BiologyISC Class 12 · Evolution

Hardy-Weinberg Principle and Numericals: ISC Class 12 Biology

Updated 15 October 2026 · Fact-checked

The Hardy-Weinberg principle states that allele and genotype frequencies in a large, randomly mating population stay constant from generation to generation unless disturbed. With allele frequencies p and q, p + q = 1 and genotype frequencies are p² + 2pq + q² = 1. Gene flow, genetic drift, mutation, recombination and natural selection disturb this equilibrium.

Random mating of gametes gives AA = p², Aa = 2pq and aa = q²; together they always add up to 1.

Illustration drawn for QuestionBuddy.

Understand Hardy-Weinberg Principle

The idea. In a population, the frequency of an allele can be measured and, under certain conditions, it stays fixed through generations. The sum of all allele frequencies of a gene is 1. This genetic equilibrium is the Hardy-Weinberg principle. A change in allele frequencies means the population is evolving; Hardy-Weinberg is the 'no-evolution' baseline.

The algebra. For a gene with alleles A (frequency p) and a (frequency q):

p + q = 1.
With random mating, the chance that an individual gets A from both parents is p × p = p² (AA); a from both, q² (aa); A from one and a from the other, 2pq (Aa).
p² + 2pq + q² = 1, the binomial expansion of (p + q)².

Conditions for equilibrium. Large population, random mating, and no gene flow, no genetic drift, no mutation, no selection.

Five factors that disturb it (NCERT).
1. Gene migration or gene flow: individuals moving in or out carry alleles; repeated migration is gene flow.
2. Genetic drift: a chance change in allele frequency, strongest in small populations. When a few individuals start a new population with chance frequencies, the original drifted population becomes the founders: founder effect. A sudden sharp fall in population size gives a similar random sample (bottleneck effect).
3. Mutation: new alleles.
4. Genetic recombination during gametogenesis: new combinations (variation).
5. Natural selection: some genotypes leave more offspring.

Working from data. Three common question types:

Given the recessive phenotype frequency: only aa shows it, so q² = that frequency; q = √(q²); p = 1 − q; then 2pq and p². This works only if the population is assumed to be in equilibrium.
Given genotype counts: count alleles directly. p = (2 × AA + Aa)/(2N); q = (2 × aa + Aa)/(2N). This needs no equilibrium assumption. Then expected numbers p²N, 2pqN, q²N show whether the population is in equilibrium.
Numbers versus frequencies: multiply frequencies by N when numbers are asked.

Carriers of a rare recessive disorder. If q² is tiny, q is small, p ≈ 1, and 2pq ≈ 2q. A disorder in 1 of 10 000 people (q = 0.01) has about 2 carriers per 100: carriers vastly outnumber affected people, so most recessive alleles hide in heterozygotes.

Selection against the recessive phenotype. If all aa individuals fail to reproduce every generation, q falls as q′ = q/(1 + q). The fall is fast while q is large and very slow when q is small, because rare alleles hide in carriers.

Flow chart for any numerical: identify what is given (phenotype frequency or genotype counts) → find q (and p) → compute p², 2pq, q² → multiply by N if numbers are asked → compare observed and expected if equilibrium is asked.

Hardy-Weinberg Principle: key facts

Allele frequencies: $p + q = 1$; p for the dominant allele A, q for the recessive allele a.
Genotype frequencies: $p^{2} (AA) + 2 pq (A a) + q^{2} (aa) = 1$; Expected only at equilibrium.
Allele frequency from counts: $p = \frac{2 N _{AA} + N _{A a}}{2 N}, q = \frac{2 N _{aa} + N _{A a}}{2 N}$; Works whether or not the population is in equilibrium.
From the recessive phenotype: $q = frequency of aa$; Assumes equilibrium.
Complete selection against aa: $q_{n} = \frac{q _{0}}{1 + n q _{0}}$; q falls slowly once the allele is rare.

How to answer Hardy-Weinberg Principle questions

ISC sets Hardy-Weinberg numericals asking frequencies and numbers, a test of equilibrium from observed counts, and a reason for a change in frequencies.

