Per disputationem veritatem quaerimusPhotosynthesis
Photosynthesis in a chloroplast 3.5.1
- Separating leaf pigments by chromatography: grind fresh leaf tissue with a small amount of solvent to extract the pigments → concentrate a small spot of the resulting pigment extract near the base of a strip of chromatography paper → stand the paper in a shallow layer of solvent (below the pigment spot) inside a covered container → as the solvent rises up the paper, each pigment travels a different distance depending on its solubility in the solvent relative to its affinity for the paper, separating into distinct coloured bands.
- Identifying the separated pigments: calculate each band's value, , and compare against known reference values (or run known pigment standards alongside) to identify which pigments (e.g. chlorophyll a, chlorophyll b, carotene, xanthophyll) were present.
- The chloroplast is bounded by a double-membrane envelope; inside, thylakoid membranes are stacked into grana (singular: granum), each stack of thylakoids joined by connecting lamellae, all surrounded by the stroma.
- The two stages of photosynthesis occur in two distinct compartments: the light-dependent reactions take place at the thylakoid membranes, and the light-independent reactions (Calvin cycle) take place in the stroma — this spatial separation is essential, since the two stages exchange specific products (ATP, reduced NADP) between the two compartments.
- Photosynthetic pigments (chlorophyll a, chlorophyll b and carotenoids) are embedded in the thylakoid membranes and absorb slightly different wavelengths of light — arranged together into photosystems, they capture more of the light spectrum than any single pigment could alone.
- Comparing the values (or relative band intensities) of pigments extracted from different plants — e.g. shade-tolerant versus shade-intolerant species, or leaves of different colours — can reveal genuine differences in pigment composition linked to their light environment, connecting this technique directly back to how a plant is adapted to the light conditions it grows in.
The light-dependent reactions 3.5.1
- Photoionisation: light absorbed by chlorophyll causes it to lose electrons.
- Photolysis of water: — replaces electrons lost from photosystem II and releases O₂ as a by-product.
- Electron transfer along the electron transfer chain (from photosystem II to photosystem I) drives active transport of H⁺ into the thylakoid lumen, building a proton gradient across the thylakoid membrane; H⁺ then flows back down this gradient through ATP synthase, driving ATP formation (chemiosmosis).
- Electrons (from photosystem I) and H⁺ combine with NADP⁺ to form reduced NADP (NADPH).
- This is structurally the same overall mechanism (light-driven electron transport building a proton gradient, then ATP synthase using that gradient to make ATP) as oxidative phosphorylation in respiration — the same chemiosmotic principle, in a different membrane, driven by a different original energy source.
- Both major products of the light-dependent reactions — ATP and reduced NADP — are used directly in the Calvin cycle in the stroma; neither is exported anywhere else in the cell for this purpose.
- PSII and PSI (photosystems II and I, named by order of discovery rather than order of use) both contain chlorophyll and other pigments; despite the numbering, electron flow proceeds through PSII before PSI in this linear pathway.
The Calvin cycle 3.5.1
- Compensation point: the light intensity (or condition) at which gross photosynthesis exactly equals respiration rate, so net (measured) O₂ production or CO₂ uptake reads zero.
- Carbon fixation: CO₂ combines with RuBP (ribulose bisphosphate, 5C), catalysed by rubisco, forming an unstable intermediate that immediately splits into two molecules of GP (glycerate 3-phosphate, 3C).
- Reduction: GP is reduced to TP (triose phosphate, 3C), using ATP and reduced NADP supplied by the light-dependent reactions (reoxidising NADP back to NADP⁺ in the process).
- RuBP regeneration: most TP (5 out of every 6 molecules produced, for 3 CO₂ fixed) is used, with further ATP, to regenerate RuBP, keeping the cycle running; the remaining TP (1 in 6) is the net output, used to make glucose and other organic compounds.
