Enzymes: Study Guide (Biology 3.1)

Why cells need catalysts

Your cells carry out thousands of chemical reactions every second. They break down food molecules, build proteins and copy DNA. Most of these reactions are energetically possible at body temperature, but on their own they would happen far too slowly to keep a cell alive. A sugar cube does not break down into carbon dioxide and water while it sits on a table, even though that reaction releases energy.

A catalyst is a substance that speeds up a reaction without being used up. Enzymes are the catalysts of living things. They are proteins, and each one catalyses one reaction, or a small group of very similar reactions. This topic explains how they speed reactions up, why each enzyme is so specific, and how you measure their effect.

Activation energy: the barrier every reaction must cross

For reactant molecules to change into products, some of their bonds must be stretched and broken before new bonds can form. That needs an input of energy at the start, even if the reaction releases more energy overall. This starting “energy hill” is the activation energy (Eₐ).

At body temperature only a tiny fraction of collisions between molecules carry enough energy to get over a high hill. Most collisions fail, and the reaction is slow.

Reading Figure 1.

• The height of the hill above the reactants is the activation energy. With the enzyme, the hill is much lower (in this model, 30 kJ mol⁻¹ instead of 80 kJ mol⁻¹, a drop of 50 kJ mol⁻¹).
• The difference between the start and the end is the free energy change, ΔG. Here the products have less free energy than the reactants (ΔG is negative), so the reaction can happen on its own; it is just slow without help.
• ΔG is the same on both curves. An enzyme changes the route, not the starting point or the destination. It cannot make an unfavourable reaction favourable, and it does not supply energy.

So an enzyme speeds up a reaction by lowering the activation energy. A much larger fraction of encounters between substrate and enzyme now lead to products, and the reaction runs millions of times faster in many cases.

Enzyme structure: the active site

An enzyme is a large protein folded into a precise three-dimensional shape (its tertiary structure, and sometimes quaternary structure, from Topic 1.7). Somewhere on its surface is a small pocket or groove called the active site. This is where the reactant molecule, called the substrate, binds and where the reaction happens.

The active site is formed by a few amino acids, often far apart in the sequence, that the folding brings close together. Their R groups line the pocket. These R groups decide two things:

• the shape of the pocket: its size, depth and outline;
• the chemical character of the pocket: which spots carry a positive or negative charge, which are polar and can form hydrogen bonds, and which are nonpolar.

For a reaction to happen, the substrate must be compatible with the active site in both shape and charge. A molecule that is the right shape but carries a positive charge where the active site also has a positive charge is repelled. A molecule with matching charges but the wrong size cannot fit. This double requirement is why most enzymes are highly specific.

The enzyme–substrate complex model

The steps of an enzyme-catalysed reaction can be written as:

E + S → ES → E + P

where E is the enzyme, S the substrate, ES the enzyme–substrate complex and P the products.

1. Binding. Substrate molecules move randomly in solution and collide with enzymes. When a substrate meets an active site in the right orientation, weak attractions (ionic attractions between opposite charges, hydrogen bonds, hydrophobic interactions) hold it in place. These are the same kinds of attraction that hold a protein’s shape together.
2. The complex. While bound, the substrate is held in a position that makes the reaction easier. The enzyme may hold two substrates close together and correctly lined up, stretch a particular bond, or place a charged R group next to the atoms that must react. Each of these lowers the activation energy.
3. Release. The products have a different shape and charge pattern from the substrate, so they no longer fit well and drift away. The enzyme is unchanged and the active site is free to bind another substrate.

Because the enzyme is reused, one enzyme molecule can catalyse the same reaction many thousands of times. A cell therefore needs only small amounts of each enzyme.

Background (not required): the active site is not perfectly rigid. When the substrate binds, the enzyme often changes shape slightly and grips the substrate more closely. This refinement is called induced fit. It still depends on shape and charge being compatible.

Enzymes build molecules as well as break them down. Some catalyse hydrolysis (splitting a polymer with water, as amylase does to starch); others catalyse dehydration synthesis (joining monomers and releasing water). The model in Figure 2 shows a breakdown reaction, but the same steps apply when two substrates are joined.

