These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: about 10% of energy passes from one trophic level to the next unless a question says otherwise; assimilated energy = ingested − egested; growth = assimilated − respired. A calculator is assumed. All organisms named with letters, places and data sets are fictional.
Question 1 (multiple choice · foundation)
Which statement about nutrient cycles is correct?
- (A) The largest reservoir of phosphorus is the atmosphere, from which plants absorb it through their leaves.
- (B) Denitrifying bacteria convert nitrate into nitrogen gas, which returns to the atmosphere.
- (C) Nitrification converts nitrogen gas into ammonia so that plants can use it.
- (D) Plants take in N₂ through their stomata and use it directly to build amino acids.
Answer and explanation
Answer: (B). Denitrification turns nitrate (NO₃⁻) back into N₂, closing the nitrogen cycle.
- (A) Phosphorus has no major atmospheric reservoir. It comes from weathered rock, and plants absorb phosphate from soil water through their roots.
- (C) describes nitrogen fixation. Nitrification converts ammonium to nitrite and then nitrate.
- (D) Plants cannot use N₂. They absorb ammonium or nitrate from the soil (assimilation).
Question 2 (multiple choice · core)
In a fictional lake, producers store 60 000 kJ m⁻² yr⁻¹ in new biomass. Assuming 10% transfer at each step, about how much energy reaches the tertiary consumers?
- (A) 6000 kJ m⁻² yr⁻¹
- (B) 600 kJ m⁻² yr⁻¹
- (C) 60 kJ m⁻² yr⁻¹
- (D) 6 kJ m⁻² yr⁻¹
Answer and explanation
Answer: (C). Producers → primary: 60 000 × 0.1 = 6000. Primary → secondary: 6000 × 0.1 = 600. Secondary → tertiary: 600 × 0.1 = 60 kJ m⁻² yr⁻¹. Three transfers, so 60 000 × 0.1³.
- (A) is the primary consumer level (one transfer).
- (B) is the secondary consumer level (two transfers).
- (D) is the quaternary level (four transfers): one transfer too many.
Question 3 (multiple choice · core)
On the dark sea floor around a hydrothermal vent, tube worms, clams and crabs live in dense communities. The water contains hydrogen sulfide and very little light reaches it. Which statement best explains how energy enters this community?
- (A) Bacteria capture energy from hydrogen sulfide and use it to build organic molecules; animals obtain energy by eating these bacteria or hosting them.
- (B) Algae photosynthesise using the faint light given off by the hot vent water.
- (C) Decomposers produce new energy by breaking down dead organisms that sink from the surface.
- (D) The animals absorb heat energy from the vent water and use it to make ATP.
Answer and explanation
Answer: (A). These bacteria are chemosynthetic autotrophs. They capture energy from a small inorganic molecule and fix carbon into organic matter, which can happen without light. They are the producers here.
- (B) There is too little light for photosynthesis to support such a large community.
- (C) Decomposers do not create energy; they release energy from dead matter that producers made elsewhere. Matter sinking from far above is too little to support communities this dense.
- (D) Animals cannot use heat to make ATP. Heat is the form in which energy leaves organisms.
Question 4 (constructed response · core)
Researchers measured the body temperature of two fictional animals, P and Q, at different air temperatures. Animal Q could move freely between sun and shade.
| Air temperature / °C | 5 | 15 | 25 | 35 |
|---|---|---|---|---|
| Animal P body temperature / °C | 37.8 | 38.0 | 38.1 | 38.4 |
| Animal Q body temperature / °C | 6.0 | 16.5 | 27.0 | 33.5 |
(a) Calculate the range of body temperature of each animal over the 30 °C range of air temperature. (b) Identify which animal is an endotherm. Justify your answer. (c) At 35 °C, animal Q’s body was cooler than the air. Suggest how, and state what this shows. (d) Predict which animal needs more food at an air temperature of 5 °C, and explain why.
Model answer and suggested Marlbridge rubric
(a) P: 38.4 − 37.8 = 0.6 °C. Q: 33.5 − 6.0 = 27.5 °C.
(b) P is the endotherm. Its body temperature stays almost constant (within 0.6 °C) while the air changes by 30 °C, so it must use heat from its own metabolism to keep its temperature steady. Q’s temperature follows the air.
(c) Q probably moved into shade or a cool burrow, where its surroundings were cooler than the open air, to avoid overheating. At lower air temperatures it was slightly warmer than the air (for example 1.0–2.0 °C above), perhaps by basking. This shows that an ectotherm can regulate its temperature by behaviour.
(d) P. At 5 °C, P must raise its metabolic rate to make enough heat to stay near 38 °C, which uses a lot of energy from food. Q’s body cools with the air, its metabolic rate falls, and it needs little food.
| Point | What earns it |
|---|---|
| 1 | Both ranges correct |
| 1 | P identified with reasoning based on the near-constant body temperature |
| 1 | Behavioural regulation by Q explained with evidence |
| 1 | P chosen for (d) and linked to metabolic heat production |
Question 5 (constructed response · core)
Over 10 days, a fictional beetle larva ate leaves containing 500 J of energy. Its faeces contained 280 J, and it released 160 J in cellular respiration.
(a) Calculate the energy assimilated and the energy available for growth. (b) Calculate the percentage of ingested energy that became new tissue. (c) In a second trial the larva received half as much food. The faeces contained the same fraction of the food energy, and respiration stayed at 160 J. Calculate the energy balance and predict the effect on the larva. (d) Explain why only a small fraction of the energy in plants ends up in animals that eat the larvae.
