Populations, evolution and ecosystems: A-level Biology AQA revision notes

3.7.1

Inheritance

Genotype, phenotype and Mendelian segregation 3.7.1

Definitions

Genotype: the alleles an organism carries at a locus. Phenotype: the resulting observable characteristic — the expression of genotype, interacting with the environment.
Homozygous: identical alleles at a locus (e.g. RR or rr). Heterozygous: differing alleles at a locus (e.g. Rr).

Key results

During meiosis, the two alleles of a gene (carried on homologous chromosomes) separate, so each haploid gamete receives just one allele of that gene.

Notes

This separation of alleles during meiosis — Mendelian segregation — is the physical, chromosomal basis for every Punnett-square calculation in this topic: a heterozygous parent (Rr) produces two gamete types (R and r) in equal proportion, precisely because the two alleles sit on different, separating homologous chromosomes.
'Alleles are versions of a gene' is the key underlying idea — every individual carries two alleles of each autosomal gene (one from each parent) at the same fixed locus, but different individuals in a population can carry different combinations of alleles at that locus.

Monohybrid and dihybrid inheritance 3.7.1

Definitions

Test cross: crossing an individual of unknown genotype (showing the dominant phenotype) with a homozygous recessive individual, to determine whether the unknown individual is homozygous or heterozygous from the resulting offspring ratio.

Key results

Monohybrid cross (one gene, Rr × Rr, complete dominance): genotype ratio 1 RR : 2 Rr : 1 rr; phenotype ratio 3 dominant : 1 recessive.
Dihybrid cross (two independently assorting genes, RrYy × RrYy, complete dominance at each locus): phenotype ratio 9 : 3 : 3 : 1 across the four possible phenotype combinations.
Test cross outcomes: Rr × rr gives ½ Rr : ½ rr (both phenotypes appear, confirming heterozygous); RR × rr gives all Rr, i.e. all dominant phenotype (confirming homozygous).

Notes

A dihybrid cross's 9:3:3:1 ratio is exactly the product of two independent monohybrid 3:1 ratios (one per gene) — it only holds when the two genes assort completely independently (different chromosomes, or far enough apart on the same chromosome that linkage is negligible).
These are EXPECTED proportions across a large number of offspring — any individual small family can deviate from the ratio purely by chance, which is exactly the kind of deviation the chi-squared test (covered later in this topic) is designed to evaluate.
A test cross is the standard way to reveal a hidden heterozygous genotype experimentally, since a dominant phenotype alone cannot distinguish RR from Rr by observation.

Codominance, multiple alleles and sex linkage 3.7.1

Definitions

Codominant alleles: both expressed distinctly in a heterozygote, neither masking the other (e.g. Iᴬ and Iᴮ in the ABO blood group system).
Sex linkage: a gene carried on a sex chromosome (usually X), producing different inheritance patterns in males and females.

Key results

ABO blood groups (multiple alleles, three allele types Iᴬ, Iᴮ, i in the population, two allele copies per individual): Iᴬ and Iᴮ are codominant with each other; i is recessive to both — group A × group B parents can produce A, B, AB or O offspring, depending on which alleles each parent actually carries.
X-linked recessive inheritance: a male (XY) has only one X chromosome, so a single copy of an X-linked recessive allele is always expressed (no second allele to mask it); a female (XX) needs two copies (homozygous) to show the recessive phenotype — this is why X-linked recessive conditions (e.g. haemophilia, red-green colour blindness) are far more common in males than females.

Notes

A gene with more than two possible alleles in the population (multiple alleles, like ABO) still gives each individual only two allele copies — 'multiple alleles' describes the population-level variation, not the number of copies any one individual carries.
Codominance is distinct from incomplete dominance (where a heterozygote shows an intermediate, blended phenotype) — in true codominance, both alleles' effects appear FULLY and DISTINCTLY, not blended, as ABO blood typing (detecting both A and B antigens on the same red blood cells in a group-AB individual) directly demonstrates.
The Y chromosome carries no corresponding allele for most X-linked genes, which is precisely why a male's phenotype for an X-linked gene is determined entirely by his single X-linked allele, with nothing on the Y to mask or complement it.

