Food chains, predator–prey cycles and field sampling (RP9) – GCSE Biology exam questions with mark scheme

1Grade 1–3Foundation4 marks

The diagram shows a food web from a grassland.

(a) Name the producer in this food web. [1]

(b) Name one secondary consumer. [1]

(a) grass(1)
(b) shrew / fox / owl(1)
(c) (number of grasshoppers) increases(1)
(c) because fewer are eaten (by shrews) / fewer predators(1)
Answer: (a) Grass (b) e.g. fox (c) Increase – fewer predators eat them.
Examiner tip: Give the effect and the reason – 'increases' alone only gets one mark.

2Grade 6–76 marks

Some students estimated the number of daisy plants in a field measuring 50 m by 80 m.

They placed 10 quadrats in the field. Each quadrat measured 0.5 m by 0.5 m. The number of daisies in each quadrat is shown.

Quadrat	1	2	3	4	5	6	7	8	9	10
Number of daisies	3	0	5	2	4	1	0	6	3	2

(a) Describe how the students should place the quadrats randomly. [2]

(b) Estimate the total number of daisy plants in the field. [3]

(a) lay out two tape measures at right angles to make a grid / axes across the field
allow divide the field into a grid of numbered squares(1)
(a) use a random number generator (or random number table) to give coordinates for each quadrat
do not accept throwing the quadrat(1)
(b) mean = 26 ÷ 10 = 2.6 daisies per quadrat
allow total of 26 daisies in 2.5 m² sampled(1)
(b) 2.6 ÷ 0.25 = 10.4 per m² (area of quadrat = 0.25 m²)
allow 26 ÷ 2.5 = 10.4 per m²; allow area of field = 4000 m² / number of quadrats in field = 16 000(1)
(b) 41 600 (daisies)
correct answer with no working scores 3(1)
(c) use more quadrats / sample a larger area
ignore 'repeat'(1)
Answer: (a) Grid from tape measures; random number coordinates (b) 41 600 (c) Use more quadrats.
Examiner tip: Students often forget the quadrat is 0.25 m², not 0.5 m² – square the side length.

3Grade 6–75 marks

A student investigated how the distribution of bluebells changes from the edge of a wood into an open field.

(a) Describe how the student could use a transect to collect data. [3]

The table shows the student's results.

Distance from wood edge in m	0	5	10	15	20
Light intensity in arbitrary units	15	30	55	80	95
Percentage cover of bluebells	60	45	20	5	0

(b) Describe the relationship between light intensity and the percentage cover of bluebells. [1]

(c) A student concludes that high light intensity kills bluebells. Give one reason why this conclusion may not be valid. [1]

(a) place a tape measure/line from inside the wood out into the field
allow from the edge of the wood into the field(1)
(a) place a quadrat at regular intervals along the line (eg every 5 m)(1)
(a) estimate percentage cover / count bluebells in each quadrat (and measure light intensity at each quadrat)
allow repeat with further transects and calculate a mean for each distance (as an alternative for any one of the three (a) marks)(1)
(b) as light intensity increases, percentage cover of bluebells decreases
allow negative correlation(1)
(c) any one from: other factors (eg soil moisture, competition from grass, grazing, soil pH) also change along the transect; correlation does not show cause; only one transect / no repeats(1)
Answer: (a) Tape from wood into field; quadrats at regular intervals; record % cover (b) Higher light, lower cover (c) Other factors also change.
Examiner tip: In (a), 'regular intervals' is the mark many miss – a transect samples at set distances, not randomly.

4Grade 4–57 marks

Weasels are predators that feed mainly on voles. Ecologists recorded the numbers of voles and weasels in a large area of grassland every year for 12 years.

The graph shows the results. Voles are read from the left-hand axis and weasels from the right-hand axis.

