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GCSE · Biology

B7 Ecology

AQA 8461/8462/8463 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Ecologists use several key terms to describe how organisms interact with each other and their surroundings.

(a)Define the term 'population'.(1)

(b)Define the term 'community'.(1)

(c)Define the term 'habitat'.(1)

(d)Define the term 'ecosystem'.(1)

(Total for Question 1 is 4 marks)

Living things can be organised into different levels within an ecosystem.

(a)Complete the sequence of levels of organisation, from smallest to largest: organism, ____, community, ecosystem.(1)

(b)Explain the difference between a population and a community.(2)

(Total for Question 2 is 3 marks)

A student measured the light intensity and counted the number of bluebell plants growing at five points along a line from the edge of a wood to the middle of the wood. Her results are shown below.
Point 1 (wood edge): light intensity 82 units, 3 bluebells
Point 2: light intensity 61 units, 9 bluebells
Point 3: light intensity 40 units, 18 bluebells
Point 4: light intensity 22 units, 27 bluebells
Point 5 (wood middle): light intensity 9 units, 34 bluebells

(a)Identify the abiotic factor that the student investigated.(1)

(b)Describe the relationship shown by the student's results.(2)

(c)Suggest a reason for the relationship you described in part (b).(2)

(Total for Question 3 is 5 marks)

A gamekeeper recorded the estimated numbers of foxes and rabbits on an estate over six years.
Year 1: rabbits 420, foxes 18
Year 2: rabbits 610, foxes 22
Year 3: rabbits 780, foxes 35
Year 4: rabbits 540, foxes 48
Year 5: rabbits 310, foxes 30
Year 6: rabbits 260, foxes 19

(a)Describe the pattern shown by the fox population compared with the rabbit population over the six years.(2)

(b)Explain why the change in the fox population lags behind the change in the rabbit population.(3)

(Total for Question 4 is 5 marks)

Required practical: a student used a 0.25 m² quadrat to sample the number of daisy plants growing in a field with a total area of 400 m^2. She placed the quadrat 8 times and recorded these counts of daisies: 3, 5, 4, 6, 2, 5, 7, 4.

(a)Calculate the mean number of daisies per quadrat.(2)

(b)Use your answer to part (a) to calculate an estimate of the total number of daisies in the whole field.(3)

(c)Describe how the student should position her quadrats to make sure her sampling is not biased.(2)

(Total for Question 5 is 7 marks)

Required practical: a student wants to estimate the population size of snails in a garden using the mark-release-recapture method. She catches 40 snails, marks each one with a small dot of non-toxic paint, then releases them back into the garden. The next day, she catches 50 snails, and finds that 8 of them are marked.

(a)State the formula used to estimate population size from mark-release-recapture data.(1)

(b)Calculate the estimated population size of snails in the garden.(3)

(c)State one assumption made when using the mark-release-recapture method.(1)

(Total for Question 6 is 5 marks)

A student ran a belt transect from the edge of a footpath into an open field to see how moss cover changed with distance. She recorded the percentage cover of moss in a quadrat every 2 metres.
Distance from path: 0 m, 2 m, 4 m, 6 m, 8 m, 10 m
Percentage cover of moss: 80%, 60%, 40%, 20%, 10%, 5%

(a)Describe the trend shown by the transect data.(2)

(b)Suggest an abiotic factor that could explain this trend, and explain your reasoning.(3)

(Total for Question 7 is 5 marks)

A food web in a hedgerow contains the following feeding relationships: grass is eaten by both rabbits and voles; rabbits are eaten by foxes and owls; voles are eaten by owls only.

(a)Identify the producer in this food web.(1)

(b)Name one secondary consumer in this food web.(1)

(c)Construct one complete food chain from this food web that includes the owl.(2)

(d)A disease sharply reduces the rabbit population. Predict the effect on the owl population, giving a reason.(2)

(Total for Question 8 is 6 marks)

A food chain in a pond has the following biomass values: producers (algae) 500 kg, primary consumers (water fleas) 60 kg, secondary consumers (small fish) 8 kg, tertiary consumers (herons) 1 kg.

