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This lesson covers the effect of substrate concentration on enzyme activity, the concept of Vmax and Km, and the mechanisms of competitive and non-competitive inhibition, as required by the Edexcel A-Level Biology B specification (9BI0), Topic 1: Biological Molecules. You need to interpret rate–concentration graphs and explain how inhibitors affect enzyme kinetics.
When enzyme concentration is held constant, increasing the substrate concentration increases the rate of reaction — but only up to a point.
| Substrate Concentration | Explanation | Rate |
|---|---|---|
| Low [S] | Many active sites are unoccupied — there are more enzyme molecules than substrate molecules. Increasing [S] increases the chance of enzyme–substrate collisions. | Rate increases proportionally with [S] |
| Moderate [S] | More active sites are occupied. The rate of increase begins to slow. | Rate increases but the curve begins to plateau |
| High [S] (saturating) | All active sites are occupied — the enzyme is working at maximum capacity. Adding more substrate has no effect because there are no free active sites. | Rate reaches a plateau — this is Vmax |
The graph of initial rate of reaction against substrate concentration produces a characteristic rectangular hyperbola (a curve that rises steeply then levels off).
| Term | Definition |
|---|---|
| Vmax | The maximum rate of reaction when all enzyme active sites are saturated with substrate |
| Km (Michaelis constant) | The substrate concentration at which the reaction rate is half of Vmax. It is a measure of the enzyme's affinity for its substrate |
| Turnover number (kcat) | The number of substrate molecules converted to product per enzyme molecule per unit time (when the enzyme is saturated). For catalase, this is approximately 40 million per second. |
Exam Tip: Km is an important concept in A-Level Biology. Remember: low Km = high affinity, high Km = low affinity. If asked to compare two enzymes, the one with the lower Km binds its substrate more effectively.
When substrate concentration is in excess (saturating), increasing the enzyme concentration increases the rate of reaction proportionally.
This is because more enzyme molecules provide more active sites, so more enzyme–substrate complexes can form per unit time. The graph of rate against enzyme concentration is a straight line through the origin (assuming substrate is not limiting).
In practice, the line may plateau at very high enzyme concentrations if the substrate becomes limiting.
| Method | Suitable For | Example |
|---|---|---|
| Measuring gas produced | Reactions that produce a gas (O₂, CO₂) | Catalase decomposing H₂O₂ — collect O₂ gas over water or with a gas syringe |
| Measuring colour change | Reactions involving a colour change | Amylase digesting starch — time for iodine to stop turning blue-black |
| Measuring change in turbidity | Reactions that clear or increase cloudiness | Protease clearing a suspension of milk protein |
| Measuring change in mass | Reactions that release gas (mass decreases) | Yeast fermenting glucose — loss of CO₂ |
When investigating the effect of substrate concentration, it is important to measure the initial rate of reaction (the rate at the very beginning, before substrate depletion or product accumulation affect the rate). This is done by:
Exam Tip: Always use initial rates when comparing enzyme activities at different substrate concentrations. As the reaction proceeds, the substrate concentration decreases and products may inhibit the enzyme, making later measurements unreliable. State this in your method to show understanding.
Enzyme inhibitors are molecules that reduce the rate of an enzyme-catalysed reaction. They are classified as competitive or non-competitive (and further as reversible or irreversible).
A competitive inhibitor is a molecule that has a similar shape to the substrate and competes with the substrate for the active site of the enzyme.
| Parameter | Effect of Competitive Inhibitor |
|---|---|
| Vmax | Unchanged — at very high substrate concentrations, the substrate outcompetes the inhibitor for the active site, so Vmax can still be reached |
| Km | Increased (apparent Km is higher) — a higher substrate concentration is needed to reach half-Vmax because some active sites are blocked by the inhibitor |
| Reversibility | Usually reversible — the inhibitor binds and leaves the active site in a dynamic equilibrium |
The competitive inhibitor can be overcome by increasing the substrate concentration — eventually, there are so many substrate molecules that they outcompete the inhibitor.
