What is the role of ATP in cellular respiration?

ATP, or adenosine triphosphate, acts as the primary energy currency in cells. During cellular respiration, energy from nutrients is transferred to ATP, which then provides energy for various cellular processes. This is crucial for maintaining cellular functions and is a key concept in the Cambridge International A Levels Biology curriculum.

How does aerobic respiration differ from anaerobic respiration?

Aerobic respiration requires oxygen and produces more ATP compared to anaerobic respiration. It involves glycolysis, the Krebs cycle, and the electron transport chain. Anaerobic respiration, on the other hand, occurs without oxygen and results in less ATP, producing lactic acid in animals or ethanol and carbon dioxide in plants and yeast. Understanding these differences is essential for students studying Energy and Respiration in Biology.

What is the significance of the Krebs cycle in respiration?

The Krebs cycle, also known as the citric acid cycle, is a crucial part of aerobic respiration. It occurs in the mitochondria and generates electron carriers NADH and FADH2, which are essential for the electron transport chain. This cycle also produces ATP and carbon dioxide as by-products. Mastery of the Krebs cycle is important for Cambridge International A Levels Biology students.

Why is the electron transport chain important in respiration?

The electron transport chain is the final stage of aerobic respiration and takes place in the inner mitochondrial membrane. It uses electrons from NADH and FADH2 to create a proton gradient that drives ATP synthesis. This process is vital for producing the majority of ATP during respiration, making it a key topic in the study of Energy and Respiration.

How does glycolysis contribute to cellular respiration?

Glycolysis is the first step of both aerobic and anaerobic respiration, occurring in the cytoplasm. It breaks down glucose into pyruvate, producing a small amount of ATP and NADH. This process is essential as it initiates the breakdown of glucose, providing substrates for further energy production in the Krebs cycle and electron transport chain. Understanding glycolysis is fundamental for students learning about respiration in Biology.

Why is "Saying that glucose enters the mitochondrion to be respired" a common mistake in Respiration ?

Why it happens: Glycolysis is often confused with mitochondrial reactions. How to avoid it: Glucose is respired in the cytoplasm to pyruvate (glycolysis). Only **pyruvate** is transported into the mitochondrion. Always state 'glycolysis in the cytoplasm; link, Krebs and OP in the mitochondrion'. Source: 9700 Examiner Reports 2023-2024

Why is "Quoting the gross ATP from glycolysis (4) rather than net (2)" a common mistake in Respiration ?

Why it happens: Students forget the 2 ATP invested in phosphorylation steps. How to avoid it: Always state 'net 2 ATP from glycolysis'. The two ATP used to phosphorylate glucose are an 'investment'; gross 4 ATP - 2 invested = net 2. Source: 9700 Examiner Reports 2024

Why is "Writing that oxygen accepts protons in the ETC" a common mistake in Respiration ?

Why it happens: Confusion between electrons and protons in chemiosmosis. How to avoid it: Oxygen accepts **electrons** (the final electron acceptor of the ETC). Electrons combine with protons (from the matrix) to form water. Get the order right: e⁻ along ETC; H⁺ pumped to intermembrane space; e⁻ + H⁺ + O₂ → H₂O at the end. Source: 9700 Examiner Reports 2024

Why is "Saying the Krebs cycle runs once per glucose" a common mistake in Respiration ?

Why it happens: Students forget glucose → 2 pyruvate → 2 acetyl-CoA. How to avoid it: Krebs runs **twice** per glucose (one for each pyruvate / acetyl-CoA). Per glucose: 4 CO₂, 6 NADH, 2 FADH₂, 2 ATP from Krebs. Source: 9700 Examiner Reports 2024

Why is "Saying anaerobic respiration in mammalian muscle produces CO₂" a common mistake in Respiration ?

Why it happens: Confusing yeast and muscle fermentation pathways. How to avoid it: Mammalian muscle produces **lactate only — no CO₂**. CO₂ is only produced in yeast (alcoholic fermentation: pyruvate → ethanal + CO₂ → ethanol). Source: 9700 Examiner Reports 2023

Why is "Writing that fermentation produces ATP" a common mistake in Respiration ?

Why it happens: Confusion about what fermentation actually does. How to avoid it: Fermentation produces **no extra ATP**. Its sole purpose is to **regenerate NAD** so that glycolysis can continue. The 2 ATP net come from glycolysis alone. Source: 9700 Examiner Reports 2024

Why is "Quoting RQ in units (e.g. 'cm³ per minute')" a common mistake in Respiration ?

Why it happens: Confusion with rate measurements. How to avoid it: RQ is a **ratio** — it has **no units**. Quote as a number (1.0, 0.7, 0.9). Source: 9700 Examiner Reports 2024

Energy and RespirationCambridge International A Levels Biology (9700)

Respiration

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Respiration — Cambridge International A Level Biology 9700 Study Notes (2025-2027 syllabus)

Aerobic respiration in four stages — glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation — plus anaerobic respiration (alcoholic fermentation in yeast, lactate fermentation in muscle), the roles of NAD and FAD, the respiratory quotient and respiratory substrates.

What you’ll learn

Mapped to the Cambridge IGCSE 9700 syllabus (2025-2027).

12.2.1 — Describe glycolysis as the phosphorylation of glucose, the splitting of fructose bisphosphate into two triose phosphates and the oxidation of triose phosphate to pyruvate, with the net production of ATP and reduced NAD.
12.2.2 — Outline the link reaction and Krebs cycle as the further oxidation of pyruvate to CO₂, with the production of reduced NAD, reduced FAD and ATP, in the mitochondrial matrix.
12.2.3 — Describe oxidative phosphorylation at the inner mitochondrial membrane as the production of ATP using energy from the oxidation of reduced NAD and reduced FAD by the electron transport chain.
12.2.4 — Explain the role of chemiosmosis in oxidative phosphorylation and the role of oxygen as the final electron acceptor.
12.2.5 — Outline anaerobic respiration in mammalian muscle (lactate fermentation) and in yeast (alcoholic fermentation), and explain why the ATP yield is lower than in aerobic respiration.
12.2.6 — Describe how the respiratory quotient (RQ) can be calculated for different respiratory substrates and used to identify the substrate being respired.
12.2.7 — Use a respirometer to investigate the rate of oxygen consumption and to calculate RQ.

Glycolysis (cytoplasm)

Glucose split into 2 pyruvate; net 2 ATP + 2 NADH per glucose.

Glycolysis ('sugar splitting') is the first stage of respiration. It takes place in the cytoplasm of all living cells and does not require oxygen — every organism on Earth uses glycolysis.

The pathway has three phases:

1. Phosphorylation (activation). Glucose (6C) is first phosphorylated by 2 ATP to form fructose-1,6-bisphosphate. The 2 ATP are an investment — they 'destabilise' the glucose so it can be split. The enzyme hexokinase catalyses the first phosphorylation.

2. Lysis (splitting). Fructose-1,6-bisphosphate is split into two molecules of triose phosphate (TP, 3-carbon each).

3. Oxidation and phosphorylation. Each TP is oxidised — hydrogen atoms are removed and accepted by NAD (NAD → NADH) — and is phosphorylated by an inorganic phosphate from the cytoplasm. The phosphate groups (one was already on the TP, and one added now) are transferred to ADP in two substrate-level phosphorylation steps, producing 4 ATP and 2 pyruvate.

