IB HL

Ecology: Populations, Communities & Energy Transfer

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Complete Study Guide

Topics Covered

  1. Populations and Communities (C4.1)
  2. Transfers of Energy and Matter (C4.2)
  3. HL Extension — Diversity Indices & Statistical Tests
  4. Exam Strategy & Common Mistakes
  5. Mixed Practice — Exam Style

Aligned to IB Biology HL 2025 syllabus — C4.1, C4.2


Watch: Energy Flow Through Ecosystems

Stephanie Castle · 10 min · Trophic levels, food chains, energy transfer efficiency, producers and consumers


Section 1: Populations and Communities (C4.1)

Ecology is the study of relationships between living organisms and their environment. Understanding how populations grow, interact, and form communities is central to IB Biology HL and appears heavily on both Paper 1 and Paper 2.

Core definitions to memorise:

TermDefinition
SpeciesA group of organisms that can interbreed and produce fertile offspring
PopulationA group of organisms of the same species living in the same area at the same time
CommunityAll the populations of different species living and interacting in the same area
EcosystemA community of organisms and its abiotic (non-living) environment
BiomeA large geographical region with a distinctive climate and community of organisms (e.g. tropical rainforest, tundra)
HabitatThe environment in which a species normally lives
NicheThe role of a species in its ecosystem, including its habitat, feeding relationships, and interactions
Carrying capacity (KK)The maximum population size that an environment can sustain indefinitely

1.1 Sampling Techniques

Ecologists cannot count every individual in a population, so they use sampling techniques to estimate population size.

Quadrats are square frames (typically 0.25 m20.25 \text{ m}^2 or 1 m21 \text{ m}^2) placed randomly within a study area. The number or percentage cover of a species is recorded in each quadrat, and the results are used to estimate population density across the whole area.

IB exam tip: Random sampling requires using a random number generator to determine quadrat positions. If you describe “throwing the quadrat over your shoulder,” examiners may not award the mark — state that coordinates are generated randomly using a table or calculator.

Transects are lines laid across an area where environmental conditions change (e.g. from shoreline to inland). Organisms are sampled at regular intervals along the line. A line transect records species touching the line; a belt transect uses quadrats placed along the line for more detailed data.

Capture-Mark-Recapture (Lincoln Index)

For mobile animals, the capture-mark-recapture method is used:

  1. Capture a sample of individuals, count them, mark them, and release them
  2. After a period (allowing marked individuals to mix with the population), capture a second sample
  3. Count the total caught and the number of marked individuals in the second sample

The Lincoln Index formula estimates population size:

N=n1×n2m2N = \frac{n_1 \times n_2}{m_2}

Where:

  • NN = estimated population size
  • n1n_1 = number captured and marked in the first sample
  • n2n_2 = total number captured in the second sample
  • m2m_2 = number of marked individuals recaptured in the second sample

Worked Example: Lincoln Index

A researcher captures 40 snails in a garden, marks them with non-toxic paint, and releases them. One week later, they capture 50 snails, of which 10 are marked.

N=n1×n2m2=40×5010=200010=200N = \frac{n_1 \times n_2}{m_2} = \frac{40 \times 50}{10} = \frac{2000}{10} = 200

Estimated population size = 200 snails

Assumptions of capture-mark-recapture (frequently tested):

  • The population is closed (no immigration, emigration, births, or deaths between samples)
  • Marking does not affect survival or behaviour (e.g. does not make individuals more visible to predators)
  • Marked individuals mix randomly with the rest of the population
  • Marks do not rub off or become unidentifiable
  • Each individual has an equal chance of being captured

If any assumption is violated, the estimate will be inaccurate. For example, if marked individuals are more easily predated, fewer will be recaptured, and NN will be overestimated.

1.2 Population Growth

Populations can grow in two characteristic patterns:

Exponential (J-curve) growth occurs when resources are unlimited. The population grows at an ever-increasing rate with no upper limit. This is rare in nature but can occur temporarily when a species colonises a new habitat or when a limiting factor is removed.

dNdt=rN\frac{dN}{dt} = rN

Where rr is the intrinsic rate of natural increase and NN is the population size.