A non-competitive inhibitor binds to a site on the enzyme that is different from the active site — called an allosteric site.
| Parameter | Effect of Non-Competitive Inhibitor |
|---|---|
| Vmax | Decreased — some enzyme molecules are permanently inactivated (their active sites are distorted), so the maximum rate is lower |
| Km | Unchanged — the affinity of the unaffected enzyme molecules for the substrate is not altered |
| Reversibility | Can be reversible (inhibitor binds and leaves) or irreversible (inhibitor permanently binds) |
| Feature | Competitive | Non-Competitive |
|---|---|---|
| Binding site | Active site | Allosteric site (not the active site) |
| Resembles substrate? | Yes — similar shape | No — different shape |
| Effect on Vmax | No change | Decreased |
| Effect on Km (apparent) | Increased | No change |
| Overcome by excess substrate? | Yes | No |
| Example | Malonate inhibiting succinate dehydrogenase | Cyanide inhibiting cytochrome c oxidase |
Exam Tip: In graph interpretation questions, look at whether Vmax changes. If the maximum rate is the same but the curve is shifted to the right → competitive inhibition. If the maximum rate is lower but the Km is the same → non-competitive inhibition. Always annotate your graphs in the exam to show the examiner your reasoning.
Some inhibitors bind permanently to the enzyme (usually by forming strong covalent bonds), permanently inactivating it. These are irreversible inhibitors.
| Irreversible Inhibitor | Target Enzyme | Mechanism |
|---|---|---|
| Organophosphates (nerve agents, some pesticides) | Acetylcholinesterase | Bind covalently to a serine residue in the active site → acetylcholine is not broken down → continuous nerve stimulation |
| Aspirin | Cyclooxygenase (COX) | Acetylates a serine residue in the active site → blocks prostaglandin synthesis → reduces inflammation and pain |
| Penicillin | Transpeptidase (bacterial enzyme) | Binds covalently to the active site → bacteria cannot form cross-links in their cell walls → cell lysis |
Many drugs and poisons work by inhibiting specific enzymes:
| Substance | Type of Inhibitor | Target | Effect |
|---|---|---|---|
| Statins (e.g. atorvastatin) | Competitive | HMG-CoA reductase | Block cholesterol synthesis → lower blood cholesterol |
| Methotrexate | Competitive | Dihydrofolate reductase | Blocks DNA synthesis → used as anticancer drug |
| Heavy metals (Hg²⁺, Pb²⁺) | Non-competitive / irreversible | Various enzymes | Bind to –SH groups, disrupting disulfide bonds and tertiary structure |
| Cyanide | Non-competitive | Cytochrome c oxidase | Blocks electron transport chain → death |
| Arsenic | Non-competitive | Pyruvate dehydrogenase | Blocks link reaction → stops aerobic respiration |
Exam Tip: The specification may ask you to apply your understanding of inhibition to unfamiliar examples. Read the question carefully — identify whether the inhibitor binds to the active site or elsewhere, whether increasing substrate concentration overcomes it, and what happens to Vmax and Km. Use these clues to classify the type of inhibition.
In metabolic pathways, the end product of the pathway often acts as a non-competitive inhibitor of an enzyme early in the pathway. This is called end-product inhibition or negative feedback.
Isoleucine is synthesised from threonine in a five-step metabolic pathway. Isoleucine (the end product) inhibits threonine deaminase (the first enzyme in the pathway) by binding to its allosteric site.
This is an efficient way of regulating metabolism — the cell only produces as much product as it needs, avoiding waste.
Exam Tip: When drawing graphs of enzyme kinetics, always label the axes clearly (initial rate of reaction vs substrate concentration), mark Vmax and Km, and draw separate curves for uninhibited, competitively inhibited, and non-competitively inhibited reactions. Annotate the key differences between the curves.