Net yield per glucose:

2 ATP (4 made − 2 invested)
2 NADH (each TP donates H to NAD)
2 pyruvate (3-carbon end product)

Pyruvate now has two possible fates:

Aerobic (O₂ present): Pyruvate is transported into the mitochondrion and enters the link reaction.
Anaerobic (no O₂): Pyruvate undergoes fermentation (see later) in the cytoplasm.

Cytoplasm; anaerobic.
Phosphorylation: glucose → fructose-1,6-bisphosphate (uses 2 ATP).
Lysis: split into 2 TP (3C each).
Oxidation: NAD → NADH; substrate-level phosphorylation → 4 ATP.
Net per glucose: 2 ATP, 2 NADH, 2 pyruvate.

See the full worked example for respiration →

The link reaction (mitochondrial matrix)

Pyruvate → acetyl-CoA + CO₂ + NADH. No ATP produced.

Under aerobic conditions, pyruvate is actively transported across the outer and inner mitochondrial membranes into the matrix. There it undergoes the link reaction, named because it links glycolysis to the Krebs cycle.

Per pyruvate molecule:

Decarboxylation. A carboxyl group is removed and released as CO₂.
Dehydrogenation. Two hydrogen atoms are removed and accepted by NAD (NAD → NADH).
Acetyl group formation. The remaining 2-carbon acetyl group is bound to coenzyme A (CoA) to form acetyl-CoA.

Because each glucose produced two pyruvate molecules in glycolysis, the link reaction runs twice per glucose. Per glucose:

2 acetyl-CoA
2 CO₂ (the first CO₂ released in aerobic respiration)
2 NADH
No ATP is produced directly in the link reaction

The enzyme complex catalysing this is pyruvate dehydrogenase.

Mitochondrial matrix; aerobic.
Pyruvate (3C) → acetyl-CoA (2C) + CO₂ + NADH.
Per glucose: 2 acetyl-CoA, 2 CO₂, 2 NADH.
No ATP made directly.

See the full worked example for respiration →

The Krebs cycle (mitochondrial matrix)

Per turn: 2 CO₂, 3 NADH, 1 FADH₂, 1 ATP. Cycle runs twice per glucose.

The Krebs cycle (also called the citric acid cycle or TCA cycle) takes place in the mitochondrial matrix. It is a cyclic pathway that completes the oxidation of the acetyl group of acetyl-CoA to CO₂, capturing the energy as reduced coenzymes (NADH, FADH₂) and a small amount of ATP.

One turn of the cycle:

Entry. The 2-carbon acetyl group from acetyl-CoA combines with a 4-carbon compound (oxaloacetate) to form a 6-carbon compound (citrate). CoA is released and recycled.
First decarboxylation + dehydrogenation. Citrate is converted via intermediates to a 5-carbon compound, with the loss of one CO₂ and the reduction of NAD → NADH.
Second decarboxylation + dehydrogenation. The 5-carbon compound is converted to a 4-carbon compound, with the loss of a second CO₂ and another NAD → NADH.
Substrate-level phosphorylation. One step is coupled to the phosphorylation of ADP, producing 1 ATP.
Two more dehydrogenations. One step reduces FAD → FADH₂; another reduces NAD → NADH. Oxaloacetate is regenerated, ready for the next turn.

Per turn: 2 CO₂, 3 NADH, 1 FADH₂, 1 ATP, oxaloacetate regenerated.

Per glucose (cycle runs twice): 4 CO₂, 6 NADH, 2 FADH₂, 2 ATP.

Note: candidates do not need to learn the names of every intermediate (citrate, isocitrate, α-ketoglutarate, succinate, etc.). What you must know is the inputs, outputs and the broad sequence (decarboxylation, dehydrogenation, substrate-level phosphorylation).

Mitochondrial matrix.
Acetyl-CoA (2C) + oxaloacetate (4C) → citrate (6C).
Per turn: 2 CO₂, 3 NADH, 1 FADH₂, 1 ATP.
Cycle runs TWICE per glucose.
Per glucose: 4 CO₂, 6 NADH, 2 FADH₂, 2 ATP.

See the full worked example for respiration →

Oxidative phosphorylation — ETC and chemiosmosis (cristae)

ETC pumps H⁺ → chemiosmosis through ATP synthase → ATP. O₂ is final e⁻ acceptor.

Oxidative phosphorylation produces the bulk of ATP in aerobic respiration. It takes place at the inner mitochondrial membrane (the cristae), which is folded to provide a large surface area for the embedded electron transport chain (ETC) and ATP synthase.

Stages:

1. Electron donation. NADH and FADH₂ produced by glycolysis, the link reaction and the Krebs cycle donate their electrons to the chain. NADH donates at the start of the chain; FADH₂ donates further along.

2. Electron transport chain. Electrons pass along a series of electron carrier proteins (NADH dehydrogenase, ubiquinone, the cytochromes b, c₁, c, a and a₃). Each carrier is alternately reduced (gains electrons) and oxidised (loses them to the next carrier). The energy released is used by three of the carriers to pump H⁺ from the matrix into the intermembrane space.

3. Proton gradient. Continuous pumping creates a proton (electrochemical) gradient across the inner membrane: H⁺ concentration in the intermembrane space becomes much higher than in the matrix, and the intermembrane space becomes positively charged relative to the matrix. The stored potential energy is the proton motive force.

4. Chemiosmosis through ATP synthase. The inner membrane is largely impermeable to H⁺, so protons can only flow back to the matrix through the F₀ portion of ATP synthase. As H⁺ flow through, the channel rotates, driving a conformational change in the F₁ catalytic head that couples ADP + Pi → ATP.

5. Oxygen — the final electron acceptor. At the end of the ETC, electrons combine with molecular oxygen and protons from the matrix to form water: $\tfrac{1}{2}\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\text{O}$ Without oxygen, electrons would back up in the chain, the gradient would collapse, and oxidative phosphorylation would stop. This is why aerobic respiration absolutely requires O₂.

6. Coenzyme recycling. NAD and FAD (released as NADH and FADH₂ donate electrons) are recycled to the matrix and cytoplasm to accept more hydrogen.

Yield. Approximately 2.5 ATP per NADH and 1.5 ATP per FADH₂. (Older textbooks quote 3 and 2 — the more precise figures reflect modern proton-stoichiometry measurements.)

Inner mitochondrial membrane (cristae).
NADH/FADH₂ → ETC carriers (redox reactions).
Energy used to pump H⁺ into intermembrane space.
Chemiosmosis: H⁺ flows back through ATP synthase → ATP.
O₂ is final electron acceptor → H₂O.
~2.5 ATP per NADH, ~1.5 ATP per FADH₂.

See the full worked example for respiration →

Total ATP yield per glucose

Glycolysis 2 + Krebs 2 + Ox phos ~28 = ~30-32 ATP per glucose.

Adding all sources of ATP from the complete aerobic respiration of one glucose:

Stage	ATP (direct)	NADH	FADH₂
Glycolysis	2 (net)	2	0
Link reaction	0	2	0
Krebs cycle	2 (2 turns × 1)	6 (2 turns × 3)	2 (2 turns × 1)
Subtotal	4	10	2

Converting reduced coenzymes via oxidative phosphorylation (≈2.5 ATP per NADH, ≈1.5 ATP per FADH₂):

Coenzyme	ATP equivalent
10 NADH × 2.5	25
2 FADH₂ × 1.5	3
Subtotal	~28

Grand total: ~30-32 ATP per glucose (range reflects the cost of shuttling cytoplasmic NADH into the mitochondrion — see going deeper).

