How does temperature affect enzyme action in biological systems?

Temperature significantly impacts enzyme activity. As temperature increases, enzyme activity typically rises due to increased kinetic energy, leading to more frequent collisions between enzymes and substrates. However, if the temperature exceeds an enzyme's optimal range, the enzyme can denature, losing its functional shape and resulting in decreased activity. This concept is crucial in Cambridge International A Levels Biology, where understanding enzyme kinetics is essential.

What role does pH play in enzyme activity?

pH levels can alter the shape and charge of an enzyme, affecting its ability to bind with substrates. Each enzyme has an optimal pH range where it functions best. Deviations from this range can lead to decreased activity or denaturation. For example, pepsin, an enzyme in the stomach, works best at a low pH, while enzymes in the small intestine prefer a higher pH. This is a key topic in the Cambridge International A Levels Biology curriculum.

How does substrate concentration influence enzyme activity?

Substrate concentration affects enzyme activity by determining the rate of reaction. At low substrate concentrations, an increase in substrate leads to a proportional increase in reaction rate. However, once all enzyme active sites are occupied (saturation point), further increases in substrate concentration do not affect the rate. This principle is fundamental in enzyme kinetics studies in Cambridge International A Levels Biology.

What is the effect of enzyme concentration on the rate of reaction?

Increasing enzyme concentration generally increases the rate of reaction, provided there is an excess of substrate. This is because more enzyme molecules mean more active sites available for substrate binding. However, if the substrate is limited, increasing enzyme concentration will have no further effect. Understanding this relationship is crucial for students studying enzyme action in Cambridge International A Levels Biology.

How do inhibitors affect enzyme activity?

Inhibitors are molecules that reduce enzyme activity. Competitive inhibitors bind to the active site, blocking substrate access, while non-competitive inhibitors bind elsewhere, altering the enzyme's shape. Both types of inhibition can decrease the rate of reaction, a concept that is important in enzyme regulation studies within the Cambridge International A Levels Biology syllabus.

Why is "Saying that the enzyme is 'killed' or 'destroyed' at high temperatures" a common mistake in Factors that affect enzyme action?

Why it happens: Students think of enzymes as alive. How to avoid it: Enzymes are proteins, not living things. They are **denatured**: the tertiary structure (active site shape) is lost. The molecule is still present. Always use 'denatured' or 'tertiary structure disrupted'. Source: 9700 Examiner Reports 2022-2024

Why is "Stating that low temperatures denature an enzyme" a common mistake in Factors that affect enzyme action?

Why it happens: Students assume any temperature extreme denatures. How to avoid it: Low temperatures only **slow** the enzyme by reducing kinetic energy. The tertiary structure is unaffected. On rewarming the enzyme regains full activity. Only high temperatures (or extreme pH) denature. Source: 9700 Examiner Reports 2023

Why is "Stating that a competitive inhibitor lowers $V_\text{max}$" a common mistake in Factors that affect enzyme action?

Why it happens: Confusion with non-competitive. How to avoid it: A competitive inhibitor is **out-competed by high [S]**, so V_max is unchanged (just reached at higher [S]). A NON-competitive inhibitor binds an allosteric site and is NOT out-competed by [S], so V_max IS reduced. Source: 9700 Examiner Reports 2024

Why is "Saying that a non-competitive inhibitor binds the active site" a common mistake in Factors that affect enzyme action?

Why it happens: Misremembering the definitions. How to avoid it: Non-competitive inhibitors bind an **allosteric site** (not the active site) and cause the active site to change shape. Competitive inhibitors bind the active site directly. Source: 9700 Examiner Reports 2023

Why is "Calculating the rate over the whole reaction instead of the initial rate" a common mistake in Factors that affect enzyme action?

Why it happens: Students take total product / total time, ignoring that the rate falls as substrate is used up. How to avoid it: Use **initial rate** = gradient of tangent at t = 0 (or gradient over the first 15-30 s). This is the most reliable measure because substrate has not been significantly depleted yet. Source: 9700 Paper 3 Examiner Reports 2023-2024

Why is "Explaining the effect of pH only by saying 'pH changes shape'" a common mistake in Factors that affect enzyme action?

Why it happens: Oversimplified explanation. How to avoid it: Mark schemes credit the mechanism: pH affects the **ionisation of R-groups** lining the active site. This disrupts **hydrogen and ionic bonds**, the tertiary structure, and therefore the **active site complementarity**. Source: 9700 Examiner Reports 2024

EnzymesCambridge International A Levels Biology (9700)

Factors that affect enzyme action

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Factors that Affect Enzyme Action — Cambridge International A Level Biology 9700 Study Notes (2025-2027 syllabus)

How temperature, pH, enzyme concentration, substrate concentration and inhibitors affect enzyme-catalysed reaction rates. Investigation methods using catalase and amylase, and graphical analysis using initial rates and $V_\text{max}$ .

