Introduction to Genetic Analysis Eleventh Edition Solution Manual

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11TheGeneTicsRevoluTionPROBLEMSIn each chapter, a set of problems tests the reader’s comprehension of the concepts in the chapter andtheir relation to concepts in previous chapters. Each problem set begins with some problems based onthe figures in the chapter, which embody important concepts. These are followed by problems of a moregeneral nature.WORKING WITH THE FIGURES1.If the white-flowered parental variety in Figure 1-3 were crossed to the first-generation hybrid plantin that figure, what types of progeny would you expect to see and in what proportions?Answer: You would get a 1:1 ratio of purple to white. This is because the first-generation hybridplant has one copy of the purple allele and one copy of the white allele, and as a result, 50 percent ofthe gametes would carry the purple allele and 50 percent of the gametes would carry the white allele.The white-flowered parental variety has two copies of the white allele, and all the gametes producedfrom the white plant will carry the white allele. Hence, a cross between the two would produce a 1:1ratio of purple to white.Hybrid plantP/p¥white plantp/pGametes50%P50%p¥100%p50%P/p: 50%p/pPurple:white2.In Mendel’s 1866 publication as shown in Figure 1-4, he reports 705 purple (violet) floweredoffspring and 224 white-flowered offspring. The ratio he obtained is 3.15:1 for purple:white. Howdo you think he explained the fact that the ratio is not exactly 3:1?Answer: This depends on the sample size. When the sample size was large, the proportions wereclose to 3:1 (e.g., for round and wrinkled seeds the ratio was 2.95:1 and the total population sizeIGA 11e SM Ch 01.indd111/12/142:05 PM

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was 7324), whereas for a small sample size such as the purple and white petal flowered plants (929plants), the ratio was not as close to 3:1.3.In Figure 1-6, the students have 1 of 15 different heights plus there are two height classes (4 ft 11 inand 5 ft 0 in) for which there are no observed students. That is a total of 17 height classes. If a singleMendelian gene can only account for two classes of a trait (such as purple or white flowers), howmany Mendelian genes would be minimally required to explain the observation of 17 height classes?Answer: If a single gene can only account for two classes of a trait, minimum of 9 genes are requiredto explain the 17 height classes.4.Figure 1-7 shows a simplified pathway for arginine synthesis inNeurospora. Suppose you have aspecial strain ofNeurosporathat makes citrulline but not arginine. Which gene(s) are likely mutant ormissing in your special strain? You have a second strain ofNeurosporathat makes neither citrullinenor arginine but does make ornithine. Which gene(s) are mutant or missing in this strain?Answer: If the mutant strain makes citrulline, that means genes A and B must be functional. Therefore,the only gene that is missing or mutant in the firstNeurosporastrain must be gene C.In the second strain, gene A must be functional since it is able to make ornithine. Gene B must bemissing or mutant since it is unable to make citrulline. However, gene C may or may not be missing/mutant. Enzyme C converts citrulline into arginine (they are in the same sequential pathway), andenzyme C is dependent on the availability of citrulline for its function.5.Consider Figure 1-8a.a.What do the small blue spheres represent?b.What do the brown slabs represent?c.Do you agree with the analogy that DNA is structured like a ladder?Answer:a.The blue ribbon represents sugar phosphate backbone (deoxyribose and a phosphate group), whilethe blue spheres signify atoms.b.Brown slabs show complementary bases (A, T, G, and C).c.Yes, it is a helical structure.6.In Figure 1-8b, can you tell if the number of hydrogen bonds between adenine and thymine is thesame as that between cytosine and guanine? Do you think that a DNA molecule with a high contentof A + T would be more stable than one with a high content of G + C?Answer: There are two hydrogen bonds between adenine and thymine; three between guanine andcytosine. No, the molecule with a high content of G-C would be more stable.2CHAPTER 1The Genetics RevolutionIGA 11e SM Ch 01.indd211/12/142:05 PM

