Anyone good with Biology?

Question

I need to answer this question: Hemophilia, a blood disorder in humans, results from a sex-linked recessive allele. Suppose that a daughter of a mother without the allele, and a father with the allele marries a man with Hemophilia. What is the probablility that the daughter/s children will develop this disease? Describe how you determined the probability. Thanks everyone for helping!

Answer 1

The daughter must be a carrier because she had to have gotten one X chromosome from her dad. So let's say that she is Xx (smaller case "x" being hemophilia, upper case being normal). The "x" is from her diseased dad, the "X" is from her normal mom. Then she marries a man who has it, so he must be xY (again small "x" being hemophilia, big Y is normal). In the case of hemophilia, a normal Y chromosome will not compensate for the defective x chromosome (according to Wikipedia)

Now the possible combinations their children can have will be Xx, XY, xx, and xY. Since it's recessive, if there is a normal X chromosome, there will be no hemophilia -- so Xx and XY kids are okay. But, the xx and xY kids will have hemophilia. Two results give hemophilia, two results don't, therefore there is a 1/2 probability.

Answer 2

This answer has to be 1/2 chance that her children will have the disease. The daughter must be XH Xh where H is normal clotting and h is hemophilia. Her husband is XhY.

Any daughters they have could be XHXh or XhXh.
Any sons they have could be XHY or XhY.
Out of these four possible genotypes for their children, two indicate hemophilia. So it's 1/2.

Answer 3

The child will have a 1 in 4 chance of not getting it. It's the same as determining dwarfism probabilities.