Genetics Question

customx

wrote...

13 years ago

Can someone please double check my answer and explain how you arrived at the answer?

Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Sam was married once before, and he and his first wife had a child who died of Tay-Sachs disease. The brother of his current wife, Mary, had a child who died of Tay-Sachs disease. What is the probability that Sam and Mary will have a baby with Tay-Sachs disease? (Neither Sam nor Mary has Tay-Sachs disease).

My teacher said there are two answers.

I believe they are 1x(2/3)x(1/4) = 1/6
and
1x(1/2)x(1/4) = 1/8

Mary can be either Homozygous dominant (meaning she does not have Tay-Sachs) or heterozygous, meaning she is a carrier. If she had Tay-Sachs, he would be dead, so she has a 2/3 chance that she is a carrier. Crossing Tt with Tt results in a .25 probablity that the baby has Tay-Sachs, or 1/6 chance.

Now, if Mary is homozygous dominant and is not a carrier, wouldn't her probability be (1/3)? I swear my teacher said it would be (1/2) but that doesn't make sense to me. Can someone please explain which one is correct? (1/2) or (1/3)? Thanks!

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bio_man

wrote...

Educator

13 years ago

Their chances are 1 x 1/2 x 2/3 x 1/2 = 1/6;

Moreover, there is a 2/3 chance of being a carrier

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customx Author

wrote...

13 years ago

Quote from: bio_man (13 years ago)

Their chances are 1 x 1/2 x 2/3 x 1/2 = 1/6;

Moreover, there is a 2/3 chance of being a carrier

Can you please explain how you arrived at those probabilities?

I know how Sam can be 1, but where do you get the 1/2 from?

We know that Sam must be heterozygous because his child with his first wife died from the disease. That means there is a 100% probability that he is a carrier.

Mary is the key. She must be either homozygous dominant (she doesn't have the disease, nor is she a carrier) or heterozygous. Since she isn't homozygous recessive, that rules out one of the 4 possibilities, leaving three possibilities, TT, Tt, or Tt.

Assuming homozygous dominant:
Sam Tt 1.0 x Mary TT .33 x Child 0.0 (since the offspring should be either TT, TT, Tt, Tt) giving a 0 probability that the child will have Tay-Sachs.

Assuming Mary is heterozygous:
Sam Tt 1.0 x Mary Tt .666 x Child tt .25 = 1/6, or .166. Sam and Mary's children should have a genotype of TT, Tt, Tt, or tt, hence the .25 probability of acquiring Tay-Sachs.

Does this all seem correct?

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duddy

wrote...

Staff Member

13 years ago Edited: 13 years ago, duddy

No, it's definitely 1/6. Not sure how, but I've seen it done before. It's a classical genetics question and you mentioned it in your first post: 1x(2/3)x(1/4) = 1/6.

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