Cambridge International AS & A Level Biology 9700: Mode of action of enzymes – Revision Notes

These notes condense 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: examined in Papers 1 and 2, with its practical skills tested in Paper 3, and assumed knowledge for Paper 4. For full explanations and worked examples, read the 3.1 study guide first.

Course pages: Cambridge A Level Biology hub and printable Biology checklist. For the whole of Topic 3, including section 3.2, see the Enzymes revision notes.

The four learning outcomes

1. State that enzymes are globular proteins acting inside cells (intracellular) or secreted to act outside cells (extracellular).
2. Explain mode of action: active site, enzyme–substrate complex, lowering of activation energy, specificity, lock-and-key and induced fit.
3. Investigate progress: rate of formation of product with catalase; rate of disappearance of substrate with amylase.
4. Outline the use of a colorimeter for reactions that involve a colour change.

Key terms

Term	Definition to learn
Enzyme	Globular protein that acts as a biological catalyst
Globular protein	Compact, roughly spherical, usually soluble; tertiary structure gives a precise shape
Intracellular enzyme	Catalyses reactions inside the cell that made it
Extracellular enzyme	Secreted from the cell; catalyses reactions outside it
Substrate	The molecule an enzyme acts on
Active site	Region of the enzyme where the substrate binds; shape and R groups complementary to the substrate
Enzyme–substrate complex	Enzyme and substrate temporarily bound together at the active site
Activation energy	Energy that must be supplied before a reaction can start
Specificity	Each enzyme catalyses one reaction (or a few very similar ones) because only a complementary substrate fits its active site

Examples to quote

Type	Example	Where it acts
Intracellular	Catalase	Inside cells, breaking down hydrogen peroxide
Intracellular	DNA polymerase	Nucleus, during DNA replication
Extracellular	Amylase	Mouth (salivary) and small intestine (pancreatic)
Extracellular	Enzymes secreted by fungi	On the food the fungus grows on

The deciding factor is where the enzyme acts. Every enzyme is made inside a cell.

Mode of action in five steps

1. Substrate collides with the enzyme and binds to the active site (complementary shape and charge).
2. Temporary hydrogen and ionic bonds form: an enzyme–substrate complex.
3. Activation energy is lowered: substrates held close in the right orientation (building reactions) or bonds strained (breaking reactions).
4. Substrate converted to product: enzyme–product complex.
5. Product released; the enzyme is unchanged and the active site is free again.

The enzyme lowers activation energy only. The energy of substrates and products, and so the overall energy change, stays the same.

Lock-and-key vs induced fit

Feature	Lock-and-key	Induced fit
Active site shape	Fixed, rigid	Flexible
Before binding	Exactly complementary	Nearly complementary
On binding	No change	Active site changes shape slightly to mould around the substrate
Explains specificity?	Yes	Yes
Explains bond strain?	No	Yes: moving R groups strain bonds in the substrate
Explains a small range of similar substrates?	No	Yes

One-line distinction: in induced fit, the active site changes shape when the substrate binds; in lock-and-key, it does not.

Must-know distinctions

• Complementary shape, not “the same” shape.
• Intracellular vs extracellular: decided by site of action.
• Lowers activation energy vs “adds energy”: enzymes never supply energy.
• Product formation vs substrate disappearance: catalase is followed by measuring O₂ formed; amylase by measuring starch remaining.
• Initial rate vs mean rate: initial rate is the gradient of a tangent at t = 0; a mean rate is change ÷ time over an interval.
• Absorbance vs transmission: as a coloured substance becomes more concentrated, absorbance rises and transmission falls.

Method box: catalase (product formed)

2H₂O₂ → 2H₂O + O₂

• Source: potato, sweet potato or mung beans, as an extract or as discs.
• Collect O₂ in a gas syringe or an inverted measuring cylinder over water; read the volume at fixed intervals.
• Alternative: time how long an extract-soaked filter-paper disc takes to rise through hydrogen peroxide.
• Plot volume against time; the curve is steepest at the start and levels off when the substrate is used up.
• Rate = volume ÷ time (cm³ s⁻¹). Initial rate = gradient of tangent at t = 0.