Logistic (S-curve) growth is the more realistic model. Growth starts exponentially but slows as the population approaches the carrying capacity (KK) of the environment. Environmental resistance (limited food, space, predation, disease) acts as a brake on growth.

dNdt=rN(KNK)\frac{dN}{dt} = rN\left(\frac{K - N}{K}\right)

Population size (N)TimeKJ-curveS-curveCarrying capacityLag phaseExponential phasePlateau phase

Phases of logistic growth:

PhaseDescription
Lag phasePopulation is small; growth is slow as individuals establish and reproduce
Exponential phaseResources abundant; population grows rapidly as birth rate exceeds death rate
Transitional phaseGrowth rate slows as environmental resistance increases
Plateau phasePopulation fluctuates around KK; birth rate approximately equals death rate

1.3 Factors Affecting Population Size

Population size changes according to four factors:

Population change=(B+I)(D+E)\text{Population change} = (B + I) - (D + E)

Where BB = births (natality), II = immigration, DD = deaths (mortality), EE = emigration.

Density-dependent factors become more intense as population density increases:

  • Competition for food, water, space, mates
  • Predation (more prey attracts more predators)
  • Disease (spreads more easily in dense populations)
  • Parasitism

Density-independent factors affect populations regardless of density:

  • Natural disasters (floods, fires, volcanic eruptions)
  • Climate change / extreme weather
  • Human activities (habitat destruction, pollution)

1.4 Community Interactions

Types of species interactions:

InteractionEffect on Species AEffect on Species BExample
Competition--Two plant species competing for light
Predation++ (predator)- (prey)Lion hunting zebra
Mutualism++++Mycorrhizae (fungi and plant roots)
Parasitism++ (parasite)- (host)Tapeworm in human intestine

Interspecific competition occurs between individuals of different species that share a resource. Intraspecific competition occurs between individuals of the same species — this is often more intense because individuals require exactly the same resources.

The Competitive Exclusion Principle (Gause’s Principle): Two species cannot occupy the same ecological niche in the same habitat at the same time. One species will outcompete the other, leading to local extinction of the inferior competitor. In practice, species may coexist through resource partitioning — they divide the niche by exploiting slightly different resources or foraging at different times.

1.5 Ecological Niches

The fundamental niche is the full range of environmental conditions and resources a species could potentially occupy in the absence of competition and predation.

The realized niche is the actual niche a species occupies in practice, which is narrower than the fundamental niche due to interspecific competition, predation, and other biotic interactions.

IB exam tip: If asked to distinguish fundamental and realized niches, always explain why the realized niche is smaller — it is competition and predation that restrict it, not simply preference.

1.6 Ecological Succession

Succession is the gradual change in species composition and community structure over time in a given area.

Primary succession starts on bare, lifeless substrates (e.g. bare rock after volcanic eruption, newly formed sand dunes). There is no pre-existing soil.

  1. Pioneer species (lichens, mosses) colonise bare rock
  2. They weather the rock and add organic matter when they die, forming a thin soil layer
  3. Small plants (grasses, ferns) can now grow in the thin soil
  4. Soil deepens over time, supporting shrubs and then trees
  5. A climax community is eventually reached — a stable, self-sustaining community in equilibrium with the climate

Secondary succession occurs in areas where a community has been disturbed but soil remains (e.g. after a forest fire, abandoned farmland). It proceeds faster because soil, seeds, and root systems are already present.

Key succession terms:

TermDefinition
Pioneer speciesFirst organisms to colonise a bare or disturbed area; they are hardy, fast-reproducing, and tolerant of extreme conditions
Seral stage (sere)Each intermediate community in the succession sequence
Climax communityThe final, stable community; species composition remains relatively unchanged unless disturbed
Deflected climax (plagioclimax)A community held below the climax by human activity (e.g. grazing, mowing)

Section 2: Transfers of Energy and Matter (C4.2)

2.1 Food Chains and Food Webs

A food chain shows a single pathway of energy transfer from one organism to the next. A food web is a network of interconnected food chains showing the feeding relationships in a community.

Trophic levels:

Trophic LevelOrganismsExample
Producer (autotroph)Photosynthetic organisms that convert light energy to chemical energyGrass, phytoplankton
Primary consumerHerbivores that feed on producersRabbit, zooplankton
Secondary consumerCarnivores that feed on primary consumersFox, small fish
Tertiary consumerTop predators that feed on secondary consumersEagle, shark
Decomposer (saprotroph)Organisms that break down dead organic matterFungi, bacteria

2.2 Energy Flow Through Ecosystems

Energy enters most ecosystems as sunlight and is converted to chemical energy by producers through photosynthesis. Energy flows through trophic levels but is not recycled — it is lost at each transfer.

The 10% Rule: On average, only about 10% of the energy at one trophic level is transferred to the next. The remaining ~90% is lost as:

  • Heat from cellular respiration (the largest loss)
  • Uneaten material (roots, bones, tough plant fibres)
  • Excretory products (urine, faeces)
  • Dead organisms not consumed before decomposition

This is why food chains rarely exceed 4-5 trophic levels — there is insufficient energy to support another level.

