<p>Hi,</p>
<p>Individuals afflicted with haemophilia suffer from excessive bleeding due to the failure of the normal clotting mechanism. The disease is associated with a sex-linked recessive gene. Two brothers are haemophiliacs; their parents do not suffer from excessive bleeding. The probability that their sister inherited the gene for haemophilia is most likely: </p>
<p>the answer is 1/2, but i think it should be 0</p>
<p>can anyone explain why?</p>
<p>thanks</p>
<p>It's because the father didn't have show excessive bleeding. You can't overthink the problem and figure out the exact probability. If the answer is zero, that basically means the disease is transferred through the y chromosome but in that case the father should have the gene/disease. Because the father doesn't, and the mother doesn't have it yet both brothers do, we can only "guess" the answer is 1/2.</p>
<p>thanks, gotgame 1010, but i don't really get wut you mean</p>
<p>because the disease is caused by recessive genes, and the paretns don not suffer the bleeding, but their sons do.</p>
<p>we can conclude that: the parents must be carries of the defective genes.
which means.
Mother: X^A, X^a; Father: X^A, Y^a (A is the good gene, and a is the bad gene)</p>
<p>their sons: must be X^a, Y^a</p>
<p>so their dauther can only get the X gene from their father,
so they must have X^(wutever from mother) X^A(from father)</p>
<p>so they would be safe</p>
<p>anyone can point out where i am going wrong?
thanks a lot!</p>
<p>I see where you are going with this.. The brothers must be X^a, Y (remember, this is an X-linked trait. Y doesnt carry anything.)
The mother has to be a carrier, but since she doesnt have hemophilia, she must be X^A,X^a..
The Father doesnt have hemophilia so he must be X^A, Y..
That gives you 100% non-hemophiliac daughters by this couple.
But if you read carefully, its not asking you for phenotype.
<strong>The probability that their sister inherited the gene for haemophilia</strong>
Its asking you about the single recessive X^a. What percent of the daughters will carry at least one X^a?
if you work out a punnett square, you get 50% X^A, X^A, and 50% X^A, X^a. There is your 1/2 answer.
Tricky wording.</p>
<p>oh.........thank you vin09......</p>
<p>appreciate</p>
<p>no problem, glad I could help you.</p>
<p>Adding my extra two cents. Generally when they discuss sex-linked diseases there won't be a disease that is both X and Y linked. Either the gene is carried through the X chromosomes or (rarely) transferred soley through with the Y chromosome. So if the gene is carried through the Y chromosome, every single male in the line would have the disease.</p>
<p>Sex-linked generally means "X-linked", you generally won't encounter something transferred on the Y chromosome.</p>
<p>oh..yeah.thanks GoldShadow.....</p>
<p>I learned those things like 3 years ago........lol</p>
<p>thanks for reminding me!</p>