AS & A Level Biology · 9700 · Topic 3 · 2025–2027 Exam

Enzymes

The biological catalysts that make life chemistry possible. Mode of action from active site to induced fit; the four kinetic factors and the inhibitor types that disrupt them; immobilised enzymes and the production of lactose-free milk.

Sub-sections 3.1–3.2 AS Level Papers 1–3 Catalysis · Kinetics · Inhibition
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Topic 3.1 · AS

Mode of action of enzymes

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Enzymes are biological catalysts: they speed up the rate of biochemical reactions without being chemically changed themselves and without altering the equilibrium position. A single enzyme molecule can convert thousands of substrate molecules per second, making enzymes essential for life chemistry to occur fast enough to sustain a living cell.

Enzymes as globular proteins

All enzymes are globular proteins (with one important exception — ribozymes are made of RNA, but these are beyond syllabus scope). Their tertiary or quaternary structure produces a precisely shaped active site on the protein surface, where catalysis takes place. Globular shape gives them three properties needed for the job:

Location 1
Intracellular enzymes

Catalyse reactions inside the cell that produced them. Examples include catalase (in peroxisomes; breaks down hydrogen peroxide), DNA polymerase (in the nucleus during replication), and the enzymes of glycolysis (in the cytoplasm).

Location 2
Extracellular enzymes

Are secreted from the cell that made them and act outside. Examples include the digestive enzymes pepsin, trypsin, lipase and pancreatic amylase, secreted into the gut lumen; and salivary amylase, secreted into the mouth.

Why secrete an enzyme?

Extracellular digestion lets cells break down food molecules outside themselves and absorb the small soluble products. Cells could not bring large polymers (starch, proteins, lipids) across their membranes intact — secreting hydrolytic enzymes solves this problem. Saprotrophic fungi take this principle to its limit, digesting their entire food supply externally before absorbing it.

Active site, ES complex, specificity

The active site is a small pocket or cleft on the surface of the enzyme, formed by a particular arrangement of amino acid R groups. It has:

The catalytic cycle
  1. Recognition — substrate(s) approach the enzyme by random collision
  2. Binding — substrate fits into the active site, forming an enzyme-substrate complex (ES)
  3. Catalysis — the enzyme catalyses the reaction (bonds in substrate broken or new bonds formed)
  4. Release — products leave the active site
  5. Reset — the enzyme returns to its original shape, ready to bind another substrate

The enzyme is unchanged at the end of the cycle, which is why a single enzyme molecule can catalyse many reactions in succession.

Enzyme specificity: most enzymes catalyse only one reaction (or one type of reaction on closely related substrates). The reason is geometric — only a substrate with shape complementary to the active site can bind well enough to undergo catalysis. This is why we need many different enzymes: roughly one per reaction type in the cell.

Lock-and-key vs induced-fit

Two models describe how the active site interacts with the substrate. The lock-and-key model came first; the induced-fit model is the modern refinement.

1894 model
Lock-and-key hypothesis

Proposed by Emil Fischer. The active site is treated as a rigid shape, and only a substrate of complementary shape (the “key”) can fit into it (the “lock”).

Strengths: simple; explains specificity; helpful first picture for students.

Limitations: some enzymes accept several similar substrates; lock-and-key cannot explain catalysis itself, only binding.

1958 model
Induced-fit hypothesis

Proposed by Daniel Koshland. The active site is treated as flexible. When the substrate begins to bind, the active site changes shape slightly to fit the substrate more precisely.

Strengths: explains catalysis (the conformational change strains substrate bonds, lowering activation energy); accounts for enzyme regulation; widely supported by structural studies.

Limitations: harder to draw; the substrate also undergoes conformational change.

Common confusion

Both models describe specificity, but only induced-fit explains why the enzyme accelerates the reaction. Exam answers asking “explain enzyme specificity” can use either; answers asking “explain how enzymes lower activation energy” should use induced-fit.

Lowering activation energy

Activation energy (Ea) is the minimum kinetic energy that reactant molecules must possess for a chemical reaction to take place — the height of the energy barrier between reactants and products. Most biochemical reactions have an activation energy too high to occur at significant rates at body temperature without a catalyst.

