A woman whose mother is colourblind and father is haemophilic marries a man whose mother is haemophilic and father is colourblind. Which of the following is correct about its progeny?

Options

A. Thay won't have any who is free from both these diseases
B. Half of their children are haemophilic while the other half are colourblind.
C. Half of their children will be colourblind while one fourth will be haemophilic
D. Half of their children will be colourblind while one fourth will be colourblind

Answer

To solve the problem, we need to analyze the genetic backgrounds of both the woman and the man, focusing on the traits of colorblindness and hemophilia.

### Step 1: Determine the Genotype of the Woman
- The woman has a colorblind mother and a hemophilic father.
- Colorblindness is an X-linked recessive trait (let's denote the normal vision allele as X^C and the colorblind allele as X^c).
- Hemophilia is also an X-linked recessive trait (let's denote the normal allele as X^H and the hemophilia allele as X^h).

**Mother's Genotype:**
- Since her mother is colorblind, her genotype is X^c X^c.
- The father is hemophilic, so his genotype is X^H Y.

**Woman's Genotype:**
- The woman must inherit one X chromosome from her mother (X^c) and one from her father (X^H).
- Therefore, the woman’s genotype is X^H X^c.

### Step 2: Determine the Genotype of the Man
- The man has a hemophilic mother and a colorblind father.
- Since the mother is hemophilic, her genotype is X^h X^h.
- The father is colorblind, so his genotype is X^c Y.

**Man's Genotype:**
- The man inherits one X chromosome from his mother (X^h) and one Y chromosome from his father.
- Therefore, the man’s genotype is X^h Y.

### Step 3: Cross the Genotypes
Now we will cross the genotypes of the woman (X^H X^c) and the man (X^h Y) to determine the possible progeny.

| | X^H (from woman) | X^c (from woman) |
|-------|-------------------|-------------------|
| X^h (from man) | X^H X^h (normal female, carrier) | X^c X^h (colorblind female) |
| Y (from man) | X^H Y (normal male) | X^c Y (colorblind male) |

### Step 4: Analyze the Progeny
From the cross, we can summarize the possible genotypes of the progeny:
1. X^H X^h - Normal female, carrier for hemophilia.
2. X^c X^h - Colorblind female, affected by hemophilia.
3. X^H Y - Normal male.
4. X^c Y - Colorblind male.

### Conclusion
- The progeny will consist of:
- 1 normal female carrier (X^H X^h)
- 1 colorblind female affected by hemophilia (X^c X^h)
- 1 normal male (X^H Y)
- 1 colorblind male (X^c Y)

None of the progeny will be free from both conditions, as at least one parent carries the alleles for both traits.

### Final Answer
The correct answer is that none of the progeny will be free from both colorblindness and hemophilia.

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