DAT Review · Biology
Hardy-Weinberg Equilibrium
On this page 7 sections
In 30 seconds
- Two equations to memorize: p + q = 1 (allele frequencies) and p² + 2pq + q² = 1 (genotype frequencies).
- p = frequency of the dominant allele; q = frequency of the recessive allele. You will be asked to calculate these from given data.
- p² = homozygous dominant; 2pq = heterozygous; q² = homozygous recessive. Know which term represents carriers vs affected individuals.
- Five conditions must be met for a population to be in Hardy-Weinberg equilibrium: large population, random mating, no migration, no mutation, no natural selection. If any condition is violated, evolution is occurring.
- HWE problems appear on every DAT — practice the three worked examples below until they are second nature.
The college version
Core Review
The Principle
The Hardy-Weinberg principle, independently derived by Godfrey Hardy and Wilhelm Weinberg in 1908, states that allele and genotype frequencies in a population remain constant from generation to generation in the absence of evolutionary influences. It provides a null model — a mathematical baseline against which to detect evolution. If observed genotype frequencies deviate significantly from Hardy-Weinberg expectations, one or more evolutionary forces are at work.
The Two Equations
(1) Allele Frequencies: For a gene with two alleles (A = dominant, a = recessive): p + q = 1
- p = frequency of the dominant allele (A)
- q = frequency of the recessive allele (a)
(2) Genotype Frequencies: p² + 2pq + q² = 1
- p² = frequency of homozygous dominant (AA)
- 2pq = frequency of heterozygous (Aa)
- q² = frequency of homozygous recessive (aa)
The Five Conditions for Equilibrium
A population is in Hardy-Weinberg equilibrium (i.e., NOT evolving) only when all five conditions hold:
| Condition | Meaning | What Happens When Violated |
|---|---|---|
| 1. Large population size | No genetic drift; allele frequencies don't fluctuate by chance | Small populations experience genetic drift |
| 2. Random mating | Individuals pair without regard to genotype | Non-random mating (assortative/inbreeding) alters genotype frequencies but not allele frequencies |
| 3. No migration (gene flow) | No alleles enter or leave the population | Migration introduces or removes alleles |
| 4. No mutation | No new alleles are created | Mutations introduce novel alleles |
| 5. No natural selection | All genotypes have equal fitness | Differential reproduction shifts allele frequencies |
In reality, no natural population perfectly meets all five conditions. HWE serves as a null hypothesis — when real populations deviate, we know evolution is occurring and can investigate why.
Worked Example 1: Finding Allele Frequency from Phenotype
Problem: In a population of 1,000 individuals, 160 show a recessive phenotype. Assuming HWE, what are the frequencies of the dominant and recessive alleles?
Solution:
- 160/1000 = 0.16 = frequency of recessive phenotype = q²
- q = √0.16 = 0.4 (frequency of recessive allele)
- p = 1 − q = 1 − 0.4 = 0.6 (frequency of dominant allele)
- Therefore: p = 0.6, q = 0.4
Worked Example 2: Calculating Carrier Frequency
Problem: Cystic fibrosis (CF) is an autosomal recessive disorder affecting approximately 1 in 2,500 newborns in a population. Assuming HWE, what proportion of the population are carriers (heterozygous)?
Solution:
- q² = 1/2500 = 0.0004
- q = √0.0004 = 0.02
- p = 1 − 0.02 = 0.98
- Carrier frequency = 2pq = 2(0.98)(0.02) = 2(0.0196) = 0.0392 ≈ 3.92%
- Approximately 1 in 25 people are carriers, far more common than those affected (1 in 2,500).
This demonstrates why recessive alleles persist in populations — most copies are hidden in heterozygous carriers.
Worked Example 3: Testing for Hardy-Weinberg Equilibrium
Problem: A population of 500 individuals has the following genotype counts for a biallelic locus: AA = 245, Aa = 210, aa = 45. Is this population in HWE?
