pierce_genetics7e_solutionsmanual_ch05

Chapter Five: Extensions and Modifications of Basic Principles COMPREHENSION QUESTIONS Section 5.1 1. What are some important characteristics of dominance? Solution: Dominance is an interaction between different alleles at a locus. Dominance does not alter the way in which genes are inherited; it only influences the way in which the genes are expressed. The type of dominance may depend on the level at which the phenotype is examined. 2. What is incomplete penetrance and what causes it? Solution: Incomplete penetrance occurs when an individual with a particular genotype does not express the expected phenotype. Environmental factors and the effects of other genes may alter phenotypic expression and produce incomplete penetrance. Section 5.2 3. What is gene interaction? What is the difference between an epistatic gene and a hypostatic gene? Solution: Gene interaction is the determination of a phenotype by genes at more than one locus, where the effect of one gene depends on the effects of another gene at a different locus. One type of gene interaction is epistasis. The alleles at the epistatic gene mask or repress the effects of alleles at another locus. The gene whose alleles are masked or repressed is called the hypostatic gene. 4. What is a recessive epistatic gene? Solution: Recessive epistasis occurs when two copies of the epistatic gene are required to mask the interacting gene or genes. In the example of coat color in Labrador retrievers, being

homozygous at the locus for deposition of color in hair shafts ( ee ) produces a yellow color regardless of which genotype ( B _ or bb ) is present at the color locus. 5. What is a complementation test and what is it used for? Solution: A complementation test is used to determine whether different recessive mutations occur at the same locus (are allelic) or whether they occur at different loci. The two mutations are introduced into the same individual by crossing homozygotes for each of the mutants. If the progeny show a mutant phenotype, then the mutations are allelic (at the same locus). If the progeny show a wild-type (dominant) phenotype, then the mutations are at different loci and are said to complement each other, because each of the mutant parents supplies a functional copy (or dominant allele) at the locus mutated in the other parent. Section 5.3 6. What characteristics are exhibited by a cytoplasmically inherited trait? 21 Solutions: Cytoplasmically inherited traits are encoded by genes in the cytoplasm. Because the cytoplasm usually is inherited from a single parent (most often the female), reciprocal crosses do not show the same results. Cytoplasmically inherited traits often show great variability because different egg cells (female gametes) may have differing proportions of cytoplasmic alleles. 7. What is genomic imprinting? Solution: Genomic imprinting refers to different expression of a gene depending on whether it was inherited from the male parent or the female parent. 8. What is the difference between genetic maternal effect and genomic imprinting? Solution: In genetic maternal effect, the phenotypes of the progeny are determined by the genotype of the mother only. The genotype of the father and the genotype of the affected individual have no effect. In genomic imprinting, the phenotype of the progeny differs based on whether a particular allele is inherited from the mother or the father. The phenotype is therefore based on both the individual’s genotype and the paternal or maternal origins of the genotype.

d phenocopy a. the percentage of individuals with a particular genotype that express the expected phenotype h pleiotrophy b. a trait determined by an autosomal gene that is more easily expressed in one sex e polygenic trait c. a trait determined by an autosomal gene that is expressed in only one sex a penetrance d. a trait that is determined by an environmental effect and has the same phenotype as a genetically determined trait c sex-limited trait e. a trait determined by genes at many loci i genetic maternal effect f. the expression of a trait is affected by the sex of the parent that transmits the gene to the offspring f genomic imprinting g. the trait appears earlier or more severely in succeeding generations b sex-influenced trait h. a gene affects more than one phenotype g anticipation i. the genotype of the maternal parent influences the phenotype of the offspring Section 5.1 *13. Palomino horses have a golden-yellow coat, chestnut horses have a brown coat, and cremello horses have a coat that is almost white. A series of crosses between the three different types of horses produced the following offspring: Cross Offspring palomino × palomino 13 palomino, 6 chestnut, 5 cremello chestnut × chestnut 16 chestnut cremello × cremello 13 cremello palomino × chestnut 8 palomino, 9 chestnut palomino × cremello 11 palomino, 11 cremello chestnut × cremello 23 palomino a. Explain the inheritance of the palomino, chestnut, and cremello phenotypes in horses. Solution: The simplest hypothesis consistent with these results is that color in the horses is an incompletely dominant characteristic. Chestnuts and cremellos are homozygotes and palominos are heterozygotes. This is supported by the observation that a cross between two cremellos produces only cremello progeny, a cross between two chestnuts produces only chestnut progeny, and a cross between two palominos produces a 1:2:1 ratio of