Genetic linkage 3.7.1

Definitions

Genetic linkage: genes at loci on the same chromosome tend to be inherited together, rather than assorting independently, because they are physically joined on the same DNA molecule.

Key results

Crossing over (at chiasmata, prophase I of meiosis) can separate linked alleles, producing recombinant gamete combinations alongside the original (parental) combinations — e.g. a parent with genotype AB/ab produces mostly parental gametes (AB and ab) plus a smaller proportion of recombinant gametes (Ab and aB), with the exact recombinant proportion depending on how far apart the two loci are.
A test cross (AB/ab × ab/ab) reveals the first parent's actual gamete types and their proportions directly from the offspring phenotype ratio, since the second (homozygous recessive) parent contributes only recessive alleles.

Notes

The further apart two linked loci are on a chromosome, the MORE likely a crossover falls somewhere between them, so the higher the proportion of recombinant gametes produced — recombination frequency is itself a rough measure of physical distance between two loci on a chromosome.
Two extremes bound this relationship: loci close enough together essentially never separate by crossing over (recombinant frequency near 0%, behaving as if completely linked); loci far enough apart on the same chromosome can approach the 50% recombination expected of genuinely independent assortment (behaving, statistically, almost as if unlinked).
Linkage and sex linkage both weaken the simple assumption behind an ordinary Punnett square (that alleles at different loci always assort completely independently) — always check whether a cross description mentions genes on the same chromosome, or an X-linked gene, before defaulting to independent assortment.

Epistasis 3.7.1

Definitions

Epistasis: the interaction between genes at two different loci in which one gene (epistatic) masks or suppresses the phenotypic expression of another gene (hypostatic) at a separate locus.

Key results

Recessive epistasis example (mouse coat colour, AaCc × AaCc, two independently assorting loci): the C locus controls whether pigment forms at all (C_ = pigment forms, cc = no pigment, regardless of the A locus); where pigment does form, the A locus determines its pattern (A_ = agouti, aa = solid black) — giving an overall phenotype ratio of 9 agouti : 3 solid black : 4 albino (the 4 albino combining the 3 aaC_-equivalent genotypes that would otherwise be solid black plus the 1 that would be agouti, all masked by cc).

Notes

Epistasis modifies the SIMPLE 9:3:3:1 dihybrid ratio precisely because one locus's effect depends entirely on the genotype at the other locus — the two genes are not simply combining independently to determine two separate features, one gene is controlling whether the other gene's effect can be seen at all.
The specific modified ratio observed (here 9:3:4, rather than the unmodified 9:3:3:1) is itself informative: different epistasis patterns (recessive, dominant, and others) produce different characteristic modified ratios, which is part of how a geneticist infers the underlying gene interaction from observed offspring proportions.
This is exactly the kind of predicted ratio the chi-squared test (covered next) is used to check observed data against.

The chi-squared test 3.7.1

Key results

Chi-squared statistic: $χ^{2} = \sum \frac{( O - E ) ^{2}}{E}$ , where $O$ is each observed count and $E$ is the count expected under the stated genetic hypothesis (predicted proportion × total).
Degrees of freedom: $df = (number of phenotype classes) - 1$ , since the model's predicted proportions were specified before observing the data.
If the calculated $χ^{2}$ value is LESS than the critical value (at the chosen significance level and correct $df$ ): do not reject the null hypothesis ( $H_{0}$ ) — the data are consistent with the predicted ratio. If it EXCEEDS the critical value: reject $H_{0}$ — the deviation is too large to plausibly be due to chance alone.

Notes

The test compares observed offspring numbers with numbers expected under a stated genetic hypothesis (e.g. a 3:1 or 9:3:3:1 ratio) — it is a general statistical tool for comparing observed and expected frequency data, not specific to genetics, but very commonly applied to genetic crosses.
'Do not reject $H_{0}$ ' is NOT the same as 'the hypothesis is proven correct' — failing to find statistical evidence against a hypothesis is not the same as positively confirming it; the test can only ever fail to reject, or reject, never definitively 'prove'.
Valid use of the test requires counts from independent observations, and expected counts of at least 5 in each class — with fewer than 5 expected per class, the chi-squared approximation becomes unreliable and the test's conclusions cannot be trusted.

