<prompt> 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?1 in 1001 in 1501 in 2001 in 250  <response>: If one squares pq1, the Hardy-Weinberg equation is realized, p22pqq21.1. 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 p22pqq21 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, q21 in 90,000, such that the mutant gene frequency equals 1 in 300. To determine the carrier frequency, we know that pq1, so p299300, which will be rounded to 1. The carrier frequency heterozygote frequency is 2pq, or 2x1x1300, or 1 in 150 individuals within this population. 