Trophic levels, pyramids and transfer of biomass – GCSE Biology exam questions with mark scheme

1Grade 6–7Triple only5 marks

The table shows the biomass at each trophic level in a food chain.

Trophic level	Biomass in g/m²
Producers	1200
Primary consumers	120
Secondary consumers	9

(a) Calculate the efficiency of biomass transfer from primary consumers to secondary consumers. [2]

(b) Explain why biomass is lost between trophic levels. [3]

(a) $\frac{9}{120}\times 100$(1)
(a) 7.5 (%)(1)
(b) not all the ingested material is absorbed – some is egested as faeces
allow not all of the organism is eaten eg bones, roots(1)
(b) some absorbed material is lost as waste eg carbon dioxide and water from respiration, or urea in urine(1)
(b) large amounts of glucose are used in respiration (eg to keep warm/move), so it is not converted into new biomass(1)
Answer: (a) 7.5% (b) Egestion of faeces; losses in respiration (CO₂, water) and urine; respiration uses glucose not stored as biomass.
Examiner tip: Check which two levels the question asks about – 120 ÷ 1200 (10%) is the wrong pair.

2Grade 1–3FoundationTriple only6 marks

This food chain is from a rocky shore.

seaweed → periwinkle → crab → herring gull

(a) Which trophic level is the crab at? [1]

(b) Periwinkles are at trophic level 2.

Give the name used for an animal at trophic level 2 that only eats plants or algae. [1]

(d) Herring gulls on this shore have no predators.

What name is given to a carnivore with no predators? [1]

(e) Which of these organisms are decomposers? Tick (✓) one box. [1]

☐ bacteria and fungi ☐ crabs and gulls ☐ seaweeds and algae ☐ viruses and protists

(a) (trophic level) 3
allow third (level)
do not accept ‘tertiary’ (tertiary consumers are at level 4)
do not accept ‘secondary consumer’ alone (the question asks for the level)(1)
(b) herbivore
allow primary consumer(1)
(c) seaweed(1)
(c) producer
allow autotroph / ‘it photosynthesises’
the two marks are independent(1)
(d) apex predator
allow top predator / top carnivore(1)
(e) bacteria and fungi (only)
more than one box ticked = 0(1)
Answer: (a) 3 (b) herbivore (primary consumer) (c) seaweed; producer (d) apex predator (e) bacteria and fungi
Examiner tip: Trophic level is a number counted from the producer (level 1); the crab is the second animal but the third organism, so it is level 3, not 2.

3Grade 4–5FoundationTriple only6 marks

Scientists measured the dry biomass at each trophic level in a pond.

Trophic level	Organism	Dry biomass in g/m²
Producers	pondweed	500
Primary consumers	pond snails	60
Secondary consumers	roach (fish)	10

The diagram shows a pyramid of biomass that a student started to draw. The bar for the producers has been drawn.

(a) Complete the pyramid of biomass.

Describe the width, in squares, and position of each bar you would draw. [3]

(b) Describe how the scientists could find the dry biomass of a sample of pondweed. [2]

(a) primary consumers bar 3 squares wide (60 ÷ 20)(1)
(a) secondary consumers bar 0.5 / half a square wide (10 ÷ 20)(1)
(a) bars stacked in order (producers at bottom, secondary consumers at top) and each bar centred on the one below
allow labelled correctly if order is shown by labels
ignore bar height if all bars are the same height(1)
(b) heat / dry the pondweed (in an oven) at about 100 °C / below the temperature at which it burns
allow leave in a warm place / dry in an oven to evaporate the water
do not accept burn(1)
(b) weigh, then reheat and reweigh until the mass stays the same / is constant
allow ‘until no more mass is lost’(1)
(c) the amount of water in organisms varies (so wet mass is not a fair / reliable comparison)
allow water is not part of the biomass / water content changes with conditions, e.g. after rain
ignore ‘more accurate’ unqualified(1)
Answer: (a) Primary consumers: bar 3 squares wide; secondary consumers: bar 0.5 square wide; both centred, stacked above producers in order (b) dry in an oven at about 100 °C; reweigh until the mass is constant (c) water content varies between organisms / over time
Examiner tip: Divide each biomass by the scale (20 g/m² per square) before drawing; students lose marks by drawing 6 squares for 60 or forgetting to centre the bars.

4Grade 4–5Triple only7 marks

Ecologists studied the transfer of energy and biomass in a grassland.

The light energy falling on the grass was 3 800 000 kJ per m² per year.

The grass transferred 41 800 kJ per m² per year into new biomass.

(a) Calculate the percentage of the light energy that the grass transferred into biomass. [2]

The table shows the dry biomass at three trophic levels in the grassland.

