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Respiration in Plants

Jharkhand Board · Class 11 · Biology

NCERT Solutions for Respiration in Plants — Jharkhand Board Class 11 Biology.

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EXERCISES — Respiration in Plants

1(a)Differentiate between Respiration and Combustion.Show solution
Given: Two processes that involve oxidation of organic molecules and release of energy.

Differences:

| Feature | Respiration | Combustion |
|---|---|---|
| Nature | Biological/enzymatic process | Physical/chemical process |
| Site | Occurs inside living cells | Occurs outside living organisms |
| Temperature | Occurs at body temperature | Requires high temperature to initiate |
| Energy release | Step-wise, controlled release; energy is trapped as ATP | Sudden, uncontrolled release; energy lost as heat and light |
| Intermediates | Several intermediate compounds formed | No intermediates; direct oxidation |
| Enzymes | Requires enzymes at each step | No enzymes involved |
| End products | CO2\mathrm{CO_2}, H2O\mathrm{H_2O}, and ATP | CO2\mathrm{CO_2}, H2O\mathrm{H_2O}, heat, and light |

Conclusion: Respiration is a controlled, enzyme-mediated, step-wise oxidation occurring in living cells, whereas combustion is an uncontrolled, rapid oxidation occurring outside living systems.
1(b)Differentiate between Glycolysis and Krebs' cycle.Show solution
Given: Two major stages of cellular respiration.

Differences:

| Feature | Glycolysis | Krebs' Cycle (TCA Cycle) |
|---|---|---|
| Location | Cytoplasm (cytosol) | Matrix of mitochondria |
| Substrate | Glucose (C6H12O6\mathrm{C_6H_{12}O_6}) | Acetyl CoA (2-carbon compound) |
| Nature | Linear (non-cyclic) pathway | Cyclic pathway |
| Oxygen requirement | Does not require O2\mathrm{O_2} (anaerobic) | Requires O2\mathrm{O_2} indirectly (aerobic) |
| End product | 2 molecules of Pyruvic acid | CO2\mathrm{CO_2}, H2O\mathrm{H_2O}, NADH, FADH2\mathrm{FADH_2} |
| Net ATP produced | 2 ATP (net) per glucose | 2 ATP (GTP) per glucose |
| CO2\mathrm{CO_2} released | No CO2\mathrm{CO_2} released | 4 molecules of CO2\mathrm{CO_2} released per glucose |
| NADH produced | 2 NADH per glucose | 6 NADH per glucose |

Conclusion: Glycolysis is the first stage occurring in the cytoplasm without O2\mathrm{O_2}, while Krebs' cycle is the second stage occurring in the mitochondrial matrix requiring aerobic conditions.
1(c)Differentiate between Aerobic respiration and Fermentation.Show solution
Given: Two types of respiration differing in oxygen requirement.

Differences:

| Feature | Aerobic Respiration | Fermentation |
|---|---|---|
| Oxygen requirement | Requires O2\mathrm{O_2} | Does not require O2\mathrm{O_2} (anaerobic) |
| Site | Cytoplasm + Mitochondria | Cytoplasm only |
| End products | CO2\mathrm{CO_2} and H2O\mathrm{H_2O} | Ethanol + CO2\mathrm{CO_2} (yeast) or Lactic acid (bacteria/muscle) |
| ATP yield | 36–38 ATP per glucose molecule | Only 2 ATP per glucose molecule |
| Completeness | Complete oxidation of glucose | Incomplete oxidation of glucose |
| Organisms | All higher organisms (eukaryotes) | Yeast, bacteria, some plant tissues |
| NADH reoxidation | Via ETS using O2\mathrm{O_2} | By reducing pyruvate to ethanol or lactate |

Conclusion: Aerobic respiration is more efficient and complete, yielding far more energy than fermentation, which is an incomplete, anaerobic process.
2What are respiratory substrates? Name the most common respiratory substrate.Show solution
Definition of Respiratory Substrates:

Respiratory substrates are the organic molecules that are oxidised during the process of cellular respiration to release energy in the form of ATP.

