Genetics: A Conceptual Approach, 6th Edition Solution Manual

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6eChapter 1Think-Pair-Share questions for the chapter opening story:Albinism occupied a special place in the Hopi culture; individuals who possessed thistrait were valued by members of the tribe. What are some examples of genetic traits that,in contrast, sometimes result in discrimination and prejudice?Possible Answers: People with many genetic traits are exposed to discrimination andprejudice. Indeed, people with albinism in cultures other than the Hopis are subjected todiscrimination for the way they look. In the past, African Americans who wereheterozygous carriers of the mutation for sickle-cell anemia were subject todiscrimination in employment and insurance, even though they were perfectly healthy.Skin color is a genetically determined characteristic and there is discrimination againstpeople with certain skin colors in some societies. Students may have witnesseddiscrimination against people with a variety of genetically determined characteristics.Albinism in humans can be caused by mutations in any one of several different genes.This situation, in which the same phenotype may result from variation in several differentgenes, is referred to as genetic heterogeneity. Is genetic heterogeneity common? Are mostgenetic traits in humans the result of variation in a single gene, or are there many genetictraits that result from variation in several genes, as albinism does?Possible Answers: Beginning genetics students often have the misconception that mosttraits are caused by a single variation at a single gene, much like the traits Mendel studiedin his pea plants. This question provides an opportunity at the beginning of the course todiscuss this misconception and the fact that the genetic basis of most traits is morecomplex.Many genetic traits and diseases exhibit genetic heterogeneity, where DNA sequences atany one of several different genes can result in the same phenotype. For example,albinism in the Hopis is caused by mutations in theOCA2gene, but mutations at severalother genes may produce albinism in other populations. Genetic heterogeneity can alsoresult from different mutations at the same gene. Both types of heterogeneity are seen in

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6ehuman albinism. Many genetic traits are also the result of combined variation at severalgenes, a situation referred to as a polygenic trait (see Chapters 5 and 24).End-of-chapter Think-Pair-Share questions:Section 1.11.Bob says that he is healthy and has no genetic diseases such as hemophilia or Downsyndrome. Therefore, genetics, he says, plays little role in his life. Do you think Bob iscorrect in his conclusion? Why or why not?Possible Answers: Bob makes a common mistake when he thinks of genetic effects onlyas classical genetic diseases, like hemophilia and Down syndrome. In fact, there aremany common disorders and diseases, as well as variations in healthy phenotypes, thatare influenced by genetics. In many of these cases, the phenotype is the result of avariation in multiple genes (the trait is polygenic) and the interaction of genes andenvironment (see Chapter 5). For example, common diseases, such as asthma, coronaryheart disease, hypertension, depression, and diabetes, are due to a combination ofvariations at multiple genes and environmental factors. Many aspects of our appearance,such as height, weight, facial features, hair color, skin color, and eye color, are alsodetermined at least to some degree by genetic variation.2.Are mutations good or bad? Explain your answer.Possible Answers: This question will likely elicit interesting responses from students.Mutations might be seen as both good and bad. On the one hand, mutations provide thenecessary genetic variation necessary for evolution, the process by which populationschange and become adapted to their environments. Adaptation and evolutionary changeallow populations to avoid extinction when environments change. So, mutations can bebeneficial to the long-term success of a population. Mutations result in individualdifferences, which makes each of us unique. Also, mutations are often useful in geneticresearch, as we will see in our journey through the field of genetics. On the other hand,when mutations result in genetic diseases, they may cause pain and suffering. Thus, many

