Communities, abiotic and biotic factors, adaptations – GCSE Biology exam questions with mark scheme

1Grade 4–55 marks

Cacti live in hot deserts.

(a) Explain how two features of a cactus adapt it to live in a hot desert. [4]

(b) Some bacteria live in hot springs at temperatures above 80 °C. What name is given to organisms that live in extreme environments? [1]

(a) first feature, any one from: leaves reduced to spines; (thick) swollen/fleshy stem; long/deep or widespread (shallow) roots; thick waxy cuticle; few / sunken stomata(1)
(a) explanation linked to that feature: spines – smaller surface area so less water lost (by transpiration) (allow deter animals from eating it for its water); stem – stores water; roots – absorb water from deep underground / from a large area (after rain); cuticle – reduces water loss by evaporation; few/sunken stomata – less water lost by transpiration(1)
(a) a second, different feature from the list(1)
(a) explanation linked to the second feature
explanation marks only awarded if linked to the feature given(1)
(b) extremophiles
allow thermophiles(1)
Answer: (a) e.g. spines – reduce water loss; swollen stem – stores water (b) Extremophiles.
Examiner tip: The explanation mark only counts if it is linked to the feature you named.

2Grade 1–3Foundation4 marks

A rock pool on a beach contains seaweed, crabs, limpets and small fish.

(a) Draw one line from each description to the correct term. [3]

Description		Term
All the crabs of one species living in the rock pool		community
All the populations of different species living in the rock pool		ecosystem
The living organisms in the rock pool interacting with the water, rocks and sand		habitat
		population

(b) The numbers of each species in the rock pool stay almost the same from year to year.

Which statement describes a stable community? Tick (✓) one box. [1]

All the species in the community have the same population size.	☐
No organism in the community ever dies.	☐
All the species and the environmental factors are in balance, so population sizes remain fairly constant.	☐
Only one species lives in the community.	☐

(a) all the crabs of one species → population(1)
(a) all the populations of different species → community(1)
(a) organisms interacting with water, rocks and sand → ecosystem (if more than one line is drawn from a description, no mark for that description)(1)
(b) third box ticked: all the species and the environmental factors are in balance (so population sizes remain fairly constant) (no mark if more than one box ticked)(1)
Answer: (a) crabs of one species → population; all populations → community; organisms + water, rocks, sand → ecosystem. (b) Species and environmental factors are in balance.
Examiner tip: ‘Habitat’ is the distractor: it is just the place where an organism lives, not the organisms themselves.

3Grade 1–3Foundation4 marks

A student studied the organisms living in a garden pond.

(a) Some factors that affect the organisms in the pond are given in the table.

Tick (✓) one box in each row to show whether the factor is abiotic or biotic. [2]

Factor	Abiotic	Biotic
Concentration of oxygen dissolved in the water	☐	☐
A heron starts visiting the pond to eat the fish	☐	☐
Temperature of the water	☐	☐
A fungal disease spreads among the frogs	☐	☐

(b) Plants growing on a windy hillside are affected by different abiotic factors.

Give two abiotic factors, other than temperature, that could affect where plants grow on the hillside. [2]

(a) any two or three rows correct (oxygen = abiotic; heron = biotic; temperature = abiotic; fungal disease = biotic)(1)
(a) all four rows correct (a row with both boxes ticked is wrong)(1)
(b) any one from: light intensity; moisture / water (in the soil); soil pH; mineral (ion) content of the soil (allow nitrates / nutrients in the soil); wind intensity / wind direction; carbon dioxide (concentration)(1)
(b) a second, different factor from the list (ignore temperature; ignore ‘weather’ / ‘climate’ unqualified; ignore ‘soil’ unqualified; do not accept biotic factors such as grazing animals)(1)
Answer: (a) Oxygen – abiotic; heron – biotic; temperature – abiotic; fungal disease – biotic. (b) Any two of: light intensity, soil moisture, soil pH, soil mineral ions, wind intensity/direction, carbon dioxide.
Examiner tip: ‘Soil’ on its own scores nothing – say which property of the soil (pH, minerals, moisture).

4Grade 4–56 marks

A student sowed different numbers of radish seeds in identical pots of compost.

