5 Normal vision in humans is dominant to color blindness and

5. Normal vision in humans is dominant to color blindness and the alleles are found on the sex chromosomes. A normal man, whose father was color blind, marries a color blind woman. Normal=C; color blind=c. Give the genotypes of those involved:

Man________________________

His Father_________________

His Wife __________________

6. In humans, the gene for hemophilia (h) is sex-linked recessive to the gene for normal blood (H). Show the genotypes of the following individuals:

a. hemophiliac woman x a normal man

b. normal (heterozygous) woman x hemophiliac man

c. normal (homozygous) woman x hemophiliac man List the expected phenotypes and ratios resulting from the above crosses:

7. What is the genotype of a man who has freckles (dominant) and is color-blind, if his mother lacked freckles?

8. A man with a widow’s peak, whose mother was color blind and whose father had a continuous hairline, is engaged to marry a woman whose color blind mother had a continuous hairline, and whose normal visioned father had a continuous hairline. a. what is the genotype of the young gentleman? b. What is the genotype of his fiancée?

9. If the couple in the preceding problem marry and have a family, what are the chances of them having:

a. A normal visioned child with a continuous hairline?

b. A color blind child with a widow’s peak?

c. A normal visioned daughter with a continuous hairline?

d. A color blind son with a widow’s peak?

Solution

5. Normal vision in humans is dominant to color blindness and the alleles are found on the sex chromosomes. A normal man, whose father was color blind, marries a colorblind woman. Normal=C; color blind=c. Give the genotypes of those involved:

Man=XCY

His Father=XcY

His Wife =XcXc

two copies of a disease allele on the X chromosome are required for an individual with two X chromosomes (a female) to be affected with an X-linked recessive disease.

6. In humans, the gene for hemophilia (h) is sex-linked recessive to the gene for normal blood (H). Show the genotypes of the following individuals:

a. hemophiliac woman x a normal man

XhXh x XHY

=XhXH (normal), XhY (hemophelia),

b. normal (heterozygous) woman x hemophiliac man

XhXH x XhY

=XhXh (hemophelia), XhY(hemophelia), XHXh (normal), XHY (normal)

c. normal (homozygous) woman x hemophiliac man

XHXH x XhY

=XHXh (normal), XHY (normal)

two copies of a disease allele on the X chromosome are required for an individual with two X chromosomes (a female) to be affected with an X-linked recessive disease.