Types of Respiratory Substrates:
- Carbohydrates (e.g., glucose, starch, sucrose)
- Fats (e.g., fatty acids, glycerol)
- Proteins (e.g., amino acids)
- Organic acids (e.g., malic acid, citric acid)

Most Common Respiratory Substrate:

The most common (favoured) respiratory substrate is Glucose (C6H12O6\mathrm{C_6H_{12}O_6}).

Reason: Glucose is readily available in cells, easily soluble, and can be directly used in glycolysis. The overall reaction is:
C6H12O6+6O26CO2+6H2O+Energy (ATP)\mathrm{C_6H_{12}O_6} + 6\mathrm{O_2} \longrightarrow 6\mathrm{CO_2} + 6\mathrm{H_2O} + \text{Energy (ATP)}
3Give the schematic representation of glycolysis.Show solution
Glycolysis (Greek: *glykos* = sugar; *lysis* = splitting) occurs in the cytoplasm and converts one molecule of glucose (6C) into two molecules of pyruvic acid (3C).

Schematic Representation of Glycolysis:

Glucose (6C)ATPADPGlucose-6-phosphate (6C)\text{Glucose (6C)} \xrightarrow{\text{ATP} \to \text{ADP}} \text{Glucose-6-phosphate (6C)}

Glucose-6-phosphate (6C)Fructose-6-phosphate (6C)\text{Glucose-6-phosphate (6C)} \longrightarrow \text{Fructose-6-phosphate (6C)}

Fructose-6-phosphate (6C)ATPADPFructose-1,6-bisphosphate (6C)\text{Fructose-6-phosphate (6C)} \xrightarrow{\text{ATP} \to \text{ADP}} \text{Fructose-1,6-bisphosphate (6C)}

Fructose-1,6-bisphosphate (6C)2×Glyceraldehyde-3-phosphate / PGAL (3C)\text{Fructose-1,6-bisphosphate (6C)} \longrightarrow 2 \times \text{Glyceraldehyde-3-phosphate / PGAL (3C)}

2×PGAL (3C)2NAD+2NADH2×1,3-bisphosphoglycerate (3C)2 \times \text{PGAL (3C)} \xrightarrow{2\,\mathrm{NAD^+} \to 2\,\mathrm{NADH}} 2 \times \text{1,3-bisphosphoglycerate (3C)}

2×1,3-bisphosphoglycerate2ADP2ATP2×3-phosphoglycerate (3C)2 \times \text{1,3-bisphosphoglycerate} \xrightarrow{2\,\mathrm{ADP} \to 2\,\mathrm{ATP}} 2 \times \text{3-phosphoglycerate (3C)}

2×3-phosphoglycerate2×2-phosphoglycerate (3C)2 \times \text{3-phosphoglycerate} \longrightarrow 2 \times \text{2-phosphoglycerate (3C)}

2×2-phosphoglycerate2H2O2×Phosphoenolpyruvate / PEP (3C)2 \times \text{2-phosphoglycerate} \xrightarrow{-2\,\mathrm{H_2O}} 2 \times \text{Phosphoenolpyruvate / PEP (3C)}

2×PEP2ADP2ATP2×Pyruvic acid (3C)2 \times \text{PEP} \xrightarrow{2\,\mathrm{ADP} \to 2\,\mathrm{ATP}} 2 \times \text{Pyruvic acid (3C)}

Net Gain per glucose molecule:
- Net ATP = 2 ATP (2 used, 4 produced)
- NADH = 2 NADH
- End product = 2 Pyruvic acid
4What are the main steps in aerobic respiration? Where does it take place?Show solution
Aerobic respiration is the process of complete oxidation of glucose in the presence of oxygen to produce CO2\mathrm{CO_2}, H2O\mathrm{H_2O}, and a large amount of ATP.