(a)Describe the shape of the pyramid of biomass drawn for this food chain.(1)

(b)Explain why the biomass decreases at each successive trophic level in a food chain.(3)

(Total for Question 9 is 4 marks)

Use the biomass data from question 9 (producers 500 kg, primary consumers 60 kg, secondary consumers 8 kg) to answer the following.

(a)Calculate the percentage of biomass transferred from the producers to the primary consumers.(2)

(b)Calculate the percentage of biomass transferred from the primary consumers to the secondary consumers. Give your answer to 1 decimal place.(2)

(c)Higher Tier only. Suggest one reason why the percentage biomass transfer between the primary and secondary consumers is not the same as the percentage transfer between the producers and the primary consumers.(2)

(Total for Question 10 is 6 marks)

Carbon is constantly cycled between living organisms, the atmosphere and the environment. Describe how carbon dioxide is removed from the atmosphere and returned to it in the carbon cycle. Your answer should refer to photosynthesis, respiration and combustion.

(Total for Question 11 is 4 marks)

Required practical: a group of students investigated the effect of temperature on the rate of decay of fresh milk. They mixed milk with a small amount of lipase enzyme and an indicator that changes colour as the pH falls, then timed how long the colour change took at different water-bath temperatures.

(a)State the independent variable in this investigation.(1)

(b)State one variable that the students should control to make this a fair test.(1)

(c)Describe how the students could use this method to compare the rate of decay at different temperatures.(3)

(Total for Question 12 is 5 marks)

The table shows the results of the milk decay investigation from question 12: the time taken for the indicator to change colour at each water-bath temperature.
10 C: 320 seconds
20 C: 180 seconds
30 C: 95 seconds
40 C: 60 seconds
50 C: 140 seconds
60 C: no colour change seen within 10 minutes

(a)Describe the pattern shown by these results.(3)

(b)Explain these results in terms of enzyme activity. Refer to the enzyme's active site in your answer.(4)

(Total for Question 13 is 7 marks)

Lichens are used as indicator species for air pollution (sulfur dioxide). A survey recorded which type of lichen was found at increasing distances from a factory.
0 m from factory: no lichen present
100 m from factory: only crusty lichen present
500 m from factory: crusty and leafy lichen present
1000 m from factory: crusty, leafy and bushy lichen present (bushy lichen is the most sensitive to pollution)

(a)State what is meant by an 'indicator species'.(1)

(b)Describe how the range of lichen species changes with distance from the factory.(2)

(c)Explain what this data suggests about air quality near the factory.(2)

(Total for Question 14 is 5 marks)

The table shows the population of a moth species and the average winter temperature over ten years in the same woodland.
Year 1: winter temp 2 C, moth population 1200
Year 3: winter temp 4 C, moth population 1650
Year 5: winter temp 1 C, moth population 950
Year 7: winter temp 6 C, moth population 2100
Year 9: winter temp 3 C, moth population 1400

(a)Describe the trend shown between winter temperature and the moth population.(2)

(b)Suggest a reason, linked to this abiotic factor, for the change in the moth population.(2)

(Total for Question 15 is 4 marks)

The human population, and the standard of living for many people, has rapidly increased over the last 200 years. Explain two ways in which this rapid increase in the human population is reducing biodiversity.

(Total for Question 16 is 4 marks)

A large area of tropical rainforest is cleared each year to plant palm oil trees.

(a)Give one reason why rainforest land is cleared for palm oil plantations.(1)

(b)Explain two negative effects of large-scale deforestation like this on the environment.(4)

(Total for Question 17 is 5 marks)

A family's car releases 180 g of carbon dioxide for every kilometre it is driven. The family drives the car a total of 12000 km in one year.

