Genetics Solutions Chapter Four

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A woman with normal chromosomes mates with a man who also has normal chromosomes. What chromosome combinations and numbers of Barr bodies would you expect to see if the chromosomes separate normally in oogenesis, but nondisjunction of the sex chromosomes takes place in meiosis I of spermatogenesis?

Eggs produced by normal meiosis: X Sperm produced by nondisjunction in meiosis I: XY and O Children: XXY (one Barr body) and XO (no Barr body)

What is the pseudoautosomal region?

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.

In Drosophila, yellow body is due to an X-linked gene that is recessive to the gene for gray body. b. A yellow female is crossed with a gray male. The F1 are intercrossed to produce the F2. Give the genotypes and phenotypes, along with the expected proportions, of the F1 and F2 progeny.

The yellow female must be homozygous XyXy because yellow is recessive, and the gray male, having only one X chromosome, must be X+Y. The F1 male progeny are all XyY (yellow) and the F1 females are all X+Xy (heterozygous gray). P XyXy (yellow female) × X+Y (gray male) F1 ½ XyY (yellow males) ½ X+Xy (gray females) F2 ¼ X+Y (gray males) ¼ XyY (yellow males) ¼ X+Xy (gray females) ¼ XyXy (yellow females)

Hypohidrotic ectodermal dysplasia (also known as anhidrotic ectodermal dysplasia) is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. This trait is usually seen in men, but women who are heterozygous carriers of the trait often have irregular patches of skin with few or no sweat glands (see the illustration below). a. Explain why women who are heterozygous carriers of a recessive gene for hypohidrotic ectodermal dysplasia have irregular patches of skin lacking sweat glands.

X-inactivation occurs randomly in each of the cells of the early embryo, and is then maintained in the mitotic progeny cells. The irregular patches of skin lacking sweat glands arose from skin precursor cells that inactivated the X chromosome with the normal allele.

Suppose that a recessive gene that produces a short tail in mice is located in the pseudoautosomal region. A short-tailed male mouse is mated with a female mouse that is homozygous for a normal tail. The F1 from this cross are intercrossed to produce the F2. Give the phenotypes, as well as their proportions, of the F1 and F2 mice.

A gene in the pseudoautosomal region must be present on both X and Y chromosomes. We will use X+ and Y+ for the normal alleles and Xs and Ys for the short-tail alleles. The short-tailed male must be XsYs, and the normal female is X+X+. Assuming no crossing over between the gene and the sex-determining factor, the expected F1 are: X+Ys males with long tails and X+Xs females with long tails. F2 from the intercross will be: ¼ X+Ys males with long tails ¼ XsYs males with short tails ¼ X+X+ females with long tails ¼ X+Xs females with long tails So all the females will have long tails, and equal proportions of the males will have short and long tails.

Red-green color-blindness in humans is due to an X-linked recessive gene. A woman whose father is color-blind possesses one eye with normal color vision and one eye with color-blindness. Would it be possible for a man to have one eye with normal color vision and one eye with color-blindness?

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.

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. b. Metaphase I of meiosis

At metaphase I of meiosis, the homologous chromosomes are paired and line up as pairs at the cell equator.

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.

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+Xc 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 Xc. Failure of sister chromatids to separate in meiosis II would then result in an egg with two copies of Xc.

Xg is an antigen found on red blood cells. This antigen is caused by an X-linked allele (Xa) 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). c. XaY × X-X- → XaX-Y (Klinefelter syndrome)

Because only the father carries the Xa allele, the child's Xa chromosome must have come from the father. The child also inherited a Y chromosome from the father, so nondisjunction must have taken place in the father. Nondisjunction must have taken place in meiosis I, where the Xa and Y failed to separate.

