Ecology and Interdependence

QUESTION 1 3 marks Criterion A

Easy

The diagram shows a simple food chain found in a meadow.

a. What is the source of all energy entering this food chain?

[1]

b. Identify the producer in this food chain.

[1]

c. Identify the tertiary consumer in this food chain, and explain how you know.

[1]

Show complete worked solution

(a)

The Sun. Grass (a producer) uses light energy from the Sun for photosynthesis to make its own food, and this stored chemical energy is what enters the food chain when the grass is eaten.

(b)

Grass is the producer — it is the only organism in the chain that makes its own food (by photosynthesis) rather than eating other organisms.

(c)

The snake is the tertiary consumer. It is the third consumer in the chain: the grasshopper is the primary consumer (eats the producer), the frog is the secondary consumer (eats the primary consumer), and the snake is the tertiary consumer (eats the secondary consumer).

QUESTION 2 3 marks Criterion A

Easy

Every food chain contains producers, consumers, and decomposers.

a. State what is meant by a ‘producer’, giving one example.

[1]

b. State what is meant by a ‘consumer’, giving one example.

[1]

c. State what is meant by a ‘decomposer’, giving one example.

[1]

Show complete worked solution

(a)

A producer is an organism that makes its own food using light energy from the Sun (photosynthesis), rather than eating other organisms. Example: grass, or any green plant.

(b)

A consumer is an organism that cannot make its own food and must obtain energy by eating other organisms. Example: a rabbit (eats plants) or a fox (eats animals).

(c)

A decomposer is an organism that breaks down dead organisms and waste material, releasing nutrients back into the soil/ecosystem. Example: bacteria, or fungi such as mushrooms.

QUESTION 3 4 marks Criterion A

Medium

A student is given four organisms found in a woodland: grass, mouse, snake, owl.

a. Arrange these four organisms into a correct food chain, using arrows to show the direction energy flows.

[2]

b. For each organism in your food chain, state its trophic level (e.g. producer, primary consumer).

[2]

Show complete worked solution

(a)

$$ \text{grass} \rightarrow \text{mouse} \rightarrow \text{snake} \rightarrow \text{owl} $$ Each arrow points from the organism being eaten to the organism eating it, showing the direction energy flows along the chain.

(b)

Grass = producer. Mouse = primary consumer (eats the producer). Snake = secondary consumer (eats the primary consumer). Owl = tertiary consumer (eats the secondary consumer).

QUESTION 4 3 marks Criterion A

Medium

The table shows four animals and the food each one eats.

Animal	Food eaten
Deer	Grass and leaves
Lion	Zebra and other animals
Bear	Berries and fish
Grasshopper	Leaves and stems

a. Classify each of the four animals as a herbivore, a carnivore, or an omnivore.

[2]

b. Explain what ‘omnivore’ means, using the bear as your example.

[1]

Show complete worked solution

(a)

Deer — herbivore (eats only plants). Lion — carnivore (eats only animals). Bear — omnivore (eats both plants and animals). Grasshopper — herbivore (eats only plants).

(b)

An omnivore is an animal that eats both plants and animals. The bear fits this because its diet includes berries (a plant food) as well as fish (an animal food).

QUESTION 5 4 marks Criterion A

Medium

The diagram shows a pyramid of energy for a grassland ecosystem, showing the energy available at each trophic level in $\text{kJ/m}^2\text{/year}$.

a. Why does a pyramid of energy always get narrower towards the top?

[1]

b. Calculate what percentage of the energy in the producers is transferred to the primary consumers.

[2]

c. State the general rule about energy transfer that this percentage illustrates.

[1]

Show complete worked solution

(a)

Energy is lost at each trophic level (mainly as heat from respiration, but also in movement and in material that is not eaten or not digested), so less energy is available to be passed to the level above — the narrowing shape of the pyramid reflects this steady decrease.

(b)

$$ \text{percentage transferred} = \frac{\text{energy in primary consumers}}{\text{energy in producers}} \times 100 = \frac{1000}{10000} \times 100 $$ Answer: $10\%$

(c)

This illustrates the ‘10% rule’ — on average, only about 10% of the energy available at one trophic level is transferred to the next, with the rest lost mainly as heat.

QUESTION 6 3 marks Criterion A

Medium

Food chains rarely contain more than four or five feeding (trophic) levels.

a. What is meant by a ‘trophic level’?

[1]

b. Using ideas about energy transfer, explain why food chains rarely have more than four or five trophic levels.

[2]

Show complete worked solution

(a)

A trophic level is the position an organism occupies in a food chain, based on the number of energy transfers needed to reach it — for example producer, primary consumer, or secondary consumer.

(b)

Only about 10% of the energy available at each trophic level is passed on to the next (the rest is lost as heat, in movement, and in undigested material). After several transfers so little energy remains that there is not enough left to support a viable population at another feeding level — for example, $10{,}000\,\text{kJ}$ at the producer level falls to only $10\,\text{kJ}$ after just three transfers.

QUESTION 7 6 marks Criterion A

Hard

In a lake ecosystem, algae (producers) fix $500{,}000\,\text{kJ/m}^2\text{/year}$ of energy from sunlight. Assume that $10\%$ of the energy at each trophic level is transferred to the next.

a. Calculate the energy available to the primary consumers (zooplankton).

[2]

b. Calculate the energy available to the secondary consumers (small fish).

[2]

c. Calculate the energy available to the tertiary consumers (large fish), and use your answer to explain why top predators like large fish exist in much smaller numbers than algae.

[2]

Show complete worked solution

(a)

$$ 500{,}000 \times 0.10 = 50{,}000\,\text{kJ/m}^2\text{/year} $$

(b)

$$ 50{,}000 \times 0.10 = 5{,}000\,\text{kJ/m}^2\text{/year} $$

(c)

$$ 5{,}000 \times 0.10 = 500\,\text{kJ/m}^2\text{/year} $$ This is only $\frac{500}{500{,}000}\times100 = 0.1\%$ of the original energy fixed by the algae. Because so little energy remains after three transfers, this tiny amount can only support a very small mass/number of large fish — which is why top predators are always far less numerous than the producers at the base of the food chain.

QUESTION 8 6 marks Criterion B

Medium

A student finds owl pellets (regurgitated balls of undigested fur and bones) below a barn owl's roost and wants to investigate what the owl has been eating, in order to build a food web for the area.

a. State a suitable method for identifying what the owl has eaten, including one piece of equipment.

[2]

b. State two variables that should be kept the same (controlled) when collecting and analysing multiple pellets, and explain why for one of them.

[2]

c. Explain why examining a large number of pellets (e.g. 30), rather than just one, gives a more reliable picture of the owl's diet.

[2]

Show complete worked solution

(a)

Carefully pull each pellet apart using tweezers and a dissecting tray/mat, then examine the bones found inside (especially skulls and jaw bones) using a magnifying glass or hand lens, and compare them to a bone identification key to identify the prey species.

(b)

Control the location the pellets are collected from (all from the same roost) and the person identifying the bones (same identification key/same observer). Why control location: pellets from a different roost could belong to a different owl with a different diet, making it unfair to combine or compare the results.

