This topic looks like the easy one and it is not. Most of the marks are for reading and drawing, not recall: constructing a chain from a paragraph of prose, tracing a knock-on effect through a web in both directions, drawing a pyramid to scale, reading a growth curve. Two sentences carry more marks than anything else. An arrow means is eaten by and shows the direction of energy transfer — reversing one is the commonest lost mark in the topic. And nutrients are recycled; energy is not. Twelve traps, six data-led walkthroughs, six lookalike pairs, a concept map and ten full practice questions below, every one aimed at a place where a sensible-sounding sentence earns nothing at all.
Twelve traps that cost marks on Topic 19 challenge papers, spread across all five sub-topics. Every one is an answer that sounds right and that mark schemes refuse.
Six challenge-level questions with real data, worked through in the order you should actually think about them. Try each part before revealing the next step.
Key: A algae · B pondweed · C water flea · D mayfly nymph · E pond snail · F dragonfly nymph · G stickleback · H perch · J heron
(a) State the letters of the two producers, and explain how you can tell. [2] (b) Construct the longest food chain in the web and state the trophic level of G in it. [3] (c) J occupies two different trophic levels in this web. Explain, using two chains. [2] (d) A disease removes all of organism E. Suggest, with reasons, the effect on B, on G and on J. [4]
You do not need to know what the organisms are. Look at the arrows: A and B have arrows leaving them and none arriving. Nothing eats energy into them, which means they must be making their own organic nutrients — they are the producers. (The key confirms it: algae and pondweed.) Everything else in the web has at least one incoming arrow, so everything else is a consumer. This test works on any web, in any habitat, even if every organism is unfamiliar.
Start at A. A is eaten by C; C is eaten by G; G is eaten by H; H is eaten by J. That is A → C → G → H → J — five organisms, so five trophic levels, which is as long as this web gets. Count organisms, not arrows: A is level 1, C level 2, G level 3. So in this chain G is a secondary consumer at the third trophic level. The commonest slip here is counting the four arrows and calling G tertiary.
Find two chains ending at J that have different lengths. In A → C → G → J, J eats G, and G eats a primary consumer — so G is secondary and J is a tertiary consumer. In A → C → G → H → J, J eats H instead, and H eats G — so H is tertiary and J is a quaternary consumer. Both answers are correct, and a full-mark answer names the chain each one belongs to.
Note what this does not apply to. G eats C, D and E, and all three of those eat producers directly, so G is a secondary consumer in every chain in this web. Before you claim that an organism sits at two levels, check the direction of every arrow into it — the claim is only true if its prey sit at different levels themselves.
B (down): E is a primary consumer that eats B, so removing E means B is grazed less and increases. G (up): G eats E, so G has lost a food source and its numbers may fall — but G also eats C and D, so it will switch rather than starve. J (sideways and above): J does not eat E directly, so the effect reaches it through G and H. If G falls, J has less food and may fall too; if instead G switches to C and D, then C and D fall and the effect passes down rather than up. A full-mark answer says which chain the effect travels along, and says that the direction of change in J is uncertain because two opposing effects act on it.
A young bullock is kept in a pen for 100 days and everything is measured. Over that period it eats grass containing 1000 MJ of energy.
| Measurement | Energy / MJ |
|---|---|
| energy in the food eaten | 1000 |
| energy in the faeces produced | 600 |
| energy released in respiration | 280 |
| energy in the urine produced | 40 |
| energy stored as new body tissue | to be calculated |
(a) Calculate the energy stored as new body tissue. [1] (b) Calculate the percentage of the energy eaten that is available to an animal that eats the bullock. [2] (c) Explain why the figure for faeces is so large in a grass-eating animal. [2] (d) A pig fed on grain shows a much smaller faeces figure and a much larger figure for new tissue. Suggest why. [2]
1000 − 600 − 280 − 40 = 80 MJ. Energy is never destroyed, so a budget question always balances, and if yours does not you have missed a row. Write the subtraction out; the working carries a mark if you slip.
The tempting wrong answer is 400 MJ — the energy the bullock absorbed. But absorbed energy that has been respired is gone as heat, and energy in urine has left the body. What a predator eats is the new tissue. So 80 ÷ 1000 × 100 = 8 %. Notice that this is not 10 %, which is exactly why you must divide rather than recite.
Plant cell walls are made of cellulose, and a mammal produces no enzyme that digests it. A large proportion of what the bullock swallows therefore cannot be digested or absorbed, passes through the alimentary canal and leaves in the faeces, taking its energy with it. That energy is not wasted from the ecosystem’s point of view — decomposers use it — but it never enters the bullock’s body, so it is not available to the next trophic level.