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7.Which of three major groups (domains) of life in Figure 1-11 is not represented by a model organism?Answer: Archaea8.Figure 1-13b shows the human chromosomes in a single cell. The green dots show the location of agene calledBAPX1. Is the cell in this figure a sex cell (gamete)? Explain your answer.Answer: It is not a sex cell. ClonedBAPXIgene has hybridized to two chromosomes in the cell,indicating there are two copies of theBAPXIgene. If it were a gamete, it would have only one copyof each chromosome and of theBAPXIgene.9.Figure 1-15 shows the family tree or pedigree for Louise Benge (Individual VI-1) who suffers fromthe disease ACDC because she has two mutant copies of theCD73gene. She has four siblings (VI-2, VI-3, VI-4, and VI-5) who have this disease for the same reason. Do all the 10 children of Louiseand her siblings have the same number of mutant copies of theCD73gene, or might this number bedifferent for some of the 10 children?Answer: All 10 children have one mutant copy of theCD73gene. Children get oneCD73copy fromtheir mom and one from their dad. Since Louise and her four siblings each carry two defective genes,all their children will get one mutantCD73copy.BASIC QUESTIONS10.Below is the sequence of a single strand of a short DNA molecule. On a piece of paper, rewrite thissequence and then write the sequence of the complementary strand below it.GTTCGCGGCCGCGAACComparing the top and bottom strands, what do you notice about the relationship between them?Answer:GTTCGCGGCCGCGAACCAAGCGCCGGCGCTTGThey are complementary to each other and run in the opposite direction (antiparallel). The sequencesare also palindromic; they read the same in either direction.11.Mendel studied atallvariety of pea plants with stems that are 20 cm long and adwarfvariety withstems that are only 12 cm long.a.Under blending theory, how long would you expect the stems of first and second hybrids to be?b.Under Mendelian rules, what would you expect to observe in the second-generation hybrids if allthe first-generation hybrids were tall?3CHAPTER 1The Genetics RevolutionIGA 11e SM Ch 01.indd311/12/142:05 PM

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Answer:a.First-generation hybrids would have stems 16 cm long because that is the average between 20 cmand 12 cm. The second generation would be a product of two 16-cm stemmed plants mating, sothey would also have stems 16 cm long.b.If the first-generation hybrids are all tall, then tall must be dominant, and you would expect a 3:1ratio of tall:dwarf in the second generation.12.If a DNA double helix that is 100 base pairs in length has 32 adenines, how many cytosines, guanines,and thymines must it have?Answer: Thymines = 32, Cytosines = 68, Guanines = 68Number of thymines is 32 because thymine and adenine are paired, so the number of adenines equalsthe number of thymines. The remaining 136 base pairs must be cytosines and guanines. The numberof cytosines equals the number of guanines because those bases are paired. Therefore, there are 136÷ 2 = 68 base pairs of each.13.The complementary strands of DNA in the double helix are held together by hydrogen bonds: G≡Cor A=T. These bonds can be broken (denatured) in aqueous solutions by heating to yield two singlestrands of DNA (see Figure 1-13a). How would you expect the relative amounts of GC versus ATbase pairs in a DNA double helix to affect the amount of heat required to denature it? How wouldyou expect the length of a DNA double helix in base pairs to affect the amount of heat required todenature it?Answer: Double-stranded DNA with a high GC content would be more stable because GC pairshave three hydrogen bonds and hence would require more heat to denature compared with a double-stranded DNA with high AT content.As the length of the double helix increases, the heat required to denature would also increase. This isbecause there would be more hydrogen bonds to break.14.The figure that follows shows the DNA sequence of a portion of one of the chromosomes from a trio(mother, father, and child). Can you spot any new point mutations in the child that are not in eitherparent? In which parent did the mutation arise?4CHAPTER 1The Genetics RevolutionIGA 11e SM Ch 01.indd411/12/142:05 PM