After 4 weeks she dried the plants in each pot and calculated the mean dry mass per plant.

The bar chart shows her results.

(a) Describe the pattern shown in the bar chart. [1]

(b) Calculate the percentage decrease in mean dry mass per plant when the number of seeds sown per pot increased from 5 to 40. [2]

(a) as the number of seeds (per pot) increases, the mean dry mass per plant decreases (allow the more seeds, the smaller / lighter the plants; ignore ‘negative correlation’ unqualified)(1)
(b) (2.4 − 0.6) ÷ 2.4 (× 100) (allow 1.8 ÷ 2.4)(1)
(b) 75 (%) (correct answer with no working = 2 marks)(1)
(c) (more) competition (between the plants) (allow plants compete with each other)(1)
for light / water / mineral ions (any one; allow nutrients from the soil; allow space; ignore food; do not accept ‘oxygen’ or ‘carbon dioxide’ alone)(1)
(c) so less photosynthesis / less glucose produced (for growth) or fewer mineral ions (e.g. nitrate) for making proteins (so less growth)(1)
Answer: (a) More seeds per pot → lower mean dry mass per plant. (b) (2.4 − 0.6)/2.4 × 100 = 75%. (c) Plants compete for light, water and mineral ions, so each gets less → less photosynthesis/glucose → less growth.
Examiner tip: In (c) ‘they compete’ gets one mark – name what plants compete for (light, water, mineral ions) and link it to less photosynthesis.

5Grade 4–55 marks

Red clover and hawthorn are two plants that grow at the edge of a field. Bumblebees feed on nectar from red clover flowers. Blackbirds eat the berries of hawthorn bushes. Each hawthorn berry contains a seed.

(a) Red clover depends on bumblebees.

Give the reason why. [1]

(b) Explain how the hawthorn bushes benefit when blackbirds eat their berries. [2]

Suggest why the population of wood mice, which eat clover seeds, may also decrease. [2]

(a) (bumblebees) pollinate the (clover) flowers (allow transfer pollen (between flowers); ignore ‘fertilise’ unqualified)(1)
(b) seeds are dispersed / spread (in the birds’ droppings) (allow seeds carried away)(1)
(b) (so seedlings grow) away from the parent plant, so less competition (with the parent) for light / water / mineral ions or so the hawthorn can colonise new areas(1)
(c) fewer (clover) flowers are pollinated (so fewer seeds are produced)(1)
(c) so less food for the mice (allow mice starve / compete more for food) (ignore ‘mice eat the pesticide’)(1)
Answer: (a) Bumblebees pollinate the flowers. (b) Seeds are dispersed away from the parent, so less competition for light, water and minerals. (c) Fewer flowers pollinated → fewer seeds → less food for mice.
Examiner tip: Interdependence chains need every link: fewer bees → less pollination → fewer seeds → less food. Missing the middle link loses a mark.

6Grade 6–7Higher5 marks

White-clawed crayfish are native to rivers in Britain. Signal crayfish were brought to Britain from North America. Signal crayfish:

grow larger and breed faster than white-clawed crayfish
eat the same food and shelter under the same stones as white-clawed crayfish
carry a pathogen that causes crayfish plague. Signal crayfish are resistant to crayfish plague but white-clawed crayfish are not.

Scientists estimated the number of each type of crayfish in one river.

Year	Number of white-clawed crayfish	Number of signal crayfish
2006	840	0
2010	420	310
2014	72	900

(a) Calculate the percentage of all the crayfish in the river in 2014 that were white-clawed crayfish.

Give your answer to 2 significant figures. [2]

(b) Use the information to explain why the population of white-clawed crayfish decreased between 2006 and 2014. [3]

(a) 72 ÷ (72 + 900) (× 100) or 72 ÷ 972 (× 100) (allow 7.407… for this mark)(1)
(a) 7.4 (%) (correct answer to 2 s.f. with no working = 2 marks; 7.41 or 7 = 1 mark only)(1)
(b) signal crayfish outcompete white-clawed crayfish for food (allow signal crayfish get more of the food because they are larger / reproduce faster)(1)
signal crayfish also outcompete white-clawed crayfish for shelter / space (under stones) (allow territory; allow so fewer white-clawed crayfish survive / more are exposed to predators)(1)
(b) signal crayfish spread crayfish plague / the pathogen to white-clawed crayfish, which have no resistance so die (allow new pathogen kills white-clawed crayfish)(1)
Answer: (a) 72 ÷ 972 × 100 = 7.4%. (b) Signal crayfish outcompete them for food and shelter, and spread crayfish plague, which kills white-clawed crayfish.
Examiner tip: In (a) the total is 72 + 900, not 900 – read what the percentage is ‘of’.

