How do temperature, pH, substrate concentration and enzyme concentration affect the rate of an enzyme-catalysed reaction?
Explain the effects of temperature, pH, substrate concentration and enzyme concentration on the rate of enzyme activity
A focused answer to the H2 Biology Energy and Equilibrium outcome on enzyme kinetics. The effect of temperature (including the optimum and denaturation), pH, substrate concentration (the plateau at saturation) and enzyme concentration, with the reasoning behind each graph.
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What this dot point is asking
SEAB wants you to describe and explain how four factors affect the rate of an enzyme-catalysed reaction: temperature, pH, substrate concentration and enzyme concentration. You need the shape of each graph and the reasoning behind it, in terms of collisions, the enzyme-substrate complex, saturation and denaturation. This is heavily tested in data and practical questions.
The answer
Temperature
As temperature rises, molecules gain kinetic energy, collide more often and with more energy, and form more enzyme-substrate complexes, so the rate rises to an optimum. Above the optimum the rate falls sharply because the enzyme denatures: increased vibration breaks the hydrogen and ionic bonds of the tertiary structure, the active site changes shape, and the substrate no longer fits.
pH
Each enzyme has an optimum pH at which its active site shape is ideal. Moving away from the optimum alters the charges on the R groups of the active site, disrupting the ionic and hydrogen bonds that hold its shape. The active site changes, the substrate binds less well, and the rate falls. Extremes of pH denature the enzyme.
Substrate concentration
At low substrate concentration the rate rises with concentration because more substrate is available to bind. The rate then plateaus because the active sites become saturated: nearly all are occupied at any moment, so adding more substrate cannot help, and enzyme concentration becomes limiting.
Enzyme concentration
Provided substrate is in excess, increasing the enzyme concentration increases the rate proportionally, because there are more active sites available to form complexes. If substrate is limited, the rate eventually plateaus when the substrate runs short.
Examples in context
Example 1. Stomach versus intestine enzymes. Pepsin works in the acidic stomach (optimum pH around 2), while intestinal enzymes work in slightly alkaline conditions (optimum pH around 8). Each enzyme's optimum pH matches the environment in which it operates, illustrating the pH effect physiologically.
Example 2. Initial rate measurement. Because product builds up and substrate is used up over time, the most reliable measure is the initial rate, taken from the steep early part of the curve. This is why experiments comparing enzyme activity measure the rate at the start, before substrate becomes limiting.
Try this
Q1. State what happens to an enzyme at a temperature well above its optimum. [1 mark]
- Cue. It denatures: its tertiary structure and active site shape are lost, so it can no longer bind substrate.
Q2. Explain why the rate of an enzyme reaction is low at a pH far from the enzyme's optimum. [2 marks]
- Cue. The change in pH alters the charges on R groups, disrupting the bonds that hold the active site shape, so the substrate binds less effectively and fewer complexes form.
Q3. A reaction has reached its plateau on a substrate-concentration graph. State what must be done to increase the rate further. [1 mark]
- Cue. Increase the enzyme concentration (add more enzyme), since active sites are saturated and enzyme is the limiting factor.
Exam-style practice questions
Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
Original5 marksDescribe and explain the effect of increasing temperature on the rate of an enzyme-catalysed reaction from low temperatures up to and beyond the optimum.Show worked answer →
The answer should describe the curve in two parts, with reasons.
As temperature rises from low values, the rate increases. This is because the enzyme and substrate molecules gain kinetic energy, move faster, and collide more frequently and with more energy, so more enzyme-substrate complexes form per unit time.
The rate reaches a maximum at the optimum temperature, where the rate of complex formation is greatest while the enzyme remains intact.
Above the optimum the rate falls sharply. The increased vibration of the molecule breaks the hydrogen and ionic bonds holding the tertiary structure, so the enzyme denatures. The active site changes shape, the substrate no longer fits, fewer complexes form, and the rate drops, eventually to zero.
Markers reward the increase due to greater kinetic energy and more frequent successful collisions, the optimum, and the fall due to denaturation with a change in active site shape.
Original4 marksExplain why the rate of an enzyme-catalysed reaction increases with substrate concentration at first but then levels off to a plateau.Show worked answer →
The answer should explain the rise and then the saturation.
At low substrate concentrations, increasing the substrate increases the rate because more substrate molecules are available to collide with and bind the enzyme active sites, so more enzyme-substrate complexes form per unit time.
As substrate concentration rises further, the rate levels off because the active sites become saturated: at any moment almost all the active sites are occupied. Adding more substrate cannot increase the rate because the enzyme concentration (the number of available active sites) is now the limiting factor.
Markers reward the increase due to more frequent binding, the plateau due to saturation of active sites, and the identification of enzyme concentration as the limiting factor at high substrate concentration.
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