Solution Manual for Concepts of Genetics 4th Edition By Robert Brooker

Page | 1 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Solution Manual for Concepts of Genetics 4th Edition By RobertBrooker Page | 2 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. CONCEPTS OF GENETICS, 4/e ANSWERS TO PROBLEM SETS Chapters 1-24 CHAPTER 1 Note: the answers to the Comprehension Questions are at the end of the chapter. Concept Check Questions (in figure legends) FIGURE 1. 1 Understanding our genes may help with diagnoses of inherited diseases. It may also lead to the development of drugs to combat diseases. Other answers are possible. FIGURE 1. 2 Many ethical issues are associated with human cloning. Is it the wrong thing to do? Does it conflict an individual’s religious views? And so on. FIGURE 1. 3 Because females mate only once, sorting out the male mosquitoes and releasing sterile males into the environment can limit mosquito reproduction. FIGURE 1. 4 DNA is a macromolecule. FIGURE 1. 5 DNA and proteins are found in chromosomes. A small amount of RNA may also be associated with chromosomes when transcription is occurring, and as discussed in Chapter 18, some non-coding RNAs may bind to chromosomes. FIGURE 1.6 The information to make a polypeptide is stored in DNA. FIGURE 1. 7 The dark-colored butterfly has a more active pigment-producing enzyme. FIGURE 1. 8 Genetic variation is the reason the frogs look different. FIGURE 1. 9 These are examples of variation in chromosome number. FIGURE 1. 10 If this girl had been given a standard diet, she would have developed the harmful symptoms of PKU, which include mental impairment and foul-smelling urine. FIGURE 1. 11 Page | 3 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. A corn gamete contains 10 chromosomes. (The leaf cells are diploid.) FIGURE 1. 12 The horse populations have become adapted to their environment, which has changed over the course of many years. FIGURE 1.13 There are several possible examples of other model organisms, including rats and frogs. End-of-chapter Questions: Conceptual Questions C1. A chromosome is a very long polymer of DNA. A gene is a specific sequence of DNA within that polymer; the sequence of bases creates a gene and distinguishes it from other genes. Genes are located in chromosomes, which are found within living cells. C2. At the molecular level, a gene (a sequence of DNA) is first transcribed into RNA. The genetic code within the RNA is used to synthesize a protein with a particular amino acid sequence. This second process is called translation. C3. A. Molecular level. This is a description of a how an allele affects protein function. B. Cellular level. This is a description of how protein function affects cell structure. C. Population level. This is a description of how the two alleles affect members of a population. D. Organism level. This is a description of how the alleles affect the traits of an individual. C4. Genetic variation is the occurrence of genetic differences within members of the same species or different species. Within any population, variation may occur in the genetic material. Variation may occur in particular genes, so some individuals carry one allele and other individuals carry a different allele. Examples include differences in coat color among mammals or flower color in plants. At the molecular level, this type of genetic variation is caused by changes in the DNA sequences of genes. There may also be variation in chromosome structure and number. C5. An extra chromosome (specifically an extra copy of chromosome 21) causes Down syndrome. C6. You can pick almost any trait. For example, flower color in petunias would be an interesting choice. Some petunias are red and others are purple. There must be different alleles in a flower color gene that affect this trait in petunias. In addition, the amount of sunlight, fertilizer, and water also affects the intensity of flower color. C7. The term diploid means that a cell has two copies of each type of chromosome. In humans, nearly all of the cells are diploid except for gametes (i.e., sperm and egg cells). Gametes usually have only one set of chromosomes. C8. A DNA sequence is a sequence of nucleotides. Each nucleotide may have one of four different bases (i.e., A, T, G, or C). When speaking of a DNA sequence, the focus is on the sequence of those bases. C9. The genetic code is the way in which the sequence of bases in RNA is read to produce a sequence of amino acids within a protein. C10. A. A gene is a segment of DNA. For most genes, the expression of the gene results in the production of a polypeptide, which is a unit of a protein. The functioning of proteins within living cells largely determines the traits of an organism. Page | 4 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. B. A gene is a segment of DNA that usually encodes the information for the production of a specific polypeptide. Genes are found within chromosomes. Many genes are found within a single chromosome. C. An allele is an alternative version of a particular gene. For example, suppose a plant has a flower color gene. One allele could produce a white flower, while a different allele could produce an orange flower. The white allele and orange allele are alleles of the flower color gene. D. A DNA sequence is a sequence of bases, which are found within nucleotides. The information within a DNA sequence (which is transcribed into an RNA sequence) specifies the amino acid sequence within a polypeptide. C11. The statement in part A is not correct. Individuals do not evolve. Populations evolve because certain individuals are more likely to survive and reproduce and pass their genes to succeeding generations. C12. A. How genes and traits are transmitted from parents to offspring. B. How the genetic material functions at the molecular and cellular levels. C. Why genetic variation exists in populations, and how it changes over the course of many generations. Application and Experimental Questions E1. There are many possible answers. Some common areas to discuss might involve the impact of genetics in the production of new medicines, the diagnosis of diseases, the production of new kinds of food, and the use of DNA fingerprinting to solve crimes. E2. A genetic cross involves breeding two different individuals. E3. This would be used to a great extent by molecular geneticists. The sequence of DNA is a molecular characteristic of DNA. In addition, the sequence of DNA is interesting to transmission and population geneticists as well. E4. You would see 47 chromosomes instead of 46. There would be three copies of chromosome 21 instead of two copies. E5. A. Transmission geneticists. Dog breeders are interested in how genetic crosses affect the traits of dogs. B. Molecular geneticists. This is a good model organism to study genetics at the molecular level. C. Both transmission geneticists and molecular geneticists. Fruit flies are easy to cross and study the transmission of genes and traits from parents to offspring. Molecular geneticists have also studied many genes in fruit flies to see how they function at the molecular level. D. Population geneticists. Most wild animals and plants would be the subject of population geneticists. In the wild, you cannot make controlled crosses. But you can study genetic variation within populations and try to understand its relationship to the environment. E. Transmission geneticists. Agricultural breeders are interested in how genetic crosses affect the outcome of traits. E6. You need to follow the scientific method. You can take a look at an experiment in another chapter to see how the scientific method is followed. Page | 5 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. CHAPTER 2 Note: the answers to Comprehension Questions are at the end of the chapter. Concept Check Questions (in figure legends) FIGURE 2. 