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Plant Physiology: questions with answers

Class 11 Biology, Chapter 11 (NCERT). 15 questions by concept, 9 with pictures. Open “Show answer” for the answer and why.

How to teach this chapter · Common misconceptions & worked visual for this chapter

Chloroplast compartmentation and light–carbon reaction dependency

Question 1

A learner says, 'Only leaves photosynthesise. In their chloroplasts the stroma traps light and makes ATP and NADPH, while grana perform the carbon reactions.' Which correction is fully correct?

  1. (a)Other green parts can photosynthesise; the stroma carries light reactions, and membranes carry carbon reactions.
  2. (b)Only leaves photosynthesise; membranes carry light reactions, and the stroma carries carbon reactions.
  3. (c)All plant parts photosynthesise; the chloroplast envelope performs both reaction phases.
  4. (d)Other green plant parts can also photosynthesise; membranes carry light reactions, and the stroma carries carbon reactions.
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(d) Other green plant parts can also photosynthesise; membranes carry light reactions, and the stroma carries carbon reactions.

The exclusive-leaf claim is too narrow because other green plant parts also photosynthesise. The site assignments are reversed: photochemical reactions use the membrane system, while enzymatic carbon reactions occur in the stroma.

Question 2

Which statement correctly interprets the conventional term 'dark reactions' for photosynthetic carbon reactions?

  1. (a)They can occur only in darkness because light prevents their enzymes from functioning.
  2. (b)They are entirely independent of light reactions and do not use ATP or NADPH.
  3. (c)They are not directly light driven, can occur in light, and depend on ATP and NADPH supplied by light reactions.
  4. (d)They are directly driven by photons in exactly the same way as the photochemical reactions.
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(c) They are not directly light driven, can occur in light, and depend on ATP and NADPH supplied by light reactions.

The term means that these reactions are not directly driven by light. They can occur in light and depend indirectly on light because they consume ATP and NADPH produced by the photochemical reactions.

Thylakoid chemiosmosis and ATP synthase

Question 3

During illumination, water splitting releases protons on the inner thylakoid side, carriers transfer protons from the stroma across the membrane, and NADP+ reduction uses stromal protons. What relative condition results?

Processes stated in the questionWater splitting releases H+ on the lumen side. Carriers move protons from stroma to lumen and NADP+ reduction uses stromal protons. No resulting concentrations or pH are shown. QUESTION DIAGRAM Processes stated in the question Only the three given processes are shown Stroma Lumen Water splitting H+ H+ Carrier NADP+ reduction H+ uses stromal H+ Compare the compartments using these processes.
  1. (a)The stroma has higher H+ concentration and lower pH than the lumen.
  2. (b)The lumen has lower H+ concentration because water splitting releases its protons into the stroma.
  3. (c)The compartments remain similar because NADP+ reduction adds protons back to the stroma.
  4. (d)The lumen has higher H+ concentration and lower pH than the stroma.
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(d) The lumen has higher H+ concentration and lower pH than the stroma.

All three contributions reinforce a proton gradient: protons are released or transferred into the lumen while stromal protons are removed. Thus the lumen has a higher proton concentration and a lower pH than the stroma.

Question 4

In the thylakoid sketch, the marker asks where NADP+ reduction changes proton abundance. Which complete account is correct?

NADP+stromalumen?
  1. (a)Water splitting adds H+ to the lumen; carriers move H+ from stroma to lumen; NADP+ reduction consumes stromal H+; the stroma becomes H+-rich.
  2. (b)Water splitting adds H+ to the stroma; carriers move H+ from stroma to lumen; NADP+ reduction consumes stromal H+; the lumen becomes H+-rich.
  3. (c)Water splitting adds H+ to the lumen; carriers move H+ from stroma to lumen; NADP+ reduction consumes stromal H+; the lumen becomes H+-rich.
  4. (d)Water splitting adds H+ to the lumen; carriers move H+ from stroma to lumen; NADP+ reduction adds H+ to stroma; the lumen becomes H+-rich.
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(c) Water splitting adds H+ to the lumen; carriers move H+ from stroma to lumen; NADP+ reduction consumes stromal H+; the lumen becomes H+-rich.

Water splitting releases protons into the lumen, an electron-linked carrier transfers protons from stroma to lumen, and reduction of NADP+ consumes stromal protons. Together these processes make the lumen proton-rich and lower its pH relative to the stroma.

Experimental evidence and the overall photosynthetic process

Question 5

A learner makes the claims below about photosynthesis experiments. Which option corrects all three?

