Study Guides
OCR A Level Biology A: Ecosystems, biomass transfer and succession (H420)
Study guide to ecosystems, biomass transfer, the nitrogen and carbon cycles, primary succession and sampling statistics for OCR A Level Biology A.
- Subject
- Biology
- Level
- A LEVELS
- Topic
- Ecosystems, biomass transfer and succession
- Author
- Marlbridge Academic Team
- Updated
Aligned to OCR A Level Biology (H420), Version 4.1 (April 2026), for first assessment in 2025. Official specification .
Syllabus page (what it covers and how it is assessed): OCR A Level Biology.
Syllabus points this page covers
H420
- 6 Genetics, evolution and ecosystems (whole topic)
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This guide covers section 6.3.1(a) to (e) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025. It sits in Module 6, Genetics, evolution and ecosystems. It teaches ecosystems, biomass transfer, the nitrogen and carbon cycles, primary succession, and measuring distribution and abundance with statistics. Section 6.3.2 is outside this unit.
Use it with the revision notes and the practice questions. The OCR A Level Biology hub, the printable checklist and the free diagnostics show where it fits. Simpson’s Index is taught in Measuring biodiversity, and general statistics and practical ideas are in Development of Practical Skills.
What this unit covers
| Spec reference | What you must be able to do |
|---|---|
| 6.3.1(a) | Describe ecosystems of different sizes, with named biotic and abiotic factors |
| 6.3.1(b) | Measure biomass transfer, calculate efficiency, explain human manipulation |
| 6.3.1(c) | Explain recycling: decomposers, the nitrogen cycle and the carbon cycle |
| 6.3.1(d) | Describe primary succession to a climax community, and deflected succession |
| 6.3.1(e) | Measure distribution and abundance, and analyse results statistically |
6.3.1(a) Ecosystems
An ecosystem is all the organisms living in an area, together with the non-living features of that area, and the interactions among them. Ecosystems vary hugely in size. A rock pool, a playing field and a single large tree are all ecosystems. A tree supports its own community of insects, lichens, fungi and birds.
Ecosystems are dynamic: populations rise and fall, species arrive and leave, and conditions change with the seasons.
- Biotic factors are caused by living things. Examples: predation, competition for light or food, disease, grazing, and the presence of a pollinator.
- Abiotic factors are the non-living conditions. Examples: temperature, light intensity, soil pH, water availability, salinity and wind speed.
In a rock pool, salinity rises on hot days as water evaporates (abiotic). Limpets grazing algae (biotic) change how much algae is left for others.
6.3.1(b) Biomass transfer through ecosystems
Measuring biomass
Fresh mass is a poor measure because water content varies, so use dry mass.
- Take a representative sample of the organisms.
- Weigh it, then dry it in an oven below the temperature at which it burns (about 80 °C is typical).
- Re-weigh at intervals. The sample is completely dry when the mass is constant.
- Scale up to the area sampled, for example g m⁻².
To measure the energy in biomass, burn a weighed, dried sample in a calorimeter. The energy released warms a known mass of water:
Energy released (J) = mass of water (g) × 4.18 × temperature rise (°C)
Worked example. A 1.20 g dried sample of algae is burned under 50 g of water. The water warms from 18.0 °C to 41.5 °C. The specific heat capacity of water is 4.18 J g⁻¹ °C⁻¹.
- Temperature rise = 41.5 − 18.0 = 23.5 °C.
- Energy = 50 × 4.18 × 23.5 = 4 911.5 J, which is 4.91 kJ.
- Energy per gram = 4.9115 / 1.20 = 4.09 kJ g⁻¹.
The true value is higher, because heat escapes and combustion may be incomplete.
Efficiency of transfer
Efficiency = (biomass transferred ÷ biomass intake) × 100.
The biomass transferred is the part built into the consumer’s own body, available to the next trophic level. The intake is the biomass the consumer takes in.
Worked example. In a pasture, the dry biomass in the plants is 1 480 g m⁻² year⁻¹. Herbivores ingest 296 g m⁻² year⁻¹ and gain 41 g m⁻² year⁻¹ of new body mass. Carnivores eat 30 g m⁻² year⁻¹ of the herbivore biomass and gain 3.6 g m⁻² year⁻¹.
- Proportion of plant biomass ingested = 296 / 1 480 × 100 = 20.0%. The rest is not eaten.
- Herbivore efficiency = 41 / 296 × 100 = 13.9%.
