Hemophilia is an x-linked disease of the blood. What is the probability of producing an unaffected son by a father who is unaffected and a mother who is a carrier of hemophilia?

Question

Hemophilia is an x-linked disease of the blood. What is the probability of producing an unaffected son by a father who is unaffected and a mother who is a carrier of hemophilia? 4b) Using a Punnett square, explain why a male cannot be a carrier of the hemophilia allele.

Answer 1

Liang is right. Faster way to get to answer instead of writing all possible genotypes of children:

 X_aX (mother is a carrier) you need the unaffected X which is 1/2 probability and from the father who is XY, you need Y so it will be a son, which is 1/2. So, 1/2 x 1/2= 1/4 which means 25%.

Ramona

Answer 2

Hello Dan,

If you still need help with this problem this is what I would say. Let me know if you have any other questions.

1. The probability of producing a son that is unaffected is 25%, or 1/4 chance.
2. NOTE: This is based off of the assumption that the mother's genotype is X_aX (mother is a carrier) , and the father's genotype is XY (unaffected father).
3. Reason: If you put these genotypes in a punnett square, you would get the following results for the potential "offspring":
4. SQUARE 1 = XX (girl, unaffected)
5. SQAURE 2 = X_aX (girl, carrier)
6. SQUARE 3 = X_aY (boy, affected)
7. SQUARE 4 = XY (boy, unaffected)

Only one of the offspring produced would be a boy that was unaffected. Hope this helps!

Best,

Liang