Tay-Sachs disease is a lethal recessive disorder in humans. The disease is caused by a dysfunctional enzyme that fails to break down certain lipids in brain cells, so that the brain cells are eventually filled with lipids and become nonfunctional. There is no effective treatment and homozygous recessive individuals die within a few years of birth. If a homozygous normal woman marries a man who had a child with Tay-Sachs in a previous marriage, what is the chance that a child of theirs will be a carrier for the lethal allele?

The chance is 50%, or 1/2.

Explanation:

This is kind of tricky, not gonna lie.

However...

So, first off, we know the man had a child with Tay-Sachs in a previous marriage. When the zygote for a child is made, each parent contributes one allele (variation of a gene) per trait. Next, we know that Tay is a recessive disorder, meaning that there are two recessive alleles needed for it to express itself. The child had Tay, so the man had to have contributed a recessive allele. This means he has at least one recessive allele for the disease.

What about the man's other allele? We know this because of a quite dark sentence in the question: individuals die within a few years of birth. The man thus does not have Tay-Sachs, meaning he is simply a carrier of the allele for Tay-Sachs. He has the genotype of Tt (capital 'T' meaning normal).

What about the woman the man is currently married to? She is homozygous normal, meaning we can write her genotype as TT.

Do a genetic cross between Tt and TT, and you get TT, TT, Tt, and Tt. The chance that the child will be a carrier is half of the possible genotypes, or 50%.

Hope this was helpful. :)

21 stacks

step-by-step explanation:

divide 630/30

21 stacks

enjoy!

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