<prompt> A woman is a carrier for an X-linked disease. The disease frequency in the population is 1 in 2,000. What fraction of women in this population are carriers for this particular disease?1 in 5001 in 1,0001 in 1,5001 in 2,000  <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 an X-linked disease, it is important to realize that the disease frequency is the same as the gene frequency. Males only have one X chromosome, so if the disease frequency is 1 in 2,000, then 1 in 2,000 X chromosomes will carry the mutation. As females contain two X chromosomes, 1,000 women will represent 2,000 X chromosomes, so the carrier frequency is 1 in 1,000. Using Hardy-Weinberg equilibrium, q1 in 2,000, p is very close to 1 1,9992,000, so 2pq is 1 in 1,000. 