- The Calvin cycle depends entirely on a continuous supply of ATP and reduced NADP from the light-dependent reactions — without light (and so without the light-dependent reactions running), the Calvin cycle cannot continue for long, even though none of its own individual reactions directly requires light.
- ADP, inorganic phosphate and NADP⁺ released by the Calvin cycle's reactions are recycled directly back to the light-dependent reactions, closing the loop between the two stages.
- Because most of the TP produced must be used to regenerate RuBP, only a small net fraction is actually available for making new organic compounds — several turns of the cycle are needed to accumulate enough net TP output for further biosynthesis.
Factors limiting the rate of photosynthesis 3.5.1
- Limiting factor: whichever of light intensity, CO₂ concentration or temperature is in shortest supply relative to demand at a given moment — increasing a non-limiting factor achieves nothing until the true limiting factor is addressed.
- Investigating the effect of a named factor (e.g. light intensity) on the rate of the light-dependent reaction, using dehydrogenase activity in chloroplast extracts as a proxy: chloroplasts are extracted from leaf tissue by homogenising and filtering (see Cell fractionation) → a blue redox indicator dye (e.g. DCPIP) is added, which accepts electrons in place of the chloroplast's own natural electron acceptor and, in doing so, changes from blue to colourless as it becomes reduced → the rate of colour loss (measured by eye against a timer, or more precisely by colorimeter) is used as a proxy for the rate of the light-dependent reaction, since a faster rate of electron release means faster reduction, and so faster decolourisation, of the dye.
- Raising light intensity increases the rate of photosynthesis only up to the point where CO₂ concentration or temperature takes over as the limiting factor — this is why two otherwise identical light-response curves, measured at different fixed CO₂ concentrations, plateau at different heights and different light intensities: each plateaus once CO₂ (not light) becomes limiting, and a higher CO₂ concentration allows a higher plateau before that happens.
- The redox indicator dye is decolourised faster under conditions that speed up the light-dependent reaction (e.g. higher light intensity, up to the point some other factor limits it) — a genuinely indirect measurement (the dye reduction rate, not photosynthesis itself), but one that tracks the underlying rate of electron transfer reliably enough to compare conditions against each other.
- Because a measurement of net O₂ production (or net CO₂ uptake) reflects gross photosynthesis minus the plant's own simultaneous respiration, any such investigation systematically understates the true gross photosynthetic rate by the plant's respiration rate — reading exactly zero at the compensation point, where the two rates are equal, even though gross photosynthesis is clearly still occurring.
Worked examples
Worked example 3.5.1 · 5 marks
| Band | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Distance travelled (cm) | 8.6 | 7.7 | 5.3 | 1.7 |
Chromatography of a leaf pigment extract produces four separated pigment bands.
The solvent front travels 9.6 cm from the origin.
Band 1 travels 8.6 cm, band 2 travels 7.7 cm, band 3 travels 5.3 cm, and band 4 travels 1.7 cm.
(a) Calculate the value of each band.
(b) Given that carotene has the highest value of typical leaf pigments and chlorophyll b has the lowest, identify which band is most likely to be carotene and which is most likely to be chlorophyll b.
Show worked solution
(a):
Band 1:
Band 2:
Band 3:
Band 4:
(b) Band 1, with the highest value, is most likely carotene.
Band 4, with the lowest value, is most likely chlorophyll b.
Mark scheme · 5 marks
- Calculates for band 1 1 mark
- Calculates for band 2 1 mark
- Calculates and for bands 3 and 4 1 mark
- Identifies band 1 as carotene (highest ) 1 mark
- Identifies band 4 as chlorophyll b (lowest ) 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Worked example 3.5.1 · 5 marks
| Light intensity | Low | Medium | High |
|---|---|---|---|
| Time to decolourise (s) | 380 | 190 | 95 |
In an investigation using a redox indicator dye (DCPIP) with chloroplast extracts, the time taken for the dye to fully decolourise was measured at three light intensities.