Because the child inherited an Xa allele, the X chromosome must have come from the father, who was XaY. Since the child has only a single X chromosome, she received no maternal X chromosome and the nondisjunction must have taken place in the mother. This could have occurred during meiosis I, when the two homologous chromosomes failed to separate, producing some gametes with no X. It could have also occurred in meiosis II, when the two chromatids of an X chromosome failed to separate, producing some gametes with no copy of X.

Craniofrontonasal syndrome (CFNS) is a birth defect in which premature fusion of the cranial sutures leads to abnormal head shape, widely spaced eyes, nasal clefts, and various other skeletal abnormalities. George Feldman and his colleagues looked at several families in which offspring had CFNS and recorded the results shown in the following table (G. J. Feldman. 1997. Human Molecular Genetics 6:1937-1941). On the basis of these results, what is the most likely mode of inheritance for CFNS? (image)

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.

A color-blind woman and a man with normal vision have three sons and six daughters. All the sons are color-blind. Five of the daughters have normal vision, but one of them is color-blind. The color-blind daughter is 16 years old, is short for her age, and has not undergone puberty. Explain how this girl inherited her color blindness.

Color-blindness is an X-linked recessive trait, so the mother is XcXc and the father must be X+Y. Normally, all the sons would be color-blind (XcY) and all the daughters should have normal vision (Xc X+). The most likely way to have a daughter who is color-blind would be for her not to have inherited an X+ from her father. The observation that the color-blind daughter is short in stature and has failed to undergo puberty is consistent with Turner syndrome (XO). The color-blind daughter would then be XcO.

A Drosophila mutation called singed (s) causes the bristles to be bent and misshapen. A mutation called purple (p) causes the fly's eyes to be purple in color instead of the normal red. Flies homozygous for singed and purple were crossed with flies that were homozygous for normal bristles and red eyes. The F1 were intercrossed to produce the F2, and the following results were obtained. Cross 1 P: male, singed bristles, purple eyes × female, normal bristles, red eyes F1: 420 female, normal bristles, red eyes 426 male, normal bristles, red eyes F2: 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 F1: 504 female, normal bristles, red eyes 498 male, singed bristles, red eyes F2: 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 Give genotypes for the parents and offspring in the P, F1, and F2 generations of Cross 2

Cross 2: The female parent with the recessive singed bristles must be homozygous XsXs, 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 F1 all have red eyes. The (2) indicates that there are twice as many of these genotypes.

How many Barr bodies would you expect to see in a human cell containing the following chromosomes?

During the process of X-inactivation, all but one X chromosome is inactivated and each inactivated X becomes a Barr body, so there should be one less Barr body than the number of X chromosomes. a. XX—1 Barr body b. XY—0 c. XO—0 d. XXY—1 e. XXYY—1 f. XXXY—2 g. XYY—0 h. XXX—2 i. XXXX—3

A woman with normal chromosomes mates with a man who also has normal chromosomes. a. Suppose that, in the course of oogenesis, the woman's sex chromosomes undergo nondisjunction in meiosis I; the man's chromosomes separate normally. Give all possible combinations of sex chromosomes that this couple's children might inherit and the number of Barr bodies you would expect to see in each of the cells of each child.

Eggs produced by nondisjunction in meiosis I: XX and O Sperm produced by normal meiosis: X and Y Children: XXX (two Barr bodies), XO (no Barr body), XXY (one Barr body). YO would be embryonic lethal, so it would not be seen in any human child.

Craniofrontonasal syndrome (CFNS) is a birth defect in which premature fusion of the cranial sutures leads to abnormal head shape, widely spaced eyes, nasal clefts, and various other skeletal abnormalities. George Feldman and his colleagues looked at several families in which offspring had CFNS and recorded the results shown in the following table (G. J. Feldman. 1997. Human Molecular Genetics 6:1937-1941). Give the most likely genotypes of the parents in families numbered 1 and 10a.

Family 1: normal father (X+Y) × CFNS mother (X+XC) Family 10a: CFNS father (XCY) × normal mother (X+X+)

How does the inheritance of traits encoded by genes in the pseudoautosomal region differ from the inheritance of other Y-linked characteristics?

