Question

An autosomal recessive disease has an incidence of 1/10,000. What is the approximate frequency of heterozygote carriers for this disease? Assume Hardy-Weinberg conditions apply

A. 1 /50, or 2%
B. 1/100, or 1%
C. 1 /1000, or 0.1%
D. 1 /10, or 10%
E. 1 /4, or 25%

Answer 1

Option B

Step-by-step explanation:

Given

Number of incidences or frequency of an autosomal recessive disease
`= (1)/(10000)`

As per Hardy Weinberg's equation, frequency of a recessive genotype is
`q^(2)`

`q = \sqrt{(1)/(10000) }\\ q = (1)/(100)`

As per first equation of Hardy Weinberg's -

`p+q=1`

so ,

`p = 1-q\\p = 1-(1)/(100)\\p = (99)/(100)`

`p^2 = ((99)/(100))^2\\p^2 =(9801)/(1000)`

As per second equation of Hardy Weinberg's -

`p^(2) + q^(2) + 2pq=1`

Substituting the given values in above equation, we get -

`(9801)/(10000) + (1)/(100) + 2pq=1\\2pq = 1-((9801)/(10000) + (1)/(100))\\2pq = 0.99%\\</p><p></p><p>OR</p><p>[tex]2pq=1`%

`2pq = (1)/(100)`

Hence, option B is correct.