OCR A Level Biology A: Ecosystems, biomass transfer and succession (H420) – Revision Notes

These notes condense 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. For full explanations and worked examples, read the study guide first. Then test yourself with the practice questions. The OCR A Level Biology hub, the checklist and the free diagnostics help you plan. For Simpson’s Index, see Measuring biodiversity.

Key definitions

Term	Meaning
Ecosystem	The organisms in an area, the non-living features, and their interactions. It can be as small as a rock pool or a large tree
Biotic factor	A living influence, such as predation, competition, grazing or disease
Abiotic factor	A non-living influence, such as temperature, light, pH, salinity or water
Biomass	Mass of living material, measured as dry mass
Trophic level	A feeding position in a food chain
Decomposer	A bacterium or fungus that digests dead material extracellularly and releases nutrients
Pioneer species	The first colonisers of a bare area
Climax community	A relatively stable community in balance with the climate and soil
Deflected succession	Succession held at an earlier stage by human activity
Distribution / abundance	Where a species is found / how many there are

Formulae and methods

Quantity	Calculation
Efficiency of transfer	(biomass transferred ÷ biomass intake) × 100
Energy in a sample	water mass (g) × 4.18 × temperature rise (°C), in joules
Population estimate (mark-release-recapture)	(first sample × second sample) ÷ marked in second sample
Spearman’s rank	rs = 1 − 6Σd² ÷ n(n² − 1)
t-test	t = (mean 1 − mean 2) ÷ √(s1²/n1 + s2²/n2); degrees of freedom = n1 + n2 − 2
Chi-squared	χ² = Σ (O − E)² ÷ E; degrees of freedom = categories − 1
Standard deviation	s = √(Σ(x − mean)² ÷ (n − 1))
Simpson’s Index of Diversity	D = 1 − Σ(n/N)²

6.3.1(b) Biomass transfer in steps

Measuring dry mass

1. Take a representative sample and weigh it.
2. Dry it in an oven at a temperature that does not burn it (around 80 °C).
3. Re-weigh until the mass is constant.
4. Scale to the area, for example g m⁻².

Measuring energy in biomass. Burn a dried, weighed sample under a known mass of water. Record the temperature rise. Heat loss to the surroundings means the result is an underestimate.

Why efficiency is low. Not all biomass is eaten. Not all that is eaten is digested (faeces). Urea is lost in excretion. Respiration releases heat and carbon dioxide, and powers movement.

Human manipulation. The aim is to cut losses and raise growth.

• Fertilisers, irrigation and greenhouses increase producer growth.
• Herbicides, pesticides, fungicides and biological control reduce competition and pest losses.
• Housing, warmth and restricted movement cut respiratory loss in livestock.
• Selective breeding gives faster growth and higher yields.
• Eating plants rather than animals avoids a trophic level.
• Costs: eutrophication, harm to non-target species, lower biodiversity and welfare concerns.
• Link to the nitrogen cycle: crops remove nitrate, so farmers add fertiliser or grow legumes, whose Rhizobium nodules fix nitrogen.

Worked reminder. Herbivores take in 900 kJ m⁻² year⁻¹ and incorporate 117 kJ m⁻² year⁻¹ into their biomass: efficiency = 117 / 900 × 100 = 13.0%.

6.3.1(c) Nutrient cycles

Nitrogen cycle

Process	From → to	Who
Fixation	N₂ → ammonia/ammonium	Azotobacter (free-living), Rhizobium (legume nodules)
Ammonification	Protein, urea → ammonium	Decomposers
Nitrification 1	Ammonium → nitrite	Nitrosomonas
Nitrification 2	Nitrite → nitrate	Nitrobacter
Denitrification	Nitrate → N₂	Denitrifying bacteria (anaerobic)

Plants take up nitrate and make amino acids, proteins and nucleotides. Animals obtain nitrogen by eating. Nitrifying bacteria need oxygen. Waterlogged soil is short of oxygen, so nitrification falls and denitrification rises.

Carbon cycle

• Removed from the air: photosynthesis; dissolving in the oceans.
• Returned to the air: respiration, decomposition, combustion, volcanic activity.
• Stores: biomass, fossil fuels, peat, limestone and shell sediments, dissolved hydrogencarbonate in the oceans.
• Chemical effect: carbon dioxide and water form carbonic acid, which releases hydrogencarbonate and hydrogen ions. Calcium carbonate shells and limestone lock carbon away.

6.3.1(d) Succession in steps

1. Bare ground: no soil, such as rock, sand or lava.
2. Pioneers (lichens, algae, mosses, marram grass) colonise.
3. They die and decompose, adding organic matter. Weathering plus humus form soil that holds water.
4. Conditions become less extreme, so new species such as grasses, then shrubs and trees colonise and outcompete earlier species.
5. Diversity, biomass and soil depth increase.
6. The climax community is stable and dominated by a few species.

Deflected succession. Mowing, grazing, burning or ploughing stops the climax forming. Examples: a mown playing field stays grass; grazed chalk grassland does not become woodland.

