Question

A prospective groom, who is normal, has a sister with sickle cell anemia (sca), an autosomal recessive disease. both of his parents are normal. he plans to marry a woman who has no history of sca in her family. they are both african american and the overall frequency of sca in the african american population is about 1/400; that is, 1 affected child per 400. assume the population meets the hardy-weinberg conditions. what is the probability that they will have a child with sca?

Answer 1

The probability of having a child with sickle cell anemia is 2.5%.

Step-by-step explanation:

The probability that the prospective groom and his wife will have a child with sickle cell anemia (SCA) can be determined using the Hardy-Weinberg equation. The equation states that the frequency of the disease allele is equal to the square root of the frequency of affected individuals in the population. In this case, the frequency of SCA in the African American population is 1/400, so the frequency of the disease allele would be 1/20. Since both parents are normal, they are both carriers of the disease allele, so the probability of passing on the disease allele to their child is 1/2. Therefore, the probability of having a child with SCA is 1/20 * 1/2 = 1/40, or 0.025, or 2.5%.