(c)

A single pellet only shows what the owl ate in one meal, which might not be typical. Examining many pellets shows the range and relative frequency of different prey species eaten over time, giving a mean/representative picture of the diet rather than one skewed by a single unusual meal.

QUESTION 9 6 marks Criterion B

Medium

A student wants to investigate whether woodlice (decomposers that feed on dead leaves) are found in greater numbers in moist leaf litter than in dry leaf litter.

a. Identify the independent and dependent variables in this investigation.

[2]

b. State two variables that should be controlled, and explain why for one of them.

[2]

c. Describe a suitable method, including equipment, for counting the woodlice in each area.

[2]

Show complete worked solution

(a)

Independent variable: the moisture level of the leaf litter (moist site vs dry site). Dependent variable: the number of woodlice found/counted.

(b)

Control the area of leaf litter sampled and the time of day sampling takes place. Why control the sampled area: a larger sampled area would naturally be expected to contain more woodlice regardless of moisture, which would make the comparison between sites unfair.

(c)

Mark out an equal area (e.g. using a $0.5\,\text{m}\times0.5\,\text{m}$ quadrat) in each of the moist and dry leaf litter sites. Search through the leaf litter within the quadrat for a fixed time (e.g. 5 minutes) and count all woodlice found, using a pooter to safely collect and count the small invertebrates. Repeat at 5 different points in each site and calculate a mean count for the moist site and the dry site.

QUESTION 10 7 marks Criterion B

Hard

A student used pitfall traps (cups sunk into the ground, level with the surface, left overnight) to sample the invertebrates living on a forest floor as part of a food web study. Rainwater collected in some of the traps overnight, and in the morning several trapped insects had escaped by climbing back out.

a. Identify two flaws in this method that would make the results unreliable.

[2]

b. Suggest an improvement for each flaw you identified, and explain how each improvement solves the problem.

[3]

c. Explain why pitfall trapping alone, even once improved, cannot capture organisms like birds or grazing mammals that also form part of the food web.

[2]

Show complete worked solution

(a)

(1) Rainwater collecting in the traps could drown or wash away some invertebrates, changing which species and how many are actually caught. (2) Insects escaping by climbing out means that active, climbing species are undercounted compared with the true population, biasing the results towards less mobile species.

(b)

For the rain: fix a small cover/roof propped just above the trap (leaving gaps around the edge for insects to fall in) to keep rain out, so conditions are kept dry and consistent between traps. For escaping insects: use smooth, steep-sided containers (e.g. plastic cups) with a few small drainage holes, so that invertebrates that fall in cannot climb back out over the smooth sides.

(c)

Pitfall traps rely on ground-dwelling invertebrates accidentally falling in as they walk across the surface. Flying or larger, mobile animals such as birds and mammals are not caught this way, since they can fly, jump, or simply step over the trap — so a complete picture of the food web needs other sampling methods too (e.g. direct observation or camera traps).

QUESTION 11 3 marks Criterion C

Easy

A student surveyed a field and counted the following numbers of organisms.

Organism	Grass tussocks	Rabbits	Foxes
Number counted	850	120	6

a. Using this data, which organism is the producer?

[1]

b. Explain what pattern you would expect if this data were drawn as a pyramid of numbers, referring to the values given.

[2]

Show complete worked solution

(a)

The grass tussocks — they are the only organism in the list capable of photosynthesis (making their own food).

(b)

A pyramid of numbers should get narrower going up: producers (grass tussocks, 850) form the wide base, primary consumers (rabbits, 120) form a narrower middle layer, and the top predator (foxes, 6) forms the narrowest layer at the top — matching the pattern in the data, since fewer individuals can be supported at each higher trophic level.

QUESTION 12 3 marks Criterion C

Easy

The diagram shows a simplified food web from a meadow habitat.

a. Using the diagram, name one organism that is eaten by both the fox and the owl.

[1]

b. Write out one complete food chain, from a producer to a top predator, that can be built using this diagram.

[2]

Show complete worked solution

(a)

The vole — the diagram shows an arrow from vole to fox and a separate arrow from vole to owl.

(b)

For example: $$ \text{Grass} \rightarrow \text{Rabbit} \rightarrow \text{Fox} $$ (Other valid chains from the diagram include Clover → Rabbit → Fox, Grass → Vole → Fox, Seeds → Vole → Owl, and Seeds → Mouse → Owl.)

QUESTION 13 5 marks Criterion C

Medium

The graph shows population data collected over 20 years for a prey species (hares) and a predator species (lynx) living in the same ecosystem.

a. Describe the relationship shown between the two populations.

[2]

b. Explain, in terms of food availability, why the predator population peaks a few years after the prey population peaks.

[3]

Show complete worked solution

(a)

Both populations rise and fall in a repeating (cyclical) pattern. The predator (lynx) population follows the same general pattern as the prey (hare) population, but its peaks and troughs occur a few years after those of the hare population — there is a time lag between them.

(b)

When the hare (prey) population is high, there is plenty of food available for the lynx, so more lynx survive and reproduce successfully, and the predator population grows over the following few years. This growing predator population then eats more hares, causing the hare population to fall; with less food now available, the predator population falls too, but again with its own delay — producing the repeating, offset cycle seen in the graph.

QUESTION 14 4 marks Criterion C

Medium

The table shows biomass measured at each trophic level of a food chain.

Trophic level	Producers	Primary consumers	Secondary consumers	Tertiary consumers
Biomass (kg)	500	60	6	40

a. Identify the anomalous value in this table.

[1]

b. Explain how you identified it.

[2]

c. Suggest one possible reason for this anomalous reading.

[1]

Show complete worked solution

(a)

The tertiary consumers value, $40\,\text{kg}$.

(b)

Biomass should decrease from one trophic level to the next, since energy (and therefore biomass) is lost at each transfer. The values fall as expected from producers ($500$) to primary consumers ($60$) to secondary consumers ($6\,\text{kg}$), but then rise sharply back up to $40\,\text{kg}$ for tertiary consumers — breaking the otherwise decreasing pattern.

(c)

The tertiary consumer biomass may have been measured or estimated incorrectly (e.g. a large, non-target animal was included by mistake, or the small sample of tertiary consumers was not representative), giving a falsely high reading.

QUESTION 15 4 marks Criterion C

Medium

The table shows the energy available at two trophic levels in a food chain.

Trophic level	Energy (kJ/m²/yr)
Producers	20,000
Primary consumers	1,800

a. Calculate the percentage of energy from the producers that is transferred to the primary consumers.

[2]

b. Compare this value to the ‘10% rule’ and suggest one reason the two values are not identical.

[2]

Show complete worked solution

(a)

$$ \frac{1800}{20000} \times 100 = 9\% $$

(b)

$9\%$ is very close to, but slightly below, the typical $10\%$ rule. Real ecosystems vary around this average value — for example, how active the primary consumers are (more energy lost as heat/movement) or how much of the plant material is actually eaten and digested (rather than left uneaten or excreted) can make the true transfer efficiency slightly higher or lower than exactly $10\%$.

QUESTION 16 4 marks Criterion C

Medium

The table shows population data for a field before and after a disease killed most of the rabbit population.

