Energy and Respiration

Energy is the price of every active process in a cell

A living cell is never at equilibrium. Even when the cell is not visibly doing anything, ions are being pumped across membranes against their concentration gradients, large molecules are being built from small ones, the genetic material is being unwound and re-wound, and chemical signals are being sent and received. Each of these processes is non-spontaneous — it does not happen on its own — and the cell has to pay for it with energy. The currency the cell uses is ATP, the same molecule introduced in §6.1 at AS, but here we see it in its A2 role: the molecule that links the energy-yielding reactions of respiration (§04–§06) to the energy-requiring processes of the cell.

The syllabus is explicit about three examples, and they cover most of the territory the examiner will test:

• Active transport (§4.5) — pumping ions or molecules across a membrane against their concentration gradient, using a carrier protein and energy from ATP hydrolysis.
• Movement — the mechanical work of contraction (muscle, flagellum, cytoplasmic streaming) and the molecular work of cell division (chromosome movement, cytokinesis, vesicle traffic).
• Anabolic reactions — building larger molecules from smaller ones. The two that recur in the mark schemes are DNA replication (joining nucleotides into a polynucleotide, with energy from ATP and the dNTPs) and protein synthesis (joining amino acids into a polypeptide, with energy from ATP and GTP). Both were covered at AS in §6; here the point is to recognise that both of these are energy-requiring.

Anything that requires the cell to do work — physical, chemical or electrical — uses ATP. The mark scheme does not require a long list; the three examples above plus one or two more are enough.

Why ATP, and not another molecule?

The cell could in principle use any high-energy molecule as its energy currency — GTP, UTP, creatine phosphate, all of which exist — but ATP has a particular combination of features that make it the universal currency. The mark schemes test four of them, all of which are reasons why ATP is a good currency (small, fast, recyclable) rather than a good store (where lipids and carbohydrates take over):

1. Releases energy on hydrolysis. The terminal phosphoanhydride bond (between the second and third phosphate groups) is broken by water; the products are ADP, inorganic phosphate (Pᵢ) and a useful amount of energy (~30.5 kJ mol⁻¹ under standard conditions, less in the cell). The energy is released in a single step, in a usable form, at a rate the cell can control.
2. The reaction is reversible. ATP can be regenerated from ADP and Pᵢ using energy from respiration. The cell therefore does not need to keep making new ATP molecules from scratch; it recycles the same pool many thousands of times per cell per second.
3. Small and soluble. ATP diffuses freely in the cytoplasm and can be moved around the cell cheaply. A bulky or insoluble energy carrier would slow every reaction that used it.
4. Releases the right amount of energy, not too much, not too little. A single hydrolysis step releases about 30 kJ mol⁻¹, which is enough to drive one reaction at a time without damaging the cell. (A molecule that released 1000 kJ mol⁻¹ in one step would be impossible to control.) The mark scheme words this as "releases energy in small, manageable amounts" or "releases the right amount of energy for cellular reactions".

A useful framing: ATP is the cash of the cell (small-denomination, instantly spendable, recycled); carbohydrates and lipids are the savings (large-denomination, stored, converted to ATP on demand). The exam tests both halves of this analogy.

Two ways the cell makes ATP

The syllabus (12.1.3) is precise about the two ways ATP is synthesised. They are not the same, and the exam will reward knowing which is which:

• Substrate-linked phosphorylation. A phosphate group is transferred from a phosphorylated substrate (a reaction intermediate that carries a phosphate) directly onto ADP, making ATP. The phosphate was put onto the substrate by an earlier reaction in the same pathway; the energy that drives the transfer comes from the substrate's own high-energy bond. The MS wording is "phosphate is transferred from a substrate molecule to ADP". Substrate-linked phosphorylation happens in glycolysis (twice) and in the Krebs cycle (once per turn, via GTP in some texts but directly to ATP in the A2 specification). It does not need a membrane.
• Chemiosmosis. Protons (H⁺ ions) are pumped across a membrane, building up a higher concentration on one side; they then flow back through the enzyme ATP synthase, and the energy of that flow is used to join ADP and Pᵢ into ATP. The MS wording is "a proton gradient across a membrane drives the synthesis of ATP via ATP synthase". Chemiosmosis happens in the inner mitochondrial membrane (during oxidative phosphorylation, §06) and in the thylakoid membrane of chloroplasts (during photophosphorylation, covered in §13 — the same mechanism, different membrane). It needs an intact, impermeable membrane.

The two mechanisms are sometimes confused because both make ATP. The test is the location: substrate-linked phosphorylation happens in the cytoplasm (glycolysis) and the matrix (Krebs cycle), and is a property of soluble enzymes; chemiosmosis happens on a membrane, and needs the membrane to be intact. The exam asks "name the type of phosphorylation that occurs in glycolysis and the Krebs cycle" — the answer is substrate-linked. The exam asks "name the type of phosphorylation that occurs on the inner mitochondrial membrane" — the answer is chemiosmosis.

