Chapter Four: Sex Determination and Sex-Linked Characteristics



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Chapter Four: Sex Determination and Sex-Linked Characteristics




COMPREHENSION QUESTIONS

1. What is the most defining difference between males and females?



Males produce relatively large gametes; females produce larger gametes.
*2. How do monoecious organisms differ from dioecious organisms?

Monoecious organisms exist as only one form, which has both male and female reproductive structures; the same organism produces both male and female gametes. Dioecious organisms exist as organisms of two distinct genders, 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?

In the XX-XO sex determination system, females have two copies of the sex-determining chromosome, whereas males have only one copy. Males must be considered 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?

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 genes in this region differ from the inheritance of other Y-linked characteristics?

The pseudoautosomal region is a region of similarity 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 inherited only from father to son.

*6. How is sex determined in insects with haplodiploid sex determination?

Diploid individuals are female, whereas haploid individuals are male. Eggs that are fertilized by a sperm develop into females, and eggs that are not fertilized develop as males.
7. What is meant by genic sex determination?

In organisms that follow this system, there is no recognizable difference in the chromosome contents 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.
8. How does sex determination in Drosophila differ from sex determination in humans?

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 determines the sex of the individual, regardless of the Y chromosome. Flies with XXY are female.
9. Give the typical sex chromosomes found in the cells of people with Turner syndrome, Klinefelter syndrome, and androgen insensitivity syndrome, as well as in poly-X females.

Turner syndrome: XO

Klinefelter syndrome: XXY (rarely XXXY or XXYY)

Androgen insensitivity: XY

Poly-X females: XXX (rarely XXXX or even XXXXX)
*10. 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 normally recessive or dominant. Males inherit X-linked traits from their mothers, pass X-linked traits to their daughters, and subsequently on to their grandsons, but not to their sons.

11. Explain how Bridges’s study of nondisjunction in Drosophila helped prove the chromosome theory of inheritance.


Bridges showed that in crosses with white-eyed flies, a sex-linked trait, exceptional progeny had abnormal inheritance of sex chromosomes. In matings of white-eyed females with red-eyed males, most of the progeny followed the expected pattern of white-eyed males and red-eyed females. Exceptional red-eyed male progeny were XO and exceptional white-eyed females were XXY. These karyotypes were exactly as Bridges predicted with his hypothesis that the exceptional red-eyed males inherited their X chromosome with the red eye allele from their red-eyed fathers and were male because they did not inherit an X chromosome from their mothers, resulting in an XO condition that is phenotypically male but sterile. Moreover, the exceptional white-eyed females inherited two X chromosomes from their white-eyed mothers, as a result of nondisjunction in meiosis I of the female, and none from their red-eyed fathers, receiving instead a Y chromosome to make them XXY females. Calvin Bridges linked exceptional inheritance of a sex-linked trait to exceptional inheritance of the X chromosome: the linked exceptions proved the rule that genes reside on chromosomes.
12. 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+.

13. What is a Barr body? How is it related to the Lyon hypothesis?


Barr bodies are darkly staining bodies in the nuclei of female mammalian cells. Mary Lyon hypothesized that Barr bodies are inactivated (condensed) X chromosomes. By inactivating all X chromosomes beyond one, female cells achieve dosage compensation for X-linked genes.
*14. What characteristics are exhibited by a Y-linked trait?

Y-linked traits appear only in males and are always transmitted from fathers to 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.
APPLICATION QUESTIONS AND PROBLEMS
*15. What is the sexual phenotype of fruit flies with the following chromosomes?





Sex

chromosomes



Autosomal

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, sterile (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 >1.0)

(h)

XX

Four haploid sets

Male (X:A = 0.5)

(i)

XXX

Four haploid sets

Intersex (X:A between 0.5 and 1.0)

(j)

XXX

Three haploid sets

Female (X:A = 1.0)

(k)

X

Three haploid sets

Metamale, sterile (X:A < 0.5)

(l)

XY

Three haploid sets

Metamale (X:A < 1.0)

(m)


XX

Three haploid sets

Intersex (X:A between 0.5 and 1.0)


In fruit flies, the X to autosome ratio determines sex: males have X:A ratios of 0.5 (metamales have X:A ratios less than 0.5) and females have X:A ratios of 1.0 (metafemales have ratios greater than 1.0).
16. For parts a through g in problem 15 what would be the human sexual phenotype (male or female) be?

In humans, the Y chromosome determines maleness. Although extra sex chromosomes may be tolerated, extra sets of autosomes are lethal.

