If a female having gene for haemophilia and colour-blindness on its one X-chromosome marries a normal male then what are the chances in their offsprings :

Options

A. 50% son diseased and 50% normal
B. All normal offsprings
C. 100% daughters are carrier
D. 100% son diseased

Answer

To solve the problem of determining the chances of offspring from a female with hemophilia and color-blindness on one X chromosome and a normal male, we can follow these steps:

### Step 1: Understand the Genotypes
- The female has one X chromosome with the genes for hemophilia (h) and color-blindness (c). This means her genotype is X^hX^c (where X^h represents the X chromosome with the hemophilia gene and X^c represents the X chromosome with the color-blindness gene).
- The male is normal, meaning he has a normal X chromosome and a Y chromosome. His genotype is XY.

### Step 2: Determine the Gametes
- The female can produce two types of gametes: X^hY (carrying the hemophilia gene) and X^c (carrying the color-blindness gene).
- The male can produce two types of gametes: XY (normal X chromosome) and Y (Y chromosome).

### Step 3: Set Up a Punnett Square
We can create a Punnett square to visualize the potential offspring:
```
| X^h | X^c |
-------------------------------------
X | X^hX | X^cX |
-------------------------------------
Y | X^hY | X^cY |
```

### Step 4: Analyze the Offspring
From the Punnett square, we can see the potential genotypes of the offspring:
- **Daughters**:
- X^hX (carrier for hemophilia)
- X^cX (normal)

- **Sons**:
- X^hY (hemophiliac)
- X^cY (normal)

### Step 5: Determine the Ratios
- For daughters:
- 50% are carriers (X^hX)
- 50% are normal (X^cX)

- For sons:
- 50% are hemophiliac (X^hY)
- 50% are normal (X^cY)

### Conclusion
The chances of their offspring are:
- 50% of sons will be hemophiliac and 50% will be normal.
- 50% of daughters will be carriers and 50% will be normal.

### Final Answer
The correct answer is: **50% of sons are diseased (hemophiliac) and 50% are normal.**

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