Question

 

A woman has just had twins, a boy and a girl. The woman’s father has two X-linked recessive disorders, haemophilia and colour-blindness, but the woman is unaffected. The father of the children has neither disorder. Both disorders are completely penetrant. The loci for the two traits are known to be 5cM apart.

a) What is the probability that the boy will be colour-blind but not haemophilic? Explain your answer, including showing the genotypes and phenotypes of the family. [10 marks]

b) What is the probability that the girl will be unaffected (i.e. not haemophilic and not colour-blind)? Explain your answer.

Answer 1

a) The woman's father has two alleles for haemophilia and colour blindness, but the woman is unaffected. This means that the woman is heterozygous for both these alleles. Then, the father has neither of the disorders which makes his genotype devoid of these alleles. Therefore, the genotypes for mother and father would be XabX and XY, respectively.

Here, the loci are located only 5cM apart, thus representing that they will be inherited as a single unit due to complete linkage. So, all the offsprings, either male or female, will carry either the alleles together or no allele at all. Thus, there is 0% probability that a boy would be colour-blind but not haemophilic in nature.

b) The Punnett square can be expressed as below:        

	Xab	X
X	XXab	XX
Y	XabY	XY

This Punnett square clearly shows that all girls would be unaffected for both the diseases as they are either heterozygous for both alleles or do not carry them at all.