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

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

Step-by-step explanation:

The frequency of homozygous recessive genotype of children is

`(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

`√(q^2)\\ = √(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