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Answer: There is a new point mutation from an A to a C. The child inherited copy F1 from the fatherwithout any de novo mutations, as shown on the bottom. The other copy (as shown on the top) musttherefore be from the mother. The mother has an SNP (A/T) at the fifth nucleotide shown, so we knowthe child inherited copy M1 (A). The A-to-C mutation in copy M1 is not observed in either of theparents, so it must be a new mutation.CHALLENGING PROBLEMS15.a.There are three nucleotides in each codon, and each of these nucleotides can have one of fourdifferent bases. How many possible unique codons are there?b.If DNA had only two types of bases instead of four, how long would codons need to be to specifyall 20 amino acids?Answer:a.There are 64 unique codons (4¥4¥4 = 64).b.In order to specify 20 amino acids using only two bases, a codon must be five bases long (2¥2¥2¥2¥2 = 32). This would give 32 unique codons that could specify 20 amino acids.16.Fathers contribute more new point mutations to their children than mothers. You may know fromgeneral biology that people have sex chromosomes—two X chromosomes in females and an X plusa Y chromosome in males. Both sexes have the autosomes (As).a.On which type of chromosome (A, X, or Y) would you expect the genes to have the greatestnumber of new mutations per base pair over many generations in a population? Why?b.On which type of chromosome would you expect the least number of new mutations per basepair? Why?c.Can you calculate the expected number of new mutations per base pair for a gene on the X and Ychromosome for every 1 new mutation in a gene on an autosome if the mutation rate in males istwice that in females?Answer:a.Y chromosome. All, or 100 percent of, Y chromosomes are passed through the male lineage viasperm. The production of sperm (spermatogenesis) continues throughout a man’s life, undergoingmany rounds of cell division. Each round of cell division include DNA replication, which providesopportunity for new mutations, so the male lineage has a higher mutation rate than the femalelineage, as explained in the text.b.X chromosome. Two out of every three X chromosomes (66.7 percent) are transmitted through thefemale lineage, which has a lower mutation rate than the male lineage. By comparison, 50 percentof autosomes are inherited through females with the lower mutation rate and 50 percent throughmales with the higher mutation rate.c.Set the male mutation rate (mM) as 2.0 per unit time and the female mutation rate (mF) as 1.0 perunit time. Since autosomes are transmitted equally through males and females, then the autosomalrate is the average of the male and female rates:(mA) = (2 + 1)/2 = 1.55CHAPTER 1The Genetics RevolutionIGA 11e SM Ch 01.indd511/12/142:05 PM

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Since genes on the Y are transmitted only through males, the Y rate equals the male rate:mY=mM= 2Since 1/3 of X chromosomes are inherited through males and 2/3 through females:mX= (1/3mM) + (2/3mF) = (1/3¥2) + (2/3¥1) = 1.33The expected number of new mutation on the Y for every 1 on an autosome is 2/1.5 = 1.33. Theexpected number of new mutation on the X for every 1 on an autosome is 1.33/1.5 = 0.889.17.For young men of age 20, there have been 150 rounds of DNA replication during sperm productionas compared with only 23 rounds for a woman of age 20. That is a 6.5-fold greater number of celldivisions and proportionately greater opportunity for new point mutations. Yet, on average, 20-year-old men contribute only about twice as many new point mutations to their offspring as do women.How can you explain this discrepancy?Answer: While experimental evidence to explain this observation are not available, one hypothesis isthat sperm cells are physiologically weak and their normal function is easily disrupted by mutations.Thus, sperm that carry deleterious new mutations are less likely to survive and form a zygote withan egg cell. Therefore, many sperm with new mutations are eliminated prezygotically.18.In computer science, a bit stores one of two states, 0 or 1. A byte is a group of 8 bits, which has 28= 256 possible states. Modern computer files are often mega-bytes (106bytes), or even giga-bytes(109bytes), in size. The human genome is approximately 3 billion base pairs in size. How manynucleotides are needed to encode a single byte? How large of a computer file would it take to storethe same amount of information as a single human genome?Answer: Four nucleotides, with four possible states (A, T, C, G) can encode 4¥4¥4¥4 = 44= 256= 1 byte. If one byte can encode four nucleotides, then 3¥109nucleotides can be encoded in (3¥109) / 4 = 7.5¥108bytes = 750¥106, or 750 megabytes.19.The human genome is approximately 3 billion base pairs in size.a.Using standard 8.5-in¥11-in paper with one-inch margins, a 12-point font size and single-spacedlines, how many sheets of paper printed on one side would be required to print out the humangenome?b.A ream of 500 sheets of paper is about 5 cm thick. How tall would the stack of paper with theentire human genome be?c.Would you want a backpack, shopping cart, or a semi trailer truck to haul around this stack?Answer:a.Assuming a single one-sided page can fit 23 lines, with 56 letters in each line, 23¥56 = 1288letters per sheet. The human genome is 3¥109base pairs, so the number of sheets required to printout the entire human genome = 3¥109/1288 = 2,329,192 sheets of paper.6CHAPTER 1The Genetics RevolutionIGA 11e SM Ch 01.indd611/12/142:05 PM