Enzymes regulate what happens in a cell

Without its enzyme, a reaction in a cell is so slow that it effectively does not happen. This turns enzymes into control points:

• A cell can switch a reaction on by making the enzyme (expressing its gene) and switch it off by not making it or by breaking the enzyme down.
• A cell can adjust how active an existing enzyme is, for example with molecules that bind to it and change its shape (you meet these in Topic 3.2).
• In a metabolic pathway, the product of one enzyme is the substrate of the next. Controlling one enzyme in the pathway controls the whole sequence.

This is why the structure of enzymes matters at the level of the whole organism. Different cell types make different sets of enzymes, so they carry out different reactions.

Measuring how fast an enzyme works

A reaction rate is the change in the amount of product (or substrate) per unit of time:

rate = change in amount ÷ time taken

Typical units are cm³ s⁻¹ (gas produced), mmol L⁻¹ min⁻¹ (concentration of product) or min⁻¹ (when you time how long a colour change takes and use 1 ÷ time as a relative rate).

On a graph of product against time, the rate is the gradient. At the start the curve is steepest: substrate is plentiful and active sites are busy. Later the curve flattens because the substrate is being used up: fewer substrate molecules are left to collide with active sites. When all the substrate has reacted, the curve is flat. The initial rate, measured over the first short period, is the fairest way to compare conditions, because every trial starts with the same substrate concentration. Topic 3.2 looks in detail at how substrate concentration and other conditions affect the rate.

Worked example 1: calculating rates with and without an enzyme

Question. Catalase is an enzyme found in most living tissues, with high levels in mammalian liver. It speeds up the breakdown of hydrogen peroxide, a harmful by-product of metabolism:

2H₂O₂ → 2H₂O + O₂

A class adds 1.0 cm³ of liver extract to 10 cm³ of hydrogen peroxide solution and collects the oxygen in an upturned measuring cylinder. A second tube receives 1.0 cm³ of water instead of extract. The fictional results are below and in Figure 3.

Time / s	0	30	60	90	120	150	180
With extract: O₂ / cm³	0.0	9.0	16.5	22.0	25.5	27.5	28.0
No extract: O₂ / cm³	0.0	0.1	0.2	0.3	0.4	0.5	0.6

(a) Calculate the initial rate with extract over the first 30 s, and the mean rate between 120 s and 180 s. (b) Calculate the initial rate without extract and compare the two initial rates. (c) Explain the shape of the curve with extract. (d) Predict the final volume of oxygen in the tube without extract if it were left for many days.

(a) Rates with extract.

1. Initial rate = (9.0 − 0.0) cm³ ÷ 30 s = 0.30 cm³ s⁻¹ (18 cm³ per minute).
2. Rate from 120 s to 180 s = (28.0 − 25.5) cm³ ÷ 60 s = 2.5 ÷ 60 = 0.042 cm³ s⁻¹, about 14% of the initial rate.

(b) Without extract.

1. Initial rate = 0.1 cm³ ÷ 30 s = 0.0033 cm³ s⁻¹.
2. Comparison: 0.30 ÷ 0.0033 ≈ 90. The extract makes the initial rate about 90 times faster. (The readings without extract are so small that this ratio is only a rough estimate.)

(c) Shape. At first the curve is steep because the peroxide concentration is high and catalase molecules bind substrate often. As peroxide is used up, collisions between substrate and active sites become less frequent, so the gradient falls. By 180 s nearly all the peroxide has been broken down and the curve levels off. Note that the catalase has not been used up: the rate falls because the substrate runs out.

(d) Prediction. About 28 cm³. The same amount of hydrogen peroxide gives the same amount of oxygen whether or not catalase is present. The enzyme only changes how fast that end point is reached, just as it changes the activation energy but not ΔG.

Check. Units: cm³ ÷ s gives cm³ s⁻¹, a rate. The gradient of the dotted line in Figure 3 matches the 0.30 cm³ s⁻¹ calculated. The true initial rate may be slightly higher, because the curve already bends a little before 30 s.

