BiologyCBSE Class 12 · Evolution

Hardy-Weinberg Principle

Updated 15 October 2026 · Fact-checked

The Hardy-Weinberg principle states that allele frequencies in a population stay constant across generations unless disturbed. If p and q are the frequencies of alleles A and a, then p + q = 1 and the genotype frequencies are p² (AA), 2pq (Aa) and q² (aa). Gene flow, genetic drift, mutation, recombination and natural selection disturb this equilibrium.

Notice the whole square has area (p+q)² = 1, so the two Aa areas make 2pq; any disturbance of these values signals evolution.

Illustration drawn for QuestionBuddy.

Understand Hardy-Weinberg Principle

Gene pool and equilibrium. The total genes and their alleles in a population form its gene pool. The Hardy-Weinberg principle says that allele frequencies in a population are stable and constant from generation to generation: the gene pool remains constant. This is genetic equilibrium (Hardy-Weinberg equilibrium). The sum of all allele frequencies is 1.

The algebra. For a gene with two alleles in a diploid population:

p = frequency of allele A; q = frequency of allele a; p + q = 1.
The chance that A appears on both homologous chromosomes is p × p = p² (AA). Similarly q² (aa), and 2pq (Aa), since a heterozygote can arise as A from the father and a from the mother, or the reverse.
p² + 2pq + q² = 1, which is the binomial expansion of (p + q)².

What a mismatch means. If the genotype frequencies measured in a population differ from the expected values, the direction and size of the difference show the extent of evolutionary change. A disturbance of genetic equilibrium, i.e. a change in allele frequencies, is interpreted as evolution.

Five factors that disturb the equilibrium.
1. Gene migration (gene flow): when a section of a population migrates to another place and population, allele frequencies change in both the old and the new population; new alleles are added to one and lost from the other. If this happens repeatedly, it is gene flow.
2. Genetic drift: a change in allele frequency that happens by chance. Sometimes the change in a new sample of a population is so great that it becomes a different species; the original drifted population acts as founders, and this is the founder effect. Drift is strongest in small populations.
3. Mutation: creates new alleles; pre-existing advantageous mutations, when selected, give new phenotypes, and over a few generations this can lead to speciation.
4. Genetic recombination: new combinations of alleles during gamete formation (crossing over, independent assortment).
5. Natural selection: heritable variations that enable better survival allow some individuals to reproduce and leave more progeny, so their alleles increase.

Solving any Hardy-Weinberg problem.

If you know the recessive phenotype frequency, that is q²: take the square root to get q.
If you know genotype counts, find allele frequency by counting alleles: p = (2 × AA + Aa) ÷ (2 × total).
Then p = 1 − q, and compute p², 2pq, q² to predict or to test equilibrium.
Carriers of a recessive disorder = 2pq.

Flow chart: phenotype data → q² (recessive phenotype) → q = √q² → p = 1 − q → p², 2pq → compare with observed → equal: equilibrium; different: evolving (find which factor).

Worked mini-example. In a population, 16% show a recessive trait: q² = 0.16 → q = 0.4 → p = 0.6 → AA = 0.36, Aa = 0.48, aa = 0.16 (sum 1.00).

Draw and label: the Hardy-Weinberg square. Checklist: a 2 × 2 Punnett-type square with gametes A (p) and a (q) on each side; boxes AA (p²), Aa (pq), aA (pq), aa (q²); below it, the sum p² + 2pq + q² = 1. Labels: allele frequencies p and q, genotype frequencies p², 2pq, q².

Hardy-Weinberg Principle: key facts

Allele frequencies: $p + q = 1$; p = frequency of A, q = frequency of a.
Genotype frequencies at equilibrium: $p^{2} + 2 pq + q^{2} = 1$; AA = p², Aa = 2pq, aa = q²; the binomial (p + q)².
Allele frequency from counts: $p = \frac{2 N _{AA} + N _{A a}}{2 N}, q = 1 - p$; Each individual carries two alleles.
From recessive phenotype: $q = frequency of recessive phenotype$; Valid only if the population is in equilibrium.
Carrier frequency: $carriers = 2 pq \approx 2 q when q is small$; For q = 0.01, 2pq = 0.0198 (about 1 in 50).