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6emutations also have detrimental effects.Section 1.23.Do you support or oppose the development of genetically engineered foods (geneticallymodified organisms, or GMOs)? Find someone who takes the opposite position anddiscuss this question with them. Think about the economic, environmental, health, andsocial effects of their use. List some reasons for and against genetically engineering thefoods we eatPossible Answers: There are many possible answers to this question.Some reasons for supporting the genetic engineering of foods include:Ability to provide more food to feed the world’s growing populationLowered use of pesticides and other chemicals that have negative health andecological effectsAbility to produce foods at lower costAbility to create crops that can grow in otherwise unsuitable habitats, such as aridregions and saline soilsSome reasons for opposing genetically engineered foods include:Unanticipated health risksPotential ecological damage caused by release of novel genetically modifiedorganisms into the environment.Negative economic effects on small farmers who cannot afford the cost ofgenetically engineered crops and animals.Section 1.34.Why do you think all organisms use nucleic acids for encoding genetic information? Whynot use proteins or carbohydrates? What advantages might DNA have as the source ofgenetic information?

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6ePossible Answers: One potential answer is that all organisms use nucleic acids becauseall living organisms are derived from a common ancestor, which just happened to usenucleic acids as its genetic information. However, there may also be reasons why nucleicacids are advantageous as information carriers. Nucleic acids have the capacity to vary intheir base sequence, but they don’t vary too much. (There are only four common basesused in DNA and RNA.) They have the ability to form long polymers, which means theycan encode a great deal of genetic information. DNA is particularly stable because of itsdouble-stranded nature and lack of any free hydroxyl groups. The double-stranded natureof DNA also provides a particularly efficient means of replicating the molecule: thecomplementary nature of the bases (G and C pair; A and T pair) means that each strandof DNA can serve as a template for the synthesis of a new strand.Chapter 2Think-Pair-Share questions for the chapter opening story:In the blind men’s riddle, two blind men must sort out 10 pairs of socks so that each mangets exactly five pairs of different colored socks. In the analogy, is it important that themen are blind? In a cell, what does the blindness represent?Possible Answers: The analogy depends on the men being blind. If they could see theycould distinguish the socks by color and easily sort the socks so that each man got twosocks of each color. While theoretically it might be possible for cells to evolve somemechanism by which they distinguish each pair of chromosomes and by thisdiscrimination ensure that one chromosome of each pair ends up in a resulting cell, cellshave not evolved such a mechanism. Through the process of mitosis, each chromosomereplicates and the two replicated copies (analogous to the two socks of a pair) get pulledin opposite directions and end up in separate cells. The cells don’t have a mechanism tosee the different types of chromosomes and ensure at the end that each daughter cell getsexactly one copy of each pair—this is what the blindness represents.

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6eEnd-of-chapter Think-Pair-Share questions:Section 2.21.A chromosome consists of two sister chromatids. Does the genetic information on thetwo sister chromatids come from only one parent or from both parents? Explain yourreasoning.Possible Answers: This questions is designed to help address the difference betweensister chromatids and homologous chromosomes, a common point of confusion for manystudents. For the most part, the genetic information on the two sister chromosomes comesfrom the same parent, because the sister chromatids are produced by replication and areexact copies of the genetic information originally present on a single chromosome, whichcomes from one parent. However, crossing over can produce information from twodifferent parents on the same chromosome. One chromosome of a homologous paircomes from one parent and the other homolog comes from the other parent. Whencrossing over takes place, information is exchanged between nonsister chromatids—chromatids from different but homologous chromosomes. So, after crossing over, theinformation on one chromatid may contain information from the other homolog (whichcomes from the other parent).2.Are homologous pairs of chromosome present in mitosis? Explain your reasoning.Possible Answers: This question addresses a common misconception among studentswho often assume that homologous pairs of chromosomes exist in meiosis but not inmitosis. Homologous pairs of chromosomes are present in mitosis. However, they don’tpair up in mitosis like they do in meiosis, and sister chromatids of each chromosomeseparate independently.3.A cell has eight chromosomes in metaphase II of meiosis. How many chromosomes andDNA molecules will be present per cell in this same organism at the following stages?a.Prophase of mitosisb.Metaphase I of meiosis