Grain is mostly starch, which mammals digest easily, and it contains far less cellulose than grass. So a much higher proportion is digested and absorbed, the faeces figure falls, and more energy is available inside the body. With more absorbed and respiration roughly similar, more is left to be stored as new tissue, so the transfer to the next level is more efficient. A second acceptable point: a pig is smaller and often less active than a bullock in a pen, so it may lose less energy as heat per unit of food.
Both diagrams describe the same oak woodland chain in the same year. (a) Explain why the pyramid of numbers has this shape. [2] (b) Explain why the pyramid of biomass does not. [2] (c) Calculate the percentage of the oak tree’s biomass present in the insects. [1] (d) State one advantage and one disadvantage of drawing a pyramid of energy for this woodland instead. [2]
The rule: a pyramid of numbers takes no account of the size of the organisms — one oak tree is counted as one, exactly like one aphid. The instance: one very large producer supports very many small primary consumers, so the base bar is the narrowest in the diagram. Give both halves. “The tree is big” is the instance without the rule and scores one.
A pyramid of biomass measures the dry mass of living material at each level. The single oak has an enormous dry mass — 5000 kg here — while the three thousand insects together weigh 80 kg. Measuring mass rather than counting individuals therefore puts the producer where you expect it, and the pyramid narrows all the way up. That is the standard answer to “state an advantage of a pyramid of biomass over a pyramid of numbers”.
80 ÷ 5000 × 100 = 1.6 %. Two things to watch. Both figures must be in the same unit — here both are kilograms, but a question may give one in grams to see whether you notice. And divide the higher level by the lower: 5000 ÷ 80 gives 62.5, which is not a percentage of anything sensible.
Advantage: a pyramid of energy measures the energy passing through each level over a period of time, so it takes account of the rate at which material is produced and it can never be inverted. It also lets you calculate the efficiency of each transfer. Disadvantage: the data are extremely difficult and slow to collect, because energy has to be measured over a whole year rather than sampled on one day.
(a) Name the process shown by the arrow labelled photosynthesis on the diagram, and state why only one arrow can point away from the carbon dioxide box. [2] (b) A student says decomposition releases carbon dioxide into the air. Correct and complete this statement. [2] (c) Explain why the carbon in coal was removed from the cycle for millions of years. [2] (d) Describe the fastest possible route by which a carbon atom fixed in a leaf this morning could be back in the air by tonight. [2]
Of the six named processes, photosynthesis alone takes carbon dioxide out of the air and fixes it into organic compounds. Three put it back: respiration, combustion and decomposition (indirectly). So on any blank carbon-cycle diagram, find the single arrow leaving the air box — it must be photosynthesis — and everything else falls into place around it.
Decomposition transfers the carbon compounds from the dead material into the decomposers. The carbon dioxide reaches the air when those decomposers respire. That is why a full answer names two processes: decomposition and respiration. Writing “decomposition releases carbon dioxide” alone is worth one mark of two.
The organisms died in conditions where decomposers could not break them down — deep water or waterlogged ground with little or no oxygen. Their carbon compounds were buried, compressed and heated over millions of years to form coal, oil and gas, and the only process that can return that carbon to the air is combustion. Until something burns it, it sits outside the cycle.
Photosynthesis fixes the carbon into glucose; the plant respires that glucose and releases the carbon dioxide back to the air within hours. Plants respire day and night, so no other organism needs to be involved at all. Every other route — being eaten first, or dying and decomposing — takes longer, and the fossil-fuel route takes millions of years.
(a) Name the phases labelled W, X, Y and Z. [2] (b) Calculate the mean rate of increase in phase X between 10 and 16 hours. [2] (c) Explain, in terms of limiting factors, why the curve levels off in phase Y. [3] (d) The experiment is repeated but fresh broth is added continuously and waste removed. Predict how the curve would differ after 18 hours, and explain. [2]
W = lag phase (flat at the start), X = exponential or log phase (rising steeply and getting steeper), Y = stationary phase (level at the maximum), Z = death phase (falling). Both names for X are accepted; write “exponential (log)” and you cannot be wrong.
At 10 hours the count is 600 cells per cm³; at 16 hours it is 3900. Rate = (3900 − 600) ÷ (16 − 10) = 3300 ÷ 6 = 550 cells per cm³ per hour. Two marks: one for the correct method, one for the answer with its unit. The classic error is dividing 3900 by 6 — you need the change, not the final value.