Experiment or conditionLearner’s claim
Starch formationDoes not need light, chlorophyll or carbon dioxide
Restoring bell-jar airAny plant tissue restores the air, even in darkness
Engelmann’s bacteriaDirectly detect carbohydrate
  1. (a)Starch formation requires light, green tissue and carbon dioxide; illuminated green tissue restores air; aerobic bacteria indicate oxygen evolution.
  2. (b)Starch formation requires the tested conditions; any tissue restores air without light; aerobic bacteria indicate oxygen evolution.
  3. (c)Starch forms without the tested requirements; illuminated green tissue restores air; aerobic bacteria indicate oxygen evolution.
  4. (d)Starch formation requires the tested conditions; illuminated green tissue restores air; aerobic bacteria directly detect carbohydrate.
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(a) Starch formation requires light, green tissue and carbon dioxide; illuminated green tissue restores air; aerobic bacteria indicate oxygen evolution.

Controlled leaf tests require light, green chlorophyll-containing tissue and carbon dioxide for starch formation. Air restoration requires sunlight and green tissue, and Engelmann's aerobic bacteria report oxygen evolution, chiefly in blue and red light.

Experiment / claimCorrect inference
Controlled starch testsLight, chlorophyll-containing tissue and CO₂ are needed for new starch
Green tissue in light restores airPhotosynthesis releases oxygen
Engelmann’s aerobic bacteriaClusters indicate oxygen evolution, not direct detection of carbohydrate

Question 6

Which interpretation fits all three observations?

ExperimentObservation
Variegated leafOnly an illuminated green region forms starch
Bell jarAn illuminated green plant restores the air; the corresponding dark setup does not
Illuminated algaAerobic bacteria cluster mainly in blue and red regions
  1. (a)The bacteria directly detect carbohydrate formation, so their clustering maps the greatest starch production.
  2. (b)Starch would form equally in covered and non-green regions; the other observations alone establish the photosynthetic requirements.
  3. (c)Any plant tissue restores bell-jar air in darkness as well as in sunlight; only the leaf test depends on illumination.
  4. (d)Light and green tissue are required for the observed photosynthetic effects, and the bacterial clusters indicate regions of greatest oxygen evolution.
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(d) Light and green tissue are required for the observed photosynthetic effects, and the bacterial clusters indicate regions of greatest oxygen evolution.

Starch formation in the leaf requires green tissue and light. Ingenhousz's comparison makes sunlight and green tissue conditions of the oxygen-releasing plant process, while Engelmann's aerobic bacteria indicate sites of greatest oxygen evolution rather than carbohydrate formation.

ObservationSupported interpretation
Illuminated green leaf region forms starchLight and chlorophyll-containing tissue support new starch formation
Green plant restores air in light, not darkLight-dependent oxygen release
Aerobic bacteria cluster in blue/red regionsMore oxygen is evolved in those regions

Photosynthetic pigments, spectra and photosystem organisation

Question 7

A learner writes, 'Because chlorophyll a is the chief pigment, it is the only photosynthetic pigment. So the action spectrum must exactly equal its absorption spectrum.' Which correction is complete?

  1. (a)Chlorophyll a is chief, but accessory pigments also contribute. So the action spectrum and the chlorophyll-a absorption spectrum broadly match without overlapping exactly.
  2. (b)Accessory pigments contribute, but the action spectrum still matches chlorophyll-a absorption exactly at every wavelength.
  3. (c)Chlorophyll a is still the only photosynthetic pigment, although the spectra do not overlap exactly.
  4. (d)Accessory pigments replace chlorophyll a as chief pigment, so the two spectra are unrelated.
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(a) Chlorophyll a is chief, but accessory pigments also contribute. So the action spectrum and the chlorophyll-a absorption spectrum broadly match without overlapping exactly.

Chief does not mean sole: chlorophyll b, xanthophylls and carotenoids also absorb light and transfer energy to chlorophyll a. Their contribution explains why the action spectrum broadly resembles but does not completely overlap chlorophyll-a absorption.

Electron flow, water splitting and photophosphorylation

Question 8

In a light reaction, loss of electrons from the 680 nm reaction centre is followed by oxygen and proton formation from water. The electrons later reach a 700 nm centre and reduce NADP+. Which inference is justified?

Reaction-centre observationsWater splitting is observed after electron loss from the 680 nm centre, forming oxygen and protons. Electrons subsequently reach a 700 nm centre and reduce NADP+. Photosystem identities and the replacement-electron mechanism are not supplied. QUESTION DIAGRAM Reaction-centre observations Observed sequence only · identities are not supplied 680 nm 700 nm electrons Electron loss Then observed: Water splitting Oxygen + protons formed NADP+ is reduced
  1. (a)Water replenishes Photosystem I, which functions after Photosystem II, and the electrons end at NADP+.
  2. (b)Water replenishes Photosystem II, but Photosystem I functions before Photosystem II, and the electrons end at NADP+.
  3. (c)Water replenishes Photosystem II, which functions before Photosystem I, and the electrons end at NADP+.
  4. (d)Water replenishes Photosystem II, which functions before Photosystem I, and the electrons then return to Photosystem II.
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(c) Water replenishes Photosystem II, which functions before Photosystem I, and the electrons end at NADP+.

