Photosynthesis in Higher Plants ISC Class 11: Light Reactions, C3 and C4

What Photosynthesis in Higher Plants is about

Weightage: Photosynthesis in Higher Plants belongs to the unit Plant Physiology, which CISCE weights at 17 of the 70 marks in the ISC Class XI Biology Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Biology syllabus, examination year 2028, Class XI unit weightage

Key concepts

Site of photosynthesis and photosynthetic pigments

Photosynthesis takes place in chloroplasts, mainly in the mesophyll cells of leaves; chloroplasts align with their flat surfaces towards the light. The grana (stacks of thylakoids) trap light and make ATP and NADPH; the stroma contains the enzymes that fix CO₂ into sugar, so the reactions there do not directly need light (though they depend on the products of the light reactions). Pigments (elementary idea): chlorophyll a (bright or blue-green in a chromatogram), the chief pigment, present at the reaction centre; chlorophyll b (yellow-green); xanthophylls (yellow); carotenoids, here meaning carotenes (yellow to yellow-orange). Chlorophyll b, xanthophylls and carotenoids are accessory pigments: they absorb wavelengths that chlorophyll a does not and pass the energy to chlorophyll a, so they widen the range of usable light, and they also protect chlorophyll a from photo-oxidation. These pigments can be separated by paper chromatography.

Worked example

Why do leaves contain several pigments instead of only chlorophyll a?

Try it first, then show the solutionSolution

1. Chlorophyll a absorbs mainly blue and red light; much of the spectrum would be wasted
2. Accessory pigments absorb other wavelengths and transfer the energy to chlorophyll a, and they protect it from damage by strong light

Answer: Accessory pigments extend the usable light range and protect chlorophyll a

Absorption spectrum, action spectrum and photosystems

An absorption spectrum is a graph of the amount of light absorbed by a pigment against wavelength. Chlorophyll a absorbs most strongly in the blue (around 430 nm) and red (around 660 nm) regions and very little green, which is reflected, so leaves look green. An action spectrum is a graph of the rate of photosynthesis against wavelength. The action spectrum has peaks in the blue and red regions and closely matches the absorption spectrum of chlorophyll a, showing that chlorophyll a is the main pigment driving photosynthesis; the match is not perfect because accessory pigments also contribute. Pigment systems: pigments are organised into two light-harvesting complexes called photosystem I (PS I) and photosystem II (PS II) (named in order of discovery, not of action). Each has an antenna of hundreds of pigment molecules bound to proteins and a reaction centre of one special chlorophyll a molecule. In PS I the reaction centre absorbs best at 700 nm (P700); in PS II at 680 nm (P680).

Worked example

If a plant is lit only with green light of the same intensity as red light, which will give a higher rate of photosynthesis, and why?

Try it first, then show the solutionSolution

1. The action spectrum shows high rates in red and blue light and low rates in green
2. Chlorophylls absorb red strongly but reflect most green light

Answer: Red light, because it is absorbed and used by chlorophyll; green is largely reflected

Light reactions: non-cyclic photophosphorylation (Z scheme)

The photochemical phase includes light absorption, splitting of water, oxygen release, and formation of ATP and NADPH. In non-cyclic photophosphorylation both photosystems work in series. (1) Light excites P680 in PS II; it loses electrons to a primary acceptor. (2) The electrons pass down an electron transport chain (plastoquinone, cytochrome b₆f complex, plastocyanin) to PS I; energy released along the way is used to pump protons. (3) Light excites P700 in PS I; its electrons go via another acceptor and ferredoxin to NADP⁺ reductase, which reduces NADP⁺ to NADPH + H⁺. (4) PS II replaces its lost electrons by splitting water (photolysis) on the lumen side of the thylakoid: $2H_{2}O\to 4H^{+}+O_2+4e^{-}$. The O₂ released comes from water, not CO₂. Because the electron path, drawn on an energy scale, looks like a letter Z, this is called the Z scheme. Products: ATP, NADPH and O₂. The electrons do not return to their starting point, hence 'non-cyclic'.