Carbohydrates, lipids and proteins differ in how much energy they release per gram

The cell has three respiratory substrates, and they are not interchangeable: each releases a different amount of energy per gram of substrate oxidised. The order is fixed, and the reason is the degree of reduction of the substrate. The more C–H (and to a lesser extent C–OH) bonds a substrate carries, the more hydrogen it can donate to the electron transport chain, and the more ATP can be made per molecule oxidised. The syllabus values are:

• Carbohydrates (e.g. glucose) → ~16 kJ g⁻¹ released on complete oxidation. The C–H content is intermediate, and the ratio of C : H : O in carbohydrates is 1 : 2 : 1.
• Lipids (e.g. triglycerides) → ~38 kJ g⁻¹ released on complete oxidation — more than twice the value per gram for carbohydrates. Triglycerides are mostly long hydrocarbon chains (the fatty acid tails), with very few oxygen atoms; the C–H content is high, so the energy yield per gram is much higher.
• Proteins → ~17 kJ g⁻¹ released on complete oxidation (slightly higher than carbohydrates). Proteins are not the cell's preferred respiratory substrate — they are used as a substrate only when carbohydrate and lipid reserves are exhausted (e.g. in prolonged starvation), because the cell's proteins are functional molecules that the cell needs.

Two MS points to memorise: (1) the order of energy values per gram is lipids > proteins > carbohydrates, with lipids roughly 2× carbohydrates; (2) the order is not the same as the order of the substrates' chemical complexity — what matters is the proportion of C–H (reducing) bonds per gram. The MS sometimes asks "explain why lipids release more energy per gram than carbohydrates" — the answer is "lipids have a higher proportion of C–H bonds (and a lower proportion of C–OH and C=O bonds) per gram, so more hydrogen is available to be oxidised in respiration, releasing more energy."

The respiratory quotient (RQ)

The respiratory quotient is a one-line ratio that tells the cell (or the experimenter) what substrate is being respired:

The RQ has a value for each substrate, and the value comes out of the stoichiometry of the balanced equation. The three the syllabus expects:

• Carbohydrate (e.g. glucose), RQ = 1.0. The complete oxidation of glucose is:
  C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O
  RQ = 6 CO₂ / 6 O₂ = 1.0.
• Lipid (e.g. tripalmitin, a typical triglyceride), RQ ≈ 0.7. The MS accepts 0.7 (the tripalmitin value). The exact value depends on the lipid, but for a triglyceride the RQ is below 1 because the substrate is highly reduced and the cell needs more O₂ to oxidise each carbon than it does for carbohydrate. For oleic acid (a common fatty acid) the RQ is closer to 0.71; for stearic acid, 0.69.
• Protein, RQ ≈ 0.9. Proteins contain nitrogen, so the carbon and hydrogen are not fully oxidised to CO₂ and H₂O; the RQ lies between 0.9 and 1.0, depending on the amino acid.

Two MS points to remember: (1) the carbohydrate equation is the one the exam expects the student to write — "C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O" — and the RQ is then read off as 6/6 = 1.0; (2) when a question gives a mixed substrate (e.g. a seed germinating and using both lipid and carbohydrate reserves), the RQ lies between the values of the two substrates and can be used to infer the proportion of each. RQ values above 1.0 indicate that CO₂ is being produced by a process other than respiration — for example, ethanolic fermentation in yeast, or the decarboxylation step of the Krebs cycle when the substrate is, e.g., pyruvate. (A malate → ethanol fermentation pathway produces 2 CO₂ per glucose without consuming O₂, so RQ → ∞ for that single step.)

How the respirometer works

A simple respirometer is a sealed container connected to a capillary tube (the manometer) and a syringe. The setup in the diagram is the one the syllabus expects; the MS wording is "a sealed container with the organism, a CO₂ absorber, and a capillary manometer with a coloured liquid, plus a control tube with an equal volume of inert material (e.g. glass beads) at the same temperature."

The principle is that, in a closed container, the organism takes up O₂ and gives out CO₂. If the CO₂ is absorbed as fast as it is produced (e.g. by soda lime or KOH at the bottom of the tube), the only gas that changes in the container is O₂ — the CO₂ is removed. The volume of the gas in the tube therefore falls as O₂ is consumed, and the coloured liquid in the capillary moves towards the tube. The rate of movement of the liquid (in mm min⁻¹) is the rate of O₂ consumption.

A respirometer measures the O₂ consumed directly. To get the RQ, the CO₂ produced must also be measured, and this is done by running a second tube without the CO₂ absorber. In the second tube, the volume change is the net gas change: O₂ consumed − CO₂ produced. With both tubes, the experimenter can solve for both O₂ and CO₂, and the RQ is CO₂/O₂. The MS sometimes simplifies this: "run two respirometers, one with CO₂ absorber and one without; the difference in manometer readings gives the CO₂ produced."

The control tube is essential. Any change in room temperature or atmospheric pressure moves the manometer liquid in both tubes; the difference in readings between the two tubes corrects for these changes and leaves only the respiration-induced movement. The control tube contains an equal volume of inert material (glass beads, dead seeds, cotton wool — anything that takes up the same space and reaches the same temperature) so that only the biological gas change differs.

A typical Paper 5 question: "describe how you would use a simple respirometer to determine the RQ of germinating seeds." A complete answer has four pieces: (1) the setup, with the CO₂ absorber and a control tube; (2) the equilibration step (leave the apparatus for 5–10 minutes before reading, so temperature and pressure stabilise); (3) the measurement (record the position of the liquid at the start and after a known time, e.g. 30 min); (4) the calculation (volume of O₂ consumed = distance moved × cross-sectional area of the capillary; rate = volume / time × mass of seeds). The RQ then comes from the parallel run without the CO₂ absorber, as above.