Worked examples

Worked example 3.7.1 · 5 marks

In a species of pea plant, seed shape (round R, dominant, over wrinkled r) and flower colour (purple P, dominant, over white p) are controlled by genes on two different chromosomes.

A plant heterozygous for both genes (RrPp) is crossed with a plant that is homozygous recessive for both (rrpp) — a standard test cross.

Calculate the expected proportion of offspring that are wrinkled-seeded with purple flowers, showing your reasoning.

Show worked solution

The two genes assort independently.

Wrinkled seeds (rr) have probability:

\frac{1}{2} \times 1 = \frac{1}{2}

Purple flowers (Pp) have probability $\frac{1}{2}$ .

Combined probability:

= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

so 25% of offspring are expected to be wrinkled-seeded with purple flowers.

Mark scheme · 5 marks

States the two genes assort independently 1 mark
Calculates probability of wrinkled seeds (rr) = 1/2 1 mark
Calculates probability of purple flowers (Pp) = 1/2 1 mark
States the combined probability is the product of the two 1 mark
Calculates the final answer as 1/4 (25%) 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.1 · 6 marks

Phenotype class	Round, purple	Round, white	Wrinkled, purple	Wrinkled, white
Observed (O)	168	62	58	32
Expected (E)	180	60	60	20

A dihybrid cross of two heterozygous pea plants (expected phenotypic ratio 9:3:3:1) produced 320 offspring with observed numbers shown in the table below.

(a) Calculate the expected number in each class.

(b) Calculate the chi-squared statistic.

(c) Given a critical value of 7.82 at the 0.05 significance level for 3 degrees of freedom, state and justify the conclusion.

Show worked solution

(a) Expected from 9:3:3:1 (16 parts, total 320): 180, 60, 60, 20. (b):

χ^{2} = Σ \frac{( O - E ) ^{2}}{E} = 0.80 + 0.067 + 0.067 + 7.20 = 8.13

(3 s.f.).

Since calculated $χ^{2}$ (8.13) exceeds the critical value (7.82), the null hypothesis is rejected — the deviation is too large to plausibly be due to chance, suggesting a factor such as genetic linkage may be affecting the results.

Mark scheme · 6 marks

Calculates expected values 180, 60, 60, 20 1 mark
Calculates $(O - E)^{2} / E$ correctly for at least two classes 1 mark
Calculates $χ^{2} = 8.13$ 1 mark
States degrees of freedom = 3 1 mark
States $χ^{2}$ exceeds the critical value, so the null hypothesis is rejected 1 mark
Suggests a factor such as genetic linkage as a possible cause of the deviation 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.1 · 4 marks

In a species of flower, gene A controls pigment production (A = pigment produced, a = no pigment produced) and gene B controls pigment colour (B = purple, b = red), but gene B's effect is only visible if pigment is produced at all.

A dihybrid cross (AaBb × AaBb) is carried out.

Predict the phenotypic ratio of the offspring, and name this type of gene interaction.

Show worked solution

This is epistasis: gene A is epistatic to gene B, since no colour is visible without pigment.

Every aa genotype (aaB_ and aabb, 3+1=4 parts) is phenotypically identical (no pigment), collapsing the usual 9:3:3:1 into 9 purple : 3 red : 4 no pigment.

Mark scheme · 4 marks

Names epistasis, identifying gene A as epistatic to gene B 1 mark
States every aa genotype produces no pigment regardless of gene B 1 mark
Combines the aaB_ and aabb classes (3+1=4 parts) into one phenotype 1 mark
States the resulting ratio is 9:3:4 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.7.2

Populations

Populations and the gene pool 3.7.2

Definitions

Gene pool: the complete set of alleles, at every gene, present across all individuals of a population at a given time.
Allele frequency: the proportion of one specific allele among all allele copies at that locus across the population.

Key results

For a diploid population of $N$ individuals, total allele copies at one autosomal locus $= 2 N$ — each individual contributes two copies (homozygous: both copies the same allele; heterozygous: one copy of each).