- Carnivore efficiency = 3.6 / 30 × 100 = 12.0%.
Why transfer is inefficient
- Not all biomass is eaten: roots, bones and shells are left, and some prey escape.
- Not all that is eaten is digested. Indigestible material such as cellulose or fur leaves as faeces.
- Biomass is lost as urea and other excretory products.
- Respiration releases the energy as heat and for movement, and carbon dioxide leaves the body.
Because so little passes on, food chains are short and each level supports less biomass than the one below.
Human manipulation of biomass transfer
Farmers raise the efficiency of conversion into human food.
- Increase the input. Fertilisers supply nitrate, phosphate and other ions so crops grow more. Irrigation and greenhouses add water and warmth.
- Reduce losses to other organisms. Herbicides remove weeds, pesticides kill pests and fungicides control disease. Biological control uses a predator or parasite.
- Reduce respiratory loss in livestock. Keeping animals indoors, in warm conditions and with limited movement means less biomass is respired to produce heat and movement, so more goes into growth. Feeding concentrated, easily digested food also raises efficiency.
- Select and shorten. Selective breeding gives faster growth and higher yields. Eating plants directly avoids the losses at an extra trophic level.
Costs include eutrophication, harm to non-target species, loss of biodiversity and animal welfare concerns.
6.3.1(c) Recycling within ecosystems
Decomposers (saprobionts, which are bacteria and fungi) secrete enzymes onto dead material and absorb the products. This extracellular digestion releases the elements in dead organisms and waste back to the environment.
The nitrogen cycle
Organisms need nitrogen for amino acids, proteins and nucleotides. Plants take up nitrate (and ammonium) ions from the soil, mostly by active transport. Animals get nitrogen by feeding.
| Process | Change | Organisms |
|---|---|---|
| Nitrogen fixation | Nitrogen gas to ammonia or ammonium | Azotobacter (free-living in soil); Rhizobium (in root nodules of legumes, with a mutualistic relationship) |
| Ammonification | Proteins and urea in dead matter and waste to ammonia or ammonium | Decomposers |
| Nitrification, step 1 | Ammonium to nitrite | Nitrosomonas |
| Nitrification, step 2 | Nitrite to nitrate | Nitrobacter |
| Denitrification | Nitrate to nitrogen gas | Denitrifying bacteria, in anaerobic soil |
Rhizobium gains carbohydrate from the legume and supplies fixed nitrogen. Nitrification needs oxygen, so it is slow in waterlogged soil. Denitrification happens in waterlogged soil, which lowers fertility. Lightning and industrial fertiliser manufacture also fix nitrogen.
The carbon cycle
| Process | Effect on atmospheric or dissolved carbon dioxide |
|---|---|
| Photosynthesis | Removes carbon dioxide and builds organic compounds |
| Respiration (plants, animals, decomposers) | Returns carbon dioxide |
| Decomposition | Releases carbon dioxide (or methane in some anaerobic conditions) |
| Combustion of wood and fossil fuels | Returns carbon dioxide quickly |
| Fossilisation | Locks carbon in coal, oil and gas, and in peat |
Physical and chemical effects also matter. Carbon dioxide dissolves in the oceans and reacts with water to form carbonic acid, hydrogencarbonate and carbonate ions. Marine organisms build calcium carbonate shells, and their remains form limestone sediment, a very long-term store. Volcanic activity and weathering of rock return carbon over long periods. Burning fossil fuels and clearing forests raise atmospheric carbon dioxide.
6.3.1(d) Primary succession
Succession is the change in the community of an ecosystem over time. Primary succession starts on newly formed land with no soil, such as bare rock, sand or volcanic lava.
- Pioneer species arrive first, for example lichens, algae and mosses, or marram grass on dunes. They tolerate extreme conditions and spread by wind-blown spores or seeds.
- They die and decompose, adding organic matter. Weathering and humus form a thin soil that holds water and nutrients.
- Conditions become less hostile (abiotic factors change), so new species such as grasses and ferns colonise. They outcompete the pioneers.
- Shrubs and then trees arrive.
- A climax community forms. It is relatively stable, with dominant species in balance with the climate and soil. In much of the UK this is deciduous woodland.
During succession, soil depth, species diversity, biomass and food web complexity increase.
Deflected succession occurs when human activity stops succession reaching the climax. Mowing, grazing, burning and ploughing keep the community at an earlier stage. Chalk grassland grazed by sheep stays grassland. If the grazing stops, shrubs and then woodland take over.
