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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 6e Chapter 1 Think-Pair-Share questions for the chapter opening story:  Albinism occupied a special place in the Hopi culture; individuals who possessed this trait 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 and prejudice. Indeed, people with albinism in cultures other than the Hopis are subjected to discrimination for the way they look. In the past, African Americans who were heterozygous carriers of the mutation for sickle-cell anemia were subject to discrimination in employment and insurance, even though they were perfectly healthy. Skin color is a genetically determined characteristic and there is discrimination against people with certain skin colors in some societies. Students may have witnessed discrimination 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 different genes, is referred to as genetic heterogeneity. Is genetic heterogeneity common? Are most genetic traits in humans the result of variation in a single gene, or are there many genetic traits that result from variation in several genes, as albinism does? Possible Answers : Beginning genetics students often have the misconception that most traits are caused by a single variation at a single gene, much like the traits Mendel studied in his pea plants. This question provides an opportunity at the beginning of the course to discuss this misconception and the fact that the genetic basis of most traits is more complex. Many genetic traits and diseases exhibit genetic heterogeneity, where DNA sequences at any one of several different genes can result in the same phenotype. For example, albinism in the Hopis is caused by mutations in the OCA2 gene, but mutations at several other genes may produce albinism in other populations. Genetic heterogeneity can also result 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 6e human albinism. Many genetic traits are also the result of combined variation at several genes, a situation referred to as a polygenic trait (see Chapters 5 and 24). End-of-chapter Think-Pair-Share questions: Section 1.1 1. Bob says that he is healthy and has no genetic diseases such as hemophilia or Down syndrome. Therefore, genetics, he says, plays little role in his life. Do you think Bob is correct in his conclusion? Why or why not? Possible Answers : Bob makes a common mistake when he thinks of genetic effects only as classical genetic diseases, like hemophilia and Down syndrome. In fact, there are many common disorders and diseases, as well as variations in healthy phenotypes, that are influenced by genetics. In many of these cases, the phenotype is the result of a variation in multiple genes (the trait is polygenic) and the interaction of genes and environment (see Chapter 5). For example, common diseases, such as asthma, coronary heart disease, hypertension, depression, and diabetes, are due to a combination of variations 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 also determined 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 the necessary genetic variation necessary for evolution, the process by which populations change and become adapted to their environments. Adaptation and evolutionary change allow populations to avoid extinction when environments change. So, mutations can be beneficial to the long-term success of a population. Mutations result in individual differences, which makes each of us unique. Also, mutations are often useful in genetic research, 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 6e mutations also have detrimental effects. Section 1.2 3. Do you support or oppose the development of genetically engineered foods (genetically modified organisms, or GMOs)? Find someone who takes the opposite position and discuss this question with them. Think about the economic, environmental, health, and social effects of their use. List some reasons for and against genetically engineering the foods we eat Possible 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 and ecological effects  Ability to produce foods at lower cost  Ability to create crops that can grow in otherwise unsuitable habitats, such as arid regions and saline soils Some reasons for opposing genetically engineered foods include:  Unanticipated health risks  Potential ecological damage caused by release of novel genetically modified organisms into the environment.  Negative economic effects on small farmers who cannot afford the cost of genetically engineered crops and animals. Section 1.3 4. Why do you think all organisms use nucleic acids for encoding genetic information? Why not use proteins or carbohydrates? What advantages might DNA have as the source of genetic information?

