BiologyISC Class 11 · Photosynthesis in Higher Plants

Calvin Cycle, C4 Pathway, Photorespiration and Factors Affecting Photosynthesis

Updated 17 October 2026 · Fact-checked

The Calvin (C3) cycle in the stroma fixes CO₂ using RuBisCO (carboxylation), reduces 3-PGA to sugar using ATP and NADPH, and regenerates RuBP; each glucose needs 6 turns, 18 ATP and 12 NADPH. C4 plants (Kranz anatomy) first fix CO₂ into oxaloacetate with PEP carboxylase, avoiding photorespiration.

Follow the arrows: CO₂ is fixed by RuBisCO, 3-PGA is reduced to G3P, and five of every six G3P rebuild RuBP, so only one leaves as sugar.

Illustration drawn for QuestionBuddy.

Understand Calvin Cycle, C4 Pathway, Photorespiration and Limiting Factors

Calvin cycle (C3 pathway). Occurs in the stroma. (1) *Carboxylation:* RuBP (5C) + CO₂ → 2 molecules of 3-phosphoglyceric acid (3-PGA, 3C), catalysed by RuBisCO. (2) *Reduction:* 3-PGA is phosphorylated by ATP and reduced by NADPH to glyceraldehyde-3-phosphate (G3P or PGAL). (3) *Regeneration:* five of every six G3P regenerate RuBP using ATP. For one CO₂ fixed the cycle uses 3 ATP and 2 NADPH; for one glucose (6 CO₂) it needs 18 ATP and 12 NADPH. RuBisCO is the most abundant enzyme on the earth.

C4 pathway (Hatch and Slack). In plants adapted to dry tropical climates (maize, sugarcane, sorghum), leaves show Kranz anatomy: large bundle sheath cells in a wreath around the vascular bundles with chloroplasts. In the mesophyll cells CO₂ is fixed by PEP carboxylase (PEPcase) to oxaloacetate (4C), converted to malate or aspartate, which moves to the bundle sheath, where it is decarboxylated to release CO₂ for the Calvin cycle by RuBisCO. Advantages: no photorespiration, higher productivity at high temperature and light; the optimum temperature is 30 to 45 °C (C3 plants: 20 to 25 °C). PEPcase has no affinity for oxygen.

Photorespiration. In C3 plants at high O₂ and low CO₂, RuBisCO acts as an oxygenase: RuBP + O₂ → one 3-PGA and one phosphoglycolate. The process spans chloroplast, peroxisome and mitochondrion, releases CO₂, and uses ATP without making ATP or NADPH, so it is wasteful.

Limiting factors. Blackman's law: if a process is controlled by several factors, its rate is limited by the factor nearest its minimal value. Factors: light (intensity, quality, duration), CO₂ concentration (about 0.03 to 0.04% in air), temperature, and water. C3 plants respond to increased CO₂ up to a limit, while C4 plants saturate at lower levels.

Calvin Cycle, C4 Pathway, Photorespiration and Limiting Factors: key facts

Calvin cycle: Carboxylation RuBP + CO₂ → 2 × 3-PGA
Reduction 3-PGA + ATP + NADPH → G3P
Regeneration G3P + ATP → RuBP (5 of every 6 G3P are used); RuBisCO catalyses carboxylation.
Cost per CO₂ and per glucose: Per CO₂: 3 ATP + 2 NADPH | Per glucose: 18 ATP + 12 NADPH; Six turns of the cycle.
C4 plants: Mesophyll: PEP + CO₂ → OAA (PEPcase) | Bundle sheath: CO₂ + RuBP (RuBisCO); Maize, sugarcane, sorghum.
Photorespiration: $R u B P + O_{2} \to 3 - P G A + p h os p h o g l y co l a t e$; No ATP or NADPH made; CO₂ released.

Carboxylation	RuBP + CO₂ → 2 × 3-PGA
Reduction	3-PGA + ATP + NADPH → G3P
Regeneration	G3P + ATP → RuBP (5 of every 6 G3P are used)

Calvin Cycle, C4 Pathway, Photorespiration and Limiting Factors in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Why sugarcane grows well in hot sun

Suppose a farmer notices that sugarcane thrives in hot, bright conditions where wheat suffers.

