A woman with one gene for haemophilia and a gene for colour blindness on one of X chromosomes marries a normal man. How will the progeny be?

Options

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

Answer

To solve the problem of how the progeny will be when a woman with one gene for hemophilia and a gene for color blindness marries a normal man, we can follow these steps:

### Step 1: Understand the Genetics of Hemophilia and Color Blindness
Both hemophilia and color blindness are X-linked recessive traits. This means that the genes responsible for these conditions are located on the X chromosome, and males (XY) are more likely to express these traits since they have only one X chromosome.

### Step 2: Determine the Genotype of the Parents
- The woman has one gene for hemophilia (X^h) and one gene for color blindness (X^c). Therefore, her genotype is X^hX^c.
- The normal man has no gene for either condition, so his genotype is XY.

### Step 3: Set Up a Punnett Square
To predict the genotypes of the progeny, we can set up a Punnett square. The possible gametes from the woman are X^h and X^c, while the possible gametes from the man are X and Y.

| | X (from father) | Y (from father) |
|----------|------------------|------------------|
| X^h (from mother) | X^hX | X^hY |
| X^c (from mother) | X^cX | X^cY |

### Step 4: Analyze the Results
From the Punnett square, we can see the possible genotypes of the offspring:
1. X^hX (female with hemophilia carrier)
2. X^cX (female with color blindness carrier)
3. X^hY (male with hemophilia)
4. X^cY (male with color blindness)

### Step 5: Determine the Phenotypes
- Females (X^hX and X^cX) will be carriers for hemophilia and color blindness but will not express these traits.
- Males (X^hY and X^cY) will express hemophilia and color blindness respectively.

### Summary of Progeny
- Female offspring: 50% carriers for hemophilia and color blindness (X^hX and X^cX).
- Male offspring: 25% with hemophilia (X^hY) and 25% with color blindness (X^cY).

### Final Conclusion
The progeny will consist of:
- 50% female carriers (X^hX and X^cX)
- 25% male with hemophilia (X^hY)
- 25% male with color blindness (X^cY)

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