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Respiration in Plants: questions with answers

Class 11 Biology, Chapter 12 (NCERT). 15 questions by concept, 4 with pictures. Open “Show answer” for the answer and why.

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

Glycolysis sequence, carbon flow and energy accounting

Question 1

The diagram shows a pathway operating when oxygen is available: one 6C compound passes through a split into two 3C branches, while ? marks the cellular compartment. Which interpretation is correct?

oxygen is available6C3C3C?
  1. (a)It is mitochondrial glycolysis; it requires oxygen directly and one glucose yields two three-carbon pyruvate molecules.
  2. (b)It is cytoplasmic glycolysis; it can operate without direct oxygen use but one glucose yields only one pyruvate molecule.
  3. (c)It is cytoplasmic glycolysis; it can operate without direct oxygen use and one glucose yields two three-carbon pyruvate molecules.
  4. (d)It is mitochondrial glycolysis; it requires oxygen directly and one glucose yields only one pyruvate molecule.
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(c) It is cytoplasmic glycolysis; it can operate without direct oxygen use and one glucose yields two three-carbon pyruvate molecules.

Glycolysis occurs in the cytoplasm and does not directly require oxygen, so it can operate in both shown conditions. One six-carbon glucose gives two three-carbon trioses, and the repeated later reactions produce two pyruvate molecules.

Question 2

A learner traces one glucose as follows: ‘Glycolysis begins only when oxygen enters the mitochondrion. The six-carbon molecule splits, but only one branch continues, so one pyruvate is formed.’ Which correction is complete?

  1. (a)Glycolysis occurs in the cytoplasm without direct oxygen use, and both three-carbon branches continue to form two pyruvate molecules.
  2. (b)Glycolysis occurs in mitochondria only when oxygen is present, and both branches form two pyruvate molecules.
  3. (c)Glycolysis occurs in the cytoplasm without direct oxygen use, but only one branch forms one pyruvate molecule.
  4. (d)Glycolysis occurs in the cytoplasm only when oxygen is present, and the branches rejoin to form one six-carbon pyruvate molecule.
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(a) Glycolysis occurs in the cytoplasm without direct oxygen use, and both three-carbon branches continue to form two pyruvate molecules.

Glycolysis takes place in the cytoplasm and does not directly require oxygen. The six-carbon intermediate splits into two three-carbon trioses, both of which continue through the later reactions, yielding two pyruvate molecules.

Respiration foundations, plant gas exchange and step-wise energy capture

Question 3

A learner says, ‘Only green plant cells respire. Therefore a plant needs a central respiratory organ to send oxygen from leaves to every non-green living part.’ Which correction is complete?

  1. (a)Green and non-green living cells respire, but a central respiratory organ must deliver gases to them.
  2. (b)Green and non-green living cells respire, and plant parts exchange gases mainly by local diffusion without a central respiratory organ.
  3. (c)Only green cells respire, but they exchange gases locally without a central respiratory organ.
  4. (d)Green and non-green cells photosynthesise, so neither tissue needs cellular respiration.
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(b) Green and non-green living cells respire, and plant parts exchange gases mainly by local diffusion without a central respiratory organ.

Respiration occurs in all living plant cells, including non-green cells. Stomata, lenticels, internal air spaces, low demand and short diffusion distances allow each plant part to exchange gases mainly for itself, so no central respiratory organ is needed.

Question 4

A learner writes, ‘The one-line respiratory equation represents a single reaction. It releases all glucose energy at once, and that energy is used directly wherever the cell performs work.’ Which correction is complete?

  1. (a)The equation summarises many enzyme-controlled steps; part of the released energy is captured in ATP, and ATP breakdown supplies cellular work.
  2. (b)The equation summarises many steps; ATP is a final waste product that removes unused energy from the cell.
  3. (c)The equation summarises many enzyme-controlled steps, but their energy is used directly for work without ATP.
  4. (d)The equation is one enzyme-controlled step; all energy is released at once and part is captured in ATP.
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(a) The equation summarises many enzyme-controlled steps; part of the released energy is captured in ATP, and ATP breakdown supplies cellular work.

The overall equation summarises many enzyme-controlled reactions. Their staged energy release allows part of the energy to be trapped in ATP, whose breakdown supplies energy for cellular work.