The four stages, located in the cell

The first thing the syllabus requires is a clean map of where the four stages of aerobic respiration happen, before any of them is described. The map is small but exact, and the MS marks it as the first sub-question of the topic:

Two MS points that the student must hit: (1) glycolysis is in the cytoplasm, not in any organelle — the reaction happens in the cytosol, dissolved in the cell's fluid; the mitochondrion's role starts with the link reaction. (2) the link reaction and the Krebs cycle are both in the mitochondrial matrix — between the inner and outer membranes; they are not on a membrane. Only the last stage, oxidative phosphorylation, is on the inner mitochondrial membrane, and the reason is that it is the only one of the four that needs a membrane to hold a proton gradient (chemiosmosis, §01).

The cell's logic is straightforward: glycolysis is the primordial pathway, the one that worked in the earliest cells before there were any mitochondria; once the cell engulfed an aerobic bacterium that became the mitochondrion, the rest of the pathway moved inside. The matrix is the inside of the mitochondrion (analogous to the inside of the bacterial cell), so reactions that were originally in the cytosol and have moved inside the bacterium are now in the matrix. The inner membrane is the descendant of the bacterium's plasma membrane, and that is where the proton gradient is held.

Structure–function relationships in the mitochondrion

The mitochondrion is the textbook example of a structure–function organelle; the MS rewards five pairings, all of which are visible in any reasonable diagram:

1. Double membrane — the outer membrane is smooth and permeable to small molecules (it has large channel proteins, porins); the inner membrane is much less permeable and is folded into cristae. The impermeability of the inner membrane to H⁺ is what allows the proton gradient to be held (§06); the permeability of the outer membrane lets substrates (pyruvate, ADP, Pᵢ) get from the cytoplasm to the intermembrane space, then through transporters in the inner membrane into the matrix.
2. Cristae (folds of the inner membrane) — these increase the surface area of the inner membrane, allowing more electron transport chains and ATP synthase complexes per mitochondrion. The MS words this as "cristae provide a large surface area for the electron transport chain and ATP synthase." Cells with high respiratory rates (liver, muscle, kidney) have more, longer, more closely-packed cristae than cells with low respiratory rates.
3. Matrix — the compartment enclosed by the inner membrane. It contains the soluble enzymes of the link reaction and Krebs cycle, the mitochondrial DNA, the mitochondrial ribosomes and tRNAs (a relic of the mitochondrion's origin as an endosymbiont), and a high concentration of substrates. The matrix is not on a membrane; the reactions that happen here (link reaction, Krebs cycle) use substrate-linked phosphorylation, not chemiosmosis.
4. Small, mobile size — mitochondria are typically 0.5–1.0 µm wide and 1–10 µm long, so they can be moved through the cytoplasm to where ATP is needed. In a muscle fibre, they line up along the myofibrils; in a sperm cell, they are wrapped around the base of the flagellum; in a dividing cell, they segregate between the daughter cells. The MS sometimes asks "why are mitochondria found in high numbers near the contractile machinery of muscle cells?" — the answer is "to supply ATP close to where it is used, so ATP does not have to diffuse over a long distance."
5. Self-replicating — mitochondria contain their own DNA and can divide by binary fission, independently of the cell cycle. A cell that needs more ATP (e.g. a muscle cell in training) signals the mitochondria to divide; a cell that needs less (e.g. a cell entering quiescence) removes mitochondria by mitophagy. The MS sometimes rewards this as an "AVP" (additional valid point).

The MS also rewards identifying these features on an electron micrograph. The signature features on an EM are the double membrane (two parallel dark lines, the outer and inner membranes, with a narrow gap), the cristae (the dark inward folds), and the matrix (the lighter, granular interior). A common MS question: "identify the features labelled X, Y and Z on the electron micrograph", with X = outer membrane, Y = inner membrane, Z = cristae or matrix.

The 6C → 3C → 3C story

Glycolysis is the first of the four stages, and the only one that does not need a mitochondrion. The whole stage happens in the cytoplasm, in nine enzymic steps, and the net effect is that one 6C glucose molecule is split into two 3C pyruvate molecules, with the production of a small amount of ATP and reduced NAD. The syllabus asks the student to outline the pathway, which is the MS's way of saying: give the carbon skeleton (6C → 3C + 3C → 2 × 3C), name the intermediates (glucose → fructose 1,6-bisphosphate → two triose phosphate → two pyruvate), and state the products (ATP and reduced NAD). The MS does not require the names of all nine enzymes.