1Read what is given: a phenotype frequency (use q² for the recessive) or genotype counts (count alleles directly).
2Find q, then p = 1 − q. Check p + q = 1.
3Compute p², 2pq and q²; check that they add to 1.
4If numbers are asked, multiply each frequency by the population size; give whole numbers where they come out whole.
5To test equilibrium, compare observed numbers with p²N, 2pqN, q²N; a large mismatch means a disturbing factor is acting.
6Name the factor that fits the story: migration (gene flow), drift (small population, founders, bottleneck), mutation, recombination or selection.

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.

Carriers of a rare recessive disorder

Suppose a health survey finds that a recessive metabolic disorder affects 1 in 10 000 newborns in a large, randomly mating town population, and parents wonder how many healthy people might carry the allele.

Estimate the carrier frequency and explain why carriers far outnumber affected people.

Try it first, then show the solutionSolution

Affected people are aa: q² = 1/10 000 = 0.0001, so q = 0.01.
p = 1 − 0.01 = 0.99.
Carriers: 2pq = 2 × 0.99 × 0.01 = 0.0198, about 1 in 50.
So for every affected newborn there are about 198 carriers; most copies of a rare recessive allele are hidden in healthy heterozygotes.

Answer: About 2% of people (1 in 50) are carriers, about 198 times as many as affected people.

Memory hook: Rare disease, common carriers.

Tricks to remember Hardy-Weinberg Principle

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

Square root the recessive
Only aa is visible, so start from q² and take the square root.
Use it when: Phenotype-frequency questions.
Count alleles, not people
Each AA gives two A, each Aa gives one A and one a, each aa two a; divide by 2N.
Use it when: Genotype-count questions and equilibrium tests.
Five spoilers: M-D-M-R-S
Migration, Drift, Mutation, Recombination, Selection.
Use it when: Factors affecting equilibrium.

Hardy-Weinberg Principle: solved examples

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

Example 1: From q² to everything · 2 marks · Short answerBoard level

In a population in Hardy-Weinberg equilibrium, 9% of individuals show a recessive trait. Find p, q and the frequencies of the three genotypes.

Try it first, then show the solutionSolution

q² = 0.09, so q = 0.3; p = 0.7 (1 mark).
AA = p² = 0.49; Aa = 2pq = 0.42; aa = 0.09 (check: total 1) (1 mark).

Answer: p = 0.7, q = 0.3; AA 0.49, Aa 0.42, aa 0.09.

Example 2: Statement and factors · 3 marks · Short answerBoard level

State the Hardy-Weinberg principle and name the five factors that affect Hardy-Weinberg equilibrium.

Try it first, then show the solutionSolution

Allele and genotype frequencies in a large, randomly mating population remain constant from generation to generation; p + q = 1 and p² + 2pq + q² = 1 (1 mark).
Factors: gene migration (gene flow), genetic drift, mutation (1 mark),
genetic recombination and natural selection (1 mark).

Answer: Frequencies stay constant (p + q = 1; p² + 2pq + q² = 1); disturbed by gene flow, drift, mutation, recombination, natural selection.

Example 3: Numbers in a population of 800 · 3 marks · Short answerHarder board level

In a population of 800 that is in Hardy-Weinberg equilibrium, 50 individuals show the recessive phenotype. Calculate the expected numbers of homozygous dominant and heterozygous individuals.

Try it first, then show the solutionSolution

q² = 50/800 = 0.0625, so q = 0.25 and p = 0.75 (1 mark).
p² = 0.5625, 2pq = 0.375 (1 mark).
AA = 0.5625 × 800 = 450; Aa = 0.375 × 800 = 300 (check: 450 + 300 + 50 = 800) (1 mark).

Answer: AA 450, Aa 300 (aa 50).

Example 4: Is it in equilibrium? · 4 marks · Short answerHarder board level

A sample of 800 individuals contains 340 AA, 320 Aa and 140 aa. (a) Calculate p and q. (b) Calculate the expected numbers of each genotype at equilibrium. (c) Is the population in equilibrium? Suggest one cause of the difference.

Try it first, then show the solutionSolution

(a) p = (2 × 340 + 320)/1600 = 1000/1600 = 0.625; q = 0.375 (1 mark).
(b) Expected AA = 0.390625 × 800 = 312.5; Aa = 2 × 0.625 × 0.375 × 800 = 375; aa = 0.140625 × 800 = 112.5 (1½ marks).
(c) No: there are fewer heterozygotes (320 versus 375) and more homozygotes than expected (1 mark).
A deficit of heterozygotes can come from non-random mating, selection against heterozygotes or migration of homozygotes into the population (½ mark).

