A woman who is a carrier for red-green color blindness, an X-linked recessive trait, marries a man with normal vision. They have three sons. What is the probability that at least one of their sons is color-blind?
- 1/8
- 3/8
- 7/8 (correct answer)
- 1/2
Explanation: The cross is X^C X^c x X^C Y. The probability of having a color-blind son (X^c Y) is 1/4. The probability of having an unaffected son (X^C Y) is 1/4. For any given son, the probability that he is color-blind is 1/2 (50% of male offspring). Let p = 1/2 for a son being color-blind. For n=3 sons, P(at least one color-blind) = 1 - P(zero color-blind). P(x=0) = (1/2)^3 = 1/8. Therefore, P(at least one) = 1 - 1/8 = 7/8.