2.3 Pyramids of Ecology

Pyramids of energy show the rate of energy flow (in kJ m2yr1\text{kJ m}^{-2} \text{yr}^{-1}) at each trophic level. They are always upright because energy is lost at each transfer.

Pyramids of biomass show the total mass of organisms at each trophic level. They are usually upright but can be inverted (e.g. in ocean ecosystems where phytoplankton biomass at any one time is less than zooplankton biomass because phytoplankton reproduce extremely rapidly).

Pyramids of numbers show the number of individuals at each trophic level. These are often irregular (e.g. one oak tree supports millions of caterpillars).

IB exam tip: If asked “which type of pyramid is always upright?”, the answer is pyramid of energy. Pyramids of biomass and numbers can both be inverted in certain ecosystems. Be prepared to explain why with a specific example.

2.4 Productivity

Gross primary productivity (GPP) is the total rate of energy fixation (photosynthesis) by producers in an ecosystem.

Net primary productivity (NPP) is the rate of energy storage in plant biomass after accounting for respiratory losses:

NPP=GPPR\text{NPP} = \text{GPP} - R

Where RR is the energy lost through cellular respiration by the producers.

NPP represents the energy available to the next trophic level (primary consumers).

Worked Example: Calculating NPP

A grassland ecosystem has a GPP of 20,000 kJ m2yr120{,}000 \text{ kJ m}^{-2} \text{yr}^{-1}. Producers use 12,000 kJ m2yr112{,}000 \text{ kJ m}^{-2} \text{yr}^{-1} for respiration.

NPP=GPPR=20,00012,000=8,000 kJ m2yr1\text{NPP} = \text{GPP} - R = 20{,}000 - 12{,}000 = 8{,}000 \text{ kJ m}^{-2} \text{yr}^{-1}

Therefore, 8,000 kJ m2yr18{,}000 \text{ kJ m}^{-2} \text{yr}^{-1} of energy is stored in plant biomass and available to herbivores.

If primary consumers assimilate 10% of this: 8,000×0.10=800 kJ m2yr18{,}000 \times 0.10 = 800 \text{ kJ m}^{-2} \text{yr}^{-1} is transferred to the secondary consumer level.

2.5 The Carbon Cycle

Carbon is recycled through ecosystems via biological, geological, and chemical processes.

CO₂ in atmosphereProducersConsumersDecomposersFossil fuelsPhotosynthesisRespirationFeedingRespirationDeathFossilisationDecompositionCombustion

Carbon cycle processes:

ProcessWhat happensEffect on atmospheric CO2\text{CO}_2
PhotosynthesisCO2\text{CO}_2 fixed into organic molecules by producersDecreases CO2\text{CO}_2
RespirationOrganic molecules broken down, releasing CO2\text{CO}_2Increases CO2\text{CO}_2
CombustionBurning of fossil fuels or biomass releases CO2\text{CO}_2Increases CO2\text{CO}_2
DecompositionSaprotrophs break down dead organic matter, releasing CO2\text{CO}_2Increases CO2\text{CO}_2
FossilisationOrganic carbon trapped in sedimentary rock / fossil fuels over millions of yearsRemoves CO2\text{CO}_2 (long-term)

2.6 The Nitrogen Cycle

Nitrogen is essential for amino acids, proteins, and nucleic acids. Although N2\text{N}_2 makes up 78% of the atmosphere, most organisms cannot use it directly — it must be fixed into usable forms.

Nitrogen cycle processes:

ProcessDescriptionOrganisms involved
Nitrogen fixationN2NH3\text{N}_2 \rightarrow \text{NH}_3 (ammonia) / NH4+\text{NH}_4^+ (ammonium)Rhizobium (in root nodules of legumes), free-living bacteria, lightning
NitrificationNH4+NO2NO3\text{NH}_4^+ \rightarrow \text{NO}_2^- \rightarrow \text{NO}_3^-Nitrifying bacteria (Nitrosomonas, Nitrobacter)
AssimilationPlants absorb NO3\text{NO}_3^- or NH4+\text{NH}_4^+ from soil; used to make amino acids and nucleotidesPlants
AmmonificationDecomposition of proteins / urea releases NH4+\text{NH}_4^+ back to soilDecomposers (saprotrophic bacteria and fungi)
DenitrificationNO3N2\text{NO}_3^- \rightarrow \text{N}_2 (returned to atmosphere)Denitrifying bacteria (anaerobic conditions, e.g. waterlogged soil)

Common exam mistake: Students confuse nitrogen fixation with nitrification. Nitrogen fixation converts N2\text{N}_2 gas into NH3\text{NH}_3/NH4+\text{NH}_4^+. Nitrification converts NH4+\text{NH}_4^+ into NO3\text{NO}_3^- (nitrate). These are carried out by different bacteria. Also note: denitrification is the reverse of fixation in terms of outcome and occurs under anaerobic conditions.

