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Respiration in Plants — NEET Biology MCQs with Solutions
Free NEET Biology Respiration in Plants MCQs with step-by-step solutions covering Cellular Respiration & Glycolysis, Fermentation (Anaerobic), Aerobic Respiration & Krebs Cycle, ETS & Oxidative Phosphorylation, Respiratory Balance Sheet & RQ. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Cellular Respiration & Glycolysis · easy · theory
Glycolysis (the EMP pathway) occurs in the ______ of the cell:
A. Mitochondrial matrix
B. Cytoplasm ✓ Correct
C. Inner mitochondrial membrane
D. Nucleus
Solution: Glycolysis takes place in the cytoplasm and is common to both aerobic and anaerobic respiration.
Q2 — Cellular Respiration & Glycolysis · easy · theory
The end product of glycolysis is:
A. Pyruvic acid (pyruvate) ✓ Correct
B. Acetyl CoA
C. Lactic acid
D. Ethanol
Solution: One glucose (6-C) is split into two molecules of pyruvate (3-C) during glycolysis.
Q3 — Fermentation (Anaerobic) · easy · theory
In yeast, anaerobic respiration (fermentation) converts pyruvate to:
A. Water
B. Ethanol and CO₂ ✓ Correct
C. Lactic acid
D. Acetyl CoA
Solution: Yeast performs alcoholic fermentation: pyruvate → acetaldehyde → ethanol + CO₂.
Q4 — Fermentation (Anaerobic) · easy · theory
In our muscle cells during vigorous exercise, pyruvate is converted to ______ under low oxygen:
A. CO₂ and water
B. Glucose
C. Ethanol
D. Lactic acid ✓ Correct
Solution: Lack of O₂ forces lactic acid fermentation in muscles, regenerating NAD⁺ for glycolysis.
Q5 — Aerobic Respiration & Krebs Cycle · easy · theory
The Krebs cycle (citric acid cycle / TCA cycle) takes place in the:
A. Cytoplasm
B. Ribosome
C. Mitochondrial matrix ✓ Correct
D. Outer mitochondrial membrane
Solution: The TCA cycle operates in the mitochondrial matrix; the ETS is on the inner membrane.
Q6 — Aerobic Respiration & Krebs Cycle · easy · theory
Before entering the Krebs cycle, pyruvate is converted to ______ by oxidative decarboxylation:
A. Lactate
B. Oxaloacetate
C. Citrate
D. Acetyl CoA ✓ Correct
Solution: Pyruvate dehydrogenase converts pyruvate to acetyl CoA, releasing CO₂ and forming NADH.
Q7 — ETS & Oxidative Phosphorylation · easy · theory
The electron transport system (ETS) is located in the:
A. Cytoplasm
B. Matrix
C. Outer membrane
D. Inner mitochondrial membrane ✓ Correct
Solution: The ETS complexes and ATP synthase are embedded in the inner mitochondrial membrane (cristae).
Q8 — ETS & Oxidative Phosphorylation · easy · theory
The final electron acceptor of the ETS in aerobic respiration is:
A. Carbon dioxide
B. Oxygen ✓ Correct
C. FAD
D. NAD⁺
Solution: O₂ accepts electrons and combines with protons to form water — the terminal acceptor.
Q9 — Respiratory Balance Sheet & RQ · easy · theory
The ratio of the volume of CO₂ evolved to O₂ consumed in respiration is called the:
A. Compensation point
B. Metabolic rate
C. Photosynthetic quotient
D. Respiratory Quotient (RQ) ✓ Correct
Solution: RQ = volume of CO₂ released / volume of O₂ used, and depends on the respiratory substrate.
Q10 — Respiratory Balance Sheet & RQ · easy · theory
When carbohydrates are the respiratory substrate, the RQ is:
A. 0.7
B. 1.0 ✓ Correct
C. Infinity
D. 0.9
Solution: For carbohydrates (e.g. glucose), CO₂ produced = O₂ consumed, so RQ = 1.
Q11 — Cellular Respiration & Glycolysis · easy · numerical
In glycolysis, two molecules of ATP are consumed in the preparatory phase before any ATP is generated. At which two steps is this ATP invested?