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b.If a ream of 500 sheets = 5 cm thick, the thickness of 2,329,192 sheets = 5 cm/500 sheets¥2,329,192 sheets = 23,291 cm = 233 meters.c.Semitrailer truck7CHAPTER 1The Genetics RevolutionIGA 11e SM Ch 01.indd711/12/142:05 PM

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82Single-geneinheritanceWORKING WITH THE FIGURES1.Intheleft-handpartofFigure2-4,theredarrowsshowselfingaspollinationwithinsingleflowersof one F1plant. Would the same F2results be produced by cross-pollinating two different F1plants?Answer: No, the results would be different. While self-pollination produces 3:1 ratio of yellow ver-sus gene phenotype, cross-pollination would result in 1:1 ratio in the F2. This is because F1yelloware heterozygous, while green are homozygous genotypes.2.In the right-hand part of Figure 2-4, in the plant showing an 11:11 ratio, do you think it would bepossibletofindapodwithallyellowpeas?Allgreen?Explain.Answer:Yes,itispossibletofindcompletepodswithonlyyellowpeasoronlygreenpeasfromthecross shown, though it would be highly unlikely. It would have the same likelihood of occurring asflippingacoinandgettingheadssixtimesinarow.3.In Table 2-1, state the recessive phenotype in each of the seven cases.Answer:wrinkledseeds;greenseeds;whitepetals;pinchedpods;yellowpods;terminalflowers;short stems4.ConsideringFigure2-8,isthesequence“pairing→replication→segregation→segregation”agood shorthand description of meiosis?Answer:No,itshouldsayeither“pairing,recombination,segregation,segregation”or“replication,pairing,segregation,segregation.”5.Point to all cases of bivalents, dyads, and tetrads in Figure 2-11.Answer: Replicate sister chromosomes or dyads are at any chromatid after the replication (S phase).A pair of synapsed dyads is called a bivalent, and it would represent two dyads together (sisterIGA 11e SM Ch 02.indd811/12/142:30 PM

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chromatids on the right), while the four chromatids that make up a bivalent are called a tetrad, andthey would be the entire square (with same or different alleles on the bivalents).6.In Figure 2-11, assume (as in corn plants) that alleleAencodes an allele that produces starch in pol-len and alleleadoes not. Iodine solution stains starch black. How would you demonstrate Mendel’sfirstlawdirectlywithsuchasystem?Answer: One would use this iodine dye to color the starch-producing corn pollen. Since pollen is aplantgametophytegeneration(haploid),itwillbeproducedbymeiosis.Mendel’sfirstlawpredictssegregationofallelesintogametes;therefore,wewouldexpect1:1ratioofstarch-producing(A) ver-sus non-starch-producing (a) pollen grains, from a heterozygous (A/a)parent/maleflower.Itwouldbe easy to color the pollen and count the observed ratio.7.Considering Figure 2-13, if you had a homozygous double mutantm3/m3m5/m5,wouldyouex-pect it to be mutant in phenotype? (Note:This line would have two mutant sites in the same codingsequence.)Answer: Yes, this double mutantm3/m3m5/m5 would be a null mutation becausem3 mutationchangestheexonsequence.Becausethem5mutation is silent, the homozygous double mutantm3/m3 m5/m5would have the same mutant phenotype as anm3/m3double mutant.8.In which of the stages of theDrosophilalifecycle(representedintheboxonpage56)wouldyoufindtheproductsofmeiosis?Answer: Meiosis happens in adult ovaries and testes to produce gametes (sperm and unfertilizedegg).Thus,theadultflyinthediagramwouldgenerategametesandparticipateinmating,andthefemale would then lay the fertilized diploid embryos (eggs).9.If you assume Figure 2-15 also applies to mice and you irradiate male sperm with X rays (known toinactivategenesviamutation),whatphenotypewouldyoulookforinprogenyinordertofindcasesof individuals with an inactivatedSRYgene?Answer: Individuals with an inactivatedSRYgenewouldbephenotypicallyfemalewithanXYsexchromosomalmakeup.Theseindividualsareoftencalled“sexreversed”andarealwayssterile.Hence,wewouldlookforfliesthatarephenotypicallyfemale,butsterile.10.In Figure 2-17, how does the 3:1 ratio in the bottom-left-hand grid differ from the 3:1 ratios obtainedby Mendel?Answer:Itdiffersbecause,inMendel’sexperiments,welearnedaboutautosomalgenes,whileinthiscase,wehaveasex-linkedgeneforeyecolor.9CHAPTER 2Single-Gene InheritanceIGA 11e SM Ch 02.indd911/12/142:30 PM