How enzymes lower Ea

Enzymes lower Ea by stabilising the transition state. Several effects contribute:

  • Substrate molecules are held in the correct orientation for reaction (no longer relying on chance collisions)
  • Substrates are positioned close together (effective concentration is dramatically increased in the active site)
  • R groups in the active site can polarise bonds in the substrate, weakening them (induced fit applies strain)
  • R groups can act as acid/base catalysts, donating or accepting protons

The net result: a much higher fraction of substrate molecules at any given temperature has enough energy to react, and the reaction proceeds many orders of magnitude faster.

What enzymes do NOT do

Enzymes do not change ΔG (the free-energy change of the reaction) or the position of equilibrium. They only change the rate at which equilibrium is reached. A reaction that is thermodynamically unfavourable cannot be made favourable by adding an enzyme.

Investigating reaction rates

The syllabus requires two specific approaches to measuring enzyme reaction rates — one based on product formation (catalase) and one based on substrate disappearance (amylase) — plus the use of a colorimeter for colour-change reactions.

Method 1
Catalase — rate of product formation

Reaction: 2 H2O2 → 2 H2O + O2. Catalase decomposes hydrogen peroxide to water and oxygen.

Setup: mix catalase with hydrogen peroxide in a sealed flask connected to a gas syringe (or inverted measuring cylinder over water).

Measure: volume of O2 produced at fixed time intervals.

Rate: initial gradient of volume-vs-time graph (cm3 s−1). Use the initial gradient because rate falls as substrate is used up.

Method 2
Amylase — rate of substrate disappearance

Reaction: starch → maltose. Amylase hydrolyses starch into maltose.

Setup: mix amylase with starch solution. At regular intervals, take a small sample and add iodine in KI solution.

Measure: time taken until iodine no longer turns blue-black (starch fully hydrolysed).

Rate: 1/time, so a shorter time corresponds to a higher rate.

Method 3
Colorimeter — for colour-change reactions

A colorimeter measures the absorbance of light passing through a solution. If the reaction produces or removes a coloured substance, absorbance changes over time, and the rate of change measures reaction rate.

Setup: select an appropriate filter (wavelength absorbed by the coloured substance), zero the meter using a blank cuvette, then read absorbance at fixed intervals.

Advantages: objective (no human colour judgement); quantitative; can be automated.

Practical accuracy points
  • Use initial rate: take the gradient at t = 0 because substrate concentration falls during the reaction, slowing the rate
  • Control variables: temperature (water bath), pH (buffer), enzyme concentration, substrate concentration, total volume
  • Multiple replicates: take 3 or more readings at each condition and report the mean
  • Standardise mixing time: start the timer the moment enzyme and substrate combine
MCQ · Topic 3.1 · Paper 1 style

Which statement best explains how enzymes increase the rate of a reaction?

  • A. They increase the kinetic energy of substrate molecules.
  • B. They alter the equilibrium constant of the reaction.
  • C. They lower the activation energy of the reaction.
  • D. They are consumed during the reaction, providing additional energy.
Answer: C — Enzymes act by stabilising the transition state, which lowers the activation energy required for reactants to convert to products. They do not change kinetic energy of molecules (A), do not affect equilibrium (B), and are not consumed (D — that's the definition of a catalyst).
Structured · Topic 3.1 · Paper 2 style · 6 marks

Compare the lock-and-key and induced-fit hypotheses of enzyme action. In your answer, explain why the induced-fit model is now generally preferred. [6]

Six creditable points (any six):

Acceptable points
  • Both models propose that the active site has a shape complementary to the substrate / explain specificity [1]
  • In lock-and-key, the active site is rigid; the substrate fits into the active site exactly like a key into a lock [1]
  • In induced-fit, the active site is flexible; it changes shape slightly when the substrate binds [1]
  • Induced-fit is preferred because it explains how enzymes catalyse the reaction, not just how they bind the substrate [1]
  • The conformational change in induced-fit puts strain on bonds in the substrate, lowering the activation energy [1]
  • Induced-fit accounts for enzymes that bind several similar substrates, which lock-and-key cannot easily explain [1]
  • Structural studies (X-ray crystallography) directly show that enzymes change shape when substrates bind, supporting induced-fit [1]

Mark scheme guidance: Both halves of the comparison must appear (rigid vs flexible) for full credit. The catalysis-vs-binding distinction is the key “preferred” mark.

Topic 3.2 · AS

Factors that affect enzyme action

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Four physical/chemical factors affect enzyme rate; two more — competitive and non-competitive inhibitors — act by binding to the enzyme. Each can be classified further as reversible or non-reversible. Industrially, enzymes can be immobilised on a support to make them reusable, with lactase in lactose-free milk as the standard syllabus example.

