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%.