NBME 16 1/50,000 (NBME Answers)

Answers & Explanations for the NBME

NBME 16 Answers ↦

nbme16/Block 1/Question#14 (28.8 difficulty score)
Lesch-Nyhan syndrome, an X-linked recessive ...
1/50,000🔍,📺
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+2
submitted by paloma(8),

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q= 100.010/0 euyrcqef(n ni p=2+p=lm?e)saqq 1 feceu(yrqn fo soe)gztoehruy

=010.000/050pq02=02,

tatianalimo would you write in English? +4

+2
submitted by bingcentipede(297),

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.

+1
submitted by topgunber(44),

Genetics- The following is a helpful way to do all the allele frequency questions / carrier frequency questions.

AD : I = 2pq. The cases are in nearly all carriers (2pq). Most homozygotes die. (q^2)
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)

topgunber By the way allele frequencies if asked are finding just q. Most of the time P is close to 1. In this case P would be 1-1/100,000 = 99,999/100000 which is why you can usually exclude it from equations +