What you need to know (from the AQA specification)
What you need to know (from the AQA specification)
In any ecosystem, plants synthesise organic compounds from atmospheric, or aquatic, carbon dioxide. Most of the sugars synthesised by plants are used by the plant as respiratory substrates. The rest are used to make other groups of biological molecules. These biological molecules form the biomass of the plants.
Plants are called producers because they make their own food through Photosynthesis.
In photosynthesis they use carbon dioxide, energy from sunlight and water to produce glucose and oxygen.
Some of this glucose can be used in respiration to produce ATP.
The remainder of this glucose can be used to make other biological molecules (e.g. cellulose). This is the biomass of plants. It’s referred to as the chemical energy store for the plant and can be transferred through the ecosystem, when one organism eats another.
Biomass
What you need to know (from the AQA specification)
What you need to know (from the AQA specification)
Biomass can be measured in terms of mass of carbon or dry mass of tissue per given area. The chemical energy store in dry biomass can be estimated using calorimetry.
How do we measure biomass?
Biomass is the total mass of living material, i.e. all the remaining glucose which has been used to make other biological molecules.
Biomass
Total mass of living material. It can be measured as mass of carbon or dry mass per given area.
Calculating Dry Mass
It’s important to calculate biomass as:
- Dry mass due to the varying amount of water in organisms
- Per given area, not individual organisms, so that biomass can be compared between different areas or ecosystems of different sizes
You can calculate dry mass by measuring the mass of the organism with water removed. To do this you can:
- Heat a sample of the organism in an oven at a low temperature, so it dries but doesn’t burn. Note the organism must not be living.
- Reweigh it at regular intervals.
- Once the mass of the sample remains constant (i.e. it doesn’t decrease anymore), this indicates that there is no more water to be lost.
This mass can then be scaled up to population or area and is usually measured in kg m⁻². For example, if the biomass of a small sample area (1 m²) was 100 kg, you could then scale this up to a larger area, e.g. 10 m² would be 1000 kg total, or 100 m² would be 10,000 kg total.
Calorimetry
Calorimetry can estimate the chemical energy store in dry mass.
It works by measuring the heat released during combustion.
A process known as bomb calorimetry involves burning a dry sample of biomass in a sealed container with oxygen. The resulting heat is transferred to a surrounding water bath (of a known volume). You can then measure the temperature change to determine the amount of chemical energy in the biomass.
It’s actually the same process you would use to measure calories in foods.
Features of a calorimeter
- Stirrer: distributes the heat evenly through the water, so the temperature reading is accurate
- Insulation / air space: reduces heat loss to (or gain from) the surroundings, so all the heat released goes into the water
- Water: has a high heat capacity, so it can absorb a lot of heat with only a small rise in temperature. This means it won’t boil and lose heat as steam (see thermal properties of water)
Calculating the energy in biomass
Question: A 2.00 g dry sample of grass was fully combusted in a calorimeter containing 200 cm³ of water. The temperature of the water rose by 15.0 °C. 4.18 J of heat energy raises the temperature of 1 cm³ of water by 1 °C. Calculate the energy released in kJ g⁻¹.
Answer:
- Energy released = volume of water × 4.18 × temperature rise
= 200 × 4.18 × 15.0 = 12 540 J - Convert J to kJ (÷ 1000) = 12.54 kJ
- Divide by the dry mass to get energy per gram = 12.54 ÷ 2.00
= 6.27 kJ g⁻¹
GPP & NPP
What you need to know (from the AQA specification)
What you need to know (from the AQA specification)
Gross primary production (GPP) is the chemical energy store in plant biomass, in a given area or volume. Net primary production (NPP) is the chemical energy store in plant biomass after respiratory losses to the environment have been taken into account, i.e. NPP = GPP – R, where GPP represents gross production and R represents respiratory losses to the environment. This net primary production is available for plant growth and reproduction. It is also available to other trophic levels in the ecosystem, such as herbivores and decomposers.
If you look at the image you should see that some of the energy that reaches producers (plants) as sunlight is never absorbed/not used for photosynthesis.