Main Steps and their Location:

Step 1: Glycolysis
- Location: Cytoplasm (cytosol)
- Glucose (6C) is broken down into 2 molecules of Pyruvic acid (3C)
- Net yield: 2 ATP + 2 NADH per glucose

Step 2: Pyruvate Oxidation (Link Reaction)
- Location: Mitochondrial matrix
- Pyruvic acid is oxidatively decarboxylated to form Acetyl CoA (2C)
Pyruvic acidPyruvate dehydrogenaseAcetyl CoA+CO2+NADH\text{Pyruvic acid} \xrightarrow{\text{Pyruvate dehydrogenase}} \text{Acetyl CoA} + \mathrm{CO_2} + \mathrm{NADH}
- Yield: 2 NADH + 2 CO2\mathrm{CO_2} per glucose

Step 3: Krebs' Cycle (TCA Cycle / Citric Acid Cycle)
- Location: Mitochondrial matrix
- Acetyl CoA (2C) combines with Oxaloacetate (4C) to form Citrate (6C)
- Complete oxidation releases CO2\mathrm{CO_2}
- Yield per glucose: 2 ATP (GTP), 6 NADH, 2 FADH2\mathrm{FADH_2}, 4 CO2\mathrm{CO_2}

Step 4: Electron Transport System (ETS) and Oxidative Phosphorylation
- Location: Inner mitochondrial membrane
- NADH and FADH2\mathrm{FADH_2} are oxidised; electrons pass through ETS
- O2\mathrm{O_2} acts as the final electron acceptor and is reduced to H2O\mathrm{H_2O}
- Large amount of ATP is synthesised (approximately 34 ATP)

Overall equation:
C6H12O6+6O26CO2+6H2O+Energy (36–38 ATP)\mathrm{C_6H_{12}O_6} + 6\mathrm{O_2} \longrightarrow 6\mathrm{CO_2} + 6\mathrm{H_2O} + \text{Energy (36–38 ATP)}
5Give the schematic representation of an overall view of Krebs' cycle.Show solution
Krebs' Cycle (Citric Acid Cycle / TCA Cycle) occurs in the mitochondrial matrix.

Schematic Representation:

Acetyl CoA (2C)+Oxaloacetate (4C)Citrate (6C)\text{Acetyl CoA (2C)} + \text{Oxaloacetate (4C)} \longrightarrow \text{Citrate (6C)}

Citrate (6C)Isocitrate (6C)\text{Citrate (6C)} \longrightarrow \text{Isocitrate (6C)}

Isocitrate (6C)CO2,  NAD+NADHα-Ketoglutarate (5C)\text{Isocitrate (6C)} \xrightarrow{-\mathrm{CO_2},\; \mathrm{NAD^+} \to \mathrm{NADH}} \alpha\text{-Ketoglutarate (5C)}

α-Ketoglutarate (5C)CO2,  NAD+NADHSuccinyl CoA (4C)\alpha\text{-Ketoglutarate (5C)} \xrightarrow{-\mathrm{CO_2},\; \mathrm{NAD^+} \to \mathrm{NADH}} \text{Succinyl CoA (4C)}

Succinyl CoA (4C)ADPATPSuccinate (4C)\text{Succinyl CoA (4C)} \xrightarrow{\mathrm{ADP} \to \mathrm{ATP}} \text{Succinate (4C)}

Succinate (4C)FADFADH2Fumarate (4C)\text{Succinate (4C)} \xrightarrow{\mathrm{FAD} \to \mathrm{FADH_2}} \text{Fumarate (4C)}

Fumarate (4C)+H2OMalate (4C)\text{Fumarate (4C)} \xrightarrow{+\mathrm{H_2O}} \text{Malate (4C)}

Malate (4C)NAD+NADHOxaloacetate (4C)\text{Malate (4C)} \xrightarrow{\mathrm{NAD^+} \to \mathrm{NADH}} \text{Oxaloacetate (4C)}

(Oxaloacetate is regenerated and the cycle continues)

Net yield per turn of cycle (per Acetyl CoA):
- CO2\mathrm{CO_2} released = 2
- NADH produced = 3
- FADH2\mathrm{FADH_2} produced = 1
- ATP (GTP) produced = 1

Per glucose molecule (2 turns): 4 CO2\mathrm{CO_2}, 6 NADH, 2 FADH2\mathrm{FADH_2}, 2 ATP
6Explain ETS (Electron Transport System).Show solution
Electron Transport System (ETS):

Definition: The Electron Transport System (ETS) is a series of electron carriers/proteins located on the inner mitochondrial membrane that transfer electrons from NADH and FADH2\mathrm{FADH_2} to molecular oxygen, releasing energy that is used to synthesise ATP.