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6ec.Anaphase of mitosisc.Anaphase II of meiosisd.Anaphase I of meiosise.After cytokinesis that follows mitosisf.After cytokinesis that follows meiosis II.Possible Answers: This question requires that students think through the different stepsof mitosis and meiosis and fully understand what is happening in each stage.Number ofNumber ofChromosomesDNA Moleculesa. Prophase of mitosis1632b. Metaphase I of meiosis1632c. Anaphase of mitosis3232c. Anaphase II of meiosis1616d. Anaphase I of meiosis1632e. After cytokinesis that follows mitosis1616f. After cytokinesis that follows meiosis II 88Section 2.34.What is the difference between sister chromatids and homologous chromosomes?Possible Answers: Students often are confused and have misconceptions about the differencebetween sister chromatids and homologous chromosomes. Sister chromatids are identicalcopies (unless crossing over takes place) of the same original chromosome. Homologouschromosomes are different chromosomes, containing information for the same traits(homologous information) but not the same genetic information. One homolog comes fromone parent; the other homolog comes from the other parent.5.List as many similarities and differences in mitosis and meiosis as you can. Which differences

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6edo you think are most important and why?Possible Answers:SimilaritiesBoth involve chromosome and cell division.Both are preceded by DNA replication.Both use spindle fibers to separate chromosomes.Both have a stage where sister chromatids separate.DifferencesMeiosis normally involves two cell divisions; mitosis usually has only a single celldivision.Chromosome reduction occurs in meiosis but not in mitosis.Resulting daughter cells are genetically different in meiosis but not in mitosis.Crossing over occurs in meiosis but does not normally take place in mitosis.Random assortment of chromosomes occurs in anaphase I of meiosis but does not occurin anaphase of mitosis.In metaphase I of meiosis, pairs of homologous chromosomes line up; in metaphase ofmitosis (and metaphase II of meiosis) individual chromosomes line up.In anaphase I of meiosis, homologous chromosomes separate; in anaphase of mitosis (andanaphase II of meiosis) sister chromatids separate.6.Describe how and where each of the following terms applies to mitosis and/or meiosis: (1)replication; (2) pairing; and (3) separation.Possible Answers:Replication, pairing, and separation are key events that take place in mitosis and meiosis.Replication: In mitosis and meiosis, DNA replication takes place during S phase precedingnuclear division.Pairing: Homologous chromosomes pair up in meiosis, but no pairing of homologouschromosomes takes place in mitosis. However, the sister chromatids of each chromosome are

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6epaired in both mitosis and meiosis.Separation: In mitosis, sister chromatids separate in anaphase. In meiosis, homologouschromosomes separate in anaphase I and sister chromatids separate in anaphase II.7.Do you know of any genetic diseases or disorders that result from errors in mitosis ormeiosis? How do errors in mitosis or meiosis bring about these diseases?Possible Answers: There are many chromosome abnormalities that result from errors inmitosis and/or meiosis, including Down syndrome (resulting from an extra copy ofchromosome 21) and Turner syndrome (resulting from a single X chromosome). Manycancers have abnormal chromosomes that result from errors in mitosis (see Chapter 23).Errors in mitosis and/or meiosis often result in abnormal separation of chromosomes, so thatcells end up with too many or too few chromosomes.Chapter 3Think-Pair-Share questions for the chapter opening story:Why is knowing about the genetic basis of a trait like blond hair important? Why wouldscientists go to the trouble to investigate the genetic basis of blond hair in the SolomonIslanders?Possible Answers: There are a number of reasons why scientists might be interested in thegenetic basis of blond hair in Solomon Islanders. First is scientific curiosity. Many scientificstudies are conducted for no obviously practical reason, but simply for the purpose oflearning the answer to an interesting question. Another reason is that science advances bygathering information about basic processes, in what is called basic science. It is oftendifficult to know how scientific information might be useful in the future and so much ofscience focuses on basic questions that have no current practical application. Often theinformation gathered in basic research provides practical applications in the future.

