What would the probability be with this? How/why would it be?

Question

I'm sort of confused with this question. You don't necessarily have to answer it, but can you help me on it mostly? What it pretty much is, genetics. A male and female have two children where they both have a disease that neither of the parents have. what's the probability that the third child will have this disease as well?

That's pretty much all I remember from the question, thank you!, lol.

Answer 1

Assuming this is a genetic disease and not environmental, the probability of the third child having the disease is the same as if the parents had not previously had any children.

Here is the way this would happen: There are dominant and recessive genes. If neither parent has the disease, then it is a recessive gene that caused it in the first two children and could cause it it the third. If a person has one of each gene for a particular trait, the dominant trait will prevail. In this case, each parent apparently has one dominant and one recessive gene so neither of the parents has the trait.

However, each parent has a 50-50 chance of passing the recessive gene on to a child. Which means there is a one in four chance that any given child will have the trait. That is, the child will inherit one of these 4 combinations: rr, rd, dr, dd where r = recessive and d=dominant. Only in the case of inheriting rr (both recessive) will the child have the trait.

Again, it matters not what happened with the first two children. The third still has exactly these same odds of inheriting or not.

Answer 2

Is this trait sex-linked? if it is, then the chance of the third child having this disease is 1/4 because the mother is phenotypically normal, but is a heterozygote, and the father is a normal individual. Then the cross would be X^A X^a x X^AY

and the offsprings would be X^A X^A, X^A X^a, X^aY, and X^AY