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3:1 ratio means that all females have red eyes (X+/–), while half the males have red (X+/Y) and halfhave white (XW/Y).CarefulsexdeterminationwhencountingF2offspringwouldpointouttoasex-linked trait.11.In Figure 2-19, assume that the pedigree is for mice, in which any chosen cross can be made. If youbredIV-1withIV-3,whatistheprobabilitythatthefirstbabywillshowtherecessivephenotype?Answer: 2/3¥2/3¥1/4 = 1/9, or 0.11The probability that IV-1 and IV-3 mice are heterozygous is 2/3. This is because both of their parentsare known heterozygotes (A/a), and since they are the dominant phenotype, they could only beA/AorA/a. Now the probability that two heterozygotes have a recessive homozygote offspring is 1/4.12.WhichpartofthepedigreeinFigure2-23inyouropinionbestdemonstratesMendel’sfirstlaw?Answer: Any part of this pedigree demonstrates the law, showing segregation of alleles into gametes.Notice how 50 percent of the children of I-1 and I-2 display the dominant trait, consistent with I-1contributing either the dominant or the recessive allele to each gamete in equal frequencies (1:1).ThemiddlepartofgenerationIImarriageshowsatypicaltestcross(expected1:1).Neitherratiointhepedigreecouldbeconfirmedbecauseofasmallsamplesizeinanygivenfamily,butalleleseg-regation is obvious.13.CouldthepedigreeinFigure2-31beexplainedasanautosomaldominantdisorder?Explain.Answer:Yes,itcouldinsomecases,butinthiscasewehavecluesthatthepedigreeisforasex-linked dominant trait. First, if fathers have a gene, only daughters would receive it; and second, ifmothers have a gene, both sons and daughters would receive it.BASICPROBLEMS14.Make up a sentence including the wordschromosome,genes,andgenome.Answer: The human genome contains an estimated 20,000–25,000 genes located on 23 differentchromosomes.15.Peas (Pisum sativum) are diploid and 2n= 14. InNeurospora,the haploid fungus,n= 7. If it werepossibletofractionategenomicDNAfrombothspeciesbyusingpulsedfieldelectrophoresis,howmany distinct DNA bands would be visible in each species?Answer:PFGEseparatesDNAmoleculesbysize.WhenDNAiscarefullyisolatedfromNeurospora(which has seven different chromosomes), seven bands should be produced using this technique.10CHAPTER 2Single-Gene InheritanceIGA 11e SM Ch 02.indd1011/12/142:30 PM