1 Compartmentalization means that cells have membrane-bound compartments. FIGURE 2. 2 The chromosomes would not be spread out very well, and would probably be overlapping. It would be difficult to see individual chromosomes. FIGURE 2. 3 Homologs are similar in size, banding pattern, and carry the same types of genes. However, the alleles of a given gene may be different. FIGURE 2. 4 FtsZ assembles into a ring at the future site of the septum and recruits to that site other proteins that produce a cell wall between the two daughter cells. FIGURE 2. 5 In the G1 phase of the cell cycle, a cell may be preparing to divide. By comparison, the G0 phase is a phase in which a cell is either not advancing through the cell cycle or has committed to never divide again. FIGURE 2. 6 Homologs are genetically similar; one is inherited from the mother and the other from the father. By comparison, chromatids are the product of DNA replication. The chromatids within a pair of sister chromatids are genetically identical. FIGURE 2.7 One end of a kinetochore microtubule is attached to a kinetochore on a chromosome. The other end is within the centrosome. FIGURE 2. 8 Anaphase FIGURE 2. 9 Ingression occurs because myosin motor proteins shorten the contractile ring, which is formed from actin proteins. FIGURE 2. 10 The end result of crossing over is that homologous chromosomes have exchanged pieces. FIGURE 2. 11 The cells at the end of meiosis are haploid, whereas the mother cell is diploid. FIGURE 2. 12 In metaphase of mitosis, each pair of sister chromatids is attached to both poles, whereas in metaphase of meiosis I, each pair of sister chromatids is attached to just one pole. FIGURE 2. 13 Polar bodies are small cells that are produced during oogenesis and then degenerate. FIGURE 2. 14 All of the nuclei in the embryo sac are haploid. The central cell has two haploid nuclei, and all of the other cells, including the egg, have just one. Page | 6 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. End-of-chapter Questions: Conceptual Questions C1. They are genetically identical, barring rare mutations, because they receive identical copies of the genetic material from the mother cell. C2. A homolog is one of the members of a chromosome pair. Homologs are usually the same size and carry the same types and order of genes. They may differ in that the genes they carry may be different alleles. C3. Sister chromatids are identical copies derived from the replication of a chromosome. They remain attached to each other at the centromere. They are genetically identical, barring rare mutations and crossing over with homologous chromosomes. C4. Metaphase is the organization phase, and anaphase is the separation phase. C5. G1, there should be six linear chromosomes. In G2, there should be 12 chromatids that are attached to each other in pairs of sister chromatids. C6. In metaphase of meiosis I, each pair of chromatids is attached to only one pole via the kinetochore microtubules. In metaphase of mitosis, there are two attachments (i.e., to both poles). If the attachment is lost, a chromosome will not migrate to a pole and may not become enclosed in a nuclear membrane after telophase. If left out in the cytosol, it would eventually be degraded. C7. A. During mitosis and meiosis II B. During meiosis I C. During mitosis, meiosis I, and meiosis II D. During mitosis and meiosis II C8. The reduction occurs because there is a single DNA replication event but two cell divisions. Because of the nature of separation during anaphase of meiosis I, each cell receives one copy of each type of chromosome. C9. C10. It means that the maternally derived and paternally derived chromosomes are randomly aligned along the metaphase plate during metaphase of meiosis I. Page | 7 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. C11. Mitosis—two diploid cells containing 10 chromosomes each (two complete sets). Meiosis—four haploid cells containing 5 chromosomes each (one complete set) C12. The number of different, random alignments equals 2n , where n equals the number of chromosomes per set. In this case, there are three per set, so the possible number of arrangements equals 2 3 , which is 8. C13. (1/2)n = (1/2)4 = 1/16 or 6.25% C14. The probability would be much lower because pieces of maternal chromosomes would be incorporated into the paternal chromosomes. Therefore, a gamete would be unlikely to carry a chromosome that was completely paternally derived. C15. Bacteria do not need to sort their chromosomes because they only have one type of chromosome. Though not discussed in the text, the attachment of the two copies of the chromosomes to the cell membrane prior to cell division also helps to ensure that each daughter cell receives one copy. C16. During interphase, the chromosomes are greatly extended. In this conformation, they might get tangled up with each other and not sort properly during meiosis and mitosis. The condensation process probably occurs so that the chromosomes easily align along the equatorial plate during metaphase without getting tangled up. C17. To produce identical quadruplets, fertilization begins with one sperm and one egg cell. This fertilized egg then could divide twice by mitosis to produce four genetically identical cells. These four cells could then separate from each other to begin the lives of four distinct individuals. Another possibility is that mitosis could produce two cells that separate from each other. These two cells could then divide by mitosis to produce two pairs of cells, which also could separate to produce four individual cells. C18. During prophase of meiosis II, your drawing should show four replicated chromosomes (i.e., four structures that look like Xs). Each chromosome is one homolog. During prophase of mitosis, there should be eight replicated chromosomes (i.e., eight Xs). During prophase of mitosis, there are pairs of homologs. The main difference is that prophase of meiosis II has a single copy of each of the four chromosomes, whereas prophase of mitosis has four pairs of homologs. At the end of meiosis I, each daughter cell has received only one copy of a homologous pair, not both. This is due to the alignment of homologs during metaphase of meiosis I and their separation during anaphase of meiosis I. C19. The products of meiosis have only one copy of each type of chromosome. For example, one human gamete may contain the paternally derived copy of chromosome 11, whereas a different gamete may contain the maternally derived copy of chromosome 11. These two homologs may carry different alleles of the same genes and therefore are not identical. In contrast, mitosis produces genetically identical daughter cells that have both copies of all the pairs of homologous chromosomes. C20. DNA replication does not take place during interphase II. The chromosomes at the end of telophase of meiosis I have already replicated (i.e., they are found in pairs of sister chromatids). During meiosis II, the sister chromatids separate from each other, yielding individual chromosomes. C21. Prophase/Prometaphase Telophase Nuclear membrane dissociates. Nuclear membrane re-forms. Mitotic spindle forms. Mitotic spindle disassembles. Page | 8 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Chromosomes condense. Chromosomes decondense. Chromosomes attach to spindle. Chromosomes detach from the spindle. C22. A. 20 B. 10 C. 30 D. 20 C23. The hybrid offspring would have 44 chromosomes (i.e., 25 + 19). The reason for infertility