At low light intensity, decolourisation took 380 seconds; at medium light intensity, 190 seconds; at high light intensity, 95 seconds.
(a) Explain what the decolourisation of DCPIP indicates is happening in the chloroplast extract.
(b) Calculate the rate of reaction (as , in s⁻¹) at each light intensity, and comment on the relationship between light intensity and rate.
Show worked solution
(a) DCPIP accepts electrons in place of the chloroplast's own natural electron acceptor, becoming colourless as it is reduced — faster decolourisation means faster electron release and transfer during the light-dependent reaction.
(b) Rate at low intensity ; medium ; high .
Doubling light intensity approximately doubles the rate at each step, consistent with light intensity being the limiting factor across this range.
Mark scheme · 5 marks
- States DCPIP accepts electrons, becoming colourless as it is reduced 1 mark
- Links faster decolourisation to faster electron transfer in the light-dependent reaction 1 mark
- Calculates rate ≈ 0.0026 s⁻¹ at low intensity 1 mark
- Calculates rate ≈ 0.0053 s⁻¹ and ≈ 0.0105 s⁻¹ at medium and high intensity 1 mark
- States doubling light intensity approximately doubles rate, consistent with light being the limiting factor over this range 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Worked example 3.5.1 · 4 marks
An investigation into the rate of photosynthesis in aquatic plant shoots is carried out at three different concentrations of dissolved carbon dioxide, at each of two light intensities (low and high).
At low light intensity, the rate of oxygen production is almost identical across all three CO₂ concentrations.
At high light intensity, the rate of oxygen production rises substantially as CO₂ concentration rises.
Explain this pattern in terms of limiting factors.
Show worked solution
At low light intensity, light itself is the limiting factor — increasing CO₂ concentration has little effect since more CO₂ is already available than the light-dependent reaction can currently supply ATP/reduced NADP for.
At high light intensity, light is no longer limiting, so CO₂ concentration becomes the limiting factor instead, and increasing it produces a genuine rise in rate.
Mark scheme · 4 marks
- States light is the limiting factor at low light intensity 1 mark
- Explains why increasing CO₂ has little effect when light is limiting 1 mark
- States CO₂ becomes the limiting factor at high light intensity 1 mark
- Explains this is why increasing CO₂ raises rate substantially at high light intensity 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Respiration
Glycolysis 3.5.2
- Glycolysis (cytoplasm, does not require oxygen): glucose (6C) is phosphorylated (using 2 ATP) and split into two triose phosphate (3C) molecules → each is oxidised to pyruvate (3C), with a net gain of 2 ATP (by substrate-level phosphorylation) and 2 NAD reduced to 2 reduced NAD, per glucose molecule.
- Investigating the effect of a named variable (e.g. temperature, or substrate concentration) on the rate of respiration of a culture of single-celled organisms (e.g. yeast): set up cultures under identical conditions except for the variable being tested, and use a redox indicator (e.g. methylene blue, or resazurin) that changes colour as it is reduced by the electron carriers respiration produces — the indicator's colour loss (or colour change) over a fixed time is used as a proxy for the rate of respiration, since faster respiration reduces the indicator faster.
- Glycolysis is the one stage of respiration common to both aerobic and anaerobic pathways, and the only stage that takes place in the cytoplasm rather than the mitochondrion — it proceeds identically whether or not oxygen is available.
- This is the same underlying redox-indicator logic used to investigate the light-dependent reaction's rate (see Factors limiting the rate of photosynthesis) — in both cases, a dye that changes colour as it accepts electrons stands in for a process that would otherwise be far harder to measure directly.
- 'Substrate-level phosphorylation' means a phosphate group is transferred directly from an intermediate substrate molecule to ADP, forming ATP — distinct from oxidative phosphorylation, which instead uses a proton gradient generated by an electron transport chain.