At a glance

Temperature: rate rises with KE up to optimum (~37 °C); above optimum H-bonds break → denatured → rate falls to zero.
pH: most enzymes optimum ~7; pepsin ~2; trypsin ~8. pH alters ionisation of R-groups → disrupts tertiary structure.
Substrate concentration: rate ∝ [S] at low [S]; plateaus at $V_\text{max}$ when active sites saturated.
Enzyme concentration: rate ∝ [E] provided substrate is in excess.
Competitive inhibitor: similar shape to S; binds active site; reversed by high [S]; $V_\text{max}$ unchanged.
Non-competitive inhibitor: binds allosteric site; changes active site shape; NOT reversed by high [S]; $V_\text{max}$ reduced.
Always use initial rate = gradient at $t=0$ .
Denatured ≠ destroyed. Low temperature does NOT denature — just slows.

What you’ll learn

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

3.2.1 — Investigate the effect of temperature on the rate of an enzyme-catalysed reaction.
3.2.2 — Investigate the effect of pH on the rate of an enzyme-catalysed reaction.
3.2.3 — Investigate the effect of enzyme and substrate concentration on the rate of an enzyme-catalysed reaction.
3.2.4 — Explain the action of competitive and non-competitive inhibitors on enzyme activity and the effect on $V_\text{max}$ .

Temperature

Rate rises with kinetic energy to an optimum, then falls sharply as enzyme denatures.

Below the optimum. As temperature rises, the kinetic energy of enzyme and substrate molecules increases. They move faster, collide more often and a greater proportion of collisions have enough energy to overcome the activation energy. More ES complexes form per unit time, so the rate increases. For most enzymes, the rate roughly doubles for every 10 °C rise — a temperature coefficient ( $Q_{10}$ ) of about 2.

At the optimum. The rate is at its maximum. For most mammalian enzymes the optimum is around 37 °C (body temperature). Plant enzymes often have lower optima (~25-30 °C); thermophilic bacterial enzymes (e.g. Taq polymerase) have optima of 60-80 °C.

Above the optimum. Increased vibration of the polypeptide chain breaks the hydrogen bonds, ionic bonds and hydrophobic interactions holding the tertiary structure together. The active site loses its complementary shape — the enzyme is denatured. ES complexes can no longer form, and the rate falls steeply.

Denaturation is usually irreversible. When the protein refolds on cooling, the bonds re-form in the wrong positions, so the original active site is not restored.

Crucially, low temperatures do NOT denature the enzyme. They simply reduce kinetic energy and slow the rate. On rewarming, full activity returns — this is why frozen meat can be safely thawed and digested.

Low T = low KE → slow rate.
Rate doubles per 10 °C rise (Q₁₀ ≈ 2) up to optimum.
Optimum ~37 °C for mammals.
Above optimum: H-bonds, ionic, hydrophobic break → denatured → rate falls to zero.
Low T does NOT denature — only slows.

See the full worked example for factors that affect enzyme action →

pH

Each enzyme has an optimum pH; extremes alter R-group ionisation and denature.

Enzymes are most active at their optimum pH, which varies by enzyme:

Enzyme	Optimum pH	Site
Pepsin	~2	Stomach
Salivary amylase	~7	Mouth
Catalase	~7	Cells / peroxisomes
Trypsin	~8	Small intestine
Lipase	~8	Small intestine
Arginase	~10	Liver

Mechanism. pH is a measure of $\text{H}^+$ ion concentration. Changing pH alters the ionisation of the acidic and basic R-groups lining the active site:

At low pH, acidic groups (—COOH) gain $\text{H}^+$ and lose their negative charge.
At high pH, basic groups (—NH₃⁺) lose $\text{H}^+$ and lose their positive charge.

These charge changes disrupt hydrogen bonds and ionic bonds holding the tertiary structure together. The active site changes shape; it is no longer complementary to the substrate. The enzyme is denatured.

Buffer use. Cambridge practicals require a buffer solution (e.g. phosphate buffer of known pH) to keep pH constant during a rate experiment, because acid or alkali produced by the reaction itself could otherwise shift the pH.

Example: pepsin vs trypsin. Pepsin's optimum pH ~2 suits the acidic stomach. When the partly-digested food enters the small intestine, bicarbonate from the pancreas raises pH to ~8 and pepsin is denatured. Trypsin (optimum ~8) takes over protein digestion.

Each enzyme has an optimum pH (often near 7, but pepsin ~2, trypsin ~8).
pH affects ionisation of R-groups → disrupts H-bonds and ionic bonds → tertiary structure / active site lost.
Buffer required in pH experiments to keep [H⁺] constant.

Enzyme concentration

Rate is directly proportional to [E] provided substrate is in excess.

When substrate is in excess, the rate of reaction is directly proportional to the enzyme concentration: doubling [E] doubles the rate. This is because more enzyme molecules mean more active sites, so more ES complexes can form simultaneously.

The rate-vs-[E] graph is therefore a straight line through the origin when [S] is saturating.

What if substrate is limiting? At low [S], the line plateaus once there is enough enzyme to use all substrate at once — adding more enzyme has no effect because substrate is the limiting factor.

Experimental application. When investigating other factors (temperature, pH, [S]), enzyme concentration must be controlled. This is usually achieved by using the same mass / surface area of an enzyme source (e.g. potato discs of identical size) and from the same batch of tissue.

