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Class 11 Biology Chapter 12 Respiration in Plants — Formulas & Key Points

CBSE Class 11 Biology Chapter 12 Respiration in Plants is a numerical and concept-heavy chapter tested heavily in board exams and NEET. Students must master the exact ATP yields at each stage—Glycolysis, Krebs cycle, and Electron Transport Chain—and calculate Respiratory Quotient for different substrates. This formula sheet collects every equation, intermediate, and value in tabular format for rapid revision, alongside memory aids and solved examples.

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Key takeaways

✓Glycolysis yields a net gain of 2 ATP and 2 NADH per glucose molecule, occurring entirely in the cytoplasm without oxygen.
✓The Krebs cycle completes two turns per glucose molecule, producing 6 NADH, 2 FADH₂, and 2 ATP in the mitochondrial matrix.
✓Electron Transport Chain oxidises NADH and FADH₂ to generate approximately 34 ATP through oxidative phosphorylation.
✓Respiratory Quotient (RQ = CO₂ released / O₂ consumed) equals 1 for carbohydrates, 0.7 for fats, and 0.9 for proteins.
✓Complete aerobic respiration of one glucose molecule theoretically yields 38 ATP, but net practical yield is about 30-32 ATP.
✓Fermentation pathways (alcoholic and lactic acid) regenerate NAD⁺ under anaerobic conditions but yield only 2 ATP per glucose.
✓Pentose phosphate pathway generates NADPH and ribose-5-phosphate without producing ATP, supporting biosynthesis.

Core Formulas and Equations in Respiration

Respiration in plants follows a sequence of chemical reactions, each with defined inputs and outputs. The overall equation for aerobic respiration summarises the complete oxidation of glucose, but board exams require you to know the step-wise ATP accounting. The table below lists the main equations tested in CBSE Class 11 Biology Chapter 12. NCERT Class 11 Biology emphasises the distinction between substrate-level phosphorylation (direct ATP synthesis in Glycolysis and Krebs cycle) and oxidative phosphorylation (ATP synthesis via ETC). Remember that the theoretical maximum of 38 ATP assumes ideal conditions; in practice, the proton gradient and transport costs reduce net yield to 30-32 ATP per glucose molecule.

Overall aerobic respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + Energy (ATP + Heat)
Glycolysis (cytoplasm): C₆H₁₂O₆ + 2NAD⁺ + 2ADP + 2Pᵢ → 2 Pyruvate + 2NADH + 2H⁺ + 2ATP + 2H₂O
Oxidative decarboxylation of pyruvate: 2 Pyruvate + 2NAD⁺ + 2CoA → 2 Acetyl-CoA + 2NADH + 2CO₂
One turn of Krebs cycle: Acetyl-CoA + 3NAD⁺ + FAD + ADP + Pᵢ + 2H₂O → 2CO₂ + 3NADH + FADH₂ + ATP + CoA
Net ATP from complete oxidation: 1 glucose → ~30-32 ATP (practical) or 38 ATP (theoretical maximum)

Glycolysis Step-by-Step Formulas

Glycolysis is the first stage of respiration, splitting one glucose (6-carbon) into two pyruvate (3-carbon) molecules. It occurs in the cytoplasm and does not require oxygen, making it common to both aerobic and anaerobic pathways. The pathway has an investment phase (consuming 2 ATP) and a payoff phase (producing 4 ATP and 2 NADH), resulting in a net gain of 2 ATP and 2 NADH. Class 11 Biology solutions often ask you to identify the enzymes catalysing each step—hexokinase, phosphofructokinase, and pyruvate kinase are the three regulatory enzymes. Exam questions may also test which steps are irreversible or where substrate-level phosphorylation occurs. The table below summarises the key steps and their ATP balance.

