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

This guide follows the AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015, with exams from June 2016 onwards. It covers section 3.1.4.2, “Many proteins are enzymes”, and Required Practical 1. The content is part of both the AS and the A-level course. The other statements in section 3.1.4 (general properties of proteins and protein structure) are taught on separate pages.

Use it with the revision notes and the practice questions. Your course page is the AQA A-level Biology hub, the printable checklist shows what to tick off, and the free diagnostics show which parts of the course need most work. For the wider topic see Biological Molecules.

What you must be able to do

Specification point	What it asks of you
Activation energy	State that each enzyme lowers the activation energy of its reaction
Induced-fit model	Describe and use the model
Active site	Link properties to the tertiary structure of the active site and the enzyme-substrate complex
Specificity	Explain why one enzyme catalyses one reaction
Factors	Enzyme concentration, substrate concentration, competitive and non-competitive inhibitors, pH, temperature
Models over time	Appreciate how models of enzyme action have changed
Wide role	Appreciate that enzymes catalyse intracellular and extracellular reactions
MS 0.5	Calculate pH from hydrogen ion concentration
Required Practical 1	Investigate the effect of a named variable on the rate of an enzyme-controlled reaction, including a tangent for the initial rate

Activation energy and how enzymes lower it

Chemical reactions need a push to start. Activation energy is the energy that must be supplied before reactants can change into products. An enzyme does not supply energy and does not change the overall energy released. It lowers the activation energy, so a larger fraction of molecules can react at body temperature.

It does this by holding substrate molecules in a position that favours the change:

• In a breakdown reaction, binding to the active site puts strain on a bond in the substrate, so the bond breaks more easily.
• In a joining reaction, the active site holds two substrates close together in the right orientation, which reduces the repulsion between them.

Enzymes are catalysts, so they are not used up. The same molecule can act on many substrate molecules in turn.

The active site and the enzyme-substrate complex

An enzyme is a globular protein. A small region on its surface, the active site, is where the substrate binds. The shape and charge pattern of the active site come from the enzyme’s tertiary structure, which is held by hydrogen bonds, ionic bonds, disulfide bridges and hydrophobic interactions between R groups. The substrate fits the active site because it is complementary to it. The substrate and enzyme form an enzyme-substrate complex. The reaction takes place, an enzyme-product complex forms briefly, and the product leaves.

Specificity. The sequence of amino acids (primary structure) fixes the tertiary structure, and that fixes the shape of the active site. Only substrates with a complementary shape bind. This is why amylase breaks down starch and does not act on a protein. A change to one amino acid in the active site can change its shape or charge, so the substrate no longer fits and the enzyme stops working.

How models of enzyme action have changed

In 1894 Emil Fischer proposed the lock-and-key model. The active site is a rigid shape and only the matching substrate fits, like a key in a lock. In 1958 Daniel Koshland proposed the induced-fit model, which is the one the specification asks you to use. The active site is not a perfect fit before the substrate arrives. On binding, the active site shifts slightly and wraps around the substrate. This puts strain on the substrate and so lowers the activation energy further.

A model explains evidence, and scientists change it when it cannot account for new observations. Lock and key pictures a rigid active site, which cannot account for strain on the substrate. Use induced fit in answers unless a question asks you to compare the two.

Enzymes inside and outside cells

Enzymes catalyse a wide range of reactions. Intracellular enzymes act inside the cell that made them. Catalase breaks down hydrogen peroxide, a harmful by-product of metabolism, and DNA polymerase copies DNA. Extracellular enzymes are secreted and act outside the cell. Amylase from the salivary glands and the pancreas, and trypsin from the pancreas, digest food in the gut. Together they link cell chemistry to whole-organism function: respiration enzymes supply energy to cells, and digestive enzymes release small molecules that the body can absorb.

Factors that affect the rate

Temperature

Raising temperature from a low value gives molecules more kinetic energy. Enzyme and substrate collide more often and with more energy, giving more enzyme-substrate complexes each second, and the rate rises. At the optimum the rate is highest. Above the optimum the enzyme vibrates more, and this breaks the hydrogen and ionic bonds holding the tertiary structure. The active site is altered, so the substrate does not fit. The enzyme is denatured. This is not reversible. A cold enzyme is slow but is not denatured.

pH

Hydrogen ions (H⁺) interact with charged R groups. A pH away from the optimum disrupts ionic and hydrogen bonds, so the tertiary structure and the active site change and the rate falls. Large changes denature the enzyme. Different enzymes have different optimum pH values, for example pepsin in the stomach and trypsin in the small intestine.

Enzyme concentration

With plenty of substrate, more enzyme means more active sites, so the rate is proportional to enzyme concentration. When the substrate runs short, extra enzyme has little to act on and the graph levels off.

Substrate concentration

At low substrate concentration the rate rises quickly, because collisions between enzyme and substrate become more frequent. The curve then flattens. Every active site is occupied at any instant, so enzyme concentration is now the limiting factor.

Competitive inhibitors

A competitive inhibitor is a molecule shaped like the substrate. It occupies the active site, so fewer enzyme-substrate complexes can form. Increasing the inhibitor concentration lowers the rate. Raising the substrate concentration reduces the effect, because substrate molecules compete better for the active sites, and the same maximum rate can be reached.

