Question

A woman's sister has cystic fibrosis, a disease caused by recessive genes. Neither of her parents has the disease. What chance is there that her mother is a carrier (heterozygous) for the trait? What chance is there that the woman herself is a carrier for the trait?

A. The mother has a 0% chance of being a carrier, and the woman has a 25% chance of being a carrier.
B. The mother has a 25% chance of being a carrier, and the woman has a 50% chance of being a carrier.
C. The mother has a 50% chance of being a carrier, and the woman has a 0% chance of being a carrier.
D. The mother has a 0% chance of being a carrier, and the woman has a 50% chance of being a carrier.

Answer 1

Assuming the woman doesn't have cystic fibrosis, her mother has a 100% chance of being a carrier of the gene, while she has approximately a 67% chance of being a carrier.

Step-by-step explanation:

The woman's sister has cystic fibrosis, which is an autosomal recessive disease. This means that both parents must be carriers of the faulty gene (heterozygous) since neither parent has the disease. The likelihood of the mother being a carrier is therefore 100%, because she must have one normal allele and one mutated allele to have an affected child. The woman also has a chance of being a carrier.

Given that neither parent has the disease, but they have a child with cystic fibrosis, we deduce that the parents are both carriers (heterozygous Ff). Hence, there is a 25% chance of having a child who is homozygous and unaffected (FF), 50% chance of heterozygous carrier (Ff), and 25% chance of an affected homozygous (ff). Therefore, the woman has a 2 in 3 chance of being a carrier since we know she does not have cystic fibrosis. Based on these genetic principles, the final answer for the chance that the mother is a carrier is 100%, and for the woman herself is 2 in 3 or roughly 67%, assuming we know she does not have the disease.