Cambridge International AS & A Level Biology 9700: Mode of action of enzymes – Study Guide

This guide teaches section 3.1, Mode of action of enzymes, of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027. It is AS Level content, so it is examined in Paper 1 (Multiple Choice) and Paper 2 (AS Level Structured Questions), and the practical skills behind it are tested in Paper 3 (Advanced Practical Skills). Paper 4 at A Level also expects you to know AS content. It goes deeper into 3.1 than the Enzymes topic overview.

Course pages: Cambridge A Level Biology hub and printable Biology checklist. When you have read this guide, use the 3.1 revision notes and the 3.1 practice questions.

What section 3.1 covers

Outcome	What you must be able to do
3.1.1	State that enzymes are globular proteins that catalyse reactions inside cells (intracellular) or are secreted to catalyse reactions outside cells (extracellular)
3.1.2	Explain mode of action: active site, enzyme–substrate complex, lowering of activation energy, specificity, lock-and-key and induced-fit hypotheses
3.1.3	Investigate the progress of reactions by measuring the rate of formation of product (catalase) and the rate of disappearance of substrate (amylase)
3.1.4	Outline the use of a colorimeter to follow reactions that involve a colour change

The syllabus assumes you know that an enzyme is a biological catalyst that is unchanged when the reaction is complete. Section 3.2 (factors affecting rate, Km, inhibitors, immobilised enzymes) is not taught here.

3.1.1 Enzymes are globular proteins

An enzyme is a globular protein. Its polypeptide chain (or chains) folds into a compact, roughly spherical tertiary structure, held by hydrogen bonds, ionic bonds, disulfide bonds and hydrophobic interactions between R groups. Hydrophilic R groups lie mostly on the outside, so most enzymes are soluble in water. A small region of the surface, the active site, has a precise shape and set of R groups. That shape comes from the tertiary structure, which depends on the primary structure.

Enzymes are grouped by where they act:

• Intracellular enzymes catalyse reactions inside the cell that made them. Examples: catalase, which breaks down hydrogen peroxide (a toxic by-product of metabolism) inside cells; the enzymes of respiration; DNA polymerase in the nucleus.
• Extracellular enzymes are secreted by the cell (by exocytosis) and catalyse reactions outside it. Examples: amylase secreted by the salivary glands and pancreas, which hydrolyses starch in the mouth and small intestine; enzymes secreted by fungi onto the food they grow on.

The test is where the enzyme acts, not where it is made.

3.1.2 How enzymes work

Active site and enzyme–substrate complex

The substrate is the molecule an enzyme acts on. It binds to the active site, where its shape and charges are complementary to the active site. Temporary bonds (hydrogen and ionic bonds) form between the substrate and R groups in the active site. The result is an enzyme–substrate complex (ES complex).

The reaction happens while the substrate is held in the active site. The substrate is converted to product, giving a brief enzyme–product complex. The product no longer fits as well, so it is released. The active site is then free to bind another substrate molecule. In summary:

E + S  →  ES  →  EP  →  E + P

Lowering activation energy

Activation energy is the energy that must be supplied to substrate molecules before a reaction can start. Enzymes lower the activation energy. On an energy diagram, the catalysed pathway has a lower “hill” between substrates and products. The energy of the substrates and of the products is the same with or without the enzyme, so the overall energy change of the reaction is not altered. The enzyme only changes how much energy is needed to get the reaction going.

Ways the ES complex lowers activation energy:

• In building reactions (for example condensation), the active site holds two substrate molecules close together in the correct orientation, so bonds form more easily.
• In breaking reactions (for example hydrolysis), binding distorts the substrate, putting strain on particular bonds so they break more easily.
• Charged R groups in the active site attract or repel parts of the substrate, helping bonds form or break.

With a lower activation energy, a much larger proportion of substrate molecules can react at a given temperature, so the rate is far higher.

Specificity

Each enzyme catalyses one reaction, or a small group of very similar reactions. This specificity follows from the active site: only a substrate with a complementary shape and charge distribution can bind and form an ES complex. For example, amylase hydrolyses the α-1,4 glycosidic bonds in starch but not the β-1,4 glycosidic bonds in cellulose, because the shape of a cellulose chain is not complementary to amylase’s active site.

Lock-and-key hypothesis

In the lock-and-key hypothesis, the active site has a fixed, rigid shape. The substrate (the key) fits exactly into the active site (the lock). Only one shape of substrate fits, which explains specificity.

Induced-fit hypothesis

In the induced-fit hypothesis, the active site is flexible. Its shape before binding is close to, but not exactly, complementary to the substrate. As the substrate binds, the active site changes shape slightly, moulding around the substrate to give a closer fit. The movement of R groups puts strain on bonds in the substrate, which helps to lower the activation energy.

Induced fit explains things the lock-and-key model does not:

• how binding itself helps catalysis, by straining bonds;
• why some enzymes can bind a small range of similar substrates;
• how a molecule binding elsewhere on an enzyme can change the shape of the active site (you will use this idea for non-competitive inhibitors in section 3.2).

The syllabus names both models. Be ready to describe each and say how they differ.

3.1.3 Following the progress of a reaction

The rate of an enzyme-catalysed reaction is the amount of product formed, or substrate used up, per unit time. You can follow progress in two ways, and the syllabus pairs each with an enzyme.