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    Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e Possible Answers : One potential answer is that all organisms use nucleic acids because all living organisms are derived from a common ancestor, which just happened to use nucleic acids as its genetic information. However, there may also be reasons why nucleic acids are advantageous as information carriers. Nucleic acids have the capacity to vary in their base sequence, but they don’t vary too much. (There are only four common bases used in DNA and RNA.) They have the ability to form long polymers, which means they can encode a great deal of genetic information. DNA is particularly stable because of its double-stranded nature and lack of any free hydroxyl groups. The double-stranded nature of DNA also provides a particularly efficient means of replicating the molecule: the complementary nature of the bases (G and C pair; A and T pair) means that each strand of DNA can serve as a template for the synthesis of a new strand. Chapter 2 Think-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 man gets exactly five pairs of different colored socks. In the analogy, is it important that the men are blind? In a cell, what does the blindness represent? Possible Answers : The analogy depends on the men being blind. If they could see they could distinguish the socks by color and easily sort the socks so that each man got two socks of each color. While theoretically it might be possible for cells to evolve some mechanism by which they distinguish each pair of chromosomes and by this discrimination ensure that one chromosome of each pair ends up in a resulting cell, cells have not evolved such a mechanism. Through the process of mitosis, each chromosome replicates and the two replicated copies (analogous to the two socks of a pair) get pulled in opposite directions and end up in separate cells. The cells don’t have a mechanism to see the different types of chromosomes and ensure at the end that each daughter cell gets exactly 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 6e End-of-chapter Think-Pair-Share questions: Section 2.2 1. A chromosome consists of two sister chromatids. Does the genetic information on the two sister chromatids come from only one parent or from both parents? Explain your reasoning. Possible Answers : This questions is designed to help address the difference between sister chromatids and homologous chromosomes, a common point of confusion for many students. For the most part, the genetic information on the two sister chromosomes comes from the same parent, because the sister chromatids are produced by replication and are exact copies of the genetic information originally present on a single chromosome, which comes from one parent. However, crossing over can produce information from two different parents on the same chromosome. One chromosome of a homologous pair comes from one parent and the other homolog comes from the other parent. When crossing over takes place, information is exchanged between nonsister chromatids— chromatids from different but homologous chromosomes. So, after crossing over, the information on one chromatid may contain information from the other homolog (which comes 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 students who often assume that homologous pairs of chromosomes exist in meiosis but not in mitosis. Homologous pairs of chromosomes are present in mitosis. However, they don’t pair up in mitosis like they do in meiosis, and sister chromatids of each chromosome separate independently. 3. A cell has eight chromosomes in metaphase II of meiosis. How many chromosomes and DNA molecules will be present per cell in this same organism at the following stages? a. Prophase of mitosis b. Metaphase I of meiosis

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    Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e c. Anaphase of mitosis c. Anaphase II of meiosis d. Anaphase I of meiosis e. After cytokinesis that follows mitosis f. After cytokinesis that follows meiosis II. Possible Answers : This question requires that students think through the different steps of mitosis and meiosis and fully understand what is happening in each stage. Number of Number of Chromosomes DNA Molecules a. Prophase of mitosis 16 32 b. Metaphase I of meiosis 16 32 c. Anaphase of mitosis 32 32 c. Anaphase II of meiosis 16 16 d. Anaphase I of meiosis 16 32 e. After cytokinesis that follows mitosis 16 16 f. After cytokinesis that follows meiosis II 8 8 Section 2.3 4. What is the difference between sister chromatids and homologous chromosomes? Possible Answers : Students often are confused and have misconceptions about the difference between sister chromatids and homologous chromosomes. Sister chromatids are identical copies (unless crossing over takes place) of the same original chromosome. Homologous chromosomes are different chromosomes, containing information for the same traits (homologous information) but not the same genetic information. One homolog comes from one 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 6e do 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 cell division.  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 occur in anaphase of mitosis.  In metaphase I of meiosis, pairs of homologous chromosomes line up; in metaphase of mitosis (and metaphase II of meiosis) individual chromosomes line up.  In anaphase I of meiosis, homologous chromosomes separate; in anaphase of mitosis (and anaphase 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 preceding nuclear division. Pairing: Homologous chromosomes pair up in meiosis, but no pairing of homologous chromosomes 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 6e paired in both mitosis and meiosis. Separation: In mitosis, sister chromatids separate in anaphase. In meiosis, homologous chromosomes 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 or meiosis? How do errors in mitosis or meiosis bring about these diseases? Possible Answers : There are many chromosome abnormalities that result from errors in mitosis and/or meiosis, including Down syndrome (resulting from an extra copy of chromosome 21) and Turner syndrome (resulting from a single X chromosome). Many cancers 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 that cells end up with too many or too few chromosomes. Chapter 3 Think-Pair-Share questions for the chapter opening story:  Why is knowing about the genetic basis of a trait like blond hair important? Why would scientists