is because each chromosome does not have a homologous partner. Therefore, the chromosomes cannot properly pair during metaphase of meiosis I, and the gametes do not receive one copy of each homolog. Gametes will be missing certain chromosomes, which makes them infertile. C24. Male gametes are usually small and mobile. Animal and some plant male gametes have flagella, which make them motile. The mobility of the male gamete makes it likely that it will come in contact with the female gamete. Female gametes are usually much larger and contain nutrients to help the growth of the embryo after fertilization occurs. C25. To produce sperm, a spermatogonial cell first goes through mitosis to produce two cells. One of these remains a spermatogonial cell and the other advances through meiosis. In this way, the testes continue to maintain a population of spermatogonial cells. C26. During oogenesis in humans, the cells are arrested in prophase of meiosis I for many years until selected primary oocytes advance through the rest of meiosis I and begin meiosis II. If fertilization occurs, meiosis II is completed. C27. There is a 1/2 chance that the mother will transmit her abnormal chromosome and a 1/2 chance that the father will. You use the product rule to calculate the chances of both outcomes happening. So the answer is 1/2 × 1/2 = 1/4, or 25%. The probability that such a child will pass both chromosomes to an offspring is also 25% because that child had a 1/2 chance of passing either chromosome. Application and Experimental Questions E1. A. G2 phase (it could not complete prophase) B. Metaphase (it could not enter anaphase) C. Telophase (it could not divide into two daughter cells) D. G2 phase (it could not enter prophase) E2. During interphase, the chromosomes are longer, thinner, and much harder to see. In metaphase, they are highly condensed, which makes them thicker and shorter. E3. You could karyotype other members of the family and see if affected members always carry the abnormal chromosome. Questions for Student Discussion/Collaboration 1. It’s not possible to give a direct answer, but the point is for students to be able to draw chromosomes in different configurations and understand the various phases. The chromosomes may or may not be: 1. In homologous pairs 2. Connected as sister chromatids 3. Associated in bivalents 4. Lined up in metaphase Page | 9 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 5. Moving toward the poles. 2. A major advantage of sexual reproduction is that it fosters genetic diversity in future populations. A major disadvantage is that to reproduce, each individual must find a mate of the opposite sex. CHAPTER 3 Note: the answers to Comprehension Questions are at the end of the chapter. Concept Check Questions (in figure legends) FIGURE 3. 2 The male gamete is found within pollen. FIGURE 3. 3 The white flower is providing the sperm and the purple flower is providing the eggs. FIGURE 3. 4 A true-breeding strain maintains the same trait over the course of many generations. FIGURE 3. 6 Segregation means that the T and t alleles separate from each other so that a haploid cell receives one of them, but not both. FIGURE 3. 7 In this hypothesis, two different genes are linked. The alleles of the same gene are not linked. FIGURE 3. 9 Independent assortment allows for new combinations of alleles among different genes to be found in future generations of offspring. FIGURE 3. 10 Such a parent could make two types of gametes, Ty and ty, in equal proportions. FIGURE 3. 12 Homologous chromosomes separate at anaphase of meiosis I. FIGURE 3. 13 chromosomes could line up in four different ways. FIGURE 3. 14 Horizontal lines connect two individuals that have offspring together, and they connect all of the offspring that produced by the same two parents. End-of-chapter Questions: Conceptual Questions C1. Mendel’s work showed that genetic determinants are inherited in a dominant/recessive manner. This was readily apparent in many of his crosses. For example, when he crossed two true-breeding plants for a trait such as height (i.e., tall versus dwarf), all the F1 plants were tall. This is inconsistent with blending. Perhaps more striking was the result obtained in the F2 generation: 3/4 of the offspring were tall and 1/4 were short. In other words, the F2 generation displayed phenotypes that were like the parental generation. There did not appear to be a blending to create an intermediate phenotype. Instead, the genetic determinants did not seem to change from one generation to the next. Page | 10 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. C2. In plants, cross-fertilization occurs when the pollen and eggs come from different plants; in selffertilization they come from the same plant. C3. The genotype is the type of genes that an individual inherits while the phenotype is the individual’s observable traits. Tall pea plants, red hair in humans, and vestigial wings in fruit flies are phenotypes. Homozygous, TT, in pea plants; a heterozygous carrier of the cystic fibrosis allele; and homozygotes for the cystic fibrosis allele are descriptions of genotypes. It is possible to have different genotypes and the same phenotype. For example, a pea plant that is TT or Tt would both have a tall phenotype. C4. A true-breeding organism is homozygote that has two copies of the same allele. C5. Conduct a cross in which the unknown individual is mated to an individual that carries only recessive alleles for the gene in question. C6. Diploid organisms contain two copies of each type of gene. When they make gametes, only one copy of each gene is found in a gamete. Two alleles cannot stay together within the same gamete. C7. B. This statement is not correct because these are alleles of different genes. C8. Genotypes: 1 Tt : 1 tt Phenotypes: 1 tall : 1 dwarf C9. The recessive phenotype must be a homozygote. The dominant phenotype could be either homozygous or heterozygous. C10. In this cross, c is the recessive allele for constricted pods; Y is the dominant allele for yellow color. The cross is ccYy × CcYy. Follow the directions for setting up a Punnett square, as described in Section 3.3. The genotypic ratio is 2 CcYY : 4 CcYy : 2 Ccyy : 2 ccYY : 4 ccYy : 2 ccyy. This 2:4:2:2:4:2 ratio can be reduced to a 1:2:1:1:2:1 ratio. The phenotypic ratio is 6 smooth pods, yellow seeds : 2 smooth pods, green seeds: 6 constricted pods, yellow seeds : 2 constricted pods, green seeds. This 6:2:6:2 ratio could be reduced to a 3:1:3:1 ratio. C11. The genotypes are 1 YY : 2 Yy : 1 yy. The phenotypes are 3 yellow : 1 green. C12. Offspring with a nonparental phenotype are consistent with the idea of independent assortment. If two different traits were always transmitted together as unit, it would not be possible to get nonparental phenotypic combinations. For example, if a true-breeding parent had two dominant traits and was crossed to a true-breeding parent having the two recessive traits, the F2 generation could not have offspring with one recessive and one dominant trait. However, because independent assortment can occur, it is possible for F2 offspring to have one dominant and one recessive trait. C13. (a) It behaves like a recessive trait because unaffected parents sometimes produce affected offspring. In such cases, the unaffected parents are heterozygous carriers. (b) It behaves like a dominant trait. An affected offspring always has an affected parent. However, recessive inheritance cannot be ruled out. C14. A. Barring a new mutation during gamete formation, the probability is 100%. They must be heterozygotes in order to produce a child with a recessive disorder. B. Construct a Punnett square. There