Notes

A single species can comprise several separate populations — the gene pool, and so the allele frequencies, of one population need not match another population of the same species, particularly if there is limited gene flow between them.
Direct counting of alleles in a real, fully-genotyped sample (as opposed to inferring frequencies from phenotype data alone) requires no equilibrium assumption at all — it is simply an exact count, valid whether or not the population is actually in Hardy-Weinberg equilibrium.

Allele frequency 3.7.2

Key results

For a two-allele locus (A and a, frequencies $p$ and $q$ ): $p + q = 1$ .
Allele frequency by direct count: (number of copies of that allele) ÷ (total allele copies at that locus).

Notes

Because $p + q = 1$ for a two-allele locus, knowing either frequency automatically gives the other — $q = 1 - p$ — a shortcut worth using directly rather than recalculating both from scratch every time.

Conditions for Hardy-Weinberg equilibrium 3.7.2

Definitions

Hardy-Weinberg principle: allele and genotype frequencies remain constant across generations, provided a specific set of conditions all hold simultaneously.

Key results

Conditions required: random mating (with respect to the locus in question); a very large population (negligible genetic drift); no mutation (alleles are not converted or newly created); no gene flow (no exchange of alleles with other populations); no natural selection (equal reproductive success of all genotypes).

Notes

This is a simple model — one autosomal locus, diploid organisms, exactly two alleles, ordinary Mendelian segregation — and every one of the five conditions must hold for the model's prediction to apply exactly; real populations typically violate at least one to some degree.
The Hardy-Weinberg model is most useful as a NULL hypothesis or baseline: comparing a real population's observed genotype frequencies against what the model predicts is a standard way to detect that evolution (in the broad sense of changing allele frequencies) is actually occurring, and to start investigating why.

The Hardy-Weinberg equation 3.7.2

Key results

Genotype frequencies under Hardy-Weinberg equilibrium: $p^{2} + 2 pq + q^{2} = 1$ , where $p^{2}$ = homozygous dominant (AA), $2 pq$ = heterozygous (Aa), $q^{2}$ = homozygous recessive (aa).
Assuming complete dominance: dominant phenotype frequency $= p^{2} + 2 pq$ ; recessive phenotype frequency $= q^{2}$ .

Notes

The $2 pq$ term for heterozygotes (rather than simply $pq$ ) exists because a heterozygote can arise in TWO distinct ways — an A-allele egg fertilised by an a-allele sperm, OR an a-allele egg fertilised by an A-allele sperm — both possibilities must be counted.
This genotype equation follows directly from random union of gametes at frequencies $p$ and $q$ (the same $p$ and $q$ in both eggs and sperm) — it is a direct mathematical consequence of the Hardy-Weinberg conditions, not a separate independent assumption.
The genotype equation itself does not require complete dominance — mapping genotype frequencies onto PHENOTYPE frequencies (as $p^{2} + 2 pq$ vs $q^{2}$ ) is the one step that specifically assumes complete dominance; with codominance or incomplete dominance, all three genotypes would instead be individually distinguishable by phenotype.
Because $q^{2}$ shrinks much faster than $2 pq$ as a recessive allele becomes rare (e.g. $q = 0.05$ gives $q^{2} = 0.0025$ but $2 pq \approx 0.095$ , roughly 38 times larger), a rare recessive condition always has a 'hidden' heterozygous carrier population many times larger than the population actually showing the condition.

Hardy-Weinberg worked example 3.7.2

Key results

Worked example: 80 of 500 plants have the recessive (white-flower) phenotype. $q^{2} = 80/500 = 0.16$ , so $q = 0.16 = 0.40$ , and $p = 1 - 0.40 = 0.60$ . Expected genotype frequencies: AA $= p^{2} = 0.36$ (180 plants), Aa $= 2 pq = 0.48$ (240 plants), aa $= q^{2} = 0.16$ (80 plants).

Method

Starting from a recessive phenotype frequency: measure the proportion of the population showing the recessive phenotype — this equals $q^{2}$ directly (assuming Hardy-Weinberg equilibrium and complete dominance) → take the square root to find $q$ → find $p = 1 - q$ → substitute both into $p^{2}$ , $2 pq$ and $q^{2}$ to find all three genotype frequencies → multiply each genotype frequency by the total population size to find expected counts.

