CH. 4 Genetics

Define the term hemizygosity.

- A situation where an individual has one copy of a set of genes in an otherwise diploid organism. Most of the genes on the X and Y chromosomes of human males are hemizygous. Exceptions are genes found in the pseudoautosomal regions which occur on the X and Y chromosomes.

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 that have more than one X chromosome. Mary Lyon correctly 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.

How many Barr bodies would you expect to see in a human cell containing the following chromosomes? a. XX b. XY c. XO d. XXY e. XXYY f. XXXY g. XYY h. XXX i. XXXX

a. 1 b. 0 c. 0 d. 1 e. 1 f. 2 g. 0 h. 2 i. 3 Human cells inactivate all X chromosomes beyond one. The Y chromosome has no effect on X-inactivation.

Anhidrotic ectodermal dysplasia is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. The 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. a. Explain why women who are heterozygous carriers of a recessive gene for anhidrotic ectodermal dysplasia have irregular patches of skin lacking sweat glands. 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?

a. X-inactivation occurs randomly in each of the cells of the early embryo, then is 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. b. 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.

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 through their daughters to their daughters' descendants, but not to their sons or their sons' descendants.

A human female with Turner syndrome also expresses the X-linked recessive trait hemophilia, as her father did. Which parent underwent non-disjuction during meiosis, giving rise to the gamete responsible for the syndrome? Assume that the mother is homoxygous for wild-type alleles.

- Non-disjunction occurred in the mother because the only X chromosome the daughter has must have come from the father. It could be either primary or secondary non-disjunction.

Bob has XXY chromosomes (Klinefelter syndrome) and is color blind. His mother and father have normal color vision, but his maternal grandfather is colorblind. 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 colorblind 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.

Red-green color blindness is a human X-linked recessive disorder. A young man with a 47, XXY karyotype (Klinefelter syndrome) is color blind. His 46, XY brother also is color blind. Both parents have normal color vision. Where did the nondisjuction that gave rise to the young man with Klinefelter syndrome take place? Assume that no crossing over took place in prophase I of meiosis.

- Because the father has normal color vision, the mother must be the carrier for color blindness. The color-blind young man with Klinefelter syndrome must have inherited two copies of the color-blind X chromosome from his mother. The nondisjunction event took place during meiosis of the egg at the second division.

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

- Colorblindness is an X-linked trait. 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 most likely 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 colorblind 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.

In chickens, congenital baldness is due to 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)

In organisms with the ZZ-ZW sex-determining system, from which of the following possibilities can a female inherit her Z chromosome?

- Her mother's mother: No Her mother's father: No Her father's mother: Yes Her father's father: Yes

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 His mother's father: Yes His father's mother: No His father's father: No 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).

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 number of X chromosomes determines sex. Flies with two X chromosomes are female and flies with one X chromosome are male. The Y chromosome is required for male fertility but does not determine sex.

What is meant by genic sex determination?

- In organisms that follow this system, there is no recognizable difference in the chromosome 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.

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.

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.

What types of offspring can arise from non-disjunction in meiosis I of a normal XY male? Assume the mother is normal and only makes gametes with an X chromosome.

- Nondisjunction in the male will produce sperm with XY and 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 XXY (Klinefelter syndrome males) and XO (Turner syndrome females).

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.

- Patches: XoX+; male: XoY; kittens: XoY, X+Y, X+Xo, XoXo

Red-green color blindness in humans is due to an X-linked recessive gene. Both John and Cathy have nomral 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 gien birth to a color-blind son, would John be justified in claiming nonpaternity?

- Since color blindness is a recessive trait, the color-blind daughter must be homozygous recessive. Because the color blindness is X-linked John has grounds for suspicion. Normally, their daughter would have inherited John's X chromosome. Because John is not color blind, he could not have transmitted a color-blind X chromosome to his daughter. However, there are other possibilities. 1) 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. 2) Cathy's egg was the result od secondary non-disjunction giving rise to an XcXc gamete. John also had non-disjunction producing a sperm with no sex chromosome. 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.