Temperature

Rate increases with temperature up to an optimum, then falls steeply.

Temperature rangeWhat happensEffect on rate
Below optimumMolecules have low kinetic energy; few successful collisionsLow rate; rate increases as temperature rises
Approaching optimumMore kinetic energy; more frequent and energetic enzyme-substrate collisions; more substrate molecules exceed EaRate roughly doubles per 10 °C rise (Q10 ≈ 2–3)
OptimumMaximum rate of ES complex formation without significant denaturationMaximum rate
Above optimumHeat energy breaks weak bonds (hydrogen, ionic) maintaining tertiary structure; active site changes shape; enzyme denaturesRate falls rapidly; eventually drops to zero
Optimum temperatures vary

Mammalian enzymes typically have optima around 37–40 °C. Enzymes from Thermus aquaticus, a hot-spring bacterium, function at over 70 °C — the source of Taq polymerase used in PCR. Cold-water fish enzymes have optima below 10 °C. There is no single “optimum”; each enzyme has its own.

Common mistake

Saying that enzymes “die” at high temperatures. Enzymes are not alive; they denature. Denaturation is the loss of tertiary structure due to broken weak bonds — covalent bonds (peptide bonds, disulfide bridges) usually remain intact, but the protein loses its functional 3D shape and so its activity.

pH

Each enzyme has an optimum pH at which its rate is highest. Either side of the optimum, rate falls; at extreme pH, the enzyme denatures.

EnzymeOptimum pHWhere it works
Pepsinaround 1.5–2Stomach (highly acidic)
Salivary amylasearound 6.7–7Mouth (near neutral)
Catalasearound 7Most cells (cytoplasm or peroxisomes)
Trypsinaround 8Small intestine (slightly alkaline)
Mechanism of pH effect

pH changes the concentration of H+ ions, which alter the charges on R groups (carboxyl groups: −COOH ↔ −COO; amino groups: −NH2 ↔ −NH3+). Changes in R-group charges disrupt the ionic and hydrogen bonds stabilising tertiary structure, altering the active site shape and reducing substrate binding.

At extreme pH, this distortion becomes severe enough to denature the enzyme. Within a few pH units of the optimum, the change is often reversible — restoring optimum pH restores activity.

Buffer solutions in experiments

The 9700 syllabus specifies pH (using buffer solutions). A buffer maintains pH constant despite small additions of acid or alkali, essential when the reaction itself releases or consumes H+ ions. Using buffers ensures that pH is a controlled (not confounding) variable.

Enzyme concentration

If substrate is present in excess, rate is directly proportional to enzyme concentration: doubling [enzyme] doubles the rate. More enzyme molecules means more active sites available to form ES complexes per unit time.

Linearity has a limit

The linear relationship breaks down once substrate becomes limiting. If [substrate] is too low to keep all active sites occupied, adding more enzyme has diminishing returns — eventually no further increase in rate.

In careful experiments, [substrate] is set high enough to ensure linear behaviour over the whole range of [enzyme] tested.

Substrate concentration

Substrate concentration has a more complex effect: linear at first, then plateauing.

Low [S]
Rate proportional to [S]

Most active sites are unoccupied. Adding more substrate produces more ES complexes per second in direct proportion. Rate increases linearly with [S].

High [S]
Plateau — saturation

Eventually, every active site is occupied at any moment (the enzyme is saturated). Adding more substrate cannot increase rate further because no free active sites are available. Rate reaches a maximum and stays there.

What sets the maximum rate?

At saturation, the bottleneck becomes the rate at which the enzyme processes substrate (catalysis + product release). The maximum rate is therefore set by the turnover number of the enzyme (how many substrate molecules each active site converts per second) and by the total enzyme concentration.

The 2025–2027 syllabus does not require the symbols Vmax or Km. Describe the behaviour in plain language: “the rate plateaus at a maximum value once all active sites are occupied”.

Competitive inhibitors

A competitive inhibitor is a molecule with shape similar to the natural substrate. It competes with substrate for the same active site.

Mechanism
  • The inhibitor has a structure that mimics the substrate
  • It binds to the active site (instead of the substrate)
  • While the inhibitor is bound, no substrate can bind → that enzyme molecule is temporarily unproductive
  • The inhibitor must dissociate before substrate can bind

Effect on rate vs [S] graph: at low [S], rate is reduced (substrate cannot easily out-compete inhibitor). At high [S], substrate dominates the active sites and the inhibition effect becomes small — the rate approaches the same maximum as the uninhibited enzyme.

