A diseased man marries a normal woman. They get three daughters and five sons. All the daughters were diseased and sons were normal. The gene of this disease is : -
- Sex linked dominant
- Autosomal dominant
- Sex linked recessive
- Sex limited character
Answer
(1) Sex linked dominant
Solution
A father passes his X-chromosome to every daughter and his Y-chromosome to every son.
∵ All daughters are diseased ⇒ the disease gene is on the father's X-chromosome.
∵ The mother is normal and still every daughter shows the disease ⇒ a single copy of the gene is enough, i.e., it is dominant.
∵ Sons get the Y-chromosome from the father and a normal X from the mother ⇒ all sons are normal.
∴ The gene is sex linked (X-linked) dominant.
Answer: (1)