Step 1: Glucose + ATP → Glucose-6-phosphate + ADP (hexokinase)
Step 2: Glucose-6-phosphate ⇌ Fructose-6-phosphate (phosphoglucose isomerase)
Step 3: Fructose-6-phosphate + ATP → Fructose-1,6-bisphosphate + ADP (phosphofructokinase, rate-limiting)
Step 6 & 7: 2(1,3-bisphosphoglycerate) + 2ADP → 2(3-phosphoglycerate) + 2ATP (substrate-level phosphorylation)
Step 10: 2(Phosphoenolpyruvate) + 2ADP → 2 Pyruvate + 2ATP (pyruvate kinase)

Krebs Cycle (Citric Acid Cycle) Intermediates and Yields

The Krebs cycle, also called the citric acid cycle or TCA cycle, runs in the mitochondrial matrix. Each acetyl-CoA (2-carbon) enters the cycle by condensing with oxaloacetate (4-carbon) to form citrate (6-carbon). Over eight enzyme-catalysed steps, two CO₂ molecules are released, regenerating oxaloacetate. One turn of the cycle yields 3 NADH, 1 FADH₂, and 1 ATP (or GTP). Because one glucose produces two acetyl-CoA, the Krebs cycle turns twice per glucose molecule, doubling all outputs. CBSE Class 11 Biology notes emphasise that the cycle does not directly use O₂ but depends on aerobic conditions because NAD⁺ and FAD must be regenerated by the Electron Transport Chain. The table below lists the eight intermediates in sequence, which is a favourite fill-in-the-blank question in board exams and NEET.

Citrate (6C) → cis-Aconitate (6C) → Isocitrate (6C) → α-Ketoglutarate (5C) → Succinyl-CoA (4C) → Succinate (4C) → Fumarate (4C) → Malate (4C) → Oxaloacetate (4C)
One turn: 3 NADH + 1 FADH₂ + 1 ATP (or GTP)
Two turns per glucose: 6 NADH + 2 FADH₂ + 2 ATP
CO₂ released: Two molecules per turn (at isocitrate → α-ketoglutarate and α-ketoglutarate → succinyl-CoA)
Enzymes tested: Citrate synthase, Isocitrate dehydrogenase, α-Ketoglutarate dehydrogenase, Succinate dehydrogenase (only membrane-bound enzyme)

Electron Transport Chain and ATP Synthesis

The Electron Transport Chain (ETC) is located on the inner mitochondrial membrane and consists of four protein complexes (I, II, III, IV) plus mobile carriers (ubiquinone and cytochrome c). NADH donates electrons to Complex I, while FADH₂ donates to Complex II. As electrons cascade down the chain, protons are pumped from the mitochondrial matrix into the intermembrane space, creating an electrochemical gradient. ATP synthase (Complex V) uses this proton-motive force to phosphorylate ADP into ATP—a process called oxidative phosphorylation or chemiosmosis. The theoretical ATP yield is 2.5-3 ATP per NADH and 1.5-2 ATP per FADH₂, though NCERT Class 11 Biology often rounds to 3 and 2 for simplicity. The final electron acceptor is molecular oxygen, which combines with protons to form water. Class 11 Biology solutions emphasise that cyanide and carbon monoxide inhibit Complex IV, halting the entire chain and ATP production.

NADH → Complex I → 10 H⁺ pumped → ~2.5-3 ATP
FADH₂ → Complex II → 6 H⁺ pumped → ~1.5-2 ATP
Total theoretical ATP from 10 NADH: 10 × 3 = 30 ATP
Total theoretical ATP from 2 FADH₂: 2 × 2 = 4 ATP
Net practical yield per glucose: ~30-32 ATP (accounting for transport and proton leak)

