IB DP Biology: Enzymes and metabolism: active sites, rates, inhibition (DP Biology) – Revision Notes

These revision notes condense IB Diploma Programme Biology section C1.1 (Enzymes and metabolism) from the IB Biology guide, first assessment 2025. C1.1.1 to C1.1.10 are SL and HL. C1.1.11 to C1.1.17 are HL only. For full explanations and worked examples, read the study guide.

Then test yourself with the practice questions. Course hub: IB DP Biology. Checklist: printable checklist. For a quick gap check, use the free 10-minute diagnostics.

Key definitions

Term	Meaning
Catalyst	Raises reaction rate and is unchanged at the end
Metabolism	The network of interdependent, interacting chemical reactions in an organism
Anabolism	Building larger molecules from smaller ones, by condensation
Catabolism	Breaking larger molecules into smaller ones, by hydrolysis or oxidation
Globular protein	Polypeptide folded into a compact three-dimensional shape
Active site	Small region, a few amino acids, where substrate binds and reacts
Induced fit	Enzyme and substrate both change shape on binding
Denaturation	Loss of three-dimensional structure; usually permanent
Activation energy	Energy needed to start a reaction
Initial rate	Rate at the start, when substrate concentration is highest

Core points (SL and HL)

Catalysts and metabolism (C1.1.1 to C1.1.3)

• Cell reactions would be too slow at body temperature without enzymes.
• Enzyme specificity means many enzymes are needed, and control of metabolism works through them.
• Anabolism examples: protein synthesis, glycogen formation, photosynthesis. Catabolism examples: hydrolysis in digestion, oxidation in respiration.

Structure and binding (C1.1.4 to C1.1.7)

• Active site properties come from a few amino acids, but the whole three-dimensional structure holds them in place.
• Induced fit: both enzyme and substrate change shape, which strains bonds in the substrate.
• A reaction needs a collision with the right orientation. One partner may be immobilised: a large substrate, or an enzyme embedded in a membrane.
• Specificity, then denaturation: shape and charge of the active site fit only certain substrates. Changing the structure changes the active site, and binding fails.

Effects on rate (C1.1.8)

Variable	Rising part of graph	Peak or plateau	Falling part
Temperature	More kinetic energy, more frequent and more energetic collisions	Optimum	Bonds break, enzyme denatured, active site lost
pH	Approach to optimum pH	Optimum pH	R group charges change, bonds disrupted, denaturation
Substrate concentration	More collisions with active sites	Plateau when all active sites are occupied	No fall (unless another factor changes)

Measuring rate (C1.1.9)

Quantity	Calculation
Rate from a change	amount of product / time
Rate from a time to finish	1 / time (or 1000 / time)
Initial rate	gradient at the start, or change over the first readings
Percentage change	(new − old) / old × 100

Activation energy (C1.1.10)

• Activation energy = energy of the peak minus energy of the reactants.
• An enzyme lowers the peak. Energies of reactants and products, and so the overall energy change, do not change.
• Bonds break in the substrate (energy needed) and form in the products (energy released).

HL only (C1.1.11 to C1.1.17)

• Intracellular: glycolysis (cytoplasm), Krebs cycle (mitochondrial matrix). Extracellular: digestion in the gut.
• Heat: metabolism is not 100% efficient, so heat is inevitable. Mammals and birds use it to hold body temperature constant.
• Pathways: glycolysis is linear; the Krebs cycle and Calvin cycle are cyclical.
• Feedback inhibition: the end product inhibits an early enzyme. In isoleucine synthesis, isoleucine inhibits threonine deaminase.

Must-know distinctions: the three inhibitor types

Feature	Competitive	Non-competitive	Mechanism-based
Binding site	Active site	Allosteric site	Active site
Resembles substrate?	Yes	No	Usually reacts with active site
Reversible?	Yes	Yes	No (irreversible)
Effect of extra substrate	Reduces the inhibition	Does not remove it	Does not remove it
Example	Statins	A reversible allosteric inhibitor	Penicillin and transpeptidase

Method in steps: deciding the inhibitor from data

1. Plot or tabulate rate against substrate concentration, with and without inhibitor.
2. Compare the gap at low and high substrate concentration, as a percentage of the uninhibited rate.
3. A gap that shrinks at high substrate means competitive inhibition.
4. A gap that stays roughly the same means non-competitive inhibition.
5. Link to the mechanism: where it binds, and why substrate does or does not outcompete it.

Worked reminder: reading an energy profile

Reactants at 30 kJ mol⁻¹. Uncatalysed peak at 70. Catalysed peak at 42. Products at 20.