Classic example: malonate competes with succinate for the active site of succinate dehydrogenase (a key enzyme in respiration). Malonate has a similar shape but cannot be processed, blocking the active site temporarily.

Non-competitive inhibitors

A non-competitive inhibitor binds to a site on the enzyme that is not the active site — usually called an allosteric site (literally “other site”). Binding causes the active site to change shape so that substrate can no longer bind effectively (or the catalytic step is disrupted).

Mechanism
  • The inhibitor has a different shape from the substrate (no competition with substrate)
  • It binds to an allosteric site, separate from the active site
  • Binding causes a conformational change in the enzyme
  • The active site is distorted so substrate either cannot bind or cannot be converted to product

Effect on rate vs [S] graph: rate is reduced at all substrate concentrations, including the highest. Adding more substrate does not overcome the inhibition because the inhibitor and substrate are not competing for the same site. The maximum rate is lower than for the uninhibited enzyme.

FeatureCompetitiveNon-competitive
Where it bindsActive siteAllosteric site (elsewhere)
ShapeSimilar to substrateDifferent from substrate
Effect on enzymePhysically blocks substrateDistorts active site shape
Effect at low [S]Marked reduction in rateMarked reduction in rate
Effect at high [S]Rate approaches uninhibited maximumRate stays below uninhibited maximum
Reversed by adding substrate?Yes (high [S] outcompetes)No (different binding site)

Reversible vs non-reversible inhibitors

The competitive/non-competitive classification describes where the inhibitor binds. The reversible/non-reversible classification is independent — it describes how strongly the inhibitor binds.

Reversible
Held by weak bonds

Reversible inhibitors bind to the enzyme using weak interactions only — hydrogen bonds, ionic attractions, hydrophobic interactions. They can dissociate from the enzyme; if the inhibitor is removed (e.g. by dialysis or dilution), enzyme activity is fully restored.

Most drugs and metabolic regulators are reversible inhibitors. They allow precise, dynamic control of metabolism.

Non-reversible
Held by covalent bonds

Non-reversible inhibitors form covalent bonds with the enzyme, permanently modifying it. The enzyme cannot recover, even if free inhibitor is removed.

Examples: heavy metal ions (Hg2+, Pb2+) bind covalently to −SH groups in cysteine residues; nerve agents (organophosphates) covalently inhibit acetylcholinesterase; some antibiotics permanently inactivate bacterial enzymes.

Two classifications, one inhibitor

The two ways of describing inhibitors are orthogonal: any inhibitor is one of competitive/non-competitive AND one of reversible/non-reversible. So malonate is a reversible competitive inhibitor; nerve agents are non-reversible non-competitive inhibitors of acetylcholinesterase. When asked about an inhibitor, both descriptors may be required.

Immobilised enzymes — lactase in lactose-free milk

Immobilised enzymes are enzymes attached to or trapped in an inert support material so that they remain in place while substrates flow through. Common methods include adsorption to a surface, covalent attachment to a support, entrapment in a gel matrix, and encapsulation in a porous membrane.

Common method
Entrapment in alginate beads

Mix the enzyme with sodium alginate solution. Drip droplets of the mixture into calcium chloride solution — the calcium ions cause the alginate to gel, trapping the enzyme inside small spherical beads (a few millimetres across). Substrate diffuses into the bead, reaches the enzyme, is converted, and the product diffuses out.

Industrial format
Continuous-flow reactor

The beads (or other immobilised enzyme support) are packed into a column. Substrate solution is pumped through one end, products flow out the other. The enzyme stays in the column and processes a continuous supply of substrate — ideal for large-scale industrial use.

AdvantagesDisadvantages
Enzyme can be reused many times — major cost savingInitial setup cost (immobilisation procedure, support material)
Product is enzyme-free, simplifying downstream purificationSome enzyme activity may be lost during the immobilisation process
More stable to changes in temperature and pH (the support stabilises tertiary structure)Substrates and products must diffuse to and from the enzyme → rate can be slower than free enzyme
Allows continuous-flow industrial processingMay not work well with very large or insoluble substrates
Easy to start and stop the reaction (just stop the substrate flow)Choice of support and method needs optimisation per enzyme
Worked example: lactose-free milk

The problem: people with lactose intolerance lack sufficient lactase in the small intestine. Undigested lactose passes to the large intestine, where bacteria ferment it — causing gas, bloating, and discomfort.

