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Important Questions: CBSE Class 11 Biology Chapter 12 Respiration in Plants
Respiration in Plants is a cornerstone chapter in CBSE Class 11 Biology, bridging biochemistry and plant physiology. The 2024-25 curriculum emphasizes understanding glycolysis, Krebs cycle, electron transport chain, and respiratory quotient through numerical and application-based problems. This question bank mirrors actual CBSE exam patterns, offering 18 important questions with detailed solutions to help you master every concept and secure full marks.
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Key takeaways
- ✓Chapter 12 Respiration in Plants typically carries 8-10 marks in CBSE Class 11 Biology term exams, distributed across MCQs and descriptive questions
- ✓Glycolysis and Krebs cycle are high-weightage topics with frequent 3-mark and 5-mark questions in board exams
- ✓The respiratory quotient concept appears in at least one numerical or application-based question every year
- ✓Electron transport chain mechanism and ATP synthesis are commonly tested through 3-mark diagram-based questions
- ✓CBSE often frames case-based questions linking respiration to real-world scenarios like fermentation or altitude effects
- ✓Common mistakes include confusing glycolysis location (cytoplasm vs mitochondria) and miscalculating net ATP yield
- ✓Practising 15-18 varied questions with model answers builds confidence for both objective and subjective sections
Chapter Overview and Marks Weightage in CBSE Exam
Respiration in Plants forms part of Unit III (Plant Physiology) in the CBSE Class 11 Biology syllabus and typically accounts for 8-10 marks in the annual examination. Questions are distributed across multiple formats: 1-mark MCQs test recall of definitions and locations of metabolic pathways, 2-mark questions assess understanding of specific steps like pyruvate oxidation, 3-mark questions demand explanation of Krebs cycle intermediates or ETC functioning, and 5-mark questions evaluate comprehensive understanding through case studies or comparative analyses. The 2023 and 2024 board papers both featured one mandatory 3-mark question on aerobic respiration and one 5-mark case-based question integrating respiratory quotient calculations. Competitive exams like NEET allocate 2-3 questions annually to this chapter, focusing on glycolysis ATP yield and FAD/NAD roles.
- 1-mark questions (2-3 per paper): definitions, enzyme names, locations of pathways
- 2-mark questions (1-2 per paper): differences between aerobic and anaerobic respiration, significance of fermentation
- 3-mark questions (1-2 per paper): glycolysis steps, Krebs cycle diagram, ETC mechanism
- 5-mark questions (1 per paper): case-based scenarios, complete ATP accounting, experimental data interpretation
1-Mark Questions: MCQs and Very Short Answers
One-mark questions in CBSE Class 11 Biology Chapter 12 test foundational knowledge of respiration terminology, enzyme names, and pathway locations. These appear as multiple-choice questions or very short answer (VSA) format requiring single-word to one-sentence responses. Mastering these ensures quick scoring and builds confidence for descriptive sections. NCERT Class 11 Biology textbook pages 224-239 contain all necessary definitions. The 2025 sample paper released by CBSE includes at least three MCQs from this chapter, emphasizing glycolysis location, respiratory substrates, and the role of oxygen in aerobic respiration. Students should focus on exact NCERT terminology—for instance, using 'EMP pathway' interchangeably with glycolysis, and remembering that the Krebs cycle is also called the TCA cycle or citric acid cycle.