Model answer and suggested Marlbridge rubric
(a) Assimilated = 500 − 280 = 220 J. Growth = 220 − 160 = 60 J.
(b) 60 ÷ 500 × 100 = 12%.
(c) Ingested = 250 J. Faeces = 250 × (280 ÷ 500) = 140 J. Assimilated = 250 − 140 = 110 J. Balance = 110 − 160 = −50 J: a net loss. The larva must use its stored energy, so it loses mass, stops growing and, if the shortage continues, will not reach adulthood or reproduce, and may die.
(d) Most of the energy at each level is not built into new biomass: much is released as heat in respiration, some is egested in faeces, and some parts are not eaten. Here only 12% of the larva’s food became larva tissue, and the predator faces the same losses again. So energy shrinks at every step.
| Point | What earns it |
|---|---|
| 1 | 220 J and 60 J |
| 1 | 12% |
| 1 | −50 J with working |
| 1 | Net loss linked to mass loss and reduced reproduction or survival |
| 1 | Losses as heat, faeces or uneaten parts at each level |
Question 6 (constructed response · core)
A fictional monitoring station beside a large temperate forest measured atmospheric CO₂ (parts per million, ppm) in four months of two years.
| Month | January | April | July | October |
|---|---|---|---|---|
| Year 1 / ppm | 421.0 | 423.5 | 416.0 | 417.5 |
| Year 2 / ppm | 423.4 | 425.9 | 418.4 | 419.9 |
(a) Calculate the fall in CO₂ from April to July in Year 1, and the rise from Year 1 to Year 2 for the same month. (b) Explain the fall from April to July using two processes of the carbon cycle. (c) Suggest a reason for the rise from one year to the next. (d) Explain why the carbon cycle depends on the nitrogen and phosphorus cycles.
Model answer and suggested Marlbridge rubric
(a) April to July, Year 1: 423.5 − 416.0 = 7.5 ppm fall. Year 1 to Year 2: 2.4 ppm in every month (for example 423.4 − 421.0).
(b) In late spring and summer the trees have full leaves, long days and warm temperatures, so photosynthesis removes CO₂ faster than cellular respiration and decomposition return it. CO₂ falls. In autumn and winter photosynthesis slows while respiration and decomposition continue, so CO₂ rises again.
(c) More CO₂ is added each year than is removed. The most likely cause is combustion of fossil fuels (and burning of forests), which releases carbon that was stored for a long time.
(d) To grow and fix carbon, producers must build enzymes and other proteins (which need nitrogen), and nucleic acids, ATP and phospholipids (which need phosphorus). If nitrogen or phosphorus is scarce, photosynthesis and growth slow, so less carbon is taken up. The cycles are interdependent.
| Point | What earns it |
|---|---|
| 1 | 7.5 ppm and 2.4 ppm |
| 1 | Photosynthesis greater than respiration and decomposition in summer |
| 1 | Combustion named as the source of the yearly rise |
| 1 | N and P needed for molecules that photosynthesis and growth depend on |
Question 7 (constructed response · stretch)
A volcanic eruption spreads ash over a fictional lake. For one summer, the sunlight reaching the water falls sharply. Net production by phytoplankton (the producers) falls from 2000 to 1200 kJ m⁻² over the season. The lake’s main consumers are water fleas (Daphnia), which eat phytoplankton, and small fish, which eat water fleas.
In normal summers, about 5% of adult female water fleas produce sexual resting eggs. In the ash summer, 35% do.
(a) Calculate the percentage fall in phytoplankton production. (b) Using 10% transfer, calculate the energy available to the water fleas and to the fish before and after the eruption. (c) Predict and explain the effect on the size of the water flea and fish populations. (d) Explain the change in water flea reproduction in terms of energy availability, and how it could help the population survive.
Model answer and suggested Marlbridge rubric
(a) (2000 − 1200) ÷ 2000 × 100 = 40% fall.
(b) Water fleas: 2000 × 0.1 = 200 before, 1200 × 0.1 = 120 kJ m⁻² after. Fish: 200 × 0.1 = 20 before, 120 × 0.1 = 12 kJ m⁻² after.
(c) Both populations should fall. Less light means less photosynthesis, so less energy enters the food web, and every level above the producers gets less. Individuals that cannot meet their energy needs lose mass, reproduce less or die, so population sizes drop. The fish, at the top, are the most at risk because their share was already small; if their energy falls too low the population may not persist, reducing the number of trophic levels.
(d) When food is plentiful, water fleas reproduce asexually, producing many offspring quickly. When energy is short, more switch to sexual reproduction. The resting eggs can survive poor conditions and hatch when food returns, and sexual reproduction gives genetically varied offspring, some of which may cope better with the changed conditions.
| Point | What earns it |
|---|---|
| 1 | 40% |
| 1 | 200 → 120 and 20 → 12 |
| 1 | Less light → less photosynthesis → less energy at all higher levels |
| 1 | Population effect explained through survival or reproduction, with the top level most affected |
| 1 | Switch to sexual reproduction linked to low energy and a survival benefit (resting eggs or variation) |
How did you do?
- Q1 wrong: re-read the nitrogen and phosphorus cycles and Figure 2 in the study guide.
- Q2 or Q7(b) wrong: rework Worked example 1; count the transfers carefully.
- Q3 wrong: re-read “Autotrophs: where energy enters”.
- Q4 or Q5 wrong: rework Worked example 2 and re-read “Energy budgets”.
- Q6 wrong: re-read the carbon cycle table.
Then tick off the topic checklist.