7Grade 4–55 marks

Organisms have adaptations that help them to survive in the conditions where they normally live.

(a) Adaptations may be structural, behavioural or functional.

Complete the table by writing structural, behavioural or functional next to each adaptation. [3]

Organism	Adaptation	Type of adaptation
Kangaroo rat (desert)	Its kidneys produce a very small volume of very concentrated urine.
Kangaroo rat (desert)	It stays in a cool burrow during the day and feeds at night.
Polar bear (Arctic)	It has a thick layer of fat under its skin.

(b) Marram grass grows on sand dunes. When the air is dry, its leaves roll up so that the stomata are on the inside of the rolled leaf.

Explain how this adaptation helps marram grass to survive on sand dunes. [2]

(a) concentrated urine – functional(1)
(a) burrow by day, feeds at night – behavioural(1)
(a) thick layer of fat – structural(1)
(b) (moist) air / water vapour is trapped inside the rolled leaf (around the stomata) (allow humid air around the stomata; allow stomata are not exposed to wind)(1)
(b) so less water is lost (by evaporation / transpiration) (allow reduces the water vapour concentration gradient / less diffusion of water vapour out of the leaf; ignore ‘stops water loss’)(1)
Answer: (a) Functional; behavioural; structural. (b) Humid air is trapped around the stomata, so less water vapour diffuses out – less water lost by transpiration in a dry, windy habitat.
Examiner tip: Functional adaptations are about processes inside the body (kidneys, enzymes); what the animal does is behavioural; a body feature you could see or dissect is structural.

8Grade 6–7Higher5 marks

Arctic foxes live in very cold places. Fennec foxes live in hot deserts.

Arctic foxes have a large, rounded body and small ears. Fennec foxes have a small body and very large ears.

A student modelled the bodies of two foxes using cubes. Cube P has sides of 2 cm. Cube Q has sides of 4 cm.

(a) Cube P has a surface area to volume ratio of 3 : 1.

Calculate the surface area to volume ratio of cube Q. [2]

(b) Use your answer to (a) to explain why a large body helps the Arctic fox to survive. [2]

Suggest how the large ears help the fennec fox to survive in a hot desert. [1]

(a) surface area = 96 (cm²) and volume = 64 (cm³) (allow 6 × 4 × 4 = 96 and 4 × 4 × 4 = 64)(1)
(a) 1.5 : 1 (allow 3 : 2; correct answer with no working = 2 marks; 96 : 64 not simplified scores the first mark only)(1)
(b) larger body has a smaller surface area to volume ratio (allow smaller surface area compared with its volume) (ignore ‘smaller surface area’ unqualified)(1)
(b) so (relatively) less heat is lost (to the cold surroundings) (allow keeps warm more easily; do not accept ‘no heat lost’; ignore ‘energy lost’)(1)
(c) (large) surface area (of the ears) so more heat is lost / transferred to the surroundings (allow helps it cool down / stops it overheating; ignore ‘hears predators’ – the question asks about the hot desert)(1)
Answer: (a) SA = 96 cm², V = 64 cm³, ratio = 1.5 : 1. (b) Larger body → smaller SA:V → less heat lost relative to volume. (c) Large surface area of ears → more heat lost, so it does not overheat.
Examiner tip: ‘Smaller surface area’ alone does not score – the marking point is the smaller surface area to volume ratio.

9Grade 8–9Higher6 marks

A factory uses river water to cool its machines. The warm water is then returned to the river through an outlet pipe.

Scientists measured some abiotic and biotic factors at three sites along the river.

