Enzymes

Activation energy is the minimum amount of energy that the reacting (substrate) molecules must have for a reaction to take place — the energy needed to start the reaction by breaking bonds.

An enzyme lowers the activation energy of the reaction. Because less energy is required, a greater proportion of substrate molecules has enough energy to react at body temperature, so the reaction proceeds faster.

Each enzyme has an active site with a specific three-dimensional shape, determined by the folding of its tertiary structure.

The shape of the active site is complementary to one particular substrate, so only that substrate can fit into the active site and form an enzyme-substrate complex.

A substrate of a different shape cannot bind to the active site, so no enzyme-substrate complex forms and the reaction is not catalysed. This makes each enzyme specific to its substrate and reaction.

As the pH increases from the optimum of 2 towards 8, the initial rate of reaction decreases and eventually falls to zero, because pepsin works fastest at its optimum and progressively more slowly away from it.

A change in pH changes the concentration of hydrogen ions surrounding the enzyme. This alters the charges on the R groups (side chains) of the amino acids in and around the active site.

The altered charges cause the ionic bonds and hydrogen bonds that maintain the enzyme's tertiary structure to break. As these bonds break, the shape of the active site changes so that it is no longer complementary to the substrate.

As a result, the substrate can no longer fit and bind, so fewer enzyme-substrate complexes form per second and the initial rate falls.

At a pH far from the optimum, such as pH 8, the change in shape is so great that the enzyme is denatured and the rate is effectively zero. This is why pepsin, which works in the acidic stomach, is inactive in the alkaline conditions of the small intestine.

In both models the substrate binds to an active site that is complementary in shape to the substrate, forming an enzyme-substrate complex, and both explain enzyme specificity.

In the lock-and-key model, the active site has a fixed, rigid shape that is already an exact match for the substrate; the active site does not change shape when the substrate binds, like a key fitting a lock.

In the induced-fit model, the active site is not a perfect fit at first. As the substrate binds, the active site changes shape and moulds around the substrate.

This moulding in the induced-fit model places strain on the bonds in the substrate, lowering the activation energy further. The induced-fit model therefore explains catalysis more completely and better fits the experimental evidence than the lock-and-key model.

Shape of the curve. At the start of the reaction the substrate (hydrogen peroxide) is at its highest concentration, so there are the most frequent collisions between substrate molecules and the active sites of catalase. Many enzyme-substrate complexes form per second, so oxygen is released rapidly and the curve is steep.

As the reaction proceeds, substrate is used up and its concentration falls. Fewer collisions occur and fewer enzyme-substrate complexes form per second, so the rate slows and the curve becomes less steep.

Eventually almost all the substrate has been converted to product, so very little further oxygen is released and the curve levels off (plateaus).

Why measure the initial rate. The initial rate, at t = 0, is the fairest measure for comparing the different temperatures because at that moment the substrate concentration is the same and at its maximum in every tube and no product has yet accumulated to slow the reaction. Therefore any difference in initial rate is due to the variable being investigated — temperature — and not to differences in how much substrate has been used up.

Determining the initial rate. She should draw a tangent to the curve at t = 0 (a straight line just touching the curve at the origin) and calculate its gradient by reading the change in volume of oxygen divided by the change in time. The initial rate should be quoted with appropriate units, for example cm³ of oxygen per second (cm³ s⁻¹).

As the temperature is increased from 10 °C towards the optimum (about 40 °C in many human enzymes), the initial rate increases. The enzyme and substrate molecules gain more kinetic energy and move faster, so there are more frequent successful collisions between substrate molecules and active sites. More enzyme-substrate complexes form per second, so the rate rises. As a guide, the rate roughly doubles for every 10 °C rise up to the optimum (a Q₁₀ of about 2).

At the optimum temperature, the active sites are forming enzyme-substrate complexes at the fastest possible rate, so the initial rate is at its maximum.

Above the optimum, increasing the temperature further causes the rate to fall. The extra vibration of the molecules breaks the hydrogen bonds and ionic bonds that hold the enzyme's tertiary structure in shape.

As these bonds break, the active site changes shape and is no longer complementary to the substrate, so the substrate can no longer bind. Fewer enzyme-substrate complexes form and the rate decreases.

By about 60 °C the enzyme is denatured — the change in the active site is permanent — and the rate falls towards zero. The overall graph of rate against temperature therefore rises to a peak at the optimum and then falls steeply.

An intracellular enzyme acts inside the cell that produced it, catalysing reactions within the cytoplasm or organelles. An example is catalase, which breaks down the toxic by-product hydrogen peroxide ($2H_2O_2 \rightarrow 2H_2O + O_2$).

An extracellular enzyme is secreted out of the cell and catalyses reactions outside it. An example is amylase (which hydrolyses starch to maltose) or trypsin (a protease in the small intestine).

The key distinction is therefore where the enzyme acts relative to the cell that makes it — inside it (intracellular) or after secretion, outside it (extracellular).

At low substrate concentrations the initial rate increases as the substrate concentration is raised. There are more substrate molecules in a given volume, so there are more frequent collisions between substrate molecules and the enzymes' active sites. More enzyme-substrate complexes form per second, so more product is formed per second and the initial rate rises.

As the substrate concentration continues to increase, the rate increases less steeply because a greater proportion of the active sites are already occupied at any instant.

At high substrate concentrations the rate reaches a plateau. Now the active sites are saturated — essentially all of them are occupied by substrate at any moment.

Because the active sites are working as fast as they can, the enzyme concentration (the number of active sites) has become the limiting factor. Adding more substrate cannot increase the rate because there are no free active sites for the extra substrate to bind to.

To raise the height of the plateau, the student would have to increase the enzyme concentration, providing more active sites.

(a) When the substrate is in excess. Increasing the enzyme concentration increases the initial rate in direct proportion. Adding more enzyme molecules provides more active sites available for substrate to bind to. Because substrate is plentiful, every additional active site can form enzyme-substrate complexes, so more product is formed per second and the rate rises proportionally. A graph of initial rate against enzyme concentration is therefore a straight line through the origin under these conditions.

(b) When the substrate is limited. Initially, increasing the enzyme concentration still increases the rate, for the same reason — more active sites form more enzyme-substrate complexes per second. However, once there is enough enzyme that the limited amount of substrate becomes the limiting factor, adding still more enzyme has no further effect: the surplus active sites cannot work because there are not enough substrate molecules to occupy them. The graph therefore rises and then levels off to a plateau.

In both cases the underlying mechanism is the same — rate depends on the number of enzyme-substrate complexes forming per second — but whether the line keeps rising or plateaus depends on whether substrate or enzyme is the limiting factor.

(a) The reaction is fastest at the start, when the substrate (hydrogen peroxide) is at its highest concentration and no product has yet accumulated to slow it. As the reaction proceeds the substrate is used up, so the rate falls and the curve levels off; a value taken later would therefore underestimate the true catalytic rate. The gradient of a tangent at t = 0 captures the rate before the substrate concentration has fallen, giving the fairest measure for comparing different conditions.

(b) The initial rate is the gradient of the tangent at t = 0:

$\text{initial rate} = \frac{\Delta V}{\Delta t} = \frac{9.6 - 0.0}{15 - 0} = \frac{9.6}{15} = 0.64\ \text{cm}^3\,\text{s}^{-1}$

So the initial rate of oxygen production is 0.64 cm³ s⁻¹.