene, which is somewhat unlikely<br /> for a large migrating population. By comparison, one might suspect<br /> that an environmental agent present in South America but not present<br /> in Central America may underlie the disease. Researchers could try to<br /> search for this environmental agent (e.g., a pathogenic organism).<br /> E6. Males I-1, II-1, II-4, II-6, III-3, III-8, and IV-5 have a normal copy of<br /> the gene. Males II-3, III-2, and IV-4 are hemizygous for an inactive<br /> mutant allele. Females III-4, III-6, IV-1, IV-2, and IV-3 have two<br /> normal copies of the gene, whereas females I-2, II-2, II-5, III-1, III-5,<br /> and III-7 are heterozygous carriers of a mutant allele.<br /> E8. A transformed cell is one that has become malignant. In a laboratory,<br /> this can be done in three ways. First, the cells could be treated with<br /> a mutagen that would convert a proto-oncogene into an oncogene.<br /> Second, cells could be exposed to the DNA from a malignant cell line.<br /> Under the appropriate conditions, this DNA can be taken up by the<br /> cells and integrated into their genome so that they become malignant.<br /> A third way to transform cells is by exposure to an oncogenic virus.<br /> E10. By comparing oncogenic viruses with strains that have lost their<br /> oncogenicity, researchers have been able to identify particular genes<br /> that cause cancer. This has led to the identification of many oncogenes.<br /> From this work, researchers have also learned that normal cells contain<br /> proto-oncogenes that usually play a role in cell division. This suggests<br /> that oncogenes exert their effects by upsetting the cell division process.<br /> In particular, it appears that oncogenes are abnormally active and keep<br /> the cell division cycle in a permanent “on” position.<br /> E12. One possible category of drugs would be GDP analogues (i.e.,<br /> compounds that resemble the structure of GDP). Perhaps one could<br /> find a GDP analogue that binds to the Ras protein and locks it in the<br /> inactive conformation.<br /> One way to test the efficacy of such a drug would be to incubate the<br /> drug with a type of cancer cell that is known to have an overactive<br /> Ras protein and then plate the cells on solid media. If the drug locked<br /> the Ras protein in the inactive conformation, it should inhibit the<br /> formation of malignant growth or malignant foci.<br /> There are possible side effects of such drugs. First, they might block<br /> the growth of normal cells, because Ras protein plays a role in normal<br /> cell proliferation. Second (if you have taken a cell biology course),<br /> there are many GTP/GDP-binding proteins in cells, and the drugs<br /> could somehow inhibit cell growth and function by interacting with<br /> these proteins.<br /> Questions for Student Discussion/Collaboration<br /> 2. There isn’t a clearly correct answer to this question, but it should<br /> stimulate a large amount of discussion.<br /> CHAPTER 23<br /> Conceptual Questions<br /> C2. A. False; the head is anterior to the tail.<br /> B. True<br /> C. False; the feet are ventral to the hips.<br /> D. True<br /> C4. A. True<br /> B. False; because gradients are also established after fertilization<br /> during embryonic development.<br /> C. True<br /> C6. A. This is a mutation in a pair-rule gene (runt).<br /> B. This is a mutation in a gap gene (knirps).<br /> C. This is a mutation in a segment-polarity gene (patched).<br /> C8. Positional information refers to the phenomenon whereby the spatial<br /> locations of morphogens and cell adhesion molecules (CAMs) provide<br /> a cell with information regarding its position relative to other cells. In<br /> Drosophila, the formation of a segmented body pattern relies initially<br /> on the spatial location of maternal gene products. These gene products<br /> lead to the sequential activation of the segmentation genes. Positional<br /> information can come from morphogens that are found within the<br /> oocyte, from morphogens secreted from cells during development,<br /> and from cell-to-cell contact. Although all three are important,<br /> morphogens in the oocyte have the greatest impact on the overall<br /> body structure.<br /> C10. The anterior portion of the antero-posterior axis is established by the<br /> action of Bicoid. During oogenesis, the mRNA for Bicoid enters the</div> <div>680 CHAPTER 24 :: POPULATION GENETICS<br /> (a) Malaria prevalence<br /> Hb S allele frequency<br /> (percent)<br /> 0–2.5<br /> 2.5–5.0<br /> 5.0–7.5<br /> 7.5–10.0<br /> 10.0–12.5<br /> > 12.5<br /> (b) Hb S allele frequency<br /> FIGURE 24.14 The geographic relationship between malaria and the frequency of the sickle-cell allele in human populations.<br /> (a) The geographic prevalence of malaria in Africa and surrounding areas. (b) The frequency of the Hb S allele in the same areas.<br /> GenesgTraits The sickle-cell allele of the β-globin gene is maintained in human populations as a balanced polymorphism. In areas where malaria is prevalent, the heterozygote<br /> carrying one copy of the Hb S allele has a greater fitness than either of the corresponding homozygotes (Hb A Hb A and Hb S Hb S ). Therefore, even though the Hb S Hb S homozygotes<br /> suffer the detrimental consequences of sickle-cell disease, this negative aspect is balanced by the beneficial effects of malarial resistance in the heterozygotes.</div> <div>778 SOLUTIONS TO EVEN-NUMBERED PROBLEMS<br /> one female) that became tetraploids. It is easy to imagine how one<br /> animal could become a tetraploid; complete nondisjunction could<br /> occur during the first cell division of a fertilized egg, thereby creating<br /> a tetraploid cell that continued to develop into a tetraploid animal.<br /> This would have to happen independently (i.e., in two individuals of<br /> opposite sex) to create a tetraploid species. If you mated a tetraploid<br /> turtle with a diploid turtle, the offspring would be triploid and<br /> probably phenotypically normal. However, the triploid offspring<br /> would be sterile because they would make highly aneuploid gametes.<br /> C32. Polyploid, triploid, and euploid should not be used.<br /> C34. The boy carries a translocation involving chromosome 21: probably<br /> a translocation in which nearly all of chromosome 21 is translocated<br /> to chromosome 14. He would have one normal copy of chromosome<br /> 14, one normal copy of chromosome 21, and the translocated<br /> chromosome that contains both chromosome 14 and chromosome 21.<br /> This boy is phenotypically normal because the total amount of genetic<br /> material is normal, although the total number of chromosomes is 45<br /> (because chromosome 14 and chromosome 21 are fused into a single<br /> chromosome). His sister has familial Down syndrome because she has<br /> inherited the translocated chromosome, but she also must have one<br /> copy of chromosome 14 and two copies of chromosome 21. She has<br /> the equivalent of three copies of chromosome 21 (i.e., two normal<br /> copies and one copy fused with chromosome 14). This is why she has<br /> familial Down syndrome. One of the parents of these two children<br /> is probably normal with regard to karyotype (i.e., the parent has 46<br /> normal chromosomes). The other parent would have a karyotype that<br /> would be like the phenotypically normal boy.<br /> C36. Nondisjunction is a mechanism whereby the chromosomes do not<br /> segregate equally into the two daughter cells. This can occur during<br /> meiosis to produce cells with altered numbers of chromosomes, or it<br /> can occur during mitosis to produce a genetic mosaic individual. A<br /> third way to alter chromosome number is by interspecies crosses that<br /> produce an alloploid.<br /> C38. A mutation occurred during early embryonic development to create<br /> the blue patch of tissue. One possibility is a mitotic nondisjunction in<br /> which the two chromosomes carrying the b allele went to one cell and<br /> the two chromosomes carrying the B allele went to the other daughter<br /> cell. A second possibility is that the chromosome carrying the B allele<br /> was lost. A third possibility is that the B allele was deleted. This would<br /> cause the recessive b allele to exhibit pseudodominance.<br /> C40. Homeologous chromosomes are chromosomes from two species that<br /> are evolutionarily related to each other. For example, chromosome 1 in<br /> the sable antelope and the roan antelope are homeologous; they carry<br /> many of the same genes.<br /> C42. In general, Turner syndrome could be due to nondisjunction during<br /> either oogenesis or spermatogenesis. However, the Turner individual<br /> with color blindness is due to nondisjunction during spermatogenesis.<br /> The sperm lacked a sex chromosome, due to nondisjunction, and the<br /> egg carried an X chromosome with the recessive color blindness allele.<br /> This X chromosome had to be inherited from the mother, because the<br /> father was not color-blind. The mother must be heterozygous for the<br /> recessive color blindness allele, and the father is hemizygous for the<br /> normal allele. Therefore, the mother must have transmitted a single<br /> X chromosome carrying the color blindness allele to her offspring,<br /> indicating that nondisjunction did not occur during oogenesis.<br /> Experimental Questions<br /> E2. Due to the mix up, the ratio of green/red fluorescence would be 0.5<br /> in regions where the cancer cells had a normal amount of DNA. If a<br /> duplication occurred on both chromosomes of cancer cells, the ratio<br /> would be 1.0. If a deletion occurred on a single chromosome, the ratio<br /> would be 0.25.<br /> E4. Colchicine interferes with the spindle apparatus and thereby causes<br /> nondisjunction. At high concentrations, it can cause complete<br /> nondisjunction and produce polyploid cells.<br /> E6. The primary purpose is to generate diploid strains that are<br /> homozygous for all of their genes. Because the pollen is haploid, it has<br /> only one copy of each gene. The strain can later be made diploid (and<br /> homozygous) by treatment with colchicine.<br /> E8. First, you would cross the two strains together. It is difficult to predict<br /> the phenotype of the offspring. Nevertheless, you would keep crossing<br /> offspring to each other and backcrossing them to the parental strains<br /> until you obtained a great-tasting tomato strain that was resistant to<br /> heat and the viral pathogen. You could then make this strain tetraploid<br /> by treatment with colchicine. If you crossed the tetraploid strain with<br /> your great-tasting diploid strain that was resistant to heat and the viral<br /> pathogen, you may get a triploid that had these characteristics. This<br /> triploid would probably be seedless.<br /> E10. A polytene chromosome is formed when a chromosome replicates<br /> many times, and the chromatids lie side by side, as shown in Figure<br /> 8.21. The homologous chromosomes also lie side by side. Therefore, if<br /> there is a deletion, there will be a loop. The loop is the segment that is<br /> not deleted from one of the two homologues.<br /> E12. Because the plant giving the pollen is heterozygous for many genes,<br /> some of the pollen grains may be haploid for recessive alleles, which<br /> are nonbeneficial or even lethal. However, some pollen grains may<br /> inherit only the dominant (beneficial) alleles and grow quite well.<br /> Questions for Student Discussion/Collaboration<br /> 2. There are many possibilities. The students could look in agriculture<br /> and botany books to find many examples. In the insect world, there<br /> are interesting examples of euploidy affecting gender determination.<br /> Among amphibians and reptiles, there are also several examples of<br /> closely related species that have euploid variation.<br /> 4. 1. Polyploids are often more robust and disease resistant.<br /> 2. Allopolyploids may have useful combinations of traits.<br /> 3. Hybrids are often more vigorous; they can be generated from<br /> monoploids.<br /> 4. Strains with an odd number of chromosome sets (e.g., triploids)<br /> are usually seedless.<br /> CHAPTER 9<br /> Conceptual Questions<br /> C2. The transformation process is described in Chapter 6.<br /> 1. A fragment of DNA binds to the cell surface.<br /> 2. It penetrates the cell wall/cell membrane.<br /> 3. It enters the cytoplasm.<br /> 4. It recombines with the chromosome.<br /> 5. The genes within the DNA are expressed (i.e., transcription and<br /> translation).<br /> 6. The gene products create a capsule. That is, they are enzymes that<br /> synthesize a capsule using cellular molecules as building blocks.<br /> C4. The building blocks of a nucleotide are a sugar (ribose or<br /> deoxyribose), a nitrogenous base, and a phosphate group. In a</div> <div>13.1 THE GENETIC BASIS FOR PROTEIN SYNTHESIS 327<br /> Coding strand<br /> DNA<br /> Transcription<br /> 5′<br /> 3′<br /> ACTGCCCATGAGCGACCACTTGGGGCTCGGGGAATAA<br /> TGACGGGTACTCGCTGGTGAACCCCGAGCCCCTTATT<br /> Template strand<br /> 3′<br /> 5′<br /> 5′ 3′<br /> ACUGCCCAUGAGCGAC<br /> CACUUGGGGCU<br /> CGGG G A A UA A<br /> mRNA<br /> Untranslated<br /> region<br /> Start<br /> codon<br /> Codons<br /> Anticodons<br /> Stop<br /> codon<br /> Translation<br /> tRNA<br /> UACUCGCUGGUGAACCCCGAGCCCCUU<br /> Polypeptide<br /> Met Ser Asp His Leu Gly Leu Gly Glu<br /> FIGURE 13.2 The relationships among the DNA coding sequence, mRNA codons, tRNA anticodons, and amino acids in a polypeptide. The<br /> sequence of nucleotides within DNA is transcribed to make a complementary sequence of nucleotides within mRNA. This sequence of nucleotides in<br /> mRNA is translated into a sequence of amino acids of a polypeptide. tRNA molecules act as intermediates in this translation process.<br /> TABLE 13.2<br /> The Genetic Code<br /> First base<br /> U<br /> C<br /> A<br /> G<br /> UUU<br /> UUC<br /> UUA<br /> UUG<br /> CUU<br /> CUC<br /> CUA<br /> CUG<br /> AUU<br /> AUC<br /> AUA<br /> AUG<br /> GUU<br /> GUC<br /> GUA<br /> GUG<br /> U<br /> Phenylalanine<br /> (Phe)<br /> Leucine (Leu)<br /> Leucine (Leu)<br /> Isoleucine (Ile)<br /> Methionine (Met);<br /> start codon<br /> Valine (Val)<br /> UCU<br /> UCC<br /> UCA<br /> UCG<br /> CCU<br /> CCC<br /> CCA<br /> CCG<br /> ACU<br /> ACC<br /> ACA<br /> ACG<br /> GCU<br /> GCC<br /> GCA<br /> GCG<br /> Second base<br /> C A G<br /> Serine (Ser)<br /> Proline (Pro)<br /> Threonine (Thr)<br /> Alanine (Ala)<br /> UAU<br /> UAC<br /> UAA<br /> UAG<br /> CAU<br /> CAC<br /> CAA<br /> CAG<br /> AAU<br /> AAC<br /> AAA<br /> AAG<br /> GAU<br /> GAC<br /> GAA<br /> GAG<br /> Tyrosine (Tyr)<br /> Stop codon<br /> Stop codon<br /> Histidine (His)<br /> Glutamine (Gln)<br /> Asparagine<br /> (Asn)<br /> Lysine (Lys)<br /> Aspartic acid<br /> (Asp)<br /> Glutamic acid<br /> (Glu)<br /> UGU<br /> UGC<br /> UGA<br /> UGG<br /> CGU<br /> CGC<br /> CGA<br /> CGG<br /> AGU<br /> AGC<br /> AGA<br /> AGG<br /> GGU<br /> GGC<br /> GGA<br /> GGG<br /> Cysteine (Cys)<br /> Stop codon<br /> Tryptophan (Trp)<br /> Arginine (Arg)<br /> Serine (Ser)<br /> Arginine (Arg)<br /> Glycine (Gly)<br /> U<br /> C<br /> A<br /> G<br /> U<br /> C<br /> A<br /> G<br /> U<br /> C<br /> A<br /> G<br /> U<br /> C<br /> A<br /> G<br /> Third base<br /> specify each type. With four types of bases in mRNA (A, U, G,<br /> and C), a genetic code containing two bases in a codon would not<br /> be sufficient because it would only have 4 2 , or 16, possible types.<br /> By comparison, a three-base codon system can specify 4 3 , or 64,<br /> different codons. Because the number of possible codons exceeds<br /> 20—which is the number of different types of amino acids—the<br /> genetic code is termed degenerate. This means that more than<br /> one codon can specify the same amino acid. For example, the<br /> codons GGU, GGC, GGA, and GGG all specify the amino acid<br /> glycine. Such codons are termed synonymous codons. In most<br /> instances, the third base in the codon is the base that varies. The<br /> third base is sometimes referred to as the wobble base. This term<br /> is derived from the idea that the complementary base in the<br /> tRNA can “wobble” a bit during the recognition of the third base<br /> of the codon in mRNA. The significance of the wobble base will<br /> be discussed later in this chapter.<br /> From the analysis of many different species, including<br /> bacteria, protists, fungi, plants, and animals, researchers have</div> <div>15.2 CHANGES IN CHROMATIN STRUCTURE 399<br /> Region that is<br /> complementary<br /> to probe<br /> +<br /> Accessible DNA<br /> cut with DNase I.<br /> Add probe DNA.<br /> Denature, then allow<br /> probe to hybridize.<br /> Region that is<br /> complementary<br /> to probe<br /> +<br /> Inaccessible DNA<br /> not cut with DNase I.<br /> Add probe DNA.<br /> Denature, then allow<br /> probe to hybridize.<br /> FIGURE 15.9 The use of DNase I and<br /> S1 nuclease to probe the chromatin structure<br /> of the �-globin gene. (a) Accessible DNA has<br /> been digested into small pieces via DNase I.<br /> These small pieces cannot hybridize with the<br /> probe DNA. When S1 nuclease is added, the<br /> single-stranded probe DNA is digested. (b) Due<br /> to compaction, the DNA is not cut with DNase<br /> I. The probe DNA can hybridize with a strand of<br /> chromosomal DNA after it has been denatured.<br /> When S1 nuclease is added, it will digest only the<br /> single-stranded regions of DNA, not the doublestranded<br /> region where the probe DNA and<br /> chromosomal DNA are bound to each other.<br /> Add S1 nuclease.<br /> No significant<br /> hybridization<br /> Probe DNA<br /> hybridizes<br /> to intact gene.<br /> Add S1 nuclease.<br /> Probe DNA<br /> is digested.<br /> Probe DNA<br /> is not<br /> digested.<br /> (a)<br /> (b)<br /> the chromosomal DNA had been in a closed conformation. This<br /> is because DNase I would not digest the chromosomal gene, and<br /> therefore, the β-globin gene would be available to hybridize to<br /> the radiolabeled DNA probe. However, if the chromosomal globin<br /> gene was in an open conformation, DNase I would digest<br /> the chromosomal DNA, preventing it from hybridizing with<br /> the radiolabeled DNA probe. In this case, the radiolabeled DNA<br /> strand would be digested by S1 nuclease. Therefore, the susceptibility<br /> of the radiolabeled DNA to S1 nuclease digestion allowed<br /> them to evaluate whether the chromosomal globin gene was in<br /> an open or closed conformation.<br /> With these concepts in mind, let’s examine the steps in<br /> their experimental procedure (Figure 15.10). They began with<br /> three cell types: reticulocytes, brain cells, and fibroblasts. Only<br /> the reticulocytes express the β-globin gene. They hypothesized<br /> that if the globin gene was expressed, the DNA would be in an<br /> open conformation and accessible to digestion. The nuclei were<br /> extracted from these cells and incubated with DNase I. The DNA<br /> was then isolated from the three types of nuclei. Following the<br /> isolation procedure, the chromosomal DNA was broken into<br /> fragments by sonication, with the average fragment size larger<br /> than the probe DNA. The DNA fragments and probe were mixed<br /> together, denatured, and then allowed to hybridize. The cooled<br /> samples were then divided into two tubes each. Into one of the<br /> two tubes, S1 nuclease was added. A key point is that the probe<br /> would not be digested by S1 nuclease if it had hybridized to a<br /> complementary strand from the globin gene and thus was double<br /> stranded. The DNA fragments were then precipitated using trichloroacetic<br /> acid. The samples were subjected to centrifugation,<br /> and the amounts of radioactivity in the pellets were determined<br /> by scintillation counting.<br /> ■ THE HYPOTHESIS<br /> A loosening of chromatin structure occurs when the β-globin<br /> gene is transcriptionally active.</div> <div>94 CHAPTER 4 :: EXTENSIONS OF MENDELIAN INHERITANCE<br /> are more beneficial for their survival in a particular environment,<br /> a phenomenon called overdominance. Overdominance may result<br /> in the prevalence of alleles, such as the sickle-cell allele, that are<br /> detrimental in the homozygous condition.<br /> With regard to single genes, we also considered that, in<br /> natural populations, multiple alleles exist for most genes. Examples<br /> include coat color in rabbits and human blood types. The<br /> ABO blood types illustrate an inheritance pattern in which two<br /> alleles are distinctly expressed in the heterozygote, a phenomenon<br /> called codominance. We also examined how sex impacts the<br /> expression and inheritance patterns of single genes. Some genes<br /> are sex linked, meaning they are found on a sex chromosome.<br /> In mammals, X-linked genes are fairly common and produce<br /> inheritance patterns in which a reciprocal cross does not produce<br /> the same result. In sex-influenced inheritance, the phenotype of<br /> the heterozygote depends on its sex. A more extreme form of sex<br /> influence involves sex-limited traits, which are found in only one<br /> sex. Finally, we considered lethal alleles that cause the death of<br /> an organism. These are often loss-of-function alleles in essential<br /> genes. Such alleles can produce unusual ratios in a genetic cross.<br /> We have also explored cases in which two different genes<br /> affect the outcome of a single trait, a concept known as a gene<br /> interaction. Epistasis is an inheritance pattern in which the alleles<br /> of one gene mask the phenotypic effects of the alleles of a different<br /> gene. Examples include comb shape in chickens, flower color<br /> in sweet peas, and coat color in rodents. Complementation is a<br /> phenomenon in which two different parents, both expressing the<br /> same or similar recessive phenotypes, produce offspring with a<br /> wild-type phenotype. In the example we explored, two different<br /> white strains of sweet peas, when crossed to each other, produced<br /> purple-flowered offspring. Modifying genes, such as the cream<br /> allele in Drosophila, alter the phenotypic outcome of the alleles<br /> of a different gene.<br /> We also considered gene interactions that are often observed<br /> in experimental organisms. When the loss of function in a single<br /> gene has no phenotypic effect, but the loss of function of two or<br /> more genes has an effect, the genes are termed redundant with<br /> each other. Intergenic suppressors are mutations that reverse the<br /> effects of a first mutation. From the analysis of mutants and their<br /> intergenic suppressors, researchers often can identify proteins<br /> that participate in a common cellular process. The proteins may,<br /> but do not always, interact directly. Other intergenic suppressors<br /> exert their effects by altering the amount of protein encoded by a<br /> mutant gene.<br /> EXPERIMENTAL SUMMARY<br /> To understand how inheritance may deviate from a simple Mendelian<br /> pattern, either a genetic or molecular approach can be<br /> taken. In a genetic approach, a researcher makes crosses and then<br /> analyzes the phenotypes of the offspring. In Chapter 4, we have<br /> seen that the experimenter must weigh many factors when designing<br /> a cross and analyzing its outcome. These include the dominant/recessive<br /> relationships of alleles (namely, dominant, recessive,<br /> incompletely dominant, overdominant, or codominant),<br /> the presence of multiple alleles in a population, the relationship<br /> between the sex of the offspring and their phenotype, and the<br /> possibility of lethal alleles. In addition, complicating factors such<br /> as incomplete penetrance, the influence of the environment,<br /> and gene interactions must be considered. Based on these factors,<br /> a researcher can construct a Punnett square to see if a pattern<br /> of inheritance fits the available data obtained from crosses.<br /> Researchers may also seek to isolate particular types of mutants<br /> such as gene knockouts and suppressor mutants.<br /> Another level of understanding in Mendelian inheritance<br /> is to appreciate how the molecular expression of genes underlies<br /> their impact on the phenotype of an organism. Using modern<br /> molecular tools that are described later in this textbook, researchers<br /> can quantitatively compare the level of gene expression<br /> between wild-type and mutant alleles. This helps us to understand<br /> how the amount of gene expression is correlated with the<br /> phenotype of the organism. Also, when gene interactions occur,<br /> an investigator may want to understand, at the molecular level,<br /> how the gene products participate in a common cellular function<br /> such as an enzy matic pathway. Therefore, the molecular investigation<br /> of gene interactions frequently involves research collaborations<br /> between geneticists, cell biologists, and biochemists.<br /> BLEM SETS & INSIGHTS<br /> PROBLEM SETS & INSIGHTS<br /> Solved Problems<br /> S1. In humans, why are X-linked recessive traits more likely to occur<br /> in males compared to females?<br /> Answer: Because a male is hemizygous for X-linked traits, the<br /> phenotypic expression of X-linked traits depends on only a single copy<br /> of the gene. When a male inherits a recessive X-linked allele, he will<br /> automatically exhibit the trait because he does not have another copy<br /> of the gene on the corresponding Y chromosome. This phenomenon is<br /> particularly relevant to the inheritance of recessive X-linked alleles that<br /> cause human disease. (Some examples will be described in Chapter 22.)<br /> S2. In Ayrshire cattle, the spotting pattern of the animals can be either<br /> red and white or mahogany and white. The mahogany and white<br /> pattern is caused by the allele M. The red and white phenotype is<br /> controlled by the allele m. When mahogany and white animals are<br /> mated to red and white animals, the following results are obtained:<br /> Genotype<br /> Phenotype<br /> Females<br /> Males<br /> MM Mahogany and white Mahogany and white<br /> Mm Red and white Mahogany and white<br /> mm Red and white Red and white<br /> Explain the pattern of inheritance.</div> <div>152 CHAPTER 6 :: GENETIC TRANSFER AND MAPPING IN BACTERIA AND BACTERIOPHAGES<br /> r103<br /> r104<br /> gene A<br /> gene A<br /> Isolate 2 different (noncomplementing)<br /> rII phage mutants, r103 and r104. Mix<br /> the 2 phages together. Coinfect E. coli B.<br /> A new population of phages will be made.<br /> The E. coli B cells will eventually lyse.<br /> Phage<br /> E. coli B<br /> Isolate this new population of phages. It will primarily<br /> contain nonrecombinant phages, but it will occasionally<br /> contain intragenic recombinants of wild type and double<br /> mutant phages (depicted in white and black, respectively).<br /> The phage preparation can contain several billion phages<br /> per milliliter.<br /> Nonrecombinant<br /> phages<br /> Wild-type<br /> phage<br /> 10 –8 10 –6 and infect E. coli K12 (λ).<br /> Double<br /> mutant<br /> phage<br /> Take some of the phage preparation,<br /> dilute it greatly (10 –8 ), and infect<br /> E. coli B. Also, take some of the phage<br /> preparation, dilute it somewhat (10 –6 ),<br /> Phage<br /> E. coli B<br /> Plate the cells and observe the<br /> number of plaques. The number<br /> of plaques observed from the<br /> E. coli B infection provides a measure<br /> of the total number of phages in the<br /> population. The number of plaques<br /> observed from the E. coli K12 (λ)<br /> infection provides a measure of<br /> the wild-type phage produced by<br /> intragenic recombination.<br /> Phage<br /> E. coli K12 (λ)<br /> FIGURE 6.18 Benzer’s method<br /> of intragenic mapping in the rII region.<br /> 66 plaques 11 plaques</div> <div>15.1 REGULATORY TRANSCRIPTION FACTORS 393<br /> tion in the forward or reverse direction. For example, if the forward<br /> orientation of an enhancer is<br /> 5'–GATA–3'<br /> 3'–CTAT–5'<br /> this enhancer will also be bound by a regulatory transcription<br /> factor and enhance transcription even when it is oriented in the<br /> reverse direction:<br /> 5'–TATC–3'<br /> 3'–ATAG–5'<br /> Striking variation is also found in the location of regulatory<br /> elements relative to a gene’s promoter. Regulatory elements<br /> are often located in a region within a few hundred base pairs<br /> upstream from the promoter site. However, they can be quite distant<br /> from the promoter, even 100,000 base pairs away, yet exert<br /> strong effects on the ability of RNA polymerase to initiate transcription<br /> at the core promoter! Regulatory elements were first discovered<br /> by Susumu Tonegawa and coworkers in the 1980s. While<br /> studying genes that play a role in immunity, they identified a<br /> region far away from the core promoter, but needed for high levels<br /> of transcription to take place. In some cases, regulatory elements<br /> are located downstream from the promoter site and may even be<br /> found within introns, the noncoding parts of genes. As you may<br /> imagine, the variation in regulatory element orientation and location<br /> profoundly complicates the efforts of geneticists to identify the<br /> regulatory elements that affect the expression of any given gene.<br /> Regulatory Transcription Factors May Exert<br /> Their Effects Through TFIID and Mediator<br /> Different mechanisms have been discovered that explain how a<br /> regulatory transcription factor can bind to a regulatory element<br /> and thereby affect gene transcription. Indeed, more than one<br /> mechanism is typically involved. The net effect of a regulatory<br /> transcription factor is to influence the ability of RNA polymerase<br /> to transcribe a given gene. However, most regulatory transcription<br /> factors do not bind directly to RNA polymerase. How then<br /> do most regulatory transcription factors exert their effects? In<br /> many cases, their mechanism of action is to influence the function<br /> of RNA polymerase by interacting with other proteins that<br /> directly bind to RNA polymerase. Two protein complexes that<br /> communicate the effects of regulatory transcription factors are<br /> TFIID and mediator.<br /> Figure 15.4 depicts how regulatory transcription factors<br /> may control transcription. In some cases, regulatory transcription<br /> factors bind to a regulatory element and then influence the<br /> function of TFIID (Figure 15.4a). As discussed in Chapter 12,<br /> TFIID is a general transcription factor that binds to the TATA<br /> box and is needed to recruit RNA polymerase to the core promoter.<br /> Activator proteins are expected to enhance the ability of<br /> TFIID to initiate transcription. One possibility is that activator<br /> proteins could help recruit TFIID to the TATA box or they could<br /> enhance the function of TFIID in a way that facilitates its ability<br /> to bind RNA polymerase. In contrast, repressors inhibit the function<br /> of TFIID. They could exert their effects by preventing the<br /> binding of TFIID to the TATA box or by inhibiting the ability of<br /> TFIID to recruit RNA polymerase to the core promoter.<br /> A second way that regulatory transcription factors control<br /> RNA polymerase is via mediator, a protein complex discovered<br /> by Roger Kornberg and colleagues in 1990 (Figure 15.4b). The<br /> term mediator refers to the observation that it mediates the<br /> interaction between RNA polymerase and regulatory transcription<br /> factors. As discussed in Chapter 12, mediator controls the<br /> ability of RNA polymerase to progress to the elongation stage of<br /> transcription. Transcriptional activators stimulate the ability of<br /> mediator to facilitate the switch between the initiation and elongation<br /> stages, whereas repressors have the opposite effect. When<br /> a repressor protein interacts with mediator, RNA polymerase<br /> cannot progress to the elongation stage of transcription.<br /> A third way that regulatory transcription factors can influence<br /> transcription is by recruiting proteins to the promoter<br /> region that affect chromatin compaction. For example, certain<br /> transcriptional activators can recruit proteins to the promoter<br /> region and thereby promote the conversion of chromatin from a<br /> closed to an open conformation. We will return to this topic later<br /> in this chapter.<br /> The Function of Regulatory Transcription Factor<br /> Proteins Can Be Modulated in Three Ways<br /> Thus far, we have considered the structures of regulatory transcription<br /> factors and the molecular mechanisms that account for<br /> their abilities to control transcription. The functions of the regulatory<br /> transcription factors themselves must also be modulated.<br /> Why is this necessary? The answer is that the genes they control<br /> must be turned on at the proper time, in the correct cell type,<br /> and under the appropriate environmental conditions. Therefore,<br /> eukaryotes have evolved different ways to modulate the functions<br /> of these proteins.<br /> The functions of regulatory transcription factor proteins<br /> are controlled in three common ways, through (1) the binding<br /> of a small effector molecule, (2) protein–protein interactions,<br /> and (3) covalent modifications. Figure 15.5 depicts these three<br /> mechanisms of modulating regulatory transcription factor function.<br /> Usually, one or more of these modulating effects are important<br /> in determining whether a transcription factor can bind to<br /> the DNA and/or influence transcription by RNA polymerase.<br /> For example, a small effector molecule may bind to a regulatory<br /> transcription factor and promote its binding to DNA (Figure<br /> 15.5a). We will see that steroid hormones function in this manner.<br /> Another important way is via protein–protein interactions<br /> (Figure 15.5b). The formation of homodimers and heterodimers<br /> is a fairly common means of controlling transcription. Finally,<br /> the function of a regulatory transcription factor can be affected<br /> by covalent modifications such as the attachment of a phosphate<br /> group (Figure 15.5c). As discussed later, the phosphorylation of<br /> activators can control their ability to stimulate transcription.<br /> Steroid Hormones Exert Their Effects by Binding<br /> to a Regulatory Transcription Factor<br /> Now that we have a general understanding regarding the structure<br /> and function of transcription factors, let’s turn our attention<br /> to specific examples that illustrate how regulatory transcription<br /> factors carry out their roles within living cells. Our first example</div> <div>22.2 GENETIC BASIS OF CANCER 623<br /> FIGURE 22.18 A comparison between chromosomes found<br /> in a normal human cell and a cancer cell from the same person. The<br /> bottom set found in a cancer cell is highly abnormal, with extra copies<br /> of some chromosomes and lost copies of others. Chromosomes made<br /> of fused pieces of chromosomes (designated mar in this figure) are also<br /> common in cancer cells.<br /> chromosome composition of a normal male cell and a tumor cell<br /> taken from the same person. The normal composition for this<br /> person is 22 pairs of chromosomes plus two sex chromosomes<br /> (X and Y). By comparison, the chromosome composition of<br /> the tumor cell is quite bizarre, including the fact that the tumor<br /> cell has two X chromosomes, which is characteristic of females.<br /> There are many cases where chromosomes are missing. If tumorsuppressor<br /> genes were on these missing chromosomes, their<br /> function is lost as well. Figure 22.18 also shows a few cases of<br /> extra chromosomes. If these chromosomes contain protooncogenes,<br /> the expression of those genes may be overactive.<br /> Finally, tumor cells often contain chromosomes that have translocations.<br /> Such translocations may create fused genes (as in the<br /> case of the Philadelphia chromosome discussed earlier in this<br /> chapter), or they may place two genes close together so that the<br /> regulatory sequences of one gene affect the expression of the<br /> other gene.<br /> DNA Microarrays Are Used to Classify Tumors<br /> As we have just seen, cancer cells are usually the result of multiple<br /> genetic alterations that cause the activation of oncogenes and<br /> the loss of function of tumor suppressor genes. Therefore, each<br /> type of tumor is characterized by a particular set of gene and<br /> chromosome alterations. Traditionally, different types of tumors<br /> have not been identified on the basis of genetic changes but<br /> instead have been classified according to their appearance under<br /> a microscope. While this approach is useful, a major drawback<br /> is that two tumors may have a very similar microscopic appearance<br /> but yet have very different underlying genetic changes and<br /> clinical outcomes. For this reason, researchers and clinicians are<br /> turning to methods that enable them to understand the molecular<br /> changes that occur in diseases such as cancer. This general<br /> approach is called molecular profiling.<br /> In cancer biology, molecular profiling involves the identification<br /> of the genes that play a role in the development of cancer.<br /> Why is this useful? First, molecular profiling can distinguish<br /> between tumors that look very similar under the microscope.<br /> Second, the classification of different tumors via molecular tools<br /> may help to predict the clinical outcome. Some tumors respond<br /> well to certain treatments, while others do not. Molecular profiling<br /> can be very effective in distinguishing such differences. Finally,<br /> researchers are optimistic that molecular profiling may lead to<br /> improved treatment options. As we gain a better understanding<br /> of the genetic changes associated with particular types of cancers,<br /> researchers may be able to develop drugs that specifically target<br /> the proteins that are encoded by cancer-causing gene mutations.<br /> As discussed earlier, the drug imatinib mesylate, which is used to<br /> treat chronic myelogenous leukemia, was developed in this way.<br /> DNA microarrays, which are described in Chapter 21, are<br /> increasingly used as a tool in the molecular profiling of tumors.<br /> The goal is to identify those genes whose pattern of expression<br /> correlates with each other—an approach called cluster analysis<br /> (see Figure 21.3). In the study of cancer, researchers can compare<br /> cancer cells to normal cells and identify groups (clusters) of<br /> genes that are turned on in the cancer cells and off in the normal<br /> cells, and other groups of genes that are turned off in the cancer<br /> cells and on in the normal cells. Likewise, researchers can compare<br /> two different types of tumors and identify groups of genes<br /> that show different patterns of expression.<br /> As an example, Figure 22.19a shows a computer-generated<br /> image that illustrates the results of a microarray analysis of 47<br /> samples, most of which came from the tumors of patients with<br /> a type of cancer called diffuse large B-cell lymphoma (DLBCL).<br /> Each column represents the expression pattern of a set of genes<br /> from a particular sample. Genes that are expressed are shown in<br /> red; those that are not expressed are shown in green. During the<br /> course of these studies, the researchers identified two different<br /> patterns of gene expression. The tumor samples on the left side<br /> showed a set of genes (next to the orange bar) that tended to be<br /> turned on in the tumor and another set of genes (next to the<br /> blue bar) that tended to be turned off in the tumor. This pattern<br /> of gene expression was similar to the pattern found in a type<br /> of B cell called germinal center B cells. In contrast, the tumors</div> <div>434 CHAPTER 16 :: GENE MUTATION AND DNA REPAIR<br /> Bulk flask<br /> Individual cultures<br /> +<br /> 1<br /> 2 3 4 5 6 7 8 9 10<br /> 11 12 13 14 15 16 17 18 19 20<br /> Remove 10 samples.<br /> Plate each tube.<br /> Plates containing T1 phage<br /> 1 2 3 4 5<br /> Relatively even distribution of ton r colonies<br /> 6 7 8 9 10<br /> 11 12 13 14 15<br /> FIGURE 16.6<br /> The Luria-Delbrück fluctuation test.