2.7 Bioaccumulation and Biomagnification

Bioaccumulation is the build-up of a persistent, non-biodegradable substance (e.g. DDT, mercury, PCBs) within an organism over its lifetime. The substance is absorbed faster than it can be metabolised or excreted.

Biomagnification is the increasing concentration of a substance at each successive trophic level in a food chain. Top predators accumulate the highest concentrations.

Classic Example: DDT and Biomagnification

DDT (dichlorodiphenyltrichloroethane) was widely used as a pesticide. It is fat-soluble, persistent, and does not break down easily:

  • Water: 0.000003 ppm0.000003 \text{ ppm}
  • Phytoplankton: 0.04 ppm0.04 \text{ ppm}
  • Zooplankton: 0.5 ppm0.5 \text{ ppm}
  • Small fish: 2 ppm2 \text{ ppm}
  • Large fish: 25 ppm25 \text{ ppm}
  • Fish-eating birds (e.g. osprey): 250+ ppm250+ \text{ ppm}

At high concentrations, DDT caused eggshell thinning in birds of prey (e.g. peregrine falcon, bald eagle), leading to population crashes. This led to the ban of DDT in many countries.

IB exam tip: If asked why biomagnification occurs, the key points are: (1) the substance is persistent (not broken down), (2) it is fat-soluble (stored in lipid tissues, not excreted), and (3) organisms at higher trophic levels consume many organisms from the level below, accumulating their stored toxins.


Section 3: HL Extensions — Diversity Indices & Statistical Tests

3.1 Simpson’s Diversity Index

Biodiversity can be quantified using the Simpson’s Diversity Index. This measures species richness (number of species) and evenness (how equally individuals are distributed among species).

D=1(nN)2D = 1 - \sum\left(\frac{n}{N}\right)^2

Where:

  • DD = diversity index (ranges from 0 to 1; higher values = greater diversity)
  • nn = number of individuals of each species
  • NN = total number of all individuals

Worked Example: Simpson’s Diversity Index

A meadow survey finds the following:

SpeciesNumber of individuals (nn)nN\frac{n}{N}(nN)2\left(\frac{n}{N}\right)^2
Daisy300.3000.0900
Buttercup250.2500.0625
Clover200.2000.0400
Dandelion150.1500.0225
Thistle100.1000.0100

Total (NN) = 100

(nN)2=0.0900+0.0625+0.0400+0.0225+0.0100=0.2250\sum\left(\frac{n}{N}\right)^2 = 0.0900 + 0.0625 + 0.0400 + 0.0225 + 0.0100 = 0.2250

D=10.2250=0.775D = 1 - 0.2250 = 0.775

Interpretation: D=0.775D = 0.775 indicates relatively high diversity. A value close to 1 means high diversity (many species, evenly distributed); a value close to 0 means low diversity (dominated by one species).

For comparison, if one species had 96 individuals and four others had 1 each: D=1(0.962+4×0.012)=10.9220=0.078D = 1 - (0.96^2 + 4 \times 0.01^2) = 1 - 0.9220 = 0.078 — very low diversity.

Common calculation errors:

  1. Forgetting to square each nN\frac{n}{N} value before summing
  2. Forgetting to subtract from 1 at the end
  3. Using the wrong version of the formula — IB uses D=1(n/N)2D = 1 - \sum(n/N)^2 (not 1/1/\sum)

Always show your working clearly in a table format as shown above — it is easy to make arithmetic mistakes and the table allows examiners to award method marks even if the final answer is wrong.

3.2 Chi-Squared Test (χ2\chi^2)

The chi-squared test is used to determine whether there is a statistically significant association between two categorical variables (e.g. species distribution and an abiotic factor like soil type).