- Q1. Where does glycolysis occur in a eukaryotic cell? A: Cytoplasm
- Q2. Name the process that converts pyruvate to ethanol in yeast. A: Alcoholic fermentation
- Q3. What is the respiratory quotient of carbohydrates? A: 1.0 (RQ = CO₂ released / O₂ consumed = 6/6)
- Q4. Which coenzyme is reduced in the Krebs cycle? A: NAD⁺ and FAD (accept either)
- Q5. In which part of mitochondria does the electron transport chain operate? A: Inner mitochondrial membrane (cristae)
2-Mark Questions with Model Answers
Two-mark questions require concise explanations spanning 30-40 words, often asking for differences, significance, or brief descriptions of a single step. CBSE marking schemes award 1 mark for correct concept identification and 1 mark for accurate explanation. Students must avoid vague statements—use specific molecules and enzyme names from NCERT Class 11 Biology Chapter 12. For instance, when explaining pyruvate oxidation, mention that pyruvate (3C) enters the mitochondrial matrix, loses one CO₂ via oxidative decarboxylation catalyzed by the pyruvate dehydrogenase complex, and forms acetyl CoA (2C), simultaneously reducing NAD⁺ to NADH. Practice writing within word limits; examiners penalize verbose answers that exceed space constraints. The 2024 CBSE marking scheme explicitly states that bullet-point answers are acceptable if each point is complete.
- Q7. Differentiate between aerobic and anaerobic respiration based on end products. A: Aerobic respiration completely oxidizes glucose to CO₂ and H₂O in the presence of oxygen, releasing ~686 kcal energy. Anaerobic respiration partially oxidizes glucose to ethanol (in yeast) or lactic acid (in muscles) without oxygen, releasing only ~50 kcal energy per glucose.
- Q8. What is the significance of the respiratory quotient? A: RQ indicates the type of respiratory substrate being oxidized. RQ=1 for carbohydrates, <1 for fats (0.7), >1 for organic acids, and ∞ for anaerobic respiration (no O₂ consumed). It helps determine metabolic state.
- Q9. Why is glycolysis called the EMP pathway? A: Named after scientists Embden, Meyerhof, and Parnas who elucidated the sequence of ten enzyme-catalyzed reactions converting glucose (6C) to two pyruvate molecules (3C each) in the cytoplasm, yielding net 2 ATP and 2 NADH.
3-Mark Questions: Glycolysis and Krebs Cycle
Three-mark questions demand structured explanations with clear sequences, often accompanied by schematic diagrams. CBSE awards 1 mark for labeling, 1 mark for pathway steps, and 1 mark for outcomes (ATP/NADH yield). When describing glycolysis, mention it is a ten-step process occurring in the cytoplasm, does not require oxygen (hence occurs in both aerobic and anaerobic organisms), begins with glucose (6C) being phosphorylated by ATP (investment phase consuming 2 ATP), splits into two 3-carbon molecules, and proceeds through oxidation and substrate-level phosphorylation to yield 4 ATP (net gain 2 ATP) and 2 NADH. For Krebs cycle questions, emphasize that it occurs in the mitochondrial matrix, begins with acetyl CoA (2C) combining with oxaloacetic acid (4C) to form citric acid (6C), proceeds through two decarboxylation steps releasing 2 CO₂, and regenerates oxaloacetic acid. Each turn produces 3 NADH, 1 FADH₂, and 1 GTP (equivalent to ATP). Since one glucose yields two acetyl CoA, multiply these figures by two.
- Q11. Describe the steps of glycolysis with net ATP gain. A: (i) Glucose is phosphorylated to glucose-6-phosphate using 1 ATP. (ii) Isomerization to fructose-6-phosphate. (iii) Second phosphorylation to fructose-1,6-bisphosphate using 1 ATP (investment phase total: 2 ATP consumed). (iv) Cleavage into two 3-carbon molecules (DHAP and G3P). (v) Oxidation and phosphorylation steps produce 4 ATP via substrate-level phosphorylation and 2 NADH. Net gain: 2 ATP, 2 NADH, 2 pyruvate.
- Q12. Draw a labeled diagram of the Krebs cycle and mention key products. A: Diagram should show acetyl CoA + oxaloacetic acid → citric acid → isocitric acid → α-ketoglutaric acid → succinyl CoA → succinic acid → fumaric acid → malic acid → oxaloacetic acid. Labels: 2 CO₂ released (at isocitrate and α-ketoglutarate steps), 3 NADH, 1 FADH₂, 1 GTP per cycle.