In practice1 in 2500 people has cystic fibrosis, a recessive condition. Find the frequency of carriers.

$q^{2} = \frac{1}{2500} = 0.0004 ⟹ q = 0.02$ the recessive phenotype frequency is $q^{2}$
$p = 1 - q = 0.98$
$2 pq = 2 (0.98) (0.02) = 0.0392$ about 1 person in 25 is a carrier

Notes

This calculation always starts from the RECESSIVE phenotype specifically, because (assuming complete dominance) it is the only phenotype class that corresponds to a single, unambiguous genotype ( $q^{2}$ ) — the dominant phenotype alone cannot distinguish AA from Aa individuals by observation.
The AA and Aa counts here are INFERRED from the model, not directly observed — flower colour alone genuinely cannot distinguish a homozygous dominant plant from a heterozygous one; the model's predictions rely on the Hardy-Weinberg conditions actually holding for this population.

Worked examples

Worked example 3.7.2 · 5 marks

In a population in Hardy-Weinberg equilibrium, 9% of individuals show a recessive condition.

Calculate the frequency of the recessive allele, the frequency of the dominant allele, and the percentage of the population expected to be heterozygous carriers, and state which of these three groups (homozygous dominant, heterozygous, homozygous recessive) is the most numerous.

Show worked solution

$q^{2} = 0.09$ , so $q = 0.3$ .

p = 1 - 0.3 = 0.7

p^{2} = 0.49 = 49%

homozygous dominant.

2 pq = 0.42 = 42%

heterozygous.

$q^{2} = 9%$ homozygous recessive.

Homozygous dominant (49%) is the largest group here.

Mark scheme · 5 marks

Calculates $q = 0.3$ 2 marks
Calculates $p = 0.7$ 1 mark
Calculates heterozygous frequency $2 pq = 42%$ 1 mark
States homozygous dominant (49%) is the largest group here 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.2 · 4 marks

A population is initially in Hardy-Weinberg equilibrium for a particular gene.

A new predator arrives, which more easily catches individuals with the recessive phenotype, significantly reducing their survival to reproductive age.

State, with reasoning, which of the five conditions required for Hardy-Weinberg equilibrium has now been violated, and predict the qualitative effect on allele frequencies in the population over subsequent generations.

Show worked solution

The 'no natural selection' condition is violated — recessive-phenotype individuals now have reduced reproductive success.

Over subsequent generations, the recessive allele's frequency ( $q$ ) would be expected to fall, while the dominant allele's frequency ( $p$ ) rises.

This is directional selection.

Mark scheme · 4 marks

States the 'no natural selection' condition is violated 1 mark
Explains recessive-phenotype individuals now have reduced reproductive success 1 mark
States $q$ would be expected to fall over subsequent generations 1 mark
Names this as directional selection 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.7.3

Evolution and speciation

Evolution and speciation 3.7.3

Definitions

Species (biological definition): a group of organisms with similar features, capable of interbreeding to produce fertile offspring — excludes cases like a mule (viable but infertile) from counting as evidence horse and donkey are the same species.
Allopatric speciation: begins with a geographical barrier separating one population into two, preventing gene flow, until independent divergence produces reproductive isolation. Sympatric speciation: reproductive isolation without geographical separation (e.g. differing breeding season or behaviour, or polyploidy in plants).

Key results

Reproductive isolation mechanisms: prezygotic (prevents mating or fertilisation, e.g. diverging mate-recognition behaviour) or postzygotic (mating occurs, but offspring are inviable or infertile).
Disruptive selection: individuals with EITHER extreme phenotype are favoured over those with the intermediate phenotype — the intermediate form is selected against, so the population's phenotype distribution splits toward the two extremes, potentially forming two distinct peaks rather than one.
Evidence for evolution converges from three independent lines: the fossil record (a chronological sequence, but incomplete, since fossilisation itself is rare), comparative anatomy (homologous structures, e.g. the pentadactyl limb, shared across superficially very different species), and molecular evidence (DNA or protein sequence similarity). No single line is individually conclusive on its own; their independent convergence is what makes the overall case for evolution robust.

