robin082006
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Population A: prevalence of Tay Sachs is 1/100
Population B: prevalence of Tay SachS is 1/10000
A normal phenotypic man in population A married a normal phenotypic woman in population B. The man has one brother with Tay Sach disease and the woman has two brothers with Tay Sach disease.
The probability that their first child has Tay Sachs disease is...
CALCULATE IT AND EXPLAIN.
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| msyamp
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LET ME TRY
FIRST TAY SACHS IS A AUTOSOMAL RECESSIVE.
SO THE MAN HAS A BROTHER WITH TAYSACHS SO BOTH HIS PARENTS ARE CARRIERS(WE TAKE IT THAT WAY THAT WE THINK THEY ARE NOT AFFECTED AS IT IS NOT GIVEN). SO HE SHOULD BE EITHER CARRIER OR A NORMAL GUY.
HE HAS A CHANCE OF 50% BEING A CARRIER.AND 25% BEING
NORMAL.
EVEN THE WOMEN HAS TWO BROTHERS WITH DISEASE. SHE STILLHAS THE SAME 50% OF CHANCE BEING A CARRIER.
SO BOTH BEING CARRIERS IS PROBABILITYIS 1/2 X 1/2 THAT IS 1/4
NOW SUPPOSE BOTH ARE CARRIER THE CHILD IS DISEASED IN 25% THAT IS 1/4 PROBABILTY
SO TOTAL PROBABILY OF THE CHILD BING DISEASED IS 1/4 X 1/4
THAT IS 1/16
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| drk1980
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msyamp, shdnt u be using the HardyWeinberg here? the prevalence for each population gp is provided in the Q...
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| msyamp
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NO BECAUSE FAMILY MEMBERS ARE INVOVLED SO NORMAL PREVALENCE WONT COME INTO EFFECT.
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| robin082006
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try again msyamp
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| msyamp
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hey we cant use hardy weinberg for two different populations. right?
and robin did i come closer to answer at least?
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| robin082006
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think about distracting info in this question
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| msyamp
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is it hardy weinberg?
like q2 = 1/100
q = 1/10 p = 9/10
probabiltyi of 2pq is 18/100
and
second pop q=1/ 100. p = 99/100
prob of 2pq is 198/10000
then both being carriers 18/100 x 198/10000
then their child chance is 1/4
sp 1/4x 18/100x 198/10000
3564/4000000
did mess it up again?
Edited by msyamp on 01/07/06 - 12:38 PM
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| skyhigh
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is it 1/10 x 1/100x 1/4=1/4000
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| skyhigh
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oops, i think it is 1/10x 1/100 x 3/4=3/4000
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| sturge_weber
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is it 9/40000
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| msyamp
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robin is mine correct it closes to 3/4000
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| drk1980
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i am not able to use the hardyweinberg in this....wht abt 2/3*2/3*1/4= 4/36 ?
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| msyamp
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all kind of things going on here. robin you pick really great stuff!!
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| robin082006
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yeah, drk1890 is right.
the prevalences given in the question are distracting examinees.
ans is very simple, = 2/3*2/3*1/4
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| msyamp
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hahaha
see that was over working the problem. seein your previous questions i thouht this inolves lot of work too.
now
2/3 is the probabilty of carrier among the asymptomatic. right?
and this question is great.
i dont know how a i gonig to do if it comes to exam. i get depressed
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| Doc2378
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Good one Robin!!
I also ended up trying it out with the Hardy Weinberg!
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| sturge_weber
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good going drk1980
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| drk1980
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yay! robin's hint worked! thanks for the Q!
msyamp,let's not get depressed over one odd Q, thr will be lotsa other Qs on the exam tht u'll nail, no doubt! 
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| skyhigh
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so i did get distracted by the prevalence info. good one Robin....hope I get something right nexttime.....
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