Solution:
- Total alleles = 500 × 2 = 1,000
- Count of A alleles = 2(245) + 210 = 700 → p = 0.7
- Count of a alleles = 2(45) + 210 = 300 → q = 0.3
- Expected under HWE:
- AA: p² × 500 = (0.49)(500) = 245
- Aa: 2pq × 500 = (0.42)(500) = 210
- aa: q² × 500 = (0.09)(500) = 45
- Observed = Expected across all three genotypes → the population is in HWE. (For statistical rigor, a χ² test would confirm, but the perfect match makes it obvious here.)
Common Traps
- q² is NOT q. The most frequent mistake: if 16% show the recessive trait, students often set q = 0.16. Wrong — q² = 0.16, so q = 0.4.
- p and q must sum to 1. If your calculated values don't add up, you made an error.
- Forgetting to double when counting alleles. For genotype data: number of A alleles = 2(AA) + 1(Aa), not just (AA) + (Aa).
- Assuming populations are in HWE without testing. Deviations often indicate inbreeding, selection, or population structure.
- X-linked traits require different HWE treatment (females: p² + 2pq + q²; males: p + q because they're hemizygous). The DAT may test this nuance.
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine a giant jar of jelly beans. Red jelly beans (A) and blue jelly beans (a) represent two versions of a gene. If you shake the jar well and the jelly beans mate randomly (red with red, red with blue, blue with blue) and there's no jelly-bean-eating monster removing certain colors, then the proportions stay exactly the same generation after generation. That's Hardy-Weinberg equilibrium — nothing changes. The math is straightforward: p is the fraction of red beans, q is the fraction of blue beans, so p + q = 100%. And p² tells you how many red-red babies there are, 2pq is mixed babies, and q² is blue-blue babies. If those proportions ever shift, you know something fishy is going on — like a monster eating all the blue ones (natural selection!)
Key takeaways
- Start with q². If given a recessive phenotype frequency, that is q² — take the square root to get q. Do NOT mistakenly set q = the phenotype frequency.
- Carrier = heterozygous = 2pq. DAT questions love asking "how many carriers?" after giving disease incidence.
- The five conditions are memorization material — be able to list them and explain what happens when each is violated.
- HWE is a null model. Deviation from HWE = evidence for evolution. Don't confuse "being in HWE" with "not evolving" — they're synonymous under the model.
Check yourself
3 review questions from the chapter. Try each one, then open the answer.
In a population at Hardy-Weinberg equilibrium, the frequency of the recessive phenotype is 0.09. What is the frequency of the heterozygous genotype?
Show answer
q² = 0.09 → q = √0.09 = 0.3. Then p = 1 − 0.3 = 0.7. Heterozygous frequency = 2pq = 2(0.7)(0.3) = 0.42, or 42% of the population.
Amish communities in Pennsylvania have a higher frequency of Ellis–van Creveld syndrome (a recessive trait) than the general population because the founding Amish settlers carried the allele. Is this population in Hardy-Weinberg equilibrium with respect to this locus? Why or why not?
Show answer
No, this population is not in HWE. It violates the "large population" condition — the founder effect (small founding population) caused genetic drift. Additionally, the Amish tend to marry within their community, violating random mating. Both factors cause allele frequencies to deviate from HWE expectations.
A student calculates p = 0.6 and q = 0.5 for a biallelic locus in a population. Assuming the data is correctly collected, what error has the student made?
Show answer
The student's p and q sum to 1.1 rather than 1.0. For a biallelic locus, p + q must equal 1. The most likely error is that the student counted genotypes incorrectly — possibly forgetting to double the homozygous count when calculating allele frequencies, or counting alleles from a sample with rounding errors.
Study tools & related lessonsYou’ll learn to · Related
You’ll learn to
- State the Hardy-Weinberg principle and write both equations, defining each term.
- Calculate allele frequencies (p, q) given genotypic or phenotypic data.
- Test whether a population is in Hardy-Weinberg equilibrium by comparing expected and observed genotype frequencies.
- List the five conditions required for Hardy-Weinberg equilibrium and predict the effect of violating each condition.
Sources & references
- OpenStax Biology 2e, Chapter 19: "The Evolution of Populations"
- NCBI Bookshelf search: "Hardy-Weinberg principle" (the cited book has been removed from Bookshelf)
This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.
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