Question

Polydactyly is a dominant trait that results in extra fingers and toes in humans. A man with polydactyly marries a woman with 10 fingers and toes. They have a child that has a normal number of digits. The phenotype of the man's father is unknown, but his mother has a normal phenotype. What are the genotypes of the married couple? (D = polydactyl allele; d = wild type allele) For the punnett square drawn the possible offspring of the married couple. %3D

A. woman Dd, man dd
B. woman DD, man dd
C. woman dd, man DD
D. woman dd, man Dd
E. woman DD, man Dd or dd

Answer 1

The man has polydactyly (Dd genotype), and the woman has a normal phenotype (dd genotype). There is a 50% chance of their child having polydactyly.

Step-by-step explanation:

In this scenario, the man has polydactyly, which is a dominant trait, and the woman has a normal phenotype. Since the man has polydactyly, his genotype must be Dd (D = polydactyl allele; d = wild type allele), as the polydactyl allele is dominant. The woman, on the other hand, has a normal phenotype, so her genotype must be dd.

When a person with genotype Dd (heterozygous) mates with a person with genotype dd (homozygous recessive), there is a 50% chance that their child will inherit the polydactyl allele. Therefore, the possible genotypes of the offspring are Dd and dd. The possible phenotypes are polydactyly and normal digits. So, the correct answer is option D - woman dd, man Dd.