Notes

The three selection types compared: directional selection shifts the population toward ONE extreme (e.g. increasing antibiotic resistance in bacteria, or increasing beak size where large seeds dominate); stabilising selection favours the INTERMEDIATE phenotype and selects against both extremes (e.g. human birth weight, where both very low and very high birth weights carry increased risk); disruptive selection favours BOTH extremes over the intermediate.
Disruptive selection is the one most directly linked to speciation of the three: if a population occupies an environment offering two distinct resource types (e.g. small and large seeds, with few intermediate-sized ones), individuals with intermediate phenotypes are at a genuine disadvantage at exploiting either resource well, while individuals at either extreme thrive — over time this can split one interbreeding population into two phenotypically (and eventually reproductively) distinct groups, without necessarily requiring the geographical separation that allopatric speciation depends on.
Speciation is the process by which one or more new species arise from an existing one, generally through accumulated genetic divergence between populations that gradually lose the ability to interbreed and produce fertile offspring.
Once reproductive isolation has genuinely developed, the two populations count as separate species, and remain so even if the original geographical barrier is later removed — the isolation, once established, does not require the original cause to persist.
Polyploidy (a sudden chromosome-number change, e.g. a diploid gamete fusing with a haploid one, producing a triploid offspring whose chromosomes cannot pair correctly at meiosis) can prevent successful interbreeding with the original population immediately, within a single generation, without any geographical separation at all — the one clear example of speciation that can occur essentially instantly rather than through gradual divergence.
Convergence of independent evidence lines is the key evaluative point examiners test — a question asking why evolution is well supported expects an answer naming multiple independent lines of evidence agreeing with each other, not just restating one line in greater detail.

Worked examples

Worked example 3.7.3 · 5 marks

A population of birds occupies an island with only two available seed sizes: small and large, with very few intermediate-sized seeds available.

Birds with intermediate beak sizes struggle to efficiently crack either seed type well, while birds with either small or large beaks feed efficiently on the matching seed size.

Explain how this situation could, over many generations, lead to the population splitting into two genetically distinct groups, naming the type of selection involved.

Show worked solution

This is disruptive selection: both extreme beak sizes are favoured, while the intermediate phenotype is selected against.

Over generations, this splits the population's phenotype distribution into two peaks.

If mating also becomes non-random with respect to beak size, this can lead to sympatric speciation, without any geographical barrier.

Mark scheme · 5 marks

Names disruptive selection 1 mark
States both extremes are favoured, the intermediate is selected against 1 mark
Explains this splits the phenotype distribution into two peaks over generations 1 mark
States non-random mating could lead to reproductive isolation 1 mark
Names sympatric speciation, noting no geographical barrier is involved 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.3 · 5 marks

Two populations of the same fish species become separated when a new waterfall forms, preventing any fish from moving between the pool above the waterfall and the river below it.

After many generations of complete isolation, fish from the two populations are tested and found unable to produce fertile offspring together.

Explain the process by which this reproductive isolation could have developed, naming the type of speciation this illustrates.

Show worked solution

This is allopatric speciation: a geographical barrier (the waterfall) prevents gene flow between the two populations.

Each population independently accumulates genetic changes (mutation, drift, selection to different environments).

Because gene flow is prevented, these differences are never shared, and can build up until the two populations can no longer interbreed.

Mark scheme · 5 marks

Names allopatric speciation 1 mark
States the waterfall is a geographical barrier preventing gene flow 1 mark
States each population independently accumulates genetic changes 1 mark
Explains the lack of gene flow allows differences to build up unshared 1 mark
Concludes sufficient divergence produces reproductive isolation 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.7.4

Populations in ecosystems

Ecosystems, communities and populations 3.7.4

Definitions

Population: individuals of one species in the same place and time. Community: populations of different species living in the same area. Ecosystem: a community together with its non-living (abiotic) environment.
Habitat: where an organism lives. Niche: its specific pattern of resource use, tolerances and interactions with other organisms.

