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

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Study Guide and Solutions Manual to accompanyEssential•GeneticsA GENOMICS PERSPECTIVEFourth Edition...V,.£»IL.Daniel L. Hartl and Elizabeth W. JonesPrepared by Elena R. Lozovsky

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Contents1The Genetic Code of Genes and Genomes.................... ............... . . 12Transmission Genetics: Heritage from Mendel ........ ........................ 43The Chromosomal Basis of Heredity............ . .......—........ 94Gene Linkage and Genetic Mapping................................................... 135Human Chromosomes and Chromosome Behavior.............. .......... 186DNA Structure, Replication, and Manipulation.......................... . ... 227The Genetics of Bacteria and Their Viruses.......... . .........................268The Molecular Genetics of Gene Expression.............................. 309Molecular Mechanisms of Gene Regulation .....................3410 Genomics, Proteomics, and Genetic Engineering..................... 3811 The Genetic Control of Development..........................4212 Molecular Mechanisms of Mutation and DNA Repair ................4513 Molecular Genetics of the Cell Cycle and Cancer... .........4814 Molecular Evolution and Population Genetics............. .................... 5215 The Genetic Basisof Complex Inheritance.. ..................................... 55Appendix: Answers.............. . .....................................58v

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PrefaceTheStudyGuideandSolutionManualwas writtenspecificallyto accompanyEssentialGenetics:AGenomicPerspective,FourthEdition,by Daniel L. Hartl and ElizabethW.Jones. It has beenorganized into chapters that parallel those of the book and provides the answers to theConcepts in Action problems from the main text. These answers have been worked in fulland the reasoning explained in detail. The objective is to help the student develop skills inproblem solving.You,the student, arc strongly urged to try to work a problem before givingup and consulting the answer.Ifthe methodofsolution is not immediately apparent, donot go to the answer immediately. Think about it for a while, consult the correspondingchapter in the text, or perhaps come back to it later. Look up the answer only after a fewtries. With this approach, you will be pleased to learn that problems that at first look verydifficult arc often within yourcapability.A willingness to be patient, to let your mind workquietly and subconsciously on a problem before looking at the answer, pays rewardsofpos-itive feedback and increased self-confidence. Even when you are unable to solve a problemafter several attempts, a delay in looking at the answer will give the problem more impact,and the methodofsolution will be less likely to be forgotten.Manyofthe problems require the understanding and applicationofonly a single con-cept, such as Mendelian segregation or polypeptide translation. Other problems requireapplication of several concepts in logical order, for example, problems that combine segrega-tion and recombination or that consider the consequencesofa mutation on the amino acidsequenceofa 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 evenifyou areunsuccessful in solving them, you will gain additional understanding of genetics when youlook at the answer.TheStudy GuideandSolutionsManualalso provides a listofthe key concepts for eachchapter 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 orin combination to account for biological observations. Other questions arc quantitative andrequire a precise numerical answer. Problems requiring calculation make use of only simplearithmetic or algebra and draw upon the generally relevant lawsofprobability and statisticsexplained in the text. I encourage you not only to give a numerical answer but also to thinka problem through, in termsof"orderofmagnitude," to verity your approacn and methodof 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 theseproblems. 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 mailat elozovsk@oeb.harvard.edu.I am grateful to my colleagues, Erica Pantages and Emily Lilly for valuable ideas andhelpful discussions. I also wish to acknowledge Rebecca Seastrong, Stephen L. Weaver, andAnne Spencer of Jones and Bartlett for their support and assistance. Special thanks go toDan Hartl for his effort in editing, correcting errors, and suggesting problems. I am gratefulfor his encouragement and support that were absolutely vital for this project.Elena R. Lozovsky, PhDHarvard University