Organism	Population before disease	Population after disease
Grass tussocks	850	1400
Rabbits	120	15
Foxes	30	9

a. Using the data, describe the change in the grass population after the rabbit population crashed, and explain why this happened.

[2]

b. Explain why the fox population also decreased, even though foxes were not directly affected by the disease.

[2]

Show complete worked solution

(a)

The grass population increased, from $850$ to $1400$ tussocks. With far fewer rabbits ($120 \rightarrow 15$) grazing on it, much less grass was eaten, so more of it survived and was able to grow, increasing the population.

(b)

Foxes rely on rabbits as a main food source. With the rabbit population crashing from $120$ to $15$, there was much less food available for foxes, so fewer foxes could survive and reproduce — causing their own population to fall from $30$ to $9$. This is an indirect (knock-on) effect passed through the food chain.

QUESTION 17 7 marks Criterion C

Hard

A student used the mark-release-recapture method to estimate the population of snails (a primary consumer) in a garden, repeating the recapture sample on three separate days to check the reliability of the estimate. The Lincoln index formula is:

$$ N = \frac{n_1 \times n_2}{n_m} $$

where $n_1$ = number marked and released, $n_2$ = total number recaptured, and $n_m$ = number of marked individuals found in the recapture sample.

Day	$n_1$ (marked & released)	$n_2$ (total recaptured)	$n_m$ (marked, recaptured)
1	40	50	8
2	40	48	7
3	40	52	20

a. Calculate the estimated total snail population for Day 1 and Day 2.

[3]

b. Calculate the Day 3 estimate, and explain why it should be considered anomalous.

[2]

c. Suggest a reason why the Day 3 estimate might be unreliable, and state what the student should do with it.

[2]

Show complete worked solution

(a)

Day 1: $$ N = \frac{40 \times 50}{8} = \frac{2000}{8} = 250 $$ Day 2: $$ N = \frac{40 \times 48}{7} = \frac{1920}{7} \approx 274 $$

(b)

$$ N = \frac{40 \times 52}{20} = \frac{2080}{20} = 104 $$ This is far lower than the Day 1 ($250$) and Day 2 ($\approx274$) estimates — a much larger proportion of the recaptured snails were marked on Day 3 than on the other days, which pulls the estimate down sharply and does not fit the otherwise consistent pattern.

(c)

The marked snails may not have mixed/dispersed fully back into the wider population before Day 3's sample was taken (e.g. they were recaptured near where they were released, before spreading out), making marked snails over-represented in that sample and giving an artificially low population estimate. This anomalous value should be excluded, and only the Day 1 and Day 2 estimates used to calculate a mean (approximately $262$).

QUESTION 18 5 marks Criterion D

Medium

DDT is a pesticide once widely used to kill insect pests on crops. It does not break down easily and can be washed into rivers and lakes, where it enters food chains. The table shows DDT concentrations measured at each trophic level of a lake ecosystem.

Trophic level	DDT concentration (ppm)
Plankton (producer)	0.04
Small fish	0.5
Large fish	2.0
Osprey (top predator)	25

Discuss one benefit and one drawback of using DDT as a pesticide, using the data to support your discussion.

Show complete worked solution

Benefit: DDT was very effective and cheap at controlling insect pests on crops, protecting harvests and increasing food production, and it also helped control disease-carrying insects such as mosquitoes, reducing the spread of diseases like malaria.

Drawback: Because DDT does not break down or get excreted easily, it builds up in each organism's body and becomes far more concentrated moving up the food chain (biomagnification) — the data show it rising more than $600$-fold, from $0.04\,\text{ppm}$ in plankton to $25\,\text{ppm}$ in ospreys ($25 \div 0.04 = 625$). At these high concentrations, DDT caused the eggshells of birds of prey like ospreys and eagles to become dangerously thin, leading to population collapses in top predators that were never directly sprayed with the pesticide.

QUESTION 19 6 marks Criterion D

Hard

Wolves, once hunted to extinction in a national park, were reintroduced as a top predator. The table shows ecological data recorded before reintroduction and 10 years after.

Measurement	Before wolves reintroduced	10 years after
Elk population	≈15,000	≈5,000
Willow/aspen tree cover	5%	25%
Beaver dams counted	1	9

Discuss one benefit and one drawback of reintroducing wolves into this ecosystem, using the data to support your answer.

Show complete worked solution

Benefit: Reintroducing wolves reduced the overgrazing elk population from about $15{,}000$ to about $5{,}000$ (a trophic cascade), allowing willow and aspen trees to recover strongly, from just $5\%$ to $25\%$ cover. This recovering vegetation provided food and building material for beavers, whose dam count rose from $1$ to $9$; beaver dams create wetland habitats that in turn support many other species, so reintroducing one top predator restored biodiversity across the whole ecosystem.

Drawback: Wolves can prey on livestock kept by nearby farmers and ranchers, causing economic losses and requiring compensation schemes or extra protective measures. The sharply reduced elk population and changed grazing patterns can also affect local hunting economies, creating ongoing human–wildlife conflict alongside the ecological benefits.

QUESTION 20 6 marks Criterion D

Hard

Anchovies are a small forage fish eaten by many larger predators, including seabirds. The table shows data collected as anchovy fishing increased over a decade.

Year	Anchovy catch (thousand tonnes)	Seabird breeding success (chicks per pair)
2000	200	1.1
2010	600	0.4

Discuss one benefit and one drawback of this increase in anchovy fishing, using the data to support your answer.

Show complete worked solution

Benefit: Anchovy fishing supplies large amounts of affordable protein and fishmeal (used for both direct food and animal feed). The catch tripled over the decade, from $200{,}000$ to $600{,}000$ tonnes, generating substantial income and jobs for fishing communities.

Drawback: Because anchovies form a key link between plankton and larger predators in the food web, removing so many reduces the food available further up the chain — seabird breeding success fell by more than half over the same period, from $1.1$ to $0.4$ chicks per breeding pair. This shows how overfishing one part of a food web can cause knock-on population declines in species that depend on it, even species that were never directly fished.

QUESTION 1 4 marks Criterion A

Easy

Ecologists use several key terms to describe the levels of organisation in nature.

a. What is meant by a ‘habitat’?

[1]

b. What is meant by a ‘population’?

[1]

c. What is meant by a ‘community’?

[1]

d. What is meant by an ‘ecosystem’?

[1]

Show complete worked solution

(a)

The place where an organism lives, e.g. a pond, a woodland, or under a rock.

(b)

All the individuals of one species living in the same area at the same time, e.g. all the frogs living in a pond.

(c)

All the populations of different species living and interacting together in the same habitat, e.g. all the frogs, pondweed, insects and fish in a pond.

(d)

A community of living organisms together with the non-living (physical and chemical) parts of their environment, all interacting as a system, e.g. a pond and everything living in and around it.

QUESTION 2 4 marks Criterion A

Easy

The environment affects organisms in two different ways: through abiotic factors and biotic factors.

a. Sort the following into abiotic and biotic factors: light intensity, predators, soil pH, disease.

[2]

b. Explain the difference between an abiotic and a biotic factor, using one example of each from part (a).

[2]

Show complete worked solution

(a)

Abiotic: light intensity, soil pH. Biotic: predators, disease.