Respiratory Quotient (RQ) Formula and Values

The Respiratory Quotient (RQ) is the ratio of the volume of CO₂ released to the volume of O₂ consumed during respiration. It helps identify which substrate—carbohydrate, fat, or protein—is being oxidised. The formula is RQ = CO₂ evolved / O₂ consumed. For complete oxidation of glucose (C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O), RQ = 6/6 = 1. Fats have more hydrogen and less oxygen per molecule, so they require more O₂ relative to CO₂ produced, yielding RQ ≈ 0.7. Proteins give RQ ≈ 0.9. Organic acids like malic acid or oxalic acid have RQ > 1 because they already contain oxygen and release more CO₂ than O₂ consumed. NCERT Class 11 Biology problems often ask you to calculate RQ from a given respiratory equation or identify the substrate based on RQ value. The table below lists standard RQ values for different respiratory substrates, which is a common 1-mark board exam question.

Carbohydrates (glucose): RQ = 1.0
Fats (tripalmitin): RQ ≈ 0.7
Proteins: RQ ≈ 0.9
Organic acids (e.g. malic acid): RQ > 1.0
Succulent plants (CAM) under certain conditions: RQ can be zero or infinite (non-steady state)

Key Definitions and Terminology

Class 11 Biology Chapter 12 introduces several technical terms that appear in definition-based questions and multiple-choice items in CBSE board exams and NEET. Aerobic respiration refers to the complete oxidation of glucose in the presence of oxygen, yielding maximum ATP. Anaerobic respiration or fermentation occurs without oxygen, producing ethanol and CO₂ (in yeast) or lactic acid (in muscle cells), with a net gain of only 2 ATP per glucose. Substrate-level phosphorylation is the direct transfer of a phosphate group from a substrate to ADP, occurring in Glycolysis and the Krebs cycle. Oxidative phosphorylation is ATP synthesis driven by the proton gradient established by the Electron Transport Chain. Chemiosmotic hypothesis, proposed by Peter Mitchell, explains how the proton-motive force across the inner mitochondrial membrane drives ATP synthesis. Amphibolic pathway means the pathway can function in both catabolism and anabolism—Krebs cycle intermediates are used to synthesise amino acids and other molecules. Understanding these definitions ensures you can tackle 1-mark 'define' questions and correctly interpret longer reasoning items.

Aerobic respiration: Complete oxidation of glucose using O₂, yielding ~30-32 ATP
Anaerobic respiration (Fermentation): Partial breakdown of glucose without O₂, yielding 2 ATP
Substrate-level phosphorylation: Direct ATP synthesis by phosphate transfer (Glycolysis, Krebs cycle)
Oxidative phosphorylation: ATP synthesis via ETC and chemiosmosis
Chemiosmotic hypothesis: ATP formation driven by proton gradient across inner mitochondrial membrane
Amphibolic pathway: Krebs cycle serves both catabolic and anabolic roles
Respiratory substrate: Organic molecule (carbohydrate, fat, protein) oxidised during respiration

Important Constants, Values and Units

Respiration in Plants involves several quantitative values that CBSE examiners test repeatedly. The net ATP yield per glucose is often asked: Glycolysis produces 2 ATP (net), the Krebs cycle produces 2 ATP (total for two turns), and the ETC produces approximately 34 ATP, summing to a theoretical maximum of 38 ATP. In practice, mitochondrial transport costs reduce this to about 30-32 ATP. Each NADH entering the ETC theoretically yields 3 ATP, and each FADH₂ yields 2 ATP, though some textbooks use 2.5 and 1.5 for more accuracy. The respiratory quotient has no units because it is a ratio of volumes. Enzymes like hexokinase (Km ≈ 0.1 mM for glucose) and phosphofructokinase (allosterically inhibited by ATP and citrate) are regulatory checkpoints. NCERT Class 11 Biology also mentions that the energy content of one mole of glucose is about 686 kcal, but only ~40-45% is captured as ATP; the rest is lost as heat. Familiarising yourself with these numbers will speed up numerical problem-solving in exams and help you cross-check your answers for errors.

