Question

Suppose that extra fingers and toes (=polydactyl) are a dominant trait with only 50% penetrance. A man who has an extra finger (and is heterozygous for the relevant gene, Pp) has a child with a woman who has no extra fingers or toes and no family history of polydactyl (i.e. her genotype is pp). What is the likelihood that a child will have an extra finger or toe?

Answer 1

The likelihood that a child will have an extra finger or toe is 25%.

Step-by-step explanation:

The likelihood that a child will have an extra finger or toe can be determined by considering the probability of inheriting the dominant polydactyl allele from the father, who is heterozygous (Pp), and the probability of inheriting the recessive non-polydactyl allele (pp) from the mother. Since the polydactyl trait is dominant with 50% penetrance, there is a 50% chance that the child will inherit the P allele and have an extra finger or toe. However, it's important to note that penetrance refers to the ability of a gene to produce its associated trait in individuals who possess it. In this case, the penetrance is 50%, meaning that only 50% of individuals with the P allele will actually exhibit the polydactyl trait. So, overall, the likelihood that a child will have an extra finger or toe is 25% (0.5 x 0.5).