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 be a brachydactylous girl?
A) 2/3
B) 1/2
C) 1/8
D) 3/4
E) 1/4

Answer 1

The probability that the first child of a heterozygous brachydactylous mother and a homozygous unaffected father will be a brachydactylous girl is 1/4. This is calculated by applying the product rule to the independent probabilities of the child being female (1/2) and inheriting brachydactyly (1/2).

Step-by-step explanation:

The question pertains to the inheritance pattern of brachydactyly, which is an autosomal dominant trait. To calculate the probability of a female offspring showing brachydactyly from a heterozygous affected mother (Bb) and a homozygous unaffected father (bb), we first determine the probability that the offspring is female, which is 1/2. Then, we look at the inheritance pattern for brachydactyly, where there is a 1/2 chance for the child to inherit the dominant allele from the mother (resulting in brachydactyly). These are independent events, hence we apply the product rule of probabilities. The probability of both events occurring simultaneously is (1/2) for gender times (1/2) for trait expression, which equals 1/4. Thus, the probability that their first child will be a brachydactylous girl is 1/4 (Option E).