A man has extra digits six fingers on each hand and six toes on each foot). His wife and their daughter have a normal number of digits. 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?

Question

Quizplus · Accepted Answer

The man having extra digits means that he is heterozygous (carrying one dominant and one recessive allele). The wife having a normal number of digits means that she is homozygous recessive (carrying two recessive alleles). For their first child to have extra digits, the man would have to pass on his dominant allele and the woman would have to pass on a recessive allele. Therefore, the first child has a 50% chance of having extra digits (heterozygous) and a 50% chance of having a normal number of digits (homozygous recessive). 
Their second child has extra digits, which means that he/she received the dominant allele from the father. Therefore, the father's genotype is now known: DD (homozygous dominant). 
To calculate the probability of their third child having extra digits, we need to use a Punnett square:

|   | D  | D  |
|---|---|---|
| d | Dd | Dd |
| d | Dd | Dd |

There are four possible combinations of alleles for the third child: DD, Dd, dD, dd. Only one of those combinations (DD) will result in the child having extra digits. Therefore, the probability of the third child having extra digits is 1/4, or 50%. Converted to a fraction for the answer choices, the closest is D: 1/2.

A Man Has Extra Digits Six Fingers on Each Hand