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 : -
A
Sex linked dominant
B
Sex linked recessive
C
Sex limited character
D
Autosomal dominant
The answer is A.
Solution
A father gives his X chromosome to all his daughters and his Y chromosome to all his sons. If the disease gene is on the X chromosome and is dominant, the affected father ($X^AY$) passes $X^A$ to every daughter, so all daughters are affected, even with a normal mother ($X^aX^a$). Sons receive the father's Y and the mother's normal X, so all sons are normal. Cross: $X^AY \times X^aX^a \rightarrow X^AX^a$ (all daughters affected) and $X^aY$ (all sons normal). This exactly matches the family. An autosomal or X-linked recessive gene would not give this pattern with a normal mother. So the gene of this disease is sex linked dominant.
ConceptFather to all daughters pattern
Principles of Inheritance and Variation in past papers
198 questions from this chapter have appeared across 27 exam years.
