Biology 11
Panayiotou
SEX LINKAGE PROBLEMS
NOTE: Haemophilia and red-green colour-blindness are both X-linked recessive alleles.
1. What would be the genotypes of the parents and the phenotype ratios of the children
resulting form the following crosses?
a. a haemophiliac woman and a normal man
b. a heterozygous normal woman and a haemophiliac man.
2. A colour-blind man marries a woman who is a carrier but does not suffer from colourblindness. State the probable phenotypes of the two sons and two daughters produced by
this marriage.
3. A mother carries the sex-linked gene for colour blindness, but does not have the disease.
The father has normal vision. What is the probability of having a colour-blind son?
4. A man who is red-green colour-blind has four normal sons by his first wife and a colour-blind
daughter by his second wife. Both wives had normal vision. What is the genotype of each
wife?
5. One twin is haemophiliac, while the twin brother is not.
a. What is the probable gender (sex) of the haemophiliac twin?
b. Are the twins monozygotic (come from the same fertilized egg)? Explain.
c. Give the genotype of both twins.
6. Red-green colour blindness is a relatively common condition that is inherited as an X-linked
recessive allele. A normal woman whose father was red-green colour-blind marries a man
who has normal vision.
a. What proportion of their sons would you expect to be red-green colour-blind?
b. If she married a man who was red-green colour blind, what proportion of their sons
would you expect to have normal vision?
c. If she married a man who was red-green colour-blind, what proportion of their
daughters would be carriers?
7. In Drosophila, miniature wings are produced by a recessive sex-linked allele on the X
chromosome. Wingless flies are produced by a recessive autosomal allele found on
Chromosome 2.
a. List the genotypes of
(i)
a female with one allele for miniature wings
(ii)
a female with one allele for winglessness
b. Use a Punnett square to compare the results of crossing a normal male with female
(i) and then female (ii) above.
c. List the phenotypes of the members of the F1 generation in each cross from (b).
d. Identify the two parents Drosophila that could produce an offspring that would be
homozygous for winglessness.
Biology 11
Panayiotou
8. A mutant sex-linked trait called notched (XN) is deadly in female Drosophila when
homozygous. Males who have a single allele (XN) will also die. The heterozygous condition
(XNXn) causes small notches on the wing. The normal condition in both males and females is
represented by the allele Xn.
a. Indicate the phenotypes of the F1 generation from the following cross: XNXn and XnY.
b. Explain why dead females are never found in the F1 generation, no matter which
parents are crossed.
c. Explain why the mating of a female XNXn and a male XNY is unlikely.
9. A woman who has unusually short fingers marries a man who has fingers of normal length.
They have four children, two of each gender (sex). One of their sons and one of their
daughters have unusually short fingers. Could the allele responsible for this trait be
located on the X chromosome?
10. A form of diabetes is caused by a recesive allele located on an autosomal chromosome. You
already know that red-green colour-blindness is caused by a recessive sex-linked allele.
Explain why the ratio of women to men with diabetes is much closer than the ratio of women
to men with red-green colour-blindness.
11. Use a Punnett square to explain how a woman who is not colour-blind, but whose father is
colour-blind, can give birth to a son who is colour-blind.
12. A woman who has Turner’s syndrome (XO) is found to have haemophilia yet neither of her
parents have the disease. How is this possible?
13. Red-green colour-blindness in females is much more common than haemophilia A, another
sex-linked disorder. Give a possible explanation.