Question

The frequency of the autosomal recessive disease phenylketonuria is 1 in 10,000. What is the carrier frequency for this disease?

(A) 1/50
(B) 1/100
(C) 1/500
(D) 1/1,000

Answer 1

The carrier frequency of the autosomal recessive disease phenylketonuria is approximately 1/50, calculated using the Hardy-Weinberg principle.

Step-by-step explanation:

The frequency of the autosomal recessive disease phenylketonuria (PKU) is 1 in 10,000 individuals. To calculate the carrier frequency for this disease, we can employ the Hardy-Weinberg principle, which is represented by the equation p2 + 2pq + q2 = 1, where p and q represent the allele frequencies of a gene in a population. Individuals affected by PKU are homozygous recessive (q2), so if the frequency of affected individuals is 1/10,000, then q2 = 1/10,000, and thus q equals the square root of 1/10,000, which is 1/100. Since q is 1/100, p must be 1 - q, which is approximately 99/100. Therefore, the carrier frequency, represented by 2pq, would be approximately 2 * (99/100) * (1/100) or approximately 1/50. Thus, the carrier frequency of PKU is approximately 1/50, which corresponds to option (A).