Ch 2 · 3.2 Factors that affect enzyme action · AS & A Level Biology 9700
Topic 2 of 2 in Enzymes — AS & A Level Biology 9700 — 68 lessons.
Introduction
The rate of an enzyme-catalysed reaction depends on the surrounding conditions and on the concentrations of enzyme, substrate and any inhibitor present. Cambridge examiners expect each effect to be explained mechanistically, in terms of molecular collisions, active site shape and enzyme-substrate complex formation, not simply described as a graph shape. (Syllabus section 3.2.)
Key ideas — Effect of temperature
As temperature rises, both enzyme and substrate molecules gain kinetic energy, move faster and collide more frequently, so more enzyme-substrate complexes form per second and the rate of reaction increases, up to an optimum temperature.
🧑🏫 Teacher's explanation
Think of an enzyme as a specialized key and the substrate as the lock. At low temperatures, the key and lock are moving very slowly in a room; they rarely bump into each other, so the door (the reaction) rarely opens. As you turn up the heat, everyone starts running around faster. They collide much more often, making the reaction happen much faster.
📖 Definitions
🔍 Worked example — Easy
An enzyme-controlled reaction is measured at 10circC and 40circC. At 10circC, the rate is 2 units/min. At 40circC, the rate is 10 units/min. Calculate the percentage increase in the rate of reaction.
🔍 Worked example — Standard
Describe the effect of increasing temperature from 20circC to 45circC on an enzyme with an optimum temperature of 37circC.
🔍 Worked example — Exam-style
A student plots a graph of reaction rate vs temperature. The graph shows a gradual rise followed by a sudden, steep drop. Explain why the drop is steeper than the rise.
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🎯 Exam question
Key ideas — Effect of pH
Every enzyme has an optimum pH at which the rate is highest. Away from this optimum, changes in H⁺ concentration disrupt the ionic bonds and hydrogen bonds between charged and polar R groups that maintain tertiary structure, altering the shape of the active site so the substrate binds less well or not at all, reducing the rate. At extremes of pH the enzyme is denatured; unlike moderate deviations, this loss of activity is not reversed by returning to the optimum pH. In investigations, pH is controlled with buffer solutions, which resist changes in pH: set up identical tubes with buffers of, say, pH 3, 5, 7, 9 and 11.
🧑🏫 Teacher's explanation
Think of an enzyme like a specialized lock and a substrate like a key. For the key to turn the lock, the internal mechanism (the active site) must be shaped perfectly. Now, imagine the lock is made of delicate magnetic parts. If you change the electrical charge of the environment around the lock, those magnetic parts might shift or pull away from each other. This is exactly what happens with pH.
📖 Definitions
🔍 Worked example — Easy
An enzyme shows maximum activity at pH 7. If the pH is lowered to pH 3, describe the effect on the enzyme's structure and its rate of reaction.
🔍 Worked example — Standard
A student investigates the effect of pH on the rate of reaction of pepsin. The rates at pH 2, 4, 6, 8, and 10 are 0.8, 0.4, 0.1, 0.05, 0.01 mumol/min. Identify the optimum pH and explain the trend.
🔍 Worked example — Exam-style
Calculate the percentage decrease in reaction rate when the pH changes from 7 (rate = 1.2 × 10^-3 mol/s) to pH 4 (rate = 4.0 × 10^-4 mol/s).
⚠️ Common errors
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🎯 Exam question
(a) Identify the optimum pH for this enzyme. [1] (b) Explain why the rate of reaction decreases as the pH increases from 7 to 9. [3] [4 marks]
Key ideas — Effect of enzyme and substrate concentration
When substrate is in excess, increasing enzyme concentration increases the rate of reaction proportionally, because more active sites are available to bind substrate at any moment. When enzyme concentration is fixed and substrate is limiting, increasing substrate concentration increases the rate, since more frequent collisions between enzyme and substrate lead to more enzyme-substrate complexes forming per second; but as substrate concentration continues to rise, an increasing proportion of active sites are occupied at any instant, and the rate levels off at a maximum once all active sites are continuously occupied (saturated).
