pierce_genetics7e_solutionsmanual_ch04

Chapter Four: Sex Determination and Sex-Linked Characteristics COMPREHENSION QUESTIONS Section 4.1 1. What is considered to be the fundamental difference between males and females of most organisms? Solution: Males produce relatively small gametes; females produce larger gametes. 2. How do monoecious organisms differ from dioecious organisms? Solution: In monoecious organisms, individuals possess both male and female reproductive structures and produce both male and female gametes. Dioecious organisms exist as organisms of two distinct sexes, one producing male gametes, the other producing female gametes. 3. Describe the XX-XO system of sex determination. In this system, which is the heterogametic sex and which is the homogametic sex? Solution: In the XX-XO sex determination system, females have two copies of the sex-determining chromosome (the X chromosome) and are the homogametic sex, whereas males have only one copy and are the heterogametic sex. Males are heterogametic because they produce two different types of gametes with respect to the sex chromosome: either containing an X or not containing an X. *4. How does sex determination in the XX-XY system differ from sex determination in the ZZ-ZW system? Solution: In the XX-XY system, males are heterogametic and produce gametes with either an X chromosome or a Y chromosome. In the ZZ-ZW system, females are heterogametic and produce gametes with either a Z or a W chromosome. 5. What is the pseudoautosomal region? How does the inheritance of traits encoded by genes in this region differ from the inheritance of other Y-linked characteristics? Solution: The pseudoautosomal region is a region of homology between the X and Y chromosomes that is responsible for pairing the X and Y chromosomes during meiotic prophase I. Genes in this region are present in two copies in both males and females and thus are

inherited like autosomal genes, whereas other Y-linked genes are passed on only from father to son. 6. What is meant by genic sex determination? Solution: In organisms that follow this system, there is no cytogenetically recognizable difference in the chromosomes of males and females. Instead of a sex chromosome that differs between males and females, alleles at one or more loci determine the sex of the individual. 7. How does sex determination in Drosophila differ from sex determination in humans? Solution: In humans, the presence of a functional Y chromosome determines maleness. People with XXY and XXXY are phenotypically male. In Drosophila , the ratio of X chromosome material to autosomes (the X:A ratio) predicts the sex of the individual, regardless of the Y chromosome. However, sex in Drosophila is actually determined by genes on the X chromosome. Flies with XXY and two haploid sets of autosomes are female, and flies with XO and two haploid sets of autosomes are sterile males. 8. Give the typical sex chromosome complement found in the cells of people with Turner syndrome, with Klinefelter syndrome, and with androgen-insensitivity syndrome. What is the sex-chromosome complement of triple-X females? Solution: Turner syndrome: XO Klinefelter syndrome: XXY (rarely XXXY, XXXXY, or XXYY) Androgen insensitivity: XY Triple-X females: XXX Section 4.2 9. What characteristics are exhibited by an X-linked trait? Solution: Males show the phenotypes of all X-linked traits, regardless of whether the X-linked allele is recessive or dominant. Males inherit X-linked traits from their mothers, pass X- linked traits to their daughters, and through their daughters to their daughters’ descendants, but not to their sons or their sons’ descendants. 10. Explain how Bridges’s study of nondisjunction in Drosophila helped prove the chromosome theory of inheritance. Solution: Bridges showed that in crosses with white-eyed flies, where white eye color is a sex-

APPLICATION QUESTIONS AND PROBLEMS Introduction 14. As discussed in the introduction to this chapter, sex in bearded dragons is determined by both sex chromosomes and temperature. What, if any, would be the effect of unusually cool temperatures during the breeding season on the sex ratio of bearded lizards? What would be the effect of unusually warm temperatures on the sex ratio? Solution: At unusually cool temperatures, there would probably be a 50:50 sex ratio, because at cool temperatures, all ZZ individuals develop as males and all ZW individuals develop as females. At warmer temperatures, some ZZ individuals develop as female, which would probably increase the proportion of females in the population. Section 4.1 *15.What is the sexual phenotype of fruit flies having the following chromosomes? Solution: In fruit flies, the X to autosome ratio predicts the sexual phenotype. Flies with an X:A ratio of 0.5 are male and flies with an X:A ratio of 1.0 are female. Flies with an X:A ratio greater than 1.0 are metafemales and flies with an X:A ratio of less than 0.5 are metamales. Flies with an X:A ratio between 0.5 and 1.0 are intersex.   Sex Autosomal     chromosomes chromosomes Sexual phenotype a. XX All normal Female (X:A = 1.0) b. XY All normal Male (X:A = 0.5) c. XO All normal Male (X:A = 0.5) d. XXY All normal Female (X:A = 1.0) e. XYY All normal Male (X:A = 0.5) f. XXYY All normal Female (X:A = 1.0) g. XXX All normal Metafemale (X:A = 1.5) h. XX Four haploid sets Male (X:A = 0.5) i. XXX Four haploid sets Intersex (X:A = 0.75) j. XXX Three haploid sets Female (X:A = 1.0) k. X Three haploid sets Metamale (X:A = 0.33) l. XY Three haploid sets Metamale (X:A = 0.33) m. XX Three haploid sets Intersex (X:A = 0.66)

