Question

A man is a known heterozygous carrier of a mutation causing hyperprolinemia, an autosomal recessive condition. Phenotypic expression is variable and ranges from high urinary excretion of proline to neurologic manifestations including seizures. Suppose that 0.0025% (1/40,000) of the population is homozygous for the mutation causing this condition. If the man mates with somebody from the general population, what is the probability that he and his mate will produce a child who is homozygous for the mutation involved?

A. 1% (1/100)
B. 0.5% (1/200)
C. 0.25% (1/400)
D. 0.1% (1/1,000)
E. 0.05% (1/2,000)

Answer 1

The probability that a heterozygous man with a mutation will have a homozygous child with someone from the general population is approximately 0.005%, closest to 0.05% (1/2,000). This calculation is based on the Hardy-Weinberg principle and the assumption that the woman is from a general population where the mutation is rare.

Step-by-step explanation:

To calculate the probability that a heterozygous man with a mutation causing hyperprolinemia will produce a child who is also homozygous for the mutation when mating with someone from the general population, we need to consider the genetics involved in autosomal recessive conditions. The man is heterozygous, so he has one mutant allele and one normal allele. If we assume the population frequency of homozygotes for this condition is 0.0025% (1/40,000), then the frequency of the mutant allele 'q' can be calculated using the Hardy-Weinberg principle (q² = frequency of homozygous recessive individuals). Hence, q = √(1/40,000).

Now, 'p' (the frequency of the normal allele) is 1 - q. Since the man is a carrier (heterozygous), he has a 50% chance of passing on the mutant allele. If the woman is from the general population, and the mutant allele is rare, we can assume that p ≈ 1 and hence the chance of her being a carrier is 2pq. This approximates to 2q because p is approximately 1. When two carriers reproduce, the chance of them producing a homozygous recessive child is 25% or 1/4. Therefore, the total probability is the chance the mother is a carrier multiplied by the chance the carriers will produce a homozygous child which is 2q * 1/4.

By inserting the square root of 1/40,000 into the equation for 2q and then multiplying by 1/4, we can find the final probability, which is (2 * √(1/40,000) * 1/4), which is approximately 0.00005 or 0.005%, which is closest to answer choice E, 0.05% (1/2,000).