- The initial investment of 2 ATP (phosphorylating glucose) is necessary before any ATP is generated — the pathway's NET yield of 2 ATP per glucose is the difference between 4 ATP generated later and the 2 ATP invested at the start.
The mitochondrion and the link reaction 3.5.2
- Coenzyme A (CoA): a carrier molecule that combines with a 2-carbon acetyl group (from pyruvate oxidation) to form acetyl CoA, delivering that acetyl group into the Krebs cycle.
- Link reaction (mitochondrial matrix, occurs twice per glucose): each pyruvate (3C, having entered the matrix by active transport) is oxidised and decarboxylated, combining with coenzyme A to form acetyl CoA (2C) — releasing CO₂ and reducing NAD to reduced NAD.
- The mitochondrion's own structure directly supports the three later stages of aerobic respiration: the matrix (bounded by the inner membrane) is where the link reaction and Krebs cycle occur, while the inner membrane itself — folded into cristae to increase its surface area — is where oxidative phosphorylation takes place.
- The link reaction happens twice for every one glucose molecule, since glycolysis produces two pyruvate molecules from each glucose — every subsequent yield figure for the link reaction and Krebs cycle should be doubled accordingly when reasoning per glucose molecule.
- Decarboxylation (loss of CO₂) and oxidation (loss of hydrogen, reducing NAD) occur together here, exactly as they do repeatedly throughout the Krebs cycle that follows.
The Krebs cycle 3.5.2
- Krebs cycle (mitochondrial matrix, two turns per glucose): each acetyl CoA (2C) combines with a 4-carbon acceptor molecule to form a 6-carbon compound → this is progressively decarboxylated and oxidised across the cycle, releasing CO₂, reducing NAD and FAD, and regenerating the original 4-carbon acceptor — each turn also produces one ATP directly, by substrate-level phosphorylation.
- Per glucose molecule (i.e. per two turns of the cycle): 4 CO₂, 6 reduced NAD, 2 reduced FAD, and 2 ATP.
- The Krebs cycle is a genuine CYCLE, not a linear pathway: the 4-carbon acceptor consumed at the start of each turn is exactly regenerated by the end of that same turn, ready to accept another acetyl CoA — this is why it can keep running continuously as long as acetyl CoA, oxidised NAD and oxidised FAD keep being supplied.
- Nearly all the energy extracted from glucose across the Krebs cycle is captured as reduced NAD and reduced FAD, not as ATP directly — only 2 ATP per glucose come from substrate-level phosphorylation in the Krebs cycle itself; the much larger ATP yield comes later, from oxidative phosphorylation using these reduced coenzymes.
Oxidative phosphorylation 3.5.2
- Reduced NAD and reduced FAD are reoxidised at the inner mitochondrial membrane, passing electrons along an electron transfer chain — this electron transfer drives active transport of H⁺ from the matrix into the intermembrane space, building a proton gradient across the inner membrane.
- H⁺ flows back down this gradient through ATP synthase (from the intermembrane space into the matrix), driving ATP formation by chemiosmosis.
- Oxygen is the final electron acceptor: .
- Oxidative phosphorylation is structurally the same chemiosmotic mechanism as the light-dependent reactions of photosynthesis — electron transport pumping H⁺ to build a gradient, then ATP synthase using that gradient to make ATP — just occurring in a different membrane, driven by a different original electron source.
- Oxygen's role in respiration reaches further than oxidative phosphorylation alone: without oxygen as the final electron acceptor, the electron transport chain backs up and stops entirely, so NAD and FAD cannot be reoxidised anywhere along the chain.
- Because the link reaction and Krebs cycle both depend on a continuing supply of OXIDISED NAD and FAD (to accept hydrogen as they oxidise their own substrates), both of those stages halt once the existing pool of oxidised coenzyme runs out — even though neither the link reaction nor the Krebs cycle directly uses oxygen itself. This is precisely why aerobic respiration as a whole depends on oxygen, despite oxygen being consumed only in this final stage.
