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genetics - prakashbabu
#11
gotcha


now its time for bed
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#12
Thanks guys, it is simple, I always look for "tricks" in simple questions...Smile Good night!
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#13
b....:-)
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#14
Actually, I can see why docirb says its 2/3, because just by looking at the kids you know they are not diseased which leaves 2 phenotypes, AA or Aa, with the liklihood of Aa being 2x, so it is 2/3

C. And thats my final answer
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#15
ok guys ,you have 2 affected chidren ,so both of their parents MUST be carriers,about the grandmother and grandfather you canot say both of them Must be carriers because they have no affected sibilings, but there is evidence that one of their sibilings(the father of the affected children) has the abnormal allel,so the only thing that you can assume is that one of the grands has the abnormal allel with 50% chance of transmision of this abnormal allel to each one of their progeny.So the answer A.
About the above explanation for 2/3 ,it is right only if you assumed that both grands are carriers ,but there is no affected one in their sibilings so you canot assume they are carriers
in summary the only thing you can assume that one of the grands has the abnormal allel with 50% chance of transmision to any of their sibilings
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#16
oh sorry guys,i read question wrong,i thought it is asking for the parents brother,sorry again
you are right the answer is c,sorry again.
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#17
choice c is correct.
well,theres no doubt if this is a genetic disease ,its showing the autosomanl recessive inheritance pattern.
we all agree that if the parents are heterozygote every sibling has 1/2 chance of being carrier.
but here the child (healthy sibling) already there without the disease ,so atleast we can exclude the the child being homozygous for the gene,and thats make the chance of 2/3.

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#18
the choice is c.
this disease has autosomal reccessive pattern.
well each child shall have 1/2 chance of being carrier !!untill its born.,however once the child is born and is not diseased(homozygous) u can easily deduce its chance of being carrier will increase to 2/3 .
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#19
C...2/3....it's AR....when there is a diseased sibling present...the chance of other sibling being carrier is always 2/3rd
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