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 F1are intercrossed to produce the F2. Give the phenotypes, as well as their expected proportions, of the F1and 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+. F1: males are Xm Y, s+s (miniature wings, red eyes) females are X+Xm, s+s (long wings, red eyes) F2: We can analyze the expected outcome of this cross with a branch diagram. F2 males: Each male will receive a Y chromosome from his father and an X chromosome from his mother (P = ½ X+ Y or ½ Xm Y) Since the sepia locus is autosomal and the cross is s+s. x s+s the expected outcome is ¾ red eyes (S_) and ¼ sepia eyes (ss). ½ X+ Y x ¾ red eyes = 3/8 long wings and red eyes ½ X+ Y x ¼ sepia eyes = 1/8 long wings and sepia eyes ½ Xm Y x ¾ red eyes = 3/8 minature wings and red eyes ½ Xm Y x ¼ sepia eyes = 1/8 minature wings and sepia eyes F2 females: Each female will receive an X chromosome from her father and mother. The father's chromosome will be Xm. There is a ½ probability that she will receive X+ or Xm from her mother. Since the sepia locus is autosomal and the cross is s+s. x s+s the expected outcome is ¾ red eyes (S_) and ¼ sepia eyes (ss). ½ XmX+ x ¾ red eyes = 3/8 long wings and red eyes ½ XmX+ x ¼ sepia eyes = 1/8 long wings and sepia eyes ½ XmXm x ¾ red eyes = 3/8 minature wings and red eyes ½ XmXm x ¼ sepia eyes = 1/8 minature wings and sepia eyes

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. a. Male Parents: Long Female Parents: Long Male offspring: 231 long, 250 miniature Female offspring: 560 long b. Male Parents: Miniature Female Parents: Long Male offspring: 610 long Female offspring: 632 long c. Male Parents: Miniature Female Parents: Long Male offspring: 410 long, 417 miniature Female offspring: 412 long, 415 miniature d. Male Parents: Long Female Parents: Miniature Male offspring: 753 miniature Female offspring: 761 long e. Male Parents: Long Female Parents: Long Male offspring: 625 long Female offsping: 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 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+

Bill and Betty have had two children with Down syndrome. Bill's brother has Down syndrome and his sister have two children with Down syndrome. On the basis of these observations, indicate which of the following statements are most likely correct and which are most likely incorrect. Explain your reasoning. a. Bill has 47 chromosomes. b. Betty has 47 chromosomes. c. Bill and Betty's children each have 47 chromosomes. d. Bill's sister has 45 chromosomes. e. Bill has 46 chromosomes. f. Betty has 45 chromosomes. g. Bill's brother has 45 chromosomes.

- The high incidence of Down syndrome in Bill's family and among Bill's relatives is consistent with familial Down syndrome, caused by a Robertsonian translocation involving chromosome 21. Bill and his sister, who are unaffected, are phenotypically normal carriers of the translocation and have 45 chromosomes. Their children and Bill's brother, who have Down syndrome, have 46 chromosomes, but one of these chromosomes is the translocation that has an extra copy of the long arm of chromosome 21. From the information given, there is no reason to suspect that Bill's wife Betty has any chromosomal abnormalities. a. Incorrect. Bill is phenotypically healthy but is a translocation carrier (with 45 chromosomes). b. Incorrect. Betty is phenotypically healthy, and thus, unlikely to have an extra chromosome. c. Incorrect. Familial Down syndrome, as evident in this family pedigree, results from an extra copy of the long arm of chromosome 21 that is translocated to another chromosome. Thus, the overall number of chromosomes in those with familial Down syndrome is still 46 (although they have three functional copies of chromosome 21). d. Correct. It is likely that Bill and his sister, who are unaffected, are phenotypically normal carriers of the translocation and have 45 chromosomes. e. Incorrect. Bill is phenotypically healthy but is a translocation carrier, with 45 chromosomes because one copy of chromosome 21 is fused with another chromosome. f. Incorrect. Betty is phenotypically healthy. The familial Down syndrome is inherited through Bill's translocation (shared with his sister), thus, Betty is unlikely to have a translocation resulting in 45 chromosomes. g. Incorrect. Because Bill's brother has familial Down syndrome, he carries a translocation with an extra copy of chromosome 21 and has a total of 46 chromosomes.