A. Phosphorylation of fructose-6-phosphate and of 1,3-bisphosphoglycerate
B. Phosphorylation of glucose (by hexokinase) and of fructose-6-phosphate (by phosphofructokinase) ✓ Correct
C. Phosphorylation of glucose and of glyceraldehyde-3-phosphate
D. Phosphorylation of glucose-6-phosphate and of phosphoenolpyruvate
Solution: ATP is spent to make glucose-6-phosphate (hexokinase) and fructose-1,6-bisphosphate (phosphofructokinase, PFK). Thus 2 ATP are used and 4 are made, giving a net of 2 ATP.
Q12 — Cellular Respiration & Glycolysis · easy · numerical
A key feature of glycolysis is that it does NOT directly require oxygen. Which statement best explains why glycolysis can proceed under anaerobic conditions only if a further reaction follows?
A. Pyruvate is toxic and must be removed by an oxygen-requiring enzyme
B. The NAD⁺ consumed during glucose oxidation must be regenerated, so a downstream reaction must reoxidise the NADH ✓ Correct
C. Glucose cannot be phosphorylated unless NADH is first oxidised by oxygen
D. ATP produced in glycolysis is unstable and must be stabilised by an oxygen-dependent step
Solution: Glycolysis reduces NAD⁺ to NADH at the glyceraldehyde-3-phosphate step. Because the cell has limited NAD⁺, it must be regenerated (by fermentation anaerobically, or the ETS aerobically) for glycolysis to continue.
Q13 — Cellular Respiration & Glycolysis · easy · numerical
Glucose entering a cell may come from sources other than free glucose. In plants, which of the following can be broken down to yield glucose (or glucose-phosphate) that then feeds into glycolysis?
A. Chlorophyll and carotenoids
B. Cellulose of the cell wall directly
C. Sucrose and starch ✓ Correct
D. Lignin of xylem vessels
Solution: Stored/transported carbohydrates such as sucrose (via invertase) and starch are broken down to glucose units that enter glycolysis. Structural lignin and pigments are not respiratory substrates in this way.
Q14 — Cellular Respiration & Glycolysis · easy · numerical
Phosphofructokinase (PFK) is regarded as the key regulatory (rate-limiting) enzyme of glycolysis. This is chiefly because it catalyses:
A. The final conversion of PEP to pyruvate
B. The oxidation step that reduces NAD⁺ to NADH
C. The first irreversible, committed step that channels the sugar specifically into glycolysis ✓ Correct
D. The substrate-level phosphorylation that yields ATP
Solution: PFK converts fructose-6-phosphate to fructose-1,6-bisphosphate — the committed, essentially irreversible step that commits the molecule to glycolysis, making it the principal control point.
Q15 — Fermentation (Anaerobic) · easy · numerical
Alcoholic fermentation in yeast converts pyruvate to ethanol in two enzymatic steps. Which pair of enzymes carries this out, in order?
A. Alcohol dehydrogenase, then pyruvate decarboxylase
B. Pyruvate dehydrogenase, then alcohol dehydrogenase
C. Lactate dehydrogenase, then pyruvate decarboxylase
D. Pyruvate decarboxylase, then alcohol dehydrogenase ✓ Correct
Solution: Pyruvate decarboxylase removes CO₂ from pyruvate to give acetaldehyde, then alcohol dehydrogenase reduces acetaldehyde to ethanol using NADH, regenerating NAD⁺.
Q16 — Fermentation (Anaerobic) · easy · numerical
Which of the following distinguishes alcoholic fermentation from lactic acid fermentation?
A. CO₂ is released in alcoholic fermentation but not in lactic acid fermentation ✓ Correct
B. ATP is produced in lactic acid fermentation but not in alcoholic fermentation
C. NADH is oxidised in alcoholic fermentation but reduced in lactic acid fermentation
D. Oxygen is required for alcoholic fermentation but not for lactic acid fermentation
Solution: Alcoholic fermentation decarboxylates pyruvate, releasing CO₂ before producing ethanol; lactic acid fermentation reduces pyruvate directly to lactate with no CO₂ release. Both are anaerobic and regenerate NAD⁺.
