NCERT Solutions for Class 11 Biology Chapter 11: Photosynthesis in Higher Plants (NCERT 2026–27)

These Class 11 Biology Chapter 11 solutions cover Photosynthesis in Higher Plants from Unit 4, Plant Physiology. The chapter explains how green plants (autotrophs) use light energy to synthesise food, the structure of the photosynthetic machinery in the chloroplast, the light reaction (Z scheme, photosystems, splitting of water, photophosphorylation), the Calvin cycle, the C4 (Hatch and Slack) pathway, photorespiration and the factors that affect the rate of photosynthesis. Every NCERT exercise question is reproduced verbatim and answered in exam-ready prose below.

Class: 11 Subject: Biology Unit: 4 — Plant Physiology Chapter: 11 Topic: Photosynthesis in Higher Plants Session: 2026–27
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Class 11 Biology Chapter 11 Solutions – Overview

Photosynthesis is the physico-chemical process by which green plants use light energy to drive the synthesis of organic compounds from carbon dioxide and water, releasing oxygen. It is the primary source of all food on Earth and of atmospheric oxygen. Early experiments by Priestley, Ingenhousz, von Sachs, Engelmann and van Niel established that light, chlorophyll and CO2 are needed and that the O2 released comes from water. Inside the chloroplast there is a clear division of labour: the thylakoid membranes (grana) carry out the light reaction—trapping light, splitting water and forming ATP and NADPH—while the stroma runs the Calvin cycle that fixes CO2 into sugar. C4 plants add a CO2-concentrating mechanism using Kranz anatomy to suppress photorespiration. Finally, light, CO2, temperature and water act as factors limiting the rate, following Blackman’s Law of Limiting Factors.

Key Concepts & Definitions

Photosynthesis: a physico-chemical process in which autotrophs use light energy to synthesise carbohydrates from CO2 and water, releasing O2.

Light reaction (photochemical phase): light absorption, splitting of water, O2 release and formation of ATP and NADPH on the thylakoid membranes.

Photosystems (PS I & PS II): pigment–protein complexes. PS I has the reaction centre P700 (absorbs 700 nm); PS II has P680 (absorbs 680 nm). They are named by order of discovery, not order of function.

Z scheme: the path of electrons from PS II, uphill to an acceptor, downhill through the electron transport chain to PS I, up again on excitation, and finally downhill to reduce NADP+ to NADPH—its shape gives it the name.

Accessory pigments: chlorophyll b, xanthophylls and carotenoids that absorb light at extra wavelengths, transfer energy to chlorophyll a, and protect it from photo-oxidation.

Chemiosmosis: ATP synthesis driven by a proton (H+) gradient across the thylakoid membrane through ATP synthase (CF0–CF1).

Calvin cycle (C3 pathway): the biosynthetic phase in the stroma—carboxylation, reduction and regeneration—in which CO2 is fixed onto RuBP by RuBisCO; first stable product is 3-PGA.

C4 (Hatch and Slack) pathway: CO2 is first fixed by PEPcase in mesophyll cells to form the 4-carbon OAA; CO2 is later released in bundle sheath cells for the Calvin cycle (Kranz anatomy).

Photorespiration: a wasteful pathway in C3 plants where RuBisCO binds O2 instead of CO2; no sugar, ATP or NADPH is made and CO2 is released.

Law of Limiting Factors (Blackman, 1905): when several factors affect a process, the rate is set by the factor nearest its minimum value.

Important Equations

6CO2 + 12H2O  ⟶[light]  C6H12O6 + 6H2O + 6O2  (overall photosynthesis; O2 comes from water)
2H2O ⟶ 4H+ + O2 + 4e−  (splitting of water at PS II)
RuBP (5C) + CO2  ⟶[RuBisCO]  2 × 3-PGA (3C)  (carboxylation, Calvin cycle)
For 1 glucose: 6 turns of the Calvin cycle → 18 ATP + 12 NADPH used (3 ATP + 2 NADPH per CO2)

NCERT Exercises – Complete Solutions

All questions below are reproduced verbatim from the NCERT textbook (Reprint 2026–27). Answers are original and written in CBSE exam-ready style.

