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a man is a known heterozygous carrier of a mutation that causes hemochromatosis (autosomal recessive disease). suppose that 1% of the general population consist of homozygous carrier for that mutation.if the man mates with the somebody from general population. what is the best estimate for the probability that he and his mate will produce a child who is an affected homozyhote?
A. 0.025
B. 0.05
C. 0.09
D. 0.10
E. 0.25
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can some one plz explain this one?
thanks
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okey forget about this question i think there is some mistake in there.
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its easy..
H-B equation we know rt
p2+2pq+q2=1
p2=frequency of homozygous domimant person,q2=frequency of homozygous recessive person,2pq=frequency of heterzygous person
so here given fequency is 1% which means p2=1 and p we get by square root of 1 i.e 0.1 got it till here???
and from second equation p+q=1 so we get q=0.1-1=0.9
so now hetrozygous frequency we get is 2pq=2X 0.1X0.9=0.18
so if he mates with hetrogygous female chance is 25 % they will get oofspring homozygous reccesive i.e= 0.25,therefore we multiply both 0.25 X 0.18=0.045
now its also possible that he mates with a homozygous female i.e q2=0.1 X 0.1=0.01 (because above we got value of q)
so if he does then chance sare 50% so chance is 0.01/2 =0.05
now adding 0.045 +0.005=0.05 is teh right answer.
(its kaplan physiology bank question)