<br /> 16 17 18 19 20<br /> Great “fluctuation” in the number of ton r colonies<br /> would have predicted that ton r bacterial mutants would occur after<br /> exposure to the selective agent. If that had been the case, the colonies<br /> would not be expected to arise in identical locations on different<br /> secondary plates but rather in random patterns.<br /> Taken together, the results of Luria and Delbrück and those<br /> of the Lederbergs supported the random mutation hypothesis,<br /> now known as the random mutation theory. According to this<br /> theory, mutations are a random process—they can occur in any<br /> gene and do not involve exposure of an organism to a particular<br /> condition that selects for specific types of mutations. In some<br /> cases, a random mutation may provide a mutant organism with<br /> an advantage, such as resistance to T1 phage. Although such<br /> mutations occur as a matter of random chance, growth conditions<br /> may select for organisms that happen to carry them.<br /> As researchers have learned more about mutation at the<br /> molecular level, the view that mutations are a totally random process<br /> has required some modification. Within the same individual,<br /> some genes mutate at a much higher rate than other genes. Why<br /> does this happen? Some genes are larger than others, which provides<br /> a greater chance for mutation. Also, the relative locations of<br /> genes within a chromosome may cause some genes to be more<br /> susceptible to mutation compared to others. Even within a single<br /> gene, hot spots are usually found—certain regions of a gene that<br /> are more likely to mutate compared to other regions (refer back<br /> to Chapter 6, Figure 6.20).<br /> Mutation Rates and Frequencies Are Ways to<br /> Quantitatively Assess Mutation in a Population<br /> Because mutations occur spontaneously among populations of<br /> living organisms, geneticists are greatly interested in learning how<br /> prevalent they are. The term mutation rate is the likelihood that<br /> a gene will be altered by a new mutation. This rate is commonly<br /> expressed as the number of new mutations in a given gene per cell<br /> generation. In general, the spontaneous mutation rate for a particular<br /> gene is in the range of 1 in 100,000 to 1 in 1 billion, or 10 –5<br /> to 10 –9 per cell generation. These numbers tell us that it is very<br /> unlikely that a particular gene will mutate due to natural causes.<br /> However, the mutation rate is not a constant number. The presence<br /> of certain environment agents, such as X-rays, can increase the rate<br /> of induced mutations to a much higher value than the spontaneous<br /> mutation rate. In addition, mutation rates vary substantially<br /> from species to species and even within different strains of the<br /> same species. One explanation for this variation is that there are<br /> many different causes of mutations (refer back to Table 16.5).</div> <div>804 SOLUTIONS TO EVEN-NUMBERED PROBLEMS<br /> C22. A. Because of their good nutrition, you may speculate that they would<br /> grow to be taller.<br /> B. If the environment is rather homogeneous, then heritability values<br /> tend to be higher because the environment contributes less to the<br /> amount of variation in the trait. Therefore, in the commune, the<br /> heritability might be higher, because they uniformly practice good<br /> nutrition. On the other hand, because the commune is a smaller<br /> size than the general population, the amount of genetic variation<br /> might be less, so this would make the heritability lower. However,<br /> because the problem states that the commune population is large,<br /> we would probably assume that the amount of genetic variation<br /> is similar to that in the general population. Overall, the best guess<br /> would be that the heritability in the commune population is higher<br /> because of the uniform nutrition standards.<br /> C. As stated in part B, the amount of variation would probably be<br /> similar, because the commune population is large. As a general<br /> answer, larger populations tend to have more genetic variation.<br /> Therefore, the general population probably has a bit more<br /> variation.<br /> C24. A natural population of animals is more likely to have a higher genetic<br /> diversity compared to a domesticated population. This is because<br /> domesticated populations have been subjected to many generations<br /> of selective breeding, which decreases the genetic diversity. Therefore,<br /> V G is likely to be higher for the natural population. The other issue<br /> is the environment. It is difficult to say which group would have a<br /> more homogeneous environment. In general, natural populations<br /> tend to have a more heterogeneous environment, but not always. If<br /> the environment is more heterogeneous, this tends to cause more<br /> phenotypic variation, which makes V E higher.<br /> Heritability � V G /V T<br /> � V G /(V G � V E )<br /> When V G is high, heritability increases. When V E is high, heritability<br /> decreases. In the natural wolf population, we would expect that V G<br /> would be high. In addition, we would guess that V E might be high as<br /> well (but that is less certain). Nevertheless, if this were the case, the<br /> heritability of the wolf population might be similar to the domestic<br /> population. This is because the high V G in the wolf population is<br /> balanced by its high V E . On the other hand, if V E is not that high in<br /> the wolf population, or if it is fairly high in the domestic population,<br /> then the wolf population would have a higher heritability for this trait.<br /> Experimental Questions<br /> E2. To calculate the mean, we add the values together and divide by the<br /> total number.<br /> Mean � 1.9 + 2(2.4) + 2(2.1) + 3(2.0) + 2(2.2) + 1.7 + 1.8 + 2(2.3) + 1.6<br /> 15<br /> Mean � 2.1<br /> The variance is the sum of the squared deviations from the mean<br /> divided by N � 1. The mean value of 2.1 must be subtracted from<br /> each value, and then the square is taken. These 15 values are added<br /> together and then divided by 14 (which is N � 1).<br /> Variance � 0.85<br /> 14<br /> � 0.061<br /> The standard deviation is the square root of the variance.<br /> Standard deviation � 0.25<br /> E4. The results are consistent with the idea that there are QTLs for this<br /> trait on chromosomes 2 and 3 but not on the X chromosome.<br /> E6. When we say an RFLP is associated with a trait, we mean that a gene<br /> that influences a trait is closely linked to an RFLP. At the chromosomal<br /> level, the gene of interest is so closely linked to the RFLP that a<br /> crossover almost never occurs between them.<br /> Note: Each plant inherits four RFLPs, but it may be homozygous for<br /> one or two of them.<br /> Small: 2,700 and 4,000 (homozygous for both)<br /> Small-medium: 2,700 (homozygous), 3,000, and 4,000; or 2,000,<br /> 2,700, and 4,000 (homozygous)<br /> Medium: 2,000 and 4,000 (homozygous for both); or 2,700 and<br /> 3,000 (homozygous for both); or 2,000, 2,700, 3,000, and 4,000<br /> Medium-large: 2,000 (homozygous), 3,000, and 4,000; or 2,000,<br /> 2,700, and 3,000 (homozygous)<br /> Large: 2,000 and 3,000 (homozygous for both)<br /> E8. Let’s assume there is an extensive molecular marker map for the<br /> rice genome. We would begin with two strains of rice, one with a<br /> high yield and one with a low yield, that greatly differ with regard to<br /> the molecular markers they carry. We would make a cross between<br /> these two strains to get F 1 hybrids. We would then backcross the F 1<br /> hybrids to either of the parental strains and then examine hundreds<br /> of offspring with regard to their rice yields and molecular markers.<br /> In this case, our expected results would be that six different markers<br /> in the high-producing strain would be correlated with offspring that<br /> produce higher yields. We might get fewer than six bands if some of<br /> these genes are closely linked and associate with the same marker. We<br /> also might get fewer than six if the two parental strains have the same<br /> marker that is associated with one or more of the genes that affect<br /> yield.<br /> E10. A. If we assume that the highly inbred strain has no genetic variance,<br /> V G (for the wild strain) � 3.2 g 2 – 2.2 g 2 � 1.0 g 2<br /> B. h<br /> 2<br /> B � 1.0 g 2 /3.2 g 2 � 0.31<br /> C. It is the same as h 2 B , so it also equals 0.31.<br /> E12. A.<br /> hN<br /> 2<br /> � X O � X<br /> X P � X<br /> 0.21 � (26.5 g � 25 g)/(27 g � 25 g)<br /> X O � 25 g � 2 g (0.21)<br /> X O � 25.42 g<br /> B. 0.21 � (26.5 g � 25 g)/(X P � 25 g)<br /> (X P � 25 g)(0.21) � 1.5 g<br /> X P � 32.14 g parents<br /> However, because this value is so far from the mean, there may<br /> not be 32.14 g parents in the population of mice that you have<br /> available.<br /> E14. We first need to calculate a and b. In this calculation, X represents the<br /> height of fathers, and Y represents the height of sons.<br /> b = 144 = 1.29<br /> 112<br /> a � 69 � (1.29)(68) = �18.7<br /> For a father who is 70 in. tall,<br /> Y � (1.29)(70) � (�18.7) � 71.6<br /> The most likely height of the son would be 71.6 in.<br /> E16. A. After six or seven generations, the selective breeding seems to<br /> have reached a plateau. This suggests that the tomato plants have<br /> become monomorphic for the alleles that affect tomato weight.<br /> B. There does seem to be heterosis, because the first generation<br /> has a weight of 1.7 lb, which is heavier than either Mary’s or<br /> Hector’s tomatoes. This partially explains why Martin has obtained<br /> tomatoes heavier than 1.5 lb. However, heterosis is not the whole<br /> story; it does not explain why Martin obtained tomatoes that weigh<br /> 2 lb. Even though Mary’s and Hector’s tomatoes were selected for<br /> heavier weight, they may not have all of the “heavy alleles” for</div> <div>368 CHAPTER 14 :: GENE REGULATION IN BACTERIA AND BACTERIOPHAGES<br /> 6. Add β-o-nitrophenylgalactoside<br /> (β-ONPG). This is a colorless<br /> compound. β-galactosidase will<br /> cleave the compound to produce<br /> galactose and o-nitrophenol (O-NP).<br /> O-nitrophenol has a yellow color.<br /> The deeper the yellow color, the<br /> more β-galactosidase was produced.<br /> β-o-nitrophenylgalactoside<br /> 1.<br /> β-ONPG<br /> Galactose<br /> +<br /> NO 2 NO 2<br /> O-NP<br /> Broken cell<br /> β-galactosidase<br /> 2.<br /> +<br /> NO 2 NO 2<br /> 7. Incubate the sonicated cells to allow<br /> β-galactosidase time to cleave<br /> β-o-nitrophenylgalactoside.<br /> 3.<br /> NO 2<br /> 1 2 3 4<br /> 8. Measure the yellow color produced<br /> with a spectrophotometer. (See<br /> the Appendix for a description<br /> of spectrophotometry.)<br /> 4.<br /> +<br /> NO 2 NO 2<br /> ■<br /> THE DATA<br /> Amount of b-Galactosidase<br /> Strain Addition of Lactose (percentage of parent strain)<br /> Mutant No 100%<br /> Mutant Yes 100%<br /> Merozygote No <1%<br /> Merozygote Yes 220%<br /> ■ INTERPRETING THE DATA<br /> As seen in the data, the yellow production in the original mutant<br /> strain was the same in the presence or absence of lactose. This<br /> result is expected because the expression of β-galactosidase in<br /> the lacI – mutant strain was already known to be constitutive. In<br /> other words, the presence of lactose was not needed to induce<br /> the operon due to a defective lacI gene. In the merozygote strain,<br /> however, a different result was obtained. In the absence of lactose,<br /> the lac operons were repressed—even the operon on the bacterial<br /> chromosome. How do we explain these results? Because the<br /> normal lacI gene on the F' factor was not physically located next<br /> to the chromosomal lac operon, this result is consistent with the<br /> idea that the lacI gene codes for a repressor protein that can diffuse<br /> throughout the cell and bind to any lac operon. The hypothesis<br /> that the lacI – mutation resulted in the synthesis of an internal<br /> inducer was rejected. If that hypothesis had been correct, the<br /> merozygote strain would have still made an internal inducer, and<br /> the lac operons in the merozygote would have been expressed in<br /> the absence of lactose. This result was not obtained.<br /> The interactions between regulatory proteins and DNA<br /> sequences illustrated in this experiment have led to the definition<br /> of two genetic terms. A trans-effect is a form of genetic</div> <div>CONCEPTUAL QUESTIONS 183<br /> S2. A maternal effect gene in Drosophila, called torso, is found as a<br /> recessive allele that prevents the correct development of<br /> anterior- and posterior-most structures. A wild-type male is<br /> crossed to a female of unknown genotype. This mating produces<br /> 100% larva that are missing their anterior- and posterior-most<br /> structures and therefore die during early development. What is the<br /> genotype and phenotype of the female fly in this cross? What are<br /> the genotypes and phenotypes of the female fly’s parents?<br /> Answer: Because this cross produces 100% abnormal offspring, the<br /> female fly must be homozygous for the abnormal torso allele. Even so,<br /> the female fly must be phenotypically normal in order to reproduce.<br /> This female fly had a mother that was heterozygous for a normal and<br /> abnormal torso allele and a father that was either heterozygous or<br /> homozygous for the abnormal torso allele.<br /> torso + torso – × torso + torso – or torso – torso –<br /> (grandmother)<br /> (grandfather)<br /> i<br /> torso – torso – (mother of 100%<br /> abnormal offspring)<br /> This female fly is phenotypically normal because its mother was<br /> heterozygous and provided the gene products of the torso + allele from<br /> the nurse cells. However, this homozygous female will produce only<br /> abnormal offspring because it cannot provide them with the normal<br /> torso + gene products.<br /> S3. An individual with Angelman syndrome produced an offspring<br /> with Prader-Willi syndrome. Why does this occur? What are the<br /> sexes of the parent with Angelman syndrome and the offspring<br /> with Prader-Willi syndrome?<br /> Answer: These two different syndromes are most commonly caused by<br /> a small deletion in chromosome 15. In addition, genomic imprinting<br /> plays a role because genes in this deleted region are differentially<br /> imprinted, depending on sex. If this deletion is inherited from the<br /> paternal parent, the offspring develops Prader-Willi syndrome.<br /> Therefore, in this problem, the individual with Angelman syndrome<br /> must have been a male because he produced a child with Prader-Willi<br /> syndrome. The child could be either a male or female.<br /> S4. In yeast, a haploid petite mutant also carries a mutant gene that<br /> requires the amino acid histidine for growth. The petite his – strain<br /> is crossed to a wild-type his + strain to yield the following tetrad:<br /> 2 cells: petite his –<br /> 2 cells: petite his +<br /> Explain the inheritance of the petite and his – mutations.<br /> Answer: The his – and his + alleles are segregating in a 2:2 ratio. This<br /> result indicates a nuclear pattern of inheritance. By comparison, all four<br /> cells in this tetrad have a petite phenotype. This is a suppressive petite<br /> that arises from a mitochondrial mutation.<br /> S5. Let’s suppose that you are a horticulturist who has recently<br /> identified an interesting plant with variegated leaves. How would<br /> you determine if this trait is nuclearly or cytoplasmically inherited?<br /> Answer: Make crosses and reciprocal crosses involving normal<br /> and variegated strains. In many species, chloroplast genomes are<br /> inherited maternally, although this is not always the case. In addition,<br /> a significant percentage of paternal leakage may occur. Nevertheless,<br /> when reciprocal crosses yield different outcomes, an organellar mode of<br /> inheritance is possibly at work.<br /> S6. A phenotype that is similar to a yeast suppressive petite was also<br /> identified in the mold Neurospora crassa. Mary and Herschel<br /> Mitchell identified a slow-growing mutant that they called poky.<br /> Unlike yeast, which are isogamous (i.e., produce one type of<br /> gamete), Neurospora is sexually dimorphic and produces male and<br /> female reproductive structures. When a poky strain of Neurospora<br /> was crossed to a wild-type strain, the results were different<br /> between reciprocal crosses. If a poky mutant was the female parent,<br /> all spores exhibited the poky phenotype. By comparison, if the<br /> wild-type strain was the female parent, all spores were wild type.<br /> Explain these results.<br /> Answer: These genetic studies indicate that the poky mutation is<br /> maternally inherited. The cytoplasm of the female reproductive cells<br /> provides the offspring with their mitochondria. Besides these genetic<br /> studies, the Mitchells and their collaborators showed that poky mutants<br /> are defective in certain cytochromes, which are iron-containing proteins<br /> that are known to be located in the mitochondria.<br /> Conceptual Questions<br /> C1. Define the term epigenetic inheritance, and describe two examples.<br /> C2. Describe the inheritance pattern of maternal effect genes. Explain<br /> how the maternal effect occurs at the cellular level. What are the<br /> expected functional roles of the proteins that are encoded by<br /> maternal effect genes?<br /> C3. A maternal effect gene exists in a dominant N (normal) allele and a<br /> recessive n (abnormal) allele. What would be the ratios of genotypes<br /> and phenotypes for the offspring of the following crosses?<br /> A. nn female × NN male<br /> B. NN female × nn male<br /> C. Nn female × Nn male<br /> C4. A Drosophila embryo dies during early embryogenesis due to a<br /> recessive maternal effect allele called bicoid. The wild-type allele is<br /> designated bicoid + . What are the genotypes and phenotypes of the<br /> embryo’s mother and maternal grandparents?<br /> C5. For Mendelian traits, the nuclear genotype (i.e., the alleles<br /> found on chromosomes in the cell nucleus) directly influences<br /> an offspring’s traits. In contrast, for non-Mendelian inheritance<br /> patterns, the offspring’s phenotype cannot be reliably predicted<br /> solely from its genotype. For the following traits, what do you need<br /> to know to predict the phenotypic outcome?<br /> A. Dwarfism due to a mutant Igf 2 allele<br /> B. Snail coiling direction<br /> C. Leber’s hereditary optic neuropathy<br /> C6. Let’s suppose a maternal effect gene exists as a normal dominant<br /> allele and an abnormal recessive allele. A mother who is<br /> phenotypically abnormal produces all normal offspring. Explain<br /> the genotype of the mother.<br /> C7. Let’s suppose that a gene affects the anterior morphology in house<br /> flies and is inherited as a maternal effect gene. The gene exists in<br /> a normal allele, H, and a recessive allele, h, which causes a small</div> <div>8.2 VARIATION IN CHROMOSOME NUMBER 201<br /> The species is diploid, having two sets of 4 chromosomes each<br /> (Figure 8.14a). A normal fruit fly is euploid because 8 chromosomes<br /> divided by 4 chromosomes per set equals two exact sets.<br /> On rare occasions, an abnormal fruit fly can be produced with<br /> 12 chromosomes, containing three sets of 4 chromosomes each.<br /> This alteration in euploidy produces a triploid fruit fly with 12<br /> chromosomes. Such a fly is also euploid because it has exactly<br /> three sets of chromosomes. Organisms with three or more sets of<br /> chromosomes are also called polyploid (Figure 8.14b). Geneticists<br /> use the letter n to represent a set of chromosomes. A diploid<br /> organism is referred to as 2n, a triploid organism as 3n, a tetraploid<br /> organism as 4n, and so on.<br /> A second way in which chromosome number can vary is by<br /> aneuploidy. Such variation involves an alteration in the number<br /> of particular chromosomes, so the total number of chromosomes<br /> is not an exact multiple of a set. For example, an abnormal fruit<br /> fly could contain nine chromosomes instead of eight because it<br /> has three copies of chromosome 2 instead of the normal two<br /> copies (Figure 8.14c). Such an animal is said to have trisomy 2 or<br /> to be trisomic. Instead of being perfectly diploid (2n), a trisomic<br /> animal is 2n + 1. By comparison, a fruit fly could be lacking a<br /> single chromosome, such as chromosome 1, and contain a total<br /> of seven chromosomes (2n – 1). This animal would be monosomic<br /> and be described as having monosomy 1.<br /> In this section, we will begin by considering several examples<br /> of aneuploidy. This is generally regarded as an abnormal<br /> condition that usually has a negative impact on phenotype. We<br /> will then examine euploid variation that occurs occasionally<br /> in animals and quite frequently in plants, and consider how it<br /> affects phenotypic variation.<br /> Aneuploidy Causes an Imbalance in Gene<br /> Expression That Is Often Detrimental<br /> to the Phenotype of the Individual<br /> The phenotype of every eukaryotic species is influenced by thousands<br /> of different genes. In humans, for example, a single set of<br /> chromosomes contains approximately 20,000 to 25,000 different<br /> genes. To produce a phenotypically normal individual, intricate<br /> coordination has to occur in the expression of thousands<br /> of genes. In the case of humans and other diploid species, evolution<br /> has resulted in a developmental process that works correctly<br /> when somatic cells have two copies of each chromosome.<br /> In other words, when a human is diploid, the balance of gene<br /> expression among many different genes usually produces a person<br /> with a normal phenotype.<br /> Aneuploidy commonly causes an abnormal phenotype.<br /> To understand why, let’s consider the relationship between gene<br /> Chromosome composition<br /> Normal<br /> fruit fly:<br /> (a)<br /> 1(X)<br /> 2 3 4<br /> Diploid; 2n (2 sets)<br /> Polyploid<br /> fruit flies:<br /> Aneuploid<br /> fruit flies:<br /> Triploid; 3n (3 sets)<br /> Trisomy 2 (2n + 1)<br /> Tetraploid; 4n (4 sets)<br /> Monosomy 1 (2n – 1)<br /> (b) Variations in euploidy<br /> (c) Variations in aneuploidy<br /> FIGURE 8.14 Types of variation in chromosome number. (a) Depicts the normal diploid number of chromosomes in Drosophila.<br /> (b) Examples of polyploidy. (c) Examples of aneuploidy.</div> <div>704 CHAPTER 25 :: QUANTITATIVE GENETICS<br /> Low environmental effect<br /> High environmental effect<br /> Number of individuals<br /> ww<br /> 1<br /> Ww<br /> 2<br /> WW<br /> 1<br /> Number of individuals<br /> ww<br /> 1<br /> Ww<br /> 2<br /> WW<br /> 1<br /> Weight<br /> Weight<br /> (a) Ww x Ww<br /> Low environmental effect<br /> Number of individuals<br /> www www<br /> 1<br /> www wwW<br /> 6<br /> www wWW<br /> 15<br /> www WWW<br /> 20<br /> wwW WWW<br /> 15<br /> wWW WWW<br /> 6<br /> WWW WWW<br /> 1<br /> Weight<br /> High environmental effect<br /> Number of individuals<br /> www www<br /> 1<br /> www wwW<br /> 6<br /> www wWW<br /> 15<br /> www WWW<br /> 20<br /> wwW WWW<br /> 15<br /> wWW WWW<br /> 6<br /> WWW WWW<br /> 1<br /> Weight<br /> (b) WWW www x WWW www<br /> FIGURE 25.4 How genotypes and phenotypes may overlap for polygenic traits. (a) Situations in which seed weight is controlled by one gene,<br /> existing in light (w) and heavy (W) alleles. (b) Situations in which seed weight is governed by three genes instead of one, each existing in light and<br /> heavy alleles. Note: The 1:2:1 and 1:6:15:20:15:6:1 ratios were derived by analyzing these crosses using a Punnett square.<br /> GenesgTraits The ability of geneticists to correlate genotype and phenotype depends on how many genes are involved and how much the environment causes the phenotype<br /> to vary. In (a), a single gene influences weight. In the graph on the left side, the environment does not cause much variation in weight. This makes it easy to distinguish the three<br /> genotypes. There is no overlap in the weights of ww, Ww, and WW individuals. In the graph on the right side, the environment causes more variation in weight. In this case, a few<br /> individuals with ww genotypes will have the same weight as a few individuals with Ww genotypes; and a few Ww genotypes will have the same weight as WW genotypes. As<br /> shown in (b), it becomes even more difficult to distinguish genotype based on phenotype when three genes are involved. The overlaps are minor when the environment does not<br /> cause much weight variation. However, when the environment causes substantial phenotypic variation, the overlaps between genotypes and phenotypes are very pronounced and<br /> greatly confound genetic analysis.</div> <div>CONCEPTUAL QUESTIONS 157<br /> this type of experiment, the recipient cell is exposed to a fairly<br /> low concentration of donor DNA, making it unlikely that the<br /> recipient bacterium will take up more than one fragment of DNA.<br /> Therefore, under these conditions, cotransformation is likely only<br /> when two genes are fairly close together and are found on one<br /> fragment of DNA.<br /> In a cotransformation experiment, a researcher has isolated<br /> DNA from an araB + and leuD + donor strain. This DNA was<br /> transformed into a recipient strain that was araB – and leuD – .<br /> Following transformation, the cells were plated on media<br /> containing arabinose and leucine. On this media, only bacteria<br /> that are araB + can grow. The bacteria can be either leuD + or<br /> leuD – because leucine is provided in the media. Colonies, which<br /> grew on this media, were then restreaked on media that contained<br /> arabinose but lacked leucine. Only araB + and leuD + cells could<br /> grow on these secondary plates. Following this protocol, a<br /> researcher obtained the following results:<br /> Number of colonies growing on arabinose plus leucine media: 57<br /> Number of colonies that grew when restreaked on arabinose media<br /> without leucine: 42<br /> What is the map distance between these two genes? Note: This<br /> problem can be solved using the strategy of a cotransduction<br /> experiment except that the researcher must determine the average<br /> size of DNA fragments that are taken up by the bacterial cells. This<br /> would correspond to the value of L in a cotransduction experiment.<br /> Answer: As mentioned, the basic principle of gene mapping via<br /> cotransformation is identical to the method of gene mapping via<br /> cotransduction described in this chapter. One way to calculate the map<br /> distance is to use the same equation that we used for cotransduction<br /> data, except that we substitute cotransformation frequency for<br /> cotransduction frequency.<br /> Cotransformation frequency = (1 – d/L) 3<br /> (Note: Cotransformation is not quite as accurate as cotransduction because<br /> the sizes of chromosomal pieces tend to vary significantly from experiment<br /> to experiment, so the value of L is not quite as reliable. Nevertheless,<br /> cotransformation has been used extensively to map the order and distance<br /> between closely linked genes along the bacterial chromosome.)<br /> The researcher needs to experimentally determine the value<br /> of L by running the DNA on a gel and estimating the average size of<br /> the DNA fragments. Let’s assume they are about 2% of the bacterial<br /> chromosome, which, for E. coli, would be about 80,000 base pairs in<br /> length. So L equals 2 minutes, which is the same as 2%.<br /> Cotransformation frequency = (1 – d/L) 3<br /> 42/57 = (1 – d/2) 3<br /> d = 0.2 minutes<br /> The distance between araB and leuD is approximately 0.2 minutes.<br /> Conceptual Questions<br /> C1. The terms conjugation, transduction, and transformation are<br /> used to describe three different natural forms of genetic transfer<br /> between bacterial cells. Briefly discuss the similarities and<br /> differences between these processes.<br /> C2. Conjugation is sometimes called “bacterial mating.” Is it a form of<br /> sexual reproduction? Explain.<br /> C3. If you mix together an equal number of F + and F – cells, how<br /> would you expect the proportions to change over time? In other<br /> S5. In our discussion of transduction via P1 or P22, the life cycle<br /> of the bacteriophage sometimes resulted in the packaging of<br /> many different pieces of the bacterial chromosome. For other<br /> bacteriophages, however, transduction may involve the transfer of<br /> only a few specific genes from the donor cell to the recipient. This<br /> phenomenon is known as specialized transduction. The key event<br /> that causes specialized transduction to occur is that the lysogenic<br /> phase of the phage life cycle involves the integration of the viral<br /> DNA at a single specific site within the bacterial chromosome. The<br /> transduction of particular bacterial genes involves an abnormal<br /> excision of the phage DNA from this site within the chromosome<br /> that would carry adjacent bacterial genes. For example, a<br /> bacteriophage called lambda (λ) that infects E. coli specifically<br /> integrates between two genes designated gal + and bio + (required<br /> for galactose utilization and biotin synthesis, respectively). Either<br /> of these genes could be packaged into the phage if an abnormal<br /> excision event occurred. How would specialized transduction be<br /> different from generalized transduction?<br /> Answer: Generalized transduction can involve the transfer of any<br /> bacterial gene, while specialized transduction can transfer only genes<br /> that are adjacent to the site where the phage integrates. As mentioned,<br /> a bacteriophage that infects E. coli cells, known as lambda (λ), provides<br /> a well-studied example of specialized transduction. In the case of phage<br /> lambda, the lysogenic life cycle results in the integration of the phage<br /> DNA at a site that is called the attachment site (described further in<br /> Chapter 17). The attachment site is located between two bacterial genes,<br /> gal + and bio + . An E. coli strain that is lysogenic for phage lambda will<br /> have the lambda DNA integrated between these two bacterial genes.<br /> On occasion, the phage may enter the lytic cycle and excise its DNA<br /> from the bacterial chromosome. When this occurs normally, the phage<br /> excises its entire viral DNA from the bacterial chromosome. The excised<br /> phage DNA is then replicated and becomes packaged into newly made<br /> phages. However, an abnormal excision does occur at a low rate (i.e.,<br /> about one in a million). In this abnormal event, the phage DNA is<br /> excised in such a way that an adjacent bacterial gene is included and<br /> some of the phage DNA is not included in the final product. For<br /> example, the abnormal excision may yield a fragment of DNA that<br /> includes the gal + gene and some of the lambda DNA but is missing part<br /> of the lambda DNA. If this DNA fragment is packaged into a virus, it is<br /> called a defective phage because it is missing some of the phage DNA.<br /> If it carries the gal + gene, it is designated λdgal (the letter d designates a<br /> defective phage). Alternatively, an abnormal excision may carry the bio +<br /> gene. This phage is designated λdbio. Defective lambda phages can then<br /> transduce the gal + or bio + genes to other E. coli cells.<br /> words, do you expect an increase in the relative proportions of F +<br /> or of F – cells? Explain your answer.<br /> C4. What is the difference between an F + and an Hfr strain? Which<br /> type of strain do you expect to transfer many bacterial genes to<br /> recipient cells?<br /> C5. What is the role of the origin of transfer during F + - and Hfrmediated<br /> conjugation? What is the significance of the direction of<br /> transfer in Hfr-mediated conjugation?</div> <div>610 CHAPTER 22 :: MEDICAL GENETICS AND CANCER<br /> 22.2 GENETIC BASIS OF CANCER<br /> Cancer is a disease characterized by uncontrolled cell division.<br /> It is a genetic disease at the cellular level. More than 100 kinds<br /> of human cancers have been identified, and they are classified<br /> according to the type of cell that has become cancerous. Though<br /> cancer is a diverse collection of many diseases, some characteristics<br /> are common to all cancers.<br /> 1. Most cancers originate in a single cell. This single cell,<br /> and its line of daughter cells, undergoes a series of genetic<br /> changes that accumulate during cell division. In this<br /> regard, a cancerous growth can be considered to be clonal<br /> in origin. A hallmark of a cancer cell is that it divides to<br /> produce two daughter cancer cells.<br /> 2. At the cellular and genetic levels, cancer usually is a<br /> multistep process that begins with a precancerous genetic<br /> change—a benign growth—and is followed by additional<br /> genetic changes that lead to cancerous cell growth (Figure<br /> 22.8).<br /> 3. When cells have become cancerous, their growth is<br /> described as malignant. Cancer cells are invasive—they<br /> can invade healthy tissues—and metastatic—they can<br /> migrate to other parts of the body and cause secondary<br /> tumors.<br /> Initial tumor cell<br /> Benign growth<br /> Malignant growth<br /> In the United States, approximately 1 million people are<br /> diagnosed with cancer each year, and about half that number will<br /> die from the disease. In 5 to 10% of all cases, a predisposition to<br /> develop the cancer is an inherited trait. We will examine some<br /> inherited forms of cancer later in this chapter. Most cancers,<br /> though, perhaps 90 to 95%, are not passed from parent to offspring.<br /> Rather, cancer is usually an acquired condition that typically<br /> occurs later in life. While some cancers are caused by spontaneous<br /> mutations and viruses, at least 80% of all human cancers<br /> are related to exposure to agents that promote genetic changes<br /> in somatic cells. These environmental agents, such as UV light<br /> and certain chemicals, are mutagens that alter the DNA in a way<br /> that affects the function of normal genes. If the DNA is permanently<br /> modified in somatic cells, such changes may be transmitted<br /> during cell division. These DNA alterations can lead to effects<br /> on gene expression that ultimately affect cell division and thereby<br /> lead to cancer. An environmental agent that causes cancer in this<br /> manner is called a carcinogen.<br /> In this section, we will begin by considering some early<br /> experimental observations that suggested genes play a role in cancer.<br /> We will then explore how genetic abnormalities, which affect<br /> the functions of particular cellular proteins, can lead to cancer.<br /> Invasion of<br /> surrounding tissue<br /> and metastasis into<br /> bloodstream<br /> FIGURE 22.8 Progression of cellular growth leading to cancer.<br /> GenesgTraits In a healthy individual, an initial gene mutation converts a normal<br /> cell into a tumor cell. This tumor cell divides to produce a benign tumor. Additional<br /> genetic changes in the tumor cells may occur, leading to malignant growth. At a<br /> later stage in malignancy, the tumor cells will invade surrounding tissues, and some<br /> malignant cells may metastasize by traveling through the bloodstream to other parts of<br /> the body where they can grow and cause secondary tumors. As a trait, cancer can be<br /> viewed as a series of genetic changes that eventually lead to uncontrolled cell growth.</div> <div>18.1 GENE CLONING 493<br /> cDNA Can Be Made from mRNA<br /> Via Reverse Transcriptase<br /> In the discussion of gene cloning described earlier in Figure 18.2,<br /> chromosomal DNA and plasmid DNA were used as the material<br /> to clone genes. Alternatively, a sample of RNA can provide<br /> a starting point to clone DNA. As mentioned in Chapter 17,<br /> the enzyme reverse transcriptase can use RNA as a template to<br /> make a complementary strand of DNA. This enzyme is encoded<br /> in the genome of retroviruses and provides a way for retroviruses<br /> to copy their RNA genome into DNA molecules that integrate<br /> into the host cell’s chromosomes. Likewise, reverse transcriptase<br /> is encoded in viral-like retroelements and is needed in the<br /> retrotransposition of such elements.<br /> Researchers can use purified reverse transcriptase in a strategy<br /> to clone genes, using mRNA as the starting material (Figure<br /> 18.4). To begin this experiment, RNA is purified from a sample<br /> of cells. The RNA is mixed with primers composed of a string<br /> of thymine-containing nucleotides. This short strand of DNA,<br /> or oligonucleotide, is called a poly-dT primer. Because eukaryotic<br /> mRNAs contain a polyA tail, poly-dT primers will be complementary<br /> to the 3' end of mRNAs. Reverse transcriptase and<br /> deoxyribonucleotides (dNTPs) are then added to make a DNA<br /> strand that is complementary to the mRNA. Next, RNaseH, DNA<br /> polymerase, and DNA ligase are added. One way to make the<br /> other DNA strand is to use RNaseH, which partially digests the<br /> RNA, generating short RNAs that are used as primers by DNA<br /> polymerase to make a second DNA strand that is complementary<br /> to the strand made by reverse transcriptase. Finally, DNA ligase<br /> will seal any nicks in this second DNA strand. When DNA is made<br /> from RNA as the starting material, the DNA is called complementary<br /> DNA (cDNA). The term originally referred to the single<br /> strand of DNA that is complementary to the RNA template. However,<br /> cDNA now refers to any DNA, whether it is single or double<br /> stranded, that is made using RNA as the starting material.<br /> Why is cDNA cloning useful? From a research perspective,<br /> an important advantage of cDNA is that it lacks introns. Because<br /> introns can be quite large, it is much simpler to insert cDNAs into<br /> vectors if researchers want to focus their attention on the coding<br /> sequence of a gene. For example, if the primary goal was to determine<br /> the coding sequence of a structural gene, a researcher would<br /> insert cDNA into a vector and then determine the DNA sequence<br /> of the insert, as described later in this chapter. Similarly, if a scientist<br /> wanted to express an encoded protein of interest in a cell that<br /> would not splice out the introns properly (e.g., in a bacterial cell),<br /> it would be necessary to make cDNA clones of the respective gene.<br /> Restriction Mapping Is Used to Locate<br /> the Restriction Sites Within a Vector<br /> As we have seen, DNA or gene cloning involves the digestion of<br /> vector and chromosomal DNA with restriction enzymes and the<br /> subsequent ligation of DNA fragments into vectors. In this type<br /> of procedure, the locations of restriction enzyme sites, or simply<br /> restriction sites, are important for the design of experiments. In<br /> the vector, for example, it is desirable to have unique restriction<br /> sites for the insertion of chromosomal DNA.<br /> A A A A A A<br /> 5′ 3′<br /> mRNA<br /> Add a poly-dT primer.<br /> 3′ 5′<br /> 5′ 3′<br /> 3′ 5′<br /> 5′ 3′<br /> 3′ 5′<br /> T T T T T T<br /> A common approach to determine the locations of restriction<br /> sites is known as restriction mapping. Figure 18.5 outlines<br /> the restriction mapping of a bacterial plasmid. To begin this experiment,<br /> the small circular plasmid DNA is isolated and purified<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> Add reverse transcriptase<br /> + dNTPs to synthesize a<br /> complementary DNA strand.<br /> Add RNaseH to<br /> cut up the RNA<br /> and generate<br /> RNA primers.<br /> Add DNA polymerase and<br /> DNA ligase to synthesize<br /> the second DNA strand.<br /> 3′ 5′<br /> 5′ 3′<br /> Double-stranded cDNA<br /> FIGURE 18.4 Synthesis of cDNA. A poly-dT primer anneals to<br /> the 3' end of mRNAs. Reverse transcriptase then catalyzes the synthesis<br /> of a complementary DNA strand (cDNA). RNaseH digests the mRNA<br /> into short pieces that are used as primers by DNA polymerase I to<br /> synthesize the second DNA strand. The 5' to 3' exonuclease function<br /> of DNA polymerase I removes all of the RNA primers except the one<br /> at the 5' end (because there is no primer upstream from this site). This<br /> RNA primer can be removed by the subsequent addition of an RNase.