The pathway is conventionally written in three phases:

1. Phosphorylation of glucose to fructose 1,6-bisphosphate (the 'investment' phase). Glucose (6C) is phosphorylated twice, using 2 ATP. The first phosphorylation makes glucose 6-phosphate (catalysed by hexokinase); the second, after isomerisation to fructose 6-phosphate, makes fructose 1,6-bisphosphate (catalysed by phosphofructokinase, PFK). The MS sometimes summarises this as "glucose is phosphorylated using 2 ATP to form fructose 1,6-bisphosphate". The 6C sugar is now committed to glycolysis — the two phosphate groups prevent it from leaving the cell, and they lower the activation energy for the next reaction.
2. Splitting to two triose phosphates (the 'splitting' phase). The 6C fructose 1,6-bisphosphate is split into two 3C triose phosphates (glyceraldehyde 3-phosphate, GAP, and dihydroxyacetone phosphate, DHAP). DHAP is rapidly converted to GAP by triose phosphate isomerase, so the net effect of the phase is one 6C sugar → two molecules of GAP. The MS wording is "fructose 1,6-bisphosphate is split into two triose phosphates".
3. Oxidation of triose phosphate to pyruvate (the 'pay-off' phase). Each GAP is oxidised (its aldehyde group is converted to a carboxyl group, releasing two hydrogen atoms that are picked up by NAD) and phosphorylated (inorganic phosphate is added). The product is 1,3-bisphosphoglycerate, which then donates one of its phosphate groups directly to ADP, making ATP. After a few more steps, the end product is pyruvate (3C). Because there are two GAPs, the phase happens twice per glucose, and the products are 2 pyruvate, 2 ATP, and 2 reduced NAD per glucose.

The MS summarises the net reaction of glycolysis as:

Glucose + 2 NAD + 2 ADP + 2 Pᵢ → 2 pyruvate + 2 reduced NAD + 2 ATP + 2 H₂O

(or, equivalently, 2 ATP + 2 reduced NAD + 2 H⁺, depending on the MS's notation). The crucial feature is that 4 ATP are made in the pay-off phase, but 2 were used in the investment phase, so the net yield is 2 ATP per glucose. The 2 ATP figure is the one the MS credits, not the gross 4.

The three things the MS wants the student to say

On a "describe glycolysis" question, the MS marks the answer against three things the student must include. (All three appear in the syllabus statement; they are the only three things the examiner is required to credit.)

1. The carbon skeleton: 6C → 3C + 3C → 2 × 3C. Glucose is split into two triose phosphates, which are then converted to pyruvate. Naming the intermediates (glucose, fructose 1,6-bisphosphate, triose phosphate, pyruvate) is the easiest way to get this mark.
2. ATP is used and made. The 2 ATP used in the phosphorylation of glucose is paid back with the 4 ATP made by substrate-linked phosphorylation in the pay-off phase, giving a net of 2 ATP per glucose.
3. Reduced NAD is made. Each GAP oxidation step makes 1 reduced NAD, and there are two GAPs, so the net yield is 2 reduced NAD per glucose.

The student does not need to name the nine enzymes, the structures of the intermediates, the steps of the pathway, or the names of the specific carbons. The MS's "outline" command is the cue: name the inputs (glucose, 2 NAD, 2 ADP, 2 Pᵢ), the path (6C → 3C → 3C), the outputs (2 pyruvate, 2 ATP, 2 reduced NAD), and stop. The MS sometimes awards a fourth mark for an AVP — usually "the reactions happen in the cytoplasm", "the pathway does not require oxygen", or "substrate-linked phosphorylation makes the ATP".

Why this matters for the rest of respiration

The two products of glycolysis — pyruvate and reduced NAD — feed into the rest of respiration. Pyruvate is the substrate for the link reaction (§05): it crosses the outer mitochondrial membrane through porins, and the inner mitochondrial membrane through a specific pyruvate carrier, before being decarboxylated and combined with coenzyme A in the matrix. Reduced NAD is the carrier of reducing equivalents: the H atoms it carries are donated to the electron transport chain at complex I, and the energy released as the electrons pass down the chain is used to pump protons and make ATP (§06). Without the 2 reduced NAD made in glycolysis, the link reaction's 2 reduced NAD, the Krebs cycle's 6 reduced NAD and 2 reduced FAD, the cell would have no reducing equivalents to feed the ETC, and the bulk of the ATP yield of respiration would be lost.

A second important feature: glycolysis does not require oxygen. The pathway was running in the earliest cells, long before oxygen was a major atmospheric gas, and the enzymes do not depend on O₂. This is why glycolysis is the only stage of respiration that can run in anaerobic conditions — and the reason anaerobic respiration (§07) works at all is that glycolysis keeps producing 2 ATP per glucose without oxygen, even though the rest of the pathway stops.

From cytoplasm to matrix — pyruvate crosses the double membrane

The link reaction is the bridge between glycolysis and the Krebs cycle. Pyruvate is made in the cytoplasm by glycolysis, but the link reaction and the Krebs cycle are both in the mitochondrial matrix, so the pyruvate has to be moved into the mitochondrion. The mechanism is straightforward: pyruvate crosses the outer mitochondrial membrane through porins (large, non-specific channels), and the inner mitochondrial membrane through a specific pyruvate carrier — a transporter that exchanges pyruvate⁻ for OH⁻ across the inner membrane, driven by the H⁺ gradient. The MS does not require the name of the carrier; the MS does require the statement "pyruvate enters the mitochondrion" before the link reaction is described.