What is the pseudoautosomal region of the human X and Y chromosomes? How does the inheritance of traits encoded by 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 passed on only from father to son.

Following secondary non-disjunction of the X chromosome during female meiosis, what types of sex chromosome aneuploidies would be expected in the offspring after fertilization with a normal sperm?

- Turner Syndrome (X0), Klinefelter Syndrome (XXY), and XXX (YO dies)

Give the typical sex chromosome complement found in the cells of people with Turner syndrome, Klinefelter syndrome, and triple-X females.

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

The following two genotypes are crossed: AaBbCcX^+ X^r x AaBBccX^+Y, where a, b, and c represent alleles of autosomal genes and X^+ and X^r represent X-linked alleles in an organism with XX-XY sex determination. What is the probability of obtaining genotype aaBbCcX^+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 aaBbCcX+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 color. 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 males 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, X+Y (gray body) and 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 X+X+ and X+Xy (all gray body). In summary: P: X+X+ (gray female) x. 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.

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.

For each of the following chromosome complements, what is the phenotypic sex of a person who has a. XY with the SRY gene deleted? b. XX with a copy of SRY gene on an autosomal chromosome? c. XO with a copy of SRY gene on an autosomal chromosome? d. XXY with the SRY gene deleted? e. XXYY with one copy of the SRY gene deleted?

a. Female b. Male c. Male d. Female e. Male (Klinefelter syndrome) 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.

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). a. XaY x X-X- gives XaO (Turner syndrome) b. XaY x XaX gives X- (Turner syndrome) c. XaY x X-X- gives XaX-Y (Klinefelter syndrome) d. XaY x XaX- gives X-X-Y (Klinefelter syndrome)

a. Since the child received no maternal X chromosome, the nondisjunction must have taken place in the mother, in either meiosis I or meiosis II. b. The child received no sex chromosome from the father, so the nondisjunction took place in the father, in either meiosis I or II. c. This child received both the Xa and Y from the father, so the nondisjunction took place in the father, in meiosis I, where the Xa and Y failed to separate. d. This child received two copies of X- from the mother. The nondisjunction took place in the mother, in meiosis II, where the sister chromatids failed to separate.

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. a. Propose an explanation for this woman's vision pattern. Assume that no new mutations have spontaneously arisen. b. Would it be possible for a man to have one eye with normal color vision and one eye with color blindness?

a. 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. b. One way would be for the man to be XXY, and the answer to part (a) would apply.

In Drosophila a cross was made between homozygous wild-type females and yellow-bodied males. All the F1 were phenotypically wild-type. In the F2 the following results were observed; 123 wild-type males, 116 yellow males, and 240 wild-type females. a. Is the yellow locus autosomal or sex-linked? b. Is the mutant gene for yellow body color recessive or dominant?

a. sex-linked b. recessive If the mutant gene is autosomal and recessive the cross would be y+y+ x yy. The F1 should all be y+y and when they are mated to themselves the F2 will exhibit a 3:1 ratio or normal body: yellow body for both sexes. If the mutant gene is autosomal and dominant all the F1 would be yellow with genotype y+y. If the mutant gene is X-linked and recessive the cross would be X+X+ x XyY. The F1 females (X+Xy) and F1 males (X+Y) would all be phenotypically wild-type. The F2 females would all be wild-type (X+Xy or X+X+) and the males would be half wild-type (X+Y) and half yellow (XyY). If the mutant allele is X-linked and dominant the F1 females (X+Xy) should all be yellow. If the mutant allele is Y-linked and dominant or recessive all of the F1 males should be yellow.