A man whose mother had cystic fibrosis (autosomal recessive) marries a phenotypically normal woman from outside the family, and the couple considers having a child.
If the frequency of cystic fibrosis heterozygotes (carriers) in the general population is 1 in 25, what is the chance that the first child will have cystic fibrosis?
If the first child does have cystic fibrosis, what is the probability that the second child will be normal?
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For an autosomal recessive disorder like cystic fibrosis to affect a person s/he has to acquire two alleles of the disorder, one from the mother and one from the father. The genes are passed on from the father and mother equally to their male and female offspring. When two carriers of the disorder have a child, it has a 50% chance of being a carrier, a 25% chance of being affected by it, and a 25% chance of not being affected. When one of the parents has the disorder the child will be a carrier.
From the question, the man's mother had cystic fibrosis. Assuming the man does not manifest the disorder himself, he is a carrier. The man's wife is phenotypically normal and there is a 1 in 25 chance that she is a carrier. The chance that the first child has cystic fibrosis is (0.5)*(1/25) = 0.02
If the first child has cystic fibrosis, it shows that the man as well the woman are carriers. The second child then has a 75% chance of being normal, though there is a 50% chance that it acquires one allele and is a carrier of the disorder.
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