cremello, palomino, and chestnut progeny. b. Assign symbols for the alleles that determine these phenotypes, and list the genotypes of all parents and offspring given in the preceding table. Solution: Let C B = chestnut, C W = cremello, C B C W = palomino. Cross Offspring palomino × palomino 13 palomino, 6 chestnut, 5 cremello C B C W × C B C W C B C W C B C B C W C W chestnut × chestnut 16 chestnut C B C B × C B C B C B C B cremello × cremello 13 cremello C W C W × C W C W C W C W palomino × chestnut 8 palomino, 9 chestnut C B C W × C B C B C B C W C B C B palomino × cremello 11 palomino, 11 cremello C B C W × C W C W C B C W C W C W chestnut × cremello 23 palomino C B C B × C W C W C B C W 14. The L M and L N alleles at the MN blood group locus exhibit codominance. Give the expected genotypes and phenotypes and their ratios in progeny resulting from the following crosses. a. L M L M × L M L N Solution: ½ L M L M (type M), ½ L M L N (type MN) b. L N L N × L N L N Solution: All L N L N (type N)

*17. When a Chinese hamster with white spots is crossed with another hamster that has no spots, approximately ½ of the offspring have white spots and ½ have no spots. When two hamsters with white spots are crossed, ⅔ of the offspring possess white spots and ⅓ have no spots. a. What is the genetic basis of white spotting in Chinese hamsters? Solution: The 2:1 ratio when two spotted hamsters are mated suggests a lethal allele, and the 1:1 ratio when spotted hamsters are mated to hamsters without spots indicates that spotted is a heterozygous phenotype. Using S to represent the allele for white spotting and s to represent the allele for solid color, spotted hamsters are Ss and solid-colored hamsters are ss . One-quarter of the progeny expected from a mating of two spotted hamsters ( Ss × Ss ) is SS , which is embryonic lethal and missing from those progeny, resulting in the 2:1 ratio of spotted to solid progeny. b. How might you go about producing Chinese hamsters that breed true for white spotting? Solution: Because spotting is a heterozygous phenotype, it would not be possible to produce Chinese hamsters that breed true for spotting. 18. In the early 1900s, Cuénot studied the genetic basis of yellow coat color in mice (discussed on p. 114). He carried out a number of crosses between two yellow mice and obtained what he thought was a 3:1 ratio of yellow to gray mice in the progeny. The following table gives Cuénot’s actual results, along with the results of a much larger series of crosses carried out by Castle and Little (W. E. Castle and C. C. Little. 1910. Science 32:868–870). Progeny Resulting from Crosses of Yellow × Yellow Mice Investigators Yellow progeny Non-yellow progeny Total progeny Cuénot 263 100 363 Castle and Little 800 435 1,235 Both combined 1,063 535 1,598 a. Using a chi-square test, determine whether Cuénot’s results are significantly different from the 3:1 ratio that he thought he observed. Are they different from a 2:1 ratio? Solution:

