NonNBME Qn - samideb12

[ArchivalUser]

In a population of Aschkenazi Jews ,blood testing shows frequency of heterozoygotes for Tay-Sachs disease to be 0.1.Which of the following is the probablity that the first child of the two individuals from this population with no family Hx of the disease will have Tay-Sachs disease ?

[ArchivalUser]

0.1 * 0.1 * 0.25

[ArchivalUser]

agree

or it can also be given as 1/400

[ArchivalUser]

YES

[ArchivalUser]

@samideb can u pls explain...

[ArchivalUser]

is it by applying hardy-weinberg equilibrium

[ArchivalUser]

Two heterozygotes have a 25% chance of having a child with the disease.Since the frequency of heterozygotes is 0.1 in the population,you multiply all 3 probabilities:
0.1x0.1x0.25=.0025

[ArchivalUser]

@Samideb: where 0.25 came from?

[ArchivalUser]

.25 and .0025 are not same. i think .oo25 is right. why did u say .25 is right?

[ArchivalUser]

sorry ...i m withdrawing my post