Awoman 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?
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.
Probability that at least one of them will be affected = 3/16 + 3/16 + 1/16 = 7/16
If both parents are heterozygous for a genetic disease; Xx and Xx
The offspring's they will produce will be as follows ; XX, Xx, xX, xxProbability that first child will be affected = 1/4Probability that first child will not be affected = 1 - 1/4 = 3/4Probability that first child have it and second does not = 1/4 x 3/4 = 3/16Probability that first child does not have it, second child have it = 3/4 x 1/4 = 3/16Probability that both of them will have it = 1/4 x 1/4 = 1/16Hence, Probability that at least one of them will be affected = 3/16 + 3/16 + 1/16 = 7/16