(b)

An abiotic factor is a non-living, physical or chemical part of the environment, such as soil pH — it affects organisms but is not itself alive. A biotic factor is a living part of the environment that affects other organisms, such as predators, which reduce prey numbers through direct interaction (being hunted and eaten).

QUESTION 3 4 marks Criterion A

Medium

Camels have several adaptations that help them survive in hot desert habitats.

Adaptation
Large, wide feet that spread the camel's weight on soft sand
Being most active at dawn, dusk and night, resting in shade during the hottest part of the day
Able to tolerate large changes in body temperature (up to several degrees) without needing to sweat

a. For each adaptation in the table, state whether it is structural, behavioural, or physiological, and give a brief reason.

[3]

b. Explain how the camel's ability to tolerate large changes in body temperature helps it survive in the desert.

[1]

Show complete worked solution

(a)

Large, wide feet — structural (a physical body feature that does not change). Being active at dawn/dusk/night — behavioural (a pattern of actions/activity the camel performs). Tolerating large body temperature changes without sweating — physiological (an internal, bodily-process adaptation).

(b)

By allowing its body temperature to rise instead of sweating to cool down, the camel avoids losing large amounts of water through sweat — conserving the scarce water available in a hot, dry habitat.

QUESTION 4 3 marks Criterion A

Medium

Robins and blue tits both live in the same woodland habitat but rarely compete directly for food.

a. What is meant by an organism's ‘niche’?

[1]

b. Using the idea of a niche, explain why robins and blue tits can share the same habitat without competing directly for food.

[2]

Show complete worked solution

(a)

The specific role an organism plays within its habitat, including what it eats, where and when it feeds, and how it interacts with other organisms.

(b)

Even though they share the same habitat, the two species likely occupy different niches — for example feeding on different types of food (such as different insects or seeds), at different heights within the trees, or at different times of day. Because they are not competing for exactly the same resources, they can coexist in the same habitat without directly competing.

QUESTION 5 4 marks Criterion A

Medium

Every population living in a habitat has a ‘carrying capacity’.

a. What is meant by the ‘carrying capacity’ of a habitat?

[2]

b. State two factors that could limit a population's carrying capacity, and briefly explain how each acts as a limit.

[2]

Show complete worked solution

(a)

The maximum population size of a species that a habitat can support long-term (sustainably), based on the resources available, such as food, water, space and shelter.

(b)

Food availability — a limited food supply means only a certain number of individuals can be fed, capping population growth once that limit is reached. Predation — more predators kill more individuals from the population, reducing survival and preventing the population from growing beyond a certain size.

QUESTION 6 4 marks Criterion A

Medium

The table describes three relationships between different species.

Relationship
(a) Oxpecker birds eat ticks and parasites off a rhino's skin; the rhino gets pest control and the oxpecker gets food.
(b) A tapeworm lives inside a dog's intestine, absorbing nutrients from the dog's food and causing the dog to become unwell.
(c) Barnacles attach to a whale's skin, gaining a place to live and access to nutrient-rich water, without noticeably affecting the whale.

a. Classify each of the three relationships as mutualism, parasitism, or commensalism.

[3]

b. For the tapeworm example, explain why this is classified as parasitism rather than mutualism.

[1]

Show complete worked solution

(a)

(a) Oxpecker and rhino — mutualism (both organisms benefit). (b) Tapeworm and dog — parasitism (the tapeworm benefits, the dog is harmed). (c) Barnacle and whale — commensalism (the barnacle benefits, the whale is unaffected).

(b)

Mutualism requires that both organisms benefit. Here, only the tapeworm benefits (it gains nutrients), while the host (the dog) is harmed — it loses nutrients and may become ill — so this is parasitism, not mutualism.

QUESTION 7 6 marks Criterion A

Hard

A student used a $1\,\text{m}\times1\,\text{m}$ quadrat to sample daisies in a field. The quadrat was placed randomly at 5 different points, and the number of daisies inside was counted each time: $8, 12, 6, 10, 9$. The whole field has an area of $250\,\text{m}^2$.

a. Calculate the mean number of daisies per quadrat.

[2]

b. Calculate the population density of daisies, in daisies per $\text{m}^2$.

[2]

c. Use your answer to estimate the total number of daisies in the whole field.

[2]

Show complete worked solution

(a)

$$ \text{mean} = \frac{8+12+6+10+9}{5} = \frac{45}{5} = 9 $$

(b)

Since each quadrat has an area of $1\,\text{m}^2$, the density is equal to the mean count per quadrat: $$ \text{density} = 9\,\text{daisies/m}^2 $$

(c)

$$ \text{total population} = \text{density} \times \text{total area} = 9 \times 250 = 2250 $$ Answer: approximately $2250$ daisies.

QUESTION 8 6 marks Criterion B

Medium

A student wants to investigate whether grass grows more densely in a sunny, open area of a field than in a shaded area under tree canopy.

a. State the independent and dependent variables.

[2]

b. State two variables that should be controlled, and explain why for one of them.

[2]

c. Describe a suitable method for this investigation, including how quadrat positions would be chosen.

[2]

Show complete worked solution

(a)

Independent variable: the light condition of the area (shaded vs sunny/open). Dependent variable: the number of grass plants (or percentage ground cover of grass) counted per quadrat.

(b)

Control the size of quadrat used and the time of year/day the sampling is carried out. Why control the time of sampling: grass growth changes with the season and time of day (e.g. after rain), so sampling both areas at a different time could make any difference due to timing rather than light, making the comparison unfair.

(c)

Lay a tape measure across each area to form a grid, and use a random number generator to select coordinates for quadrat placement (to avoid bias from choosing spots by eye). Place a $0.5\,\text{m}\times0.5\,\text{m}$ quadrat at each randomly chosen coordinate, count (or estimate percentage cover of) the grass plants inside, and repeat at 10 random points in each area (shaded and sunny), then calculate a mean for each area to compare.

QUESTION 9 6 marks Criterion B

Medium

A student wants to investigate how species distribution changes along a transect from the edge of a pond into a nearby woodland, and whether this relates to an abiotic factor.

a. State a suitable independent variable and dependent variable for this investigation.

[2]

b. State two pieces of equipment needed for this investigation, and what each is used for.

[2]

c. Describe how the transect method would be carried out.

[2]

Show complete worked solution

(a)

Independent variable: distance along the transect from the pond edge (e.g. 0 m, 5 m, 10 m …). Dependent variable: the number of individuals of a chosen species (e.g. moss plants) counted in each quadrat, and/or a reading of an abiotic factor such as light intensity.

(b)

A tape measure, to lay out the transect line and mark regular distance intervals. A quadrat, placed at each interval to count the species. A light meter (lux meter) could also be used to measure light intensity at each point along the transect.

(c)

Lay a tape measure in a straight line from the pond edge into the woodland. At regular intervals along the tape (e.g. every 2 m), place a quadrat and record the number of the chosen species present, along with a light intensity reading at that point using a light meter. Repeat along the full length of the transect to see how species distribution changes with the abiotic factor measured.

QUESTION 10 7 marks Criterion B

Hard

A student investigated dandelion distribution in a field by placing a quadrat in 5 spots that ‘looked like they had a lot of dandelions’, and recorded these counts: $20, 22, 18, 25, 19$.

a. Identify a flaw in how the student chose the quadrat locations, and explain the effect this would have on the results.