(a)Calculate the total mass of carbon dioxide, in kilograms, released by the car in one year.(3)

(b)Carbon dioxide and methane are greenhouse gases. Explain how increasing levels of these gases in the atmosphere lead to global warming.(3)

(Total for Question 18 is 6 marks)

Human activity is reducing biodiversity around the world. Evaluate the methods that can be used to maintain biodiversity, using named examples to support your answer.

(Total for Question 19 is 6 marks)

Quick-fire recap questions on key ecology ideas.

(a)Which of these is the correct definition of a population?(1)

A) All the different species living in a habitat
B) All the organisms of one species living in the same area at the same time
C) All the living and non-living components of an area
D) A place where an organism lives

(b)Which of the following is an abiotic factor?(1)

A) Predation
B) Competition for food
C) Availability of light
D) Disease

(c)What term describes an organism that makes its own food using light energy?(1)

A) Consumer
B) Producer
C) Decomposer
D) Predator

(d)Which process in the carbon cycle releases carbon dioxide into the atmosphere?(1)

A) Photosynthesis
B) Respiration
C) Absorption
D) Transpiration

(e)Biodiversity is best described as...(1)

A) The total number of individual organisms in a habitat
B) The range of different species living in an area, or on Earth
C) The total biomass of producers in an ecosystem
D) The number of trophic levels in a food chain

(Total for Question 20 is 5 marks)

Mark scheme · B7 Ecology

Question 1

(a) B1 all the organisms of one species living in the same area at the same time (oe)
(a) Answer: All the organisms of one species living in the same area at the same time.
(b) B1 all the populations of different species living together in the same area (oe)
(b) Answer: All the populations of different species living together in the same area.
(c) B1 the place where an organism lives (oe)
(c) Answer: The place where an organism lives.
(d) B1 the interaction of a community of living organisms (biotic) with the non-living (abiotic) parts of their environment (oe)
(d) Answer: The interaction of a community of living organisms with the non-living parts of their environment.

Question 2

(a) B1 population
(a) Answer: Population.
(b) B1 a population is all the organisms of one species in an area (oe)
(b) B1 a community is all the populations of different species in an area (oe)
(b) Answer: A population is all the organisms of a single species in an area, whereas a community is all the different populations (of different species) living together in that area.

Question 3

(a) B1 light intensity
(a) Answer: Light intensity.
(b) B1 as light intensity decreases (from the wood edge to the middle), the number of bluebells increases (oe)
(b) B1 correct use of at least two data points/values to support the trend (e.g. 82 units/3 bluebells compared with 9 units/34 bluebells)
(b) Answer: As light intensity decreases from the wood edge to the middle of the wood, the number of bluebell plants increases.
(c) B1 bluebells flower and grow in early spring, before the tree canopy is in full leaf, so they can photosynthesise in the brighter conditions near the wood edge, but face little competition in the shadier middle of the wood (oe)
(c) B1 in the middle of the wood, taller trees/other plants may out-compete other species for light, so bluebells (which are adapted to lower light) face less competition there than at the brighter, more crowded wood edge (oe)
(c) Answer: Bluebells are adapted to grow well in shadier conditions and face less competition from other plants (e.g. grasses) in the middle of the wood, whereas the brighter wood edge favours other species that out-compete bluebells there.

Question 4

(a) B1 the fox population rises and falls in the same general pattern as the rabbit population (oe)
(a) B1 the change in fox numbers lags behind (happens slightly after) the change in rabbit numbers, e.g. rabbits peak in year 3 but foxes peak in year 4
(a) Answer: The fox population follows the same rise-and-fall pattern as the rabbit population, but the change in fox numbers lags about a year behind the change in rabbit numbers.
(b) B1 foxes are predators of rabbits (their prey), so when rabbit numbers rise there is more food available for foxes (oe)
(b) B1 this greater food supply means more foxes survive and reproduce successfully, so the fox population increases, but only after the rabbit population has already increased
(b) B1 as the fox population rises, predation increases and reduces the rabbit population; the falling rabbit numbers then leave less food for foxes, so the fox population later falls too (oe)
(b) Answer: Foxes rely on rabbits for food, so a rise in rabbit numbers only leads to a rise in fox numbers once the extra food has allowed more foxes to survive and breed; the resulting increase in predation then reduces rabbit numbers, causing fox numbers to fall shortly afterwards, so predator numbers always change slightly after prey numbers.

