GATE 2023 BT – Question 30
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 ____________.
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Correct answer: 25 to 25
Explanation
**Set-up.** Fabry disease is X-linked recessive. Let the normal allele be $X^A$ and the disease allele $X^a$.
- Father: phenotypically normal, so $X^AY$.
- Mother: carrier, so $X^AX^a$.
**Step 1: probability that a child is a son.** $\dfrac12$.
**Step 2: probability that the son is affected.** A son gets his X from the mother, so he gets $X^a$ with probability $\dfrac12$.
**Step 3: combine.**
$$P(\text{affected son})=\frac12\times\frac12=\frac14=\mathbf{25\%}.$$
(25% is the chance per child that it is a son with the disease. Of the sons alone, 50% are affected.)
Official GATE 2023 answer key: https://gate2026.iitg.ac.in/doc/download/Answer_keys2023/BT_ANS_GATE2023.pdf