6.3.1(e) Distribution and abundance
Distribution is where a species is found. Abundance is how many there are. Both are measured by sampling, because counting every organism is not practical.
- Random sampling uses random coordinates (for example from a random number table or generator) to place quadrats. It removes bias.
- Systematic sampling uses a line or belt transect, with quadrats at regular intervals. It suits a gradient, such as shore to inland.
- Take enough samples to be representative.
Quadrats. Record abundance as a count (density per m²), frequency (the proportion of quadrats containing the species), or percentage cover. Use percentage cover where individual plants are hard to count. An abundance scale gives a quick estimate.
Other methods. Use pitfall traps for ground invertebrates, sweep nets for vegetation and kick sampling in streams.
Mark-release-recapture estimates the population of mobile animals. Capture a sample, mark it harmlessly, release it, then capture a second sample.
N = (number in first sample × number in second sample) ÷ number of marked animals in second sample
Example: 48 marked, a second catch of 55 with 11 marked. N = 48 × 55 / 11 = 240.
It assumes the marked animals mix evenly, marking does not affect survival or capture, no births, deaths or movement occur, and marks are not lost.
Record abiotic factors with a light meter, thermometer, soil moisture meter or pH probe, so you can look for links.
Statistics
| Question | Test |
|---|---|
| Are two means different? | Student’s t-test |
| Are two variables correlated? | Spearman’s rank correlation |
| Do observed frequencies differ from expected? | Chi-squared |
State a null hypothesis, such as “there is no significant difference”. Look up the critical value at P = 0.05. A calculated value at least as large as the critical value means you reject the null hypothesis.
Spearman worked example. Light intensity (arbitrary units) and ground cover (%) at eight sites along a woodland path:
| Light | Cover | Rank light | Rank cover | d | d² |
|---|---|---|---|---|---|
| 12 | 64 | 8 | 1 | 7 | 49 |
| 18 | 52 | 7 | 3 | 4 | 16 |
| 25 | 58 | 6 | 2 | 4 | 16 |
| 31 | 41 | 5 | 4 | 1 | 1 |
| 40 | 30 | 4 | 6 | −2 | 4 |
| 46 | 35 | 3 | 5 | −2 | 4 |
| 58 | 18 | 2 | 7 | −5 | 25 |
| 70 | 12 | 1 | 8 | −7 | 49 |
Σd² = 164 and n = 8.
rs = 1 − (6 × Σd²) / (n(n² − 1)) = 1 − (6 × 164) / (8 × 63) = 1 − 984/504 = −0.952
For n = 8 the critical value at P = 0.05 is 0.738 (ignoring the sign). Since 0.952 is above it, reject the null hypothesis: there is a significant negative correlation. Correlation does not show cause.
t-test worked example. Ground beetles per trap in eight traps in each of two habitats. A: 14, 11, 16, 9, 13, 15, 12, 10. B: 8, 6, 10, 7, 5, 9, 6, 7.
- Means: A = 12.5, B = 7.25.
- Standard deviations, s = √(Σ(x − mean)² / (n − 1)): A = 2.45, B = 1.67. The standard deviation shows spread and is less affected by one extreme value than the range.
- t = (mean A − mean B) / √(sA²/nA + sB²/nB) = 5.25 / √(0.75 + 0.348) = 5.25 / 1.048 = 5.01
- Degrees of freedom = nA + nB − 2 = 14. The critical value at P = 0.05 is 2.145.
5.01 is greater than 2.145, so reject the null hypothesis: the difference in means is significant.
Chi-squared. χ² = Σ (O − E)² / E, with degrees of freedom = number of categories − 1. Expected values come from the null hypothesis, such as an even spread, or a spread in proportion to habitat area. The revision notes include a worked case.
Common errors
- Using fresh mass as biomass.
- Dividing by the wrong figure: efficiency uses intake, not the amount available in the level below.
- Saying respiration “uses up” energy rather than transferring it as heat.
- Confusing nitrification with nitrogen fixation, or saying Nitrosomonas makes nitrate.
- Saying a test “proves” something, instead of rejecting the null hypothesis.
Official syllabus
OCR, A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025: section 6.3.1(a) to (e), Ecosystems.
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Condensed revision notes on ecosystems, biomass efficiency, nitrogen and carbon cycles, succession and sampling statistics for OCR A Level Biology A.
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