χ2=(OE)2E\chi^2 = \sum \frac{(O - E)^2}{E}

Where:

  • OO = observed frequency
  • EE = expected frequency (calculated assuming no association)

Worked Example: Chi-Squared Test

A student investigates whether the distribution of a moss species is associated with the direction a wall faces (north vs south). They sample 100 quadrats:

Moss presentMoss absentRow total
North-facing351550
South-facing203050
Column total5545100

Step 1: Calculate expected values

E=row total×column totalgrand totalE = \frac{\text{row total} \times \text{column total}}{\text{grand total}}

Moss present (EE)Moss absent (EE)
North-facing50×55100=27.5\frac{50 \times 55}{100} = 27.550×45100=22.5\frac{50 \times 45}{100} = 22.5
South-facing50×55100=27.5\frac{50 \times 55}{100} = 27.550×45100=22.5\frac{50 \times 45}{100} = 22.5

Step 2: Calculate χ2\chi^2

χ2=(3527.5)227.5+(1522.5)222.5+(2027.5)227.5+(3022.5)222.5\chi^2 = \frac{(35-27.5)^2}{27.5} + \frac{(15-22.5)^2}{22.5} + \frac{(20-27.5)^2}{27.5} + \frac{(30-22.5)^2}{22.5}

=56.2527.5+56.2522.5+56.2527.5+56.2522.5= \frac{56.25}{27.5} + \frac{56.25}{22.5} + \frac{56.25}{27.5} + \frac{56.25}{22.5}

=2.045+2.500+2.045+2.500=9.09= 2.045 + 2.500 + 2.045 + 2.500 = 9.09

Step 3: Determine significance

Degrees of freedom = (r1)(c1)=(21)(21)=1(r-1)(c-1) = (2-1)(2-1) = 1

Critical value at p=0.05p = 0.05 with 1 degree of freedom = 3.8413.841

Since χ2=9.09>3.841\chi^2 = 9.09 > 3.841, we reject the null hypothesis. There is a statistically significant association between wall direction and moss distribution at the 5% significance level.

IB exam tip: Always state the null hypothesis before calculating. For example: “There is no significant association between wall direction and moss distribution.” You must compare the calculated χ2\chi^2 to the critical value from the table and state whether you reject or fail to reject the null hypothesis at p=0.05p = 0.05.

3.3 Mark-Recapture Assumptions and Limitations

HL extension — evaluating mark-recapture:

The Lincoln Index provides an estimate only. Its reliability depends on the validity of the assumptions:

AssumptionIf violatedEffect on estimate
Closed population (no births, deaths, immigration, emigration)Individuals leave or enterOverestimate (if emigration) or underestimate (if immigration)
Marking does not affect survivalMarked individuals predated more easilyFewer recaptured, so NN is overestimated
Marks do not rub offMarks lost between samplesFewer marked recaptured, so NN is overestimated
Random mixingMarked individuals remain clusteredIf second sample is from the cluster, NN is underestimated; if from elsewhere, NN is overestimated
Equal probability of captureSome individuals are trap-shy or trap-happyTrap-shy: overestimate; trap-happy: underestimate

To improve reliability: increase sample sizes, allow adequate mixing time, use durable non-toxic marking methods, and verify assumptions where possible.


Section 4: Exam Strategy & Common Mistakes

Top mistakes in ecology exams:

  1. Confusing biomass and energy pyramids — only energy pyramids are always upright
  2. Saying “energy is recycled” — energy flows through ecosystems; only matter (carbon, nitrogen) is recycled
  3. Forgetting units — productivity must be in kJ m2yr1\text{kJ m}^{-2} \text{yr}^{-1}; diversity index is unitless
  4. Using “adapted” loosely — say “the species has adaptations that are suited to…” not “the species adapted to its environment” (which implies Lamarckian evolution)
  5. Confusing food chains and food webs — a food chain is a single linear pathway; a food web shows multiple interconnected chains
  6. Saying “top predators have the most energy” — they have the least energy; they have the highest concentration of persistent toxins via biomagnification
  7. Simpson’s Index errors — forgetting to square, forgetting to subtract from 1, or using the reciprocal formula instead of the complement formula

IB Exam-Style Questions

Question 1 (3 marks)

Explain why food chains rarely have more than four or five trophic levels.

Markscheme
  • Energy is lost at each trophic level, mainly as heat through cellular respiration; [1]
  • Only approximately 10% of the energy at one trophic level is transferred to the next; [1]
  • By the fourth or fifth trophic level, there is insufficient energy to sustain a viable population of organisms; [1]

Reject “energy is used up” without reference to heat loss through respiration. Reject “the animals at the top are too big” — size is not the limiting factor; energy availability is.


Question 2 (4 marks)

A population of rabbits is introduced to an island with abundant vegetation and no predators. Describe and explain the expected changes in population size over time.