1. By looking at a plant externally, can you tell whether a plant is C3 or C4? Why and how?

ANSWER No, it is not possible to tell whether a plant is C3 or C4 simply by looking at it externally. The difference between the two types lies in their internal leaf anatomy and in the biochemical pathway of CO2 fixation, not in any external feature such as colour, shape or size of the leaf. Both C3 and C4 plants have green leaves and may look quite similar from outside. To distinguish them we must examine a vertical section of the leaf under a microscope. C4 plants (e.g. maize, sorghum) show the characteristic Kranz anatomy—large bundle sheath cells with many chloroplasts surrounding the vascular bundles—while C3 plants do not. Hence the test is internal, not external.

2. By looking at which internal structure of a plant you can tell whether a plant is C3 or C4? Explain.

ANSWER We can tell a C4 plant from a C3 plant by examining the internal anatomy of the leaf, specifically the presence or absence of Kranz anatomy around the vascular bundles. In C4 plants, the vascular bundles are surrounded by large bundle sheath cells arranged in a wreath-like (Kranz) pattern. These cells have a large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces. The Calvin cycle here occurs only in the bundle sheath cells, while the initial CO2 fixation occurs in the mesophyll. In C3 plants such specialised bundle sheath cells are absent, and the Calvin cycle takes place in all mesophyll cells. Therefore, the presence of well-developed bundle sheath cells with chloroplasts (Kranz anatomy) confirms a C4 plant.

3. Even though a very few cells in a C4 plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?

ANSWER In C4 plants the Calvin cycle is restricted to the bundle sheath cells, which are relatively few. Despite this, C4 plants are highly productive because they possess an efficient CO2-concentrating mechanism that keeps the Calvin cycle working at its maximum rate. In the mesophyll cells, the enzyme PEPcase fixes CO2 into the C4 acid OAA. This is carried as malic/aspartic acid to the bundle sheath cells, where it is broken down to release CO2. As a result, the concentration of CO2 around RuBisCO in the bundle sheath cells becomes very high. This high CO2 level ensures that RuBisCO acts almost entirely as a carboxylase and not as an oxygenase, so photorespiration is virtually eliminated. Because no fixed carbon is wasted in photorespiration, and because these plants also tolerate high temperatures and high light intensities, the few Calvin-cycle cells achieve very high productivity.

4. RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C4 plants?

ANSWER RuBisCO can bind both CO2 and O2 at its active site, and this binding is competitive—whichever gas is present in higher relative concentration is favoured. RuBisCO has a greater affinity for CO2, so a high CO2 concentration pushes it towards carboxylation. In C4 plants, the C4 acid transported from the mesophyll is broken down in the bundle sheath cells to release CO2, raising the intracellular CO2 concentration at the site of RuBisCO to a high level. With abundant CO2 and very little O2 available to compete, RuBisCO functions mainly as a carboxylase. This is why RuBisCO carries out more carboxylation (and minimal oxygenation) in C4 plants, avoiding photorespiration.

5. Suppose there were plants that had a high concentration of Chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?

ANSWER No, such a plant would not be able to carry out photosynthesis. Chlorophyll a is the chief (primary) pigment of photosynthesis and forms the reaction centre (P700 in PS I and P680 in PS II) where the light-driven reactions actually begin. Without chlorophyll a there is no reaction centre, so the trapped light energy cannot be converted into chemical energy and photosynthesis cannot proceed. Plants still have chlorophyll b and other accessory pigments (xanthophylls, carotenoids) for two reasons: (i) they absorb light at wavelengths that chlorophyll a cannot, thus widening the range of light usable for photosynthesis, and they pass this absorbed energy on to chlorophyll a; and (ii) they protect chlorophyll a from photo-oxidation (damage caused by excess light). So accessory pigments make photosynthesis more efficient and safer, but they cannot replace chlorophyll a.

6. Why is the colour of a leaf kept in the dark frequently becomes yellow, or pale green? Which pigment do you think is more stable?

ANSWER A leaf kept in the dark gradually turns yellow or pale green because chlorophyll is broken down and cannot be re-synthesised in the absence of light (chlorophyll formation requires light). As the green chlorophyll degrades, the yellow accessory pigments—the carotenoids (xanthophylls and carotenes)—that were earlier masked by the green colour become visible, giving the leaf a yellow or pale appearance. This shows that carotenoids are more stable than chlorophyll. Chlorophyll degrades quickly in the dark, whereas carotenoids persist for longer, which is why their yellow colour remains after the green has faded.

7. Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?

ANSWER The leaves on the shady side (and the plants kept in the shade) are usually darker green than the leaves on the sunny side. This is because leaves growing in shade receive less light, so they produce a larger amount of chlorophyll to capture as much of the available light as possible for photosynthesis; the higher chlorophyll content makes them appear deeper green. In contrast, leaves in bright sunlight receive plenty of light and do not need as much chlorophyll; moreover, very intense light can cause some breakdown of chlorophyll. As a result, sun-exposed leaves tend to be lighter or yellowish-green.

8. Figure 11.10 shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions:

(a) At which point/s (A, B or C) in the curve light is a limiting factor?

ANSWER Light is a limiting factor at point A. In the region around A the curve is still rising steeply—the rate of photosynthesis increases as light intensity increases—which shows that light is in short supply and is controlling (limiting) the rate. At B and C the curve has levelled off (light saturation), so light is no longer limiting there.

(b) What could be the limiting factor/s in region A?

ANSWER In region A, where the curve rises linearly with light, light intensity itself is the main limiting factor—CO2, temperature and other factors are adequate, and only an increase in light raises the rate.

(c) What do C and D represent on the curve?

ANSWER C represents the point/region where the rate of photosynthesis has reached its maximum and the curve has become flat—light saturation has occurred, so further increase in light intensity does not increase the rate (some other factor such as CO2 is now limiting). D represents the light intensity (x-axis value) at which this light saturation / plateau begins, i.e. the saturating light intensity beyond which the rate no longer rises.

9. Give comparison between the following:

(a) C3 and C4 pathways

ANSWER
FeatureC3 pathwayC4 pathway
Primary CO2 acceptorRuBP (5-carbon)PEP (3-carbon)
Enzyme of first fixationRuBisCOPEP carboxylase (PEPcase)
First stable product3-PGA (3-carbon)OAA (4-carbon)
Site of Calvin cycleMesophyll cellsBundle sheath cells only
Leaf anatomyNormal (no Kranz)Kranz anatomy present
PhotorespirationPresent (wasteful)Absent / negligible
Optimum temperature & productivityLower; lower productivityHigher; higher productivity
ExamplesRice, wheatMaize, sorghum, sugarcane

(b) Cyclic and non-cyclic photophosphorylation

ANSWER
FeatureCyclic photophosphorylationNon-cyclic photophosphorylation
Photosystem(s) involvedOnly PS IBoth PS II and PS I (in series)
Path of electronsElectron cycles back to PS I (closed)Open Z scheme; electrons end on NADP+
ProductsOnly ATPATP, NADPH + H+ and O2
Splitting of water (O2 release)Does not occurOccurs at PS II
SiteStroma lamellae (lack PS II & NADP reductase)Grana thylakoids
Occurs whenOnly light > 680 nm available; or extra ATP neededNormal light with both photosystems active

(c) Anatomy of leaf in C3 and C4 plants

ANSWER
FeatureC3 leaf anatomyC4 leaf anatomy
Kranz anatomyAbsentPresent
Bundle sheath cellsSmall, few or undeveloped chloroplastsLarge, in a wreath around vascular bundles, many chloroplasts
Wall of bundle sheathThin, permeableThick, impervious to gases; no intercellular spaces
Cells fixing CO2One type (mesophyll)Two types (mesophyll + bundle sheath)
Location of RuBisCOMesophyll cellsBundle sheath cells only

Extra Practice Questions

Short Answer Type Questions

Q1. Why is the dark reaction wrongly named? Suggest a better name.

ANSWERThe dark reaction does not require darkness and is not independent of light—it depends on ATP and NADPH made in the light reaction and stops soon after light is removed. A more accurate name is the biosynthetic phase or carbon-fixing (carbon) reactions.

Q2. Why are twelve molecules of water shown on the substrate side of the photosynthesis equation?

ANSWERIn the balanced equation, water appears both as a reactant and a product (6H2O is released). Showing 12H2O on the left makes it clear that water is the source of all the O2 evolved (6O2), as proved by radioisotope studies, while 6 water molecules are also regenerated.