Components of ETS (in order):
1. Complex I – NADH dehydrogenase (accepts electrons from NADH)
2. Ubiquinone (CoQ) – mobile electron carrier
3. Complex II – Succinate dehydrogenase (accepts electrons from FADH2\mathrm{FADH_2})
4. Complex III – Cytochrome bc₁ complex
5. Cytochrome c – mobile electron carrier
6. Complex IV – Cytochrome c oxidase (transfers electrons to O2\mathrm{O_2})

Process:
- NADH donates 2 electrons to Complex I; FADH2\mathrm{FADH_2} donates 2 electrons to Complex II.
- Electrons pass through the carriers in a downhill manner (from higher to lower energy).
- As electrons move through the complexes, protons (H+\mathrm{H^+}) are pumped from the matrix to the intermembrane space, creating a proton gradient.
- O2\mathrm{O_2} acts as the final electron acceptor and is reduced to water:
12O2+2H++2eH2O\frac{1}{2}\mathrm{O_2} + 2\mathrm{H^+} + 2e^- \longrightarrow \mathrm{H_2O}
- The proton gradient drives ATP synthase (Complex V) to synthesise ATP from ADP + Pi — this is called oxidative phosphorylation or chemiosmosis.

ATP yield:
- Each NADH yields approximately 2.5 ATP
- Each FADH2\mathrm{FADH_2} yields approximately 1.5 ATP

Significance: ETS is the major site of ATP production in aerobic respiration, accounting for about 34 of the ~36–38 ATP produced per glucose.
7(a)Distinguish between Aerobic respiration and Anaerobic respiration.Show solution
Differences between Aerobic and Anaerobic Respiration:

| Feature | Aerobic Respiration | Anaerobic Respiration |
|---|---|---|
| Oxygen requirement | Requires O2\mathrm{O_2} | Does not require O2\mathrm{O_2} |
| Site | Cytoplasm + Mitochondria | Cytoplasm only |
| End products | CO2\mathrm{CO_2} + H2O\mathrm{H_2O} | Ethanol + CO2\mathrm{CO_2} (yeast) or Lactic acid (bacteria) |
| ATP yield | 36–38 ATP per glucose | 2 ATP per glucose |
| Oxidation of glucose | Complete | Incomplete |
| Stages | Glycolysis + Krebs' cycle + ETS | Glycolysis + Fermentation |
| Organisms | Most eukaryotes and aerobes | Yeast, anaerobic bacteria, muscle cells (temporarily) |
| NADH fate | Oxidised via ETS | Oxidised by reducing pyruvate |

Conclusion: Aerobic respiration is far more efficient in energy production compared to anaerobic respiration.
7(b)Distinguish between Glycolysis and Fermentation.Show solution
Differences between Glycolysis and Fermentation:

| Feature | Glycolysis | Fermentation |
|---|---|---|
| Definition | Breakdown of glucose to pyruvate | Anaerobic breakdown of pyruvate to ethanol/lactic acid |
| Location | Cytoplasm | Cytoplasm |
| Oxygen | Not required | Not required |
| Substrate | Glucose (6C) | Pyruvic acid (3C) |
| End product | Pyruvic acid | Ethanol + CO2\mathrm{CO_2} or Lactic acid |
| ATP produced | 2 ATP (net) | No additional ATP produced |
| NADH | 2 NADH produced | NADH is reoxidised to NAD+\mathrm{NAD^+} |
| Purpose | Energy production (ATP) | Regeneration of NAD+\mathrm{NAD^+} to allow glycolysis to continue |
| Universality | Occurs in all living organisms | Occurs only in anaerobic organisms or under anaerobic conditions |

