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-influenced disease
B. Blood group inheritance disease
C. Sex-linked disease
D. Sex-limited disease

Answer

To solve the question regarding the inheritance pattern of a disease in a family where a diseased man marries a normal woman, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parents' Genotypes**:
- The man is diseased, and the woman is normal.
- Since the man is diseased, we need to determine if the disease is sex-linked or autosomal.

2. **Analyze the Offspring**:
- The couple has three daughters and five sons, all of whom are normal.
- If the disease were autosomal dominant, at least some of the children would be expected to inherit the disease. Since all children are normal, this suggests that the disease is not autosomal dominant.

3. **Consider X-Linked Inheritance**:
- In the case of X-linked inheritance, the man has one X chromosome (which carries the disease) and one Y chromosome.
- Daughters inherit one X chromosome from their father and one from their mother. If the father has the disease, he would pass his affected X chromosome to all daughters. Therefore, all daughters would be carriers (heterozygous) but not affected if the disease is recessive.

4. **Evaluate Sons' Inheritance**:
- Sons inherit their X chromosome from their mother and the Y chromosome from their father. Since all sons are normal, the mother must have two normal X chromosomes (XX) and cannot pass on the disease.

5. **Conclusion**:
- Since all daughters are carriers and all sons are normal, the disease must be X-linked recessive. The affected father cannot pass the disease to his sons, as they inherit the Y chromosome from him. Therefore, the disease is indeed sex-linked.

6. **Final Answer**:
- The gene of this disease is **X-linked disease**.