A-Level Revision Aid: This resource is designed to support your revision. Content may contain errors. If you find a discrepancy with your class teaching, your teacher is correct — please let us know at alevelrevise@scott.scottrix.co.uk.
A-Level Biology revision on biological molecules and respiration: carbohydrates, lipids and proteins, enzyme kinetics (Km, Vmax, inhibitors), glycolysis, the Krebs cycle, chemiosmosis and anaerobic pathways, with multi-board exam practice.
🎯 Learning Objectives
Describe the structure of monosaccharides, disaccharides, polysaccharides, triglycerides and phospholipids, and state the biochemical tests (AQA 3.1.1-3.1.3; OCR 2.1.2)
Explain protein structure levels and enzyme action via the induced fit model (AQA 3.1.4; OCR 2.1.4)
Interpret Km, Vmax and the effects of competitive and non-competitive inhibitors from rate graphs
Detail glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation with locations and yields (AQA 3.5.2; OCR 5.2.1)
Explain chemiosmosis and calculate theoretical vs actual ATP yield per glucose; calculate RQ values
Compare anaerobic respiration in animals (lactate) and yeast (ethanol) and explain the lower yield
📋 Key Definitions and Core Concepts
Monomer: A small single molecular unit that can join to others in repeating chains; biological monomers include monosaccharides, amino acids and nucleotides.
Condensation reaction: A reaction joining two molecules with the formation of a covalent bond and the release of a molecule of water.
Hydrolysis reaction: A reaction breaking a covalent bond using a molecule of water; digestion and polymer breakdown use hydrolysis.
Glycosidic bond: A covalent bond formed between two monosaccharides in a condensation reaction (e.g. 1,4-bonds in amylose, 1,6-bonds at glycogen branch points).
Ester bond: The bond formed between glycerol and a fatty acid in a condensation reaction; three ester bonds join each triglyceride.
Induced fit model: Model of enzyme action in which substrate binding induces a conformational change in the active site that stresses bonds and lowers activation energy; explains enzyme specificity.
Km (Michaelis constant): The substrate concentration at which reaction velocity is half Vmax; a lower Km indicates higher enzyme-substrate affinity.
Vmax: The maximum rate of an enzyme-catalysed reaction, reached when all active sites are saturated with substrate.
Chemiosmosis: ATP synthesis driven by the flow of protons down an electrochemical gradient across a membrane (inner mitochondrial membrane or thylakoid) through ATP synthase.
Respiratory quotient (RQ): RQ = CO₂ produced ÷ O₂ consumed; ~1.0 for carbohydrate, ~0.7 for lipid, ~0.9 for protein, and can exceed 1 with anaerobic fermentation.
📌 Key Points
Glycolysis (cytoplasm): glucose → 2 pyruvate; net 2 ATP, 2 NADH. Link reaction (matrix): pyruvate → acetyl CoA + CO₂ + NADH. Krebs (matrix): per acetyl CoA → 2CO₂ + 3NADH + FADH₂ + ATP
Oxidative phosphorylation: NADH/FADH₂ donate electrons to the ETC; proton pumping builds the gradient; H⁺ flows through ATP synthase (chemiosmosis); O₂ is the terminal electron acceptor
Theoretical yield ~32 ATP per glucose; actual 30-32 due to proton leak, shuttle costs and ATP spent transporting pyruvate/ADP
Carbohydrates, lipids and biochemical tests (AQA 3.1.1-3.1.3; OCR 2.1.2; Edexcel A T1)
Monosaccharides (glucose, galactose, fructose) are monomers joined by condensation reactions releasing water to form glycosidic bonds; hydrolysis reverses this. α-glucose monomers join into starch (amylose: unbranched 1,4-bonds, coiled helix; amylopectin: branched 1,4- and 1,6-bonds) and glycogen (more highly branched, animals). The storage-function link is a guaranteed four-marker: insoluble so it does not affect water potential or draw water into cells; compact so lots of glucose fits into a small volume; branched so many terminal ends allow rapid simultaneous enzymatic hydrolysis when glucose demand is high; α-glycosidic bonds are easily hydrolysed by amylase/maltase. Cellulose contrasts: β-glucose in unbranched straight chains with hydrogen bonds between chains forming microfibrils - structural, not storage.
