Factors that affect enzyme activity

1. Overview

The rate of an enzyme-catalysed reaction depends on how often substrate molecules successfully collide with, and bind to, an enzyme's active site to form an enzyme-substrate complex. Five factors change this rate: temperature, pH, enzyme concentration, substrate concentration and inhibitor concentration. Each factor works by changing either the frequency of successful collisions or the shape of the active site. This topic also covers how the maximum rate ($V_{max}$) and the Michaelis-Menten constant ($K_m$) are used to compare enzymes, and the use of immobilised enzymes in industry.

Key Definitions

• Active site: the region of an enzyme with a specific three-dimensional shape that is complementary to its substrate, where the substrate binds.
• Enzyme-substrate complex: the temporary structure formed when a substrate binds to the active site of an enzyme.
• Denaturation: a permanent change to the tertiary (three-dimensional) structure of an enzyme so that its active site is no longer complementary to the substrate and it can no longer catalyse the reaction.
• Optimum temperature: the temperature at which an enzyme catalyses its reaction at the maximum rate.
• Maximum rate (Vmax): the highest rate of reaction an enzyme can reach, achieved when all of its active sites are saturated with substrate.
• Michaelis-Menten constant (Km): the substrate concentration that gives a rate equal to half of Vmax; a lower Km means a higher affinity of the enzyme for its substrate.
• Competitive inhibitor: a molecule with a shape similar to the substrate that binds to the active site and competes with the substrate, so its effect is reduced by raising the substrate concentration.
• Non-competitive inhibitor: a molecule that binds to a site other than the active site and changes the shape of the active site, so its effect is not reduced by raising the substrate concentration.
• Immobilised enzyme: an enzyme attached to or trapped within an inert material such as alginate beads, so it is not free in solution and can be reused.

Content

Temperature

As temperature rises, enzyme and substrate molecules gain kinetic energy and move faster, so there are more frequent successful collisions and more enzyme-substrate complexes form per second. The rate therefore increases up to the optimum temperature. As a rough guide the rate roughly doubles for each 10 °C rise up to the optimum (a temperature coefficient $Q_{10}$ of about 2).

Above the optimum, the extra vibration breaks the hydrogen bonds and other interactions that hold the enzyme's tertiary structure together. The active site changes shape and is no longer complementary to the substrate, so fewer complexes form and the rate falls steeply. This permanent change is denaturation — the enzyme is denatured, not "killed".

pH

Every enzyme has an optimum pH at which the active site is the correct shape and the rate is highest. Moving away from this optimum disrupts the enzyme in two linked steps:

• Cause: the change in pH alters the concentration of H⁺ and OH⁻ ions. These ions interfere with the hydrogen and ionic bonds that hold the tertiary structure together, and they change the charges on the R-groups in the active site.
• Consequence: the active site changes shape so it is less complementary to the substrate. Fewer enzyme-substrate complexes form and the rate falls. At extreme pH the change becomes permanent and the enzyme is denatured.

In experiments a buffer solution is used to hold the pH constant at each value being tested.

Enzyme concentration

When substrate is in excess, increasing the enzyme concentration provides more active sites, so the rate increases in direct proportion to enzyme concentration. If substrate is limited, adding more enzyme has no further effect because there is not enough substrate to occupy the extra active sites.

Substrate concentration, Vmax and Km

At low substrate concentrations many active sites are empty, so adding substrate increases the rate. As substrate concentration rises, the active sites become fully occupied — the enzyme is saturated — and the rate levels off at the maximum rate, $V_{max}$. As the graph of rate against substrate concentration below shows, the curve climbs steeply at first and then flattens into a plateau. At that plateau the enzyme concentration (the number of active sites) is the limiting factor; the enzyme is not "used up", because enzymes are catalysts and are reused.

Reading the active-site picture from the curve. It helps to think about how many active sites are empty at each stage:

• At the very start, no substrate has bound yet, so almost all active sites are empty.
• As substrate is added, more sites fill, so the number of empty sites falls and the rate rises.
• On the plateau, sites fill as fast as they empty, so a constant number are occupied and a constant number stay empty — a steady state. The reaction is still going (substrate is still being converted); it has not stopped and the enzyme has not run out.

The Michaelis-Menten constant, $K_m$. This is the substrate concentration that gives a rate of exactly $\tfrac{1}{2}V_{max}$. Because it is a concentration, $K_m$ is read off the x-axis of the curve shown above. It tells you how strongly an enzyme binds its substrate:

$K_m$ value	Substrate needed to reach $\tfrac{1}{2}V_{max}$	Affinity for substrate
Low $K_m$	small amount	high affinity
High $K_m$	large amount	low affinity

So $K_m$ lets you compare different enzymes: the one with the lower $K_m$ reaches half its maximum rate with less substrate, meaning it binds its substrate more readily.

Worked example

Exam-style question: The rate of an enzyme-controlled reaction is measured at substrate concentrations from low to high, with temperature and pH kept constant. The rate rises and then levels off. Explain this pattern. [3]

Model answer:

• At low substrate concentration many active sites are empty, so adding substrate forms more enzyme-substrate complexes per second and the rate rises.
• As substrate concentration increases, a greater proportion of the active sites become occupied.
• The rate plateaus when the active sites are saturated, so the enzyme concentration (the number of active sites) becomes the limiting factor — the enzyme is not "used up".

