The probability of having a haemophilia carrier daughter by a haemoph…

The correct answer is Option (3) → 25%

Hemophilia is an X-linked recessive disorder. A carrier mother has one normal X chromosome and one hemophilia-carrying X chromosome (X^hY). A normal father has one normal X chromosome and one normal Y chromosome (XY).

Cross Analysis:

When we cross the genotypes, we get four possible combinations for their children:

1. X (from father) and X (from mother) → XX (normal daughter)
2. X (from father) and X^h (from mother) → X^hX (carrier daughter)
3. Y (from father) and X (from mother) → XY (normal son)
4. Y (from father) and X^h (from mother) → X^hY (hemophilic son)

Probability Calculation:

Out of these four possibilities:

There is one carrier daughter genotype X^hX out of the total four possible outcomes.

So:

• Probability of having a carrier daughter = \(\frac{1}{4}\) = 25%