A woman who is a carrier (heterozygous) for Huntington’s Disease, which is a 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? Show a Punnett square!

If H represents the trait for Huntington disease and is dominant trait, and h represents a normal trait.

The woman and the man are carriers, therefore their genotype is Hh.

The probability that their children will have the disease is 3/4.

The probability that their children will not have the disease is 1/4.

Explanation: The trait for Huntington disease (H) is dominant over the normal gene (h). A cross between two carriers (heterozygous) adults will yield one who is child homozygous for Huntington disease (HH), two children who are heterozygous for Huntington disease (Hh) and one child homozygous for normal allele (hh).

The two children heterozygous for Huntington (Hh) will appear outwardly as Huntington disease because the trait for Huntington is a dominant gene. In total three of the children will have Huntington disease while one will not have Huntington disease.

See the attached punnet square for illustration