<br /> After the double-stranded cDNA is made, it can then be inserted into<br /> vectors as described in Figure 18.2. Prior to vector insertion, short<br /> pieces of DNA, called linkers, are covalently attached to both ends of<br /> the cDNA. The DNA sequence of the linkers contains unique restriction<br /> sites, making it easy to cut the cDNA at both ends and insert it into<br /> a vector.<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A<br /> T<br /> A</div> <div>17.1 HOMOLOGOUS RECOMBINATION 459<br /> increase the level of genetic exchange. When cells are exposed<br /> to these types of mutagens, the technique of harlequin staining<br /> has revealed a substantial increase in the frequency of SCEs. In<br /> addition, certain genetic disorders that result in higher levels of<br /> chromosome breakage also show elevated SCEs. For example,<br /> Bloom syndrome is a rare autosomal recessive disorder characterized<br /> by short stature, skin abnormalities, and a predisposition<br /> for developing certain forms of cancer. The defect is associated<br /> with a gene that is involved with DNA replication. In the cells<br /> of Bloom syndrome patients, chromosome breaks are more frequent<br /> during DNA replication. Likewise, SCE is typically 10- to<br /> 15-fold more frequent in Bloom syndrome patients compared to<br /> unaffected individuals.<br /> A self-help quiz involving this experiment can be found at<br /> the Online Learning Center.<br /> The Holliday Model Describes a Molecular<br /> Mechanism for the Recombination Process<br /> We now turn our attention to genetic exchange that occurs<br /> between homologous chromosomes. Perhaps it is surprising that<br /> the first molecular model of homologous recombination did<br /> not come from a biochemical analysis of DNA or from electron<br /> microscopy studies. Instead, it was deduced from the outcome of<br /> genetic crosses in fungi.<br /> As discussed in Chapter 5, geneticists have learned a great<br /> deal from the analysis of fungal asci, because an ascus is a sac<br /> that contains the products of a single meiosis. When two haploid<br /> fungi that differ at a single gene are crossed to each other, the<br /> ascus is expected to contain an equal proportion of each genotype.<br /> For example, if a pigmented strain of Neurospora producing<br /> orange spores is crossed to an albino strain producing white<br /> spores, the resulting group of eight cells, or octad, should contain<br /> four orange spores and four white spores (refer to Figure 5.14).<br /> As early as 1934, H. Zickler noticed that unequal proportions<br /> of the spores sometimes occurred within asci. He occasionally<br /> observed octads with six orange spores and two white spores, or<br /> six white spores and two orange spores.<br /> Zickler used the term gene conversion to describe the<br /> phenomenon in which one allele is converted to the allele on<br /> the homologous chromosome. Subsequent studies by several<br /> researchers confirmed this phenomenon in yeast and Neurospora.<br /> Gene conversion occurred at too high a rate to be explained by<br /> new mutations. In addition, research showed that gene conversion<br /> often occurs in a chromosomal region where a crossover has<br /> taken place.<br /> Based on studies involving gene conversion, Robin Holliday<br /> proposed a model in 1964 to explain the molecular steps that<br /> occur during homologous recombination. We will first consider<br /> the steps in the Holliday model and then consider more recent<br /> models. Later, we will examine how the Holliday model can<br /> explain the phenomenon of gene conversion.<br /> The Holliday model is shown in Figure 17.4a. At the<br /> beginning of the process, two homologous chromatids are<br /> aligned with each other. According to the model, a break or nick<br /> occurs at identical sites in one strand of each of the two homologous<br /> chromatids. The strands then invade the opposite helices<br /> and base pair with the complementary strands. This event is followed<br /> by a covalent linkage to create a Holliday junction. The<br /> cross in the Holliday junction can migrate in a lateral direction.<br /> As it does so, a DNA strand in one helix is swapped for a DNA<br /> strand in the other helix. This process is called branch migration<br /> because the branch connecting the two double helices migrates<br /> laterally. Because the DNA sequences in the homologous chromosomes<br /> are similar but not identical, the swapping of the DNA<br /> strands during branch migration may produce a heteroduplex,<br /> a region in the double-stranded DNA that contains base mismatches.<br /> In other words, because the DNA strands in this region<br /> are from homologous chromosomes, their sequences are not perfectly<br /> complementary, yielding mismatches.<br /> The final two steps in the recombination process are collectively<br /> called resolution because they involve the breakage<br /> and rejoining of two DNA strands to create two separate chromosomes.<br /> In other words, the entangled DNA strands become<br /> resolved into two separate structures. The bottom left side of<br /> Figure 17.4a shows the Holliday junction viewed from two different<br /> planes. If breakage occurs in the same two DNA strands<br /> that were originally nicked at the beginning of this process, the<br /> subsequent joining of strands produces nonrecombinant chromosomes<br /> with a heteroduplex region. Alternatively, if breakage<br /> occurs in the strands that were not originally nicked, the rejoining<br /> process results in recombinant chromosomes, also with a<br /> heteroduplex region.<br /> The Holliday model can account for the general properties<br /> of recombinant chromosomes formed during eukaryotic meiosis.<br /> As mentioned, the original model was based on the results<br /> of crosses in fungi where the products of meiosis are contained<br /> within a single ascus. Nevertheless, molecular research in many<br /> other organisms has supported the central tenets of the model.<br /> Particularly convincing evidence came from electron microscopy<br /> studies in which recombination structures could be visualized. Figure<br /> 17.4b shows an electron micrograph of two DNA fragments in<br /> the process of recombination. This structure has been called a chi<br /> form because its shape is similar to the Greek letter χ (chi).<br /> More Recent Models Have Refined the Molecular<br /> Steps of Homologous Recombination<br /> As more detailed studies of homologous recombination have<br /> become available, certain steps in the Holliday model have been<br /> reconsidered. In particular, more recent models have modified the<br /> initiation phase of recombination. Researchers now suggest that<br /> homologous recombination is not likely to involve nicks at identical<br /> sites in one strand of each homologous chromatid. Instead, it<br /> is more likely for a DNA helix to incur a break in both strands<br /> of one chromatid or a single nick. Both of these types of changes<br /> have been shown to initiate homologous recombination. Therefore,<br /> newer models have tried to incorporate these experimental</div> <div>174 CHAPTER 7 :: NON-MENDELIAN INHERITANCE<br /> PWS<br /> AS<br /> PWS gene<br /> AS gene<br /> PWS<br /> AS<br /> AS gene<br /> silenced<br /> in sperm<br /> Fertilized<br /> egg<br /> Angelman syndrome<br /> The offspring does not carry an<br /> active copy of the AS gene.<br /> 15<br /> 15<br /> PWS gene<br /> silenced<br /> in egg<br /> Fertilized<br /> egg<br /> Silenced allele<br /> Transcribed allele<br /> Deleted region that<br /> includes both the AS<br /> and the PWS genes<br /> PWS<br /> AS<br /> PWS<br /> AS<br /> Prader-Willi syndrome<br /> The offspring does not carry an<br /> active copy of the PWS gene.<br /> FIGURE 7.12 The role of imprinting in<br /> the development of Angelman and Prader-Willi<br /> syndromes.<br /> Genes g Traits A small region on chromosome 15 contains two different genes<br /> designated the AS gene and PWS gene in this figure. If a chromosome 15 deletion<br /> is inherited from the maternal parent, Angelman syndrome occurs because the<br /> offspring does not inherit an active copy of the AS gene (left). Alternatively, the<br /> chromosome 15 deletion may be inherited from the male parent, leading to<br /> Prader-Willi syndrome. The phenotype of this syndrome occurs because the offspring<br /> does not inherit an active copy of the PWS gene (right).<br /> movements, and exhibit mental deficiencies. Most commonly,<br /> both PWS and AS involve a small deletion in human chromosome<br /> 15. If this deletion is inherited from the maternal parent, it<br /> leads to Angelman syndrome; if inherited from the father, it leads<br /> to Prader-Willi syndrome (Figure 7.12).<br /> Researchers have discovered that this region contains<br /> closely linked but distinct genes that are maternally or paternally<br /> imprinted. AS results from the lack of expression of a single gene<br /> (UBE3A) that codes for a protein called E6-AP, which functions<br /> to transfer small ubiquitin molecules to certain proteins to target<br /> their degradation. Both copies of this gene are active in many of<br /> the body’s tissues. In the brain, however, only the copy inherited<br /> from a person’s mother (the maternal copy) is active. The paternal<br /> allele of UBE3A is silenced. Therefore, if the maternal allele<br /> is deleted, as in the left side of Figure 7.12, the individual will<br /> develop AS because he or she will not have an active copy of the<br /> UBE3A gene.<br /> The gene(s) responsible for PWS has not been definitively<br /> determined, although five imprinted genes in this region of chromosome<br /> 15 are known. One possible candidate involved in PWS<br /> is a gene designated SNRPN. The gene product is part of a small<br /> nuclear ribonucleoprotein polypeptide N, which is a complex that<br /> controls RNA splicing and is necessary for the synthesis of critical<br /> proteins in the brain. The maternal allele of SNRPN is silenced,<br /> and only the paternal copy is active.<br /> 7.3 EXTRANUCLEAR INHERITANCE<br /> Thus far, we have considered several types of non-Mendelian<br /> inheritance patterns. These include maternal effect genes, dosage<br /> compensation, and genomic imprinting. All of these inheritance<br /> patterns involve genes found on chromosomes in the cell nucleus.<br /> Another cause of non-Mendelian inheritance patterns concerns<br /> genes that are not located in the cell nucleus. In eukaryotic species,<br /> the most biologically important example of extranuclear<br /> inheritance involves genetic material in cellular organelles. In<br /> addition to the cell nucleus, the mitochondria and chloroplasts<br /> contain their own genetic material. Because these organelles are<br /> found within the cytoplasm of the cells, the inheritance of organellar<br /> genetic material is called extranuclear inheritance (the prefix<br /> extra- means outside of) or cytoplasmic inheritance. In this<br /> section, we will examine the genetic composition of mitochondria<br /> and chloroplasts and explore the pattern of transmission<br /> of these organelles from parent to offspring. We will also consider<br /> a few other examples of inheritance patterns that cannot be<br /> explained by the transmission of nuclear genes.<br /> Mitochondria and Chloroplasts Contain<br /> Circular Chromosomes with Many Genes<br /> In 1951, Y. Chiba was the first to suggest that chloroplasts contain<br /> their own DNA. He based his conclusion on the staining properties<br /> of a DNA-specific dye known as Feulgen. Researchers later developed<br /> techniques to purify organellar DNA. In addition, electron<br /> microscopy studies provided interesting insights into the organization<br /> and composition of mitochondrial and chloroplast chromosomes.<br /> More recently, the advent of molecular genetic techniques<br /> in the 1970s and 1980s has allowed researchers to determine the<br /> genome sequences of organellar DNAs. From these types of studies,<br /> the chromosomes of mitochondria and chloroplasts were<br /> found to resemble smaller versions of bacterial chromosomes.<br /> The genetic material of mitochondria and chloroplasts<br /> is located inside the organelle in a region known as the nucleoid<br /> (Figure 7.13). The genome is a single circular chromosome<br /> (composed of double-stranded DNA), although a nucleoid contains<br /> several copies of this chromosome. In addition, a mitochondrion<br /> or chloroplast often has more than one nucleoid. In<br /> mice, for example, each mitochondrion has one to three nucleoids,<br /> with each nucleoid containing two to six copies of the circular<br /> mitochondrial genome. However, this number is variable<br /> and depends on the type of cell and the stage of development.<br /> In comparison, the chloroplasts of algae and higher plants tend<br /> to have more nucleoids per organelle. Table 7.3 describes the<br /> genetic composition of mitochondria and chloroplasts for a few<br /> selected species.</div> <div>ARIS ™<br /> (Assessment Review<br /> and Instruction System)<br /> Explore this dynamic site for<br /> a variety of study tools:<br /> •<br /> •<br /> •<br /> •<br /> Interactive genetic problems<br /> Self-quizzes<br /> Animations with quizzing<br /> Flashcards<br /> • Tweaking the Experiment quizzes<br /> Go to aris.mhhe.com or directly<br /> to this book’s ARIS site at<br /> www.mhhe.com/brookergenetics3e<br /> to learn more.<br /> Student Study Guide/Solutions Manual ISBN: 978-0-07-299283-0 MHID: 0-07-299283-2<br /> Written to support the concepts presented in Genetics: Analysis<br /> & Principles, Third Edition, this manual includes solutions<br /> to the end-of-chapter problems and questions. Solutions are<br /> presented in a step-by-step method to help develop your problem-<br /> solving skills. The Study Guide follows the order of sections<br /> and subsections in the textbook and summarizes the main<br /> points in the text, figures, and tables. It also contains conceptbuilding<br /> exercises, self-help quizzes, and practice exams.</div> <div>8.2 VARIATION IN CHROMOSOME NUMBER 205<br /> Variations in Euploidy Can Occur<br /> in Certain Tissues Within an Animal<br /> Thus far, we have considered variations in chromosome number<br /> that occur at fertilization, so all the somatic cells of an individual<br /> contain this variation. In many animals, certain tissues of the<br /> body display normal variations in the number of sets of chromosomes.<br /> Diploid animals sometimes produce tissues that are<br /> polyploid. For example, the cells of the human liver can vary to a<br /> great degree in their ploidy. Liver cells contain nuclei that can be<br /> triploid, tetraploid, and even octaploid (8n). The occurrence of<br /> polyploid tissues or cells in organisms that are otherwise diploid<br /> is known as endopolyploidy. What is the biological significance<br /> of endopolyploidy? One possibility is that the increase in chromosome<br /> number in certain cells may enhance their ability to produce<br /> specific gene products that are needed in great abundance.<br /> An unusual example of natural variation in the ploidy of<br /> somatic cells occurs in Drosophila and some other insects. Within<br /> certain tissues, such as the salivary glands, the chromosomes<br /> undergo repeated rounds of chromosome replication without<br /> cellular division. For example, in the salivary gland cells of Drosophila,<br /> the pairs of chromosomes double approximately nine<br /> times (2 9 = 512). Figure 8.19a illustrates how repeated rounds of<br /> chromosomal replication produce a bundle of chromosomes that<br /> lie together in a parallel fashion. This bundle, termed a polytene<br /> chromosome, was first observed by E. G. Balbiani in 1881. Later,<br /> in the 1930s, Theophilus Painter and colleagues recognized that<br /> the size and morphology of polytene chromosomes provided<br /> geneticists with unique opportunities to study chromosome<br /> structure and gene organization.<br /> Figure 8.19b shows a micrograph of a polytene chromosome.<br /> The structure of polytene chromosomes is different from<br /> other forms of endopolyploidy because the replicated chromosomes<br /> remain attached to each other. Prior to the formation of<br /> polytene chromosomes, Drosophila cells contain eight chromosomes<br /> (two sets of four chromosomes each; see Figure 8.14a). In<br /> the salivary gland cells, the homologous chromosomes synapse<br /> with each other and replicate to form a polytene structure. During<br /> this process, the four types of chromosomes aggregate to form<br /> a single structure with several polytene arms. The central point<br /> where the chromosomes aggregate is known as the chromocenter.<br /> Each of the four types of chromosome is attached to the chromocenter<br /> near its centromere. The X and Y and chromosome 4<br /> are telocentric, and chromosomes 2 and 3 are metacentric. Therefore,<br /> chromosomes 2 and 3 have two arms that radiate from the<br /> chromocenter, while the X and Y and chromosome 4 have a single<br /> arm projecting from the chromocenter (Figure 8.19c).<br /> Because of their considerable size, polytene chromosomes<br /> lend themselves to an easy microscopic examination. Ordinarily,<br /> we use light microscopy to visualize the highly condensed<br /> metaphase chromosomes seen during mitosis or meiosis, as<br /> in Figure 8.1a. Because polytene chromosomes are so large, we<br /> (a) Repeated chromosome replication produces<br /> polytene chromosome.<br /> L<br /> 2<br /> 4<br /> 3<br /> L<br /> R<br /> x<br /> R<br /> Each polytene arm<br /> is composed of<br /> hundreds of<br /> chomosomes<br /> aligned side by side.<br /> Chromocenter<br /> (b) A polytene chromosome<br /> (c) Relationship between a polytene chromosome and<br /> regular Drosophila chromosomes<br /> FIGURE 8.19 Polytene chromosomes in Drosophila. (a) A schematic illustration of the formation of polytene chromosomes. Several rounds<br /> of repeated replication without cellular division result in a bundle of sister chromatids that lie side by side. Both homologues also lie parallel to each<br /> other. This replication does not occur in highly condensed, heterochromatic DNA near the centromere. (b) A photograph of a polytene chromosome.<br /> (c) This drawing shows the relationship between the four pairs of chromosomes and the formation of a polytene chromosome in the salivary gland.<br /> The heterochromatic regions of the chromosomes aggregate at the chromocenter, and the arms of the chromosomes project outward.</div> <div>17.1 HOMOLOGOUS RECOMBINATION 463<br /> 5′ 3′<br /> G<br /> C<br /> G C C<br /> C G G<br /> 3′ 5′<br /> Branch migration<br /> 5′<br /> 3′<br /> 3′<br /> 5′<br /> The top chromosome carries the recessive<br /> b allele, while the bottom chromosome<br /> carries the dominant B allele.<br /> Gene region<br /> b<br /> B<br /> b<br /> 3′<br /> 5′<br /> 5′<br /> 3′<br /> A double-strand break occurs<br /> within the recessive gene.<br /> 5′ 3′<br /> C C<br /> G A<br /> G G<br /> C T<br /> 3′ 5′<br /> Branch migration travels<br /> over a region that has a<br /> minor sequence difference.<br /> Heteroduplexes<br /> 5′<br /> 3′<br /> G<br /> G<br /> C C<br /> C<br /> G G<br /> T<br /> 3′ 5′<br /> B<br /> B<br /> A region adjacent to the doublestrand<br /> break is digested away,<br /> which eliminates the recessive b allele.<br /> Strand invasion causes<br /> D-loop formation.<br /> 5′ 3′<br /> A C C<br /> G<br /> G G<br /> C C<br /> 3′ 5′<br /> DNA mismatch repair yields 4<br /> possible combinations.<br /> B<br /> B<br /> Gap repair synthesis uses the strands<br /> from the dominant B allele to fill in the<br /> region.<br /> G<br /> C<br /> G C C<br /> C G G<br /> G<br /> C<br /> G C C<br /> C G G<br /> G<br /> C<br /> A<br /> T<br /> C C<br /> G G<br /> G<br /> C<br /> A<br /> T<br /> C C<br /> G G<br /> B<br /> The intertwined<br /> strands are resolved.<br /> G<br /> C<br /> G C C<br /> C G G<br /> G<br /> C<br /> A<br /> C C<br /> T G G<br /> G<br /> C<br /> G C C<br /> C G G<br /> G<br /> C<br /> A<br /> T<br /> C C<br /> G G<br /> B<br /> Gene<br /> conversion<br /> No gene<br /> conversion<br /> No gene<br /> conversion<br /> Gene<br /> conversion<br /> FIGURE 17.6 Gene conversion by DNA mismatch repair.<br /> A branch migrates past a homologous region that contains slightly<br /> different DNA sequences. This produces two heteroduplexes—DNA<br /> double helices with mismatches. The mismatches can be repaired in<br /> four possible ways by the mismatch repair system described in Chapter<br /> 16. Two of these ways result in gene conversion. The repaired base is<br /> shown in red.<br /> B<br /> Both chromosomes carry the B allele.<br /> FIGURE 17.7 Gene conversion by gap repair synthesis in the<br /> double-strand break model. A gene is found in two alleles, designated<br /> B and b. A double-strand break occurs in the DNA encoding the b allele.<br /> Both of these DNA strands are digested away, thereby eliminating the b<br /> allele. A complementary DNA strand encoding the B allele migrates to<br /> this region and provides the template to synthesize a double-stranded<br /> region. Following resolution, both DNA double helices carry the B allele.</div> <div>9.1 IDENTIFICATION OF DNA AS THE GENETIC MATERIAL 227<br /> 6. The heavy bacterial cells sediment to the<br /> pellet, while the lighter phages remain in<br /> the supernatant. (See Appendix for<br /> explanation of centrifugation.)<br /> 7. Count the amount of radioisotope in<br /> the supernatant with a scintillation<br /> counter (see Appendix). Compare it<br /> with the starting amount.<br /> Supernatant<br /> with<br /> 35 S-labeled<br /> empty phage<br /> Pellet with<br /> unlabeled<br /> DNA in<br /> infected<br /> E. coli cells<br /> Supernatant<br /> with<br /> unlabeled<br /> empty phage<br /> Pellet with<br /> 32 P-labeled<br /> DNA in<br /> infected<br /> E. coli cells<br /> (labeled)<br /> Sheared<br /> empty<br /> phage<br /> Sheared empty<br /> phages (labeled)<br /> Sheared<br /> empty<br /> phages<br /> (unlabeled)<br /> ■<br /> THE DATA<br /> Total isotope in supernatant (%)<br /> 100<br /> 80<br /> 60<br /> Extracellular 35 S<br /> 80%<br /> Blending removes 80%<br /> of 35 S from E. coli cells.<br /> 40<br /> 20<br /> Extracellular 32 P<br /> 35%<br /> Most of the 32 P (65%)<br /> remains with intact<br /> E. coli cells.<br /> 0<br /> 0 1 2 3 4 5 6 7 8<br /> Agitation time in blender (min)<br /> ■<br /> INTERPRETING THE DATA<br /> As seen in the data, most of the 35 S isotope was found in the<br /> supernatant. Because the shearing force was expected to remove<br /> the phage coat, this result indicates that the empty phages contain<br /> primarily protein. By comparison, only about 35% of the<br /> 32<br /> P was found in the supernatant following shearing. Therefore,<br /> most of the DNA was located within the bacterial cells in the<br /> pellet. These results are consistent with the idea that the DNA<br /> is injected into the bacterial cytoplasm during infection, which<br /> would be the expected result if DNA is the genetic material.<br /> By themselves, the results described in Figure 9.6 were not<br /> conclusive evidence that DNA is the genetic material. For exam-<br /> ple, you may have noticed that less than 100% of the phage protein<br /> was found in the supernatant. Therefore, some of the phage<br /> protein could have been introduced into the bacterial cells (and<br /> could function as the genetic material). Nevertheless, the results<br /> of Hershey and Chase were consistent with the conclusion that<br /> the genetic material is DNA rather than protein. Overall, their<br /> studies of the T2 phage were quite influential in convincing the<br /> scientific community that DNA is the genetic material.<br /> A self-help quiz involving this experiment can be found at<br /> the Online Learning Center.<br /> RNA Functions as the Genetic<br /> Material in Some Viruses<br /> We now know that bacteria, archaea, protists, fungi, plants, and<br /> animals all use DNA as their genetic material. As mentioned,<br /> viruses also have their own genetic material. Hershey and Chase<br /> concluded from their experiments that this genetic material is<br /> DNA. In the case of T2 bacteriophage, that is the correct conclusion.<br /> However, many viruses use RNA, rather than DNA, as their<br /> genetic material. In 1956, Alfred Gierer and Gerhard Schramm</div> <div>584 CHAPTER 21 :: GENOMICS II: FUNCTIONAL GENOMICS, PROTEOMICS, AND BIOINFORMATICS<br /> The two common types of protein microarray analyses are<br /> antibody microarrays and functional protein microarrays. The<br /> purpose of an antibody microarray is to study protein expression.<br /> As discussed in Chapter 17, antibodies are proteins that recognize<br /> antigens. One type of antigen that an antibody can recognize is a<br /> short peptide sequence found within another protein. Therefore,<br /> an antibody can specifically recognize a cellular protein. Researchers<br /> can produce thousands of different antibodies, each one recognizing<br /> a different peptide sequence. These can be spotted onto<br /> a microarray. Cellular proteins can then be isolated, fluorescently<br /> labeled, and exposed to the antibody microarray. When a given<br /> protein is recognized by an antibody on the microarray, it will be<br /> captured by the antibody and remain bound to that spot. The level<br /> of fluorescence at a given spot indicates the amount of a cellular<br /> protein that is recognized by a particular antibody.<br /> The other type of array is a functional protein microarray.<br /> To make this type of array, researchers must purify cellular<br /> proteins and then spot them onto a microarray. The microarray<br /> can then be analyzed with regard to specific kinds of protein<br /> function. In 2000, for example, Heng Zhu, Michael Snyder,<br /> and colleagues purified 119 proteins from yeast that were known<br /> to function as protein kinases. These kinds of proteins attach<br /> phosphate groups onto other cellular proteins. A microarray was<br /> made consisting of different possible proteins that may or may<br /> not be phosphorylated by these 119 kinases, and then the array<br /> was exposed to each of the kinases in the presence of radiolabeled<br /> ATP. By following the incorporation of phosphate into the<br /> array, they determined the protein specificity of each kinase. On<br /> a much larger scale, the same group of researchers purified 5,800<br /> different yeast proteins and spotted them onto a microarray.<br /> The array was then exposed to fluorescently labeled calmodulin,<br /> which is a regulatory protein that binds calcium ions. Several<br /> proteins in the microarray were found to bind calmodulin. While<br /> some of these were already known to be regulated by calmodulin,<br /> other proteins in the array that had not been previously known<br /> to bind calmodulin were identified.<br /> 21.3 BIOINFORMATICS<br /> Geneticists use computers to collect, store, manipulate, and analyze<br /> data. Molecular genetic data is particularly amenable to<br /> computer analysis because it comes in the form of a sequence,<br /> such as a DNA, RNA, or amino acid sequence. The ability of<br /> computers to analyze data at a rate of millions or even billions<br /> of operations per second has made it possible to solve problems<br /> concerning genetic information that were thought intractable a<br /> few decades ago.<br /> In recent years, the marriage between genetics and computational<br /> tools and approaches has yielded an important branch<br /> of science known as bioinformatics. Several scientific journals<br /> are largely devoted to this topic. Computer analysis of genetic<br /> sequences relies on three basic components: a computer, a computer<br /> program, and some type of data. In genetic research, the<br /> data consist of a particular genetic sequence or several sequences<br /> that a researcher or clinician wants to study. For example, this<br /> could be a DNA sequence derived from a cloned DNA fragment.<br /> A sequence of interest may be relatively short or thousands to<br /> millions of nucleotides in length. Experimentally, DNA sequences<br /> and related data are obtained using the techniques described in<br /> Chapters 18 and 20.<br /> In this section, we will first consider the fundamental concepts<br /> that underlie the analysis of genetic sequences. We will<br /> then explore how these methods are used to provide insights<br /> regarding functional genomics and proteomics. Chapter 26 will<br /> describe applications of bioinformatics in the area of evolutionary<br /> biology. In addition, you may wish to actually run computer<br /> programs, which are widely available at university and government<br /> websites (for example, see www.ncbi.nlm.nih.gov/). This<br /> type of hands-on learning will help you to see how the computer<br /> has become a valuable tool to analyze genetic data.<br /> Sequence Files Are Analyzed by Computer Programs<br /> Most people are familiar with computer programs, which consist<br /> of a defined series of operations that can manipulate and<br /> analyze data in a desired way. For example, a computer program<br /> might be designed to take a DNA sequence and translate it into<br /> an amino acid sequence. A first step in the computer analysis of<br /> genetic data is the creation of a computer data file to store the<br /> data. This file is simply a collection of information in a form<br /> suitable for storage and manipulation on a computer. In genetic<br /> studies, a computer data file might contain an experimentally<br /> obtained DNA, RNA, or amino acid sequence. For example, a file<br /> could contain the DNA sequence of one strand of the lacY gene<br /> from Escherichia coli, as shown here. The numbers to the left represent<br /> the base number in the sequence file.<br /> 1 ATGTACTATT<br /> 51 CTTTTACTTT<br /> 101 TACATGACAT<br /> 151 GCTATTTCTC<br /> 201 TGACAAACTC<br /> 251 TAGTCATGTT<br /> 301 TACAACATTT<br /> 351 TTTTAACGCC<br /> 401 GTCGCAGTAA<br /> 451 TGGGCGCTGT<br /> 501 GTTTGTTTTC<br /> 551 TCTTTTTCGC<br /> 601 GTAGGTGCCA<br /> 651 CAGACAGCCA<br /> 701 GCACCTACGA<br /> 751 TTTGCTACCG<br /> 801 GGGCGCAATTA<br /> 851<br /> 901<br /> 951<br /> 1,001<br /> 1,051<br /> 1,101<br /> 1,151<br /> ATCGCATCGG<br /> CTACGTATTA<br /> TCTGAAAACG<br /> TTAAATATAT<br /> CTGGTCTGTT<br /> ACTGGCGGGC<br /> TGCTGGGTCT<br /> 1,201 AGCGGCCCCG<br /> 1,251 TTAA<br /> TAAAAAACAC<br /> TTTATCATGG<br /> CAACCATATC<br /> TGTTCTCGCT<br /> GGGCTGCGCA<br /> TGCGCCGTTC<br /> TAGTAGGATC<br /> GGTGCGCCAG<br /> TTTCGAATTT<br /> GTGCCTGAT<br /> TGGCTGGGCT<br /> CAAAACGGAT<br /> ACCATTCGGC<br /> AAACTGTGGT<br /> TGTTTTTGAC<br /> GTGAACAGGG<br /> CTTAACGCCT<br /> TGGGAAAAAC<br /> TTGGCTCATC<br /> CTGCATATGT<br /> TACCAGCCAG<br /> TCTGCTTCTT<br /> AATATGTATG<br /> GGTGGCGCTG<br /> AAACTTTTGG<br /> GAGCCTACTT<br /> AGCAAAGTG<br /> ATTATTCCAA<br /> AATACCTGCT<br /> TTTATTTTTA<br /> GATTGTTGGT<br /> CAGTAGAGGC<br /> GGTCGCGCGC<br /> TGTCGGCATC<br /> CTGGCTGTGC<br /> GCGCCCTCTT<br /> ATTTAGCCTT<br /> TTTGTCTACT<br /> CAACAGTTTG<br /> TACGCGGGTA<br /> CGATTATCTT<br /> GCCCTGCTGC<br /> GTTCGCCACC<br /> TTGAAGTACC<br /> TTTGAAGTGC<br /> TAAGCAACTG<br /> AAAGCATCGG<br /> GGCTTCACCT<br /> GCCCGCTTTC CCTGCTGCGT<br /> ATGTTCGGTT<br /> CCCGTTTTTC<br /> ATACGGGTAT<br /> CCGCTGTTTG<br /> GTGGATTATT<br /> TCTTCGGGCC<br /> GGTATTTATC<br /> ATTTATTGAG<br /> GGATGTTTGG<br /> ATGTTCACCA<br /> ACTCATCCTC<br /> CTGCCACGGT<br /> AAGTGGCAC<br /> GTATGTTATT<br /> CTAATTTCTT<br /> TTTGGCTACG<br /> CTTGCGCCA<br /> TGGCTGGCAC<br /> TCAGCGCTGG<br /> GTTCCTGCTG<br /> GTTTTTCAGC<br /> GCATGATTT<br /> TTTCCAGGGC<br /> TAATTTCCGT<br /> TATTCTTTTT<br /> CCGATTTGGC<br /> TATTTTGCC<br /> GTCTGCTTC<br /> ACGGCATGT<br /> ACTGTTACAA<br /> TAGGCTTTTG<br /> AAAGTCAGCC<br /> CTGTGTTGGC<br /> TCAATAATCA<br /> GCCGTTTTAC<br /> TGCCAATGCG<br /> TGGAACTGTT<br /> GGCGTTTCCT<br /> TACTTCGTTC<br /> TAACGACAAT<br /> CTGATCATTA<br /> TATTATCTCT<br /> AAGTGGTTAT<br /> GTGGGCTGCT<br /> GACGATTTAT<br /> TTATGTCTGT<br /> GCTTATCTGG<br /> GTTCACGCTT<br /> CGTCAGGTGA ATGAAGTCGC<br /> To store data in a computer data file, a scientist creates the file<br /> and enters the data, either by hand or, what is now more common,<br /> by laboratory instruments such as densitometers and<br /> fluorometers. These instruments have the capability to read data,<br /> such as a sequencing ladder, and enter the DNA sequence information<br /> directly into a computer file.<br /> The purpose of making a computer file that contains a<br /> genetic sequence is to take advantage of the swift speed with<br /> which computers can analyze this information. Genetic sequence</div> <div>208 CHAPTER 8 :: VARIATION IN CHROMOSOME STRUCTURE AND NUMBER<br /> Nondisjunction<br /> in meiosis I<br /> Normal<br /> meiosis I<br /> Normal<br /> meiosis II<br /> Nondisjunction<br /> in meiosis II<br /> n + 1<br /> n + 1 n – 1 n – 1 n + 1<br /> n – 1 n n<br /> (a) Nondisjunction in meiosis I<br /> (b) Nondisjunction in meiosis II<br /> FIGURE 8.22 Nondisjunction during meiosis I and II. The chromosomes shown in purple are behaving properly during meiosis I and II,<br /> so each cell receives one copy of this chromosome. The chromosomes shown in blue are not disjoining correctly. In (a), nondisjunction occurred in<br /> meiosis I, so the resulting four cells receive either two copies of the blue chromosome or zero copies. In (b), nondisjunction occurred during meiosis<br /> II, so one cell has two blue chromosomes and another cell has zero. The remaining two cells are normal.<br /> and two normal haploid cells. If a gamete that is missing a chromosome<br /> is viable and participates in fertilization, the resulting<br /> offspring is monosomic for the missing chromosomes. Alternatively,<br /> if a gamete carrying an extra chromosome unites with a<br /> normal gamete, the offspring will be trisomic.<br /> In rare cases, all of the chromosomes can undergo nondisjunction<br /> and migrate to one of the daughter cells. The net result of<br /> complete nondisjunction is a diploid cell and a cell without any<br /> chromosomes. While the cell without chromosomes is nonviable,<br /> the diploid cell might participate in fertilization with a normal<br /> haploid gamete to produce a triploid individual. Therefore, complete<br /> nondisjunction can produce individuals that are polyploid.<br /> Mitotic Nondisjunction or Chromosome<br /> Loss Can Produce a Patch of Tissue<br /> with an Altered Chromosome Number<br /> Abnormalities in chromosome number occasionally occur after<br /> fertilization takes place. In this case, the abnormal event happens<br /> during mitosis rather than meiosis. One possibility is that the sister<br /> chromatids separate improperly, so one daughter cell has three<br /> copies of that chromosome while the other daughter cell has only<br /> one (Figure 8.23a). Alternatively, the sister chromatids could separate<br /> during anaphase of mitosis, but one of the chromosomes<br /> could be improperly attached to the spindle, so that it would not<br /> migrate to a pole (Figure 8.23b). A chromosome will be degraded<br /> if it is left outside the nucleus when the nuclear membrane reforms.<br /> In this case, one of the daughter cells would have two copies<br /> of that chromosome, while the other would have only one.<br /> When genetic abnormalities occur after fertilization, the<br /> organism will contain a subset of cells that are genetically different<br /> from those of the rest of the organism. This condition is<br /> referred to as mosaicism. The size and location of the mosaic<br /> region depend on the timing and location of the original abnormal<br /> event. If a genetic alteration happens very early in the<br /> embryonic development of an organism, the abnormal cell will<br /> be the precursor for a large section of the organism. In the most<br /> extreme case, an abnormality could take place at the first mitotic<br /> division. As a bizarre example, consider a fertilized Drosophila<br /> egg that is XX. One of the X chromosomes may be lost during<br /> the first mitotic division, producing one daughter cell that is XX<br /> and one that is X0. Flies that are XX develop into females, and<br /> X0 flies develop into males. Therefore, in this example, one-half<br /> of the organism will become female and one-half will become<br /> male! This peculiar and rare individual is referred to as a bilateral<br /> gynandromorph (Figure 8.24).</div> <div>17.3 TRANSPOSITION 469<br /> c sh Wx<br /> 5. Conduct crosses to determine the<br /> location of Ds in the plants derived from<br /> the solid red kernels. Note: This can be<br /> done using the chromosomal markers<br /> and the strategy that was described at<br /> the beginning of this experiment. When<br /> observing the results of these crosses,<br /> determine whether the C gene is still<br /> mutable. (Is red sectoring occurring?)<br /> Make crosses and<br /> observe kernels.<br /> ■<br /> THE DATA<br /> Kernel<br /> Location<br /> Strain Phenotype of Ds Mutability?<br /> From parental Colorless Within the Yes, red sectorcross<br /> background C gene ing occurred in<br /> (see step 2) with red sectoring strains containing<br /> Ac.<br /> From red Red kernels Ds had No, the C gene<br /> kernels moved out was stable; no<br /> (see step 3)<br /> of the C gene sectoring was<br /> to another observed.<br /> location.<br /> ■ INTERPRETING THE DATA<br /> By conducting the appropriate crosses, McClintock found that in<br /> the progeny of a solid red kernel, the Ds locus had moved out<br /> of the C gene to another location (see the data of Figure 17.11).<br /> In addition, the “restored” C gene behaved normally. In other<br /> words, the C gene was no longer highly mutable, and the kernels<br /> did not show a sectoring phenotype. Taken together, the results<br /> are consistent with the hypothesis that the Ds locus can move<br /> around the corn genome by transposition.<br /> When McClintock published these results in 1951, they<br /> were met with great skepticism. Some geneticists of that time<br /> were unable to accept the idea that the genetic material was susceptible<br /> to frequent rearrangement. Instead, they believed that<br /> the genetic material was always very stable and permanent in its<br /> structure. Over the next several decades, the scientific community<br /> came to realize that transposable elements are a widespread phenomenon.<br /> Much like Gregor Mendel and Charles Darwin, Barbara<br /> McClintock was clearly ahead of her time. She was awarded<br /> the Nobel Prize in 1983, more than 30 years after her original<br /> discovery.<br /> A self-help quiz involving this experiment can be found at<br /> the Online Learning Center.<br /> Transposable Elements and Retroelements Move<br /> Via One of Three Transposition Pathways<br /> Since the pioneering studies of Barbara McClintock, many different<br /> transposable elements (TEs) have been found in bacteria,<br /> fungi, plants, and animals. Three general types of transposition<br /> pathways have been identified (Figure 17.12). In simple, or conservative<br /> transposition, the TE is removed from its original site<br /> and transferred to a new target site. This mechanism is also called<br /> a cut-and-paste mechanism because the element is cut out of its<br /> original site and pasted into a new site. Transposable elements<br /> that move via simple transposition are widely found in bacterial<br /> and eukaryotic species. By comparison, replicative transposition<br /> involves the replication of the TE and insertion of the newly<br /> made copy into a second site. In this case, one of the TEs remains<br /> in its original location, and the other is inserted at another location.<br /> Replicative transposition is relatively uncommon and found<br /> only in bacterial species.<br /> The first two categories of TEs we have considered thus<br /> far move as a DNA molecule from one site to another. Such TEs<br /> are called transposons. A third category of elements move via<br /> an RNA intermediate. This form of transposition, termed retrotransposition,<br /> is found in eukaryotic species, where it is very<br /> common. Transposable elements that move via retrotransposition<br /> are known as retroelements, retrotransposons, or retroposons.<br /> In retrotransposition, the element is transcribed into RNA.<br /> An enzyme called reverse transcriptase uses the RNA as a template<br /> to synthesize a DNA molecule that is integrated into a new<br /> region of the genome. Like replicative transposons, retroelements<br /> increase in number during retrotransposition.