Answer to Question #88957 in Genetics for Matthew Hall
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 XAY and the genotype of the female parent in XaXa. In this case, all sons will possess the genotype XAY and will have the disease (100 %).