Site	Position	Water temperature in °C	Dissolved oxygen in mg/dm³	Number of mayfly larvae per sample	Number of different invertebrate species found
A	100 m upstream of the outlet	14	10.2	46	12
B	50 m downstream of the outlet	24	6.1	3	4
C	2 km downstream of the outlet	16	9.4	38	10

Mayfly larvae live on the river bed and are eaten by trout.

Explain how the warm water from the factory affects the community of organisms in the river.

Use data from the table in your answer. [6]

temperature is higher at site B (24 °C) than upstream at site A (14 °C)(1)
dissolved oxygen is lower at B (6.1 mg/dm³) than at A (10.2 mg/dm³) – warmer water holds less dissolved oxygen(1)
oxygen is an abiotic factor for aquatic animals: it is needed for aerobic respiration (to release energy)(1)
so fewer mayfly larvae survive at B: 46 → 3, a decrease of about 93%(1)
higher temperature may also be above the optimum for (enzymes in) some animals / increase their oxygen demand(1)
fewer invertebrate species at B (4 vs 12) so lower biodiversity / the community is less stable(1)
biotic effect / interdependence: fewer mayfly larvae means less food for trout, so trout numbers may fall or trout move away(1)
species that tolerate low oxygen may increase at B because there is less competition(1)
at C the water has cooled (16 °C) and oxygen has risen again (9.4 mg/dm³), so mayfly numbers (38) and species (10) have almost recovered(1)
limitations: only one sample / one occasion; outlet water may contain other pollutants, so cannot be certain temperature is the only cause(1)

Level	Marks	What it looks like
Level 3	5–6	A detailed, logically linked explanation of the effect of the warm water on both abiotic and biotic factors, from higher temperature through lower oxygen and less respiration to fewer invertebrates and an effect on other organisms such as trout. Data from the table are used to support the points, including the recovery at site C.
Level 2	3–4	An explanation that links the higher temperature or lower oxygen to the decrease in mayfly larvae or species, with some data quoted. Links may be incomplete, or the effect on the wider community or the recovery downstream is not covered.
Level 1	1–2	Simple statements describing differences between the sites, e.g. the water is warmer or there are fewer mayflies at B, with little or no explanation.
No relevant content	0	No relevant content.

Answer: Warm water at B (24 °C) holds less dissolved oxygen (6.1 vs 10.2 mg/dm³). Animals need oxygen for aerobic respiration, so mayfly larvae fall from 46 to 3 and species from 12 to 4; less food for trout. Downstream at C the water cools and re-oxygenates, so the community largely recovers.
Examiner tip: Top-level answers link the chain: temperature ↑ → oxygen ↓ → less aerobic respiration → fewer invertebrates → less food for trout, quoting the numbers as they go.

10Grade 6–75 marks

Scientists measured the rate of growth of two species of bacteria, A and B, at different temperatures.

Species B was collected from a hydrothermal vent on the deep sea floor.

The graph shows the results. Species A did not grow at any temperature above 50 °C.

(a) Give the optimum temperature for the growth of species B. [1]

(b) Species B is an extremophile. What is an extremophile? [1]

(d) Give one condition, other than high temperature, that species B must be adapted to at a deep-sea vent. [1]

(a) 105 (°C) (allow 104–106)(1)
(b) an organism that lives in (an environment with) extreme conditions (allow very high or very low temperature / pressure / salt concentration as examples of extreme conditions; ignore ‘lives in harsh places’)(1)
(c) (the high temperature) denatures its enzymes / proteins (allow changes the shape of the active site; do not accept ‘enzymes are killed’)(1)
(c) so (the substrate no longer fits the active site and) reactions such as respiration cannot take place / metabolism stops (allow so it cannot respire / release energy)(1)
(d) high pressure (allow lack of light / darkness; allow high concentration of (dissolved) minerals / salt / toxic chemicals such as hydrogen sulfide; ignore ‘no oxygen’; ignore ‘cold’)(1)
Answer: (a) 105 °C. (b) An organism that lives in extreme conditions (e.g. very high temperature, pressure or salt concentration). (c) Its enzymes are denatured at 90 °C, so reactions such as respiration stop. (d) High pressure.
Examiner tip: Enzymes are denatured, not ‘killed’ – they are molecules, not living things.

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