</div> <div>748 CHAPTER 26 :: EVOLUTIONARY GENETICS<br /> Chromosome 1<br /> 3<br /> 2<br /> 1<br /> 1<br /> 2<br /> 3<br /> 4<br /> 6<br /> 5<br /> 4<br /> 3<br /> 2<br /> .33<br /> .23<br /> .22<br /> .21<br /> .13<br /> .12<br /> .11<br /> .3<br /> .2<br /> .1<br /> .3<br /> .2<br /> .1<br /> .32 .31<br /> .3<br /> .2<br /> .1<br /> .3<br /> 1<br /> .2<br /> .13<br /> .12<br /> .11<br /> 2<br /> 1<br /> .3<br /> .2<br /> .1<br /> .3<br /> .2<br /> .1<br /> .3<br /> 3 .2<br /> .1<br /> 2 .2<br /> 1 .1<br /> 1 .1<br /> .2<br /> 2<br /> .1<br /> .21<br /> .22<br /> 1 .23<br /> .3<br /> 2 .1<br /> 3 .2<br /> .3<br /> .1<br /> 4 .2<br /> .3<br /> .1<br /> 5 .2<br /> .3<br /> 1<br /> .1<br /> .2<br /> .3<br /> 2<br /> 1<br /> .1<br /> .2<br /> .3<br /> .11<br /> .12<br /> .13 2<br /> .2<br /> .3<br /> 3<br /> 4<br /> Human<br /> Chimpanzee<br /> Gorilla<br /> Orangutan<br /> Chromosome 2<br /> 7 .2<br /> .3<br /> .1<br /> 3 3<br /> Chromosome 3<br /> 7<br /> 6<br /> .1<br /> .2<br /> .3<br /> 4<br /> 5<br /> 6<br /> .1<br /> .2<br /> .3<br /> 5<br /> 2<br /> .1 1<br /> 1 .2<br /> .3<br /> 2<br /> .11<br /> .12<br /> .13<br /> 3<br /> 2 .2<br /> .31<br /> 4<br /> .32<br /> .33<br /> 8<br /> 9<br /> .1<br /> .1<br /> 2 .2<br /> .3<br /> 2 .1<br /> 3 .2<br /> .3<br /> .1<br /> .2<br /> 4<br /> .3<br /> 1<br /> .1<br /> .2<br /> .3<br /> 1<br /> 1<br /> 2<br /> 1<br /> 3<br /> 4<br /> .1<br /> .2<br /> .3<br /> 5<br /> 4<br /> 1 .3<br /> 3 .2 .1<br /> .3<br /> 2 .2<br /> .1<br /> .23<br /> .22<br /> 1 .21<br /> .1<br /> .1<br /> .21<br /> 1 .23<br /> .23<br /> .1<br /> 2 .2<br /> .3<br /> 1 3<br /> 3<br /> 1<br /> 2<br /> 4<br /> 1<br /> 4<br /> 2 3<br /> 2<br /> 6<br /> .3<br /> .2<br /> .1<br /> 1<br /> 5<br /> 2<br /> .3<br /> .2<br /> .1<br /> 6<br /> .3<br /> .2 5<br /> .1<br /> .3<br /> 4 .2<br /> .1<br /> 2 .3<br /> 3 .2<br /> .1<br /> Human<br /> Chimpanzee<br /> Gorilla<br /> Orangutan<br /> .2<br /> .31<br /> .32<br /> .33<br /> .1<br /> .2<br /> .3<br /> .1<br /> .1 .2 .31<br /> .32<br /> .33<br /> .2<br /> .3<br /> .1 .2 .3<br /> .2<br /> .31<br /> .32<br /> .33<br /> .1<br /> .2<br /> .3<br /> .11<br /> .12<br /> .13<br /> .1<br /> .3<br /> .2<br /> .1<br /> .2<br /> Inversion<br /> .2<br /> .1<br /> .2<br /> .1<br /> .3<br /> .31<br /> .3<br /> .2<br /> .1<br /> .33<br /> .32<br /> .3<br /> .2<br /> .1<br /> .3<br /> .2<br /> .1<br /> .3<br /> .2<br /> .1<br /> Human<br /> Chimpanzee<br /> Gorilla<br /> Orangutan<br /> FIGURE 26.18 A comparison of banding<br /> patterns among the three largest human<br /> chromosomes and the corresponding chromosomes<br /> in apes. This is a schematic drawing of late prophase<br /> chromosomes. The conventional numbering system<br /> of the banding patterns is shown next to the human<br /> chromosomes.<br /> Rice Wheat Maize Foxtail millet Sugarcane Sorghum<br /> R1a R5a R12a R1a R1a R1a R1a<br /> 1<br /> 3<br /> F<br /> 1<br /> R1b 1 R10 R1a R5a R1b R1b R1b<br /> 8 F<br /> 2 R2<br /> R5b R1b R5b R5a<br /> R5a R5a<br /> R3a<br /> R4a<br /> R1b R5b I<br /> 2<br /> R6a<br /> R5b R5b<br /> 3 R3b<br /> 2 R7<br /> R6b R6a R6a<br /> 6<br /> H<br /> R6a R6a<br /> R4b<br /> R3c<br /> R5a R6b R6b H<br /> 3<br /> 9<br /> R6b R6b<br /> R4a<br /> R1a<br /> R5b R8<br /> B R8<br /> 4<br /> 3<br /> R4b<br /> R1b R3c<br /> R3c<br /> E R8 4 R8<br /> R3c<br /> R3b<br /> R8 R2<br /> A R3c R3c<br /> R5a<br /> R10<br /> 4<br /> 5<br /> R5b<br /> R3c<br /> 1 R10 R10 C<br /> R10<br /> 5 R3b<br /> R10 5<br /> R3b R3b<br /> R3b<br /> R6a<br /> R12a<br /> R3a<br /> C R3b<br /> R3a R3a<br /> R3a<br /> R3a<br /> 6<br /> R6b<br /> R11a<br /> G R2<br /> R11a R11a<br /> R2<br /> R11b<br /> G R2<br /> 6<br /> 4<br /> 7 R7 5<br /> R12b<br /> R2 R9 7<br /> A R12a<br /> R12a<br /> 7 R12a<br /> 8 R8<br /> R9<br /> R4a<br /> R7 D R11a<br /> J R12b 8 R11a<br /> R3a R4b R4a<br /> 9 R9<br /> 2<br /> I<br /> R9 R11b<br /> R9<br /> 6 R2<br /> R9 R4b 10 R7<br /> 9<br /> 10 R10<br /> R11a R7<br /> R7 R12a<br /> R6a<br /> R4a D R9<br /> R11a<br /> E<br /> R4a<br /> 11<br /> 7 R8<br /> R4b<br /> R7 10<br /> R11b<br /> R4b<br /> R6b<br /> R4a<br /> 12<br /> R12a<br /> B<br /> R4b<br /> R12b<br /> FIGURE 26.19 Synteny groups in cereal<br /> grasses. The synteny groups are named according to<br /> their relative arrangement in rice, which is shown on<br /> the left. In some cases, a synteny group or portion of<br /> a group may have incurred an inversion, but these<br /> are not shown in this figure. Also, the genome of<br /> maize contains two copies of most synteny groups,<br /> suggesting that it is derived from an ancestral<br /> polyploid species.</div> <div>5.1 LINKAGE AND CROSSING OVER 103<br /> higher proportion of parental combinations was that all three<br /> genes are located on the X chromosome and, therefore, tend to<br /> be transmitted together as a unit.<br /> However, to fully account for the data shown in Figure 5.3,<br /> Morgan needed to interpret two other key observations. First,<br /> he needed to explain why a significant proportion of the F 2 generation<br /> had nonparental combinations of alleles. Along with the<br /> two parental phenotypes, five other phenotypic combinations<br /> appeared that were not found in the parental generation. Second,<br /> he needed to explain why a quantitative difference was observed<br /> between nonparental combinations involving body color and eye<br /> color versus eye color and wing length. This quantitative difference<br /> is revealed by reorganizing the data from Morgan’s cross by<br /> pairs of genes.<br /> Gray body, red eyes 1,159<br /> Yellow body, white eyes 1,017<br /> Gray body, white eyes 17 Nonparental<br /> Yellow body, red eyes 12 offspring<br /> Total 2,205<br /> Red eyes, long wings 770<br /> White eyes, miniature wings 716<br /> Red eyes, miniature wings 401 Nonparental<br /> White eyes, long wings 318 offspring<br /> Total 2,205<br /> Morgan found a substantial difference between the numbers<br /> of nonparental offspring when pairs of genes were considered<br /> separately. Nonparental combinations involving only eye color<br /> and wing length were fairly common—401 + 318 nonparental<br /> offspring. In sharp contrast, nonparental combinations for<br /> body color and eye color were quite rare—17 + 12 nonparental<br /> offspring.<br /> How did Morgan explain these data? He considered the<br /> studies conducted in 1909 of the Belgian cytologist F. A. Janssens,<br /> who observed chiasmata under the microscope and proposed<br /> that crossing over involves a physical exchange between homologous<br /> chromosomes. Morgan shrewdly realized that crossing over<br /> between homologous X chromosomes was consistent with his<br /> data. He assumed that crossing over did not occur between the<br /> X and Y chromosome and that these three genes are not found<br /> on the Y chromosome. With these ideas in mind, he made three<br /> important hypotheses to explain his results:<br /> 1. The genes for body color, eye color, and wing length<br /> are all located on the same chromosome, namely, the X<br /> chromosome. Therefore, the alleles for all three traits are<br /> most likely to be inherited together.<br /> 2. Due to crossing over, the homologous X chromosomes<br /> (in the female) can exchange pieces of chromosomes and<br /> create new (nonparental) combinations of alleles.<br /> 3. The likelihood of crossing over depends on the distance<br /> between two genes. If two genes are far apart from each<br /> other, crossing over is more likely to occur between them.<br /> With these ideas in mind, Figure 5.4 illustrates the possible<br /> events that occurred in the F 1 female flies of Morgan’s experiment.<br /> One of the X chromosomes carried all three dominant<br /> alleles, while the other had all three recessive alleles. During<br /> oogenesis in the F 1 female flies, crossing over may or may not<br /> have occurred in this region of the X chromosome. If no crossing<br /> over occurred, the parental phenotypes were produced in<br /> the F 2 offspring. Alternatively, a crossover sometimes occurred<br /> between the eye color gene and the wing length gene to produce<br /> nonparental offspring with gray bodies, red eyes, and miniature<br /> wings or yellow bodies, white eyes, and long wings. According<br /> to Morgan’s proposal, such an event is fairly likely because these<br /> two genes are far apart from each other on the X chromosome.<br /> In contrast, he proposed that the body color and eye color genes<br /> are very close together, which makes crossing over between them<br /> an unlikely event. Nevertheless, it occasionally occurred, yielding<br /> offspring with gray bodies, white eyes, and miniature wings,<br /> or with yellow bodies, red eyes, and long wings. Finally, it was<br /> also possible for two homologous chromosomes to cross over<br /> twice. This double crossover is very unlikely. Among the 2,205<br /> offspring Morgan examined, he found only one fly with a gray<br /> body, white eyes, and long wings that could be explained by this<br /> phenomenon.<br /> A Chi Square Analysis Can Be Used to Distinguish<br /> Between Linkage and Independent Assortment<br /> Now that we have an appreciation for linkage and the production<br /> of recombinant offspring, let’s consider how an experimenter can<br /> objectively decide whether two genes are linked or assort independently.<br /> In Chapter 2, we used chi square analysis to evaluate<br /> the goodness of fit between a genetic hypothesis and observed<br /> experimental data. This method can similarly be employed to<br /> determine if the outcome of a dihybrid cross is consistent with<br /> linkage or independent assortment.<br /> To conduct a chi square analysis, we must first propose a<br /> hypothesis. In a dihybrid cross, the standard hypothesis is that<br /> the two genes are not linked. This hypothesis is chosen even<br /> if the observed data suggest linkage, because an independent<br /> assortment hypothesis allows us to calculate the expected number<br /> of offspring based on the genotypes of the parents and the<br /> law of independent assortment. In contrast, for two linked genes<br /> that have not been previously mapped, we cannot calculate the<br /> expected number of offspring from a genetic cross because we<br /> do not know how likely it is for a crossover to occur between<br /> the two genes. Without expected numbers of recombinant and<br /> parental offspring, we cannot conduct a chi square test. Therefore,<br /> we begin with the hypothesis that the genes are not linked.<br /> Recall from Chapter 2 that the hypothesis we are testing is called<br /> a null hypothesis, because it assumes there is no real difference<br /> between the observed and expected values. The goal is to determine<br /> whether or not the data fit the hypothesis. If the chi square<br /> value is low and we cannot reject the null hypothesis, we infer that<br /> the genes assort independently. On the other hand, if the chi square<br /> value is so high that our hypothesis is rejected, we will accept the<br /> alternative hypothesis, namely, that the genes are linked.</div> <div>724 CHAPTER 25 :: QUANTITATIVE GENETICS<br /> Experimental Questions<br /> E1 . Here are data for height and weight among 10 male college<br /> students.<br /> Height (cm)<br /> Weight (kg)<br /> 159 48<br /> 162 50<br /> 161 52<br /> 175 60<br /> 174 64<br /> 198 81<br /> 172 58<br /> 180 74<br /> 161 50<br /> 173 54<br /> A. Calculate the correlation coefficients for this group.<br /> B. Is the correlation coefficient statistically significant? Explain.<br /> E2. The abdomen length (in millimeters) was measured in 15 male<br /> Drosophila, and the following data were obtained: 1.9, 2.4, 2.1, 2.0,<br /> 2.2, 2.4, 1.7, 1.8, 2.0, 2.0, 2.3, 2.1, 1.6, 2.3, and 2.2. Calculate the<br /> mean, standard deviation, and variance for this population of male<br /> fruit flies.<br /> E3. You will need to understand solved problem S5 before answering<br /> this question. The average weights for two varieties of cattle were<br /> 514 kg and 621 kg. The genetic variance for weight calculated for<br /> both strains was 382 kg 2 . What is the minimum number of genes<br /> that affect weight variation in these two varieties of cattle?<br /> E4. Using the same strategy as the experiment of Figure 25.6, the<br /> following data are the survival of F 2 offspring obtained from<br /> backcrosses to insecticide-resistant and control strains:<br /> 1<br /> 2<br /> 3<br /> 4<br /> 5<br /> 6<br /> 7<br /> 8<br /> 9<br /> 10<br /> 11<br /> 12<br /> 13<br /> 14<br /> 15<br /> 16<br /> Chromosomes<br /> Percent survival<br /> X 2 3 0 1 5 10 20 40 60 80<br /> Chromosome from<br /> control strain<br /> Chromosome from<br /> resistant strain<br /> Interpret these results with regard to the locations of QTLs.<br /> E5. In one strain of cabbage, you conduct an RFLP analysis with<br /> regard to head weight; you determine that seven QTLs affect this<br /> trait. In another strain of cabbage, you find that only four QTLs<br /> affect this trait. Note that both strains of cabbage are from the<br /> same species, although they may have been subjected to different<br /> degrees of inbreeding. Explain how one strain can have seven<br /> QTLs and another strain four QTLs for exactly the same trait. Is<br /> the second strain missing three genes?<br /> E6. From an experimental viewpoint, what does it mean to say that<br /> an RFLP is associated with a trait? Let’s suppose that two strains<br /> of pea plants differ in two RFLPs that are linked to two genes<br /> governing pea size. RFLP-1 is found in 2,000 bp and 2,700 bp<br /> bands, and RFLP-2 is found in 3,000 bp and 4,000 bp bands. The<br /> plants producing large peas have RFLP-1 (2,000 bp) and RFLP-2<br /> (3,000 bp); those producing small peas have RFLP-1 (2,700 bp)<br /> and RFLP-2 (4,000 bp). A cross is made between these two strains,<br /> and the F 1 offspring are allowed to self-fertilize. Five phenotypic<br /> classes are observed: small peas, small-medium peas, medium<br /> peas, medium-large peas, and large peas. We assume that each<br /> of the two genes makes an equal contribution to pea size and<br /> that the genetic variance is additive. Draw a gel and explain what<br /> RFLP banding patterns you would expect to observe for these five<br /> phenotypic categories. Note: Certain phenotypic categories may<br /> have more than one possible banding pattern.<br /> E7. Let’s suppose that two strains of pigs differ in 500 RFLPs. One<br /> strain is much larger than the other. The pigs are crossed to each<br /> other, and the members of the F 1 generation are also crossed<br /> among themselves to produce an F 2 generation. Three distinct<br /> RFLPs are associated with F 2 pigs that are larger. How would you<br /> interpret these results?<br /> E8. Outline the steps you would follow to determine the number of<br /> genes that influence the yield of rice. Describe the results you<br /> might get if rice yield is governed by six different genes.<br /> E9. A researcher has two highly inbred strains of mice. One strain is<br /> susceptible to infection by a mouse leukemia virus, while the other<br /> strain is resistant. Susceptibility/resistance is a polygenic trait.<br /> The two strains were crossed together, and all the F 1 mice were<br /> resistant. The F 1 mice were then allowed to interbreed, and 120<br /> F 2 mice were obtained. Among these 120 mice, 118 were resistant<br /> to the viral pathogen, and 2 were sensitive. Discuss how many<br /> different genes may be involved in this trait. How would your<br /> answer differ if none of the F 2 mice had been susceptible to the<br /> leukemia virus? Hint: You should assume that the inheritance of<br /> one viral-resistance allele is sufficient to confer resistance.<br /> E10. In a wild strain of tomato plants, the phenotypic variance for<br /> tomato weight is 3.2 g 2 . In another strain of highly inbred<br /> tomatoes raised under the same environmental conditions, the<br /> phenotypic variance is 2.2 g 2 . With regard to the wild strain,<br /> A. Estimate V G .<br /> B. What is h 2 B ?<br /> C. Assuming all the genetic variance is additive, what is h 2 N ?<br /> E11. The average thorax length in a Drosophila population is 1.01 mm.<br /> You want to practice selective breeding to make larger Drosophila.<br /> To do so, you choose 10 parents (five males and five females) of<br /> the following sizes: 0.97, 0.99, 1.05, 1.06, 1.03, 1.21, 1.22, 1.17, 1.19,<br /> and 1.20. You mate them and then analyze the thorax sizes of 30<br /> offspring (half male and half female):<br /> 0.99, 1.15, 1.20, 1.33, 1.07, 1.11, 1.21, 0.94, 1.07, 1.11, 1.20, 1.01,<br /> 1.02, 1.05, 1.21, 1.22, 1.03, 0.99, 1.20, 1.10, 0.91, 0.94, 1.13, 1.14,<br /> 1.20, 0.89, 1.10, 1.04, 1.01, 1.26<br /> Calculate the realized heritability in this group of flies.</div> <div>IR<br /> IR<br /> IR<br /> IR<br /> 478 CHAPTER 17 :: RECOMBINATION AND TRANSPOSITION AT THE MOLECULAR LEVEL<br /> Answer: These results can be explained by gene conversion. The gene<br /> conversion took place in a limited region of the chromosome (within<br /> the pdx-1 gene), but it did not affect the flanking genes (pyr-1 and<br /> col-4) located on either side of the pdx-1 gene. In the asci containing<br /> two pdx-1 alleles and six pdx-1 + alleles, a crossover occurred during<br /> meiosis I in the region of the pdx-1 gene. Gene conversion changed<br /> the pdx-1 allele into the pdx-1 + allele. This gene conversion could have<br /> occurred by two mechanisms. If branch migration occurred across<br /> the pdx-1 gene, a heteroduplex may have formed, and this could be<br /> repaired by mismatch DNA repair, as described in Figure 17.6. In the<br /> aberrant asci with two pdx-1 and six pdx-1 + alleles, the pdx-1 allele<br /> was converted to pdx-1 + . Alternatively, gene conversion of pdx-1 into<br /> pdx-1 + could have taken place via gap repair synthesis, as described in<br /> Figure 17.7. In this case, the pdx-1 allele would have been digested away,<br /> and the DNA encoding the pdx-1 + allele would have migrated into the<br /> digested region and provided a template to make a copy of the pdx-1 +<br /> allele. (Note: Since this pioneering work, additional studies have shed<br /> considerable light concerning the phenomenon of gene conversion.<br /> It occurs at a fairly high rate in fungi, approximately 0.1 to 1% of the<br /> time. It is not due to new mutations occurring during meiosis.)<br /> S2. Recombination involves the pairing of identical or similar<br /> sequences, followed by crossing over and the resolution of the<br /> intertwined helices. On rare occasions, the direct repeats or the<br /> inverted repeats within a single transposable element can align and<br /> undergo homologous recombination. What are the consequences<br /> when the direct repeats recombine? What are the consequences<br /> when the inverted repeats recombine?<br /> Answer:<br /> IR<br /> Transposase<br /> DR<br /> DR<br /> DR<br /> DR<br /> The sequence within the transposable element has been inverted.<br /> Note that the transposase gene has changed to the opposite direction.<br /> (b) Recombination between inverted repeats<br /> S3. A schematic drawing of an uncrossed Holliday junction is shown<br /> here. One chromatid is shown in red, and the homologous<br /> chromatid is shown in blue. The red chromatid carries a dominant<br /> allele labeled A and a recessive allele labeled b, whereas the blue<br /> chromatid carries a recessive allele labeled a and a dominant allele<br /> labeled B.<br /> A<br /> IR<br /> b<br /> Transposase<br /> DR<br /> IR<br /> Transposase<br /> IR<br /> DR<br /> 2<br /> 1<br /> 3<br /> Transposase<br /> Transposase<br /> B<br /> 4<br /> a<br /> DR<br /> DR<br /> IR<br /> IR<br /> DR<br /> Most of the transposable element has been excised.<br /> (a) Recombination between direct repeats<br /> DR<br /> Where would the breakage of the crossed strands have to occur to<br /> get recombinant chromosomes? Would it have to occur at sites 1<br /> and 3, or at sites 2 and 4? What would be the genotypes of the two<br /> recombinant chromosomes?<br /> Answer:<br /> Breakage would have to occur at the arrows labeled 2 and 4. This<br /> would connect the A allele with the B allele. The a allele in the other<br /> homologue would become connected with the b allele. In other words,<br /> one chromosome would be AB, and the homologue would be ab.<br /> Conceptual Questions<br /> C1. Describe the similarities and differences between homologous<br /> recombination involving sister chromatid exchange (SCE) versus<br /> that involving homologues. Would you expect the same types of<br /> proteins to be involved in both processes? Explain.<br /> C2. The molecular mechanism of SCE is similar to homologous<br /> recombination between homologues except that the two segments<br /> of DNA are sister chromatids instead of homologous chromatids.<br /> If branch migration occurs during SCE, will a heteroduplex be<br /> formed? Explain why or why not. Can gene conversion occur<br /> during sister chromatid exchange?</div> <div>5.1 LINKAGE AND CROSSING OVER 107<br /> ■ TESTING THE HYPOTHESIS — FIGURE 5.6 Experimental correlation between genetic recombination and<br /> crossing over.<br /> Starting materials: Two different strains of corn. One strain, referred to as parent A, had an abnormal chromosome 9 (knobbed/translocation)<br /> with a dominant C allele and a recessive wx allele. It also had a cytologically normal copy of chromosome 9 that carried the recessive<br /> c allele and the dominant Wx allele. Its genotype was Cc Wxwx. The other strain (referred to as parent B) had two normal versions of<br /> chromosome 9. The genotype of this strain was cc Wxwx.<br /> Experimental level<br /> Conceptual level<br /> 1. Cross the two strains described. The<br /> tassel is the pollen-bearing structure, and<br /> the silk (equivalent to the stigma and<br /> style) is connected to the ovary. After<br /> fertilization, the ovary will develop into<br /> an ear of corn.<br /> Tassel<br /> x<br /> Silk<br /> C<br /> wx<br /> c<br /> Wx<br /> x<br /> c<br /> Wx<br /> c<br /> wx<br /> Parent A<br /> Cc Wxwx<br /> Parent B<br /> cc Wxwx<br /> 2. Observe the kernels from this cross.<br /> F 1 ear of corn<br /> Each kernel is a separate seed that has<br /> inherited a set of chromosomes from<br /> each parent.<br /> F 1 kernels<br /> 3. Microscopically examine chromosome 9<br /> in the kernels.<br /> Colored/waxy<br /> From parent B<br /> Colorless/waxy<br /> C<br /> c<br /> A recombinant<br /> chromosome<br /> c<br /> c<br /> Microscope<br /> wx<br /> wx<br /> wx<br /> wx<br /> From parent A<br /> This illustrates only 2 possible<br /> outcomes in the F 1 kernels. The<br /> recombinant chromosome on the right is<br /> due to crossing over during meiosis in<br /> parent A. As shown in The Data, there are<br /> several possible outcomes.</div> <div>88 CHAPTER 4 :: EXTENSIONS OF MENDELIAN INHERITANCE<br /> A surprising result came in an experiment where they<br /> crossed two different varieties of white-flowered plants (Figure<br /> 4.20). All of the F 1 generation plants had purple flowers! Bateson<br /> and Punnett then allowed the F 1 offspring to self-fertilize.<br /> The F 2 generation resulted in purple and white flowers in a ratio<br /> of 9 purple : 7 white. From this result, Bateson and Punnett<br /> deduced that two different genes were involved, with the following<br /> relationship:<br /> C (one purple-color-producing) allele is dominant to c<br /> (white)<br /> P (another purple-color-producing) allele is dominant to<br /> p (white)<br /> cc or pp masks the P or C alleles, producing white color<br /> If possible, geneticists use the wild-type phenotype as<br /> their reference phenotype when describing an epistatic interaction.<br /> In this case, purple flowers are wild type. Homozygosity for<br /> the white allele of one gene masks the expression of the purpleproducing<br /> allele of another gene. In other words, the cc genotype<br /> is epistatic to a purple phenotype, and the pp genotype is<br /> also epistatic to a purple phenotype. At the level of genotypes,<br /> cc is epistatic to PP or Pp, and pp is epistatic to CC or Cc. This<br /> is another example of recessive epistasis. As seen in Figure 4.20,<br /> this epistatic interaction produces only two phenotypes—purple<br /> or white flowers—in a 9:7 ratio.<br /> Epistatis often occurs because two (or more) different<br /> proteins participate in a common function. For example, two<br /> or more proteins may be part of an enzymatic pathway leading<br /> to the formation of a single product. To illustrate this idea, let’s<br /> consider the formation of a purple pigment in the sweet pea.<br /> Colorless<br /> Enzyme C<br /> Colorless<br /> Enzyme P<br /> Purple<br /> precursor intermediate pigment<br /> In this example, a colorless precursor molecule must be acted on<br /> by two different enzymes to produce the purple pigment. Gene<br /> C encodes a functional protein called enzyme C, which converts<br /> the colorless precursor into a colorless intermediate. Two copies<br /> of the recessive allele (cc) result in a lack of production of this<br /> enzyme in the homozygote. Gene P encodes a functional enzyme<br /> P, which converts the colorless intermediate into the purple pigment.<br /> Like the c allele, the recessive p allele encodes a defective<br /> enzyme P. If an individual is homozygous for either recessive<br /> allele (cc or pp), it will not make any functional enzyme C or<br /> enzyme P, respectively. When one of these enzymes is missing,<br /> purple pigment cannot be made, and the flowers remain white.<br /> The parental cross shown in Figure 4.20 illustrates another<br /> genetic phenomenon called complementation. This term refers<br /> to the production of offspring with a wild-type phenotype from<br /> parents that both display the same or similar recessive phenotype.<br /> In this case, purple-flowered F 1 offspring were obtained<br /> from two white-flowered parents. Complementation typically<br /> occurs because the recessive phenotype in the parents is due to<br /> homozgyosity at two different genes. In our sweet pea example,<br /> one parent is CCpp and the other is ccPP. In the F 1 offspring, the<br /> C and P alleles, which are wild-type and dominant, complement<br /> White variety #1<br /> (CCpp )<br /> CP<br /> Cp<br /> cP<br /> cp<br /> F 1 generation<br /> F 2 generation<br /> CP Cp<br /> x<br /> All purple<br /> (CcPp)<br /> White variety #2<br /> (ccPP )<br /> Self-fertilization<br /> cP<br /> Complementation: Each<br /> recessive allele (c and p)<br /> is complemented by a<br /> wild-type allele (C and P).<br /> This phenomenon indicates<br /> that the recessive alleles<br /> are in different genes.<br /> Epistasis: Homozygosity<br /> for the recessive allele<br /> of either gene results in<br /> a white phenotype, thereby<br /> masking the purple<br /> (wild-type) phenotype.<br /> Both gene products<br /> encoded by the wild-type<br /> alleles (C and P) are<br /> needed for a purple<br /> phenotype.<br /> the c and p alleles, which are recessive. The offspring needs one<br /> wild-type allele of both genes to display the wild-type phenotype.<br /> Why is complementation an important experimental observation?<br /> When geneticists observe complementation in a genetic<br /> cross, the results suggest that the recessive phenotype in the two<br /> parent strains is caused by mutant alleles in two different genes.<br /> A Cross Involving a Two-Gene Interaction Can<br /> Produce Three Distinct Phenotypes Due to Epistasis<br /> Thus far, we have observed two epistatic interactions, one producing<br /> four phenotypes and the other producing only two. Coat color<br /> in rodents provides an example that produces three phenotypes.<br /> If a true-breeding black rat is crossed to a true-breeding albino<br /> rat, the result is a rat with agouti coat color. Animals with agouti<br /> coat color have black pigmentation at the tips of each hair that<br /> changes to orange pigmentation near the root. If two agouti animals<br /> of the F 1 generation are crossed to each other, they produce<br /> agouti, black, and albino offspring in a 9:3:4 ratio (Figure 4.21).<br /> cp<br /> CCPP CCPp CcPP CcPp<br /> Purple Purple Purple Purple<br /> CCPp CCpp CcPp Ccpp<br /> Purple White Purple White<br /> CcPP CcPp ccPP ccPp<br /> Purple Purple White White<br /> CcPp Ccpp ccPp ccpp<br /> Purple White White White<br /> FIGURE 4.20 A cross between two different<br /> white varieties of the sweet pea.<br /> Genes g Traits The color of the sweet pea flower is controlled<br /> by two genes, which are epistatic to each other and show complementation. Each<br /> gene is necessary for the production of an enzyme required for pigment synthesis.<br /> The recessive allele of either gene encodes a defective enzyme. If an individual is<br /> homozygous recessive for either of the two genes, the purple pigment cannot be<br /> synthesized. This results in a white phenotype.</div> <div>CONCEPTUAL QUESTIONS 41<br /> and valine. One of the symptoms is that the urine smells like<br /> maple syrup. An unaffected couple produced six children in the<br /> following order: unaffected daughter, affected daughter, unaffected<br /> son, unaffected son, affected son, and unaffected son. The youngest<br /> unaffected son marries an unaffected woman and has three<br /> children in the following order: affected daughter, unaffected<br /> daughter, and unaffected son. Draw a pedigree that describes this<br /> family. What type of inheritance (dominant or recessive) would<br /> you propose to explain maple syrup urine disease?<br /> C24. Marfan syndrome is a rare inherited human disorder characterized<br /> by unusually long limbs and digits plus defects in the heart<br /> (especially the aorta) and the eyes. Following is a pedigree for<br /> this disorder. Affected individuals are shown with filled (black)<br /> symbols. What type of inheritance pattern do you think is the<br /> most likely?<br /> I-1 I-2<br /> II-1 II-2 II-3 II-4<br /> II-5<br /> II-6<br /> III-1 III-2 III-3 III-4 III-5<br /> IV-1 IV-2 IV-3 IV-4<br /> C25. A true-breeding pea plant with round and green seeds was crossed<br /> to a true-breeding plant with wrinkled and yellow seeds. Round<br /> and yellow seeds are the dominant traits. The F 1 plants were<br /> allowed to self-fertilize. What are the following probabilities for the<br /> F 2 generation?<br /> A. An F 2 plant with wrinkled, yellow seeds.<br /> B. Three out of three F 2 plants with round, yellow seeds.<br /> C. Five F 2 plants in the following order: two have round, yellow<br /> seeds; one has round, green seeds; and two have wrinkled, green<br /> seeds.<br /> D. An F 2 plant will not have round, yellow seeds.<br /> C26. A true-breeding tall pea plant was crossed to a true-breeding dwarf<br /> plant. What is the probability that an F 1 individual will be truebreeding?<br /> What is the probability that an F 1 individual will be a<br /> true-breeding tall plant?<br /> C27. What are the expected phenotypic ratios from the following cross:<br /> Tt Rr yy Aa × Tt rr YY Aa, where T � tall, t � dwarf, R � round,<br /> r � wrinkled, Y � yellow, y � green, A � axial, a � terminal; T,<br /> R, Y, and A are dominant alleles. Note: See solved problem S3 for<br /> help in answering this problem.<br /> C28. When an abnormal organism contains three copies of a gene<br /> (instead of the normal number of two copies), the alleles for the<br /> gene usually segregate so that a gamete will contain one or two<br /> copies of the gene. Let’s suppose that an abnormal pea plant has<br /> three copies of the height gene. Its genotype is TTt. The plant is<br /> also heterozygous for the seed color gene, Yy. How many types of<br /> gametes can this plant make, and in what proportions? (Assume<br /> that it is equally likely that a gamete will contain one or two copies<br /> of the height gene.)<br /> C29. Honeybees are unusual in that male bees (drones) have only one<br /> copy of each gene, while female bees have two copies of their<br /> genes. That is because drones develop from eggs that have not<br /> been fertilized by sperm cells. In bees, the trait of long wings is<br /> dominant over short wings, and the trait of black eyes is dominant<br /> over white eyes. If a drone with short wings and black eyes was<br /> mated to a queen bee that is heterozygous for both genes, what<br /> are the predicted genotypes and phenotypes of male and female<br /> offspring? What are the phenotypic ratios if we assume an equal<br /> number of male and female offspring?<br /> C30. A pea plant that is dwarf with green, wrinkled seeds was crossed to<br /> a true-breeding plant that is tall with yellow, round seeds. The F 1<br /> generation was allowed to self-fertilize. What types of gametes, and<br /> in what proportions, would the F 1 generation make? What would<br /> be the ratios of genotypes and phenotypes of the F 2 generation?<br /> C31. A true-breeding plant with round and green seeds was crossed to a<br /> true-breeding plant with wrinkled and yellow seeds. The F 1 plants<br /> were allowed to self-fertilize. What is the probability of obtaining<br /> the following plants in the F 2 generation: two that have round,<br /> yellow seeds; one with round, green seeds; and two with wrinkled,<br /> green seeds? (Note: See solved problem S7 for help.)<br /> C32. Wooly hair is a rare dominant trait found in people of<br /> Scandinavian descent in which the hair resembles the wool of a<br /> sheep. A male with wooly hair, who has a mother with straight<br /> hair, moves to an island that is inhabited by people who are not<br /> of Scandinavian descent. Assuming that he never leaves the island<br /> and that no other Scandinavians immigrate to the island, what is<br /> the probability that a great-grandchild of this male will have wooly<br /> hair? (Hint: You may want to draw a pedigree to help you figure<br /> this out.) If this wooly-haired male has eight great-grandchildren,<br /> what is the probability that one out of eight will have wooly hair?<br /> C33. Huntington disease is a rare dominant trait that causes<br /> neurodegeneration later in life. A man in his thirties, who already<br /> has three children, discovers that his mother has Huntington<br /> disease though his father is unaffected. What are the following<br /> probabilities?<br /> A. That the man in his thirties will develop Huntington disease.<br /> B. That his first child will develop Huntington disease.<br /> C. That one out of three of his children will develop Huntington<br /> disease.<br /> C34. A woman with achondroplasia (a dominant form of dwarfism)<br /> and a phenotypically unaffected man have seven children, all of<br /> whom have achondroplasia. What is the probability of producing<br /> such a family if this woman is a heterozygote? What is the<br /> probability that the woman is a heterozygote if her eighth child<br /> does not have this disorder?</div> <div>EXPERIMENTAL QUESTIONS 567<br /> In this study, the researchers restriction mapped this region using<br /> two enzymes, NotI and XhoI. The locations of sites are shown here:<br /> MET D7S122 IRP CF gene D7S8<br /> hA h B h C h D h E h F h<br /> N X X X X X N<br /> N = NotI, X = XhoI<br /> How would the researchers know they were walking toward the CF<br /> gene and not away from it?<br /> Answer: The general answer is that they had to determine if they were<br /> walking toward the IRP marker or toward the MET marker.<br /> As depicted in the previous diagram, NotI would produce a large<br /> DNA fragment (including regions A–F) that contained both the D7S122<br /> and IRP markers. Because the D7S122 and IRP markers were available<br /> as probes, they could confirm that this region contained both markers<br /> by Southern blotting of this NotI fragment to both markers. XhoI cuts<br /> this region into smaller pieces. The piece labeled B carried the D7S122<br /> marker, and the piece labeled D carried the IRP marker. Again, this<br /> could be confirmed by hybridization. This map enabled the researchers<br /> to determine which direction to walk. They began at fragment B<br /> (which contained the D7S122 marker) and walked toward the adjacent<br /> fragment labeled C. They then walked to the fragment labeled D, which<br /> contains the IRP marker. If they had walked toward the fragment<br /> labeled A, they would then have walked to the MET marker. This is the<br /> incorrect direction, because they knew from their linkage mapping that<br /> IRP is closer to the CF gene compared to MET.<br /> Conceptual Questions<br /> C1. A person with a rare genetic disease has a sample of her<br /> chromosomes subjected to in situ hybridization using a probe that<br /> is known to recognize band p11 on chromosome 7. Even though<br /> her chromosomes look cytologically normal, the probe does not<br /> bind to this person’s chromosomes. How would you explain these<br /> results? How would you use this information to positionally clone<br /> the gene that is related to this disease?<br /> C2. For each of the following, decide if it could be appropriately<br /> described as a genome:<br /> A. The E. coli chromosome<br /> B. Human chromosome 11<br /> C. A complete set of 10 chromosomes in corn<br /> D. A copy of the single-stranded RNA packaged into human<br /> immunodeficiency virus (HIV)<br /> C3. Which of the following statements are true about molecular<br /> markers?<br /> A. All molecular markers are segments of DNA that carry specific<br /> genes.<br /> B. A molecular marker is a segment of DNA that is found at a<br /> specific location in a genome.<br /> C. We can follow the transmission of a molecular marker<br /> by analyzing the phenotype (i.e., the individual’s bodily<br /> characteristics) of offspring.<br /> D. We can follow the transmission of molecular markers using<br /> molecular techniques such as gel electrophoresis.<br /> E. An STS is a molecular marker.<br /> Experimental Questions<br /> E1. Would the following methods be described as linkage, cytogenetic,<br /> or physical mapping?<br /> A. Fluorescence in situ hybridization (FISH)<br /> B. Conducting dihybrid crosses to compute map distances<br /> C. Chromosome walking<br /> D. Examination of polytene chromosomes in Drosophila<br /> E. Use of RFLPs in crosses<br /> F. Using BACs and cosmids to construct a contig<br /> E2. In an in situ hybridization experiment, what is the relationship<br /> between the sequence of the probe DNA and the site on the<br /> chromosomal DNA where the probe binds?<br /> E3. Describe the technique of in situ hybridization. Explain how it can<br /> be used to map genes.<br /> E4. The cells from a malignant tumor were subjected to in situ<br /> hybridization using a probe that recognizes a unique sequence<br /> on chromosome 14. The probe was detected only once in each of<br /> these cells. Explain these results and speculate on their significance<br /> with regard to the malignant characteristics of these cells.<br /> E5. Figure 20.3 describes the technique of FISH. Why is it necessary to<br /> “fix” the cells (and the chromosomes inside of them) to the slides?<br /> What does it mean to fix them? Why is it necessary to denature the<br /> chromosomal DNA?<br /> E6. Explain how the use of DNA probes with different fluorescence<br /> emission wavelengths can be used in a single FISH experiment to<br /> map the locations of two or more genes. This method is called<br /> chromosome painting. Explain why this is an appropriate term.<br /> E7. A researcher is interested in a gene found on human chromosome<br /> 21. Describe the expected results of a FISH experiment using a<br /> probe that is complementary to this gene. How many spots would<br /> you see if the probe was used on a sample from an individual with<br /> 46 chromosomes versus an individual with Down syndrome?<br /> E8. What is a contig? Explain how you would determine that two<br /> clones in a contig are overlapping.<br /> E9. Contigs are often made using BAC or cosmid vectors. What are the<br /> advantages and disadvantages of these two types of vectors? Which<br /> type of contig would you make first, a BAC or cosmid contig?<br /> Explain.