is a 50% chance of heterozygous offspring. C. Use the product rule. The chance of being phenotypically unaffected is 0.75 (i.e., 75%), so the answer is 0.75 × 0.75 × 0.75 = 0.422, which is 42.2%. D. Use the binomial expansion equation, where n = 3, x = 2, p = 0.75, q = 0.25. The answer is 0.422, or 42.2%. Page | 11 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. C15. A. 100% because they are genetically identical. B. Construct a Punnett square. You know the parents are heterozygotes because they produced a blue-eyed child. The fraternal twin is not genetically identical, but she or he has the same parents as its twin. The answer is 25%. C. The probability that an offspring inherits the allele is 50% and the probability that this offspring will pass it on to his/her offspring is also 50%. You use the product rule: (0.5)(0.5) = 0.25, or 25%. D. Barring a new mutation during gamete formation, the chance is 100% because they must be heterozygotes in order to produce a child with blue eyes. C16. First construct a Punnett square. The chances are 75% of producing a solid pup and 25% of producing a spotted pup. A. Use the binomial expansion equation, where n = 5, x = 4, p = 0.75, q = 0.25. The answer is 0.396, or 39.6%. B. You can use the binomial expansion equation for each litter. For the first litter, n = 6, x = 4, p = 0.75, q = 0.25; for the second litter, n = 5, x = 5, p = 0.75, q = 0.25. Because the litters are in a specified order, you use the product rule and multiply the probability of the first litter times the probability of the second litter. The answer is 0.070, or 7.0%. C. To calculate the probability of the first litter, you use the product rule and multiply the probability of the first pup (0.75) times the probability of the remaining four. You use the binomial expansion equation to calculate the probability of the remaining four, where n = 4, x = 3, p = 0.75, q = 0.25. The probability of the first litter is 0.316. To calculate the probability of the second litter, you use the product rule and multiply the probability of the first pup (0.25) times the probability of the second pup (0.25) times the probability of the remaining five. To calculate the probability of the remaining five, you use the binomial expansion equation, where n = 5, x = 4, p = 0.75, q = 0.25. The probability of the second litter is 0.025. To get the probability of these two litters occurring in this order, you use the product rule and multiply the probability of the first litter (0.316) times the probability of the second litter (0.025). The answer is 0.008, or 0.8%. D. Because this is a specified order, you use the product rule and multiply the probability of the firstborn (0.75) times the probability of the second born (0.25) times the probability of the remaining four. You use the binomial expansion equation to calculate the probability of the remaining four pups, where n = 4, x = 2, p = 0.75, q = 0.25. The answer is 0.040, or 4.0%. C17. If B is the black allele, and b is the white allele, the male is bb, the first female is probably BB, and the second female is Bb. You are uncertain of the genotype of the first female. She could be Bb, although it is unlikely because she didn’t produce any white pups out of a litter of eight. C18. A. Use the product rule: (1/4)(1/4)=1/16 B. Use the binomial expansion equation: n = 4, p = ¼, q = ¾, x = 2 P = 0.21, or 21% C. Use the product rule: (1/4)(3/4)(3/4) = 0.14, or 14% C19. The parents must be heterozygotes, so the probability is 1/4. C20. A. 1/4 B. 1, or 100% Page | 12 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. C. (3/4)(3/4)(3/4) = 27/64 = 0.42, or 42% D. Use the binomial expansion equation, where n = 7, p = 3/4, q = 1/4, x = 3 P = 0.058, or 5.8% E. The probability that the first plant is tall is 3/4. To calculate the probability that among the next four, any two will be tall, you use the binomial expansion equation, where n = 4, p = 3/4, q = 1/4, x = 2. The probability P equals 0.21. To calculate the overall probability of these two outcomes: (3/4)(0.21) = 0.16, or 16% C21. A. T Y R, T y R, T Y r, T y r B. T Y r, t Y r C. T Y R, T Y r, T y R, T y r, t Y R, t Y r, t y R, t y r D. t Y r, t y r C22. The gamete violates the law of segregation because two copies of one gene are in the gamete. The two alleles for the A gene did not segregate from each other. C23. It is recessive inheritance. The pedigree is shown here. Affected individuals are shown with filled (black) symbols. The mode of inheritance appears to be recessive. Unaffected parents (who must be heterozygous) produce affected children. C24. Based on this pedigree provided, the disease is likely to be a dominant trait because an affected child always has an affected parent. In fact, Marfan syndrome is a dominant disorder. C25. A. 3/16 B. You can either use the binomial expansion equation where n = 3, x = 3, and p = 9/16 or you could use the product rule. The answer is 0.18. C. Use the multinomial expansion equation, where n = 5, a = 2, b = 1, c = 2, p = 9/16, q = 3/16, and r = 1/16. The answer is 0.007. D. Another way of looking at this is that the probability it will have round, yellow seeds is 9/16. Therefore, the probability that it will not is 1 – 9/16 = 7/16. C26. It is impossible for the F1 individuals to be true-breeding because they are all heterozygotes. Page | 13 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. C27. This problem is a bit unwieldy, but you can solve it using the multiplication method. For height, the ratio is 3 tall : 1 dwarf. For seed texture, the ratio is 1 round : 1 wrinkled. For seed color, they are all yellow. For flower location, the ratio is 3 axial : 1 terminal. Thus, the product is (3 tall + 1 dwarf)(1 round + 1 wrinkled)(1 yellow)(3 axial + 1 terminal) Multiplying this out, the answer is 9 tall, round, yellow, axial 9 tall, wrinkled, yellow, axial 3 tall, round, yellow, terminal 3 tall, wrinkled, yellow, terminal 3 dwarf, round, yellow, axial 3 dwarf, wrinkled, yellow, axial 1 dwarf, round, yellow, terminal 1 dwarf, wrinkled, yellow, terminal C28. 2 TY, tY, 2 Ty, ty, TTY, TTy, 2 TtY, 2 Tty This question is a bit tricky, but you get 2 TY and 2 Ty because either of the two T alleles could combine with Y or y. Also, you get 2 TtY and 2 Tty because either of the two T alleles could combine with t and then combine with Y or y. C29. The drone is sB and the queen is SsBb. According to the laws of segregation and independent assortment, the male can make only sB gametes, while the queen can make SB, Sb, sB, and sb, in equal proportions. Therefore, male offspring will be SB, Sb, sB, and sb, and female offspring will be SsBB, SsBb, ssBB, and ssBb. The phenotypic ratios, assuming an equal number of males and females, will be: Males Females 1 normal wings/black eyes 2 normal wings, black eyes 1 normal wings/white eyes 2 short wings, black eyes 1 short wings/black eyes 1 short wings/white eyes C30. The genotype of the F1 plants is Tt Yy Rr. According to the laws of segregation and independent assortment, the alleles of each gene will segregate from each other, and the alleles of different genes will randomly assort into gametes. A Tt Yy Rr individual could make eight types of gametes: TYR, TyR, Tyr, TYr, tYR, tyR, tYr, and tyr, in equal proportions (i.e., 1/8 of the gametes will be of each type). To determine genotypes and phenotypes, you could make a large Punnett square that contains 64 boxes. You would need to line up the eight possible gametes across the top and along the side, and then fill in the 64 boxes. Alternatively, you could use the multiplication method or the forkedline method. The genotypes and phenotypes are as follows: 1 TT YY RR 2 TT Yy RR 