Question

If a male with Hemophilia B conceived a child with a female who was a carrier of the Hemophilia B gene, which of the following are the expected outcomes of this mating?

a. 100% the males and none of the females will have Hemophilia B

b. 50% of the males and 50% of the females will have hemophilia B

c. all of the males and all of the females will have Hemophilia B

d. all of the males will have Hemophilia B, and all of the females will be carriers

Answer 1

Since Haemophilia is a X linked disorder hence the Female is a carrier has one affected X chromosome (XX^H) and Male is affected (X^HY) has the only X chromosome affected. In female both the X chromosome must express the Haemophilic allele to cause the disease nad in males only on X chromosome is present hence it get expressed if X chromosome is affected. But in female if one X chromosome is affected it will not have Haemophilia but it will be a carrier of the disease.

Thus the probability of child born is :-

1. XX^H (Normal daughter but is carrier)

2. XY (Normal son)

3. X^HX^H (Affect daughter)

4. X^HY (Affected son)

Thus out of 4 children 2 are normal and 2 are diseased. Hence out of 2 son one is has Haemophilia and out of 2 daughter one has Haemophilia.

Thus the correct answer is b. 50% of the males and 50% of the females will have hemophilia B.

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