This is 15-16 times more efficient than anaerobic respiration (2 ATP per glucose). Aerobic respiration captures the energy of the hydrogen atoms removed in glycolysis, the link reaction and Krebs; anaerobic respiration cannot.

Direct ATP (substrate-level phosphorylation): 4 per glucose (2 glycolysis + 2 Krebs).
Indirect ATP (oxidative phosphorylation): ~28 per glucose.
Total: ~30-32 ATP per glucose.
Anaerobic: only 2 ATP per glucose — 15-16× less.

Anaerobic respiration

Glycolysis only → 2 ATP. Yeast → ethanol + CO₂; muscle → lactate.

When oxygen is absent (or limiting), the ETC cannot operate (no final electron acceptor) and oxidative phosphorylation stops. NADH cannot be re-oxidised, and without NAD, glycolysis would also stop.

To survive, cells regenerate NAD by fermentation — diverting pyruvate from the Krebs cycle into a short pathway that re-oxidises NADH to NAD. Two routes exist:

Alcoholic fermentation (yeast and many plants). $\text{pyruvate} \xrightarrow{\text{decarboxylase}} \text{ethanal} + \text{CO}_2$ $\text{ethanal} + \text{NADH} \xrightarrow{\text{alcohol dehydrogenase}} \text{ethanol} + \text{NAD}$

Products: ethanol + CO₂.
NAD regenerated so glycolysis continues.
Net yield: 2 ATP per glucose (from glycolysis alone).
Used industrially in brewing, baking, and bioethanol production.

Lactate fermentation (mammalian muscle, some bacteria). $\text{pyruvate} + \text{NADH} \xrightarrow{\text{lactate dehydrogenase}} \text{lactate} + \text{NAD}$

Product: lactate. No CO₂.
NAD regenerated so glycolysis continues.
Net yield: 2 ATP per glucose.
Lactate accumulates, lowers muscle pH and contributes to fatigue.
Lactate is later transported via the blood to the liver, where it is oxidised back to pyruvate (or converted to glucose by gluconeogenesis). Extra O₂ is needed for this — the oxygen debt.

Why anaerobic yield is low. The H atoms on NADH (and any FADH₂) are never passed to the ETC, so their chemical energy is never captured as ATP. Aerobic respiration is ~15-16× more efficient.

Anaerobic = no O₂; glycolysis still runs.
Yeast: pyruvate → ethanal + CO₂ → ethanol (NADH → NAD).
Muscle: pyruvate → lactate (NADH → NAD); no CO₂.
Both regenerate NAD → glycolysis continues.
Net 2 ATP per glucose (no oxidative phosphorylation).
Lactate → oxygen debt (oxidised in liver post-exercise).

See the full worked example for respiration →

Respiratory substrates and RQ

RQ = CO₂/O₂. Carbohydrate 1.0; lipid 0.7; protein 0.9.

Cells can respire carbohydrates, lipids and proteins, but they differ in energy per gram and in the ratio of CO₂ released to O₂ consumed.

Respiratory quotient (RQ). $\text{RQ} = \frac{\text{volume of CO}_2 \text{ produced}}{\text{volume of O}_2 \text{ consumed}}$ RQ is a ratio, so it has no units.

RQ values by substrate:

Substrate	Typical RQ	Energy per gram
Carbohydrate (glucose)	1.0	~17 kJ g⁻¹
Protein	~0.9	~17 kJ g⁻¹
Lipid (triglyceride)	~0.7	~38 kJ g⁻¹

Why the differences arise. The more hydrogen a molecule contains (relative to its oxygen content), the more O₂ is needed to oxidise it fully to CO₂ and H₂O. Lipids are highly reduced (many C-H bonds, little internal O) so they need much more O₂ per CO₂ produced — RQ is low. Carbohydrates already contain some O (e.g. glucose $\text{C}_6\text{H}_{12}\text{O}_6$ ) so less external O₂ is needed — RQ is higher.

Practical use. RQ can be used to:

Identify the substrate being respired by a tissue.
Detect switching between substrates — a germinating fatty seed starts at RQ ~0.7 and rises towards 1.0 as oil reserves run out and carbohydrate respiration takes over.
Detect anaerobic respiration — if cells are respiring partly anaerobically, CO₂ continues to be produced (from yeast fermentation or from buffering of lactate in muscle) while O₂ consumption stops or falls, so the apparent RQ rises above 1.0.

Measuring RQ with a respirometer. A respirometer measures O₂ consumption when KOH or soda lime absorbs the CO₂ released. To find CO₂ produced, run a parallel experiment without KOH; the difference between O₂ consumed and the (smaller) net gas change equals CO₂ released. Then divide.

RQ = CO₂ produced / O₂ consumed (no units).
Carbohydrate 1.0; protein 0.9; lipid 0.7.
Lipid is the most reduced (most H) → needs most O₂ → lowest RQ.
RQ > 1.0 suggests anaerobic respiration too.
Respirometer + soda lime measures O₂; subtract to find CO₂.

See the full worked example for respiration →

Practical: respirometer (Paper 5)

Soda lime absorbs CO₂; movement of coloured liquid in capillary gives O₂ consumption rate.

A simple respirometer consists of a sealed boiling tube containing the respiring organism (typically germinating seeds, woodlice or maggots) separated from a small mass of soda lime (or KOH solution) by gauze. The tube is connected to a horizontal graduated capillary containing a coloured liquid (e.g. eosin / methylene blue).

Setting up.

Place the organisms above gauze; place soda lime / KOH at the bottom to absorb CO₂.
Seal the tube with a bung connected to the capillary.
Immerse in a water bath at constant temperature for ~5 min to equilibrate.
Open the screw clip / vent to atmosphere, then close it.
Record the position of the coloured liquid at $t = 0$ , $t = 5$ , $t = 10$ , $t = 15$ minutes.

What happens. As the organism respires, it consumes O₂ and releases CO₂. The CO₂ is absorbed by soda lime, so the gas volume in the tube falls. The coloured liquid moves towards the organism.

Calculating O₂ consumption rate. Distance moved × cross-sectional area of the capillary = volume of O₂ consumed in that time. Divide by time to get rate (cm³ min⁻¹). Normalise by the mass of organism to get specific rate (cm³ min⁻¹ g⁻¹).

Control tube. A parallel respirometer containing dead (boiled) seeds — or a mass equivalent of glass beads — corrects for any volume change caused by changes in atmospheric pressure or temperature. The control value is subtracted from the experimental value.

Calculating RQ. Run two experiments:

Tube A — with soda lime: measures O₂ consumed.
Tube B — without soda lime: measures (O₂ consumed − CO₂ produced).
CO₂ produced = (Tube A reading) − (Tube B reading).
RQ = CO₂ produced ÷ O₂ consumed.

Independent variables that can be investigated: temperature, type of organism, age of seed, presence of inhibitor (e.g. cyanide on the ETC).

Soda lime / KOH absorbs CO₂.
Capillary with coloured liquid shows volume change.
Distance × area = volume of O₂ consumed.
Control: boiled seeds / glass beads.
Two tubes (with and without soda lime) → RQ.

