Question

10. In humans Wiskott–Aldrich syndrome (WAS) is a rare sex-linked recessive disease characterized by eczema, thrombocytopenia (low platelet count), immune deficiency, and bloody diarrhea. This disease is due to a sex-linked recessive allele.

a. Examine the statement below. Is this statement true or false? Why? Explain. Boys usually inherit sex-linked recessive disorders from their fathers. Sarah is healthy and but her brother has WAS. Her mother and father both have healthy phenotypes. Sarah marries Fred, a healthy man without WAS.

b. What is the probability that Sarah is a carrier of WAS? c. If Sarah is a carrier of WAS; then, what is the probability that Fred and Sarah’s first son inherits the WAS disease?

Answer 1

Sex-linked diseases whose genes are located on sex chromosomes either X or Y. Since Y is present only in males and that too in a single copy, there is no recessive dominant nature associated with genes on Y chromosome. On the other hand, females have two copies of X chromosome and thus, each chromosome can have a different allele of a particular gene. And the protein product that will be expressed depends on the dominant alleles or the contributing allele. In males, however, only a single copy of X chromosome is present thereby leading to forced expression of the allele without any mechanism to suppress the expression of a recessive mutant.

The Wiskott-Aldrich syndrome (WAS) is a sex-linked recessive disorder. Hence the gene that is responsible for this disease is present on the X chromosome. Let the gene by W. It has two alleles W and w, wherein W is dominant over w. The disease is present only when W is absent in the individual. In females, this can happen only in ww and not in WW or Ww scenarios. Males, on the other hand, only have one X chromosome. Hence the presence of a single mutant allele as in wY will lead to the disease phenotype.

a) For a male child, the Y chromosome comes from the father, the X chromosome comes from the mother. Hence if the disease phenotype is present in the child, the disease is transmitted from the X chromosome of the mother and hence the boy inherits the sex-linked recessive disorders from the mother. The statement, therefore, holds False.

Sarah is healthy but her brother has WAS. Being a male his sex-chromosome composition has to be wY. He has inherited the Y chromosome from his father and the X chromosome from his mother. But his mother is healthy. Hence she must be heterozygote for the locus and her genotype would be Ww. The father is also healthy and his genotype must be WY.

b) Thus, Sarah being a girl inherits one X chromosome from her father which would obviously be W and one X chromosome from her mother which can be either of W or w with equal probability. Therefore, the probability that Sarah is a carrier of the disease is 50%.

c) Now, she married a man named Fred who is healthy. Hence as per our argument, his genotype would be WY. And if Sarah is a carrier her genotype would be Ww.

Thus, we can generate a Punnett square to examine their offsprings.

	W	Y
W	WW	WY
w	Ww	wY

The son has a Y chromosome. The possible genotypes for the son are WY and wY each with 50% probability. Of these two genotypes, WY will give a healthy phenotype while wY will give WAS disease. Thus the probability of the son being diseased is 50%.