explain that the maximum rate of reaction (Vmax) is used to derive the Michaelis–Menten constant (Km), which is used to compare the affinity of different enzymes for their substrates

Learning Objective

Explain how the maximum rate of reaction (Vmax) is used to derive the Michaelis–Menten constant (Km), and show how Km can be used to compare the affinity of different enzymes for their substrates. In doing so, discuss quantitatively how substrate concentration, enzyme concentration, temperature, pH, inhibitors and immobilisation affect Vmax and/or Km.

1. Introduction to Enzyme Kinetics

  • Enzymes are biological catalysts that lower the activation energy of a reaction.
  • The rate of an enzyme‑catalysed reaction depends on:

    • the physical‑chemical environment (temperature, pH, etc.)
    • the interaction between enzyme and substrate (affinity, saturation)

  • When every enzyme molecule is bound with substrate the reaction proceeds at its theoretical maximum, Vmax. The relationship between Vmax, substrate concentration ([S]) and the Michaelis–Menten constant (Km) is described by the Michaelis–Menten equation (see Section 3).

2. Quantitative Effect of the Six Factors on Vmax and Km

2.1 Substrate concentration ([S])

  • Increasing [S] increases the initial rate (v) because more enzyme‑substrate (ES) complexes form.
  • When [S] ≫ Km, the enzyme is saturated and v approaches Vmax.
  • Effect on kinetic parameters: no change in either Vmax or Km; they are intrinsic properties of the enzyme–substrate pair.

2.2 Enzyme concentration ([E]total)

  • More enzyme molecules mean more active sites are available.
  • Mathematically, Vmax = kcat [E]total, where kcat (turnover number) is the number of substrate molecules converted per enzyme per second.
  • Effect on kinetic parameters:

    • ↑ Vmax (proportional to [E]total)
    • Km unchanged (affinity is an intrinsic property).

2.3 Temperature

  • Rising temperature increases kinetic energy → more frequent, energetic collisions → higher reaction rate.
  • Up to the optimum temperature:

    • ↑ Vmax (greater catalytic turnover)
    • Km may fall slightly (enhanced binding) or remain unchanged, depending on the enzyme.

  • Beyond the optimum, thermal denaturation distorts the active site:

    • ↓ Vmax (loss of catalytic activity)
    • ↑ Km (lower apparent affinity).

2.4 pH

  • Each enzyme has an optimum pH at which ionisable side‑chains in the active site are correctly charged.
  • Deviation from the optimum:

    • ↓ Vmax (reduced catalytic efficiency)
    • ↑ Km (weaker substrate binding) – the effect is usually more pronounced for acids or bases that alter the charge of key residues.

2.5 Inhibitors

Inhibition typeEffect on VmaxEffect on KmKey kinetic feature
CompetitiveNo changeIncreases (apparent Km larger)Line of LB plot intersect at the y‑axis; slope ↑
Non‑competitiveDecreasesNo changeLB plot intersect at the x‑axis; y‑intercept ↑
Uncompetitive* (A‑Level)DecreasesDecreasesLB plot produces parallel lines (same slope, higher y‑intercept).

*Uncompetitive inhibition is included in the A‑Level specification (Topic 3, sub‑topic 3.2).

2.6 Immobilised Enzymes

  • Enzyme molecules are physically trapped (e.g., entrapment in calcium alginate beads, covalent attachment to a solid support).
  • Typical kinetic consequences:

    • ↓ Vmax – diffusion of substrate into the matrix limits the rate at which ES complexes can form.
    • ↑ Km (apparent) – the effective substrate concentration at the active site is lower because of mass‑transfer resistance.

  • Advantages (relevant for industry and practical work):

    • Easy separation of enzyme from product → reusable catalyst.
    • Often increased thermal and pH stability because the support matrix protects the protein.
    • Facilitates continuous‑flow processes.