16. If nondisjunction of the sex chromosomes takes place in meiosis I in the male in Figure 4.5 , what sexual phenotypes and proportions of offspring will be produced? Solution: Nondisjunction in the male will produce ½ sperm with XY and ½ sperm with O (no sex chromosomes). Normal separation of the sex chromosomes in the female will produce all eggs with a single X chromosome. These gametes will combine to produce ½ XXY (Klinefelter syndrome males) and ½ XO (Turner syndrome females). *17. For each of the following chromosome complements, what is the phenotypic sex of a person who has: a. XY with the SRY gene deleted? Recall that in humans a single functional copy of the SRY gene, normally located on the Y chromosome, determines phenotypic maleness by causing gonads to differentiate into testes. In the absence of a functional SRY gene, gonads differentiate into ovaries and the individual is phenotypically female. Solution: Female b. XX with a copy of SRY gene on an autosomal chromosome? Solution: Male c. XO with a copy of SRY gene on an autosomal chromosome? Solution: Male d. XXY with the SRY gene deleted? Solution: Female e. XXYY with one copy of the SRY gene deleted Solution: Male (Klinefelter syndrome)

Females inherit the W chromosome from the mother and the Z chromosome from the father. Thus, no Z chromosome can be inherited from the mother’s parents. Males inherit one copy of the Z chromosome from each parent, so females have equal probability of inheriting the Z chromosome from the father’s mother or the father’s father. Section 4.2 *22. When Bridges crossed white-eyed females with red-eyed males, he obtained a few red- eyed males and white-eyed females (see Figure 4.13 ). What types of offspring would be produced if these red-eyed males and white-eyed females were crossed with each other? Solution: Bridges’ exceptional white-eyed females were X w X w Y and the red-eyed males were X + Y. Meiosis in the white-eyed females generates 45% X w Y, 45% X w , 5% X w X w , and 5% Y gametes. Meiosis in the red-eyed male generates 50% X + and 50% Y gametes. When these two flies are crossed, the following progeny are produced. X + Y X w Y X w X + Y red-eyed females X w YY white-eyed males X w X w X + red-eyed females X w Y white-eyed males X w X w X w X w X + metafemales X w X w Y white-eyed females Y X + Y red-eyed males YY dies diez *23.Joe has classic hemophilia, an X-linked recessive disease. Could Joe have inherited the gene for this disease from the following persons? Solution: Yes No a. His mother’s mother X b. His mother’s father X c. His father’s mother X d. His father’s father X Males always inherit their X chromosome from their mother (they inherit the Y chromosome from their father), and thus, X-linked traits are passed on from mother to son. Females inherit an X chromosome from both their mother and father, so X-linked traits can be passed from mother or father to females. Joe must have inherited the hemophilia trait from his mother. His mother could have inherited the trait from either her mother (a) or her father, because both contribute an X chromosome to their daughters (b). Joe could not have inherited the trait from his father (Joe inherited the Y chromosome

