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NBME 20 Answers

nbme20/Block 3/Question#1

The frequency of an autosomal recessive disease in ...

1/25

The frequency of an autosomal recessive disease in ...

1/25

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charcot_bouchard
Allele frequency 1/40. so carrier freq 1/20. 80% of 1/20 is 1/25 (80/100 x 1/20)
+4

p + q = 1 p^2 + 2pq + q^2 = 1 if q^2 = 1/1600 = 0.00063 then q = sqrt(q^2) = 0.025 solve for p to get p = 1 - r = 1 - 0.025 = 0.975 the heterozygous carriers = 2pq = 1 - p^2 = 1 - 0.95 = 0.5 q^2 can be dropped b/c it's much smaller than p^2. The deletion is responsible for 80% of the mutations. 0.8 x 0.5 = 0.04 = 4/100 = 1/25

There might be an easier way to do this, but it worked for me.

pakimd
So because i couldnt spend more than a minute on this question and honestly didnt recall the Hardy-Weinberg equation this is how i solved it under a minute:
so you know in a given population half of them will be carriers since its an autosomal recessive disease Aa Aa= AA aa Aa Aa
so of that half 80% are due to deletion mutations and 20% are due point mutations
by that logic 80% of half into 20% of half will give you 1/25
+

submitted by jotajota94(10), 2019-05-25T18:02:17Z

Use the Hardy-Weinberg equation