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AS 3

Enzymes

AS Level · Topic 3 · Papers 1, 2 and 3

🎯What you need to be able to do

  • State that enzymes are globular proteins acting inside cells (intracellular) or secreted (extracellular).
  • Explain enzyme action in terms of the active site, the enzyme–substrate complex, lowered activation energy and specificity, using both the lock-and-key and induced-fit models.
  • Investigate reaction progress by measuring product formation with catalase and substrate disappearance with amylase.
  • Outline the use of a colorimeter to follow reactions involving a colour change.
  • Investigate and explain the effects of temperature, pH, enzyme concentration, substrate concentration and inhibitor concentration.
  • Explain how Vmax is used to derive Km, and what Km tells you about affinity.
  • Explain the effects of competitive and non-competitive reversible inhibitors, and state the advantages of immobilised enzymes.

📚The biology

What an enzyme does

Enzymes are globular proteins that act as biological catalysts. Some work inside the cell that made them (intracellular — catalase, the respiratory enzymes); some are secreted to act outside it (extracellular — amylase, protease).

An enzyme has an active site, a depression whose shape and charge distribution are complementary to one particular substrate. Substrate binds to form an enzyme–substrate complex; the reaction occurs; product is released and the enzyme is unchanged. The catalytic effect comes from lowering the activation energy — the energy barrier the reaction must cross — so a far greater proportion of collisions at a given temperature are successful.

Enzymes do not “provide energy” or change the equilibrium position. They lower the activation energy only. The overall energy change of the reaction is identical with or without the enzyme; only the rate changes. Writing that an enzyme “gives energy to the substrate” scores nothing.

Lock-and-key and induced fit

The lock-and-key hypothesis treats the active site as a rigid shape that only the correct substrate fits, which explains specificity simply. The induced-fit hypothesis is the modern refinement: the active site is flexible, and binding of the substrate causes the site to change shape slightly so that it moulds around the substrate. That conformational change strains the bonds in the substrate, which is a better account of how the activation energy falls, and it explains why some enzymes accept a small range of related substrates.

Both models explain specificity; only induced fit explains the mechanism. Say both if a question asks you to compare.

Following the rate of a reaction

Two directions, both examinable:

  • Product formation — catalase. Catalase breaks hydrogen peroxide into water and oxygen. Collect the oxygen in a gas syringe or by displacement and record volume against time.
  • Substrate disappearance — amylase. Amylase digests starch. Remove a drop at fixed intervals into iodine solution on a spotting tile; the blue-black fades and eventually the iodine stays orange-brown, at which point no starch remains.

A colorimeter converts a colour change into a numerical absorbance, removing the subjectivity of judging colour by eye and letting you plot a proper graph.

Whichever method, the curve of product against time is steep at first and flattens. Always take the initial rate, the gradient of the tangent at time zero, because that is the only point at which substrate concentration is the value you set. Later the substrate is being used up and the rate falls for a reason that has nothing to do with the variable you are testing.

The five factors

Temperature. Rate rises with temperature because molecules have more kinetic energy, so there are more frequent collisions with sufficient energy, more enzyme–substrate complexes per second. Above the optimum, increasing vibration breaks the hydrogen and ionic bonds holding the tertiary structure; the active site changes shape, is no longer complementary, and the enzyme is denatured. The fall is steep and, unlike the rise, irreversible.

pH. A narrow optimum. Away from it, excess H+ or OH ions interfere with the ionic bonds and hydrogen bonds of the tertiary structure and with the charges on R groups in the active site, so substrate no longer binds. Small deviations are reversible; large ones denature. This is why pH must be held with a buffer solution, not by adding acid at the start and hoping.

Enzyme concentration. With substrate in excess, rate is directly proportional to enzyme concentration — more active sites, more complexes per second. The line only levels off if substrate becomes limiting.

Substrate concentration. Rate rises steeply, then levels off at Vmax. At low concentrations the active sites are the limiting factor's opposite — there is spare capacity, so adding substrate adds rate. At high concentrations every active site is occupied as soon as it is free: the enzyme is saturated, and adding more substrate changes nothing.