go to the trouble to investigate the genetic basis of blond hair in the Solomon Islanders? Possible Answers : There are a number of reasons why scientists might be interested in the genetic basis of blond hair in Solomon Islanders. First is scientific curiosity. Many scientific studies are conducted for no obviously practical reason, but simply for the purpose of learning the answer to an interesting question. Another reason is that science advances by gathering information about basic processes, in what is called basic science. It is often difficult to know how scientific information might be useful in the future and so much of science focuses on basic questions that have no current practical application. Often the information 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 6e There may also be practical reasons for studying the genetic basis of blond hair. One of the hypotheses explaining blond hair in Solomon Islanders is that the population has some European genes, so the study of this trait provides insight into the history and evolution of Solomon 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 UV radiation (see the introduction to Chapter 1 about albinism). So the underlying genetics of blond hair may lead to a better understanding of the biosynthesis of melanin and disorders caused 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 your reasoning. Possible Answer : Most likely, none of the offspring would have blond hair, because different genes are coding for blond hair in Europeans and Solomon Islanders. The European’s genotype might be aa B + B +, where aa codes for blond hair. The Solomon islander might be A + A + bb , where bb codes for blond hair. A mating between these two individuals would be: aa B + B + × A + A + bb  A + a B + b , which would have nonblond hair. End-of-chapter Think-Pair-Share questions: Introduction 1. About 40% of Solomon Islanders carry a gene for blond hair, and yet only 5%–10% of these people 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 phenotypic frequencies and how, with recessive traits, people can carry a gene for the trait (be heterozygous) and yet not have the trait. The allele for blond hair ( b ) is recessive to the allele for dark hair. Most of the people who carry a blond allele are heterozygotes ( Bb ); because the blond allele is recessive, these individuals 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 6e random, then the percent of people with blond hair is expected to be q 2 , where q equals the frequency of the blond gene. If the percent of blond people is 10%, then q 2 = 0.10. The square root of 0.10 is 10 . 0 = 0.32 = q , the frequency of the blond gene. The frequency of the dark-haired gene, p , will be 1 – q = 0.68. According to the Hardy-Weinberg law, which we will learn in Chapter 25, the probability of a person being heterozygous is expected to be 2 pq = 2(0.68)(0.32) = 0.44, which is about the frequency of dark-haired people on the Solomon Islands who carry a blond gene. 2. Why was Mendel’s success dependent on his studying characteristics that exhibit only two easily distinguished phenotypes, such as white versus gray seed coats and round versus wrinkled seeds? Would he have been less successful if he had instead studied traits like seed weight or length of the leaves, which vary much more in their phenotypes? Explain your answer. Possible Answer : This question is useful for discussing some of the factors that led to Mendel’s success and the differences between discontinuous and continuous traits (discussed in more detail in Chapter 24). Probably Mendel would have been less successful if he had studied characteristics that vary more in phenotypes, such as seed weight or length of the leaves. Because his characteristics differed in only two easily distinguishable phenotypes, he was able to infer their genotypes. Many traits that vary continuously in phenotype are encoded by multiple loci and often the phenotypes of these traits are influenced by environmental effects (see Chapter 24). Section 3.2 3. Geneticists often carry out reciprocal crosses when they are studying the inheritance of traits. Why do geneticists use reciprocal crosses? Possible Answer : Reciprocal crosses provide information about whether the sex of a parent in a cross affects the types and proportions of progeny from the cross. For a simple autosomal trait, reciprocal crosses usually give the same result. However, reciprocal crosses 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 6e sex-limited, sex-influenced, cytoplasmically inherited traits, genetic maternal effects, and genomically imprinted traits (see Chapters 4 and 5). When these factors are present, the inheritance of a trait may differ from that of simple autosomal traits, so it is important to know 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’t genetically possible for two black-haired people to have ¾ red-haired children, and he accuses Sarah of infidelity. Sarah says Bill is a homozygous dominant idiot and knows nothing about genetics. Who is correct and why? Possible Answer : This question helps to emphasize that the ratio of progeny in a genetic cross often differs by chance from the expected ratio, especially when the number of progeny is small. Assume black hair is determined by a dominant allele R + and red hair by a recessive allele r . If Bill and Sarah both have black hair and have one or more red-haired children, then Bill and Sarah are probably both heterozygous for red hair ( R + r ). Mating between Bill and Sarah would produce: R + r × R + r  ¾ R + _ (black hair) and ¼ rr (red hair), so the expectation is that only one out of four children would have red hair. However, just by chance, they might have three red-haired children. The probability of this can be determined by the expansion of the binomial: ( a + b ) 4 = a 4 + 4 a 3 b + 6 a 2 b 2 + 4 ab 3 + b 4 , where a = the probability of black hair (¾) and b = the probability of red hair (¼). In this case, the term 4 ab 3 equals the probability of having four children, one with black hair and three with red hair, or 4(¾)(¼) 3 , which equals 3 / 64 or 0.047. Thus, there is only about a 5% chance (one in 20) of getting three children out of four with red hair when both parents are heterozygous. This is not very likely but not impossible. Section 3.3 5. Are Mendel’s principles of segregation and independent assortment even relevant today in the age 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 6e of its genetic information? Why is it important to study these principles and how can they be used? Possible Answer : This question is useful for addressing a common misconception that Mendel’s principles are outdated and no longer relevant because we can determine the entire sequence of a genome. These principles are still relevant and important today because they allow us to make predictions about the outcome of genetic crosses, which are of great practical value in agriculture, biotechnology, and human heredity. Having the DNA sequence information does not, by itself, provide us with information about individual genes, the dominance relationships of alleles, or whether genes assort independently, characteristics that are 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 genotype Ll Aa mates with a cat that is Ll aa and they produce a litter of six kittens. 