Why do you think this is? (2 mark exam question)
Why do you think this is? (2 mark exam question)
- (Light is) reflected;
- (Light is) wrong wavelength;
- (Light) misses chlorophyll/chloroplasts/photosynthetic tissue;
- CO₂ concentration or temperature is a limiting factor.
GPP
Gross Primary Production (GPP) is the total chemical energy store in plant biomass, in a given area or volume. Essentially it’s the total energy captured by the plants (producers) during photosynthesis.
Not all the GPP can be passed on to the next trophic level. The plants use some of the energy in respiration, so this has to be taken off (R for respiration).
NPP
Net Primary Production (NPP) is the chemical energy store in plant biomass, once respiration has been accounted for. NPP is available for plant growth, reproduction and can be passed on to other trophic levels, e.g. consumers and decomposers.
NPP
NPP = GPP - R
Net Production of Consumers
What you need to know (from the AQA specification)
What you need to know (from the AQA specification)
The net production of consumers (N), such as animals, can be calculated as: N = I – (F + R) where I represents the chemical energy store in ingested food, F represents the chemical energy lost to the environment in faeces and urine and R represents the respiratory losses to the environment.
Not all the chemical energy stored in the food consumers ingest can be transferred to the next trophic level. In fact, it’s really inefficient, with often only about 10% of the energy available to the next trophic level.
Give 3 reasons for the low efficiency of energy transfer from secondary consumers to tertiary consumers in an ecosystem. (3 marks from exam question)
Give 3 reasons for the low efficiency of energy transfer from secondary consumers to tertiary consumers in an ecosystem. (3 marks from exam question)
- Heat (loss) from respiration;
- Some of the food is not digested (lost in faeces) or not all eaten;
- Excretion (lost in urine).
The energy that is stored in the consumer’s biomass and available to the next trophic level is called ‘net production’, see equation below.
Net production of consumers
N = I − (F + R)
- N = Net production of consumers
- I = Chemical energy from ingested food
- F = Energy lost in faeces and urine
- R = Energy lost through respiration
Productivity (Rate)
What you need to know (from the AQA specification)
What you need to know (from the AQA specification)
Primary and secondary productivity is the rate of primary or secondary production, respectively. It is measured as biomass in a given area in a given time eg kJ ha⁻¹ year⁻¹.
- Net production of consumers can also be called secondary productivity, when it’s expressed as a rate, in a given area in a given time, e.g. kJ ha⁻¹ year⁻¹.
- Similarly, primary production (GPP or NPP) can be expressed as productivity when it’s expressed as a rate in a given area in a given time, e.g. kJ ha⁻¹ year⁻¹.
Example: Calculating Net Production
Calculating net production
Question: A cow ingests 5500 kJ of chemical energy per day. It loses 3200 kJ in faeces and urine, and 1800 kJ through respiration. Calculate the net production of the cow.
Answer:
N = I − (F + R)
N = 5500 − (3200 + 1800)
N = 5500 − 5000
N = 500 kJ per day
Example: Calculating Efficiency of Energy Transfer
Calculating efficiency
Question: The net production of producers in an ecosystem is 40 000 kJ m⁻² year⁻¹. The net production of the primary consumers is 4400 kJ m⁻² year⁻¹. Calculate the percentage efficiency of energy transfer from the producers to the primary consumers.
Answer:
$$\text{Efficiency} = \frac{\text{Net production of the next trophic level}}{\text{Net production of the previous trophic level}} \times 100$$
$$\text{Efficiency} = \frac{4400}{40,000} \times 100$$
Efficiency = 11%
Farming Practices
What you need to know (from the AQA specification)
What you need to know (from the AQA specification)
Students should be able to appreciate the ways in which production is affected by farming practices designed to increase the efficiency of energy transfer by:
- simplifying food webs to reduce energy losses to non-human food chains
- reducing respiratory losses within a human food chain.
Farmers want to increase the efficiency of energy transfer to maximise the amount of energy (and therefore biomass) available for human consumption (in farming products). They can do this by:
Simplifying food webs
This is aimed at reducing the energy lost to other organisms (e.g. pests) in the food chain.
- Using pesticides/herbicides: to remove organisms that would otherwise take energy away from the crop (pests eat it, and weeds compete with it for light, water and mineral ions).