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6eThere may also be practical reasons for studying the genetic basis of blond hair. One of thehypotheses explaining blond hair in Solomon Islanders is that the population has someEuropean genes, so the study of this trait provides insight into the history and evolution ofSolomon Islanders. Blond hair results from reduced amounts of the melanin in hair.Melanin plays a vital role in many processes, including vision and protection from UVradiation (see the introduction to Chapter 1 about albinism). So the underlying genetics ofblond hair may lead to a better understanding of the biosynthesis of melanin and disorderscaused by reduced pigment.If a blond-haired person from northern Europe mated with a blond Solomon Islander,what proportion of their offspring would be expected to have blond hair? Explain yourreasoning.Possible Answer: Most likely, none of the offspring would have blond hair, becausedifferent genes are coding for blond hair in Europeans and Solomon Islanders. TheEuropean’s genotype might beaa B+B+, whereaacodes for blond hair. The Solomonislander might beA+A+bb, wherebbcodes for blond hair. A mating between these twoindividuals would be:aa B+B+ ×A+A+bbA+a B+b, which would have nonblond hair.End-of-chapter Think-Pair-Share questions:Introduction1.About 40% of Solomon Islanders carry a gene for blond hair, and yet only 5%–10% of thesepeople actually have blond hair. Why is the number of people with blond hair only 5%–10%when so many people carry genes for blond hair?Answer: This question focuses on the difference between genotypic and phenotypicfrequencies and how, with recessive traits, people can carry a gene for the trait (beheterozygous) and yet not have the trait.The allele for blond hair (b) is recessive to the allele for dark hair. Most of the peoplewho carry a blond allele are heterozygotes (Bb); because the blond allele is recessive, theseindividuals have dark hair even though they carry a blond allele. If mating for hair color is

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6erandom, then the percent of people with blond hair is expected to beq2, whereqequals thefrequency of the blond gene. If the percent of blond people is 10%, thenq2= 0.10. Thesquare root of 0.10 is10.0= 0.32 =q, the frequency of the blond gene. The frequency ofthe dark-haired gene,p, will be 1 –q= 0.68. According to the Hardy-Weinberg law, whichwe will learn in Chapter 25, the probability of a person being heterozygous is expected to be2pq= 2(0.68)(0.32) = 0.44, which is about the frequency of dark-haired people on theSolomon Islands who carry a blond gene.2.Why was Mendel’s success dependent on his studying characteristics that exhibit only twoeasily distinguished phenotypes, such as white versus gray seed coats and round versuswrinkled seeds? Would he have been less successful if he had instead studied traits like seedweight or length of the leaves, which vary much more in their phenotypes? Explain youranswer.Possible Answer: This question is useful for discussing some of the factors that led toMendel’s success and the differences between discontinuous and continuous traits(discussed in more detail in Chapter 24). Probably Mendel would have been less successfulif he had studied characteristics that vary more in phenotypes, such as seed weight or lengthof the leaves. Because his characteristics differed in only two easily distinguishablephenotypes, he was able to infer their genotypes. Many traits that vary continuously inphenotype are encoded by multiple loci and often the phenotypes of these traits areinfluenced by environmental effects (see Chapter 24).Section 3.23.Geneticists often carry out reciprocal crosses when they are studying the inheritance oftraits. Why do geneticists use reciprocal crosses?Possible Answer: Reciprocal crosses provide information about whether the sex of aparent in a cross affects the types and proportions of progeny from the cross. For a simpleautosomal trait, reciprocal crosses usually give the same result. However, reciprocalcrosses do not give the same results if sex interacts with heredity, as it does for sex-linked,