How it’s examined

Cambridge 9700 examines this heavily on Paper 4 (A Level structured questions, often 7-12 marks per part). Frequent questions: (a) describe the four stages with locations and ATP yields (8 marks); (b) explain chemiosmosis / oxidative phosphorylation (6-7 marks); (c) compare aerobic and anaerobic ATP yield (5-6 marks); (d) compare yeast and muscle fermentation (4-5 marks); (e) calculate RQ for a given substrate or experimental data (3-4 marks). Paper 5 (PAE) regularly tests respirometer practicals — calculation of O₂ consumption rate, design of controls, and interpretation of RQ values.

Sources: Cambridge International AS & A Level Biology 9700 syllabus (2025-2027); 9700 Examiner Reports 2022-2024; 9700/42 May/Jun 2024 question paper and mark scheme; 9700/42 Oct/Nov 2024 question paper and mark scheme; 9700/52 Practical Paper 5 (PAE) — respirometer tasks. Last reviewed 2026-05-12.

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Step-by-step worked examples — Respiration

Step-by-step solutions to past-paper-style questions on respiration , written exactly the way a tutor would explain them at the board.

1Glycolysis — net yield (5 marks)
Extended• Adapted from 9700/42 May/Jun 2024• glycolysis, ATP, NADH, Paper 4
Question
Describe the main events of glycolysis and state the net yield of ATP and NADH per molecule of glucose. (5 marks)
Step-by-step solution
1. Step 1
  Location and phosphorylation. Glycolysis occurs in the cytoplasm. Glucose (6C) is phosphorylated by 2 ATP to form fructose-1,6-bisphosphate.
2. Step 2
  Lysis (splitting). Fructose-1,6-bisphosphate is split into two triose phosphates (TP, 3C each).
3. Step 3
  Oxidation and phosphorylation. Each TP is oxidised — H atoms are removed and accepted by NAD → NADH — and is phosphorylated by inorganic phosphate. The phosphate groups are then transferred to ADP, producing ATP.
4. Step 4
  Product. Two molecules of pyruvate (3C) are formed per glucose. Each TP yields 2 ATP and 1 NADH, so per glucose: 4 ATP gross + 2 NADH.
5. Step 5
  Net yield. 4 ATP gross − 2 ATP invested = net 2 ATP per glucose, plus 2 NADH and 2 pyruvate. Glycolysis is anaerobic — no oxygen required.
Answer
Cytoplasm; glucose phosphorylated by 2 ATP → fructose-1,6-bisphosphate → split into 2 TP → oxidised + phosphorylated → 2 pyruvate. Net: 2 ATP, 2 NADH, 2 pyruvate per glucose.
Examiner tip
Mark scheme: (1) cytoplasm; (2) phosphorylation by 2 ATP / activation step; (3) splitting into 2 triose phosphates; (4) NAD → NADH / oxidation; (5) net 2 ATP + 2 NADH + 2 pyruvate. 9700 Examiner Reports flag candidates who write '4 ATP made' without subtracting the 2 ATP invested.
2Link reaction (4 marks)
Extended• Adapted from 9700/42 Oct/Nov 2024• link reaction, acetyl-CoA, Paper 4
Question
Describe the link reaction and explain why it is essential for aerobic respiration. (4 marks)
Step-by-step solution
1. Step 1
  Location. The link reaction occurs in the mitochondrial matrix after pyruvate has been actively transported from the cytoplasm into the mitochondrion.
2. Step 2
  Decarboxylation and oxidation. Pyruvate (3C) is decarboxylated (CO₂ removed) and dehydrogenated (H removed, accepted by NAD → NADH). The resulting 2-carbon acetyl group is then bound to coenzyme A to form acetyl-CoA.
3. Step 3
  Stoichiometry. Per glucose, two pyruvate molecules undergo the link reaction, producing 2 acetyl-CoA, 2 CO₂ and 2 NADH. No ATP is made directly in the link reaction.
4. Step 4
  Significance. The link reaction converts pyruvate (the end product of glycolysis) into acetyl-CoA — the substrate for the Krebs cycle. Without the link reaction, pyruvate cannot enter the Krebs cycle and aerobic ATP yield is limited to the 2 ATP from glycolysis.
Answer
Mitochondrial matrix. Pyruvate decarboxylated + dehydrogenated → acetyl group + CO₂ + NADH; acetyl group joins CoA → acetyl-CoA. Per glucose: 2 acetyl-CoA, 2 CO₂, 2 NADH (no ATP). Essential to provide Krebs substrate.
Examiner tip
Mark scheme: (1) matrix; (2) decarboxylation (CO₂); (3) dehydrogenation (NAD → NADH); (4) acetyl-CoA formed / substrate for Krebs. Common error: candidates write 'glucose enters mitochondrion' — glucose is too large; pyruvate does.
3Krebs cycle (6 marks)
Extended• Krebs cycle, Paper 4
Question
Describe the main events of one turn of the Krebs cycle. State the products formed per turn and per molecule of glucose. (6 marks)
Step-by-step solution
1. Step 1
  Entry of acetyl-CoA. The 2-carbon acetyl group from acetyl-CoA combines with a 4-carbon compound (oxaloacetate) to form a 6-carbon compound (citrate). CoA is released and recycled.
2. Step 2
  Two decarboxylations. Citrate is converted through a series of intermediates back to oxaloacetate. Along the way, two CO₂ molecules are released (decarboxylation steps).
3. Step 3
  Dehydrogenations. Hydrogen atoms are removed in four steps and accepted by coenzymes: 3 NAD → 3 NADH and 1 FAD → 1 FADH₂.
4. Step 4
  Substrate-level phosphorylation. One step is coupled to the phosphorylation of ADP, producing 1 ATP directly per turn.
5. Step 5
  Per turn yield. From one acetyl-CoA: 2 CO₂, 3 NADH, 1 FADH₂, 1 ATP and oxaloacetate is regenerated.
6. Step 6
  Per glucose yield. Each glucose yields 2 acetyl-CoA, so the cycle turns twice: 4 CO₂, 6 NADH, 2 FADH₂, 2 ATP per glucose. The Krebs cycle takes place in the mitochondrial matrix.
Answer
Acetyl-CoA (2C) + oxaloacetate (4C) → citrate (6C). Two decarboxylations (2 CO₂), three NAD→NADH, one FAD→FADH₂, one ATP per turn. Oxaloacetate regenerated. Per glucose × 2: 4 CO₂, 6 NADH, 2 FADH₂, 2 ATP.
Examiner tip
Mark scheme: (1) acetyl + 4C → 6C; (2) two decarboxylations / 2 CO₂; (3) 3 NADH + 1 FADH₂ per turn; (4) 1 ATP per turn; (5) oxaloacetate regenerated; (6) cycle runs twice per glucose. Examiner Reports note many students forget the cycle runs TWICE per glucose.
4Oxidative phosphorylation and chemiosmosis (7 marks)