Conclusion: Glycolysis produces pyruvate and NADH; fermentation uses pyruvate to reoxidise NADH so that glycolysis can continue under anaerobic conditions.
7(c)Distinguish between Glycolysis and Citric acid Cycle (Krebs' cycle).Show solution
Differences between Glycolysis and Citric Acid Cycle:

| Feature | Glycolysis | Citric Acid Cycle (Krebs' Cycle) |
|---|---|---|
| Location | Cytoplasm (cytosol) | Mitochondrial matrix |
| Nature of pathway | Linear (non-cyclic) | Cyclic |
| Substrate | Glucose (6C) | Acetyl CoA (2C) |
| Oxygen requirement | Not required | Required (aerobic) |
| End product | 2 Pyruvic acid | CO2\mathrm{CO_2}, NADH, FADH2\mathrm{FADH_2}, ATP |
| CO2\mathrm{CO_2} released | None | 4 molecules per glucose (2 per turn) |
| Net ATP | 2 ATP per glucose | 2 ATP (GTP) per glucose |
| NADH produced | 2 NADH per glucose | 6 NADH per glucose |
| FADH2\mathrm{FADH_2} produced | None | 2 FADH2\mathrm{FADH_2} per glucose |
| Universality | Occurs in all organisms | Only in aerobic organisms |

Conclusion: Glycolysis is the first, universal, anaerobic stage, while the Krebs' cycle is the second, aerobic, cyclic stage occurring in mitochondria.
8What are the assumptions made during the calculation of net gain of ATP?Show solution
Given: Calculation of net ATP gain from complete aerobic oxidation of one glucose molecule.

Assumptions made:

1. Sequential and non-overlapping reactions: It is assumed that glycolysis, Krebs' cycle, and ETS occur in a sequential and orderly manner without any overlap.

2. NADH from glycolysis enters mitochondria: The 2 NADH produced in the cytoplasm during glycolysis are assumed to be transported into the mitochondria for oxidation via ETS. (In reality, this transport may cost energy.)

3. Fixed ATP yield per NADH and FADH2\mathrm{FADH_2}:
- Each NADH produced in the mitochondria yields 2.5 ATP (earlier assumed 3 ATP).
- Each FADH2\mathrm{FADH_2} yields 1.5 ATP (earlier assumed 2 ATP).
- Each cytoplasmic NADH yields 1.5–2.5 ATP depending on the shuttle used.

4. No ATP is used for other cellular processes: All ATP produced is assumed to be available for cellular work.

5. The P/O ratio is constant: It is assumed that a fixed number of ATP molecules are produced per atom of oxygen consumed.

6. Complete oxidation: It is assumed that glucose is completely oxidised to CO2\mathrm{CO_2} and H2O\mathrm{H_2O} without any intermediates being diverted for biosynthesis.

7. Proton gradient is used only for ATP synthesis: The proton gradient generated during ETS is assumed to be used exclusively for ATP synthesis by ATP synthase.

Note: Due to these assumptions, the theoretical yield is approximately 36–38 ATP per glucose, but the actual yield in living cells may be lower.
9Discuss 'The respiratory pathway is an amphibolic pathway.'Show solution
Definition of Amphibolic Pathway:

A metabolic pathway that involves both catabolism (breakdown of complex molecules) and anabolism (synthesis of complex molecules) is called an amphibolic pathway.

Respiration as an Amphibolic Pathway:

Respiration is primarily a catabolic process — it breaks down glucose and other organic molecules to release energy. However, the intermediates of the respiratory pathway are also used for biosynthetic (anabolic) reactions. Hence, it is called an amphibolic pathway.

Examples of how respiratory intermediates are used in anabolism:

| Intermediate | Anabolic Use |
|---|---|
| Glucose-6-phosphate | Synthesis of polysaccharides (starch, cellulose) |
| Pyruvic acid | Synthesis of alanine (amino acid) |
| Acetyl CoA | Synthesis of fatty acids and steroids |
| α\alpha-Ketoglutarate | Synthesis of glutamic acid (amino acid) |
| Oxaloacetate | Synthesis of aspartic acid (amino acid) |
| PEP (Phosphoenolpyruvate) | Synthesis of aromatic amino acids |

Catabolic role: Fats, proteins, and carbohydrates are broken down and fed into the respiratory pathway:
- Fats → Fatty acids → Acetyl CoA
- Proteins → Amino acids → Keto acids → enter Krebs' cycle
- Polysaccharides → Glucose → Glycolysis

Conclusion: Since the respiratory pathway serves both as a source of energy (catabolism) and as a source of carbon skeletons for biosynthesis (anabolism), it is rightly called an amphibolic pathway.
10Define RQ. What is its value for fats?Show solution
Definition of Respiratory Quotient (RQ):

The Respiratory Quotient (RQ) is defined as the ratio of the volume of carbon dioxide released to the volume of oxygen consumed during respiration in a given time.