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The Genetic Code of Genesand GenomesKey Concepts•Inherited traits arc affected by genes.•Genes are composedofthe chemical deoxyribonucleic acid (DNA).•DNAreplicates to form (usually identical) copiesofitself.•DNAcontains a code specifying what typesofenzymes and other proteins are madein cells.•DNAoccasionally 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.•Traitsare affected by environment as well as by genes.• Organisms change genetically through generations in the processofbiologicalevolution.•Because of their common descent, organisms share many featuresoftheir geneticsand biochemistry.Key Terms1.complementary base pairing2.antiparallel3.central dogma4.transcription5.ribosome6.transfer RNA (tRNA)7. proteome8.substrate9.alkaptonuria10.phenylketonuria11.pleiotropic effect or pleiotropy12.prokaryoteConcepts in Action1.1.(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 duringcell division, and the observation that each species has a characteristic chromosomenumber 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 beavery complex molecule.Proteins were chemically the most complex macromolecules known at the time.DNA was thought to bea monotonouspolymer composed of a simple repeating unit.1.4.Because thematureT2 phage contains only DNA andprotein,the labeled RNA wasleft behind in material released by the burst cells.1.5.Watson and Cricknoted that the nucleotide sequence of the DNA molecule could bereplicated if each of the strands were used as a template for the formation ofanewdaughter strand having a complementary sequence of bases. They also noted thatgenetic 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, theynoted that changes in genetic information could result from errors in replication,and the altered nucleotide sequence could then beperpetuated.1.6.RNA differs from DNA in that the sugar-phosphatc backbone contains ribose ratherthan deoxyribose. RNA contains the base uracil (U) instead of thymine (T), and RNAusually exists as a single strand (although any particular molecule of RNA may containshort 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 thatpercentA = 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 eitherstrand contains 5'-GT-3', the other contains 3'-CA-5'.1.10.The repeating Asn results from translation in the reading frame5'-AAUAAUAAUAAU...-3', and the repeating He results from translation in thereading frame 5'-AUAAUAAUAAUA...-3'. There is no product corresponding tothe third reading frame (5'-UAAUAAUAAUAA...-3'), because 5'-UAA-3' is a stopcodon.1.11.Most likely themutant 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 ofthree nucleotides would shift the reading frame (phase) in which the mRNA istranslated. 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 is5'- UGUCGUAUUUGCAAG-3'.1.15.Cys-Arg-Ile-Cys-Lys or, using the single-letter abbreviations, CRICK.1.16.Cys-His-Ile-Cys-Lys, CHICK1.17.The codon 5'-UGG-3' codes for Trp, and in this random polymer the Trp codon isexpected with a frequency of 1/4 x 3/4 x 3/4 = 9/64. The amino acid Vai could bespecified by either 5'-GUU-3' or 5'-GUG-3'; the former has an expected frequencyof 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 firstAUG; translation will start with the next AUG farther along the mRNA or, if this istoo distant, not at all. (c) Met-Ser-Thr-AJa-Val-Leu-Glu-Asn-Pro-Gly; there is nochange, 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 because5'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, Yin excess.Study Guide to accompany Essential Genetics: A Genomics Perspective, Fourth Edition2

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1.20.The finding that the cells can grow in the presenceofYimplies that step C is func-tional. The finding that the cells cannot grow in the presenceof Ximplies that stepB is blocked. The results with W imply that a downstream step is blocked, but donot reveal which one.Study Questions1.S1.When the base composition of double-strandedDNAfrom a new species of bacteriawas determined, 15% of the bases were found to be cytosine. What is the percent-ageofadenine in theDNAof this organism?A)15%fSZCB)25%/S7.D)45%3 5 7 . TX1.S2.A duplexDNAmolecule contains a random sequence of the four nucleotides inequal proportions. What is the average spacing between consecutive occurrences ofthe sequence 5'-GGGG-3'?CACMmxcUjrticU.ocean"Q24D)421.53.The sequenceofone strand ofDNAis5'-CCATATGC-3'.The sequenceofthe com-plementary strand would be what?-AT5'-CCATATGC-3'((JBf5 -GGTATACG-3'GCA'T ft'VGG’3£j)5'-GCATATGG-3'-~&r5'-AAGCGCTA-3'5'-GCAUAUGC-3'£ N A1.54.AnRNAmolecule folds back uponitselfto form a "hairpin" structure held togetherby a region of base pairing. One segmentofthe molecule in the paired region hasthe 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 oftRNAmet5'-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 calledreverse transcriptasecan produce a complementary DNA strandfrom antemplate.1.58.If DNA from Baker's yeast has a guanine content of 25%, then what is the contentof the adenine?{1.S9.JIf a particular piece of RNA has a uracil content of 25%, then what is its guaninecontent?NIdA1A-(1.S10.How many different DNA sequences can encode the amino acid sequencePro-Met-Arg?CHAPTER 1 The Genetic Code of Genes and Genomes3

