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

Option E,
`0.64`

Step-by-step explanation:

The frequency of x-linked recessive disease is equal to

`(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 = √(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