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Solution Manual for Essential Genetics: A Genomic Perspective , 4th Edition

Solution Manual for Essential Genetics: A Genomic Perspective, 4th Edition makes textbook problem-solving simple, with detailed answers that make learning fun.

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    Study Guide and Solutions Manual to accompany Essential • Genetics A GENOMICS PERSPECTIVE Fourth Edition ... V,. £»IL. Daniel L. Hartl and Elizabeth W. Jones Prepared by Elena R. Lozovsky

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    Contents 1 The Genetic Code of Genes and Genomes .................... ............... . . 1 2 Transmission Genetics: Heritage from Mendel ........ ........................ 4 3 The Chromosomal Basis of Heredity............ . ....... — ........ 9 4 Gene Linkage and Genetic Mapping................................................... 13 5 Human Chromosomes and Chromosome Behavior.............. .......... 18 6 DNA Structure, Replication, and Manipulation.......................... . ... 22 7 The Genetics of Bacteria and Their Viruses.......... . ........................ .26 8 The Molecular Genetics of Gene Expression. ............................. 30 9 Molecular Mechanisms of Gene Regulation ..................... 34 10 Genomics, Proteomics, and Genetic Engineering........ ............. 38 11 The Genetic Control of Development.......................... 42 12 Molecular Mechanisms of Mutation and DNA Repair ................ 45 13 Molecular Genetics of the Cell Cycle and Cancer... ......... 48 14 Molecular Evolution and Population Genetics............. .................... 52 15 The Genetic Basis of Complex Inheritance.. ..................................... 55 Appendix: Answers .............. . ..................................... 58 v

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    Preface The Study Guide and Solution Manual was written specifically to accompany Essential Genetics: A Genomic Perspective, Fourth Edition, by Daniel L. Hartl and Elizabeth W. Jones. It has been organized into chapters that parallel those of the book and provides the answers to the Concepts in Action problems from the main text. These answers have been worked in full and the reasoning explained in detail. The objective is to help the student develop skills in problem solving. You, the student, arc strongly urged to try to work a problem before giving up and consulting the answer. If the method of solution is not immediately apparent, do not go to the answer immediately. Think about it for a while, consult the corresponding chapter in the text, or perhaps come back to it later. Look up the answer only after a few tries. With this approach, you will be pleased to learn that problems that at first look very difficult arc often within your capability. A willingness to be patient, to let your mind work quietly and subconsciously on a problem before looking at the answer, pays rewards of pos- itive feedback and increased self-confidence. Even when you are unable to solve a problem after several attempts, a delay in looking at the answer will give the problem more impact, and the method of solution will be less likely to be forgotten. Many of the problems require the understanding and application of only a single con- cept, such as Mendel ian segregation or polypeptide translation. Other problems require application of several concepts in logical order, for example, problems that combine segrega- tion and recombination or that consider the consequences of a mutation on the amino acid sequence of a polypeptide. Although some chapters contain a few problems that are admit- tedly difficult, they nevertheless do not require any concepts or principles beyond those dis- cussed in the main text. Some problems are intended to be challenging, and even if you are unsuccessful in solving them, you will gain additional understanding of genetics when you look at the answer. The Study Guide and Solutions Manual also provides a list of the key concepts for each chapter for quick review and the answers to the main text's Key Terms and Concepts fill-in- the-blank questions. In addition, each chapter includes a set of Study Questions. These con- sist of a few multiple -choice questions as practice in the genre. Some call for verbal profi- ciency, asking for a very short essay explaining how relevant concepts can be used alone or in combination to account for biological observations. Other questions arc quantitative and require a precise numerical answer. Problems requiring calculation make use of only simple arithmetic or algebra and draw upon the generally relevant laws of probability and statistics explained in the text. I encourage you not only to give a numerical answer but also to think a problem through, in terms of "order of magnitude," to verity your approacn and method of solution as well as to catch any howling mistakes caused by miscalculation. I will be glad to have any feedback from students or instructors who make use of these problems. I welcome comments and suggestions, corrections, and especially additional prob- lems for inclusion in later renditions of this collection. I can be contacted by electronic mail at elozovsk@oeb.harvard.edu. I am grateful to my colleagues, Erica Pantages and Emily Lilly for valuable ideas and helpful discussions. I also wish to acknowledge Rebecca Seastrong, Stephen L. Weaver, and Anne Spencer of Jones and Bartlett for their support and assistance. Special thanks go to Dan Hartl for his effort in editing, correcting errors, and suggesting problems. I am grateful for his encouragement and support that were absolutely vital for this project. Elena R. Lozovsky, PhD Harvard University

