Question

The frequency of carriers for a rare autosomal recessive genetic condition is 0.04 in a population. Assuming this population is in Hardy-Weinberg equilibrium, what is the allele frequency of the recessive allele?
a. 0.2
b. 0.4
c. 0.64
d. 0.8
e. Not enough information is provided.

Answer 1

Answer:

The right answer is option A

Explanation:

Hardy-Weinberg equilibrium states that the variations in genetic distribution will always be constant as long as there are no external factors to disturb it across generations like mutations, genetic drift and gene flow among others. This principle is an ideal situation and is not really applicable in reality.

A simple formula for the Hardy-Weinberg principle is

$p^{2}$+ 2pq + $q^{2}$ = 1 and p + q = 1 where p is the frequency of the dominant allele and q is the frequency of the recessive allele. While $p^{2}$ and $q^{2}$  are the frequencies of the dominant and recessive genes respectively.

The frequency of a recessive genetic condition of 0.04 in a population means 4% of the population has a recessive genetic condition.

$q^{2$= 0.04 Genetic condition

q = square root of 0.04 = 0.2 Since q is the frequency of the allele then the frequency is 20%

Answer 2

Answer:

E

Explanation:

Hardy-Weinberg principle states that frequency of alelle and genotype will remain constant in the absence of genetic disturbances such as mutation, no change in the DNA sequence, the population must be large and others

the principle is defined by the equation

p² + 2pq + q² = 1

where the frequency of the dominant allele is p, and the frequency of the recessive allele is q. Going by the what is given which is for the carriers is 0.04 = 2pq

substitute the value into the expression

p² + 0.04  +  q² = 1;  The information is not enough to calculate either p or q going by the equation generated.