Water splitting is associated with Photosystem II and replaces electrons lost from P680. Electron transfer then reaches Photosystem I and ends in reduction of NADP+, so this is a non-cyclic route from Photosystem II to Photosystem I.

Question 9

The neutral electron-path diagram emphasises the terminal electron-transfer segment. Which account correctly describes the complete non-cyclic route?

680 nm700 nmH2ONADP+
  1. (a)Flow starts at Photosystem II; water supplies replacement electrons there; electrons pass through Photosystem I and finally reduce NADP+.
  2. (b)Flow starts at Photosystem I; water supplies replacement electrons to Photosystem II; electrons finally reduce NADP+.
  3. (c)Flow starts at Photosystem II; water supplies replacement electrons to Photosystem I; electrons finally reduce NADP+.
  4. (d)Flow starts at Photosystem II; water supplies replacement electrons there; electrons pass through Photosystem I and return to Photosystem II.
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(a) Flow starts at Photosystem II; water supplies replacement electrons there; electrons pass through Photosystem I and finally reduce NADP+.

Non-cyclic electron flow starts when P680 in Photosystem II is excited. Water splitting associated with Photosystem II supplies replacement electrons, and electrons then pass through Photosystem I before reducing NADP+; they do not return to the starting reaction centre.

Calvin-cycle fixation, stages and energy budget

Question 10

The neutral Calvin-cycle redraw emphasises the first stable three-carbon products without naming its molecules or stages. Which account correctly interprets the cycle?

1C5C6C3C3CATP
  1. (a)A two-carbon acceptor receives CO2, two three-carbon PGA molecules form, and ATP-dependent regeneration restores the acceptor.
  2. (b)Five-carbon RuBP accepts CO2, two three-carbon PGA molecules form, and ATP-dependent regeneration restores RuBP.
  3. (c)Three-carbon PGA accepts CO2, a five-carbon product forms, and ATP-dependent regeneration restores PGA.
  4. (d)Five-carbon RuBP accepts CO2, two three-carbon PGA molecules form, and RuBP is consumed without regeneration.
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(b) Five-carbon RuBP accepts CO2, two three-carbon PGA molecules form, and ATP-dependent regeneration restores RuBP.

Five-carbon RuBP accepts carbon dioxide through RuBisCO, and the unstable six-carbon intermediate yields two three-carbon PGA molecules. Reduction follows, and ATP-dependent regeneration restores RuBP so carbon fixation is cyclic rather than linear.

Question 11

A learner writes: ‘PGA accepts carbon dioxide in the Calvin cycle. The pathway consumes its acceptor without restoring it, and it is absent from C4 plants.’ Which correction is complete?

  1. (a)PGA accepts CO2; RuBP is the first stable product; RuBP is regenerated; the cycle occurs in both C3 and C4 plants.
  2. (b)RuBP accepts CO2; PGA is the first stable product; RuBP is regenerated; the cycle occurs only in C3 plants.
  3. (c)RuBP accepts CO2; PGA is the first stable product; RuBP is regenerated; the cycle occurs in both C3 and C4 plants.
  4. (d)RuBP accepts CO2; PGA is the first stable product; RuBP is not regenerated; the cycle occurs in both C3 and C4 plants.
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(c) RuBP accepts CO2; PGA is the first stable product; RuBP is regenerated; the cycle occurs in both C3 and C4 plants.

RuBP, not PGA, is the carbon-dioxide acceptor; PGA is the first stable three-carbon product. ATP-dependent regeneration restores RuBP, and the Calvin cycle occurs in both C3 and C4 plants, though its cellular location differs.

C4 spatial pathway and suppression of photorespiration

Question 12

A schematic shows a four-carbon compound moving from mesophyll to bundle-sheath cells and a three-carbon compound returning to mesophyll. Which account correctly completes the C4 carbon route?

