A diseased man marries a normal woman.They get three daughters and five sons were normal . The gene of this disease is :

Options

A. Sex linked dominant
B. Sex linked recessive
C. Sex limited character
D. Autosomal dominant

Answer

To solve the question regarding the inheritance pattern of a disease in a family where a diseased man marries a normal woman and they have three daughters and five sons, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parents' Genotypes**:
- The diseased man has a disease-causing gene. Assuming this disease is sex-linked, he has the genotype X^D Y (where X^D represents the diseased allele).
- The normal woman has the genotype X^N X^N (where X^N represents the normal allele).

2. **Determine the Offspring's Genotypes**:
- The man can pass on either his X^D or his Y chromosome. Since he has daughters, he must pass on his X^D chromosome to each daughter. Therefore, all daughters will have the genotype X^D X^N (diseased).
- The sons will inherit the Y chromosome from their father and one X chromosome from their mother. Since the mother is normal (X^N), all sons will have the genotype X^N Y (normal).

3. **Analyze the Offspring**:
- The three daughters (X^D X^N) are diseased, and the five sons (X^N Y) are normal. This confirms that the disease is passed from father to daughter.

4. **Determine the Nature of the Disease**:
- Since the father (X^D Y) can only pass on the X^D allele to daughters, and all daughters are affected, this indicates that the disease is dominant. If it were recessive, we would expect some daughters to be normal if the mother is normal.

5. **Conclusion**:
- The disease is a sex-linked dominant disorder. Therefore, the answer to the question is that the gene of this disease is sex-linked dominant.

### Final Answer:
The gene of this disease is **sex-linked dominant**.

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