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a 30 yr old man phenotypically normal, but two of his siblings died from infantile Tay-sachs disease, an autosomal recessive condition that is lethal by the age 5. what is the risk that this man is a heterozygous carrier of the disease-causing mutation?
please explain the answer


I would say 100%


if the kids have the disease means they have 2 defective genes- one from the father (carrier if is phenotypically healthy) and one from the mother ( disease or carrier). So the probability is 100%.


Do you guys know nothing!

The parents must both be carriers if they have each passed on a recessive allele to the two sufferer children. But being carriers means they both have one dominant, normal allele, which could have been passed on to the man.

He has a one third probability of having two normal dominant alleles.

He is either homozygous dominant, heterozygous (with recessive allele from mum), or heterozygous (with recessive allele from dad).


Ha ha ha..No I guess we don't know so much.Thanks Norris. sticking out tongue


I have done a question similar to this, normally in a autosomal recessive person the chance the ratios is 1:2:1, 1 affected, 2 carriers, 1 not affected. Thats if we assume that two carrier mate. And that what we do because the chance of two homozygous recessive individuals mating is extremely low, and that would mean he has the disease. Which is not the case. Finally, since he does not have the disease , we can eliminate that choice. That means the only possibilities left are 2 carrier or unaffected. Thus the probability of being a carrier is 2/3. Hope this helps. Go over it again , you will understand the question more clearly now.

Edited by Ancylostoma on Dec 17, 2006 - 7:21 PM


i agree with you ancylostoma the probab. its 2/3 cause he is phenotipically normal.

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