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.

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 dryness 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 inheritance for hypohidrotic ectodermal dysplasia? What evidence in Darwin's description supports your conclusion? (See also Problem 47 and Think-Pair- Share Question 8.)

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.

In Drosophila, yellow body is due to an X-linked gene that is recessive to the gene for gray body. c. A yellow female is crossed with a gray male. The F1 females are backcrossed with gray males. Give the genotypes and phenotypes, along with the expected proportions, of the F2 progeny.

If the F1 X+Xy females are backcrossed to X+Y gray males, then: F2 ¼ X+Y (gray males) ¼ XyY (yellow males) ¼ X+X+ (gray females) ¼ X+Xy (gray females)

What characteristics are exhibited by an X-linked trait?

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.

Joe has classic hemophilia, an X-linked recessive disease. Could Joe have inherited the gene for this disease from the following persons? His father's mother

No (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).

Joe has classic hemophilia, an X-linked recessive disease. Could Joe have inherited the gene for this disease from the following persons? His father's father

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?

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).

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

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?

Red-green color-blindness is due to an X-linked allele (Xc) that is recessive to the allele for color vision (X+). The color-blind daughter must be homozygous for the color-blind allele (XcXc); 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+Xc). 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 XcO and have Turner syndrome. 2) John has Klinefelter syndrome (and is heterozygous for the color-blind allele (X+XcY). 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 (XcY), 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+Xc).

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 (Xo) encodes orange fur. Females can be black (X+X+), orange (XoXo), or a mixture of orange and black called tortoiseshell (X+Xo). Males are either black (X+Y) or orange (XoY). 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 XoX+. The genotypes of the kittens are: orange male: XoY black male: X+Y tortoiseshell females: X+Xo orange female: XoXo Because the orange female has two orange alleles, the father must be carrying an Xo chromosome and a Y chromosome. Thus, the male cat's genotype must be XoY.

For each of the following chromosome complements, what is the phenotypic sex of a person who has: d. XXY with the SRY gene deleted?

Solution: Female

For each of the following chromosome complements, what is the phenotypic sex of a person who has: b. XX with a copy of SRY gene on an autosomal chromosome?

Solution: Male

For each of the following chromosome complements, what is the phenotypic sex of a person who has: c. XO with a copy of SRY gene on an autosomal chromosome?

Solution: Male

For each of the following chromosome complements, what is the phenotypic sex of a person who has: e. XXYY with one copy of the SRY gene deleted

Solution: Male (Klinefelter syndrome)

Hypohidrotic ectodermal dysplasia (also known as anhidrotic ectodermal dysplasia) is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. This trait is usually seen in men, but women who are heterozygous carriers of the trait often have irregular patches of skin with few or no sweat glands (see the illustration below). b. Why does the distribution of the patches of skin lacking sweat glands differ among the females depicted in the illustration, even between the two identical twins?

The X-inactivation event occurs randomly in each of the cells of the early embryo. Even in identical twins, the different ectodermal precursor cells will inactivate different X chromosomes, resulting in different distributions of patches lacking sweat glands.

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.

The child must have received the X- chromosome from the mother, as the father is XaY and carries no X- chromosome. Thus, the child received no X chromosome from the father, so the nondisjunction took place in the father. Nondisjunction of the sex chromosomes could have occurred in meiosis I, when the two sex chromosomes failed to separate, producing some gametes with no sex chromosome. Alternatively, it could have occurred in meiosis II, when the two chromatids of either the X or Y chromosome failed to separate, producing some gametes with no sex chromosome.

Miniature wings in Drosophila are due to an X-linked gene (Xm) that is recessive to an allele for long wings (X+). Sepia eyes are produced by an autosomal gene (s) that is recessive to an allele for red eyes (s+). A female fly that is homozygous for normal wings and has sepia eyes is crossed with a male that has miniature wings and is homozygous for red eyes. The F1 are intercrossed to produce the F2. Give the phenotypes, as well as their expected proportions, of the F1 and F2 flies.