Pyruvate fates, fermentation and aerobic contrast

Question 5

A learner claims that whenever pyruvate is processed without oxygen, ethanol and carbon dioxide form and extra ATP is produced after glycolysis. Which comparison correctly evaluates that claim? Consider yeast and oxygen-limited muscle.

Claim to testLearner’s statement
ProductsEthanol and carbon dioxide in every case
ATP after glycolysisAdditional ATP is produced during the conversion
  1. (a)Yeast forms ethanol and carbon dioxide while muscle can form lactate, but both routes add ATP after glycolysis.
  2. (b)Yeast and muscle both form lactate, consume NAD+ and stop glycolysis after using its ATP.
  3. (c)All anaerobic cells form ethanol and carbon dioxide, but these routes only regenerate NAD+ and add no ATP after glycolysis.
  4. (d)Yeast forms ethanol and carbon dioxide, muscle can form lactate, and both routes regenerate NAD+ without adding ATP after glycolysis.
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(d) Yeast forms ethanol and carbon dioxide, muscle can form lactate, and both routes regenerate NAD+ without adding ATP after glycolysis.

The product depends on the organism or tissue: yeast forms ethanol and carbon dioxide, whereas oxygen-limited muscle and some bacteria form lactate. Both fermentations regenerate NAD+ and add no ATP beyond the net two ATP formed in glycolysis.

Fermentation exampleProducts from pyruvateCommon roleATP added after glycolysis
YeastEthanol and CO₂Regenerates NAD⁺None in this fermentation step
Oxygen-limited muscleLactateRegenerates NAD⁺None in this fermentation step

Question 6

A learner writes, ‘Glycolysis occurs only without oxygen. When oxygen is present, cells skip glycolysis; when it is absent, fermentation completely oxidises glucose.’ Which correction is complete?

  1. (a)Glycolysis forms pyruvate in both conditions; oxygen affects the later pyruvate fate, and fermentation incompletely oxidises glucose.
  2. (b)Glycolysis forms pyruvate in both conditions, but fermentation completely oxidises glucose when oxygen is absent.
  3. (c)Glycolysis occurs only without oxygen; fermentation is incomplete, while aerobic cells skip pyruvate formation.
  4. (d)Glycolysis occurs only with oxygen, and fermentation precedes glycolysis to make glucose from pyruvate.
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(a) Glycolysis forms pyruvate in both conditions; oxygen affects the later pyruvate fate, and fermentation incompletely oxidises glucose.

Glycolysis forms pyruvate in both aerobic and anaerobic conditions. Oxygen availability and cellular context influence the fate of pyruvate, while fermentation is incomplete and retains substantial energy in lactate or ethanol.

Pyruvate oxidation and TCA-cycle regeneration

Question 7

A student writes: “Pyruvate enters the cycle directly, where molecular oxygen converts it to citrate.” Which correction exactly matches the source?

  1. (a)Replace acetyl CoA with carbon dioxide as the compound that condenses with oxaloacetate.
  2. (b)First form acetyl CoA from pyruvate, and place molecular oxygen only at the terminal ETS step.
  3. (c)Keep molecular oxygen as a direct matrix reactant; only change the number of carbon dioxide molecules released.
  4. (d)Keep direct pyruvate entry into citrate formation; only change the name of the reduced carrier.
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(b) First form acetyl CoA from pyruvate, and place molecular oxygen only at the terminal ETS step.

Pyruvate must be converted to acetyl CoA before the acetyl group enters the TCA cycle. Carbon dioxide release and reduced-carrier formation occur in matrix reactions without direct use of molecular oxygen, whose direct role is terminal in the ETS.

Question 8

A student describes one TCA-cycle turn as follows: “Oxaloacetate is used up, and the only products are carbon dioxide and one ATP equivalent.” Which correction repairs every error in that statement?

  1. (a)Keep oxaloacetate as a one-use reactant and change only the carbon dioxide coefficient.
  2. (b)Keep carbon dioxide and one ATP equivalent as the only outputs; reduced carriers are not products of the cycle.
  3. (c)Replace the three NADH + H+ with three FADH2 and the one FADH2 with one NADH + H+.
  4. (d)Restore oxaloacetate and include 3 NADH + H+ and 1 FADH2 along with 2 carbon dioxide and 1 GTP/ATP equivalent.
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(d) Restore oxaloacetate and include 3 NADH + H+ and 1 FADH2 along with 2 carbon dioxide and 1 GTP/ATP equivalent.