</div> <div>24<br /> POPULATION<br /> GENETICS<br /> ::<br /> Until now, we have primarily focused our attention on genes within<br /> individuals and their related family members. In this chapter and Chapters 25<br /> and 26, we will turn to the study of genes in a population or species. The field<br /> of population genetics is concerned with changes in genetic variation within a<br /> group of individuals over time. Population geneticists want to know the extent<br /> of genetic variation within populations, why it exists, and how it changes over<br /> the course of many generations. The field of population genetics emerged as a<br /> branch of genetics in the 1920s and 1930s. Its mathematical foundations were<br /> developed by theoreticians who extended the principles of Gregor Mendel and<br /> Charles Darwin by deriving formulas to explain the occurrence of genotypes<br /> within populations. These foundations can be largely attributed to three scientists:<br /> Sir Ronald Fisher, Sewall Wright, and J. B. S. Haldane. As we will see, support<br /> for their mathematical theories was provided by several researchers who<br /> analyzed the genetic composition of natural and experimental populations. More<br /> recently, population geneticists have used techniques to probe genetic variation<br /> at the molecular level. In addition, staggering advances in computer technology<br /> have aided population geneticists in the analysis of their genetic theories and<br /> data. In this chapter, we will explore the genetic variation that occurs in populations<br /> and consider the reasons why the genetic composition of populations may<br /> change over the course of several generations.<br /> CHAPTER OUTLINE<br /> 24.1 Genes in Populations, and the Hardy-<br /> Weinberg Equation<br /> 24.2 Factors That Change Allele and<br /> Genotype Frequencies in Populations<br /> 24.3 Sources of New Genetic Variation<br /> 24.1 GENES IN POPULATIONS, AND<br /> THE HARDY-WEINBERG EQUATION<br /> Population genetics may seem like a significant departure from other topics in<br /> this textbook, but it is a direct extension of our understanding of Mendel’s laws<br /> of inheritance, molecular genetics, and the ideas of Darwin. In the field of population<br /> genetics, the focus shifts away from the individual and toward the population<br /> of which the individual is a member. Conceptually, all of the alleles of every<br /> gene in a population make up the gene pool. In this regard, each member of the<br /> population is viewed as receiving its genes from its parents, which, in turn, are<br /> members of the gene pool. Furthermore, individuals that reproduce contribute<br /> to the gene pool of the next generation. Population geneticists study the genetic<br /> variation within the gene pool and how such variation changes from one generation<br /> to the next. The emphasis is often on allelic variation. In this introductory<br /> section, we will examine some of the general features of populations and<br /> gene pools.<br /> The African cheetah. This species has a<br /> relatively low level of genetic variation because<br /> the population was reduced to a small size<br /> approximately 10,000 to 12,000 years ago.</div> <div>672 CHAPTER 24 :: POPULATION GENETICS<br /> In Small Populations, Allele Frequencies<br /> Can Be Altered by Random Genetic Drift<br /> In the 1930s, geneticist Sewall Wright played a key role in developing<br /> the concept of random genetic drift, which refers to changes<br /> in allele frequencies in a population due to chance fluctuations.<br /> As a matter of random chance, the frequencies of alleles found in<br /> gametes that unite to form zygotes will vary from generation to<br /> generation. Over the long run, genetic drift favors either the loss<br /> of an allele or its fixation at 100% in the population. The rate<br /> at which this occurs depends on the population size and on the<br /> initial allele frequencies. Figure 24.6 illustrates the potential consequences<br /> of genetic drift in one large (N � 1,000) and five small<br /> (N � 20) populations. At the beginning of this hypothetical simulation,<br /> all of these populations have identical allele frequencies:<br /> A � 0.5 and a � 0.5. In the five small populations, this allele<br /> frequency fluctuates substantially from generation to generation.<br /> Eventually, one of the alleles is eliminated and the other is fixed<br /> at 100%. At this point, the allele has become monomorphic and<br /> cannot fluctuate any further. By comparison, the allele frequencies<br /> in the large population fluctuate much less, because random<br /> sampling error is expected to have a smaller impact. Nevertheless,<br /> genetic drift will lead to homozygosity even in large populations,<br /> but this will take many more generations to occur.<br /> Now let’s ask two questions:<br /> 1. How many new mutations do we expect in a natural<br /> population?<br /> 2. How likely is it that any new mutation will be either<br /> fixed in, or eliminated from, a population due to random<br /> genetic drift?<br /> With regard to the first question, the average number of new mutations<br /> depends on the mutation rate (m) and the number of individuals<br /> in a population (N). If each individual has two copies of the<br /> gene of interest, the expected number of new mutations in this gene<br /> is<br /> Expected number of new mutations � 2Nm<br /> From this, we see that a new mutation is more likely to occur in<br /> a large population than in a small one. This makes sense, because<br /> the larger population has more copies of the gene to be mutated.<br /> With regard to the second question, the probability of fixation of<br /> a newly arising allele due to genetic drift is<br /> Probability of fixation �<br /> 1<br /> 2N<br /> (assuming equal numbers of<br /> males and females contribute<br /> to the next generation)<br /> In other words, the probability of fixation is the same as the<br /> initial allele frequency in the population. For example, if N � 20,<br /> the probability of fixation of a new allele equals 1/(2 � 20), or<br /> 2.5%. Conversely, a new allele may be lost from the population.<br /> Probability of elimination � 1 � probability of fixation<br /> 1<br /> � 1 �<br /> 2N<br /> If N � 20, the probability of elimination equals 1 � 1/(2 � 20),<br /> or 97.5%. As you may have noticed, the value of N has opposing<br /> effects with regard to new mutations and their eventual fixation in<br /> a population. When N is very large, new mutations are much more<br /> likely to occur. Each new mutation, however, has a greater chance<br /> of being eliminated from the population due to random genetic<br /> drift. On the other hand, when N is small, the probability of new<br /> mutations is also small, but if they occur, the likelihood of fixation<br /> is relatively large.<br /> Now that we have an appreciation for the phenomenon of<br /> genetic drift, we can ask a third question:<br /> 3. If fixation of a new allele does occur, how many<br /> generations is it likely to take?<br /> 1.0<br /> N = 20<br /> N = 20<br /> Fixation of allele A<br /> Frequency of A<br /> 0.5<br /> N = 1,000<br /> N = 20<br /> N = 20<br /> N = 20<br /> 0 Loss of allele A<br /> Generations<br /> FIGURE 24.6 A hypothetical simulation of random genetic drift. In all cases, the starting allele frequencies are A � 0.5 and<br /> a � 0.5. The colored lines illustrate five populations in which N � 20; the black line shows a population in which N � 1,000.</div> <div>OVERVIEW<br /> OF GENETICS<br /> 1<br /> ::<br /> Hardly a week goes by without a major news story involving a<br /> genetic breakthrough. The increasing pace of genetic discoveries has become<br /> staggering. The Human Genome Project is a case in point. This project began<br /> in the United States in 1990, when the National Institutes of Health and the<br /> Department of Energy joined forces with international partners to decipher the<br /> massive amount of information contained in our genome—the DNA found<br /> within all of our chromosomes (Figure 1.1). Working collectively, a large group<br /> of scientists from around the world has produced a detailed series of maps that<br /> help geneticists navigate through human DNA. Remarkably, in only a decade,<br /> they determined the DNA sequence (read in the bases of A, T, G, and C) covering<br /> over 90% of the human genome. The first draft of this sequence, published<br /> in 2001, is nearly 3 billion nucleotide base pairs in length. The completed<br /> sequence, published in 2003, has an accuracy greater than 99.99%; fewer than<br /> one mistake was made in every 10,000 base pairs!<br /> Studying the human genome allows us to explore fundamental details<br /> about ourselves at the molecular level. The results of the project are expected<br /> to shed considerable light on basic questions, like how many genes we have,<br /> how genes direct the activities of living cells, how species evolve, how single cells<br /> develop into complex tissues, and how defective genes cause disease. Furthermore,<br /> such understanding may lend itself to improvements in modern medicine<br /> by leading to better diagnoses of diseases and the development of new medicines<br /> to treat them.<br /> As scientists have attempted to unravel the mysteries within our genes,<br /> this journey has involved the invention of many new technologies. This textbook<br /> emphasizes a large number of these modern approaches. Paradoxically,<br /> one could argue that these advancements may have a greater impact on our<br /> lives than the original discoveries upon which they were based. For example,<br /> the technology of gene cloning originally arose from discoveries concerning<br /> bacterial enzymes that “cut and paste” DNA fragments together. These scientific<br /> discoveries have made it possible to create medicines that would otherwise<br /> be difficult or impossible to make. An example is human recombinant insulin,<br /> termed Humulin, which is synthesized in a strain of Escherichia coli bacteria that<br /> has been genetically altered by the addition of the gene for human insulin. The<br /> bacteria are grown in a laboratory and make large amounts of human insulin.<br /> As discussed in Chapter 19, this insulin is purified and administered to many<br /> people with insulin-dependent diabetes.<br /> CHAPTER OUTLINE<br /> 1.1 The Relationship Between Genes<br /> and Traits<br /> 1.2 Fields of Genetics<br /> Copycat, the first cloned pet. In 2002, the cat<br /> shown here, called Copycat, was produced by<br /> cloning, a procedure described in Chapter 19.<br /> PART I<br /> Introduction</div> <div>236 CHAPTER 9 :: MOLECULAR STRUCTURE OF DNA AND RNA<br /> reason, DNA sequences that have a high proportion of G and C<br /> tend to form more stable double-stranded structures.<br /> The AT/GC rule implies that we can predict the sequence<br /> in one DNA strand if the sequence in the opposite strand is<br /> known. For example, if one strand has the sequence of 5'–ATG-<br /> GCGGATTT–3', then the opposite strand must be 3'–TACC-<br /> GCCTAAA–5'. In genetic terms, we would say that these two<br /> sequences are complementary to each other or that the two<br /> sequences exhibit complementarity. In addition, you may have<br /> noticed that the sequences are labeled with 5' and 3' ends. These<br /> numbers designate the direction of the DNA backbones. The<br /> direction of DNA strands is depicted in the inset to Figure 9.17.<br /> When going from the top of this figure to the bottom, one strand<br /> is running in the 5' to 3' direction, while the other strand is 3' to<br /> 5'. This opposite orientation of the two DNA strands is referred<br /> to as an antiparallel arrangement. An antiparallel structure was<br /> initially proposed in the models of Watson and Crick.<br /> Figure 9.18a is a schematic model that emphasizes certain<br /> molecular features of DNA structure. The helical structure<br /> is formed by the DNA backbone. The bases in this model are<br /> depicted as flat rectangular structures that hydrogen bond in<br /> pairs. (The hydrogen bonds are the dotted lines.) Although the<br /> bases are not actually rectangular, they do form flattened planar<br /> structures. Within DNA, the bases are oriented so that the flattened<br /> regions are facing each other, an arrangement referred to<br /> as base stacking. In other words, if you think of the bases as flat<br /> plates, these plates are stacked on top of each other in the doublestranded<br /> DNA structure. Along with hydrogen bonding, base<br /> stacking is a structural feature that stabilizes the double helix.<br /> By convention, the direction of the DNA double helix<br /> shown in Figure 9.18a spirals in a direction that is called “righthanded.”<br /> To understand this terminology, imagine that a double<br /> helix is laid on your desk; one end of the helix is close to you,<br /> and the other end is at the opposite side of the desk. As it spirals<br /> away from you, a right-handed helix turns in a clockwise<br /> direction. By comparison, a left-handed helix would spiral in a<br /> counterclockwise manner. Both strands in Figure 9.18a spiral in<br /> a right-handed direction.<br /> Figure 9.18b is a space-filling model for DNA in which<br /> the atoms are represented by spheres. This model emphasizes the<br /> surface features of DNA. Note that the backbone—composed of<br /> sugar and phosphate groups—is on the outermost surface. In a<br /> living cell, the backbone has the most direct contact with water. In<br /> contrast, the bases are more internally located within the doublestranded<br /> structure. Biochemists use the term grooves to describe<br /> the indentations where the atoms of the bases are in contact with<br /> the surrounding water. As you travel around the DNA helix, the<br /> structure of DNA has two grooves: the major groove and the<br /> minor groove. As will be discussed in later chapters, certain proteins<br /> can bind within these grooves and interact with a particular<br /> sequence of bases, thereby altering DNA structure and regulating<br /> gene expression.<br /> Minor<br /> groove<br /> Major<br /> groove<br /> Minor<br /> groove<br /> Major<br /> groove<br /> (a) Ball-and-stick model of DNA<br /> (b) Space-filling model of DNA<br /> FIGURE 9.18 Two models of the double helix. (a) Ball-and-stick model of the double helix. The ribose–phosphate backbone is<br /> shown in detail, while the bases are depicted as flattened rectangles. (b) Space-filling model of the double helix.</div> <div>196 CHAPTER 8 :: VARIATION IN CHROMOSOME STRUCTURE AND NUMBER<br /> may prompt a physician to request a microscopic examination of<br /> the individual’s chromosomes. In this way, phenotypically normal<br /> individuals may discover they have a chromosome with an<br /> inversion. Next, we will examine why an individual carrying an<br /> inversion may produce offspring with phenotypic abnormalities.<br /> Inversion Heterozygotes May Produce Abnormal<br /> Chromosomes Due to Crossing Over<br /> An individual that carries one copy of a normal chromosome and<br /> one copy of an inverted chromosome is known as an inversion<br /> heterozygote. Such an individual, though possibly phenotypically<br /> normal, may have a high probability of producing haploid cells<br /> that are abnormal in their total genetic content. This likelihood<br /> depends on the size of the inverted segment. During meiosis, an<br /> inversion heterozygote with a fairly large inverted segment may<br /> produce a sizable fraction of abnormal haploid cells, perhaps 1/3<br /> or even higher.<br /> The underlying cause of abnormality is the phenomenon<br /> of crossing over within the inverted region. During meiosis I,<br /> pairs of homologous sister chromatids synapse with each other.<br /> Figure 8.10 illustrates how this occurs in an inversion heterozygote.<br /> For the normal chromosome and inversion chromosome<br /> to synapse properly, an inversion loop must form to permit<br /> the homologous genes on both chromosomes to align next to<br /> each other despite the inverted sequence. If a crossover occurs<br /> within the inversion loop, highly abnormal chromosomes are<br /> produced.<br /> Normal:<br /> Replicated chromosomes<br /> A B C D E F G H I<br /> Normal:<br /> Replicated chromosomes<br /> A B C D E F G H I<br /> With<br /> inversion:<br /> A<br /> B<br /> C<br /> D<br /> E<br /> a b c g f e d h i<br /> F<br /> G<br /> H<br /> I<br /> With<br /> inversion:<br /> A B C D E F G H I<br /> a e d c b f g h i<br /> A<br /> a<br /> B<br /> b c g f e d h i<br /> Homologous pairing<br /> during prophase<br /> Crossover site<br /> E<br /> F<br /> D<br /> e f<br /> C<br /> d e f G H I<br /> d g g<br /> a e d c b f g h i<br /> A<br /> B<br /> C<br /> c<br /> b<br /> Homologous pairing<br /> during prophase<br /> D<br /> d<br /> e<br /> E<br /> Crossover site<br /> F G H I<br /> a<br /> b<br /> c<br /> h<br /> i<br /> a<br /> f<br /> g<br /> h<br /> i<br /> Products after crossing over<br /> Products after crossing over<br /> A B C D E F G H I<br /> A B C D E F<br /> G H I<br /> A B C D E f g c b a<br /> I<br /> H<br /> G<br /> F e d h i<br /> a b c g f e d h i<br /> Duplicated/<br /> deleted<br /> Acentric<br /> fragment<br /> Dicentric<br /> chromosome<br /> A B C d e a<br /> I H G F E D c b f g h i<br /> a<br /> Dicentric bridge<br /> e d c b f g h i<br /> (a) Pericentric inversion<br /> (b) Paracentric inversion<br /> FIGURE 8.10 The consequences of crossing over in the inversion loop. (a) Crossover within a pericentric inversion. (b) Crossover within a<br /> paracentric inversion.</div> <div>25.1 QUANTITATIVE TRAITS 701<br /> Mother’s Offspring’s<br /> Weight (kg) Weight (kg) X�X Y�Y (X�X) (Y�Y)<br /> 570 568 �26 �30 780<br /> 572 560 �24 �38 912<br /> 599 642 3 44 132<br /> 602 580 6 �18 �108<br /> 631 586 35 �12 �420<br /> 603 642 7 44 308<br /> 599 632 3 34 102<br /> 625 580 29 �18 �522<br /> 584 605 �12 7 �84<br /> 575 585 �21 �13 273<br /> X 5 596 Y 5 598 S51,373<br /> SD X � 21.1 SD Y � 30.5<br /> S3(X 2 X)(Y 2 Y)4<br /> CoV (X,Y) 5<br /> N 2 1<br /> CoV (X,Y) 5 1,373<br /> 10 2 1<br /> CoV (X,Y) � 152.6<br /> r value differs from zero due only to random sampling error. We<br /> followed a similar approach in the chi square analysis described<br /> in Chapter 2. Like the chi square value, the significance of the<br /> correlation coefficient is directly related to sample size and the<br /> degrees of freedom (df). In testing the significance of correlation<br /> coefficients, df equals N � 2, which is one less than the degrees<br /> of freedom of variance (i.e., df for variance equals N � 1). Table<br /> 25.2 shows the relationship between the r values and degrees of<br /> freedom at the 5% and 1% confidence intervals.<br /> The use of Table 25.2 is valid only if several assumptions<br /> are met. First, the values of X and Y in the study must have been<br /> obtained by an unbiased sampling of the entire population. In<br /> addition, this approach assumes that the scores of X and Y follow<br /> a normal distribution, like that of Figure 25.1, and that the relationship<br /> between X and Y is linear.<br /> TABLE 25.2<br /> Values of r at the 5% and 1% Confidence Intervals<br /> Degrees of<br /> Freedom (df) 5% 1%<br /> Degrees of<br /> Freedom (df ) 5% 1%<br /> After we have calculated the covariance, we can evaluate<br /> the strength of the association between the two variables by calculating<br /> a correlation coefficient (r). This value, which ranges<br /> between �1 and �1, indicates how two factors vary in relation<br /> to each other. The correlation coefficient is calculated as<br /> r (X,Y) 5 CoV (X,Y)<br /> SD X SD Y<br /> A positive r value means that two factors tend to vary in the<br /> same way relative to each other; as one factor increases, the other<br /> will increase with it. A value of zero indicates that the two factors<br /> do not vary in a consistent way relative to each other; the values<br /> of the two factors are not related. Finally, an inverse correlation,<br /> in which the correlation coefficient is negative, indicates that the<br /> two factors tend to vary in opposite ways to each other; as one<br /> factor increases, the other will decrease.<br /> Let’s use the data of 5-year weights for mother and offspring<br /> to calculate a correlation coefficient.<br /> 1 .997 1.000<br /> 2 .950 .990<br /> 3 .878 .959<br /> 4 .811 .917<br /> 5 .754 .874<br /> 6 .707 .834<br /> 7 .666 .798<br /> 8 .632 .765<br /> 9 .602 .735<br /> 10 .576 .708<br /> 11 .553 .684<br /> 12 .532 .661<br /> 13 .514 .641<br /> 14 .497 .623<br /> 24 .388 .496<br /> 25 .381 .487<br /> 26 .374 .478<br /> 27 .367 .470<br /> 28 .361 .463<br /> 29 .355 .456<br /> 30 .349 .449<br /> 35 .325 .418<br /> 40 .304 .393<br /> 45 .288 .372<br /> 50 .273 .354<br /> 60 .250 .325<br /> 70 .232 .302<br /> 80 .217 .283<br /> r (X,Y) 5<br /> r (X, Y) � 0.237<br /> 152.6<br /> (21.1)(30.5)<br /> The result is a positive correlation between the 5-year weights of<br /> mother and offspring. In other words, the positive correlation<br /> value suggests that heavy mothers tend to have heavy offspring<br /> and that lighter mothers have lighter offspring.<br /> How do we evaluate the value of r ? After a correlation<br /> coefficient has been calculated, one must consider whether the<br /> r value represents a true association between the two variables<br /> or whether it could be simply due to chance. To accomplish<br /> this, we can test the hypothesis that there is no real correlation<br /> (i.e., the null hypothesis, r � 0). The null hypothesis is that the<br /> 15 .482 .606<br /> 16 .468 .590<br /> 17 .456 .575<br /> 18 .444 .561<br /> 19 .433 .549<br /> 20 .423 .537<br /> 21 .413 .526<br /> 22 .404 .515<br /> 23 .396 .505<br /> 90 .205 .267<br /> 100 .195 .254<br /> 125 .174 .228<br /> 150 .159 .208<br /> 200 .138 .181<br /> 300 .113 .148<br /> 400 .098 .128<br /> 500 .088 .115<br /> 1,000 .062 .081<br /> Note: df equals N � 2.<br /> From Spence, J. T. et al. (1976). Elementary Statistics. Prentice-Hall, Englewood Cliffs, New Jersey.</div> <div>318 CHAPTER 12 :: GENE TRANSCRIPTION AND RNA MODIFICATION<br /> 5′<br /> AAUAAA<br /> AAUAAA<br /> Polyadenylation sequence<br /> 5′ 3′<br /> Endonuclease cleavage occurs<br /> about 20 nucleotides downstream<br /> from the AAUAAA sequence.<br /> PolyA-polymerase adds<br /> adenine nucleotides<br /> to the 3′ end.<br /> (toward the 3' end) from the stop codon in the pre-mRNA.<br /> An endonuclease recognizes the polyadenylation sequence and<br /> cleaves the pre-mRNA at a location that is about 20 nucleotides<br /> beyond the 3' end of the AAUAAA sequence. The fragment<br /> beyond the 3' cut is degraded. Next, an enzyme known as polyApolymerase<br /> attaches many adenine-containing nucleotides. The<br /> length of the polyA tail varies among different mRNAs, from a<br /> few dozen to several hundred adenine nucleotides. As discussed<br /> in Chapter 15, a long polyA tail increases the stability of mRNA<br /> and plays a role during translation.<br /> AAAAAAAAAAAA....<br /> 5′ 3′<br /> AAUAAA<br /> PolyA tail<br /> FIGURE 12.24 Attachment of a polyA tail. First, an<br /> endonuclease cuts the RNA at a location that is 11 to 30 nucleotides<br /> after the AAUAAA polyadenylation sequence, making the RNA shorter<br /> at its 3' end. Adenine-containing nucleotides are then attached, one at a<br /> time, to the 3' end by the enzyme polyA-polymerase.<br /> CONCEPTUAL SUMMARY<br /> Transcription is the process of RNA synthesis from a DNA template.<br /> Different types of RNA transcripts can be made. These<br /> include mRNA, which specifies the sequence of amino acids<br /> within a polypeptide, and tRNA and rRNA, which are necessary<br /> to translate mRNA. In addition, several other small RNA molecules,<br /> such as microRNA, 7S RNA, scRNA, the RNA component<br /> of RNaseP, snRNA, telomerase RNA, and snoRNA have various<br /> functions in the cell.<br /> The synthesis of RNA during gene transcription involves<br /> three steps: initiation, elongation, and termination. During<br /> the initiation stage, proteins that bind to DNA recognize specific<br /> sequences within the promoter region. In the case of many<br /> bacterial promoters, the –35 and –10 sequences are recognized<br /> by s factor, which is part of RNA polymerase holoenzyme. To<br /> begin the synthesis or elongation of a transcript, the DNA must<br /> be converted to an open complex by denaturing a small doublestranded<br /> region. At this point, RNA polymerase can slide down<br /> the DNA, synthesizing a complementary RNA molecule. At the<br /> end of the gene, a termination signal is found. In bacteria, termination<br /> signals can promote transcriptional termination in<br /> two different ways. In r-dependent termination, the r protein is<br /> responsible for dissociating the RNA transcript when a stem-loop<br /> structure causes RNA polymerase to pause. In r-independent termination,<br /> the formation of a stem-loop within the RNA causes<br /> RNA polymerase to stall and eventually fall off the DNA, but this<br /> occurs without the help of r protein.<br /> Transcription in eukaryotes is similar but more complex<br /> compared to the process in bacteria. In eukaryotes, three RNA<br /> polymerases designated I, II, and III transcribe different types<br /> of genes. Transcriptional initiation of structural genes requires<br /> five general transcription factors that enable RNA polymerase<br /> II to recognize the TATA box in the –25 region. Transcription<br /> is initiated by an assembly process that leads to the formation<br /> of the open complex. A large protein complex termed mediator<br /> plays a role in the switch between initiation and elongation.<br /> For transcription to occur, the chromatin structure must<br /> be loosened so that RNA polymerase can transcribe the template<br /> DNA. This involves two mechanisms: covalent modifications<br /> of histone proteins and ATP-dependent chromatin<br /> remodeling.<br /> Following transcription, a newly made RNA transcript<br /> may be modified in various ways. For example, tRNA transcripts<br /> are processed to smaller forms by endo- and exonucleases.<br /> Some genes contain introns that must be removed from<br /> the RNA via splicing. Group I and Group II introns are selfsplicing,<br /> which means they catalyze their own removal. These<br /> types of introns are found in the rRNA genes of Tetrahymena<br /> and other simpler eukaryotes, in organelle DNA, and occasionally<br /> in bacterial genes. In contrast, introns are commonly found<br /> in eukaryotic pre-mRNA, where they are removed by a multicomponent<br /> structure known as a spliceosome. Many eukaryotic<br /> pre-mRNAs undergo alternative splicing, with the result that<br /> a single gene can produce more than one type of polypeptide.<br /> Besides splicing, most eukaryotic pre-mRNA is also capped with<br /> 7-methylguanosine at its 5' end, and a polyA tail is attached to<br /> its 3' end.</div> <div>40 CHAPTER 2 :: MENDELIAN INHERITANCE<br /> C13. For the following pedigrees, describe what you think is the most<br /> likely inheritance pattern (dominant versus recessive). Explain your<br /> reasoning. Filled (black) symbols indicate affected individuals.<br /> I-1 I-2<br /> II-1 II-2 II-3<br /> II-4<br /> III-1 III-2 III-3 III-4 III-5<br /> (a)<br /> (b)<br /> IV-1 IV-2 IV-3<br /> I-1 I-2<br /> IV-1 IV-2 IV-3<br /> II-5<br /> III-6<br /> II-1 II-2 II-3 II-4 II-5<br /> III-7<br /> III-1 III-2 III-3 III-4 III-5<br /> C14. Ectrodactyly, also known as “lobster claw syndrome,” is a<br /> recessive disorder in humans. If a phenotypically unaffected<br /> couple produces an affected offspring, what are the following<br /> probabilities?<br /> A. Both parents are heterozygotes.<br /> B. An offspring is a heterozygote.<br /> C. The next three offspring will be phenotypically unaffected.<br /> D. Any two out of the next three offspring will be phenotypically<br /> unaffected.<br /> C15. Identical twins are produced from the same sperm and egg (which<br /> splits after the first mitotic division), whereas fraternal twins<br /> are produced from separate sperm and separate egg cells. If two<br /> parents with brown eyes (a dominant trait) produce one twin boy<br /> with blue eyes, what are the following probabilities?<br /> A. If the other twin is identical, he will have blue eyes.<br /> B. If the other twin is fraternal, he/she will have blue eyes.<br /> C. If the other twin is fraternal, he/she will transmit the blue eye<br /> allele to his/her offspring.<br /> D. The parents are both heterozygotes.<br /> C16. In cocker spaniels, solid coat color is dominant over spotted coat<br /> color. If two heterozygous dogs were crossed to each other, what<br /> would be the probability of the following combinations of<br /> offspring?<br /> A. A litter of five pups, four with solid fur and one with spotted fur.<br /> B. A first litter of six pups, four with solid fur and two with spotted<br /> fur, and then a second litter of five pups, all with solid fur.<br /> C. A first litter of five pups, the firstborn with solid fur, and then<br /> among the next four, three with solid fur and one with spotted<br /> fur, and then a second litter of seven pups in which the firstborn<br /> is spotted, the second born is spotted, and the remaining five are<br /> composed of four solid and one spotted animal.<br /> D. A litter of six pups, the firstborn with solid fur, the second born<br /> spotted, and among the remaining four pups, two with spotted<br /> fur and two with solid fur.<br /> C17. A cross was made between a white male dog and two different<br /> black females. The first female gave birth to eight black pups, while<br /> the second female gave birth to four white and three black pups.<br /> What are the likely genotypes of the male parent and the two<br /> female parents? Explain whether you are uncertain about any of<br /> the genotypes.<br /> C18. In humans, the allele for brown eye color (B) is dominant to blue<br /> eye color (b). If two heterozygous parents produce children, what<br /> are the following probabilities?<br /> A. The first two children have blue eyes.<br /> B. A total of four children, two with blue eyes and the other two<br /> with brown eyes.<br /> C. The first child has blue eyes, and the next two have brown eyes.<br /> C19. Albinism, a condition characterized by a partial or total lack<br /> of skin pigment, is a recessive human trait. If a phenotypically<br /> unaffected couple produced an albino child, what is the probability<br /> that their next child will be albino?<br /> C20. A true-breeding tall plant was crossed to a dwarf plant. Tallness is<br /> a dominant trait. The F 1 individuals were allowed to self-fertilize.<br /> What are the following probabilities for the F 2 generation?<br /> A. The first plant is dwarf.<br /> B. The first plant is dwarf or tall.<br /> C. The first three plants are tall.<br /> D. For any seven plants, three are tall and four are dwarf.<br /> E. The first plant is tall, and then among the next four, two are tall<br /> and the other two are dwarf.<br /> C21. For pea plants with the following genotypes, list the possible<br /> gametes that the plant can make:<br /> A. TT Yy Rr<br /> B. Tt YY rr<br /> C. Tt Yy Rr<br /> D. tt Yy rr<br /> C22. An individual has the genotype Aa Bb Cc and makes an abnormal<br /> gamete with the genotype Aa Bc. Does this gamete violate the law<br /> of independent assortment and/or the law of segregation? Explain<br /> your answer.<br /> C23. Maple syrup urine disease is a disease found in humans in which<br /> the body is unable to metabolize the amino acids leucine, isoleucine,</div> <div>14.1 TRANSCRIPTIONAL REGULATION 363<br /> To investigate this phenomenon, Jacob and Monod focused<br /> their attention on lactose metabolism in E. coli. Key experimental<br /> observations that led to an understanding of this genetic system<br /> are listed here.<br /> 1. The exposure of bacterial cells to lactose increased the<br /> levels of lactose-utilizing enzymes by 1,000- to 10,000-fold.<br /> 2. Antibody and labeling techniques revealed that the<br /> increase in the activity of these enzymes was due to the<br /> increased synthesis of the enzymes.<br /> 3. The removal of lactose from the environment caused an<br /> abrupt termination in the synthesis of the enzymes.<br /> 4. Mutations that prevented the synthesis of particular<br /> enzymes involved with lactose utilization showed that a<br /> separate gene encoded each enzyme.<br /> These critical observations indicated to Jacob and Monod that<br /> enzyme adaptation is due to the synthesis of specific cellular<br /> proteins in response to lactose in the environment. As described<br /> next, we will examine how Jacob and Monod discovered that this<br /> phenomenon is due to the interactions between genetic regulatory<br /> proteins and small effector molecules. In other words, we<br /> will see that enzyme adaptation is due to the transcriptional regulation<br /> of genes.<br /> The lac Operon Encodes Proteins Involved<br /> in Lactose Metabolism<br /> In bacteria, it is common for a few structural genes to be arranged<br /> together in an operon—a group of two or more genes under the<br /> transcriptional control of a single promoter. An operon encodes<br /> a polycistronic mRNA, an mRNA that contains the coding<br /> sequence for two or more structural genes. Why do operons occur<br /> in bacteria? One biological advantage of an operon organization<br /> is that it allows a bacterium to coordinately regulate a group of<br /> genes that encode proteins with a common functional goal; the<br /> expression of the structural genes occurs as a single unit. For<br /> transcription to take place, an operon is flanked by a promoter<br /> that signals the beginning of transcription and a terminator that<br /> specifies the end of transcription. Two or more structural genes<br /> are found between these two sequences.<br /> Figure 14.3a shows the organization of the genes involved<br /> with lactose utilization and their transcriptional regulation. Two<br /> distinct transcriptional units are present. The first unit, known<br /> i promoter<br /> Regulatory gene<br /> lac operon<br /> lacI lacZ lacY lacA<br /> Encodes β-galactosidase<br /> Encodes lactose<br /> CAP site<br /> Operator site (lacO)<br /> permease<br /> lac promoter (lacP)<br /> Encodes<br /> galactoside<br /> transacetylase<br /> lac<br /> terminator<br /> E.coli<br /> chromosome<br /> (a) Organization of DNA sequences in the lac region of the E.coli chromosome<br /> CH 2 OH<br /> HO O<br /> Lactose H<br /> O<br /> OH<br /> H<br /> H<br /> H<br /> H<br /> CH 2 OH<br /> HO<br /> H<br /> OH<br /> H<br /> Lactose<br /> H OH<br /> Galactose<br /> CH 2 OH<br /> H<br /> H<br /> OH<br /> CH 2 OH<br /> H<br /> H<br /> OH<br /> HO<br /> O OH<br /> H<br /> H<br /> H OH<br /> Glucose<br /> (b) Functions of lactose permease and �-galactosidase<br /> OH<br /> β-galactosidase<br /> O OH<br /> +<br /> H<br /> H<br /> H + H +<br /> H<br /> Lactose permease<br /> O OH<br /> H<br /> H<br /> OH<br /> β-galactosidase<br /> side reaction<br /> CH 2 OH<br /> HO O<br /> H<br /> OH<br /> H<br /> H<br /> H<br /> H<br /> OH<br /> β-galactosidase<br /> Cytoplasm<br /> O CH 2<br /> H<br /> H<br /> OH<br /> HO<br /> H<br /> O OH<br /> H<br /> H<br /> OH<br /> Allolactose<br /> FIGURE<br /> 14.3 Organization<br /> of the lac operon and<br /> other genes involved with lactose<br /> metabolism in E. coli. (a) The CAP<br /> site is the binding site for CAP; the<br /> operator site provides a binding site for<br /> the lac repressor. The promoter (lacP)<br /> is responsible for the transcription of<br /> the lacZ, lacY, and lacA genes as a single<br /> unit, which ends at the lac terminator.<br /> The i promoter is responsible for the<br /> transcription of the lacI gene. (b)<br /> Lactose permease is a membrane protein<br /> that allows the uptake of lactose into<br /> the bacterial cytoplasm. It cotransports<br /> lactose with H + . Because bacteria<br /> maintain an H + gradient across their<br /> cytoplasmic membrane, this cotransport<br /> permits the active accumulation<br /> of lactose against a gradient. β-<br /> galactosidase is a cytoplasmic enzyme<br /> that cleaves lactose and related<br /> compounds into galactose and glucose.<br /> As a minor side reaction, β-galactosidase<br /> also converts lactose into allolactose.<br /> As shown here, allolactose can also be<br /> broken down into galactose and glucose.</div> <div>652 CHAPTER 23 :: DEVELOPMENTAL GENETICS<br /> Peripheral zone<br /> Leaf bud<br /> Differentiation<br /> Stem cells<br /> Central zone<br /> CLV3<br /> WUS<br /> Organizing<br /> center<br /> Peripheral zone<br /> Differentiation<br /> Leaf bud<br /> FIGURE 23.22 Organization of a shoot meristem. The<br /> organization of a shoot meristem is controlled by the WUS and CLV3<br /> genes, which are acronyms for Wuschel and clavata, respectively. The<br /> WUS gene is expressed in the organizing center and induces the cells<br /> in the central zone to become undifferentiated stem cells. The red<br /> arrow indicates that the WUS protein induces these stem cells to turn<br /> on the CLV3 gene, which encodes a secreted protein that binds to<br /> receptors in the cells of the peripheral zone. The black lines with a<br /> vertical slash indicate that the CLV3 protein prevents the cells in the<br /> peripheral zone from expressing the WUS gene. This limits the area of<br /> WUS gene expression to the underlying organizing center and thereby<br /> maintains a small population of stem cells at the growing tip. The cells<br /> in the peripheral zone are allowed to grow and differentiate into lateral<br /> structures such as leaves.<br /> of actively dividing stem cells. The central zone is an area where<br /> undifferentiated stem cells are always maintained. The peripheral<br /> zone contains dividing cells that will eventually differentiate into<br /> plant structures. For example, the peripheral zone may form a<br /> bud, which will produce a leaf or flower.<br /> In Arabidopsis, the organization of a shoot meristem is controlled<br /> by two critical genes termed WUS and CLV3. The WUS<br /> gene encodes a transcription factor that is expressed in the organizing<br /> center (Figure 23.22). The expression of the WUS gene<br /> induces the adjacent cells in the central zone to become undifferentiated<br /> stem cells. These stem cells then turn on the CLV3<br /> gene, which encodes a secreted protein. The CLV3 protein binds<br /> to receptors in the cells of the peripheral zone, preventing them<br /> from expressing the WUS gene. This limits the area of WUS<br /> gene expression to the underlying organizing center and thereby<br /> maintains a small population of stem cells at the growing tip. A<br /> shoot meristem in Arabidopsis contains only about 100 cells. The<br /> inhibition of WUS expression in the peripheral cells also allows<br /> them to embark on a path of cell differentiation so they can produce<br /> structures such as leaves and flowers.<br /> In the seedling shown earlier in Figure 23.21e, three main<br /> regions are observed. The apical region produces the leaves and<br /> flowers of the plant. The central region (not to be confused with<br /> the central zone) creates the stem. The radial pattern of cells in<br /> the central region causes the radial growth observed in plants.<br /> Finally, the basal region produces the roots. Each of these three<br /> regions develops differently as indicated by their unique cell division<br /> patterns and distinct morphologies. In addition, by analyzing<br /> mutants that disrupt the developmental process, researchers<br /> have discovered that these three regions express different sets of<br /> genes. Plant biologists have identified a category of genes, known<br /> as apical-basal-patterning genes, that are important in the early<br /> stages of development. As described in Table 23.2, defects in<br /> apical-basal-patterning genes cause dramatic effects in each of<br /> the three main regions. For example, the gurke gene is necessary<br /> for apical development. When it is defective, the embryo lacks<br /> apical structures. Currently, a great amount of effort is directed<br /> toward the identification of genes that govern pattern formation<br /> in the three regions of Arabidopsis and other plants.<br /> TABLE 23.2<br /> Examples of Arabidopsis Apical-Basal-Patterning Genes That Affect<br /> the Development of the Apical, Central, or Basal Region<br /> Gene<br /> Apical<br /> Gurke<br /> Pin1<br /> Aintegumenta<br /> Central<br /> Fackel<br /> Scarecrow<br /> Basal<br /> Monopterous<br /> Hobbit<br /> Overall Organization<br /> Gnom<br /> Description<br /> Minor loss-of-function alleles in the Gurke gene<br /> produce seedlings with highly reduced or no cotyledons.<br /> Complete loss-of-function eliminates the entire shoot.<br /> Encodes a putative auxin exporter that is expressed in<br /> the peripheral zone. It plays a role in the location of<br /> bud sites, which give rise to leaves and flowers.