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Similarly, the pea has seven different chromosomes and will produce seven bands (homologouschromosomes will co-migrate as a single band).16.The broad bean (Vicia faba) is diploid and 2n=18.Eachhaploidchromosomesetcontainsapproxi-mately 4 m of DNA. The average size of each chromosome during metaphase of mitosis is 13mm.What is the average packing ratio of DNA at metaphase? (Packing ratio = length of chromosome/length of DNA molecule therein.) How is this packing achieved?Answer: There is a total of 4 m of DNA and nine chromosomes per haploid set. On average, each is4/9m long. At metaphase, their average length is 13mm, so the average packing ratio is 13¥10–6m:4.4¥10–1m, or roughly 1:34,000! This remarkable achievement is accomplished through theinteraction of the DNA with proteins. At its most basic, eukaryotic DNA is associated with histonesin units called nucleosomes, and during mitosis, coils into a solenoid. As loops, it associates with andwinds into a central core of nonhistone protein called the scaffold.17.If we call the amount of DNA per genome “x,”nameasituationorsituationsindiploidorganismsinwhich the amount of DNA per cell is:a.xb.2xc.4xAnswer:BecausetheDNAlevelsvaryfour-fold,therangecoverscellsthatarehaploid(gametes)tocells that are dividing (after DNA has replicated but prior to cell division). The following cells wouldfittheDNAmeasurements:a.xhaploid cellsb.2xdiploid cells in G1or cells after meiosis I but prior to meiosis IIc.4xdiploid cells after S but prior to cell division18.Name the key function of mitosis.Answer: The key function of mitosis is to generate two daughter cells that are genetically identicalto the original parent cell.19.Name two key functions of meiosis.Answer:TwokeyfunctionsofmeiosisaretohalvetheDNAcontentandtoreshufflethegeneticcontent of the organism to generate genetic diversity among the progeny.20.Design a different nuclear-division system that would achieve the same outcome as that of meiosis.Answer: It’s pretty hard to beat several billion years of evolution, but it might be simpler if DNAdid not replicate prior to meiosis. The same events responsible for halving the DNA and producinggenetic diversity could be achieved in a single cell division if homologous chromosomes paired,recombined, randomly aligned during metaphase, and separated during anaphase, etc. However, youwould lose the chance to check and repair DNA that replication allows.11CHAPTER 2Single-Gene InheritanceIGA 11e SM Ch 02.indd1111/12/142:30 PM

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21.In a possible future scenario, male fertility drops to zero, but, luckily, scientists develop a way forwomen to produce babies by virgin birth. Meiocytes are converted directly (without undergoingmeiosis) into zygotes, which implant in the usual way. What would be the short-term and long-termeffects in such a society?Answer:Inlargepart,thisquestionisasking:Whysex?Parthenogenesis(theabilitytoreproducewithout fertilization—in essence, cloning) is not common among multicellular organisms. Parthe-nogenesisoccursinsomespeciesoflizardsandfishesandseveralkindsofinsects,butitistheonlymeans of reproduction in only a few of these species. In plants, about 400 species can reproduceasexuallybyaprocesscalledapomixis.Theseplantsproduceseedswithoutfertilization.However,themajorityofplantsandanimalsreproducesexually.Sexualreproductionproducesawidevarietyof different offspring by forming new combinations of traits inherited from both the father and themother.Despitethenumericaladvantagesofasexualreproduction,mostmulticellularspeciesthathaveadopteditastheironlymethodofreproducinghavebecomeextinct.However,thereisnoagreed-uponexplanationofwhythelossofsexualreproductionusuallyleadstoearlyextinction,orconversely,whysexualreproductionisassociatedwithevolutionarysuccess.Ontheotherhand,theimmediate effects of such a scenario are obvious. All offspring would be genetically identical to theirmothers,andmaleswouldbeextinctwithinonegeneration.22.In what ways does the second division of meiosis differ from mitosis?Answer: As cells divide mitotically, each chromosome consists of identical sister chromatids that areseparated to form genetically identical daughter cells. Although the second division of meiosis ap-pearstobeasimilarprocess,the“sister”chromatidsarelikelytobedifferent.Recombinationduringearlier meiotic stages has swapped regions of DNA between sister and nonsister chromosomes suchthat the two daughter cells of this division typically are not genetically identical.23.MakeupmnemonicsforrememberingthefivestagesofprophaseIofmeiosisandthefourstagesofmitosis.Answer:Thefourstagesofmitosisareprophase,metaphase,anaphase,andtelophase.Thefirstlet-ters, PMAT, can be remembered by a mnemonic such as Playful Mice Analyze Twice.ThefivestagesofprophaseIareleptotene,zygotene,pachytene,diplotene,anddiakinesis.Thefirstletters, LZPDD, can be remembered by a mnemonic such as Large Zoos Provide Dangerous Distrac-tions.24.Inanattempttosimplifymeiosisforthebenefitofstudents,madscientistsdevelopawayofpre-venting premeiotic S phase and making do with having just one division, including pairing, crossingover, and segregation. Would this system work, and would the products of such a system differ fromthose of the present system?Answer: Yes, it could work, but certain DNA repair mechanisms (such as postreplication recombina-tion repair) could not be invoked prior to cell division. There would be just two cells as products ofthis meiosis, rather than four.12CHAPTER 2Single-Gene InheritanceIGA 11e SM Ch 02.indd1211/12/142:30 PM

