AQA · A-Level · Biology
Course version: 7402 · Version from Sept 2016
The Hardy–Weinberg principle provides a mathematical model, which predicts…
Full specification wording
Populations (A-level only): The Hardy–Weinberg principle provides a mathematical model, which predicts that allele frequencies will not change from generation to generation.
The Hardy-Weinberg principle is a mathematical model, not a law of nature.
Checked against the AQA 7402 specification (opens in a new tab) on .
- Point 4 of 5
- 2 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
The Hardy-Weinberg principle is a mathematical model, not a law of nature. It predicts that in a population meeting a set of conditions the allele frequencies stay the same from one generation to the next, so no evolution takes place. Two equations carry it: p + q = 1 for the frequencies of a dominant and a recessive allele, and p squared + 2pq + q squared = 1 for the genotype frequencies, where p squared is the homozygous dominant, 2pq the heterozygous and q squared the homozygous recessive. Because only the homozygous recessive phenotype reveals its own genotype, almost every calculation begins by setting q squared equal to the proportion showing the recessive characteristic, taking the square root to find q, then finding p as 1 - q. Its real value is as a null model.
What earns the marks
- Stating that allele frequencies remain constant from generation to generation when the conditions of the model are met.
- Using q squared as the frequency of the homozygous recessive phenotype to find q, then p = 1 - q.
- Calculating 2pq as the frequency of heterozygotes from p and q rather than measuring it directly.
- Explaining that the principle does not hold when any of its conditions is broken, for example when individuals carrying an allele are selected against, so allele frequency changes.
- Listing the conditions required for the principle to hold: large population, random mating, no migration, no mutation and no selection.
Common mistakes and how to put them right
Giving q when the question asks for q squared.
Put it right: Check the symbol and the units requested.
Forgetting to take the square root, or taking the square root of the wrong quantity.
Put it right: q squared is the homozygous recessive frequency, so take the square root to find q.
Treating everyone showing the dominant characteristic as p squared, when that group also contains the heterozygotes, 2pq.
Put it right: p squared is only the homozygous dominant.
Giving the frequency of carriers as pq rather than 2pq.
Put it right: Heterozygotes are represented by 2pq.
Saying the principle is simply wrong for a real population instead of naming the condition that has been broken.
Put it right: Identify which condition, such as selection or non-random mating, is violated.
Exam tips
- Write down which value the question has given you, and which symbol it belongs to, before you calculate anything.
- When asked whether the principle would hold, answer yes or no and then justify your answer by referring to the relevant conditions of the model.
- Keep frequencies as decimals to at least three places until the final answer, then round as the question instructs.
Try the lesson
The start of the Populations (A-level only): The Hardy–Weinberg principle provides a mathematical model, which predicts that allele frequencies will not change from generation to generation. lesson, made from checked teaching notes for this specification point.
Have a go first
In the Hardy–Weinberg equation p + q = 1, what does q represent?
Listen
The Hardy-Weinberg principle is a mathematical model, not a law of nature. It predicts that in a population meeting a set of conditions the allele frequencies stay the same from one generation to the next, so no evolution takes place. Two equations carry it: p + q = 1 for the frequencies of a dominant and a recessive allele, and p squared + 2pq + q squared = 1 for the genotype frequencies, where p squared is the homozygous dominant, 2pq the heterozygous and q squared the homozygous recessive. Because only the homozygous recessive phenotype reveals its own genotype, almost every calculation begins by setting q squared equal to the proportion showing the recessive characteristic, taking the square root to find q, then finding p as 1 - q. Its real value is as a null model. A strong answer earns marks for these points. Stating that allele frequencies remain constant from generation to generation when the conditions of the model are met. Using q squared as the frequency of the homozygous recessive phenotype to find q, then p = 1 - q. Calculating 2pq as the frequency of heterozygotes from p and q rather than measuring it directly. Explaining that the principle does not hold when any of its conditions is broken, for example when individuals carrying an allele are selected against, so allele frequency changes. A common slip: Giving q when the question asks for q squared; read whether it wants an allele frequency or a genotype frequency. Instead, check the symbol and the units requested. Exam tip: Write down which value the question has given you, and which symbol it belongs to, before you calculate anything.
Key words
Say what each word means before you turn the card.
p + q = 1
Try answering aloud or in your head before turning the card.
p squared + 2pq + q squared = 1
Try answering aloud or in your head before turning the card.
See every definition
- p + q = 1
- An equation for the frequencies of a dominant and a recessive allele.
- p squared + 2pq + q squared = 1
- An equation for the genotype frequencies, where p squared is the homozygous dominant, 2pq the heterozygous and q squared the homozygous recessive.
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