Testing Cuénot’s data for a 3:1 ratio: Obs Expected (3:1) O – E (O – E) 2 /E Yellow 263 272.25 –9.25 0.314 Non-yellow 100 90.75 9.25 0.943 Total 363 363 1.257 = χ 2 df = 2 – 1 = 1; 0.1 < p < 0.5 Cannot reject hypothesis of 3:1 ratio Now test for 2:1 ratio: Obs Expected (2:1) O – E (O – E) 2 /E Yellow 263 242 21 1.82 Non-yellow 100 121 –21 3.64 Total 363 363 5.46 = χ 2 df = 1; p < 0.025 The observations are inconsistent with a 2:1 ratio. b. Determine whether Castle and Little’s results are significantly different from a 3:1 ratio. Are they different from a 2:1 ratio? Solution: Obs Expected (3:1) O – E (O – E) 2 /E Yellow 800 926.25 –126.25 17.2 Non-yellow 435 308.75 126.25 51.6 Total 1,235 1,235 68.8 = χ 2 df = 1; p << 0.005 Reject 3:1 ratio Obs Expected (2:1) O – E (O – E) 2 /E Yellow 800 823.3 –23.3 0.66 Non-yellow 435 411.7 23.3 1.32 Total 1,235 1,235 1.98 = χ 2 df = 1; 0.1 < p < 0.5 Cannot reject 2:1 ratio c. Combine the results of Castle and Cuénot and determine whether they are significantly different from a 3:1 ratio and a 2:1 ratio.

Solution: ¼ Rp 1 / Rp 2 (red), ¼ Rp 1 / rp (red), ¼ Rp 2 / rp (purple), ¼ rp/rp (green), for overall phenotypic ratio of ½ red, ¼ purple, ¼ green. c. Rp 1 / Rp 2 × Rp 1 / Rp 2 Solution: This cross is equivalent to a two-allele cross of heterozygotes, so the expected phenotypic ratio is ¾ red, ¼ purple. d. Rp 2 / rp × rp / rp Solution: This is another two-allele cross of a heterozygote with a homozygous recessive. Phenotypic ratio is ½ purple, ½ green. e. rp / rp × Rp 1 / Rp 2 Solution: ½ Rp 1 /rp (red), ½ Rp 2 /rp (purple) 20. If there are five alleles at a locus, how many genotypes may there be at this locus? How many different kinds of homozygotes will there be? How many genotypes and homozygotes may there be with eight alleles at a locus? Solution: The number of genotypes possible will be n ( n +1)/2, where n equals the number of different alleles at a locus. Thus, the number of genotypes possible with 5 alleles is 5(5+1)/2 = 15 There is one homozygote possible for each allele, so there will be five homozygotes. For eight alleles, there are 8(8+1)/2 = 36 possible genotypes and eight homozygotes. 21. Turkeys have black, bronze, or black-bronze plumage. Examine the results of the following crosses:

Parents Offspring Cross 1: black and bronze all black Cross 2: black and black ¾ black, ¼ bronze Cross 3: black-bronze and black-bronze all black-bronze Cross 4: black and bronze ½ black, ¼ bronze, ¼ black-bronze Cross 5: bronze and black-bronze ½ bronze, ½ black-bronze Cross 6: bronze and bronze ¾ bronze, ¼ black-bronze Do you think these differences in plumage arise from incomplete dominance between two alleles at a single locus? If yes, support your conclusion by assigning symbols to each allele and providing genotypes for all turkeys in the crosses. If your answer is no, provide an alternative explanation and assign genotypes to all turkeys in the crosses. Solution: A 3:1 ratio in the progeny (¾ of one phenotype and ¼ of the other) results from a cross between two heterozygotes for a recessive trait. With incomplete dominance, we may get a 1:2:1 ratio, but not a 3:1 ratio. Thus, the results of Cross 2 tell us that black is dominant to bronze. Similarly, the results of Cross 6 tell us that bronze is dominant to black-bronze. Thus, it appears that color in turkeys is the result of multiple alleles. We can use B L for black, B R for bronze, and b for black-bronze. The dominance relationships among these alleles are B L > B R > b. Parents Offspring Cross 1: black ( B L B L ) × bronze ( B R B R ) All black ( B L B R ) Cross 2: black ( B L B R ) × black ( B L B R ) ¾ black ( B L  ) , ¼ bronze ( B R B R ) Cross 3: black-bronze ( bb ) × black-bronze ( bb ) All black-bronze ( bb ) Cross 4: black ( B L b ) × bronze ( B R b ) ½ black ( B L _), ¼ bronze ( B R b ), ¼ black- bronze ( bb ) Cross 5: bronze ( B R b ) × black-bronze ( bb ) ½ bronze ( B R b ), ½ black-bronze ( bb ) Cross 6: bronze ( B R b ) × bronze ( B R b ) ¾ bronze ( B R _) , ¼ black-bronze ( bb ) 22. In rabbits, an allelic series helps to determine coat color: C (full color), c ch (chinchilla, gray color), c h (Himalayan, white with black extremities), and c (albino, all white). The C allele is dominant over all others, c ch is dominant over c h and c, c h is dominant over c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > c ch > c h > c . The rabbits in the following list are crossed and produce the progeny shown. Give the genotypes of the parents for each cross. a. full color × albino → ½ full color, ½ albino