Worked example

How many molecules of water must be split to release six molecules of O₂, and how many electrons are released?

Try it first, then show the solutionSolution

1. $2H_{2}O\to 4H^{+}+O_2+4e^{-}$: two water molecules per O₂
2. For 6 O₂: 12 H₂O split, releasing 6 × 4 = 24 electrons

Answer: 12 water molecules; 24 electrons

Cyclic photophosphorylation and the chemiosmotic hypothesis

In cyclic photophosphorylation only PS I works. Excited electrons from P700 pass to an acceptor and then back to PS I through the electron transport chain, so they cycle. Along the way protons are pumped and ATP is made, but no NADPH and no O₂ are formed (water is not split). It probably occurs in the stroma lamellae, which lack PS II and NADP reductase, and when only light of wavelengths beyond 680 nm is available. It supplies the extra ATP the Calvin cycle needs. Chemiosmotic hypothesis (Peter Mitchell): ATP synthesis is linked to a proton gradient across the thylakoid membrane. Protons build up in the lumen because (1) water is split on the lumen side, releasing H⁺; (2) as electrons move through the chain, plastoquinone carries protons from the stroma into the lumen; (3) NADP reductase on the stroma side uses H⁺ from the stroma to form NADPH, lowering stromal H⁺. The lumen therefore becomes acidic. Protons flow back to the stroma down this gradient only through the CF₀ channel of ATP synthase; this flow changes the shape of the CF₁ head (facing the stroma), which joins ADP and Pi into ATP. So ATP is made in the stroma, exactly where the Calvin cycle needs it.

Worked example

Give two differences between cyclic and non-cyclic photophosphorylation in terms of products.

Try it first, then show the solutionSolution

1. Non-cyclic forms ATP, NADPH and O₂
2. Cyclic forms only ATP; no NADPH, and no O₂ because water is not split

Answer: Cyclic: ATP only; non-cyclic: ATP, NADPH and O₂

Biosynthetic phase: the Calvin (C3) cycle

Melvin Calvin traced the path of carbon using radioactive ¹⁴CO₂ in algae. In C3 plants the first stable product is the 3-carbon 3-phosphoglycerate (PGA). The cycle runs in the stroma in three stages (graphic outline): (1) Carboxylation: CO₂ combines with the 5-carbon acceptor ribulose-1,5-bisphosphate (RuBP), catalysed by RuBisCO (RuBP carboxylase–oxygenase), forming two molecules of PGA. (2) Reduction (glycolytic reversal): PGA is converted to the triose sugar glyceraldehyde-3-phosphate (G3P) using 2 ATP and 2 NADPH per CO₂ fixed; these steps reverse part of glycolysis. Some triose phosphate leaves the cycle to make glucose, sucrose and starch. (3) Regeneration: the remaining triose phosphate is rearranged to re-form RuBP, using 1 ATP per CO₂. Totals: each CO₂ fixed needs 3 ATP and 2 NADPH; one glucose (6 CO₂) needs 6 turns, 18 ATP and 12 NADPH. The extra ATP beyond the 3:2 ratio of non-cyclic flow comes partly from cyclic photophosphorylation.

Worked example

How many ATP and NADPH molecules are needed to make 3 molecules of glucose by the Calvin cycle?