• Activation energy without enzyme: 70 − 30 = 40 kJ mol⁻¹.
• Activation energy with enzyme: 42 − 30 = 12 kJ mol⁻¹.
• Overall change: 20 − 30 = −10 kJ mol⁻¹ with or without enzyme.

Worked reminder: percentage change from a rate table

A catalase-type experiment gives these rates in arbitrary units: 10 °C, 6; 20 °C, 12; 30 °C, 21; 40 °C, 15.

• 10 to 20 °C: (12 − 6) / 6 × 100 = 100% increase.
• 20 to 30 °C: (21 − 12) / 12 × 100 = 75% increase.
• 30 to 40 °C: (15 − 21) / 21 × 100 = about −29%, a fall.
• The optimum is above 20 °C and below 40 °C. Rate rises less steeply at the top of the rising part, and the fall after 30 °C suggests denaturation.

Describing and explaining graphs

Use the same three steps for every rate graph.

1. Describe with data: give the variable, the direction and one pair of values, for example “rate rises from 12 to 21 units between 20 and 30 °C”.
2. Explain with theory: collision theory for a rise, denaturation for a fall, saturation of active sites for a plateau.
3. Evaluate if asked: a smooth sketch graph is a model. Real data have scatter, limited readings and a coarse temperature interval, so the optimum can only be placed in a range.

Model answer: why the rate falls above the optimum

The enzyme gains more kinetic energy and vibrates more. Hydrogen and ionic bonds that hold the tertiary structure break. The active site changes shape, so substrate cannot bind. More enzyme molecules are denatured as temperature rises, so the rate falls.

Model answer: competitive against non-competitive

A competitive inhibitor has a shape like the substrate and binds reversibly to the active site, so the two compete and extra substrate reduces inhibition. A non-competitive inhibitor binds reversibly to an allosteric site and causes a conformational change to the active site, so substrate cannot be converted whatever its concentration.

Link to other topics

• Digestion and respiration are the catabolic examples; protein synthesis and photosynthesis are the anabolic ones.
• The pH effect depends on the properties of amino acid R groups, so it links to protein structure.
• The Krebs and Calvin cycles are the cyclical-pathway examples, so keep them in mind when you meet cell respiration and photosynthesis.

Quick self-test

1. Name the two types of reaction that make up metabolism and give one example of each.
2. State what is meant by induced fit.
3. Explain why a rise in temperature below the optimum increases rate.
4. A reaction produces 18 cm³ of gas in 45 s. Calculate the mean rate.
5. A reaction takes 25 s to turn a solution clear. Calculate the rate as 1000 / time.
6. A rate rises from 15 to 21 units. Calculate the percentage increase.
7. State why the rate levels off at high substrate concentration.
8. Outline why a denatured enzyme cannot catalyse its reaction.
9. (HL only) Distinguish how extra substrate affects a competitive and a non-competitive inhibitor.
10. (HL only) Explain how penicillin kills bacteria, and how resistance arises.

Answers

1. Anabolism, for example protein synthesis; catabolism, for example hydrolysis in digestion.
2. Both enzyme and substrate change shape slightly on binding.
3. Molecules have more kinetic energy, so collide more often and more collisions have enough energy to react.
4. 18 / 45 = 0.4 cm³ s⁻¹.
5. 1000 / 25 = 40 arbitrary units.
6. (21 − 15) / 15 × 100 = 40%.
7. Every active site is already occupied nearly all the time, so more substrate cannot raise the rate.
8. High temperature or extreme pH breaks bonds, the active site changes shape, and substrate no longer binds.
9. More substrate outcompetes a competitive inhibitor for the active site, so inhibition falls. A non-competitive inhibitor binds elsewhere, so extra substrate does not remove it.
10. It binds irreversibly to the active site of transpeptidase, which makes cross-links in the cell wall, so the wall is weakened. Resistance arises when a changed transpeptidase cannot bind penicillin, and these bacteria survive and reproduce.

Where marks are usually lost

• Writing that enzymes are “killed” or “die” rather than denatured.
• Saying the substrate “does not fit” without stating that the active site has changed shape.
• Describing only the rising part of a rate graph and ignoring the fall or the plateau.
• Explaining a plateau as “enzymes used up” rather than active sites saturated.
• Quoting a rate from the whole curve instead of the initial rate.
• Forgetting units, or forgetting to convert a time into a rate.
• Saying a competitive inhibitor is “blocked” by more substrate, instead of outcompeted.
• Claiming an enzyme changes the energy of the products.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. This page is a paraphrase aligned to section C1.1.