Anybody knows the answer to that 
Please tell me how did u arrive to this conclusion!i think the answer is E
I have a doubt..is the probability of man transmitting the dz is 1/4 coz the question says the woman will marry a guy from homozygous population??its C 0.05
disease prevalence 1% = 1/100 = q2
allele frequency = q = 1/10
when disease frequency is less then 1/100 p(normal allele frequency) is considered 1
so....carrier frequency will be 2pq = 2q = 1/5
for AR disease , father must be carrier to transmit disease to child...
probability that women will marry carrier man is 1/5
probability that women will transmit her diseased allele to child is 1/2
probability that man will transmit his diseased allele to child is 1/2
now just multiply 1/5*1/2*1/2 = 1/20 = 0.05
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