Hardy–Weinberg Principle — Genetic Equilibrium & the Equation — Class 12 Biology Notes

Frequencies That Stay Put

In any population you can measure how common each version of a gene is. For a gene with two alleles, some fraction of all the copies in the population will be one allele and the rest the other; these fractions are called allele frequencies. The interesting claim, worked out independently by Hardy and Weinberg, is that these frequencies do not have to change on their own. Left undisturbed, they stay stable and constant from one generation to the next.

What this really says is that the gene pool — the total collection of all the genes and their alleles in the population — remains constant over time. When a population holds its allele frequencies steady like this, it is said to be in genetic equilibrium. Hardy and Weinberg expressed this idea not in words but in a compact set of algebraic equations, and those equations are what make the principle so useful.

Naming the Frequencies

Start with the simplest case: a single gene that has just two alleles, a dominant $A$ and a recessive $a$ . Give the frequency of $A$ the symbol $p$ and the frequency of $a$ the symbol $q$ . Since these are the only two alleles, together they must account for everything, so their frequencies add up to one:

$p + q = 1$

More generally, the sum of all the allele frequencies for a gene is always 1. That single fact is the anchor for everything that follows — if you know one frequency, you immediately know the other, because $q = 1 - p$ .

The Genotype Equation

Alleles come together in pairs to make genotypes, and the frequencies of those genotypes follow from $p$ and $q$ by simple probability. The chance that an allele $A$ (frequency $p$ ) turns up on both chromosomes of a diploid individual is the product of the probabilities, $p \times p$ , so the frequency of $AA$ individuals is $p^2$ . In the same way the frequency of $aa$ individuals is $q^2$ , and the frequency of heterozygous $Aa$ individuals is $2pq$ — the factor of 2 appears because the $A$ can come from either parent.

Since every individual must be one of these three genotypes, the three frequencies add up to one:

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

This is nothing more than the binomial expansion of $(p+q)^2$ . Because $p + q = 1$ , squaring both sides gives $(p+q)^2 = 1$ , which is exactly the equation above. Two small formulas, $p + q = 1$ and $p^2 + 2pq + q^2 = 1$ , capture the whole principle.

Reading the Numbers

The equations become powerful once you feed real data into them. Usually the easiest thing to count in a population is the recessive phenotype, the $aa$ individuals, because that frequency equals $q^2$ directly. Take the square root to get $q$ , subtract from one to get $p$ , and every other frequency follows.

Suppose a recessive condition shows up in 1 out of every 100 people, so the frequency of $aa$ is $q^2 = 0.01$ . Then $q = 0.1$ and $p = 1 - 0.1 = 0.9$ . The frequency of unaffected carriers, the heterozygotes, is $2pq = 2 \times 0.9 \times 0.1 = 0.18$ — that is, 18 out of every 100 people carry the allele without showing the condition. The frequency of homozygous dominants is $p^2 = 0.81$ . As a check, $0.81 + 0.18 + 0.01 = 1$ , exactly as the equation demands.

When the Equation Fails — a Sign of Evolution

The real payoff is what happens when the numbers do not fit. The Hardy–Weinberg equation predicts the genotype frequencies you should see if a population is genuinely at equilibrium. If you go out, measure the actual frequencies, and find they differ from those expected values, that difference is telling you something: the population is not at equilibrium.

Since equilibrium means allele frequencies are holding constant, a departure from it means allele frequencies are changing — and a change in allele frequency across generations is exactly what evolution is. So the size and direction of the gap between the measured frequency and the expected frequency indicate the extent of evolutionary change taking place. A disturbance of Hardy–Weinberg equilibrium is read as evolution in action.

Quick Recap

Hardy–Weinberg principle: allele frequencies in a population are stable and constant from generation to generation; the gene pool stays constant. This steady state is genetic equilibrium.
The sum of all allele frequencies for a gene is 1. For two alleles, $p + q = 1$ , where $p$ is the frequency of $A$ and $q$ of $a$ .
Genotype frequencies: $AA = p^2$ , $aa = q^2$ , $Aa = 2pq$ , and $p^2 + 2pq + q^2 = 1$ — the binomial expansion of $(p+q)^2$ .
To use it: the recessive phenotype gives $q^2$ ; take the root for $q$ , then $p = 1 - q$ , and carriers = $2pq$ .
When measured frequencies differ from the expected equilibrium values, allele frequencies are changing — the difference signals that evolution is occurring.

Solved Examples — Section 10

Q1. State the Hardy–Weinberg principle in one sentence.

Answer: In a population that is undisturbed, allele frequencies remain stable and constant from generation to generation, so the gene pool stays constant — a state called genetic equilibrium.

Q2. For a gene with two alleles of frequency $p$ and $q$ , write the two equations of the principle and explain each term.

Answer: $p + q = 1$ says the two allele frequencies add to one; $p^2 + 2pq + q^2 = 1$ gives the genotype frequencies, where $p^2$ is $AA$ , $2pq$ is $Aa$ and $q^2$ is $aa$ . The second is the binomial expansion of $(p+q)^2$ .

Q3. In a population the frequency of the recessive allele $a$ is $q = 0.2$ . Find the frequencies of $AA$ , $Aa$ and $aa$ .

Answer: $p = 1 - 0.2 = 0.8$ . So $AA = p^2 = 0.64$ , $Aa = 2pq = 2 \times 0.8 \times 0.2 = 0.32$ , and $aa = q^2 = 0.04$ . They add to $0.64 + 0.32 + 0.04 = 1$ .

Q4. A recessive disorder affects 4 people in every 100. What fraction of the population are unaffected carriers?

Answer: The affected are $aa$ , so $q^2 = 0.04$ , giving $q = 0.2$ and $p = 0.8$ . Carriers are the heterozygotes, $2pq = 2 \times 0.8 \times 0.2 = 0.32$ — that is, 32 in every 100.

Q5. Why is the term $2pq$ multiplied by 2 while $p^2$ and $q^2$ are not?

Answer: A heterozygote can receive the $A$ from one parent and the $a$ from the other, or the reverse, so there are two ways to form $Aa$ ; the homozygotes $AA$ and $aa$ can each form in only one way.

Q6. How does the Hardy–Weinberg equation let us detect that evolution is happening?

Answer: The equation predicts the genotype frequencies expected at equilibrium. If the frequencies actually measured differ from these expected values, allele frequencies are changing — and a change in allele frequency across generations is evolution.