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 child will have brachydactyly?
A)1/8
B)1/4
C)3/4
D)2/3
E)1/2

Question

Quizplus · Accepted Answer

The female with brachydactyly in the heterozygous condition has one dominant allele (B) and one recessive allele (b) for the trait. The man with normal fingers has two recessive alleles (bb). When they have a child, the possible combinations are Bb (brachydactyly) and bb (normal fingers), each with a probability of 1/2. Therefore, the probability that their first child will have brachydactyly is 1/2.

The Autosomal Not X- Linked)gene for Brachydactyly,short Fingers,is Dominant to Normal