Question

Cystic fibrosis is a genetic disorder in homozygous recessives that causes death during the teenage years. If 9 in 10,000 newborn babies have the disease,
what are the expected frequencies of the dominant (A1) and recessive (A2) alleles according to the Hardy-Weinberg equation?
(a) f(A1) = 0.9997, f(A2) = 0.0003(b) f(A1) = 0.9800, f(A2) = 0.0200(c) f(A1) = 0.9700, f(A2) = 0.0300(d) f(A1) = 0.9604, f(A2) = 0.0392

Answer 1

Answer:

f(A1) = 0.9700, f(A2) = 0.0300

Explanation:

The frequency of homozygous recessive genotype of children is  

$\frac{9}{10000} \\= 0.0009$

As per Hardy weinberg's equation, frequency of homozygous recessive genotype is represented by  

$q^2$

Here $q^2$ $= 0.0009$

Thus, frequency of recessive allele would be  

$\sqrt{q^2}\\ = \sqrt{0.0009}\\ = 0.03$

as per first law of Hardy weinberg

$p+q = 1\\p = 1- 0.03\\p = 0.97$

Hence, option C is correct