Notes

These four terms nest inside one another in scale, from smallest to largest: population → community → ecosystem — keeping this hierarchy straight matters, since exam questions often specifically test whether a described scenario is being analysed at the correct level.
A species' adaptations determine how it responds to both biotic (living) and abiotic (non-living) conditions in its environment — niche and habitat together describe not just WHERE a species lives, but HOW it lives there, distinctly from any other species sharing the same physical space.

Factors limiting population size 3.7.4

Definitions

Carrying capacity (K): the maximum population size a particular environment's resources can sustain.

Key results

Abiotic factors limiting population size: light, temperature, water, pH, mineral ions.
Biotic interactions limiting population size: intraspecific competition (within a species, for the same limited resources), interspecific competition (between different species, for overlapping resources), predation (a predator kills and eats prey).

Notes

In the fixed-K model, population size grows but levels off as it approaches carrying capacity — though real populations can fluctuate around K, or temporarily exceed it before falling back, rather than settling smoothly at a constant value.
Carrying capacity is not a fixed, universal constant for a species — it depends on the specific environment's conditions and resources, and changes if those conditions change (e.g. a drought lowering K by reducing available water and food).
Intraspecific competition is very often the dominant factor limiting a population's own size as it approaches its own carrying capacity, since individuals of the same species have near-identical resource requirements and so compete most directly with one another.

Sampling populations: quadrats and transects 3.7.4

Definitions

Frequency (quadrat sampling): the percentage of quadrats in which the target species is present at all. Percentage cover: the percentage of a quadrat's area actually covered by the target species.

Method

Random quadrats: choose sampling coordinates at random (e.g. using random number coordinates across a marked grid) to avoid observer bias, keep quadrat area fixed, and use enough representative samples for a reliable population estimate — population estimate = mean density per quadrat × total habitat area.
Interrupted belt transect: place quadrats at regular intervals along a line crossing an environmental gradient (e.g. increasing light intensity), recording the target species' abundance (frequency or percentage cover) and the relevant environmental factor at each station — Required practical 12: investigating the relationship between an environmental factor and species distribution.

In practiceTen random 0.25 m² quadrats in a 50 m² field contain 3, 5, 2, 0, 4, 6, 3, 1, 4 and 2 plants.

$mean = \frac{30}{10} = 3 per quadrat$
$density = \frac{3}{0.25} = 12 per m^{2}$
$estimate = 12 \times 50 = 600 plants$

Notes

Random quadrat placement is specifically designed to give an unbiased ABUNDANCE estimate across a habitat; a transect instead deliberately samples systematically along a gradient to investigate HOW abundance changes with an environmental factor — the two techniques answer genuinely different questions and are not interchangeable.
Repeating transects, and using consistent recording methods at every station, helps assess natural variation and improves confidence that an observed pattern is real rather than a sampling artefact.
A correlation observed between an environmental factor and species distribution along a transect does not, by itself, establish causation — the environmental factor measured might not be the actual cause of the distribution pattern, even if the two vary together consistently.

Mark-release-recapture 3.7.4

Key results

Lincoln index: estimated population size $N \approx \frac{M \times C}{R}$ , where $M$ = number marked in the first catch, $C$ = total number in the second catch, $R$ = number of marked individuals recaptured in the second catch.

Notes

The method's core assumptions all exist to protect the marked-recapture proportion from being distorted: a closed population (no births, deaths, immigration or emigration during the study), full mixing (marked individuals redistribute randomly and have the same capture probability as unmarked ones), and marks that are harmless (do not affect survival or capture probability) and remain identifiable throughout.
Because recaptured marked individuals ( $R$ ) sit in the Lincoln index's denominator, anything that selectively reduces that count (e.g. marking that increases predation risk, or reduces catchability) INFLATES the population estimate above the true value — a small, easily overlooked source of systematic error worth stating explicitly when evaluating the method.
This is fundamentally an ESTIMATE, not an exact count — a small number of recaptures increases the uncertainty in the resulting estimate, since the calculation then rests on a small, more variable sample.

Ecological succession 3.7.4

Definitions

Succession: the directional change in a community's species composition over time. Primary succession begins on bare substrate with no developed soil (e.g. bare rock); secondary succession begins where soil and some biological legacy already exist (e.g. after fire), so it proceeds faster from a more advanced starting point.

