Cambridge International AS & A Level Biology 9700: Factors that affect enzyme action – Revision Notes

These notes condense 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), set on Papers 1 and 2 and assumed knowledge for Paper 4. For full explanations and worked examples, use the Factors that affect enzyme action study guide. For the mechanism of catalysis (subtopic 3.1), see the Enzymes revision notes.

The four learning outcomes

1. Investigate and explain the effects of temperature, pH (using buffers), enzyme concentration, substrate concentration and inhibitor concentration on rate.
2. Explain that Vmax is used to derive Km, and that Km compares enzymes’ affinity for their substrates.
3. Explain the effects of reversible inhibitors: competitive and non-competitive.
4. Investigate immobilised (alginate) vs free enzyme activity; state the advantages of immobilisation.

Measuring rate

Method	When to use it	How
Initial rate	You have a progress curve (product or substrate against time)	Gradient of the tangent at t = 0
1/t	You timed the reaction to a fixed end point	Rate ∝ 1/t (often written 1000/t)
Colorimeter	The reaction involves a colour change	Change in absorbance or transmission per unit time

Use initial rate because substrate concentration is the same in every tube at t = 0. Later, the rate falls as substrate is used up, not because the enzyme is used up.

The five factors at a glance

Factor	Graph of rate against the factor	Key explanation
Temperature	Rises to an optimum, then falls steeply	Below optimum: more kinetic energy, more collisions, more E–S complexes. Above: hydrogen and ionic bonds break, active site changes shape, enzyme denatured
pH	Peak at the optimum, falling either side	H⁺ concentration changes the charge on R groups; ionic and hydrogen bonds disrupted; active site shape and charges change
Enzyme concentration	Straight line through origin (substrate in excess); levels off if substrate is limited	More active sites, more E–S complexes per second
Substrate concentration	Rises steeply, then levels off at Vmax	Low: substrate is limiting. High: all active sites occupied (saturated); enzyme concentration is limiting
Inhibitor concentration	Rate falls as inhibitor concentration rises	A greater proportion of enzyme molecules is inhibited at any moment

Investigation checklist

• Independent variable: the one factor you change, with at least five values.
• Dependent variable: what you measure (volume of oxygen, time for iodine colour to go, absorbance).
• Standardised variables: everything else. For enzyme work that means temperature (water bath), pH (buffer), enzyme concentration, substrate concentration and volumes.
• Equilibrate enzyme and substrate separately at the test temperature before mixing.
• Replicates: at least three at each value so you can calculate a mean and spot anomalies.
• Control: boiled (denatured) enzyme in place of active enzyme shows the change is caused by enzyme activity.

Buffers. In a pH investigation, each tube gets a different buffer. In every other investigation, every tube gets the same buffer.

Dilutions

Type	Formula or pattern	Gives
Proportional	volume of stock = (wanted ÷ stock) × total volume; top up with distilled water	Evenly spaced concentrations
Serial	Each solution is diluted by the same factor (often ×½ or ×⅒) from the previous one	A wide range of concentrations

Method in steps: Km from Vmax

1. Plot initial rate (y) against substrate concentration (x); draw a smooth curve.
2. Estimate Vmax from where the curve levels off (not just the highest point plotted).
3. Calculate ½Vmax.
4. Read across from ½Vmax to the curve, then down: that x-value is Km.

Km = substrate concentration at which rate = ½Vmax. Low Km → high affinity (E–S complexes form readily, so half the maximum rate is reached at low substrate concentration). High Km → low affinity.

Small reminder: Vmax = 64 units, so ½Vmax = 32 units. If the curve passes 32 units at 1.8 mmol dm⁻³, then Km = 1.8 mmol dm⁻³.

Reversible inhibitors

	Competitive	Non-competitive
Shape	Similar to substrate	Not similar to substrate
Binds to	Active site	A site other than the active site
Effect on active site	Occupies it; substrate cannot enter	Changes its shape
More substrate?	Reduces the inhibition	No effect on the inhibition
Vmax	Unchanged	Lowered
Km	Increased (lower apparent affinity)	Unchanged
Inhibition at high vs low substrate	Much smaller at high substrate	Same proportion at both

Reversible means the inhibitor binds by weak interactions and can leave, so activity returns when it is removed.

Data test. Calculate the inhibited rate as a percentage of the uninhibited rate at a low and a high substrate concentration. Percentage much higher at high substrate → competitive. About the same → non-competitive.