Net ATP from one glucose (practical): ~30-32 ATP
Theoretical maximum ATP: 38 ATP (or 36 ATP if NADH from Glycolysis costs 2 ATP to shuttle into mitochondria)
ATP per NADH: ~3 ATP (or 2.5 ATP in some texts)
ATP per FADH₂: ~2 ATP (or 1.5 ATP in some texts)
Energy content of glucose: ~686 kcal/mol; ~40-45% captured as ATP
RQ values: Carbohydrates = 1.0, Fats = 0.7, Proteins = 0.9, Organic acids > 1.0
Glycolysis net gain: 2 ATP + 2 NADH per glucose
Krebs cycle per turn: 3 NADH + 1 FADH₂ + 1 ATP

Memory Tricks and Mnemonics

Remembering the eight intermediates of the Krebs cycle in order is a common challenge. Use the mnemonic 'Can I Keep Selling Seashells For Money, Officer?' to recall: Citrate, Isocitrate, α-Ketoglutarate (often written as Ketoglutarate), Succinyl-CoA, Succinate, Fumarate, Malate, Oxaloacetate. For Glycolysis steps, 'Good Grapes Ferment Directly, Producing Great Beverages' helps: Glucose, Glucose-6-P, Fructose-6-P, Fructose-1,6-bisP, DHAP/G3P, 1,3-BPG, 3-PG, 2-PG, PEP, Pyruvate. To remember RQ values, think 'Fats are lighter (0.7), Carbs are one-to-one (1.0), Proteins in between (0.9)'. For the three irreversible steps of Glycolysis, recall 'Hexokinase, PFK, Pyruvate Kinase' using the acronym HPP. These mnemonics save precious minutes during board exams and reduce silly errors on fill-in-the-blank questions. Many Class 11 Biology notes and NEET coaching centres recommend creating your own visual or rhyme-based memory aids for the multi-step pathways, as active engagement strengthens recall far better than passive reading.

Krebs cycle intermediates: 'Can I Keep Selling Seashells For Money, Officer?'
Glycolysis: 'Good Grapes Ferment Directly, Producing Great Beverages'
RQ values: Fats 0.7 (lighter), Carbs 1.0 (one-to-one), Proteins 0.9 (in between)
Irreversible Glycolysis steps: HPP (Hexokinase, PFK, Pyruvate Kinase)
ETC complexes: 'I, II, III, IV' plus 'Q' for ubiquinone and 'Cyt c' for cytochrome c
ATP accounting: 'Glyco-2, Krebs-2, ETC-34' to quickly sum to 38 theoretical

Common Sign, Unit and Notation Mistakes

Students often lose marks on respiration numericals due to small notation errors. First, RQ is dimensionless—never write 'RQ = 0.7 units'. Second, ATP counts must be integers or simple fractions; avoid writing '2.73 ATP' unless the question explicitly asks for decimal precision. Third, when writing chemical equations, balance atoms and charges: the overall equation C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O must show equal carbons (6 on each side), hydrogens (12), and oxygens (18). Fourth, distinguish NADH from NAD⁺ and FADH₂ from FAD—incorrect subscripts or charges signal conceptual confusion. Fifth, when calculating net ATP, remember to subtract the ATP invested in Glycolysis (2 ATP); writing gross ATP (4) instead of net ATP (2) is a frequent mistake. Sixth, Krebs cycle intermediates are often misspelled—Isocitrate, not 'Iso-citrate'; α-Ketoglutarate, not 'Ketogluterate'. Seventh, state the location: Glycolysis in cytoplasm, Krebs cycle in mitochondrial matrix, ETC on inner mitochondrial membrane. Class 11 Biology solutions that omit these details lose marks in descriptive answers. Finally, when asked for 'total ATP from one glucose', clarify whether the question wants theoretical maximum (38) or practical yield (~30-32).