Q3. What is the role of the proton gradient in ATP synthesis in chloroplasts?

ANSWERSplitting of water and electron transport pump protons into the thylakoid lumen, creating a proton gradient. When these protons diffuse back to the stroma through the CF0 channel of ATP synthase, the energy released drives a conformational change in CF1 that catalyses ATP formation (chemiosmosis).

Q4. Where exactly are PS I and PS II located within the thylakoid system?

ANSWERThe grana thylakoid membranes contain both PS I and PS II, whereas the stroma lamellae contain only PS I and lack PS II and the NADP reductase enzyme; hence cyclic photophosphorylation occurs in the stroma lamellae.

Q5. State Blackman’s Law of Limiting Factors with an example.

ANSWERIf a process is affected by more than one factor, its rate is determined by the factor that is nearest its minimum value. For example, even with a green leaf, ample light and CO2, a plant will not photosynthesise if the temperature is very low; raising the temperature then allows photosynthesis to begin.

Long Answer Type Questions

Q1. Describe the Calvin cycle under its three stages and state the ATP and NADPH needed to make one glucose.

ANSWERThe Calvin cycle operates in the stroma in three stages. (1) Carboxylation: CO2 is fixed onto the 5-carbon RuBP by the enzyme RuBisCO, forming two molecules of 3-carbon 3-PGA—the most crucial step. (2) Reduction: a series of reactions using 2 ATP and 2 NADPH per CO2 fixed converts 3-PGA to triose phosphate (sugar). (3) Regeneration: RuBP, the CO2 acceptor, is regenerated using 1 ATP so the cycle continues. For each CO2 fixed, 3 ATP and 2 NADPH are used. Since 6 turns of the cycle (6 CO2) are required to make one glucose, the total is 18 ATP and 12 NADPH per glucose molecule.

Q2. Explain the Z scheme of the light reaction, including the splitting of water.

ANSWERIn PS II, the reaction centre P680 absorbs red light (680 nm); electrons are excited and picked up by a primary acceptor, then passed downhill through an electron transport chain of cytochromes to PS I. In PS I, P700 absorbs 700 nm light; its electrons are excited and transferred to another acceptor of higher redox potential, then moved downhill to reduce NADP+ to NADPH + H+. The electrons lost by PS II are replaced by the splitting of water (2H2O → 4H+ + O2 + 4e−), which also releases O2 into the lumen. Plotted on a redox scale, the path—up, down, up, down—forms a characteristic ‘Z’ shape, hence the name Z scheme. This non-cyclic flow produces ATP, NADPH and O2.

Q3. Explain photorespiration and why it does not occur in C4 plants.

ANSWERPhotorespiration occurs in C3 plants when RuBisCO, whose active site binds both CO2 and O2, binds O2 instead of CO2 under high O2/low CO2 conditions. RuBP then forms one molecule of phosphoglycerate and one of phosphoglycolate (2-carbon). In this pathway there is no synthesis of sugar, ATP or NADPH; instead CO2 is released and ATP is consumed, so it is wasteful. C4 plants avoid photorespiration because their CO2-concentrating mechanism (C4 acid broken down in bundle sheath cells) keeps CO2 very high around RuBisCO, ensuring it acts as a carboxylase. This higher efficiency gives C4 plants greater productivity and tolerance to high temperatures.

MCQs & Assertion–Reason

1. The reaction centre of Photosystem I is:

(a) P680    (b) P700    (c) P870    (d) P960

2. The O2 released during photosynthesis comes from:

(a) CO2    (b) glucose    (c) water    (d) RuBP

3. The first stable product of CO2 fixation in the C4 pathway is:

(a) 3-PGA    (b) RuBP    (c) OAA    (d) PEP

4. The primary CO2 acceptor in the Calvin cycle is:

(a) PEP    (b) RuBP    (c) OAA    (d) PGA

5. Number of ATP and NADPH required to make one glucose in the Calvin cycle:

(a) 12 ATP, 18 NADPH    (b) 18 ATP, 12 NADPH    (c) 6 ATP, 6 NADPH    (d) 9 ATP, 6 NADPH

6. Kranz anatomy is a characteristic feature of:

(a) C3 plants    (b) C4 plants    (c) all algae    (d) fungi

7. Cyclic photophosphorylation produces:

(a) ATP and NADPH    (b) only NADPH    (c) only ATP    (d) ATP, NADPH and O2

8. The enzyme that fixes CO2 in mesophyll cells of C4 plants is:

(a) RuBisCO    (b) PEP carboxylase    (c) ATP synthase    (d) NADP reductase

9. ATP synthesis in chloroplasts is explained by the:

(a) lock-and-key model    (b) chemiosmotic hypothesis    (c) induced-fit model    (d) Z scheme