Explain with reference to C4 photosynthesis.

Try it first, then show the solutionSolution

Sugarcane is a C4 plant: PEP carboxylase in mesophyll cells fixes CO₂ efficiently even when stomata are partly closed.
The CO₂ is concentrated in bundle sheath cells, so RuBisCO works as a carboxylase and photorespiration is nearly absent.
Its optimum temperature is higher (30 to 45 °C), unlike wheat, a C3 plant (20 to 25 °C).

Answer: C4 pathway and Kranz anatomy suit hot, bright conditions.

Memory hook: CO₂ concentration in the bundle sheath prevents photorespiration.

Tricks to remember Calvin Cycle, C4 Pathway, Photorespiration and Limiting Factors

Short memory aids. Say each one aloud once, then use it on a question the same day.

5 + 1 = 2 × 3 for carbons
RuBP (5C) + CO₂ (1C) → two 3-PGA (3C each): 5 + 1 = 2 × 3.
Use it when: Carboxylation.
3 and 2 per CO₂
Three ATP and two NADPH for each CO₂ fixed.
Use it when: Energy accounting.

Calvin Cycle, C4 Pathway, Photorespiration and Limiting Factors: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: First stable product · 1 mark · MCQBoard level

The first stable product of CO₂ fixation in the C3 cycle is:
(a) oxaloacetic acid
(b) 3-phosphoglyceric acid
(c) RuBP
(d) glucose

Try it first, then show the solutionSolution

RuBP + CO₂ gives two molecules of 3-PGA.

Answer: (b) 3-phosphoglyceric acid

Example 2: Steps of the Calvin cycle · 3 marks · Short answerBoard level

Name the three stages of the Calvin cycle with the enzyme of the first stage and the energy inputs.

Try it first, then show the solutionSolution

Carboxylation: RuBP + CO₂ → 3-PGA, catalysed by RuBisCO. (1 mark)
Reduction: 3-PGA is converted to G3P using ATP and NADPH. (1 mark)
Regeneration of RuBP from G3P, using ATP. (1 mark)

Answer: Carboxylation, reduction, regeneration.

Example 3: Energy accounting · 4 marks · Short answerHarder board level

Calculate the ATP and NADPH needed to make one glucose molecule in the Calvin cycle and the number of oxygen molecules released through the light reactions that supply them.

Try it first, then show the solutionSolution

Per CO₂: 3 ATP and 2 NADPH; for 6 CO₂: 18 ATP and 12 NADPH. (1.5 marks)
Making 12 NADPH requires 24 electrons from water. (1 mark)
Each O₂ comes from 4 electrons: 24/4 = 6 O₂. (1 mark)
This agrees with 6CO₂ + 12H₂O → C₆H₁₂O₆ + 6O₂ + 6H₂O. (0.5 mark)

Answer: 18 ATP, 12 NADPH; 6 molecules of O₂.

Example 4: C3 and C4 plants · 4 marks · Short answerHarder board level

Differentiate C3 and C4 plants on the first product, enzyme, leaf anatomy, photorespiration and optimum temperature.

Try it first, then show the solutionSolution

First product: 3-PGA (C3); oxaloacetic acid (C4). (0.8 mark)
Enzyme for fixation: RuBisCO (C3); PEPcase in mesophyll (C4). (0.8 mark)
Anatomy: normal mesophyll in C3; Kranz anatomy in C4. (0.8 mark)
Photorespiration: present in C3; negligible in C4. (0.8 mark)
Optimum temperature: 20 to 25 °C (C3); 30 to 45 °C (C4). (0.8 mark)

Answer: Five points of difference.

Example 5: Photorespiration and limiting factors · 5 marks · Long answerHOTS / competency

Explain photorespiration, why it is wasteful, and why it does not occur in C4 plants. A plant is in strong light with ample water but at a low CO₂ concentration. Name the limiting factor by Blackman's law and state what would happen if the light were increased further.

