AQA A-level Biology: Enzymes and Required Practical 1 (7402) – Revision Notes

Short recall notes for the AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015, section 3.1.4.2 and Required Practical 1. The study guide gives the full explanations and worked examples. Then test yourself with the practice questions. Tick off topics on the printable checklist, find your course at the AQA A-level Biology hub, and use the free diagnostics to find weak spots. A wider overview is in Biological Molecules.

Key definitions

Term	Meaning
Enzyme	A globular protein that acts as a biological catalyst
Activation energy	The energy needed to start a reaction
Active site	The region of the enzyme where the substrate binds
Enzyme-substrate complex	Enzyme and substrate bound together at the active site
Complementary	Having a shape (and charge pattern) that matches
Denaturation	Loss of the tertiary structure, so the active site changes shape
Specificity	Each enzyme binds one substrate, or a very small group, because of its active site shape
Intracellular enzyme	Works inside the cell that made it
Extracellular enzyme	Secreted and works outside the cell

How enzymes work

1. Substrate collides with the active site.
2. Substrate binds, forming the enzyme-substrate complex.
3. The active site changes shape slightly (induced fit), straining the substrate bonds or bringing two substrates together.
4. The reaction takes place at a lower activation energy.
5. Product leaves. The enzyme is unchanged and can be used again.

Tertiary structure fixes the shape of the active site. A different amino acid sequence can change the tertiary structure and the active site. That is why specificity is a property of the active site.

Models over time

Model	Date and scientist	Active site	Limit
Lock and key	Emil Fischer, 1894	Rigid, fixed shape	Does not explain strain on the substrate
Induced fit	Daniel Koshland, 1958	Flexible, moulds around the substrate	The model to use in answers

Intracellular and extracellular enzymes

• Intracellular: catalase breaks down hydrogen peroxide; DNA polymerase copies DNA.
• Extracellular: amylase (salivary glands and pancreas) and trypsin (pancreas) digest food in the gut.
• Enzymes control reactions from the cell level (respiration) to the whole organism (digestion).

Effects on rate

Factor	Effect on rate	Reason
Temperature, up to optimum	Rises	More kinetic energy, more collisions, more complexes
Temperature, above optimum	Falls sharply	Bonds in tertiary structure break, active site changes, denatured
Low temperature	Slow, not denatured	Little kinetic energy
pH away from optimum	Falls	Ionic and hydrogen bonds disrupted, active site changes
Enzyme concentration	Rises (excess substrate)	More active sites
Substrate concentration	Rises, then plateaus	Plateau: all active sites occupied, enzyme limits

Inhibitors

Feature	Competitive	Non-competitive
Shape	Similar to the substrate	Not like the substrate
Binds at	The active site	Another site on the enzyme
Effect on enzyme	Blocks the active site	Changes the tertiary structure, so the active site changes
Raise substrate concentration	Effect reduced, same maximum reached	No recovery, lower maximum
Raise inhibitor concentration	Rate falls	Rate falls

Method in steps: pH calculation (MS 0.5)

pH = -log₁₀ [H⁺] with [H⁺] in mol dm⁻³.

• From concentration to pH: [H⁺] = 4.0 × 10⁻⁴ gives pH = 3.40.
• From pH to concentration: [H⁺] = 10⁻ᵖᴴ. pH 3.5 gives 10⁻³·⁵ = 3.2 × 10⁻⁴ mol dm⁻³.
• One unit lower in pH means ten times more H⁺. Two units means 100 times.

Method in steps: Required Practical 1

1. Choose the independent variable (temperature, pH, enzyme concentration, substrate concentration or inhibitor concentration).
2. Control everything else: volumes, concentrations, pH, temperature, time.
3. Bring solutions to temperature in a water bath before mixing.
4. Measure a rate: volume of product against time, or time taken for a change (rate = 1 ÷ time).
5. Repeat and calculate a mean. Look for anomalies.
6. Plot rate against the independent variable.
7. Find an initial rate from a tangent at time 0.

Tangent reminder (MS 3.6)

Product concentration rises from 0 and the tangent at time 0 passes through (0, 0) and (20 s, 5.0 mmol dm⁻³). Initial rate = 5.0 ÷ 20 = 0.25 mmol dm⁻³ s⁻¹. Quote the unit as quantity per unit time. The mean rate over a longer period is lower, because the curve flattens.

Uncertainty and dilution

• Percentage uncertainty = (absolute uncertainty ÷ measured value) × 100. A reading of 25.0 cm³ with ±0.5 cm³ gives 2.0%.
• Dilution: stock volume = (required concentration ÷ stock concentration) × final volume. For 20 cm³ at 0.8% from a 2.5% stock: (0.8 ÷ 2.5) × 20 = 6.4 cm³ stock plus 13.6 cm³ water.
• Graphs: a line graph or scatter graph for a continuous variable, a bar chart for categories.
• Safety: eye protection, and take care with allergens and hydrogen peroxide.

