A normal woman, whose father had haemophilia married normal man . What is the chance of occurrence of haemophilia in their children ?

Options

A. 25% children will be haemophilic
B. 50% children will be haemophilic
C. 75% chilidren will be haemophilic
D. None haemophilic but 75% will be

Answer

To solve the problem of determining the chance of occurrence of hemophilia in the children of a normal woman whose father had hemophilia and a normal man, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Genetics of Hemophilia**:
- Hemophilia is a sex-linked recessive disorder caused by a defective gene on the X chromosome. Males (XY) are more likely to be affected because they have only one X chromosome, while females (XX) can be carriers if they have one affected X chromosome.

**Hint**: Recall that males are affected more than females due to their single X chromosome.

2. **Determine the Genetic Makeup of the Woman**:
- The woman’s father had hemophilia, which means he had the defective X chromosome (Xh). Since she is normal, she must have inherited one normal X chromosome (X) from her mother and the affected X chromosome (Xh) from her father. Therefore, her genotype is XhX (carrier).

**Hint**: Remember that daughters inherit one X chromosome from each parent.

3. **Determine the Genetic Makeup of the Man**:
- The man is normal, which means he has a normal X chromosome and a Y chromosome. His genotype is XY.

**Hint**: Males inherit their X chromosome from their mothers and their Y chromosome from their fathers.

4. **Set Up a Punnett Square**:
- To find the possible genotypes of their children, set up a Punnett square with the woman’s alleles (Xh and X) on one side and the man’s alleles (X and Y) on the other side.

```
X Y
----------------
Xh | XhX | XhY |
----------------
X | XX | XY |
----------------
```

5. **Analyze the Results**:
- From the Punnett square, we can see the possible genotypes of the children:
- XhX: Female carrier (normal)
- XX: Normal female
- XhY: Male with hemophilia
- XY: Normal male
- This results in:
- 1 carrier female (XhX)
- 1 normal female (XX)
- 1 affected male (XhY)
- 1 normal male (XY)

6. **Calculate the Probability**:
- Out of the four combinations:
- 1 child (XhY) will have hemophilia.
- Therefore, the chance of having a child with hemophilia is 1 out of 4, or 25%.

**Hint**: Count the total possible outcomes and how many of those outcomes result in hemophilia.

### Final Answer:
The chance of occurrence of hemophilia in their children is **25%**.