RQ is a dimensionless ratio—never add 'units'
ATP counts are whole numbers or simple fractions (2.5, 3), not arbitrary decimals
Balance all chemical equations for C, H, O atoms
Distinguish NADH/NAD⁺ and FADH₂/FAD; incorrect oxidation states lose marks
Glycolysis net ATP = 4 produced − 2 invested = 2 net
Spell intermediates correctly: Isocitrate, α-Ketoglutarate, Succinyl-CoA, etc.
State cellular location: cytoplasm, mitochondrial matrix, or inner mitochondrial membrane
Clarify theoretical (38) vs. practical (~30-32) ATP yield when asked

Solved Mini-Examples Applying Formulas

Worked examples consolidate formula application. Example 1: Calculate the RQ when 9 mL of O₂ is consumed and 9 mL of CO₂ is released. Solution: RQ = CO₂/O₂ = 9/9 = 1.0, indicating carbohydrate respiration. Example 2: A student isolates mitochondria and adds succinate plus ADP and Pi. Which Krebs cycle enzyme is directly involved, and what coenzyme is reduced? Solution: Succinate dehydrogenase oxidises succinate to fumarate, reducing FAD to FADH₂. Example 3: If a plant cell produces 10 NADH and 2 FADH₂ during complete glucose oxidation, estimate the ATP from oxidative phosphorylation alone, using the 3:2 ratio. Solution: (10 × 3) + (2 × 2) = 30 + 4 = 34 ATP from the ETC. Add substrate-level ATP (2 from Glycolysis + 2 from Krebs cycle) for a grand total of 38 ATP theoretical maximum. These mini-examples mirror the style of CBSE Class 11 Biology Chapter 12 board questions and NEET items. Practising them builds speed and confidence.

One-Glance Last-Minute Revision Box

Use this condensed table the night before your exam to verify you have memorised every critical formula, value, and sequence. Glycolysis: Cytoplasm, 2 ATP net, 2 NADH. Krebs cycle (per glucose, two turns): Matrix, 6 NADH, 2 FADH₂, 2 ATP. ETC: Inner mitochondrial membrane, ~34 ATP. Total theoretical: 38 ATP; practical: ~30-32 ATP. RQ: Carbs 1.0, Fats 0.7, Proteins 0.9. Krebs intermediates: Citrate, Isocitrate, α-Ketoglutarate, Succinyl-CoA, Succinate, Fumarate, Malate, Oxaloacetate. Irreversible Glycolysis enzymes: Hexokinase, PFK, Pyruvate Kinase. Chemiosmotic hypothesis: Proton gradient drives ATP synthase. Amphibolic: Krebs cycle serves catabolism and anabolism. Keep this box on your phone or print it on a flashcard for quick review during study breaks. Many toppers revise this single page an hour before entering the exam hall to refresh the exact numbers and spellings that examiners love to test.

Glycolysis: Cytoplasm, net 2 ATP, 2 NADH
Krebs cycle (2 turns): Matrix, 6 NADH, 2 FADH₂, 2 ATP, 4 CO₂
ETC: Inner mitochondrial membrane, ~34 ATP (10 NADH × 3 + 2 FADH₂ × 2)
Total ATP: Theoretical 38, Practical ~30-32
RQ: Carbs = 1.0, Fats = 0.7, Proteins = 0.9, Organic acids > 1.0
Krebs intermediates mnemonic: 'Can I Keep Selling Seashells For Money, Officer?'
Irreversible Glycolysis steps: Hexokinase, PFK, Pyruvate Kinase
Chemiosmotic hypothesis: H⁺ gradient → ATP synthase → ATP

