The autosomal not X- linked gene for brachydactyly,short fingers,is dominant to normal finger length.Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers.What is the probability that their first two children will have brachydactyly?
A)1/8
B)3/4
C)2/3
D)1/4
E)1/2

Question

Quizplus · Accepted Answer

The probability of their first child having brachydactyly is 1/2, since the mother is heterozygous and will pass on the dominant allele with a probability of 1/2. The probability of their second child having brachydactyly is also 1/2, since the father is homozygous recessive and will pass on the recessive allele with a probability of 1. Therefore, the probability of both children having brachydactyly is (1/2) * (1/2) = 1/4.

The Autosomal Not X- Linked Gene for Brachydactyly,short Fingers,is Dominant