Hardy-Weinberg Equilibrium

Allele frequencies and the null model

A population's gene pool is all the alleles of all its members. The allele frequency is the fraction of the gene pool made up of one allele. For a gene with two alleles, we call the dominant allele frequency $p$ and the recessive allele frequency $q$.

Every allele in the pool is either one or the other, so the frequencies must add up to 1:

Evolution is a change in allele frequencies over generations. To detect change, you need a baseline: what would the frequencies look like if nothing were changing? The Hardy-Weinberg equilibrium is that baseline. A population stays in equilibrium only if all five conditions hold:

1. No mutation (no new alleles appear or change).

2. No gene flow (no migration of individuals or gametes in or out).

3. Random mating (no preference for particular genotypes).

4. No natural selection (all genotypes survive and reproduce equally well).

5. Very large population size (no genetic drift from chance).

The Hardy-Weinberg equation

Think of each parent as drawing one gamete from the gene pool. The chance of drawing $A$ is $p$ and the chance of drawing $a$ is $q$. Two gametes combine at random, so we multiply probabilities:

• $AA$: $p \times p = p^2$

• $Aa$: $p \times q$ and $q \times p$, so $2pq$

• $aa$: $q \times q = q^2$

All genotypes together make up the whole population:

Working from the recessive phenotype

Here is the key trick. You can only see phenotypes. A dominant phenotype could be $AA$ or $Aa$, so you cannot count $p^2$ or $2pq$ directly. But the recessive phenotype can only be $aa$, so it tells you $q^2$. Start there.

This answers our opening question: most copies of a rare recessive allele are hidden in healthy carriers.

Testing whether a population is in equilibrium

If you know the genotype counts, you can calculate allele frequencies by counting alleles, predict the genotype counts with Hardy-Weinberg, and compare. Each individual has 2 alleles, so a population of $N$ individuals has $2N$ alleles.

In practice, real counts rarely match perfectly. To decide whether a difference is just chance, AP Biology uses the chi-square test from the exam formula sheet:

Here $o$ is the observed count and $e$ is the expected count (from Hardy-Weinberg) for each genotype. Add the terms for all genotypes. A larger $\chi^2$ means the observed counts are farther from the expected counts. If $\chi^2$ is bigger than the critical value for your degrees of freedom (number of genotypes minus 1), the difference is unlikely to be chance alone. A large gap like the one above is a clear signal.

When a condition is violated: selection against a recessive

Suppose the recessive genotype is lethal before reproduction. Then selection is violated, and $q$ changes.

As $q$ gets smaller, carriers ($2pq$) vastly outnumber $aa$ individuals ($q^2$), so selection removes fewer and fewer $a$ alleles each generation. That is why $q$ declines slowly over many generations.

Summary

• Hardy-Weinberg equilibrium describes a non-evolving population: no mutation, no gene flow, random mating, no selection, very large population.

• $p + q = 1$ for allele frequencies; $p^2 + 2pq + q^2 = 1$ for genotype frequencies.

• Start from the recessive phenotype: $q^2 \to q \to p \to 2pq$.

• Compare observed counts with expected counts to detect evolution; the model does not say which condition is violated.

• Selection against a recessive allele is slow because most copies hide in carriers.

Back to the start: for a rare recessive condition, carriers far outnumber affected people, and Hardy-Weinberg lets you estimate them from $q^2$ alone.