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25.TheodorBoverisaid,“Thenucleusdoesn’tdivide;itisdivided.”Whatwashegettingat?Answer: The nucleus contains the genome and separates it from the cytoplasm. However, duringcell division, the nuclear envelope dissociates (breaks down). It is the job of the microtubule-basedspindle to actually separate the chromosomes (divide the genetic material) around which nucleireform during telophase. In this sense, it can be viewed as a passive structure that is divided by thecell’s cytoskeleton.26.Francis Galton, a geneticist of the pre-Mendelian era, devised the principle that half of our geneticmakeup is derived from each parent, one-quarter from each grandparent, one-eighth from each great-grandparent,andsoforth.Washeright?Explain.Answer: Yes, half of our genetic makeup is derived from each parent, each parent’s genetic makeupisderivedhalffromeachoftheirparents,etc.Theprocessisabitmorecomplexwhenoneconsidersthe recombination of homologous chromosomes in prophase I, as is discussed in later chapters.27.If children obtain half their genes from one parent and half from the other parent, why aren’t siblingsidentical?Answer:Becausethe“half”inheritedisveryrandom,thechancesofreceivingexactlythesamehalfis vanishingly small. Ignoring recombination and focusing just on which chromosomes are inheritedfrom one parent, there are 223=8,388,608possiblecombinations!28.Statewherecellsdividemitoticallyandwheretheydividemeioticallyinafern,amoss,afloweringplant,apinetree,amushroom,afrog,abutterfly,andasnail.Answer:UMitosisUMeiosisfernsporophyte gametophyte(sporangium)mosssporophytegametophytesporophyte (antheridium and archegonium)plantsporophyte gametophytesporophyte (anther and ovule)pine treesporophyte gametophytesporophyte (pine cone)mushroomsporophyte gametophytesporophyte (ascus or basidium)frogsomatic cellsgonadsbutterflysomatic cellsgonadssnailsomatic cellsgonads29.Humancellsnormallyhave46chromosomes.Foreachofthefollowingstages,statethenumberofnuclear DNA molecules present in a human cell:a.Metaphase of mitosisb.Metaphase I of meiosisc.Telophase of mitosis13CHAPTER 2Single-Gene InheritanceIGA 11e SM Ch 02.indd1311/12/142:30 PM

