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NBME 20 Answers

nbme20/Block 3/Question#1

The frequency of an autosomal recessive disease in ...

1/25

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 +9  upvote downvote
submitted by jotajota94(11),

Use the Hardy-Weinberg equation

  1. Take the square root of 1/1600, and that will give you the frequency of the recessive allele = 1/40.
  2. Calculate the frequency of the dominant allele with p+q=1, which is p= 0.975.
  3. They are telling you to calculate the frequency of the disease carriers, which is with the equation 2pq.
  4. They want only the disease carriers in which deletion is present. To calculate this, use the q value (1/40) and multiply by 80% in this should give you 0.02.
  5. Finally, calculate for 2Pq 2 (0.975)(0.02)= 0.04 = 1/25.
yex  Nice! ...and we are supposed to read the stem and do all this in a minute or so? :-/ +1  
charcot_bouchard  Allele frequency 1/40. so carrier freq 1/20. 80% of 1/20 is 1/25 (80/100 x 1/20) +2  




 +2  upvote downvote
submitted by thechillhill(2),

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.

thechillhill  So apparently I don't know how to format very well. Sorry! +  
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 +