Rate ∝ [E] when substrate is in excess.
Graph is a straight line through origin (substrate saturating).
At low [S], line plateaus when substrate becomes limiting.

Substrate concentration

Rate ∝ [S] at low [S]; plateaus at Vmax when active sites saturate.

At a constant enzyme concentration, the relationship between rate and substrate concentration is a rectangular hyperbola:

At low [S] — substrate molecules are rare. Active sites spend most of the time empty. The rate is directly proportional to [S]: doubling [S] doubles the rate.
At intermediate [S] — more substrate molecules are available. Active sites are occupied more of the time. The rate continues to rise, but with diminishing returns as active sites begin to become saturated.
At high [S] — the rate plateaus at $V_\text{max}$ . Every active site is occupied at any moment: the enzyme is saturated. Adding more substrate has no effect because the rate-limiting step is the turnover of ES to product.

At $V_\text{max}$ , the limiting factor is enzyme concentration, not substrate. To increase $V_\text{max}$ , you must add more enzyme.

$K_m$ . Cambridge does not formally require $K_m$ at AS, but it is useful to know: $K_m$ is the substrate concentration at which the rate is half of $V_\text{max}$ . A low $K_m$ means the enzyme binds substrate tightly; a high $K_m$ means weak binding.

Rate ∝ [S] at low [S].
Curve plateaus at $V_\text{max}$ .
At saturation every active site occupied; enzyme conc. now limiting.
Increase $V_\text{max}$ by adding more enzyme, not more substrate.

See the full worked example for factors that affect enzyme action →

Inhibitors — competitive vs non-competitive

Competitive: binds active site, overcome by high [S], Vmax unchanged. Non-competitive: binds allosteric site, NOT overcome by high [S], Vmax reduced.

Enzyme inhibitors are molecules that reduce the rate of enzyme-catalysed reactions. There are two main types:

Competitive inhibitors

Have a shape similar to the substrate.
Bind reversibly to the active site, blocking the substrate.
Fewer ES complexes form, so the rate falls.
Increasing [S] outcompetes the inhibitor (mass-action), so the inhibitor's effect is reduced at high [S].
$V_\text{max}$ is unchanged — it is just reached at higher [S].
Example: malonate competes with succinate for the active site of succinate dehydrogenase.

Non-competitive inhibitors

Have a shape unlike the substrate.
Bind at an allosteric site (away from the active site).
Binding causes a conformational change that changes the shape of the active site, so the substrate can no longer bind (or, if it does, no catalysis occurs).
Increasing [S] does NOT help, because substrate and inhibitor bind at different sites — they do not compete.
$V_\text{max}$ is reduced — a constant fraction of enzyme molecules are inactivated at all [S].
Example: heavy-metal ions such as $\text{Hg}^{2+}$ , $\text{Pb}^{2+}$ or $\text{Ag}^+$ bind covalently to $-\text{SH}$ groups of cysteine, often irreversibly.

Graphical comparison. On a rate-vs-[S] plot:

Competitive: same $V_\text{max}$ , reached at higher [S].
Non-competitive: lower $V_\text{max}$ at all [S].

Reversibility. Competitive inhibition is usually reversible (remove inhibitor / flood with substrate). Non-competitive inhibition is often irreversible because the inhibitor binds covalently.

Competitive: similar shape to S, binds active site, reversed by high [S], $V_\text{max}$ unchanged.
Non-competitive: binds allosteric site, changes active-site shape, NOT reversed by high [S], $V_\text{max}$ reduced.
Both reduce rate, but only competitive can be overcome by more substrate.

See the full worked example for factors that affect enzyme action →

Investigation methods (catalase & amylase)

Catalase → measure O₂; amylase → measure starch disappearance (iodine) or maltose appearance (Benedict's).

Cambridge 9700 expects you to be able to plan, carry out and analyse enzyme experiments. Two model systems are standard:

System 1 — Catalase + hydrogen peroxide

$2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$

Source: potato discs, yeast suspension, or liver homogenate.
Measurement: collect O₂ in a gas syringe or inverted measuring cylinder; record volume vs time. Alternatives: foam height (with detergent), time to fixed volume ( $1/t$ ), mass loss on a balance.
Controlled variables: mass / surface area of tissue, volume and pH of buffer, temperature, total volume.

System 2 — Amylase + starch

$\text{starch} \rightarrow \text{maltose}$

Source: salivary amylase or bacterial $\alpha$ -amylase.
Measurement (method A — disappearance of starch): at fixed time intervals, take a small sample and add to iodine solution on a spotting tile. Note the time at which the blue-black colour disappears (i.e. all starch hydrolysed).
Measurement (method B — appearance of maltose): at intervals, sample and test with Benedict's reagent; brick-red precipitate intensity (or colorimeter absorbance) gives the maltose concentration.
Controlled variables: enzyme concentration, volume and pH of buffer, temperature.

Calculating rate

Always use the initial rate = the gradient of the tangent to the product-vs-time curve at $t = 0$ (or the slope over the first 15-30 s).
Plot initial rate (y-axis) against the independent variable (e.g. [S], [E], temperature, pH).
Repeat each measurement at least 3 times and calculate the mean; calculate standard deviation or range bars for reliability.

