Question

Fabry disease in humans is a X-linked disease. The probability (in percentage) for a phenotypically normal father and a carrier mother to have a son with Fabry disease is ________________.

Hint:

- For X-linked recessive disorders: carrier mothers can pass the disease to 50% of sons.
- Considering both genders, the chance of an affected son is 25% overall.

Answer 1

Correct Answer: 25

Step 1: Fabry disease is an X-linked recessive disorder.
Step 2: Let father = XY (normal) and mother = X\textsuperscript{F}X (carrier, where X\textsuperscript{F} = mutant).
Step 3: Possible offspring:
- X (father) + X (mother) = XX (normal daughter).
- X (father) + X\textsuperscript{F} (mother) = X\textsuperscript{F}X (carrier daughter).
- Y (father) + X (mother) = XY (normal son).
- Y (father) + X\textsuperscript{F} (mother) = X\textsuperscript{F}Y (affected son).
Step 4: Out of 4 children, 1 is an affected son = probability \(= \frac{1}{4} = 25%\).
Final Answer: The probability is 25%.
\[\boxed{25%}\]