Method box: amylase (substrate disappearing)

• Starch + iodine = blue-black; no starch = iodine stays orange-brown.
• Drops of iodine in a spotting tile; add a drop of reaction mixture at fixed intervals.
• Record the first time the iodine stays orange-brown.
• Rate = 1/time (s⁻¹).
• Limits: sampling interval sets the uncertainty in the end point; colour is judged by eye.

Method box: colorimeter

1. Filter of the complementary colour (red for blue-black starch–iodine).
2. Zero with a blank (everything except the coloured substance).
3. Calibration curve: absorbance of known concentrations, plotted against concentration.
4. Sample the reaction at set times; treat each sample the same way; read absorbance.
5. Convert absorbance to concentration with the calibration curve; plot concentration against time.

Coloured substrate disappearing → absorbance falls. Coloured product forming → absorbance rises.

Worked reminders

• Mean rate: 12.0 cm³ of oxygen in the first 40 s → 12.0 ÷ 40 = 0.30 cm³ s⁻¹.
• Rate from end-point time: iodine stays orange-brown at 250 s → 1 ÷ 250 = 0.0040 s⁻¹.
• Calibration: if absorbance = 0.75 × concentration (g dm⁻³), an absorbance of 0.30 means 0.30 ÷ 0.75 = 0.40 g dm⁻³ of starch.

Quick self-test

1. Name the level of protein structure that gives an enzyme its active site shape.
2. Is salivary amylase intracellular or extracellular? Give a reason.
3. What is an enzyme–substrate complex?
4. An enzyme lowers the activation energy of a reaction from 64 kJ mol⁻¹ to 48 kJ mol⁻¹. Calculate the percentage reduction.
5. Does an enzyme change the overall energy change of the reaction it catalyses?
6. State two ways in which formation of an enzyme–substrate complex lowers activation energy.
7. Give one thing the induced-fit hypothesis explains that the lock-and-key hypothesis does not.
8. What product is measured when following catalase activity, and with what apparatus?
9. A student records that iodine stops turning blue-black at 250 s. Calculate the rate.
10. Why is a red filter used in a colorimeter for the starch–iodine colour?
11. What should a colorimeter be zeroed with when following the disappearance of starch?
12. Why is initial rate used to compare reactions, rather than total oxygen collected?

Answers

1. Tertiary structure (which depends on the primary structure).
2. Extracellular: it is secreted from the salivary glands and acts in the mouth, outside the cells that made it.
3. The enzyme and substrate temporarily bound together at the active site, held by temporary bonds such as hydrogen and ionic bonds.
4. (64 − 48) ÷ 64 × 100 = 25%.
5. No. Substrates and products have the same energy with or without the enzyme; only the activation energy changes.
6. Any two: holds substrates close together in the correct orientation; strains bonds in the substrate; charged R groups attract or repel parts of the substrate.
7. Any one: bond strain caused by the active site changing shape; how an enzyme can bind a small range of similar substrates; how binding elsewhere can change the active site.
8. Oxygen; a gas syringe (or an inverted measuring cylinder over water).
9. 1 ÷ 250 = 0.0040 s⁻¹ (4.0 × 10⁻³ s⁻¹).
10. The blue-black solution absorbs red light most strongly, so absorbance changes most with starch concentration.
11. A blank containing everything except the coloured starch–iodine complex, for example iodine solution and water.
12. The total oxygen depends on how much substrate was present; the initial rate shows how fast the enzyme works before substrate concentration falls.

Where marks are usually lost

• Writing “same shape” instead of complementary shape for active site and substrate.
• Saying the substrate changes shape in induced fit, instead of the active site changing shape slightly.
• Stating that enzymes “provide energy” or “reduce the energy of the reaction” rather than lower the activation energy.
• Classifying catalase as extracellular because it was extracted from potato tissue.
• Naming the enzyme and method but not the quantity measured: oxygen volume for catalase, starch remaining for amylase.
• Giving a mean rate when the question asks for an initial rate from a tangent.
• Missing units: cm³ s⁻¹ for gas volume per time, s⁻¹ for 1/time.
• Saying iodine “turns orange-brown” instead of stays orange-brown.
• Forgetting the calibration curve, so absorbance readings cannot be turned into concentrations.

Next steps

• 3.1 practice questions
• 3.1 study guide
• Enzymes study guide and Enzymes practice questions
• 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.