NCERT grounding
NCERT Class 12 Biology, Chapter 6 (Evolution), §6.7, fixes the syllabus boundary. The book states that allele frequencies in a population are stable and constant from generation to generation, calls this state genetic equilibrium, and identifies the gene pool as the total of all genes and their alleles in a population. NIOS Lesson 1, §1.2.5, supplements this with the term panmictic population — a sexually reproducing population in which mating is random — and credits the principle to G. H. Hardy and W. Weinberg working independently in 1908.
"Allele frequencies in a population are stable and constant from generation to generation. The gene pool remains a constant. This is called genetic equilibrium."
NCERT Class 12, Chapter 6, §6.7
The principle and its equation
Consider one autosomal locus in a sexually reproducing diploid population with two alleles: A (dominant) at frequency p and a (recessive) at frequency q. Because every allele in the population must be either A or a, the two frequencies must sum to one:
Allele-frequency identity
The sum of all allelic frequencies at a locus is always 1. For two alleles this gives q = 1 − p, a step NEET problems demand before squaring.
Now ask: under random mating, with what probability do two A alleles meet in a zygote? Each gamete carries an A with probability p. Sampling two gametes independently, the probability of an AA zygote is p × p = p². The probability of an aa zygote is q × q = q². A heterozygote Aa can arise in two ways — an A egg meeting an a sperm (p × q) or an a egg meeting an A sperm (q × p) — giving a combined probability of 2pq. The three genotype frequencies must again exhaust the population:
Hardy–Weinberg expansion
The genotype frequencies are simply the binomial expansion of (p + q)². NCERT highlights this explicitly: "This is a binomial expansion of (p + q)²."
Figure 1. Random union of gametes treats A and a as samples drawn with probabilities p and q. The four cells partition the offspring gene pool into the three genotype classes, recovering Hardy–Weinberg's algebra.
Reading the equation backwards
Most NEET problems give you one observable — typically the frequency of the recessive phenotype — and demand the rest. Because aa is the only genotype that displays the recessive phenotype (AA and Aa share the dominant phenotype), the trick is to set the observed recessive frequency equal to q², take the square root to recover q, and then back-fill p, p², and 2pq. This single workflow handles the 2019 PYQ, the standard sickle-cell carrier estimate, and any "X% albinos — what fraction are carriers?" question the examiner can write.
Five assumptions of equilibrium
Hardy–Weinberg's prediction holds only when an idealised set of conditions is met. Memorise these as a checklist — every NEET trap is built by flipping exactly one of them.
Large population
Effectively infinite size so sampling chance is negligible.
Violation → genetic drift.
Random mating
Population is panmictic — no genotype-based mate choice.
Violation → assortative mating skews 2pq.
No mutation
No new alleles A→a or a→A created.
Violation → mutation adds raw variation.
No gene flow
No migration in or out of the population.
Violation → allele frequencies shift in both donor and recipient.
No selection
All genotypes have equal fitness; survival and reproduction are independent of genotype.
Violation → favoured alleles rise in frequency.
When all five assumptions are met simultaneously, the gene pool is locked in place and the population is, by definition, not evolving. This is the inverted way population geneticists define evolution: any allele-frequency change across generations is microevolution. NCERT writes it bluntly — "Disturbance in genetic equilibrium, or Hardy–Weinberg equilibrium, i.e. change of frequency of alleles in a population would then be interpreted as resulting in evolution."
Five factors that disturb equilibrium
NCERT names five forces capable of changing allele frequencies, and the NEET 2024 paper directly asked students to identify the one that does not disturb equilibrium. Lock the list down.
- 01
Gene migration / gene flow
Movement of individuals between populations transfers alleles. Repeated migration = gene flow.
- 02
Genetic drift
Random change in allele frequency in small populations. Sub-cases: founder effect, bottleneck effect.
- 03
Mutation
Heritable change in DNA introduces a new allele. Rate ≈ 10⁻⁵–10⁻⁸ per locus per generation.