Answer: (a) p = 0.625, q = 0.375. (b) 312.5 : 375 : 112.5. (c) Not in equilibrium (heterozygote deficit).

Example 5: Carriers of a rare disorder · 2 marks · Short answerHarder board level

A recessive disorder affects 1 in 10 000 people of a large, randomly mating population. What is the frequency of carriers? Express it as roughly 1 in how many people.

Try it first, then show the solutionSolution

q² = 0.0001, q = 0.01, p = 0.99 (1 mark).
2pq = 2 × 0.99 × 0.01 = 0.0198, about 1 in 50 (1 mark).

Answer: 0.0198, about 1 in 50.

Example 6: Selection against the recessive · 4 marks · Short answerHOTS / competency

Suppose a recessive allele has q = 0.5, and every generation all aa individuals die before reproducing (other genotypes reproduce equally, mating is random). (a) Find q after one, two and three generations using q′ = q/(1 + q). (b) How many generations are needed to bring q down to 0.1, and to 0.01? (c) Why does selection fail to remove the allele quickly when it becomes rare?

Try it first, then show the solutionSolution

(a) 0.5 → 0.5/1.5 = 0.333 → 0.333/1.333 = 0.25 → 0.25/1.25 = 0.2 (1½ marks).
(b) qₙ = q₀/(1 + nq₀): for 0.1, n = (0.5/0.1 − 1)/0.5 = 8 generations; for 0.01, n = (50 − 1)/0.5 = 98 generations (1½ marks).
(c) When q is small, almost all copies of the allele are in heterozygotes (2pq ≫ q²), which look normal and are not removed by selection (1 mark).

Answer: (a) 0.333, 0.25, 0.2. (b) 8 and 98 generations. (c) Rare alleles hide in carriers.

Where students lose marks in Hardy-Weinberg Principle

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

Mistake: Assuming equilibrium when counts are given
In 2025 examiners noted that candidates assumed a rabbit population was in Hardy-Weinberg equilibrium and used 2pq, instead of reading the heterozygote frequency from the counts.
Do this: If genotype counts are given, the observed frequency is count/N; use p² and 2pq only for expected values.
Usually costs: 1 to 2 marks.
Mistake: Giving frequencies when numbers are asked
In 2024 examiners reported percentages instead of numbers of flies, and wrong frequencies carried into later parts.
Do this: Multiply each frequency by N; check that the numbers add to N.
Usually costs: 1 mark per part.
Mistake: Subtracting instead of using p², 2pq, q²
Examiners (2024) saw answers such as 40 or 60 out of 200 obtained by subtraction, and carrier numbers guessed without calculation.
Do this: Write p² N, 2pq N, q² N explicitly.
Usually costs: Each part.
Mistake: Misnaming the principle
Examiners (2025) found 'Mendelism', 'Darwinism' or 'Hindsberg principle'.
Do this: Write Hardy-Weinberg principle (genetic equilibrium).
Usually costs: 1 mark.

Exam tips

Write p + q = 1 and p² + 2pq + q² = 1 at the start of every numerical.
Keep at least two decimal places until the end; round only final numbers.
Check that genotype frequencies sum to 1 and numbers sum to N.
For 'reason for change' parts, name a specific factor and link it to the data.
Learn the five factors in NCERT's order: gene flow, drift, mutation, recombination, selection.

