Energy and ecosystems - AQA A-Level Biology (Paper 2) Exam Questions

Practise 19 exam-style questions on Energy and ecosystems for AQA A-Level Biology (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

19 questions
1–8 marks each
Instant marking
No sign-up needed

19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

The plants in a grassland have a gross primary production (GPP) and a net primary production (NPP).

Explain why the NPP of the plants, and not their GPP, is the chemical energy available to the herbivores that feed on them.

Model answer

GPP is the chemical energy stored in the sugars made by the plants, but the plants use some of these sugars as respiratory substrates. The energy released in respiration is lost to the environment as heat, so it is not in the plants' biomass (NPP = GPP - respiratory losses). Herbivores can only ingest the chemical energy that remains stored in the plant biomass, which is the NPP.

Explanation

Gross primary production is the chemical energy store in plant biomass before the plant's own respiratory losses are taken into account. The plant respires much of the sugar it makes, and that energy is lost as heat, so only the net primary production is left in the plant's tissues for herbivores and decomposers. Answers must link NPP to biomass that can be eaten; avoid saying energy is 'produced' or 'made' in respiration.

Mark scheme

1 mark
GPP includes chemical energy (in sugars) that the plants use in (their own) respiration / NPP = GPP minus respiratory losses
Max 2. Accept NPP = GPP - R if R is identified as respiration or respiratory losses. 'NPP is GPP minus energy lost' with no reference to respiration is insufficient.
1 mark
Energy released in (plant) respiration is lost (to the environment) as heat / is no longer in the plants' biomass
Reject 'energy is produced in respiration'. Accept 'respiration releases energy that is not stored in the plant'.
1 mark
Only chemical energy stored in plant biomass/organic molecules (the NPP) can be eaten/ingested by herbivores
Accept 'NPP is the energy in the plant biomass, which is what the herbivores eat'. Ignore references to the herbivores' own respiration.

Question 2

3 marks

A student estimated the biomass of brown seaweed on a rocky shore. At low tide, she removed all the seaweed from inside a 0.25 m² quadrat. She rinsed the seaweed in fresh water and dried it in an oven at 80 °C until its mass was constant. The final dry mass was 186 g.

Calculate the dry biomass of the seaweed in kg m⁻².

Explain why she dried the seaweed at 80 °C rather than at 200 °C.

Explain why biomass is measured as dry mass rather than fresh mass.

Model answer

Dry biomass = 186 ÷ 0.25 = 744 g m⁻² = 744 ÷ 1000 = 0.744 kg m⁻²

At 200 °C the organic compounds in the seaweed would start to burn (be oxidised to carbon dioxide), so the mass would fall because of the loss of biomass as well as water, and the dry mass would be underestimated.

The amount of water in fresh seaweed varies (for example, it depends on how much rinsing water is left on it), and water is not part of the biomass and contains no chemical energy, so only the dry mass is a reliable measure of the organic material.

Explanation

Biomass is the mass of organic material, so water must be removed: the sample is heated at a temperature high enough to evaporate water but too low to burn organic compounds, until its mass stays constant. To convert to a mass per square metre, divide by the area sampled (0.25 m²), then by 1000 to convert grams to kilograms. A common error is to multiply by the quadrat area instead of dividing.

Mark scheme

1 mark
0.744 (kg m⁻²) (accept 0.74), from 186 ÷ 0.25 ÷ 1000; a correct answer with no working scores this mark
Point 1 is the calculation: a correct answer of 0.744 (accept 0.74) with no working scores 1 mark. Do not accept 744 (not converted to kg), 46.5 or 0.0465 (multiplied by 0.25 instead of divided), or 0.186 (not converted to per m²).
1 mark
At 200 °C (organic compounds in) the seaweed would burn/combust/decompose/be oxidised, so mass other than water would be lost / the dry mass would be too low
Accept 'carbon would be lost as carbon dioxide'. 'It would burn' alone is enough only if linked to losing mass/biomass. Ignore 'to save energy' and 'for safety'.
1 mark
The water content (of fresh seaweed) varies, e.g. with how wet it is from rinsing or the tide, so fresh mass does not reliably measure the organic material / water is not part of the biomass / water contains no chemical energy
Accept 'dry mass measures only the organic matter'. 'Dry mass is more accurate' alone is insufficient.

Question 3

3 marks

Septoria is a fungus that infects the leaves of wheat plants and feeds on them. A farmer estimated that the yield of grain from his wheat field would be 11.0 tonnes per hectare if the crop had no disease. He sprayed most of the field with a fungicide but left one strip unsprayed. The sprayed crop yielded 9.6 tonnes of grain per hectare and the unsprayed strip yielded 7.4 tonnes per hectare.

Calculate the yield of the sprayed crop as a percentage of the yield expected with no disease.

Explain, in terms of energy transfer, why spraying the crop with fungicide increased the yield of grain.

