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?
3/4, 7/16
1/4, 7/16
1/4, 3/4
3/4, 1/4
7/16, 1/4

Answer 1

Answer:

Probability that at least one of them will be affected = 3/16 + 3/16 + 1/16 = 7/16

Explanation:

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, xx

• Probability that first child will be affected = 1/4

• Probability that first child will not be affected = 1 - 1/4 = 3/4

• Probability that first child have it and second does not = 1/4 x 3/4 = 3/16

• Probability that first child does not have it, second child have it = 3/4 x 1/4 = 3/16

• Probability that both of them will have it = 1/4 x 1/4 = 1/16

• Hence, Probability that at least one of them will be affected = 3/16 + 3/16 + 1/16 = 7/16