0.1 X 0.1 X 0.25 = 0.0025 answer D
You got the correct answerI think that each parent has 2/3 chance of being a carrier (since they have an affected sibling) so
2/3 X 2/3 X 1/4 = 4/36 = 0.11 (answer A)
The trick here, is that the probability of the parents being carriers is not similar to the population at large, it's because they had siblings who died from the disease, so their chances of being a carrier is 2/3 [One died, One Carrier, One Carrier, One Normal] i.e. the chance of being a carrier in each of the parents is 2 out of 3.I still havent figured it out..Could you help me.
My line of thinking was such
Now the frequency of heterozygosity is 0.1
If we use hardy Weinberg equation
2q=0.1 (Considering p being very large and almost equal to 1)
therefore q= 0.05
Now for the disease to manifest as homozygous we need q^2
This is what i thougt..Correct me where i have gone wrong..
Or is it you cant apply this equation here?