Markscheme
  • Initially the population grows slowly (lag phase) as there are few individuals reproducing and becoming established; [1]
  • Population then grows exponentially / rapidly (exponential phase) because resources (food, space) are abundant and there is no predation; birth rate greatly exceeds death rate; [1]
  • Growth rate slows (transitional phase) as intraspecific competition for resources increases and environmental resistance builds; [1]
  • Population levels off at / fluctuates around the carrying capacity (KK) where birth rate approximately equals death rate; resources are limiting; [1]

Award marks for description of the S-curve shape with appropriate biological explanation for each phase. Reject “the population keeps growing forever” — all populations are eventually limited by resources.


Question 3 (3 marks)

Distinguish between bioaccumulation and biomagnification.

Markscheme
  • Bioaccumulation is the build-up of a persistent, non-biodegradable substance within a single organism over its lifetime; the substance is absorbed faster than it is excreted or metabolised; [1]
  • Biomagnification is the increasing concentration of a substance at successively higher trophic levels in a food chain; [1]
  • Biomagnification occurs because organisms at higher trophic levels consume many organisms from lower levels, each containing accumulated toxins; the substance is fat-soluble and persistent, so it becomes more concentrated at each transfer; [1]

Reject confusion between the two terms. Award marks for a named example (e.g. DDT, mercury) with correct application of both terms.


Question 4 (4 marks)

Outline the role of bacteria in the nitrogen cycle.

Markscheme
  • Nitrogen-fixing bacteria (e.g. Rhizobium in root nodules of legumes) convert atmospheric N2\text{N}_2 into NH3\text{NH}_3 / NH4+\text{NH}_4^+ (ammonium); [1]
  • Nitrifying bacteria (e.g. Nitrosomonas, Nitrobacter) convert NH4+\text{NH}_4^+ to NO2\text{NO}_2^- (nitrite) and then to NO3\text{NO}_3^- (nitrate) in aerobic soil; [1]
  • Decomposing bacteria carry out ammonification — breaking down organic nitrogen (dead organisms, urea) into NH4+\text{NH}_4^+; [1]
  • Denitrifying bacteria convert NO3\text{NO}_3^- back to N2\text{N}_2 gas under anaerobic conditions (e.g. waterlogged soil), returning nitrogen to the atmosphere; [1]

Reject vague statements like “bacteria cycle nitrogen.” Specific roles with correct chemical conversions are required for full marks. Award marks for correct naming of genera where appropriate.


Mixed Practice — Exam Style

How to use this section: Unlike topic-specific practice, these questions are interleaved — they mix all topics from this guide in random order. Before answering, identify which concept or topic area the question is testing. This is exactly the skill you need on Paper 1 and Paper 2, where you don’t know in advance which topic each question covers.

  1. [Energy Flow] In a food chain, which trophic level contains the most energy?

    A. Tertiary consumers (top predators)

    B. Secondary consumers

    C. Primary consumers (herbivores)

    D. Producers (autotrophs)

  2. [Population Growth] A population of bacteria is growing in a flask with limited nutrients. The growth curve is best described as:

    A. Exponential growth only, continuing indefinitely

    B. Logistic growth — initially exponential, then levelling off at carrying capacity

    C. Linear growth at a constant rate

    D. Immediate plateau — growth rate is constant from the start

  3. [Nitrogen Cycle] Which process converts atmospheric N2\text{N}_2 into a form usable by plants?

    A. Nitrification

    B. Denitrification

    C. Nitrogen fixation

    D. Ammonification

  4. [Ecological Pyramids] Which type of ecological pyramid is always upright?

    A. Pyramid of numbers

    B. Pyramid of biomass

    C. Pyramid of energy

    D. All three types are always upright

  5. [Simpson’s Diversity Index] A habitat has 4 species with the following numbers: 90, 5, 3, 2. Another habitat has 4 species with: 25, 25, 25, 25. Which statement is correct?

    A. The first habitat has greater diversity because it has more total individuals

    B. The second habitat has greater diversity because individuals are more evenly distributed among species

    C. Both habitats have equal diversity because they have the same number of species

    D. Diversity cannot be compared without knowing the species names

  6. [Succession] Which of the following best describes primary succession?

    A. Regrowth of a forest after a fire, using existing soil and seed bank

    B. Colonisation of bare rock by pioneer species, leading gradually to a climax community

    C. A farmer abandoning a field, which gradually returns to forest

    D. The replacement of one dominant species by another after a disease outbreak

  7. [Bioaccumulation] A persistent pesticide is found at highest concentrations in:

    A. Producers (phytoplankton)

    B. Primary consumers (zooplankton)

    C. Secondary consumers (small fish)

    D. Tertiary consumers (fish-eating birds)

  8. [Carbon Cycle] Which process removes CO2\text{CO}_2 from the atmosphere?

