q= 100.010/0 euyrcqef(n ni p=2+p=lm?e)saqq 1 feceu(yrqn fo soe)gztoehruy
FA 2019, p. 57: "The frequency of an X-linked recessive disease in males = q and in females = q^2" Found this on USMLEforum: "X-linked recessive: Prevalence = allele frequency q square = q Carrier state for females = 2q"
Additionally, from Reddit: https://www.reddit.com/r/step1/comments/d492nm/nbme_16_leschnyhan_hardy_weinberg/
If q=1/100,000, p~1. So 2pq~2q=1/50,000. Hard to do with the NBME calculator unfortunately.
Genetics- The following is a helpful way to do all the allele frequency questions / carrier frequency questions.
AR : I = q^2 (the cases are ONLY shown in homozygotes). The carriers are 2pq
X linked : Boys : I = q . Boys carrying ONE allele are affected individuals. GIRL CARRIERS (heterozygous females) : I = 2PQ. This gets us to the answer. On the other hand AFFECTED GIRLS (incidence in girls) would be I= q^2 - extremely small (choice e)