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    The Genetic Code of Genes and Genomes Key Concepts • Inherited traits arc affected by genes. • Genes are composed of the chemical deoxyribonucleic acid (DNA). • DNA replicates to form (usually identical) copies of itself. • DNA contains a code specifying what types of enzymes and other proteins are made in cells. • DNA occasionally mutates, and the mutant forms specify altered proteins. • A mutant enzyme is an "inborn error of metabolism" that blocks one step in a bio- chemical pathway for the metabolism of small molecules. • Traits are affected by environment as well as by genes. • Organisms change genetically through generations in the process of biological evolution. • Because of their common descent, organisms share many features of their genetics and biochemistry. Key Terms 1. complementary base pairing 2. antiparallel 3. central dogma 4. transcription 5. ribosome 6. transfer RNA (tRNA) 7. proteome 8. substrate 9. alkaptonuria 10. phenylketonuria 11. pleiotropic effect or pleiotropy 12. prokaryote Concepts in Action 1.1. (a) false; (b) true; (c) true; (d) false. 1.2. The importance of the nucleus in inheritance was implied by its prominence in fer- tilization. The discovery of chromosomes inside the nucleus, their behavior during cell division, and the observation that each species has a characteristic chromosome number made it likely that chromosomes were the carriers of the genes. Micro- scopic studies showed that DNA and proteins are both present in chromosomes, but whereas nearly all cells of a given species contain a constant amount of DNA, the amount and kinds of proteins differ greatly in different cell types. 1