The female parent (homozygous for normal wings, sepia eyes) is X+X+, ss The male parent (miniature wings, homozygous red eyes) is XmY, s+s+ The F1 males are X+Y, s+s (normal wings, red eyes) The F1 females are X+Xm, s+s (normal wings, red eyes) At the eye color locus, we expect the following proportions in the F2. At the locus that determines wing length, we expect the following proportions in the F2. Now we can apply the branch method to combine the probability of obtaining both characteristics together in the F2.

Miniature wings in Drosophila are due to an X-linked gene (Xm) that is recessive to an allele for long wings (X+). Sepia eyes are produced by an autosomal gene (s) that is recessive to an allele for red eyes (s+). A female fly that has miniature wings and sepia eyes is crossed with a male that has normal wings and is homozygous for red eyes. The F1 are intercrossed to produce the F2. Give the phenotypes, as well as their expected proportions, of the F1 and F2 flies.

The female parent (miniature wings, sepia eyes) must be XmXm, ss. The male parent (normal wings, homozygous red eyes) is X+Y, s+s+. The F1 males are XmY, s+s (miniature wings, red eyes) The F1 females are X+Xm, s+s (long wings, red eyes) T he proportions expected in the F2 can be obtained by breaking the cross into two separate crosses, one for each locus, and then combining the probabilities with the branch method. First, let's determine the outcome for the locus that codes for eye color. Next, let's consider the locus that determines wing length. Now we can use the branch method to combine the probability of obtaining both characteristics together in the F2.

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 and normal fingers and toes. If Bill and Martha marry, what types and proportions of children can they produce?

The first step is to deduce the genotypes of Martha and Bill. Because the two traits are on different chromosomes, they are independent, and we can deal with just one trait at a time. Starting with the X-linked color-blind trait, Bill must be XcY because he is color-blind. Bill's mother has normal color vision, and she passed a color-blind allele to Bill, so she must be a carrier (X+Xc). Martha must be X+Xc, a carrier for color-blindness, because she has normal color vision and her father is color-blind (XcY). For polydactyly, Bill must be Dd (D denotes the dominant polydactyly allele). Because his mother has normal fingers (dd), he cannot be homozygous DD. Martha, with normal fingers, must be dd. If Martha (dd, X+Xc) marries Bill (Dd, XcY), then we can predict the types and probability ratios of children they could produce. For polydactyly, the cross is dd × Dd. ½ of the children will be polydactylous, and ½ will have normal fingers. For color-blindness, the cross is X+Xc × XcY , so ¼ of children will be color-blind girls, ¼ will be girls with normal vision but carrying the color-blindness allele, ¼ will be color-blind boys, and ¼ will be boys with normal vision. Combining both traits, the probability of the different phenotypes is: color-blind girls (1/4) with normal fingers (1/2) = ¼ × ½ = 1/8 color-blind girls (1/4) with polydactyly (1/2) = ¼ × ½ = 1/8 girls with normal vision (1/4) and normal fingers (1/2) = ¼ × ½ = 1/8 girls with normal vision (1/4) and polydactyly (1/2) = ¼ × ½ = 1/8 color-blind boys (1/4) with normal fingers (1/2) = ¼ × ½ = 1/8 color-blind boys (1/4) with polydactyly (1/2) = ¼ × ½ = 1/8 boys with normal vision (1/4) and normal fingers (1/2) = ¼ × ½ = 1/8 boys with normal vision (1/4) and polydactyly (1/2) = ¼ × ½ = 1/8 This analysis can also be carried out with a Punnett square.

Miniature wings (Xm) 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.

