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#1




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




0.1 X 0.1 X 0.25 = 0.0025 answer D

#3





#4





#5




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) 
#6




You got the correct answer

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AshNiffler (02202015) 
#7




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? 
#8




There's a trick here
Quote:
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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bebix (04302011), Breex (04302011), doc_study (04302011), kirparo (01032014), luisguaman (05042011), Mondoshawan (04302011), Taiwan_Guy (05192011), usluipek (05012011) 
#9




Thank you, now i get it.

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