- The overall ATP yield from aerobic respiration varies somewhat with cell conditions (e.g. how reduced coenzymes' hydrogen is shuttled into the mitochondrion), so quoted total ATP-per-glucose figures are typically given as an approximate range rather than one fixed exact number.
Anaerobic respiration and other respiratory substrates 3.5.2
- Animals (lactic acid fermentation): pyruvate is reduced directly to lactate, reoxidising reduced NAD back to NAD⁺.
- Yeast and many plant tissues (alcoholic fermentation): pyruvate is decarboxylated to ethanal (releasing CO₂), then ethanal is reduced to ethanol, reoxidising reduced NAD back to NAD⁺.
- Both anaerobic routes yield a net total of only about 2 ATP per glucose — from glycolysis alone — since converting pyruvate to lactate or ethanol adds no further ATP itself.
- The entire purpose of both anaerobic routes is to regenerate oxidised NAD, NOT to produce more ATP — without regenerating NAD⁺, glycolysis itself would stall (since it requires oxidised NAD as a reactant), so anaerobic respiration is really 'a way of keeping glycolysis running without oxygen', not an independent energy-yielding pathway in its own right.
- Respiratory substrates other than glucose all enter the same aerobic pathway at different points, rather than requiring wholly separate machinery: glycerol (from lipids) enters via glycolysis; fatty acids (from lipids) are converted to acetyl-CoA and enter directly into the Krebs cycle; amino acid carbon skeletons (after deamination, removing the nitrogen-containing amine group) enter as pyruvate, acetyl-CoA, or a Krebs-cycle intermediate, depending on the specific amino acid.
Worked examples
Worked example 3.5.2 · 5 marks
| Temperature (°C) | 15 | 25 | 35 |
|---|---|---|---|
| Time to decolourise (s) | 640 | 210 | 95 |
In an investigation into the effect of temperature on the rate of respiration in a yeast culture, methylene blue (a redox indicator, blue when oxidised, colourless when reduced) was added to identical yeast cultures held at 15 °C, 25 °C and 35 °C.
Time to full decolourisation was 640 s, 210 s, and 95 s respectively.
(a) Explain, in terms of respiration, why methylene blue decolourises in this experiment.
(b) Calculate the rate (as ) at each temperature, and comment on what the pattern suggests about the effect of temperature on respiration rate across this range.
Show worked solution
(a) Respiration produces reduced coenzymes as electron carriers are reduced during glycolysis, the link reaction and the Krebs cycle; methylene blue accepts electrons similarly, becoming reduced (colourless) — faster respiration decolourises it faster.
(b) Rate at 15°C ≈0.00156 s⁻¹, at 25°C ≈0.00476 s⁻¹, at 35°C ≈0.01053 s⁻¹.
The rate rises with temperature over this range, consistent with increased kinetic energy raising collision frequency.
Mark scheme · 5 marks
- States respiration produces reduced coenzymes that reduce methylene blue similarly 1 mark
- States faster respiration decolourises the indicator faster 1 mark
- Calculates rate ≈ 0.00156 s⁻¹ at 15°C 1 mark
- Calculates rate ≈ 0.00476 s⁻¹ and ≈ 0.01053 s⁻¹ at 25°C and 35°C 1 mark
- Explains the rise via increased kinetic energy and collision frequency with temperature 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Worked example 3.5.2 · 3 marks
During a short period of intense exercise, when oxygen delivery to a muscle cannot meet demand, the muscle respires anaerobically, consuming glucose at 15 mmol per minute, with lactic acid fermentation providing a net yield of 2 ATP molecules per glucose molecule from glycolysis alone.
Calculate the net rate of ATP production from this anaerobic respiration, and explain qualitatively why this rate is far lower than the muscle could achieve respiring aerobically at the same rate of glucose consumption.