2 TT YY Rr 2 Tt YY RR 4 TT Yy Rr Page | 14 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 4 Tt Yy RR 4 Tt YY Rr 8 Tt Yy Rr= 27 tall, yellow, round 1 TT yy RR 2 Tt yy RR 2 TT yy Rr 4 Tt yy Rr = 9 tall, green, round 1 TT YY rr 2 TT Yy rr 2 Tt YY rr 4 Tt Yy rr = 9 tall, yellow, wrinkled 1 tt YY RR 2 tt Yy RR 2 tt YY Rr 4 tt Yy Rr = 9 dwarf, yellow, round 1 TT yy rr 2 Tt yy rr = 3 tall, green, wrinkled 1 tt yy RR 2 tt yy Rr = 3 dwarf, green, round 1 tt YY rr 2 tt Yy rr = 3 dwarf, yellow, wrinkled 1 tt yy rr = 1 dwarf, green, wrinkled C31. Construct a Punnett square to determine the probability of these three phenotypes. The probabilities are 9/16 for round, yellow; 3/16 for round, green; and 1/16 for wrinkled, green. Use the multinomial expansion, where n = 5, a = 2, b = 1, c = 2, p = 9/16, q = 3/16, r = 1/16. The answer is 0.007, or 0.7%, of the time. C32. The woolly haired male is a heterozygote, because he has the trait and his mother did not. (He must have inherited the normal allele from his mother.) Therefore, he has a 50% chance of passing the woolly allele to his offspring; his offspring have a 50% of passing the allele to their offspring; and these grandchildren have a 50% chance of passing the allele to their offspring (the woolly haired man’s great-grandchildren). Because this is an ordered sequence of independent outcomes, you use the product rule: 0.5 × 0.5 × 0.5 = 0.125, or 12.5%. Because no other Scandinavians are on the island, the chance is 87.5% for the offspring being normal (because they could not inherit the woolly hair allele from anyone else). You use the binomial expansion equation to determine the likelihood that one out of eight great-grandchildren will have woolly hair, where n = 8, x = 1, p = 0.125, q = 0.875. The answer is 0.393, or 39.3%, of the time. C33. A. Construct a Punnett square. Because it is a rare disease, you would assume that the mother is a heterozygote and the father is normal. The chances are 50% that the man in his thirties will have the allele. B. Use the product rule: 0.5 (chance that the man has the allele) times 0.5 (chance that he will pass it to his offspring), which equals 0.25, or 25%. Page | 15 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. C. You use the binomial expansion equation. From part B, you calculated that the probability of an affected child is 0.25. Therefore, the probability of an unaffected child is 0.75. For the binomial expansion equation, n = 3, x = 1, p = 0.25, q = 0.75. The answer is 0.422 or 42.2%. C34. Use the product rule. If the woman is heterozygous, there is a 50% chance of having an affected offspring: (0.5)7 = 0.0078, or 0.78%, of the time. This is a pretty small probability. If the woman has an eighth child who is unaffected, however, she has to be a heterozygote, because it is a dominant trait. She would have to pass a normal allele to an unaffected offspring. The answer is 100%. Application and Experimental Questions E1. Pea plants are relatively small and hardy. They produce both pollen and eggs within the same flower. Because a keel covers the flower, self-fertilization is quite easy. In addition, crossfertilization is possible by the simple manipulation of removing the anthers in an immature flower and later placing pollen from another plant. Finally, peas exist in several variants. E2. The difference lies in where the pollen comes from. In self-fertilization, the pollen and eggs come from the same plant. In cross-fertilization, they come from different plants. E3. Two generations would take two growing seasons. About 1 and 1/2 years. E4. According to Mendel’s law of segregation, the genotypic ratio should be 1 homozygote dominant : 2 heterozygotes : 1 homozygote recessive. The data table considers only the plants with a dominant phenotype. The genotypic ratio should be 1 homozygote dominant : 2 heterozygotes. The homozygote dominants would be true-breeding while the heterozygotes would not be true-breeding. This 1:2 ratio is very close to what Mendel observed. E5. In a single-factor experiment, the experimenter is only concerned with the outcome of a single trait. In a two-factor experiment, the experimenter follows the pattern of inheritance for two different traits. E6. All three offspring had black fur. The ovaries from the albino female could only produce eggs with the dominant black allele (because they were obtained from a true-breeding black female). The actual phenotype of the albino mother does not matter. Therefore, all offspring were heterozygotes (Bb) with black coats. E7. The data are consistent with two genes (let’s call them gene 22 and gene 24) that exist in two alleles each, a susceptible allele and a resistant allele. The observed data approximate a 9:3:3:1 ratio. This is the expected ratio if two genes are involved, and if resistance is dominant to susceptibility. E8. If you construct a Punnett square according to Mendel’s laws, you expect a 9:3:3:1 ratio. Because a total of 556 offspring were observed, the expected number of offspring are 556 × 9/16 = 313 round, yellow 556 × 3/16 = 104 wrinkled, yellow 556 × 3/16 = 104 round, green 556 × 1/16 = 35 wrinkled, green If you plug the observed and expected values into the chi square equation, you get a value of 0.51. With four categories, the degrees of freedom are n – 1, or 3. If you look up the value of 0.51 in the chi square table (see Table 3.2), you see that it falls between the P values of 0.80 and 0.95. This means that the probability is a 80% to 95% that a value equal to or greater than 0.51 is expected to occur due to random sampling error. Therefore, you accept the hypothesis. In other words, the results are consistent with the law of independent assortment. Page | 16 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. E9. No, the law of independent assortment applies to transmission patterns of two or more genes. In a single-factor experiment, you are monitoring only the transmission pattern of a single gene. E10. A. Let c + represent straight wings and c represent curved wings, and e + represents gray body and e represents ebony body: Parental cross: cc e + e +  c + c + ee. F1 generation is heterozygous: c + c e + e An F1 offspring crossed to a fly with curved wings and ebony body is c + c e + e  ccee The F2 offspring would have this genotypic ratio: c + c e + e : c + cee : cc e + e : ccee B. The phenotypic ratio of the F2 flies is 1:1:1:1, as follows: straight wings, gray body : straight wings, ebony bodies : curved wings, gray bodies : curved wings, ebony bodies C. From part B, you expect 1/4 of each category. There are a total of 444 offspring. The expected number of each category is 1/4  444, which equals 111.  