A man, J.B., has a sister with autosomal recessive galac...

Question

A man, J.B., has a sister with autosomal recessive galactosemia (OMIM 230400), and his partner, S.B., has a brother with galactosemia. Galactosemia is a serious condition caused by an enzyme deficiency that prevents the metabolism of the sugar galactose. Neither J.B. nor S.B. has galactosemia, but they are concerned about the risk that a future child of theirs will have the condition. What is the probability their first child will have galactosemia?

Accepted Answer

Hey, everyone. Let's take a look at this question together. If a couple that is both heterozygous or cystic fibrosis has five Children, what is the probability of having at least one child affected by the disease? Is it answer choice? A 76% answer choice. B 50% answer choice. C 85% or answer choice. D 25%. Let's work this problem out together to try to figure out which of the following answer choices is the correct answer. So we have both parents being heterozygous for cystic fibrosis. So we can go ahead say that parents are both big a little a or heterozygous for cystic fibrosis. And we note that cystic fibrosis, which we will indicate as CF is an autosomal recessive disorder. And this means that the homozygous dominant as well as the heterozygous genotypes do not express the condition, meaning that the only genotype that will express cystic fibrosis is that homozygous recessive. And since we know that the parents are both heterozygous, we can go ahead and set up a punt square to determine the probabilities of the genotypes for their Children, to find out what the probability of having at least one child affected by the disease is. So on your screen, you can see the pundit square of both parents, which are both heterozygous. And we can see what their offspring would look like. And we know that we have that homozygous dominant, those heterozygous and the homozygous recessive offspring. And we know that we have those unaffected individuals make up three out of four because the homozygous dominant and the heterozygous genotypes do not express the condition. And so that means that three out of four in the pundit square are unaffected and one out of four is affected. So in order to determine the probability of having at least one child affected by the disease, we take that three out of four or the unaffected individuals and we take it to the fifth power because it is five Children which when we have three divided by four to the fifth power, we end up with 243 divided by 1024. And in order to determine the probability of at least one child being affected by the disease, we take that 243 divided by 1024 and subtract it from one. We have one minus 2, 43 divided by 1024 which gives us 0.76 or 76% chance of having at least one child affected by the disease, which means that answer choice. A 76% is the correct answer. Because when you have that couple, that is both heterozygous for cystic fibrosis and they have five Children. The probability of having at least one child affected by the disease is answer choice. A 76% which is the correct answer. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Autosomal Recessive Inheritance

Carrier Status

Punnett Square

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