Show worked solution
Net ATP production rate:
This is far lower than aerobic respiration could achieve because, without oxygen, pyruvate cannot enter the link reaction or Krebs cycle, and no reduced NAD/FAD from these stages is available to oxidative phosphorylation — by far the largest ATP-yielding stage.
Mark scheme · 3 marks
- Calculates 30 mmol ATP min⁻¹ 1 mark
- States pyruvate cannot enter the link reaction/Krebs cycle without oxygen 1 mark
- Explains oxidative phosphorylation (the largest ATP-yielding stage) is unavailable, so far less ATP is produced 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Energy and ecosystems
Energy and ecosystems 3.5.3
- Gross primary productivity (GPP): the total rate at which producers fix light energy into chemical energy. Net primary productivity (NPP): what is left over for new plant biomass or transfer to consumers, after the producers' own respiratory losses (R).
- Energy transfer between trophic levels is typically only ~10% efficient — lost as heat (respiration), in material never eaten, and in material eaten but not digested (faeces) — a hundredfold reduction across just two transfers, which is why food chains rarely extend beyond four or five levels.
- Herbivores typically lose more energy in faeces than carnivores, since plant material has a higher proportion of tough, poorly digestible material (e.g. cellulose) — successive transfers in one food chain need not have identical efficiency.
- Net production of a consumer: , where is the chemical energy store in food ingested, is chemical energy lost to the environment in faeces and urine, and is the consumer's own respiratory loss.
- . Only energy converted into an organism's own growth is available to the level above, which is also why producers can support a far greater total biomass than top predators.
- is the animal (consumer) equivalent of for a plant — both express the same underlying idea (energy actually available for growth = energy taken in, minus what is lost before it can be used for growth), but a consumer has an extra loss term () that a plant does not, since a plant does not ingest and then egest undigested material in the same way.
- Reducing the number of trophic levels between the sun and human food consumption — eating producers directly, or farming practices reducing an animal's own energy losses (limiting movement or heat loss) — increases overall food-energy yield efficiency per unit input.
- This is a strong general tendency rather than an absolute rule: some land can support grazing livestock but cannot efficiently be farmed for crops, and some nutrients are more readily obtained from animal products, so energy-transfer efficiency is not the only criterion that matters to a real food system.
Worked examples
Worked example 3.5.3 · 4 marks
A forest has a gross primary productivity of 60 000 kJ m⁻² year⁻¹.
The trees respire at a rate of 25 000 kJ m⁻² year⁻¹.
Primary consumers in the forest have 3 500 kJ m⁻² year⁻¹ available in their biomass.
Calculate the forest's net primary productivity, and the percentage efficiency of energy transfer from net primary productivity to primary consumers.
Show worked solution
Efficiency of transfer:
NPP, not GPP, is the correct denominator, since NPP is the energy actually available to pass on to consumers.
Mark scheme · 4 marks
- Calculates NPP = 35 000 kJ m⁻² year⁻¹ 2 marks
- Calculates transfer efficiency = 10% 1 mark
- Explains NPP, not GPP, is the correct denominator 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Worked example 3.5.3 · 4 marks
A population of herbivores ingests food containing 8 400 kJ m⁻² year⁻¹ of chemical energy.
Of this, 5 200 kJ m⁻² year⁻¹ is lost in faeces and urine, and a further 2 100 kJ m⁻² year⁻¹ is lost as heat through respiration.
Calculate the net production available to this population for growth and reproduction, and calculate what percentage of the originally ingested energy this represents.
Show worked solution
As a percentage of ingested energy:
Mark scheme · 4 marks
- Substitutes correctly into 1 mark
- Calculates kJ m⁻² year⁻¹ 1 mark
- Calculates the percentage as ≈13.1% 1 mark
- Comments this is broadly consistent with the ~10% rule-of-thumb transfer efficiency 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Nutrient cycles
The nitrogen cycle 3.5.4
- Saprobiotic nutrition: decomposers (saprobionts: bacteria and fungi) secrete hydrolytic enzymes onto or into dead organic matter and waste, breaking large molecules into smaller, soluble ones that are then absorbed — external digestion followed by absorption, unlike the internal digestion typical of most animals.