[2]

b. Suggest an improvement to the method that would remove this bias, and explain how it works.

[2]

c. The student only used 5 quadrats. Explain why increasing the number of quadrat samples (e.g. to 20) would improve the reliability of the population estimate, and state one further improvement to make the field survey more representative.

[3]

Show complete worked solution

(a)

The quadrat locations were not chosen randomly — the student deliberately picked spots that looked dandelion-rich. This biases the sample towards high counts and would give a mean that overestimates the true average dandelion density across the whole field, since low-density areas were deliberately excluded.

(b)

Use a random number generator to generate random coordinates (e.g. using two tape measures laid at right angles as x- and y-axes across the field) for quadrat placement. This gives every point in the field an equal chance of being sampled, removing the observer's bias in choosing locations.

(c)

More quadrats sample a larger proportion of the field, reducing the effect of chance variation (e.g. one quadrat happening to fall on an unusually dense or sparse patch) on the calculated mean — a mean from 20 quadrats is far less likely to be skewed by a single unusual reading than a mean from just 5. A further improvement would be to spread sampling evenly across the whole field, not just one area, and possibly repeat the survey at a different time of year, since dandelion numbers can change with season.

QUESTION 11 3 marks Criterion C

Easy

The table shows soil moisture measured at four points in a garden.

Point	1	2	3	4
Soil moisture (%)	10	35	60	85

a. Which point has the driest soil?

[1]

b. A cactus is adapted to survive in very dry soil, while a fern needs consistently moist soil. At which point would each be most likely found, and why?

[2]

Show complete worked solution

(a)

Point 1 (10% soil moisture).

(b)

The cactus would be most likely found at Point 1 (10%, the driest), matching its drought-tolerant adaptations. The fern would be most likely found at Point 4 (85%, the wettest), matching its need for consistently moist conditions.

QUESTION 12 3 marks Criterion C

Easy

The table shows the number of buttercups counted in 5 quadrats, each of area $1\,\text{m}^2$.

Quadrat	1	2	3	4	5
Number of buttercups	4	6	5	7	3

a. Calculate the mean number of buttercups per quadrat.

[2]

b. State the population density of buttercups.

[1]

Show complete worked solution

(a)

$$ \text{mean} = \frac{4+6+5+7+3}{5} = \frac{25}{5} = 5 $$

(b)

Since each quadrat has an area of $1\,\text{m}^2$, the population density is $5$ buttercups per $\text{m}^2$.

QUESTION 13 5 marks Criterion C

Medium

The table shows grass plant counts recorded in 5 quadrats in a sunny area and 5 quadrats in a shaded area of the same field.

Area	Quadrat 1	Quadrat 2	Quadrat 3	Quadrat 4	Quadrat 5
Sunny	18	22	20	25	15
Shaded	6	4	9	5	7

a. Calculate the mean number of grass plants per quadrat in the sunny area.

[2]

b. Calculate the mean number of grass plants per quadrat in the shaded area.

[2]

c. What does this data suggest about the effect of light on grass growth?

[1]

Show complete worked solution

(a)

$$ \text{mean} = \frac{18+22+20+25+15}{5} = \frac{100}{5} = 20 $$

(b)

$$ \text{mean} = \frac{6+4+9+5+7}{5} = \frac{31}{5} = 6.2 $$

(c)

Grass grows much more densely in sunny areas (mean $20$ per quadrat) than in shaded areas (mean $6.2$ per quadrat), suggesting light availability strongly affects grass growth/density — likely because grass needs light for photosynthesis.

QUESTION 14 5 marks Criterion C

Medium

The graph shows how a population's size changes over time after it is introduced into a new habitat with limited resources.

a. Describe how the population growth rate changes from the start to the end of the graph.

[2]

b. What does the flat, level section of the graph represent?

[2]

c. State one resource that might become limiting and cause the population to level off.

[1]

Show complete worked solution

(a)

Growth starts slowly (few individuals present), then increases rapidly for a period (plentiful resources allow fast reproduction), before finally slowing down and levelling off towards the end of the graph.

(b)

The population has reached the carrying capacity ($K$) of the habitat. Resources such as food, space or water have become limiting, so births and deaths are roughly balanced, and population size stays approximately constant.

(c)

Food (other acceptable answers: space, water, or shelter).

QUESTION 15 4 marks Criterion C

Medium

The table shows the number of clover plants counted in 5 quadrats.

Quadrat	1	2	3	4	5
Number of clover plants	10	12	9	45	11

a. Identify the anomalous reading in this table.

[1]

b. Explain how you identified it.

[2]

c. Calculate the mean, excluding the anomalous value.

[1]

Show complete worked solution

(a)

Quadrat 4 (45 clover plants).

(b)

The other four readings are all close together, between $9$ and $12$. The value of $45$ is far higher than any of the others and does not fit this otherwise consistent pattern.

(c)

$$ \text{mean} = \frac{10+12+9+11}{4} = \frac{42}{4} = 10.5 $$

QUESTION 16 5 marks Criterion C

Medium

The table shows data collected along a transect from a pond edge into woodland.

Distance from pond edge (m)	0	5	10	15	20
Light intensity (lux, thousands)	2	6	12	18	20
Number of moss plants	18	11	5	2	0

a. Describe the relationship between distance from the pond edge and the number of moss plants found.

[2]

b. Suggest an explanation for this pattern, referring to moss's habitat requirements.

[2]

c. Identify the abiotic factor being investigated as the main variable along this transect.

[1]

Show complete worked solution

(a)

As distance from the pond edge increases, light intensity increases, and the number of moss plants decreases — moss is most abundant close to the pond edge ($18$ at $0\,\text{m}$) and is completely absent further away ($0$ at $20\,\text{m}$).

(b)

Moss typically needs shaded, moist conditions to grow well. Near the pond edge, light intensity is low ($2000\,\text{lux}$, likely shaded/moist conditions) and moss thrives, whereas further into the open woodland edge where light intensity is high ($20{,}000\,\text{lux}$, drier/more exposed conditions), moss struggles to survive, so numbers fall to zero.

(c)

Light intensity (measured at each point along the transect).

QUESTION 17 7 marks Criterion C

Hard

A student compared woodlice counts from 5 quadrats in two different sites: Site A (moist leaf litter near a fallen log) and Site B (dry, open ground).

Site	Quadrat 1	Quadrat 2	Quadrat 3	Quadrat 4	Quadrat 5
A (moist)	14	16	12	15	13
B (dry)	3	2	38	4	3

a. Calculate the mean woodlice count for Site A.

[2]

b. One reading in Site B looks anomalous. Identify it, and calculate the mean for Site B excluding this reading.

[2]

c. Using your calculated means, evaluate whether this data supports the conclusion that woodlice prefer moist, sheltered conditions (Site A) over dry, open conditions (Site B). Refer to the size of the difference between the means.