Reading enzyme graphs

Graph	What to say
Rate against temperature	Rises to an optimum, then falls steeply. Explain both halves: kinetic energy first, denaturation second
Rate against pH	Rises to an optimum and falls on both sides. Name the bonds affected: ionic and hydrogen bonds
Rate against enzyme concentration	A straight line through the origin while substrate is in excess, then it levels off
Rate against substrate concentration	Steep at first, then a plateau at a maximum. Name the limiting factor on the plateau
Product against time	Steepest at the start, then flattens as substrate is used up
Substrate against time	Falls steeply, then levels at zero when the substrate has gone

Describe first with figures (for example “the rate doubled from 20 to 30 units”), then explain using the enzyme-substrate complex.

Worked reminders

Comparing rates from times. A suspension clears in 60 s at one temperature and in 30 s at another. Rate = 1 ÷ time, so 1 ÷ 60 = 0.017 s⁻¹ and 1 ÷ 30 = 0.033 s⁻¹. The second rate is twice the first.

Percentage change. A rate rises from 4.0 to 6.0 units: (6.0 - 4.0) ÷ 4.0 × 100 = +50%. A rate falls from 8.0 to 2.0 units: (2.0 - 8.0) ÷ 8.0 × 100 = -75%.

Percentage inhibition. The rate is 6.0 without inhibitor and 4.2 with it: (6.0 - 4.2) ÷ 6.0 × 100 = 30%. If this value stays at 30% at every substrate concentration, suspect a non-competitive inhibitor. If it falls as substrate rises, suspect a competitive one.

Answer frames

• Explain the effect of temperature: more kinetic energy, more frequent collisions, more enzyme-substrate complexes. Above the optimum, bonds holding the tertiary structure break, the active site is altered, the substrate is no longer complementary, the enzyme is denatured.
• Explain the effect of pH: change in H⁺ concentration, ionic and hydrogen bonds disrupted, tertiary structure and active site change, fewer complexes.
• Explain inhibition: say where the inhibitor binds, what happens to the active site, and whether more substrate helps.
• Explain specificity: primary structure, tertiary structure, shape of the active site, complementary to one substrate.

Must-know distinctions

• Lock and key and induced fit: rigid versus flexible active site.
• Denatured and inactive: cold gives a slow enzyme that recovers on warming. Heat gives permanent change.
• Competitive and non-competitive: location of binding, and whether more substrate helps.
• Mean rate and initial rate: the mean averages over a period, the initial rate is the gradient at time 0.
• Substrate-limited and enzyme-limited: the plateau on a substrate graph is enzyme-limited.

Quick self-test

1. State what is meant by activation energy.
2. Why is induced fit regarded as a better model than lock and key?
3. Calculate the pH of a solution with [H⁺] = 4.0 × 10⁻⁴ mol dm⁻³.
4. Calculate [H⁺] at pH 6.5.
5. How do you make 20 cm³ of 0.8% enzyme from a 2.5% stock?
6. A tangent at time 0 passes through (0, 0) and (20 s, 5.0 mmol dm⁻³). What is the initial rate?
7. How does raising substrate concentration affect a competitive inhibitor?
8. Why is an enzyme at 5 °C slow but not denatured?
9. An enzyme clears a suspension in 25 s. What is the rate?
10. Give one intracellular and one extracellular enzyme.

Answers

1. The energy that must be supplied before a reaction can happen.
2. The active site changes shape as the substrate binds, which strains the substrate bonds and lowers activation energy further. Lock and key treats it as rigid.
3. 3.40.
4. 10⁻⁶·⁵ = 3.2 × 10⁻⁷ mol dm⁻³.
5. 6.4 cm³ stock plus 13.6 cm³ water.
6. 5.0 ÷ 20 = 0.25 mmol dm⁻³ s⁻¹.
7. It reduces the inhibition, because substrate molecules out-compete the inhibitor for the active sites. The same maximum rate can be reached.
8. Molecules have little kinetic energy, so few collisions occur, but the tertiary structure is intact.
9. 1 ÷ 25 = 0.04 s⁻¹.
10. Catalase (intracellular), amylase (extracellular).

Where marks are usually lost

• Saying an enzyme “dies” or is “killed” by heat, when the word needed is denatured.
• Describing an active site as having the “same shape” as the substrate without saying complementary.
• Failing to mention tertiary structure when explaining denaturation or non-competitive inhibition.
• Saying a high temperature “kills” the collisions, instead of explaining bond breakage.
• Mixing up inhibitors: a non-competitive inhibitor does not bind the active site.
• Calling every plateau “denaturation” when it is a limiting factor.
• Giving a tangent gradient without units or with points too close together.
• Describing a control variable without saying how it is kept constant.

Official syllabus

AQA AS and A-level Biology specification (7401, 7402), for first teaching 2015, published by AQA. Section 3.1.4.2 and Required Practical 1.