Worked example 2: choosing and justifying controls

Question. Germinating seeds break down their stored starch for energy. A student wants to test the claim that an extract of germinating barley seeds contains amylase, an enzyme that hydrolyses starch. She can use iodine solution, which turns blue-black with starch and stays yellow-brown when no starch is left. She sets up four tubes, each with 5 cm³ of 1% starch solution at 25 °C, and samples a drop every 30 s onto iodine:

Tube	Added (1 cm³)	Time until iodine stays yellow-brown / min (3 trials)
A	barley extract	5.5, 6.0, 6.5
B	barley extract boiled for 5 min, then cooled	still blue-black at 30 min (all trials)
C	distilled water	still blue-black at 30 min (all trials)
D	purchased amylase solution	3.5, 4.0, 4.5

(a) Identify the independent and dependent variables. (b) Explain the purpose of tubes B, C and D. (c) Calculate the mean time and a relative rate (1 ÷ mean time) for tubes A and D. (d) State a conclusion.

(a) Variables.

1. Independent variable: what is added to the starch (fresh extract, boiled extract, water or known amylase).
2. Dependent variable: the time taken for starch to disappear, judged by the iodine test.
3. Controlled variables: starch concentration and volume, volume added, temperature (25 °C), pH, and the sampling interval.

(b) Controls, and why each one is there.

1. Tube C, water (negative control). It shows whether starch disappears without any extract, for example by slow breakdown on its own or by dilution. Starch is still present at 30 min, so any change in tube A must come from the extract.
2. Tube B, boiled extract (negative control). Boiling destroys the shape of proteins (Topic 3.2 explains how). If the extract still worked after boiling, the starch breakdown might be caused by something other than an enzyme, such as acid in the extract. Boiled extract has no effect, so the active ingredient is heat-sensitive, as an enzyme would be.
3. Tube D, known amylase (positive control). It shows that the method can detect amylase activity: the starch, iodine and timing all work. Without it, a negative result in tube A could just mean the test had failed.

(c) Calculations.

1. Tube A mean = (5.5 + 6.0 + 6.5) ÷ 3 = 6.0 min. Relative rate = 1 ÷ 6.0 = 0.17 min⁻¹.
2. Tube D mean = (3.5 + 4.0 + 4.5) ÷ 3 = 4.0 min. Relative rate = 1 ÷ 4.0 = 0.25 min⁻¹.

A longer time means a slower rate, which is why 1 ÷ time is used.

(d) Conclusion. The data support the claim. Fresh extract removes starch, water does not, and boiling removes the effect, which points to a protein catalyst. The positive control shows the test works. The experiment does not prove the enzyme is amylase specifically; a further test could check that the product is maltose, the sugar amylase releases.

Common misconceptions

• “Enzymes are used up in the reaction.” The enzyme leaves each reaction unchanged and is reused. Reactions slow down because substrate runs out.
• “Enzymes provide the energy for a reaction.” They lower the energy barrier; they do not add energy, and ΔG stays the same.
• “An enzyme can make any reaction happen.” It only speeds up reactions that are already possible. It changes how fast equilibrium or completion is reached, not where the reaction ends up.
• “Only shape matters for binding.” Charge and polarity matter too. A substrate of the right shape is repelled if its charges clash with the active site.
• “The active site is the whole enzyme.” It is a small region; the rest of the protein holds the active site in the right shape.
• “Enzymes only break molecules down.” Many enzymes join molecules together, for example in dehydration synthesis.
• “Enzymes are alive.” They are molecules. They can work outside cells, for example in a test tube or in a biological washing powder.
• “A negative control and a positive control do the same job.” A negative control shows what happens without the factor being tested; a positive control shows that the method can detect a real effect.

Where this leads

Because an enzyme’s activity depends on its shape and charge, anything that changes the shape (temperature, pH, chemicals) or competes for the active site (inhibitors) changes the rate. That is Topic 3.2, Environmental Impacts on Enzyme Function. Test yourself with the practice questions, then use the revision notes and the checklist to consolidate.