A-Level · Biology · AQA · Mark scheme decoded

AQA A-Level Biology: Population Genetics and Hardy-Weinberg Principle — mark scheme explained

Machine-verifiedchecked against the AQA A-Level Biology specificationlast verified 3 July 2026

The short answer

In the study of genetics, populations, evolution, and ecosystems, understanding how genetic variation is maintained within a population is crucial. This section focuses on the concepts of species, populations, gene pools, allele frequencies, and the Hardy-Weinberg principle. We will explore these ideas in detail to provide a comprehensive understanding of population genetics.

The question

In a population of 100 individuals, 36 are homozygous dominant (AA), 48 are heterozygous (Aa), and 16 are homozygous recessive (aa). Calculate the allele frequencies p and q.

[Paraphrased for study — not reproduced from any exam paper.]

3 marks

Mark scheme, decoded

How the examiner actually awards the marks on this topic.

Gradora's own decode of the marking approach — not the exam board's published mark scheme.

How marks are awarded

For calculation questions, marks are typically awarded for correct substitution of values into formulas, accurate arithmetic, and providing the final answer with appropriate units. For conceptual questions, marks are given for clear and concise explanations that demonstrate a deep understanding of the topic.

What the command words demand

Calculate: Perform a numerical calculation using given data and appropriate formulas.
Explain: Provide a detailed description of the concept, including relevant principles and relationships.
Determine: Find or derive a specific value or quantity based on given information.
Analyze: Examine and interpret data or graphs to draw conclusions about the genetic structure of a population.

Model answer

A full-mark response to the question above, worked through step by step.

Timing: Allocate approximately 5-7 minutes per question to ensure you have enough time to carefully read the problem, perform calculations, and check your work.

Determine the total number of alleles in the population.1 mark
Total alleles = 2 × 100 = 200
Count the number of each allele.1 mark
Number of A alleles = (36 × 2) + 48 = 72 + 48 = 120Number of a alleles = (16 × 2) + 48 = 32 + 48 = 80
Calculate the frequency of each allele.1 mark
p = 120 / 200 = 0.6q = 80 / 200 = 0.4

Final answer: p = 0.6, q = 0.4

Work through every step correctly and you earn all 3 marks.

Another worked example

In a population of 500 individuals, the frequency of the recessive allele (a) is 0.2. Calculate the expected number of homozygous recessive individuals (aa).

3 marks

Identify the given values.0 marks
q = 0.2Total individuals = 500
Use the Hardy-Weinberg equation to find q 2 .1 mark
q 2 = (0.2) 2 = 0.04
Calculate the expected number of homozygous recessive individuals.2 marks
Expected number of aa = 0.04 × 500 = 20

Final answer: 20

Work through every step correctly and you earn all 3 marks.

Common mistakes

Confusing the total number of alleles with the total number of individuals.
Why it happens: Students often forget that each individual has two alleles for a given gene, leading to incorrect calculations.
Fix: Always remember to multiply the number of individuals by 2 to get the total number of alleles.
Forgetting to use p + q = 1 when calculating allele frequencies.
Why it happens: Students may overlook this simple relationship, leading to incorrect values for p and q.
Fix: Always check that the sum of p and q equals 1 after calculating each frequency.
Incorrectly applying the Hardy-Weinberg equation.
Why it happens: Students may substitute values incorrectly or forget to square terms, leading to wrong genotype frequencies.
Fix: Practice using the Hardy-Weinberg equation step-by-step. Ensure that p 2 , 2pq, and q 2 are calculated correctly and sum to 1.
Misinterpreting the conditions for Hardy-Weinberg equilibrium.
Why it happens: Students may not fully understand the implications of each condition, leading to incorrect conclusions about a population's genetic structure.
Fix: Review and memorize the five conditions required for Hardy-Weinberg equilibrium. Practice identifying which conditions are met or not met in given scenarios.
Failing to check if the sum of genotype frequencies equals 1.
Why it happens: Students may perform calculations but forget to verify that p 2 + 2pq + q 2 = 1, leading to incorrect results.
Fix: Always double-check your calculations by ensuring that the sum of genotype frequencies equals 1.
Incorrectly interpreting allele and genotype frequencies in a population.
Why it happens: Students may struggle to understand how allele frequencies relate to genotype frequencies, leading to confusion in problem-solving.
Fix: Practice converting between allele and genotype frequencies using the Hardy-Weinberg equation. Understand that p 2 , 2pq, and q 2 represent specific genotypes in a population.

Where the marks go

The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.

Question type	What you’re asked to do	Marks
Calculate Allele Frequencies	Work out the frequencies of the dominant and recessive alleles from genotype counts.	3
Hardy-Weinberg Calculation	Use the Hardy-Weinberg equation to find the expected number of homozygous recessive individuals.	3
Hardy-Weinberg Allele Frequencies	Use the recessive genotype frequency to calculate the two allele frequencies p and q.	2
Calculate Allele Frequencies	Use the Hardy-Weinberg equation to find dominant and recessive allele frequencies from a genotype frequency.	2
Hardy-Weinberg Calculation	Use the heterozygous frequency to calculate the dominant and recessive allele frequencies.	5
Total across these question types		15

Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.