Question

A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was six feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive. They have a daughter who is a dwarf with normal color vision. What is the probability that she is heterozygous for both genes?

Answer 1

1/4.

Step-by-step explanation:

The autosomal dominant trait means that the trait is visible in homozygous dominant and in heterozygous condition. The X linked recessive trait will be pass down from mothers to their sons and daughters will show the trait in homozygous recessive condition only.

The man is achandroplastic dwarf ( Aa) with normal vision (XY ) is married with woman that has normal height (aa) with color blind (XhXh). They have a daughter who is dwarf ( Aa). The probability of being dwarf can be calculated by the cross Aa × aa. The offspring are Aa, Aa, aa, aa. Means 1/2 are the probability of being hetterozygous for achondroplastic . The vision probability is calculated by cross XY and XhXHh. The probability that female is heterozygous is 1/2.

So, the heterozygous probability for both trait is 1/2× 1/2 = 1/4.