The industrial solution:

  1. Lactase is immobilised on alginate beads (or similar support)
  2. The beads are packed into a column reactor
  3. Milk is passed through the column at controlled temperature
  4. Lactase hydrolyses lactose into glucose and galactose: lactose + H2O → glucose + galactose
  5. The output is lactose-free milk — safe for lactose-intolerant consumers

Why immobilisation matters here specifically:

  • Lactase is expensive; immobilisation allows reuse
  • Free lactase would contaminate the milk; immobilised lactase stays in the column — the milk is enzyme-free
  • Continuous flow allows large-scale, industrial-volume production
  • The product is sweeter than ordinary milk (glucose and galactose taste sweeter than lactose) — a useful side effect
MCQ · Topic 3.2 · Paper 1 style

A non-competitive inhibitor is added to a reaction mixture in which substrate is at low concentration. The substrate concentration is then gradually increased while the inhibitor concentration stays constant. Which graph correctly describes the rate compared to a control reaction without the inhibitor?

  • A. Rate eventually reaches the same maximum as the control because excess substrate outcompetes the inhibitor.
  • B. Rate is unaffected at all substrate concentrations because non-competitive inhibitors do not bind the active site.
  • C. Rate is reduced at all substrate concentrations and the maximum rate is lower than the control.
  • D. Rate is initially zero, then rises to match the control as inhibitor is displaced.
Answer: C — Non-competitive inhibitors bind at an allosteric site (not the active site), so adding more substrate cannot displace them. Rate is reduced at every [S], and the maximum rate is below the uninhibited maximum. Option A describes competitive inhibition; B is wrong because non-competitive inhibitors still affect the rate (via shape change to the active site).
Structured · Topic 3.2 · Paper 2 style · 8 marks

Lactase is used industrially to produce lactose-free milk. The enzyme is immobilised on an inert support and milk is passed over it.

(a) Explain three advantages of using immobilised lactase rather than free lactase in this process. [3]
(b) Suggest one disadvantage of immobilising lactase. [1]
(c) Lactose intolerance results from a deficiency of lactase. Explain why undigested lactose causes the symptoms (gas, bloating, diarrhoea) experienced by people with lactose intolerance. [4]

(a) Three advantages [3 marks]
Acceptable points (one mark each, max 3)
  • The enzyme can be reused many times instead of being discarded with each batch of milk — reduces cost
  • The milk is not contaminated with enzyme; the enzyme stays in the column / on the beads
  • Immobilised enzymes are more stable to temperature and pH changes — less denaturation, longer working life
  • Allows continuous-flow processing of large milk volumes
  • Easy to control: stop the milk flow to stop the reaction
(b) One disadvantage [1 mark]

Substrate (lactose) must diffuse to the enzyme inside the support, and product (glucose, galactose) must diffuse out, so the rate may be slower than free lactase. OR: Initial cost of immobilisation. OR: Some enzyme activity may be lost during the immobilisation process.

(c) Why undigested lactose causes symptoms [4 marks]
Acceptable points (any four)
  • Without lactase, lactose is not hydrolysed in the small intestine [1]
  • Lactose passes into the large intestine [1]
  • Bacteria in the large intestine ferment the lactose, producing gas (e.g. methane, hydrogen) — causing flatulence and bloating [1]
  • Lactose and fermentation products lower the water potential of the gut contents [1]
  • Water moves from surrounding tissues into the gut by osmosis, increasing fluid in the colon — causing diarrhoea [1]

Mark scheme guidance: Part (c) tests synoptic understanding linking enzymes (Topic 3) with osmosis (Topic 4) and gut function. Both the “gas from fermentation” and the “water by osmosis” routes are required for full marks.

Exam Prep

Topic 3 Practice — Comprehensive

Mixed practice covering both sub-sections in 9700 P1/P2 style. Try each before revealing the answer.

MCQ · Mode of action · Paper 1

Which property of an enzyme determines which substrate it catalyses?

  • A. The total number of amino acids in the polypeptide
  • B. The size of the enzyme molecule
  • C. The 3D shape of the active site
  • D. Whether the enzyme is intracellular or extracellular
Answer: C — Enzyme specificity arises from the precise 3D shape of the active site, which only complementary substrates can fit into. Total length, overall size, and intracellular/extracellular location do not determine which substrate is catalysed.
MCQ · Temperature · Paper 1

An enzyme is heated from 10 °C to 80 °C and the rate is recorded throughout. Which best describes the rate change?