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    1.3. It was generally believed that the genetic material must be a very complex molecule. Proteins were chemically the most complex macromolecules known at the time. DNA was thought to be a monotonous polymer composed of a simple repeating unit. 1.4. Because the mature T2 phage contains only DNA and protein, the labeled RNA was left behind in material released by the burst cells. 1.5. Watson and Crick noted that the nucleotide sequence of the DNA molecule could be replicated if each of the strands were used as a template for the formation of a new daughter strand having a complementary sequence of bases. They also noted that genetic information could be coded by the sequence of bases along the DNA mole- cule, analogous to letters of the alphabet printed on a strip of paper. Finally, they noted that changes in genetic information could result from errors in replication, and the altered nucleotide sequence could then be perpetuated. 1.6. RNA differs from DNA in that the sugar-phosphatc backbone contains ribose rather than deoxyribose. RNA contains the base uracil (U) instead of thymine (T), and RNA usually exists as a single strand (although any particular molecule of RNA may contain short regions of complementary base pairs that can come together to form duplexes). 1.7. Percent C = 37.5%, so percent G = 37.5% also. Percent A + T = 1 - 0.375 - 0.375 = 25%, but because percent A = percent T, it must be that percent A = 12.5%. 1.8. Because A A T and G A C in this DNA, it seems likely that the DNA molecule pres- ent in this particular virus is single-stranded. 1.9. 3'-CA-5', because the dinucleotide 3'-CA-5' pairs with 5'-GT-3', so where either strand contains 5'-GT-3', the other contains 3'-CA-5'. 1.10. The repeating Asn results from translation in the reading frame 5'-AAUAAUAAUAAU...-3', and the repeating He results from translation in the reading frame 5'-AUAAUAAUAAUA...-3'. There is no product corresponding to the third reading frame (5'-UAAUAAUAAUAA...-3'), because 5'-UAA-3' is a stop codon. 1.11. Most likely the mutant protein does not fold properly and is degraded by proteases. 1.12. It is the case because each codon is exactly three nucleotides in length. In a protein- coding region, an insertion or deletion of anything other than an exact multiple of three nucleotides would shift the reading frame (phase) in which the mRNA is translated. All amino acids downstream of the site of the mutation would be trans- lated incorrectly. 1.13. 5'-TGTCGTATTTGCAAG-3' 1.14. Transcription takes place from left to right, and the mRNA sequence is 5'- UGUCGUAUUUGCAAG-3'. 1.15. Cys-Arg-Ile-Cys-Lys or, using the single-letter abbreviations, CRICK. 1.16. Cys-His-Ile-Cys-Lys, CHICK 1.17. The codon 5'-UGG-3' codes for Trp, and in this random polymer the Trp codon is expected with a frequency of 1/4 x 3/4 x 3/4 = 9/64. The amino acid Vai could be specified by either 5'-GUU-3' or 5'-GUG-3'; the former has an expected frequency of 3/4 x 1/4 x 1/4 = 3/64, and the latter of 3/4 x 1/4 x 3/4 = 9/64, totaling 12/64. The amino acid Phe could be specified only by 5'-UUU-3' in this random polymer, so Phe would have an expected frequency of 1/4 x 1/4 x 1/4 =1/64. 1.18. (a) Mct-Ser-Thr-Ala-Val-Leu-Glu-Asn-Pro-Gly. (b) The mutation changes the initia- tion codon into a noninitiation codon, so translation will not start with the first AUG; translation will start with the next AUG farther along the mRNA or, if this is too distant, not at all. (c) Met-Ser-Thr-AJa-Val-Leu-Glu-Asn-Pro-Gly; there is no change, because both 5'-UCC-3' and 5'-UCG-3' code for serine, (d) Met-Ser-Thr- Ala-Ala-Leu-Glu-Asn-Pro-Gly; there is a Vai-*Ala amino acid replacement because 5'GUC-3' codes for Vai, whereas 5'GCC-3' codes for Ala. (e) Met-Ser-Thr-Ala-Val- Leu; translation is terminated at UAA because 5'-UAA-3' is a "stop" (termination) code. 1.19. (a) X,Y, and Z missing, W in excess; (b) Y and Z missing, X in excess; (c) Z missing, Y in excess. Study Guide to accompany Essential Genetics: A Genomics Perspective, Fourth Edition 2

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    1.20. The finding that the cells can grow in the presence of Y implies that step C is func- tional. The finding that the cells cannot grow in the presence of X implies that step B is blocked. The results with W imply that a downstream step is blocked, but do not reveal which one. Study Questions 1.S1. When the base composition of double-stranded DNA from a new species of bacteria was determined, 15% of the bases were found to be cytosine. What is the percent- age of adenine in the DNA of this organism? A) 15% fSZ C B) 25% /S7. D) 45% 3 5 7 . T X 1.S2. A duplex DNA molecule contains a random sequence of the four nucleotides in equal proportions. What is the average spacing between consecutive occurrences of the sequence 5'-GGGG-3'? CACM mxcUjrticU. ocean "Q 2 4 D) 4 2 1.53. The sequence of one strand of DNA is 5'-CCATATGC-3'. The sequence of the com- plementary strand would be what? -AT5'-CCATATGC-3' ( ( JBf 5 -GGTATACG-3' GCA'T ft'VGG’ 3 £j)5'-GCATATGG-3' -~&r 5'-AAGCGCTA-3' 5'-GCAUAUGC-3' £ N A 1.54. An RNA molecule folds back upon itself to form a "hairpin" structure held together by a region of base pairing. One segment of the molecule in the paired region has the base sequence 5'-UCAAUGC-3'. What is the base sequence with which this seg- ment is paired? ®5'-GCAUUGA-3< „ „ --------------8) r - i s e A u u c A - j ' 1 --------------------- -------------------------------------------------------------------------------- C) 5'-TCAATGC-3' D) 5'-GCATTGA-3' What codon would pair with the anticodon of tRNA met 5'-UAG-3'? 5'-AUC-S’ 1.S5. A) 5'-CAU-3' B) 5'-GAU-3' C) 5'-UAG-3' D) 5'-ATG-3' (tp5'-CUA-3' 1.56. Traits are affected by environment as well as by . 1.57. An enzyme called reverse transcriptase can produce a complementary DNA strand from an template. 1.58. If DNA from Baker's yeast has a guanine content of 25%, then what is the content of the adenine? {1.S9. J If a particular piece of RNA has a uracil content of 25%, then what is its guanine content? N Id A 1A- ( 1.S10. How many different DNA sequences can encode the amino acid sequence Pro- Met-Arg? CHAPTER 1 The Genetic Code of Genes and Genomes 3