Controls. Always include a control with no enzyme (e.g. boiled enzyme — denatured) to confirm that any observed change is due to enzyme activity and not background hydrolysis.

Catalase: collect O₂ vs time.
Amylase: time for iodine to lose colour (disappearance) OR Benedict's intensity (appearance).
Use initial rate = gradient at $t=0$ .
Repeats + mean + control (boiled enzyme).

See the full worked example for factors that affect enzyme action →

Summary of rate-vs-factor graphs

Recognise and sketch the four classical enzyme-rate graphs.

Four canonical graphs are tested repeatedly on Paper 2:

Rate vs Temperature — bell-shaped curve. Rises gradually to optimum (~37 °C), then falls sharply as denaturation begins. Falls to zero at high temperatures.
Rate vs pH — bell-shaped curve, narrower than temperature. Peaks at the optimum pH; falls steeply either side as R-group ionisation alters tertiary structure.
Rate vs Substrate concentration — rectangular hyperbola. Linear at low [S], plateaus at $V_\text{max}$ as active sites saturate.
Rate vs Enzyme concentration — straight line through the origin (provided [S] is saturating). Plateaus only if substrate becomes limiting.

Inhibitor overlays (on the rate-vs-[S] curve):

Competitive inhibitor: same $V_\text{max}$ but reached at higher [S] (curve shifted right).
Non-competitive inhibitor: lower $V_\text{max}$ at all [S].

You should be able to draw each graph with labelled axes, mark the optimum or $V_\text{max}$ , and annotate the molecular reason for each region (low KE / saturation / denaturation / R-group ionisation).

Temperature: bell-shaped, optimum ~37 °C.
pH: bell-shaped (narrower), optimum varies (pepsin 2, amylase 7, trypsin 8).
[S]: hyperbola → $V_\text{max}$ plateau.
[E]: straight line through origin (saturating [S]).
Inhibitors: competitive shifts curve right; non-competitive lowers $V_\text{max}$ .

How it’s examined

Cambridge 9700 examines this on Paper 2 (AS, structured 5-7 mark questions), Paper 3 (AS practical — collecting and analysing rate data, drawing calibration curves and tangents) and Paper 4 (A2 synthesis with metabolism). Most-tested questions: (a) describe and explain the effect of temperature (5-6 marks); (b) compare competitive and non-competitive inhibition (5-6 marks); (c) plan a method to investigate one factor (6 marks, Paper 3); (d) interpret a rate-vs-[S] graph and identify $V_\text{max}$ / limiting factor (4-5 marks).

Sources: Cambridge International AS & A Level Biology 9700 syllabus (2025-2027); 9700 Examiner Reports 2022-2024; 9700/22 May/Jun 2024 question paper and mark scheme; 9700/42 May/Jun 2024 question paper and mark scheme; 9700/32 May/Jun 2024 practical paper and mark scheme; 9700/12 Oct/Nov 2024 question paper and mark scheme. Last reviewed 2026-05-12.

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Step-by-step worked examples — Factors that affect enzyme action