  • A. Rate increases steadily over the whole range.
  • B. Rate stays constant until 60 °C, then falls sharply.
  • C. Rate increases up to an optimum, then falls steeply due to denaturation.
  • D. Rate is highest at 10 °C because cold preserves enzyme structure.
Answer: C — Rate rises with temperature because of increased kinetic energy, reaches a peak at the optimum, then falls sharply because heat breaks the weak bonds maintaining tertiary structure (denaturation).
MCQ · Inhibitors · Paper 1

Mercury ions (Hg2+) bind covalently to the −SH groups of cysteine residues in many enzymes, causing permanent loss of activity. The binding is not at the active site. Hg2+ is best classified as a:

  • A. Reversible competitive inhibitor
  • B. Reversible non-competitive inhibitor
  • C. Non-reversible competitive inhibitor
  • D. Non-reversible non-competitive inhibitor
Answer: D — “Covalently” means non-reversible; “not at the active site” means non-competitive. Both descriptors are needed to fully classify an inhibitor.
Structured · Mixed factors · Paper 2 · 8 marks

A student investigates the effect of substrate concentration on the rate of an enzyme-catalysed reaction. Initial rate is plotted against substrate concentration.

(a) Explain the shape of the curve: linear at low [S], plateauing at high [S]. [4]
(b) The student then repeats the experiment in the presence of a competitive inhibitor at fixed concentration. Predict and explain how the curve would differ from the control. [4]

(a) Linear-then-plateau shape [4 marks]
Acceptable points
  • At low [S], most active sites are unoccupied [1]
  • Adding more substrate increases the chance of substrate-active site collisions / increases ES complex formation in proportion, so rate is proportional to [S] [1]
  • At high [S], all active sites are occupied at any one moment / the enzyme is saturated [1]
  • Adding more substrate cannot increase rate further because there are no free active sites; the rate is now limited by enzyme turnover, so the curve plateaus [1]
(b) With competitive inhibitor [4 marks]
Acceptable points
  • At low [S], the inhibitor binds the active site and competes successfully with substrate; the rate is much lower than the control [1]
  • The curve initially rises more slowly with [S] than the control [1]
  • At high [S], substrate is in large excess and outcompetes the inhibitor for the active site; the proportion of active sites occupied by inhibitor falls to a low level [1]
  • The curve eventually plateaus at the same maximum rate as the control / approaches the control value at very high [S] [1]
Structured · Synoptic Topic 2 + Topic 3 · Paper 2 · 6 marks

The active site of an enzyme is part of its tertiary structure. Explain how (a) extreme heat and (b) very low pH could cause an enzyme to lose activity, with reference to the bonds involved. [6]

(a) Heat [3 marks]
Acceptable points
  • High temperature increases the vibrational energy of atoms in the protein [1]
  • Weak bonds maintaining tertiary structure are broken — especially hydrogen bonds and ionic bonds (and weak hydrophobic interactions) [1]
  • The active site shape changes; substrate can no longer bind; the enzyme is denatured (active site no longer complementary to substrate) [1]
  • (Bonus point: peptide bonds and disulfide bridges may remain intact; the protein doesn't break apart but loses its functional fold)
(b) Very low pH [3 marks]
Acceptable points
  • At very low pH, high H+ concentration alters the charges on R groups (e.g. −COO → −COOH; −NH2 → −NH3+) [1]
  • This disrupts ionic and hydrogen bonds between R groups in the tertiary structure [1]
  • The active site changes shape and is no longer complementary to the substrate — the enzyme is denatured [1]

Synoptic note: This question links Topic 2.3 (R-group bonds in tertiary structure) with Topic 3.2 (effect of T and pH on enzyme rate). The structure-based explanation is the key to full marks.

Exam Prep

Topic 3 — Common Mistakes

Topic 3 strategy

Topic 3 is the foundation for every metabolic and physiological topic that follows: respiration (12), photosynthesis (13), digestion, signalling, immunology (11), all rely on enzymes. Highest-yield items: induced-fit hypothesis with explicit catalysis benefit, all four kinetic factors with reasons, the orthogonal classification of inhibitors (competitive/non-competitive × reversible/non-reversible), the lactase + alginate immobilisation case study, and recognising rate-vs-[S] graph shapes. The 2025–2027 syllabus does not require Vmax/Km — describe behaviour in plain language.

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