(a) Give the year in which the number of voles was highest. [1]

(b) The numbers of both animals rise and fall in a regular pattern. How many years are there between the two peaks in the number of voles? [1]

(d) Explain why the number of voles decreased between year 2 and year 5. [2]

(e) Suggest one factor, other than predation, that could have caused the number of voles to fall. [1]

(a) year 8(1)
(b) 6 (years)
(peaks at year 2 and year 8)(1)
(c) more voles / more prey
allow more food (for weasels)(1)
(c) so more weasels survive and / or (more weasels) reproduce / breed / have more young
ignore ‘weasels grow’(1)
(d) more weasels / more predators (in years 2–3)(1)
(d) so more voles are eaten / killed
allow (voles are) eaten faster than they can reproduce
ignore voles ‘die’ unqualified(1)
(e) any one from:
• shortage of food / plants / grass (for voles)
• disease
• competition (for food / space) with other voles or other species
• harsh / cold winter or flooding
• habitat loss (e.g. ploughing)
allow lack of water
ignore weather unqualified / ‘pollution’ unqualified
do not accept another predator (question says other than predation)(1)
Answer: (a) year 8 (b) 6 years (c) more voles = more food, so more weasels survive and breed (d) more weasels, so more voles eaten (e) e.g. less food for voles / disease / competition
Examiner tip: In (c) and (d) the second mark needs the consequence — say the weasels survive and breed, or the voles are eaten; ‘there is more food’ on its own only earns 1 mark.

5Grade 6–76 marks

A park has two areas of grass. One area is mown every week. The other area is never mown.

A student wants to find out whether mowing affects the population of buttercup plants.

She has: 0.5 m × 0.5 m quadrats, two 30 m tape measures and a random number generator.

Plan an investigation to compare the number of buttercup plants in the two areas. [6]

lay the two tape measures at right angles (along two edges) to make a grid / axes in the first area(1)
use the random number generator to give coordinates (e.g. 7 m along and 12 m across)(1)
place a quadrat at each pair of coordinates (corner of quadrat at the point)(1)
count the number of buttercup plants in each quadrat(1)
decide (before starting) what counts as ‘in’ the quadrat, e.g. count a plant if more than half of it / its stem base is inside(1)
use a large number of quadrats, e.g. at least 10 / 20, in each area(1)
use the same number of quadrats and the same size quadrat in both areas(1)
repeat the same method in the second area(1)
calculate the mean number of buttercups per quadrat for each area (allow per m², mean × 4)(1)
compare the means of the two areas to see whether mowing affects the population(1)
control variables: sample both areas on the same day / same time of year; areas with similar light / soil / slope(1)
(optional) estimate each total population: mean per m² × area of each region(1)

Level	Marks	What it looks like
Level 3	5–6	A logical method that could be followed to produce valid results: random placement explained (tape-measure grid and random coordinates), what is counted in each quadrat, a suitable number of quadrats in both areas, and use of the means to compare the two areas.
Level 2	3–4	The main steps are present — quadrats placed in both areas and buttercups counted — with some indication of random placement or of using means, but some detail or order is missing so the method could not be followed without extra thought.
Level 1	1–2	Simple, relevant statements such as ‘put quadrats down and count the buttercups’ or ‘do it in both areas’, without a clear method of random sampling or comparison.
No marks	0	No relevant content.

Answer: Lay tape measures at right angles to form a grid in each area; use random numbers as coordinates; place ≥10 quadrats per area at the coordinates; count buttercups in each; same number and size of quadrats in both areas, same day; calculate the mean per quadrat (or per m²) for each area and compare.
Examiner tip: ‘Throw the quadrat over your shoulder’ is not random sampling in a mark scheme — describe the tape-measure grid and random-number coordinates to reach Level 3.

6Grade 6–76 marks

A footpath crosses a meadow. A student placed a 1 m × 1 m quadrat at 1 m intervals along a line from the centre of the footpath into the meadow. She estimated the percentage cover of two plants in each quadrat: greater plantain (a short plant with flat leaves) and tall grasses.

The diagram shows the quadrat she placed 1 m from the centre of the path.

(a) Use the diagram to estimate the percentage cover of plantain at 1 m. [1]

The table shows the student's results.

Distance from centre of path in m	0	1	2	3	4	5
Percentage cover of plantain	46	X	21	9	3	0
Percentage cover of tall grasses	0	12	37	58	79	88

(b) Give one reason why the student measured percentage cover rather than counting the number of plants. [1]

Suggest an explanation for the distribution of plantain shown in the table. [3]

(d) Give one way the student could make her results more representative of the whole meadow. [1]

