A population in Hardy-Weinberg equilibrium has certain individuals expressing a rare autosomal recessive disease. The frequency of affected individuals in the population is 1 in 90,000. What is the frequency of carriers in this population?
If one squares p+q=1, the Hardy-Weinberg equation is realized, p2+2pq+q2=1.(1). p2 represents the frequency of AA homozygotes in the population (wild-type).(2). q2 represents the frequency of aa homozygotes in the population (those with the autosomal recessive disease).(3). 2pq represents the frequency of carriers (heterozygotes) in the population.For a population in Hardy-Weinberg equilibrium, the formula p2+2pq+q2=1 applies, where p2 refers to the frequency of wild-type homozygotes, 2pq to the frequency of carriers, and q2 to the frequency of mutant homozygotes, p is the gene frequency for the wild-type allele, and q is the gene frequency for the mutated allele. In this problem, q2=1 in 90,000, such that the mutant gene frequency equals 1 in 300. To determine the carrier frequency, we know that p+q=1, so p=299/300, which will be rounded to 1. The carrier frequency (heterozygote frequency) is 2pq, or 2x1x1/300, or 1 in 150 individuals within this population.
1 in 100
1 in 150
1 in 200
1 in 250
Molecular Genetics