from his father); thus, he could not have inherited hemophilia from either his father’s mother (c) or his father’s father (d). *24.In Drosophila, yellow body is due to an X-linked gene that is recessive to the gene for gray body. a. A homozygous gray female is crossed with a yellow male. The F 1 are intercrossed to produce F 2 . Give the genotypes and phenotypes, along with the expected proportions, of the F 1 and F 2 progeny. Solution: We will use X + as the symbol for the dominant gray body color and X y for the recessive yellow body color. The homozygous gray female parent is thus X + X + and the yellow male parent is X y Y. Male progeny always inherit the Y chromosome from the male parent and either of the two X chromosomes from the female parent. Female progeny always inherit the X chromosome from the male parent and either of the two X chromosomes from the female parent. F 1 males inherit the Y chromosome from their father and X + from their mother; hence, their genotype is X + Y and they have gray bodies. F 1 females inherit X y from their father and X + from their mother; hence, they are X + X y and also have gray bodies. When the F 1 progeny are intercrossed, the F 2 males again inherit the Y from the F 1 male, and they inherit either X + or X y from their mother. Therefore, we should get ½ X + Y (gray body) and ½ X y Y (yellow body). The F 2 females will all inherit the X + from their father and either X + or X y from their mother. Therefore, we should get ½ X + X + and ½ X + X y (all gray body). In summary: P X + X + (gray female) × X y Y (yellow male) F 1 ½ X + Y (gray males) ½ X + X y (gray females) F 2 ¼ X + Y (gray males) ¼ X y Y (yellow males) ¼ X + X y (gray females) ¼ X + X + (gray females) The net F 2 phenotypic ratios are ½ gray females, ¼ gray males, and ¼ yellow males. The F 2 progeny can also be predicted using a Punnett square.

Male gametes: ½ Y, ¼ X + , ¼ X y Half the F 2 females in part (b) are homozygous X y X y , so all their gametes will be X y : ½ X y . The other half are heterozygous and will produce equal proportions of X + and X y gametes: ½(½X + ) = ¼ X + ; ½(½ X y ) = ¼ X y . Female gametes: ¼ X + , ¾ X y Now using a Punnett square: 1/2 Y 1/4 X + 1/4 X y 1/4 X + 1/8 X + Y 1/16 X + X + 1/16 X + X y 3/4 X y 3/8 X y Y 3/16 X + X y 3/16 X y X y Overall genotypic ratios are 1/8 X + Y, 3/8 X y Y, 1/16 X + X + , 4/16 X + X y , and 3/16 X y X y . Overall phenotypic ratios are 1/8 gray males, 3/8 yellow males, 5/16 gray females, and 3/16 yellow females. 25. Coat color in cats is determined by genes at several different loci. At one locus on the X chromosome, one allele (X + ) codes for black fur; another allele (X o ) encodes orange fur. Females can be black (X + X + ), orange (X o X o ), or a mixture of orange and black called tortoiseshell (X + X o ). Males are either black (X + Y) or orange (X o Y). Bill has a female tortoiseshell cat named Patches. One night, Patches escapes from Bill’s house, spends the night out, and mates with a stray male. Patches later gives birth to the following kittens: one orange male, one black male, two tortoiseshell females, and one orange female. Give the genotypes of Patches, her kittens, and the stray male with which Patches mated. Solution: Because Patches is tortoiseshell, she must be X o X + . The genotypes of the kittens are: orange male: X o Y black male: X + Y tortoiseshell females: X + X o orange female: X o X o Because the orange female has two orange alleles, the father must be carrying an X o chromosome and a Y chromosome. Thus, the male cat’s genotype must be X o Y. *26. Red-green color-blindness in humans is due to an X-linked recessive gene. Both John and Cathy have normal color vision. After 10 years of marriage to John, Cathy gave birth to a color-blind daughter. John filed for divorce, claiming he is not the father of the child. Is John justified in his claim of nonpaternity? Explain why. If Cathy had given birth to a color-blind son, would John be justified in claiming nonpaternity?

Solution: Red-green color-blindness is due to an X-linked allele (X c ) that is recessive to the allele for color vision (X + ). The color-blind daughter must be homozygous for the color-blind allele (X c X c ); thus, she must have inherited a color-blind allele from each parent. She could have inherited a color-blind allele from her mother, as the mother (who had normal color vision) could have been heterozygous (X + X c ). However, her father also had normal color vision, with genotype X + Y, and thus could not have passed on a color-blind allele. John would therefore be justified in being suspicious about the paternity of the child. A few remote alternative possibilities could be considered. 1) The daughter is XO, inheriting a color-blind allele from her mother and no sex chromosome from her father. In that case, the daughter could be X c O and have Turner syndrome. 2) John has Klinefelter syndrome (and is heterozygous for the color-blind allele (X + X c Y). In this way, he could have passed a color-blind allele to his daughter. However, most people with Klinefelter syndrome are sterile. 3) The daughter is heterozygous, but the X with the normal vision allele was inactivated in both eyes. 4) The daughter inherited a new X-linked color-blind mutation on the X chromosome she inherited from her father. If Cathy had a color-blind son (X c Y), then John would have no grounds for suspicion. The son would have inherited John’s Y chromosome and the color-blind X chromosome from Cathy, who could have been heterozygous (X + X c ). 27. Maria has red– green color blindness; Juan has normal color vision. Together, Maria and Juan have two children, Maggi and Daniel. Draw the sex chromosomes of Maggi and Daniel at the following stages, and label on these chromosomes the alleles for color blindness and normal color vision (see Chapter 2 for stages of meiosis). Assume that no crossing over has occurred. a. Metaphase of mitosis b. Metaphase I of meiosis c. Metaphase II of meiosis Solution: The allele for colorblindness (c) is located on the maternal X (yellow) and the allele for normal color vision (C) is located on the paternal X (blue). The Y chromosome (gray) has no allele for color vision. The chromosomes consist of two chromatids. a. Maggi has two X chromosomes, the one from her mother Maria with the allele for colorblindness and the other one from her father Juan with the normal allele. Daniel possesses only one X chromosome, which he received from Maria. In mitosis, the homologous chromosomes do not pair when they line up at the equator of the cell.