How to answer Hardy-Weinberg Principle questions

Follow the same five steps for every Hardy-Weinberg numerical and show each line; marks are given for the steps.

1Write the given data and decide which quantity it is: an allele frequency (p or q), a genotype frequency (p², 2pq, q²) or counts.
2If the recessive phenotype frequency is given, set it equal to q² and take the square root. Never take the square root of the dominant phenotype frequency (it is p² + 2pq).
3If counts are given, count alleles: p = (2 × AA + Aa)/(2 × total).
4Find the other allele: p = 1 − q (or q = 1 − p).
5Compute p², 2pq, q² and check they add to 1. Multiply by population size for numbers.
6To test equilibrium, compare observed genotype frequencies with p², 2pq, q² from the same p and q; a clear difference means the population is evolving.

Hardy-Weinberg Principle in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Carrier frequency of a recessive disorder

Suppose a health survey in a large town finds that 1 in every 10,000 newborns has a particular autosomal recessive disorder, and the town's population is assumed to be in Hardy-Weinberg equilibrium for this gene.

What is the frequency of the disease allele, and roughly what fraction of people are healthy carriers? In a town of 50,000, how many carriers would you expect?

Try it first, then show the solutionSolution

Affected = aa = q² = 1/10,000 = 0.0001, so q = √0.0001 = 0.01.
p = 1 − 0.01 = 0.99.
Carriers = 2pq = 2 × 0.99 × 0.01 = 0.0198, about 2% (roughly 1 in 50).
Expected carriers in 50,000 = 0.0198 × 50,000 = 990.

Answer: q = 0.01; carriers about 1.98% (1 in 50); about 990 carriers among 50,000 people.

Memory hook: Rare disease, common carriers: q² is tiny, but 2pq is about 2q.

Tricks to remember Hardy-Weinberg Principle

Short memory aids. Say each one aloud once, then use it on a question the same day.

Square the binomial
(p + q)² = p² + 2pq + q²: the genotype equation is just the allele equation squared.
Use it when: Recalling the equation.
Root the recessive
Only the recessive phenotype gives q² directly; take its root.
Use it when: Starting a numerical.
Five disturbers: Flow, Drift, Mutation, Recombination, Selection
'Few Dogs Make Real Sounds'.
Use it when: Listing factors that upset equilibrium.

Hardy-Weinberg Principle: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Heterozygote frequency · 1 mark · MCQBoard level

At a locus, the frequency of allele A is 0.8 and of allele a is 0.2. What is the frequency of heterozygotes at equilibrium?

Try it first, then show the solutionSolution

Heterozygotes = 2pq.
2 × 0.8 × 0.2 = 0.32.

Answer: 0.32.

Example 2: Five factors · 3 marks · Short answerBoard level

List the five factors that affect Hardy-Weinberg equilibrium and explain genetic drift and the founder effect.

Try it first, then show the solutionSolution

Gene migration (gene flow), genetic drift, mutation, genetic recombination, natural selection (1 mark).
Genetic drift: a change in allele frequencies that happens by chance, strongest in small populations (1 mark).
Founder effect: when a small group drifts away and starts a new population, its allele frequencies can differ so much from the original that it may become a different species; the drifted group are the founders (1 mark).

Answer: Flow, drift, mutation, recombination, selection; drift = chance change; founder effect = drifted founders.

Example 3: Red and black beetles · 3 marks · Short answerHarder board level

In a beetle population in Hardy-Weinberg equilibrium, black (B) is dominant over red (b), and 40% of beetles are red. (a) Find the frequencies of b and B. (b) What percentage are heterozygous? (c) What percentage are homozygous black?

Try it first, then show the solutionSolution

(a) q² = 0.40, so q = √0.40 ≈ 0.632; p = 1 − 0.632 = 0.368 (1 mark).
(b) 2pq = 2 × 0.368 × 0.632 ≈ 0.465, about 46.5% (1 mark).
(c) p² = 0.368² ≈ 0.135, about 13.5%. Check: 13.5 + 46.5 + 40 = 100 (1 mark).

Answer: q ≈ 0.63, p ≈ 0.37; Bb ≈ 46.5%; BB ≈ 13.5%.