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
In a population at Hardy-Weinberg equilibrium, p = 0.8. The frequency of heterozygotes is:
(a) 0.16 (b) 0.32 (c) 0.64 (d) 0.04
Hint
2pq with q = 0.2.
Answer
(b) 0.32.
Q2 · 2 marks · Short answerBoard level
Define genetic drift and the founder effect.
Hint
Chance; a few founders.
Answer
Genetic drift is a change in allele frequency by chance, strongest in small populations. Founder effect: when a few individuals found a new population, its allele frequencies differ by chance from the original population; these individuals are the founders.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In a large, randomly mating population with no mutation, migration or selection, allele frequencies remain constant.
Reason (R): p² + 2pq + q² = 1 is the binomial expansion of (p + q)².
(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
R is algebra; does it explain why frequencies stay constant?
Answer
(b). Both are true, but the expansion only describes genotype frequencies; constancy follows from the absence of disturbing factors.
Q4 · 3 marks · Short answerHarder board level
84% of a population in equilibrium shows the dominant phenotype. Find q, p, the frequency of heterozygotes, and the fraction of dominant-phenotype individuals who are heterozygous.
Hint
Recessive phenotype = 16%.
Answer
q² = 0.16, q = 0.4, p = 0.6. 2pq = 0.48. Among the dominant phenotype: 0.48/0.84 ≈ 0.57, so about 57% are heterozygous.
Q5 · 3 marks · Short answerHarder board level
In a sample of 2000 people, 720 are MM, 960 MN and 320 NN for the MN blood group (alleles co-dominant). Calculate the allele frequencies and state whether the population is in equilibrium.
Hint
Count alleles; then compare with p²N, 2pqN, q²N.
Answer
p(M) = (1440 + 960)/4000 = 0.6; q(N) = 0.4. Expected: 0.36 × 2000 = 720, 0.48 × 2000 = 960, 0.16 × 2000 = 320. Observed equals expected, so the population is in equilibrium.
Q6 · 2 marks · Short answerHarder board level
Why does a change in allele frequencies indicate evolution? Give an example of how gene flow changes allele frequencies.
Hint
Equilibrium = no evolution.
Answer
Hardy-Weinberg equilibrium describes a population that is not evolving; a change in frequencies shows that a disturbing factor (and so evolution) is acting. Example: if individuals carrying a rare allele repeatedly migrate into a population, that allele's frequency rises there.
Q7 · 3 marks · Short answerHOTS / competency
Suppose, in a population of 1200 beetles, 300 are green (recessive, gg) and the rest brown. A study then shows that birds eat green beetles much more often. (a) Assuming equilibrium before predation, find the numbers of GG and Gg beetles. (b) Predict the direction of change in q over generations, and name the factor. (c) Why will q not reach zero quickly?
Hint
q² = 0.25.
Answer
(a) q² = 300/1200 = 0.25, q = 0.5, p = 0.5: GG = 0.25 × 1200 = 300, Gg = 0.5 × 1200 = 600. (b) q will fall; natural selection (directional). (c) Many g alleles are carried by brown heterozygotes, which birds do not pick out.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a storm carries 10 birds from a mainland population to an empty island. On the mainland a rare allele b has q = 0.01, but the 10 founders carry 4 copies of b among their 20 alleles.
(a) What is q in the island population? (1)
(b) Name the phenomenon. (1)
(c) If the island birds mate randomly and nothing else acts, what fraction of the next generation is expected to be bb? (1)
Hint
Count alleles among the founders.
Answer
(a) q = 4/20 = 0.2. (b) Founder effect (a form of genetic drift). (c) q² = 0.04, i.e. 4%, compared with 0.01% (q² = 0.0001) on the mainland.

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 2pq represent?Show answer
The frequency of heterozygotes (Aa).
2.Name the five factors affecting Hardy-Weinberg equilibrium.Show answer
Gene migration (gene flow), genetic drift, mutation, genetic recombination and natural selection.
3.If q = 0.1, what is p²?Show answer
0.81.
4.What is the founder effect?Show answer
Chance difference in allele frequencies of a new population started by a few individuals.
5.How do you find p from genotype counts?Show answer
p = (2 × AA + Aa)/(2N).
6.What does a change in allele frequency over generations indicate?Show answer
That evolution is occurring.

Hardy-Weinberg Principle: frequently asked questions

What is the Hardy-Weinberg principle?

In a large, randomly mating population with no gene flow, drift, mutation, recombination or selection, allele and genotype frequencies stay constant from generation to generation.

What are p and q?

The frequencies of the dominant and recessive alleles of a gene; p + q = 1.

How do you find carriers from a disease frequency?

Take q = √(disease frequency), p = 1 − q, and carriers = 2pq.

How can you test whether a population is in equilibrium?

Calculate p and q from observed counts, compute expected p²N, 2pqN and q²N, and compare them with the observed numbers.

Why is Hardy-Weinberg called a null model of evolution?

It describes a population that is not evolving, so any change in frequencies shows that some evolutionary force is acting.