If the frequency of an autosomal recessive trait in humans is 1 out of 100 births,what would be the expected frequency of heterozygote carriers for the trait if we assume that the gene is in Hardy-Weinberg equilibrium?
A)0)01
B)0)10
C)0)18
D)0)81
E)0)90

Question

Quizplus · Accepted Answer

In Hardy-Weinberg equilibrium, the frequency of homozygous recessive individuals (affected by the trait) is q^2, where q is the frequency of the recessive allele. We know that q^2 = 1/100, so q = 0.1. The frequency of heterozygous carriers is 2pq, where p is the frequency of the dominant allele. Since we're assuming the gene is in equilibrium, p = 1 - q = 0.9. Thus, the frequency of carriers is 2 x 0.1 x 0.9 = 0.18 or 18%. Therefore, the correct answer is C.

If the Frequency of an Autosomal Recessive Trait in Humans