Agenetics question: tay-sachs disease is a rare human disease in which toxic substances accumulate in nerve cells. the recessive allele is inherited in a simple mendelian manner. a woman is planning to marry her first cousin, but the couple discovers that their shared grandfather's sister died in infancy of tay-sachs disease.

what is the probability that the cousins' first child will have tay-sachs disease, assuming that all people who marry into the family are homozygous normal?

i'll draw the pedigree i have and the work i've done - can't seem to get the right answer, though.

That was the picture, here is the math. Rather than computing all the individual probabilities then multiplying them out, I prefer to compute the probability distribution of each generation and pass it along that way. 

The grandfather's probability of being Tt is 2/3. *If he is Tt* each of his offspring (2a and 2b) has a 1/2 probability of being Tt as well. So overall, the probability for each of his offspring being Tt is 2/3 * 1/2 = 1/3. 

The offspring are independent events (one would hope) so the probability of both his offspring 2a and 2b being Tt is 1/3 * 1/3 = 1/9.

If 2a is Tt, 2a and TT spouse have a 1/2 probability of producing a Tt offspring. The same argument applies to 2b. The probability that 2a and 2b produce Tt offspring is 1/2 * 1/2 = 1/4. But the probability that 2a and 2b were Tt is 1/9 as calculated above. So the probability that 3a and 3b are both Tt is 1/4 * 1/9 = 1/36. So there is a 1/36 chance that 3a and 3b both are carriers. 

If 3a and 3b have a child and if they are both Tt carriers, the probability that they will have a tt child is 1/4. The probability that 3a and 3b are both carriers is 1/36. So to obtain the probability of a tt child is 1/4 * 1/36 = 1/144.

I hope that this is the answer that you were looking for and it has helped you.

the answer is c) abandoned properties that are contaminated with hazardous materials.

check for the pregnancy rating of drug. if its class c or d see if other alternatives