<br /> Encodes a transcription factor that is also expressed<br /> in the peripheral zone. Its expression maintains the<br /> proliferative cell state during the growth of lateral buds.<br /> Encodes an enzyme involved in plant sterol synthesis.<br /> Loss-of-function alleles show a severe defect in stem<br /> growth.<br /> Encodes a transcription factor protein that plays a role<br /> in the asymmetric division that produces the radial<br /> pattern of growth in the stem. Note: The scarecrow<br /> protein also affects cell division patterns in roots and<br /> plays a role in sensing gravity.<br /> Encodes a transcription factor. When this gene is<br /> defective, the plant embryo cannot initiate the<br /> formation of root structures, but root structures<br /> can be formed postembryonically under the correct<br /> growth conditions. This gene seems to be required for<br /> organizing root formation in the embryo but is not<br /> required for root formation per se.<br /> Encodes a subunit of a protein that functions as a cell<br /> cycle checkpoint during anaphase. Hobbit function may<br /> be required to couple cell division to cell differentiation<br /> in the root meristem or to restrict the response to<br /> growth hormones such as auxin. Loss-of-function alleles<br /> result in plants that are incapable of forming a root.<br /> Plays a role in the stable fixation of the apical-basal<br /> axis of the Arabidopsis embryo.</div> <div>EXPERIMENTAL QUESTIONS 725<br /> E12. In a strain of mice, the average 6-week body weight is 25 g and the<br /> narrow-sense heritability for this trait is 0.21.<br /> A. What would be the average weight of the offspring if parents<br /> with a mean weight of 27 g were chosen?<br /> B. What weight of parents would you have to choose to obtain<br /> offspring with an average weight of 26.5 g?<br /> E13. Two tomato strains, A and B, both produce fruit that weighs, on<br /> average, 1 lb each. All of the variance is due to V G . When these two<br /> strains are crossed to each other, the F 1 offspring display heterosis<br /> with regard to fruit weight, with an average weight of 2 lb. You<br /> take these F 1 offspring and backcross them to strain A. You then<br /> grow several plants from this cross and measure the weights of<br /> their fruit. What would be the expected results for each of the<br /> following scenarios?<br /> A. Heterosis is due to a single overdominant gene.<br /> B. Heterosis is due to two dominant genes, one in each strain.<br /> C. Heterosis is due to two overdominant genes.<br /> D. Heterosis is due to dominance of several genes each from strains<br /> A and B.<br /> E14. You will need to understand solved problem S4 before answering<br /> this question. With regard to height, the variance for fathers (in<br /> square inches) was 112, the variance for sons was 122, and the<br /> covariance was 144. The mean height for fathers was 68 in., and<br /> the mean height for sons was 69 in. If a father had a height of<br /> 70 in., what is the most probable height of his son?<br /> E15. A danger in computing heritability values from studies involving<br /> genetically related individuals is the possibility that these<br /> individuals share more similar environments than do unrelated<br /> individuals. In the experiment of Figure 25.10, which data are<br /> the most compelling evidence that ridge count is not caused by<br /> genetically related individuals sharing common environments?<br /> Explain.<br /> E16. A large, genetically heterogeneous group of tomato plants was used<br /> as the original breeding stock by two different breeders, named<br /> Mary and Hector. Each breeder was given 50 seeds and began an<br /> artificial selection strategy, much like the one described in Figure<br /> 25.13. The seeds were planted, and the breeders selected the 10<br /> plants with the highest mean tomato weights as the breeding stock<br /> for the next generation. This process was repeated over the course<br /> of 12 growing seasons, and the following data were obtained:<br /> Mean Weight of Tomatoes (lb)<br /> Year Mary’s Tomatoes Hector’s Tomatoes<br /> 1 0.7 0.8<br /> 2 0.9 0.9<br /> 3 1.1 1.2<br /> 4 1.2 1.3<br /> 5 1.3 1.3<br /> 6 1.4 1.4<br /> 7 1.4 1.5<br /> 8 1.5 1.5<br /> 9 1.5 1.5<br /> 10 1.5 1.5<br /> 11 1.5 1.5<br /> 12 1.5 1.5<br /> A. Explain these results.<br /> B. Another tomato breeder, named Martin, got some seeds from<br /> Mary’s and Hector’s tomato strains (after 12 generations), grew<br /> the plants, and then crossed them to each other. The mean<br /> weight of the tomatoes in these hybrids was about 1.7 lb. For a<br /> period of five years, Martin subjected these hybrids to the same<br /> experimental strategy that Mary and Hector had followed, and<br /> he obtained the following results:<br /> Mean Weight of Tomatoes (lb)<br /> Year<br /> Martin’s Tomatoes<br /> 1 1.7<br /> 2 1.8<br /> 3 1.9<br /> 4 2.0<br /> 5 2.0<br /> Explain Martin’s data. Is heterosis occurring? Why was Martin<br /> able to obtain tomatoes heavier than 1.5 lb, while Mary’s and<br /> Hector’s strains appeared to plateau at this weight?<br /> E17. The correlations for height were determined for 15 pairs of<br /> individuals with the following genetic relationships:<br /> Mother/daughter: 0.36<br /> Mother/granddaughter: 0.17<br /> Sister/sister: 0.39<br /> Sister/sister (fraternal twins): 0.40<br /> Sister/sister (identical twins): 0.77<br /> What is the average heritability for height in this group of females?<br /> E18. An animal breeder had a herd of sheep with a mean weight of<br /> 254 lb at 3 years of age. He chose animals with mean weights of<br /> 281 lb as parents for the next generation. When these offspring<br /> reached 3 years of age, their mean weights were 269 lb.<br /> A. Calculate the narrow-sense heritability for weight in this herd.<br /> B. Using the heritability value that you calculated in part A, what<br /> weight of animals would you have to choose to get offspring<br /> that weigh 275 lb (at 3 years of age)?<br /> E19. The trait of blood pressure in humans has a frequency distribution<br /> that is similar to a normal distribution. The following graph shows<br /> the ranges of blood pressures for a selected population of people.<br /> The red line depicts the frequency distribution of the systolic<br /> pressures for the entire population. Several individuals with high<br /> blood pressure were identified, and the blood pressures of their<br /> relatives were determined. This frequency distribution is depicted<br /> (continued)</div> <div>444 CHAPTER 16 :: GENE MUTATION AND DNA REPAIR<br /> H<br /> N<br /> O<br /> O<br /> N<br /> Thymine dimer<br /> CH 3 H 3 C<br /> H H<br /> O<br /> N<br /> H<br /> N<br /> O<br /> O 6 -methylguanine<br /> N<br /> H<br /> N<br /> O<br /> N<br /> CH 3<br /> N<br /> NH 2<br /> SH<br /> CH 2<br /> Alkyltransferase<br /> DNA<br /> backbone<br /> H<br /> O<br /> N<br /> O<br /> N<br /> CH 3<br /> H<br /> DNA photolyase<br /> cleaves the 2<br /> bonds between<br /> the thymine dimer.<br /> H 3 C<br /> H<br /> O<br /> N<br /> N<br /> O<br /> H<br /> DNA<br /> backbone<br /> Guanine<br /> N<br /> H<br /> N<br /> O<br /> N<br /> N<br /> Alkyltransferase catalyzes<br /> the removal of the methyl<br /> group onto itself.<br /> H<br /> NH 2<br /> CH 2<br /> CH 3<br /> S<br /> The normal structure of the 2 thymines is restored.<br /> (a) Direct repair of a thymine dimer<br /> The normal structure of guanine is restored.<br /> (b) Direct repair of a methylated base<br /> FIGURE 16.17 Direct repair of damaged bases in DNA. (a) The repair of thymine dimers by photolyase. (b) The repair of<br /> methylguanine by the transfer of the methyl group to alkyltransferase.<br /> 5′ 3′<br /> G<br /> C<br /> C G<br /> T<br /> A<br /> C<br /> G<br /> A<br /> T<br /> A<br /> U<br /> U<br /> C<br /> G<br /> G<br /> C<br /> A<br /> T<br /> C<br /> G<br /> G<br /> C<br /> 3′ 5′<br /> 5′ 3′<br /> G<br /> C<br /> C G<br /> T<br /> A<br /> C<br /> G<br /> A<br /> T<br /> A<br /> T<br /> C<br /> G<br /> G<br /> C<br /> A<br /> T<br /> C<br /> G<br /> G<br /> C<br /> 3′ 5′<br /> A<br /> T<br /> A<br /> T<br /> C<br /> G<br /> C<br /> G<br /> C<br /> G<br /> C<br /> G<br /> T<br /> A<br /> N-glycosylase recognizes an abnormal<br /> base and cleaves the bond between the<br /> base and the sugar.<br /> 5′ 3′<br /> G<br /> C T C<br /> C G<br /> A<br /> C C C C T<br /> A G A G A<br /> A G G G G G A<br /> T C T C T<br /> 3′ 5′<br /> Apyrimidinic<br /> nucleotide<br /> AP endonuclease recognizes a missing<br /> base and cleaves the DNA backbone on<br /> the 5′ side of the missing base.<br /> 5′ 3′<br /> G<br /> C T C<br /> C G<br /> A<br /> C C C C T<br /> A G A G A<br /> A G G G G G A<br /> T C T C T<br /> 3′ 5′<br /> DNA polymerase uses its 5′ 3′ exonuclease<br /> Nick activity to remove the damaged region<br /> and then fills in the region with normal DNA.<br /> DNA ligase seals the region.<br /> Nick-translated region<br /> T<br /> A<br /> template for resynthesis of a normal complementary strand. NER<br /> is found in all eukaryotes and prokaryotes, although its molecular<br /> mechanism is better understood in prokaryotic species.<br /> In E. coli, the NER system requires four key proteins, designated<br /> UvrA, UvrB, UvrC, and UvrD, plus the help of DNA polymerase<br /> and DNA ligase. UvrA, B, C, and D recognize and remove<br /> a short segment of a damaged DNA strand. They are named Uvr<br /> because they are involved in Ultraviolet light repair of pyrimidine<br /> dimers, although the UvrA–D proteins are also important<br /> in repairing chemically damaged DNA.<br /> Figure 16.19 outlines the steps involved in the E. coli NER<br /> system. A protein complex consisting of two UvrA molecules<br /> and one UvrB molecule tracks along the DNA in search of damaged<br /> DNA. Such DNA will have a distorted double helix, which is<br /> sensed by the UvrA/UvrB complex. When a damaged segment is<br /> identified, the two UvrA proteins are released, and UvrC binds to<br /> the site. The UvrC protein makes cuts in the damaged strand on<br /> both sides of the damaged site. Typically, the damaged strand is<br /> cut four to five nucleotides away from the 3' side of the damage<br /> and eight nucleotides from the 5' end. After this process, UvrD,<br /> which is a helicase, recognizes the region and separates the two<br /> FIGURE 16.18 Base excision repair. An abnormal base is<br /> initially recognized by an enzyme known as N-glycosylase, which<br /> cleaves the bond between the base and the sugar. Depending on<br /> whether a purine or pyrimidine is removed, this creates an apurinic or<br /> apyrimidinic site, respectively. In either case, the region is recognized<br /> by AP endonuclease, which makes a cut at the 5' side of the abnormal<br /> nucleotide. DNA polymerase then removes the abnormal region and fills<br /> it in with normal DNA. Finally, DNA ligase seals the repaired strand.</div> <div>232 CHAPTER 9 :: MOLECULAR STRUCTURE OF DNA AND RNA<br /> The experimental protocol of Chargaff is described in Figure<br /> 9.14. He began with various types of cells as starting material.<br /> The chromosomes were extracted from cells and then treated<br /> with protease to separate the DNA from chromosomal proteins.<br /> The DNA was then subjected to a strong acid treatment that<br /> cleaved the bonds between the sugars and bases. Therefore, the<br /> strong acid treatment would release the individual bases from the<br /> DNA strands. This mixture of bases was then subjected to paper<br /> chromatography to separate the four types. The amounts of the<br /> four bases were determined spectroscopically.<br /> ■ THE GOAL<br /> An analysis of the base composition of DNA in different organisms<br /> may reveal important features about the structure of DNA.<br /> ■ ACHIEVING THE GOAL — FIGURE 9.14 An analysis of base composition among different DNA samples.<br /> Starting material: The following types of cells were obtained: Escherichia coli, Streptococcus pneumoniae (type III), yeast, turtle red blood<br /> cells, salmon sperm cells, chicken red blood cells, and human liver cells.<br /> Experimental level<br /> Conceptual level<br /> 1. For each type of cell, extract the<br /> chromosomal material. This can be<br /> done in a variety of ways, including the<br /> use of high salt, detergent, or mild alkali<br /> treatment. Note: The chromosomes<br /> contain both DNA and protein.<br /> Solution of<br /> chromosomal<br /> extract<br /> DNA +<br /> proteins<br /> 2. Remove the protein. This can be done in<br /> several ways, including treament with<br /> protease.<br /> Protease<br /> DNA<br /> 3. Hydrolize the DNA to release the bases<br /> from the DNA strands. A common way<br /> to do this is by strong acid treatment.<br /> Acid<br /> A<br /> G<br /> A<br /> C<br /> G<br /> C<br /> G<br /> T<br /> T<br /> C<br /> Individual<br /> bases</div> <div>QUESTIONS FOR STUDENT DISCUSSION/COLLABORATION 631<br /> E7. Let’s suppose a cell line has become malignant because it has<br /> accumulated mutations that inactivate two different tumorsuppressor<br /> genes. A researcher followed the protocol described<br /> in the experiment of Figure 22.10 and isolated DNA from<br /> this mutant cell line. The DNA was used to transform normal<br /> fibroblast (NIH3T3) cells. What results would you expect? Would<br /> you expect a high number of malignant foci or not? Explain your<br /> answer.<br /> E8. What is a transformed cell? Describe three different methods to<br /> transform cells in a laboratory.<br /> E9. In the experiment of Figure 22.10, what would be the results if<br /> the DNA sample had been treated with either DNase, RNase, or<br /> protease prior to the treatment with calcium and phosphate ions?<br /> E10. Explain how the experimental study of cancer-causing viruses has<br /> increased our understanding of cancer.<br /> E11. Discuss ways to distinguish whether a particular form of<br /> cancer involves an inherited predisposition or is due strictly to<br /> (postzygotic) somatic mutations. In your answer, consider that<br /> only one mutation may be inherited, but the cancer might develop<br /> only after several somatic mutations.<br /> E12. The codon change (Gly-12 to Val-12) in human rasH that converts<br /> it to oncogenic rasH has been associated with many types of<br /> cancers. For this reason, researchers would like to develop drugs to<br /> inhibit oncogenic rasH. Based on your molecular understanding of<br /> the Ras protein, what types of drugs might you develop? In other<br /> words, what would be the structure of the drugs, and how would<br /> they inhibit Ras protein? How would you test the efficacy of the<br /> drugs? What might be some side effects?<br /> E13. Describe how normal mammalian cells grow on solid media in a<br /> laboratory and how cancer cells grow. Is a malignant focus derived<br /> from a single cancer cell or from many independent cancer cells<br /> that happen to be in the same region of a tissue culture plate?<br /> Questions for Student Discussion/Collaboration<br /> 1. Make a list of the benefits that may arise from genetic testing as<br /> well as possible negative consequences. Discuss the items on your<br /> list.<br /> 2. Our government has finite funds to devote to cancer research.<br /> Discuss which aspects of cancer biology you would spend the most<br /> money pursuing.<br /> A. Identifying and characterizing oncogenes and tumor-suppressor<br /> genes<br /> B. Identifying agents in our environment that cause cancer<br /> C. Identifying viruses that cause cancer<br /> D. Devising methods aimed at killing cancer cells in the body<br /> E. Informing the public of the risks involved in exposure to<br /> carcinogens<br /> In the long run, which of these areas would you expect to be the<br /> most effective in decreasing mortality due to human cancer?<br /> Note: All answers appear at the website for this textbook; the answers to<br /> even-numbered questions are in the back of the textbook.<br /> www.mhhe.com/brookergenetics3e<br /> Visit the Online Learning Center for practice tests, answer keys, and other learning aids for this chapter. Enhance your understanding of genetics with<br /> our interactive exercises, quizzes, animations, and much more.</div> <div>812 GLOSSARY<br /> phage. Cosmid vectors can accept fragments of DNA<br /> that are typically tens of thousands of base pairs in<br /> length.<br /> C 0 t curve a plot of C/C 0 versus C 0 t.<br /> cotransduction the phenomenon in which bacterial<br /> transduction transfers a piece of DNA carrying two<br /> closely linked genes.<br /> cotransformation the phenomenon in which bacterial<br /> transformation transfers a piece of DNA carrying two<br /> closely linked genes.<br /> cotranslational events that occur during translation.<br /> cotranslational import during the synthesis of certain<br /> eukaryotic proteins, translation begins in the<br /> cytosol and then is temporarily halted by the signal<br /> recognition particle (SRP). Translation resumes when<br /> the ribosome has become bound to the ER membrane<br /> and the polypeptide is synthesized into the ER lumen<br /> or ER membrane.<br /> cotranslational sorting refers to the sorting of proteins<br /> into the ER. The protein is actually translated into the<br /> ER lumen or ER membrane.<br /> covariance a statistic that describes the degree of<br /> variation between two variables within a group.<br /> cpDNA an abbreviation for chloroplast DNA.<br /> CpG island a group of CG sequences that may be<br /> clustered near a promoter region of a gene. The<br /> methylation of the cytosine bases usually inhibits<br /> transcription.<br /> CREB protein (cAMP response element-binding<br /> protein) a regulatory transcription factor that<br /> becomes activated in response to specific cellsignaling<br /> molecules that cause the synthesis of cAMP.<br /> cross a mating between two distinct individuals. An analysis<br /> of their offspring may be conducted to understand how<br /> traits are passed from parent to offspring.<br /> cross-fertilization same meaning as cross. It requires that<br /> the male and female gametes come from separate<br /> individuals.<br /> crossing over a physical exchange of chromosome pieces<br /> that most commonly occurs during prophase of<br /> meiosis I.<br /> C-terminus see carboxyl terminus.<br /> cyclic AMP (cAMP) in bacteria, a small effector molecule<br /> that binds to CAP (catabolite activator protein). In<br /> eukaryotes, cAMP functions as a second messenger in<br /> a variety of intracellular signaling pathways; in some<br /> cases, it binds to transcription factors such as the<br /> CREB protein.<br /> cyclin a type of protein that plays a role in the regulation<br /> of the eukaryotic cell cycle.<br /> cyclin-dependent protein kinases (CDKs) enzymes that<br /> are regulated by cyclins and can phosphorylate other<br /> cellular proteins by covalently attaching a phosphate<br /> group.<br /> cytogeneticist a scientist who studies chromosomes<br /> under the microscope.<br /> cytogenetic mapping the mapping of genes or genetic<br /> sequences using microscopy.<br /> cytogenetics the field of genetics that involves the<br /> microscopic examination of chromosomes.<br /> cytokinesis the division of a single cell into two cells. The<br /> two nuclei produced in M phase are segregated into<br /> separate daughter cells during cytokinesis.<br /> cytological mapping see cytogenetic mapping.<br /> cytoplasmic inheritance (also known as extranuclear<br /> inheritance) refers to the inheritance of genetic<br /> material that is not found within the cell nucleus.<br /> cytosine a pyrimidine base found in DNA and RNA. It<br /> base-pairs with guanine in DNA.<br /> DDam see DNA adenine methyltransferase.<br /> Darwinian fitness the relative likelihood that a<br /> phenotype will survive and contribute to the gene<br /> pool of the next generation as compared with other<br /> phenotypes.<br /> database a computer storage facility that stores many data<br /> files such as those containing genetic sequences.<br /> daughter strand in DNA replication, the newly made<br /> strand of DNA.<br /> deamination the removal of an amino group from a<br /> molecule. For example, the removal of an amino<br /> group from cytosine produces uracil.<br /> decoding function the ability of the 16S rRNA to detect<br /> when an incorrect tRNA is bound at the A site and<br /> prevent elongation until the mispaired tRNA is<br /> released from the A site.<br /> decondensed refers to chromosomes forming a less<br /> compact structure.<br /> deficiency condition in which a segment of chromosomal<br /> material is missing.<br /> degeneracy in genetics, this term means that more<br /> that one codon specifies the same amino acid. For<br /> example, the codons GGU, GGC, GGA, and GGG all<br /> specify the amino acid glycine.<br /> degrees of freedom in a statistical analysis, the number of<br /> categories that are independent of each other.<br /> deleterious mutation a mutation that is detrimental with<br /> regard to its effect on phenotype.<br /> deletion condition in which a segment of DNA is<br /> missing.<br /> deletion mapping the use of strains carrying deletions<br /> within a defined region to map a mutation of<br /> unknown location.<br /> deme see local population.<br /> de novo methylation the methylation of DNA that has<br /> not been previously methylated. This is usually a<br /> highly regulated event.<br /> deoxyribonucleic acid (DNA) the genetic material. It<br /> is a double-stranded structure, with each strand<br /> composed of repeating units of deoxyribonucleotides.<br /> deoxyribose the sugar found in DNA.<br /> depurination the removal of a purine base from DNA.<br /> determined cell a cell that is destined to differentiate into<br /> a specific cell type.<br /> developmental genetics the area of genetics concerned<br /> with the roles of genes in orchestrating the changes<br /> that occur during development.<br /> diakinesis the fifth stage of prophase of meiosis I.<br /> diauxic growth the sequential use of two sugars by a<br /> bacterium.<br /> dicentric describes a chromosome with two centromeres.<br /> dicentric bridge the region between the two centromeres<br /> in a dicentric chromosome.<br /> dicer an endonuclease that makes a cut in doublestranded<br /> RNA.<br /> dideoxyribonucleotide a nucleotide used in DNA<br /> sequencing that is missing the 3'—OH group. If a<br /> dideoxyribonucleotide is incorporated into a DNA<br /> strand, it stops any further growth of the strand.<br /> dideoxy sequencing a method of DNA sequencing that<br /> uses dideoxyribonucleotides to terminate the growth of<br /> DNA strands.<br /> differential centrifugation a form of centrifugation<br /> involving a series of centrifugation steps in which<br /> the supernatant or pellet is used in each subsequent<br /> centrifugation step.<br /> differentially methylated region (DMR) in the case<br /> of imprinting, a site that is methylated during<br /> spermatogenesis or oogenesis, but not both.<br /> differentiated cell a cell that has become a specialized<br /> type of cell within a multicellular organism.<br /> dihybrid cross a cross in which an experimenter follows<br /> the outcome of two different traits.<br /> dihybrid testcross a cross in which an experimenter<br /> crosses an individual that is heterozygous for two<br /> genes to an individual that is homozygous recessive<br /> for the same two genes.<br /> dimeric DNA polymerase a complex of two DNA<br /> polymerase proteins that move as a unit during DNA<br /> replication.<br /> dioecious in plants, a species in which male and female<br /> gametophytes are produced on a single (sporophyte)<br /> individual.<br /> diploid an organism or cell that contains two copies of<br /> each type of chromosome.<br /> diplotene the fourth stage of prophase of meiosis I.<br /> directionality in DNA and RNA, refers to the 5' to 3'<br /> arrangement of nucleotides in a strand; in proteins,<br /> refers to the linear arrangement of amino acids from<br /> the N-terminal to C-terminal ends.<br /> directional selection natural selection that favors an<br /> extreme phenotype. This usually leads to the fixation<br /> of the favored allele.<br /> direct repeat (DR) short DNA sequences that flank<br /> transposable elements in which the DNA sequence is<br /> repeated in the same direction.<br /> discontinuous trait a trait in which each offspring can be<br /> put into a particular phenotypic category.<br /> discovery-based science experimentation that does not<br /> require a preconceived hypothesis. In some cases,<br /> the goal is to collect data to be able to formulate a<br /> hypothesis.<br /> disequilibrium in population genetics, refers to<br /> a population that is not in Hardy-Weinberg<br /> equilibrium.<br /> dispersive model an incorrect model for DNA replication<br /> in which segments of parental DNA and newly made<br /> DNA are interspersed in both strands following the<br /> replication process.<br /> disruptive selection natural selection that favors both<br /> extremes of a phenotypic category. This results in a<br /> balanced polymorphism.<br /> dizygotic twins also known as fraternal twins; twins<br /> formed from separate pairs of sperm and egg cells.<br /> DNA the abbreviation for deoxyribonucleic acid.<br /> DNA adenine methyltransferase an enzyme in bacteria<br /> that attaches methyl groups to the adenine base in<br /> DNA that is found within the sequence GATC.<br /> DnaA box sequence serves as a recognition site for the<br /> binding of the DnaA protein, which is involved in the<br /> initiation of bacterial DNA replication.<br /> DnaA protein a protein that binds to the dnaA box<br /> sequence at the origin of replication in bacteria and<br /> initiates DNA replication.<br /> DNA fingerprinting a technology to identify a particular<br /> individual based on the properties of their DNA.<br /> DNA gap repair synthesis the synthesis of DNA in a<br /> region where a DNA strand has been previously<br /> removed, usually by a DNA repair enzyme or by an<br /> enzyme involved in homologous recombination.<br /> DNA gyrase also known as topoisomerase II; an enzyme<br /> that introduces negative supercoils into DNA using<br /> energy from ATP. Gyrase can also relax positive<br /> supercoils when they occur.<br /> DNA helicase an enzyme that separates the two strands<br /> of DNA.<br /> DNA library a collection of many hybrid vectors, each<br /> vector carrying a particular fragment of DNA from<br /> a larger source. For example, each hybrid vector<br /> in a DNA library might carry a small segment of<br /> chromosomal DNA from a particular species.<br /> DNA ligase an enzyme that catalyzes a covalent bond<br /> between two DNA fragments.<br /> DNA methylation the phenomenon in which an enzyme<br /> covalently attaches a methyl group (—CH 3 ) to a base<br /> (usually adenine or cytosine) in DNA.<br /> DNA methyltransferase the enzyme that attaches methyl<br /> groups to adenine or cytosine bases.<br /> DNA microarray a small silica, glass, or plastic slide that<br /> is dotted with many different sequences of DNA,<br /> corresponding to short sequences within known<br /> genes.<br /> DNA-N-glycosylase an enzyme that can recognize an<br /> abnormal base and cleave the bond between it and the<br /> sugar in the DNA backbone.<br /> DNA polymerase an enzyme that catalyzes the covalent<br /> attachment of nucleotides together to form a strand<br /> of DNA.<br /> DNA primase an enzyme that synthesizes a short RNA<br /> primer for DNA replication.<br /> DNA probe in a hybridization experiment, a singlestranded<br /> DNA fragment with a base sequence that is<br /> complementary to a gene of interest. The DNA probe<br /> is labeled to detect a gene of interest.<br /> DNA profiling see DNA fingerprinting.<br /> DNA replication the process in which original DNA<br /> strands are used as templates for the synthesis of new<br /> DNA strands.<br /> DNA replication licensing in eukaryotes, occurs when<br /> MCM helicase is bound at an origin, enabling the<br /> formation of two replication forks.<br /> DNase an enzyme that cuts the sugar–phosphate<br /> backbone in DNA.<br /> DNase I an endonuclease that cleaves DNA.<br /> DNase I footprinting a method to study protein–DNA<br /> interactions in which the binding of a protein to DNA<br /> protects the DNA from digestion by DNase I.<br /> DNase I sensitivity the phenomenon in which DNA is<br /> in an open conformation and able to be digested by<br /> DNase I.</div> <div>CONCEPTUAL QUESTIONS 723<br /> A. Discuss the possible reasons for the differences in variance.<br /> B. If you were a potato farmer, would you rather raise a variety<br /> with a low or high variance? Explain your answer from a<br /> practical point of view.<br /> C. If you were a potato breeder and you wanted to develop<br /> potatoes with a heavier weight, would you choose the variety<br /> with a low or high variance? Explain your answer.<br /> C8. If an r value equals 0.5 and N � 4, would you conclude a positive<br /> correlation is found between the two variables? Explain your<br /> answer. What if N � 500?<br /> C9. What does it mean when a correlation coefficient is negative? Can<br /> you think of examples?<br /> C10. When a correlation coefficient is statistically significant, what do<br /> you conclude about the two variables? What do the results mean<br /> with regard to cause and effect?<br /> C11. What is polygenic inheritance? Discuss the issues that make<br /> polygenic inheritance difficult to study.<br /> C12. What is a quantitative trait locus (QTL)? Does a QTL contain<br /> one gene or multiple genes? What technique is commonly used to<br /> identify QTLs?<br /> C13. Let’s suppose that weight in a species of mammals is polygenic,<br /> and each gene exists as a heavy and light allele. If the allele<br /> frequencies in the population were equal for both types of allele<br /> (i.e., 50% heavy alleles and 50% light alleles), what percentage<br /> of individuals would be homozygous for the light alleles at all of<br /> the genes affecting this trait, if the trait was determined by the<br /> following number of genes?<br /> A. Two<br /> B. Three<br /> C. Four<br /> C14. The broad-sense heritability for a trait equals 1.0. In your own<br /> words, explain what this value means. Would you conclude that<br /> the environment is unimportant in the outcome of this trait?<br /> Explain your answer.<br /> C15. Compare and contrast the dominance and overdominance<br /> hypotheses. Based on your knowledge of mutations and genetics,<br /> which do you think tends to be the more common explanation for<br /> heterosis?<br /> C16. What is hybrid vigor (also known as heterosis)? Give examples that<br /> you might find in a vegetable garden.<br /> C17. From an agricultural point of view, discuss the advantages and<br /> disadvantages of selective breeding. It is common for plant<br /> breeders to take two different, highly inbred strains, which are the<br /> product of many generations of selective breeding, and cross them<br /> to make hybrids. How does this approach overcome some of the<br /> disadvantages of selective breeding?<br /> C18. Many beautiful varieties of roses have been produced, particularly<br /> in the last few decades. These newer varieties often have very<br /> striking and showy flowers, making them desirable as horticultural<br /> specimens. However, breeders and novices alike have noticed that<br /> some of these newer varieties do not have very fragrant flowers<br /> compared to the older, more traditional varieties. From a genetic<br /> point of view, suggest an explanation why some of these newer<br /> varieties with superb flowers are not as fragrant.<br /> C19. In your own words, explain the meaning of the term heritability.<br /> Why is a heritability value valid only for a particular population of<br /> individuals raised in a particular environment?<br /> C20. What is the difference between broad-sense heritability and<br /> narrow-sense heritability? Why is narrow-sense heritability such a<br /> useful concept in the field of agricultural genetics?<br /> C21. The heritability for egg weight in a group of chickens on a farm in<br /> Maine is 0.95. Are the following statements regarding heritability<br /> true or false? If a statement is false, explain why.<br /> A. The environment in Maine has very little impact on the<br /> outcome of this trait.<br /> B. Nearly all of the phenotypic variation for this trait in this group<br /> of chickens is due to genetic variation.<br /> C. The trait is polygenic and likely to involve a large number of<br /> genes.<br /> D. Based on the observation of the heritability in the Maine<br /> chickens, it is reasonable to conclude that the heritability for egg<br /> weight in a group of chickens on a farm in Montana is also very<br /> high.<br /> C22. In a fairly large population of people living in a commune in the<br /> southern United States, everyone cares about good nutrition. All<br /> the members of this population eat very nutritional foods, and<br /> their diets are very similar to each other. With regard to height,<br /> how do you think this commune population would compare to the<br /> general population in the following categories?<br /> A. Mean height<br /> B. Heritability for height<br /> C. Genetic variation for alleles that affect height<br /> C23. When artificial selection is practiced over many generations, it is<br /> common for the trait to reach a plateau in which further selection<br /> has little effect on the outcome of the trait. This phenomenon is<br /> illustrated in Figure 25.13. Explain why.<br /> C24. Discuss whether a natural population of wolves or a domesticated<br /> population of German shepherds is more likely to have a higher<br /> heritability for the trait of size.<br /> C25. With regard to heterosis, would the following statements be<br /> consistent with the dominance hypothesis, the overdominance<br /> hypothesis, or both?<br /> A. Strains that have been highly inbred have become<br /> monomorphic for one or more recessive alleles that are<br /> somewhat detrimental to the organism.<br /> B. Hybrid vigor occurs because highly inbred strains are<br /> monomorphic for many genes, while hybrids are more likely to<br /> be heterozygous for those same genes.<br /> C. If a gene exists in two alleles, hybrids are more vigorous<br /> because heterozygosity for the gene is more beneficial than<br /> homozygosity of either allele.</div> <div>PREFACE<br /> xv<br /> Joel Chandlee, University of Rhode Island<br /> Henry Chang, Purdue University<br /> Karen Cichy, Michigan State University<br /> Brian Condie, University of Georgia<br /> Erin Cram, Northeastern University<br /> Jeff DeJong, University of Texas at Dallas<br /> Rodney Dyer, Virginia Commonwealth University<br /> Kevin Edwards, Illinois State University<br /> Aboubaker Elkharroubi, Johns Hopkins University<br /> Cedric Feschotte, UT–Arlington<br /> David Foltz, Louisiana State University<br /> Michael Foster, Eastern Kentucky University<br /> Thomas Fowler, Southern Illinois University<br /> Anne Galbraith, UW–Lacrosse<br /> Jack Girton, Iowa State University<br /> Elliott Goldstein, ASU<br /> Christine Gray, University of Puget Sound<br /> Tara Harmer Luke, The Richard Stockton College of NJ<br /> Marilyn Hart, Mankato State University<br /> Daryl Henderson, Stony Brooker University<br /> Adam Hrincevich, LSU<br /> Michael Ibba, Ohio State University<br /> Margaret Jefferson, CSU–LA<br /> David Kass, Eastern Michigan University<br /> Cindy Malone, CSU–Northridge<br /> John Mishler, Delaware Valley College<br /> James Morris, Clemson University<br /> Sang-Chul Nam, Baylor University<br /> Daniel Odom, California State University–Northridge<br /> Michael Polymenis, Texas A&M University<br /> David Reed, University of Mississippi<br /> Laurie Russell, St. Louis University<br /> Mark Seeger, Ohio State University<br /> Theresa Spradling, University of Northern Iowa<br /> JD Swanson, University of Central Arkansas<br /> James Thompson, University of Oklahoma<br /> John Tomkiel, U of NC–Greensboro<br /> Carol Trent, Western Washington University<br /> Fyodor Urnov, University of California–Berkeley<br /> Harald Vaessin, Ohio State University<br /> Alan Waldman, University of South Carolina<br /> Daniel Wang, University of Miami<br /> Sarah Ward, Colorado State University<br /> Matt White, Ohio University<br /> Dwayne Wise, Mississippi State University<br /> Yang Yen, South Dakota State University<br /> Malcolm Zellars, Georgia State University<br /> Jianzhi Zhang, University of Michigan</div> <div>20 CHAPTER 2 :: MENDELIAN INHERITANCE<br /> stimulates the growth of a pollen tube. This enables sperm cells<br /> to enter the stigma and migrate toward an ovule. Fertilization<br /> occurs when a sperm enters the micropyle, an opening in the<br /> ovule wall, and fuses with an egg cell. The term gamete is used<br /> to describe haploid reproductive cells that can unite to form a<br /> zygote. It should be emphasized, however, that the process that<br /> produces gametes in animals is quite different from the way that<br /> gametes are produced in plants and fungi. These processes are<br /> described in greater detail in Chapter 3.<br /> In some experiments, Mendel wanted to carry out selffertilization,<br /> which means that the pollen and egg are derived<br /> from the same plant. In peas, a modified petal known as the<br /> keel covers the reproductive structures of the plant. Because<br /> of this covering, pea plants naturally reproduce by selffertilization.<br /> In fact, pollination occurs even before the flower<br /> opens. In other experiments, however, Mendel wanted to make<br /> crosses between different plants. How did he accomplish this<br /> goal? Fortunately, pea plants contain relatively large flowers<br /> that are easy to manipulate, making it possible to make crosses<br /> between two particular plants and study their outcomes. This<br /> process, known as cross-fertilization, requires that the pollen<br /> from one plant be placed on the stigma of another plant. This<br /> procedure is shown in Figure 2.3. Mendel was able to pry open<br /> immature flowers and remove the anthers before they produced<br /> pollen. Therefore, these flowers could not self-fertilize. He would<br /> then obtain pollen from another plant by gently touching its<br /> mature anthers with a paintbrush. Mendel applied this pollen to<br /> the stigma of the flower that already had its anthers removed. In<br /> this way, he was able to cross-fertilize his pea plants and thereby<br /> obtain any type of hybrid he wanted.<br /> Mendel Studied Seven Traits That Bred True<br /> When he initiated his studies, Mendel obtained several varieties<br /> of peas that were considered to be distinct. These plants were<br /> different with regard to many morphological characteristics.<br /> Such characteristics of an organism are called characters, or<br /> traits. Over the course of two years, Mendel tested the strains<br /> to determine if their characteristics bred true. This means that<br /> a trait did not vary in appearance from generation to generation.<br /> For example, if the seeds from a pea plant were yellow,<br /> the next generation would also produce yellow seeds. Likewise,<br /> if these offspring were allowed to self-fertilize, all of their offspring<br /> would also produce yellow seeds, and so on. A variety<br /> that continues to produce the same characteristic after several<br /> generations of self-fertilization is called a true-breeding line,<br /> or strain.<br /> Mendel next concentrated his efforts on the analysis of<br /> characteristics that were clearly distinguishable between different<br /> true-breeding lines. Figure 2.4 illustrates the seven traits that<br /> Parental<br /> generation<br /> Firstgeneration<br /> offspring<br /> Remove anthers<br /> from purple flower.<br /> Purple<br /> Anthers<br /> Transfer pollen<br /> from anthers of<br /> white flower to<br /> the stigma of a<br /> purple flower.<br /> Cross-pollinated flower<br /> produces seeds.<br /> White<br /> Plant the seeds.<br /> FIGURE 2.3 How Mendel cross-fertilized two different pea<br /> plants. This illustration depicts a cross between a plant with purple<br /> flowers and another plant with white flowers. The offspring from this<br /> cross are the result of pollination of the purple flower using pollen<br /> from a white flower.<br /> Mendel eventually chose to follow in his breeding experiments.