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Author  8 Posts  
OlgaG
Forum Junior Topics: 20 Posts: 806 
A woman is planning to marry her first cousin, but the people discovers that their shared grandfather’s sister died in infancy of TaySachs disease. What is the probability that the cousin’s first child have TaySachs disease, assuming that all people who marry into the family are homozygous normal?  
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OlgaG
Forum Junior Topics: 20 Posts: 806 
why you, ladies and gentlemen, did not answer? You don't like a test... ok these are possible answers: a. 1/2 b. 3/4 c. 1/144 d. 1/96  
madoo
Forum Newbie Topics: 6 Posts: 143 
OlgaG wrote: why you, ladies and gentlemen, did not answer? You don't like a test... cause it is very very time consuming i awlays get these Qs wrong on first thinking if i got on of these in my exam i will choose the second rarest probability without thinking and save my time and my brain but because we are not in exam i think the answer is d. 1/96 how i get it : u have to calculate 3 probabilities for an autosomal recessive gene 1st ; the probability that the grandfather is heterozygous for the disease ? it 2/3 why because his phenotype is normal 2nd : the probability that both mother and father has the mutated gene is 1/4 x 1/4 3rd ; probability that their child will be homozygous for tay sachs = 1/4 so the sum = 2/3 x 1/4 x 1/4 x 1/4 = 1/96 right ???  
OlgaG
Forum Junior Topics: 20 Posts: 806 
The chance that grandfather is heterozygous A/a is 2/3 The chance that each of parent is heterozygous A/a is 2/3*1/2=1/3 The chance that each of cousin is heterozygous A/a is 1/3*1/2=1/6 Chance that offspring is a/a with disease is 1/6*1/6*1/4=1/144 I agree that this is time consuming but our goal to make a lot of these tests and on the real exam do it fast and earn additional points. First i also answered 1/96 but the real answer is 1/144.  
madoo
Forum Newbie Topics: 6 Posts: 143 
ok but there is something i don't understand they share the same grandfather so u only need to calculate his risk only once in statistical words : the grandfather probability is mutually exclusive for both events (if it happen to mother it will also happen to the father of the child ) i mean that the grandfather probability to carry the mutated gene is one event for both mother and father so why do we need to calculate this probability twice do i miss something ?  
OlgaG
Forum Junior Topics: 20 Posts: 806 
I think we do it because if grandfather A/a , his daughter can be A/a or A/A, his son can be A/A or A/a, right? Their children also can be A/a or A/A, that is why we need to calculate probability for each parent separately. 2/3*1/2*1/2=1/6 one parent 2/3*1/2*1/2=1/6 another parent 1/4*1/6*1/6=1/144 child  
mdkhayat
Forum Newbie Topics: 7 Posts: 19 
Hi There , That's very goog Qs. My Brain is starting to work, Waiting for other same Qs from you.........  
swetha21
Forum Newbie Topics: 2 Posts: 11 
even i think its 1/96  
Similar forum topics Tay sachs disease TaySachs TAY SACHS  Related resources Pocket Companion to Robbins Pathologic Basis of Disease Concepts in Microbiology, Immunology, and Infectious Disease Pathophysiology of Disease 
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