IB DP Environmental Systems and Societies – Ecosystems, energy flow and biogeochemical cycles Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set is for IB Diploma Programme Environmental Systems and Societies, syllabus sections 2.1 to 2.3 (individuals, populations, communities and ecosystems; energy and biomass in ecosystems; biogeochemical cycles). It follows the IB ESS subject brief for first assessment 2026 – the course examined in the May and November 2026, 2027 and 2028 sessions. All questions suit both SL and HL. All data sets are fictional and written for practice.

Before you start, work through the study guide and revision notes. Course hub: IB DP ESS. Checklist: IB DP ESS checklist. For broader Topic 2 questions, see the Topic 2 Ecology practice set.

Questions

1. Distinguish between the fundamental niche and the realised niche of a species. [2]

2. Distinguish between gross primary productivity (GPP) and net primary productivity (NPP). [2]

3. Ten 0.25 m² quadrats were placed at random in a 1200 m² field. The numbers of a clover species counted were: 3, 5, 2, 4, 6, 1, 3, 4, 5, 7. Estimate the number of clover plants in the field. [3]

4. In a pond, 48 water beetles were caught, marked and released. Two days later 60 were caught, of which 16 were marked.

(a) Calculate the estimated population size. [2] (b) State two assumptions this method makes. [2]

5. A population of flour beetles was kept in a container with a fixed daily food supply.

Week	0	1	2	3	4	5	6	7	8	9	10	11
Beetles	10	14	21	33	52	80	112	138	151	156	158	157

(a) Estimate the carrying capacity of the container. [1] (b) Explain, using the idea of negative feedback, why the population stops increasing. [4]

6. A population of grazing sheep on an upland pasture has the following energy budget (kJ m⁻² yr⁻¹): food eaten 2400; faecal loss 1560; respiration 690.

(a) Calculate the gross secondary productivity (GSP). [2] (b) Calculate the net secondary productivity (NSP). [2] (c) Explain why a large proportion of the energy eaten by the sheep is lost in faeces. [2]

7. The table shows productivity at each trophic level of a lake ecosystem.

Trophic level	Productivity / kJ m⁻² yr⁻¹
Producers (NPP)	25 000
Primary consumers	2250
Secondary consumers	180
Tertiary consumers	12.6

(a) Calculate the ecological efficiency of each of the three transfers. [3] (b) Calculate the percentage of producer NPP that reaches the tertiary consumers. [1] (c) Explain two reasons why energy is lost between trophic levels. [2] (d) Use the data to explain why more people can be fed per hectare from crops than from livestock. [2]

8. Explain why a pyramid of numbers for an oak woodland may be inverted, while its pyramid of productivity cannot be. [2]

9. In the nitrogen cycle:

(a) Name the process by which N₂ is converted to ammonium in legume root nodules. [1] (b) Name the process by which ammonium is converted to nitrate. [1] (c) Name the process by which nitrate is converted to N₂ in waterlogged soil. [1] (d) Explain why a farmer might grow clover in rotation with wheat. [2]

10. Carbon flows were measured in two forest plots (units: tonnes of carbon per hectare per year). Plot P is intact forest; plot Q was logged two years earlier.

Flow	Plot P	Plot Q
GPP	22.0	9.0
Plant respiration	11.5	4.8
Decomposer (soil) respiration	8.3	10.6

(a) Calculate the NPP of each plot. [2] (b) Calculate the net carbon exchange of each plot (NPP − decomposer respiration). [2] (c) Using your answers, state whether each plot is a carbon sink or source, and explain why decomposer respiration is higher in plot Q. [3]

11. Discuss how human activities have altered the carbon and nitrogen cycles, and evaluate one strategy for reducing the impact on each cycle. [9]

Answers

1. The fundamental niche is the full range of conditions and resources a species could use [1]; the realised niche is the smaller part it actually occupies because of competition, predation or other interactions [1]. [2] Examiner insight: “Distinguish” needs both terms defined and the contrast clear; a definition of only one term earns at most one mark.

2. GPP is the total energy (or biomass) fixed by producers in photosynthesis per unit area per unit time [1]; NPP is what remains after the producers’ own respiration, NPP = GPP − R [1]. [2] Examiner insight: Leaving out “per unit area per unit time” often costs the first mark, because productivity is a rate, not a stock.

3. Mean = 40 ÷ 10 = 4.0 per quadrat [1]; density = 4.0 ÷ 0.25 = 16 per m² [1]; 16 × 1200 = 19 200 clover plants [1]. [3] Examiner insight: Method marks are available for each step, so write the density line even if the final number is wrong.

4. (a) N = (M × C) / R = (48 × 60) / 16 [1] = 180 beetles [1]. (b) Any two: no births, deaths or migration between samples [1]; marks are not lost and do not affect survival or catchability [1]; marked beetles mix fully with the population. [4] Examiner insight: Only the first two assumptions given are normally credited, so list your two strongest and do not pad with a third.

5. (a) About 157 beetles (accept 155-158) [1]. (b) As density rises, food per beetle falls because the supply is fixed [1]; competition for food (and build-up of waste) increases [1]; birth rate falls and/or death rate rises [1]; when births equal deaths the population levels off at carrying capacity – the change reduces itself, which is negative feedback [1]. [5] Examiner insight: The feedback mark needs the loop closed: say that the rise in numbers causes the change that stops the rise.

