7.2Genetics, Populations, Evolution and Ecosystems (A Level only)
Populations (A Level only) — AQA A Level Biology 7402 Revision Notes & Topical Questions
5 min read 12 parts Lesson 5 of 25 Unit quiz
Populations, Gene Pools and the Hardy-Weinberg Principle — AQA A LEVEL BIOLOGY 7402 notes: Define population and gene pool, calculate phenotype frequencies and use the Hardy-Weinberg equations to find allele and genotype frequencies.A species is a group of similar organisms that …
- Allele frequency
- p2 + 2pq + q2 = 1
- Worked example 1
- Worked example 2
- Worked example 3
- q = √0.10 = 0.32
Section 1 of 12Populations and species
A species is a group of similar organisms that can reproduce to give fertile offspring.
Section 2 of 12Gene pools and frequencies
A gene pool is the collection of genes within an interbreeding population at a particular time; since genes have different alleles, it is the sum of all the alleles of the genes of a population of a single species at that time. Allele frequency is the proportion of a particular allele in the gene pool.
Phenotype frequency = (number of individuals with the phenotype ÷ total number of individuals) × 100. Worked example: in a population of 9 pea plants, 7 have purple flowers and 2 white. Purple: 7 ÷ 9 × 100 = 77.8 %. White: 2 ÷ 9 × 100 = 22.2 %.
Section 3 of 12The Hardy-Weinberg principle
The Hardy-Weinberg principle states that allele frequencies in a population stay constant from one generation to the next if certain conditions are met: no selection, no mutation, no migration (no gene flow), a large population (no genetic drift) and random mating. Frequencies are given as proportions between 0 and 1: if every individual is homozygous dominant (BB), its frequency is 1; if half the population show this genotype the frequency is 0.5.
For a gene with two alleles B and b, p is the frequency of the dominant allele and q is the frequency of the recessive allele:
p + q = 1 and p2 + 2pq + q2 = 1
where p2 = frequency of homozygous dominant, 2pq = frequency of heterozygotes, q2 = frequency of homozygous recessive.
Section 4 of 12Worked example: feather colour
Dominant F and recessive f control feather colour. 10 % of the individuals have the recessive phenotype, which must be homozygous recessive ff, so q2 = 0.10.
Section 5 of 12Worked examples
In a population of 9 pea plants 7 are purple and 2 white. Calculate the percentage with white flowers. [1]1 mark
2 ÷ 9 × 100 = 22.2 %.
10 % of birds have the recessive phenotype. Calculate the frequency of heterozygotes. [3]3 marks
q2 = 0.10, q = 0.32, p = 0.68, 2pq = 0.44.
State two conditions for the Hardy-Weinberg principle to hold. [2]2 marks
Large population (no genetic drift); random mating; no mutation, migration or selection.
Section 6 of 12Step 1: start with q²
Always begin by identifying the frequency of the recessive phenotype. This is the only genotype you can determine directly, since it must be homozygous recessive (bb), so its frequency equals q2. From q2, q = √(q2), then p = 1 − q.
Section 7 of 12Step 2: identify known and unknown values
Check what the question gives you (for example q2 or p) and work out what it asks for (for example 2pq). Then use the appropriate equation or equations to connect them:
| Given | Find | Use |
|---|---|---|
| q2 | q | square root |
| q | p | p = 1 − q |
| p and q | carriers (heterozygotes) | 2pq |
| p | homozygous dominant | p2 |
Check by adding p2 + 2pq + q2; the total must be 1.
Section 8 of 12Step 3: do not confuse the equations with the principle
The equations estimate allele and genotype frequencies. The principle assumes no change in allele frequencies between generations (genetic equilibrium). A question about whether a population is in equilibrium needs the principle's conditions, not just the equations. If frequencies calculated in successive generations differ, then one of the conditions (no selection, mutation, migration or drift) has been broken.
Section 9 of 12Common slips to avoid
Examiners often see the following
- giving the answer as a percentage when the question asks for a frequency (a value between 0 and 1);
- writing p when q is meant;
- stating that 2pq is the frequency of the dominant allele;
- and forgetting that a carrier frequency in a population is 2pq, not q.
- Always label every value with its symbol, show the square root step, and finish with the check that the three genotype frequencies sum to 1.
- Also remember the population must be large, randomly mating and free from selection, mutation and migration for the equations to give valid estimates.
Section 10 of 12Worked examples
The recessive phenotype of a disease has a frequency of 1 in 2500. Calculate the carrier frequency. [4]4 marks
q2 = 0.0004, q = 0.02, p = 0.98, 2pq = 2 × 0.98 × 0.02 = 0.039 (about 3.9 %).
Explain why only the recessive phenotype can be used as the starting point. [2]2 marks
It has only one possible genotype (homozygous recessive), whereas the dominant phenotype may be homozygous or heterozygous.
Section 11 of 12Key terms
| Term | Meaning |
|---|---|
| Population | Group of organisms of the same species in a place at a time that can interbreed. |
| Gene pool | All the alleles of all genes in an interbreeding population. |
| Allele frequency | Proportion of a given allele in the gene pool. |
| Hardy-Weinberg principle | Allele frequencies stay constant between generations if specific conditions are met. |
| p and q | Frequencies of the dominant and recessive allele, p + q = 1. |
| Term | Meaning |
|---|---|
| q² | Frequency of the homozygous recessive genotype. |
| 2pq | Frequency of heterozygotes. |
| Genetic equilibrium | Allele frequencies remain unchanged between generations. |
| Carrier | Heterozygous individual. |
Section 12 of 12Exam technique
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