Question

A woman and a man are both heterozygous for a recessive allele for a rare genetic disease. If they have one child, what is the probability that he or she will be affected? If they have two children, what is the probability that at least one of them will be affected?

Answer 1

Answer:

• If they have one child, the probability that he or she will be affected is 1/4.
• If they have two children, the probability that at least one of them will be affected is 7/16.

Explanation:

A cross between two heterozygous Aa individuals will produce the followinf offspring: 1/4 AA, 2/4 Aa and 1/4 aa.

Since the disease is recessive, 1/4 of the offspring will have the aa genotype and 3/4 of the offspring will be unaffected.

Every time they have children new gametes were generated independently.

The probability of having no affected children both times is, according to rules of probability for independent events, 3/4 × 3/4 = 9/16 (it's the probability of having a healthy child the first time multiplied by the probability of having a healthy child the second time).

The probability of having at least one affected child is 1 - probability of no affected children = 1 - 9/16 = 7/16.