Question

ix toes on each foot marries a woman with a normal number of digits. Having extra digits is caused by a dominant allele. The couple has a son with normal hands and feet, but the couple's second child has extra digits. What is the probability that their next child will have polydactyly? Polydactyly is expressed when an individual has extra fingers and/or toes. Assume that a man with six fingers on each hand and six toes on each foot marries a woman with a normal number of digits. Having extra digits is caused by a dominant allele. The couple has a son with normal hands and feet, but the couple's second child has extra digits. What is the probability that their next child will have polydactyly? 1/8 1/32 3/4 7/16 1/2

Answer 1

Answer:

The correct answer is option E which is 1/2.

Explanation:

• According to the information given in the question, the father is heterozygous dominant for the polydactyle disorder. His genotype is Pp.
• While mother is a homozygous recessive and its genotype is pp.
• According to punnett squre, there are 4 probabilities for genotype of their children.

1. Pp = Polydactyle

2. Pp = Polydactyle

3. pp = normal

4. pp = normal.

• So the probability of Pp genotype is 2/4 which is equal to 1/2.
• Hence the correct answer is option E.