Extended• Adapted from 9700/42 May/Jun 2024• ETC, chemiosmosis, Paper 4
Question
Describe how ATP is produced by oxidative phosphorylation in the mitochondrion. (7 marks)
Step-by-step solution
1. Step 1
  Location. Oxidative phosphorylation occurs at the inner mitochondrial membrane (the cristae), where the electron transport chain and ATP synthase are embedded.
2. Step 2
  Electron donation. NADH and FADH₂ produced by glycolysis, the link reaction and the Krebs cycle donate their electrons (and H⁺) to the electron transport chain. NAD and FAD are regenerated for reuse.
3. Step 3
  Electron transport chain. Electrons pass along a series of electron carrier proteins (e.g. NADH dehydrogenase, ubiquinone, cytochromes) by sequential redox reactions. Each carrier is alternately reduced (gains electrons) then oxidised (loses electrons).
4. Step 4
  Proton pumping. Energy released by electron transfer is used to pump H⁺ from the matrix into the intermembrane space. This creates a proton (electrochemical) gradient across the inner membrane.
5. Step 5
  Chemiosmosis. Protons flow back through ATP synthase down their electrochemical gradient. The flow rotates part of ATP synthase, driving the condensation of ADP + Pi → ATP.
6. Step 6
  Final electron acceptor. At the end of the chain, electrons (together with H⁺) combine with molecular oxygen to form water: ½O₂ + 2H⁺ + 2e⁻ → H₂O. This is why aerobic respiration requires O₂.
7. Step 7
  Yield. Approximately 2.5 ATP per NADH and 1.5 ATP per FADH₂. With 10 NADH and 2 FADH₂ per glucose, oxidative phosphorylation produces ~25 + 3 = ~28 ATP. Total ATP per glucose ~30-32.
Answer
Inner mitochondrial membrane / cristae. NADH/FADH₂ → ETC; redox carriers pump H⁺ into intermembrane space; chemiosmosis through ATP synthase → ATP; O₂ is final e⁻ acceptor → H₂O. Yield ~2.5 ATP/NADH, ~1.5 ATP/FADH₂. Total ~30-32 ATP per glucose.
Examiner tip
Mark scheme awards: (1) cristae / inner membrane; (2) NADH/FADH₂ donate e⁻; (3) ETC carriers + redox; (4) H⁺ pumped to intermembrane space; (5) proton gradient / chemiosmosis; (6) ATP synthase; (7) O₂ = final e⁻ acceptor, forms H₂O. Examiner Reports flag candidates who say 'O₂ accepts protons' — O₂ accepts ELECTRONS (and combines with protons).
5Anaerobic respiration in yeast and muscle (5 marks)
Extended• anaerobic, fermentation, lactate, Paper 4
Question
Compare anaerobic respiration in yeast with anaerobic respiration in mammalian muscle. (5 marks)
Step-by-step solution
1. Step 1
  Common starting point. In both cases, glycolysis proceeds normally in the cytoplasm to produce pyruvate, 2 ATP and 2 NADH per glucose. After this, the pathways diverge.
2. Step 2
  Yeast — alcoholic fermentation. Pyruvate is decarboxylated (CO₂ released) to form ethanal. Ethanal is then reduced by NADH (NADH → NAD) to form ethanol. Net product: ethanol + CO₂.
3. Step 3
  Mammalian muscle — lactate fermentation. Pyruvate is reduced directly by NADH (NADH → NAD) to form lactate. No CO₂ is produced. Lactate accumulates and contributes to muscle fatigue; it is later oxidised back to pyruvate in the liver (oxygen debt).
4. Step 4
  Common purpose. Both pathways regenerate NAD from NADH. This is essential because glycolysis requires NAD; without regeneration, glycolysis would stop and no ATP could be produced anaerobically.
5. Step 5
  Yield. Both yield only 2 ATP per glucose (the original glycolytic 2). Far less than aerobic respiration (~30-32 ATP) because the H atoms on NADH are not used by the ETC.
Answer
Both: glycolysis → 2 pyruvate + 2 ATP + 2 NADH. Yeast: pyruvate → ethanal + CO₂ → ethanol (NADH oxidised). Muscle: pyruvate → lactate (NADH oxidised), no CO₂. Both regenerate NAD so glycolysis continues. Net 2 ATP per glucose either way.
Examiner tip
Mark scheme: (1) glycolysis common to both; (2) yeast → ethanal → ethanol + CO₂; (3) muscle → lactate, no CO₂; (4) regeneration of NAD essential; (5) only 2 ATP per glucose / oxygen debt. Examiner Reports note candidates often write 'alcohol' instead of 'ethanol' — both credited but ethanol preferred.
6Respiratory quotient (4 marks)
Extended• RQ, respiratory substrate, Paper 4
Question
Define the respiratory quotient (RQ). Explain why the RQ for carbohydrate respiration is 1.0 and for lipid respiration is about 0.7. (4 marks)
Step-by-step solution
1. Step 1
  Definition. The respiratory quotient is the ratio of carbon dioxide produced to oxygen consumed during respiration: $\text{RQ} = \dfrac{\text{CO}_2 \text{ produced}}{\text{O}_2 \text{ consumed}}$ .
2. Step 2
  Carbohydrate (glucose) RQ = 1.0. From the overall equation $\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}$ , 6 mol of CO₂ are released per 6 mol of O₂ consumed. $\text{RQ} = 6/6 = 1.0$ .
3. Step 3
  Lipid RQ ≈ 0.7. A typical triglyceride has formula approximately $\text{C}_{55}\text{H}_{104}\text{O}_6$ . The equation is roughly $\text{C}_{55}\text{H}_{104}\text{O}_6 + 78\text{O}_2 \rightarrow 55\text{CO}_2 + 52\text{H}_2\text{O}$ . $\text{RQ} = 55/78 ≈ 0.7$ .
4. Step 4
  Reason. Lipids contain more hydrogen (and proportionally less oxygen) than carbohydrates, so more O₂ is needed to oxidise the hydrogen to water. More O₂ is consumed relative to the CO₂ produced, so the RQ is lower. This is also why lipids release more energy per gram (~38 kJ g⁻¹ vs ~17 kJ g⁻¹ for carbohydrates).
Answer
RQ = CO₂ produced / O₂ consumed. Carbohydrate: glucose + 6 O₂ → 6 CO₂ → RQ = 1.0. Lipid: more H per molecule needs more O₂ → less CO₂ per O₂ → RQ ≈ 0.7. Lipids release more energy per gram.
Examiner tip
Mark scheme: (1) RQ = CO₂/O₂ definition; (2) carbohydrate RQ 1.0 from balanced equation; (3) lipid ratio quoted (~0.7); (4) explanation — more H in lipid needs more O₂. Protein RQ ≈ 0.9 (sometimes asked).

Key Formulae — Respiration

The formulae you need to memorise for respiration on the Cambridge International A Level 9700 paper, with every variable defined in plain English and a note on when to use it.

