SPEC 3.7.2 - GENETICS, POPULATIONS, EVOLUTION & ECOSYSTEMS
Populations: Gene Pool, Allele Frequency & the Hardy-Weinberg Equation
The mathematical framework for measuring evolution — and the five conditions that stop it
Population & Gene Pool
Key Definition
Population — a group of organisms of the same species occupying a particular place at the same time, capable of interbreeding.
Key Definition
Gene pool — the total of all alleles of all genes in a population at a given time.
Allele frequency — the proportion of all alleles at a given locus that are a particular allele; expressed as a decimal (0–1) or as a percentage.
Why Allele Frequency Matters
Evolution is defined as a change in allele frequency in a population over time. A population is evolving when the proportion of one allele increases or decreases relative to others — even before any visible change in phenotype occurs.
Allele frequencies are calculated from population data — not from individuals.
Calculating Allele Frequency
Allele frequency is calculated from genotype counts in the population. For a gene with two alleles (A and a):
frequency of A = [(2 × number of AA) + (number of Aa)] / (2 × total individuals)
Worked Example
In a population of 200 individuals: 84 are AA, 88 are Aa, 28 are aa.Total alleles = 2 × 200 = 400A alleles = (2 × 84) + 88 = 168 + 88 = 256a alleles = (2 × 28) + 88 = 56 + 88 = 144Frequency of A (p) = 256/400 = 0.64Frequency of a (q) = 144/400 = 0.36Check: p + q = 0.64 + 0.36 = 1.00 ✓
Examiner Alert
Allele frequency is NOT the same as genotype frequency. The frequency of the dominant allele is NOT equal to the frequency of the dominant phenotype. Heterozygotes carry both alleles — always count each allele separately.
The Hardy-Weinberg Principle
Key Definition
Hardy-Weinberg equilibrium — the principle that allele frequencies in a population will remain constant from generation to generation, provided certain conditions are met.
The Hardy-Weinberg equations describe the relationship between allele frequencies and genotype frequencies in a non-evolving population:
p + q = 1 allele frequencies sum to 1
p² + 2pq + q² = 1
genotype frequencies sum to 1
SymbolMeaningGenotype
pFrequency of dominant allele (A)—
qFrequency of recessive allele (a)—
p²Frequency of homozygous dominantAA
2pqFrequency of heterozygousAa
q²Frequency of homozygous recessiveaa
The Five Conditions for Equilibrium
Hardy-Weinberg equilibrium only holds when all five of these conditions are met simultaneously. If any one is violated, allele frequencies or genotype frequencies will change. Note: violating random mating changes genotype frequencies but not allele frequencies; all other violations change allele frequencies and therefore constitute evolution.
- 1Large population size — prevents genetic drift (random sampling errors that change allele frequency by chance in small populations)
- 2Random mating — every individual has an equal chance of mating with any other; no sexual selection or assortative mating
- 3No mutation — no new alleles created; no existing alleles converted to others
- 4No migration — no gene flow into or out of the population; no immigration or emigration of individuals carrying different allele frequencies
- 5No natural selection — all genotypes have equal survival and reproductive success; no allele confers a fitness advantage
Memory Aid
The conditions can be remembered as: Large population, Random mating, No mutation, No migration, No selection — LR-NNN.
Worked Example — Cystic Fibrosis
Cystic fibrosis (CF) is autosomal recessive. In the UK, approximately 1 in 2500 newborns is affected.
Use Hardy-Weinberg to estimate the carrier frequency.
- 1Identify q²: affected individuals are homozygous recessive (aa).q² = 1/2500 = 0.0004
- 2Calculate q: q = √q² = √0.0004q = 0.02
- 3Calculate p: p = 1 − q = 1 − 0.02p = 0.98
- 4Calculate carrier frequency: 2pq = 2 × 0.98 × 0.022pq = 0.0392 ≈ 1 in 26
Interpretation
Although only 1 in 2500 people are affected with CF, approximately 1 in 26 people in the UK carry one copy of the CF allele. Carriers vastly outnumber affected individuals.