A TCA-cycle turn regenerates oxaloacetate and forms reduced carriers as major products. The complete per-acetyl ledger is 2 carbon dioxide, 3 NADH + H+, 1 FADH2 and 1 GTP/ATP equivalent, with oxaloacetate restored.

Mitochondrial ETS, oxygen and oxidative phosphorylation

Question 9

The question mark in the diagram identifies one ATP-synthase component. Which option correctly names that component, its function and the proton direction during ATP synthesis?

Intermembrane spaceInner mitochondrial membraneMatrixH+H+?
  1. (a)F1 is the integral proton channel, while F0 is the peripheral catalytic site that forms ATP.
  2. (b)F1 is the peripheral matrix-side catalytic headpiece; proton passage through F0 drives ATP formation at F1.
  3. (c)F1 is the catalytic headpiece, but it is driven by protons passing through F0 from matrix to intermembrane space.
  4. (d)Both F0 and F1 are soluble matrix enzymes, and protons are chemically consumed to make ATP.
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(b) F1 is the peripheral matrix-side catalytic headpiece; proton passage through F0 drives ATP formation at F1.

F0 is the integral inner-membrane proton channel, while the peripheral F1 headpiece on the matrix side contains the catalytic site. Protons flow down their electrochemical gradient through F0 from the intermembrane space to the matrix, coupling that passage to ATP formation at F1.

Question 10

A model claims: “F1 is the proton channel, F0 is catalytic, and protons move from matrix to intermembrane space.” Which is the minimal complete correction?

The stated ATP-synthase modelThis diagram reproduces the learner’s stated component assignments or proton route for checking. QUESTION DIAGRAM The stated ATP-synthase model Learner’s model Intermembrane space Matrix F1 F0 Catalytic site H+ Claimed route
  1. (a)Make F0 the channel and F1 the catalytic headpiece, and direct protons from intermembrane space through F0 into the matrix.
  2. (b)Retain proton movement through ATP synthase from matrix to intermembrane space during ATP production.
  3. (c)Make both components soluble in the matrix and remove the inner membrane from the model.
  4. (d)Retain F1 as the proton channel and F0 as the catalytic ATP-forming site.
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(a) Make F0 the channel and F1 the catalytic headpiece, and direct protons from intermembrane space through F0 into the matrix.

F0 is the integral inner-membrane proton channel and F1 is the peripheral catalytic headpiece. During ATP synthesis protons flow through F0 from intermembrane space to matrix; they are not sent in the reverse direction.

Respiratory balance-sheet assumptions and ATP accounting

Question 11

Which comparison correctly relates glucose breakdown and ATP formation in fermentation and aerobic respiration?

  1. (a)All ATP in both pathways is formed by oxidative phosphorylation; fermentation nevertheless stops after gaining two ATP.
  2. (b)Because both pathways begin with glycolysis, fermentation and aerobic respiration have similar total ATP yields.
  3. (c)Fermentation incompletely breaks down glucose and nets two ATP; aerobic respiration completely degrades it and can yield many more ATP under the stated assumptions.
  4. (d)Fermentation completely degrades glucose, whereas aerobic respiration retains most of the substrate energy in its end products.
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(c) Fermentation incompletely breaks down glucose and nets two ATP; aerobic respiration completely degrades it and can yield many more ATP under the stated assumptions.

Fermentation only partially breaks down glucose and gives a net gain of two ATP through glycolytic substrate-level phosphorylation. Aerobic respiration completely degrades the substrate and adds extensive oxidative phosphorylation, so it can yield many more ATP under the theoretical assumptions.

Question 12

A learner writes: ‘All ATP in respiration is formed by the ETS. Fermentation and aerobic respiration therefore give about the same ATP yield.’ Which correction is complete?

  1. (a)Respiration includes substrate-level and oxidative phosphorylation, but fermentation and aerobic respiration have similar total ATP yields.
  2. (b)All respiratory ATP is formed by oxidative phosphorylation, but fermentation nets two ATP while aerobic respiration can yield many more.
  3. (c)Respiration includes substrate-level and oxidative phosphorylation; fermentation nets two ATP, while aerobic respiration can yield many more under the stated assumptions.
  4. (d)Only fermentation includes substrate-level phosphorylation, and only aerobic respiration includes any ATP formation.
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(c) Respiration includes substrate-level and oxidative phosphorylation; fermentation nets two ATP, while aerobic respiration can yield many more under the stated assumptions.

