Question

Haemolytic jaundice is caused by a dominant gene but only 10% of the people actually develop it. What proportion of the children would be expected to develop the disease, if a heterozygous man marries a homozygous normal woman?

• A. 1/5
• B. 1/10
• C. 1/15
• D. 1/20

Answer 1

The correct option is D 1/20

Genotype of the man --> A${}^H$A (where A${}^H$ is the chromosome carrying dominant gene for haemolytic jaundice and A is the homologous chromosome carrying the normal allele).
Genotype of woman --> AA
P generation : A${}^H$A X AA

F${}_1$ generation : A${}^H$A A${}^H$A AA AA
(Diseased:Normal) = (1:1)
Only 10% of the people having diseased genotype actually suffer from haemolytic jaundice, this turns the ratio to (0.1:1).
Thus, proportion of children expected to develop the disease for the given cross= 0.1/2 (0.1 children are affected out of two children) = 1/20