- Biological control: this involves introducing natural predators of pest species to reduce energy losses (e.g. ladybirds, parasitic wasps).
Reducing respiratory losses
This is aimed at controlling the conditions that livestock are in, to reduce the losses due to respiration.
- Restricting movement of livestock (e.g. battery farming, indoor housing) so less energy is used in muscle contraction.
- Controlling temperature of the environment for livestock: keeping animals warm reduces the energy they need to use to maintain body temperature, lowering respiratory losses.
How this topic is tested
This analysis is based on past paper data from 2017 to 2025. It is intended for interest only and is not predictive of what will appear in future papers.
- Tested in 8 of 9 years (2017–2025): 21 question parts worth 42 marks.
- 25th most-examined topic overall by marks, 4th in Unit 5.
Marks by year
Most-tested spec points
- Gross & Net Primary Production: tested in 7 parts (16 marks)
- Biomass & Its Measurement: tested in 5 parts (10 marks)
- Net Production of Consumers: tested in 4 parts (7 marks)
- Measuring Dry Mass & Energy of Biomass: tested in 2 parts (4 marks)
- Farming Practices & Energy Transfer: tested in 1 part (2 marks)
Also links to: Fertilisers & Environmental Issues, Limiting Factors & Agricultural Practice.
Maths and practical skills
- Percentages, ratios and fractions (MS 0.3): e.g. 2022 P2 Q2.3
- Units and standard form (MS 0.1): e.g. 2025 P2 Q3.2
- Using equations (MS 2.3, 2.4): e.g. 2022 P3 Q5.4
- Logarithms and log scales (MS 2.5): e.g. 2021 P3 Q4.4
- Powers, exponentials and logs on a calculator (MS 0.5): e.g. 2017 P2 Q5.5
- Significant figures (MS 1.1): e.g. 2019 P2 Q1.2
Tips from examiner reports
What students commonly get wrong
- Faeces aren't excretionWatch out: faeces are undigested food that was never absorbed, so they're not excretion. Urine is excretion. 2023 P2 Q7.3
- Link heat loss to respirationWatch out: respiration releases some energy as heat, which is lost to the surroundings, so it isn't passed to the next trophic level. "Energy is used in respiration" or "energy is produced" doesn't explain the loss. 2023 P2 Q7.3 · 2022 P2 Q2.4
- Say energy is lost between trophic levelsWatch out: when explaining why eating animals is less efficient than eating crops, first say energy is lost at each trophic level, then say how (respiration, faeces, urine). 2022 P2 Q2.4
- Dry mass: heat until the mass is constantWatch out: heat the sample (e.g. in an oven) and reweigh it until the mass is constant. Don't burn it: burning is calorimetry, not drying. 2025 P2 Q3.1
- Calorimetry: kJ and per gramWatch out: in calorimetry calculations, convert J to kJ (÷ 1000), and divide by the sample's dry mass when the answer is per gram. 2025 P2 Q3.2 · 2020 P2 Q3.2
- Productivity units: mass, per area, per timeWatch out: productivity units need mass (or energy), per area, per time, e.g. kg m⁻² h⁻¹ or kJ ha⁻¹ year⁻¹. Start the units with the mass or energy unit. 2022 P3 Q5.4
- GPP and NPP are for plantsWatch out: GPP and NPP describe producers (plants). For animals (consumers), use N = I − (F + R). 2021 P3 Q4.1
Exam Question Practice
A student wanted to find the dry mass of a sample of grass cuttings. She had an oven and a balance.
Describe how the student could use this equipment to find the dry mass of the grass cuttings.
(2 marks)
Hint
How do you remove all the water without burning the grass? How do you know when all the water has gone?
Mark Scheme
- Heat the grass in the oven, then weigh it (1 mark)
- Repeat until the mass stays constant (1 mark)
Also accepted / rejected
Tips
- Heat the sample in the oven; don’t burn it, or the biomass is lost
- Keep reweighing until the mass is constant, which shows all the water has gone
- Subtracting the final mass from the starting mass isn’t what’s asked
The student burned 2.00 g of dried grass completely in a calorimeter.
- The calorimeter contained 200 cm³ of water.
- The temperature of the water rose by 15.6 °C.