Catalase: rate of formation of product

Catalase catalyses the breakdown of hydrogen peroxide:

2H₂O₂  →  2H₂O + O₂

The product you measure is oxygen gas. Plant sources include potato, sweet potato and mung beans. Ways to measure the oxygen:

• a gas syringe, read at set intervals;
• downward displacement of water in an inverted measuring cylinder;
• timing how long a filter-paper disc soaked in extract takes to rise through hydrogen peroxide solution.

Worked example 1. Potato extract is added to hydrogen peroxide and the oxygen is collected in a gas syringe.

Time / s	0	20	40	60	80	100	120
Volume of O₂ / cm³	0.0	9.0	16.0	21.0	24.5	26.5	27.0

(a) Mean rate over the first 20 s:

rate = change in volume ÷ time = 9.0 ÷ 20 = 0.45 cm³ s⁻¹

(b) Mean rate from 60 s to 120 s:

(27.0 − 21.0) ÷ (120 − 60) = 6.0 ÷ 60 = 0.10 cm³ s⁻¹

(c) Initial rate. Plot volume against time and draw a tangent to the curve at t = 0. Suppose your tangent passes through (0, 0) and (20, 10.0):

gradient = 10.0 ÷ 20 = 0.50 cm³ s⁻¹

(d) Explanation. The rate is highest at the start, when substrate concentration is highest. As hydrogen peroxide is used up, fewer ES complexes form per second, so the curve flattens, then levels off when all the substrate has gone.

The initial rate is the best value to compare between experiments, because later rates are lowered by the falling substrate concentration.

Amylase: rate of disappearance of substrate

Amylase hydrolyses starch to maltose. Starch gives a blue-black colour with iodine solution; when no starch is left, iodine stays orange-brown. To follow the reaction:

1. Put a drop of iodine solution in each well of a spotting tile.
2. Mix amylase and starch suspension (both at the same temperature, with a buffer) and start a timer.
3. At fixed intervals, transfer a drop of the mixture to the next well.
4. Record the first time at which the iodine stays orange-brown. This is when the starch has disappeared.

The time taken is inversely related to rate, so you can calculate rate = 1/time.

Worked example 2. Samples are taken every 20 s. The 160 s sample is still blue-black; the 180 s sample stays orange-brown.

rate = 1 ÷ 180 = 0.0056 s⁻¹  (5.6 × 10⁻³ s⁻¹)

The real end point lies somewhere between 160 s and 180 s, an uncertainty of up to 20 s. As a percentage of 180 s this is about 11%. Shorter sampling intervals reduce it. Judging the colour by eye is also subjective, which leads to the colorimeter.

3.1.4 Using a colorimeter

A colorimeter measures how much light of a chosen colour a solution absorbs. Light from a lamp passes through a coloured filter, then through the sample in a cuvette, and reaches a photocell. The meter shows absorbance (or transmission, which falls as absorbance rises). For a coloured substance, absorbance rises with its concentration.

How to use it to follow an enzyme-catalysed reaction:

1. Choose a filter of the complementary colour to the solution, because the solution absorbs that colour most. For the blue-black starch–iodine colour, a red filter is used.
2. Zero (calibrate) the colorimeter with a blank cuvette that contains everything except the coloured substance, such as water or iodine solution with no starch.
3. Make a calibration curve: measure the absorbance of solutions of known concentration and plot absorbance against concentration.
4. Take samples of the reaction mixture at set times, treat them in the same way (for amylase, add each to the same volume of iodine solution) and read the absorbance.
5. Read each concentration from the calibration curve and plot concentration against time.

If a coloured substrate disappears, absorbance falls. If a coloured product forms, absorbance rises. The readings are numerical, not a judgement by eye.

Worked example 3. Calibration with starch and iodine:

Starch / g dm⁻³	0.0	0.2	0.4	0.6	0.8	1.0
Absorbance	0.00	0.16	0.32	0.48	0.64	0.80

The line is straight, so absorbance = 0.80 × concentration. Samples from an amylase reaction gave:

Time / min	0	2	4	6	8
Absorbance	0.80	0.56	0.38	0.26	0.18
Starch / g dm⁻³ (= A ÷ 0.80)	1.00	0.70	0.475	0.325	0.225

Mean rate of starch disappearance from 0 to 2 min:

(1.00 − 0.70) ÷ 2 = 0.15 g dm⁻³ min⁻¹

From 6 to 8 min: (0.325 − 0.225) ÷ 2 = 0.050 g dm⁻³ min⁻¹. The rate falls as starch is used up. After 8 min, 0.225 ÷ 1.00 × 100 = 22.5% of the starch remains.

Common errors

• Calling catalase extracellular because you extract it from potato. It acts inside the cells that make it.
• Writing that the substrate and active site have the “same shape”. They are complementary.
• Describing induced fit as the substrate changing shape. It is the active site that changes shape slightly as the substrate binds.
• Calling the total oxygen collected the rate. The final volume depends on how much substrate there was.
• For amylase, saying the iodine “turns orange-brown”. Iodine is orange-brown to start with; it stays orange-brown when starch has gone.

Next steps

• 3.1 revision notes for condensed recall and a self-test
• 3.1 practice questions with marked answers
• Enzymes practice questions and Enzymes revision notes for the whole topic, including 3.2
• AS Level Biology 9700 diagnostic

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, published by Cambridge Assessment International Education (Cambridge University Press & Assessment, September 2022). Topic 3, Enzymes, section 3.1 Mode of action of enzymes, learning outcomes 1 to 4.