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Transmission Genetics:Heritage from MendelKey Concepts•Inherited traits are determined by the genes present in the reproductive cells unitedin fertilization.•Genes are usually inherited in pairs, one from the mother and one from the father.•The genes in a pair may differ inDNAsequence and in their effect on the expressionofa particular inherited trait.•The maternally and paternally inherited genes are not changed by being together inthe same organism.• hi the formationofreproductivecells,the paired genes separate again into different cells.• Random combinationsofreproductive cells containing different genes result inMendel's ratiosoftraits appearing among the progeny.• The ratios actually observed for any traits are determined by the types of dominanceand gene interaction.Key Terms1.allele2.wildtype3.genotype4.segregation5.testcross6.independent assortment7.incomplete dominance8.pedigree9.sibling10.epistasis11.complementation test12.variable expressivityConcepts in Action2.1.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 fromthe mother and one from the father.4

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2.2.ProgenyParentsAAAaaaAAXAA100AAXAa1/21/20AA Xaa010AaXAa1/41/21/4AaXaa01/21/2aaXaa0012.3.Ww; 1/4 wrinkled seeds are expected.2.4.The birth of an affected offspring implies that parents are at risk of another affectedoffspring. Once the genotypes of the parents have been deduced from their ownphenotypes and the fact that they have an affected offspring, the recurrence risk iscalculated as the probability of an affected offspring in the next birth, (a) The matingmust beAaxaa,and so the risk of anActoffspring in any birth (and therefore therecurrence risk) is 1/2. (b) The mating must beAaxAa,and so the risk of anaaoff-spring in any birth is 1/4. (c) The mating must beAaxAa,and so the risk of anaaoffspring 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 beaa.(b)1-1 and 1-2 are the—parents of the affected person and because neither is affected, their genotypes mustboth beAa.(c)II-I and II-3 are siblings of the affected person. Because they are notaffected, and because they result from the matingAaxAa,their possible genotypesareAAandAa.(d)From the matingAaxAa,the ratio ofAA.Aais 1:2; hence, theprobability that II-3 is a carrier (genotypeAa)equals 2/3.2.7.(2/3)2=4/9;(1/3)2=1/9; 1 - (1/3)2= 8/92.8.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 a2with the 1-kb deletion results in a band at3 kb-1 kb = 2 kb. DNA from each homozygous genotypes produces only one band,whereas that from each heterozygous genotype produces two bands.AAaia2a2AaiAa2aia25kb3kbIkb2.9.The recessive mutations are in different genes. They are not alleles. This situation isan example of complementation in human beings.2.10.The existing data enable us to group the mutants into three complementationgroups as follows:[a, c,d}, {b,f},and{e}.The missing entries are shown in theaccompanying table.a©G©©a©©©©©©©©©0©C2006 by Jones and Bartlett PublishersCHAPTER2Transmission Genetics: Heritage from Mendel5