Try it first, then show the solutionSolution

1. One glucose needs 6 CO₂: 18 ATP and 12 NADPH
2. Three glucose: 3 × 18 = 54 ATP and 3 × 12 = 36 NADPH

Answer: 54 ATP and 36 NADPH

C4 (Hatch–Slack) pathway and Kranz anatomy

C4 plants (maize, sugarcane, sorghum, *Amaranthus*) are adapted to hot, dry, bright conditions. Kranz anatomy ('wreath'): the vascular bundles are surrounded by large bundle sheath cells arranged like a wreath; these cells have many large chloroplasts, thick walls that are nearly impermeable to gases, and no intercellular spaces. Hatch–Slack pathway: (1) In mesophyll cells, the CO₂ acceptor phosphoenolpyruvate (PEP) (3C) combines with CO₂ (as bicarbonate) under the enzyme PEP carboxylase (PEPcase), which has no affinity for O₂, forming the 4-carbon oxaloacetic acid (OAA), the first stable product. (2) OAA is converted to other 4-carbon acids such as malic acid or aspartic acid, which move into bundle sheath cells. (3) There they are decarboxylated, releasing CO₂ and a 3-carbon acid (pyruvate). (4) The released CO₂ enters the Calvin cycle, which runs in bundle sheath cells (they are rich in RuBisCO; mesophyll cells lack it). (5) The 3-carbon molecule returns to the mesophyll and is converted back to PEP, using ATP. Because CO₂ is pumped into the bundle sheath, its concentration around RuBisCO is high, so photorespiration is almost absent. The cost is 2 extra ATP per CO₂, i.e. 30 ATP per glucose instead of 18.

Worked example

Why is RuBisCO placed only in the bundle sheath cells of C4 plants and not in the mesophyll?

Try it first, then show the solutionSolution

1. Mesophyll cells are exposed to air spaces with normal O₂ levels, where RuBisCO would also bind O₂
2. Bundle sheath cells are gas-tight and receive concentrated CO₂ from 4-carbon acids, so RuBisCO works mainly as a carboxylase

Answer: To keep RuBisCO in a high-CO₂, low-O₂ space and avoid photorespiration

Photorespiration: RuBisCO as oxygenase

RuBisCO can bind both CO₂ and O₂ at the same active site; which one wins depends on their relative concentrations. Its affinity for CO₂ is much higher, but when O₂ is high and CO₂ low (e.g. hot, dry days with closed stomata), it acts as an oxygenase. Pathway (in brief): in the chloroplast, RuBP + O₂ gives one PGA and one 2-carbon phosphoglycolate; glycolate passes to the peroxisome, where it is converted to glycine; in the mitochondrion, two glycines give serine and CO₂ is released; the products return through the peroxisome to the chloroplast as PGA. Significance: it is a wasteful process: no sugar, no ATP and no NADPH are made; instead ATP is used and fixed carbon is lost as CO₂, lowering net photosynthesis in C3 plants. It is absent or negligible in C4 plants. A possible benefit is that it uses up excess energy and reducing power in strong light when CO₂ is short, which may reduce damage to the photosynthetic machinery.

Worked example

On a hot, dry afternoon a wheat plant closes its stomata. Explain why its net photosynthesis falls more than that of maize.

Try it first, then show the solutionSolution

1. Closed stomata cause CO₂ inside the leaf to fall and O₂ from photolysis to build up
2. In wheat (C3), RuBisCO then acts as an oxygenase, causing photorespiration and CO₂ loss; maize (C4) concentrates CO₂ in the bundle sheath and avoids this

Answer: Photorespiration rises in the C3 plant but is suppressed in the C4 plant

Factors affecting photosynthesis and Blackman's law

Blackman's law of limiting factors (1905): when a chemical process is affected by more than one factor, its rate is determined by the factor that is nearest its minimal value; only a change in that factor alters the rate directly. In other words, the factor in shortest supply limits the rate, and increasing the others has no effect until it is raised. Light: rate rises linearly with intensity at low light; light saturation is reached at about 10% of full sunlight, so, except for shaded plants, light is rarely limiting in nature; very strong light can damage chlorophyll. CO₂: at about 0.03–0.04% of air, CO₂ is often the major limiting factor. C3 plants respond to higher CO₂ concentrations with increased rates, so greenhouse crops such as tomatoes and bell peppers are grown in CO₂-enriched air; C4 plants saturate at lower CO₂ (about 360 µL L⁻¹). Temperature: the biosynthetic phase is enzyme-controlled, so it is more temperature-sensitive than the light phase; C4 plants have a higher optimum (about 30–40 °C) than C3 plants. Water: its effect is mainly indirect: water stress closes stomata (cutting CO₂ supply) and makes leaves wilt, reducing surface area and metabolic activity. Internal factors include the number, size, age and orientation of leaves, mesophyll cells, chloroplasts and the internal CO₂ concentration.