Example 4: Is the population in equilibrium? · 5 marks · Long answerHarder board level

A population of 1000 has 490 AA, 420 Aa and 90 aa individuals. A second population of 1000 has 500 AA, 300 Aa and 200 aa. Test whether each is in Hardy-Weinberg equilibrium.

Try it first, then show the solutionSolution

Population 1: p = (2 × 490 + 420)/2000 = 1400/2000 = 0.7; q = 0.3 (1 mark).
Expected: p² = 0.49 (490), 2pq = 0.42 (420), q² = 0.09 (90): identical to observed, so it is in equilibrium (1 mark).
Population 2: p = (2 × 500 + 300)/2000 = 1300/2000 = 0.65; q = 0.35 (1 mark).
Expected: p² = 0.4225 (about 423), 2pq = 0.455 (455), q² = 0.1225 (about 123); observed 500, 300, 200 (1 mark).
Population 2 has too few heterozygotes and too many homozygotes, so it is not in equilibrium: its allele or genotype frequencies are being changed (e.g. by selection, non-random mating or drift), indicating evolution (1 mark).

Answer: Population 1 is in equilibrium (p = 0.7); population 2 is not (p = 0.65; excess homozygotes).

Example 5: Founder population · 3 marks · Short answerHOTS / competency

Suppose 20 birds from a large mainland population (where q = 0.5 for a colour allele) are blown to an island, and by chance 15 of them are AA, 5 are Aa and none are aa. (a) Find q in the island population. (b) Name the process. (c) What fraction of the island chicks would be aa in the next generation if mating is random?

Try it first, then show the solutionSolution

(a) a alleles = 5 (from Aa); total alleles = 40; q = 5/40 = 0.125 (1 mark).
(b) Founder effect, a form of genetic drift: allele frequency changed by chance in a small group (1 mark).
(c) q² = 0.125² = 0.015625, about 1.6% aa, compared with 25% on the mainland (1 mark).

Answer: (a) q = 0.125 (b) founder effect (genetic drift) (c) about 1.6%.

Example 6: Using population size · 2 marks · Very short answerHarder board level

In a population of 2000 people in equilibrium, the frequency of a recessive allele is 0.3. How many people are expected to be AA, Aa and aa?

Try it first, then show the solutionSolution

q = 0.3, p = 0.7; p² = 0.49, 2pq = 0.42, q² = 0.09 (1 mark).
AA = 0.49 × 2000 = 980; Aa = 840; aa = 180 (total 2000) (1 mark).

Answer: 980 AA, 840 Aa, 180 aa.

Where students lose marks in Hardy-Weinberg Principle

The slips examiners see most often, and what to write instead.

Mistake: Rooting the wrong number
Students take √(dominant %).
Wrong working
75% dominant → p = √0.75 = 0.87.
Right working
25% recessive → q = √0.25 = 0.5; p = 0.5.
Do this: Start from the recessive phenotype (q²).
Usually costs: Most of the numerical.
Mistake: Forgetting the 2 in 2pq
Students write pq for heterozygotes.
Wrong working
Heterozygotes = 0.8 × 0.2 = 0.16.
Right working
Heterozygotes = 2pq = 0.32.
Do this: Two ways to be Aa: A from the father or from the mother.
Usually costs: 1 mark.
Mistake: Using percentages and fractions together
Students mix 40 and 0.4.
Wrong working
q² = 40, q = 6.3.
Right working
q² = 0.40, q ≈ 0.632.
Do this: Convert percentages to decimals first.
Usually costs: 1 mark.
Mistake: Counting individuals instead of alleles
Students divide by N instead of 2N.
Wrong working
p = (490 + 420)/1000 = 0.91.
Right working
p = (2 × 490 + 420)/2000 = 0.7.
Do this: Each diploid individual has two alleles.
Usually costs: 1-2 marks.

Exam tips

Write both equations at the top of every numerical: p + q = 1 and p² + 2pq + q² = 1.
Show a check line where p², 2pq and q² add up to 1.
State the conclusion in words: 'in equilibrium' or 'not in equilibrium, so evolution is occurring'.
Learn the five disturbing factors in order and give a one-line meaning for each.
Round to 2-3 decimal places, but keep the full value in intermediate steps.

