Here is the trap that costs marks on Hardy-Weinberg questions: students write p² + q² = 1 and forget the 2pq heterozygote term entirely. They then solve for p or q, get a wrong number, pick a distractor, and lose 5 marks (1 lost + 4 forfeited by negative marking). The full equation is p² + 2pq + q² = 1, and dropping 2pq is the single most reliable way to get the wrong answer on this topic.
What the principle states. In an idealised, large population with no evolutionary forces acting, allele frequencies and genotype frequencies remain constant across generations (NCERT Class 12 Biology Chapter 6, page 120). The two equations are:
The five conditions for equilibrium. Hardy-Weinberg holds only when ALL five conditions are met simultaneously: (1) no mutation, (2) no gene flow (migration), (3) no genetic drift (large population), (4) no natural selection, (5) random mating. Violation of any single condition pushes the population out of equilibrium — this is a common trap in NEET questions that ask "which factor disturbs/maintains Hardy-Weinberg equilibrium."
NEET application. Questions typically give you the frequency of one genotype (often the homozygous recessive aa = q²) and ask you to calculate carrier frequency (2pq) or dominant homozygote frequency (p²). The workflow: q² given → take square root → get q → compute p = 1 − q → plug into p², 2pq as needed. Every step requires you to remember that genotype frequencies have three terms, not two.
Watch-out. If a question asks what happens when one of the five conditions is violated, the answer is always "allele frequencies change" — the population evolves. Random mating alone does not cause evolution; it is the absence of all disturbing forces that maintains equilibrium.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following is NOT a condition required for Hardy-Weinberg equilibrium?
Answer: D. Hardy-Weinberg equilibrium requires a large population (no genetic drift). Small population size promotes drift, which violates equilibrium (NCERT Class 12 Biology Chapter 6, page 121).
Why A is wrong: A is wrong because no mutation IS a required condition for Hardy-Weinberg equilibrium — mutations introduce new alleles, changing frequencies.
Why B is wrong: B is wrong because random mating IS a required condition — non-random (assortative) mating alters genotype frequencies.
Why C is wrong: C is wrong because no natural selection IS a required condition — selection favours certain alleles, changing their frequencies.
The Hardy-Weinberg equation for genotype frequencies in a population is:
Answer: C. The Hardy-Weinberg genotype frequency equation is p² + 2pq + q² = 1, representing homozygous dominant (p²), heterozygous (2pq), and homozygous recessive (q²) frequencies (NCERT Class 12 Biology Chapter 6, pages 120–121).
Why A is wrong: A is wrong because p + q = 1 is the allele frequency equation, not the genotype frequency equation. It describes the sum of allele frequencies, not genotype proportions.
Why B is wrong: B is wrong because p² + q² = 1 omits the heterozygote term 2pq — this is the most common mistake on this topic (trap: forgetting heterozygotes).
Why D is wrong: D is wrong because this equation has incorrect coefficients — the correct expansion of (p + q)² yields p² + 2pq + q², not 2p² + pq + 2q².
In the Hardy-Weinberg equation, the term '2pq' represents the frequency of:
Answer: A. In p² + 2pq + q² = 1, the term 2pq represents the frequency of heterozygous (Aa) individuals in the population (NCERT Class 12 Biology Chapter 6, page 121).
Why B is wrong: B is wrong because homozygous recessive (aa) frequency is represented by q², not 2pq.
Why C is wrong: C is wrong because homozygous dominant (AA) frequency is represented by p², not 2pq.
Why D is wrong: D is wrong because total allele frequency is described by p + q = 1, which is a separate equation from genotype frequencies.
In a population in Hardy-Weinberg equilibrium, the frequency of the homozygous recessive genotype (aa) is 0.09. What is the frequency of the heterozygous genotype (Aa)?
Answer: A. q² = 0.09, so q = 0.3. Then p = 1 − 0.3 = 0.7. Heterozygote frequency = 2pq = 2 × 0.7 × 0.3 = 0.42 (NCERT Class 12 Biology Chapter 6, page 121).
Why B is wrong: B is wrong because 0.21 equals pq (single product, not doubled) — the heterozygote term requires the factor of 2: 2pq, not pq (trap: dropping the coefficient 2 from the heterozygote term).
Why C is wrong: C is wrong because 0.49 equals p² (homozygous dominant frequency), not the heterozygous frequency. This answer confuses AA with Aa.
Why D is wrong: D is wrong because 0.30 equals q (the allele frequency), not the genotype frequency 2pq. This conflates allele frequency with genotype frequency (trap: allele vs genotype confusion).
In a population, the frequency of allele 'a' is 0.4. Assuming Hardy-Weinberg equilibrium, what is the frequency of the homozygous dominant genotype (AA)?
Answer: D. q = 0.4, so p = 1 − 0.4 = 0.6. AA frequency = p² = 0.6² = 0.36 (NCERT Class 12 Biology Chapter 6, page 121).
Why A is wrong: A is wrong because 0.16 equals q² = 0.4² — that is the homozygous recessive (aa) frequency, not AA. This confuses the two homozygous classes.
Why B is wrong: B is wrong because 0.60 equals p (the dominant allele frequency), not p². This conflates allele frequency with genotype frequency (trap: allele vs genotype confusion).
Why C is wrong: C is wrong because 0.48 equals 2pq = 2 × 0.6 × 0.4 — that is the heterozygote (Aa) frequency, not homozygous dominant.
Which one of the following factors, if acting alone, would cause a population to deviate from Hardy-Weinberg equilibrium?