[3]

Show complete worked solution

(a)

$$ \text{mean} = \frac{14+16+12+15+13}{5} = \frac{70}{5} = 14 $$

(b)

The reading of $38$ (Quadrat 3) is anomalous. Excluding it: $$ \text{mean} = \frac{3+2+4+3}{4} = \frac{12}{4} = 3 $$

(c)

Yes, the data strongly supports this conclusion — the mean count in Site A ($14$) is more than four times higher than the corrected mean in Site B ($3$), a large difference unlikely to be due to chance alone. This matches the expectation that woodlice, which lose water easily through their exoskeleton, prefer the moist, sheltered, shaded conditions found in Site A's leaf litter/near the log, over the dry, exposed conditions of Site B.

QUESTION 18 5 marks Criterion D

Medium

A wetland was drained to create new farmland. The table shows data collected before and after draining.

Measurement	Before draining	After draining
Farmland area (hectares)	0	450
Wetland bird species recorded	32	9

Discuss one benefit and one drawback of draining this wetland, using the data to support your answer.

Show complete worked solution

Benefit: Draining the wetland created $450$ hectares of new farmland from previously unusable land, allowing significantly increased food production and providing farming income for the local community.

Drawback: The loss of wetland habitat caused the number of wetland bird species recorded in the area to fall drastically, from $32$ to just $9$ species. This shows that draining wetlands can severely reduce biodiversity, since many bird (and other) species depend entirely on wetland conditions — shallow water, reeds, and mud — that no longer exist once the land is drained.

QUESTION 19 6 marks Criterion D

Hard

As global temperatures rise, many species are shifting where they live. Over 30 years, the northern range boundary of a particular butterfly species has shifted $120\,\text{km}$ further north, and a mountain-dwelling species has been recorded $200\,\text{m}$ higher up mountain slopes than previously.

Discuss one way this range shift could benefit a species, and one way it could put a species at greater risk, using the data given.

Show complete worked solution

Benefit: As temperatures rise, species able to shift their range — such as moving $120\,\text{km}$ further north over 30 years, or $200\,\text{m}$ further up a mountain — can track (follow) the cooler conditions they are adapted to, allowing them to continue finding suitable habitat rather than declining or going locally extinct as their old habitat warms.

Risk: Not all species can shift range fast enough or far enough. Mountain species already found near a summit have nowhere higher left to go once they run out of mountain — a shift of $200\,\text{m}$ could use up most of the remaining altitude available. Species shifting into new areas may also end up competing with different species for the same resources, or arrive before the plants or prey they depend on have also shifted, putting them at serious risk of population decline or extinction.

QUESTION 20 6 marks Criterion D

Hard

A national park was created to protect a threatened elephant population. The table shows data recorded before protection began and 15 years later.

Measurement	Before protection	After 15 years of protection
Elephant population	800	2200
Local farmland available to residents (hectares)	5000	3200

Discuss one benefit and one drawback of creating this protected area, using the data to support your answer.

Show complete worked solution

Benefit: Protecting the area allowed the elephant population to recover strongly, nearly tripling from $800$ to $2200$ over 15 years, showing that reducing human pressures (such as hunting and habitat clearance) allowed a threatened species' population to grow substantially. Protected areas like this can also generate income for the country through ecotourism, as visitors pay to see the wildlife.

Drawback: Protecting the land reduced the farmland available to local residents from $5000$ to $3200$ hectares, a loss of $1800$ hectares. This could reduce farming income and food production for local communities, and may cause conflict if residents are not consulted or fairly compensated for the reduced access to land they previously depended on.

QUESTION 1 3 marks Criterion A

Easy

Human activity can cause pollution, which affects ecosystems in many ways.

a. What is meant by ‘pollution’?

[1]

b. Give one example each of air pollution and water pollution, and state how each harms an ecosystem.

[2]

Show complete worked solution

(a)

The introduction of harmful substances or waste products into the environment by humans, causing damage to ecosystems.

(b)

Air pollution: burning fossil fuels releases sulfur dioxide, which causes acid rain that damages trees and acidifies lakes, harming aquatic life. Water pollution: fertiliser washed into a river causes eutrophication, which can lead to fish dying from lack of oxygen.

QUESTION 2 3 marks Criterion A

Easy

Deforestation is a major cause of habitat loss worldwide.

a. What is meant by ‘deforestation’?

[1]

b. State two causes of deforestation.

[2]

Show complete worked solution

(a)

The large-scale clearing or removal of forest/trees, usually so the land can be used for another purpose.

(b)

Clearing land for agriculture (e.g. crop farming or cattle grazing) and logging for timber or paper. (Other acceptable causes: clearing land for roads/urban development, or for mining.)

QUESTION 3 4 marks Criterion A

Medium

Eutrophication is a process that can occur when fertiliser is washed into rivers and lakes. The five events below describe how it happens, but they are in the wrong order.

(i) Fish and other aquatic animals die from lack of oxygen.
(ii) Fertiliser is washed off farmland into a river.
(iii) Decomposers (bacteria) break down the dead algae, using up oxygen in the water as they respire.
(iv) The extra nutrients cause algae to grow rapidly (an ‘algal bloom’).
(v) The dense algal bloom blocks light, causing algae deeper in the water to die.

a. Put the five events into the correct order.

[3]

b. Explain why fish die as a direct result of step (iii).

[1]

Show complete worked solution

(a)

The correct order is: (ii) → (iv) → (v) → (iii) → (i).

(b)

Because decomposers use up dissolved oxygen in the water as they respire while breaking down the dead algae, oxygen levels fall so low that fish (which need dissolved oxygen to survive) suffocate.

QUESTION 4 4 marks Criterion A

Medium

Human activities are increasing the concentration of greenhouse gases in the atmosphere.

a. What is a greenhouse gas, and name two examples.

[2]

b. Explain what is meant by the ‘enhanced greenhouse effect’, and state one human activity that contributes to it.

[2]

Show complete worked solution

(a)

A greenhouse gas is a gas in the atmosphere that traps heat, by absorbing infrared radiation given off by the Earth's surface and re-radiating some of it back down, keeping the planet warmer than it would otherwise be. Examples: carbon dioxide and methane (water vapour and nitrous oxide are also acceptable).

(b)

The enhanced greenhouse effect is the additional warming caused by human activities increasing the concentration of greenhouse gases beyond natural levels, trapping more heat than before and raising global temperatures. Example activity: burning fossil fuels (coal, oil, gas) for energy and transport, which releases extra carbon dioxide.

QUESTION 5 3 marks Criterion A

Medium

Cane toads were deliberately introduced to Australia in the 1930s to control a crop pest, but they have since become a major ecological problem.

a. What is meant by an ‘invasive species’?

[1]

b. Explain how an invasive species like the cane toad can harm a native ecosystem.

[2]

Show complete worked solution

(a)

A non-native species introduced (deliberately or accidentally) to a new ecosystem, which spreads rapidly and causes ecological or economic harm.

(b)

Cane toads have no natural predators in Australia and are poisonous, so native predators that try to eat them are often killed. Cane toads also compete with native species for food and can prey on native insects and small animals, disrupting the existing food web and reducing native biodiversity.

QUESTION 6 3 marks Criterion A

Medium

Some pollutants, such as certain pesticides and heavy metals like mercury, build up in living organisms rather than being broken down or excreted.

a. Explain what is meant by ‘bioaccumulation’, using an example of a pollutant.