Try it first, then show the solutionSolution

RuBisCO fixes O₂ instead of CO₂ at high O₂ and low CO₂, giving one 3-PGA and one phosphoglycolate, which is metabolised through peroxisome and mitochondrion, releasing CO₂. (1.5 marks)
It is wasteful because no ATP or NADPH is made and carbon is lost. (1 mark)
C4 plants concentrate CO₂ in the bundle sheath, where RuBisCO acts only as a carboxylase. (1 mark)
The limiting factor is CO₂ (the factor nearest its minimum). (1 mark)
Increasing the light would not increase the rate. (0.5 mark)

Answer: Photorespiration is wasteful; CO₂ is limiting, so more light would not help.

Where students lose marks in Calvin Cycle, C4 Pathway, Photorespiration and Limiting Factors

The slips examiners see most often, and what to write instead.

Mistake: Mixing the enzymes
Giving RuBisCO for mesophyll cells in C4 plants.
Do this: PEP carboxylase in mesophyll; RuBisCO in bundle sheath cells.
Usually costs: 1 to 2 marks.
Mistake: Wrong ATP count
Using 2 ATP per CO₂.
Do this: The cycle needs 3 ATP and 2 NADPH per CO₂.
Usually costs: 1 mark.
Mistake: Calling photorespiration useful
Missing that it wastes energy.
Do this: It consumes energy and releases CO₂ without making sugar.
Usually costs: 1 mark.

Exam tips

Draw the Calvin cycle with carbon numbers.
Compare C3 and C4 in a table.
State the stoichiometry per CO₂ and per glucose.
Use Blackman's law in factor questions.

Practice questions on Calvin Cycle, C4 Pathway, Photorespiration and Limiting Factors

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The enzyme that fixes CO₂ in the mesophyll cells of C4 plants is:
(a) RuBisCO
(b) PEP carboxylase
(c) catalase
(d) hexokinase
Hint
It has no affinity for oxygen.
Answer
(b) PEP carboxylase
Q2 · 2 marks · Short answerBoard level
What is Kranz anatomy?
Hint
Wreath-like arrangement.
Answer
The arrangement of large chloroplast-containing bundle sheath cells around the vascular bundles in a wreath, found in C4 leaves.
Q3 · 3 marks · Short answerHarder board level
How many ATP and NADPH molecules are needed for fixing 3 CO₂ into one triose phosphate?
Hint
Per CO₂: 3 ATP and 2 NADPH.
Answer
9 ATP and 6 NADPH.
Q4 · 3 marks · Short answerHarder board level
State Blackman's law of limiting factors with an example.
Hint
It deals with a factor at minimum.
Answer
When a process depends on several factors, its rate is limited by the factor present nearest its minimum value. At low CO₂ the rate cannot be raised by increasing light.
Q5 · 3 marks · Assertion-reasonHarder board level
Assertion (A): C4 plants show little photorespiration. Reason (R): They concentrate CO₂ around RuBisCO in the bundle sheath cells. Choose: (a) both true and R explains A (b) both true, R does not explain A (c) A true, R false (d) A false, R true.
Hint
Think about the CO₂ to O₂ ratio.
Answer
(a) Both true and R explains A.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student compares maize and wheat leaves in a hot afternoon.
(i) Name the type of pathway in each. (1 mark)
(ii) Name the first product in maize. (1 mark)
(iii) Which has Kranz anatomy? (1 mark)
(iv) Which shows higher photorespiration? (1 mark)
Hint
Maize is C4 and wheat C3.
Answer
(i) C4 in maize; C3 in wheat. (ii) Oxaloacetic acid. (iii) Maize. (iv) Wheat.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Enzyme for carboxylation in C3 plants?Show answer
RuBisCO.
2.C4 pathway discovered by?Show answer
Hatch and Slack.
3.ATP per glucose in the Calvin cycle?Show answer
18.
4.First product in C4 plants?Show answer
Oxaloacetic acid.

Calvin Cycle, C4 Pathway, Photorespiration and Limiting Factors: frequently asked questions

Why is the Calvin cycle called the dark reaction?

It does not need light directly, though it depends on the ATP and NADPH made in the light and runs mainly in the day.

Why do C4 plants do well at high temperatures?

Because they avoid photorespiration, which rises with temperature in C3 plants.

What is the role of PEPcase?

It fixes CO₂ into the four-carbon acid oxaloacetate in the mesophyll and has no affinity for O₂.