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2.11.A 15:1 ratio is characteristic of two genes with independent assortment in whichthe minority trait is due to a double -homozygous recessive. Suppose that the geno-typeaa bbresults in ovoid-shaped fruit and that all other genotypes result in trian-gular fruit. The true-breeding parents would then have beenaa BBandAA bb.Onetest of this hypothesis is that the ovoid F2plants should be true-breeding.Furthermore, crossing the Fj triangular plants(Aa Bb)with the F2ovoid plants(aa bb)should yield progeny with the phenotypic ratio3triangular:1 ovoid.2.12.We expecta3:1 ratio of thedominantto the recessivephenotypewhen two het-erozygous genotypes are crossed, but in this case we observe 2:1. A cross of twoCy!+heterozygotes is expected to yield a progeny genotypic ratio of1CICy.2 Cy /+:1+/+. Because we only see two curly-winged flics for every wildtype fly, one possibleexplanation is that theCylCyhomozygotes die. In other words,Cyisdominantwithrespect to wing phenotype but recessive with respect to lethality.2.13.(a)Two phenotypic classes arc expected for each of theAaandBbpairs ofalleles, and three are expected for theRrpair, yielding a total number of 12.(b) 1/64. (c) 1/8.2.14.Let AABBrepresent the wildtype genotype with a phenotypeof a disc shape. Weare told that genotypesaa BBandAa bbboth have spherical fruit and thataa bbhaselongated fruit. The F2progeny are expected in the ratio9 A-B-:3A-bb:3 aa B-t1aa 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-erozygousHD/hd,wherehdrepresents the normal allele. Half of the father'sgametes contain themutantHDallele, so the probability is 1/2 that the son receivedthe allele and will later develop the disorder, (b) We do not knowwhetherthe sonis heterozygousHD/hd,but the probability is 1/2 that he is; if the son is heterozy-gous, half of his gametes will contain theHDallele. Therefore, the overall probabil-ity that the grandchild has theHDallele 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 oneboy include all families except those with all girls. Therefore, the simplest way toobtain the answer is to calculate the proportion of families with four girls and thensubtract this from 1. The probability of having four girls is (1/2)“, which equals0.0625; the rest of the families are those with at least one boy, which account forthe proportion1-0.0625 = 0.9375 of all families,(b)Because a particular birthorder is specified, the answer is (1/2)4= 1/16. To look at the problem in anotherway, note that although 6/16 of all families with four children have two boys andtwo girls, only 1/6 of such families have the specific birth order FMFM; hence, theanswer is, as before, 1/16.2.18.In the functional female gametes, the ratio ofAtais l / 2 : l / 2 because of Mendeliansegregation, In males, the products of meiosis in anAaindividual 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 ofAtaamong functional male gametes is therefore1:2 or, converting to proportions, 1/3 A:2/3a.The Punnet square shown here indi-cates that the F2ratio ofAAtAataais l/6:3/6:2/6 (or, reducing the fractions,l / 6 : l / 2 : l / 3 ) .Eggs1/2 A1/2 a1/3 A1/6 AA1/6 AaPollen---------------------2/3 a2/6 Aa2/6 aa2.19.(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 indicatesconsanguineous mating,(c)III-1 and III-2 are first cousins,(d)Either 1-1 or 1-2 areStudy Guide to accompany Essential Genetics: A Genomics Perspective, Fourth Edition6