How CBSETUTOR.ai Helps Master Respiration in Plants

Respiration in Plants is dense with multi-step pathways, enzyme names, and ATP accounting that many students find overwhelming. CBSETUTOR.ai offers a 24×7 AI tutor that lets you photograph any diagram—whether it is the Krebs cycle wheel from your textbook or a half-finished Glycolysis flowchart from your notebook—and receive instant, step-by-step explanations in simple language. Stuck on why FADH₂ yields fewer ATP than NADH? Upload your doubt and get a clear answer referencing Complex II entry into the ETC. Preparing for a unit test on RQ calculations? The platform generates practice problems at varying difficulty levels, complete with worked solutions that match CBSE marking schemes. At just ₹999 per month for all subjects across Classes 6–12, it is far more affordable than hiring separate Biology tutors or joining expensive coaching centres. The 3-day free trial lets your child explore the respiration module risk-free, making last-minute revision efficient and stress-free. Thousands of CBSE students already rely on CBSETUTOR.ai to clarify tricky metabolic pathways, ensuring they walk into the exam hall confident about every intermediate, every enzyme, and every ATP molecule.

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Frequently asked questions

What is the net ATP yield from Glycolysis per glucose molecule?+

Glycolysis produces 4 ATP by substrate-level phosphorylation but consumes 2 ATP in the investment phase, giving a net gain of 2 ATP. It also generates 2 NADH molecules that can yield additional ATP via the Electron Transport Chain.

How many turns of the Krebs cycle occur per glucose molecule?+

One glucose molecule produces two pyruvate molecules, each forming one acetyl-CoA. Since one acetyl-CoA completes one turn of the Krebs cycle, two turns occur per glucose, yielding 6 NADH, 2 FADH₂, and 2 ATP in total.

Why does FADH₂ produce less ATP than NADH in the ETC?+

FADH₂ donates electrons to Complex II of the Electron Transport Chain, bypassing Complex I. This results in fewer protons being pumped across the inner mitochondrial membrane, yielding approximately 2 ATP per FADH₂ compared to 3 ATP per NADH.

What is the formula for calculating Respiratory Quotient (RQ)?+

RQ = Volume of CO₂ released / Volume of O₂ consumed. It is a dimensionless ratio that identifies the respiratory substrate: 1.0 for carbohydrates, 0.7 for fats, 0.9 for proteins, and greater than 1.0 for organic acids.

Which steps of Glycolysis are irreversible?+

The three irreversible steps are catalysed by hexokinase (glucose → glucose-6-phosphate), phosphofructokinase (fructose-6-phosphate → fructose-1,6-bisphosphate), and pyruvate kinase (phosphoenolpyruvate → pyruvate). These are key regulatory points.

Where does the Krebs cycle occur in the cell?+

The Krebs cycle takes place in the mitochondrial matrix. All eight enzymes of the cycle are located in the matrix except succinate dehydrogenase, which is embedded in the inner mitochondrial membrane.

What is the mnemonic to remember Krebs cycle intermediates?+

Use 'Can I Keep Selling Seashells For Money, Officer?' to recall the sequence: Citrate, Isocitrate, α-Ketoglutarate, Succinyl-CoA, Succinate, Fumarate, Malate, Oxaloacetate. This mnemonic helps in fill-in-the-blank and diagram-labelling questions.

How much ATP is produced theoretically from one glucose molecule?+

Theoretical maximum is 38 ATP: Glycolysis (2 ATP + 2 NADH → 6 ATP), Oxidative decarboxylation (2 NADH → 6 ATP), Krebs cycle (2 ATP + 6 NADH + 2 FADH₂ → 24 ATP). Practically, cells obtain about 30-32 ATP due to transport costs and proton leakage.

What is the difference between substrate-level and oxidative phosphorylation?+

Substrate-level phosphorylation directly transfers a phosphate group from a high-energy substrate to ADP, occurring in Glycolysis and Krebs cycle. Oxidative phosphorylation generates ATP via the proton gradient established by the Electron Transport Chain and ATP synthase.

Why is the Krebs cycle called an amphibolic pathway?+

The Krebs cycle is amphibolic because it participates in both catabolism (breaking down acetyl-CoA for energy) and anabolism (providing intermediates like α-ketoglutarate and oxaloacetate for biosynthesis of amino acids, nucleotides, and other molecules).

Related resources

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