Question

A genetics 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.

Answer 1

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.