10. The major limiting factor for photosynthesis in nature is usually:

(a) light    (b) CO2 concentration    (c) chlorophyll    (d) water in the soil

Answer key: 1-(b), 2-(c), 3-(c), 4-(b), 5-(b), 6-(b), 7-(c), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: C4 plants do not show photorespiration.

Reason: C4 plants have a mechanism that raises the CO2 concentration around RuBisCO, keeping it acting as a carboxylase.

A-R 2. Assertion: Chlorophyll a is the chief pigment of photosynthesis.

Reason: Accessory pigments cannot absorb any light at all.

A-R 3. Assertion: The dark reaction can also occur in the presence of light.

Reason: The dark reaction depends on ATP and NADPH supplied by the light reaction.

A-R 4. Assertion: Cyclic photophosphorylation produces both ATP and NADPH.

Reason: In cyclic flow only PS I is functional and the electron returns to PS I.

A-R 5. Assertion: A leaf kept in the dark gradually turns yellow.

Reason: Carotenoids are more stable than chlorophyll, which degrades and is not re-synthesised in the dark.

Answer key: 1-(A), 2-(C), 3-(A), 4-(D), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Saying the O2 in photosynthesis comes from CO2 — it comes from the splitting of water.
  • Calling the dark reaction “light-independent” — it needs ATP and NADPH from the light reaction; better called the biosynthetic phase.
  • Mixing up PS I (P700) with PS II (P680), and forgetting they are named by order of discovery, not function.
  • Writing that cyclic photophosphorylation makes NADPH — it makes only ATP.
  • Confusing the primary CO2 acceptor (RuBP in C3; PEP in C4) with the first stable product (3-PGA in C3; OAA in C4).
  • Stating wrong totals — one glucose needs 18 ATP and 12 NADPH, i.e. 3 ATP + 2 NADPH per CO2.
  • Saying C4 plants do not use the Calvin cycle — they do, but only in bundle sheath cells.

How to score full marks in this chapter

Learn the four landmark scientists and what each proved (Priestley – air is restored; Ingenhousz – sunlight needed; von Sachs – glucose/chloroplast; Engelmann – action spectrum; van Niel – O2 from water). Be able to draw and label the Z scheme and the Calvin cycle with its three stages. Memorise the key numbers (P680/P700, 3 ATP + 2 NADPH per CO2, 18 ATP + 12 NADPH per glucose) and the C3 vs C4 comparison table—these are frequent exam questions. For factor-based questions, always quote Blackman’s Law of Limiting Factors.

Frequently Asked Questions

What is Class 11 Biology Chapter 11 about?

Chapter 11, Photosynthesis in Higher Plants, explains how green plants use light energy to make food—the early experiments, the structure of the chloroplast, the light reaction (Z scheme, photosystems, water splitting, photophosphorylation, chemiosmosis), the Calvin cycle, the C4 (Hatch and Slack) pathway, photorespiration and the factors affecting the rate of photosynthesis.

How many ATP and NADPH are needed to make one glucose in the Calvin cycle?

Three ATP and two NADPH are used per CO2 molecule fixed. Since six turns of the cycle (six CO2) are needed for one glucose, the totals are 18 ATP and 12 NADPH per glucose molecule.

What is the difference between C3 and C4 plants?

C3 plants fix CO2 directly with RuBisCO forming 3-PGA in mesophyll cells and show photorespiration. C4 plants first fix CO2 with PEPcase to form the 4-carbon OAA in mesophyll cells, have Kranz anatomy, run the Calvin cycle only in bundle sheath cells, avoid photorespiration and are more productive.

Are these Class 11 Biology Chapter 11 solutions free?

Yes. All solutions are free and follow the official NCERT Biology textbook for session 2026–27, with every exercise question reproduced verbatim and answered.

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