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    Transmission Genetics: Heritage from Mendel Key Concepts • Inherited traits are determined by the genes present in the reproductive cells united in fertilization. • Genes are usually inherited in pairs, one from the mother and one from the father. • The genes in a pair may differ in DNA sequence and in their effect on the expression of a particular inherited trait. • The maternally and paternally inherited genes are not changed by being together in the same organism. • hi the formation of reproductive cells, the paired genes separate again into different cells. • Random combinations of reproductive cells containing different genes result in Mendel's ratios of traits appearing among the progeny. • The ratios actually observed for any traits are determined by the types of dominance and gene interaction. Key Terms 1. allele 2. wildtype 3. genotype 4. segregation 5. testcross 6. independent assortment 7. incomplete dominance 8. pedigree 9. sibling 10. epistasis 11. complementation test 12. variable expressivity Concepts in Action 2.1. Any number of different alleles may exist at the same time in the population. A sin- gle diploid organism can have only two alleles of the same gene, one inherited from the mother and one from the father. 4

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    2.2. Progeny Parents AA Aa aa AAXAA 1 0 0 AA X Aa 1/2 1/2 0 AA X aa 0 1 0 Aa X Aa 1/4 1/2 1/4 Aa X aa 0 1/2 1/2 aa X aa 0 0 1 2.3. Ww; 1/4 wrinkled seeds are expected. 2.4. The birth of an affected offspring implies that parents are at risk of another affected offspring. Once the genotypes of the parents have been deduced from their own phenotypes and the fact that they have an affected offspring, the recurrence risk is calculated as the probability of an affected offspring in the next birth, (a) The mating must be Aa x aa, and so the risk of an Act offspring in any birth (and therefore the recurrence risk) is 1/2. (b) The mating must be Aa x Aa, and so the risk of an aa off- spring in any birth is 1/4. (c) The mating must be Aa x Aa, and so the risk of an aa offspring in any birth is 1/2. 2.5. Sixteen possible types of gametes [all possible combinations with one each of (A, a), " ---- (B, b), (C, c), and (D, d), with an expected proportion of 1/16 each. 2.6. (a) 112 is the affected person, and so the genotype must be aa. (b) 1-1 and 1-2 are the — parents of the affected person and because neither is affected, their genotypes must both be Aa. (c) II-I and II-3 are siblings of the affected person. Because they are not affected, and because they result from the mating Aa x Aa, their possible genotypes are AA and Aa. (d) From the mating Aa x Aa, the ratio of AA.Aa is 1:2; hence, the probability that II-3 is a carrier (genotype Aa) equals 2/3. 2.7. (2/3) 2 = 4/9; (1/3) 2 = 1/9; 1 - ( 1/3) 2 = 8/9 2.8. The DNA fragment from a, with the 2-kb insertion results in a band at 3 kb + 2 kb = 5 kb, and the DNA fragment from a 2 with the 1-kb deletion results in a band at 3 kb-1 kb = 2 kb. DNA from each homozygous genotypes produces only one band, whereas that from each heterozygous genotype produces two bands. AA a i a 2 a 2 A a i Aa2 a i a 2 5kb 3kb Ikb 2.9. The recessive mutations are in different genes. They are not alleles. This situation is an example of complementation in human beings. 2.10. The existing data enable us to group the mutants into three complementation groups as follows: [a, c, d}, {b,f}, and {e}. The missing entries are shown in the accompanying table. a © G © © a © © © © © © © © © 0 © C 2006 by Jones and Bartlett Publishers CHAPTER 2 Transmission Genetics: Heritage from Mendel 5