The link reaction is run once per pyruvate, so it runs twice per glucose. Each pyruvate (3C) is converted to acetyl-CoA (2C) in three steps:

1. Decarboxylation — one carbon is removed as CO₂. The carboxyl group of pyruvate is released as CO₂; this is one of the CO₂ molecules that eventually leaves the cell as a waste product of respiration.
2. Oxidation — the remaining 2C fragment is oxidised (the H atoms are removed), and the electrons are passed to NAD, making reduced NAD.
3. Combination with coenzyme A — the 2C fragment (acetyl group) is joined to coenzyme A (CoA), making acetyl-CoA. CoA is a vitamin-derived cofactor with a thiol (–SH) group that forms a high-energy thioester bond with the acetyl group; the bond's energy is what drives the next step (the condensation with oxaloacetate).

Per glucose (2 pyruvates), the link reaction makes:

• 2 acetyl-CoA
• 2 CO₂ (one per pyruvate)
• 2 reduced NAD (one per pyruvate)

No ATP is made in the link reaction. The MS will mark the answer "2 acetyl-CoA, 2 CO₂, 2 reduced NAD, no ATP" as a one-shot list.

The Krebs cycle — citrate synthase, then eight small steps back

The Krebs cycle (citric acid cycle, tricarboxylic acid cycle) is run once per acetyl-CoA, so it runs twice per glucose. The cycle has eight enzymic steps in textbook treatments, but the syllabus asks for the outline, and the MS marks only the high-level structure: 4C + 2C → 6C, then six small steps that regenerate the 4C oxaloacetate, with decarboxylation, dehydrogenation and the reduction of NAD and FAD along the way.

The cycle's net reaction per turn is:

Acetyl-CoA + 3 NAD + FAD + ADP + Pᵢ + 2 H₂O → 2 CO₂ + CoA + 3 reduced NAD + reduced FAD + ATP + 3 H⁺

Or, per glucose (2 turns of the cycle):

• 2 ATP (substrate-linked phosphorylation; sometimes written as GTP)
• 6 reduced NAD (3 per turn × 2)
• 2 reduced FAD (1 per turn × 2)
• 4 CO₂ (2 per turn × 2)

The cycle's logic: oxaloacetate (4C) is regenerated at the end of each turn, so the molecule that turns over is catalytic — the same oxaloacetate can be used turn after turn. The acetyl group is the fuel; it is fully oxidised to 2 CO₂. The reducing equivalents (6 NADH + 2 FADH₂ per glucose) are the cycle's output — they go to the electron transport chain and yield the bulk of the ATP.

The MS marks the following points as the must-haves for an "outline the Krebs cycle" answer:

1. Oxaloacetate (4C) + acetyl-CoA (2C) → citrate (6C). The condensation is catalysed by citrate synthase and releases CoA. The 6C citrate then goes through a series of small steps.
2. The 6C citrate is converted back to the 4C oxaloacetate in a series of small steps. The MS does not require the names of all eight intermediates; the wording "in a series of small steps" is sufficient.
3. The cycle involves decarboxylation and dehydrogenation. Decarboxylation is the release of CO₂ (twice per turn, so 4 CO₂ per glucose). Dehydrogenation is the removal of H atoms, which reduce the coenzymes NAD and FAD. The MS marks the words "decarboxylation" and "dehydrogenation" as the two distinct points.
4. NAD and FAD are reduced. Per turn, 3 NAD and 1 FAD. Per glucose, 6 NAD and 2 FAD. The MS sometimes asks for the explicit numbers; the answer "3 reduced NAD and 1 reduced FAD per turn" is the safest.

The role of decarboxylation and dehydrogenation

Two chemical processes drive the Krebs cycle, and the MS marks both explicitly. Each happens in different steps of the cycle, and each has a different fate for the atoms removed.

• Decarboxylation is the removal of a carboxyl group (–COO⁻) as CO₂. It happens twice per turn (once at the 6C → 5C step, once at the 5C → 4C step). The MS wording is "decarboxylation releases CO₂". The 4 CO₂ made per glucose (2 from the link reaction, 2 from each turn of the cycle × 2 turns) account for all the carbon in the original glucose — the carbon skeleton is fully oxidised.
• Dehydrogenation is the removal of hydrogen atoms. It happens in four steps per turn (three reducing NAD, one reducing FAD). The MS wording is "dehydrogenation reduces NAD and FAD". The reducing equivalents (reduced NAD and reduced FAD) are then re-oxidised by the electron transport chain, and the energy released is used to make the bulk of the ATP yield of respiration.

The MS sometimes asks "what is the role of NAD and FAD in the Krebs cycle?" The answer is: "to accept hydrogen atoms removed by dehydrogenation, becoming reduced NAD and reduced FAD; these then carry the hydrogen to the electron transport chain on the inner mitochondrial membrane, where the energy is released to make ATP." The coenzymes are carriers of reducing equivalents, not stores — they are constantly being reduced in the Krebs cycle and re-oxidised in the ETC.

NAD and FAD are carriers of hydrogen — but the hydrogen is split before it reaches the chain

The reduced NAD and reduced FAD made by glycolysis, the link reaction and the Krebs cycle are the output of those stages; they are the form in which the energy of glucose is packaged for the next stage. The next stage happens on the inner mitochondrial membrane, and it begins with the carriers donating their hydrogen to the first complex of the electron transport chain. But the hydrogen is not passed on as a single atom — it is split. The H atom is separated into a proton (H⁺, which is released into the matrix) and an electron (e⁻, which is the part that carries the energy). The MS wording is precise: "the hydrogen atoms are split into protons and electrons at the first complex of the ETC".