[2]

b. What is ‘biomagnification’?

[1]

Show complete worked solution

(a)

Bioaccumulation is the build-up of a substance (e.g. a pesticide such as DDT, or a heavy metal such as mercury) within an individual organism's body over its lifetime, because the substance is absorbed or eaten faster than the organism's body can excrete or break it down.

(b)

The increase in concentration of a pollutant at each successive trophic level in a food chain, as each predator eats many contaminated prey and accumulates their combined pollutant load.

QUESTION 7 6 marks Criterion A

Hard

A country had $80{,}000\,\text{km}^2$ of rainforest in 2000. By 2020, deforestation had reduced this to $52{,}000\,\text{km}^2$.

a. Calculate the total area of forest lost between 2000 and 2020.

[2]

b. Calculate the mean rate of deforestation, in $\text{km}^2$ per year, over this period.

[2]

c. If this mean rate continued, calculate how many more years it would take for all the remaining forest (as of 2020) to be lost.

[2]

Show complete worked solution

(a)

$$ 80{,}000 - 52{,}000 = 28{,}000\,\text{km}^2 $$

(b)

$$ \frac{28{,}000}{20} = 1400\,\text{km}^2\text{/year} $$

(c)

$$ \frac{52{,}000}{1400} \approx 37.1 $$ Answer: approximately $37$ more years.

QUESTION 8 6 marks Criterion B

Medium

A student wants to investigate the effect of fertiliser concentration on the growth of algae in pond water, to model the process of eutrophication.

a. State the independent and dependent variables.

[2]

b. State two variables that should be controlled, and explain why for one of them.

[2]

c. Describe a suitable method for this investigation.

[2]

Show complete worked solution

(a)

Independent variable: fertiliser concentration added to the pond water (e.g. $0, 1, 2, 4\,\text{g/L}$). Dependent variable: algae growth, e.g. measured as % light transmission through the water, or algae mass/coverage after a fixed time.

(b)

Control the volume of pond water used in each container, and the light and temperature conditions all containers are kept under. Why control light: algae need light to photosynthesise and grow — if some containers received more light than others, this alone could cause differences in growth unrelated to the fertiliser concentration, making the comparison unfair.

(c)

Set up several containers with the same volume and starting amount of pond water/algae. Add a different concentration of fertiliser solution to each container, and place all containers under identical light and temperature conditions. After a fixed time (e.g. 7 days), measure algae growth in each container — for example using a light meter to measure how much light passes through the water, since more algae blocks more light. Repeat each concentration 3 times and calculate a mean.

QUESTION 9 6 marks Criterion B

Medium

Lichens are very sensitive to air pollution (particularly sulfur dioxide) and are often used as ‘bioindicators’ of air quality. A student wants to investigate how air quality changes with distance from a busy road, using lichens.

a. State the independent and dependent variables.

[2]

b. State two pieces of equipment needed, and what each is used for.

[2]

c. Describe a suitable method for this investigation.

[2]

Show complete worked solution

(a)

Independent variable: distance of the sampling site from the road. Dependent variable: the number of different lichen species present (or the percentage of the tree trunk covered by lichen).

(b)

A quadrat, placed against the tree trunk to mark a fixed sampling area. A lichen identification key/chart, used to identify and count the different lichen species present within the quadrat. A tape measure could also be used to accurately measure distance from the road at each site.

(c)

At several sites at increasing distance from the road, select trees of a similar type and age, and place a quadrat on the trunk at a fixed height (e.g. $1.5\,\text{m}$). Record the number of different lichen species (or percentage cover) present, using the identification key. Repeat on several trees at each site and calculate a mean, since fewer and different lichen species are expected to grow where air quality is poor.

QUESTION 10 7 marks Criterion B

Hard

To investigate the effect of oil pollution on plant growth, a student grew one bean plant in soil with no oil added, and one bean plant in soil with oil added, then compared their heights after 2 weeks.

a. Identify a flaw in this method and explain why it makes the conclusion unreliable.

[2]

b. Suggest an improvement to this flaw, and explain how it helps.

[2]

c. State two further variables that should be controlled in this investigation, and explain why controlling one of them matters.

[3]

Show complete worked solution

(a)

Only one plant was used per condition. Any difference in height between the two plants could simply be due to natural variation between individual plants, rather than being caused by the oil — without repeats, it is impossible to tell whether the result is a real effect or just chance/individual variation.

(b)

Grow several plants (e.g. 10) in each condition, and compare the mean height for each group instead of a single plant. Using a larger sample and taking a mean reduces the effect of individual variation, making it much clearer whether any height difference is really caused by the oil.

(c)

Control the amount of water given to each plant, the light each plant receives, and the type/amount of soil and starting size of each plant. Why control watering: if the oil-treated plants also happened to receive less water than the untreated plants, any reduced growth observed could be caused by the lack of water rather than by the oil itself, making it impossible to isolate the oil as the true cause of any difference.

QUESTION 11 3 marks Criterion C

Easy

The table shows measured atmospheric carbon dioxide concentration over several decades.

Year	1960	1980	2000	2020
CO₂ concentration (ppm)	317	339	369	414

a. What was the atmospheric CO₂ concentration in 2000?

[1]

b. Calculate the increase in CO₂ concentration between 1960 and 2020.

[2]

Show complete worked solution

(a)

$369\,\text{ppm}$.

(b)

$$ 414 - 317 = 97\,\text{ppm} $$

QUESTION 12 3 marks Criterion C

Easy

The table shows the area of forest lost to deforestation each year in a region.

Year	2015	2016	2017	2018	2019
Forest lost (km²)	120	150	180	140	160

a. Calculate the total area of forest lost over these 5 years.

[2]

b. In which year was the most forest lost?

[1]

Show complete worked solution

(a)

$$ 120+150+180+140+160 = 750\,\text{km}^2 $$

(b)

2017 ($180\,\text{km}^2$).

QUESTION 13 5 marks Criterion C

Medium

The graph shows the global temperature anomaly (the difference from the 1960 average temperature) recorded between 1960 and 2020.

a. Describe the trend shown in the graph.

[2]

b. Calculate the mean rate of temperature increase, in $^\circ\text{C}$ per decade, over the full 60-year period shown.

[3]

Show complete worked solution

(a)

Global temperature anomaly has risen fairly steadily since 1960, from $0.0^\circ\text{C}$ to about $0.9^\circ\text{C}$ above the 1960 baseline by 2020, with the rate of increase appearing to speed up slightly in more recent decades.

(b)

$$ \text{total change} = 0.9 - 0.0 = 0.9^\circ\text{C over } 60 \text{ years } (6 \text{ decades}) $$ $$ \text{rate} = \frac{0.9}{6} = 0.15^\circ\text{C per decade} $$

QUESTION 14 4 marks Criterion C

Medium

The table shows mercury concentration measured at each trophic level of a marine food chain.

Trophic level	Phytoplankton	Zooplankton	Small fish	Tuna (top predator)
Mercury concentration (ppm)	0.002	0.02	0.2	2.0

a. Calculate how many times more concentrated the mercury is in tuna compared with phytoplankton.

[2]

b. Describe the pattern shown in this data and name the process responsible.