Worked example

A plant in dim light shows no increase in photosynthesis when CO₂ is doubled, but responds when light is increased. Explain using Blackman's law.

Try it first, then show the solutionSolution

1. In dim light, light is the factor nearest its minimum and so limits the rate
2. Raising CO₂ (not limiting) has no effect; raising light removes the limit until another factor (CO₂ or temperature) becomes limiting

Answer: Light is the limiting factor

Photosynthesis in Higher Plants in real life

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

Topics in this chapter

• Calvin Cycle, C4 Pathway, Photorespiration and Limiting Factors
• Photosynthetic Pigments and Light Reactions

Photosynthesis in Higher Plants: key facts

Overall equation

$6C{O}_2+12H_{2}O\underset{\text{chlorophyll}}{\overset{\text{light}}{\to }}{C}_6{H}_{12}{O}_6+6H_{2}O+6O_2$

O₂ comes from water.

Photolysis of water

$2H_{2}O\to 4H^{+}+O_2+4e^{-}$

Linked to PS II, on the lumen side.

Calvin cycle cost

$\text{per }C{O}_2:3\text{ATP}+2\text{NADPH};\text{per glucose}:18\text{ATP}+12\text{NADPH}$

Six turns per glucose.

C4 cost

$\text{per glucose}:18+12=30\text{ATP};12\text{NADPH}$

2 extra ATP per CO₂ to regenerate PEP.

First steps of fixation

$C_3:RuBP+C{O}_2\stackrel{\text{RuBisCO}}{\to }2PGA;C_4:PEP+C{O}_2\stackrel{\text{PEPcase}}{\to }OAA$

First stable products: PGA (3C) and OAA (4C).

Photosynthesis in Higher Plants: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Photosynthesis in Higher Plants

The question patterns to expect, and the method that works for each.

Tricks to remember Photosynthesis in Higher Plants

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

• Two before one

  Electrons go PS II first, then PS I: the systems are numbered by discovery, not by order of action.

  Use it when: Z scheme.

• Cyclic = Cheap (ATP only)

  Cyclic gives only ATP; Non-cyclic gives ATP, NADPH and O₂.

  Use it when: Photophosphorylation types.

• CRR: Carboxylate, Reduce, Regenerate

  The three stages of the Calvin cycle.

  Use it when: C3 cycle.

• 3-2 per CO₂, 18-12 per glucose

  3 ATP + 2 NADPH per CO₂; ×6 for glucose. C4 adds 2 ATP per CO₂ (30 total).

  Use it when: Energy accounting.

• PEP picks, RuBisCO fixes

  In C4, PEP carboxylase picks up CO₂ in mesophyll; RuBisCO fixes it in bundle sheath.

  Use it when: C4 pathway.

Photosynthesis in Higher Plants: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Photosynthesis in Higher Plants

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

• ✗Mistake: Saying O₂ comes from CO₂

  The equation is read as CO₂ → O₂.

  Wrong working

  The oxygen given off comes from carbon dioxide.

  Right working

  It comes from photolysis of water; that is why 12 H₂O appear on the left and 6 H₂O on the right.

  ✓Do this: Write the balanced equation with 12 H₂O.

  Usually costs: 1 mark.

• ✗Mistake: Calling the dark reaction light-independent in every sense

  The name suggests it happens in darkness.

  Wrong working

  The Calvin cycle runs at night.

  Right working

  It does not need light directly but depends on ATP and NADPH from the light reactions, so it slows quickly in darkness.

  ✓Do this: Use 'biosynthetic phase'.

  Usually costs: 1 mark.

• ✗Mistake: Wrong order of photosystems

  Numbers suggest order.

  Wrong working

  Electrons flow from PS I to PS II.

  Right working

  In non-cyclic flow electrons go from PS II to PS I to NADP⁺.

  ✓Do this: Remember: two before one.

  Usually costs: 1-2 marks.

• ✗Mistake: Putting RuBisCO in C4 mesophyll

  C3 pattern applied to C4 plants.

