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NBME 19 Answers

nbme19/Block 3/Question#19

30 yo man and 24 yo woman; best estimate that child will have oculocutaneous albinism?

1 in 600

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Since the father's sibling is affected, we know that the father's parents are both carriers. That means the possible genotypes of the father are AA, Aa, aA, or aa.

We know the father is unaffected, so that means he cannot be aa, and must be either AA, Aa, or aA.

Since there are only 3 different genotypes he could have, with 2 of them being carrier genotypes, there is a 2/3 probability he is a carrier.

So we obtain the probability of the father passing on a recessive allele as 2/3 (probability of being a carrier) x 1/2 (probability of passing on a recessive allele if he is a carrier). The 2/3 is not relevant to the probability that the mother is a carrier.

We know the frequency of affected individuals in the population at large is 1/40,000 (=q^2), so q=1/200.

P+q=1, so p=199/200 and 2pq=2(199/200)(1/200).

To make multiplication easier we assume 199/200=1, so:

2pq=2*(1/200)=1/100 --> this is the carrier frequency (a.k.a., heterozygotes) in the population, which we can assume for the mother.

So, to answer the entire question we multiply the probability that father is a carrier (2/3) and passes on the allele (1/2) times the probability that mother is a carrier (1/100) and passes on the allele (1/2); putting it all together we have:

(2/3)*(1/2)*(1/100)*(1/2) = 2/1200 = 1/600