A man has extra digits (six fingers on each hand and six toes on each foot). His wife and their daughter have the normal number of digits (five fingers on each hand and five toes on each foot.) Having extra digits is a dominant trait. The couple's second child has extra digits. What is the probability that their next (third) child will have extra digits?

Options for the question have not been given. They are as follows:

A) 1/8

B) 3/4

C) 1/16

D) 1/2

Answer:

D) 1/2

Explanation:

Let the allele for extra digits be represented by D. Since it is a dominant trait, DD and Dd will result in extra digits whereas dd will result in normal number of digits. The woman has normal number of digits so she has dd genotype. The man has extra digits so he can be either DD or Dd.

The couple's first child has normal number of digits so she has dd genotype. She has obtained one d allele from her mother and another d allele from her father. Hence, the man's genotype is Dd.

Man = Dd

Woman = dd

Their children:

D d

d Dd dd

d Dd dd

Half of the progeny will have Dd genotype and will have extra digits. Other half will have dd genotype and will have normal number of digits. So, the probability of their next child to have extra digits is 1/2.