(a) female

(b) male


(c) female (Turner syndrome)

(d) male (Klinefelter syndrome)

(e) male

(f) male (Klinefelter syndrome)

(g) female (triple-X syndrome)
*17. Joe has classic hemophilia, which is an X-linked recessive disease. Could Joe have inherited the gene for this disease from the following persons?

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
X-linked traits are passed on from mother to son. Therefore, 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 (b). Because Joe could not have inherited the trait from his father (Joe inherited the Y chromosome from his father), he could not have inherited hemophilia from either (c) or (d).

*18. 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. 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.

This problem can also be analyzed using a Punnett square.






X+

Y


X+

X+X+ (gray females)

X+Y (gray males)

Xy

X+Xy (gray females)

XyY (yellow males)

(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)
(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)
(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. Since there are equal numbers of males with either the 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: ½Y, 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:





½ Y

¼ X+

¼ Xy

¼ X+

1/8 X+Y

1/16 X+X+

1/16 X+Xy

¾ Xy

3/8 XyY

3/16 X+Xy

3/16 XyXy


Overall genotypic ratios are 1/8 X+Y, 3/8 XyY, 1/16 X+X+, 4/16 X+Xy, 3/16 XyXy. Overall phenotypic ratios are 1/8 gray males, 3/8 yellow males, 5/16 gray females, and 3/16 yellow females.

*19. Both John and Cathy have normal color vision. After 10 years of marriage to John, Cathy gives birth to a color-blind daughter. John files 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?


Assuming the color-blindness is a recessive trait, the color-blind daughter must be homozygous recessive. If the color-blindness trait here is not X-linked, then John has no justification, because both John and Cathy could be carriers. If the color-blindness is the X-linked red-green color-blindess, then John has grounds for suspicion. Normally, their daughter would have inherited John’s X chromosome. Since John is not color-blind, he could not have transmitted a color-blind X chromosome to the daughter.

A remote alternative possibility is that the daughter is XO, having inherited a recessive color-blind allele from her mother and no sex chromosome from her father. In that case, the daughter would have Turner syndrome.

If Cathy had a color-blind son, 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.
20. Red-green color-blindness in humans is due to an X-linked recessive gene. A woman whose father was color-blind possesses one eye with normal color vision and one eye with color-blindness.

(a) Propose an explanation for this woman’s vision pattern.



One of the two X chromosomes in the woman’s zygotes must have been inactivated during early embryogenesis. If one eye derived exclusively from a progenitor cell that inactivated the normal X, then that eye would be color-blind, whereas the other eye may be derived from a progenitor cell 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?



The only way would be for the man to be XXY.

*21. 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 occur? Explain your answer.


Since 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.
22. In certain salamanders, it is possible to alter the sex of a genetic female, making her into a functional male; these salamanders are called sex-reversed males. When a sex-reversed male is mated with a normal female, approximately ⅔ of the offspring are female and ⅓ are male. How is sex determined in these salamanders? Explain the results of this cross.

The 2:1 ratio of females to males could be explained if a quarter of the progeny are embryonic lethals. The sex-reversed male has the same chromosome complement as normal females. If females were homogametic, then matings between sex-reversed males and normal females must result in all females: XX (sex-reversed male) crossed to XX (normal female) results in all XX (female) progeny. However, if females are heterogametic (ZW), then ZW (sex-reversed male) crossed to ZW (normal female) results in ¼ ZZ (male), ½ ZW (female), and ¼ WW (embryonic lethal). The net result is a 2:1 ratio of females to males.
23. In some mites, males pass genes to their grandsons, but they never pass genes to male offspring. Explain.

A system in which males are haploid and females are diploid would explain these results. Haploid males pass genes only to female progeny. The female progeny can then generate haploid male grandsons that contain the grandfather’s genes.

24. 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. Is this religious law consistent with sound genetic principles? Explain your answer.

Yes. If a woman has a son with hemophilia, then she is a carrier. Any of her sons have a 50% chance of having hemophilia. Her sisters may also be carriers. Her brothers, if they do not themselves have hemophilia (since they survived circumcision, they most likely do not have hemophilia), cannot be carriers, and therefore there is no risk of passing hemophilia on to their children.
*25. 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:




Male parent


Female parent


Male offspring



Female offspring

(a)

Long

Long

231 long,

250 miniature

560 long


(b)

Miniature

Long

610 long

632 long

(c)

Miniature

Long

410 long,

417 miniature



412 long,

415 miniature



(d)

Long

Miniature

753 miniature

761 long

(e)

Long

Long

625 long

630 long


The genotype of the male parent is the same as his phenotype for an X-linked trait. Because the male progeny get their X chromosomes from their mother, the phenotypes of the male progeny give us the genotypes of the female parents.

