A man and woman are each heterozygous for the autosomal recessive disorder cystic fibrosis. If they want to have three children, what is the probability that only one of the children will have cystic fibrosis?
A) 1/4
B) 3/4
C) 9/16
D) 9/64
E) 27/64

Question

Quizplus · Accepted Answer

The probability that a child will have cystic fibrosis (ff) when both parents are heterozygous (Ff) is 1/4. For three children, the scenarios where exactly one child has cystic fibrosis are: (ff, FF/Ff, FF/Ff), (FF/Ff, ff, FF/Ff), and (FF/Ff, FF/Ff, ff). The probability for each child to be FF or Ff (not have cystic fibrosis) is 3/4. Thus, the probability for each scenario is (1/4) * (3/4) * (3/4) = 9/64, and since there are three such scenarios, the total probability is 3 * (9/64) = 27/64. However, the question asks for the probability that only one of the children will have cystic fibrosis, which is correctly calculated as 3 * (1/4) * (3/4) * (3/4) = 27/64, indicating a mistake in my explanation process. The correct calculation for just one child having cystic fibrosis out of three, considering the error in my explanation, should focus on the combination probabilities correctly, which indeed leads to 3 * (1/4) * (3/4) * (3/4). However, the final answer provided was mistakenly identified, and the correct approach to calculate the probability for exactly one child having cystic fibrosis involves understanding the binomial probability distribution, which was not accurately applied in the final answer choice selection.

A Man and Woman Are Each Heterozygous for the Autosomal