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likely to be heterozygousAa(but not both, because the trait is said to be rare), andallofII-2, II-3,HI-1, and HI-2 are likely to beAa.On the other hand,II-1 and II-4are most likely to beAA.2.20.The accompanying Punnet squares indicate that the expected ratios of genotypes in(a)are 9/16DD:6/16Dd:1/16dd,an in(b)are 3/8DD:4/8Dd:1/8dd.(a)Eggs3/4D\ / d d(b)Eggs3/4D\ / A d3/4DSperm1/4d9/16DD3/\f>Dd1/2DSperm1/23/8 DD1/8Dd3/16Dd1/16dd3/8Dd1/8ddStudy Questions2.51.Twoparents with blood types A and B haveachild who has the O blood type.What is the chance that their next child will be O?A)0B)1/2C)1/4D)12.52.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 homozygousdominant for all four genes?A)1/2B)1/5C)(1/2)5D)(1/4)4E)(1/4)52.53.Among sibships consistingoffive children, and assuming a sex ratio of 1:1, what isthe proportion with no girls?A)1/2B)(1/2)5C)(1/5)5D)1-(1/2)52.54.Assuming equal sex ratios,ifa mating has already produced 5 boys, what is theprobability that the noxt child will be a boy?A)0B)1C)1/2D)(1/2)3E)1-(1/2)32.55.Assuming independent assortment, how many different gametes can be formed byan organism that is homozygous for 7 and heterozygous for 2 genes?A)2B)4C)32D)5E)62.56.In genetic analysis, the complementation test is used to determine whether tworecessive mutations that cause similar phenotype areof the same gene.2.57.Consider a family with six children, and remember that each birth is equally likelyto result in a boy or a girl. What proportion of sibships will include at least one girl?CHAPTER 2Transmission Genetics: Heritage from Mendel7

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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 isheterozygous for the gene?2.S9.In a testcross of Aa BBCcDd EeFF,where the genes show independent assortment:a.What is the expected frequencyofaa bb Cc dd ceFfprogeny?b.What is the expected frequencyofprogeny that are heterozygous for all six genes?2.S10.Assuming sex ratioof1:1,a.What is the probability that a couple will have eight boys?b.Ifthey already have seven boys, what is the probability that the eighth child willbe a boy?Study Guide to accompany Essential Genetics: A Genomics Perspective, Fourth Edition8

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The Chromosomal Basisof HeredityKey Concepts•Chromosomes in eukaryotic cells are usually present in pairs.•The chromosomesofeach pair separate in meiosis, one going to each gamete.•In meiosis, the chromosomesofdifferent pairs undergo independent assortment.•Chromosomes consist largely ofDNAcombined with histone proteins.• In many animals, sex is determined by a special pairofchromosomes, theXand Y.•Irregularities in the inheritanceofan X-linked gene inDrosophilagave experimentalproof ofthe chromosomal theoryofheredity.• The progenyofgenetic crosses follow the binomial probability formula.• The chi-square statistical test is used to determine how well observed genetic dataagree with expectations from a hypothesis.Key Terms1.synapsis2.kinetochore3.diplotete4.chiasma5.telomerase6.anaphase II7.histone8.chromosome territory9.heterochromatin10.nondisjunction11.chi-square12.significantConcepts in Action3.1.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;ifthere is cross-ing over between a gene and the centromere, they are separated at anaphase II. Ineither 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 beexpected 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 thesame 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 cellcycle. The reason is thatDNAreplication cannot begin at the extreme 3' endofatemplate strand. The function of telomerase is to restore the unreplicated region byelongating the 3' end after each round ofDNAreplication.3.7.Because female chickens areZWand males areZZ,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 beSs(silver plumage) and the female prog-eny will bes(gold plumage), and these are easily distinguished.3.8.Panel B is anaphaseofmitosis, because the homologous chromosomes are notpaired. Panel A is anaphase I of meiosis, because the homologous chromosomes arepaired. Panel C is anaphase IIofmeiosis, because the chromosome number hasbeen reduced by half.3.9.Assuming that she happens upon cells in each stage according to the lengthoftimethat an average cell spends in that stage, prophase takes up 320/2000 = 16.0%ofthetotal time, metaphase 150/2000-7.5%, anaphase 80/2000 = 4.0%, and telophase120/2000 = 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)cbrcbs;(b)cbrcbr;(c)cb9cb9.3.12.(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, hergenotype must beHh,wherehdenotes the hemophilia mutation. The most likelyexplanation of the 47,XXYhemophiliac son is that in meiosis in the mother, the-bearing chromatids underwent nondisjunction and produced an XX-bearing eggwith genotypehh.Fertilization by a normalY-bearingsperm resulted in the XXYzygoteofgenotypehh,which is the causeofthe 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 eitherofthe firsttwo is 5/7. Continuing in this manner, we find that the overall chance is 6/7 x 5/7 x4/7 x 3/7 x 2/7 x 1/7 = 0.00612, or less than 1 percent. The likelihoodofhavingevery gene on a different chromosome is quite small, and for this reason Mendel hasbeen accused of deliberately choosing unlinked genes. In fact, it is now known thatnot allofMendel's genes are on different chromosomes. Threeofthe genes(fa,deter-mining axial vs. terminal flowers;le,determining tall vs. short plants; andv,determin-ing smooth vs. constricted pods) arc all located on chromosome 4. Theleandvgenesundergo recombination at the rale of about 12 percent, but Mendel apparently didnot study this particular combination for independent assortment. Thefagene is verydistant from the other two and shows independent assortment with them.3.15.These are typical characteristics ofX-linked inheritance. Affected males transmit themutant X chromosome only to their daughters. A carrier female will have bothaffected and unaffected sons as well as both carrier and noncarrier daughters. Thecarrier daughters (sisters of the affected sons) will transmit the mutant gene, butthe 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 + (178-186)2/186 = 2.57. It has 3 degrees of freedom (four classes ofdata), and thePvalue 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 increasingchi-square value needed for significance docs not imply that one is less likely toobtain a statistically significant chi-square with increasing degrees of freedom. Theprobability of obtaining a statistically significant chi-square value, given that thegenetic hypothesis is true, is 5%, no matter what the number of degrees of freedom.10Study Guide to accompany Essential Genetics: Genomics Perspective, Fourth Edition

