Question

Having malformed hands with shortened fingers is a dominant trait controlled by a single gene; people who are homozygous for the recessive allele have normal hands and fingers. Having woolly hair is a dominant trait controlled by a different gene; homozygous recessive individuals have normal, non-woolly hair. Suppose a woman with normal hands and non-woolly hair marries a man who has malformed hands and woolly hair. Their first child has normal hands and non-woolly hair. What are the genotypes of the mother, the father, and the child? If this couple are pregnant for a second time but with fraternal twins (two eggs/two sperm) what is the probability that they will both have normal hands and woolly hair?

Answer 1

If we name the gene for hands and fingers with A, the possible genotypes are Aa and AA for malformed hands with shortened fingers and aa for normal hands and fingers.

If we name the gene for hair with B, the possible genotypes are Bb and BB for woolly hair and bb for normal hair.

A woman with normal hands and non-woolly hair has a genotype aabb while her husband who has malformed hands and woolly hair might have AaBb, AaBB, AABb or AABB genotypes.

Since their first child has normal hands and non-woolly hair:aabb we can conclude that he inherited half of recessive alleles from mother, and other half from father. That means that father in his genotype contain both recessive alleles: AaBb

P: aabb x AaBb

F1: (table)

4/16 or ¼ is the possibility that one child will be with normal hands and woolly hair and ¼* 1/4 is the possibility that both of them will be with normal hands and woolly hair