Practice questions on Hardy-Weinberg Principle

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of the following does NOT disturb Hardy-Weinberg equilibrium?
(a) Gene flow (b) Genetic drift (c) Random mating in a large population (d) Natural selection
Hint
Equilibrium assumes this condition.
Answer
(c) Random mating in a large population.
Q2 · 2 marks · Very short answerBoard level
If 9% of a population shows a recessive trait, find p, q and the percentage of carriers.
Hint
q² = 0.09.
Answer
q = √0.09 = 0.3; p = 0.7; carriers 2pq = 2 × 0.7 × 0.3 = 0.42 = 42%.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A change in allele frequency in a population indicates evolution.
Reason (R): Hardy-Weinberg equilibrium is maintained when gene flow, drift, mutation, recombination and selection are absent.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Does R explain why a change means evolution?
Answer
(a). Allele frequencies stay constant without these factors, so a change shows that evolutionary forces are acting.
Q4 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a large, randomly mating population of flowers in equilibrium has 64% red (RR and Rr) and 36% white (rr) flowers.
(a) Find q and p. (1)
(b) What fraction of red flowers are heterozygous? (1)
(c) Suppose all white flowers are removed before they set seed, for one generation. Find the new q among the breeding plants and say why equilibrium no longer holds. (2)
Hint
Breeding plants: only RR and Rr remain.
Answer
(a) q² = 0.36 → q = 0.6, p = 0.4. (b) RR = 0.16, Rr = 2pq = 0.48; fraction of red that are Rr = 0.48/0.64 = 0.75. (c) Breeders: RR 0.16 and Rr 0.48 (total 0.64). r alleles = 0.48 out of 2 × 0.64 = 1.28 alleles, so new q = 0.48/1.28 = 0.375. Allele frequency changed because of selection against rr, so the population is no longer in Hardy-Weinberg equilibrium: it has evolved.
Q5 · 2 marks · Very short answerHarder board level
Explain how gene flow differs from genetic drift. Which of them would you expect to change allele frequencies more in a small island population cut off from the mainland?
Hint
Migration vs chance.
Answer
Gene flow is the change in allele frequencies through migration of individuals (and their alleles) between populations, repeated over time; genetic drift is a change in allele frequencies by chance. In a small isolated island population, there is little gene flow, so genetic drift (chance) would have the larger effect.
Q6 · 2 marks · Very short answerHarder board level
Why is it wrong to take the square root of the dominant phenotype frequency to find p? Show with a population in which 75% show the dominant phenotype.
Hint
Dominant phenotype = p² + 2pq.
Answer
The dominant phenotype includes both AA and Aa, so its frequency is p² + 2pq, not p². With 75% dominant, q² = 0.25 → q = 0.5 and p = 0.5. Wrongly, √0.75 ≈ 0.87 would be taken as p, giving the wrong answer.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does p² represent?Show answer
The frequency of the homozygous dominant genotype (AA).
2.If q = 0.2, what is p?Show answer
0.8.
3.Frequency of heterozygotes if p = 0.6?Show answer
2pq = 2 × 0.6 × 0.4 = 0.48.
4.What is the founder effect?Show answer
A large change in allele frequencies when a small drifted group starts a new population.
5.Name the five factors that affect Hardy-Weinberg equilibrium.Show answer
Gene flow, genetic drift, mutation, genetic recombination, natural selection.
6.What is a gene pool?Show answer
The total genes and their alleles in a population.

Hardy-Weinberg Principle: frequently asked questions

What is the Hardy-Weinberg principle?

It states that allele frequencies in a population remain constant from generation to generation in the absence of disturbing factors, giving genotype frequencies p², 2pq and q².

What is the Hardy-Weinberg equation?

p² + 2pq + q² = 1, where p and q are the frequencies of the two alleles (p + q = 1).

What disturbs Hardy-Weinberg equilibrium?

Gene migration (gene flow), genetic drift, mutation, genetic recombination and natural selection.

How do you find q from phenotype data?

Take the frequency of the recessive phenotype as q² and find its square root.

Why does a change in allele frequency mean evolution?

Without disturbing factors, allele frequencies stay constant; a change shows that forces such as selection or drift are acting, which is evolution.