Glycolysis — net yield per glucose
$\text{glucose} \rightarrow 2\,\text{pyruvate} + 2\,\text{ATP}_{\text{net}} + 2\,\text{NADH}$
$2\,\text{ATP}_{\text{net}}$
4 ATP made − 2 ATP invested = net 2 ATP
$2\,\text{NADH}$
Two reduced NAD per glucose
$2\,\text{pyruvate}$
Two 3-carbon pyruvate molecules enter the link reaction
When to use
Cytoplasm; anaerobic up to this point — no O₂ needed. Always state the net ATP (2), not the gross (4).
Link reaction — yield per pyruvate (per glucose: × 2)
$\text{pyruvate} + \text{CoA} + \text{NAD} \rightarrow \text{acetyl-CoA} + \text{CO}_2 + \text{NADH}$
$\text{acetyl-CoA}$
2-carbon acetyl group bound to coenzyme A
$\text{CO}_2$
First CO₂ release of aerobic respiration
$\text{NADH}$
Reduced coenzyme for the ETC
When to use
Mitochondrial matrix. Per glucose, the link reaction runs twice: 2 acetyl-CoA, 2 CO₂, 2 NADH. No ATP directly.
Krebs cycle — yield per turn (per glucose: × 2)
$\text{acetyl-CoA} \rightarrow 2\,\text{CO}_2 + 3\,\text{NADH} + 1\,\text{FADH}_2 + 1\,\text{ATP}$
$2\,\text{CO}_2$
Two decarboxylation steps
$3\,\text{NADH}$
Three dehydrogenation steps with NAD
$1\,\text{FADH}_2$
One dehydrogenation step with FAD
$1\,\text{ATP}$
Substrate-level phosphorylation
When to use
Mitochondrial matrix. Cycle runs twice per glucose. Per glucose: 4 CO₂, 6 NADH, 2 FADH₂, 2 ATP.
Oxidative phosphorylation — ATP per reduced coenzyme
$\text{NADH} \rightarrow \sim 2.5\,\text{ATP} \quad ; \quad \text{FADH}_2 \rightarrow \sim 1.5\,\text{ATP}$
$\sim 2.5$
ATP yield per NADH; older textbooks quote 3
$\sim 1.5$
ATP yield per FADH₂; older textbooks quote 2 (FADH₂ enters ETC further down so fewer H⁺ pumped)
When to use
Cristae (inner mitochondrial membrane). Yields are approximate because some protons leak back across the membrane and NADH from the cytoplasm must be shuttled in (a process that consumes some energy).
Example
Per glucose: 10 NADH × 2.5 + 2 FADH₂ × 1.5 = 25 + 3 = ~28 ATP from oxidative phosphorylation.
Total ATP yield per glucose (aerobic)
$\text{Total ATP} = 2_{\text{glycolysis}} + 2_{\text{Krebs}} + \sim 28_{\text{ox phos}} \approx 30\text{–}32$
$2 (glycolysis)$
Substrate-level phosphorylation in cytoplasm
$2 (Krebs)$
Substrate-level phosphorylation in matrix (1 per turn × 2 turns)
$~28 (ox phos)$
Chemiosmosis from 10 NADH + 2 FADH₂
When to use
Aerobic respiration of one molecule of glucose. Anaerobic respiration yields only 2 ATP (from glycolysis alone) — far less efficient. The range 30-32 accounts for the cost of NADH shuttling between cytoplasm and mitochondrion.
Respiratory quotient (RQ)
$\text{RQ} = \dfrac{\text{CO}_2 \text{ produced}}{\text{O}_2 \text{ consumed}}$
$Carbohydrate (glucose)$
RQ = 1.0 (6 CO₂ / 6 O₂)
$Lipid (e.g. triglyceride)$
RQ ≈ 0.7 (more H, more O₂ needed)
$Protein (amino acid)$
RQ ≈ 0.9
$Anaerobic respiration in yeast$
RQ → ∞ (no O₂ consumed but CO₂ still released)
When to use
Measure with a respirometer: KOH / soda lime absorbs CO₂ to give O₂ consumption; then a parallel respirometer without KOH gives net (O₂ − CO₂). Subtraction yields CO₂ produced.

Model Answers — Respiration

High-scoring sample answers for respiration on the Cambridge International A Level 9700 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question
9700/42 May/Jun 2024 Q5 (adapted)8 marks
Q (8 marks). Outline the four stages of aerobic respiration of glucose, naming the location, the main products and the ATP yield of each stage.
Model answer
Stage 1 — Glycolysis (cytoplasm). Glucose (6C) is phosphorylated by 2 ATP and split into 2 triose phosphate (3C). The triose phosphates are oxidised to 2 pyruvate, with NAD reduced to NADH. Net per glucose: 2 ATP, 2 NADH, 2 pyruvate. Glycolysis is anaerobic — no O₂ required.

Stage 2 — Link reaction (mitochondrial matrix). Each pyruvate is actively transported into the matrix, decarboxylated (CO₂ removed) and dehydrogenated (NAD → NADH); the remaining 2C acetyl group binds coenzyme A to form acetyl-CoA. Per glucose (cycle runs twice): 2 acetyl-CoA, 2 CO₂, 2 NADH; no ATP.

Stage 3 — Krebs cycle (mitochondrial matrix). Acetyl-CoA combines with 4C oxaloacetate to form 6C citrate. The cycle then regenerates oxaloacetate via two decarboxylation and four dehydrogenation steps. Per turn: 2 CO₂, 3 NADH, 1 FADH₂, 1 ATP (substrate-level phosphorylation). Per glucose (× 2): 4 CO₂, 6 NADH, 2 FADH₂, 2 ATP.

Stage 4 — Oxidative phosphorylation (inner mitochondrial membrane / cristae). NADH and FADH₂ from the previous stages donate electrons to the electron transport chain. Energy released by electron transfer pumps H⁺ from the matrix into the intermembrane space, building a proton gradient. Protons flow back through ATP synthase (chemiosmosis), driving ADP + Pi → ATP. Oxygen is the final electron acceptor, forming water. Per glucose: ~28 ATP (≈2.5 per NADH, ≈1.5 per FADH₂), all 6 H₂O.

Total per glucose: ~30-32 ATP, 6 CO₂, 6 H₂O.
Why this scores
Why this scores 8/8. Mark scheme: 2 marks per stage — (a) location + key event named, (b) yield stated. Maximum 8 in total. Examiner Reports flag candidates who confuse the locations: glycolysis is cytoplasm, the link reaction and Krebs are matrix, oxidative phosphorylation is the inner membrane / cristae.
Question
9700/42 Oct/Nov 2024 Q6(b) (adapted)7 marks
Q (7 marks). Explain how a proton gradient across the inner mitochondrial membrane is established and used to produce ATP in oxidative phosphorylation.
Model answer
Oxidative phosphorylation takes place at the inner mitochondrial membrane (the cristae), which contains an electron transport chain (ETC) and ATP synthase embedded in it.

Establishing the gradient. NADH and FADH₂ produced by glycolysis, the link reaction and the Krebs cycle donate their electrons to the chain. Electrons pass along a series of electron carrier proteins (e.g. NADH dehydrogenase, ubiquinone, cytochromes) by alternating redox reactions — each carrier is reduced and then re-oxidised as it passes electrons on. The energy released by these electron transfers is used by three of the carriers to pump H⁺ from the matrix into the intermembrane space, against their concentration gradient.

This pumping generates an electrochemical (proton) gradient across the inner membrane: H⁺ concentration is higher in the intermembrane space than in the matrix, and the intermembrane space is more positive. The gradient stores potential energy — known as the proton motive force.

Using the gradient — chemiosmosis. The inner membrane is largely impermeable to H⁺. Protons can therefore only flow back to the matrix through a special channel — the F₀ portion of ATP synthase. As H⁺ flow through, the channel rotates, and this rotation drives a conformational change in the F₁ catalytic head. The conformational change couples ADP + Pi → ATP. About 2.5 ATP are produced per NADH and 1.5 ATP per FADH₂ (FADH₂ enters the chain further along, so fewer protons are pumped per pair of electrons).

