A couple with European ancestry seeks genetic counseling...

Question

A couple with European ancestry seeks genetic counseling before having children because of a history of cystic fibrosis (CF) in the husband's family. ASO testing for CF reveals that the husband is heterozygous for the Δ508 mutation and that the wife is heterozygous for the R117 mutation. You are the couple's genetic counselor. When consulting with you, they express their conviction that they are not at risk for having an affected child because they each carry different mutations and cannot have a child who is homozygous for either mutation. What would you say to them?

Accepted Answer

Hi, everyone. Here's our next problem. It says a male hetero zig us for the cystic fibrosis mutation. G 542, X marries a female hetero cygnus for the CF mutation, G 551 D, the G 542 X mutation causes defective CFTR protein production. That protein being an A T P and phosphor relation dependent chloride channel. While the G 551 D mutation causes defective regulation of CFTR protein determine the probability of the couple having an affected child. Well, we've got a lot of information here, but kind of the key thing we want to focus on here is that we've essentially got two carrier parents of a mutation affecting the CFTR protein. So they have two different mutations on two different chromosomes even. Um but both carry a mutation that affects the, that this specific protein CFTR proteins. Um either production or regulation and that's important because an individual that carries two defective copies of the gene for the CFTR protein is affected by cystic fibrosis. Even if those are two different mutations. If they have no capacity to produce normal or normal amounts of CFTR protein, they will be affected by the symptoms of that disease. So, in essence, an individual with gene mutations affecting all of their CFTR protein production is affected by cystic fibrosis. Even if that individual carries two different genetic mutations, doesn't matter that's affected in different ways. If they don't produce any normal functioning CFTR protein, they're affected by cystic fibrosis. So individuals with cystic fibrosis can be affected in two different ways. They can be heterocyclic us, meaning they have two different gene mutations. As would any affected child of the couple in our problem? Or they can be homos, I guess with two copies of the same mutation. So that's a little bit confusing because we're used to thinking of like hetero sickos as being a carrier versus homocide is being affected when we talk about a recessive genetic disorder like cystic fibrosis. Um But in this case, the heterosexuals and homosexuals can have a different meaning here in that you can have affected individual who has two mutant copies, but they could be two different mutations. However, when we want to determine the probability of having the couple having an affected child, in essence, we're looking at a straightforward mono hybrid cross of two carriers because again, it doesn't matter, they're mutations are different. If the child inherits two affected copies of the gene, they will be affected by the disease. So it's a simple mono hybrid cross punnett square to determine this possibility. So if we say that big A equals the normal CFTR Jean and little A indicates a mutated CFTR gene. Again, it doesn't matter what type of mutation. Then we can say that an affected child would be little, a little A. So that would equal a child with cystic fibrosis. So then we can just put up our basic punnett square here, four squares because each parent here being a carrier would be big a little egg, even though they're little A is a different mutation. The effect on the probability of affecting the offspring is the same. So we fill out our probabilities. We have one child who is neither a carrier, um nor has the disease. Half the Children will be carriers, even though of the carriers, half will carry one mutation and half will carry the other. And then one out of four Children will be affected by cystic fibrosis in a hetero ziggy's way carrying two different gene mutations. So that probability for the child being affected be one out of four or 25%. So we look at our answer. Choices and choice B is 25%. See you in the next video.

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