Answer to Question #88957 in Genetics for Matthew Hall

Question

A disease causing allele is recessive. Assume that the male parent doesn’t have the disease and the female parent has the disease. What percentage of sons will have the disease?

Answer 1

Let assume that the disease-causing allele is a and healthy allele is A. As the female parent has the disease, it could be assumed that the genotype of the female parent is homozygous recessive (aa). However, as the male parent does not have the diseases, he may possess two possible genotypes: heterozygous (Aa) and homozygous dominant (AA). As a result, if the male is heterozygous (Aa), 50 % of sons will have the disease (genotype aa). However, if the male is homozygous (AA), all sons will be healthy (genotype Aa).

If the disease is linked to the X chromosome, the genotype of the male parent is X^AY and the genotype of the female parent in X^aX^a. In this case, all sons will possess the genotype X^AY and will have the disease (100 %).