- 4.18 J of energy raises the temperature of 1 cm³ of water by 1 °C.
Calculate the energy released per gram of dried grass, in kJ g⁻¹.
Show your working.
Answer = ________ kJ g⁻¹
(2 marks)
Hint
Energy = volume × 4.18 × temperature rise. Then convert to kJ and divide by the mass.
Mark Scheme
6.52 kJ g⁻¹ (2 marks)
Working:
- Energy released: 200 × 4.18 × 15.6 = 13 041.6 J
- In kJ: 13 041.6 ÷ 1000 = 13.04 kJ
- Per gram: 13.04 ÷ 2.00 = 6.52 kJ g⁻¹
If the answer is wrong, 1 mark for any of:
- the digits 652 with the decimal point in the wrong place (wrong conversion)
- 13.0 (not divided by 2.00)
- 0.0326 (not multiplied by 200)
- 1.56 (not multiplied by 4.18)
- 0.418 (not multiplied by 15.6)
- 26.1 (multiplied by 2.00 instead of dividing)
Also accepted / rejected
Also accepted
- 6.5 or 7
Tips
- Convert joules to kilojoules by dividing by 1000; this is the most common slip
- Divide by the mass of grass to get kJ per gram
- Show your working, so one error can still earn a mark
A farmer grew lettuces in a glasshouse. During a heatwave, the temperature rose well above the optimum for the lettuces, and they gained less biomass.
Use your knowledge of net primary production to explain why.
(2 marks)
Hint
Write the equation for net primary production. Which part of it would a high temperature change, and in which direction?
Mark Scheme
- NPP = GPP − R (gross primary production minus respiratory losses) (1 mark)
- At the higher temperature, respiration increases, or photosynthesis (GPP) decreases because its enzymes denature (1 mark)
Also accepted / rejected
Not accepted
- Enzymes for both photosynthesis and respiration denaturing (point 2)
- “More energy is used in respiration” (point 2)
Tips
- Give the equation, NPP = GPP − R
- Then link the temperature to one part of the equation. Either respiration increases or photosynthesis decreases. Don’t say both decrease
- If you mention denaturing, say it’s the enzymes for photosynthesis
Here is a woodland food chain.
oak leaves → caterpillars → blue tits → sparrowhawks
Only a small percentage of the energy in the blue tits (secondary consumers) is transferred to the sparrowhawks (tertiary consumers).
Give three reasons why.
(3 marks)
Hint
What happens to the energy the blue tits release in respiration? Does a sparrowhawk eat and digest every part of a blue tit? What else leaves the body?
Mark Scheme
- Energy is lost as heat from respiration by the blue tits (1 mark)
- Not all of each blue tit is eaten (e.g. feathers and bones), or some is not digested and lost in faeces (1 mark)
- Some is lost in excretion (urine or urea) (1 mark)
Also accepted / rejected
Also accepted
- “Faeces” for not digested (point 2)
Tips
- Link heat loss to respiration. Write “heat from respiration”, not just “energy lost as heat”
- Parts not eaten and food not digested count as one reason
- Faeces aren’t excretion. Excretion is urine or urea
- “Energy lost in waste” is too vague
The wheat from a field can be used in two ways.
- People eat the wheat themselves.
- The wheat is fed to chickens, and people eat the chickens.
The same field feeds more people when they eat the wheat themselves.
Explain why.
(2 marks)
Hint
How many trophic levels are there in each case? What happens to energy at each trophic level?
Mark Scheme
- Energy is lost at each trophic level, and feeding the wheat to chickens adds an extra trophic level (1 mark)
- The chickens lose energy as heat in respiration, and in faeces and excretion (1 mark)
Also accepted / rejected
Also accepted
- Energy lost in movement or in keeping warm (point 2)
- A description of a longer food chain (point 1)
Not accepted
- “Energy used in respiration” (point 2)
- Energy “produced” or “made” (point 2)
Tips
- Say energy is lost at each trophic level, and eating the chickens adds an extra level compared with eating the wheat
- Give how it’s lost, such as heat from respiration, movement, faeces or excretion
- Say energy is “lost” or “released” in respiration, not “used in respiration”. Energy is never “produced”
Also accepted
Not accepted