2 = (114 −111)2 + (105 −111)2 + (111−111)2 + (114 −111)2  2 = 0.49 With 3 degrees of freedom, a value of 0.49 or greater is likely to occur between 80% and 95% of the time. Therefore, you accept our hypothesis. E11. You would expect a ratio of 3 normal : 1 long neck. In other words, there should be 1/3 as many long-necked mice as normal mice. If you multiply 522 times 1/3, the expected value is 174. However, only 62 were observed. Therefore, it appears that 174 – 62, or 112, mice died during early embryonic development; 112 divided by 174 gives us the percentage that died, which equals 0.644, or 64.4%. E12. Follow the basic chi square strategy. You expect a 3:1 ratio, or 3/4 of the dominant phenotype and 1/4 of the recessive phenotype. The observed and expected values are as follows (rounded to the nearest whole number): Observed* Expected (O − E) 2 E 5,474 5,493 0.066 1,850 1,831 0.197 6,022 6,017 0.004 2,001 2,006 0.012 705 697 0.092 224 232 0.276 882 886 0.018 299 295 0.054 428 435 0.113 152 145 0.338 Page | 17 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 651 644 0.076 207 215 0.298 787 798 0.152 277 266 0.455 2 = 2.15 *Due to rounding, the observed and expected values may not add up to precisely the same number. Because n = 14, there are 13 degrees of freedom. If you look up this value in the chi square table, you have to look between 10 and 15 degrees of freedom. In either case, you expect a value of 2.15 or greater to occur more than 99% of the time. Therefore, you accept the hypothesis. E13. This means that a deviation value of 1.005 or greater (between the observed and expected data) would occur 80% of the time. In other words, it is fairly likely to obtain this value due to random sampling error. Therefore, you accept our hypothesis. E14. The dwarf parent with terminal flowers must be homozygous for both genes, because it is expressing these two recessive traits: ttaa, where t is the recessive dwarf allele, and a is the recessive allele for terminal flowers. The phenotype of the other parent is dominant for both traits. Because this parent was able to produce dwarf offspring with axial flowers, it must have been heterozygous for both genes: TtAa. E15. Your hypothesis is that disease sensitivity and herbicide resistance are dominant traits and they are governed by two genes that assort independently. According to this hypothesis, the F2 generation should yield a ratio of 9 disease sensitive, herbicide resistant : 3 disease sensitive, herbicide sensitive : 3 disease resistant, herbicide resistant : 1 disease resistant, herbicide sensitive. Because there are a total of 288 offspring produced, the expected numbers would be 9/16  288 = 162 disease sensitive, herbicide resistant 3/16  288 = 54 disease sensitive, herbicide sensitive 3/16  288 = 54 disease resistant, herbicide resistant 1/16  288 = 18 disease resistant, herbicide sensitive  2 = (157 −162)2 + (57 − 54)2 + (54 − 54)2 + (20 −18)2  2 = 0.54 If you look up this value in the chi square table under 3 degrees of freedom, the value lies between the 0.95 and 0.80 probability values. Therefore, you expect a value equal to or greater than 0.54, at least 80% of the time, due to random sampling error. Therefore, you accept the hypothesis. E16. You need to make crosses to understand the pattern of inheritance of traits (determined by genes) from parents to offspring. And you need to microscopically examine cells to understand the pattern of transmission of chromosomes. The correlation between the pattern of transmission of chromosomes during meiosis, and Mendel’s laws of segregation and independent assortment, is what led to the chromosome theory of inheritance. Questions for Student Discussion/Collaboration 1. The methods for making a Punnett square and using the multiplication method are described in the chapter. Presumably, the multiplication method (or the forked-line method) will work a lot faster. Page | 18 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2. If you construct a Punnett square, the following probabilities will be obtained: tall with axial flowers: 3/8 dwarf with terminal flowers: 1/8 The probability of being tall with axial flowers or dwarf with terminal flowers: 3/8 + 1/8 = 4/8 = 1/2 You use the product rule to calculate the probability of the ordered outcome of the first three offspring being tall/axial or dwarf/terminal, and fourth offspring being tall/axial: (1/2)(1/2)(1/2)(3/8) = 3/64 = 0.047 = 4.7% 3. Ignore the other genes. The cross is Rr × RR. All of the offspring will have round seeds. The probability is 100% CHAPTER 4 Note: the answers to the Comprehension Questions are at the end of the chapter. Concept Check Questions (in figure legends) FIGURE 4. 1 In the X-Y system, the presence of the Y chromosome causes maleness, whereas in the X-0 system, it is the ratio between the number of X chromosomes and number of sets of autosomes that determines sex. A ratio of 0.5 results in a male and a ratio of 1.0 produces a female. FIGURE 4. 2 A male bee is not produced by sexual reproduction because it is produced from an unfertilized egg. Sexual reproduction involves the union of gametes. FIGURE 4. 3 A higher average temperature would favor a high percentage of male alligators. Assuming that males are not limiting for alligator reproduction, fewer females would be expected to decrease the population size of alligators. FIGURE 4. 4 The sporophytes are the opposite sexes in dioecious plants. FIGURE 4. 5 The Barr body is more brightly stained because it is very compact. FIGURE 4. 6 X-chromosome inactivation first occurs during embryonic development. It is then maintained into adulthood. FIGURE 4. 7 Only the maintenance phase occurs in an adult female. FIGURE 4. 9 In the F2 generation, only the males had white eyes. FIGURE 4. 10 The key pedigree feature that points to X-linked inheritance is that only males are affected with the disorder. Also, carrier females often have affected brothers. FIGURE 4. 11 Page | 19 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. The reason why the reciprocal cross yields a different result is because females carry two copies of an X-linked gene whereas males have only one. End-of-chapter Questions: Conceptual Questions C1. In the X-Y system the male is heterogametic whereas in the Z-W system the female is heterogametic. C2. A. Dark males and light females; reciprocal: all dark offspring B. All dark offspring; reciprocal: dark females and light males C. All dark offspring; reciprocal: dark females and light males D. All dark offspring; reciprocal: dark females and light males C3. A. The fly is a male because the ratio of X chromosomes to sets of autosomes is 1/2, or 0.5. B. The fly is female because the ratio of X chromosomes to sets of autosomes is 1.0. C. The fly is male because the ratio of X chromosomes to sets of autosomes is 0.5. D. The fly is female because the ratio of X chromosomes to sets of autosomes is 1.0. C4. A. Female; there are no Y chromosomes. B. Female; there are no Y chromosomes. C. Male; the Y chromosome determines maleness. D. Male; the Y chromosome determines maleness. C5. Dosage compensation refers to the phenomenon that the level of expression of genes on the sex chromosomes is similar in males and females, even though they have different numbers of sex chromosomes. In many species it seems necessary so that the balance of gene expression between the autosomes and sex chromosomes is similar between the two sexes. C6. A Barr body is a mammalian X chromosome that is highly condensed. It is found in somatic cells with two or more X chromosomes. Most genes on the Barr body are inactive. C7. In mammals, one of the X chromosomes is inactivated in females; in Drosophila, the level of transcription on the X chromosome in males is doubled; in C. elegans, the level of transcription of the X chromosome in hermaphrodites is decreased by 50% of that of males. C8. X-chromosome inactivation in heterozygous females produces a mosaic pattern of gene expression. During early embryonic development, some cells have the maternal X chromosome inactivated and other cells have the paternal X chromosome inactivated; these embryonic cells will divide and produce billions of cells. In the case of a female that is heterozygous for a gene that affects pigmentation of the fur, this produces a variegated pattern of coat color. Because it is a random process in any given animal, two female cats will vary as to where the orange and black patches occur. A variegated coat pattern could not occur in female marsupials due to X-chromosome inactivation because the paternal X chromosome is always inactivated in the somatic cells of females. C9. X-chromosome inactivation begins