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+Xm. 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 XmY. Because the male offspring are all long, the female parent must be X+X+. c. Male parent is XmY; female parent is X+Xm. d. Male parent is X+Y; female parent is XmXm. e. Male parent is X+Y; female parent is X+X+.

In Drosophila, yellow body is due to an X-linked gene that is recessive to the gene for gray body. d. If the F2 flies in part b mate randomly, what are the expected phenotypic proportions of flies in the F3?

The outcome of F2 flies from b mating randomly should be equivalent to random union of the male and female gametes. We need to predict the overall male and female gamete types and their frequencies. As a result of meiosis, half of the male gametes will have the Y chromosome. Because there are equal numbers of males with either X+ or Xy, the X-bearing male gametes will be split equally: (½ with X+)(½ with X+) = ¼ X+; (½ with X+)(½ with Xy) = ¼ Xy Male gametes: ½ Y, ¼ X+, ¼ Xy Half the F2 females in part (b) are homozygous XyXy, so all their gametes will be Xy: ½ Xy. The other half are heterozygous and will produce equal proportions of X+ and Xy gametes: ½(½X+) = ¼ X+; ½(½ Xy) = ¼ Xy. Female gametes: ¼ X+, ¾ Xy Now using a Punnett square: Overall genotypic ratios are 1/8 X+Y, 3/8 XyY, 1/16 X+X+, 4/16 X+Xy, and 3/16 XyXy. Overall phenotypic ratios are 1/8 gray males, 3/8 yellow males, 5/16 gray females, and 3/16 yellow females.

Orange coat color in cats is an X- linked trait that is recessive to black. A black female cat had a father who was orange. How many copies of the orange allele will be present in the following cells in oogenesis of this black female? See Figure 2.20 for a review of oogenesis in animals. a. primary oocyte b. first polar body c. ovum

The primary oocyte will have one orange allele as well as one black allele. Assuming that there is no nondisjunction of the X chromosomes and no crossing over involving this locus, the first polar body will have either the orange allele or the black allele. There is a 50% chance that the polar body will receive the orange allele. As for part b, there is a 50% chance that the ovum will have the orange allele. However, if the first polar body receives the X with the orange allele, then the ovum has no chance of having the orange allele unless there was a nondisjunction event.

Red-green color-blindness in humans is due to an X-linked recessive gene. A woman whose father is color-blind possesses one eye with normal color vision and one eye with color-blindness. Propose an explanation for this woman's vision pattern. Assume that no new mutations have spontaneously arisen.

The woman is heterozygous, with one X chromosome bearing the allele for normal vision and one X chromosome with the allele for color-blindness. One of the two X chromosomes is inactivated at random during early embryogenesis. If one eye 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.

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.

Explain why tortoiseshell cats are almost always female and why they have a patchy distribution of orange and black fur.

Tortoiseshell cats have two different alleles of an X-linked gene: X+ (non-orange, or black) and Xo (orange). The patchy distribution results from X-inactivation during early embryo development. Each cell of the early embryo randomly inactivates one of the two X chromosomes, and the inactivation is maintained in all of the daughter cells. So each patch of black fur arises from a single embryonic cell that inactivated the Xo, and each patch of orange fur arises from an embryonic cell that inactivated the X+. Normal male cats have only one X chromosome, so they cannot have patches of black and orange fur.

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?

Turner syndrome: XO Klinefelter syndrome: XXY (rarely XXXY, XXXXY, or XXYY) Androgen insensitivity: XY Triple-X females: XXX

We define Xs 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 F1 males all have normal bristles, so the female parent is homozygous for normal bristles: X+X+. The singed male parent is XsY. 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.