Start with the balance: the rate of reproduction now equals the death rate, so the number stays constant. Then give the causes: the food supply is running short, toxic waste products have accumulated, and competition for the remaining nutrients and space is intense. Whichever of those runs short first is the limiting factor — the same idea you met with limiting factors in photosynthesis, where the factor in shortest supply sets the rate and increasing anything else changes nothing.
Adding fresh broth continuously keeps the food supply high; removing waste prevents toxic products accumulating. Those were the two things that ended the exponential phase, so the population would continue to rise for much longer and there would be no death phase while the supply lasted. It would still level off eventually, because space in the flask is finite and competition for it would become the new limiting factor. Saying that last part is what turns a good answer into a full one.
A farmer compares two fields of the same size, given the same fertiliser and the same rainfall. Field A is ploughed and has drains beneath it. Field B is flat and holds standing water for weeks after rain. After three years the soil is tested.
| Field A | Field B | |
|---|---|---|
| nitrate in soil / mg per kg | 42 | 9 |
| ammonium in soil / mg per kg | 6 | 21 |
| crop yield / tonnes per hectare | 8.4 | 4.1 |
(a) Calculate how many times greater the nitrate concentration is in field A. [1] (b) Explain the difference in nitrate, naming the bacteria involved. [4] (c) Explain why field B has more ammonium than field A even though it has less nitrate. [2] (d) Suggest why the yield in field B is only about half that of field A. [2]
42 ÷ 9 = 4.7 times greater (2 significant figures). Ten seconds, one mark, and it cannot be argued with. Calculation parts are the cheapest marks on any data question — never leave them until the end.
Same size, same fertiliser, same rainfall — the question has controlled those deliberately, which is the examiner pointing at the one thing left. Drainage and ploughing put air, and therefore oxygen, into the soil. In field A there is plenty of oxygen, so nitrifying bacteria are active and convert ammonium ions through nitrite to nitrate ions. In waterlogged field B the water fills the air spaces, oxygen is short, nitrification slows, and denitrifying bacteria, which thrive where oxygen is scarce, convert nitrate back to nitrogen gas, which is lost to the air.
Ammonium ions are the input to nitrification. In field B nitrification is slow because of the shortage of oxygen, so ammonium is produced by decomposition but is not converted onwards into nitrate, and it accumulates. This row is the strongest single piece of evidence that the problem is nitrification and not, say, a shortage of decomposers — if decomposition had failed, the ammonium would be low too.
Plant roots absorb nitrate ions, by active transport; they cannot absorb ammonium as their main nitrogen source in this syllabus, and they certainly cannot use nitrogen gas. With only 9 mg per kg available, the crop in field B can make fewer amino acids and less protein, so it grows less and yields less. A second acceptable point: waterlogged soil is also short of oxygen for root respiration, and without energy from respiration the roots cannot carry out active transport at all — which makes the nitrate shortage worse still.
Six pairs that look almost identical and have different answers. In this topic the distinction is nearly always where the marks live.
Click each node. The whole topic is three stories: energy going one way, atoms going round, and numbers rising until something stops them.
Six answers of the kind that read fluently and score badly. Find the fault before you reveal it.
Ten Cambridge-style challenge questions, each drawing on more than one sub-topic. Write your answer first, then reveal the model answer and the examiner’s notes.
Key: A grass · B acacia tree · C grasshopper · D zebra · E impala · F giraffe · G lizard · H mongoose · J cheetah · K eagle · L lion · M hyena
(a) State the letters of the producers and explain how the diagram alone tells you. [2] (b) Construct a food chain containing four organisms and ending in M, and state the trophic level of the organism you have placed second. [3] (c) A disease removes all of E. Suggest, with reasons, the effect on B, on J and on A. [3]
(a) Name the four phases W, X, Y and Z. [2] (b) Explain the shape of the curve in phase W. [2] (c) Explain, referring to limiting factors, why the curve behaves as it does in phase Y. [3] (d) Sketch in words how the curve would differ if fresh nutrient broth were supplied continuously and waste removed. [1]
(a) Name the six processes shown in Fig. 4.1 that Cambridge includes in the carbon cycle. [3] (b) Explain why only one process can remove carbon dioxide from the air. [2] (c) Describe, naming two processes, how carbon in a dead animal reaches the air. [2] (d) State one reason why carbon locked in coal was out of the cycle for millions of years. [1]
Fig. 6.1 shows what happens to 1000 kJ of energy eaten by a bullock. (a) Calculate the energy stored as new tissue and the percentage efficiency of the transfer. [2] (b) Explain why the faeces figure is the largest single loss for this animal. [2] (c) A student says this proves that carnivores transfer energy more efficiently than herbivores. Evaluate that claim. [3]