    A. Respiration

    B. Combustion of fossil fuels

    C. Photosynthesis

    D. Decomposition

  9. [Community Interactions — Distractor] Two species of finch on an island both eat medium-sized seeds. Over time, one species evolves a larger beak and shifts to eating larger seeds. This is an example of:

    A. Mutualism — both species benefit

    B. Parasitism — one species exploits the other

    C. Resource partitioning driven by interspecific competition — the competitive exclusion principle leads to niche differentiation

    D. Predation — one species preys on the other

  10. [Mark-Recapture — Distractor] In a mark-recapture study, a researcher marks 50 fish and releases them. In the second sample, they catch 60 fish, of which 5 are marked. The estimated population size is:

    A. 110

    B. 300

    C. 600

    D. 3000

Show Answers
  1. D — Producers. Energy is lost at each trophic level, so producers always contain the most energy. The 10% rule means each successive level has approximately one-tenth of the previous level’s energy.

  2. B — Logistic growth. With limited nutrients, the population initially grows exponentially when resources are abundant but eventually levels off as nutrients become limiting and the carrying capacity is reached. A (exponential only) ignores resource limitation.

  3. C — Nitrogen fixation. This is the conversion of N2\text{N}_2 gas into NH3\text{NH}_3 / NH4+\text{NH}_4^+ by nitrogen-fixing bacteria (e.g. Rhizobium) or lightning. Nitrification (A) converts NH4+\text{NH}_4^+ to NO3\text{NO}_3^-. Denitrification (B) converts NO3\text{NO}_3^- back to N2\text{N}_2.

  4. C — Pyramid of energy. Energy is always lost at each trophic level, so energy pyramids must be upright. Pyramids of numbers can be inverted (e.g. one tree supporting many insects). Pyramids of biomass can be inverted (e.g. ocean ecosystems with fast-reproducing phytoplankton).

  5. B — The second habitat has greater diversity. Simpson’s Diversity Index accounts for both species richness AND evenness. The first habitat is heavily dominated by one species (90 out of 100), giving low evenness and a low DD value. The second habitat has perfectly even distribution, giving the maximum DD for 4 species. Total number of individuals does not directly determine diversity.

  6. B — Primary succession starts on bare substrates (rock, lava, new sand) where no soil exists. Pioneer species (lichens, mosses) colonise first, gradually building soil and enabling later species. A and C describe secondary succession (soil already present). D describes a single species replacement, not full succession.

  7. D — Tertiary consumers. Biomagnification means persistent, fat-soluble toxins become more concentrated at each successive trophic level. Top predators accumulate the highest concentrations because they consume many organisms that have themselves accumulated the toxin.

  8. C — Photosynthesis. Producers absorb CO2\text{CO}_2 and fix it into organic compounds (C6H12O6\text{C}_6\text{H}_{12}\text{O}_6). Respiration (A), combustion (B), and decomposition (D) all release CO2\text{CO}_2 into the atmosphere.

  9. C — Resource partitioning. When two species compete for the same resource, the competitive exclusion principle predicts one will be displaced unless they differentiate their niches. Character displacement (evolution of differing beak sizes) reduces competition by allowing each species to exploit different food sources. This is not mutualism (both species don’t benefit from competition) or predation.

  10. C — Using the Lincoln Index: N=n1×n2m2=50×605=30005=600N = \frac{n_1 \times n_2}{m_2} = \frac{50 \times 60}{5} = \frac{3000}{5} = 600. Note the common distractor D (3000) is obtained by multiplying numerator values without dividing. Always apply the formula carefully.


May 2026 Prediction Questions

These are NOT official IB questions. These are trend-based practice questions written to reflect the topic areas and question styles most likely to appear on the May 2026 IB Biology HL Paper 2. Based on recent exam patterns (2022—2025), expect heavy weighting on: energy flow and trophic efficiency (including calculations), the nitrogen cycle, and human impacts on ecosystems (pollution, invasive species, habitat fragmentation).


Question 1 — Energy Flow and Ecological Efficiency [8 marks]

The table shows the energy content (kJ m2^{-2} yr1^{-1}) at each trophic level in a grassland ecosystem.