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  ¾ L _ (short) and ¼ ll (long). The cross for stripes is Aa × aa  ½ Aa (stripes) and ½ aa (solid). Using the multiplication rule, the probability of a kitten having long hair and strips ( ll Aa ) is: ¼ × ½ = 1 / 8 . The probability of a kitten not having long hair and stripes is 1 – 1 / 8 = 7 / 8 . We can use the binomial expansion ( a + b ) 6 to determine the probability that in a litter of six kittens, four will have long hair and stripes, where a = the probability of having long hair and stripes and b = the probability of not having long hair and stripes. The binomial expansion is: a 6 + 6 a 5 b + 15 a 4 b 2 + 20 a 3 b 3 + 15 a 2 b 4 + 6 ab 5 + b 6 . The term 15 a 4 b 2 represents the probability of four kittens with long hair and stripes in a litter of six. Substituting in the values of a and b , we obtain: 15( 1 / 8 ) 4 ( 7 / 8 ) 2 = 0.0028. Section 3.4 7. In corn, purple kernels ( P ) are dominant to yellow kernels ( p ) and starchy kernels ( Su ) are dominant to sugary kernels ( su ). A corn plant grown from a purple and starchy kernel is crossed 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 6e Phenotype Number purple starchy 150 purple, sugary 142 yellow, starchy 161 yellow, sugary 115 Formulate a hypothesis about the genotypes of the parents and offspring in this cross. Perform a chi-square goodness-of-fit test comparing the observed progeny with the numbers expected based on your genetic hypothesis. What conclusion can you draw based on the results of your chi-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 of yellow and sugary kernels in the progeny tells us that the purple and starchy parent is heterozygous for both purple and starchy: Pp Susu × pp susu . This cross is expected to produce a 1:1:1:1 progeny ratio: Pp Susu × pp susu ¼ Pp Susu purple, starchy ¼ Pp susu purple, sugary ¼ pp Susu yellow, starchy ¼ pp susu yellow, sugary The observed and expected progeny are: Phenotype Observed Expected purple starchy 150 ¼ × 568 = 142 purple, sugary 142 ¼ × 568 = 142 yellow, starchy 161 ¼ × 568 = 142 yellow, sugary 115 ¼ × 568 = 142 total 568

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    Answers to Chapter Opening and End-of-Chapter Think-Pair-Share Questions: Pierce Genetics 6e The chi-square value is   2 2 observed exp ected exp ected     . Substituting the observed and expected values, we obtain: 12 . 8 142 ) 142 115 ( 142 ) 142 161 ( 142 ) 142 142 ( 142 ) 142 150 ( 2 2 2 2         The degrees of freedom are n – 1 = 4 – 1 = 3 From 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 expected by chance alone is low and we reject the hypothesis that this represents a 1:1:1:1 ratio. All of the observed numbers are close to those expected with the exception of the yellow and sugary phenotype. There are far fewer yellow sugary progeny than are expected. One possible explanation 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 appear among the progeny. Chapter 4 Think-Pair-Share questions for the chapter opening story:  Climate warming could lead to all-female populations of bearded dragons, resulting in extinction of the species. What are some other potential effects of climate warming on natural 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 are frequently genetically adapted to function at a particular temperature. A change in temperature may make them unable to survive and reproduce, leading to population declines and eventual extinction. For example, breeding in many amphibians is triggered by early spring temperatures. A rise in temperature may cause them to breed earlier, when food and sufficient rainfall may not be available for the development of tadpoles. Many organisms such as trout are adapted to cooler temperatures and may not be able to survive and reproduce 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 6e less rainfall to some areas, leading to the local extinction of many populations of plants and animals. Melting of ice may lead to sea-level rise, which may make it difficult for some organisms to survive. Warmer temperatures may increase the prevalence of pathogens and vectors of disease, leading to more disease. Some organisms may be able to genetically adapt 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 to develop as females instead of males. Imagine that temperature also caused sex reversal in ZW individuals, so that high temperatures caused ZW individuals to develop as males instead of females (this doesn’t actually happen in the bearded dragons). What would be the 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 32 o C, the ZZ individuals would develop as male and the ZW would develop as female. The W chromosome is like the Y chromosome, with little genetic information. A WW zygote, like a YY zygote, would probably be nonviable. Therefore, the sex ratio of the offspring would become 1 / 3 ZZ males and 2 / 3 ZW females. At higher temperatures, some ZW individuals might develop as males, skewing the sex ratio towards more males. End-of-chapter Think-Pair-Share questions: Section 4.1 1. The duck-billed platypus has a unique mechanism of sex determination: females have five pairs of X chromosomes (X 1 X 1 X 2 X 2 X 3 X 3 X 4 X 4 X 5 X 5 ) and males have five pairs of X and Y chromosomes (X 1 Y 1 X 2 Y 2 X 3 Y 3 X 4 Y 4 X 5 Y 5 ). Do you think each of the X and Y chromosome pairs 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 not separate independently. If they did separate independently, gametes with differing numbers of X and Y chromosomes would result, which would produce offspring with abnormal numbers of sex chromosomes. For example, independent separation of X and Ys in males would

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