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NCERT Solutions · Class 11 Biology · Chapter 12

Chapter 12: Respiration in Plants (Plant Physiology)

Answers to all NCERT exercises of Chapter 12, Respiration in Plants (NCERT Class 11 Biology, 2026-27): glycolysis, fermentation, Krebs cycle, electron transport system and respiratory quotient. Each answer is written in full, the way it is expected in school and board exams. All 12 questions are answered, with the key answer highlighted.

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Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-11-biology/chapter-12-respiration-in-plants

Exercises

1
Differentiate: (a) respiration and combustion (b) glycolysis and Krebs cycle (c) aerobic respiration and fermentation.
Solution

(a)

respirationcombustion
biochemical, inside living cellsphysico-chemical, non-living
stepwise, enzyme-controlled, at body temperaturesingle uncontrolled step at high temperature
energy trapped as ATP, little heatenergy released as heat and light
intermediates used for biosynthesisno intermediates

(b)

glycolysisKrebs cycle
cytoplasm; all organismsmitochondrial matrix; aerobes
no O₂ neededworks only when O₂ is available
glucose → 2 pyruvate; linearacetyl CoA oxidised to CO₂; cyclic
no CO₂; net 2 ATP, 2 NADHCO₂ released; NADH, FADH₂, ATP (GTP)

(c)

aerobic respirationfermentation
O₂ needed; mitochondriano O₂; cytoplasm
complete oxidation to CO₂ + H₂Oincomplete: ethanol + CO₂ or lactic acid
38 ATP2 ATP
NADH oxidised vigorously by ETSNADH oxidised slowly by the product step
2
What are respiratory substrates? Name the most common one.
Solution

Respiratory substrates are the organic compounds oxidised in respiration to release energy: carbohydrates, fats, proteins and organic acids. The most common one is glucose (carbohydrates).

3
Give the schematic representation of glycolysis.
Solution

See the glycolysis flow chart in the concept notes: Glucose → (−ATP, hexokinase) glucose-6-phosphate → fructose-6-phosphate → (−ATP) fructose-1,6-bisphosphate → splits into PGAL + DHAP (DHAP → PGAL) → 2 × 1,3-BPGA (+2 NADH) → 2 × 3-PGA (+2 ATP) → 2 × 2-phosphoglycerate → 2 × PEP (−H₂O) → 2 × pyruvic acid (+2 ATP). Net: 2 ATP + 2 NADH per glucose.

4
What are the main steps in aerobic respiration? Where do they take place?
Solution

(i) Glycolysis: glucose → 2 pyruvate, in the cytoplasm. (ii) Oxidative decarboxylation (link reaction): pyruvate → acetyl CoA + CO₂ + NADH, in the mitochondrial matrix. (iii) Krebs cycle: acetyl CoA oxidised to CO₂ forming NADH, FADH₂ and ATP, in the matrix. (iv) ETS and oxidative phosphorylation: NADH and FADH₂ oxidised, electrons to O₂ forming water, ATP made by ATP synthase, on the inner mitochondrial membrane.

5
Give a schematic overview of the Krebs cycle.
Solution

See the Krebs cycle diagram in the concept notes: Acetyl CoA (2C) + OAA (4C) → citric acid (6C) → isocitrate → α-ketoglutaric acid (5C) [CO₂, NADH] → succinyl-CoA (4C) [CO₂, NADH] → succinic acid [GTP → ATP] → fumaric acid [FADH₂] → malic acid → OAA [NADH]. Per turn: 2 CO₂, 3 NADH, 1 FADH₂, 1 ATP (with the link reaction: 3 CO₂, 4 NADH per pyruvate).

6
Explain ETS.
Solution

The electron transport system is the chain of electron carriers in the inner mitochondrial membrane through which electrons from NADH and FADH₂ pass to O₂. Complex I (NADH dehydrogenase) passes electrons from NADH to ubiquinone; complex II passes electrons from FADH₂ to ubiquinone; reduced ubiquinone passes them via complex III (cytochrome bc₁) to the mobile carrier cytochrome c, then to complex IV (cytochrome c oxidase: cytochromes a, a₃, two Cu centres), which gives them to O₂, forming water. The energy released pumps protons to the intermembrane space; their return through ATP synthase (complex V: F₀ + F₁) makes ATP (oxidative phosphorylation): 3 ATP per NADH, 2 per FADH₂.

7
Distinguish: (a) aerobic and anaerobic respiration (b) glycolysis and fermentation (c) glycolysis and citric acid cycle.
Solution

(a) Aerobic: uses O₂; complete oxidation to CO₂ + H₂O; in cytoplasm and mitochondria; 38 ATP. Anaerobic: no O₂; incomplete oxidation to ethanol + CO₂ or lactic acid; only in cytoplasm; 2 ATP. (b) Glycolysis: glucose → pyruvic acid, the first step of both aerobic and anaerobic respiration; produces 2 ATP and 2 NADH. Fermentation: glycolysis followed by conversion of pyruvate to ethanol + CO₂ or lactic acid, using the NADH (reoxidised to NAD⁺); no extra ATP. (c) As in Q1(b).

8
What assumptions are made in calculating the net gain of ATP?
Solution

(i) There is a sequential, orderly pathway: glycolysis, TCA cycle and ETS follow one after another. (ii) The NADH made in glycolysis is transferred into mitochondria and undergoes oxidative phosphorylation. (iii) None of the intermediates is used to synthesise other compounds. (iv) Only glucose is respired; no other substrate enters at intermediate stages. (Under these, 38 ATP per glucose.)

9
Discuss: "The respiratory pathway is an amphibolic pathway."
Solution

Respiration was considered catabolic (breakdown). But other substrates enter the pathway at particular points (fatty acids as acetyl CoA, glycerol as PGAL, amino acids as pyruvate, acetyl CoA or Krebs intermediates), and the same intermediates are withdrawn from it to synthesise these substances: acetyl CoA for fatty acids, α-ketoglutarate and OAA for amino acids. Since the pathway takes part in both catabolism and anabolism, it is better called amphibolic.

10
Define RQ. What is its value for fats?
Solution

RQ (respiratory quotient) = volume of CO₂ evolved / volume of O₂ consumed in respiration. For fats it is less than 1: for tripalmitin, 2C₅₁H₉₈O₆ + 145O₂ → 102CO₂ + 98H₂O, RQ = 102/145 = 0.7.

11
What is oxidative phosphorylation?
Solution

The synthesis of ATP from ADP and Pi using the energy of oxidation–reduction released as electrons from NADH and FADH₂ pass through the ETS to O₂ on the inner mitochondrial membrane. The energy pumps protons across the membrane; their flow back through ATP synthase (F₀ channel, F₁ catalytic site) drives ATP formation. It is called "oxidative" because redox energy, not light (as in photophosphorylation), creates the proton gradient.

12
What is the significance of stepwise release of energy in respiration?
Solution

(i) If glucose were oxidised in one step (combustion), most energy would be lost as heat, which could damage cells. (ii) Releasing energy in small enzyme-controlled steps lets some steps be just large enough to be coupled to ATP synthesis, so energy is stored efficiently. (iii) It allows control of the rate. (iv) The many intermediates can be used for synthesising other molecules (amphibolic role) and other substrates can enter at different points.

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