7.5 Hardy-Weinberg Equilibrium

Hardy-Weinberg equilibrium is a model that predicts allele and genotype frequencies in a population that is not evolving. If the real frequencies in a population do not match what the model predicts, that mismatch is evidence the population is evolving. For AP Biology, you need to use the equations, identify which value is given, and explain which evolutionary condition may be changing the population.

Hardy-Weinberg in AP Bio

Hardy-Weinberg equilibrium is the AP Bio model for predicting allele and genotype frequencies in a population that is not evolving. Use $p + q = 1$ for allele frequencies and $p^2 + 2pq + q^2 = 1$ for genotype frequencies.

Most Hardy-Weinberg practice problems start with the recessive phenotype. If the recessive phenotype frequency is given, treat it as $q^2$, take the square root to find $q$, use $p = 1 - q$, then calculate $p^2$ and $2pq$.

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Why This Matters for the AP Biology Exam

This topic gives you a mathematical tool for thinking about evolution at the population level. On the AP Biology exam you will likely calculate allele and genotype frequencies from data, then use those numbers to decide whether a population is in equilibrium or evolving. Because Hardy-Weinberg works as a null hypothesis (it describes what you would expect if nothing were changing), it connects directly to the data analysis and explanation skills the exam rewards. Expect to plug values into the equations, interpret what the results mean, and explain in writing which conditions a real population is breaking.

Key Takeaways

• Hardy-Weinberg equilibrium describes a population that is not evolving, so allele frequencies stay constant across generations.
• Five conditions must hold: large population size, no migration, no new mutations, random mating, and no natural selection. These conditions are never fully met in nature.
• Use $p + q = 1$ for allele frequencies and $p^2 + 2pq + q^2 = 1$ for genotype frequencies.
• $p^2$ is the homozygous dominant frequency, $2pq$ is the heterozygous frequency, and $q^2$ is the homozygous recessive frequency.
• The homozygous recessive group ($q^2$) is often your starting point because that phenotype tells you the genotype directly.
• When real frequencies differ from predicted ones, the population is evolving, which makes the model a useful null hypothesis.

Understanding the Hardy-Weinberg Model

Hardy-Weinberg equilibrium is a model that describes and predicts allele frequencies in a population that is not evolving. The core idea is that if certain conditions are met, allele frequencies stay the same from one generation to the next.

That stable scenario almost never happens in real life, which is exactly why the model is useful. It gives you a baseline, or null hypothesis, for what a non-evolving population should look like. When you compare actual population data to the model's predictions and they do not match, you have evidence that the population is evolving.

The Five Conditions

Allele frequencies stay constant only if all five of these conditions hold:

1. Large population size: the population is large enough that random changes in allele frequency (genetic drift) do not occur.
2. No migration: no individuals move in or out, so gene flow does not add or remove alleles.
3. No new mutations: no new alleles are added to the gene pool.
4. Random mating: mating is not influenced by genotype or any other trait.
5. No natural selection: no allele is favored over another, so survival and reproduction are not tied to specific traits.

These conditions are never all met in a real population. Mutations happen, populations are finite, and mating is rarely fully random. That is why Hardy-Weinberg is a model for calculation rather than a description of any actual population.

If any condition is broken, the population can evolve, meaning allele frequencies can shift across generations. For example, mutation introduces new alleles, migration adds or removes alleles, natural selection increases the frequency of alleles that improve reproductive success, genetic drift changes frequencies in small populations, and nonrandom mating shifts genotype frequencies.

Working With the Equations

You only need two equations, and they share the same two variables:

• $p + q = 1$
• $p^2 + 2pq + q^2 = 1$

Here, p is the frequency of one allele and q is the frequency of the other allele. Both are decimals, and since they are the only two alleles for that gene, they add to 1.

In the genotype equation:

• $p^2$ is the frequency of the homozygous dominant genotype
• $2pq$ is the frequency of the heterozygous genotype
• $q^2$ is the frequency of the homozygous recessive genotype

A Reliable Solving Order

Many AP Biology problems start with the recessive phenotype because individuals showing it must be homozygous recessive, which gives you q^2 directly. From there:

1. Take the square root of $q^2$ to find $q$.

2. Use $p = 1 - q$ to find $p$.

3. Plug $p$ and $q$ back into $p^2 + 2pq + q^2 = 1$ to find the other genotype frequencies.

Some problems instead hand you allele frequencies or genotype frequencies directly, so read carefully to see what you are actually given.

Worked Example: Bird Feather Color

Imagine a population of 100 birds where 84 have white feathers (dominant phenotype) and 16 have grey feathers (recessive phenotype).