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

1Effect of temperature on enzyme rate (6 marks)
Extended• Adapted from 9700/22 May/Jun 2024• temperature, denaturation, Paper 2
Question
Describe and explain the effect of temperature on the rate of an enzyme-catalysed reaction. (6 marks)
Step-by-step solution
1. Step 1
  Low temperatures. Molecules have low kinetic energy. Enzyme and substrate molecules move slowly, so successful collisions are rare and few enzyme-substrate complexes form per unit time. The rate is low.
2. Step 2
  As temperature rises (up to the optimum), kinetic energy increases, so substrate molecules collide with active sites more often and with sufficient energy to react. The rate increases — typically doubling for every 10 °C rise (Q₁₀ ≈ 2) up to the optimum.
3. Step 3
  At the optimum temperature (~37 °C for most mammalian enzymes), the rate is maximum. Enzyme and substrate collide frequently and most ES complexes form.
4. Step 4
  Above the optimum. Vibrations of the polypeptide become so violent that the hydrogen bonds, ionic bonds and hydrophobic interactions holding the tertiary structure are broken.
5. Step 5
  Active site loses shape. The enzyme is denatured — the active site is no longer complementary to the substrate. ES complexes cannot form.
6. Step 6
  Rate falls sharply above the optimum and reaches zero when all enzyme molecules are denatured. Denaturation is generally irreversible because hydrogen and ionic bonds re-form in the wrong positions.
Answer
Rate increases with KE / collisions up to optimum (~37 °C); above optimum, tertiary structure denatured, active site no longer complementary, rate falls to zero.
Examiner tip
Mark scheme: (1) low T = low KE / few collisions; (2) rate increases with T up to optimum; (3) optimum named; (4) above optimum H-bonds break / tertiary structure disrupted; (5) active site no longer complementary / denatured; (6) rate decreases / no ES complex. 9700 examiners flag candidates who write 'enzyme is killed' — this is wrong and scores 0.
2Competitive inhibition (5 marks)
Extended• competitive, inhibitor, Paper 2
Question
Explain how a competitive inhibitor reduces the rate of an enzyme-catalysed reaction. (5 marks)
Step-by-step solution
1. Step 1
  Similar shape. A competitive inhibitor has a shape and chemistry similar to the substrate, so it can fit into the active site.
2. Step 2
  Binds active site. The inhibitor binds reversibly at the active site, blocking the substrate from binding.
3. Step 3
  Fewer ES complexes. While inhibitor is bound, the active site cannot bind substrate, so fewer enzyme-substrate complexes form per unit time. The reaction rate falls.
4. Step 4
  Effect reduced by high [S]. At high substrate concentrations, substrate molecules outcompete the inhibitor for the active site (mass-action / probability of binding). At very high [S], the inhibitor has little effect.
5. Step 5
  $V_\text{max}$ unchanged. Because the inhibitor can be displaced by substrate, $V_\text{max}$ is the same as without inhibitor (just reached at higher [S]). The Lineweaver-Burk plot shows the same y-intercept but a steeper gradient.
Answer
Inhibitor similar shape to substrate; binds active site reversibly; blocks substrate; fewer ES complexes; rate falls; effect reversed by high [S]; Vmax unchanged.
Examiner tip
Mark scheme: (1) similar shape to substrate; (2) binds active site / reversibly; (3) blocks substrate binding; (4) effect reduced by high [S] / outcompeted; (5) Vmax unchanged. Candidates who confuse competitive with non-competitive on the Vmax point lose marks 4-5.
3Effect of substrate concentration (5 marks)
Extended• substrate concentration, Vmax, Paper 2
Question
Describe and explain the effect of substrate concentration on the rate of an enzyme-catalysed reaction at a constant enzyme concentration. (5 marks)
Step-by-step solution
1. Step 1
  At low [S], substrate molecules are rare. Most active sites are empty for most of the time. The rate is directly proportional to [S] — doubling [S] doubles the rate.
2. Step 2
  As [S] increases, more substrate molecules are available to collide with active sites. More ES complexes form per unit time, and the rate continues to rise.
3. Step 3
  The curve plateaus when nearly all active sites are occupied at any moment. Substrate concentration is no longer the limiting factor — enzyme concentration is.
4. Step 4
  At $V_\text{max}$ . The enzyme is saturated with substrate: every active site is occupied as soon as the previous product is released. Adding more substrate has no further effect.
5. Step 5
  Limiting factor. Above saturation, the rate-limiting step is the conversion of ES to product (or the turnover number of the enzyme). $V_\text{max}$ can only be increased by adding more enzyme.
Answer
Rate ∝ [S] at low [S]; curve plateaus as active sites saturate; at Vmax all active sites occupied; enzyme concentration becomes limiting.
Examiner tip
Mark scheme: (1) rate proportional to [S] at low values; (2) idea of collisions / ES complexes; (3) plateau / Vmax; (4) saturation / all active sites occupied; (5) enzyme concentration is limiting factor at Vmax.
4Investigating the effect of [S] on catalase (6 marks)
Extended• Adapted from 9700/32 May/Jun 2024 (practical)• catalase, method, Paper 3
Question
Describe how you would investigate the effect of hydrogen peroxide concentration on the rate of decomposition catalysed by catalase. Refer to apparatus, variables and how the rate is calculated. (6 marks)
Step-by-step solution
1. Step 1
  Prepare substrate dilutions. Make a serial dilution of hydrogen peroxide using distilled water — for example 0, 1, 2, 5, 10 and 20 vol (or 0–6% w/v).
2. Step 2
  Source of catalase. Use a standard mass of fresh potato discs (or yeast suspension), cut to the same surface area each time, to keep enzyme concentration constant.
3. Step 3
  Apparatus. Attach a delivery tube to a sealed flask containing the enzyme + buffer. Run the delivery tube into an inverted measuring cylinder over water (or a gas syringe) to collect the oxygen produced.
4. Step 4
  Run the reaction. Add the chosen volume of H₂O₂ to the flask, immediately seal, and record the volume of O₂ collected at fixed intervals (e.g. every 15 s for 2 minutes). Repeat each concentration at least three times for reliability.
5. Step 5
  Controlled variables. Same mass / surface area of potato; same volume and pH of buffer; same temperature (water bath); same total volume by topping up with water; same person observing.
6. Step 6
  Calculate rate. Plot O₂ volume against time for each concentration. Calculate the initial rate as the gradient of the first ~30 s of each curve. Plot initial rate (y-axis) against [H₂O₂] (x-axis). Expect a curve that rises proportionally then plateaus at $V_\text{max}$ .
Answer
Serial dilution of H₂O₂; same mass/surface area of potato (constant [E]); gas syringe / displaced water collects O₂; control temperature, pH, total volume; calculate initial rate = gradient; plot rate vs [H₂O₂].
Examiner tip
Mark scheme: (1) named substrate range / dilution method; (2) named catalase source + control of [E]; (3) apparatus to collect O₂; (4) named controlled variables (at least 3); (5) repeats for reliability; (6) initial rate calculated as gradient. Examiner reports stress 'initial rate' — using rate over the full reaction loses a mark because rate is non-linear over long times.