derived exclusively from progenitor cells that inactivated the normal X, then that eye would be color-blind, whereas the other eye may be derived from progenitor cells that inactivated the color-blind X, or is a mosaic with sufficient normal retinal cells to permit color vision. b. Would it be possible for a man to have one eye with normal color vision and one eye with color-blindness? Solution: An XY male would not be able to have one eye with color vision and one eye with color-blindness through X-inactivation, as males only have a single X chromosome. If the man were XXY (Klinefelter syndrome), then he might have one eye with color vision and one eye with color-blindness through X inactivation, as explained in part (a). Another rare possibility would be a somatic mutation in the progenitor cells for one retina but not the other. 28. In 1875, Charles Darwin, author of On the Origin of Species , wrote of a peculiar family from Sind, a province in northwest India (C. Darwin. The Variation of Animals and Plants under Domestication , 2nd ed. 1875. London: John Murray, London. p. 319 ): . . . in which ten men, in the course of four generations, were furnished in both jaws taken together, with only four small and weak incisor teeth and with eight posterior molars. The men thus affected have little hair on the body, and become bald early in life. They also suffer much during hot weather from excessive dry- ness of the skin. It is remarkable that no instance has occurred of a daughter being thus affected. . . . Though daughters in the above family are never affected, they transmit the tendency to their sons; and no case has occurred of a son transmitting it to his sons. The men in this family had a condition that is now known as hypohidrotic ectodermal dysplasia. Darwin’s account is an accurate description of the symptoms and inheritance of this disorder. Based on Darwin’s description, what is the most likely mode of inheri- tance for hypohidrotic ectodermal dysplasia? What evidence in Darwin’s description sup- ports your conclusion? (See also Problem 47 and Think-Pair- Share Question 8.) Solution: Hypohidrotic dysplasia (HED) is an X-linked recessive trait. Three observations provide evidence for this mode of inheritance: 1) this trait is expressed only in males of a family and not in females, 2) unaffected daughters in families with HED transmit to sons and not to their daughters, and 3) there is no father to son transmission. The latter is highly supportive of X-linked inheritance. The absence of the trait in females, who are presumably heterozygous, indicates that the trait is recessive. *30. Bob has XXY chromosomes (Klinefelter syndrome) and is color-blind. His mother and father have normal color vision, but his maternal grandfather is color-blind. Assume that Bob’s chromosome abnormality arose from nondisjunction in meiosis. In which parent

