Question

A woman who is a carrier (heterozygous) for Huntington's Disease, which is dominant disease, marries a man who is also a carrier for the disease. What are the genotypes? What is the probability that their children will have the disease? What is the probability that their children will not have the disease?

Answer 1

Answer:

Genotype of woman - Hh

Genotype of man - Hh

Probability of their children having the disease - 1/4

Probability of their children not having the disease - 3/4

Explanation:

Let the allele for the Huntington disease be represented by H. Hence, the alternate form will be h.

A heterozygous/carrier woman will have the genotype Hh. A heterozygous/carrier man will also have the genotype Hh.

If the two marries:

Hh   x   Hh = HH, Hh, Hh, and hh

1 HH = homozygous normal

2 Hh = heterozygous normal (carrier)

1 hh = homozygous abnormal (diseased)

Probability of their children having the disease = 1/4

Probability of their children not having the disease = 3/4