Triglycerides are one glycerol + three fatty acids joined by three ester bonds in condensation reactions. Function links: hydrophobic energy store releasing more energy per gram than carbohydrate (more C-H bonds to oxidise); insoluble storage; metabolic water source (e.g. camel hump, desert eggs); low mass. Phospholipids have one fatty acid replaced by a hydrophilic phosphate group - hydrophilic head, hydrophobic tails - explaining bilayer formation in membranes. Saturated fatty acids have no C=C double bonds (straight chains, pack tightly, solid at room temperature); unsaturated have double bonds (kinked chains, lower melting point, oils).
Tests are free marks - quote the sequence and positive result: Benedict's: heat with Benedict's reagent, brick-red precipitate = reducing sugar (green→yellow→red grades concentration with a colorimeter); non-reducing sugar: boil with dilute HCl to hydrolyse glycosidic bonds, neutralise with sodium hydrogencarbonate, re-test with Benedict's - positive means non-reducing sugar was present; iodine/potassium iodide: blue-black = starch; Biuret: NaOH then copper sulfate, purple = protein (peptide bonds); emulsion test: dissolve in ethanol, add to water, white emulsion = lipid. AQA RP6 and OCR PAG6 test the colorimetric calibration-curve technique.
Proteins and enzyme kinetics (AQA 3.1.4; OCR 2.1.4)
Amino acids have an amino group, carboxyl group, hydrogen and variable R-group on a central carbon; peptide bonds join amino acids in condensation reactions. Know the four structure levels with the bonds that stabilise each: primary - unique amino-acid sequence (peptide bonds); secondary - hydrogen bonding into α-helix or β-pleated sheet; tertiary - 3D folding via hydrogen, ionic and disulfide bonds between R-groups, determining active-site shape; quaternary - multiple polypeptide chains (haemoglobin: 4 chains + haem prosthetic groups). Tertiary structure is what enzymes and inhibitors attack - link any denaturation question to disruption of hydrogen/ionic bonds changing active-site shape.
Induced fit beats lock-and-key in mark schemes: the active site is close to complementary but changes shape slightly as the substrate binds, putting the enzyme-substrate complex under stress that lowers activation energy. Enzymes are biological catalysts, globular proteins, specific to substrates. Rate factors: temperature - more kinetic energy, more successful collisions, rate rises to an optimum (~37 °C in humans), then hydrogen/ionic bonds vibrate too much and break, denaturing the active site so rate falls sharply; pH - deviation from the optimum changes H⁺/OH⁻ concentrations, disrupting ionic and hydrogen bonds, denaturing the enzyme (often irreversibly at extremes); substrate concentration - rate rises to Vmax when all active sites are saturated; enzyme concentration - proportional rise if substrate is non-limiting.
Km and Vmax interpretation: Km is the substrate concentration at half-Vmax; lower Km = higher affinity. Competitive inhibitors bind the active site reversibly, competing with substrate: raising substrate concentration overcomes inhibition, so Vmax unchanged, Km increased. Non-competitive inhibitors bind an allosteric site, changing tertiary conformation so the active site is no longer complementary: substrate concentration cannot help, so Vmax decreased, Km effectively unchanged (for the fraction of uninhibited enzyme). Statins (competitive, HMG-CoA reductase) and cyanide (non-competitive, cytochrome c oxidase) are the stock examples. Metabolic pathways are regulated by feedback inhibition: the end product inhibits the first enzyme of the pathway when it accumulates (e.g. isoleucine inhibiting threonine deaminase), and by allosteric effectors that switch enzymes between active/inactive conformations - link these to homeostatic control of respiration rate.
ATP, glycolysis and the link reaction (AQA 3.1.6, 3.5.2; OCR 5.2.1)
ATP is a nucleotide derivative: adenine + ribose + three phosphates. Energy is released in small, manageable packets when the terminal phosphate bond is hydrolysed by ATPase to ADP + Pi (30.6 kJ mol⁻¹), and ATP is resynthesised from ADP + Pi using energy from respiration. Properties that suit its role: small (moves easily), water-soluble, releases energy in small amounts, quickly resynthesised, and couples energy-releasing to energy-requiring reactions via transfer of the terminal phosphate. Phosphorylation of substrates (e.g. glucose → glucose phosphate in glycolysis) also makes them more reactive.
Glycolysis occurs in the cytoplasm and is anaerobic-compatible: (1) glucose is phosphorylated using 2 ATP to fructose bisphosphate (6C); (2) it splits into two triose phosphate (3C) molecules; (3) each TP is oxidised (losing H to NAD) and converted to pyruvate, producing 4 ATP by substrate-level phosphorylation. Net yield per glucose: 2 ATP, 2 NADH, 2 pyruvate. Substrate-level phosphorylation = direct transfer of a phosphate from an intermediate to ADP (contrast with oxidative phosphorylation via the proton gradient).