- Q13. Explain oxidative decarboxylation of pyruvate. A: Pyruvate (3C) formed in glycolysis enters mitochondrial matrix. The pyruvate dehydrogenase complex catalyzes removal of one CO₂ (decarboxylation) and oxidation (removal of hydrogen captured by NAD⁺ forming NADH), yielding acetyl CoA (2C) which enters the Krebs cycle. This is an irreversible, link reaction connecting glycolysis to the Krebs cycle.
3-Mark Questions: Electron Transport Chain and ATP Synthesis
Questions on the electron transport chain (ETC) and oxidative phosphorylation require understanding of chemiosmosis and the role of electron carriers. The ETC is located on the inner mitochondrial membrane (cristae) and consists of four protein complexes (I, II, III, IV) plus mobile carriers (ubiquinone and cytochrome c). NADH donates electrons at Complex I, FADH₂ at Complex II; electrons cascade through carriers of decreasing energy, and the released energy pumps protons (H⁺) from the mitochondrial matrix into the intermembrane space, creating an electrochemical gradient. ATP synthase (Complex V) harnesses this proton-motive force to phosphorylate ADP to ATP—this mechanism is called chemiosmosis, proposed by Peter Mitchell in 1961. Each NADH generates approximately 2.5 ATP, each FADH₂ approximately 1.5 ATP. Total theoretical yield from one glucose is 38 ATP (glycolysis 2 ATP + 2 NADH→5 ATP, Krebs cycle 2 GTP + 6 NADH→15 ATP + 2 FADH₂→3 ATP, link reaction 2 NADH→5 ATP), but actual yield in eukaryotes is 30-32 ATP due to energy cost of transporting cytoplasmic NADH into mitochondria.
- Q15. Explain the chemiosmotic hypothesis of ATP synthesis. A: (i) Electron transport through Complexes I, III, IV pumps H⁺ from mitochondrial matrix to intermembrane space. (ii) This creates a proton gradient (high [H⁺] in intermembrane space, low in matrix). (iii) Protons flow back through ATP synthase (F₀F₁ particle) down the gradient. (iv) The energy released drives phosphorylation of ADP + Pi → ATP. This coupling of electron transport to ATP synthesis is chemiosmosis.
- Q16. Why is oxygen called the terminal electron acceptor? A: At the end of the ETC (Complex IV, cytochrome c oxidase), electrons are passed to molecular oxygen (O₂), which combines with protons (H⁺) to form water (H₂O). Without oxygen, the ETC halts because electrons accumulate, NADH and FADH₂ cannot be reoxidized, and ATP synthesis stops. Hence O₂ is 'terminal' or final acceptor.
- Q17. Calculate ATP yield from one glucose molecule in aerobic respiration. A: Glycolysis: 2 ATP (substrate-level) + 2 NADH (→5 ATP). Pyruvate oxidation: 2 NADH (→5 ATP). Krebs cycle (×2): 2 GTP (→2 ATP) + 6 NADH (→15 ATP) + 2 FADH₂ (→3 ATP). Total: 2+5+5+2+15+3 = 32 ATP (considering malate-aspartate shuttle cost). Theoretical maximum is 38 ATP; actual yield is 30-32 ATP in eukaryotes.
5-Mark Questions and Case-Based Problems
Five-mark questions test integrative understanding, often presenting experimental data, graphs, or real-world scenarios. CBSE case-based questions introduced in the 2021 curriculum reform typically describe a phenomenon—like muscle fatigue during sprinting or fermentation in bread-making—and ask 3-4 sub-questions totaling 5 marks. Students must read the case carefully, extract relevant data, apply concepts from NCERT Class 11 Biology Chapter 12, and write logically sequenced answers. For instance, a case on altitude effects might describe reduced oxygen availability at 3000 m elevation and ask why climbers experience breathlessness (insufficient O₂ for ETC), how cells compensate (anaerobic respiration, lactic acid formation), and long-term adaptations (increased RBC count). Award marks as per sub-question breakdown. Always conclude case answers by linking back to the biological principle—here, the absolute requirement of O₂ for complete glucose oxidation and maximum ATP yield.
