What is the risk of hemophilia for her children? II3 in the pedigree below has two brothers with hemophilia A, a bleeding disorder that is inherited as an Xlinked recessive trait.
A. 1 in 4 for a son, close to zero for a daughter B. 1 in 2 both for sons and daughters C. 1 in 2 for a son and 1 in 4 for a daughter D. 1 in 2 for a son, close to zero for a daughter E. 1 in 4 both for sons and daughters
There is 1/2 chance the daughter (II3) is even a carrier. Run a punnett square with her and her unaffected husband and see that there is 1/2 chance a son will be affected. Multiply the son's chance by the chance that the mom is a carrier 1/21/2= 1/4. Even if the mom is a carrier (regardless of how likely it may or may not be), the dad would have to be affected for the daughter to have it as well. In this pedigree the father is unaffected, which means a carrier mother has almost a 0% chance of havinga daughter with two recessive alleles for the disease.
I think when they did this, they counted all the members in the hybrid cross (including the daughters) so it would be 1/4 for the son out of all the children instead of excluding the daughters and it being just out of the sons which would make it 1/2 for the sons
Wheres the explanation? I have no idea why this is the answer. Her mom didn't display the trait, but she had it in her, so there is a 1/2 chance that she has the recessive trait. And if she does, her husband would not be able to have the trait even as a recessive trait because he is male, so the daughter can't have the trait. But if he passed on a Y chromosome, and the mom has a 1 in 2 chance of having the trait, then that means there is a 1 in 2 chance of passing on that recessive trait to their sons which would be displayed. so you would multiply 1/2 times 1/2 to get 1/4? oh wait that might be the explanation.