The link reaction occurs in the mitochondrial matrix: pyruvate is actively transported into the matrix, oxidatively decarboxylated - each pyruvate loses CO₂, loses H to NAD (forming NADH) and combines with coenzyme A to form acetyl CoA (2C). Per glucose: 2 CO₂ + 2 NADH. The 2C acetyl group then feeds the Krebs cycle, also in the matrix: acetyl CoA (2C) + oxaloacetate (4C) → citrate (6C); citrate is decarboxylated and dehydrogenated in a series of redox reactions regenerating oxaloacetate. Per turn: 2 CO₂, 3 NADH, 1 FADH₂, 1 ATP (by substrate-level phosphorylation); per glucose (two turns): 4 CO₂, 6 NADH, 2 FADH₂, 2 ATP. Coenzymes NAD and FAD carry H atoms (electrons) to the electron transport chain; remember that dehydrogenation is oxidation. When asked to state where each stage occurs and its products, a quick table of stage/location/yields is the fastest full-mark format.
Oxidative phosphorylation, chemiosmosis and ATP yield (AQA 3.5.2; OCR 5.2.1)
Oxidative phosphorylation happens on the folded inner mitochondrial membrane: NADH and FADH₂ deliver electrons (from their H atoms) to the electron transport chain of protein complexes I-IV. As electrons pass along the carriers of decreasing energy, energy released pumps H⁺ from the matrix into the intermembrane space, building a steep electrochemical gradient. Protons then flow back into the matrix through ATP synthase channels, and this flow (chemiosmosis) drives phosphorylation of ADP to ATP. Oxygen is the terminal electron acceptor: it combines with electrons and protons to form water (½O₂ + 2H⁺ + 2e⁻ → H₂O). Without oxygen, electrons stop flowing, the proton gradient dissipates, NADH cannot be reoxidised, and oxidative phosphorylation ceases - explaining both cyanide poisoning (inhibits complex IV) and the far lower anaerobic yield.
ATP yield per glucose: glycolysis 2 ATP + 2 NADH; link 2 NADH; Krebs 2 ATP + 6 NADH + 2 FADH₂. Each NADH yields ~2.5 ATP and each FADH₂ ~1.5 ATP via chemiosmosis, giving a theoretical ~32 ATP. Actual yields (30-32) are lower because: some protons leak back across the inner membrane without making ATP; energy is spent shuttling NADH from the cytoplasm into the mitochondrion (glycerol-3-phosphate vs malate-aspartate shuttles yield different returns); and ATP is used to actively transport pyruvate and ADP across the mitochondrial membranes. Exam questions may ask you to calculate yield from NADH/FADH₂ numbers or to explain why the theoretical maximum is not reached - learn both directions.
Respirometers measure oxygen consumption (a manometer fluid moves as O₂ is consumed and CO₂ is absorbed by soda lime): link rate of oxygen uptake to rate of respiration, and control temperature with a water bath. RQ calculations: for tripalmitin C₅₁H₉₈O₆ + 72.5 O₂ → 51 CO₂ + 49 H₂O, RQ = 51 ÷ 72.5 = 0.70. RQ near 1.0 = carbohydrate substrate; below 1 = lipid/protein; above 1 = some anaerobic respiration occurring. These interpretation questions appear regularly on OCR (5.2.1) and WJEC C1 papers.
When oxygen is limiting, glycolysis continues but the link reaction, Krebs cycle and oxidative phosphorylation stop because NADH cannot unload its electrons. Anaerobic pathways regenerate NAD⁺ so glycolysis can keep producing its net 2 ATP. In animals: pyruvate + NADH → lactate + NAD⁺, catalysed by lactate dehydrogenase in muscle during intense exercise. Lactate accumulates, lowering pH; it is transported in the blood to the liver and oxidised back to pyruvate (or converted to glucose via gluconeogenesis) - the oxygen debt. Blood lactate assays and muscle fatigue questions expect this. In yeast (fermentation): pyruvate → acetaldehyde + CO₂ (pyruvate decarboxylase), then acetaldehyde + NADH → ethanol + NAD⁺ (ethanol dehydrogenase) - the basis of brewing and of AQA RP-style investigations into yeast respiration rate under different variables.