Solution Manual for Essential Genetics: A Genomic Perspective , 4th Edition - Page 15 preview image

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3.18.The chi-square value equals (462 - 450)2/450 + (167 - 150)2/150 + (127 -150)2/l 50 + (44 - 50)2/50 = 6.49. It has 3 degrees of freedom (four classes of data),and thePvalue 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. ThePvalue isabout 0.035, which is less than 0.05, hence there is reason to reject, at the 5% levelof 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 suggestsX-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 bandpresent in the mother and one in the father (which may be identical if the daughteris homozygous). The genotypes are: 1-1, A2A2; 1-2, AjY; H-l, A2Y; II-2, AjA2; II-3, A2Y;II-4, A.Y; U-5, A, A • H-6, A.Y; III-l, A.Y; III-2, A.AIII-3, A,A,; IH-4, A.Y.XI X11I XI XXStudy Questions3.51.Which of the following types of inheritance have the feature that an affected malehas all affected sons but no affected daughters?A)Autosomal recessiveB)Autosomal dominantC)Y-linkedD)X-linked recessiveE)X-linked dominant3.52.The ribosome incorporates an essential RNA molecule called what?A)tRNAB)mRNAC)rRNAD)Circular RNAE)Guide RNA3.53.If an organism has 12 chromosomes in each body cell, how many chromosomeswould you expect to be present in the nucleus in telophase II of meiosis?A)48B)24C)18______________________________________________________________________D)12E)63.54.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)1B)2/3C)1/2D)1/16E)1/643.55.Among the F2progeny of a monohybrid cross, what is the number ofA-andaaprogeny whose probability of occurrence is given by the expression [601/(40120!)] x(3/4)40x (1/4)20?A)60 and 0B)30 and 30C)40 and 20D)20 and 40E)0 and 60CHAPTER 3 The Chromosomal Basis of Heredity11

Solution Manual for Essential Genetics: A Genomic Perspective , 4th Edition - Page 16 preview image

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3.56.In, the centromeres divide longitudinally and the sister chromatids ofeach chromosome, move toward opposite poles of the spindle,3.57.Each nucleosome is composed of a, linker DNA, and one molecule ofHl 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 insibships of size 5?Study Guide to accompany Essential Genetics: Genomics Perspective, Fourth Edition12

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