The reason for the split is that the electron transport chain is, in fact, an electron transport chain — it is the electrons that move from one carrier to the next down the chain, releasing energy at each step, and it is the energy of those electrons that is captured and used to pump protons across the membrane. The protons do not move along the chain at all; they are released into the matrix and re-used at the end, when they flow back through ATP synthase to make ATP.

The four complexes of the ETC and the two mobile carriers (coenzyme Q / ubiquinone, and cytochrome c) are the apparatus. The MS does not require the names of the four complexes; the MS does require the order (electrons enter at complex I from reduced NAD, or at complex II from reduced FAD; they pass to coenzyme Q; then to complex III; then to cytochrome c; then to complex IV; then to oxygen). The energy released as the electrons move down the chain (from higher to lower redox potential) is used to pump H⁺ from the matrix to the intermembrane space at complexes I, III and IV.

The five-step story of chemiosmosis

The MS asks the student to explain oxidative phosphorylation as a five-step story, and the same five points appear in the mark scheme every session. The student who knows these five points in order will mark well on the most heavily examined sub-topic of the topic.

1. Hydrogen atoms split into protons (H⁺) and electrons (e⁻). This happens at the first complex of the ETC. The protons are released into the matrix; the electrons carry the energy.
2. Electrons release energy as they pass along the electron transport chain. The carriers in the chain have progressively higher reduction potentials (or, equivalently, progressively lower energy levels for the electrons). The energy released at each step is captured by the complexes.
3. The released energy is used to pump protons from the matrix to the intermembrane space. This happens at complexes I, III and IV. The result is a higher [H⁺] in the intermembrane space than in the matrix, and a difference in electrical charge (the intermembrane space is more positive). The two together constitute the proton-motive force.
4. Protons return to the matrix by facilitated diffusion through ATP synthase. The only route back for protons is through ATP synthase (the inner membrane is otherwise impermeable to H⁺). The energy of the proton flow is used by ATP synthase to join ADP and Pᵢ into ATP. The MS wording: "protons flow down their electrochemical gradient through ATP synthase, providing energy for the synthesis of ATP."
5. Oxygen is the final electron acceptor. At complex IV, the electrons (which have now given up most of their energy) are passed to O₂. The O²⁻ combines with 2 H⁺ (from the matrix) to form H₂O. The MS wording: "oxygen acts as the final electron acceptor to form water". Without O₂, the chain stalls: the carriers remain reduced, no protons can be pumped, no gradient forms, no ATP is made, and the cell dies (or falls back on glycolysis alone — see §07).

The student should know the story in the order above. The MS marks the five points in order, and a "describe" answer that names all five gets 5 marks.

The role of NAD and FAD: the same idea, two entries

The two coenzymes carry hydrogen to the ETC, but they enter at different points, and the difference matters for the ATP yield:

• Reduced NAD donates to complex I. Two electrons are passed along the chain, and complex I pumps 4 H⁺ per pair of electrons. The full chain runs, so the electron reaches complex IV and combines with O₂. Net yield per reduced NAD: roughly 2.5 ATP (the modern figure; the A2 specification accepts "many ATP" or the per-electron count of ~3 H⁺ per ATP).
• Reduced FAD donates to complex II. Complex II does not pump protons; the electrons are passed to coenzyme Q and continue down the chain, but the chain is shorter. Net yield per reduced FAD: roughly 1.5 ATP (the A2 specification accepts "fewer ATP than reduced NAD").

The MS does not require the specific 2.5 / 1.5 figures. The MS does require the qualitative difference: "reduced NAD yields more ATP than reduced FAD", or "the electrons from reduced NAD enter the chain at an earlier complex than those from reduced FAD". The per-glucose totals are 10 reduced NAD (2 from glycolysis, 2 from the link reaction, 6 from the Krebs cycle) and 2 reduced FAD (from the Krebs cycle), making roughly 28 ATP from oxidative phosphorylation per glucose, depending on the shuttle used to move the glycolytic reduced NAD into the mitochondrion.

A small but testable point: the 2 reduced NAD from glycolysis are made in the cytoplasm, but the ETC is in the mitochondrion. The cell has two shuttles for moving the reducing equivalents in. The glycerol 3-phosphate shuttle transfers the electrons to FAD in the inner membrane, costing one ATP per reduced NAD (so each glycolytic reduced NAD yields ~1.5 ATP). The malate–aspartate shuttle transfers the electrons to NAD in the matrix, preserving the full ~2.5 ATP per reduced NAD. The A2 specification does not require the shuttle names, but the MS sometimes asks "why is the ATP yield of aerobic respiration often given as 30 or 32, not 38?" — the answer is "because of the cost of moving the cytoplasmic reduced NAD into the mitochondrion".

Two fermentation pathways that regenerate NAD⁺

When oxygen is scarce, the ETC stalls, the proton gradient is not maintained, and the cell cannot regenerate NAD⁺ by re-oxidising reduced NAD at complex I. Without NAD⁺, the link reaction and the Krebs cycle stop, and glycolysis — the only pathway that does not require O₂ — would also stop at the GAP-dehydrogenase step, because the reaction reduces NAD to reduced NAD and the cell has run out of NAD. To keep glycolysis running (and so keep making 2 ATP per glucose), the cell needs a way to re-oxidise reduced NAD back to NAD without using the ETC. That way is fermentation — a small, terminal pathway that takes the pyruvate made by glycolysis and converts it into a different product, with the oxidation of reduced NAD back to NAD as a side effect.