- Nitrogen fixation: free-living soil bacteria and root-nodule bacteria convert atmospheric N₂ into ammonia/ammonium.
- Ammonification: saprobionts decompose dead organisms and waste, releasing organic nitrogen as ammonium (NH₄⁺).
- Nitrification (aerobic nitrifying bacteria): ammonium is oxidised to nitrite (NO₂⁻), then nitrite is oxidised to nitrate (NO₃⁻) — the form most plants take up and assimilate into organic nitrogen.
- Denitrification (bacteria, anaerobic conditions): nitrate is converted back to atmospheric N₂, completing the cycle.
- The nitrogen cycle exists specifically because atmospheric N₂ is chemically very unreactive and directly unusable by almost all organisms, including plants — nitrogen fixation is the essential step converting it into a chemically usable form.
- None of this is automatic: the atmosphere's roughly 78% nitrogen content does not by itself guarantee adequate nitrogen availability to a plant's roots — soil aeration, waterlogging, leaching and the level of bacterial activity all determine how much usable nitrogen is actually present in the soil at a given time, which is exactly why nitrogen deficiency (and fertiliser use to correct it) are both common in practice.
- Ammonification feeds directly into nitrification, which is what keeps the cycle genuinely cyclical — without decomposition releasing nitrogen from dead organic matter, nitrogen would simply accumulate one-way, locked in dead biomass.
- The carbon cycle works on the same general principle of continuous cycling between atmosphere, living organisms and (over long timescales) geological stores: CO₂ is fixed by photosynthesis, released again by respiration, and can be locked away for very long periods as fossil fuels before combustion returns it to the atmosphere — decomposition plays the same essential 'unlocking' role for carbon as it does for nitrogen, releasing it from dead organic matter for recycling rather than leaving it permanently sequestered.
The phosphorus cycle 3.5.4
- Unlike the nitrogen and carbon cycles, the phosphorus cycle has no major gaseous reservoir — phosphorus moves between rock, soil solution, organisms, and sediment, but essentially never through the atmosphere.
- Phosphate is released from rock by weathering into soil solution, taken up by plant roots, passed to animals by feeding, and returned to the soil (as organic phosphorus) by death and decomposition — the same organism-to-decomposer-to-inorganic-form pattern as the nitrogen cycle, run by saprobionts in both cases.
- Phosphate can also be lost from the readily-available cycle for very long periods: runoff and drainage carry it to rivers and seas, where sedimentation and burial can lock it into sedimentary rock, only returning to active cycling much later via geological uplift and weathering — phosphorus cycling operates on both a fast (biological) and an extremely slow (geological) timescale simultaneously.
Saprobionts and mycorrhizae 3.5.4
- Mycorrhiza: a mutualistic association between a plant's roots and a fungus, in which the fungus's hyphae extend into the soil and the plant's roots together form one functional unit.
- Saprobionts (bacteria and fungi) secrete extracellular hydrolytic enzymes directly onto dead organic matter, breaking down large molecules into small, soluble products which the saprobiont then absorbs — releasing NH₄⁺ and phosphate into the surrounding soil as a by-product, exactly the ammonification and mineralisation steps that keep the nitrogen and phosphorus cycles running.
- In a mycorrhizal association, the fungus receives organic carbon from the plant (a product of the plant's own photosynthesis), while the plant receives water and inorganic ions (especially phosphate) via the fungal hyphae — a genuine two-way, mutually beneficial exchange, not one organism simply parasitising the other.
- Fungal hyphae are far finer and more extensively branching than plant roots alone, so a mycorrhizal association dramatically increases the effective surface area available for water and ion uptake — the practical reason many plants form these associations at all.