This lesson sits in Edexcel 9BI0 Topic 1 — Biological Molecules, on the quantitative behaviour of enzymes: the Michaelis–Menten rectangular-hyperbola relating initial rate to [S] at fixed [E], pH and T; Vmax as the saturation plateau; Km as [S] at Vmax/2, reporting active-site affinity; and the kinetic signatures of competitive, non-competitive and irreversible inhibition (refer to the official Pearson Edexcel 9BI0 specification for exact wording). Content statements paraphrase to: define Vmax and Km with units (Km in μmol dm⁻³ or mmol dm⁻³); interpret rate-vs-[S] graphs; recognise that doubling [E] doubles Vmax but leaves Km unchanged (Km is intrinsic to the E–S pair); state that competitive inhibitors raise apparent Km but leave Vmax unchanged (substrate outcompetes inhibitor at high [S]); state that non-competitive inhibitors lower Vmax but leave Km unchanged (allosteric distortion removes functional enzyme); recognise irreversible inhibition as covalent active-site modification; and apply these principles to drug action and metabolic regulation. The lesson is examined directly on Paper 1 and reactivates synoptically across Topic 5 (PFK inhibited by ATP; cyanide on cytochrome oxidase), Topic 6 (most drugs are enzyme inhibitors), Topic 7 (orlistat on lipase) and Topic 8 (PKU — defective phenylalanine hydroxylase).
Question (8 marks): Initial rate is measured across [S] at fixed [E] = 1.0 nmol dm⁻³, pH 7.4 and 37 °C. The rate-vs-[S] curve plateaus at Vmax = 80 μmol dm⁻³ s⁻¹, with rate 40 μmol dm⁻³ s⁻¹ at [S] = 0.50 mmol dm⁻³.
(a) State Km with units and explain what it tells you about the enzyme. (3)
(b) With inhibitor X added, Vmax is unchanged at 80 μmol dm⁻³ s⁻¹ but Vmax/2 is now reached at [S] = 1.5 mmol dm⁻³. Identify the inhibition type, justify, and predict the effect of further raising [S]. (3)
(c) With inhibitor Y, Vmax falls to 40 μmol dm⁻³ s⁻¹ while apparent Km remains 0.50 mmol dm⁻³. Identify the inhibition type and explain in molecular terms. (2)
Solution with mark scheme:
(a) M1 (AO2) — Km = [S] at Vmax/2 = 40 μmol dm⁻³ s⁻¹, so Km = 0.50 mmol dm⁻³.
A1 (AO1) — units must be a concentration; Km is not a pure number.
A1 (AO3) — Km reports active-site affinity: lower Km = higher affinity. Half-saturation at sub-millimolar [S] indicates moderately tight substrate binding.
(b) M1 (AO2) — Vmax unchanged + apparent Km tripled → competitive inhibition.
A1 (AO1) — inhibitor X is substrate-like and binds reversibly to the active site, competing for the same pocket; more [S] is needed to half-saturate.
A1 (AO3) — at sufficiently high [S], substrate outcompetes inhibitor by mass-action, so the original Vmax is still reached; the curve shifts right but reaches the same plateau.
(c) M1 (AO2) — Vmax halved with Km unchanged identifies non-competitive inhibition.
A1 (AO1) — inhibitor Y binds an allosteric site; the conformational change distorts the active site of the bound enzyme fraction. The unaffected fraction retains the original Km, but fewer functional active sites lowers Vmax. Excess [S] cannot rescue — substrate cannot displace an inhibitor that is not at the active site.
Total: 8 marks (a: M1 A1 A1; b: M1 A1 A1; c: M1 A1). A* candidates state Km units explicitly, derive Km graphically, and explain why the kinetic signatures differ.
Question (6 marks): A drug is being trialled as a treatment that targets a specific bacterial enzyme. Initial-rate measurements show that the drug raises the apparent Km of the enzyme but does not alter Vmax. Explain the molecular basis of this kinetic signature, and evaluate whether such a drug could in principle be defeated by raising the substrate concentration in vivo.
Mark scheme decomposition by AO:
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