with the counting of Xics. If there are two X chromosomes, in the process of initiation, one is targeted for inactivation. During embryogenesis, this inactivation begins at the Xic locus and spreads to both ends of the X chromosome until it becomes a highly condensed Barr body. The Xist gene, which is located in the Xic region, remains transcriptionally active on the inactivated X chromosome. It is thought to play an important role in X-chromosome inactivation by coating the inactive X chromosome. After X-chromosome inactivation is Page | 20 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. established, it is maintained in the same X chromosome in somatic cells during subsequent cell divisions. In germ cells, however, the X chromosomes are not inactivated, so that an egg can transmit either copy of an active (noncondensed) X chromosome. C10. The male is XXY. The person is male due to the presence of the Y chromosome. Because two X chromosomes are counted, one of the X chromosomes is inactivated to produce a Barr body. C11. A. One B. Zero C. Two D. Zero C12. A. In females, one of the X chromosomes is inactivated. When the X chromosome that is inactivated carries the normal allele, only the defective color-blindness allele will be expressed. Therefore, on average, about half of a female’s eye cells are expected to express the common (not color blind) allele. Depending on the relative amounts of cells expressing the common versus the color-blindness allele, the end result may be partial color blindness. B. In this female, as a matter of chance, X-chromosome inactivation in the right eye always, or nearly always, inactivated the X chromosome carrying the normal allele. The opposite occurred in the left eye. In the left eye, the chromosome carrying the color-blindness allele was primarily inactivated. C13. The offspring inherited XB from its mother and XO and Y from its father. It is an XXY animal, which is male (but somewhat feminized). C14. The spreading phase is when the X chromosome is inactivated (i.e., condensed) in a wave that spreads outward from the X-inactivation center (Xic). The condensation spreads from Xic to the rest of the X chromosome. The Xist gene is transcribed from the inactivated X chromosome. It encodes an RNA that coats the X chromosome, which subsequently attracts proteins that are responsible for the compaction. C15. The outcome is that all the daughters will be affected and all the sons will be unaffected. The ratio is 1:1. C16. First set up the following Punnett square: A. 1/4 B. (3/4)(3/4)(3/4)(3/4) = 81/256 C. 3/4 Page | 21 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. D. The probability of an affected offspring is ¼, and the probability of an unaffected offspring is 3/4. For this problem, you use the binomial expansion equation, where x = 2, n = 5, p = 1/4, and q = 3/4. The answer is 0.26, or 26%, of the time. C17. 1 affected daughter : 1 unaffected daughter : 1 affected son : 1 unaffected son Application and Experimental Questions E1. The first type of observation was based on cytological studies. The presence of the Barr body in female cells was consistent with the idea that one of the X chromosomes was highly condensed. The second type of observation was based on the phenotypes of heterozygous females. A variegated phenotype that is found only in females is consistent with the idea that certain patches express one allele and other patches express the other allele. This variegated phenotype would occur only if the inactivation of one X chromosome happened at an early stage of embryonic development and was inherited permanently thereafter. E2. In general, you cannot distinguish between autosomal and pseudoautosomal inheritance from a pedigree analysis. Mothers and fathers have an equal probability of passing the alleles to sons and daughters. However, if an offspring had a chromosomal abnormality, you might be able to tell. For example, in a family tree involving the Mic2 allele, an offspring that was X0 would have less of the gene product and an offspring that was XXX or XYY or XXY would have an extra amount of the gene products. This may lead you to suspect that the gene is located on the sex chromosomes. E3. Perhaps the most convincing observation was that all of the white-eyed flies of the F2 generation were males. This suggests a link between sex determination and the inheritance of this trait. Because sex determination in fruit flies is determined by the number of X chromosomes, this suggests a relationship between the inheritance of the X chromosome and the inheritance of this trait. E4. Actually, his data are consistent with this hypothesis. To rule out a Y-linked allele, he could have crossed an F1 female with a red-eyed male rather than an F1 male. The same results would be obtained. Because the red-eyed male would not have a white allele, this would rule out Y-linkage. E5. The basic strategy is to set up a pair of reciprocal crosses. The phenotype of sons is usually the easiest way to discern the two patterns. If it is Y linked, the trait will be passed only from father to son. If it is X linked, the trait will be passed from mother to son. E6. To be a white-eyed female, a fly must inherit two X chromosomes and both must carry the whiteeye allele. This could occur only if both X chromosomes in the female stayed together and the male gamete contained a Y chromosome. The white-eyed females would be XXY. To produce a red-eyed male, a female gamete lacking any sex chromosomes could unite with a normal male gamete carrying the X w+ . This would produce an X0, red-eyed male. E7. If you use the data from the F1 mating (i.e., F2 results), there were 3,470 red-eyed flies. You would expect a 3:1 ratio between red- and white-eyed flies. Therefore, assuming that all red-eyed offspring survived, there should have been about 1,157 (i.e., 3,470/3) white-eyed flies. However, there were only 782. If you divide 782 by 1,157, you get a value of 0.676, or a 67.6% survival rate. E8. The rare female flies would be XXY. Both X chromosomes would carry the white-eye allele and the miniature allele. These female flies would have miniature wings because they would have inherited both X chromosomes from their mother. Page | 22 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. E9. Originally, individuals who had abnormalities in their composition of sex chromosomes provided important information. In mammals, X0 individuals are females, while in flies, X0 individuals are males. In mammals, XXY individuals are males, while in flies, XXY individuals are females. These results indicate that the presence of the Y chromosome causes maleness in mammals, but it does not in flies. A further analysis of flies with abnormalities in the number of sets of autosomes revealed that it is the ratio between the number of X chromosomes and the number of sets of autosomes that determines sex in flies. Questions for Student Discussion/Collaboration 1. The wide variety of sex determination mechanisms suggests that such mechanisms may have arisen independently on multiple occasions. It would seem that it may have arisen relatively recently after the major groups of animals diverged from each other. 2. One possibility is fertilization of an abnormal female gamete. The white-eyed male parent could make sperm carrying the X chromosome that could fertilize a female gamete without any sex chromosomes. This would produce an X0, white-eyed male. This is the most likely explanation. Another possibility is that the flies in Morgan’s lab were not completely true-breeding. The allele creating the white phenotype may have occurred several generations earlier in a female fly. Perhaps, among the many red-eyed sisters, one of them could have already been a heterozygote. The rare sister could produce white-eyed male offspring. New mutations are a third, but unlikely possibility, because you already know that the mutation rate is very low. It took Morgan 2 years to get one white-eyed male. CHAPTER 5 Note: the answers to the Comprehension Questions are at the end of the chapter. Concept Check Questions (in figure legends) FIGURE 5. 