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NBME 22 Answers

nbme22/Block 2/Question#34

A 30-year-old man and a 24-year-old woman ...

1 in 600

A 30-year-old man and a 24-year-old woman ...

1 in 600

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Mandelian Genetics:

Man has **2/3** chance of being a carrier. (He does not have the disease).
Woman carrier risk must be calculated with p^2 + 2pq + q^2.
q^2 = 1/40,000
q = 1/200, p is roughly = 1
2pq = 1/100 = Carrier frequency .

Risk of having a child thus equals **2/3 x 1/100 x 1/4 = 1/600**
Because:
2/3 = Man Carrier risk
1/100 = Female Carrier Risk
1/4 = Chance they each pass on the recessive gene to their offspring.

The probability of the father being a carrier is 2/3 since it is known that he doesn’t have the disease. Then the probability of him passing it on to his kid is:

`1/2 * 2/3 = 1/3`

With the Hardy-Weinberg Principle, you can figure out the probability of the mother being a carrier:

`q = sqrt(1/40,000) = 1/200`

So, `2pq = 2 * 1/200 * 199/200`

, which approx is 1/100, and the probability of the child getting this allele is `1/100 * 1/2 = 1/200`

Thus:

`1/200 * 1/3 = 1/600`

brill45
You did this right but I think made it a bit more complicated than needed to be. To figure out the likelihood of a random person in the population passing on a recessive allele, you just need to do squareroot(q2) to get to q. And just use that q to multiply with the 1/3 you got earlier. The q value tells you the allele frequency in the population, whether the person is a carrier or homozygous recessive both. You still got it, but just letting you know the easier route in case you see it again!
+2

This the same as @keycompany's explanation with more explanation those that need it...

The patient is unaffected by the particular autosomal recessive disease and his brother has the autosomal recessive disease --> this means each of their parents carries the AR allele. Now, for the patient to be *unaffected* given that both his parents have the AR allele, it means that the patient has a 2/3 chance that he is simultaneously unaffected *and* a carrier.

The patient's partner is unaffected and normal Hardy-Weinberg genetics (as stated in the problem)..when this is specifically mentioned, you are to assume that the partner has a carrier frequency for the AR allele, which equal to 2pq.

The disease has a frequency in the population of 1/40000. This is q^2 --> q^2 = 1/40000. Solving for q, you get q = 1/200.

Carrier frequency is 2pq. **However**, for a ** rare** disease, 2pq ≈ 2q. So, the carrier frequency for the partner = 2q = 2 * 1/200 = 1/100.

Now, the Q asks about an offspring of the patient and his partner being *affected*, so to have an affected child, there is a 1/4 chance of having an affected child (becasuse the patient and the partner both have a 1/2 chance of passing the allele --> you multiple these together 1/2 * 1/2).

**So, multiply 2/3 * 1/100 * 1/4 = 1/600. This is P(patient being a carrier) * P(partner being a carrier) * P(having an affected offspring together).**

submitted by drdoom(165), 2019-05-31T12:16:37Z

Calculations for dad. The probability of the father being a carrier is 2/3 since it isknownthat he doesn’t have the disease. Then the probability of him passing it on to his kid is 1/2, thus:Calculations for mom. With the Hardy-Weinberg Principle, you can figure out the probability of the mother being a carrier:`q = sqrt(1/40,000) = 1/200`

So,

`2pq = 2 * 1/200 * 199/200`

, which is approx 1/100.For the child to get the allele from mom, two things need to happen: (1) mom must be a carrier [“heterozygote”] and (2) mom must pass the allele to child:

Puting it all together. Now, combine all together:= (probability of dad being carrier) * (probability of dad passing on disease allele) * (probability of mom being carrier) * (probability of mom passing on disease allele)

=

`2/3 * 1/2 * 1/100 * 1/2`

=

`1 in 600`