and in which meiotic division did nondisjunction take place? Assume no crossing over has taken place. Explain your answer. Solution: Because Bob must have inherited the Y chromosome from his father, and his father has normal color vision, there is no way a nondisjunction event from the paternal lineage could account for Bob’s genotype. Bob’s mother must be heterozygous X + X c because she has normal color vision, and she must have inherited a color-blind X chromosome from her color-blind father. For Bob to inherit two color-blind X chromosomes from his mother, the egg must have arisen from a nondisjunction in meiosis II. In meiosis I, the homologous X chromosomes separate, so one cell has the X + and the other has X c . Failure of sister chromatids to separate in meiosis II would then result in an egg with two copies of X c . 31. Xg is an antigen found on red blood cells. This antigen is caused by an X-linked allele (X a ) that is dominant over an allele for the absence of the antigen (X – ). The inheritance of these X-linked alleles was studied in children with chromosome abnormalities to determine where nondisjunction of the sex chromosomes took place. For each type of mating in parts a through d , indicate whether nondisjunction took place in the mother or in the father and, if possible, whether it took place in meiosis I or meiosis II (assume no crossing over). a. X a Y  X – X –  X a (Turner syndrome) b. X a Y  X a X –  X – (Turner syndrome) c. X a Y  X – X –  X a X – Y (Klinefelter syndrome) d. X a Y  X a X –  X – X – Y (Klinefelter syndrome) Solution: This child received two copies of X – , which must have come from the mother. Thus, nondisjunction took place in the mother. This must have occurred in meiosis II, where the sister chromatids of the X – chromosome failed to separate. 32. The Talmud, an ancient book of Jewish civil and religious laws, states that if a woman bears two sons who die of bleeding after circumcision (removal of the foreskin from the penis), any additional sons that she has should not be circumcised. (The bleeding is most likely due to the X-linked disorder hemophilia.) Furthermore, the Talmud states that the sons of her sisters must not be circumcised, while the sons of her brothers should be. Is this religious law consistent with sound genetic principles? Explain your answer. Solution: Yes. If a woman has a son with hemophilia, then she carries at least one X h allele for hemophilia. If she is a carrier (X + X h ), any of her sons have a 50% chance of inheriting the X h allele and having hemophilia. If she is homozygous X h X h (unlikely), then all sons will inherit the X h allele and have hemophilia. Thus, it is prudent that her sons not be circumcised. Her sisters may also be carriers and thus their sons should also not be

  Parents Offspring Family number Father Mother Normal male Normal female CFNS male CFNS female 1 normal CFNS 1 0 2 1 5 normal CFNS 0 2 1 2 6 normal CFNS 0 0 1 2 8 normal CFNS 1 1 1 0 10a CFNS normal 3 0 0 2 10b normal CFNS 1 1 2 0 12 CFNS normal 0 0 0 1 13a normal CFNS 0 1 2 1 13b CFNS normal 0 0 0 2 7b CFNS normal 0 0 0 2 a. On the basis of these results, what is the most likely mode of inheritance for CFNS? Solution: Children with CFNS are born in families where one parent has CFNS. Thus, CFNS is most likely a dominant trait. Moreover, we see that in families where the father has CFNS and the mother does not, CFNS is transmitted only to girls, not to boys. If the mother has CFNS, then both boys and girls may have CFNS. These data are consistent with an X-linked dominant mode of inheritance for CFNS. b. Give the most likely genotypes of the parents in families numbered 1 and 10a. Solution: Family 1: normal father (X + Y)  CFNS mother (X + X C ) Family 10a: CFNS father (X C Y)  normal mother (X + X + ) 34. Miniature wings (X m ) in Drosophila result from an X-linked allele that is recessive to the allele for long wings (X + ). Give the genotypes of the parents in the following crosses.   Male parent Female parent Male offspring Female off- spring a . Long Long 231 long, 250 minia- ture 560 long b . Miniature Long 610 long 632 long c . Miniature Long 410 long, 417 minia- ture 412 long, 415 minia- ture d . Long Miniature 753 miniature 761 long e . Long Long 625 long 630 long

Solution: The genotype of the male parent is the same as his phenotype for an X-linked trait. Because the male progeny inherit their X chromosomes from their mother, the phenotypes of the male progeny give us the genotypes of the female parent. a. Male parent is X + Y. Because the male offspring are 1:1 long:miniature, the female parent must be X + X m . You can use a Punnett square to verify that all the female progeny from such a cross will have long wings (they get the dominant X + from the father). b. Male parent is X m Y. Because the male offspring are all long, the female parent must be X + X + . c. Male parent is X m Y; female parent is X + X m . d. Male parent is X + Y; female parent is X m X m . e. Male parent is X + Y; female parent is X + X + . *35. In chickens, congenital baldness is due to a Z-linked recessive gene. A bald rooster is mated with a normal hen. The F 1 from this cross are interbred to produce the F 2 . Give the genotypes and phenotypes, along with their expected proportions, among the F 1 and F 2 progeny. Solution: For species with the ZZ-ZW sex-determination system, the females are heterogametic ZW. So a bald rooster must be Z b Z b (where Z b denotes the recessive allele for baldness), and a normal hen must be Z + W. P Z b Z b × Z + W F 1 ½ Z b Z + (normal males) ½ Z b W (bald females) F 2 Using a Punnett square: 1 Z b Z b W W Z + Z + Z b (normal roosters) Z + W (normal hens) Z b Z b Z b (bald roosters) Z b W (bald hens) 36. If the blue F 1 females in Figure 4.15b are backcrossed to the blue males in the P generation, what types and proportions of offspring will be produced? Solution: The F 1 females are Z Ca + W. If they are backcrossed to blue males in the P generation (Z Ca + Z Ca + ), the resulting progeny will be ½ Z Ca + Z Ca + (blue males) and ½ Z Ca + W (blue females). So all of the offspring will be blue. 37. Red-green color-blindness is an X-linked recessive trait in humans. Polydactyly (extra fingers and toes) is an autosomal dominant trait. Martha has normal fingers and toes and normal color vision. Her mother is normal in all respects, but her father is color-blind and polydactylous. Bill-is color-blind and polydactylous. His mother has normal color vision