Question 5

(a) M1 total = 36 (sum of the 8 counts), divided by 8
(a) A1 4.5 (daisies per quadrat) cao
(a) Answer: 4.5 daisies per quadrat.
(b) M1 density = mean per quadrat / quadrat area, e.g. 4.5 / 0.25 (= 18 daisies per m²)
(b) M1 18 x 400 (total field area)
(b) A1 7200 daisies, allow ecf from part (a)
(b) Answer: Approximately 7200 daisies.
(c) B1 lay out two tape measures at right angles (or mark out a grid) across the field to create a set of coordinates
(c) B1 use a random number generator/table to pick coordinates and place the quadrat at each randomly generated point, repeating for all 8 quadrats (oe: prevents the student choosing areas that look like they have more/fewer daisies)
(c) Answer: Lay two tape measures at right angles across the field to form a grid of coordinates, then use a random number generator to pick coordinate pairs and place the quadrat at each randomly chosen point, so the sampling positions are not chosen by the student and are not biased.

Question 6

(a) B1 estimated population size = (number caught and marked in first sample x total number caught in second sample) / number of marked individuals recaptured in second sample (oe, accept in symbols)
(a) Answer: Estimated population size = (number marked in the first sample x total caught in the second sample) / number of marked individuals recaptured in the second sample.
(b) M1 correct substitution: (40 x 50) / 8
(b) M1 40 x 50 = 2000
(b) A1 250 snails, allow ecf
(b) Answer: 250 snails.
(c) B1 any one of: no immigration, emigration, births or deaths occur between the two samples; the marks do not fade, wash off or affect survival; marked individuals redistribute/mix randomly and fully with the rest of the population before the second sample is taken
(c) Answer: The population does not change between the two samples (no immigration, emigration, births or deaths), and the marked snails mix back in randomly with the rest of the population.

Question 7

(a) B1 percentage cover of moss decreases with increasing distance from the footpath (oe)
(a) B1 correct use of at least two data values to support the trend, e.g. 80% at 0 m falling to 5% at 10 m
(a) Answer: The percentage cover of moss decreases steadily as the distance from the footpath increases, from 80% at the path edge to just 5% at 10 m.
(b) B1 moisture/water availability (e.g. the soil near the footpath may be more compacted and retain more surface water, or be shadier and damper) (oe, allow light intensity with suitable linked reasoning)
(b) B1 moss needs a constantly moist surface to grow well/to reproduce, so areas with more moisture support more moss growth
(b) B1 further from the path the ground may be drier/more exposed, so conditions are less suitable for moss and cover decreases
(b) Answer: Moisture is likely to explain the trend: the compacted, shadier ground near the footpath probably stays damper, and moss needs a constantly moist surface to grow and reproduce, so cover is high there; further into the field the ground is likely drier and less suitable for moss, so cover falls.

Question 8

(a) B1 grass
(a) Answer: Grass.
(b) B1 fox or owl (either accepted)
(b) Answer: The fox (or the owl).
(c) B1 correct three organisms in the chain, e.g. grass -> vole -> owl or grass -> rabbit -> owl
(c) B1 arrows correctly drawn/described in the direction of energy transfer (from grass to consumer to owl)
(c) Answer: grass -> vole -> owl (or grass -> rabbit -> owl).
(d) B1 the owl population is unlikely to fall very much / may stay fairly stable at first, because owls can eat voles instead of rabbits (oe)
(d) B1 owls may increase predation on voles to compensate for the loss of rabbits as a food source, which could then reduce the vole population (oe, allow any logically linked consequence)
(d) Answer: The owl population is unlikely to collapse immediately, because owls can switch to eating more voles instead of rabbits; however, this increased predation pressure could then cause the vole population to fall too.