Model Answers — Factors that affect enzyme action

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

Question
9700/22 May/Jun 2024 Q3(b) (adapted)6 marks
Q (6 marks). Describe and explain the effect of temperature on the rate of an enzyme-catalysed reaction.
Model answer
At low temperatures, enzyme and substrate molecules have low kinetic energy. They move and collide slowly, so few successful collisions occur per second and few enzyme-substrate complexes form. The reaction rate is therefore low.

As temperature rises towards the optimum, kinetic energy increases. Molecules move faster and collide more frequently, and a greater proportion of collisions have enough energy to overcome the (already-lowered) activation energy. More ES complexes form per second, so the rate rises — for many enzymes the rate roughly doubles for every 10 °C rise, giving a temperature coefficient ( $Q_{10}$ ) of about 2.

At the optimum temperature (around 37 °C for most mammalian enzymes; closer to 60-70 °C for thermophilic bacteria), the rate reaches a maximum.

Above the optimum, the increased vibration of the polypeptide chain begins to break the hydrogen bonds, ionic bonds and hydrophobic interactions that hold the enzyme's tertiary structure together. The polypeptide unfolds and the active site loses its complementary shape — the enzyme is denatured.

Once denatured, the active site can no longer bind the substrate, so ES complexes do not form and the rate falls sharply. At high enough temperatures all enzyme molecules are denatured and the rate falls to zero. Denaturation is generally irreversible — re-cooling does not restore the original tertiary structure because the bonds re-form in the wrong positions.

Note that low temperatures do NOT denature an enzyme; they merely slow it down. On rewarming, the rate is restored.
Why this scores
Why this scores 6/6. Mark scheme awards: (1) low T → low KE / few collisions; (2) rate increases with T up to optimum; (3) optimum identified (or named for mammalian enzymes); (4) above optimum H-bonds / ionic / hydrophobic break OR vibration disrupts tertiary structure; (5) active site no longer complementary / denatured; (6) rate falls / ES complexes do not form. Additional credit available for: low T does not denature; denaturation irreversible. Examiner reports repeatedly flag 'enzyme is killed' — wrong term, 0 marks.
Question
9700/42 May/Jun 2024 Q4(c) (adapted)5 marks
Q (5 marks). Explain how a competitive inhibitor reduces the rate of an enzyme-catalysed reaction. Refer to the effect of substrate concentration in your answer.
Model answer
A competitive inhibitor has a shape and chemistry similar to the substrate, so it can fit into the active site of the enzyme.

The inhibitor binds reversibly at the active site, occupying it temporarily. While the inhibitor is bound, the substrate cannot enter the active site, so an enzyme-substrate complex cannot form. Fewer ES complexes form per unit time, and the reaction rate is reduced.

Because substrate and inhibitor compete for the same site, the effect of the inhibitor depends on their relative concentrations. At high substrate concentrations, the probability that the active site encounters a substrate rather than an inhibitor increases — substrate molecules outcompete the inhibitor.

For this reason, at sufficiently high $[\text{S}]$ , the maximum rate of reaction ( $V_\text{max}$ ) is unchanged by a competitive inhibitor — it is simply reached at a higher substrate concentration. On a rate-vs- $[\text{S}]$ curve, the inhibited curve approaches the same plateau but more slowly; on a Lineweaver-Burk plot, the y-intercept is unchanged but the gradient (and $K_m$ ) increases.

A classic example is malonate, which competes with succinate for the active site of succinate dehydrogenase in the Krebs cycle.
Why this scores
Why this scores 5/5. Mark scheme: (1) inhibitor shape similar to substrate; (2) binds active site (reversibly); (3) blocks substrate binding / no ES complex; (4) effect reduced by increasing [S] / substrate outcompetes inhibitor; (5) Vmax unchanged. Recent examiner reports note that candidates who write 'Vmax decreases' for a competitive inhibitor lose mark 5 — this is the non-competitive case.
Question
9700/32 May/Jun 2024 (practical, adapted)6 marks
Q (6 marks). Describe how you would investigate the effect of substrate concentration on the rate of the catalase-catalysed decomposition of hydrogen peroxide.
Model answer
Substrate range. Prepare a serial dilution of hydrogen peroxide solution at known concentrations — for example 0, 1, 2, 5, 10 and 20 vol — by diluting a stock solution with distilled water. The total volume in each tube should be the same so that only $[\text{H}_2\text{O}_2]$ varies.

Enzyme source. Use a fixed mass of fresh potato discs (or a fixed volume of yeast suspension). Cut the discs to the same size and surface area each time, since catalase concentration must remain constant.

Apparatus. Connect the reaction flask, via a delivery tube and bung, to a gas syringe (or an inverted measuring cylinder over water) to collect the oxygen produced as the reaction proceeds.

Procedure. Place the potato in the flask with phosphate buffer of pH 7. Add a known volume of the chosen $[\text{H}_2\text{O}_2]$ , immediately stopper the flask and start the timer. Record the volume of O₂ collected every 15 s for two minutes. Repeat each concentration three times and calculate the mean.