RQ=Volume of CO2 releasedVolume of O2 consumed\mathrm{RQ} = \frac{\text{Volume of } \mathrm{CO_2} \text{ released}}{\text{Volume of } \mathrm{O_2} \text{ consumed}}

RQ for Fats:

When fats are used as respiratory substrates, the RQ is less than 1 (approximately 0.7).

Reason: Fats contain a higher proportion of hydrogen and carbon relative to oxygen compared to carbohydrates. Therefore, more oxygen is required to oxidise fats completely, resulting in more O2\mathrm{O_2} consumed than CO2\mathrm{CO_2} released.

Example — Tripalmitin (a fat):
2(C51H98O6)+145O2102CO2+98H2O+Energy2(\mathrm{C_{51}H_{98}O_6}) + 145\mathrm{O_2} \longrightarrow 102\mathrm{CO_2} + 98\mathrm{H_2O} + \text{Energy}

RQ=102CO2145O2=0.7\mathrm{RQ} = \frac{102\,\mathrm{CO_2}}{145\,\mathrm{O_2}} = \mathbf{0.7}

RQ values for different substrates:
- Carbohydrates: RQ = 1.0
- Fats: RQ = 0.7
- Proteins: RQ ≈ 0.9
11What is oxidative phosphorylation?Show solution
Definition:

Oxidative phosphorylation is the process by which ATP is synthesised from ADP and inorganic phosphate (Pi\mathrm{P_i}) using the energy released during the transfer of electrons through the Electron Transport System (ETS), coupled with the reduction of molecular oxygen to water.

Location: Inner mitochondrial membrane

Mechanism (Chemiosmosis):

1. NADH and FADH2\mathrm{FADH_2} (produced in glycolysis and Krebs' cycle) donate electrons to the ETS.
2. As electrons pass through the protein complexes (Complex I → CoQ → Complex III → Cyt c → Complex IV), energy is released.
3. This energy is used to pump protons (H+\mathrm{H^+}) from the mitochondrial matrix to the intermembrane space, creating a proton gradient (electrochemical gradient).
4. Protons flow back into the matrix through ATP synthase (Complex V) — this flow drives the synthesis of ATP:
ADP+PiATP synthaseATP\mathrm{ADP} + \mathrm{P_i} \xrightarrow{\text{ATP synthase}} \mathrm{ATP}
5. O2\mathrm{O_2} acts as the final electron acceptor and is reduced to water:
12O2+2H++2eH2O\frac{1}{2}\mathrm{O_2} + 2\mathrm{H^+} + 2e^- \longrightarrow \mathrm{H_2O}

Significance: Oxidative phosphorylation accounts for the majority of ATP produced during aerobic respiration (~34 out of ~36–38 ATP per glucose molecule).
12What is the significance of step-wise release of energy in respiration?Show solution
Step-wise Release of Energy in Respiration:

During cellular respiration, glucose is not oxidised in a single step. Instead, it is broken down through a series of enzyme-controlled reactions, releasing energy gradually in small amounts.

Significance:

1. Efficient energy capture: Step-wise release allows the cell to trap the released energy efficiently in the form of ATP at each step. If all energy were released at once, most of it would be lost as heat.

2. Prevention of cellular damage: A sudden, explosive release of all the energy stored in glucose would generate excessive heat, which would denature enzymes and damage cellular structures. Gradual release prevents this.

3. Regulation and control: Each step is catalysed by a specific enzyme, allowing the cell to regulate the rate of respiration according to its energy needs.

4. Versatility: Intermediates formed at various steps can be diverted for biosynthetic (anabolic) reactions, making respiration an amphibolic pathway.

5. Maintenance of body temperature: The small amounts of heat released at each step help maintain a relatively constant body temperature in organisms.

6. Higher efficiency: The step-wise process achieves approximately 40% efficiency in converting chemical energy of glucose into ATP, which is much higher than what would be possible with a single-step oxidation.

Conclusion: The step-wise release of energy ensures maximum conservation of energy in the form of ATP, prevents cellular damage, and allows metabolic regulation — making it far more advantageous than a single-step combustion-like reaction.

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