- Q18. Case Study: During strenuous exercise, muscles switch from aerobic to anaerobic respiration. Lactic acid accumulates, causing cramps. (a) Why does this switch occur? (2 marks) (b) What is the RQ during this phase? (1 mark) (c) How is normal respiration restored post-exercise? (2 marks). A: (a) During intense activity, oxygen supply to muscles cannot meet the high ATP demand of contracting muscle fibers. Cells switch to anaerobic glycolysis, converting pyruvate to lactic acid via lactate dehydrogenase, yielding only 2 ATP per glucose (vs 32 in aerobic). (b) RQ is ∞ (infinity) because no O₂ is consumed, only CO₂ is produced from partial oxidation. (c) Post-exercise, deep breathing repays the 'oxygen debt'. Accumulated lactic acid is transported to the liver, converted back to pyruvate, and oxidized aerobically via the Krebs cycle and ETC, fully restoring ATP and clearing metabolic waste.
- Q19. A student measures RQ of germinating seeds and finds it to be 0.7. (a) Identify the respiratory substrate. (2 marks) (b) Write the equation for respiration of this substrate. (2 marks) (c) Why do seeds use this substrate initially? (1 mark). A: (a) RQ of 0.7 indicates fats (lipids) are the respiratory substrate, as they are rich in hydrogen and consume more O₂ relative to CO₂ released. (b) 2(C₅₁H₉₈O₆) + 145 O₂ → 102 CO₂ + 98 H₂O. Simplified: Fat + O₂ → CO₂ + H₂O (RQ = 102/145 ≈ 0.7). (c) Seeds store energy as oils (e.g., castor, groundnut). Fats yield more ATP per gram (9 kcal/g) than carbohydrates (4 kcal/g), providing sustained energy for germination until photosynthesis begins.
- Q20. Explain how cyanide poisoning affects cellular respiration and leads to death. (5 marks). A: (i) Cyanide (CN⁻) irreversibly binds to cytochrome c oxidase (Complex IV) in the electron transport chain. (ii) This blocks electron transfer to oxygen, halting the ETC and stopping ATP synthesis via oxidative phosphorylation. (iii) NADH and FADH₂ accumulate in reduced form; the Krebs cycle and glycolysis slow down due to lack of NAD⁺/FAD. (iv) Cells switch to anaerobic respiration, but 2 ATP per glucose is insufficient for vital organs like the brain and heart. (v) Rapid ATP depletion causes cellular dysfunction, organ failure, and death within minutes. Antidote: hydroxocobalamin binds cyanide, restoring ETC function.
How CBSE Frames Questions from Respiration in Plants
Understanding CBSE question-framing patterns helps you anticipate exam questions and tailor your preparation. The board typically sources 60-70% of questions directly from NCERT Class 11 Biology textbook exercises (in-text and end-of-chapter questions), 20-30% from NCERT Exemplar, and 10-15% from applied or integrated scenarios. Glycolysis and Krebs cycle are tested almost every year through 3-mark 'Describe the pathway' or 'Draw and label' questions. The respiratory quotient appears as a 2-mark calculation or 1-mark MCQ, often integrated into a case study (e.g., identifying substrate from given RQ value). Electron transport chain questions favor 3-mark 'Explain the mechanism' format, sometimes asking for a schematic diagram of the inner mitochondrial membrane. CBSE also tests conceptual clarity through 'Why' and 'What would happen if' questions—such as 'Why does aerobic respiration yield more ATP than anaerobic?' or 'What happens to the Krebs cycle if NAD⁺ is unavailable?' The shift to competency-based questions since 2021 means more application and analysis, fewer rote recall items.
