Question

An x-linked recessive disease, caused by a single gene, affects in 1/25 men (1 in 25 0.04). What is the frequency of affected females (assume Hardy-Weinberg can be applied).
A. 0.0016
B. 0.04
C. 0.2
D. 0.32
E. 0.64

Answer 1

Answer:

Option E, $0.64$

Explanation:

The frequency of x-linked recessive disease is equal to

$\frac{1}{25}\\ = 0.04$

As per Hardy-Weinberg equation, the frequency of recessive genotype is represented by "$q^{2}$"

Thus, here $q^{2}$$= 0.04$

Frequency of recessive allele $= \sqrt{q^{2} }$

Substituting the given value, we get -

$q = \sqrt{0.4} \\q=0.2$

As per hardy Weinberg equation-

$p+q=1\\p+0.2+1\\p=0.8$

Now "p" signifies frequency of dominant allele. Thus, frequency of dominant genotype would be

$p^{2} \\0.8^2\\0.64$

Hence, option E is correct