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    2.11. A 15:1 ratio is characteristic of two genes with independent assortment in which the minority trait is due to a double -homozygous recessive. Suppose that the geno- type aa bb results in ovoid-shaped fruit and that all other genotypes result in trian- gular fruit. The true-breeding parents would then have been aa BB and AA bb. One test of this hypothesis is that the ovoid F 2 plants should be true-breeding. Furthermore, crossing the Fj triangular plants (Aa Bb) with the F 2 ovoid plants (aa bb) should yield progeny with the phenotypic ratio 3 triangular: 1 ovoid. 2.12. We expect a 3:1 ratio of the dominant to the recessive phenotype when two het- erozygous genotypes are crossed, but in this case we observe 2:1. A cross of two Cy!+ heterozygotes is expected to yield a progeny genotypic ratio of 1 CICy.2 Cy /+: 1 +/+. Because we only see two curly-winged flics for every wildtype fly, one possible explanation is that the CylCy homozygotes die. In other words, Cy is dominant with respect to wing phenotype but recessive with respect to lethality. 2.13. (a) Two phenotypic classes arc expected for each of the Aa and Bb pairs of alleles, and three are expected for the Rr pair, yielding a total number of 12. (b) 1/64. (c) 1/8. 2.14. Let AA BB represent the wildtype genotype with a phenotype of a disc shape. We are told that genotypes aa BB and Aa bb both have spherical fruit and that aa bb has elongated fruit. The F 2 progeny are expected in the ratio 9 A-B-:3 A- bb:3 aa B-t 1 aa bb, which implies that the expected ration of fruit-shaped phenotypes is 9 disc: 6 sphere:! elongate. 2.15. 1 / 2 x 1 / 2 x 1 / 4 = 1 / 1 6 . 2.16. (a) Because the trait is rare, it is reasonable to assume that the affected father is het- erozygous HD/hd, where hd represents the normal allele. Half of the father's gametes contain the mutant HD allele, so the probability is 1/2 that the son received the allele and will later develop the disorder, (b) We do not know whether the son is heterozygous HD/hd, but the probability is 1/2 that he is; if the son is heterozy- gous, half of his gametes will contain the HD allele. Therefore, the overall probabil- ity that the grandchild has the HD allele is (1/2) x (1/2) = 1/4. 2.17. (a) In approaching this kind of problem, note first that the families with at least one boy include all families except those with all girls. Therefore, the simplest way to obtain the answer is to calculate the proportion of families with four girls and then subtract this from 1. The probability of having four girls is (1/2)“, which equals 0.0625; the rest of the families are those with at least one boy, which account for the proportion 1 - 0.0625 = 0.9375 of all families, (b) Because a particular birth order is specified, the answer is (1/2) 4 = 1/16. To look at the problem in another way, note that although 6/16 of all families with four children have two boys and two girls, only 1/6 of such families have the specific birth order FMFM; hence, the answer is, as before, 1/16. 2.18. In the functional female gametes, the ratio of Ata is l / 2 : l / 2 because of Mendelian segregation, In males, the products of meiosis in an Aa individual also consist of A + A + a + a, but as stated in the problem, half of the A-bearing products are nonfunc- tional. Hence, each male meiosis produces, on the average, three functional prod- ucts, namely A + a + a. The ratio of Ata among functional male gametes is therefore 1:2 or, converting to proportions, 1/3 A:2/3 a. The Punnet square shown here indi- cates that the F 2 ratio of AAtAataa is l/6:3/6:2/6 (or, reducing the fractions, l / 6 : l / 2 : l / 3 ) . Eggs 1/2 A 1/2 a 1/3 A 1/6 AA 1/6 Aa Pollen --------------------- 2/3 a 2/6 Aa 2/6 aa 2.19. (a) The trait is more likely to be due to a recessive allele because there is consan- guinity (mating between relatives) in the pedigree, (b) The double line indicates consanguineous mating, (c) III-1 and III-2 are first cousins, (d) Either 1-1 or 1-2 are Study Guide to accompany Essential Genetics: A Genomics Perspective, Fourth Edition 6

