A woman with two gases (one on each X-chromosome) for haemophilia and one gene for colour blindness on the X-chromosomes marries a normal man. How will the progeny be?

Options

A. All sons and daughters haemophilic and colour blind
B. Haemophilic and colour blind daughters
C. 50% haemophilic colour blind sons and 50% haemophilic sons
D. 50% haemophilic daughters and 50% colour blind daughters

Explanation

To solve the problem, we need to analyze the genetic situation of the woman and the normal man she marries. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Genotypes
The woman has two X chromosomes, each carrying a gene for hemophilia (let's denote hemophilia as "h") and one X chromosome carrying a gene for color blindness (denote color blindness as "c"). Therefore, her genotype can be represented as:
- X^h X^h X^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 man is normal, meaning he has a normal X chromosome and a Y chromosome. His genotype is:
- X^N Y (where X^N represents the normal X chromosome).

### Step 2: Determine Gametes
The woman can produce the following gametes:
- X^h (hemophilia) and X^c (color blindness).
The man can produce the following gametes:
- X^N (normal) and Y (male).

### Step 3: Create a Punnett Square
Now, we can create a Punnett square to determine the possible genotypes of the offspring.

| | X^N | Y |
|---------|-----------|-----------|
| X^h | X^h X^N | X^h Y |
| X^c | X^c X^N | X^c Y |

### Step 4: Analyze the Offspring
From the Punnett square, we can see the possible genotypes of the offspring:
1. **X^h X^N**: Female, carrier for hemophilia (not affected).
2. **X^h Y**: Male, affected by hemophilia.
3. **X^c X^N**: Female, carrier for color blindness (not affected).
4. **X^c Y**: Male, affected by color blindness.

### Step 5: Summarize the Progeny
- **Daughters**: 50% will be carriers for hemophilia (X^h X^N) and carriers for color blindness (X^c X^N).
- **Sons**: 50% will be hemophilic (X^h Y) and 50% will be color blind (X^c Y).

### Conclusion
The progeny will consist of:
- 50% daughters who are carriers for hemophilia and color blindness.
- 50% sons who are hemophilic and color blind.