There are two fermentation pathways the syllabus requires, and they are organism-specific:

• Lactate fermentation (mammals, especially muscle). Pyruvate (3C) is reduced directly to lactate (3C) by the enzyme lactate dehydrogenase. The H atoms that reduce pyruvate come from reduced NAD, which is re-oxidised to NAD. The lactate is released into the blood and taken up by the liver, where it is converted back to glucose by the Cori cycle. The MS wording: "pyruvate is reduced to lactate, oxidising reduced NAD to NAD, so glycolysis can continue."
• Ethanol fermentation (yeast, and some plant cells under waterlogging — §08). Pyruvate (3C) is decarboxylated to acetaldehyde (2C), releasing CO₂; the acetaldehyde is then reduced to ethanol (2C) by the enzyme alcohol dehydrogenase, with the oxidation of reduced NAD to NAD. The CO₂ is what makes bread rise and what makes beer fizzy. The MS wording: "pyruvate is decarboxylated to acetaldehyde (releasing CO₂), then reduced to ethanol, oxidising reduced NAD to NAD."

Why the ATP yield is so much lower in anaerobic conditions

The MS rewards a precise explanation, and the explanation has three parts. The student who names all three gets the mark.

1. Only glycolysis runs. The link reaction, Krebs cycle and oxidative phosphorylation all need oxygen (or, in the link reaction, the NAD⁺ that is regenerated only by the ETC). Without O₂, the only ATP-making stage is glycolysis.
2. Glycolysis makes only 2 ATP per glucose by substrate-linked phosphorylation. The bulk of the ATP from a glucose molecule (around 28 ATP, depending on the shuttle) comes from the reduced NAD made by glycolysis, the link reaction, and the Krebs cycle being re-oxidised in the ETC. Without the ETC, none of that reduced NAD can be re-oxidised by chemiosmosis, so none of that ATP can be made.
3. The fermentation steps do not make ATP. The reduction of pyruvate to lactate (or to ethanol) is a redox step that regenerates NAD⁺; it does not phosphorylate ADP. The only ATP comes from the substrate-linked phosphorylation of glycolysis, so the total is 2 ATP per glucose — versus ~30–32 in aerobic conditions.

The MS will sometimes ask the student to put a number on the difference. The answer is "about 15× more ATP in aerobic conditions": 30–32 ATP per glucose aerobic, 2 ATP per glucose anaerobic. The 15× figure is a useful memorisable number; the exact figure depends on the shuttle and the textbook, but the order of magnitude is right.

The MS also rewards a comment on the fate of the lactate or ethanol. In mammals, lactate is taken up by the liver and converted back to glucose by the Cori cycle, but at a cost of 6 ATP per glucose (the gluconeogenesis pathway). The ethanol made by yeast is toxic to the cell at high concentrations; this is why the alcohol content of fermenting yeast cultures is limited to about 12–15% before the yeast dies. The MS occasionally asks: "explain why anaerobic respiration is a useful short-term emergency but cannot be sustained for long" — the answer is "the lactate (or ethanol) accumulates because it is not further metabolised, and the cell's 2 ATP per glucose is not enough to cover its energy needs."

Three adaptations to life with the roots underwater

Rice is unusual among crops: it is grown in flooded paddy fields, where the roots are submerged in water with very little dissolved O₂. Most flowering plants cannot survive this, because their roots need O₂ for the ETC and would quickly run out of ATP without it. Rice has evolved three adaptations that allow it to grow in these conditions. The MS marks all three.

1. Aerenchyma in the roots. Aerenchyma is a tissue with large, air-filled intercellular spaces that run continuously from the leaves, down the stems, and into the roots. The air spaces act as a plumbing system for gases: O₂ produced by photosynthesis in the leaves (or dissolved in the water around the leaves) diffuses down the aerenchyma to the roots, where it can be used by the mitochondria of root cells for the ETC. CO₂ produced by the roots' respiration diffuses back up the aerenchyma to the leaves and is released or used in photosynthesis. The MS wording: "aerenchyma are air-filled spaces that allow the diffusion of O₂ from the leaves to the roots." The adaptation is structural, and it is the only one of the three that can be seen in a cross-section of the root under a microscope.
2. Ethanol fermentation in the roots. Even with aerenchyma, the O₂ supply to the deep root tips is limited. Rice roots therefore supplement aerobic respiration with ethanol fermentation: pyruvate is decarboxylated to acetaldehyde (releasing CO₂), and acetaldehyde is reduced to ethanol. The pathway is the same as in yeast (§07), and the effect is the same — NAD⁺ is regenerated so glycolysis can keep making 2 ATP per glucose anaerobically. The MS wording: "rice roots carry out ethanol fermentation to generate ATP anaerobically when O₂ is in short supply." Rice tolerates the ethanol that accumulates because its root cells have high levels of alcohol dehydrogenase and can re-metabolise the ethanol when O₂ returns.
3. Faster growth of the stems. In anaerobic conditions, ATP is in short supply and the root system is limited; the plant responds by spending its limited ATP on rapid stem elongation, so that the leaves reach the water surface and the shoot can start photosynthesising in air. This is controlled by the hormone ethylene, which accumulates in submerged tissues and promotes stem elongation. The MS wording: "faster stem growth allows the leaves to reach the surface, where O₂ is more abundant, so that aerobic respiration can resume in the leaves and photosynthesis can support the plant."