  Wrong working

  In maize, RuBisCO fixes CO₂ in mesophyll cells.

  Right working

  PEPcase works in mesophyll; RuBisCO is in bundle sheath cells.

  ✓Do this: PEP picks, RuBisCO fixes.

  Usually costs: 1 mark.

• ✗Mistake: Writing the full photorespiration chemistry

  Over-preparation.

  Wrong working

  Every intermediate of the glycolate pathway.

  Right working

  The syllabus asks for the photorespiration pathway in brief: RuBisCO as oxygenase, three organelles, CO₂ loss and significance.

  ✓Do this: Learn a four-line outline.

  Usually costs: Time.

Practice questions on Photosynthesis in Higher Plants

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

1. Q1 · 1 mark · MCQExam level

   The reaction centre of PS II absorbs best at:
   (a) 680 nm
   (b) 700 nm
   (c) 430 nm
   (d) 560 nm

   Hint

   P680.

   Answer

   (a)

2. Q2 · 1 mark · MCQExam level

   Photorespiration does not occur significantly in:
   (a) wheat
   (b) rice
   (c) maize
   (d) soybean

   Hint

   C4.

   Answer

   (c)

3. Q3 · 1 mark · Very short answerExam level

   Name the primary CO₂ acceptor in C3 plants.

   Hint

   5-carbon.

   Answer

   Ribulose-1,5-bisphosphate (RuBP)

4. Q4 · 1 mark · Very short answerExam level

   Name the pigment present at the reaction centre of both photosystems.

   Hint

   Chief pigment.

   Answer

   Chlorophyll a

5. Q5 · 2 marks · Short answerExam level

   Give two functions of accessory pigments.

   Hint

   Range and protection.

   Answer

   They absorb wavelengths not absorbed by chlorophyll a and pass the energy to it, widening the usable spectrum; they protect chlorophyll a from photo-oxidation.

6. Q6 · 2 marks · Short answerHarder exam level

   Distinguish between PS I and PS II.

   Hint

   Reaction centres and water.

   Answer

   PS I: reaction centre P700; takes part in both cyclic and non-cyclic flow; passes electrons to NADP⁺. PS II: reaction centre P680; only in non-cyclic flow; linked with photolysis of water and O₂ release.

7. Q7 · 2 marks · Short answerHarder exam level

   Why is RuBisCO called a carboxylase–oxygenase? When does it act as an oxygenase?

   Hint

   Two substrates.

   Answer

   Its active site can bind either CO₂ (carboxylation of RuBP) or O₂ (oxygenation of RuBP). It acts as an oxygenase when the O₂ concentration is high relative to CO₂, e.g. on hot, dry days with stomata closed.

8. Q8 · 3 marks · Short answerHarder exam level

   Describe the Hatch–Slack pathway.

   Hint

   Mesophyll to bundle sheath.

   Answer

   In mesophyll, PEP + CO₂ → OAA by PEP carboxylase; OAA → malate or aspartate, which moves to bundle sheath cells. There it is decarboxylated to CO₂ and pyruvate; CO₂ enters the Calvin cycle with RuBisCO. Pyruvate returns to the mesophyll and is converted to PEP using ATP.

9. Q9 · 3 marks · Short answerHarder exam level

   Explain how temperature and water affect photosynthesis.

   Hint

   Enzymes, stomata.

   Answer

   The biosynthetic phase is enzyme-controlled, so rate rises with temperature to an optimum and then falls; C4 plants have a higher optimum than C3. Water acts mainly indirectly: water stress closes stomata, reducing CO₂ supply, and leaves wilt, reducing surface area and metabolism.

10. Q10 · 3 marks · Short answerHarder exam level

    How many turns of the Calvin cycle, ATP and NADPH are needed to make one molecule of sucrose (a 12-carbon sugar) in a C3 plant?

    Hint

    12 CO₂.

    Answer

    12 CO₂ are needed: 12 turns, 12 × 3 = 36 ATP and 12 × 2 = 24 NADPH.

11. Q11 · 5 marks · Long answerHarder exam level

    (a) Describe non-cyclic photophosphorylation with the help of the Z scheme. [3]
    (b) What is the source of electrons for PS II? Write the equation. [2]

    Hint

    PS II → ETC → PS I → NADP⁺.