As discussed in the text, Barry had blood type A and her child had blood type B. Barry’s genotype therefore must be I A i , and the child inherited the i allele from Barry. The child must have inherited the I B allele from the father. Thus, the father could have been B or AB. b. If Chaplin had possessed one of these blood types, would that prove that he fathered Barry’s child? Solution: No. Many other men have these blood types. The results would have meant only that Chaplin cannot be eliminated as a possible father of the child. *24. A woman has blood-type A M. She has a child with blood-type AB MN. Which of the following blood types could not be that of the child’s father? Explain your reasoning. George O N Tom AB MN Bill B MN Claude A NN Henry AB M Solution: The child’s blood type has a B allele ( I B ) and an N allele ( L N ) that could not have come from the mother and must have come from the father. George, Claude, and Henry are eliminated as possible fathers because they lack either an I B allele or an L N . Section 5.2 *25. In chickens, comb shape is determined by alleles at two loci ( R , r and P , p ). A walnut comb is produced when at least one dominant allele R is present at one locus and at least one dominant allele P is present at a second locus (genotype R_ P_ ). A rose comb is produced when at least one dominant allele is present at the first locus and two recessive alleles are present at the second locus (genotype R_ pp ). A pea comb is produced when two recessive alleles are present at the first locus and at least one dominant allele is present at the second (genotype rr P_ ). If two recessive alleles are present at the first and at the second loci ( rr pp ), a single comb is produced. Progeny with what types of combs and in what proportions will result from the following crosses?

a. RR PP × rr pp Solution: All walnut ( Rr Pp ) b. Rr Pp × rr pp Solution: ¼ walnut ( Rr Pp ) , ¼ rose ( Rr pp ) , ¼ pea ( rr Pp ) , ¼ single ( rr pp ) c. Rr Pp × Rr Pp Solution: 9 / 16 walnut ( R_ P_ ), 3 / 16 rose ( R_ pp ), 3 / 16 pea ( rr P_ ), 1 / 16 single ( rr pp ) d. Rr pp × Rr pp Solution: ¾ rose ( R_pp ) , ¼ single ( rr pp ) e. Rr pp × rr Pp Solution: ¼ walnut ( Rr Pp ) , ¼ rose ( Rr pp ) , ¼ pea ( rr Pp ) , ¼ single ( rr pp ) f. Rr pp × rr pp Solution: ½ rose ( Rr pp ) , ½ single ( rr pp ) *26. Tatuo Aida investigated the genetic basis of color variation in the Medaka ( Aplocheilus latipes ), a small fish found in Japan (T. Aida. 1921. Genetics 6:554–573). Aida found that genes at two loci ( B, b and R, r ) determine the color of the fish: fish with a dominant allele at both loci ( B _ R _) are brown, fish with a dominant allele at the B locus only ( B _ rr ) are blue, fish with a dominant allele at the R locus only ( bb R _) are red, and fish with recessive alleles at both loci ( bb rr ) are white. Aida crossed a homozygous brown fish with a homozygous white fish. He then backcrossed the F 1 with the homozygous white parent and obtained 228 brown fish, 230 blue fish, 237 red fish, and 222 white fish.