Cross 1 P : male, singed bristles, purple eyes  female, normal bristles, red eyes F 1 : 420 female, normal bristles, red eyes 426 male, normal bristles, red eyes F 2 : 337 female, normal bristles, red eyes 113 female, normal bristles, purple eyes 168 male, normal bristles, red eyes 170 male, singed bristles, red eyes 56 male, normal bristles, purple eyes 58 male, singed bristles, purple eyes Cross 2 P : female, singed bristles, purple eyes  male, normal bristles, red eyes F 1 : 504 female, normal bristles, red eyes 498 male, singed bristles, red eyes F 2 : 227 female, normal bristles, red eyes 223 female, singed bristles, red eyes 225 male, normal bristles, red eyes 225 male, singed bristles, red eyes 78 female, normal bristles, purple eyes 76 female, singed bristles, purple eyes 74 male, normal bristles, purple eyes 72 male, singed bristles, purple eyes a. What are the modes of inheritance of singed and purple ? Explain your reasoning. Solution: We examine each trait separately. The singed mutation is recessive because the F 1 for Cross 1 all have normal bristles. It is also X-linked because the reciprocal crosses (Cross 1 and Cross 2) give different F 1 progeny: Cross 2 yields F 1 singed males and normal females. The purple eye-color mutation appears recessive because the F 1 progeny of both crosses all have red eyes. It appears autosomal because there is no difference in the progeny of the reciprocal crosses, and also because there is no significant difference between male and female progeny with respect to eye color. b. Give genotypes for the parents and offspring in the P, F 1 , and F 2 generations of Cross 1 and Cross 2. Solution: We define X s as the singed allele, X + as the normal bristles allele, p as the purple allele, and P as the red-eyed allele. Cross 1: The F 1 males all have normal bristles, so the female parent is homozygous for normal bristles: X + X + . The singed male parent is X s Y.

The purple-eyed male parent must be homozygous recessive pp . The red-eyed female parent must be homozygous PP because all the progeny have red eyes. P male, singed bristles, purple eyes  female, normal bristles, red eyes X s Y, pp  X + X + , PP F 1 420 female, normal bristles, red eyes X + X s , Pp 426 male, normal bristles, red eyes X + Y, Pp F 2 337 female, normal bristles, red eyes X + X + , PP X + X s , PP X + X + , Pp (2) X + X s , Pp (2) 113 female, normal bristles, purple eyes X + X + , pp X + X s , pp 168 male, normal bristles, red eyes X + Y, PP X + Y, Pp (2) 170 male, singed bristles, red eyes X s Y, PP X s Y, Pp (2) 56 male, normal bristles, purple eyes X + Y, pp 58 male, singed bristles, purple eyes X s Y, pp Cross 2: The female parent with the recessive singed bristles must be homozygous X s X s , and the male parent must be X + Y. The purple-eyed parent is pp , and the red-eyed parent must be homozygous PP because the F 1 all have red eyes. P female, singed bristles, purple eyes  male, normal bristles, red eyes X s X s , pp  X + Y, PP F 1 504 female, normal bristles, red eyes X + X s , Pp 498 male, singed bristles, red eyes X s Y, Pp F 2 227 female, normal bristles, red eyes X + X s , PP X + X s , Pp (2) 223 female, singed bristles, red eyes X s X s , PP X s X s , Pp (2) 225 male, normal bristles, red eyes X + Y, PP X + Y, Pp (2) 225 male, singed bristles, red eyes X s Y, PP X s Y, Pp (2) 78 female, normal bristles, purple eyes X + Xs, pp 76 female, singed bristles, purple eyes X s X s , pp 74 male, normal bristles, purple eyes X + Y, pp 72 male, singed bristles, purple eyes X s Y, pp The (2) indicates that there are twice as many of these genotypes.