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6esex-limited, sex-influenced, cytoplasmically inherited traits, genetic maternal effects, andgenomically imprinted traits (see Chapters 4 and 5). When these factors are present, theinheritance of a trait may differ from that of simple autosomal traits, so it is important toknow whether these factors are present.4.Red hair in humans is inherited as a recessive trait. Bill and Sarah both have black hair.They marry and have four children, three of whom have red hair. Bill says it isn’tgenetically possible for two black-haired people to have ¾ red-haired children, and heaccuses Sarah of infidelity. Sarah says Bill is a homozygous dominant idiot and knowsnothing about genetics. Who is correct and why?Possible Answer: This question helps to emphasize that the ratio of progeny in a geneticcross often differs by chance from the expected ratio, especially when the number ofprogeny is small.Assume black hair is determined by a dominant alleleR+and red hair by a recessive alleler. If Bill and Sarah both have black hair and have one or more red-haired children, then Billand Sarah are probably both heterozygous for red hair (R+r). Mating between Bill and Sarahwould produce:R+r×R+r¾R+_ (black hair) and ¼rr(red hair), so the expectation isthat only one out of four children would have red hair. However, just by chance, they mighthave three red-haired children. The probability of this can be determined by the expansionof the binomial: (a+b)4=a4+ 4a3b+ 6a2b2+ 4ab3+b4, wherea= the probability of blackhair (¾) andb= the probability of red hair (¼). In this case, the term 4ab3equals theprobability of having four children, one with black hair and three with red hair, or 4(¾)(¼)3,which equals3/64or 0.047. Thus, there is only about a 5% chance (one in 20) of gettingthree children out of four with red hair when both parents are heterozygous. This is not verylikely but not impossible.Section 3.35.Are Mendel’s principles of segregation and independent assortment even relevant today in theage of genomics, where it is possible to sequence an organism’s entire DNA and determine all

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6eof its genetic information? Why is it important to study these principles and how can they beused?Possible Answer: This question is useful for addressing a common misconception thatMendel’s principles are outdated and no longer relevant because we can determine the entiresequence of a genome. These principles are still relevant and important today because theyallow us to make predictions about the outcome of genetic crosses, which are of greatpractical value in agriculture, biotechnology, and human heredity. Having the DNA sequenceinformation does not, by itself, provide us with information about individual genes, thedominance relationships of alleles, or whether genes assort independently, characteristics thatare essential for making predictions about heredity.6.In cats, short hair (L) is dominant to long hair (l) and stripes (A) are dominant to solid color(a). A cat with genotypeLl Aamates with a cat that isLl aaand they produce a litter of sixkittens. What is the probability that four of six kittens will have both long hair and stripes?Answer: The cross for hair length is:Ll×Ll¾L_ (short) and ¼ll(long). The cross forstripes isAa×aa½Aa(stripes) and ½aa(solid). Using the multiplication rule, theprobability of a kitten having long hair and strips (ll Aa) is: ¼ × ½ =1/8. The probability of akitten not having long hair and stripes is 1 –1/8=7/8. We can use the binomial expansion (a+b)6to determine the probability that in a litter of six kittens, four will have long hair andstripes, wherea= the probability of having long hair and stripes andb= the probability of nothaving long hair and stripes. The binomial expansion is:a6+ 6a5b+ 15a4b2+ 20a3b3+ 15a2b4+ 6ab5+b6. The term 15a4b2represents the probability of four kittens with long hair andstripes in a litter of six. Substituting in the values ofaandb, we obtain: 15(1/8)4(7/8)2= 0.0028.Section 3.47.In corn, purple kernels (P) are dominant to yellow kernels (p) and starchy kernels (Su) aredominant to sugary kernels (su). A corn plant grown from a purple and starchy kernel iscrossed with a plant grown from a yellow and sugary kernel and the following progeny(kernels) are produced:

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6ePhenotypeNumberpurple starchy150purple, sugary142yellow, starchy161yellow, sugary115Formulate a hypothesis about the genotypes of the parents and offspring in this cross. Performa chi-square goodness-of-fit test comparing the observed progeny with the numbers expectedbased on your genetic hypothesis. What conclusion can you draw based on the results of yourchi-square test? Can you suggest an explanation for the observed results?Answer: The cross is purple and starchy kernels × yellow and sugary kernels. The presence ofyellow and sugary kernels in the progeny tells us that the purple and starchy parent isheterozygous for both purple and starchy:Pp Susu×pp susu. This cross is expected toproduce a 1:1:1:1 progeny ratio:Pp Susu×pp susu¼Pp Susupurple, starchy¼Pp susupurple, sugary¼pp Susuyellow, starchy¼pp susuyellow, sugaryThe observed and expected progeny are:PhenotypeObservedExpectedpurple starchy150¼ × 568 = 142purple, sugary142¼ × 568 = 142yellow, starchy161¼ × 568 = 142yellow, sugary115¼ × 568 = 142total568