27. A variety of opium poppy ( Papaver somniferum ) with lacerate leaves was crossed with a variety that has normal leaves. All the F 1 had lacerate leaves. Two F 1 plants were interbred to produce the F 2 . Of the F 2 , 249 had lacerate leaves and 16 had normal leaves. Give genotypes for all the plants in the P, F 1 , and F 2 generations. Explain how lacerate leaves are determined in the opium poppy. Solution: The F 1 progeny tell us that lacerate is dominant over normal leaves. In the F 2 , 249:16 does not come close to a 3:1 ratio. Let’s see if these numbers fit a dihybrid ratio. Dividing 265 total progeny by 16 (because dihybrid ratios are based on 16ths), we see that 1 / 16 of 265 is 16.56. When we divide 249 (the number of progeny with lacerate leaves) by 16.56, we get 15, which tells us that approximately 15 / 16 of the progeny have lacerate leaves and 1 / 16 have normal leaves, a modified dihybrid ratio. If we symbolize the two genes as A and B , then: F 1 Aa Bb × Aa Bb all lacerate F 2 9 / 16 A_ B_ lacerate (like F 1 ) 3 / 16 A_ bb lacerate 3 / 16 aa B_ lacerate 1 / 16 aa bb normal A dominant allele at either gene A or gene B , or both, results in lacerate leaves. Finally, the parents must have been AA BB lacerate × aa bb normal. Note that only AA BB for the lacerate parent would result in F 1 that are Aa Bb. 28. E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescent-white seedlings, and 260 yellow seedlings (E. W. Lindstrom. 1921. Genetics 6:91–110). a. Give the genotypes for the green, virescent-white, and yellow progeny. Solution: The ratio does not fit the 1:2:1 ratio expected with incomplete dominance for a single locus, suggesting that multiple loci with gene interactions may be involved. The simplest case is two loci, so we look for a fit to a ratio based on a dihybrid cross. A dihybrid ratio cross produces progeny in ratios based on 16ths; 1 / 16 of the total ( 4696 / 16 ) is 293.5. Dividing the number of each phenotype by 293.5 gives a 12:3:1 ratio of green:virescent- white:yellow, a modified 9:3:3:1 ratio. Let G and g represent alleles at one locus and Y and y alleles at a second locus. 9 / 16 G_ Y_ + 3/16 G_ yy = 12/16 green 3 / 16 gg Y_ = 3/16 virescent-white 1 / 16 gg yy = 1/16 yellow

b. Explain how color is determined in these seedlings. Solution: The green arises when the G locus is dominant, regardless of the alleles at the other Y locus. Yellow requires that both loci be recessive, and virescent-white arises when the G locus is homozygous recessive and the Y locus has a dominant allele. c. Does epistasis occur among the genes that determine color in the corn seedlings? If so, which gene is epistatic and which is hypostatic? Solution: As defined above, the G locus is the epistatic locus. It is an example of dominant epistasis, because a dominant allele at this locus masks the effect of the Y locus. The Y locus is hypostatic, and its effect revealed only when the epistatic locus is homozygous recessive. *29. A dog breeder liked yellow and brown Labrador retrievers. In an attempt to produce yellow and brown puppies, he mated a yellow Labrador male and a brown Labrador female. Unfortunately, all the puppies produced in this cross were black. (See text for a discussion of the genetic basis of coat color in Labrador retrievers.) a. Explain this result. Solution: Labrador retrievers vary in two loci, B and E . Black dogs have dominant alleles at both loci ( B_ E_ ), brown dogs have bb E_ , and yellow dogs have B_ ee or bb ee . Because all the puppies were black, they must all have inherited a dominant B allele from the yellow parent and a dominant E allele from the brown parent. The brown female parent must have been bb EE , and the yellow male must have been BB ee . The black puppies were all Bb Ee . b. How might the breeder go about producing yellow and brown Labradors? Solution: Simply mating yellow with yellow will produce all yellow Labrador puppies. Mating two brown Labradors will produce either all brown puppies, if at least one of the parents is homozygous EE , or ¾ brown and ¼ yellow, if both parents are heterozygous Ee . 30. When a yellow female Labrador retriever was mated with a brown male, half of the