A Drosophila mutation called singed (s) causes the bristles to be bent and misshapen. A mutation called purple (p) causes the fly's eyes to be purple in color instead of the normal red. Flies homozygous for singed and purple were crossed with flies that were homozygous for normal bristles and red eyes. The F1 were intercrossed to produce the F2, and the following results were obtained. Cross 1 P: male, singed bristles, purple eyes × female, normal bristles, red eyes F1: 420 female, normal bristles, red eyes 426 male, normal bristles, red eyes F2: 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 F1: 504 female, normal bristles, red eyes 498 male, singed bristles, red eyes F2: 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 What are the modes of inheritance of singed and purple? Explain your reasoning.

We examine each trait separately. The singed mutation is recessive because the F1 for Cross 1 all have normal bristles. It is also X-linked because the reciprocal crosses (Cross 1 and Cross 2) give different F1 progeny: Cross 2 yields F1 singed males and normal females. The purple eye-color mutation appears recessive because the F1 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.

The following two genotypes are crossed: Aa Bb Cc X+ Xr × Aa BB cc X+Y, where a, b, and c represent alleles of autosomal genes and X+ and Xr represent X-linked alleles in an organism with XX-XY sex determination. What is the probability of obtaining genotype aa Bb Cc X+ X+ in the progeny?

We have to assume that the autosomal genes a, b, and c assort independently of each other as well as of the sex chromosomes. Given independent assortment, we can calculate the probability of the genotype for each gene separately, and then multiply the probabilities to calculate the probability of the combined genotype for all four genes. For gene a: Aa × Aa → ¼ aa For gene b: Bb × BB → ½ Bb For gene c: Cc × cc → ½ Cc For the sex-linked gene r: X+Xr × X+Y → ¼ X+X+ Combined probability of genotype aa Bb Cc X+X+ = ¼ × ½ × ½ × ¼ = 1/64

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 F1 are intercrossed to produce F2. Give the genotypes and phenotypes, along with the expected proportions, of the F1 and F2 progeny.

We will use X+ as the symbol for the dominant gray body color and Xy for the recessive yellow body color. The homozygous gray female parent is thus X+X+ and the yellow male parent is XyY. 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. F1 males inherit the Y chromosome from their father and X+ from their mother; hence, their genotype is X+Y and they have gray bodies. F1 females inherit Xy from their father and X+ from their mother; hence, they are X+Xy and also have gray bodies. When the F1 progeny are intercrossed, the F2 males again inherit the Y from the F1 male, and they inherit either X+ or Xy from their mother. Therefore, we should get ½ X+Y (gray body) and ½ XyY (yellow body). The F2 females will all inherit the X+ from their father and either X+ or Xy from their mother. Therefore, we should get ½ X+X+ and ½ X+Xy (all gray body). In summary: P X+X+ (gray female) × XyY (yellow male) F1 ½ X+Y (gray males) ½ X+Xy (gray females) F2 ¼ X+Y (gray males) ¼ XyY (yellow males) ¼ X+Xy (gray females) ¼ X+X+ (gray females) The net F2 phenotypic ratios are ½ gray females, ¼ gray males, and ¼ yellow males. The F2 progeny can also be predicted using a Punnett square.

What characteristics are exhibited by a Y-linked trait?

Y-linked traits appear only in males and are always transmitted from fathers to all sons, thus following a strict paternal lineage. Autosomal male-limited traits also appear only in males, but they can be transmitted to boys through their mothers.

Joe has classic hemophilia, an X-linked recessive disease. Could Joe have inherited the gene for this disease from the following persons? His mother's mother

Yes 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

Joe has classic hemophilia, an X-linked recessive disease. Could Joe have inherited the gene for this disease from the following persons? His mother's father

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.

Yes. If a woman has a son with hemophilia, then she carries at least one Xh allele for hemophilia. If she is a carrier (X+Xh), any of her sons have a 50% chance of inheriting the Xh allele and having hemophilia. If she is homozygous XhXh (unlikely), then all sons will inherit the Xh 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 circumcised. Males always pass on their Y chromosome to their sons, so her brothers cannot pass on the hemophilia allele to their sons.

Genetics Solutions Chapter Four

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