Organism	Dry biomass in g/m²
grass	2500
grasshoppers	230
shrews	19

(b) Calculate the efficiency of biomass transfer from the grass to the grasshoppers. [2]

(d) Suggest one reason why the percentage you calculated in part (a) is so low. [1]

(a) 41 800 ÷ 3 800 000 (× 100)(1)
(a) 1.1 (%)
an answer of 1.1 scores 2 marks
an answer of 0.011 scores 1 mark(1)
(b) 230 ÷ 2500 (× 100)(1)
(b) 9.2 (%)
an answer of 9.2 scores 2 marks
an answer of 0.092 scores 1 mark
an answer of 1087 (2500 ÷ 230, inverted) scores 0 marks(1)
(c) 19 ÷ 2500 (× 100)
do not accept 19 ÷ 230 for this mark (that is grasshoppers to shrews)(1)
(c) 0.76 (%)
an answer of 0.76 scores 2 marks
allow 0.8
an answer of 8.3 (19 ÷ 230) scores 0 marks(1)
(d) any one from:
• some light is reflected (from the leaves)
• some light is the wrong wavelength / colour (green) for chlorophyll to absorb
• some light passes through the leaves / is transmitted
• some light misses the leaves / falls on bare ground
• photosynthesis is limited by another factor, e.g. carbon dioxide / temperature
• some of the glucose made is used in respiration (so not stored as biomass)
ignore ‘energy lost as heat’ unqualified(1)
Answer: (a) 1.1% (b) 9.2% (c) 0.76% (d) e.g. most light is reflected, transmitted, the wrong wavelength or misses the leaves
Examiner tip: In (c) divide by the producer biomass, not by the level just below — ‘percentage of the grass biomass that ends up in shrews’ means shrews ÷ grass.

5Grade 6–7HigherTriple only8 marks

A farmer measured what happened to the grass eaten by a young bull (bullock) over 6 months.

	Dry biomass in kg
grass eaten	2900
egested in faeces	1560
lost as urea in urine	75
increase in body biomass (growth)	145

The rest of the biomass was lost through respiration.

(a) Calculate the efficiency of transfer of biomass from the grass to the bullock. [2]

(b) Calculate the percentage of the biomass eaten that was lost through respiration. Give your answer to 2 significant figures.

Name the two products in which this biomass leaves the bullock. [3]

Explain why the fox egests a smaller percentage of the biomass it eats than the bullock. [3]

(a) 145 ÷ 2900 (× 100)(1)
(a) 5 (%)
an answer of 5 scores 2 marks
allow 0.05 for 1 mark(1)
(b) 2900 − (1560 + 75 + 145) = 1120 (kg)(1)
(b) 1120 ÷ 2900 × 100 = 39 (%)
allow ecf from an incorrect mass lost in respiration
38.6 (not given to 2 s.f.) does not score this mark
an answer of 39 scores 2 marks(1)
(b) carbon dioxide and water
both needed(1)
(c) grass contains cellulose / fibre that the bullock cannot (fully) digest
allow plant cell walls are hard to digest(1)
(c) meat is mainly protein / fat (and has no cell walls), which is easily digested
allow the fox has enzymes / proteases / lipases to digest all of these(1)
(c) so a greater proportion of the fox’s food is absorbed (into the blood) and less passes out as faeces
allow less undigested food is egested
ignore the fox eats less / the fox is smaller(1)
Answer: (a) 5% (b) 1120 kg; 39%; carbon dioxide and water (c) grass contains indigestible cellulose; meat (protein/fat) is easily digested, so more of the fox’s food is absorbed and less egested
Examiner tip: In (b) subtract all three named routes (faeces, urine and growth) from the biomass eaten first — students often subtract only the faeces.

6Grade 4–5Triple only6 marks

The diagram shows a pyramid of biomass for a hedgerow. It is drawn to scale.

(a) Use the diagram to determine the biomass of the caterpillars. [1]

(b) Calculate the efficiency of biomass transfer from the hawthorn to the caterpillars. [2]

(d) Sparrowhawks eat blue tits. The biomass of sparrowhawks in the hedgerow was 0.5 g/m².

Calculate how many squares wide the bar for sparrowhawks would be on this pyramid. [1]

(a) 40 g/m²
unit required
(4 squares × 10)(1)
(b) hawthorn = 400 g/m² (40 squares) and 40 ÷ 400 (× 100)
allow ecf from (a)(1)
(b) 10 (%)
an answer of 10 scores 2 marks(1)
(c) blue tits = 5 g/m² (half a square) and 5 ÷ 40 (× 100)
allow ecf from (a)(1)
(c) 12.5 (%)
allow 13
an answer of 12.5 scores 2 marks(1)
(d) 0.05 (squares)
(0.5 ÷ 10)
allow 1/20 of a square(1)
Answer: (a) 40 g/m² (b) 10% (c) 12.5% (d) 0.05 squares
Examiner tip: Count squares and multiply by the scale — the blue tit bar is only half a square, so read it carefully as 5 g/m², not 10 g/m².