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    likely to be heterozygous Aa (but not both, because the trait is said to be rare), and all of II-2, II-3, HI- 1, and HI-2 are likely to be Aa. On the other hand, II- 1 and II-4 are most likely to be A A. 2.20. The accompanying Punnet squares indicate that the expected ratios of genotypes in (a) are 9/16 DD: 6/16 Dd: 1/16 dd, an in (b) are 3/8 DD: 4/8 Dd: 1/8 dd. (a) Eggs 3/4 D \ / d d (b) Eggs 3/4 D \ / A d 3/4 D Sperm 1/4 d 9/16 DD 3/\f>Dd 1/2 D Sperm 1/2 3/8 DD 1/8 Dd 3/16 Dd 1/16 dd 3/8 Dd 1/8 dd Study Questions 2.51. Two parents with blood types A and B have a child who has the O blood type. What is the chance that their next child will be O? A) 0 B) 1/2 C) 1/4 D) 1 2.52. In the cross Aa Bb Cc Dd Be x Aa Bb Cc Dd Ee, in which all genes undergo inde- pendent assortment, what proportion of offspring are expected to be homozygous dominant for all four genes? A) 1/2 B) 1/5 C) (1/2) 5 D) (1/4) 4 E) (1/4) 5 2.53. Among sibships consisting of five children, and assuming a sex ratio of 1:1, what is the proportion with no girls? A) 1/2 B) (1/2) 5 C) (1/5) 5 D) 1 - ( 1/2) 5 2.54. Assuming equal sex ratios, if a mating has already produced 5 boys, what is the probability that the noxt child will be a boy? A) 0 B) 1 C) 1/2 D) (1/2) 3 E) 1 - (1/2) 3 2.55. Assuming independent assortment, how many different gametes can be formed by an organism that is homozygous for 7 and heterozygous for 2 genes? A) 2 B) 4 C) 3 2 D) 5 E) 6 2.56. In genetic analysis, the complementation test is used to determine whether two recessive mutations that cause similar phenotype are of the same gene. 2.57. Consider a family with six children, and remember that each birth is equally likely to result in a boy or a girl. What proportion of sibships will include at least one girl? CHAPTER 2 Transmission Genetics: Heritage from Mendel 7

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    2.S8. A normal woman has a brother, who is albino (a trait determined by a rare reces- sive autosomal allele). What is the probability that her phenotypically normal son is heterozygous for the gene? 2.S9. In a testcross of Aa BB Cc Dd Ee FF, where the genes show independent assortment: a. What is the expected frequency of aa bb Cc dd ce Ff progeny? b. What is the expected frequency of progeny that are heterozygous for all six genes? 2.S10. Assuming sex ratio of 1:1, a. What is the probability that a couple will have eight boys? b. If they already have seven boys, what is the probability that the eighth child will be a boy? Study Guide to accompany Essential Genetics: A Genomics Perspective, Fourth Edition 8

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    The Chromosomal Basis of Heredity Key Concepts • Chromosomes in eukaryotic cells are usually present in pairs. • The chromosomes of each pair separate in meiosis, one going to each gamete. • In meiosis, the chromosomes of different pairs undergo independent assortment. • Chromosomes consist largely of DNA combined with histone proteins. • In many animals, sex is determined by a special pair of chromosomes, the X and Y. • Irregularities in the inheritance of an X-linked gene in Drosophila gave experimental proof of the chromosomal theory of heredity. • The progeny of genetic crosses follow the binomial probability formula. • The chi-square statistical test is used to determine how well observed genetic data agree with expectations from a hypothesis. Key Terms 1. synapsis 2. kinetochore 3. diplotete 4. chiasma 5. telomerase 6. anaphase II 7. histone 8. chromosome territory 9. heterochromatin 10. nondisjunction 11. chi-square 12. significant Concepts in Action 3.1. It means that, if there is no crossing over between a gene and the centromere, homologous alleles are separated from one another at anaphase I; if there is cross- ing over between a gene and the centromere, they are separated at anaphase II. In either case, they undergo segregation (separation} at one of the anaphase divisions. 3.2. After one cell cycle carried out in the presence of colchicine, a human cell would be expected to have 46 x 2 = 92 chromosomes. 3.3. It means that nonhomologous chromosomes have no influence on each other's ori- entation as they align on the metaphase plate at metaphase I. Hence, genes on non- homologous chromosomes are independent in whether or not they proceed to the same anaphase pole, which is equivalent to independent assortment. 3.4. After the centromeres have split, each former sister chromatid is counted as a chro- mosome in its own right, so at anaphase there are 48 chromosomes. 9