(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 dad).

(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+.
*26. In chickens, congenital baldness results from a Z-linked recessive gene. A bald rooster is mated with a normal hen. The F1 from this cross are interbred to produce the F2. Give the genotypes and phenotypes, along with their expected proportions, among the F1 and F2 progeny.

For species with the ZZ-ZW sex-determination system, the females are heterogametic ZW. So a bald rooster must be ZbZb (where Zb denotes the recessive allele for baldness), and a normal hen must be Z+W.

P ZbZb × Z+W

F1 ½ ZbZ+ (normal males)

½ ZbW (bald females)

F2 Using a Punnett square:





Zb

W


Z+

Z+Zb (normal roosters)

Z+W (normal hens)

Zb

ZbZb (bald roosters)

ZbW (bald hens)

27. In the eastern mosquito fish (Gambusia affinis holbrooki), which has XX-XY sex determination, spotting is inherited as a Y-linked trait. The trait exhibits 100% penetrance when the fish are raised at 22C, but the penetrance drops to 42% when the fish are raised at 26C. A male with spots is crossed with a female without spots and the F1 are intercrossed to produce the F2. If all the offspring are raised at 22C, what proportion of the F1 and F2 will have spots? If all the offspring are raised at 26C, what proportion of the F1 and F2 will have spots?


Y-linked genes are passed on from father to son, and females are always unaffected. Genotypes and ratios:

P XYs (spotted male) × XX (female without spots)

F1 ½ XX (females without spots)

½ XYs (spotted males)

F2 same as F1

At 22 (100% penetrance), all males will have spots in the F1 and F2.

At 26 (42% penetrance), 42% of males will have spots in the F1 and F2.
*28. How many Barr bodies would you expect to see in human cells containing the following 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



Human cells inactivate all X chromosomes beyond one. The Y chromosome has no effect on X-inactivation.
29. 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 independent, 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 must be a carrier (X+Xc). Martha must be X+Xc, a carrier for color blindness because her father is color blind (XcY).

For polydactyly, Bill must be Dd (D denotes the dominant polydactyly allele). Since his mother has normal fingers (dd), he can’t 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, ½ of children will be polydactylous, and ½ will have normal fingers.

For color-blindness, 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, then:

1/8 color-blind girls with normal fingers

1/8 color-blind girls with polydactyly

1/8 girls with normal vision and normal fingers

1/8 girls with normal vision and polydactyly

1/8 color-blind boys with normal fingers

1/8 color-blind boys with polydactyly

1/8 boys with normal vision and normal fingers

1/8 boys with normal vision and polydactyly

This analysis can also be carried out with a Punnett square.

*30. Miniature wings in Drosophila melanogaster result from an X-linked gene (Xm) that is recessive to an allele for long wings (Xm+). Sepia eyes are produced by an autosomal gene (s) that is recessive to an allele for red eyes (s+).

(a) 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 and their proportions expected in the F1 and F2 flies from this cross.



The female parent (miniature wings, sepia eyes) must be XmXm,ss.

The male parent (normal wings, homozygous red eyes) is Xm+Y,s+s+.

F1 males are XmY,s+s (miniature wings, red eyes)

females are Xm+Xm,s+s (long wings, red eyes)

F2: We can analyze the expected outcome of this cross with either a branch diagram or with a Punnett square.

First, the branch diagram:

½ male (Y)½ Xm+ normal wings ¾ red (¾ s+) = ½ × ½ × ¾ = 3/16 male, normal, red

¼ sepia (¼ ss) = ½ × ½ × ¼ = 1/16 male, normal, sepia

½ Xm miniature ¾ red = ½ × ½ × ¾ = 3/16 male, miniature, red

¼ sepia = ½ × ½ × ¼ = 1/16 male, miniature, sepia


½ female (Xm) ½ Xm+ normal ¾ red = ½ × ½ × ¾ = 3/16 female, normal, red

¼ sepia = ½ × ½ × ¼ = 1/16 female, normal, sepia

½ Xm miniature ¾ red = ½ × ½ × ¾ = 3/16 female, miniature, red

¼ sepia = ½ × ½ × ¼ = 1/16 female, mini, sepia
Explanation: ½ of the F2 progeny are males because they inherit Y from the F1 male. The other ½ are females that inherit the Xm from the F1 male. In each case, ½ of the F2 males and females inherit Xm+ from the F1 female and have normal wings, whereas the other ½ inherit Xm and have miniature wings. Finally, ¾ of the progeny will have the dominant red eyes, and ¼ will have sepia eyes.