Model answer

Percentage yield = 9.6 ÷ 11.0 × 100 = 87.3%

The fungus feeds on the wheat leaves, so some of the chemical energy in the wheat's biomass is transferred to the fungus, a non-human food chain, rather than being stored in the grain. Infected leaves also have less living tissue and chlorophyll, so the wheat photosynthesises less and has a lower net primary production. Spraying kills the fungus, so more of the energy fixed by the wheat ends up in the grain harvested by humans.

Explanation

Percentage yield = actual yield ÷ expected (maximum) yield × 100. Pests and pathogens are consumers in non-human food chains: controlling them simplifies the food web, so a greater proportion of the crop's net primary production is available to humans. Infection also reduces the area of leaf that can photosynthesise. A common error is to calculate the percentage increase caused by spraying instead of the percentage of the expected yield.

Mark scheme

1 mark
87.3 (%) (accept 87 to 87.3), from 9.6 ÷ 11.0 × 100; a correct answer with no working scores this mark
Point 1 is the calculation: a correct answer of 87.3 (accept 87 to 87.3) with no working scores 1 mark. Do not accept 67.3 (the unsprayed strip) or 29.7 (percentage increase from spraying) for this point.
1 mark
The fungus feeds on the wheat, so (chemical) energy/biomass from the wheat passes to the fungus / to a non-human food chain (instead of to the grain harvested by humans)
Accept 'fungus uses sugars/organic compounds from the wheat (as respiratory substrates)'.
1 mark
Infected leaves have less (living) leaf area/chlorophyll, so the wheat photosynthesises less / has a lower GPP/NPP
Accept 'damaged leaves cannot photosynthesise as much'. Ignore 'the plant is unhealthy'.

Question 4

5 marks

In most ecosystems, the efficiency of energy transfer from producers to primary consumers is lower than the efficiency of energy transfer from primary consumers to secondary consumers.

Describe the ways in which chemical energy is lost at each transfer between two trophic levels. Explain why the transfer from producers to primary consumers is usually less efficient.

Model answer

Chemical energy is lost at each transfer because not all of the organisms at the lower trophic level are eaten; some parts pass to decomposers. Not all of the food that is eaten is digested and absorbed, so some chemical energy is lost in faeces, and some is lost in urine (urea). Much of the energy in the absorbed food is released in respiration and lost to the environment as heat, so it is not stored in the consumer's biomass.

The transfer from producers to primary consumers is less efficient because plant cell walls contain cellulose and lignin, which are difficult to digest. Herbivores therefore lose a greater proportion of the energy they ingest in their faeces than carnivores do, because animal tissue is mostly protein and fat, which are easily digested and absorbed. Also, a larger proportion of plant biomass than of animal biomass, such as roots and woody stems, is never eaten.

Explanation

Energy transfer between trophic levels is inefficient because not all biomass is eaten, not all that is eaten is absorbed (losses in faeces), some is lost in urine, and much of what is absorbed is released in respiration and lost as heat. The first two losses are larger for herbivores, because plant cell walls are made of cellulose and lignin that are hard to digest. Make the comparison clear by saying what happens in herbivores and in carnivores, and never say that energy is 'produced' in respiration.

Mark scheme

1 mark
Not all of the organisms/biomass at the lower trophic level is eaten / some parts are not eaten (and pass to decomposers)
Max 5. Points 1 to 4 (ways chemical energy is lost): max 3. Points 5 to 8 (why the transfer from producers to primary consumers is less efficient): max 3. Ignore 'energy is lost' with no route of loss.
1 mark
Not all of the food eaten is digested/absorbed, so some chemical energy is lost in faeces
Accept 'lost in egestion'.
1 mark
Some chemical energy is lost in urine / urea / excretion
Accept 'lost in excretory products'. Do not award point 3 for faeces.
1 mark
Chemical energy released in respiration is lost (to the environment) as heat / is used for movement or maintaining body temperature, so it is not stored in biomass
Reject 'energy is produced/made in respiration'.
1 mark
Plant material/cell walls contain cellulose/lignin, which is difficult to digest/hydrolyse
Accept 'herbivores cannot digest cellulose without microorganisms'.
1 mark
So herbivores lose a greater proportion of the chemical energy they ingest in faeces / absorb a smaller proportion of their food than carnivores
A comparison with carnivores is needed for point 6.
1 mark
Animal tissue is mostly protein/fat, which is more easily digested and absorbed
Ignore 'meat has more energy'.
1 mark
A larger proportion of plant biomass than of animal biomass is not eaten / is inedible, e.g. roots, woody stems or bark
The comparison with animals (or an example of an inedible plant part compared with animal tissue) is needed for point 8; it is a separate point from point 1.

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

Study Helper Hints first, answers after you try

Stuck on this one? Ask for a hint and I’ll point you in the right direction, without giving the answer away. Once you’ve checked your answer, I’ll show you how the marks are awarded.