Question

Hemolytic jaundice is due to the dominant gene but only 20% of the people develop this disease. A heterozygous man marries a homozygous normal woman. What proportion of the children in population would be expected to have this disorder?

• A. $\frac{1}{5}$
• B. $\frac{1}{20}$
• C. $\frac{1}{10}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$\frac{1}{10}$

	A	a
a	Aa	aa
a	Aa	aa

Genes present in man: Aa

Genes present in a woman: aa

The children would have genes: Aa, Aa, aa, aa.

Aa: diseased, aa: normal

Proportion expected to have the disorder: 2/4= 1/2

Since only 20% of people develop this disease, therefore 1/2*0.20= 1/10

So the correct option is '$\frac{1}{10}$'.