Worked Example — Sickle Cell Anaemia
In a population, 9% of individuals show sickle cell anaemia (homozygous recessive). Calculate the expected frequencies of all genotypes.
q² = 0.09 (9% affected)
q = √0.09 = 0.30p = 1 − 0.30 = 0.70
p² (HbAHbA) = 0.70² = 0.49 (49%)
2pq (HbAHbS) = 2 × 0.70 × 0.30 = 0.42 (42%)
q² (HbSHbS) = 0.09 = 0.09 (9%)
Check: 0.49 + 0.42 + 0.09 = 1.00 ✓
Exam Tip — Show All Steps
AQA awards marks for each step separately. Always write: q² = ..., therefore q = ..., p = 1 − q = ..., then 2pq = ... Do not skip from q² to 2pq. Each step can score independently.
Using Hardy-Weinberg to Detect Evolution
If observed genotype frequencies differ significantly from Hardy-Weinberg predictions, one or more of the five conditions is being violated — and the population is evolving.
Observed = Expected
Population is at Hardy-Weinberg equilibrium — no evolution detected. The five conditions are likely being met.
Observed ≠ Expected
Population deviates from equilibrium — one or more conditions are violated. Evolution is occurring. Chi-squared test can assess significance.
What Violating Each Condition Does
Condition violatedWhat changesDirection
Small populationGenetic drift — random allele frequency changesUnpredictable — could go either way
Non-random matingGenotype frequencies shift; allele frequencies unchangedIncreases homozygosity if inbreeding
MutationNew alleles appear; existing alleles convertSlow — increases variation
MigrationGene flow imports or exports allelesDepends on incoming allele frequencies
Natural selectionFavoured alleles increase; disfavoured alleles decreaseDirectional — adaptive change
Examiner Alert
A common trap: non-random mating changes genotype frequencies but does NOT change allele frequencies. The gene pool composition stays the same — only the pairing of alleles shifts. Inbreeding increases homozygotes but does not change p or q.
Working from Phenotype Data
In many questions, you are only given phenotype frequencies (not genotype counts). Use this approach:
- 1Identify the recessive phenotype frequency — this equals q² (only homozygous recessives show the recessive phenotype)
- 2Take the square root to find q = √q²
- 3Find p = 1 − q
- 4Calculate carrier frequency = 2pq
- 5Calculate homozygous dominant frequency = p²
Critical — Start with q²
You can only begin a Hardy-Weinberg calculation from the recessive phenotype frequency — because only recessive homozygotes (q²) can be unambiguously identified from phenotype alone. You cannot distinguish AA from Aa by looking at the dominant phenotype.
Examiner Traps
Trap 1 — p is NOT the dominant phenotype frequency
p² is the frequency of AA. The dominant phenotype frequency = p² + 2pq (includes heterozygotes). Students often confuse p with the proportion showing the dominant phenotype — these are different values.
Trap 2 — Skipping steps in calculations
AQA mark schemes award marks at each step: finding q² → q → p → 2pq. If you jump to the final answer without showing working, you can score 0 even if the final answer is correct. Always write every step explicitly.
Trap 3 — Conditions vs consequences
The question may ask "what would happen to allele frequency if...". Know that selection causes directional change, drift causes unpredictable random change, mutation is slow, migration depends on the incoming frequencies. Do not give vague answers like "the frequency would change."
Trap 4 — Hardy-Weinberg assumes the condition IS met
When you are asked to calculate carrier frequency using Hardy-Weinberg, you are ASSUMING all five conditions are met. The calculation gives an estimate of what frequency should be, not a measurement of what it actually is in a real (always evolving) population.
