Cambridge International AS & A Level Biology 9700: Factors that affect enzyme action – Study Guide

This study guide teaches subtopic 3.2, Factors that affect enzyme action, from the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027. It is AS Level content (Topic 3, Enzymes). The syllabus bases Paper 1 (Multiple Choice) and Paper 2 (AS Level Structured Questions) on AS content, and Paper 4 expects you to bring AS knowledge with you. The investigations also build skills for Papers 3 and 5, which can set contexts outside the syllabus content.

How catalysis works (subtopic 3.1) is in the Enzymes study guide.

What this subtopic covers

Syllabus 3.2	What you must be able to do	Stage
1	Investigate and explain the effects of temperature, pH (using buffer solutions), enzyme concentration, substrate concentration and inhibitor concentration on the rate of enzyme-catalysed reactions	AS
2	Explain that Vmax is used to derive the Michaelis–Menten constant (Km), and that Km is used to compare the affinity of different enzymes for their substrates	AS
3	Explain the effects of reversible inhibitors, both competitive and non-competitive, on enzyme activity	AS
4	Investigate the difference in activity between an enzyme immobilised in alginate and the same enzyme free in solution; state the advantages of immobilised enzymes	AS

Measuring the rate first

Every factor in 3.2 is judged by its effect on rate.

• Initial rate. Plot product formed against time, draw a tangent at t = 0 and find its gradient. At t = 0 substrate has not yet been used up, so only your independent variable differs between tubes.
• Rate as 1/t. If you time a reaction to a fixed end point (such as starch no longer giving a blue-black colour with iodine), rate is proportional to 1/t.

Worked example 1: rate from a tangent

Catalase from potato extract is added to hydrogen peroxide and the oxygen is collected in a gas syringe. A tangent drawn at t = 0 passes through (0 s, 0 cm³) and (25 s, 16.0 cm³). A second tangent drawn at t = 60 s passes through (40 s, 22.0 cm³) and (80 s, 30.0 cm³).

initial rate  = 16.0 ÷ 25            = 0.64 cm³ s⁻¹
rate at 60 s  = (30.0 − 22.0) ÷ (80 − 40) = 8.0 ÷ 40 = 0.20 cm³ s⁻¹

The rate at 60 s is about 31% of the initial rate. The enzyme is not used up; there is less hydrogen peroxide left to collide with active sites.

Temperature

As temperature rises towards the optimum, enzyme and substrate molecules have more kinetic energy. They collide more often, so more enzyme–substrate complexes form per second and the rate increases.

Above the optimum, the enzyme molecule vibrates more strongly. Hydrogen bonds (and some ionic bonds) holding the tertiary structure break. The shape of the active site changes, so it is no longer complementary to the substrate. Fewer enzyme–substrate complexes form and the rate falls steeply. This loss of shape is denaturation, and it is usually permanent.

Investigating it. Use a thermostatically controlled water bath. Equilibrate enzyme and substrate in separate tubes before mixing. Standardise pH (buffer), concentrations and volumes.

pH

Each enzyme has an optimum pH. A change in hydrogen ion concentration changes the charges on the R groups of amino acids, for example whether a carboxyl group is –COOH or –COO⁻ and whether an amine group is –NH₂ or –NH₃⁺. This has two effects:

• ionic bonds and hydrogen bonds in the tertiary structure are disrupted, so the active site changes shape
• charged groups in the active site that normally attract the substrate may no longer do so

Either way, fewer enzyme–substrate complexes form. Extreme pH denatures the enzyme.

Buffer solutions. The syllabus states that pH is investigated using buffer solutions. A buffer keeps pH constant even if the reaction produces acidic or alkaline substances. For a pH investigation, use a different buffer (for example pH 4, 5, 6, 7, 8) in each tube. For every other factor, add the same buffer to every tube so pH is standardised.

Enzyme concentration

If substrate is in excess, the initial rate is directly proportional to enzyme concentration: doubling the number of active sites doubles the number of enzyme–substrate complexes formed per second. The graph is a straight line through the origin.

If the amount of substrate is limited, the line levels off at high enzyme concentrations, because now there are free active sites with no substrate to occupy them. Substrate concentration has become the limiting factor.

Worked example 2: making the concentrations

You have a 2.0% enzyme stock solution and want 10 cm³ of each of five concentrations. A proportional dilution uses volume of stock = (wanted ÷ stock) × total volume.

Concentration / %	Stock / cm³	Distilled water / cm³
0.4	2.0	8.0
0.8	4.0	6.0
1.2	6.0	4.0
1.6	8.0	2.0
2.0	10.0	0.0

A serial dilution (each step half the previous one) from the same stock gives 2.0, 1.0, 0.5, 0.25 and 0.125%. Proportional dilution gives evenly spaced values; serial dilution covers a wide range, which suits a curve.

Substrate concentration

At low substrate concentration, most active sites are empty. Adding substrate increases the collision rate, so the rate rises steeply: substrate concentration is the limiting factor.

As substrate concentration increases, more active sites are occupied at any instant. Eventually all active sites are occupied as fast as products are released. The enzyme is saturated, and adding more substrate has no further effect. The curve levels off at the maximum rate, Vmax. At this plateau, enzyme concentration is the limiting factor.

Vmax and the Michaelis–Menten constant, Km

The syllabus asks you to explain that Vmax is used to derive Km. The method:

1. Measure the initial rate at a range of substrate concentrations, enzyme concentration fixed.
2. Plot initial rate (y-axis) against substrate concentration (x-axis) and draw a smooth curve.
3. Estimate Vmax from the level the curve approaches.
4. Halve it to get ½Vmax.
5. Read across from ½Vmax to the curve, then down to the x-axis. That substrate concentration is Km.