Now the Punnett square:





¼ Xm s+

¼ Xm s

¼ Y s+

¼ Y s

¼ Xm+ s+

Xm+ Xm s+s+

Long wings,

red eyes

Xm+ Xm s+s

Long wings,

red eyes

Xm+ Y s+s+

Long wings, red eyes


Xm+ Y s+s

Long wings, red eyes

¼ Xm+ s

Xm+ Xm s+s

Long wings, red eyes

Xm+ Xm ss

Long wings, sepia eyes

Xm+ Y s+s

Long wings, red eyes

Xm+ Y ss

Long wings, sepia eyes

¼ Xm s+

XmXm s+s+

Mini wings, red eyes

XmXm s+s

Mini wings, red eyes

XmY s+s+

Mini wings, red eyes

XmY s+s

Mini wings, red eyes

¼ Xm s

XmXm s+s

Mini wings, red eyes

XmXm ss

Mini wings, sepia eyes

XmY s+s

Mini wings, red eyes

XmY ss

Mini wings, sepia eyes

(b) 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 and proportions expected in the F1 and F2 flies from this cross.


Parents Xm+ Xm+ ,ss and XmY,s+s+

F1 Xm+ Xm,s+s and Xm+ Y,s+s

F2

Sex chromosome inherited Autosomal Combined

From male From female phenotype phenotype




½ Y male ½ Xm+ ¾ red 3/16 long wings, red eyes

¼ sepia 1/16 long wings, sepia eyes

½ Xm ¾ red 3/16 mini wings, red eyes

¼ sepia 1/16 mini wings, sepia eyes

½ Xm+ female Xm+ or Xm ¾ red 3/8 long wings, red eyes

¼ sepia 1/8 long wings, sepia eyes
Note that in this case, the X chromosome the F2 females inherit from the mother does not affect their phenotype because they all have a dominant X+ for long wings from their father.

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


If the gene is in the pseudoautosomal region, then it must be present on both the X and Y chromosome. 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+. The expected F1 would then be:

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.
*32. A color-blind female and a male 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 never undergone puberty. Propose an explanation for how this girl inherited her color-blindness.

The trivial explanation for these observations is that this form of color-blindness is an autosomal recessive trait. In that case, the father would be a heterozygote, and we would expect equal proportions of color-blind and normal children, of either sex.

If, on the other hand, we assume that this is an X-linked trait, then the mother is XcXc and the father must be X+Y. Normally, all the sons would be color-blind, and all the daughters should have normal vision. The only way to have a daughter that 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.


CHALLENGE QUESTIONS

33. On average, what proportion of the X-linked genes in the first individual is the same as that in the second individual?



  1. A male and his mother

A male inherits his single X chromosome from his mother, so 100% of his X-

linked genes are the same as those in his mother.

  1. A female and her mother.

A female inherits one of her two X chromosomes from her mother, so 50% of her

X-linked genes are the same as those in her mother.

  1. A male and his father

A male inherits only the Y chromosome from his father, so none of his X-linked

genes are the same as those in his father.

  1. A female and her father

A female inherits one of her two X chromosomes from her father, so 50%.



  1. A male and his brother

The chance that a male’s brother inherited the same X chromosome of the two

possible maternal X chromosomes is ½, or 50%.

  1. A female and her sister

A female and her sister share at least 50% because they inherited the same X

chromosome from their father. In addition, there is another ½ chance that they

inherited the same X chromosome from their mother, so overall it is 50% +

½(50%) = 75%.

  1. A male and his sister

The chance that a sister inherited the same X chromosome from her mother as


her brother did is ½, or 50%.


  1. A female and her brother

Of the two X chromosomes in the female, the one inherited from the father

cannot be present in her brother. Then there is a ½, or 50%, chance that the

second X chromosome is shared.
34. A geneticist discovers a male mouse in his laboratory colony with greatly enlarged testes. He suspects that this trait results from a new mutation that is either Y-linked or autosomal dominant. How could he determine if the trait is autosomal dominant or Y-linked?

Since testes are present only in males, enlarged testes could either be a sex-limited autosomal dominant trait or a Y-linked trait. Assuming the male mouse with enlarged testes is fertile, mate it with a normal female. If the trait is autosomal dominant, and the parental male is heterozygous, only half the male progeny will have enlarged testes. If the trait is Y-linked, all the male progeny will have enlarged testes.