DEEP Chain — Extended Response
"A population of moths shows two phenotypes: pale (dominant, PP or Pp) and dark (recessive, pp). In a sample of 400 moths, 64 are dark. Estimate the frequency of carriers in this population and explain what conditions must be met for your estimate to be valid."
DEEP Answer Framework
D — Define
"The Hardy-Weinberg equation states that p² + 2pq + q² = 1, where p = frequency of the dominant allele, q = frequency of the recessive allele, 2pq = frequency of carriers (heterozygotes)."
E — Explain the Mechanism
"Dark moths are homozygous recessive (pp = q²). q² = 64/400 = 0.16. Therefore q = √0.16 = 0.4. p = 1 − 0.4 = 0.6. Carrier frequency = 2pq = 2 × 0.6 × 0.4 = 0.48, or 48% of the population."
E — Evidence / Example
"This means 48% of moths carry one copy of the dark allele but appear pale — carriers vastly outnumber affected individuals when allele frequencies are in this range."
P — Practical / Conditions
"The estimate is valid only if: the population is large (no genetic drift), mating is random, there is no mutation, no migration, and no natural selection. If any condition is violated — for example if pale moths are selectively predated — the actual allele frequencies will deviate from Hardy-Weinberg predictions."
Synoptic Links
Link to Natural Selection (4.05 / 7.04)
Natural selection violates Hardy-Weinberg condition 5. When selection operates, allele frequencies shift directionally. Hardy-Weinberg provides the baseline — deviation from it quantifies how much selection (or drift, or migration) is acting. Covered in detail in Lesson 7.04.
Link to Chi-Squared (7.01)
Chi-squared can be used to test whether observed genotype frequencies are significantly different from Hardy-Weinberg expectations. If χ² exceeds the critical value, the null hypothesis (equilibrium) is rejected — the population is evolving. This is a direct AQA exam application.
Link to Genetic Drift (7.04)
Genetic drift is the violation of condition 1 (large population). In small populations — e.g. after a bottleneck or founder effect — random changes in allele frequency can eliminate alleles entirely, reducing genetic diversity regardless of selective pressure.
Practice Question
In a large population of cats, the allele for black coat (B) is dominant over the allele for ginger coat (b). A census of 500 cats finds that 45 cats have ginger coats.
(a) Using the Hardy-Weinberg equation, calculate the expected frequency of carrier cats (Bb) in this population. Show all working. [4 marks]
(b) State two conditions that must be met for the Hardy-Weinberg principle to apply, and for each condition, explain what would happen to allele frequencies if it were not met. [4 marks]
D
q² = frequency of ginger cats = 45/500 = 0.09
E
q = √0.09 = 0.3p = 1 − 0.3 = 0.7
E
Carrier frequency 2pq = 2 × 0.7 × 0.3 = 0.42 (42%) → expected 210 carriers in 500 cats
P
(b) Example answers: (1) No natural selection — if selection operated against one coat colour, that allele would decrease in frequency over generations. (2) Large population — in a small population, genetic drift could randomly increase or decrease the ginger allele frequency, independent of any selective advantage.
Lesson Summary
Core Content — Spec 3.7.2
Gene pool = all alleles of all genes in a population at a given time
Allele frequency = proportion of one allele at a locus; expressed as decimal 0–1
Hardy-Weinberg equations: p + q = 1 and p² + 2pq + q² = 1
Five conditions : large population, random mating, no mutation, no migration, no selection
Start from q² (recessive phenotype frequency) — the only unambiguously identifiable class from phenotype data
Show all steps: q² → q → p → 2pq (each step independently marked)
Key HW Steps
q² = recessive phenotype frequency
q = √q²
p = 1 − q
2pq = carrier frequency
p² = homozygous dominant frequency
If a Condition Is Violated...
Selection → directional change in allele freq.
Drift → random change (small populations)
Migration → allele freq. shifts by gene flow
Mutation → slow introduction of new alleles
Non-random mating → genotype freq. changes, allele freq. unchanged
Next: Population & Gene Pool