🧑🏫 Teacher's explanation
Think of an enzyme like a worker in a factory and the substrate as the raw material they need to process. If you have only one worker but a mountain of raw material, the worker can only work so fast; adding more workers (increasing enzyme concentration) directly increases the factory's output. However, if you have 100 workers but only 5 pieces of raw material, adding more workers won't help—the workers will just stand around waiting. This is because the raw material (substrate) is the limiting factor.
📖 Definitions
🔍 Worked example — Easy
An enzyme-controlled reaction is measured at different substrate concentrations. At 0.1 mol dm^-3, the rate is 2× 10^-3 mol dm^-3 s^-1. At 0.2 mol dm^-3, the rate is 4× 10^-3 mol dm^-3 s^-1. Calculate the percentage increase in rate.
🔍 Worked example — Standard
A graph shows the rate of reaction against substrate concentration. The rate increases linearly until it reaches a plateau at V_max = 0.8 mg s^-1. Describe the effect of increasing substrate concentration when the rate has reached V_max.
🔍 Worked example — Exam-style
In an experiment, the enzyme concentration is doubled while the substrate concentration is kept constant at a very low level. Predict and explain the effect on the rate of reaction.
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🎯 Exam question
Key ideas — Competitive and non-competitive inhibitors
A competitive inhibitor has a molecular shape similar to the substrate and binds reversibly to the active site itself, physically blocking the substrate from binding; because it competes directly with the substrate, its effect can be overcome by increasing substrate concentration sufficiently, so the maximum rate can still be reached, though a higher substrate concentration is needed to approach it. A non-competitive inhibitor binds to an allosteric site elsewhere on the enzyme, distinct from the active site; binding causes a conformational change that distorts the active site itself, so the substrate can no longer bind properly. Because this effect does not depend on competing with the substrate for the active site, increasing substrate concentration cannot overcome it, and the maximum possible rate is permanently reduced while the inhibitor is present.
🧑🏫 Teacher's explanation
Think of an enzyme as a lock and a substrate as the key. A competitive inhibitor is like a 'dummy key' that fits into the lock but won't turn; it sits in the keyhole, blocking the real key. However, if you have a massive pile of real keys, the chance of the dummy key getting in becomes much smaller. A non-competitive inhibitor is like someone gluing the lock's internal mechanism or changing the shape of the keyhole from the back. No matter how many real keys you bring, the lock is fundamentally broken and won't accept them.
📖 Definitions
🔍 Worked example — Easy
A reaction reaches a maximum rate (V_max) of 100 mumol/min. In the presence of a competitive inhibitor, the substrate concentration required to reach 50 mumol/min increases from 2 mM to 8 mM. Calculate the change in K_m.
🔍 Worked example — Standard
An enzyme-catalyzed reaction has a V_max of 50 units. When a non-competitive inhibitor is added, the reaction rate at saturating substrate concentrations drops to 20 units. Calculate the percentage decrease in V_max.
🔍 Worked example — Exam-style
Compare the effect of a competitive inhibitor and a non-competitive inhibitor on the V_max and K_m of an enzyme-catalyzed reaction.
⚠️ Common errors
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🎯 Exam question
(i) State the effect of the non-competitive inhibitor on the V_max of the reaction. (ii) Explain why increasing the substrate concentration does not reverse the effect of a non-competitive inhibitor. [3 marks]
Key ideas — Vmax and Km
Vmax is the maximum rate of reaction reached when substrate concentration is so high that all the enzyme's active sites are continuously occupied (saturated); it depends on the amount of enzyme present. Km, the Michaelis constant, is the substrate concentration at which the rate of reaction is exactly half of Vmax. A low Km indicates that the enzyme reaches half its maximum rate at a low substrate concentration, meaning it has a high affinity for its substrate; a high Km indicates a lower affinity, since a higher substrate concentration is needed to reach half-maximum rate. To find Km, first find Vmax from where the rate levels off, halve it, and read the substrate concentration at that rate from the graph.