- 04
Genetic recombination
Crossing-over during gametogenesis shuffles existing alleles into new combinations.
- 05
Natural selection
Differential reproductive success of genotypes. Outcomes: stabilising, directional, disruptive.
How each factor moves the algebra
Genetic drift
Random
non-directional
- Allele-frequency change due to sampling error.
- Strongest in small populations.
- Can fix or lose alleles even if neutral.
- Founder & bottleneck effects are special cases.
Natural selection
Directional
fitness-driven
- Allele-frequency change due to differential fitness.
- Operates in populations of any size.
- Tends to increase fit alleles, decrease unfit ones.
- Three modes: stabilising, directional, disruptive.
Genetic drift: founder & bottleneck
Two well-named sub-cases of drift sit on the high-yield list. NEET 2021 directly asked which factor produces the founder effect — the answer is genetic drift, not selection or recombination.
Figure 2. Both founder and bottleneck effects are forms of genetic drift, but the trigger differs. Founder effect — a small group leaves the parent population. Bottleneck effect — a large population is reduced in place. In both, the post-event gene pool is a non-random sample of the original.
NCERT closes the discussion with a striking line: "Sometimes the change in allele frequency is so different in the new sample of population that they become a different species. The original drifted population becomes founders and the effect is called founder effect." Drift can therefore, in principle, drive speciation without any role for selection — a key point because students reflexively credit speciation to natural selection alone.
Worked examples
Q. A gene locus has two alleles A and a. If the frequency of dominant allele A is 0.4, what are the frequencies of homozygous dominant, heterozygous and homozygous recessive individuals?
Solution. Given p(A) = 0.4, so q(a) = 1 − 0.4 = 0.6. Apply p² + 2pq + q² = 1. AA frequency = p² = (0.4)² = 0.16. Aa frequency = 2pq = 2 × 0.4 × 0.6 = 0.48. aa frequency = q² = (0.6)² = 0.36. Check: 0.16 + 0.48 + 0.36 = 1.00. This is the exact algebra of NEET 2019 Q.61.
Q. In a population of 10,000, 1% express a recessive phenotype (e.g. an autosomal recessive disorder). Assuming Hardy–Weinberg equilibrium, how many heterozygous carriers exist?
Solution. The recessive phenotype is shown only by aa, so q² = 0.01. Therefore q = √0.01 = 0.1, and p = 1 − 0.1 = 0.9. Carrier (heterozygote) frequency = 2pq = 2 × 0.9 × 0.1 = 0.18. In a population of 10,000, expected carriers = 0.18 × 10,000 = 1,800 individuals. Notice the asymmetry: 100 affected, but eighteen times that number silently carry the allele — the standard "hidden reservoir" argument used in medical-genetics MCQs.
Q. A bird population is in Hardy–Weinberg equilibrium for a wing-colour locus with q = 0.3. A storm wipes out 80% of the population at random. Predict the new q immediately after the storm and explain.
Solution. If the kill is genuinely random with respect to genotype, the expected q remains 0.3 — selection has not acted. But because the population has been bottlenecked to a small size, the observed q in the survivors will deviate from 0.3 by sampling chance. This is the bottleneck effect, a form of genetic drift. The smaller the surviving population, the larger the expected deviation — and the alleles that happen to be over-represented may even be neutral or mildly deleterious.
Q. In a Hardy–Weinberg population the frequency of heterozygotes is 0.42. What is the maximum possible value of 2pq, and at what allele frequencies is it attained?
Solution. The function 2pq with p + q = 1 reaches its maximum when p = q = 0.5, giving 2pq = 2 × 0.5 × 0.5 = 0.5. So at most half the population can be heterozygous at a two-allele locus. The given 0.42 is achievable; setting 2p(1 − p) = 0.42 gives p² − p + 0.21 = 0, with p ≈ 0.7 or p ≈ 0.3 — both valid solutions, mirror images of each other.