- Direct NCERT lifts: Definitions, pathway steps, product yields often match textbook language verbatim
- Diagram-based questions: Glycolysis flowchart, Krebs cycle, mitochondrial structure with ETC complexes labeled
- Numerical/calculation: ATP accounting, RQ determination from given substrate, efficiency calculation (ATP yield / total energy)
- Comparison tables: Aerobic vs anaerobic, glycolysis vs Krebs cycle, substrate-level vs oxidative phosphorylation
- Application scenarios: Fermentation in industry (bread, alcohol), altitude physiology, muscle fatigue, seed germination
- Assertion-Reason format (introduced 2023): Statement pairs testing causal understanding, e.g., 'Assertion: Oxygen is essential for aerobic respiration. Reason: Oxygen is the final electron acceptor in ETC.'
Common Mistakes Students Make and How to Avoid Them
Even well-prepared students lose marks due to recurring errors in CBSE Class 11 Biology Chapter 12 questions. The most frequent mistake is incorrect ATP accounting—forgetting to double Krebs cycle outputs (since one glucose produces two acetyl CoA), or ignoring the energy cost of shuttling cytoplasmic NADH into mitochondria (2 ATP equivalent). Many students confuse glycolysis location, writing 'mitochondria' instead of 'cytoplasm', costing 1 mark in MCQs or VSAs. Another common error is mixing up NADH and FADH₂ counts: the Krebs cycle produces 3 NADH and 1 FADH₂ per turn, not 3 of each. In RQ calculations, students sometimes invert the formula (writing O₂/CO₂ instead of CO₂/O₂), yielding incorrect values. Diagram labeling errors include mislabeling cristae as matrix, or omitting key enzymes like ATP synthase in ETC diagrams. Finally, vague language—'energy is produced' rather than 'ATP is synthesized'—shows conceptual weakness. CBSE marking schemes reward precise scientific terminology from NCERT Class 11 Biology text.
- Mistake 1: Writing 'glycolysis occurs in mitochondria'. Correction: Glycolysis occurs in cytoplasm; only pyruvate oxidation and Krebs cycle occur in mitochondria.
- Mistake 2: Calculating net ATP as 4 in glycolysis. Correction: Gross ATP is 4, but 2 are consumed in phosphorylation steps, so net gain is 2 ATP.
- Mistake 3: Stating RQ of fats as 1.0. Correction: Fat RQ is approximately 0.7 (they consume more O₂ relative to CO₂ produced).
- Mistake 4: Drawing Krebs cycle in cytoplasm. Correction: Krebs cycle (TCA cycle) occurs in mitochondrial matrix.
- Mistake 5: Confusing substrate-level and oxidative phosphorylation. Correction: Substrate-level (glycolysis, Krebs) directly transfers phosphate to ADP; oxidative (ETC) uses proton gradient.
- Mistake 6: Omitting CO₂ release points in diagrams. Correction: Mark decarboxylation steps clearly—pyruvate to acetyl CoA (1 CO₂), isocitrate to α-ketoglutarate (1 CO₂), α-ketoglutarate to succinyl CoA (1 CO₂).
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Frequently asked questions
How many marks does Chapter 12 Respiration in Plants carry in the CBSE Class 11 Biology annual exam?+
Respiration in Plants typically carries 8-10 marks in the CBSE Class 11 Biology theory paper, distributed across 2-3 MCQs (1 mark each), 1-2 short answer questions (2-3 marks each), and 1 long answer or case-based question (5 marks). The exact distribution varies by paper set, but expect at least one mandatory question on glycolysis or Krebs cycle pathways and one on respiratory quotient or ATP yield calculations.
What is the net ATP yield from one glucose molecule in aerobic respiration?+
The theoretical maximum ATP yield is 38 ATP per glucose, but the actual yield in eukaryotic cells is 30-32 ATP. This difference arises because transporting cytoplasmic NADH (from glycolysis) into mitochondria consumes energy. The breakdown is: glycolysis (2 ATP + 2 NADH→5 ATP), pyruvate oxidation (2 NADH→5 ATP), Krebs cycle (2 GTP + 6 NADH→15 ATP + 2 FADH₂→3 ATP), totaling approximately 32 ATP.