Vmax and Km

Km, the Michaelis–Menten constant, is the substrate concentration at which the rate is half of Vmax. You read it off the graph: find Vmax from the plateau, halve it, go across to the curve, drop down to the x-axis.

\[ \text{at } [S] = K_{\mathrm{m}}, \quad v = \tfrac{1}{2}V_{\max} \]

The interpretation is the part that gets tested: a low Km means a high affinity for the substrate, because only a little substrate is needed to reach half maximum rate. A high Km means low affinity. Km therefore lets you compare two enzymes acting on the same substrate, or one enzyme's affinity for different substrates.

Inhibitors

  • Competitive. A molecule similar in shape to the substrate binds to the active site and blocks it. Substrate and inhibitor compete, so the effect depends on their relative concentrations. Raising substrate concentration far enough out-competes the inhibitor, so Vmax is still reached — it just takes more substrate. On the graph: the curve rises more slowly, Km appears higher, but the plateau is unchanged.
  • Non-competitive. The inhibitor binds elsewhere — at an allosteric site — and changes the shape of the enzyme, so the active site is no longer complementary. Substrate cannot out-compete it because they are not competing for the same place. Vmax is reduced and no amount of extra substrate restores it.

Both of these are reversible inhibitors, which is what the syllabus asks for. The distinguishing experiment is always the same: raise the substrate concentration and see whether the original Vmax comes back.

Immobilised enzymes

Enzymes can be trapped in beads of sodium alginate, made by dripping an enzyme–alginate mixture into calcium chloride solution. Compared with free enzyme in solution, an immobilised enzyme usually has a slightly lower rate, because the substrate must diffuse into the bead to reach the active sites. That cost buys real advantages:

  • the enzyme is easily recovered and reused, so it is cheaper;
  • the product is not contaminated with enzyme, so downstream purification is simpler — important for food and medical products;
  • the enzyme is more stable to changes in temperature and pH, because the matrix supports its tertiary structure;
  • the reaction can be run continuously by passing substrate through a column of beads.

✏️Worked example

An enzyme was assayed at a range of substrate concentrations. The initial rate rose to a plateau of 40 µmol min−1, and the rate was 20 µmol min−1 at a substrate concentration of 2.0 mmol dm−3. The assay was repeated with inhibitor P and with inhibitor Q. With P, the plateau was still 40 µmol min−1 but half-maximal rate needed 6.0 mmol dm−3. With Q, the plateau was 16 µmol min−1. (a) State Vmax and Km for the uninhibited enzyme. (b) Identify P and Q, with reasons. (c) A second enzyme acting on the same substrate has Km = 0.4 mmol dm−3. Which enzyme has the greater affinity?

(a) Vmax is the plateau, so Vmax = 40 µmol min−1. Km is the substrate concentration giving half of that, and half of 40 is 20, which the data say occurs at Km = 2.0 mmol dm−3.

(b) With P, Vmax is unchanged at 40 but Km has risen from 2.0 to 6.0 — three times as much substrate is now needed for half maximum rate. Unchanged Vmax means the inhibition can be overcome by adding substrate, so P is competitive: it binds the active site, substrate and inhibitor compete for it, and enough substrate wins.

With Q, Vmax has fallen from 40 to 16. No amount of substrate restores the original maximum, so Q is not competing for the active site. Q is non-competitive: it binds at an allosteric site and alters the shape of the active site, so those enzyme molecules are permanently out of action while it is bound.

(c) The second enzyme, with Km = 0.4 mmol dm−3, reaches half its maximum rate at one fifth of the substrate concentration our enzyme needs. A lower Km means a higher affinity, so the second enzyme binds the substrate more readily.