(a) 38 (%)(1)
(b) it is difficult to tell where one (grass) plant ends and the next begins / individual plants are hard to count
allow the plants grow in mats / clumps / spread by runners
allow it is quicker
ignore ‘more accurate’ unqualified(1)
(c) near the path the tall grasses are damaged / killed by trampling (so there is little grass)(1)
so plantain has less competition (from the grasses) near the path
allow further from the path the tall grasses outcompete the plantain(1)
for light (allow space / water / mineral ions)
allow the tall grasses shade the plantain(1)
(d) repeat the transect at other places along the path / do more transects (and calculate a mean)
ignore sampling at smaller intervals along the same line
ignore ‘repeat’ unqualified(1)
Answer: (a) 38% (b) individual grass plants are hard to count / quicker (c) trampling kills tall grasses near the path, so plantain has less competition for light and space; further away tall grasses out-compete / shade it (d) carry out more transects at different places along the path
Examiner tip: In (c) don’t just describe the pattern — the marks are for linking trampling to the grasses and then to less competition for light for the plantain.

7Grade 8–9Higher8 marks

Aphids are small insects that feed on the sap of pepper plants. Ladybirds eat aphids.

A grower released ladybirds into a greenhouse of pepper plants in week 2. The table shows the mean numbers of aphids and ladybirds on the plants.

Week	0	1	2	3	4	5	6	7	8	9	10
Mean number of aphids per plant	12	30	65	110	140	95	48	20	14	22	35
Mean number of ladybirds per 10 plants	0	0	5	9	16	24	27	21	12	8	7

(a) Calculate the percentage decrease in the mean number of aphids per plant from week 4 to week 8. [2]

(b) The number of ladybirds was highest two weeks after the number of aphids was highest.

Explain why. [2]

(d) The grower concluded that the ladybirds caused the fall in the number of aphids.

Suggest how the grower could have improved the investigation to support this conclusion. [1]

(a) (140 − 14) ÷ 140 (× 100)
allow 126 ÷ 140(1)
90 (%)
an answer of 90 scores 2 marks
an answer of 0.9 scores 1 mark
do not credit 900 (divided by the final value)(1)
(b) ladybirds had plenty of food (aphids) so more survived and reproduced / laid eggs(1)
(b) it takes time for ladybirds to reproduce / for eggs and larvae to develop into adults
or ladybird numbers only fell once there were too few aphids (to feed them all), which happened after week 6
allow there were still enough aphids for ladybird numbers to rise until week 6(1)
(c) aphids increase (from 14 to 35 per plant)
ignore descriptions of ladybird data alone for this mark(1)
(c) because the number of ladybirds decreased (27 → 7 per 10 plants) as there were too few aphids / less food (so ladybirds starved / left / bred less)(1)
(c) so fewer aphids were eaten and the aphids could reproduce faster than they were eaten / more aphids survived to breed
max 2 if no reference to the fall in ladybirds(1)
(d) use a second (identical) greenhouse with no ladybirds released as a control, and compare the aphid numbers
allow measure other factors that could affect aphids (e.g. temperature / plant age / disease) and keep them the same
ignore ‘repeat’ unqualified(1)
Answer: (a) (140 − 14)/140 × 100 = 90% (b) many aphids → ladybirds survive and reproduce, but breeding/development takes time, so their peak lags; they fall only when aphids become scarce (c) aphids rise from 14 to 35 because ladybirds have fallen (lack of food), so fewer aphids eaten and aphids reproduce (d) a control greenhouse with no ladybirds
Examiner tip: Percentage change always divides by the starting value (140), not the final value — dividing by 14 gives 900%, a common slip.

8Grade 1–3Foundation6 marks

This is a food chain from a pond.

algae → mayfly larvae → stickleback → heron

(a) Name the producer in this food chain. [1]

(b) Complete the sentence. Choose the answer from the box.

[ biomass | minerals | oxygen only | water ]

Producers are photosynthetic organisms that make ____________ for the rest of the food chain. [1]

(d) Name the organism that is both a predator and prey in this food chain. [1]

(e) What do the arrows in a food chain show? [1]

(f) What is the original source of energy for this food chain? [1]

(a) algae
allow alga
do not accept pond / water / plants (not in the chain)(1)
(b) biomass(1)
(c) stickleback
allow phonetic spelling(1)
(d) stickleback
do not accept mayfly larvae (they eat algae, so are not predators)
do not accept heron (it is not eaten)(1)
(e) the transfer of energy / biomass (from one organism to the next)
allow what is eaten by what / the direction of feeding, e.g. the stickleback eats the mayfly larvae
do not accept ‘what eats what’ written the wrong way round, e.g. ‘the algae eats the mayfly’(1)
(f) the Sun
allow (sun)light / light energy
ignore heat
do not accept algae / photosynthesis alone(1)
Answer: (a) algae (b) biomass (c) stickleback (d) stickleback (e) transfer of energy / biomass (direction of feeding) (f) the Sun (light)
Examiner tip: Count the steps from the producer: the first animal is the primary consumer, the second is the secondary consumer — students often name the heron because it is the ‘biggest’ predator.