Why anaerobic yields far less ATP: glucose is only partially oxidised - the energy-rich hydrogen in pyruvate/lactate is never transferred to the ETC, so only the 2 ATP from substrate-level phosphorylation in glycolysis are made, versus ~32 aerobically; the large chemiosmotic yield from oxidising 10 NADH + 2 FADH₂ is lost. In animals the lactate pathway also costs ATP later (liver conversion), and lactate acidosis impairs muscle contraction - an exam staple linking respiration to muscle physiology (Edexcel T7, WJEC).
Regulation of respiration completes the topic: phosphofructokinase (an early glycolytic enzyme) is allosterically inhibited by high ATP and citrate, and activated by high AMP/ADP, so glycolysis slows when energy charge is high - a concrete feedback-inhibition example worth quoting. Mitochondrial diseases that impair the ETC reduce ATP yield and raise dependence on anaerobic pathways; uncouplers (e.g. DNP) dissipate the proton gradient, so respiration continues without ATP production, releasing energy as heat - questions use these to test whether you really understand chemiosmosis rather than recite stages.
💡 Worked Example
Exam-style: AQA-style (3.5.2, 5 marks): Explain why respiration of one glucose molecule in anaerobic conditions produces far less ATP than in aerobic conditions.
Mark-scheme logic: Model answer logic (1 mark per point, 5 total): (1) Anaerobically glycolysis is the only stage operating - glucose is only partially oxidised to pyruvate, then to lactate (animals) or ethanol + CO₂ (yeast). (2) The link reaction, Krebs cycle and oxidative phosphorylation cannot run because NADH cannot be reoxidised via the electron transport chain. (3) Only the net 2 ATP from substrate-level phosphorylation in glycolysis are produced. (4) Aerobically, oxidation of 10 NADH and 2 FADH₂ pumps protons and drives chemiosmosis, yielding ~30 more ATP. (5) Without oxygen as the terminal electron acceptor the proton gradient cannot be maintained, so oxidative phosphorylation stops.
❓ Practice Questions & Mark Schemes
State where glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation occur, and give the net yield of glycolysis. (4 marks)
Answer: Glycolysis: cytoplasm - net 2 ATP, 2 NADH, 2 pyruvate per glucose. Link reaction: mitochondrial matrix - per glucose 2 CO₂, 2 NADH, 2 acetyl CoA. Krebs cycle: mitochondrial matrix - per glucose 4 CO₂, 6 NADH, 2 FADH₂, 2 ATP. Oxidative phosphorylation: inner mitochondrial membrane via the electron transport chain and ATP synthase (chemiosmosis).
Explain how the structure of glycogen is related to its function as an energy store. (4 marks)
Answer: Polymer of α-glucose - easily hydrolysed to release glucose for respiration. Highly branched with 1,6-glycosidic bonds - many terminal ends allow rapid simultaneous enzymatic hydrolysis. Insoluble - does not affect the water potential of cells so water is not drawn in. Compact coiled structure - stores a large amount of energy in a small volume in liver and muscle.
A non-competitive inhibitor is added to an enzyme-catalysed reaction. Explain its effect on Vmax and contrast it with a competitive inhibitor. (5 marks)
Answer: Non-competitive inhibitor binds at an allosteric site away from the active site, altering the tertiary structure so the active site is no longer complementary to the substrate. Vmax decreases and cannot be restored by adding substrate because the enzyme shape is changed. A competitive inhibitor binds the active site reversibly, competing with the substrate; increasing substrate concentration overcomes it, so Vmax is unchanged but apparent Km increases.
Explain the role of oxygen in aerobic respiration. (3 marks)
Answer: Oxygen acts as the terminal electron acceptor at the end of the electron transport chain (complex IV), combining with electrons and protons to form water. This maintains the flow of electrons along the chain, sustaining proton pumping and the proton gradient needed for chemiosmosis. Without it, oxidative phosphorylation stops and NADH accumulates.
Calculate the RQ for a substrate respired as: C₃H₆O₃ + 3 O₂ → 3 CO₂ + 3 H₂O, and state what the value indicates. (3 marks)
Answer: RQ = CO₂ produced ÷ O₂ consumed = 3 ÷ 3 = 1.0. An RQ of 1.0 indicates the substrate is a carbohydrate being respired aerobically; values below 1 suggest lipid or protein use, and values above 1 indicate some anaerobic respiration.
Explain the induced fit model of enzyme action. (3 marks)
Answer: The active site is nearly complementary to the substrate; as the substrate binds, the enzyme changes shape slightly to mould around it. This forms the enzyme-substrate complex and places strain on substrate bonds. The induced fit lowers the activation energy, allowing the reaction to proceed, and explains the specificity of enzyme-substrate binding.