Allele frequencies in population equilibrium. Five conditions required.

Hardy-Weinberg Equation — p² + 2pq + q² = 1 Explained

Question

A population of 10,000 individuals has 9% showing a recessive phenotype (homozygous recessive). Assuming Hardy-Weinberg equilibrium, what are the allele frequencies and the expected number of heterozygous carriers?

Solution — Step by Step

The recessive phenotype means individuals are homozygous recessive, so their frequency is $q^2$ .

q^2 = 0.09

We always start here because $q^2$ is the only genotype we can directly read from phenotype data for recessive traits.

q = \sqrt{0.09} = 0.3

Since $p + q = 1$ , we get $p = 1 - 0.3 = \mathbf{0.7}$ .

Here $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele.

Using $p^2 + 2pq + q^2 = 1$ :

Genotype	Formula	Value
Homozygous dominant (AA)	$p^2 = (0.7)^2$	0.49
Heterozygous carriers (Aa)	$2pq = 2(0.7)(0.3)$	0.42
Homozygous recessive (aa)	$q^2 = (0.3)^2$	0.09

Quick check: $0.49 + 0.42 + 0.09 = 1.00$ ✓

Heterozygous carriers = $2pq \times N = 0.42 \times 10{,}000 = \mathbf{4200}$

This is the expected number of carriers — individuals who carry the recessive allele but don’t express the recessive phenotype.

Why This Works

The Hardy-Weinberg equation comes directly from expanding $(p + q)^2 = 1$ . Think of it as a Punnett square for an entire population — if allele frequencies are $p$ and $q$ , random mating produces genotypes in exact $p^2 : 2pq : q^2$ proportions.

The equation holds only when five conditions are met: large population size, random mating, no mutation, no natural selection, and no gene flow (no migration). These conditions define an ideal population — real populations deviate, and that deviation is what drives evolution.

In NEET, the five conditions appear as a direct 1-mark question almost every year. Memorise them as: Large Mating No Selection No Migration — LMNSM. No shortcuts needed, just five concepts.

For board exams, understanding what violates equilibrium is just as important as knowing the conditions. A small island population (violates large size), selective mating (violates random mating), antibiotic resistance spreading (selection) — all of these shift allele frequencies and break the equilibrium.

Alternative Method

If a question gives you carrier frequency (2pq) instead of recessive phenotype frequency, work backwards:

Suppose 42% are carriers → $2pq = 0.42$

We know $p + q = 1$ , so $p = 1 - q$ . Substituting:

2(1-q)q = 0.42

2q - 2q^2 = 0.42

2q^2 - 2q + 0.42 = 0

q^2 - q + 0.21 = 0

Using the quadratic formula, $q = 0.3$ , which gives the same answer. This approach appeared in NEET 2024 to test whether students can work the equation in both directions.

Common Mistake

Most students write $q = 0.09$ directly from “9% show recessive phenotype.” That is wrong. The 9% is $q^2$ (genotype frequency), not $q$ (allele frequency). You must take the square root: $q = \sqrt{0.09} = 0.3$ . Forgetting to square root is the single most common error in this topic — it appeared in NEET 2024 answer key discussions as the most-selected wrong option.

A second trap: some questions ask for allele frequency and others ask for genotype frequency. Read the question twice. Allele frequency = $p$ or $q$ . Genotype frequency = $p^2$ , $2pq$ , or $q^2$ . Mixing these up costs the mark even if all your calculations are correct.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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