The genetic frequencies of two separate populations are: $$\begin{array} { | c c c c | }
\hline & A A & A a & a a \
\hline \text { Population 1 } & .36 & .48 & .16 \
\text { Population 2 } & .33 & .33 & .33 \
\text { Popllation 3 } & .55 & .10 & .35 \
\hline
\end{array}$$ Which of these populations are in Hardy-Weinberg equilibrium?

Question

The genetic frequencies of two separate populations are: $$\begin{array} { | c c c c | } 
\hline & A A & A a & a a \
\hline \text { Population 1 } & .36 & .48 & .16 \
\text { Population 2 } & .33 & .33 & .33 \
\text { Popllation 3 } & .55 & .10 & .35 \
\hline
\end{array}$$ Which of these populations are in Hardy-Weinberg equilibrium?

Quizplus · Accepted Answer

For a population to be in Hardy-Weinberg equilibrium, the genotype frequencies must satisfy $p^2 + 2pq + q^2 = 1$ where $p$ and $q$ are the allele frequencies. In Population 1, $p = 0.6$ and $q = 0.4$, and the genotype frequencies $0.36, 0.48, 0.16$ satisfy $p^2, 2pq, q^2$. Population 2 and 3 do not satisfy this condition.

The Genetic Frequencies of Two Separate Populations Are Which of These Populations Are in Hardy-Weinberg Equilibrium? A)Population 1

The Genetic Frequencies of Two Separate Populations Are Which of These Populations Are in Hardy-Weinberg Equilibrium?
A)Population 1