7Grade 8–9HigherTriple only8 marks

A pair of ospreys lives by a lake. Ospreys are apex predators that eat only fish.

The food chain is:

algae → invertebrates → fish → ospreys

The pair of ospreys eats 220 kg of fish each year. The algae in the lake produce 1.1 kg of new biomass per m² each year.

(a) Assume that 10% of the biomass at each trophic level is transferred to the next level.

Calculate the area of lake, in m², needed to produce enough algae to support the fish eaten by the ospreys. [3]

(b) The real area of lake needed is much larger than your answer to (a).

Give two reasons why. [2]

(c) The light energy reaching the lake surface is 3.5 × 10⁶ kJ per m² per year. Each gram of algae biomass contains 20 kJ of energy.

Calculate the percentage of the light energy that is transferred to the algae biomass. Give your answer to 2 significant figures. [3]

(a) biomass of invertebrates needed = 220 × 10 = 2200 (kg)
allow 220 ÷ 0.1(1)
(a) biomass of algae needed = 2200 × 10 = 22 000 (kg)
allow ecf(1)
(a) 22 000 ÷ 1.1 = 20 000 (m²)
allow ecf from their algae biomass
an answer of 20 000 scores 3 marks
an answer of 2000 (only one factor of 10 used) scores 2 marks(1)
(b) any two from:
• not all the algae are eaten by invertebrates / not all invertebrates are eaten by fish
• the ospreys do not catch all the fish / other predators (e.g. herons, other fish) also eat the fish
• the efficiency of transfer may be less than 10%
• the fish / invertebrates also eat other food sources not counted, so the chain is a simplification (accept: food webs are more complex)
• some biomass goes to decomposers / dead organisms are not eaten
ignore ‘the ospreys eat more’(1)
(b) second reason from the list(1)
(c) energy in algae = 1100 × 20 = 22 000 (kJ per m² per year)
allow 1.1 × 20 000(1)
(c) 22 000 ÷ 3 500 000 × 100
allow ecf from their energy in algae(1)
(c) 0.63 (%)
an answer of 0.63 scores 3 marks
0.629 / 0.6286 scores 2 marks (not 2 s.f.)
an answer of 0.0063 / 6.3 × 10⁻³ (not × 100) scores 2 marks
an answer of 0.00063 (1.1 kg not converted to g) or 0.031 (not multiplied by 20 kJ per g) scores 1 mark(1)
Answer: (a) 220 kg fish ← 2200 kg invertebrates ← 22 000 kg algae; 22 000 ÷ 1.1 = 20 000 m² (b) e.g. not all organisms at each level are eaten; other predators also eat the fish; efficiency may be below 10% (c) 1100 g × 20 kJ = 22 000 kJ; 22 000 ÷ 3.5 × 10⁶ × 100 = 0.63%
Examiner tip: Work backwards up the chain: each step down towards the producers multiplies the biomass needed by 10 — and convert kg to g before using the 20 kJ per gram value.

8Grade 6–7Triple only6 marks

Most food chains have no more than four or five trophic levels.

Explain why.

In your answer refer to how energy and biomass are transferred in a food chain. [6]

producers transfer only about 1% of the light energy falling on them into biomass (so the energy entering the chain is limited)(1)
only about 10% of the biomass at one trophic level is transferred to the next level(1)
not all of an organism is eaten, e.g. roots, bones, shells, fur(1)
not all of the material eaten is digested / absorbed; it is egested as faeces (e.g. cellulose in plants)(1)
biomass is lost as carbon dioxide and water from respiration(1)
respiration releases energy for movement / keeping warm (in mammals and birds), so the glucose used is not stored as biomass(1)
biomass is lost as urea in urine / in excretion(1)
some organisms die without being eaten and are broken down by decomposers(1)
so the biomass (and energy) available decreases at each trophic level(1)
e.g. 10 000 → 1000 → 100 → 10 kg: after four or five levels there is too little biomass left to support another (viable) population(1)
organisms at higher levels / apex predators are few in number / need to hunt over a large area to get enough food(1)
a pyramid of biomass gets narrower towards the top for this reason(1)

Level	Marks	What it looks like
Level 3	5–6	A detailed, logically linked explanation: identifies that only about 10% of biomass passes between levels, explains at least two named ways biomass is lost (egestion, respiration products, urea) and links the cumulative decrease to too little biomass/energy remaining to support a further trophic level.
Level 2	3–4	A partial explanation: states that biomass or energy decreases at each level and gives at least one correct reason for the loss, but the link to the limited number of trophic levels is weak or missing.
Level 1	1–2	Simple statements, e.g. ‘energy is lost at each level’ or a single loss route, with little or no explanation.
No relevant content	0	Nothing worthy of credit.

