OCR · A-Level · Biology

    Course version: H420 · Version from Sept 2025

    (f) the use of the Hardy–Weinberg principle to calculate allele…

    Full specification wording

    Patterns of inheritance: (f) the use of the Hardy–Weinberg principle to calculate allele frequencies in populations

    The Hardy–Weinberg principle models allele and genotype frequencies in an idealised population that is large, randomly mating, with no mutation, no migration and no selection.

    Checked against the OCR H420 specification (opens in a new tab) on .

    • Point 8 of 10
    • 4 key terms

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    The Hardy–Weinberg principle models allele and genotype frequencies in an idealised population that is large, randomly mating, with no mutation, no migration and no selection. For a gene with two alleles, let the dominant allele frequency be p and the recessive allele frequency be q, so p + q = 1. Genotype frequencies are then p² (homozygous dominant), 2pq (heterozygous) and q² (homozygous recessive), summing to 1. To calculate allele frequencies, start from the observable recessive phenotype frequency, which equals q²; take its square root to obtain q, then p = 1 − q. For example, if 9% of a population shows the recessive phenotype, q² = 0.09, so q = 0.3 and p = 0.7. Allele frequencies are proportions of the gene pool, not counts of individuals, and results are estimates valid only while the assumptions hold.

    What earns the marks

    • States that for two alleles p + q = 1, where p and q are the frequencies of the dominant and recessive alleles.
    • States that genotype frequencies are p² for homozygous dominant, 2pq for heterozygous and q² for homozygous recessive, summing to 1.
    • Identifies that the frequency of the recessive phenotype equals q², so q is the square root of that frequency.
    • Calculates p as 1 − q after finding q, and can convert a proportion to a percentage or vice versa.
    • Recognises that the calculation gives estimated allele frequencies and depends on assumptions such as a large population, random mating and no selection, mutation or migration.

    Common mistakes and how to put them right

    • Treating the recessive phenotype frequency directly as q rather than q².

      Put it right: Take the square root first.

    • Assuming the dominant phenotype frequency equals p² when it actually equals p² + 2pq.

      Put it right: Use the recessive phenotype to find q².

    • Reporting allele frequencies as whole numbers of individuals.

      Put it right: Frequencies are proportions of the gene pool between 0 and 1 (or percentages).

    • Forgetting that p + q must equal 1 and producing values that do not sum correctly.

      Put it right: Check the sum after calculating.

    Exam tips

    • Write down p + q = 1 and p² + 2pq + q² = 1 before substituting numbers, so the algebra is visible.
    • Identify which phenotype frequency is given and decide whether it corresponds to q² or to p² + 2pq before calculating.
    • Keep frequencies as decimals through the working and only convert to a percentage at the end, checking that p + q = 1.

    Try the lesson

    The start of the Patterns of inheritance: (f) the use of the Hardy–Weinberg principle to calculate allele frequencies in populations lesson, made from checked teaching notes for this specification point.

    Have a go first

    In a Hardy–Weinberg population the dominant allele frequency p is 0.6. What is the frequency of the recessive allele q?

    Listen

    The Hardy–Weinberg principle models allele and genotype frequencies in an idealised population that is large, randomly mating, with no mutation, no migration and no selection. For a gene with two alleles, let the dominant allele frequency be p and the recessive allele frequency be q, so p + q = 1. Genotype frequencies are then p² (homozygous dominant), 2pq (heterozygous) and q² (homozygous recessive), summing to 1. To calculate allele frequencies, start from the observable recessive phenotype frequency, which equals q²; take its square root to obtain q, then p = 1 − q. For example, if 9% of a population shows the recessive phenotype, q² = 0.09, so q = 0.3 and p = 0.7. Allele frequencies are proportions of the gene pool, not counts of individuals, and results are estimates valid only while the assumptions hold. A strong answer earns marks for these points. States that for two alleles p + q = 1, where p and q are the frequencies of the dominant and recessive alleles. States that genotype frequencies are p² for homozygous dominant, 2pq for heterozygous and q² for homozygous recessive, summing to 1. Identifies that the frequency of the recessive phenotype equals q², so q is the square root of that frequency. Calculates p as 1 − q after finding q, and can convert a proportion to a percentage or vice versa. A common slip: Treating the recessive phenotype frequency directly as q rather than q². Instead, take the square root first. Exam tip: Write down p + q = 1 and p² + 2pq + q² = 1 before substituting numbers, so the algebra is visible.

    See it

    A flowchart running from top to bottom. The first box states that the recessive phenotype frequency equals q squared. An arrow leads to a box saying to take the square root to find q. An arrow leads to a box giving p equals 1 minus q. An arrow leads to a box saying to check that p plus q equals 1. An arrow leads to a box listing the genotype frequencies: p squared for homozygous dominant, 2pq for heterozygous and q squared for homozygous recessive. A final arrow leads to a box stating that p squared plus 2pq plus q squared equals 1.
    Shows the steps for using the Hardy–Weinberg principle to go from the recessive phenotype frequency to q, then p, and on to the genotype frequencies.

    Put it in order

    Put these steps in the order they happen.

    1. 1. p = 1 − q
    2. 2. Take square root to find q
    3. 3. Recessive phenotype frequency = q²
    4. 4. p² + 2pq + q² = 1
    5. 5. Genotype frequencies: p² homozygous dominant, 2pq heterozygous, q² homozygous recessive
    6. 6. Check p + q = 1

    Key words

    Say what each word means before you turn the card.

    Hardy–Weinberg principle

    Try answering aloud or in your head before turning the card.

    Allele frequencies

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    Genotype frequencies

    Try answering aloud or in your head before turning the card.

    Recessive phenotype frequency

    Try answering aloud or in your head before turning the card.

    See every definition
    Hardy–Weinberg principle
    Models allele and genotype frequencies in an idealised population that is large, randomly mating, with no mutation, no migration and no selection.
    Allele frequencies
    Proportions of the gene pool, where for two alleles p + q = 1, with p the dominant and q the recessive allele frequency.
    Genotype frequencies
    p² for homozygous dominant, 2pq for heterozygous and q² for homozygous recessive, summing to 1.
    Recessive phenotype frequency
    Equals q², so q is the square root of that frequency and p is then found as 1 − q.

    Label the diagram

    Some boxes have lost their names. Pick the right name for each number.

    The diagram again, with 3 boxes numbered instead of named. A flowchart running from top to bottom. The first box states that the recessive phenotype frequency equals q squared. An arrow leads to a box saying to take the square root to find q. An arrow leads to a box giving p equals 1 minus q. An arrow leads to a box saying to check that p plus q equals 1. An arrow leads to a box listing the genotype frequencies: p squared for homozygous dominant, 2pq for heterozygous and q squared for homozygous recessive. A final arrow leads to a box stating that p squared plus 2pq plus q squared equals 1.

    • 1Box 1

    • 2Box 2

    • 3Box 3

    0 of 3 labelled

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