2.7 Summary Table – Direction of Change for Each Factor

FactorEffect on VmaxEffect on Km
Substrate concentrationnone (parameter)none (parameter)
Enzyme concentration↑ (proportional)none
Temperature (≤ optimum)≈ none or slight ↓
Temperature (≫ optimum)↓ (denaturation)↑ (lower affinity)
pH (optimal)maximalminimal
pH (away from optimum)
Competitive inhibitornone
Non‑competitive inhibitornone
Uncompetitive inhibitor
Immobilisation↓ (diffusion limit)↑ (apparent)

3. Michaelis–Menten Kinetics

The relationship between initial velocity (v) and substrate concentration ([S]) for many enzymes is:

v = Vmax [S] / (Km + [S])

  • Vmax – rate when every enzyme molecule is saturated with substrate.
  • Km – substrate concentration at which v = ½ Vmax. It provides a quantitative measure of affinity: lower Km → higher affinity.
  • From the definition of Vmax we have Vmax = kcat [E]total, where kcat is the turnover number (s⁻¹).

4. Determining Vmax and Km from Experimental Data

4.1 Michaelis–Menten Plot (v vs [S])

  1. Measure initial rates (v) for a series of substrate concentrations, keeping [E]total, temperature and pH constant.
  2. Plot v (y‑axis) against [S] (x‑axis). The curve approaches an asymptote – read this as Vmax.
  3. The substrate concentration at which the curve reaches half‑maximal height is Km.

4.2 Lineweaver–Burk (double‑reciprocal) Plot – Worked Example

Given the following data for an enzyme assay (enzyme concentration constant, temperature = 37 °C, pH = 7.0):

[S] (µM)v (µmol min⁻¹ mg⁻¹)
20.95
51.80
102.70
203.55
504.30
1004.80

  1. Calculate the reciprocals 1/[S] (µM⁻¹) and 1/v (min mg µmol⁻¹).
  2. Plot 1/v (y‑axis) against 1/[S] (x‑axis). Fit a straight line (least‑squares).
  3. From the linear equation 1/v = (Km/Vmax)·(1/[S]) + 1/Vmax obtain:

    • y‑intercept = 1/Vmax → Vmax = 1 / (y‑intercept)
    • x‑intercept = –1/Km → Km = –1 / (x‑intercept)

Using the data above (rounded to two decimal places):

1/[S] (µM⁻¹)1/v (min mg µmol⁻¹)
0.501.05
0.200.56
0.100.37
0.050.28
0.020.23
0.010.21

Linear regression gives the equation 1/v = 0.84·(1/[S]) + 0.20.

  • y‑intercept = 0.20 → Vmax = 1 / 0.20 = 5.0 µmol min⁻¹ mg⁻¹
  • x‑intercept = –0.84⁻¹ = –1.19 → Km = 1 / 1.19 ≈ 0.84 µM

Interpretation: The enzyme reaches half‑maximal velocity at a very low substrate concentration (0.84 µM), indicating high affinity.

Limitations of the Lineweaver–Burk Plot

  • Both axes are reciprocals; any experimental error at low [S] (large 1/[S] values) is magnified, giving disproportionate weight to the least reliable points.
  • Modern practice prefers non‑linear regression of the Michaelis–Menten equation, but the LB plot remains a required skill for the Cambridge A‑Level exam.

4.3 Eadie–Hofstee Plot (optional)

Plots v (y‑axis) against v/[S] (x‑axis). The slope equals –Km and the y‑intercept equals Vmax. It reduces the weighting problem of the LB plot because the axes are not reciprocals.

5. Inhibition – Kinetic Signatures

  • Competitive inhibition: inhibitor competes with substrate for the active site.

    • Vmax unchanged.
    • Apparent Km increases (requires more substrate to reach ½ Vmax).
    • On a LB plot the lines intersect on the y‑axis.

  • Non‑competitive inhibition: inhibitor binds a separate site, reducing the number of functional enzymes.

    • Vmax decreases.
    • Km unchanged.
    • LB lines intersect on the x‑axis.

  • Uncompetitive inhibition (A‑Level): inhibitor binds only to the ES complex.

    • Both Vmax and Km decrease proportionally.
    • LB plots give a set of parallel lines (same slope, higher y‑intercept).

6. Immobilised Enzymes – Practical Implications

  • Typical experiment: compare free catalase with alginate‑entrapped catalase.

    1. Prepare equal enzyme activities (e.g., same protein concentration) in free and immobilised forms.
    2. Measure initial rates at a range of H₂O₂ concentrations.
    3. Plot Michaelis–Menten and Lineweaver–Burk curves for each preparation.