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d.Telophase I of meiosise.Telophase II of meiosisAnswer:Thisproblemistrickybecausetheanswersdependonhowacellisdefined.Ingeneral,geneticists consider the transition from one cell to two cells to occur with the onset of anaphase inboth mitosis and meiosis, even though cytoplasmic division occurs at a later stage.a.46chromosomes,eachwithtwochromatids=92chromatidsb.46chromosomes,eachwithtwochromatids=92chromatidsc.46physicallyseparatechromosomesineachoftwoabout-to-be-formedcellsd.23 chromosomes in each of two about-to-be-formed cells, each with two chromatids =46chromatidse.23 chromosomes in each of two about-to-be-formed cells30.Four of the following events are part of both meiosis and mitosis, but only one is meiotic. Whichone? (1) chromatid formation, (2) spindle formation, (3) chromosome condensation, (4) chromo-some movement to poles, (5) synapsisAnswer: (5) chromosome pairing (synapsis)31.Incorn,thealleleƒ´causesflouryendospermandtheallelef´´causesflintyendosperm.Inthecrossƒ´/ƒ´♀¥ƒ´´/ƒ´´♂,alltheprogenyendospermsarefloury,butinthereciprocalcross,alltheprogenyendospermsareflinty.Whatisapossibleexplanation?(CheckthelegendforFigure2-7.)Answer:First,examinethecrossesandtheresultinggenotypesoftheendosperm:FemaleMalePolar nucleiSpermEndospermƒ´/ƒ´ƒ´´/ƒ´´ƒ´ and ƒ´ƒ´´/ƒ´´ƒ´/ƒ´/ƒ´´(floury)ƒ´´/ƒ´´ƒ´/ƒ´ƒ´´ and ƒ´´ƒ´/ƒ´ƒ´´/ƒ´´/ƒ´(flinty)As can be seen, the phenotype of the endosperm correlates to the predominant allele present.32.WhatisMendel’sfirstlaw?Answer:Mendel’sfirstlawstatesthatallelessegregateintogametesduringmeiosis.Thisdiscoverycamefromhismonohybridexperimentalcrosses.33.Ifyouhadafruitfly(Drosophila melanogaster) that was of phenotypeA,what test would you maketodetermineifthefly’sgenotypewasA/AorA/a?Answer: Do a testcross (cross toa/a).IftheflywasA/A, all the progeny will be phenotypicallyA; iftheflywasA/a, half the progeny will beAand half will bea.14CHAPTER 2Single-Gene InheritanceIGA 11e SM Ch 02.indd1411/12/142:30 PM

Introduction to Genetic Analysis Eleventh Edition Solution Manual - Page 16 preview image

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34.Inexaminingalargesampleofyeastcoloniesonapetridish,ageneticistfindsanabnormal-lookingcolony that is very small. This small colony was crossed with wild type, and products of meiosis(ascospores) were spread on a plate to produce colonies. In total, there were 188 wild-type (normal-size) colonies and 180 small ones.a.What can be deduced from these results regarding the inheritance of the small-colony pheno-type? (Invent genetic symbols.)b.What would an ascus from this cross look like?Answer:a.Adiploidmeiocytethatisheterozygousforonegene(forexample,s+/s,wheresis the allele thatconfers the small colony phenotype) will, after replication and segregation, give two meioticproducts of genotypes+and two ofs. If the random spores of many meiocytes are analyzed, youwouldexpecttofindabout50percentnormal-sizecoloniesand50percentsmallcoloniesiftheabnormal phenotype is the result of a mutation in a single gene. Thus, the actual results of 188normal-size and 180 small-size colonies support the hypothesis that the phenotype is the resultof a mutation in a single gene.b.The following represents an ascus with four spores. The important detail is that two of the sporesaresand will generate small colonies, and two ares+and will generate normal colonies.35.Two black guinea pigs were mated and over several years produced 29 black and 9 white offspring.Explaintheseresults,givingthegenotypesofparentsandprogeny.Answer:Theprogenyratioisapproximately3:1,indicatingclassicheterozygous-by-heterozygousmating. Since black (B) is dominant to white (b):Parents:B/b¥B/bProgeny:3 black:1 white (1B/B:2B/b:1b/b)This ratio indicates that black parents were probably heterozygous and that black is dominant overwhite.36.In a fungus with four ascospores, a mutant allelelys-5causes the ascospores bearing that allele to bewhite, whereas the wild-type allelelys-5+results in black ascospores. (Ascospores are the spores thatconstitute the four products of meiosis.) Draw an ascus from each of the following crosses:a.lys-5¥lys-5+b.lys-5¥lys-5c.lys-5+¥UlysU-5+15CHAPTER 2Single-Gene InheritanceIGA 11e SM Ch 02.indd1511/12/142:30 PM

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