Photosynthesis in Higher Plants — the Z-Scheme, Calvin Cycle and C₄ Advantage
The Calvin Cycle and Its Cost
| Phase | What happens | Cost per CO₂ |
|---|---|---|
| Carboxylation | RuBP (5C) + CO₂ → two molecules of 3-PGA, catalysed by RuBisCO — the most crucial step | No ATP or NADPH |
| Reduction | 3-PGA is reduced to sugar | 2 ATP + 2 NADPH |
| Regeneration | RuBP is rebuilt so the cycle can continue | 1 ATP |
| Totals | For one glucose, six turns of the cycle are needed | 18 ATP + 12 NADPH per glucose; 3 ATP + 2 NADPH per CO₂ |
| Feature | C₃ plants | C₄ plants |
|---|---|---|
| First CO₂ acceptor | RuBP (5 carbons) | PEP (3 carbons), by PEP carboxylase |
| First stable product | 3-PGA (3 carbons) | OAA (4 carbons) — hence the name |
| Anatomy | Ordinary mesophyll; no special sheath | Kranz anatomy — large, chloroplast-rich bundle sheath cells |
| Where RuBisCO acts | In the mesophyll, exposed to atmospheric O₂ | In the bundle sheath, where CO₂ is concentrated |
| Photorespiration | Significant, and worse in heat | Absent — that is the whole point of the pathway |
| Examples and optimum | Rice, wheat, most temperate crops | Maize, sugarcane, sorghum, *Amaranthus* — higher temperature optimum, higher productivity |
Key takeaways
- Light reactions on the thylakoid give ATP, NADPH and O₂ only; the Calvin cycle in the stroma spends them to fix CO₂.
- Van Niel showed O₂ comes from water, not CO₂; Engelmann gave the first action spectrum, peaking in blue and red.
- Chlorophyll a is the reaction centre; the rest are accessory pigments that widen the spectrum and protect it. P680 and P700 are both chlorophyll a.
- Cyclic photophosphorylation uses PS I only, in stroma lamellae, and makes ATP alone — no NADPH, no O₂.
- Protons accumulate in the thylakoid lumen and flow out through ATP synthase, making ATP in the stroma.
- Calvin cycle costs 3 ATP + 2 NADPH per CO₂, hence 18 ATP + 12 NADPH per glucose.
- Photorespiration makes no sugar, no ATP and no NADPH — it only consumes ATP and releases CO₂.
- C₄ concentrates CO₂ at RuBisCO via PEP carboxylase and Kranz anatomy, abolishing photorespiration at a cost of ~5 ATP per CO₂. Light saturates at ~10% of sunlight; CO₂ is the real limiting factor.
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Which of these is produced during photorespiration?
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Answer: A — None of sugar, ATP or NADPH — only phosphoglycolate, and CO₂ downstream
Photorespiration is pure loss: RuBisCO binds O₂ instead of CO₂, giving one 3-PGA and one two-carbon phosphoglycolate, with no sugar, no ATP and no NADPH produced and ATP actually consumed. Every distractor here offers one of the three products that the process specifically fails to make.Cyclic photophosphorylation differs from the non-cyclic form in that it:
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Answer: A — Involves PS I only and produces ATP without NADPH or oxygen
It runs on PS I alone in the stroma lamellae, which lack PS II and NADP reductase, so no water is split and no NADPH forms — only ATP. Producing oxygen requires PS II and the photolysis of water, which is exactly what the cyclic pathway does not include.The first stable product of CO₂ fixation in a C₄ plant such as maize is:
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Answer: A — Oxaloacetic acid, a four-carbon acid
PEP carboxylase adds CO₂ to the three-carbon PEP to give the four-carbon OAA, which is where the name C₄ comes from. 3-PGA is the C₃ answer, RuBP is the acceptor rather than a product, and phosphoglycolate belongs to photorespiration — which C₄ plants avoid altogether.Van Niel's work with purple and green sulphur bacteria established that:
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Answer: A — The oxygen released in photosynthesis comes from water, not from CO₂
Those bacteria use H₂S as the hydrogen donor and deposit sulphur instead of releasing oxygen, showing that the donor determines the by-product — so in plants, where water is the donor, the O₂ must come from water. Light being essential and only green parts working were Ingenhousz's conclusions, and starch was von Sachs'.How many ATP and NADPH molecules does the Calvin cycle consume to produce one molecule of glucose?
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Answer: A — 18 ATP and 12 NADPH
Each CO₂ costs 3 ATP and 2 NADPH, and six CO₂ are needed for one glucose, giving 18 and 12. The third option is the per-CO₂ figure rather than the per-glucose one, and the second simply swaps the two numbers — both are traps for a half-remembered ratio.In a C₄ plant, the mesophyll cells characteristically lack:
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Answer: A — RuBisCO
The division of labour is strict: mesophyll cells carry PEP carboxylase and no RuBisCO, while bundle sheath cells carry RuBisCO and no PEP carboxylase. That is precisely what keeps RuBisCO away from atmospheric oxygen and prevents photorespiration, so reversing the two enzymes destroys the whole mechanism.During photophosphorylation, protons accumulate in the:
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Answer: A — Thylakoid lumen, making it acidic
Water is split on the inner face of the thylakoid membrane and the carriers pump H⁺ inward, so the lumen turns acidic and the stroma alkaline; protons then return through ATP synthase, making ATP in the stroma where the Calvin cycle needs it. Option two reverses the gradient, which would put the ATP in the wrong compartment.For a leaf in the open on a clear day, the factor most likely to be limiting photosynthesis is:
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Answer: A — Carbon dioxide concentration
Light saturates at roughly 10% of full sunlight, so on a clear day it is well above the limiting level, whereas CO₂ is only about 0.03–0.04% of air — which is why enriching CO₂ in a greenhouse raises yield. By Blackman's law the rate is set by whichever factor is nearest its minimum, and that is almost always CO₂ in the open.