  • Interpretation:

    • A right‑shift of the curve (higher apparent Km) indicates diffusion limitation.
    • A lower asymptote (lower Vmax) shows that the maximum catalytic turnover is reduced, often because not all immobilised enzyme molecules are accessible.

7. Practical Investigation – Mapping to the Syllabus

The following activity satisfies the Cambridge requirement to plan an experiment, collect data, analyse and evaluate (Topic 3, sub‑topic 3.2):

  1. Planning

    • Define the research question: “How does immobilisation affect the kinetic parameters of catalase?”
    • Identify variables:

      • Independent: enzyme form (free vs. immobilised), substrate concentration.
      • Dependent: initial rate (Δ[O₂] / Δt).
      • Controlled: temperature (25 °C), pH (7.0), enzyme concentration (standardised by protein assay).

    • Choose appropriate substrate concentrations (e.g., 0.5–10 mM H₂O₂) to cover the region below and above the expected Km.

  2. Data collection

    • Measure the volume of O₂ evolved in a fixed time interval for each substrate concentration.
    • Repeat each measurement at least three times to obtain an average and standard deviation.

  3. Data analysis

    • Convert O₂ volume to µmol min⁻¹ mg⁻¹ (using gas‑law conversion).
    • Construct Michaelis–Menten and Lineweaver–Burk plots for both enzyme forms.
    • Calculate Vmax and Km from the LB plots (or by non‑linear regression if software is available).

  4. Evaluation checklist

    • Sources of random error – timing of gas collection, temperature fluctuations.
    • Systematic error – inaccurate calibration of the gas syringe, incomplete mixing of substrate.
    • Specific to immobilisation – diffusion limitation within the bead, possible enzyme leakage.
    • Repeatability – have you performed replicates and obtained consistent values?
    • Improvement suggestions – use a stirring bath to minimise diffusion gradients, test different bead sizes.

8. Using Km to Compare Enzyme Affinity

Because Km is the substrate concentration required to reach half‑maximal velocity, it provides a direct, quantitative way to compare how tightly different enzymes bind the same substrate, or how one enzyme behaves with different substrates.

EnzymeSubstrateKm (µM)Vmax (µmol min⁻¹ mg⁻¹)Affinity interpretation
HexokinaseGlucose0.105.2Very high affinity (low Km)
GlucokinaseGlucose8.07.8Low affinity (high Km)
Lactate dehydrogenasePyruvate0.3012.4High affinity
Alginate‑entrapped catalase (immobilised)H₂O₂1.53.6Moderate affinity; Vmax lower because of diffusion barrier
Free catalaseH₂O₂0.96.8Higher affinity and higher Vmax than immobilised form

9. Summary (Key Points for Revision)

  • Vmax = kcat [E]total – the theoretical maximum rate when the enzyme is fully saturated.
  • Km = [S] at which v = ½ Vmax. Lower Km → higher substrate affinity.
  • Quantitative effects of the six factors:

    • Substrate concentration – no change in kinetic parameters.
    • Enzyme concentration – ↑ Vmax, Km unchanged.
    • Temperature – ↑ Vmax up to optimum; beyond optimum ↓ Vmax and ↑ Km.
    • pH – optimum gives maximal Vmax and minimal Km; deviations cause ↓ Vmax and ↑ Km.
    • Inhibitors – competitive (↑ Km), non‑competitive (↓ Vmax), uncompetitive (↓ both).
    • Immobilisation – typically ↓ Vmax and ↑ apparent Km because of diffusion limits.

  • Lineweaver–Burk plots allow extraction of Vmax and Km, but the double‑reciprocal transformation amplifies errors at low [S]; students should comment on this limitation.
  • Uncompetitive inhibition, though optional in some curricula, is part of the A‑Level specification and must be recognised.
  • Practical skills: design experiments, collect reliable data, construct kinetic plots, calculate Vmax and Km, and critically evaluate sources of error.
  • Comparing Km values provides a quantitative method for judging which enzyme has the stronger affinity for a given substrate.

Suggested diagram: Michaelis–Menten curve showing the relationship between substrate concentration and reaction velocity, with markers for Km (half‑maximal velocity) and Vmax (asymptote). A second curve for an immobilised enzyme should be drawn shifted to the right (higher apparent Km) and with a lower asymptote (lower Vmax).