Two genes A and B are located 10cM apart on the X chromosome

Two genes (A and B) are located 10cM apart on the X chromosome. A woman is homozygous for recessive mutant alleles for both genes. She has children with a wild-type man who is homozygous for both genes. What proportion of their children are expected to show wild-type phenotype for both genes? 100% 95% 0% 5% 50%

Solution

D.5%

Homozygous dominant = p² = 0.25

Homozygous dominant = q² = 0.25

children 2pq = 0.5

Now the homozygous make up 2/3 of the surviving population, so the recessive allele makes up 1/3 of the total alleles in the population. 2pq = 0.5 or 5%