Question 9

(a) B1 a true pyramid shape: each bar is narrower/shorter than the bar below it, getting smaller from the producers up to the herons
(a) Answer: A true pyramid shape, narrowing at each level from the wide producer bar at the bottom to the narrow heron bar at the top.
(b) B1 not all the biomass/material of an organism is eaten by the next organism in the chain (e.g. bones, roots), or is not digested and is egested as faeces (oe)
(b) B1 some absorbed biomass/energy is used in respiration, which releases energy that is eventually lost to the surroundings/environment as heat
(b) B1 some biomass/energy is used for movement and other life processes rather than being converted into new growth, so less biomass is available to pass on to the next trophic level
(b) Answer: Not everything is eaten or fully digested, and much of what is absorbed is used for respiration (releasing energy lost as heat) or for movement and other life processes rather than growth, so only a small proportion of biomass is passed on to the next trophic level.

Question 10

(a) M1 (60 / 500) x 100
(a) A1 12% cao
(a) Answer: 12%.
(b) M1 (8 / 60) x 100
(b) A1 13.3% awrt
(b) Answer: 13.3%.
(c) B1 different types of organism lose different proportions of biomass/energy, e.g. active animals (such as fish) use more energy for movement and to maintain body temperature than water fleas do, so less biomass is available to pass on (oe)
(c) B1 or: the proportion of inedible/indigestible material differs between trophic levels (e.g. algae may be more fully consumed/digested by water fleas than water fleas are by fish), so the transfer efficiency is not fixed at every stage of a food chain
(c) Answer: Transfer efficiency is not fixed because different organisms lose different amounts of biomass to processes such as movement, maintaining body temperature and egestion; for example, more active secondary consumers may use more energy for movement than primary consumers do, so a different percentage of biomass is transferred at each stage.

Question 11

B1 carbon dioxide is removed from the atmosphere by green plants and algae during photosynthesis, which converts it into carbon compounds/biomass (oe)
B1 carbon is passed along food chains as organisms eat other organisms/plants
B1 carbon dioxide is returned to the atmosphere through respiration, carried out by plants and animals (and other organisms)
B1 carbon dioxide is also returned to the atmosphere by combustion (burning) of wood or fossil fuels formed from the remains of dead organisms (oe)
Answer: Carbon dioxide is taken from the atmosphere by photosynthesis in plants and algae, which converts it into carbon compounds passed along food chains; it is returned to the atmosphere through the respiration of living organisms and through the combustion of wood and fossil fuels.

Question 12

(a) B1 temperature (of the water bath/milk mixture)
(a) Answer: Temperature.
(b) B1 any one of: volume/concentration of milk used, volume/concentration of lipase used, volume/amount of indicator used, type/make of indicator, starting pH
(b) Answer: The volume and concentration of lipase used (or an equivalent controlled variable such as the volume of milk or indicator).
(c) B1 set up several identical test tubes of milk, lipase and indicator, and place each in a water bath at a different temperature (e.g. 10 C, 20 C, 30 C, 40 C, 50 C)
(c) B1 start a timer as soon as the enzyme is added/mixed in, and record the time taken for the indicator to change colour at each temperature
(c) B1 repeat each temperature (and calculate a mean time) to improve reliability; a shorter time indicates a faster rate of decay/reaction
(c) Answer: Set up identical tubes of milk, lipase and indicator, place each at a different water-bath temperature, time how long the colour change takes at each temperature, and repeat to find a mean time; a shorter time shows a faster rate of decay.