🧑🏫 Teacher's explanation
Imagine you are running a busy coffee shop. The 'reaction rate' is how many coffees you serve per hour. The 'Vmax' is the absolute maximum number of coffees you can serve when your machines are running at full capacity and every barista is working—you simply cannot go faster because you have reached the limit of your equipment. The 'substrate' is the number of customers waiting in line.
📖 Definitions
🔍 Worked example — Easy
An enzyme-catalyzed reaction reaches a maximum velocity (V_max) of 200 \ μ mol \ min^-1. Calculate the substrate concentration required to reach half of this maximum velocity (K_m).
🔍 Worked example — Standard
A researcher observes that an enzyme has a V_max of 80 \ mg \ s^-1 and a K_m of 2 \ mM. If a competitive inhibitor is added, how will the new V_max and K_m compare to the original values?
🔍 Worked example — Exam-style
A non-competitive inhibitor is added to a reaction. The original V_max was 100 \ units/min and K_m was 0.5 \ mM. After adding the inhibitor, the new V_max is 50 \ units/min. What is the new K_m?
⚠️ Common errors
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🎯 Exam question
(a) Predict and explain the effect of the non-competitive inhibitor on the V_max and the K_m of the reaction. [3] (b) A competitive inhibitor is added instead. State how the V_max and K_m would change compared to the original reaction (without inhibitor). [2] [5 marks]
Key ideas — Immobilised enzymes
An immobilised enzyme is held in place on or inside an inert material, so it does not mix freely with the product. The standard school method uses alginate beads:
🧑🏫 Teacher's explanation
Imagine you are a chef trying to cook pasta. If you throw the pasta into a giant pot of boiling water, it's easy to cook, but once it's done, you have to strain it to separate the pasta from the water. Now, imagine if you cooked the pasta inside little mesh bags. It's slightly harder for the water to reach the pasta through the mesh, so it might take a little longer, but when you're finished, you just lift the bags out. The pasta is perfectly clean, and you can use those same bags again and again.
📖 Definitions
🔍 Worked example — Easy
A student uses lactase immobilised in calcium alginate beads to hydrolyse lactose into glucose. State two advantages of using this method compared to using free lactase in a solution.
🔍 Worked example — Standard
A reaction rate for a free enzyme is 0.5 mol L^-1 s^-1. When the enzyme is immobilised, the rate drops to 0.35 mol L^-1 s^-1. Calculate the percentage decrease in the rate of reaction.
🔍 Worked example — Exam-style
Explain why the rate of reaction is typically lower for an immobilised enzyme compared to a free enzyme at the same substrate concentration.
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🎯 Exam question
Worked example
Question. A gas syringe experiment measured the volume of oxygen produced by catalase acting on hydrogen peroxide at a range of substrate concentrations. At very high substrate concentrations, the rate of oxygen production reached a constant maximum of 40 mm³ s⁻¹. At a hydrogen peroxide concentration of 8 mmol dm⁻³, the rate was measured as 20 mm³ s⁻¹. State the value of Km for this reaction and explain your reasoning. [3]
🔍 Worked example — Easy
An enzyme-catalysed reaction has a V_max of 100 mumol dm^-3 s^-1. Calculate the substrate concentration required to reach a rate of 50 mumol dm^-3 s^-1.
🔍 Worked example — Standard
Enzyme A has a K_m of 2 mM for substrate X. Enzyme B has a K_m of 10 mM for substrate X. Compare the affinity of Enzyme A and Enzyme B for substrate X.
🔍 Worked example — Exam-style
A gas syringe experiment measured the volume of oxygen produced by catalase acting on hydrogen peroxide at a range of substrate concentrations. At very high substrate concentrations, the rate of oxygen production reached a constant maximum of 40 mm^3 s^-1. At a hydrogen peroxide concentration of 8 mmol dm^-3, the rate was measured as 20 mm^3 s^-1. State the value of K_m for this reaction and explain your reasoning.
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