[2]

Show complete worked solution

(a)

$$ \frac{2.0}{0.002} = 1000 $$ Answer: $1000$ times more concentrated.

(b)

The mercury concentration increases roughly ten-fold at each successive trophic level (from $0.002 \rightarrow 0.02 \rightarrow 0.2 \rightarrow 2.0\,\text{ppm}$). This process is called biomagnification (bioaccumulation increasing up the food chain).

QUESTION 15 4 marks Criterion C

Medium

The table shows cod catch data recorded over 20 years.

Year	2000	2005	2010	2015	2020
Cod catch (thousand tonnes)	300	250	200	310	100

a. Identify the anomalous value in this table.

[1]

b. Explain how you identified it.

[1]

c. Excluding the anomalous value, calculate the overall percentage decline in cod catch from 2000 to 2020.

[2]

Show complete worked solution

(a)

2015 ($310$ thousand tonnes).

(b)

The catch decreases fairly steadily each period ($300 \rightarrow 250 \rightarrow 200$), except in 2015, where it rises sharply to $310$ before falling again to $100$ in 2020 — breaking the otherwise consistent declining trend.

(c)

$$ \text{decrease} = 300 - 100 = 200 $$ $$ \text{\% decline} = \frac{200}{300} \times 100 \approx 66.7\% $$

QUESTION 16 4 marks Criterion C

Medium

The table shows dissolved oxygen concentration measured at increasing distances from a sewage outflow pipe.

Distance from outflow (m)	0	100	200	400	800
Dissolved oxygen (mg/L)	1.2	3.5	6.0	8.2	8.8

a. Describe the pattern shown between distance from the sewage outflow and dissolved oxygen concentration.

[2]

b. Explain why dissolved oxygen is so low close to the sewage outflow.

[2]

Show complete worked solution

(a)

Dissolved oxygen is very low close to the outflow ($1.2\,\text{mg/L}$ at $0\,\text{m}$) and increases with distance, reaching $8.8\,\text{mg/L}$ by $800\,\text{m}$ — oxygen levels recover the further away the water is from the pollution source.

(b)

Sewage adds excess nutrients to the water, causing rapid growth of algae and bacteria. Decomposers breaking down this organic waste and any dead algae use up dissolved oxygen through respiration faster than it can be replaced, sharply lowering oxygen levels close to the outflow (eutrophication).

QUESTION 17 7 marks Criterion C

Hard

A student investigated the effect of fertiliser concentration on algae growth, repeating each concentration in 3 separate containers. The table shows the increase in algae mass after 7 days.

Fertiliser conc. (g/L)	Trial 1 (g)	Trial 2 (g)	Trial 3 (g)
0	0.5	0.6	0.4
1	1.8	2.0	1.9
2	3.5	3.6	9.0
4	4.9	5.1	5.0

a. Calculate the mean algae mass increase for the $2\,\text{g/L}$ fertiliser concentration, excluding any anomalous reading.

[2]

b. Identify the anomalous reading and suggest one possible cause for it.

[2]

c. Using the corrected means, evaluate whether the data supports the conclusion that increasing fertiliser concentration increases algae growth. Refer to values in your answer.

[3]

Show complete worked solution

(a)

The $9.0\,\text{g}$ reading is anomalous and should be excluded: $$ \text{mean} = \frac{3.5+3.6}{2} = 3.55\,\text{g} $$

(b)

The $9.0\,\text{g}$ reading at $2\,\text{g/L}$ is anomalous. Possible cause: contamination (extra algae or nutrients accidentally introduced), a measurement error, or that container receiving extra light or warmth by mistake.

(c)

The corrected means show a clear, consistent increasing trend: $0.5\,\text{g}$ at $0\,\text{g/L}$, $1.9\,\text{g}$ at $1\,\text{g/L}$, $3.55\,\text{g}$ at $2\,\text{g/L}$, and $5.0\,\text{g}$ at $4\,\text{g/L}$. Algae growth increases every time fertiliser concentration increases, and the differences between means are much larger than the small spread within each set of 3 trials, so the data strongly supports the conclusion that higher fertiliser concentration causes greater algae growth — consistent with real eutrophication, where nutrient-rich runoff causes algal blooms.

QUESTION 18 5 marks Criterion D

Medium

The table shows global plastic production and an estimate of plastic entering the oceans each year.

Year	Global plastic production (million tonnes/year)
1970	35
2020	370

An estimated $8$ million tonnes of plastic enters the world's oceans every year.

Discuss one benefit and one drawback of plastic use, using the data to support your discussion.

Show complete worked solution

Benefit: Plastic is cheap, lightweight, durable and versatile, which is why production has grown enormously, from $35$ million tonnes in 1970 to $370$ million tonnes in 2020. It is used in countless important applications, from food packaging that reduces spoilage and waste to sterile, single-use medical equipment, improving daily life and health worldwide.

Drawback: Plastic's durability means it does not easily break down in the environment; an estimated $8$ million tonnes enters the oceans every year, where it can entangle or be ingested by marine animals such as turtles and seabirds, and gradually breaks into microplastics that enter food chains — with effects on marine ecosystems, and potentially human health, that are still being studied.

QUESTION 19 6 marks Criterion D

Hard

A country is deciding whether to replace some of its coal power stations with wind farms. The table shows the estimated carbon dioxide emitted per unit of electricity generated by different energy sources.

Energy source	CO₂ emitted (g per kWh)
Coal	820
Natural gas	490
Solar	41
Wind	11

Discuss one benefit and one drawback of this change, using the data to support your answer.

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Benefit: Switching from coal ($820\,\text{g CO}_2\text{/kWh}$) to wind ($11\,\text{g CO}_2\text{/kWh}$) would cut carbon dioxide emissions per unit of electricity by about $75$-fold ($820 \div 11 \approx 74.5$), significantly reducing the country's contribution to the enhanced greenhouse effect and climate change, and reducing the air pollution associated with burning coal.

Drawback: Wind power is intermittent — it only generates electricity when the wind is blowing, so on its own it cannot supply a constant, reliable electricity supply the way a coal power station can. Building enough wind farms (plus back-up power or storage) to reliably replace coal requires large upfront investment and significant land or sea space, and some communities may object to wind farms being built near them.

QUESTION 20 6 marks Criterion D

Hard

A farming region introduced regular pesticide use on its crops. The table shows data recorded before and after this change.

Measurement	Before pesticide use	After regular pesticide use
Crop yield (tonnes/hectare)	2.5	6.0
Wild bee species recorded in the area	18	6

Discuss one benefit and one drawback of this pesticide use, using the data to support your answer.

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Benefit: Regular pesticide use more than doubled crop yield, from $2.5$ to $6.0$ tonnes per hectare, by killing insect pests that would otherwise damage or eat the crop. This allows more food to be produced from the same area of farmland, which is important for feeding a growing global population.

Drawback: Pesticides do not just kill pest insects — the number of wild bee species recorded in the surrounding area fell sharply, from $18$ to just $6$, since many pesticides are also toxic to pollinators. Bees are essential for pollinating many food crops and wild plants, so this decline threatens both biodiversity and, ironically, future crop yields that depend on pollination — showing the wider ecological cost of pesticide use beyond its intended target.

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