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?

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 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, q=1 in 2,000, p is very close to 1 (1,999/2,000), so 2pq is 1 in 1,000.