Now we see that purple requires dominant alleles for both genes, so the purple parent must have been AA BB , and the white parent must have been aa bb to give all purple F 1 . b. Draw a hypothetical biochemical pathway to explain the production of purple and white flowers in sweet peas. Solution: *33. Refer to the chapter for a discussion of how coat color and pattern are determined in dogs. a. Why are Irish setters not black in color? Solution: According to the information in Table 5.3, Irish setters are BB ee SS . The B permits expression of black pigment, but the ee genotype prevents black color on the body coat, resulting in a reddish color except on the nose and in the eyes. The S prevents spotting, resulting in a uniform coat color. b. Can a poodle crossed with any other breed produce spotted puppies? Why or why not? Solution: Poodles are SS . Because the dominant S allele prevents spotting, no puppies from matings with poodles will have spotting. c. If a St. Bernard is crossed with a Doberman, what will be the coat color of the offspring: solid, yellow, saddle, or bicolor? Solution: St. Bernards are a y a y BB , and Dobermans are a t a t EE SS . The offspring will be of genotype a y a t B _ E_ S_ . Because a y specifying yellow is dominant over a t , and the E allele allows expression of the A genotype throughout, the offspring will have yellow coats.

d. If a Rottweiler is crossed with a Labrador retriever, what will be the coat color of the offspring: solid, yellow, saddle, or bicolor? Solution: Rottweilers are a t a t BB EE SS , and Labrador Retrievers are A s A s SS . The offspring will be A s a t B_E_SS . The combination of the dominant A s and E alleles should create solid coats. Section 5.3 34. Male-limited precocious puberty results from a rare, sex-limited autosomal allele ( P ) that is dominant over the allele for normal puberty ( p ) and is expressed only in males. Bill undergoes precocious puberty, but his brother Jack and his sister Beth underwent puberty at the usual time, between the ages of 10 and 14. Although Bill’s mother and father underwent normal puberty, two of his maternal uncles (his mother’s brothers) underwent precocious puberty. All of Bill’s grandparents underwent normal puberty. Give the most likely genotypes for all the relatives mentioned in this family. Solution: Since precocious puberty is dominant, all the males who experienced normal puberty, such as Jack, Bill’s father, and Bill’s grandfathers, must be pp . Bill and his two maternal uncles, who all experienced precocious puberty, are Pp . We know they are heterozygotes not only because P is a rare allele but also because these individuals all had fathers that are pp . This means Bill inherited P from his mother, who must have been Pp . Bill’s sister Beth could be either Pp or pp . *35. In some goats, the presence of horns is produced by an autosomal gene that is dominant in males and recessive in females. A horned female is crossed with a hornless male. The F 1 offspring are intercrossed to produce the F 2 . What proportion of the F 2 females will have horns? Solution: Let H + represent the allele for the presence of horns and H – represent the allele for hornlessness. Since H + is recessive in females, the horned female parent must be H + H + . The hornless male is H – H – because the absence of horns is recessive in males. Then their F 1 progeny must be all heterozygous H + H – . An intercross of the F 1 would produce both male and female progeny in the ratio of 1 H + H + , 2 H + H – , and 1 H – H – . Again, remembering that H + is recessive in females, we would expect a ratio of 3:1 hornless to horned females. 36. In goats, a beard is produced by an autosomal allele that is dominant in males and recessive in females. We’ll use the symbol B b for the beard allele and B + for the beardless allele. Another independently assorting autosomal allele that produces a black coat ( W ) is