<br /> All seven were found in two variants. The term variant refers to a<br /> particular trait that may be found in two or more versions within<br /> a single species. For example, one trait he followed was height,<br /> which was found in two variants—tall and dwarf plants. Mendel<br /> studied this trait by crossing the variants to each other. A cross<br /> in which an experimenter is observing only one trait is called a<br /> single-factor cross, also called a monohybrid cross. When the<br /> two parents are different variants for a given trait, this type of<br /> cross produces single-trait hybrids, also known as monohybrids.</div> <div>::<br /> 13<br /> TRANSLATION<br /> OF mRNA<br /> CHAPTER OUTLINE<br /> 13.1 The Genetic Basis for Protein Synthesis<br /> 13.2 Structure and Function of tRNA<br /> 13.3 Ribosome Structure and Assembly<br /> 13.4 Stages of Translation<br /> The synthesis of cellular proteins occurs via the translation of the<br /> sequence of codons within mRNA into a sequence of amino acids of a polypeptide.<br /> The general steps that occur in this process have already been outlined<br /> in Chapter 1. In this chapter, we will explore the current state of knowledge<br /> regarding translation, with an eye toward the specific molecular interactions<br /> responsible for this process. During the past few decades, the concerted efforts<br /> of geneticists, cell biologists, and biochemists have profoundly advanced our<br /> understanding of translation. Even so, many questions remain unanswered, and<br /> this topic continues to be an exciting area of research investigation.<br /> We will begin by considering the classic experiments that revealed the<br /> purpose of some genes is to encode proteins that function as enzymes. Next, we<br /> will examine how the genetic code is used to decipher the information within<br /> mRNA to produce a polypeptide with a specific amino acid sequence. The rest<br /> of this chapter is devoted to a molecular understanding of translation as it<br /> occurs in living cells. This will involve an examination of the cellular components—including<br /> many different proteins, RNAs, and small molecules—needed<br /> for the translation process. We will consider the structure and function of tRNA<br /> molecules, which act as the translators of the genetic information within mRNA,<br /> and then examine the composition of ribosomes. Finally, we will explore the<br /> differences between translation in bacterial cells and eukaryotic cells.<br /> A molecular model for the structure of a<br /> ribosome. This is a model of ribosome structure<br /> based on X-ray crystallography. Ribosomes are<br /> needed to synthesize polypeptides, using mRNA<br /> as a template.<br /> 13.1 THE GENETIC BASIS<br /> FOR PROTEIN SYNTHESIS<br /> Proteins are critically important as active participants in cell structure and function.<br /> The primary role of DNA is to store the information needed for the synthesis<br /> of all the proteins that an organism makes. As we discussed in Chapter 12,<br /> genes that encode an amino acid sequence are known as structural genes. The<br /> RNA transcribed from structural genes is called messenger RNA (mRNA). The<br /> main function of the genetic material is to encode the production of cellular<br /> proteins in the correct cell, at the proper time, and in suitable amounts. This<br /> is an extremely complicated task because living cells make thousands of different<br /> proteins. Genetic analyses have shown that a typical bacterium can make a<br /> few thousand different proteins, and estimates for eukaryotes range from several<br /> thousand in simple eukaryote organisms, such as yeast, to tens of thousands in<br /> plants and animals.</div> <div>268 CHAPTER 10 :: CHROMOSOME ORGANIZATION AND MOLECULAR STRUCTURE<br /> D. If you had coated your double helix with rubber cement and<br /> allowed the cement to dry before making the three additional<br /> right-handed turns, would the rubber cement make it more or<br /> less likely for the three turns to create supercoiling? Would a<br /> pair of cemented strings be more or less like a real DNA double<br /> helix compared to an uncemented pair of strings? Explain your<br /> answer.<br /> C10. Try to explain the function of DNA gyrase with a drawing.<br /> C11. How are two topoisomers different from each other? How are they<br /> the same?<br /> C12. On rare occasions, a chromosome can suffer a small deletion<br /> that removes the centromere. When this occurs, the chromosome<br /> usually is not found within subsequent daughter cells. Explain<br /> why a chromosome without a centromere is not transmitted very<br /> efficiently from mother to daughter cells. (Note: If a chromosome<br /> is located outside the nucleus after telophase, it is degraded.)<br /> C13. What is the function of a centromere? At what stage of the cell<br /> cycle would you expect the centromere to be the most important?<br /> C14. Describe the characteristics of highly repetitive DNA.<br /> C15. Describe the structures of a nucleosome and a 30 nm fiber.<br /> C16. Beginning with the G 1 phase of the cell cycle, describe the level of<br /> compaction of the eukaryotic chromosome. How does the level of<br /> compaction change as the cell progresses through the cell cycle?<br /> Why is it necessary to further compact the chromatin during<br /> mitosis?<br /> C17. If you assume the average length of linker DNA is 50 bp,<br /> approximately how many nucleosomes are found in the haploid<br /> human genome, which contains 3 billion bp?<br /> C18. Draw the binding between the nuclear matrix and MARs.<br /> C19. Compare heterochromatin and euchromatin. What are the<br /> differences between them?<br /> C20. Compare the structure and cell localization of chromosomes<br /> during interphase and M phase.<br /> C21. What types of genetic activities occur during interphase? Explain<br /> why these activities cannot occur during M phase.<br /> C22. Let’s assume the linker DNA averages 54 bp in length. How many<br /> molecules of H2A would you expect to find in a DNA sample that<br /> is 46,000 bp in length?<br /> C23. In Figure 10.15, what are we looking at in part b? Is this an 11 nm<br /> fiber, a 30 nm fiber, or a 300 nm fiber? Does this DNA come from<br /> a cell during M phase or interphase?<br /> C24. What are the roles of the core histone proteins compared to the<br /> role of histone H1 in the compaction of eukaryotic DNA?<br /> C25. A typical eukaryotic chromosome found in humans contains about<br /> 100 million bp of DNA. As described in Chapter 9, 1 bp of DNA<br /> has a linear length of 0.34 nm.<br /> A. What is the linear length of the DNA for a typical human<br /> chromosome in micrometers?<br /> B. What is the linear length of a 30 nm fiber of a typical human<br /> chromosome?<br /> C. Based on your calculation of part B, would a typical human<br /> chromosome fit inside the nucleus (with a diameter of 5 µm)<br /> if the 30 nm fiber were stretched out in a linear manner? If<br /> not, explain how a typical human chromosome fits inside the<br /> nucleus during interphase.<br /> C26. Which of the following terms should not be used to describe a<br /> Barr body?<br /> A. Chromatin<br /> B. Euchromatin<br /> C. Heterochromatin<br /> D. Chromosome<br /> E. Genome<br /> C27. Discuss the differences in the compaction levels of metaphase<br /> chromosomes compared to interphase chromosomes. When would<br /> you expect gene transcription and DNA replication to take place,<br /> during M phase or interphase? Explain why.<br /> C28. Explain two ways that histone modifications may affect the level of<br /> chromatin compaction.<br /> C29. What is an SMC protein? Describe two examples.<br /> Experimental Questions<br /> E1. Two circular DNA molecules, which we will call molecule A and<br /> molecule B, are topoisomers of each other. When viewed under the<br /> electron microscope, molecule A appears more compact compared<br /> to molecule B. The level of gene transcription is much lower<br /> for molecule A. Which of the following three possibilities could<br /> account for these observations?<br /> First possibility: Molecule A has 3 positive supercoils, and<br /> molecule B has 3 negative supercoils.<br /> Second possibility: Molecule A has 4 positive supercoils, and<br /> molecule B has 1 negative supercoil.<br /> Third possibility: Molecule A has 0 supercoils, and molecule B<br /> has 3 negative supercoils.<br /> E2. Explain how a renaturation experiment can provide quantitative<br /> information about genome sequence complexity.<br /> E3. In a renaturation experiment, does the copy number affect<br /> only the rate of renaturation, or does it also affect the rate of<br /> denaturation? Explain your answer.<br /> E4. Let’s suppose that you have isolated DNA from a cell and<br /> have viewed it under a microscope. It looks supercoiled. What<br /> experiment would you perform to determine if it is positively or<br /> negatively supercoiled? In your answer, describe your expected<br /> results. You may assume that you have purified topoisomerases at<br /> your disposal.<br /> E5. We seem to know more about the structure of eukaryotic<br /> chromosomal DNA than bacterial DNA. Discuss why you think<br /> this is so, and list several experimental procedures that have<br /> yielded important information concerning the compaction of<br /> eukaryotic chromatin.</div> <div>INDEX 831<br /> Environmental factors<br /> in creation of overlaps between genotypes and<br /> phenotypes, 702–3, 704<br /> in induced mutations, 438–40<br /> norm of reaction and, 9<br /> in phenotypic variation, 710<br /> in sex determination, 63–64<br /> in trait variation, 8–9, 76–77<br /> in transgenic plant introduction, 534–35<br /> Enzyme adaptation, 361–63<br /> Enzyme(s)<br /> allosteric, 377<br /> anabolic, 4, 337<br /> catabolic, 4, 337<br /> feedback inhibition, 377<br /> function of, 337<br /> genes in encoding of, 325<br /> one gene-one enzyme hypothesis, 325–26<br /> processive, 281<br /> restriction, 484, 485, 546<br /> Eosin allele, 89–91<br /> Ephrussi, Boris, 178<br /> Epidermal growth factor, 336<br /> Epigenetic inheritance, 164–74<br /> dosage compensation in, 164–66<br /> Episomes, 142<br /> Epistasis pattern, 86<br /> recessive, 87<br /> three distinct phenotypes due to, 88–89<br /> two distinct phenotypes due to, 87–88<br /> Epitopes, 501<br /> Epstein-Barr virus, 228<br /> ER retention signal, 354<br /> Erythropoietin, 518<br /> Escherichia coli<br /> base pairs of, 248<br /> binary fission in, 48<br /> conjugation in genetic mapping of, 139–42<br /> cotransduction in, 145<br /> DNA repair in<br /> mismatch system, 446<br /> nucleotide excision system, 444<br /> DNA replication in<br /> mutants involved in, 288<br /> proteins involved in, 277<br /> sequence of oriC, 276<br /> enzyme adaptation in, 363<br /> F factor in, 136<br /> genetic transfer during conjugation in, 135<br /> homologous recombination in, 461<br /> in human insulin synthesis, 1<br /> lactose metabolism in, 363<br /> loop domains in, 249<br /> as model organism, 11, 12<br /> negative supercoiling in, 250<br /> osmoregulation in, 376<br /> RNA processing in, 309<br /> transcription in, 301<br /> “Essay on the Principle of Population”<br /> (Malthus), 728<br /> Essential gene, 84–85<br /> Ethical issues<br /> genetic testing and screening, 608<br /> human cloning, 3, 528<br /> stem cell research, 530<br /> Ethyl methanesulfonate, 441<br /> Euchromatin, 260, 261<br /> Eukaryotes<br /> cell cycle in, 48–50, 49<br /> crossing over in haploid<br /> first- and second-division segregation in, 121<br /> in ordered octad arrangement of cells, 120<br /> in production of parental ditype, tetratype,<br /> and nonparental ditype, 123<br /> site of, in separation of allele from<br /> centromere, 121<br /> defined, 44–45<br /> gene regulation in, 390–417<br /> genetic mapping in haploid, 118–24<br /> gene variation in, 685<br /> organelles, 45<br /> ribosome composition in, 345<br /> RNA polymerase in, 305–6<br /> transcription in, 305–9<br /> translation in, 349<br /> Eukaryotic chromosomes, 251–66<br /> banding patterns of, 188, 189<br /> centromeres of, 252<br /> compaction<br /> histone proteins in, 264<br /> leading to the metaphase chromosome, 239<br /> 30 nm fiber in, 259–61<br /> scaffolding proteins in, 264<br /> condensation of, 262–64<br /> crossing over between, 456<br /> DNA polymerases in, 289–90, 289<br /> DNA replication in, 288–93<br /> telomerase in, 290–92<br /> DNA sequences in, 252–53, 253<br /> repetitive, 253–54<br /> folding of, 255<br /> karyotype preparation, 45–47, 46<br /> nucleosomes<br /> 30 nm fiber, 258–59, 259<br /> structure of, 255–58, 256<br /> origins of replication in, 252<br /> physical mapping of, 556–58, 557<br /> renaturation experiment in sequence complexity<br /> evaluation, 254–55<br /> sets of, 47<br /> sizes of, 252<br /> symbiotic infective particles in, 181<br /> telomeres of, 253<br /> translocation between, 197–99<br /> Eumelanin, 79<br /> Euploid organism, 200<br /> Euploidy<br /> alterations in<br /> by autopolyploidy, alloploidy, and<br /> allopolyploidy, 209<br /> in certain animal tissue, 205–6<br /> naturally occurring, 204<br /> in plants, 206–7<br /> European Conditional Mouse Mutagenesis<br /> Program (EUCOMM), 523<br /> Evolution<br /> biological, 10<br /> formation of gene families, 191–93<br /> fundamental principles of, 728<br /> gene transfer during, 181<br /> of globin gene family in humans, 192<br /> horizontal gene transfer in, 742–44<br /> of horse, 11<br /> macro-, 727<br /> micro-, 670, 670–71, 727<br /> of mitochondria and chloroplast genomes,<br /> 180–81<br /> molecular, 727, 734–49<br /> mutations in, 424<br /> natural selection in. See Natural selection<br /> neutral theory of, 736<br /> non-Darwinian, 736–37<br /> of paralogs, 192<br /> population genetics and, 13<br /> of primates, 739<br /> transposable elements influence on, 474–75<br /> vertical, 742<br /> web of life, 743, 744<br /> Evolutionary developmental biology, 749–52<br /> Evolutionary genetics, 727–53<br /> ancient DNA analysis in, 744–47<br /> molecular evolution, 734–49<br /> origin of species, 728–34<br /> phylogenetic trees, 734–49<br /> Evolutionary species concept, 729, 730<br /> Exit site (E site), 347<br /> Exons, 309<br /> alternative, 408<br /> alternative splicing in regulation of, 407–9<br /> constitutive, 408<br /> Exon shuffling, 475, 685, 685, 686<br /> Exon skipping, 409<br /> Exonucleases, 253, 309<br /> “Experiments on Plant Hybrids” (Mendel), 18<br /> Expressed sequence tag (EST) library, 572<br /> Expressivity, 75<br /> Extragenic suppressor, 92<br /> Extranuclear inheritance, 174–81<br /> in Chlamydomonas, 177–79, 179<br /> genetic material in mitochondria and<br /> chloroplasts, 174–76, 175<br /> human diseases caused by mitochondrial<br /> mutations, 179–80, 180<br /> non-Mendelian results in reciprocal crosses,<br /> 176–77<br /> pattern of inheritance of mitochondria and<br /> chloroplasts, 179<br /> in yeast, 177–79<br /> Ex vivo approach, 537<br /> eyeless gene, 752<br /> Eyes<br /> color<br /> binomial expansion equation in predicting,<br /> 32–33<br /> OCA2 gene in, 47<br /> Drosophila<br /> cream-eye gene in, 89–91<br /> eyeless gene in, 752<br /> position effect in eye color alteration, 431<br /> protein function in eye color, 337<br /> temperature in eye development, 76, 77<br /> white-eye trait in, 64–66<br /> evolution of, 727<br /> Pax6 gene in, 751–52<br /> F<br /> Facioscapulohumeral muscular dystrophy, 2<br /> Factor IX, 526, 604<br /> Factor VIII, 518, 604<br /> Factor XI deficiency, 2<br /> Facultative heterochromatin, 260–61<br /> Familial Down syndrome, 198, 198–99<br /> Familial fatal insomnia, 609<br /> Familial hypercholesterolemia, 603<br /> Fasman, Gerald, 594<br /> Fearon, Eric, 622<br /> Feedback inhibition, 377<br /> Fertility<br /> in allotetraploids, 209–12<br /> chromosome pairing and, 210–11<br /> Feulgen, 174<br /> F factor, 142<br /> in conjugation, 136<br /> in Hfr strains, 138–39<br /> transfer of, 137<br /> F factors, 366<br /> F' factors, 138, 366<br /> F 1 generation, 21<br /> F 2 generation, 21<br /> Ficedula albicollis, 678<br /> Fidelity, in DNA synthesis, 282<br /> Fine structure mapping, 148</div> <div>CREDITS<br /> ::<br /> Photographs<br /> Chapter 1<br /> Opener: © Photo courtesy of the College of Veterinary Medicine, Texas A&M<br /> University/CORBIS; 1.2(left): © Roslin Institute; 1.3a(right): Advanced Cell<br /> Technology, Inc., Worcester, Massachusetts; 1.3b: Photo taken by Flaminia<br /> Catteruccia, Jason Benton, and Andrea Crisanti, and assembled by www.<br /> luciariccidesign.com; 1.4: © Biophoto Associates/Photo Researchers; 1.5:<br /> © CNRI/SCIENCE PHOTO LIBRARY/Photo Researchers; 1.8(top left):<br /> © Edmund D. Brodie III; 1.9a(top right): © Joseph Sohm/ChromoSohm Inc./<br /> CORBIS; 1.9b: © Paul Edmondson/Photodisc Red/Getty Images; 1.10: © March<br /> of Dimes Birth Defects Foundation; 1.13a: © George Musil/Visuals Unlimited; 1.13b:<br /> © SciMAT/Photo Researchers, Inc.; 1.13c: © Steve Hopkin/ARDEA LONDON;<br /> 1.13d: © Brad Mogen/Visuals Unlimited; 1.13e: © Mark Smith/Photo Researchers,<br /> Inc.; 1.13f: © J-M. Labat/Photo Researchers, Inc.; 1.13g: © Wally Eberhart/Visuals<br /> Unlimited.<br /> Chapter 2<br /> Opener: © CHRIS MARTIN BAHR/SCIENCE PHOTO LIBRARY; 2.1: © SPL/<br /> Photo Researchers, Inc.; 2.2: © Nigel Cattlin/Photo Researchers, Inc.<br /> Chapter 3<br /> Opener: © Photomicrographs by Dr. Conly L. Rieder, Wadsworth Center, Albany,<br /> New York 12201-0509; 3.2b(right top): © Burger/Photo Researchers; 3.2c(right<br /> bottom): © Leonard Lessin/Peter Arnold; 3.6a(left): © Leonard Lessin/Peter<br /> Arnold; 3.6a(right): © Biophoto Associates/Photo Researchers, Inc.; 3.8a-f:<br /> © Photomicrographs by Dr. Conly L. Rieder, Wadsworth Center, Albany, New<br /> York 12201-0509; 3.9a(1): © Dr. David M. Phillips/Visuals Unlimited; 3.9a(2):<br /> © Carolina Biological Supply Company/Phototake.com; 3.11: © Diter von Wettstein.<br /> Chapter 4<br /> Opener: © Robert Calentine/Visuals Unlimited; 4.1: © Blickwinkel/Alamy; 4.5b:<br /> © Bob Shanley/The Palm Beach Post; 4.6a(1): © PETER WEIMANN/Animals<br /> Animals-Earth Scenes; 4.6a(2): © Tom Walker/Visuals Unlimited; 4.6b: © Sally<br /> Haugen/Virginia Schuett, www.pkunews.org; 4.6c: © PHOTOTAKE Inc./Alamy;<br /> 4.7a(left), 4.7b(right): © Stan Flegler/Visuals Unlimited; 4.9b: © John T. Fowler/<br /> Alamy; 4.9c: © Wegner, P./Peter Arnold, Inc.; 4.9d: © Gary Randall/Visuals<br /> Unlimited; 4.9a(top left): © Zig Leszcynski/AnimalsAnimals-Earth Scenes;<br /> 4.10(top right): © Alan & Sandy Carey/Photo Researchers; 4.13a: © AP Images;<br /> 4.15a(top left), 4.15b(top right), 4.15c(bottom left): National Parks Service,<br /> Adams National Historical Park; 4.15d(bottom right): © Bettmann/CORBIS ;<br /> 4.17a(left), 4.17b(right): © Robert Maier/AnimalsAnimals-Earth Scenes; 4.18a:<br /> © Jane Burton/naturepl.com.<br /> Chapter 5<br /> Opener: © Carolina Biological/Visuals Unlimited.<br /> Chapter 6<br /> Opener: © David Scharf/Peter Arnold; 6.4b: © Dr. L. Caro/Science Photo Library/<br /> Photo Researchers; 6.15a(left), 6.15b(right): © Carolina Biological Supply/<br /> Phototake.<br /> Chapter 7<br /> Opener: © John Mendenhall, Institute for Cellular and Molecular Biology,<br /> University of Texas at Austin; 7.3a(1), 7.3a(2): Courtesy of I. Solovei, University<br /> of Munich (LMU); 7.6: © G.W. Willis, MD/Visuals Unlimited; TA 7.1: Ronald<br /> G. Davidson, Harold M. Nitowsky, and Barton Childs. “Demonstration of<br /> Two Populations of Cells in the Human Female Heterozygous for Glucose-6-<br /> Phosphate Dehydrogenase Variants.” PNAS 50 (1963) f. 2, p. 484. Courtesy Harold<br /> M. Nitowsky; 7.9: Courtesy of Dr. Argiris Efstratiadis; 7.13a(left): Reproduced<br /> with permission from The Journal of Cell Biology, 1977, 72:687-694. Copyright<br /> 1977 The Rockefeller University Press; 7.13b(right): Gibbs, SP., Mak, R., Ng, R., &<br /> Slankis, T. “The chloroplast nucleoid in Ochromonas danica, II. Evidence for an<br /> increase in plastid DNA during greening.” J Cell Sci. 1974 Dec;16(3):579-91. Fig. 1.<br /> By permission of the Company of Biologists Limited.<br /> Chapter 8<br /> Opener: © BSIP/Phototake, Inc.; 8.1a(top left): © Scott Camazine/Photo<br /> Researchers; 8.1a(center): © Michael Abbey/Photo Researchers; 8.1a(top right):<br /> © Carlos R Carvalho/Universidade Federal de Viçosa; 8.1c(bottom left):<br /> © C.N.R.I./Phototake; 8.4a(left): © Biophoto Associates/Science Source/Photo<br /> Researchers; 8.4b(right): © Jeff Noneley; 8.12b(left): © Paul Benke/University<br /> of Miami School of Medicine; 8.12c(right): © Will Hart/PhotoEdit; 8.18a(top),<br /> 8.18b(bottom): © A. B. Sheldon; 8.19b: © David M. Phillips/Visuals Unlimited;<br /> 8.20a(left): © James Steinberg/Photo Researchers; 8.20a(top right), 8.20b(bottom<br /> right): © Biophoto Associates/Science Source/Photo Researchers; 8.26: Robinson,<br /> T.J. & Harley, E.H. “Absence of geographic chromosomal variation in the roan and<br /> sable antelope and the cytogenetics of a naturally occurring hybrid.” Cytogenet<br /> Cell Genet. 1995; 71(4):363-9. Permission granted by S. Karger AG, Basel.<br /> Chapter 9<br /> Opener: © Ken Eward/Photo Researchers; 9.4(left): © Omikron/Photo<br /> Researchers; 9.12b(left): © Pictorial Parade/Getty Images; 9.16a(left):<br /> © Barrington Brown/Photo Researchers; 9.16b(right): © Hulton Archive by Getty<br /> Images; 9.18b: © Michael Freeman/Phototake; 9.24b: © Alfred Pasieka/Photo<br /> Researchers, Inc.<br /> Chapter 10<br /> Opener: © Dr. Gopal Murti/Visuals Unlimited; 10.3: ©American Society for<br /> Microbiology; 10.10b(top): © Simpson’s Nature Photography; 10.10c(bottom):<br /> © William Leonard; 10.13c: Image courtesy of Timothy J. Richmond. Figure 1<br /> from” Crystal structure of the nucleosome core particle at 2.8 Å resolution.”<br /> Nature 1997 Sept 18; 389(6648): 251-60. Reprinted by permission from<br /> Macmillan Publishers Ltd.; TA 10.14: Reprinted by permission from Macmillan<br /> Publishers Ltd. Nature Subunit structure of chromatin. Markus Noll. 251: 5472,<br /> 249-251. 1974.; 10.17b, 10.17c: Nickerson et al. “The nuclear matrix revealed by<br /> eluting chromatin from a cross-linked nucleus.” PNAS 94: 4446-4450. Figure 2a<br /> &b. © 1997 National Academy of Sciences, U.S.A.; 10.18a(left), 10.18b(right):<br /> Reprinted by permission from Macmillan Publishers Ltd. Nature Reviews/<br /> Genetics. Chromosome territories, nuclear architecture and gene regulation in<br /> mammalian cells. Cremer, T. & Cremer, C., 2: 4, 292-301, 2001; 10.21a(2nm):<br /> © Dr. Gopal Murit/Visuals Unlimited; 10.21a(11nm): © Olins and Olins/<br /> Biological Photo Service; 10.21b(30nm): Jerome Rattner/University of Calgary;<br /> 10.21c(300nm): This article was published in Cell. Nov. 12(3). Paulson, JR. &<br /> Laemmli, UK. “The structure of histone-depleted metaphase chromosomes.”<br /> 817-2 8. f. 5. Copyright Elsevier, 1977; 10.21d(700nm), 10.21d(1400nm), 10.22a:<br /> © Peter Engelhardt/Department of Virology, Haartman Institute.; 10.22b:<br /> © Dr. Donald Fawcett/Visuals Unlimited.<br /> Chapter 11<br /> Opener: © Clive Freeman, The Royal Institution/Photo Researchers; TA 11.1:<br /> Meselson M, Stahl, F. “The Replication of DNA in Escherichia Coli.” PNAS Vol.<br /> 44, 1958. f. 4a, p. 673. Courtesy of M. Meselson; 11.4b: From Cold Spring Harbor<br /> Symposia of Quantitative Biology, 28, p. 43 (1963). Copyright holder is Cold<br /> Spring Habour Laboratory Press; 11.8b: Reprinted by permission from Macmillan<br /> Publishers Ltd. The Embo Journal. “Crystal structures of open and closed forms<br /> of binary and ternary complexes of the large fragment of Thermus aquaticus<br /> DNA polymerase I: structural basis for nucleotide incorporation.” Ying Li et al.<br /> 17: 24, 7514-7525, 1998.; 11.20: This article was published in Journal of Molecular<br /> Biology. Mar 14; 32(2). Huberman JA. Riggs AD. “Links On the mechanism of<br /> DNA replication in mammalian chromosomes”:327-41. Copyright Elsevier,<br /> 1968; 11.21b: Henry J. Kriegstein and David S. Hogness “Mechanism of DNA<br /> Replication in Drosophila Chromosomes: Structure of Replication Forks and<br /> Evidence for Bidirectionality.” PNAS. January 1, 1974. vol. 71, no. 1. 135-139.<br /> Permission courtesy D. Hogness.<br /> Chapter 12<br /> Opener: From Patrick Cramer, David A. Bushnell, Roger D. Kornberg. “Structural<br /> Basis of Transcription: RNA Polymerase II at 2.8 Angstrom Resolution.” Science,<br /> 823</div> <div>QUESTIONS FOR STUDENT DISCUSSION/COLLABORATION 513<br /> E38. Let’s suppose you want to use site-directed mutagenesis to<br /> investigate a DNA sequence that functions as a response element<br /> for hormone binding. From previous work, you have narrowed<br /> down the response element to a sequence of DNA that is 20 bp in<br /> length with the following sequence:<br /> 5'–GGACTGACTTATCCATCGGT–3'<br /> 3'–CCTGACTGAATAGGTAGCCA–5'<br /> As a strategy to pinpoint the actual response element sequence,<br /> you decide to make 10 different site-directed mutants and then<br /> analyze their effects by a gel retardation assay. What mutations<br /> would you make? What results would you expect to obtain?<br /> E39. Site-directed mutagenesis can also be used to explore the structure<br /> and function of proteins. For example, changes can be made to<br /> the coding sequence of a gene to determine how alterations in<br /> the amino acid sequence affect the function of a protein. Let’s<br /> suppose that you are interested in the functional importance of<br /> one glutamic acid residue within a protein you are studying. By<br /> site-directed mutagenesis, you make mutant proteins in which this<br /> glutamic acid codon has been changed to other codons. You then<br /> test the encoded mutant proteins for functionality. The results are<br /> as follows:<br /> Functionality<br /> Normal protein 100%<br /> Mutant proteins containing<br /> Tyrosine 5%<br /> Phenylalanine 3%<br /> Aspartic acid 94%<br /> Glycine 4%<br /> From these results, what would you conclude about the functional<br /> significance of the glutamic acid residue within the protein?<br /> Questions for Student Discussion/Collaboration<br /> 1. Discuss and make a list of some of the reasons why it would be<br /> informative for a geneticist to determine the amount of a gene<br /> product. Use specific examples of known genes (e.g., β-globin and<br /> other genes) when making your list.<br /> 2. Make a list of all the possible genetic questions that could be<br /> answered using site-directed mutagenesis.<br /> Note: All answers appear at the website for this textbook; the answers to<br /> even-numbered questions are in the back of the textbook.<br /> www.mhhe.com/brookergenetics3e<br /> Visit the Online Learning Center for practice tests, answer keys, and other learning aids for this chapter. Enhance your understanding of genetics with<br /> our interactive exercises, quizzes, animations, and much more.</div> <div>206 CHAPTER 8 :: VARIATION IN CHROMOSOME STRUCTURE AND NUMBER<br /> can see them during interphase, when normal chromosomes are<br /> not readily visible. Remarkably, a polytene chromosome during<br /> interphase is 100 to 200 times larger than the average metaphase<br /> chromosome. As shown in Figure 8.19b, polytene chromosomes<br /> exhibit a characteristic banding pattern. Each dark band is known<br /> as a chromomere. The structure of the genetic material within a<br /> dark band is more compact than the interband region. More than<br /> 95% of the DNA is found within these dark bands. The banding<br /> patterns of polytene chromosomes are much more detailed<br /> than those observed in metaphase chromosomes. Cytogeneticists<br /> have identified approximately 5,000 bands along polytene chromosomes.<br /> At one time, each chromomere was thought to correspond<br /> to one gene. However, this idea was found to be incorrect<br /> because the sequencing of the entire Drosophila genome has<br /> revealed an approximate gene number of 14,000.<br /> Polytene chromosomes have allowed geneticists to study<br /> the organization and functioning of interphase chromosomes in<br /> great detail. When a gene is deleted or duplicated via mutation,<br /> researchers can map the change if it results in a microscopically<br /> visible alteration in the structure of a polytene chromosome. In<br /> addition, because polytene chromosomes can be observed during<br /> interphase, the expression of particular genes in salivary cells can<br /> be correlated with changes in the compaction of certain bands in<br /> the polytene chromosome.<br /> Variations in Euploidy Are Common in Plants<br /> We now turn our attention to variations of euploidy that occur<br /> in plants. Compared to animals, plants more commonly exhibit<br /> polyploidy. Among ferns and flowering plants, at least 30 to 35%<br /> of species are polyploid. Polyploidy is also important in agriculture.<br /> Many of the fruits and grains we eat are produced from<br /> polyploid plants. For example, the species of wheat that we use<br /> to make bread, Triticum aestivum, is a hexaploid (6n) that arose<br /> from the union of diploid genomes from three closely related<br /> species (Figure 8.20a).<br /> In many instances, polyploid strains of plants display outstanding<br /> agricultural characteristics. They are often larger in size<br /> and more robust. These traits are clearly advantageous in the<br /> production of food. In addition, polyploid plants tend to exhibit<br /> a greater adaptability, which allows them to withstand harsher<br /> environmental conditions. Also, polyploid ornamental plants<br /> often produce larger flowers than their diploid counterparts (Figure<br /> 8.20b).<br /> Polyploid plants having an odd number of chromosome sets,<br /> such as triploids (3n) or pentaploids (5n), usually cannot reproduce.<br /> Why are they sterile? The sterility arises because they produce<br /> highly aneuploid gametes. During prophase of meiosis I, homologous<br /> pairs of sister chromatids will form bivalents. However,<br /> Tetraploid<br /> (a) Cultivated wheat, a hexaploid species<br /> Diploid<br /> (b) A comparison of diploid and tetraploid petunias<br /> FIGURE 8.20 Examples of polyploid plants. (a) Cultivated wheat, Triticum aestivum, is a hexaploid. It was derived from three different<br /> diploid species of grasses that originally were found in the Middle East and were cultivated by ancient farmers in that region. (b) Differences in<br /> euploidy may exist in two closely related petunia species. The flower at the bottom is diploid, whereas the large one at the top is tetraploid.<br /> Genes g Traits An increase in chromosome number from diploid to tetraploid or hexaploid affects the phenotype of the individual. In the case of many plant species, a polyploid<br /> individual is larger and more robust than its diploid counterpart. This suggests that having additional copies of each gene is somewhat better than having two copies of each gene.<br /> This phenomenon in plants is rather different from the situation in animals. Tetraploidy in animals may have little effect (as in Figure 8.18b), and it is also common for polyploidy<br /> in animals to be detrimental.</div> <div>ABOUT THE AUTHOR<br /> Robert J. Brooker is a Professor in the Department of<br /> Genetics, Cell Biology, and Development at the University of<br /> Minnesota–Minneapolis. He received his B.A. in Biology from<br /> Wittenberg University in 1978 and his Ph.D. in Genetics from<br /> Yale University in 1983. At Harvard, he conducted post-doctoral<br /> studies on the lactose permease, which is the product of the lacY<br /> gene of the lac operon. He continues his work on transporters at<br /> the University of Minnesota. Dr. Brooker’s laboratory primarily<br /> investigates the structure, function, and regulation of iron<br /> transporters found in bacteria and C. elegans. At the University of<br /> Minnesota, he teaches undergraduate courses in biology, genetics,<br /> and cell biology.<br /> DEDICATION<br /> To my wife, Deborah, and our children,<br /> Daniel, Nathan, and Sarah<br /> iii</div> <div>23.3 PLANT DEVELOPMENT 651<br /> and a genome size of 14 � 10 7 bp, which is similar to Drosophila<br /> and C. elegans. A flowering Arabidopsis plant produces a large<br /> number of seeds and is small enough to be grown in the laboratory.<br /> Like Drosophila, Arabidopsis can be subjected to mutagens<br /> to generate mutations that alter developmental processes. The<br /> small genome size of this organism makes it relatively easy to<br /> map these mutant alleles and eventually clone the relevant genes<br /> (as described in Chapters 18 and 20).<br /> The morphological patterns of growth are markedly different<br /> between animals and plants. As described previously, animal<br /> embryos become organized along antero-posterior, dorsoventral,<br /> and right-left axes, and then they subdivide into segments.<br /> By comparison, the form of plants has two key features.<br /> The first is the root-shoot axis. Most plant growth occurs via cell<br /> division near the tips of the shoots and the bottoms of the roots.<br /> Second, this growth occurs in a well-defined radial pattern.<br /> For example, early in Arabidopsis growth, a rosette of leaves or<br /> flowers is produced from buds that emanate in a spiral pattern<br /> directly from the main shoot (see Figure 23.20). Later, the shoot<br /> generates branches that will also produce leaf buds as they grow.<br /> Overall, the radial pattern in which a plant shoot gives off the<br /> buds that produce leaves, flowers, and branches is an important<br /> mechanism that determines much of the general morphology of<br /> the plant.<br /> At the cellular level, too, plant development differs markedly<br /> from animal development. For example, cell migration does<br /> not occur during plant development. In addition, the development<br /> of a plant does not rely on morphogens that are deposited<br /> asymmetrically in the oocyte. In plants, an entirely new individual<br /> can be regenerated from many types of somatic cells. In other<br /> words, many plant cells are totipotent, meaning that they have<br /> the ability to differentiate into every cell type and to produce an<br /> entire individual. By comparison, animal development typically<br /> relies on the organization within an oocyte as a starting point for<br /> development.<br /> In spite of these apparent differences, the underlying<br /> molecular mechanisms of pattern development in plants still<br /> share some similarities with those in animals. In this section, we<br /> will consider a few examples in which the genes encoding transcription<br /> factors play a key role in plant development.<br /> Plant Growth Occurs from Meristems Formed<br /> During Embryonic Development<br /> Figure 23.21 illustrates a common sequence of events that takes<br /> place in the development of seed plants such as Arabidopsis.<br /> After fertilization, the first cellular division is asymmetrical and<br /> produces a smaller cell, called the apical cell, and a larger basal<br /> cell (Figure 23.21a). The apical cell will give rise to most of the<br /> embryo, and it will later develop into the shoot of the plant. The<br /> basal cell will give rise to the root, along with the suspensor that<br /> produces extraembryonic tissue required for seed formation.<br /> At the heart stage, which is composed of only about 100 cells,<br /> the basic organization of the plant has been established (Figure<br /> 23.21d). The shoot meristem will arise from a group of cells<br /> located between the cotyledons. These cells are the precursors<br /> that will produce the shoot of the plant, along with lateral structures<br /> such as leaves and flowers. The root meristem is located at<br /> the opposite side and will create the root.<br /> A meristem contains an organized group of actively dividing<br /> stem cells. As discussed in Chapter 19, stem cells retain the<br /> ability to divide and differentiate into multiple cell types. As they<br /> grow, meristems produce offshoots of proliferating cells. On a<br /> shoot meristem, for example, these offshoots or buds give rise to<br /> structures such as leaves and flowers. The organization of a shoot<br /> meristem is shown in Figure 23.22. It is organized into three areas<br /> called the organizing center, the central zone, and the peripheral<br /> zone. The role of the organizing center is to ensure the proper<br /> organization of the meristem and preserve the correct number<br /> Cotyledon<br /> primordia<br /> Shoot<br /> meristem<br /> Cotyledon<br /> Shoot<br /> meristem<br /> Apical<br /> region<br /> Apical<br /> cell<br /> Pro-embryo<br /> Suspensor<br /> Apical region<br /> Central region<br /> Basal region<br /> Suspensor<br /> Epidermis (E)<br /> Ground<br /> tissue (G)<br /> Basal<br /> Vascular<br /> cell<br /> primordia (V)<br /> Root<br /> meristem<br /> (a) (b) (c) (d) (e)<br /> (E)<br /> (G)<br /> (V)<br /> Root<br /> meristem<br /> Central<br /> region<br /> Basal<br /> region<br /> FIGURE 23.21 Developmental steps in the formation of a plant embryo. (a) The two-cell stage consists of the apical cell and basal cell.<br /> (b) The eight-cell stage consists of a pro-embryo and a suspensor. The suspensor gives rise to extraembryonic tissue, which is needed for seed<br /> formation. (c) At this stage of embryonic development, the three main regions of the embryo (i.e., apical, central, and basal) have been determined.<br /> (d) At the heart stage, all of the plant tissues have begun to form. Note that the shoot meristem is located between the future cotyledons, and the root<br /> meristem is on the opposite side. (e) A seedling.</div> <div>EXPERIMENTAL QUESTIONS 69<br /> C17. Nine-banded armadillos almost always give birth to four offspring<br /> that are genetically identical quadruplets. Explain how you think<br /> this happens.<br /> C18. A diploid species contains four chromosomes per set for a total<br /> of eight chromosomes in its somatic cells. Draw the cell as it<br /> would look in late prophase of meiosis II and prophase of mitosis.<br /> Discuss how prophase of meiosis II and prophase of mitosis differ<br /> from each other, and explain how the difference originates.<br /> C19. Explain why the products of meiosis may not be genetically<br /> identical while the products of mitosis are.<br /> C20. The period between meiosis I and meiosis II is called<br /> interphase II. Does DNA replication take place during<br /> interphase II? Explain your answer.<br /> C21. List several ways in which telophase appears to be the reverse of<br /> prophase and prometaphase.<br /> C22. In corn, there are 10 chromosomes per set and the sporophyte of<br /> the species is diploid. If you performed a karyotype, what is the<br /> total number of chromosomes that you would expect to see in the<br /> following types of cells?<br /> A. A leaf cell<br /> B. The sperm nucleus of a pollen grain<br /> C. An endosperm cell after fertilization<br /> D. A root cell<br /> C23. The arctic fox has 50 chromosomes (25 per set), and the common<br /> red fox has 38 chromosomes (19 per set). These species can<br /> interbreed to produce viable but infertile offspring. How many<br /> chromosomes would the offspring have? What problems do you<br /> think may occur during meiosis that would explain the offspring’s<br /> infertility?<br /> C24. Let’s suppose that a gene affecting pigmentation is found on the<br /> X chromosome (in mammals or insects) or the Z chromosome<br /> (in birds) but not on the Y or W chromosome. It is found on<br /> an autosome in bees. This gene is found in two alleles, D (dark),<br /> which is dominant to d (light). What would be the phenotypic<br /> results of crosses between a true-breeding dark female and<br /> true-breeding light male, and the reciprocal crosses involving a<br /> true-breeding light female and true-breeding dark male, in the<br /> following species? Refer back to Figure 3.18 for the mechanism of<br /> sex determination in these species.<br /> A. Birds<br /> B. Drosophila<br /> C. Bees<br /> D. Humans<br /> C25. Describe the cellular differences between male and female gametes.<br /> C26. At puberty, the testes contain a finite number of cells and produce<br /> an enormous number of sperm cells during the life span of a male.<br /> Explain why testes do not run out of spermatogonial cells.<br /> C27. Describe the timing of meiosis I and II during human oogenesis.<br /> C28. Three genes (A, B, and C) are found on three different<br /> chromosomes. For the following diploid genotypes, describe all the<br /> possible gamete combinations.