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6eThe chi-square value is22observedexp ectedexp ected . Substituting the observed andexpected values, we obtain:12.8142)142115(142)142161(142)142142(142)142150(2222The degrees of freedom are n – 1 = 4 – 1 = 3From Table 3.7, we find the probability associated with a chi-square value of 8.12 and 3 d.f.is less than 0.05. Therefore, the probability that the observed would deviate from the expectedby chance alone is low and we reject the hypothesis that this represents a 1:1:1:1 ratio. All ofthe observed numbers are close to those expected with the exception of the yellow and sugaryphenotype. There are far fewer yellow sugary progeny than are expected. One possibleexplanation is that the yellow sugary progeny exhibits lower probability of developing.Perhaps some of them died at an early developmental stage and therefore do not appearamong the progeny.Chapter 4Think-Pair-Share questions for the chapter opening story:Climate warming could lead to all-female populations of bearded dragons, resulting inextinction of the species. What are some other potential effects of climate warming onnatural populations of organisms?Possible Answers: There are many possible answers to this question. For many organisms,temperature plays an important role in many aspects of their ecology, and organisms arefrequently genetically adapted to function at a particular temperature. A change intemperature may make them unable to survive and reproduce, leading to population declinesand eventual extinction. For example, breeding in many amphibians is triggered by earlyspring temperatures. A rise in temperature may cause them to breed earlier, when food andsufficient rainfall may not be available for the development of tadpoles. Many organismssuch as trout are adapted to cooler temperatures and may not be able to survive andreproduce in water that has warmed because of climate change. Climate warming will bring

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Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6eless rainfall to some areas, leading to the local extinction of many populations of plants andanimals. Melting of ice may lead to sea-level rise, which may make it difficult for someorganisms to survive. Warmer temperatures may increase the prevalence of pathogens andvectors of disease, leading to more disease. Some organisms may be able to geneticallyadapt to these changes (see Chapters 25 and 26), but others may not and might go extinct.High incubation temperatures reverse the sex of ZZ bearded dragons, causing them todevelop as females instead of males. Imagine that temperature also caused sex reversal inZW individuals, so that high temperatures caused ZW individuals to develop as malesinstead of females (this doesn’t actually happen in the bearded dragons). What would bethe result of a mating between a normal ZW female and a sex-reversed ZW male?Answer: A cross between a ZW female and sex reversed ZW male would produce: ZW ×ZW¼ ZZ, ½ ZW, ¼ WW. At 32oC, the ZZ individuals would develop as male and theZW would develop as female. The W chromosome is like the Y chromosome, with littlegenetic information. A WW zygote, like a YY zygote, would probably be nonviable.Therefore, the sex ratio of the offspring would become1/3ZZ males and2/3ZW females. Athigher temperatures, some ZW individuals might develop as males, skewing the sex ratiotowards more males.End-of-chapter Think-Pair-Share questions:Section 4.11.The duck-billed platypus has a unique mechanism of sex determination: females have fivepairs of X chromosomes (X1X1X2X2X3X3X4X4X5X5) and males have five pairs of X and Ychromosomes (X1Y1X2Y2X3Y3X4Y4X5Y5). Do you think each of the X and Y chromosomepairs in males assort independently of other X and Y pairs during meiosis? Why or why not?Possible Answer: The individual pairs of X and Y chromosomes in male platypuses must notseparate independently. If they did separate independently, gametes with differing numbers ofX and Y chromosomes would result, which would produce offspring with abnormal numbersof sex chromosomes. For example, independent separation of X and Ys in males would

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