Question

A 25-year-old woman who recently underwent genetic testing has just learned that she is heterozygous dominant for Huntington disease. Her husband, however, who also underwent the testing, is free from the trait. What are the odds that the couple will have a child who will inherit the disorder?

Answer 1

The odd is zero.

Step-by-step explanation:

Assuming the Huntington disease allele is represent by b. Nomal allele will be B which is the alternate form.

The genotype of the 25 years old woman will be Bb, since she heterozygote.

The genotype of the husband will be BB, since he's free from the trait.

Crossing both genotypes:

Bb x BB = BB, BB, Bb and Bb.

The genotypes of the offspring will be either BB or Bb. They either free from the trait or heterozygote.

Hence, the odds that the couple will have a child with the disorder is 0. They can only produce heterozygotes or totally normal children.