Hardy-Weinberg Equilibrium

AP Biology· AP Biology CED — Natural Selection· 14 min read

1. What Is Hardy-Weinberg Equilibrium?★★☆☆☆⏱ 3 min

Hardy-Weinberg Equilibrium (HWE, also called the Hardy-Weinberg principle or law) is a null model in population genetics that describes a population that is not evolving. In a population at HWE, allele and genotype frequencies remain constant across generations, in the absence of any evolutionary forces.

According to the AP Biology CED, HWE is a key topic within Unit 7 (Natural Selection), making up approximately 10-15% of the unit's exam weight. It appears in both multiple choice (MCQ) and free response (FRQ) sections of the AP exam.

📘 Definition

Hardy-Weinberg Equilibrium

HWE

A null population genetics model describing constant allele and genotype frequencies across generations when no evolutionary forces are acting. $p$ = dominant allele frequency, $q$ = recessive allele frequency for bi-allelic systems.

Example:

Used as a baseline to detect when evolution is occurring in a natural population.

Exam tip:

Expect 1-2 MCQ and at least a partial FRQ question testing HWE concepts on your AP exam.

2. Core Hardy-Weinberg Equations and Frequency Calculation★★☆☆☆⏱ 4 min

HWE is built on two core equations for bi-allelic loci, which is the only case AP Biology regularly tests. The first comes from the fact that the sum of all allele frequencies at a locus must equal 1 (100% of all alleles):

p + q = 1

The second equation for genotype frequencies comes directly from the product rule of probability. For a diploid individual, to get a homozygous dominant genotype (AA), you must inherit one A allele from each parent. The probability of inheriting A from each parent is $p$ , so the probability of AA is $p \times p = p^{2}$ . For homozygous recessive (aa), the same logic gives $q^{2}$ . For heterozygotes, there are two possible combinations: A from mother + a from father, or a from mother + A from father. This gives $pq + q p = 2 pq$ . Summing these gives the second core equation:

p^2 + 2pq + q^2 = 1

📐 Worked Example

In a population of wild sunflowers, 25% of individuals have yellow seeds (recessive trait), while brown seeds are dominant. Calculate the frequency of heterozygous brown-seeded individuals in the population, assuming HWE.

1
Only recessive phenotypes have a known genotype, so yellow seeds = aa, meaning $q^{2} = 0.25$ .
2
Solve for $q$ :
$q = \sqrt{0.25} = 0.5$
3
Solve for $p$ using the allele frequency rule:
$p = 1 - q = 1 - 0.5 = 0.5$
4
Calculate heterozygote frequency:
$2pq = 2(0.5)(0.5) = 0.5$
5
Check your work: $0. 5^{2} + 0.5 + 0. 5^{2} = 0.25 + 0.5 + 0.25 = 1$ , which adds up correctly.

Exam tip:

Always start with $q^{2}$ when you know the recessive phenotype frequency. Only recessive individuals have a known genotype from their phenotype—dominant phenotypes can be either homozygous or heterozygous, so you cannot use their frequency directly to find $p$ .

3. Hardy-Weinberg Equilibrium Core Assumptions★★★☆☆⏱ 3 min

HWE only holds if five core assumptions are met. If any assumption is violated, allele and genotype frequencies will change across generations, meaning the population is evolving and is not in HWE. The key purpose of these assumptions is to define the null case: any deviation from HWE predictions means at least one assumption is broken, and evolution is occurring.

No mutation: No new alleles are created or modified at the locus.
No genetic drift: The population is infinitely large, so random sampling of alleles does not cause frequency changes.
No gene flow: No individuals enter or leave the population, so no alleles are gained or lost.
Random mating: Individuals choose mates without reference to their genotype at the locus.
No natural selection: All genotypes have equal survival and reproductive fitness (no genotype is more likely to reproduce than another).

📐 Worked Example

A biologist studying a population of wild goats finds that males with longer horns (genotype LL or Ll) win 8x more fights for access to females than males with short horns (genotype ll), so they father far more offspring. All other conditions (no mutation, no migration, large population size) are met. Which HWE assumption is violated, and how will genotype frequencies deviate from HWE predictions?

1
First, match the scenario to the assumptions: differential reproductive success based on genotype means the assumption of no natural selection is violated.
2
The L allele for long horns confers a strong reproductive advantage, so it will be passed to the next generation at a higher rate than the l allele for short horns.
3
Expected deviation: The frequency of the ll genotype will be lower than predicted by HWE, and the frequency of LL and Ll genotypes will be higher than HWE predictions.
4
Confirm no other assumption is broken, per the problem constraints.

Exam tip:

When asked to identify a violated assumption, match the scenario directly: mating preference = non-random mating, migration = gene flow, new trait = mutation, small population = genetic drift, differential survival/reproduction = natural selection.

4. Chi-Square Testing for HWE Deviation★★★★☆⏱ 4 min

To formally test whether observed genotype frequencies differ significantly from HWE predictions, we use a chi-square goodness-of-fit test. The null hypothesis ( $H_{0}$ ) is that the population is in HWE (no significant deviation, no evolution). The alternative hypothesis ( $H_{A}$ ) is that the population is not in HWE (significant deviation, evolution is occurring). For bi-allelic HWE tests, degrees of freedom are always 1 (3 genotypes minus 2 estimated parameters: p and q). If your calculated chi-square value is larger than the critical value (usually 3.84 for p=0.05, df=1), you reject the null hypothesis and conclude the population is not in HWE.

📐 Worked Example

A population of clover has the following observed genotype counts: AA = 40, Aa = 40, aa = 20. Total population size N=100. Test whether the population is in HWE, using a critical value of 3.84 for df=1, α=0.05.