Key results

Typical temperate primary succession sequence: bare rock (little organic matter, no developed soil) → pioneer species (e.g. lichens and mosses, tolerating harsh conditions) → grasses and herbs (more soil and organic matter, new species establish) → shrubs (greater shading, community changes) → woodland (an example climax community, relatively stable composition).

Notes

Pioneer species (e.g. lichens) begin soil formation through weathering and the accumulation of litter and their own decomposition — generally the slowest stage of the whole sequence, since almost no soil exists at the start.
Each successive seral stage changes local conditions (more soil, more organic matter, more shading) in ways that make the environment more suitable for the NEXT community and, often, less suitable for the current one — the community that colonises frequently ends up outcompeted and replaced by the one it helped enable.
Rates and endpoints depend on local conditions — woodland is a common temperate climax community but is not the universal endpoint everywhere, and disturbance (fire, grazing, human management) can redirect or reset succession rather than letting it proceed to a 'natural' climax.

Conservation of habitats and ecosystems 3.7.4

Notes

Conservation can deliberately involve MANAGING succession rather than simply leaving an area undisturbed — e.g. to retain species-rich, open grassland (which would otherwise progress toward shrub and woodland stages), timed grazing or mowing limits woody plant growth, and removing cuttings limits nutrient build-up that would otherwise favour more competitive, less diverse vegetation.
Effective conservation management requires ongoing monitoring of species and vegetation, with management adjusted based on that evidence — a fixed, one-off intervention is not enough to maintain a deliberately-managed community state against succession's constant underlying pressure.
Conservation decisions weigh ecological evidence (target-species abundance and diversity, habitat condition and its change over time) against human needs (food production and livelihoods, recreation, access and cost) — sustainable management requires balancing both sets of priorities, not ecological evidence alone.
The appropriate management approach depends on the specific site and its particular conservation goals — there is no single universally 'correct' management strategy that applies to every habitat regardless of what that habitat is actually being conserved for.

Worked examples

Worked example 3.7.4 · 5 marks

In an investigation into the abundance of a particular plant species in a field, 20 quadrats of 1 m² each were placed at random locations.

The plant was present in 14 of the 20 quadrats, and its percentage cover was recorded in each quadrat where present, giving a mean percentage cover of 35% across those 14 quadrats.

Calculate the plant's frequency (as a percentage of quadrats in which it was present), and explain why frequency and percentage cover are both useful but give genuinely different information about a population.

Show worked solution

Frequency:

= \frac{14}{20} \times 100 = 70%

Frequency describes how widespread a species is; percentage cover describes how dense/dominant it is where it occurs.

A species could have high frequency but low cover, or the reverse — using only one measure gives an incomplete picture.

Mark scheme · 5 marks

Calculates frequency = 70% 1 mark
States frequency describes how widespread the species is 1 mark
States percentage cover describes density/dominance where present 1 mark
Explains the two measures can diverge (high frequency, low cover, or vice versa) 1 mark
States both are needed for a complete picture of abundance 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.7.4 · 5 marks

A newly exposed area of bare rock, created by a retreating glacier, is observed over several decades.

Lichens and mosses colonise first, followed eventually by grasses and small herbaceous plants, then shrubs, and finally trees, forming a stable woodland community.

Explain, in terms of how each successional stage changes the environment, why this particular sequence of species occurs, and name the type of succession this describes.

Show worked solution

This is primary succession, starting from bare, previously lifeless substrate.

Lichens and mosses (pioneers) weather the rock and, decomposing, form a thin soil layer suitable for grasses.

Grasses further build soil depth and organic content, eventually permitting shrubs and then trees.

Each stage alters conditions to favour the next.

The final stable community (woodland) is the climax community.

Mark scheme · 5 marks

Names primary succession, justified by bare, previously lifeless substrate 1 mark
States lichens/mosses are pioneer species, weathering rock and beginning soil formation 1 mark
States decomposing pioneers build soil suitable for the next stage 1 mark
Explains each stage alters conditions to favour the next stage's species 1 mark
Names the final stable community as the climax community 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.