9Grade 4–5Foundation6 marks

Some students counted the number of limpets in ten 0.5 m × 0.5 m quadrats on a rocky shore.

Quadrat	1	2	3	4	5	6	7	8	9	10
Number of limpets	12	0	7	9	7	31	8	6	7	5

(a) Calculate the mean number of limpets per quadrat. [2]

(b) Give the median number of limpets per quadrat. [1]

(d) Quadrat 6 was placed on a large rock covered in limpets.

Explain why the median may represent the typical number of limpets per quadrat better than the mean does. [2]

(a) 92 ÷ 10
allow (12 + 0 + 7 + 9 + 7 + 31 + 8 + 6 + 7 + 5) ÷ 10(1)
(a) 9.2
an answer of 9.2 scores 2 marks
allow 9(1)
(b) 7
(ordered: 0, 5, 6, 7, 7, 7, 8, 9, 12, 31; (7 + 7) ÷ 2)(1)
(c) 7(1)
(d) quadrat 6 / 31 is an extreme value / outlier / much higher than the others(1)
which increases / distorts the mean (but has little effect on the median)
allow the mean is higher than most of the counts / only 2 quadrats are above the mean
ignore ‘anomalous’ unqualified for this mark(1)
Answer: (a) 9.2 (b) 7 (c) 7 (d) the 31 in quadrat 6 is an extreme value that raises the mean but hardly affects the median
Examiner tip: For the median, put the numbers in order first and, with an even number of values, take the mean of the middle two — using the table order gives the wrong answer.

10Grade 6–7Higher6 marks

A student estimated the population of dandelions on a school field. The field is a rectangle measuring 120 m by 75 m.

The student used a square quadrat with sides of 40 cm. She placed 20 quadrats at random and counted a total of 72 dandelions.

(a) Estimate the number of dandelions on the field.

Give your answer in standard form to 2 significant figures. [4]

(b) Another student placed all 20 of his quadrats in the sunniest corner of the field, where dandelions grow best.

Explain how this would affect his estimate of the dandelion population. [2]

(a) mean per quadrat = 72 ÷ 20 = 3.6
or total area sampled = 20 × 0.16 = 3.2 (m²)(1)
(a) area of quadrat = 0.4 × 0.4 = 0.16 (m²) and number per m² = 3.6 ÷ 0.16 = 22.5
or 72 ÷ 3.2 = 22.5 (per m²)
or area of quadrat = 0.16 (m²) and number of quadrats that fit in the field = 9000 ÷ 0.16 = 56 250(1)
(a) area of field = 9000 (m²) and 22.5 × 9000 = 202 500 or 3.6 × 56 250 = 202 500
allow ecf from an incorrect number per m²(1)
2.0 × 10⁵
an answer of 2.0 × 10⁵ scores 4 marks
an answer of 202 500 / 200 000 / 2 × 10⁵ (not in standard form or not to 2 s.f.) scores 3 marks
allow ecf: their population correctly given in standard form to 2 s.f., e.g. quadrat area taken as 1 m² → 32 400 → 3.2 × 10⁴ scores 3 marks(1)
(b) the estimate would be too high / an overestimate(1)
(b) because the sample is not representative of the whole field / the quadrats were not placed randomly, so (sunny) areas with more dandelions are over-sampled
allow (sampling) bias
ignore ‘unreliable’ / ‘inaccurate’ unqualified(1)
Answer: (a) 72 ÷ 20 = 3.6 per quadrat; 3.6 ÷ 0.16 = 22.5 per m²; 22.5 × 9000 = 202 500 ≈ 2.0 × 10⁵ (b) overestimate — the sample is biased / not representative because the quadrats were not placed at random
Examiner tip: Convert the quadrat side to metres before squaring: 40 cm = 0.4 m, so the area is 0.16 m², not 0.4 m² or 1600 m².

Still losing marks on this topic?

Book a free introductory session and we'll find exactly which step is costing you.

Book a free intro