Factors Affecting Enzyme Activity
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You should understand the induced fit model of enzyme-substrate interaction and be comfortable with basic thermodynamic concepts, specifically the distinction between activation energy and Gibbs free energy.
Why this matters
Enzymes are not static machines; their catalytic power is exquisitely sensitive to the physical and chemical environment.
In the previous lecture, we established that enzymes lower the activation energy barrier (\(E_a\)) to accelerate reactions. However, this catalytic efficiency is not an intrinsic, unchangeable property. It is a dynamic state dependent on temperature, pH, substrate concentration, and the presence of inhibitors. Understanding these factors is critical for predicting how enzymes behave in vivo, where conditions fluctuate, and in industrial applications where conditions are manipulated to optimise yield. This lecture quantifies how these environmental variables modulate the rate constant \(k_{\text{cat}}\) and the overall reaction velocity \(v\).
The Dynamic Nature of Catalysis
Think of an enzyme not as a rigid lock, but as a flexible spring. Its ability to perform work—converting substrate to product—depends on its structural integrity and the kinetic energy of the molecules around it. If the temperature is too low, the molecules move too slowly to collide effectively. If it is too high, the delicate non-covalent bonds holding the enzyme’s tertiary structure together break, causing the active site to lose its specific shape. Similarly, the ionisation state of amino acid side chains in the active site is dictated by the surrounding pH. If the pH shifts, the charges on these residues change, potentially disrupting the electrostatic interactions necessary for substrate binding or catalysis. Therefore, enzyme activity is a balance between sufficient molecular motion and structural stability.
Temperature and the Arrhenius Relationship
The effect of temperature on reaction rate is governed by the Arrhenius equation. As temperature increases, the fraction of substrate molecules with sufficient kinetic energy to overcome the activation energy barrier increases exponentially. This leads to a rise in the rate constant \(k\). However, enzymes are proteins, and their structure is maintained by weak intermolecular forces (hydrogen bonds, ionic interactions, hydrophobic effects). Above a certain threshold, known as the optimal temperature, these bonds break, leading to denaturation. The active site distorts, and the enzyme loses its catalytic ability. The net result is a bell-shaped curve: activity rises with temperature until the rate of denaturation begins to outpace the rate of catalysis, causing a sharp decline in activity.
Where \(A\) is the frequency factor, \(R\) is the gas constant, and \(T\) is the absolute temperature. Enzymes lower \(E_a\) but do not alter the overall \(\Delta G^{\circ}\) of the reaction. The temperature dependence remains, but the magnitude of the rate increase is amplified by the lower \(E_a\).
pH and Ionisation States
Enzymes have an optimal pH at which their active site residues are in the correct protonation state for catalysis. Deviations from this pH alter the charge distribution on the enzyme and substrate. For example, if a catalytic histidine residue must be unprotonated to act as a base, a drop in pH will protonate it, rendering it inactive. This effect is often described by a pH-rate profile, which is typically bell-shaped. The maximum activity occurs at the pH where the necessary ionisation states are simultaneously present. The sensitivity to pH varies among enzymes; pepsin, which functions in the stomach, has an optimal pH of approximately 2, whereas trypsin, functioning in the small intestine, has an optimal pH of approximately 8. This reflects the specific microenvironments in which these enzymes evolved to operate.
Substrate Concentration and Saturation Kinetics
At low substrate concentrations \([S]\), the rate of reaction \(v\) is directly proportional to \([S]\). This is because most enzyme active sites are empty, and the rate-limiting step is the collision between enzyme and substrate. As \([S]\) increases, more active sites become occupied, and the rate increases. However, at very high \([S]\), all enzyme molecules are saturated with substrate. The enzyme is working at its maximum capacity, and the rate reaches a plateau known as \(V_{\max}\). This relationship is described by the Michaelis-Menten equation. The parameter \(K_m\), the Michaelis constant, represents the substrate concentration at which the reaction rate is half of \(V_{\max}\). A low \(K_m\) indicates high affinity between the enzyme and substrate, as saturation is reached at low concentrations.
The concept
Enzymes and activation energy
Biological catalysts that speed reactions without being consumed.
A common misconception is that \(V_{\max}\) is a fixed property of an enzyme. In fact \(V_{\max} = k_{\text{cat}}[\mathrm{E}]_{\text{total}}\), so it grows in proportion to the amount of enzyme present; what characterise the enzyme itself (under given conditions of temperature and pH) are \(k_{\text{cat}}\) and \(K_m\). Separately, any real reaction slows over time as substrate is used up, which is why kinetics are measured as initial rates. C
Worked example
Problem. An enzyme has a \(K_m\) of \(2.0 \times 10^{-4}\ \mathrm{mol\,dm^{-3}}\) and a \(V_{\max}\) of \(5.0 \times 10^{-5}\ \mathrm{mol\,dm^{-3}\,s^{-1}}\). Calculate the initial reaction velocity \(v\) when the substrate concentration \([S]\) is \(1.0 \times 10^{-4}\ \mathrm{mol\,dm^{-3}}\).
- Identify the given values: \(K_m = 2.0 \times 10^{-4}\ \mathrm{mol\,dm^{-3}}\), \(V_{\max} = 5.0 \times 10^{-5}\ \mathrm{mol\,dm^{-3}\,s^{-1}}\), \([S] = 1.0 \times 10^{-4}\ \mathrm{mol\,dm^{-3}}\).
- Substitute these values into the Michaelis-Menten equation: \(v = \frac{V_{\max} [S]}{K_m + [S]}\).
- Calculate the numerator: \(V_{\max} [S] = (5.0 \times 10^{-5}) \times (1.0 \times 10^{-4}) = 5.0 \times 10^{-9}\ \mathrm{mol^2\,dm^{-6}\,s^{-1}}\).
- Calculate the denominator: \(K_m + [S] = (2.0 \times 10^{-4}) + (1.0 \times 10^{-4}) = 3.0 \times 10^{-4}\ \mathrm{mol\,dm^{-3}}\).
- Divide the numerator by the denominator: \(v = \frac{5.0 \times 10^{-9}}{3.0 \times 10^{-4}} = 1.67 \times 10^{-5}\ \mathrm{mol\,dm^{-3}\,s^{-1}}\).
Units check. The numerator has units of \(\mathrm{mol^2\,dm^{-6}\,s^{-1}}\) and the denominator has units of \(\mathrm{mol\,dm^{-3}}\), resulting in \(\mathrm{mol\,dm^{-3}\,s^{-1}}\), which is consistent with a rate of concentration change per time.
Three quick tests
- Why does the reaction rate plateau at high substrate concentrations?
Answer
All active sites are occupied (saturated), so the enzyme is working at its maximum turnover rate \(V_{\max}\). - If the pH is lowered significantly below the optimum, what happens to the ionisation state of a catalytic histidine residue that needs to be unprotonated?
Answer
It becomes protonated, losing its ability to act as a base and thus reducing catalytic activity. - What is the physical significance of \(K_m\) in terms of enzyme-substrate affinity?
Answer
A lower \(K_m\) indicates a higher affinity, as it represents the substrate concentration required to reach half of \(V_{\max}\).
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Every lecture and every concept in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.