Trophic LevelEnergy (kJ m2^{-2} yr1^{-1})
Producers40,000
Primary consumers4,000
Secondary consumers480
Tertiary consumers52

(a) Calculate the trophic efficiency between producers and primary consumers. Show your working. [2 marks]

(b) Trophic efficiency between primary and secondary consumers differs from your answer in (a). Suggest two reasons why energy transfer efficiency varies between trophic levels. [2 marks]

(c) Explain why food chains rarely exceed five trophic levels. [2 marks]

(d) Evaluate the claim that “eating lower on the food chain is always more energy-efficient for humans.” [2 marks]

Model Answer:

(a) Trophic efficiency = (energy at next level / energy at current level) x 100 = (4000 / 40000) x 100 = 10%. (2)

(b) Any two from: different respiratory losses at each level (ectotherms vs. endotherms); different proportions of biomass that are indigestible (e.g. more cellulose in plants than in animal prey); different feeding efficiency or foraging behaviours; differences in reproductive strategy affecting proportion of energy used for growth. (1 mark each, max 2)

(c) Energy is lost at each trophic level — approximately 90% is lost as heat (respiration), waste, and indigestible material (1); so by the time you reach a fifth or sixth trophic level, the available energy is too low to support a viable population of predators (1).

(d) In favour: consuming plants directly (first trophic level) avoids the 90% energy loss at each trophic step, so theoretically more people can be fed from the same land area (1). However, not all land is suitable for crop production — some is only suitable for grazing, so livestock can convert non-human-edible plant matter into food; also, human nutritional requirements include specific amino acids and micronutrients that may not all be efficiently obtained from plants alone (1). Accept any well-reasoned balanced evaluation.


Question 2 — Nitrogen Cycle and Human Impact [7 marks]

(a) Draw and annotate a diagram of the nitrogen cycle, including the following processes: nitrogen fixation, nitrification, denitrification, decomposition/ammonification, and assimilation by plants. [4 marks]

(b) Excessive use of nitrogen fertiliser in agriculture can lead to eutrophication in nearby water bodies. Outline the sequence of events by which this occurs and its ecological consequences. [3 marks]

Model Answer:

(a) Award 1 mark each for any four of: Nitrogen fixation — N2\text{N}_2 in atmosphere converted to NH3\text{NH}_3 / NH4+\text{NH}_4^+ by Rhizobium bacteria in root nodules or free-living bacteria like Azotobacter / by lightning (1); Nitrification — NH4+\text{NH}_4^+ converted to NO2\text{NO}_2^- then NO3\text{NO}_3^- by nitrifying bacteria (1); Assimilation — plants absorb NO3\text{NO}_3^- / NH4+\text{NH}_4^+ through roots and incorporate nitrogen into amino acids and nucleotides (1); Decomposition/ammonification — decomposers break down organic nitrogen in dead organisms into NH4+\text{NH}_4^+ (1); Denitrification — NO3\text{NO}_3^- converted to N2\text{N}_2 by denitrifying bacteria in anaerobic conditions, returning N2\text{N}_2 to the atmosphere (1). Max 4.

(b) Nitrates leach from fields into waterways (1); algae and cyanobacteria grow rapidly in response to the nutrient influx (algal bloom) (1); algal bloom reduces light penetration, causing submerged plants to die; decomposers break down dead organic matter using aerobic respiration, depleting dissolved oxygen (BOD increases); fish and invertebrates die from oxygen depletion (1).


Question 3 — Biodiversity and Invasive Species [6 marks]

(a) Define biodiversity, distinguishing between species richness and species evenness. [2 marks]

(b) Explain how the introduction of an invasive species can reduce native biodiversity, using a named example. [2 marks]

(c) A researcher uses Simpson’s Diversity Index (DD) to compare two habitats before and after the introduction of an invasive species. Predict how DD would change and explain your reasoning. [2 marks]

Model Answer:

(a) Biodiversity is the variety of life in an area (1). Species richness is the total number of different species present. Species evenness refers to how equally individuals are distributed among those species — a community where all species are equally abundant has greater evenness (1).

(b) Any named invasive species with correct explanation, e.g. cane toads (Rhinella marina) introduced to Australia killed native predators (quolls, goannas) that attempted to eat them because the toads are toxic — native predators lacked evolutionary adaptations to recognise or tolerate the toxin (1, 1). Accept: grey squirrels outcompeting red squirrels in the UK for food and nest sites, also spreading squirrelpox virus; water hyacinth (Eichhornia crassipes) blocking light and oxygen in waterways. Must name the species and explain the mechanism (2).

(c) DD would decrease (become closer to 0) (1); because the invasive species typically dominates numerically (reducing evenness) and may cause local extinction of native species (reducing richness) — both of which lower the DD value in Simpson’s index (1).


IB Biology HL — Ecology: Populations, Communities & Energy Transfer — Complete Study Guide — 2025 Syllabus — Good luck!