MesophyllBundle sheath3-C acceptor4-C carrierCO23-C return
  1. (a)Mesophyll PEPCase fixes carbon dioxide into OAA. A four-carbon acid is decarboxylated in bundle sheath. Carbon dioxide enters the Calvin cycle there. A three-carbon molecule returns to regenerate PEP.
  2. (b)Bundle-sheath PEPCase produces OAA, which moves to mesophyll where the complete Calvin cycle occurs.
  3. (c)Mesophyll PEPCase produces OAA or malate, and that four-carbon acid itself is the final sugar product retained in bundle sheath.
  4. (d)Mesophyll RuBisCO fixes carbon dioxide into PGA from RuBP; PGA then moves to bundle sheath as the initial four-carbon product.
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(a) Mesophyll PEPCase fixes carbon dioxide into OAA. A four-carbon acid is decarboxylated in bundle sheath. Carbon dioxide enters the Calvin cycle there. A three-carbon molecule returns to regenerate PEP.

In mesophyll, three-carbon PEP accepts carbon dioxide through PEPCase to form four-carbon OAA, which may form malate or aspartate. A four-carbon acid moves to bundle sheath and releases carbon dioxide for the Calvin cycle; the remaining three-carbon molecule returns to mesophyll and regenerates PEP.

Question 13

A transverse section of a leaf shows large bundle-sheath cells arranged around the vascular bundles. Which conclusion and cell-role assignment is supported by the chapter?

Leaf section observedA transverse section shows large bundle-sheath cells arranged around the vascular bundles. QUESTION DIAGRAM Leaf section observed Observed structure Large bundle-sheath cells Vascular bundle
  1. (a)The internal feature supports a C4 identity; initial PEPCase fixation is in mesophyll and the Calvin cycle is in bundle-sheath cells.
  2. (b)The internal feature supports a C4 identity; the entire Calvin cycle occurs in mesophyll and bundle-sheath cells only surround the vein.
  3. (c)The plant should be classified from its external habit instead, because internal bundle-sheath anatomy cannot distinguish C3 from C4.
  4. (d)The internal feature supports a C4 identity; mesophyll RuBP is carboxylated by RuBisCO to give PGA as the initial C4 fixation product.
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(a) The internal feature supports a C4 identity; initial PEPCase fixation is in mesophyll and the Calvin cycle is in bundle-sheath cells.

This feature—large bundle-sheath cells arranged around the vascular bundles—is part of Kranz anatomy and supports identifying the leaf as C4 from internal anatomy. In a C4 leaf, mesophyll cells perform initial PEPCase fixation, while bundle-sheath cells contain RuBisCO and carry out the Calvin cycle.

Limiting factors and photosynthetic response curves

Question 14

A tropical plant and a temperate plant are exposed to the same temperature and both become water-stressed. The diagram shows the narrowed stomatal pore of one stressed leaf. Which explanation is valid for interpreting their photosynthetic rates?

  1. (a)Water stress can reduce carbon-dioxide entry and leaf activity, and the same temperature may affect the two plants differently because their habitat-linked optimum temperatures differ.
  2. (b)Water stress affects only water splitting, so stomatal narrowing, wilting and carbon-dioxide entry are irrelevant.
  3. (c)The same temperature must affect both plants identically because every plant has one common temperature optimum.
  4. (d)Adding carbon dioxide outside the leaf must fully overcome water stress even when the stomatal pore remains narrowed.
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(a) Water stress can reduce carbon-dioxide entry and leaf activity, and the same temperature may affect the two plants differently because their habitat-linked optimum temperatures differ.

Water stress mainly lowers photosynthesis indirectly: stomata close and reduce carbon-dioxide entry, while wilting reduces leaf area and metabolic activity. The same temperature also need not be equally favourable to both plants because temperature optimum depends on habitat adaptation.

Question 15

The question mark marks the high-light plateau of a photosynthetic light-response curve. Which interpretation is supported?

Light intensityPhotosynthetic rate?
  1. (a)Light is still the limiting factor at this high intensity, so still more light is the required correction.
  2. (b)The shape shows that photosynthesis has become independent of every internal and external factor.
  3. (c)The rate must continue to increase whenever light is increased; a plateau or decline cannot occur.
  4. (d)More light no longer raises the rate because another factor has become limiting.
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(d) More light no longer raises the rate because another factor has become limiting.

At low light, photosynthetic rate rises approximately linearly with incident light. For the marked plateau: More light no longer raises the rate because another factor has become limiting.

Common misconception in Plant Physiology

Open full misconception guide →

Does doubling light always double the rate of photosynthesis?

Wrong

If light intensity doubles, the rate of photosynthesis must double at every light level.

Why it feels right. More light helps under dim conditions, so I extend that relationship.

Right

Another resource or process can limit photosynthesis when light is abundant.

Worked example

With low CO₂ the schematic rate curve rises in dim light and then plateaus. More light alone brings little change on that plateau.
  1. Imagine a leaf at constant temperature and low CO₂.
  2. Raising dim light initially increases the rate.
  3. After the rate plateaus, more light alone has little effect.
  4. Raising CO₂ can increase the rate if CO₂ was limiting.