Check it. Test each conclusion by imagining the opposite experiment. For P, predict that at a very high substrate concentration the inhibited and uninhibited curves converge — and they do, since both plateau at 40. For Q, predict that they never converge, and they do not: 16 against 40 at every high concentration. If a data set showed both a lower Vmax and an unchanged Km, non-competitive is the only reading; if it showed a raised Km and unchanged Vmax, competitive is.
Reading Km as ‘the rate’ and treating high Km as good. Km is a substrate concentration, so its units are mmol dm−3, never µmol min−1. Quoting “Km = 20” here confuses it with the half-maximal rate and loses the mark outright. And the affinity relationship runs backwards from intuition: the smaller number is the better binder, because it takes less substrate to get halfway to maximum.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Explain why the rate of an enzyme-controlled reaction falls sharply above the optimum temperature but rises steadily below it.
Below the optimum, increasing temperature increases the kinetic energy of enzyme and substrate molecules, so collisions are more frequent and a greater proportion have the activation energy. More enzyme–substrate complexes form per second, so the rate rises. Above the optimum, the extra energy causes the enzyme molecule to vibrate enough to break the hydrogen and ionic bonds maintaining its tertiary structure. The active site changes shape and is no longer complementary to the substrate, so no complexes form. This is denaturation, and because the bonds do not re-form correctly on cooling it is effectively irreversible — which is why the fall is steeper than the rise.
2. A student measures the volume of oxygen released by catalase and hydrogen peroxide over 5 minutes, then divides the total volume by 5 to obtain the rate. Explain why this underestimates the true rate at the stated substrate concentration, and what they should do instead.
The curve of oxygen volume against time is steep at the start and flattens, because the substrate is used up as the reaction proceeds, so the rate falls progressively. Dividing the total by 5 gives a mean rate over the whole period, which is lower than the rate at the beginning and does not correspond to the substrate concentration the student set up. They should measure the initial rate: plot volume against time, draw a tangent to the curve at time zero, and take its gradient. Only at t = 0 is the substrate concentration the intended value, so only the initial rate is a fair comparison between different concentrations.
3. Distinguish between competitive and non-competitive inhibition in terms of binding site, effect of substrate concentration, and effect on Vmax and Km.
Competitive: binds the active site; is structurally similar to the substrate; increasing substrate concentration reduces the inhibition because the two compete for the same site; Vmax unchanged (reached at higher substrate concentration); Km increased. Non-competitive: binds an allosteric site away from the active site; changes the shape of the enzyme so the active site is no longer complementary; increasing substrate concentration has no effect on the degree of inhibition; Vmax reduced; Km unchanged. The single distinguishing test is whether excess substrate restores the original maximum rate.
4. Describe how you would investigate the effect of pH on the rate of the amylase–starch reaction, naming the variables you would control.
Set up a series of tubes each containing the same volume and concentration of starch, buffered to a different pH (for example 4, 5, 6, 7, 8) using buffer solutions, and equilibrate them all in a water bath at a fixed temperature. Add the same volume and concentration of amylase to each and start a timer. At 10-second intervals, remove a drop into iodine solution on a spotting tile. Record the time for the blue-black colour to disappear, which is the point at which no starch remains, and calculate rate as 1/time. Controlled variables: temperature, starch concentration and volume, enzyme concentration and volume, iodine volume, sampling interval, and the same person judging the end point. Improvements: repeat at each pH and take a mean; narrow the pH interval around the apparent optimum; use a colorimeter rather than the eye to fix the end point.
5. Alginate beads containing lactase are packed into a column and milk is run through. Give three advantages of this over adding free lactase to a tank of milk.
(i) The enzyme is not lost with the product — it stays in the column, so it can be reused many times, which is far cheaper than adding fresh enzyme to every batch. (ii) The product is uncontaminated by enzyme, so no purification step is needed to remove it and the lactose-free milk is not affected by residual protein. (iii) The enzyme is more stable to variation in temperature and pH, because the alginate matrix supports its tertiary structure, so the process tolerates less precise control. A fourth advantage: the process can run continuously rather than batch by batch. The trade-off is a somewhat lower rate, because substrate must diffuse into the beads to reach the active sites.
6. Two enzymes act on the same substrate. Enzyme A has Km = 0.5 mmol dm−3 and Vmax = 30 µmol min−1; enzyme B has Km = 5.0 mmol dm−3 and Vmax = 90 µmol min−1. Which would you choose for a reaction where substrate is scarce, and why?
Enzyme A. Km is the substrate concentration at which the enzyme runs at half its maximum rate, so A reaches half-maximal rate at one tenth of the concentration B needs — A has the higher affinity for the substrate. Where substrate is scarce, the working concentration sits far below B’s Km, so B operates at a small fraction of its 90 µmol min−1 and its high Vmax is never realised. B would be the better choice only where substrate is plentiful enough to approach saturation, since its ceiling is three times higher. The general point: Vmax tells you the ceiling, Km tells you how much substrate it takes to get near it, and you need both to choose.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • PhET — simulations that let you vary substrate concentration and inhibitor and watch the rate curve respond in real time
  • Khan Academy — Michaelis–Menten kinetics, for a slower derivation of where Km comes from than the syllabus requires
  • Nuffield Foundation practical biology — protocols for immobilising enzymes in alginate beads and for the catalase and amylase rate experiments