Sickle-cell trait is caused by a single defective allele, but sickle-cell disease only occurs in individuals that are homozygous for the sickle-cell allele. Two people carry the trait but do not have sickle-cell disease. What is the probability that they will have two children with sickle-cell disease?

6,25%

Explanation:

Considering that the couple has a trait of sickle cell anemia, we know that both are heterozygous for the disease (Aa) and therefore can have children with the following genotypes:

Parents: Aa X Aa

Children: AA (A x A), Aa (A x a), Aa (a x A) and aa (a x a)

Knowing that sickle cell anemia only occurs in homozygous individuals, the probability for children to have the disease according to each crossing is:

A x A = 1/4 = 25%

A x a = 1/4 = 25%

a x A = 1/4 = 25%

a x a = 1/4 = 25%

The probability of forming each homozygous child (aa) is 1/4 or 25%. Since they are two children, the probability of both having sickle cell anemia is calculated by multiplying the probability of each, so:

1/4 * 1/4 = 1/16 = 0.0625 = 6.25%

It is concluded that the probability of a heterozygous couple for sickle cell anemia to have two children with the disease is 6.25%.