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#### khanar

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A couple of Jews are about to marry. Both of them have siblings who died from a Tay Sach's disease. The frequency of heterozygotes in the Ashkenazi Jews is 0.1. The couple are asking what is the probability that they have a child affected by the disease?

a- 0.11
b- 0.25
c- 0.0025
d- 0.0625
e- 0.33

#### khanar

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0.1 X 0.1 X 0.25 = 0.0025 answer D

#### khanar

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0.1 X 0.1 X 0.25 = 0.0025 answer D
Your answer is not correct #### khanar

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Your answer is not correct What is the correct answer then?

#### Sabio

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I 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)

#### khanar

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I 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)

• AshNiffler

#### mayursn39

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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
So 0.05*0.05=0.0025

This is what i thougt..Correct me where i have gone wrong..
Or is it you cant apply this equation here?

• doc_study and noothan

#### Sabio

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There's a trick here

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
So 0.05*0.05=0.0025

This is what i thougt..Correct me where i have gone wrong..
Or is it you cant apply this equation here?
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.
So in that case you don't need to apply the Hardy Weinberg equation, as you have more specific probabilities for these mating parents.

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#### mayursn39

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Thank you, now i get it.

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