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    3.5. The hybrid has 14 + 7 = 21 chromosomes. 3.6. The chromosome tips (telomeres) would be expected to become shorter in each cell cycle. The reason is that DNA replication cannot begin at the extreme 3' end of a template strand. The function of telomerase is to restore the unreplicated region by elongating the 3' end after each round of DNA replication. 3.7. Because female chickens are ZW and males are ZZ, females receive their Z chromo- some from their father. Therefore, the answer is to mate a gold male (ss) with a sil- ver female (X). The male progeny will be Ss (silver plumage) and the female prog- eny will be s (gold plumage), and these are easily distinguished. 3.8. Panel B is anaphase of mitosis, because the homologous chromosomes are not paired. Panel A is anaphase I of meiosis, because the homologous chromosomes are paired. Panel C is anaphase II of meiosis, because the chromosome number has been reduced by half. 3.9. Assuming that she happens upon cells in each stage according to the length of time that an average cell spends in that stage, prophase takes up 320/2000 = 16.0% of the total time, metaphase 150/2000 - 7.5%, anaphase 80/2000 = 4.0%, and telophase 120/2000 = 6.0%. The remaining time (66.5%) must be spent in interphase. 3.10. (a) Ratio X/A = 0.5, phenotype male, (b) Ratio X/A = 1.0, phenotype female. (c) Ratio X/A > 1.0, phenotype female, (d) Ratio X/A = 0.66, phenotype intersex. 3.11. (a) cb r cb s ; (b) cb r cb r ; (c) cb 9 cb 9 . 3.12. (a) 1/2 x 1/2 = 1/4; (b) 1/2 x 1/2 = 1/4. 3.13. The mother must be heterozygous, because her father was affected. Therefore, her genotype must be Hh, where h denotes the hemophilia mutation. The most likely explanation of the 47, XXY hemophiliac son is that in meiosis in the mother, the -bearing chromatids underwent nondisjunction and produced an XX-bearing egg with genotype hh. Fertilization by a normal Y-bearing sperm resulted in the XXY zygote of genotype hh, which is the cause of the hemophilia. 3.14. The chance that the second gene is on a different chromosome than the first is 6/7; the chance that the third gene is on a different chromosome than either of the first two is 5/7. Continuing in this manner, we find that the overall chance is 6/7 x 5/7 x 4/7 x 3/7 x 2/7 x 1/7 = 0.00612, or less than 1 percent. The likelihood of having every gene on a different chromosome is quite small, and for this reason Mendel has been accused of deliberately choosing unlinked genes. In fact, it is now known that not all of Mendel's genes are on different chromosomes. Three of the genes (fa, deter- mining axial vs. terminal flowers; le, determining tall vs. short plants; and v, determin- ing smooth vs. constricted pods) arc all located on chromosome 4. The le and v genes undergo recombination at the rale of about 12 percent, but Mendel apparently did not study this particular combination for independent assortment. The fa gene is very distant from the other two and shows independent assortment with them. 3.15. These are typical characteristics of X- linked inheritance. Affected males transmit the mutant X chromosome only to their daughters. A carrier female will have both affected and unaffected sons as well as both carrier and noncarrier daughters. The carrier daughters (sisters of the affected sons) will transmit the mutant gene, but the unaffected sons (brothers of the affected sons) will not. 3.16. The chi-square value equals (188 - 186) 2 /186 + (203 - 186) 2 / 186 + (175 - 186) 2 / 186 + (178 - 1 86) 2 / 1 86 = 2.57. It has 3 degrees of freedom (four classes of data), and the P value is about 0.5. We therefore conclude that there is no reason, on the basis of these data, to reject the hypothesis of 1:1:1 :1 segregation. 3.17. 3.84 for 1 df, 5.99 for 2 df, 7.82 for 3 df, 9.49 for 4 df, and 11.07 for 5 df (these val- ues are from statistical tables; values read from the graph will not be so accurate). These values correspond to two, three, four, five, and six classes of data, respectively, for the type of chi-square tests in this chapter. Perhaps surprisingly, the increasing chi-square value needed for significance docs not imply that one is less likely to obtain a statistically significant chi-square with increasing degrees of freedom. The probability of obtaining a statistically significant chi-square value, given that the genetic hypothesis is true, is 5%, no matter what the number of degrees of freedom. 10 Study Guide to accompany Essential Genetics: Genomics Perspective, Fourth Edition