Answer: Only ~1% of light energy is transferred by producers and only ~10% of biomass passes to each next level, because parts are not eaten, food is egested, biomass is lost as CO₂ and water in respiration and as urea in urine; so after 4–5 levels there is not enough biomass to support another population.
Examiner tip: To reach Level 3 you must link the losses (egestion, respiration, urine) to the falling biomass at each level and then say why that limits the number of levels — just listing the losses stops at Level 2.

9Grade 8–9HigherTriple only9 marks

The table shows the dry biomass at each trophic level in a meadow food chain.

Trophic level	Organism	Dry biomass in g/m²
1	grasses	X
2	leafhoppers	315
3	spiders	27
4	shrews	2.4

(a) The efficiency of biomass transfer from the grasses to the leafhoppers was 9%.

Calculate biomass X. [2]

(b) Calculate the efficiency of biomass transfer from the spiders to the shrews. Give your answer to 2 significant figures. [2]

(c) A student said: ‘These data support the idea that about 10% of biomass is transferred from one trophic level to the next.’

Evaluate the student’s statement. Use calculations in your answer. [2]

(d) Spiders are ectotherms: their body temperature changes with the temperature of their surroundings. Shrews are mammals that keep a constant body temperature.

A spider and a shrew each eat 10 g (dry biomass) of insects.

Explain why the shrew is likely to turn a smaller mass of this food into new body tissue than the spider. [3]

(a) 315 ÷ 0.09
allow 315 × 100 ÷ 9(1)
(a) 3500 (g/m²)
an answer of 3500 scores 2 marks
an answer of 28.35 (315 × 0.09) scores 0 marks(1)
(b) 2.4 ÷ 27 (× 100)(1)
(b) 8.9 (%)
an answer of 8.9 scores 2 marks
8.89 / 8.889 scores 1 mark (not 2 s.f.)(1)
(c) efficiency from leafhoppers to spiders = 27 ÷ 315 × 100 = 8.6% (allow 8.57)
and all three efficiencies (9%, 8.6%, 8.9%) are close to / slightly below 10%, so the data support the statement(1)
(c) but only one food chain / one meadow / one set of measurements was studied (so this cannot be generalised)
allow biomass may vary with season / time of year, or other food chains may differ(1)
(d) shrews respire more / use more of their food in respiration (than spiders do)
allow a greater proportion of the biomass shrews eat is lost as carbon dioxide and water(1)
(d) to release energy to keep their body temperature constant / to keep warm (higher than surroundings)
allow energy transferred to the surroundings by heating
ignore ‘to make heat’(1)
(d) so less of the biomass eaten / less glucose is available for growth / is stored as new tissue
allow so less biomass is available to the next trophic level(1)
Answer: (a) X = 315 ÷ 0.09 = 3500 g/m² (b) 8.9% (c) 8.6% leafhoppers→spiders; 9%, 8.6% and 8.9% are all close to 10%, so supported, but only one meadow/one sample (d) the shrew uses more of its food in respiration to keep its body temperature constant, so more biomass is lost as CO₂ and water and less is built into new body tissue
Examiner tip: When you are given an efficiency and the higher level, divide by the decimal (÷ 0.09) to go back down to the producers — multiplying gives a producer biomass smaller than the consumers, which should warn you it is wrong.

10Grade 1–3FoundationTriple only7 marks

A sheep ate 100 g of grass (dry biomass).

The bar chart shows how much of this biomass was lost from the sheep in three different ways.

(a) Use the bar chart. What mass of biomass was lost in faeces? [1]

(b) The rest of the biomass was used for growth of the sheep.

Calculate the mass of biomass used for growth. [2]

Name the two waste products. [2]

(d) Name the substance in urine that contains biomass. [1]

(e) Suggest why so much of the grass biomass is lost in faeces. [1]

(a) 45 g
allow 44–46 g(1)
(b) 45 + 40 + 5 = 90 or 100 − 45 − 40 − 5
allow ecf from (a)(1)
(b) 10 (g)
an answer of 10 scores 2 marks(1)
(c) carbon dioxide
allow CO₂(1)
(c) water
allow H₂O
ignore oxygen / glucose / energy (energy is not a substance)(1)
(d) urea
do not accept urine / uric acid / ammonia(1)
(e) not all of the grass can be digested (by the sheep)
allow grass contains cellulose / fibre (which is hard to digest)
allow not all the food is absorbed
ignore the sheep eats too much(1)
Answer: (a) 45 g (b) 100 − 90 = 10 g (c) carbon dioxide and water (d) urea (e) some of the grass (e.g. cellulose) cannot be digested so it passes out in the faeces
Examiner tip: Read every bar carefully against the scale (each small gridline is 5 g) and subtract all three losses from 100 g — egestion in faeces is not the same as excretion of urea in urine.

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