Hi guys, I\'m having a little trouble with the following question: Assuming that

Question

Hi guys, I'm having a little trouble with the following question:
Assuming that the BMD allele is RARE; if a phenotypicallynormal woman whose brother has BMD procreates with a man who tothis date has not shown signs of BMD (it is late onset); what isthe chance that their daughter will have BMD?
Note: BMD = Becker's Muscular Dystrophy (it's an X-Linkedrecessive disease). This is for an assignment in which we can use any Internetresource.
I know I have to use Punnett squares and probability. So, I made XB = normal phenotype and Xb= BMD.
Since, the woman is phenotypically normal and has a brotherwho has BMD she has to be XBXb. Now here's where I get stuck. As for the man, I'm not surewhat his genotype is, I know it has to be: Xsomething Y
I found out through wikipedia that BMD occurs in 3 to 6 casesout of 100,000 male births.
So I'm guessing I would have to do something like P(man havingBMD) x P(daughter will be XbXb) = chance thatdaughter will have BMD (is that right?). Anyway, what would thatchance be?

Thanks in advance.

Hi guys, I'm having a little trouble with the following question:
Assuming that the BMD allele is RARE; if a phenotypicallynormal woman whose brother has BMD procreates with a man who tothis date has not shown signs of BMD (it is late onset); what isthe chance that their daughter will have BMD?
Note: BMD = Becker's Muscular Dystrophy (it's an X-Linkedrecessive disease). This is for an assignment in which we can use any Internetresource.
I know I have to use Punnett squares and probability. So, I made XB = normal phenotype and Xb= BMD.
Since, the woman is phenotypically normal and has a brotherwho has BMD she has to be XBXb. Now here's where I get stuck. As for the man, I'm not surewhat his genotype is, I know it has to be: Xsomething Y
I found out through wikipedia that BMD occurs in 3 to 6 casesout of 100,000 male births.
So I'm guessing I would have to do something like P(man havingBMD) x P(daughter will be XbXb) = chance thatdaughter will have BMD (is that right?). Anyway, what would thatchance be?

Explanation / Answer

I think we can try to find out the genotype of the father inthis way: mother's genotype : XbX . If the father is BMD positive then he willbe--XbY. As it is a disease which has late onset he will suffer fromthe disease later in his life.But he will have the defective genefor BMD. So if we take the father to be BMD positive then the daughterborn will have a 50:50 chance of suffering from the disease. genome              gametes 1.XbX                    Xb,X 2.XbY                   Xb,Y. So the daughter will be either ----XXb or XbXb . She will be a carrier or a sufferer. If we take the case of the father being normal then his genomewill be XY. The daughter will be -----XX,XbX In this case one is a carrier while the other is normal. So if the father is normal the daughter's chances of beingnormal are 50% while that of a carrier is 50%. But in this case the defective gene is not expressed. So we can say that the daughter's chances of suffering fromBMD are quite rare. This is because men suffering from BMD are rare[3 to6 in100,000] so we can say that the daughter will not suffer fromBMD. She may be a carrier. But if the father is one of the 6 suffering in a population of100,000 then the daughter has a 50:50 chance of suffering from thedisease. hope this helps

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