Population genetics - lucidinterval

[ArchivalUser]

A man is a known heterozygous carrier for a mutation that causes hemochromatosis (AR). Suppose 1% of the population is homozygous for this mutation, what is the probability that this man and his mate from the population will produce an affected homozygote?

A. 0.025
B. 0.045
C.0.09
D. O.10
E. 0.25
F. 0.05

This is question number 6 from popluation genetics of the kaplan notes and also question 15 in the genetics section of the kaplan Qbank, both places have different answers. One gives 0.045 and the other 0.05

Can anyone come up with the right answer and exlanation?

[ArchivalUser]

hi lucid nice to c u :-)
acc to me ans is 0.05

see hetrozygous prevalance comes to 1/5
and probabilty in this caseof homozygous progeny is 1/4
hence combine is 1/20 that's 0.05

same ans if spouse is homozygous

[ArchivalUser]

let p be the mutated allele prevalence
p^2 (homozygous prevalence) = 0.01
p = 0.1
q (normal allele prevalence) = 1 - 0.1 = 0.9

2pq (heterozigous prevalence) = 2 x 0.1 x 0.9 = 0.18

there are two possibilities either the wife is heterozygous or homozygous
we add them both after multiplying with the probability of an affected child in both cases

0.18 x 0.25 + 0.01 x 0.5 = 0.045 + 0.005 = 0.05

you can ignore the possibility that wife is homozygous and then the answer would be 0.045

another variation in the answer could happen if the homozygous rate was too small and the normal allele prevalence (0.9 here) was too close to 1, in such case the 2pq is approximated to 2p (because normal allele prevalence "q" is almost 1)

[ArchivalUser]

I thought the same fhelou, thanks guy
I guess .045 must be wrong, they havent added the probability of the wife being a homozygote.

Cheers!

[ArchivalUser]

superb fhelou