    Answer

    (a) Light excites P680 (PS II); electrons pass via acceptor, plastoquinone, cytochrome b₆f and plastocyanin to PS I, forming ATP; light excites P700 (PS I); electrons pass via ferredoxin to NADP⁺ reductase, forming NADPH. Diagram shaped like a Z on an energy scale. (b) Water: $2H_{2}O\to 4H^{+}+O_2+4e^{-}$.

12. Q12 · 5 marks · Long answerHarder exam level

    (a) State Blackman's law of limiting factors. [2]
    (b) Explain with a graph how light intensity and CO₂ concentration interact to limit photosynthesis. [3]

    Hint

    Plateaus at two CO₂ levels.

    Answer

    (a) The rate of a process affected by several factors is limited by the factor nearest its minimal value; changing that factor changes the rate. (b) Graph of rate vs light intensity at low and high CO₂: both rise together at low light (light limiting), then the low-CO₂ curve flattens early (CO₂ limiting) while the high-CO₂ curve rises further before flattening; raising CO₂ helps only when light is not limiting.

13. Q13 · 2 marks · Short answerHarder exam level

    Why do leaves appear green, and what would happen to the photosynthesis of a plant kept under pure green light for a week?

    Hint

    Reflection.

    Answer

    Chlorophylls absorb blue and red light and reflect most green light, so leaves look green. Under pure green light the rate of photosynthesis would be very low; the plant would use up its reserves and weaken.

14. Q14 · 3 marks · Short answerHOTS / competency

    A C3 plant is placed in a sealed, illuminated chamber. Over hours, CO₂ falls and O₂ rises. Predict how the balance between photosynthesis and photorespiration changes and what happens to net carbon gain.

    Hint

    RuBisCO's competition.

    Answer

    As CO₂ falls and O₂ rises, RuBisCO increasingly acts as an oxygenase, so photorespiration rises while carboxylation falls. Eventually CO₂ fixation equals CO₂ released (by photorespiration and respiration), the CO₂ compensation point, and net carbon gain becomes zero.

15. Q15 · 3 marks · Short answerHOTS / competency

    Explain why the number of ATP needed per glucose is higher in C4 plants, yet C4 plants are often more productive in tropical climates.

    Hint

    Cost vs loss.

    Answer

    C4 plants spend 2 extra ATP per CO₂ to regenerate PEP (30 vs 18 ATP per glucose). But in hot, bright conditions C3 plants lose much fixed carbon to photorespiration and must partly close stomata. C4 plants concentrate CO₂ at RuBisCO, avoid photorespiration and fix CO₂ even at low internal CO₂, so their net gain is higher; ATP is not scarce in bright light.

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. 1.Site of light reactions?Show answer

   Thylakoid membranes (grana).

2. 2.Site of Calvin cycle in C3 plants?Show answer

   Stroma of mesophyll chloroplasts.

3. 3.Reaction centre of PS I?Show answer

   P700.

4. 4.Products of cyclic photophosphorylation?Show answer

   ATP only.

5. 5.Who proposed the chemiosmotic hypothesis?Show answer

   Peter Mitchell.

6. 6.Enzyme fixing CO₂ in C4 mesophyll?Show answer

   PEP carboxylase.

7. 7.ATP per glucose in C4 plants?Show answer

   30.

8. 8.Organelles in photorespiration?Show answer

   Chloroplast, peroxisome, mitochondrion.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you balance the overall equation?
• ☐Can you draw an action spectrum and explain it?
• ☐Can you draw the Z scheme with all carriers?
• ☐Can you explain three causes of the proton gradient?
• ☐Can you state the cost of one glucose in C3 and C4 plants?
• ☐Can you describe Kranz anatomy?
• ☐Can you explain why photorespiration is wasteful?
• ☐Can you state Blackman's law with an example?

Photosynthesis in Higher Plants: frequently asked questions