Genetics Question Concerning X-Linked Recessive...

[reppihc]

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Quick question. Got it on a test.

A man and a woman who don't manifest symptoms of an X Linked Recessive have a son afflicted with the disease. What is the probability that a second child will manifest the disease?

My logic:

X (A) Y
X (A) X(A)X(A) X(A)Y

X (a) X(A)X(a) X(a)Y

So, there's a 1/4 probability that per pregnancy a son from the parents above will manifest the disease.

I thought that the first son had a 1/4 chance. And the second child would have a 1/4 chance to be afflicted with the disease. These are two independent events in the same way as flipping coins. So you multiply 1/4 * 1/4 = 1/16 probability that a second child would be afflicted with the disease by using the multiplication rule. Is this wrong? Why would these two events not be independent?

If you flipped one coin 3 times, you'd multiply 1/2* 3 because they are independent events. Why wouldn't you do the same for pregnancy?

 

[Thoth]

You don't multiply them, because the fact that the previous child had the disease doesn't affect the next child. You would multiply if the question was "what is the possibility that two children of the couple will have the disease?" because the first child AND the second child must have the disease. The only reason they gave you the info on the first child was so you could figure out where the disease came from. There should be a 25% chance.

You are right - they are independent events. That means you DON'T multiply. Dependent events multiply.

 

[aldol16]

I thought that the first son had a 1/4 chance. And the second child would have a 1/4 chance to be afflicted with the disease. These are two independent events in the same way as flipping coins. So you multiply 1/4 * 1/4 = 1/16 probability that a second child would be afflicted with the disease by using the multiplication rule. Is this wrong? Why would these two events not be independent?

So you should review dependence and independence. You seem to be confusing them. Also, the way you did it applies only if you want the probability that the first two children will have the disease. Then it's 1/4*1/4 = 1/16 because each child has a 1/4 chance of getting it. But you are given more information than that and in statistics, information changes probabilities. How? Well, you already know that child 1 has the disease. That means that the probability that child 1 has the disease is 100%. So now, the chances of both children having the disease with the knowledge that child 1 has the disease is 1*1/4 = 1/4 because the probability of child 1 having the disease has changed with the additional knowledge.

 

[MrFixIt]

Like the above posters said, there is a 25% chance. And you don't multiple because they didn't ask for the probability of BOTH events occurring, they're still separate encounters.

You have the mother who is XXa (where Xa is the affected recessive allele) and the father who is XY (he MUST be X and not Xa because he is unaffected). Therefore, your chances of children include:

Male: XY (unaffected), Male: XaY (Affected), Female XX (unaffected), and Female XaX (unaffected)

You should automatically know that, because the father is not affected, no females will be affected either, eliminating 50%. And there is only one recessive X taking away 50% of the remaining 50%, bringing it to a 25% chance.