With either outcome, however, the results from this first cross will not be conclusive. If all the male progeny do express the trait, the trait may still be autosomal dominant if the parental male was homozygous. If only some of the male progeny express the trait, the possibility still remains that the trait is Y-linked but incompletely penetrant. In either case more conclusive evidence is needed.

Mate the female progeny (F1 females) with normal males. If the trait is autosomal dominant, some of the male F2 progeny will have enlarged testes, proving that the trait can be passed through a female. If the trait is Y-linked, all the male F2 progeny will have normal testes, like their normal male father.

35. Amanda is a genetics student at a small college in Connecticut. While counting her fruit flies in the laboratory one afternoon, she observed a strange species of fly in the room. Amanda captured several of the flies and began to raise them. After having raised the flies for several generations, she discovered a mutation in her colony that produces yellow eyes, in contrast with normal red eyes, and Amanda determined that this trait is definitely X-linked recessive. Because yellow eyes are X-linked, she assumed that either this species has the XX-XY system of sex determination with genic balance similar to Drosophila or it has the XX-XO system of sex determination.

How can Amanda determine whether sex determination in this species is XX-XY or XX-XO? The chromosomes of this species are very small and hard for Amanda to see with her student microscope so she can only conduct crosses with flies having the yellow-eye mutation. Outline the crosses Amanda should conduct and explain how they will prove XX-XY or XX-XO sex determination in this species.


To distinguish between these two possibilities, Amanda has to look for situations where the presence of the Y chromosome makes a difference in the outcome of a cross. The XX-XY and XX-XO systems behave similarly in normal matings. However, the outcomes can be very different in exceptional circumstances, such as nondisjunction of the X chromosomes in female meiosis.

Amanda should cross yellow-eyed females with red-eyed males. With either sex determination system, the progeny of such a cross should be all yellow-eyed males and all red-eyed females. However, she should be able to observe rare (~1/1000) progeny that are yellowed-eyed females or red-eyed males, as a result of meiotic nondisjunction, as Calvin Bridges observed in a similar cross with Drosophila melanogaster white-eyed mutants. Nondisjunction at either meiosis I or II will produce female gametes with two Xm (bearing the yellow-eye mutant allele) or nullo-X. Upon fertilization of the nullo-X egg by an X+-bearing sperm, exceptional red-eyed X+O males will be produced with either sex determination system. However, fertilization of the XmXm egg with a Y-bearing sperm will result in exceptional yellow-eyed females with XmXmY in the case of XX-XY sex determination. For the XX-XO sex determination system, fertilization of the XmXm egg by a nullo-X bearing sperm will produce XmXm females.

If these exceptional yellow-eyed females are now mated again to normal red-eyed males, then the two systems give different predicted outcomes. Meiosis in XmXmY females will be complicated by the presence of three sex chromosomes and should yield relatively high proportions of XmY and Xm gametes along with XmXm and Y. In contrast, meiosis in XmXm females should proceed normally, and produce just Xm gametes, excepting the rare nondisjunction events. Therefore, the progeny produced from mating these yellow-eyed exceptional females to normal red-eyed males will be very different, as shown in the following table:



P: XmXm × X+Y

F1: X+Xm females, XmY males

P: XmXm × X+O

F1: X+Xm females, XmO males

Rare: XmXmY females

X+O males



Cross XmXmY female × X+Y male
Rare: XmXm females

X+O males



Cross XmXm females × X+O male

If sex determination is XX-XY

If sex detemination is XX-XO

Gametes from

Xm Xm Y



X+Y male gametes

Gametes from

XmXm



X+O male gametes

X+

Y

X+



O

Xm

X+Xm

red female


XmY

yellow male

Xm

X+Xm


red female

XmO yellow male

XmY



X+XmY

red female

XmYY

yellow male



XmXm

XmXmX+


lethal

XmXmY yellow female



Y

X+Y

red male

YY

lethal


If the cross of the exceptional yellow-eyed females results in a relatively high frequency (at least several percent, rather than 1/1000) of red-eyed males and yellow-eyed females, then these flies use an XX-XY system. If the matings produce the usual phenotypic ratios, then the flies use an XX-XO system.
36. Occasionally, a mouse X chromosome is broken into two pieces and each piece becomes attached to a different autosomal chromosome. In this event, only the genes on one of the two pieces undergo X-inactivation. What does this observation indicate about the mechanism of X-chromosome inactivation?

The X-inactivation mechanism must require or recognize a specific region or locus on the X-chromosome and must inactivate chromatin attached to this center of inactivation. When the X-chromosome breaks, the fragment containing the X-inactivation locus or center becomes inactivated. The other fragment escapes inactivation because it is no longer attached.






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