Final electron acceptor. At the end of the chain, electrons (and the protons in the matrix) combine with molecular oxygen to form water: $\tfrac{1}{2}\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\text{O}$ . Without oxygen, electrons would back up in the chain and the gradient would collapse — this is why aerobic respiration requires O₂.

NAD and FAD released from NADH / FADH₂ are recycled to glycolysis, the link reaction and the Krebs cycle to accept further hydrogen.
Why this scores
Why this scores 7/7. Mark scheme: (1) inner mitochondrial membrane / cristae; (2) NADH/FADH₂ donate e⁻; (3) chain of e⁻ carriers with redox; (4) energy used to pump H⁺ into intermembrane space; (5) chemiosmosis through ATP synthase; (6) ATP synthase phosphorylates ADP; (7) O₂ = final e⁻ acceptor / H₂O formed. 9700 Examiner Reports note this is the most frequent 7-8 mark question on Paper 4.
Question
Practice question — recurrent Paper 4 style6 marks
Q (6 marks). Compare the products and ATP yield of aerobic and anaerobic respiration of one molecule of glucose.
Model answer
Aerobic respiration. A complete oxidation of glucose: $\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}$ . Glucose is oxidised through glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation. Products: 6 CO₂, 6 H₂O, and approximately 30-32 ATP per glucose.

Anaerobic respiration in yeast (alcoholic fermentation). Glycolysis proceeds as normal, producing 2 pyruvate, 2 NADH and 2 ATP. The pyruvate is then decarboxylated to ethanal (releasing CO₂), and ethanal is reduced by NADH to ethanol. Products: ethanol + CO₂, and just 2 ATP per glucose. The NADH is re-oxidised to NAD so glycolysis can continue.

Anaerobic respiration in mammalian muscle (lactate fermentation). Glycolysis again proceeds as normal. Pyruvate is then reduced directly by NADH to lactate — no CO₂ is released. Products: lactate, and 2 ATP per glucose. The NADH is re-oxidised to NAD so glycolysis can continue. Lactate accumulates, lowering pH and causing muscle fatigue; the oxygen debt must be repaid by oxidising lactate back to pyruvate (in the liver) once oxygen is available.

Yield comparison. Aerobic ~30-32 ATP per glucose; anaerobic only 2 ATP per glucose — about 15-16× more efficient with oxygen. The reason is that anaerobic respiration cannot use the NADH/FADH₂ in oxidative phosphorylation (no O₂ to accept electrons), so the chemical energy of the H atoms is not captured as ATP.
Why this scores
Why this scores 6/6. Mark scheme: (1) aerobic products (CO₂ + H₂O); (2) aerobic yield ~30-32 ATP; (3) yeast anaerobic: ethanol + CO₂; (4) muscle anaerobic: lactate (no CO₂); (5) both yield 2 ATP; (6) reason for difference — no ETC / oxidation of NADH without O₂.
Question
9700/42 May/Jun 2024 Q7 (adapted)6 marks
Q (6 marks). Describe how a respirometer could be used to measure the rate of respiration of germinating seeds, and explain how the respiratory quotient could be calculated.
Model answer
A respirometer consists of a sealed tube containing the respiring organism (e.g. germinating peas) connected to a graduated capillary tube containing a coloured liquid (e.g. methylene blue dye). The seeds are separated from the coloured liquid by gauze and a small mass of soda lime (or KOH solution) that absorbs all CO₂ produced.

Procedure. The apparatus is left to equilibrate at constant temperature in a water bath. The position of the coloured liquid in the capillary is recorded. As the seeds respire, they consume O₂ and release CO₂; the CO₂ is absorbed by the soda lime, so the gas volume falls. The liquid moves towards the seeds. The distance moved per unit time gives a rate of O₂ consumption (after multiplying by the capillary cross-sectional area).

Controls. A second respirometer identical except containing dead (boiled) seeds is set up as a control to correct for atmospheric pressure / temperature changes. Volume changes here are subtracted from the experimental tube.

Calculating RQ. Run two experiments:

With soda lime — all CO₂ absorbed, so the volume change measures O₂ consumed.

Without soda lime — the volume change measures (O₂ consumed − CO₂ produced).

CO₂ produced = (O₂ consumed) − (volume change without soda lime). Then $\text{RQ} = \dfrac{\text{CO}_2 \text{ produced}}{\text{O}_2 \text{ consumed}}$ .

An RQ of 1.0 indicates pure carbohydrate respiration; ~0.7 indicates lipid; ~0.9 protein. Germinating seeds rich in oil (e.g. sunflower) start at ~0.7 and rise towards 1.0 as oil reserves are exhausted and carbohydrate metabolism takes over.
Why this scores
Why this scores 6/6. Mark scheme: (1) respirometer with graduated capillary and coloured liquid; (2) soda lime / KOH absorbs CO₂; (3) control tube (boiled seeds or empty); (4) measure distance moved by liquid; (5) repeat without soda lime; (6) RQ = CO₂ produced / O₂ consumed. 9700 Paper 5 (Practical PAE) often probes this experiment.
Question
9700/42 Oct/Nov 2024 Q5(c) (adapted)5 marks
Q (5 marks). Compare carbohydrates, lipids and proteins as respiratory substrates.
Model answer
Carbohydrates are the default respiratory substrate. Glucose enters glycolysis directly; glycogen / starch are first hydrolysed to glucose. Carbohydrate respiration yields approximately 17 kJ g⁻¹ of energy and has an RQ of 1.0.

Lipids are the highest energy-density substrate, yielding approximately 38 kJ g⁻¹ — about twice as much per gram as carbohydrate. This is because triglycerides contain a much higher proportion of C–H bonds (and almost no oxygen), so more reduced coenzymes (NADH, FADH₂) are generated per molecule when they are oxidised. Lipid respiration has an RQ of ~0.7 because more O₂ is needed to oxidise all the hydrogen. Lipids enter respiration as glycerol (which feeds into glycolysis) and fatty acids (which undergo β-oxidation in the matrix to produce acetyl-CoA).

Proteins are not a preferred respiratory substrate — they are saved for biosynthesis. When carbohydrate and lipid stores are exhausted (e.g. starvation), amino acids are deaminated in the liver: the amino group is removed (and excreted as urea), and the remaining keto-acid is fed into the link reaction or the Krebs cycle. Protein yields ~17 kJ g⁻¹ with RQ ≈ 0.9.

Summary. Lipid > protein ≈ carbohydrate in energy per gram, but carbohydrate is the fastest-mobilised substrate. Migratory birds and hibernating mammals rely on lipid stores precisely because of the high energy density.
Why this scores
Why this scores 5/5. Mark scheme: (1) carbohydrate ~17 kJ g⁻¹ / RQ 1.0 / glucose enters glycolysis; (2) lipid ~38 kJ g⁻¹ / RQ 0.7; (3) reason lipids have more energy (more H, more reduced); (4) protein ~17 kJ g⁻¹ / RQ 0.9 / deaminated; (5) protein used as last resort.

Key Definitions and Keywords — Respiration

Definitions to memorise and the exact keywords mark schemes credit for respiration answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9700 sitting.