1
Calculate allele frequencies: Total alleles = 200. Number of A alleles = (2×40) + 40 = 120, so $p = 120/200 = 0.6$ , $q = 1 - 0.6 = 0.4$ .
2
Calculate expected counts:
$E(AA) = p^2 N = 0.36 × 100 = 36, E(Aa) = 2pq N = 0.48 × 100 = 48, E(aa) = q^2 N = 0.16 × 100 = 16$
3
Calculate the chi-square statistic:
$\chi^2 = \frac{(40-36)^2}{36} + \frac{(40-48)^2}{48} + \frac{(20-16)^2}{16} = 0.44 + 1.33 + 1 = 2.77$
4
Compare to critical value: 2.77 < 3.84, so we fail to reject the null hypothesis. The population is consistent with HWE.

Exam tip:

Always use df = number of genotypes - 2 for HWE chi-square tests, not the default df = number of categories - 1 used for other chi-square tests. This is one of the most commonly tested mistakes on the AP exam.

5. AP Style Concept Check★★★☆☆⏱ 3 min

✓ Quick check

Test your understanding with these AP-style practice questions:

In a population of lizards, the ability to produce toxin in their skin is controlled by a recessive allele. 4% of the lizard population produces toxin. What is the frequency of the dominant non-toxin producing allele in this population?
- 0.02
- 0.2
- 0.8
- 0.96
Reveal answer
0.8 —
Correct: Toxin production is recessive, so $q^{2} = 0.04$ , $q = 0.2$ , $p = 1 - 0.2 = 0.8$ . Other options correspond to common student mistakes.
Sickle cell anemia is an autosomal recessive disorder that affects approximately 1 in 1600 individuals in a population. Assuming HWE, what is the expected frequency of heterozygous carriers?
Reveal answer
~4.9% —
Correct: $q^{2} = 1/1600 = 0.000625$ , $q = 0.025$ , $2 pq = 2 (0.975) (0.025) = 0.04875$ or ~4.9%.

6. Common Pitfalls

Wrong move:

Calculating $p$ as $1 - q^{2}$ , instead of $1 - q$ .

Why:

Students confuse the frequency of the recessive allele ( $q$ ) with the frequency of the recessive genotype ( $q^{2}$ ).

Correct move:

Always write down and label both equations at the start of every problem, clearly separating allele frequencies from genotype frequencies.

Wrong move:

Forgetting to multiply $pq$ by 2 when calculating heterozygote frequency.

Why:

Students forget there are two distinct ways to inherit a heterozygous genotype, one from each parent.

Correct move:

Always write the full formula $p^{2} + 2 pq + q^{2} = 1$ before plugging in values, so you do not miss the coefficient 2.

Wrong move:

Assuming that any deviation from HWE must be caused by natural selection.

Why:

Students associate evolution with natural selection, but any broken HWE assumption causes deviation.

Correct move:

Always match the scenario description to the specific broken assumption, do not default to natural selection automatically.

Wrong move:

Calculating degrees of freedom for HWE chi-square as $n - 1$ instead of $n - 2$ .

Why:

Most chi-square tests use $df = c a t e g or i es - 1$ , but HWE requires an extra correction because we estimate two parameters (p and q) from the data.

Correct move:

Memorize that for all standard two-allele HWE chi-square tests, df is always 1.

Wrong move:

Assuming dominant alleles are always more common than recessive alleles.

Why:

Students associate "dominant" inheritance with "dominant (common) frequency" from basic Mendelian genetics.

Correct move:

Always calculate frequencies from the given data—recessive alleles can be far more common than dominant alleles, so never assume $p > q$ .

Wrong move:

Using observed genotype frequencies to calculate expected HWE frequencies, instead of counting alleles first.

Why:

Students skip the step of calculating p from total alleles and use observed p directly.

Correct move:

Always count all alleles in the population (2 per diploid individual) to get p and q before calculating expected frequencies.

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Allele frequency sum (bi-allelic)	$p + q = 1$	$p$ = dominant, $q$ = recessive allele frequency
Genotype frequency sum (HWE)	$p^{2} + 2 pq + q^{2} = 1$	$p^{2}$ = AA, $2 pq$ = Aa, $q^{2}$ = aa frequency
Allele frequency calculation	$p = \frac{2 ( AA ) + A a}{2 N}$	N = total diploid individuals, count all alleles
Chi-square goodness-of-fit	$χ^{2} = \sum \frac{( O - E ) ^{2}}{E}$	O = observed, E = expected genotype count
Degrees of freedom (HWE)	$df = genotypes - 2$	df = 1 for all standard 2-allele tests
No mutation	Core assumption	No new alleles introduced
No genetic drift	Core assumption	Large population, no random frequency changes
No gene flow	Core assumption	No migration in/out of population
Random mating	Core assumption	Mating not biased by genotype
No natural selection	Core assumption	All genotypes have equal fitness
Null hypothesis for HWE	Population is in HWE	Reject null = evolution is occurring

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Calculate dominant allele frequency
2022 · FRQ
Test for HWE deviation
2021 · MCQ
Identify violated HWE assumption

What's Next

Hardy-Weinberg Equilibrium is the foundational null model for all studies of microevolution in Unit 7 Natural Selection. Without mastering HWE calculations and null hypothesis testing, you cannot quantify the magnitude or direction of evolutionary change, making follow-up topics like measuring selection strength and genetic drift impossible to apply. HWE also provides the baseline for comparing allele frequencies between populations, core to studying speciation and macroevolution, and translates to real applications in human genetic counseling.