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

Answer:

Option B

Explanation:

Given

Number of incidences or frequency of an autosomal recessive disease $= \frac{1}{10000}$

As per Hardy Weinberg's equation, frequency of a recessive genotype is $q^{2}$

$q = \sqrt{\frac{1}{10000} }\\ q = \frac{1}{100}$

As per first equation of Hardy Weinberg's -

$p+q=1$

so ,

$p = 1-q\\p = 1-\frac{1}{100}\\p = \frac{99}{100}$

$p^2 = (\frac{99}{100})^2\\p^2 =\frac{9801}{1000}$

As per second equation of Hardy Weinberg's -

$p^{2} + q^{2} + 2pq=1$

Substituting the given values in above equation, we get -

$\frac{9801}{10000} + \frac{1}{100} + 2pq=1\\2pq = 1-(\frac{9801}{10000} + \frac{1}{100})\\2pq = 0.99%\\OR[tex]2pq=1$%

$2pq = \frac{1}{100}$

Hence, option B is correct.