Km is the substrate concentration at which the rate is half of Vmax. It compares how readily enzymes bind their substrate:

• low Km = high affinity — the enzyme reaches half its maximum rate at a low substrate concentration, so it forms enzyme–substrate complexes readily
• high Km = low affinity — far more substrate is needed to reach half the maximum rate

Worked example 3: finding Km

An enzyme gives these initial rates (arbitrary units):

Substrate concentration / mmol dm⁻³	0.5	1	2	4	8	16	32
Initial rate	6.7	11.4	17.8	24.6	30.5	34.6	37.1

The curve is still rising slowly at 32 mmol dm⁻³. Extending the curve, it levels off at a rate of about 40, so Vmax ≈ 40. Then ½Vmax = 20. A rate of 20 lies between the readings at 2 and 4 mmol dm⁻³; reading from the curve gives Km ≈ 2.5 mmol dm⁻³.

An enzyme with Km = 9 mmol dm⁻³ for the same substrate has a lower affinity.

Inhibitor concentration and reversible inhibitors

An inhibitor reduces the rate of an enzyme-catalysed reaction. The syllabus deals with reversible inhibitors: they bind to the enzyme by weak interactions and can leave again, so the enzyme recovers if the inhibitor is removed. As inhibitor concentration increases, a greater proportion of enzyme molecules is inhibited at any moment, so the rate falls.

Competitive inhibition

A competitive inhibitor has a shape similar to the substrate. It binds to the active site, preventing the substrate from binding while it is there. Inhibitor and substrate compete, so the effect depends on their relative concentrations:

• at low substrate concentration the inhibitor occupies a large share of active sites and the rate falls a lot
• at high substrate concentration substrate molecules out-compete the inhibitor and the rate approaches the uninhibited rate

Vmax is unchanged (it is still reached, but only at a much higher substrate concentration). Km increases, because more substrate is needed to reach ½Vmax, so the enzyme appears to have a lower affinity.

Non-competitive inhibition

A non-competitive inhibitor binds to the enzyme at a site other than the active site. This changes the shape of the enzyme, including the active site, so the substrate cannot bind or cannot be converted to product. Adding more substrate does not help because the inhibitor is not competing for the active site.

In effect, some enzyme molecules are taken out of action, so Vmax is lowered. The enzyme molecules that are not inhibited work normally, so Km is unchanged.

Worked example 4: telling them apart from data

Without inhibitor, a reaction runs at 30 units at low substrate concentration. With inhibitor P it runs at 12 units. At high substrate concentration, the uninhibited rate is 55 units and the rate with P is 50 units.

low substrate:  12 ÷ 30 × 100 = 40% of the uninhibited rate (a 60% reduction)
high substrate: 50 ÷ 55 × 100 ≈ 91% of the uninhibited rate

The inhibition is much smaller at high substrate concentration, so P is competitive. A non-competitive inhibitor would cut the rate by the same proportion at both.

Immobilised enzymes in alginate

Making the beads. Mix the enzyme with sodium alginate solution and drip it from a syringe into calcium chloride solution. Each drop forms a bead of calcium alginate gel with enzyme trapped inside. Rinse the beads with distilled water.

The comparison investigation. Set up the free enzyme and the immobilised enzyme so that the same quantity of enzyme acts on the same volume and concentration of substrate, at the same temperature and pH. For example, with amylase: add beads to buffered starch solution, and add the equivalent volume of free amylase solution to an identical tube. Take samples at intervals and test with iodine, recording the time for the blue-black colour to disappear. Or pack the beads in a column and run substrate through it, testing what comes out.

Worked example 5: comparing rates

Free amylase clears the starch in 50 s; the same quantity of amylase in alginate beads takes 125 s.

free:        1000 ÷ 50  = 20.0 s⁻¹ (×10⁻³)
immobilised: 1000 ÷ 125 = 8.0 s⁻¹ (×10⁻³)
immobilised rate = 8.0 ÷ 20.0 = 0.4 of the free rate

The immobilised enzyme is slower at first. Substrate has to diffuse into the bead and product has to diffuse out, and some active sites are less accessible inside the gel.

Advantages of immobilised enzymes (the syllabus asks you to state these):

• the enzyme is easily separated from the product, so the product is not contaminated with enzyme
• the enzyme can be recovered and reused, which reduces cost
• immobilised enzymes are usually more stable: they keep their activity over a wider range of temperature and pH, because the gel holds each molecule and limits unfolding
• substrate can flow continuously through a column, allowing a continuous process

Common errors

• Saying enzymes are “killed” at high temperature. Enzymes are not alive: say the active site changes shape and the enzyme is denatured.
• Forgetting to use a buffer in every tube when pH is not the independent variable.
• Reading Km at half the highest measured rate when the curve has not yet levelled off. Estimate Vmax from the plateau first.
• Writing that a competitive inhibitor lowers Vmax. It raises Km; Vmax is unchanged.
• Explaining non-competitive inhibition as “blocking” the active site. It binds elsewhere and changes the active site’s shape.
• Comparing free and immobilised enzyme without using the same quantity of enzyme in each.

Where next

• Condensed recall: Factors that affect enzyme action revision notes
• Data and planning questions: Factors that affect enzyme action practice questions
• The whole topic: Enzymes revision notes and Enzymes practice questions
• Check your AS knowledge with the 9700 AS diagnostic
• Course hub: Cambridge A Level Biology and the printable checklist

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, published by Cambridge International: Topic 3 Enzymes, subtopic 3.2 Factors that affect enzyme action, learning outcomes 1–4.