Where does glycolysis occur, and does it require oxygen?+
Glycolysis occurs in the cytoplasm of the cell and does not require oxygen (it is an anaerobic process). This pathway converts one glucose (6-carbon) molecule into two pyruvate (3-carbon) molecules, yielding a net gain of 2 ATP and 2 NADH. Because glycolysis is anaerobic, it proceeds in both aerobic and anaerobic organisms.
What is respiratory quotient (RQ), and how is it calculated?+
Respiratory quotient (RQ) is the ratio of the volume of CO₂ evolved to the volume of O₂ consumed during respiration: RQ = CO₂ released / O₂ consumed. For carbohydrates RQ=1, for fats RQ≈0.7, for proteins RQ≈0.8, and for organic acids RQ>1. In anaerobic respiration, RQ is infinity because no oxygen is consumed. RQ helps identify which substrate is being oxidized.
Why is the Krebs cycle called an amphibolic pathway?+
The Krebs cycle is termed amphibolic because it participates in both catabolism (breaking down acetyl CoA to CO₂ and H₂O, releasing energy) and anabolism (providing intermediates like α-ketoglutaric acid and oxaloacetic acid for biosynthesis of amino acids, nucleotides, and other molecules). This dual role makes it a central hub linking carbohydrate, fat, and protein metabolism.
What is the role of oxygen in the electron transport chain?+
Oxygen acts as the final (terminal) electron acceptor in the electron transport chain at Complex IV (cytochrome c oxidase). Electrons passing through the ETC are transferred to O₂, which combines with protons (H⁺) to form water (H₂O). Without oxygen, electrons cannot be removed, the ETC halts, NADH and FADH₂ accumulate, and ATP synthesis via oxidative phosphorylation stops.
How many NADH and FADH₂ molecules are produced in one turn of the Krebs cycle?+
One turn of the Krebs cycle produces 3 NADH molecules (at isocitrate→α-ketoglutarate, α-ketoglutarate→succinyl CoA, and malate→oxaloacetate steps), 1 FADH₂ molecule (at succinate→fumarate step), and 1 GTP (equivalent to ATP). Since one glucose yields two acetyl CoA molecules, the totals per glucose are 6 NADH, 2 FADH₂, and 2 GTP from the Krebs cycle.
What is the difference between substrate-level phosphorylation and oxidative phosphorylation?+
Substrate-level phosphorylation directly transfers a phosphate group from a high-energy substrate molecule to ADP, forming ATP (occurs in glycolysis and Krebs cycle). Oxidative phosphorylation synthesizes ATP using energy from the proton gradient created by the electron transport chain (chemiosmosis) on the inner mitochondrial membrane. Oxidative phosphorylation yields far more ATP (26-28) than substrate-level (4 total).
Why is aerobic respiration more efficient than anaerobic respiration?+
Aerobic respiration completely oxidizes glucose to CO₂ and H₂O using oxygen, extracting maximum energy and producing 30-32 ATP per glucose. Anaerobic respiration only partially oxidizes glucose to lactic acid or ethanol (no O₂), releasing minimal energy and yielding just 2 ATP per glucose. The presence of oxygen enables the Krebs cycle and electron transport chain, which account for the bulk of ATP synthesis.
What happens to pyruvate under aerobic and anaerobic conditions?+
Under aerobic conditions, pyruvate enters the mitochondrial matrix and undergoes oxidative decarboxylation to form acetyl CoA, which then enters the Krebs cycle. Under anaerobic conditions in muscles, pyruvate is reduced to lactic acid by lactate dehydrogenase. In yeast and some plant tissues (like waterlogged roots), pyruvate is converted to ethanol and CO₂ via alcoholic fermentation.
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