Glycolysis and the succinyl-CoA step include substrate-level phosphorylation, so respiratory ATP is not all made by the ETS. Fermentation nets two ATP after incomplete breakdown, whereas aerobic respiration can produce many more under the theoretical assumptions.

Amphibolic respiration and alternative substrate entry

Question 13

A learner draws fat splitting into glycerol and fatty acids, then sends both products to the glucose step before respiration. Which correction is complete?

The learner’s fat-entry mapFat splits into glycerol and fatty acids; both are sent to the glucose step in the learner’s map. The map is reproduced for audit, without corrected entry points. QUESTION DIAGRAM The learner’s fat-entry map Learner’s proposed route · not the accepted pathway Fat Glycerol Fatty acids Glucose step
  1. (a)Keep both routes at glucose because all respiratory substrates must enter at the first step.
  2. (b)Send glycerol to PGAL and fatty acids to acetyl CoA; neither needs to enter at glucose.
  3. (c)Send glycerol into the Krebs' cycle and fatty acids to pyruvate without any prior conversion.
  4. (d)Move both routes together to acetyl CoA because glycerol and fatty acids share one entry point.
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(b) Send glycerol to PGAL and fatty acids to acetyl CoA; neither needs to enter at glucose.

Glycerol enters the respiratory pathway after conversion to PGAL, while fatty acids are degraded to acetyl CoA before entry. Thus neither product must return to glucose, and the two products do not share one entry point.

Question 14

A metabolic map contains two separate links: fatty acids are broken down to acetyl CoA before entering respiration, and acetyl CoA is withdrawn when fatty acids are synthesised. Which interpretation is correct?

  1. (a)Acetyl CoA can participate only in synthesis, so its entry into respiration cannot represent catabolism.
  2. (b)The two links mean that one individual reaction must simultaneously break down and synthesise fatty acids.
  3. (c)Only the breakdown link matters for respiration, so the respiratory pathway remains purely catabolic.
  4. (d)The two links show catabolic entry and anabolic removal, so the respiratory pathway is amphibolic at the network level.
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(d) The two links show catabolic entry and anabolic removal, so the respiratory pathway is amphibolic at the network level.

The first link is catabolic and the second is anabolic. Their connection through acetyl CoA shows that the respiratory pathway functions in both roles as a network, so it is amphibolic rather than only catabolic.

Respiratory quotient calculation and substrate inference

Question 15

A learner claims: ‘All respiratory substrates have RQ 1.0. Also, whenever an organism has RQ 1.0, that value proves it is respiring pure carbohydrate alone.’ Which review is complete?

  1. (a)Only the second claim is faulty; every substrate does have RQ 1.0, but the value cannot identify carbohydrate.
  2. (b)Both claims are correct because RQ is independent of substrate and uniquely identifies carbohydrate at 1.0.
  3. (c)Only the first claim is faulty; substrate benchmarks differ, so RQ 1.0 proves that pure carbohydrate alone is used.
  4. (d)Both claims are faulty: RQ depends on substrate, and one observed value does not necessarily identify a single pure substrate.
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(d) Both claims are faulty: RQ depends on substrate, and one observed value does not necessarily identify a single pure substrate.

The first claim wrongly generalises carbohydrate's equal gas volumes: fat has RQ below one and protein about 0.9. The second claim ignores mixed substrate use, so an observed value need not identify one pure substrate with certainty.

Common misconception in Respiration in Plants

Open full misconception guide →

Do plants only respire at night?

Wrong

Photosynthesis replaces respiration in daylight.

Why it feels right. The leaf releases oxygen overall in bright light, so I think none is being used.

Right

A lit leaf can photosynthesise and respire at the same time.

Worked example

Photosynthesis
Light-driven reactions→O₂ released
Respiration
Mitochondria→O₂ consumed
Net in bright light
Release exceeds use→O₂ leaves the leaf
Both pathways operate in a lit leaf. Net release depends on the difference between them.
  1. Mitochondria in a lit leaf keep using oxygen during aerobic respiration.
  2. If photosynthesis produces more oxygen than respiration consumes, the leaf still releases oxygen overall.