Question

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

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

Answer 1

The Correct Option is A

The question pertains to genetics, specifically, it is about inheritance patterns related to genetic diseases. In this scenario, a diseased man marries a normal woman, resulting in diseased daughters and normal sons. Let's analyze this using the understanding of genetic inheritance:

Explanation

To determine the type of genetic inheritance, consider the following:

1. Sex-Linked Genes: These are genes located on the sex chromosomes. Humans have two sex chromosomes, X and Y. Typically, females have two X chromosomes (XX), while males have one X and one Y (XY).
2. Dominant vs Recessive:
   • A dominant gene manifests in the phenotype even if only one copy is present.
   • A recessive gene requires two copies to manifest in the phenotype.

Determining the Gene Type

Given the scenario:

• Diseased Daughters: All daughters are diseased, which hints that they inherited a gene from their father. Since they have only one X chromosome from the father, it suggests the disease is dominant.
• Normal Sons: None of the sons are diseased. Sons receive the Y chromosome from their father, and since they do not manifest the disease, it further suggests the disease is linked to the X chromosome and is expressed dominantly only when the diseased X is present.

Thus, the gene for the disease must be a sex-linked dominant gene. This means the gene is located on the X chromosome and is dominant.

Conclusion

Based on the inheritance pattern where all daughters but no sons are affected, the correct answer is:

Sex linked dominant