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    3.18. The chi-square value equals (462 - 450) 2 /450 + (167 - 1 50) 2 / 1 50 + (127 - 1 50) 2 /l 50 + (44 - 50) 2 /50 = 6.49. It has 3 degrees of freedom (four classes of data), and the P value is about 0.08. Although this is close to the 5% level of significance, it is not sufficient reason to reject the hypothesis of: 9:3:3:1 segregation. 3.19. The chi-square value equals 8.79 with three degrees of freedom. The P value is about 0.035, which is less than 0.05, hence there is reason to reject, at the 5% level of statistical significance, the hypothesis of 1:1:1:1 segregation. 3.20. DNA from males exhibits one band and that from females two bands, which suggests X-linked inheritance. This can be verified further by examining the pattern of inheri- tance. Sons exhibit a band present in the mother, and daughters exhibit one band present in the mother and one in the father (which may be identical if the daughter is homozygous). The genotypes are: 1-1, A 2 A 2 ; 1-2, AjY; H-l, A 2 Y; II-2, AjA 2 ; II-3, A 2 Y; II-4, A.Y; U-5, A, A • H-6, A.Y; III-l, A.Y; III-2, A.A III-3, A, A,; IH-4, A.Y. X I X 1 1 I X I X X Study Questions 3.51. Which of the following types of inheritance have the feature that an affected male has all affected sons but no affected daughters? A) Autosomal recessive B) Autosomal dominant C) Y-linked D) X-linked recessive E) X-linked dominant 3.52. The ribosome incorporates an essential RNA molecule called what? A) tRNA B) mRNA C) rRNA D) Circular RNA E) Guide RNA 3.53. If an organism has 12 chromosomes in each body cell, how many chromosomes would you expect to be present in the nucleus in telophase II of meiosis? A) 48 B) 24 C) 18______________________________________________________________________ D) 12 E) 6 3.54. Hemophilia is an X-linked trait. A woman whose father is affected married a nor- mal man and they have one son. What is the probability that the son has hemo- philia? A) 1 B) 2/3 C) 1/2 D) 1/16 E) 1/64 3.55. Among the F 2 progeny of a monohybrid cross, what is the number of A- and aa progeny whose probability of occurrence is given by the expression [601/(40120!)] x (3/4) 40 x (1/4) 20 ? A) 60 and 0 B) 30 and 30 C) 40 and 20 D) 20 and 40 E) 0 and 60 CHAPTER 3 The Chromosomal Basis of Heredity 11

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    3.56. In , the centromeres divide longitudinally and the sister chromatids of each chromosome, move toward opposite poles of the spindle, 3.57. Each nucleosome is composed of a , linker DNA, and one molecule of Hl that binds to the histone octamer and to the linker DNA. 3.58. A conventional measure of . between a set of observed numbers and the- oretical expectations is known as chi-square (% 2 ) , 3.59. Calculate the probability of having three boys and one girl in a family of five, 3310. Assuming a sex ratio of 1:1, what is the expected distribution of the sexes in sibships of size 5? Study Guide to accompany Essential Genetics: Genomics Perspective, Fourth Edition 12

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