Read the core explanations from The Hardy-Weinberg Principle before testing yourself from memory.
1. Why Allele Frequencies Matter
Here is a question that sits at the heart of evolutionary biology: if a dominant allele gives an advantage, why does the recessive allele not disappear over time? The Hardy-Weinberg principle gives us the mathematical answer — and it turns out that, under the right conditions, allele frequencies stay perfectly stable from one generation to the next.
This stability is not obvious. Godfrey Hardy and Wilhelm Weinberg independently worked it out in 1908, and their model remains one of the most powerful tools in population genetics. At A Level, you need to know both why the model works and how to use it to calculate allele, genotype, and phenotype frequencies.
2. The Conditions for Hardy-Weinberg Equilibrium
The principle only holds when a population meets a strict set of conditions. Think of these as the assumptions baked into the model — if any one of them is violated, allele frequencies will shift, and evolution is occurring.
- The population is very large (large enough that random sampling errors — genetic drift — are negligible)
- Mating is random (no sexual selection or mate preference based on genotype)
- There is no mutation (no new alleles are being introduced)
- There is no migration (no individuals entering or leaving the population, which would alter allele frequencies)
- There is no natural selection (all genotypes have equal survival and reproductive success — no allele confers an advantage)
No real population satisfies all five conditions perfectly. That is precisely what makes the principle useful: when you measure a real population and find its allele frequencies do not match the Hardy-Weinberg prediction, you have evidence that at least one of these forces — selection, drift, migration, or mutation — is acting on that gene.
The conditions collectively describe a population where chance and choice play no role in which alleles are passed on. When those conditions hold, the model predicts that allele frequencies will remain unchanged indefinitely — this stable state is called Hardy-Weinberg equilibrium.
3. The Hardy-Weinberg Equations
The principle uses two linked equations. The first tracks allele frequencies; the second translates those into genotype frequencies. You need both to solve exam questions.
Notice that $p^2 + 2pq + q^2 = 1$ is the expansion of $(p + q)^2 = 1$. The three terms represent the three possible genotypes for a gene with two alleles. Because every individual must have one of these genotypes, the three frequencies must sum to 1.
The term $2pq$ appears because heterozygotes can arise in two ways: inheriting the dominant allele from the mother and the recessive from the father, or vice versa. That factor of 2 is easy to forget under exam pressure — keep it in mind.
4. Where the Equations Come From — The Logic Behind the Maths
Before working through calculations, it helps to see why these equations are correct. Picture a very large gene pool. The frequency of allele A is $p$ and the frequency of allele a is $q$. When random mating occurs, each parent contributes one allele to an offspring at random.
The probability that an offspring receives A from both parents is $p \times p = p^2$. The probability of receiving a from both parents is $q \times q = q^2$. The probability of receiving A from one parent and a from the other is $p \times q + q \times p = 2pq$. Add all three together and you get $p^2 + 2pq + q^2 = 1$ — every individual is accounted for.
This derivation shows that the equation is not an empirical observation but a mathematical certainty, given random mating. Understanding this logic helps you justify the model in evaluation questions, which are common at A Level.
5. Worked Calculation — Finding Allele and Genotype Frequencies
Exam questions almost always give you one piece of information — usually the frequency of the recessive phenotype — and ask you to work backwards to find allele frequencies, then forwards to find genotype or heterozygote frequencies. The entry point is always $q^2$, because homozygous recessive individuals are the only group whose genotype you can identify from phenotype alone.
Worked example: Hardy-Weinberg Calculation: Cystic Fibrosis Carrier Frequency
In a population of 10,000 people, 100 individuals are affected by cystic fibrosis (homozygous recessive, genotype aa). Calculate: (a) the frequency of the recessive allele $q$; (b) the frequency of the dominant allele $p$; (c) the expected number of carriers (heterozygotes, Aa) in the population. Show all working. [4 marks]
- $q^2 = \frac{\text{number affected}}{\text{total population}} = \frac{100}{10000} = 0.01$ — The frequency of the homozygous recessive phenotype equals $q^2$. This is always your starting point. [M1]
- $q = \sqrt{q^2} = \sqrt{0.01} = 0.1$ — Check: $0.1 \times 0.1 = 0.01$ — confirmed. $q = 0.1$ means 10% of alleles at this locus are the recessive allele. [A1]
- $p = 1 - q = 1 - 0.1 = 0.9$ — The two allele frequencies must sum to 1. [M1]
- $2pq = 2 \times 0.9 \times 0.1 = 0.18$ — 18% of individuals in the population are expected to be carriers.