Respiration
Examiner keyword
The enzyme-controlled oxidation of organic substrates (mainly glucose) within cells, releasing energy used to synthesise ATP from ADP + Pi.
Glycolysis
Examiner keyword
The cytoplasmic pathway in which one glucose (6C) is converted to two pyruvate (3C), with a net production of 2 ATP and 2 NADH. Anaerobic — no O₂ required.
Link reaction
Examiner keyword
The pathway in the mitochondrial matrix in which pyruvate is decarboxylated and dehydrogenated to form acetyl-CoA, releasing CO₂ and producing NADH.
Krebs cycle (citric acid cycle / TCA cycle)
Examiner keyword
The cycle in the mitochondrial matrix in which acetyl-CoA combines with oxaloacetate, and a series of oxidation, decarboxylation and dehydrogenation steps regenerate oxaloacetate, producing 3 NADH, 1 FADH₂, 1 ATP and 2 CO₂ per turn.
Oxidative phosphorylation
Examiner keyword
The production of ATP using energy from the oxidation of reduced coenzymes (NADH, FADH₂) by the electron transport chain at the inner mitochondrial membrane.
Electron transport chain (ETC)
Examiner keyword
A series of electron carrier proteins in the inner mitochondrial membrane that pass electrons by sequential redox reactions; energy released is used to pump H⁺ into the intermembrane space.
Chemiosmosis
Examiner keyword
The flow of protons (H⁺) down their electrochemical gradient through ATP synthase, driving the phosphorylation of ADP to ATP.
Decarboxylation
Examiner keyword
The removal of a carboxyl group (CO₂) from a substrate. Occurs once in the link reaction and twice in each Krebs cycle.
Dehydrogenation
Examiner keyword
The removal of hydrogen atoms from a substrate, with their transfer to a coenzyme (NAD or FAD), reducing it to NADH or FADH₂.
NAD (nicotinamide adenine dinucleotide)
Examiner keyword
A coenzyme that accepts hydrogen during dehydrogenation steps, becoming reduced to NADH. NADH carries hydrogen / electrons to the ETC, where each donates enough energy for ~2.5 ATP.
FAD (flavin adenine dinucleotide)
Examiner keyword
A coenzyme that accepts hydrogen at one step of the Krebs cycle (and in β-oxidation of fatty acids), becoming reduced to FADH₂. Each FADH₂ donates enough energy for ~1.5 ATP.
Respiratory quotient (RQ)
Examiner keyword
The ratio of CO₂ produced to O₂ consumed during respiration: $\text{RQ} = \dfrac{\text{CO}_2 \text{ produced}}{\text{O}_2 \text{ consumed}}$ . Carbohydrate 1.0; lipid 0.7; protein 0.9.
Anaerobic respiration
Examiner keyword
Respiration in the absence of oxygen. Limited to glycolysis followed by fermentation (ethanol + CO₂ in yeast; lactate in mammalian muscle). Net yield 2 ATP per glucose.
Oxygen debt (excess post-exercise oxygen consumption)
The extra oxygen needed after vigorous exercise to oxidise the lactate that built up during anaerobic respiration in muscle, restore creatine phosphate and reoxygenate myoglobin / haemoglobin.

Common Mistakes and Misconceptions — Respiration

The traps other students keep falling into on respiration questions — taken from recent Cambridge International A Level 9700 examiner reports and mark schemes — and how to avoid them.

✕Saying that glucose enters the mitochondrion to be respired
9700 Examiner Reports 2023-2024
Why it happens
Glycolysis is often confused with mitochondrial reactions.
How to avoid it
Glucose is respired in the cytoplasm to pyruvate (glycolysis). Only pyruvate is transported into the mitochondrion. Always state 'glycolysis in the cytoplasm; link, Krebs and OP in the mitochondrion'.
✕Quoting the gross ATP from glycolysis (4) rather than net (2)
9700 Examiner Reports 2024
Why it happens
Students forget the 2 ATP invested in phosphorylation steps.
How to avoid it
Always state 'net 2 ATP from glycolysis'. The two ATP used to phosphorylate glucose are an 'investment'; gross 4 ATP - 2 invested = net 2.
✕Writing that oxygen accepts protons in the ETC
9700 Examiner Reports 2024
Why it happens
Confusion between electrons and protons in chemiosmosis.
How to avoid it
Oxygen accepts electrons (the final electron acceptor of the ETC). Electrons combine with protons (from the matrix) to form water. Get the order right: e⁻ along ETC; H⁺ pumped to intermembrane space; e⁻ + H⁺ + O₂ → H₂O at the end.
✕Saying the Krebs cycle runs once per glucose
9700 Examiner Reports 2024
Why it happens
Students forget glucose → 2 pyruvate → 2 acetyl-CoA.
How to avoid it
Krebs runs twice per glucose (one for each pyruvate / acetyl-CoA). Per glucose: 4 CO₂, 6 NADH, 2 FADH₂, 2 ATP from Krebs.
✕Saying anaerobic respiration in mammalian muscle produces CO₂
9700 Examiner Reports 2023
Why it happens
Confusing yeast and muscle fermentation pathways.
How to avoid it
Mammalian muscle produces lactate only — no CO₂. CO₂ is only produced in yeast (alcoholic fermentation: pyruvate → ethanal + CO₂ → ethanol).
✕Writing that fermentation produces ATP
9700 Examiner Reports 2024
Why it happens
Confusion about what fermentation actually does.
How to avoid it
Fermentation produces no extra ATP. Its sole purpose is to regenerate NAD so that glycolysis can continue. The 2 ATP net come from glycolysis alone.
✕Quoting RQ in units (e.g. 'cm³ per minute')
9700 Examiner Reports 2024
Why it happens
Confusion with rate measurements.
How to avoid it
RQ is a ratio — it has no units. Quote as a number (1.0, 0.7, 0.9).

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Respiration — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Respiration

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Glycolysis — net yield (5 marks)

2Link reaction (4 marks)

3Krebs cycle (6 marks)

4Oxidative phosphorylation and chemiosmosis (7 marks)

5Anaerobic respiration in yeast and muscle (5 marks)

6Respiratory quotient (4 marks)

Glycolysis — net yield per glucose

Link reaction — yield per pyruvate (per glucose: × 2)

Krebs cycle — yield per turn (per glucose: × 2)

Oxidative phosphorylation — ATP per reduced coenzyme

Total ATP yield per glucose (aerobic)

Respiratory quotient (RQ)

Respiration

Glycolysis

Link reaction

Krebs cycle (citric acid cycle / TCA cycle)

Oxidative phosphorylation

Electron transport chain (ETC)

Chemiosmosis

Decarboxylation

Dehydrogenation

NAD (nicotinamide adenine dinucleotide)

FAD (flavin adenine dinucleotide)

Respiratory quotient (RQ)

Anaerobic respiration

Oxygen debt (excess post-exercise oxygen consumption)

✕Saying that glucose enters the mitochondrion to be respired

✕Quoting the gross ATP from glycolysis (4) rather than net (2)

✕Writing that oxygen accepts protons in the ETC

✕Saying the Krebs cycle runs once per glucose

✕Saying anaerobic respiration in mammalian muscle produces CO₂

✕Writing that fermentation produces ATP

✕Quoting RQ in units (e.g. 'cm³ per minute')

Practice questions

Past paper style quiz

Assess your understanding

Respiration — frequently asked questions