Answer: B. Natural selection is one of the five forces that disturb Hardy-Weinberg equilibrium. Any one of mutation, gene flow, drift, selection, or non-random mating acting alone is sufficient to change allele frequencies (NCERT Class 12 Biology Chapter 6, page 121).
Why A is wrong: A is wrong because large population size is a requirement FOR equilibrium — it prevents genetic drift. Large size maintains, not disturbs, HW (trap: confusing conditions for equilibrium with conditions that disrupt it).
Why C is wrong: C is wrong because absence of mutation is a requirement FOR equilibrium — mutations introduce new alleles and change frequencies.
Why D is wrong: D is wrong because random mating is a requirement FOR equilibrium — non-random mating (assortative mating) is the disrupting factor.
A researcher finds that genotype frequencies in a population do not match Hardy-Weinberg predictions. She then discovers the population has been experiencing immigration from a neighbouring population. Which Hardy-Weinberg condition has been violated?
Answer: C. Immigration introduces alleles from another population, which is gene flow (migration). This violates the "no gene flow" condition of Hardy-Weinberg equilibrium (NCERT Class 12 Biology Chapter 6, page 121).
Why A is wrong: A is wrong because mutation refers to the origin of new alleles within the population's own DNA, not the introduction of alleles from another population through migration.
Why B is wrong: B is wrong because genetic drift refers to random fluctuations in allele frequency due to small population size, not the directional introduction of alleles from immigrants.
Why D is wrong: D is wrong because random mating concerns mate choice within the population (assortative vs random), not the arrival of individuals from outside (trap: confusing gene flow with non-random mating).
In a large randomly mating population in Hardy-Weinberg equilibrium, 16% of individuals show the recessive phenotype. What percentage of the population is expected to be carriers (heterozygous)?
Answer: B. Recessive phenotype frequency = q² = 0.16, so q = 0.4. Then p = 1 − 0.4 = 0.6. Carrier (heterozygote) frequency = 2pq = 2 × 0.6 × 0.4 = 0.48 = 48% (NCERT Class 12 Biology Chapter 6, page 121).
Why A is wrong: A is wrong because 24% equals pq without the factor of 2 — heterozygote frequency is 2pq, not pq (trap: forgetting the 2 in the heterozygote term).
Why C is wrong: C is wrong because 36% equals p² = 0.6² — that is the homozygous dominant (AA) frequency, not the carrier frequency.
Why D is wrong: D is wrong because 32% does not correspond to any standard Hardy-Weinberg term for these allele frequencies. This may result from an arithmetic error such as computing 2 × 0.4 × 0.4 = 0.32, confusing p and q.
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Given
In a population in Hardy-Weinberg equilibrium, 4% of individuals are affected by a certain autosomal recessive condition (genotype aa).
Required
Find (a) the frequency of carriers (Aa), and (b) the frequency of unaffected homozygous dominant individuals (AA).
Concept
Hardy-Weinberg principle: in an ideal population, genotype frequencies are determined by allele frequencies via p² + 2pq + q² = 1.
Formula
• p + q = 1• p² + 2pq + q² = 1
Substitution
• q² = 0.04 (given: 4% show recessive phenotype)• q = √0.04 = 0.2• p = 1 − 0.2 = 0.8
Calculation
• Carrier frequency: 2pq = 2 × 0.8 × 0.2 = 0.32 = 32%• Homozygous dominant frequency: p² = 0.8² = 0.64 = 64%• Verification: p² + 2pq + q² = 0.64 + 0.32 + 0.04 = 1.00 ✓
Final answer
Carrier frequency (Aa) = 32%. Homozygous dominant frequency (AA) = 64%.
Note: the integers 2 in "2pq" and the value 1 in "p + q = 1" are exact mathematical constants and do not affect significant-figure reasoning.
Common trap
Computing p² + q² = 0.64 + 0.04 = 0.68 and calling the remaining 0.32 "other" without recognising it as the heterozygote 2pq. Or, worse, writing p² + q² = 1, which gives p = √0.96 ≈ 0.98 — a completely wrong allele frequency because the heterozygote term was omitted.
Similar NEET-style question
"In a Hardy-Weinberg population, the frequency of allele 'a' is 0.3. What fraction of the population is heterozygous?" (Answer: 2 × 0.7 × 0.3 = 0.42 = 42%.)
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p² + 2pq + q² = 1 (genotype frequency); p + q = 1 (allele frequency). Disequilibrium → evolution. Five forces: mutation, gene flow (migration), genetic drift, natural selection, non-random mating.
-- NCERT Class 12 Biology, Ch. 6, p. 121In an idealised population (no mutation, drift, selection, gene flow, random mating), allele and genotype frequencies remain constant.
| Symbol | Quantity | SI Unit |
|---|---|---|
| p | freq of dominant allele A | - |
| q | freq of recessive allele a | - |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Negative Marking
Five forces disturb HW: mutation, gene flow, drift, selection, non-random mating. ANY of these violates equilibrium.
Question asks which factor maintains/disturbs HW.
Random mating + no other forces → equilibrium. ANY of mutation/migration/drift/selection/assortative mating → disequilibrium.
Root cause: formula misuse
Genotype frequencies: p² (AA) + 2pq (Aa) + q² (aa) = 1. Allele frequencies: p + q = 1. Don't drop 2pq.
More in Genetics and Evolution: 19 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
2 questions from NEET 2024, 2026. Answers verified against NTA official keys.
Which one of the following factors will not affect the Hardy-Weinberg equilibrium?
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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