Non-competitive inhibitors

A non-competitive inhibitor attaches elsewhere on the enzyme, away from the active site. That alters the tertiary structure, so the active site is distorted and the substrate cannot bind. Adding more substrate does not overcome the effect, and the maximum rate falls. The higher the inhibitor concentration, the more enzymes are affected.

Worked example 1: effect of an inhibitor

An enzyme was tested with and without inhibitor X. Rates are in arbitrary units.

Substrate concentration	No inhibitor	Inhibitor X
1	2.5	1.0
2	4.0	1.8
4	5.7	3.1
8	7.3	4.7
16	8.4	6.4
32	9.1	7.8

Percentage reduction at concentration 2: (4.0 - 1.8) ÷ 4.0 × 100 = 55% (to 2 significant figures). At concentration 32: (9.1 - 7.8) ÷ 9.1 × 100 = 14%. The effect shrinks as substrate increases and the two curves are heading for the same maximum, so X is a competitive inhibitor. A non-competitive inhibitor would give the same percentage reduction at every concentration and a lower plateau.

pH calculations (MS 0.5)

pH = -log₁₀ [H⁺], where [H⁺] is the hydrogen ion concentration in mol dm⁻³. Each fall of one pH unit is a tenfold rise in [H⁺].

Worked example 2: a solution has [H⁺] = 2.5 × 10⁻³ mol dm⁻³. pH = -log₁₀(2.5 × 10⁻³) = 2.60.

Worked example 3: a buffer has pH 8.2. [H⁺] = 10⁻⁸·² = 6.3 × 10⁻⁹ mol dm⁻³.

Worked example 4: a medium changes from pH 7 to pH 5. [H⁺] rises from 1 × 10⁻⁷ to 1 × 10⁻⁵ mol dm⁻³, which is 100 times greater.

Required Practical 1

Task: investigate the effect of a named variable on the rate of an enzyme-controlled reaction. The variable can be temperature, pH, enzyme concentration, substrate concentration or inhibitor concentration.

Method A, measuring product (catalase). Mix hydrogen peroxide with a catalase source such as a yeast suspension. Collect the oxygen in a gas syringe or an inverted measuring cylinder and record the volume every 10 seconds. To vary temperature, bring both solutions to each temperature in a water bath before mixing.

Method B, timing a change. Time how long a protease takes to clear a cloudy milk suspension. The rate is 1 ÷ time.

Variables to control: volumes and concentrations of enzyme and substrate, pH (use a buffer unless pH is the variable), temperature (a water bath, unless temperature is the variable), and time allowed for equilibration. Repeat each value and calculate a mean.

Safety: wear eye protection, because hydrogen peroxide can irritate eyes and skin and enzymes can cause allergic reactions.

Worked example 5: initial rate from a tangent (MS 3.6)

The volume of oxygen collected from a catalase reaction was:

Time (s)	0	10	20	30	45	60	90	120	180
Volume (cm³)	0.0	3.7	6.8	9.4	12.7	15.2	18.6	20.8	22.8

The curve flattens because substrate is used up, so the gradient falls with time. The rate is greatest at time 0, when substrate concentration is highest and little product has built up. To find it:

1. Plot volume against time and draw a smooth curve.
2. Draw a straight line touching the curve at time 0, with the same slope as the curve there. This is the tangent.
3. Choose two points far apart on the tangent, for example (0, 0) and (30, 12.0).
4. Gradient = 12.0 ÷ 30 = 0.40 cm³ s⁻¹.

The mean rate over the first 60 s is 15.2 ÷ 60 = 0.25 cm³ s⁻¹. It is lower because the rate falls during the 60 s. The initial rate is the better value to compare between runs.

Worked example 6: timing and uncertainty

Time to clear the milk suspension: 240 s at 20 °C, 130 s at 30 °C, 70 s at 40 °C and 55 s at 50 °C. Rates are 1 ÷ time: 0.0042, 0.0077, 0.0143 and 0.0182 s⁻¹. The rate at 40 °C is 240 ÷ 70 = 3.4 times the rate at 20 °C. The enzyme did not clear the suspension at 60 °C within 10 minutes, which suggests denaturation.

(PS 3.3) A reading from a syringe has an uncertainty of ±0.5 cm³. A volume of 12.0 cm³ measured as a difference between two readings has an uncertainty of ±1.0 cm³, which is 1.0 ÷ 12.0 × 100 = 8.3%. Larger volumes give smaller percentage uncertainties.

Graphs (MS 3.2)

Plot the rate (dependent variable) on the y-axis and the variable you changed on the x-axis. For a continuous variable such as temperature or pH, use a line graph or scatter graph with a smooth curve. Use a bar chart if the x-axis shows separate categories.

Making enzyme dilutions

To make 10 cm³ of enzyme at 1.0% from a 4.0% stock: stock volume = (1.0 ÷ 4.0) × 10 = 2.5 cm³, plus 7.5 cm³ of water.

Common errors

• Writing “the enzyme is killed” instead of “denatured”. Enzymes are not alive.
• Saying the substrate “fits the active site exactly” with no mention of induced fit.
• Explaining a plateau at high temperature as “all active sites are full”. It is denaturation.
• Saying a non-competitive inhibitor “blocks the active site”. It binds elsewhere.

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.