Q17 — Fermentation (Anaerobic) · easy · numerical
Both lactic acid and alcoholic fermentation yield less than 7% of the energy available in glucose and do not release all of it. Which statement correctly explains this?
A. Fermentation destroys the NADH so no energy can be captured
B. The end products (lactic acid or ethanol) still contain most of the chemical energy of glucose ✓ Correct
C. Fermentation converts glucose entirely to CO₂ and water
D. All the energy of glucose is lost as heat during fermentation
Solution: Fermentation stops at lactate or ethanol, both energy-rich molecules; only the net 2 ATP of glycolysis are captured, so less than 7% of glucose energy is released.
Q18 — Fermentation (Anaerobic) · easy · numerical
During strenuous exercise, when muscle cells run short of oxygen, the immediate benefit of converting pyruvate to lactate is that it:
A. Reoxidises NADH to NAD⁺ so that glycolysis can keep producing ATP ✓ Correct
B. Generates additional ATP directly from lactate
C. Produces oxygen needed for the electron transport system
D. Releases CO₂ that stimulates faster breathing
Solution: Lactate dehydrogenase transfers electrons from NADH to pyruvate, regenerating NAD⁺. This keeps glycolysis (and its ATP output) running when oxygen is scarce, even though no extra ATP comes from the reduction itself.
Q19 — Aerobic Respiration & Krebs Cycle · easy · numerical
The link reaction connecting glycolysis to the Krebs cycle converts pyruvate to acetyl CoA. This conversion involves all of the following EXCEPT:
A. Reduction of NAD⁺ to NADH
B. Catalysis by the pyruvate dehydrogenase complex
C. Reduction of FAD to FADH₂ ✓ Correct
D. Release of one molecule of CO₂
Solution: Oxidative decarboxylation of pyruvate by the pyruvate dehydrogenase complex releases CO₂ and reduces NAD⁺ to NADH. No FADH₂ is produced at this step.
Q20 — Aerobic Respiration & Krebs Cycle · easy · numerical
Where in a plant cell does the pyruvate produced in the cytoplasm undergo its further oxidation during aerobic respiration?
A. It is transported into the mitochondrial matrix ✓ Correct
B. It moves into the chloroplast stroma
C. It remains in the cytoplasm throughout
D. It enters the nucleus
Solution: Pyruvate formed by glycolysis in the cytoplasm is moved into the mitochondrial matrix, where the link reaction and Krebs cycle occur.
Q21 — Aerobic Respiration & Krebs Cycle · easy · numerical
The Krebs cycle is also described as the citric acid cycle. Which enzyme catalyses the very first, condensation step of the cycle?
A. Succinate dehydrogenase
B. Citrate synthase ✓ Correct
C. Aconitase
D. Malate dehydrogenase
Solution: Citrate synthase condenses the two-carbon acetyl group of acetyl CoA with the four-carbon oxaloacetate to form the six-carbon citrate, starting the cycle.
Q22 — Aerobic Respiration & Krebs Cycle · easy · numerical
The respiratory pathway is described as amphibolic because it:
A. Uses only fats as respiratory substrates
B. Operates only in the presence of oxygen
C. Serves both to break down molecules for energy and to provide intermediates for the synthesis (anabolism) of other compounds ✓ Correct
D. Occurs simultaneously in the cytoplasm and the nucleus
Solution: Respiratory intermediates (e.g. acetyl CoA, α-ketoglutarate, oxaloacetate) are withdrawn to build fatty acids, amino acids, etc. — so the pathway is catabolic and anabolic, i.e. amphibolic.
Q23 — ETS & Oxidative Phosphorylation · easy · numerical
In the electron transport system, electrons from FADH₂ enter at a different point from those of NADH. Where does FADH₂ feed its electrons?
A. At Complex I (NADH dehydrogenase), the same as NADH
B. Directly at Complex IV (cytochrome c oxidase)
C. At ATP synthase, skipping the cytochromes
D. At Complex II (succinate dehydrogenase), bypassing Complex I ✓ Correct
Solution: FADH₂ transfers electrons at Complex II, downstream of Complex I. Because it pumps fewer protons than the NADH route, each FADH₂ yields less ATP than each NADH.