The three adaptations are coordinated: aerenchyma delivers some O₂ to the roots; ethanol fermentation keeps glycolysis running in the root cells that are still short of O₂; the fast stem growth gets the leaves to the surface so that more O₂ can be taken up. The MS sometimes asks "explain how rice is adapted to growing in water" — the answer is "rice has three adaptations: aerenchyma in the roots to deliver O₂; ethanol fermentation in the roots to generate ATP anaerobically; and faster stem growth to reach the water surface." All three are marks.

Redox indicators: a colour change as a proxy for respiration

DCPIP (2,6-dichlorophenolindophenol) and methylene blue are redox indicators — they change colour when they are reduced. DCPIP is blue in its oxidised form and colourless when reduced; methylene blue is blue in its oxidised form and colourless when reduced. The principle of the investigation is that, in a respiring cell, substrates (e.g. glucose) are oxidised and NAD is reduced; the reduced NAD then reduces the indicator, and the time taken for the colour to change is a measure of the rate of respiration. A faster colour change means a faster rate of respiration.

The standard setup is:

• A suspension of yeast cells in a buffer, with a known concentration of substrate (e.g. glucose).
• DCPIP (or methylene blue) added to the suspension.
• The mixture is incubated at a known temperature (often in a water bath for temperature-control experiments).
• The time taken for the blue colour to disappear is recorded. This is the end-point.

The rate of respiration is inversely proportional to the time taken for the colour to change: rate ∝ 1/t (per minute). The MS wording is "the time taken for DCPIP to be decolourised is inversely proportional to the rate of respiration."

To investigate the effect of temperature: repeat the experiment at a series of temperatures (e.g. 10, 20, 30, 40, 50 °C). The rate is highest at the optimum temperature (around 30–40 °C for yeast) and falls off at higher temperatures (denaturation of enzymes) and at lower temperatures (kinetic energy limit). A control tube without substrate (e.g. with water instead of glucose) checks that the colour change is due to respiration and not to some other reaction. A control tube without yeast checks that the substrate does not reduce DCPIP on its own.

To investigate the effect of substrate concentration: vary the concentration of glucose (e.g. 0, 0.1, 0.5, 1.0, 2.0 mol dm⁻³) and keep the temperature and other conditions constant. The rate rises with substrate concentration up to a saturation point, beyond which all the active sites of the rate-limiting enzyme are occupied and the rate plateaus (Michaelis–Menten kinetics; the MS calls this the saturation point).

The MS rewards the explicit statement of the control in each experiment. The standard controls are:

• No substrate (replace glucose with water): shows that the colour change is due to respiration, not to spontaneous reduction of DCPIP.
• No yeast (boiled or autoclaved yeast suspension): shows that the colour change requires living cells (i.e. enzymes).
• No DCPIP: shows that the colour change is not a property of the yeast itself.

The MS also rewards a comment on why a control is needed: "to check that the colour change is due to respiration and not to some other reaction".

The respirometer: a quantitative version of the same idea

The respirometer is the more quantitative version of the same investigation, and it is the apparatus that gives the RQ (covered in §02). The MS rewards the same idea — the apparatus measures O₂ consumed directly, with CO₂ absorbed by soda lime or KOH — but the respirometer measures a continuous rate, not a single end-point.

To investigate the effect of temperature on the rate of respiration with a respirometer:

• Set up a respirometer with germinating seeds (or small invertebrates) and a CO₂ absorber.
• Place the respirometer in a water bath at a known temperature (e.g. 20 °C); allow the apparatus to equilibrate for 5–10 min.
• Record the position of the manometer liquid at the start, and again after a known time (e.g. 30 min).
• The volume of O₂ consumed = distance moved × cross-sectional area of the capillary. Divide by the time and the mass of seeds to get the rate in mm³ O₂ g⁻¹ min⁻¹.
• Repeat at a series of temperatures (e.g. 10, 20, 30, 40, 50 °C). The rate rises to a peak at the optimum and then falls (denaturation).

A control tube (with glass beads or dead seeds) is essential. The MS rewards the statement "a control tube is used to correct for changes in temperature and atmospheric pressure." The control tube must be at the same temperature as the experimental tube (so the gas expansion due to temperature change is the same in both), and the manometer reading is the difference between the two tubes — only this difference is the respiration-induced change.

The MS sometimes asks "what are the limitations of the respirometer?" The answer has three parts:

• The capillary is narrow, so any small change in temperature or pressure is amplified as a large movement of the manometer liquid. A drift in room temperature can give a false reading; the control tube corrects for this.
• The CO₂ absorber (soda lime) is not selective. It absorbs any acid gas (e.g. SO₂), and it absorbs CO₂ from the respiration of microorganisms in the seeds as well as from the seeds themselves. This is usually not a problem in a well-controlled experiment, but it is a source of error.
• The respirometer is closed. Once set up, no O₂ enters; the experiment must be finished before the O₂ runs out (or the rate slows because of O₂ limitation).