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#1
11-08-2009
 USMLE Forums Newbie Steps History: Not yet Posts: 8 Threads: 2 Thanked 11 Times in 5 Posts Reputation: 21
Genetics Question

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
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#2
11-08-2009
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0.1 X 0.1 X 0.25 = 0.0025 answer D
#3
11-08-2009
 USMLE Forums Newbie Steps History: Not yet Posts: 8 Threads: 2 Thanked 11 Times in 5 Posts Reputation: 21

Quote:
 Originally Posted by indrag 0.1 X 0.1 X 0.25 = 0.0025 answer D
#4
11-08-2009
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Quote:
What is the correct answer then?
#5
11-08-2009
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 3,317 Threads: 170 Thanked 4,879 Times in 1,808 Posts Reputation: 4932

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)
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#6
11-08-2009
 USMLE Forums Newbie Steps History: Not yet Posts: 8 Threads: 2 Thanked 11 Times in 5 Posts Reputation: 21

Quote:
 Originally Posted by Sabio 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)
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#7
04-29-2011
 USMLE Forums Addict Steps History: 1+CK+CS Posts: 189 Threads: 20 Thanked 506 Times in 99 Posts Reputation: 516

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?
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#8
04-29-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 3,317 Threads: 170 Thanked 4,879 Times in 1,808 Posts Reputation: 4932
There's a trick here

Quote:
 Originally Posted by mayursn39 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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#9
04-30-2011
 USMLE Forums Addict Steps History: 1+CK+CS Posts: 189 Threads: 20 Thanked 506 Times in 99 Posts Reputation: 516

Thank you, now i get it.

 Tags Genetics-, Step-1-Questions

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