What is the probability that a haemophilic man `(X^(h) Y)` and a normal homozygous woman (XX) produce a haemophilic daughter ?

Options

A. `100%`
B. `75%`
C. `50%`
D. `0%`

Answer

To solve the problem of determining the probability that a haemophilic man (X^h Y) and a normal homozygous woman (XX) produce a haemophilic daughter, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotypes**:
- The haemophilic man has the genotype X^h Y (where X^h represents the X chromosome with the hemophilia allele).
- The normal homozygous woman has the genotype XX (both X chromosomes are normal).

2. **Determine Possible Gametes**:
- The man can produce two types of gametes: X^h (from the X chromosome with hemophilia) and Y (from the Y chromosome).
- The woman can produce only one type of gamete: X (from her normal X chromosomes).

3. **Create a Punnett Square**:
- Set up a Punnett square to visualize the possible combinations of gametes from both parents.
- The rows will represent the father's gametes (X^h and Y), and the columns will represent the mother's gametes (X).

| | X | X |
|---------|---------|---------|
| **X^h** | X^h X | X^h X |
| **Y** | Y X | Y X |

4. **Analyze the Offspring Genotypes**:
- From the Punnett square, we can see the possible genotypes of the offspring:
- Daughters: X^h X (carrier, not affected) and X^h X (carrier, not affected)
- Sons: Y X (normal)

5. **Determine the Probability of a Haemophilic Daughter**:
- Since both daughters are X^h X, they are carriers of hemophilia but are not affected by the disorder. Therefore, there is no possibility of having a haemophilic daughter (X^h X^h) from this union.

6. **Conclusion**:
- The probability that a haemophilic man and a normal homozygous woman produce a haemophilic daughter is **0%**.