6.3.1(e) Sampling and statistics

Choosing a sampling approach

Situation	Method
Even habitat, unbiased estimate	Random quadrats using random coordinates
Change along a gradient	Line or belt transect, quadrats at regular intervals
Plants too small or tangled to count	Percentage cover
Ground invertebrates	Pitfall traps
Stream invertebrates	Kick sampling
Mobile animals	Mark-release-recapture

Which test?

Aim	Test	Reject the null hypothesis when
Compare two means	Student’s t-test	calculated t is at least the critical value
Test correlation	Spearman’s rank	rs (ignoring sign) is at least the critical value
Observed against expected frequencies	Chi-squared	calculated χ² is at least the critical value

Use P = 0.05 unless told otherwise.

Chi-squared worked reminder. 120 beetles are counted in four habitats that make up 40%, 30%, 20% and 10% of the area. The null hypothesis is that beetles are spread in proportion to area, so the expected numbers are 48, 36, 24 and 12. The observed numbers are 38, 41, 28 and 13.

• (38 − 48)²/48 = 2.083; (41 − 36)²/36 = 0.694; (28 − 24)²/24 = 0.667; (13 − 12)²/12 = 0.083.
• χ² = 3.53, degrees of freedom = 3, and the critical value at P = 0.05 is 7.81.
• 3.53 is smaller, so do not reject the null hypothesis. There is no significant evidence that the beetles prefer particular habitats.

Planning a transect investigation. Lay a tape from the shore (or other starting point) in a straight line. Place a quadrat at fixed intervals, for example every 5 m. In each quadrat, record the species and the percentage cover or count, and measure the abiotic factors, such as light, soil moisture and pH, using the same instrument each time. Repeat the transect in several places and use a Spearman test to look for an association. State the null hypothesis before collecting data.

Standard deviation worked reminder. Data: 6, 8, 9, 11, 16. Mean = 10. Squared differences: 16, 4, 1, 1, 36, which sum to 58. Divide by n − 1 = 4 to get 14.5. The standard deviation is √14.5 = 3.81. A single extreme value (16) raises it, but it shows the spread of all the data, unlike the range, which uses only the two extremes.

Simpson’s Index worked reminder. A sample has 10, 6 and 4 individuals of three species, N = 20. Σ(n/N)² = 0.25 + 0.09 + 0.04 = 0.38. D = 1 − 0.38 = 0.62. A higher D means greater diversity.

Must-know distinctions

• Nitrification and fixation. Nitrification makes nitrite and nitrate from ammonium. Fixation makes ammonium from nitrogen gas.
• Nitrosomonas and Nitrobacter. Nitrosomonas makes nitrite. Nitrobacter makes nitrate.
• Azotobacter and Rhizobium. Free-living and symbiotic.
• Primary succession and deflected succession. Natural progression, and progression blocked by humans.
• Frequency and percentage cover. Presence in quadrats, and the area covered.
• Correlation and difference. Spearman tests an association. The t-test tests a difference in means.

Quick self-test

1. Why is dry mass used rather than fresh mass?
2. A calorimeter holds 25 g of water that warms by 12 °C. How much energy was released?
3. Herbivores take in 900 kJ and incorporate 117 kJ. What is the efficiency?
4. Which bacteria convert nitrite to nitrate?
5. A first catch of 24 animals is marked. A second catch of 35 contains 7 marked animals. Estimate the population.
6. How many degrees of freedom has a chi-squared test with five categories?
7. In a Spearman test, Σd² = 30 and n = 10. Calculate rs.
8. Name two pioneer species.
9. Give one example of deflected succession.
10. Which test compares the mean of two groups?
11. Name three processes that return carbon dioxide to the atmosphere.

Answers

1. Water content varies, so fresh mass is not a consistent measure of biological material.
2. 25 × 4.18 × 12 = 1 254 J.
3. 117 / 900 × 100 = 13.0%.
4. Nitrobacter.
5. 24 × 35 / 7 = 120.
6. 5 − 1 = 4.
7. rs = 1 − (6 × 30) / (10 × 99) = 1 − 180/990 = 0.818.
8. For example, lichens and mosses (or algae, or marram grass on dunes).
9. For example, grazing keeps chalk grassland from becoming woodland.
10. Student’s t-test.
11. Any three from respiration, decomposition, combustion and volcanic activity.

Where marks are usually lost

• Writing “biomass” but measuring fresh mass.
• Using the wrong denominator in efficiency calculations.
• Writing that energy is “lost” without saying it is lost as heat from respiration, or in faeces and urine.
• Mixing up the four bacteria, or the order of nitrification.
• Forgetting that nitrifying bacteria need oxygen.
• Describing a climax community as unchanging, when it is stable but still dynamic.
• Not stating a null hypothesis, or not comparing the calculated and critical value.
• Using the wrong degrees of freedom.
• Describing a sampling method without saying how the positions are chosen.
• Stating that a correlation shows one factor causes the other.

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.