Question 13

(a) B1 the time taken decreases (the rate of decay increases) as temperature rises from 10 C to 40 C
(a) B1 the time taken then increases again at 50 C (the rate of decay decreases)
(a) B1 at 60 C no colour change/reaction occurs at all within the time given
(a) Answer: The time taken falls steadily from 10 C to 40 C (the reaction speeds up), then rises again at 50 C (the reaction slows down), and at 60 C no colour change happens at all.
(b) B1 as temperature increases towards 40 C, particles/molecules have more kinetic energy, so the enzyme and substrate collide more frequently and successfully
(b) B1 this increases the rate of the enzyme-catalysed reaction, so less time is needed for the colour change, up to an optimum temperature (around 40 C here)
(b) B1 above the optimum temperature, the enzyme begins to denature: the shape of its active site changes/is destroyed by the heat
(b) B1 the substrate can no longer fit/bind to the active site (as well), so the enzyme-substrate complex cannot form and the reaction slows down or, once fully denatured (as at 60 C), stops altogether
(b) Answer: Up to about 40 C, rising temperature gives particles more kinetic energy, so the enzyme and its substrate collide more often and successfully, speeding up the reaction; above this optimum temperature the enzyme starts to denature, its active site changes shape, and the substrate can no longer fit it, so the reaction slows down and eventually stops completely once the enzyme is fully denatured, as seen at 60 C.

Question 14

(a) B1 a species whose presence, absence or abundance shows the level of pollution (or another environmental condition) in an area (oe)
(a) Answer: A species whose presence or absence indicates the level of pollution (or another environmental condition) in a habitat.
(b) B1 the number/range of lichen species present increases with distance from the factory
(b) B1 correct reference to the data, e.g. no lichen at all at 0 m, but all three types (including the most sensitive, bushy lichen) present by 1000 m
(b) Answer: The range of lichen species increases with distance from the factory: no lichen grows right next to it, but by 1000 m away all three types, including the most pollution-sensitive bushy lichen, are present.
(c) B1 air quality is likely to be poor/more polluted (with sulfur dioxide) close to the factory, and improves further away
(c) B1 because only the most pollution-tolerant lichen (crusty) can survive close to the factory, whereas the most pollution-sensitive lichen (bushy) can only survive where sulfur dioxide levels are low, i.e. far from the factory
(c) Answer: Air quality is likely to be worse (more sulfur dioxide) close to the factory, since only the most pollution-tolerant lichen can survive there, while the most sensitive species only grows further away, where pollution levels are lower.

Question 15

(a) B1 as average winter temperature increases, the moth population increases (a positive correlation) (oe)
(a) B1 correct use of data, e.g. the lowest temperature (1 C, year 5) matches the lowest population (950), and the highest temperature (6 C, year 7) matches the highest population (2100)
(a) Answer: The moth population rises as the average winter temperature rises: the coldest winter (1 C) has the smallest population (950) and the mildest winter (6 C) has the largest population (2100).
(b) B1 milder winters mean more moths (or their eggs/pupae) survive the winter, e.g. fewer die from cold stress or from a lack of energy reserves (oe)
(b) B1 more survivors in spring means more moths are available to reproduce, increasing the population further (oe, allow reference to food availability)
(b) Answer: Milder winters allow more moths, eggs or pupae to survive the cold, so more individuals are left to reproduce in spring, leading to a larger population overall.

Question 16

B1 more land is used for housing, farming or quarrying, which destroys/reduces natural habitats (named point 1)
B1 this means fewer species have suitable places to live, so biodiversity falls (linked explanation for point 1)
B1 more waste (e.g. household or industrial waste) is produced and sent to landfill, or more pollutants are released into water, land or air (named point 2)
B1 this pollution/waste can poison, harm or destroy the habitats of organisms, reducing biodiversity (linked explanation for point 2)
Answer: More land is being used for housing, farming and quarrying, destroying natural habitats and leaving fewer species with somewhere to live; and more waste and pollution are being produced, which can harm organisms and damage habitats, both of which reduce biodiversity.