6. (a) GSP = food eaten − faecal loss = 2400 − 1560 [1] = 840 kJ m⁻² yr⁻¹ [1]. (b) NSP = GSP − R = 840 − 690 [1] = 150 kJ m⁻² yr⁻¹ [1]. (c) Grass contains a lot of cellulose and lignin, which are hard to digest [1]; undigested material leaves as faeces, so is never assimilated into sheep tissue (here 65% of intake) [1]. [6] Examiner insight: An answer with the correct number but no unit usually loses the accuracy mark in (a) or (b).

7. (a) 2250 ÷ 25 000 × 100 = 9.0% [1]; 180 ÷ 2250 × 100 = 8.0% [1]; 12.6 ÷ 180 × 100 = 7.0% [1]. (b) 12.6 ÷ 25 000 × 100 = 0.0504% [1]. (c) Any two: most assimilated energy is respired and lost as heat [1]; parts of the lower level are not eaten or not digested and pass to decomposers [1]; energy is used for movement and other life processes. (d) Only about 9% of producer energy reaches primary consumers [1]; eating crops directly uses the producer level, so roughly ten times more food energy is available per hectare than from livestock [1]. [8] Examiner insight: In (d) the mark depends on quoting or using the data; a general statement about “energy loss” without a figure from the table scores less.

8. In the woodland one oak tree supports thousands of insects, so the producer bar is narrower than the consumer bar [1]; productivity measures energy flow over time, and each level receives only part of the energy from the level below, so it can never be larger [1]. [2] Examiner insight: Both halves of the question carry a mark; explaining only the inverted pyramid of numbers caps the answer at one.

9. (a) Nitrogen fixation [1]. (b) Nitrification [1]. (c) Denitrification [1]. (d) Clover is a legume with Rhizobium in its root nodules that fixes nitrogen [1]; when it is ploughed in and decomposes, ammonium and then nitrate are added to the soil for the wheat, reducing the need for artificial fertiliser [1]. [5] Examiner insight: For “name” questions, one-word answers are enough, but misspellings that change the meaning (nitrification for denitrification) score zero.

10. (a) P: 22.0 − 11.5 = 10.5 [1]; Q: 9.0 − 4.8 = 4.2 t C ha⁻¹ yr⁻¹ [1]. (b) P: 10.5 − 8.3 = +2.2 [1]; Q: 4.2 − 10.6 = −6.4 t C ha⁻¹ yr⁻¹ [1]. (c) Plot P is a carbon sink and plot Q a carbon source [1]; logging leaves large amounts of dead wood, roots and litter for decomposers [1]; removing the canopy warms the soil and exposes it, which speeds up decomposition [1]. [7] Examiner insight: The sink/source mark must follow from the sign of your calculated values; an answer that contradicts its own figures loses the mark even if the idea is right.

11. Indicative creditworthy points:

• Carbon: fossil-fuel combustion moves carbon from long-term storage to the atmosphere; deforestation removes a sink and adds a source through burning and decay [1].
• Result: atmospheric CO₂ has risen from about 280 ppm to above 420 ppm, and more CO₂ dissolving in the ocean lowers pH [1].
• Nitrogen: the Haber process fixes large amounts of N₂ for fertiliser [1].
• Excess nitrate leaches into water, causing eutrophication [1].
• Combustion also releases nitrogen oxides that contribute to acid deposition.
• Carbon strategy (e.g. reforestation) increases photosynthesis and builds a sink [1]; but it is slow, needs land, and stored carbon can be lost to fire [1].
• Nitrogen strategy (e.g. precision fertiliser use or legume rotation) cuts leaching [1]; but may reduce yields or need investment and training [1].
• Reasoned conclusion: reducing emissions or inputs at the source is more effective than managing them afterwards, and both need economic support to be adopted [1]. [9] Examiner insight: An extended ESS response is judged on balance; strategies described without any limitation rarely reach the top of the mark range. These points are indicative: the real exam judges extended responses as a whole against the IB’s own markscheme, not point by point, so ask your teacher how they are assessed.

Where marks are usually lost

• Productivity answers without units, or with units that miss the time component.
• Subtracting respiration before faecal loss when calculating consumer productivity.
• Ecological efficiency divided the wrong way up.
• Lincoln index answers left as non-integers without comment.
• Explaining a population plateau with “resources run out” but no feedback loop.
• Naming a nitrogen process without the bacteria or the oxygen condition when “explain” is asked.
• Carbon budgets where the sink/source conclusion does not match the calculated sign.
• Evaluation answers that list strategies with no limitation or judgement.

Next steps

• Revision notes for this unit
• Study guide for this unit
• IB DP ESS course hub
• Printable IB DP ESS checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

International Baccalaureate Organization, Diploma Programme Subject Brief, Environmental systems and The brief lists subtopics for Topic 1 only; the subtopic numbering on this page follows the printable ESS checklist. societies, first assessment 2026.

The brief gives Topic 2 Ecology 22 teaching hours at SL and 35 at HL. It does not list the subtopics or learning outcomes for this topic, so the numbered subtopics and outcomes on this page follow the syllabus numbering used in the printable ESS checklist, not the brief itself.