Controlled variables. Mass and surface area of potato; volume and pH of buffer; temperature (water bath at 25 °C); total volume in flask; same observer.

Analysis. Plot O₂ volume against time for each concentration. The initial rate (the gradient of the tangent to the curve over the first 30 s) is the most reliable measure because the reaction slows as substrate is used up. Plot initial rate (y-axis) against $[\text{H}_2\text{O}_2]$ (x-axis): the expected curve rises in direct proportion at low $[\text{S}]$ and plateaus at $V_\text{max}$ when active sites are saturated.
Why this scores
Why this scores 6/6. Mark scheme: (1) range of named substrate concentrations / dilution method; (2) named catalase source + constant mass/surface area; (3) apparatus to collect O₂ named correctly; (4) at least 3 controlled variables stated; (5) repeats + mean for reliability; (6) initial rate calculated as gradient of tangent / first 30 s. 'Calculate rate over whole reaction' does not score — examiner reports flag this consistently as it ignores that rate falls as [S] depletes.
Question
9700/12 Oct/Nov 2024 Q6 (adapted)4 marks
Q (4 marks). Pepsin is a protein-digesting enzyme that works in the stomach but is inactive in the small intestine. Explain this observation.
Model answer
Pepsin has an optimum pH of approximately 2. In the stomach, hydrochloric acid is secreted by parietal cells, giving a pH of about 1.5 to 2.0. Pepsin's active site has its correctly folded tertiary structure under these acidic conditions, so it is fully active and hydrolyses proteins to shorter peptides.

In the small intestine, however, bicarbonate secreted in pancreatic juice raises the pH to about 8. At this high pH, the ionisation of the R-groups lining the active site of pepsin changes — many acidic groups become deprotonated and basic groups become uncharged — disrupting the hydrogen and ionic bonds that maintain the tertiary structure.

The active site loses its complementary shape, and pepsin is denatured. It can no longer bind substrate or catalyse hydrolysis.

In the small intestine, protein digestion continues using trypsin (and chymotrypsin), which have an optimum pH of about 8 and are perfectly active in the alkaline conditions there.
Why this scores
Why this scores 4/4. Mark scheme: (1) pepsin optimum pH ~2 / acidic; (2) small intestine pH high / ~8 due to bicarbonate; (3) high pH alters ionisation of R-groups / disrupts H- and ionic bonds → tertiary structure / active site no longer complementary / denatured; (4) ORA — trypsin replaces pepsin in small intestine at higher pH. 9700 mark schemes commonly use 'ORA' (or reverse argument) for pH explanations — answers may flip stomach ↔ intestine without losing marks.
Question
Practice question — recurring Paper 2 style5 marks
Q (5 marks). Compare the effects of a competitive inhibitor and a non-competitive inhibitor on the rate of an enzyme-catalysed reaction.
Model answer
A competitive inhibitor has a shape similar to the substrate and binds reversibly to the active site, blocking substrate binding. A non-competitive inhibitor has a shape unlike the substrate and binds to an allosteric site away from the active site.

When a non-competitive inhibitor binds, the enzyme's tertiary structure changes and the active site changes shape so that the substrate no longer fits. The enzyme is functionally inactivated even though the active site itself is unoccupied.

The effect on rate vs $[\text{S}]$ differs sharply:

For a competitive inhibitor, increasing $[\text{S}]$ dilutes out the inhibitor by mass-action — substrate molecules outcompete the inhibitor for the active site. Therefore $V_\text{max}$ is unchanged; it is just reached at a higher $[\text{S}]$ (higher $K_m$ ).

For a non-competitive inhibitor, increasing $[\text{S}]$ does not help: substrate cannot displace the inhibitor because they bind at different sites. A constant fraction of enzyme remains inactivated at all $[\text{S}]$ . Therefore $V_\text{max}$ is reduced.

Competitive inhibition can sometimes be reversed by removing the inhibitor or flooding with substrate; non-competitive inhibition often involves covalent binding (e.g. heavy-metal ions such as Hg²⁺ binding to cysteine -SH groups) and may be irreversible.
Why this scores
Why this scores 5/5. Mark scheme: (1) competitive = similar shape, binds active site; (2) non-competitive = different shape, binds allosteric / non-active site; (3) non-competitive changes shape of active site; (4) competitive overcome by increasing [S], non-competitive not; (5) Vmax unchanged (competitive) vs Vmax reduced (non-competitive). This is one of the most frequent Paper 2 comparison questions — examiners want clear comparison points, not two separate descriptions.

