Question

hardy-weinberg equilibrium - formula
$p + q = 1$
$p$ is the frequency of one allele in the population
$q$ is the frequency of the other allele in the population
therefore $p + q$ must equal $100\\%$ of the population or $1$
value: 10
if the frequency of $p$ is $0.6$, what is the frequency of $q$?
o a. $0.6$
o b. $0.4$
o c. $0.2$
o d. $0.5$
check answer

Explanation:

Step1: Substitute the value of \(p\) into the formula

Given \(p + q=1\) and \(p = 0.6\).

Step2: Solve for \(q\)

We can rewrite the formula as \(q=1 - p\). Substituting \(p = 0.6\) into it, we get \(q=1-0.6\).

Answer:

B. \(0.4\)