<br /> A. Aa Bb Cc<br /> B. AA Bb CC<br /> C. Aa BB Cc<br /> D. Aa bb cc<br /> C29. A phenotypically normal woman with an abnormally long<br /> chromosome 13 (and a normal homologue of chromosome 13)<br /> marries a phenotypically normal man with an abnormally short<br /> chromosome 11 (and a normal homologue of chromosome 11).<br /> What is the probability of producing an offspring that will have both<br /> a long chromosome 13 and a short chromosome 11? If such a child<br /> is produced, what is the probability that this child would eventually<br /> pass both abnormal chromosomes to one of his/her offspring?<br /> C30. Assuming that such a fly would be viable, what would be the sex of<br /> a fruit fly with the following chromosomal composition?<br /> A. One X chromosome and two sets of autosomes<br /> B. Two X chromosomes, one Y chromosome, and two sets of<br /> autosomes<br /> C. Two X chromosomes and four sets of autosomes<br /> D. Four X chromosomes, two Y chromosomes, and four sets of<br /> autosomes<br /> C31. What would be the sex of a human with the following numbers of<br /> sex chromosomes?<br /> A. XXX<br /> B. X (also described as X0)<br /> C. XYY<br /> D. XXY<br /> Experimental Questions<br /> E1. When studying living cells in a laboratory, researchers sometimes<br /> use drugs as a way to make cells remain at a particular stage of<br /> the cell cycle. For example, aphidicolin inhibits DNA synthesis<br /> in eukaryotic cells and causes them to remain in the G 1 phase<br /> because they cannot replicate their DNA. In what phase of the cell<br /> cycle—G 1 , S, G 2 , prophase, metaphase, anaphase, or telophase—<br /> would you expect somatic cells to stay if the following types of<br /> drug were added?<br /> A. A drug that inhibits microtubule formation<br /> B. A drug that allows microtubules to form but prevents them<br /> from shortening<br /> C. A drug that inhibits cytokinesis<br /> D. A drug that prevents chromosomal condensation<br /> E2. In Morgan’s experiments, which result do you think is the most<br /> convincing piece of evidence pointing to X-linkage of the eye color<br /> gene? Explain your answer.<br /> E3. In his original studies of Figure 3.19, Morgan first suggested that the<br /> original white-eyed male had two copies of the white-eye allele. In<br /> this problem, let’s assume that he meant the fly was X w Y w instead<br /> of X w Y. Are his data in Figure 3.19 consistent with this hypothesis?<br /> What crosses would need to be made to rule out the possibility that<br /> the Y chromosome carries a copy of the eye color gene?</div> <div>386 CHAPTER 14 :: GENE REGULATION IN BACTERIA AND BACTERIOPHAGES<br /> C5. An operon is repressible—a small effector molecule turns off<br /> transcription. Which combinations of small effector molecules and<br /> regulatory proteins could be involved?<br /> A. An inducer plus a repressor<br /> B. A corepressor plus a repressor<br /> C. An inhibitor plus an activator<br /> D. An inducer plus an activator<br /> C6. Some mutations have a cis-effect, whereas others have a transeffect.<br /> Explain the molecular differences between cis- and<br /> trans-mutations. Which type of mutation (cis or trans) can be<br /> complemented in a merozygote experiment?<br /> C7. What is enzyme adaptation? From a genetic point of view, how<br /> does it occur?<br /> C8. In the lac operon, how would gene expression be affected if one of<br /> the following segments were missing?<br /> A. lac operon promoter<br /> B. Operator site<br /> C. lacA gene<br /> C9. If an abnormal repressor protein could still bind allolactose but<br /> the binding of allolactose did not alter the conformation of the<br /> repressor protein, how would this affect the expression of the lac<br /> operon?<br /> C10. What is diauxic growth? Explain the roles of cAMP and the<br /> catabolite activator protein in this process.<br /> C11. Mutations may have an effect on the expression of the lac operon,<br /> the ara operon, and the trp operon. Would the following mutations<br /> have a cis- or trans-effect on the expression of the structural genes<br /> in the operon?<br /> A. A mutation in the operator site that prevents the lac repressor<br /> from binding to it<br /> B. A mutation in the lacI gene that prevents the lac repressor from<br /> binding to DNA<br /> C. A mutation in the araC gene that prevents two AraC proteins<br /> from binding to each other and forming a loop<br /> D. A mutation in trpL that prevents attenuation<br /> C12. Would a mutation that inactivated the lac repressor and prevented<br /> it from binding to the lac operator site result in the constitutive<br /> expression of the lac operon under all conditions? Explain. What<br /> is the disadvantage to the bacterium of having a constitutive lac<br /> operon?<br /> C13. Describe the function of the AraC protein. How does it positively<br /> and negatively regulate the ara operon?<br /> C14. Explain how a mutation would affect the regulation of the ara<br /> operon if the mutation prevented AraC protein from binding to<br /> the following sites:<br /> A. araO 2<br /> B. araO 1<br /> C. araI<br /> D. araO 2 and araI<br /> C15. What is meant by the term attenuation? Is it an example of gene<br /> regulation at the level of transcription or translation? Explain your<br /> answer.<br /> C16. As described in Figure 14.14, four regions within the trpL gene can<br /> form stem-loop structures. Let’s suppose that mutations have been<br /> previously identified that prevent the ability of a particular region<br /> to form a stem-loop structure with a complementary region. For<br /> example, a region 1 mutant cannot form a 1–2 stem-loop structure,<br /> but it can still form a 2–3 or 3–4 structure. Likewise, a region 4<br /> mutant can form a 1–2 or 2–3 stem-loop but not a 3–4 stem-loop.<br /> Under the following conditions, would attenuation occur?<br /> A. Region 1 is mutant, tryptophan is high, and translation is not<br /> occurring.<br /> B. Region 2 is mutant, tryptophan is low, and translation is<br /> occurring.<br /> C. Region 3 is mutant, tryptophan is high, and translation is not<br /> occurring.<br /> D. Region 4 is mutant, tryptophan is low, and translation is not<br /> occurring.<br /> C17. As described in Chapter 13, enzymes known as aminoacyl-tRNA<br /> synthetases are responsible for attaching amino acids to tRNAs.<br /> Let’s suppose that tryptophanyl-tRNA synthetase was partially<br /> defective at attaching tryptophan to tRNA; its activity was only<br /> 10% of that found in a normal bacterium. How would that affect<br /> attenuation of the trp operon? Would it be more or less likely to be<br /> attenuated? Explain your answer.<br /> C18. The 3–4 stem-loop and U-rich attenuator found in the trp operon<br /> (see Figure 14.14) is an example of ρ-independent termination.<br /> The function of ρ-independent terminators is described in Chapter<br /> 12. Would you expect attenuation to occur if the tryptophan levels<br /> were high and mutations at the end of the trpL gene changed the<br /> UUUUUUUU sequence to UGGUUGUC? Explain why or why not.<br /> C19. Mutations in tRNA genes can create tRNAs that recognize stop<br /> codons. Because stop codons are sometimes called nonsense<br /> codons, these types of mutations that affect tRNAs are called<br /> nonsense suppressors. For example, a normal tRNA gly has an<br /> anticodon sequence CCU that recognizes a glycine codon in<br /> mRNA (GGA) and puts in a glycine during translation. However,<br /> a mutation in the gene that encodes tRNA gly could change the<br /> anticodon to ACU. This mutant tRNA gly would still carry glycine,<br /> but it would recognize the stop codon UGA. Would this mutation<br /> affect attenuation of the trp operon? Explain why or why not.<br /> Note: To answer this question, you need to look carefully at Figure<br /> 14.14 and see if you can identify any stop codons that may exist<br /> beyond the UGA stop codon that is found after region 1.<br /> C20. Translational control is usually aimed at preventing the initiation<br /> of translation. With regard to cellular efficiency, why do you think<br /> this is the case?<br /> C21. What is antisense RNA? How does it affect the translation of a<br /> complementary mRNA?<br /> C22. A species of bacteria can synthesize the amino acid histidine so<br /> that it does not require histidine in its growth medium. A key<br /> enzyme, which we will call histidine synthetase, is necessary for<br /> histidine biosynthesis. When these bacteria are given histidine in<br /> their growth media, they stop synthesizing histidine intracellularly.<br /> Based on this observation alone, propose three different<br /> regulatory mechanisms to explain why histidine biosynthesis<br /> ceases when histidine is in the growth medium. To explore this<br /> phenomenon further, you measure the amount of intracellular<br /> histidine synthetase protein when cells are grown in the presence</div> <div>84 CHAPTER 4 :: EXTENSIONS OF MENDELIAN INHERITANCE<br /> I-1 I-2<br /> II-1<br /> II-2<br /> II-3<br /> II-4 II-5 II-6 II-7 II-8<br /> III-1 III-2 III-3 III-4 III-5 III-6 III-7 III-8 III-9 III-10<br /> IV-1 IV-2 IV-3 IV-4 IV-5 IV-6 IV-7 IV-8 IV-9 IV-10 IV-11 IV-12 IV-13 IV-14<br /> (a) A pedigree for human pattern baldness<br /> Bb<br /> �<br /> Bb<br /> B<br /> Sperm<br /> b<br /> B<br /> BB<br /> Bald male<br /> Bald female<br /> Bb<br /> Bald male<br /> Nonbald female<br /> Egg<br /> b<br /> Bb<br /> Bald male<br /> Nonbald female<br /> bb<br /> Nonbald male<br /> Nonbald female<br /> (b) Example of an inheritance pattern involving baldness<br /> FIGURE 4.16 Inheritance of pattern<br /> baldness, a sex-influenced trait involving an<br /> autosomal gene. (a) A family pedigree. Bald individuals<br /> are shown in black. (b) The predicted offspring from two heterozygous<br /> parents.<br /> genotypic ratios from this cross would be 1 BB bald son : 1 BB<br /> bald daughter : 2 Bb bald sons : 2 Bb nonbald daughters : 1 bb<br /> nonbald son : 1 bb nonbald daughter. The predicted phenotypic<br /> ratios would be 3 bald sons : 1 bald daughter : 3 nonbald daughters<br /> : 1 nonbald son. The ratio of bald to nonbald offspring is<br /> 4:4, which is the same as 1:1.<br /> Another example in which sex affects an organism’s phenotype<br /> is provided by sex-limited inheritance, in which the trait<br /> occurs in only one of the two sexes. In humans, for example,<br /> breast development is a trait that is normally limited to females,<br /> whereas beard growth is limited to males. Among many types<br /> of birds, the male of the species has more ornate plumage than<br /> the female. As shown in Figure 4.17, roosters have a larger comb<br /> and wattles and longer neck and tail feathers than do hens. These<br /> are sex-limited features that may be found in roosters but never<br /> in normal hens. However, some varieties of chickens have males<br /> that are hen-feathered. In these species, hen-feathering is controlled<br /> by a dominant allele (H) that is expressed both in males<br /> and females, while cock-feathering is controlled by a recessive<br /> allele (h) that is expressed only in males. These alleles are located<br /> on an autosome; they are not on a sex chromosome.<br /> Genotype<br /> Phenotype<br /> Females<br /> Males<br /> hh Hen-feathered Cock-feathered<br /> Hh Hen-feathered Hen-feathered<br /> HH Hen-feathered Hen-feathered<br /> Like baldness in humans, hen- feathering depends on the<br /> production of sex hormones. If a newly hatched female with an<br /> hh genotype has her single ovary removed surgically, she will<br /> develop cock-feathering and look indistinguishable from a male.<br /> Mutations That Cause a Loss of Function<br /> in an Essential Gene Result in a Lethal Phenotype<br /> Let’s now turn our attention to alleles that have the most detrimental<br /> effect on phenotype—those that result in death. An allele<br /> that has the potential to cause the death of an organism is called<br /> a lethal allele. These are usually inherited in a recessive manner.<br /> When the absence of a specific protein results in a lethal phenotype,<br /> the gene that encodes the protein is considered an essential</div> <div>17.1 HOMOLOGOUS RECOMBINATION 461<br /> observations. In 1975, a model proposed by Matthew Meselson and<br /> Charles Radding hypothesized that a single nick in one DNA strand<br /> initiates recombination. A second model, proposed by Jack Szostak,<br /> Terry Orr-Weaver, Rodney Rothstein, and Franklin Stahl, suggests<br /> that a double-strand break initiates the recombination process.<br /> This is called the double-strand break model. Though recombination<br /> may occur via more than one mechanism, recent evidence<br /> suggests that double-strand breaks commonly promote homologous<br /> recombination during meiosis and during DNA repair.<br /> Figure 17.5 shows the general steps in the double-strand<br /> break model. As seen here, the top chromosome has experienced<br /> A<br /> 5′<br /> 3′<br /> 3′<br /> 5′<br /> a<br /> A<br /> a<br /> A<br /> Double-strand<br /> break<br /> Z<br /> 3′<br /> 3′<br /> z<br /> Strand degradation occurs at the<br /> double-strand break site to<br /> yield single-stranded ends.<br /> Strand invasion causes<br /> D-loop formation.<br /> D-loop<br /> Z<br /> z<br /> Z<br /> 5′<br /> 5′<br /> Homologous<br /> chromosomes<br /> a double-strand break. A small region near the break is degraded,<br /> which generates a single-stranded DNA segment that can invade<br /> the intact double helix. The strand displaced by the invading<br /> segment forms a structure called a displacement loop (D-loop).<br /> After the D-loop is formed, two regions have a gap in the DNA.<br /> How is the problem fixed? DNA synthesis occurs in the relatively<br /> short gaps where a DNA strand is missing. This DNA synthesis<br /> is called DNA gap repair synthesis. Once this is completed, two<br /> Holliday junctions are produced. Depending on the way these are<br /> resolved, the end result is nonrecombinant or recombinant chromosomes<br /> containing a short heteroduplex. In eukaryotes such<br /> as yeast, recent evidence suggests that certain proteins bound to<br /> Holliday junctions may regulate the resolution step in a way that<br /> favors the formation of recombinant chromosomes rather than<br /> nonrecombinant chromosomes.<br /> Various Proteins Are Necessary to Facilitate<br /> Homologous Recombination<br /> The homologous recombination process requires the participation<br /> of many proteins that catalyze different steps in the recombination<br /> pathway. Homologous recombination is found in all<br /> species, and the types of proteins that participate in the steps<br /> outlined in Figure 17.5 are very similar. The cells of any given<br /> species may have more than one molecular mechanism to carry<br /> out homologous recombination. This process is best understood<br /> in Escherichia coli. Table 17.1 summarizes some of the E. coli proteins<br /> that play critical roles in one recombination pathway found<br /> in this species. Though it is beyond the scope of this textbook, E.<br /> TABLE 17.1<br /> coli has other pathways to carry out homologous recombination.<br /> A<br /> a<br /> Nonrecombinant<br /> chromosomes<br /> a<br /> A<br /> a<br /> Gap repair synthesis fills in<br /> the vacant regions.<br /> Z<br /> z<br /> or<br /> z<br /> Branch migration and resolution<br /> can produce recombinant or<br /> nonrecombinant chromosomes.<br /> A<br /> a<br /> Z<br /> z<br /> Recombinant<br /> chromosomes<br /> Crossover site<br /> FIGURE 17.5 A simplified version of the double-strand break<br /> model. For simplicity, this illustration does not include the formation<br /> of heteroduplexes. The dashed arrow indicates that the pathway to the<br /> left may be less favored.<br /> z<br /> Z<br /> TABLE 16.6<br /> TABLE 17.1<br /> E. coli Proteins That Play a Role in Homologous Recombination<br /> Protein<br /> RecBCD<br /> Single-stranded binding<br /> protein<br /> RecA<br /> RuvABC<br /> RecG<br /> Description<br /> A complex of three proteins that tracks along the<br /> DNA and recognizes double-strand breaks. The<br /> complex partially degrades the double-stranded<br /> regions to generate single-stranded regions that<br /> can participate in strand invasion. RecBCD is also<br /> involved in loading RecA onto single-stranded DNA.<br /> In addition, RecBCD can create single-strand breaks<br /> at chi sequences.<br /> Coats broken ends of chromosomes and prevents<br /> excessive strand degradation.<br /> Binds to single-stranded DNA and promotes strand<br /> invasion, which enables homologous strands to find<br /> each other. It also promotes the displacement of the<br /> complementary strand to generate a D-loop.<br /> This protein complex binds to Holliday junctions.<br /> RuvAB promotes branch migration. RuvC is an<br /> endonuclease that cuts the crossed or uncrossed<br /> strands to resolve Holliday junctions into separate<br /> chromosomes.<br /> RecG protein can also promote branch migration of<br /> Holliday junctions.</div> <div>102 CHAPTER 5 :: LINKAGE AND GENETIC MAPPING IN EUKARYOTES<br /> P generation<br /> P generation<br /> x<br /> x<br /> Purple flowers,<br /> long pollen (PPLL)<br /> Red flowers,<br /> round pollen (ppll )<br /> X ywm X ywm<br /> X y + w + m + Y<br /> F 1 offspring<br /> F 2 offspring<br /> Purple flowers,<br /> long pollen (PpLl )<br /> Self-fertilization<br /> Observed<br /> Expected<br /> number Ratio number Ratio<br /> F 1 generation<br /> X y + w + m + X ywm<br /> x<br /> X ywm Y<br /> F 1 generation<br /> contains wild-type<br /> females and<br /> yellow-bodied,<br /> white-eyed,<br /> miniature-winged<br /> males.<br /> Purple flowers, long pollen<br /> Purple flowers, round pollen<br /> Red flowers, long pollen<br /> Red flowers, round pollen<br /> 296<br /> 19<br /> 27<br /> 85<br /> 15.6<br /> 1.0<br /> 1.4<br /> 4.5<br /> 240<br /> 80<br /> 80<br /> 27<br /> FIGURE 5.2 An experiment of Bateson and Punnett with<br /> sweet peas, showing that independent assortment does not always<br /> occur. Note: The expected numbers are rounded to the nearest whole<br /> number.<br /> Genes gTraits Two genes that govern flower color and pollen shape are found on<br /> the same chromosome. Therefore, the offspring tend to inherit the parental<br /> combinations of alleles (PL or pl). Due to occasional crossing over, a lower<br /> percentage of offspring inherit nonparental combinations of alleles (Pl or pL).<br /> a strain with red flowers (pp) and round pollen (ll). This yielded<br /> an F 1 generation of plants that all had purple flowers and long<br /> pollen (PpLl). An unexpected result came from the F 2 generation.<br /> Even though the F 2 generation had four different phenotypic<br /> categories, the observed numbers of offspring did not<br /> conform to a 9:3:3:1 ratio. Bateson and Punnett found that the<br /> F 2 generation had a much greater proportion of the two phenotypes<br /> found in the parental generation—purple flowers with<br /> long pollen and red flowers with round pollen. Therefore, they<br /> suggested that the transmission of these two traits from the<br /> parental generation to the F 2 generation was somehow coupled<br /> and not easily assorted in an independent manner. However,<br /> Bateson and Punnett did not realize that this coupling was due<br /> to the linkage of the flower color gene and the pollen shape gene<br /> on the same chromosome.<br /> 9<br /> 3<br /> 3<br /> 1<br /> F 2 generation<br /> Gray body, red eyes, long wings<br /> Gray body, red eyes, miniature wings<br /> Gray body, white eyes, long wings<br /> Gray body, white eyes, miniature wings<br /> Yellow body, red eyes, long wings<br /> Yellow body, red eyes, miniature wings<br /> Yellow body, white eyes, long wings<br /> Yellow body, white eyes, miniature wings<br /> Females Males Total<br /> 439<br /> 208<br /> 1<br /> 5<br /> 7<br /> 0<br /> 178<br /> 365<br /> 319<br /> 193<br /> 0<br /> 11<br /> 5<br /> 0<br /> 139<br /> 335<br /> FIGURE 5.3 Morgan’s trihybrid cross involving three<br /> X-linked traits in Drosophila.<br /> 758<br /> 401<br /> 1<br /> 16<br /> 12<br /> 0<br /> 317<br /> 700<br /> Genes gTraits Three genes that govern body color, eye color, and wing length are<br /> all found on the X chromosome. Therefore, the offspring tend to inherit the parental<br /> combinations of alleles (y + w + m + or y w m). Figure 5.4 explains how single and<br /> double crossovers can create nonparental combinations of alleles.<br /> Morgan Provided Evidence for the Linkage<br /> of X-Linked Genes and Proposed That Crossing<br /> Over Between X Chromosomes Can Occur<br /> The first direct evidence that different genes are physically located<br /> on the same chromosome came from the studies of Thomas<br /> Hunt Morgan in 1911, who investigated the inheritance pattern<br /> of different traits that had been shown to follow an X-linked pattern<br /> of inheritance. Figure 5.3 illustrates an experiment involving<br /> three traits that Morgan studied. His parental crosses were<br /> wild-type male fruit flies mated to females that had yellow bodies<br /> (yy), white eyes (ww), and miniature wings (mm). The wild-type<br /> alleles for these three genes are designated y + (gray body), w + (red<br /> eyes), and m + (long wings). As expected, the phenotypes of the F 1<br /> generation were wild-type females, and males with yellow bodies,<br /> white eyes, and miniature wings. The linkage of these genes was<br /> revealed when the F 1 flies were mated to each other and the F 2<br /> generation examined.<br /> Instead of equal proportions of the eight possible phenotypes,<br /> Morgan observed a much higher proportion of the combinations<br /> of traits found in the parental generation. He observed<br /> 758 flies with gray bodies, red eyes, and long wings, and 700 flies<br /> with yellow bodies, white eyes, and miniature wings. The combination<br /> of gray body, red eyes, and long wings was found in the<br /> males of the parental generation, while the combination of yellow<br /> body, white eyes, and miniature wings was the same as the<br /> females of the parental generation. Morgan’s explanation for this</div> <div>TABLE OF CONTENTS<br /> ::<br /> Preface<br /> ix<br /> A Visual Guide to Genetics: Analysis<br /> and Principles xvi<br /> 1 OVERVIEW<br /> PART I<br /> INTRODUCTION 1<br /> OF GENETICS 1<br /> 1.1 The Relationship Between Genes<br /> and Traits 4<br /> 1.2 Fields of Genetics 10<br /> PART II<br /> PATTERNS OF INHERITANCE 17<br /> 2 MENDELIAN<br /> INHERITANCE 17<br /> 2.1 Mendel’s Laws of Inheritance 18<br /> Experiment 2A Mendel Followed the<br /> Outcome of a Single Trait for Two<br /> Generations 21<br /> Experiment 2B Mendel Also Analyzed<br /> Crosses Involving Two Different<br /> Traits 25<br /> 2.2 Probability and Statistics 30<br /> 3 REPRODUCTION<br /> AND<br /> CHROMOSOME TRANSMISSION 44<br /> 3.1 General Features of<br /> Chromosomes 44<br /> 3.2 Cell Division 48<br /> 3.3 Sexual Reproduction 54<br /> 3.4 The Chromosome Theory<br /> of Inheritance and Sex<br /> Chromosomes 60<br /> Experiment 3A Morgan’s Experiments<br /> Showed a Connection Between a<br /> Genetic Trait and the Inheritance of a<br /> Sex Chromosome in Drosophila 64<br /> 4 EXTENSIONS<br /> OF MENDELIAN<br /> INHERITANCE 71<br /> 4.1 Inheritance Patterns<br /> of Single Genes 71<br /> 4.2 Gene Interactions 86<br /> Experiment 4A Bridges Observed an<br /> 8:4:3:1 Ratio Because the Cream-Eye<br /> Gene Can Modify the X-Linked Eosin<br /> Allele but Not the Red or<br /> White Alleles 89<br /> 5 LINKAGE<br /> AND GENETIC MAPPING<br /> IN EUKARYOTES 100<br /> 5.1 Linkage and Crossing Over 100<br /> Experiment 5A Creighton and<br /> McClintock Showed That Crossing<br /> Over Produced New Combinations of<br /> Alleles and Resulted in the Exchange<br /> of Segments Between Homologous<br /> Chromosomes 106<br /> 5.2 Genetic Mapping in Plants<br /> and Animals 109<br /> Experiment 5B Alfred Sturtevant Used<br /> the Frequency of Crossing Over in<br /> Dihybrid Crosses to Produce the First<br /> Genetic Map 113<br /> 5.3 Genetic Mapping in Haploid<br /> Eukaryotes 118<br /> 6 GENETIC<br /> TRANSFER AND<br /> MAPPING IN BACTERIA AND<br /> BACTERIOPHAGES 133<br /> 6.1 Genetic Transfer and Mapping<br /> in Bacteria 134<br /> Experiment 6A Conjugation<br /> Experiments Can Map Genes Along<br /> the E. coli Chromosome 139<br /> 6.2 Intragenic Mapping in<br /> Bacteriophages 148<br /> 7 NON-MENDELIAN<br /> INHERITANCE 161<br /> 7.1 Maternal Effect 161<br /> 7.2 Epigenetic Inheritance 164<br /> Experiment 7A In Adult Female<br /> Mammals, One X Chromosome Has<br /> Been Permanently Inactivated 166<br /> 7.3 Extranuclear Inheritance 174<br /> v</div> <div>C G<br /> T A<br /> T A<br /> T A<br /> 2 CHAPTER 1 :: OVERVIEW OF GENETICS<br /> DNA, the molecule of life<br /> Trillions of cells<br /> Each cell contains:<br /> Chromosomes<br /> Cell<br /> • 46 human chromosomes,<br /> found in 23 pairs<br /> • 2 meters of DNA<br /> Gene<br /> • Approximately 3 billion<br /> DNA base pairs per set<br /> of chromosomes,<br /> containing the bases A,<br /> T, G, and C<br /> • Approximately 20,000 to<br /> 25,000 genes code for<br /> proteins that perform<br /> most life functions<br /> C G<br /> C G<br /> T A<br /> T A<br /> A T<br /> A T<br /> DNA<br /> G<br /> Amino acid<br /> (a) The genetic composition of humans<br /> Protein (composed<br /> of amino acids)<br /> Chromosome 4<br /> p<br /> q<br /> 16<br /> 15<br /> 1<br /> 13<br /> 1<br /> 13<br /> 21<br /> 24<br /> 2<br /> 26<br /> 28<br /> 31<br /> 32<br /> 3<br /> 35<br /> Huntington disease<br /> Wolf-Hirschhorn syndrome<br /> PKU due to dihydropteridine<br /> reductase deficiency<br /> Dentinogenesis imperfecta-1<br /> C3b inactivator deficiency<br /> Aspartylglucosaminuria<br /> Williams-Beuren syndrome, type II<br /> Sclerotylosis<br /> Anterior segment<br /> mesenchymal dysgenesis<br /> Pseudohypoaldosteronism<br /> Hepatocellular carcinoma<br /> Glutaric acidemia type IIC<br /> Factor XI deficiency<br /> Fletcher factor deficiency<br /> MPS 1 (Hurler and Scheie syndromes)<br /> Mucopolysaccharidosis I<br /> Periodontitis, juvenile<br /> Dysalbuminemic hyperzincemia<br /> Dysalbuminemic hyperthyroxinemia<br /> Analbuminemia<br /> Hereditary persistence of alpha-fetoprotein<br /> AFP deficiency, congenital<br /> Piebaldism<br /> Polycystic kidney disease, adult, type II<br /> Mucolipidosis II<br /> Mucolipidosis III<br /> Severe combined immunodeficiency due<br /> to IL2 deficiency<br /> Rieger syndrome<br /> Dysfibrinogenemia, gamma types<br /> Hypofibrinogenemia, gamma types<br /> Dysfibrinogenemia, alpha types<br /> Amyloidosis, hereditary renal<br /> Dysfibrinogenemia, beta types<br /> Facioscapulohumeral muscular dystrophy<br /> (b) Genes on one human chromosome that are associated with disease when mutant<br /> FIGURE 1.1 The Human Genome Project. (a) The human genome is a complete set of human chromosomes. People have two sets of<br /> chromosomes, one from each parent. Collectively, each set of chromosomes is composed of a DNA sequence that is approximately 3 billion nucleotide<br /> base pairs long. Estimates suggest that each set contains about 20,000 to 25,000 different genes. This figure emphasizes the DNA found in the cell<br /> nucleus. Humans also have a small amount of DNA in their mitochondria, which has also been sequenced. (b) An important outcome of this work is<br /> the identification of genes that contribute to human diseases. This illustration depicts a map of a few genes that are located on human chromosome 4.<br /> When these genes carry certain rare mutations, they can cause the diseases designated in this figure.</div> <div>732 CHAPTER 26 :: EVOLUTIONARY GENETICS<br /> During parapatric speciation, hybrid zones may exist<br /> where two populations can interbreed. For speciation to occur,<br /> the amount of gene flow within the hybrid zones must become<br /> very limited. In other words, there must be selection against the<br /> offspring produced in the hybrid zone. One way that this can<br /> happen is that each of the two parapatric populations may accumulate<br /> different chromosomal rearrangements, such as inversions<br /> and balanced translocations. How do chromosomal rearrangements,<br /> such as inversions, prevent interbreeding? As discussed<br /> in Chapter 8, if a hybrid individual has one chromosome with a<br /> large inversion and one that does not carry the inversion, crossing<br /> over during meiosis can lead to the production of grossly<br /> abnormal chromosomes. Therefore, such an individual is substantially<br /> less fertile. By comparison, an individual homozygous<br /> for two normal chromosomes or for two chromosomes carrying<br /> the same inversion will be fertile, because crossing over can proceed<br /> normally.<br /> Sympatric speciation (from the Greek sym, together)<br /> occurs when a new species arises in the same geographic area<br /> as the species from which it was derived. In plants, a common<br /> way for sympatric speciation to occur is the formation of polyploids.<br /> As discussed in Chapter 8, complete nondisjunction of<br /> chromosomes during gamete formation can increase the number<br /> of chromosome sets within a single species (autopolyploidy)<br /> or between different species (allopolyploidy). Polyploidy is so<br /> frequent in plants that it is a major form of speciation. In ferns<br /> and flowering plants, at least 30% of the species are polyploid.<br /> By comparison, polyploidy is much less common in animals,<br /> but it can occur. For example, roughly 30 species of reptiles and<br /> amphibians have been identified that are polyploids derived from<br /> diploid relatives.<br /> The formation of a polyploid can abruptly lead to reproductive<br /> isolation. As an example, let’s consider the probable<br /> events that led to the formation of a natural species of common<br /> hemp nettle known as Galeopsis tetrahit. This species is thought<br /> to be an allotetraploid derived from two diploid species, Galeopsis<br /> pubescens and Galeopsis speciosa. As shown in Figure 26.3a,<br /> G. tetrahit has 32 chromosomes, whereas the two diploid species<br /> contain 16 chromosomes each (2n � 16). Figure 26.3b illustrates<br /> what would happen in crosses between the allotetraploid<br /> Galeopsis tetrahit G. pubescens G. speciosa<br /> (a) Chromosomal composition of 3 Galeopsis species<br /> Fertile<br /> G. tetrahit x G. tetrahit<br /> Infertile<br /> G. tetrahit x G. pubescens<br /> Infertile<br /> G. tetrahit x G. speciosa<br /> (b) Outcome of intraspecies and interspecies crosses<br /> FIGURE 26.3 A comparison of crosses between three natural species of hemp nettle with different ploidy levels. Galeopsis tetrahit is an<br /> allotetraploid that is thought to be derived from Galeopsis pubescens and Galeopsis speciosa. If G. tetrahit is mated with the other two species, the F 1<br /> hybrid offspring will be monoploid for one chromosome set and diploid for the other set. The F 1 offspring are likely to be sterile, because they will<br /> produce highly aneuploid gametes.</div> <div>160 CHAPTER 6 :: GENETIC TRANSFER AND MAPPING IN BACTERIA AND BACTERIOPHAGES<br /> ability to synthesize proline and confer streptomycin resistance,<br /> respectively.) A recipient strain is proA – and strC – and is<br /> resistant to tetracycline. After transformation, the bacteria were<br /> first streaked on media containing proline, streptomycin, and<br /> tetracycline. Colonies were then restreaked on media containing<br /> streptomycin and tetracycline. (Note: Both types of media had<br /> carbon and nitrogen sources for growth.) The following results<br /> were obtained:<br /> 70 colonies grew on media containing proline, streptomycin,<br /> and tetracycline, while only 2 of these 70 colonies grew when<br /> restreaked on media containing streptomycin and tetracycline<br /> but lacking proline.<br /> A. If we assume the average size of the DNA fragments is 2<br /> minutes, how far apart are these two genes?<br /> B. What would you expect the cotransformation frequency to be<br /> if the average size of the DNA fragments was 4 minutes and the<br /> two genes are 1.4 minutes apart?<br /> E17. If you took a pipette tip and removed a phage plaque from a petri<br /> plate, what would it contain?<br /> E18. As shown in Figure 6.16, phages with rII mutations cannot<br /> produce plaques in E. coli K12(λ), but wild-type phages can. From<br /> an experimental point of view, explain why this observation is so<br /> significant.<br /> E19. In the experimental strategy described in Figure 6.18, explain why<br /> it was necessary to dilute the phage preparation used to infect<br /> E. coli B so much more than the phage preparation used to infect<br /> E. coli K12(λ).<br /> E20. Here are data from several complementation experiments,<br /> involving rapid-lysis mutations in genes rIIA and rIIB. The strain<br /> designated L51 is known to have a mutation in rIIB.<br /> Phage Mixture<br /> Complementation<br /> L91 and L65<br /> No<br /> L65 and L62<br /> No<br /> L33 and L47<br /> Yes<br /> L40 and L51<br /> No<br /> L47 and L92<br /> No<br /> L51 and L47<br /> Yes<br /> L51 and L92<br /> Yes<br /> L33 and L40<br /> No<br /> L91 and L92<br /> Yes<br /> L91 and L33<br /> No<br /> List which groups of mutations are in the rIIA gene and which<br /> groups are in the rIIB gene.<br /> E21. A researcher has several different strains of T4 phage with<br /> single mutations in the same gene. In these strains, the mutations<br /> render the phage temperature sensitive. This means that<br /> temperature-sensitive phages can propagate when the bacterium<br /> (E. coli) is grown at 32°C but cannot propagate themselves when<br /> E. coli is grown at 37°C. Think about Benzer’s strategy for<br /> intragenic mapping and propose an experimental strategy to map<br /> the temperature-sensitive mutations.<br /> E22. Explain how Benzer’s results indicated that a gene is not an<br /> indivisible unit.<br /> E23. Explain why deletion mapping was used as a step in the intragenic<br /> mapping of rII mutations.<br /> Questions for Student Discussion/Collaboration<br /> 1. Discuss the advantages of the genetic analysis of bacteria and<br /> bacteriophages. Make a list of the types of allelic differences among<br /> bacteria and phages that are suitable for genetic analyses.<br /> 2. Complementation occurs when two defective alleles in two<br /> different genes are found within the same organism and produce a<br /> normal phenotype. What other examples of complementation have<br /> we encountered in previous chapters of this textbook?<br /> Note: All answers appear at the website for this textbook; the answers to<br /> even-numbered questions are in the back of the textbook.<br /> www.mhhe.com/brookergenetics3e<br /> Visit the Online Learning Center for practice tests, answer keys, and other learning aids for this chapter. Enhance your understanding of genetics with<br /> our interactive exercises, quizzes, animations, and much more.</div> <div>650 CHAPTER 23 :: DEVELOPMENTAL GENETICS<br /> At the molecular level, certain key features enable myogenic<br /> bHLH proteins to promote muscle cell differentiation. The basic<br /> domain binds to a muscle-cell-specific enhancer sequence; this<br /> sequence is adjacent to genes that are expressed only in muscle<br /> cells (Figure 23.19). Therefore, when myogenic bHLH proteins<br /> are activated, they can bind to these enhancers and activate the<br /> expression of many different muscle-cell-specific genes. They may<br /> exert their effects via alterations in chromatin structure or via the<br /> activation of RNA polymerase to a transcriptionally active state.<br /> In this way, myogenic bHLH proteins function as master switches<br /> that activate the expression of many muscle-specific genes. When<br /> the encoded proteins are synthesized, they change the characteristics<br /> of an undifferentiated cell into those of a highly specialized<br /> skeletal muscle cell.<br /> Another important aspect of myogenic bHLH proteins<br /> is that their activity is regulated by dimerization, which occurs<br /> via the helix-loop-helix domain. As shown in Figure 23.19, heterodimers—dimers<br /> formed from two different proteins—may<br /> be activating or inhibitory. When a heterodimer forms between<br /> a myogenic bHLH protein and an E protein, which also contains<br /> a basic domain, the heterodimer binds to the DNA and activates<br /> gene expression (Figure 23.19a). However, when a heterodimer<br /> forms between a myogenic bHLH protein and a protein called<br /> Id (for inhibitor of differentiation), the heterodimer cannot<br /> bind to DNA because the Id protein lacks a basic domain (Figure<br /> 23.19b). Two basic domains are needed for the heterodimer<br /> to bind to the DNA. The Id protein is produced during early<br /> stages of development and prevents myogenic bHLH proteins<br /> from promoting muscle differentiation too soon. At later stages<br /> of development, the amount of Id protein falls, and myogenic<br /> bHLH proteins can then combine with E proteins to induce<br /> muscle-cell differentiation.<br /> 23.3 PLANT DEVELOPMENT<br /> In developmental plant biology, the model organism for genetic<br /> analysis is Arabidopsis thaliana (Figure 23.20). Unlike most<br /> flowering plants, which have long generation times and large<br /> genomes, Arabidopsis has a generation time of about two months<br /> Muscle-cell-specific<br /> enhancer sequence<br /> Basic<br /> domain<br /> Myogenic<br /> bHLH<br /> E<br /> HLH<br /> domain<br /> Binding of<br /> heterodimer<br /> activates the gene.<br /> (a) Action of bHLH–E heterodimer<br /> Muscle-cell-specific<br /> enhancer sequence<br /> Myogenic<br /> bHLH<br /> Id<br /> Heterodimer is<br /> unable to bind<br /> –gene is not<br /> activated.<br /> (b) Action of bHLH–Id heterodimer<br /> FIGURE 23.19 Regulation of muscle-cell-specific genes by<br /> myogenic bHLH proteins. (a) A heterodimer formed from a myogenic<br /> bHLH protein and an E protein can bind to a muscle-cell-specific<br /> enhancer sequence and activate gene expression. (b) When a myogenic<br /> bHLH protein forms a heterodimer with an Id protein, it cannot bind to<br /> the DNA and therefore does not activate gene transcription.<br /> FIGURE 23.20 The model organism Arabidopsis. The plant is<br /> relatively small, making it easy to grow many of them in the laboratory.</div> <div>180 CHAPTER 7 :: NON-MENDELIAN INHERITANCE<br /> TABLE 7.5<br /> Examples of Human Mitochondrial Diseases<br /> Disease<br /> Mitochondrial Gene Mutated<br /> Cyanobacterium<br /> Purple bacterium<br /> Leber’s hereditary<br /> optic neuropathy<br /> Neurogenic muscle<br /> weakness<br /> Mitochondrial<br /> encephalomyopathy,<br /> lactic acidosis, and<br /> strokelike episodes<br /> Mitochondrial<br /> myopathy<br /> Maternal myopathy<br /> and cardiomyopathy<br /> A mutation in one of several mitochondrial genes<br /> that encode respiratory chain proteins: ND1, ND2, CO1,<br /> ND4, ND5, ND6, and cytb<br /> A mutation in the ATPase6 gene that encodes a<br /> subunit of the mitochondrial ATP-synthetase, which is<br /> required for ATP synthesis<br /> A mutation in genes that encode tRNAs for leucine<br /> and lysine<br /> A mutation in a gene that encodes a tRNA<br /> for leucine<br /> A mutation in a gene that encodes a tRNA<br /> for leucine<br /> Primordial<br /> eukaryotic cells<br /> Evolution<br /> Myoclonic epilepsy<br /> with ragged-red<br /> muscle fibers<br /> A mutation in a gene that encodes a tRNA<br /> for lysine<br /> Data from: Wallace, D. C. (1993). Mitochondrial diseases: genotype versus phenotype. Trends<br /> Genet. 9, 128–33.<br /> loss of vision in one or both eyes. LHON can be caused by a<br /> defective mutation in one of several different mitochondrial<br /> genes. Researchers are still investigating how a defect in these<br /> mitochondrial genes produces the symptoms of this disease.<br /> Extranuclear Genomes of Mitochondria<br /> and Chloroplasts Evolved from an<br /> Endosymbiotic Relationship<br /> The idea that the nucleus, mitochondria, and chloroplasts contain<br /> their own separate genetic material may at first seem puzzling.<br /> Wouldn’t it be simpler to have all of the genetic material<br /> in one place in the cell? The underlying reason for distinct<br /> genomes of mitochondria and chloroplasts can be traced back to<br /> their evolutionary origin, which is thought to involve a symbiotic<br /> association.<br
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