Fertilisers and eutrophication 3.5.4
- Eutrophication: the process by which excess nutrients (especially nitrate and phosphate) entering a body of water lead to algal overgrowth and, ultimately, oxygen depletion.
- Fertilisers (natural, e.g. manure, or artificial, e.g. mineral salts) replace nitrogen and phosphorus removed from soil by harvesting crops or livestock.
- Eutrophication sequence: nutrient enrichment (runoff/leaching carrying nitrate and phosphate into water) → algal bloom → reduced light reaching submerged plants, which then die → decomposition of dead algae and plants increases microbial (saprobiont) respiration → dissolved oxygen falls → hypoxia, in which aquatic animals may die from lack of oxygen.
- Fertiliser use and eutrophication are directly connected consequences of the same nutrient cycles: fertilisers are applied specifically because harvesting removes nitrogen and phosphorus from the natural cycle faster than natural processes replace them, but any excess applied (beyond what crops actually take up) can run off or leach into watercourses instead.
- The chain of harm in eutrophication is indirect: it is not the excess nutrient itself that kills aquatic animals, but the resulting oxygen depletion, driven by decomposer respiration after the algae and shaded-out plants die — the nutrient enrichment is the root cause, but oxygen depletion is the proximate cause of death.
- This is exactly the same saprobiont-driven decomposition process covered above, simply occurring at a scale and rate that overwhelms the water body's normal oxygen supply, rather than the steady, small-scale decomposition that keeps a nutrient cycle in balance under normal conditions.
Worked examples
Worked example 3.5.4 · 5 marks
A gardener adds fresh manure, containing organic nitrogen compounds largely as proteins, to a well-aerated garden bed.
Trace the sequence of processes, and name the bacteria responsible for each, by which the nitrogen in this manure could eventually become available to a growing plant as nitrate, and explain what would instead happen to that nitrate if the bed later became waterlogged.
Show worked solution
Decomposers break down proteins, releasing ammonia — ammonification.
Nitrifying bacteria oxidise ammonium to nitrite, then to nitrate — nitrification, requiring aerobic conditions.
The nitrate becomes available to plant roots.
If waterlogged, anaerobic conditions favour denitrifying bacteria, converting nitrate back to atmospheric N₂, removing it from the soil.
Mark scheme · 5 marks
- States ammonification: decomposers release ammonia from organic nitrogen 1 mark
- States nitrification: ammonium oxidised to nitrite, then to nitrate 1 mark
- States nitrification requires aerobic conditions 1 mark
- States waterlogging favours denitrifying bacteria 1 mark
- States denitrification converts nitrate back to atmospheric N₂, removing it from the soil 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.
Worked example 3.5.4 · 5 marks
A lake receives runoff from surrounding farmland that has been heavily treated with nitrate and phosphate fertiliser.
Over the following months, the lake develops a dense algal bloom on its surface, and shortly afterward a large number of fish are found dead.
Explain the full sequence of events linking the fertiliser runoff to the fish deaths, making clear that it is not the nutrients themselves that directly kill the fish.
Show worked solution
Excess nitrate/phosphate cause explosive algal growth (a bloom).
Dense surface growth blocks light from reaching deeper plants/algae, which die.
Decomposers break down this dead organic matter in large numbers, and their aerobic respiration depletes dissolved oxygen.
It is this fall in dissolved oxygen, not the nutrients directly, that kills the fish.
Mark scheme · 5 marks
- States excess nutrients cause an algal bloom 1 mark
- States the bloom blocks light, killing deeper plants/algae 1 mark
- States decomposers break down the resulting dead organic matter 1 mark
- States decomposer respiration depletes dissolved oxygen 1 mark
- States it is oxygen depletion, not the nutrients directly, that kills the fish 1 mark
Do not count matching words alone — ask whether your answer actually makes the same claim.