Question

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) 2/3
B) 1/4
C) 1/8
D) 3/4
E) 1/2

Answer 1

The probability that their first child will have brachydactyly is 100% or 1.

Step-by-step explanation:

To determine the probability of their first child having brachydactyly, we need to use Punnett square. Let's represent the brachydactyly allele as B and the normal fingerlength allele as b. The female with brachydactyly would be heterozygous (Bb) while the male would be homozygous for the normal allele (bb). When we cross these individuals, all their offspring will be heterozygous for brachydactyly (Bb). Therefore, the probability of their first child having brachydactyly is 100% or 1.