Immobilised enzymes

Making beads: enzyme + sodium alginate solution, dripped from a syringe into calcium chloride solution → calcium alginate beads with enzyme trapped inside; rinse with distilled water.

Fair comparison with free enzyme: same quantity of enzyme, same substrate volume and concentration, same temperature and pH; measure rate the same way (for example, time for starch to stop giving a blue-black colour with iodine).

Typical findings and reasons

• Initial rate is often lower for beads: substrate must diffuse into the gel and product out; some active sites are less accessible.
• Beads keep more activity after exposure to high temperature or extreme pH: the gel holds each molecule and limits unfolding.
• Beads can be reused many times with little loss of activity.

Advantages to state: product not contaminated with enzyme; enzyme recovered and reused (cheaper); greater stability over a wider range of temperature and pH; allows a continuous process in a column.

Must-know distinctions

• Denatured vs inhibited: denaturation is a permanent change in tertiary structure; reversible inhibition ends when the inhibitor leaves.
• Vmax vs Km: Vmax is a rate; Km is a substrate concentration.
• Substrate-limited vs enzyme-limited: the rising part of a substrate curve is substrate-limited; the plateau is enzyme-limited.
• Initial rate vs mean rate: the tangent at t = 0 is not the same as total product ÷ total time.
• Serial vs proportional dilution: serial multiplies by a fixed factor each step; proportional mixes set volumes of stock and water.

Quick self-test

1. A reaction takes 80 s to reach its end point. Calculate the rate as 1/t, with units.
2. An enzyme has Vmax = 64 arbitrary units. What rate do you read across from to find Km?
3. Adding an inhibitor reduces a rate from 18 to 12 units. Calculate the percentage decrease.
4. How would you make 5.0 cm³ of 0.6% enzyme solution from a 1.5% stock?
5. List the concentrations from a serial dilution of 10% substrate, halving four times.
6. Enzyme X has Km 0.8 mmol dm⁻³; enzyme Y has Km 3.0 mmol dm⁻³ for the same substrate. Which has the higher affinity?
7. A graph of rate against enzyme concentration levels off. Suggest why.
8. State the effect of a competitive inhibitor on Vmax and on Km.
9. Explain why the rate falls above the optimum temperature even though collisions are more frequent.
10. Why is a buffer used when temperature is the independent variable?
11. Give one reason why alginate beads may give a lower initial rate than the free enzyme.
12. Name the two solutions used to make alginate beads.

Answers

1. 1/80 = 0.0125 s⁻¹ (or 1000/t = 12.5).
2. 32 units (½Vmax).
3. (18 − 12) ÷ 18 × 100 = 33% decrease.
4. Volume of stock = (0.6 ÷ 1.5) × 5.0 = 2.0 cm³ stock + 3.0 cm³ distilled water.
5. 10, 5, 2.5, 1.25, 0.625%.
6. Enzyme X (lower Km).
7. Substrate has become the limiting factor: there are free active sites but not enough substrate.
8. Vmax unchanged; Km increased.
9. Hydrogen and ionic bonds break, the active site changes shape (denaturation), so fewer E–S complexes form.
10. To keep pH constant, so any change in rate is caused only by temperature.
11. Substrate must diffuse into the bead (or some active sites are less accessible).
12. Sodium alginate and calcium chloride.

Where marks are usually lost

• Writing “the enzyme is killed” instead of “the active site changes shape so it is no longer complementary to the substrate”.
• Stating “rate increases with temperature” without the mechanism: kinetic energy, collision frequency, more E–S complexes.
• Explaining pH effects without mentioning the charges on R groups and the bonds holding tertiary structure.
• Saying the substrate curve levels off “because the enzyme is used up”. Active sites are saturated; enzymes are not used up.
• Reading Km at half the highest plotted rate when the curve has not levelled off.
• Defining Km as “half the maximum rate”. It is a substrate concentration, not a rate.
• Saying competitive inhibitors lower Vmax, or that non-competitive inhibitors change Km.
• Describing non-competitive inhibitors as blocking the active site.
• Comparing free and immobilised enzyme using different quantities of enzyme.
• Listing “cheaper” as an advantage of immobilisation without explaining that the enzyme can be reused.

Next: test yourself on data and planning questions in the Factors that affect enzyme action practice questions, or try the 9700 AS diagnostic. Course hub: Cambridge A Level Biology; 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.