- $0.18 \times 10000 = 1800 \text{ individuals}$ — Multiply the frequency by the total population size to get the expected number of carriers. [A1]
Answer: (a) $q = 0.1$; (b) $p = 0.9$; (c) 1800 carriers expected in the population.
Notice how many carriers there are compared to affected individuals: 1800 versus 100. This is a key biological insight from the Hardy-Weinberg model — recessive alleles are often far more common in heterozygous carriers than in affected homozygotes, which is why recessive conditions can persist in populations at relatively stable frequencies across generations.
6. Interpreting Deviations from Hardy-Weinberg Equilibrium
The real power of the Hardy-Weinberg model is not just in calculating frequencies — it is in using those predictions as a baseline. If you measure allele frequencies in a real population across several generations and find they are changing, that deviation is direct evidence that one or more of the five conditions is being violated.
For example, if the frequency of a recessive allele is declining faster than the model predicts, natural selection against the homozygous recessive phenotype is a likely explanation. If allele frequencies fluctuate randomly in a small isolated population, genetic drift is the probable cause. The model gives you the null hypothesis; the data tells you whether evolution is happening.
At A Level, evaluation questions often ask you to assess whether a population is likely to be in Hardy-Weinberg equilibrium. Your answer should address the specific conditions, link each to the biological context given, and reach a justified conclusion — not list the conditions in the abstract.
7. A Second Worked Example — Working from Phenotype Ratios
Some exam questions give you phenotype counts rather than a simple frequency. The method is identical, but you must calculate $q^2$ yourself from the raw numbers before proceeding. Here is a typical example.
Worked example: Hardy-Weinberg Calculation: Coat Colour in a Mouse Population
In a population of 400 mice, coat colour is controlled by a single gene. Black coat (B) is dominant over white coat (b). 64 mice have white coats. Calculate the expected frequency of heterozygous black mice (Bb) in this population. [3 marks]
- $q^2 = \frac{64}{400} = 0.16$ — White coat mice are homozygous recessive (bb), so their frequency equals $q^2$. [M1]
- $q = \sqrt{0.16} = 0.4$ — Check: $0.4 \times 0.4 = 0.16$ — confirmed.
- $p = 1 - q = 1 - 0.4 = 0.6$ — Using $p + q = 1$.
- $2pq = 2 \times 0.6 \times 0.4 = 0.48$ — 48% of the population are expected to be heterozygous (Bb). [A1]
- $p^2 + 2pq + q^2 = (0.6)^2 + 0.48 + (0.4)^2 = 0.36 + 0.48 + 0.16 = 1.00$ — Check: $0.6 \times 0.6 = 0.36$ and $0.4 \times 0.4 = 0.16$. Total $= 1.00$ — the calculation is internally consistent. [A1]
Answer: The expected frequency of heterozygous black mice (Bb) is 0.48, meaning 48% of the population.
8. Exam Technique and Common Mistakes
Hardy-Weinberg questions are reliable mark-earners if you follow a consistent method. The most common reason students lose marks is not misunderstanding the biology — it is making arithmetic errors or using the wrong term as the starting point. The guidance below addresses the mistakes that appear most frequently in mark schemes.
The Hardy-Weinberg principle is, at its core, a null model for evolution: it describes what a population looks like when nothing is changing. Every spec requirement — the conditions, the two equations, and the interpretation of deviations — connects back to that central idea. Master the calculation method, understand the logic behind the equations, and you will be equipped to handle both the numerical and the evaluative questions this topic generates.