• The grey birds are homozygous recessive, so $q^2 = 16/100 = 0.16$.
• $q = \sqrt{0.16} = 0.4$.
• $p = 1 - q = 1 - 0.4 = 0.6$.
• Homozygous dominant: $p^2 = 0.6^2 = 0.36$, or 36 birds.
• Heterozygous: $2pq = 2(0.6)(0.4) = 0.48$, or 48 birds.
• Homozygous recessive: $q^2 = 0.16$, or 16 birds.

The genotype frequencies add to 1 (0.36 + 0.48 + 0.16 = 1), which is a quick way to check your work.

Example Problem

A rare genetic disorder is caused by a recessive allele, "a," in a population. The frequency of that allele is 0.02. Calculate the frequency of the dominant allele "A," along with the frequency of homozygous recessive (aa) and heterozygous (Aa) individuals.

Assign the recessive allele frequency as $q$ and the dominant allele frequency as $p$. Since A and a are the only two alleles at this locus, $p + q = 1$. (Hardy-Weinberg does not require the alleles to be codominant.)

Given $q = 0.02$:

• $p = 1 - q = 1 - 0.02 = 0.98$
• Homozygous recessive (aa): $q^2 = 0.02^2 = 0.0004$, or 0.04%
• Heterozygous (Aa): $2pq = 2(0.98)(0.02) = 0.0392$, or 3.92%

This calculation assumes all five Hardy-Weinberg conditions: no mutation, no selection, no gene flow, a large population, and random mating. Remember that this equilibrium is theoretical, and real populations almost never meet every condition.

How to Use This on the AP Biology Exam

Problem Solving

• Identify what the question gives you first. If it provides the recessive phenotype count, start with $q^2$. If it gives an allele frequency, you can go straight to $p + q = 1$.
• Keep $p$ and $q$ as decimals during the math, then convert to percentages or counts only at the end if asked.
• Check that $p^2 + 2pq + q^2$ adds to 1 to catch arithmetic mistakes.

Free Response

• When asked whether a population is evolving, compare observed genotype frequencies to the frequencies the model predicts. A clear mismatch supports the conclusion that the population is evolving.
• If you argue a population is evolving, name the specific condition being broken (for example, selection or gene flow) instead of just saying "it is not in equilibrium."
• State your assumptions when you use the equations. Explaining that you assumed random mating and no selection shows you understand the model's limits.

Common Trap

• Do not confuse allele frequencies with genotype frequencies. $p$ and $q$ are allele frequencies; $p^2$, $2pq$, and $q^2$ are genotype frequencies.
• Remember that $2pq$ counts heterozygotes only, not all carriers plus homozygotes. The heterozygous group is its own term.

Common Misconceptions

• Hardy-Weinberg proves a population is not evolving. It does the opposite job. It predicts what a non-evolving population would look like so you can test real data against that baseline.
• If a population is in equilibrium, allele frequencies are static forever. The model describes conditions under which frequencies stay constant, but it does not mean frequencies cannot change. The point is to identify what changes them.
• p always means the dominant allele. $p$ and $q$ just label two alleles. Which one you call $p$ depends on the problem; always read which allele or phenotype you are given.
• $q^2$ is the frequency of carriers. $q^2$ is the homozygous recessive frequency. Heterozygous carriers are counted by $2pq$.
• The five conditions are realistic. They are never all met in nature. That is intentional, since the unrealistic ideal is what makes the model a useful null hypothesis.

zygous dominant, $2pq$ is heterozygous, and $q^2$ is homozygous recessive.

What does p represent in Hardy-Weinberg?

$p$ represents the frequency of one allele in the population. It is often used for the dominant allele in AP Bio problems, but the problem should tell you which allele is being labeled as $p$.

What does q squared represent?

$q^2$ represents the frequency of the homozygous recessive genotype. If the recessive phenotype is visible in the population, that phenotype frequency usually gives you $q^2$ directly.

What are the five Hardy-Weinberg conditions?

The five conditions are a large population, no migration, no new mutations, random mating, and no natural selection. If any condition is broken, allele frequencies can change and the population can evolve.

How do you solve Hardy-Weinberg practice problems?

Start by identifying what the problem gives you. If it gives the recessive phenotype frequency, treat it as $q^2$, take the square root to find $q$, use $p = 1 - q$, then calculate genotype frequencies with $p^2 + 2pq + q^2 = 1$.

Related AP Biology Guides

• 7.1 Introduction to Natural Selection
• Unit 7 Overview: Natural Selection
• 7.2 Natural Selection
• 7.4 Population Genetics
• 7.3 Artificial Selection
• 7.6 Evidence of Evolution