Key Definitions and Keywords — Factors that affect enzyme action

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

Optimum temperature
Examiner keyword
The temperature at which the rate of an enzyme-catalysed reaction is maximum. For most mammalian enzymes this is about 37 °C; thermophilic bacterial enzymes can have optima of 60-80 °C.
Optimum pH
Examiner keyword
The pH at which the rate of an enzyme-catalysed reaction is maximum. Pepsin ~2; salivary amylase ~7; trypsin ~8. Outside this range, ionisation of R-groups disrupts tertiary structure.
Denaturation
Examiner keyword
Loss of the three-dimensional tertiary structure of a protein (and therefore the active-site shape of an enzyme) caused by heat or extreme pH. Hydrogen bonds, ionic bonds and hydrophobic interactions break; the active site is no longer complementary to the substrate.
$V_\text{max}$
Examiner keyword
The maximum rate of an enzyme-catalysed reaction, reached when all active sites are saturated with substrate. Can be increased by adding more enzyme but not by adding more substrate.
Saturation
Examiner keyword
The state in which every enzyme active site is occupied by substrate at any given moment. Above this point, adding more substrate does not increase the rate; the rate-limiting step becomes the turnover of ES to product.
Limiting factor
Examiner keyword
The factor that determines the rate of a reaction at a particular set of conditions. For an enzyme reaction this may be substrate concentration (at low [S]), enzyme concentration (at saturating [S]), temperature, pH or inhibitor presence.
Competitive inhibitor
Examiner keyword
A molecule of similar shape to the substrate that binds reversibly to the active site, blocking substrate binding. Its effect is reduced by increasing [S]; $V_\text{max}$ is unchanged.
Non-competitive inhibitor
Examiner keyword
A molecule that binds at a site away from the active site (allosteric site). This binding changes the shape of the active site so substrate cannot bind. Its effect is NOT reduced by increasing [S]; $V_\text{max}$ is reduced.
Allosteric site
Examiner keyword
A binding site on an enzyme distinct from the active site. Binding of a non-competitive inhibitor (or a regulatory molecule) at this site changes the conformation of the active site.
Initial rate
Examiner keyword
The rate of an enzyme-catalysed reaction at the very start, before the substrate concentration falls appreciably. Calculated as the gradient of the tangent to a product-vs-time curve at $t = 0$ .

Common Mistakes and Misconceptions — Factors that affect enzyme action

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

✕Saying that the enzyme is 'killed' or 'destroyed' at high temperatures
9700 Examiner Reports 2022-2024
Why it happens
Students think of enzymes as alive.
How to avoid it
Enzymes are proteins, not living things. They are denatured: the tertiary structure (active site shape) is lost. The molecule is still present. Always use 'denatured' or 'tertiary structure disrupted'.
✕Stating that low temperatures denature an enzyme
9700 Examiner Reports 2023
Why it happens
Students assume any temperature extreme denatures.
How to avoid it
Low temperatures only slow the enzyme by reducing kinetic energy. The tertiary structure is unaffected. On rewarming the enzyme regains full activity. Only high temperatures (or extreme pH) denature.
✕Stating that a competitive inhibitor lowers $V_\text{max}$
9700 Examiner Reports 2024
Why it happens
Confusion with non-competitive.
How to avoid it
A competitive inhibitor is out-competed by high [S], so $V_\text{max}$ is unchanged (just reached at higher [S]). A NON-competitive inhibitor binds an allosteric site and is NOT out-competed by [S], so $V_\text{max}$ IS reduced.
✕Saying that a non-competitive inhibitor binds the active site
9700 Examiner Reports 2023
Why it happens
Misremembering the definitions.
How to avoid it
Non-competitive inhibitors bind an allosteric site (not the active site) and cause the active site to change shape. Competitive inhibitors bind the active site directly.
✕Calculating the rate over the whole reaction instead of the initial rate
9700 Paper 3 Examiner Reports 2023-2024
Why it happens
Students take total product / total time, ignoring that the rate falls as substrate is used up.
How to avoid it
Use initial rate = gradient of tangent at $t = 0$ (or gradient over the first 15-30 s). This is the most reliable measure because substrate has not been significantly depleted yet.
✕Explaining the effect of pH only by saying 'pH changes shape'
9700 Examiner Reports 2024
Why it happens
Oversimplified explanation.
How to avoid it
Mark schemes credit the mechanism: pH affects the ionisation of R-groups lining the active site. This disrupts hydrogen and ionic bonds, the tertiary structure, and therefore the active site complementarity.

Practice questions

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

Past paper style quiz

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

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Factors that affect enzyme action — frequently asked questions

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

Factors that affect enzyme action

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Effect of temperature on enzyme rate (6 marks)

2Competitive inhibition (5 marks)

3Effect of substrate concentration (5 marks)

4Investigating the effect of [S] on catalase (6 marks)

Optimum temperature

Optimum pH

Denaturation

VmaxV_\text{max}Vmax​

Saturation

Limiting factor

Competitive inhibitor

Non-competitive inhibitor

Allosteric site

Initial rate

✕Saying that the enzyme is 'killed' or 'destroyed' at high temperatures

✕Stating that low temperatures denature an enzyme

✕Stating that a competitive inhibitor lowers VmaxV_\text{max}Vmax​

✕Saying that a non-competitive inhibitor binds the active site

✕Calculating the rate over the whole reaction instead of the initial rate

✕Explaining the effect of pH only by saying 'pH changes shape'

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Factors that affect enzyme action — frequently asked questions

$V_\text{max}$

✕Stating that a competitive inhibitor lowers $V_\text{max}$