1 Both of these colors are considered wild type because both are prevalent in natural populations. FIGURE 5. 2 Yes. The PP homozygote probably makes twice the amount of protein that is needed for purple pigment formation. FIGURE 5. 3 Individual III-2 shows incomplete penetrance. FIGURE 5. 4 Genes and the environment determine an organism’s traits. FIGURE 5. 5 50% of the functional protein is not enough to give a red color. FIGURE 5. 6 It is often easier to observe incomplete dominance at the molecular or cellular level. FIGURE 5. 7 In this case, the heterozygote is resistant to malaria. Page | 23 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. FIGURE 5. 8 The scenario in part (a) explains the effects of the sickle-cell allele. FIGURE 5. 9 The i allele is a loss-of-function allele. FIGURE 5. 10 A heterozygous female does not have scurs. FIGURE 5. 11 Certain traits are expressed only in males or females, possibly due to differences in the levels of sex hormones or other factors that differ between the sexes. FIGURE 5. 12 The heterozygote has one normal copy of the gene, which allows for development to proceed in a way that is not too far from normal. Having two mutant copies of the gene probably adversely affects development to a degree that is incompatible with survival. FIGURE 5. 14 Epistasis means that the alleles of one gene mask the phenotypic effects of the alleles of a different gene. Complementation occurs when two strains exhibiting the same recessive trait produce offspring that show the dominant (wild-type) trait. This usually means that the alleles for the recessive trait are in two different genes. FIGURE 5. 15 The two genes are redundant. Having one functional copy of either gene produces a triangular capsule. If both genes are nonfunctional, an ovate capsule is produced. End-of-chapter Questions: Conceptual Questions C1. Dominance occurs when one allele completely exerts its phenotypic effects over another allele. Incomplete dominance is a situation in which two alleles in the heterozygote have an intermediate phenotype. In codominance, both alleles exert their effects independently in the heterozygote. Overdominance is a case in which the heterozygote has a phenotype that is superior to either homozygote. C2. Sex-influenced traits are affected by the sex of the individual even though the gene that governs the trait is autosomally inherited. Scurs in cattle is an example. The expression of a sex-limited trait is limited to one sex. For example, colorful plumage in certain species of birds is limited to the male sex. Sex-linked inheritance are traits whose genes are found on the sex chromosomes. Examples in humans include hemophilia and color blindness. C3. The term gene interaction refers to the phenomenon that two or more different genes can have an impact on the same trait. This can occur, for example, if two genes encode enzymes in the same metabolic pathway. If the pathway is disrupted by a mutation in either of these two genes, the same net result may occur. C4. If the functional allele is dominant, then one copy of the gene produces a sufficient amount of the protein. Having twice as much of this protein, as in the normal homozygote, does not alter the phenotype. If the allele is incompletely dominant, this means that one copy of the normal allele does not produce the same trait as the homozygote. Page | 24 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. C5. Recessive alleles are often loss-of-function alleles. It would generally be more likely for a recessive allele to eliminate a trait or function rather than create one. Therefore, the peach carries the dominant allele, whereas the nectarine has a loss-of-function allele that prevents fuzz formation. The recessive allele is in a gene that is necessary for fuzz formation. C6. The ratio would be 1 normal : 2 star-eyed individuals. C7. If individual 1 is ii, individual 2 could be I A i, I A I A , I B i, I B I B , or I A I B . If individual 1 is I A i or I A I A , individual 2 could be I B i, I B I B , or I A I B . If individual 1 is I B i or I B I B , individual 2 could be I A i, I A I A , or I A I B . Assuming individual 1 is the parent of individual 2: If individual 1 is ii, individual 2 could be I A i or I B i. If individual 1 is I A i, individual 2 could be I B i or I A I B . If individual 1 is I A I A , individual 2 could be I A I B . If individual 1 is I B i, individual 2 could be I A i or I A I B . If individual 1 is I B I B , individual 2 could be I A I B . C8. Types O and AB provide an unambiguous genotype. Type O can only be ii, and type AB can only be I A I B . It is possible for a couple to produce children with all four blood types. The couple would have to be I A i and I B i. If you construct a Punnett square, you will see that they can produce children with AB, A, B, and O blood types. C9. The father could not be I A I B , I B I B , or I A I A . He is contributing the O allele to his offspring. Genotypically, he could be I A i, I B i, or ii and have type A, B, or O blood, respectively. C10. A. 1/4 B. 0 C. (1/4)(1/4)(1/4) = 1/64 D. Use the binomial expansion equation: P = n! x!(n − x)! p x q ( n− x ) n = 3, p =1/ 4, q =1/ 4, x = 2 P = 3/ 64 = 0.047, or 4.7% C11. Perhaps it should be called codominant at the “hair level” because one or the other allele is dominant with regard to a single hair. However, this is not the same as codominance in blood types, in which every cell can express both alleles. C12. All of the F1 generation will be white because they have inherited the dominant white allele from their Leghorn parent. Construct a Punnett square. Let W and w represent one gene, where W is dominant and causes a white phenotype. Let A and a represent the second gene, where the recessive allele causes a white phenotype in the homozygous condition. The genotype of the F1 birds is WwAa. The phenotypic ratio of the F2 generation will be 13 white : 3 brown. The only brown birds will be 2 wwAa and 1 wwAA. C13. X-linked recessive (unaffected mothers transmit the trait to sons) C14. You know that the parents must be heterozygotes for both genes. The genotypic ratio of their offspring is 1 ScP ScP : 2 ScP ScA : 1 ScA ScA The phenotypic ratio depends on sex: 1 ScP ScP male with scurs : 1 ScP ScP female with scurs : 2 ScP ScA males with scurs : 2 ScP ScA females without scurs : 1 ScA ScA male without scurs : 1 ScA ScA female without scurs. Page | 25 Copyright 2022 © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. A. 50% B. 1/8, or 12.5% C. (3/8)(3/8)(3/8)= 27/512 = 0.05, or 5% C15. A. Could be. B. No, because an unaffected father has an affected daughter. C. No, because two unaffected parents have affected children. D. No, because an unaffected father has an affected daughter. E. No, because both sexes exhibit the trait. F. Could be. C16. You would look at the pattern within families over the course of many generations. For a recessive trait, 25% of the offspring within a family are expected to be affected if both parents are unaffected carriers, and 50% of the offspring are expected to be affected if one parent is affected. You could look at many families and see if these 25% and 50% values are approximately true. Incomplete penetrance would not necessarily