Q24 — ETS & Oxidative Phosphorylation · easy · numerical
Which mobile carrier shuttles electrons from Complex III to Complex IV of the electron transport chain?
A. Ubiquinone (coenzyme Q)
B. NAD⁺
C. Plastocyanin
D. Cytochrome c ✓ Correct
Solution: Cytochrome c is a small mobile protein that carries electrons from Complex III to Complex IV. (Ubiquinone shuttles between Complexes I/II and III; plastocyanin belongs to photosynthesis.)
Q25 — ETS & Oxidative Phosphorylation · easy · numerical
Oxidative phosphorylation is so named because it couples:
A. The phosphorylation of glucose to its subsequent oxidation
B. The oxidation of reduced coenzymes (via the ETS) to the phosphorylation of ADP to ATP ✓ Correct
C. The oxidation of glucose to the reduction of NAD⁺
D. The direct transfer of a phosphate from a substrate to ADP
Solution: In oxidative phosphorylation the energy released as NADH and FADH₂ are oxidised through the ETS drives ATP synthase to phosphorylate ADP — distinct from substrate-level phosphorylation.
Q26 — ETS & Oxidative Phosphorylation · easy · numerical
The role of oxygen as the terminal acceptor of the electron transport system is described as vital and indispensable because it:
A. Splits glucose into two pyruvate molecules
B. Directly phosphorylates ADP to ATP
C. Removes hydrogen (as water) and keeps electrons flowing, allowing the coenzymes to be reoxidised ✓ Correct
D. Provides the protons that are pumped across the membrane
Solution: O₂ accepts the spent electrons and combines with protons to form water. This continual removal keeps the chain flowing and regenerates NAD⁺/FAD; without it the whole chain and Krebs cycle back up.
Q27 — Respiratory Balance Sheet & RQ · easy · numerical
The theoretical net gain of 38 ATP per glucose in the standard respiratory balance sheet is arrived at by adding which three contributions?
A. 8 (glycolysis) + 6 (Krebs) + 24 (oxidative phosphorylation)
B. 2 (glycolysis) + 6 (Krebs) + 30 (oxidative phosphorylation)
C. 2 (net glycolysis) + 2 (Krebs, substrate-level) + 34 (oxidative phosphorylation) ✓ Correct
D. 4 (glycolysis) + 4 (Krebs) + 30 (oxidative phosphorylation)
Solution: Net substrate-level ATP is 2 (glycolysis) + 2 (Krebs), and oxidative phosphorylation contributes 34, giving the conventional 38 ATP per glucose.
Q28 — Respiratory Balance Sheet & RQ · easy · numerical
The respiratory quotient (RQ) for the oxidation of a pure carbohydrate such as glucose is:
A. Greater than 1.0, because CO₂ released exceeds O₂ consumed
B. About 0.9, intermediate between fats and carbohydrates
C. Exactly 1.0, because the volumes of CO₂ released and O₂ consumed are equal ✓ Correct
D. About 0.7, because more O₂ is consumed than CO₂ released
Solution: For C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, six CO₂ are produced for six O₂ used, so RQ = 6/6 = 1.0.
Q29 — Respiratory Balance Sheet & RQ · easy · numerical
When proteins are used as the respiratory substrate, the RQ is:
A. About 0.9 ✓ Correct
B. About 0.7
C. Greater than 1.0
D. Exactly 1.0
Solution: Protein respiration gives an RQ of roughly 0.9, reflecting that slightly less CO₂ is evolved than O₂ consumed.
Q30 — Respiratory Balance Sheet & RQ · easy · numerical
A germinating castor bean seed, rich in stored oil, is respiring. What RQ would you expect, and why?
A. Less than 1 (around 0.7), because fatty acids are highly reduced and consume much O₂ relative to the CO₂ produced ✓ Correct
B. Exactly 0.5, because half the oxygen is stored
C. Exactly 1.0, because all seeds respire carbohydrates
D. Greater than 1.0, because fats release extra CO₂ without using O₂
Solution: Fats are highly reduced, so their oxidation needs a large amount of O₂ compared with the CO₂ released, giving an RQ well below 1 (about 0.7) — typical of fat-storing germinating seeds.