Question 17

(a) B1 to grow palm oil for use in food, cosmetics or biofuel (or to make money/for the economic benefit of the crop) (oe)
(a) Answer: Palm oil is a valuable crop used in food, cosmetics and biofuel, so clearing the land to grow it generates income.
(b) B1 loss of habitat for the species that lived in the rainforest (named effect 1)
(b) B1 this reduces biodiversity, as species may become endangered or extinct if they cannot survive elsewhere (linked explanation)
(b) B1 less carbon dioxide is removed from the atmosphere by photosynthesis (fewer trees), and burning/decaying cleared trees releases stored carbon dioxide (named effect 2)
(b) B1 this contributes to an increase in atmospheric carbon dioxide levels, adding to the greenhouse effect/global warming (linked explanation)
(b) Answer: Deforestation destroys the habitat of rainforest species, reducing biodiversity as some species cannot survive elsewhere; it also removes trees that would otherwise take in carbon dioxide by photosynthesis, and burning or decaying the cleared trees releases stored carbon dioxide, adding to the greenhouse effect.

Question 18

(a) M1 180 x 12000 (= 2 160 000)
(a) M1 correct conversion from grams to kilograms (divide by 1000)
(a) A1 2160 kg cao
(a) Answer: 2160 kg of carbon dioxide.
(b) B1 short-wavelength radiation from the Sun passes through the atmosphere and reaches/warms the Earth's surface
(b) B1 the Earth re-radiates this energy as longer-wavelength (infrared) radiation
(b) B1 greenhouse gases absorb this infrared radiation and re-radiate some of it back towards the Earth, trapping heat in the atmosphere; higher levels of these gases trap more heat, increasing the Earth's average temperature
(b) Answer: Short-wavelength radiation from the Sun passes through the atmosphere and warms the Earth's surface, which re-radiates the energy as longer-wavelength infrared radiation; greenhouse gases absorb this infrared radiation and radiate some of it back towards the Earth, trapping heat, so higher concentrations of these gases trap more heat and raise the Earth's average temperature.

Question 19

Level 1 (1-2): Simple statements are made about ways to maintain biodiversity, with limited or no named examples. The answer lacks a logical structure.
Level 2 (3-4): Relevant methods of maintaining biodiversity are described, supported by at least one named example. There is some evaluation of how effective the method(s) might be. The answer shows some structure.
Level 3 (5-6): A range of methods of maintaining biodiversity are clearly described, supported by named examples throughout. The answer includes a clear evaluation of the effectiveness and limitations of the methods described, with a logical, well-structured argument.
Indicative content:
- Breeding programmes for endangered species in captivity (e.g. in zoos or wildlife parks), followed by reintroduction into the wild
- Protection and regeneration of rare habitats, e.g. protecting coral reefs or replanting mangrove forests
- Reintroduction of hedgerows and field margins around agricultural fields, providing habitats and food for wildlife such as birds and insects
- Reducing the rate of deforestation and running afforestation/reforestation schemes to replace lost woodland
- Government and international schemes to reduce over-exploitation of forests and fishing grounds, e.g. fishing quotas to prevent overfishing
- Recycling resources to reduce the amount of waste sent to landfill and reduce demand for raw materials taken from natural habitats
- Evaluation: breeding programmes can be expensive, slow, and reintroduced animals may struggle to survive in the wild
- Evaluation: hedgerow and field margin schemes are relatively low cost but only protect small areas of land
- Evaluation: international agreements (e.g. fishing quotas) can be difficult to enforce consistently across different countries

Question 20

(a) B1 B
(a) Answer: B) All the organisms of one species living in the same area at the same time.
(b) B1 C
(b) Answer: C) Availability of light.
(c) B1 B
(c) Answer: B) Producer.
(d) B1 B
(d) Answer: B) Respiration.
(e) B1 B
(e) Answer: B) The range of different species living in an area, or on Earth.

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