Worked example

Exam-style question: The table shows the substrate concentration at which two enzymes, P and Q, each reach half of their maximum rate.

Enzyme	$[S]$ at $\tfrac{1}{2}V_{max}$ / mmol dm⁻³
P	2.0
Q	8.0

State the $K_m$ of each enzyme and identify which enzyme has the higher affinity for its substrate, giving a reason. [3]

Model answer:

• $K_m$ is the substrate concentration that gives $\tfrac{1}{2}V_{max}$, read on the x-axis, so $K_m$ of P = 2.0 mmol dm⁻³ and $K_m$ of Q = 8.0 mmol dm⁻³.
• Enzyme P has the higher affinity for its substrate.
• P has the lower $K_m$, so it reaches half its maximum rate at a lower substrate concentration, meaning it binds its substrate more readily.

Worked example

Exam-style question: An enzyme-controlled reaction has a rate of 8 arbitrary units at 20 °C and 16 arbitrary units at 30 °C. Calculate $Q_{10}$ and state whether the enzyme is below its optimum temperature, giving a reason. [2]

Model answer:

• $Q_{10} = \dfrac{\text{rate at }30\ °\text{C}}{\text{rate at }20\ °\text{C}} = \dfrac{16}{8} = 2$.
• The enzyme is still below its optimum, because the rate increased when the temperature was raised (above the optimum the rate would fall as the enzyme denatured).

Inhibitors and inhibitor concentration

Reversible inhibitors slow enzymes without permanently damaging them. They fall into two types:

	Competitive inhibitor	Non-competitive inhibitor
Where it binds	the active site	a different site (not the active site)
Effect on active site	blocks substrate by competing for the site	changes the active site shape
Raising substrate concentration	overcomes the inhibitor; $V_{max}$ can still be reached	does not overcome it; $V_{max}$ is lowered

The concentration of inhibitor also matters, because more inhibitor molecules block or alter more active sites:

• More competitive inhibitor lowers the rate at any given substrate concentration, because more inhibitor molecules are competing for the active sites. However, because they compete, a high enough substrate concentration still outcompetes them, so $V_{max}$ can still be reached — it just takes more substrate to get there (the apparent $K_m$ rises).
• More non-competitive inhibitor binds more sites away from the active site, distorting more active sites and taking them out of action. Adding substrate cannot win these sites back, so increasing the inhibitor concentration progressively lowers $V_{max}$.

Immobilised enzymes

An immobilised enzyme is trapped in or bound to an inert support such as alginate beads.

Investigating the difference in activity. To compare an immobilised enzyme with the same enzyme free in solution:

• Mix the enzyme with sodium alginate solution, then drip the mixture into calcium chloride solution. The drops set into firm alginate beads with enzyme trapped inside.
• Pass substrate over (or mix it with) the beads and measure the rate of product formation over time.
• Repeat with the same mass of enzyme free in solution, keeping temperature, pH, substrate concentration and volume the same so it is a fair comparison.

Why the immobilised rate is often lower at first. The enzyme is held inside the bead, so substrate must diffuse into the bead to reach the active sites, and product must diffuse out. This diffusion barrier means fewer substrate molecules reach the active sites per second, so the initial rate is usually a little lower than for free enzyme, whose active sites are fully exposed in solution.

Advantages of immobilised enzymes. Despite the slightly lower initial rate, immobilising enzymes brings large practical benefits:

• The enzyme can be recovered and reused, lowering cost.
• The product is not contaminated with enzyme, so less purification is needed.
• The enzyme is more stable to changes in temperature and pH, allowing continuous-flow industrial processes.

Key Equations

Temperature coefficient (rate change per 10 °C rise): $$Q_{10} = \frac{\text{rate at } (T+10)}{\text{rate at } T}$$

Michaelis-Menten relationship (rate against substrate concentration $[S]$): $$\text{rate} = \frac{V_{max}[S]}{K_m + [S]} \qquad\text{so when } [S] = K_m,\ \text{rate} = \tfrac{1}{2}V_{max}$$

Common Mistakes to Avoid

• Saying the plateau happens because the enzyme is "used up". The rate levels off because every active site is occupied — the enzyme is saturated — so the number of active sites (enzyme concentration) becomes the limiting factor. Enzymes are catalysts and are not consumed.
• Reading $K_m$ off the y-axis. $K_m$ is a substrate concentration, so it is found on the x-axis — the concentration that gives half of $V_{max}$.
• Writing that the enzyme is "killed" at high temperature. Enzymes are not alive; the correct term is denatured, meaning the tertiary structure (and so the active site) has permanently changed shape.
• Saying "the enzyme changes shape" without naming the active site. State specifically that the active site changes shape so it is no longer complementary to the substrate.
• Explaining a temperature rise only as "more collisions". Add that molecules gain kinetic energy and move faster, giving more frequent successful collisions that form enzyme-substrate complexes.
• Choosing extreme or too few temperatures when planning an experiment. Use a sensible range of at least five temperatures around the expected optimum (for example 20–60 °C) and control the other variables.
• Confusing the two types of inhibitor. Raising substrate concentration overcomes a competitive inhibitor (so $V_{max}$ is still reached) but not a non-competitive inhibitor (which lowers $V_{max}$).
• Misreading the "empty active sites" curve. The number of empty active sites is highest at the very start (no substrate has bound yet) and highest again at the very end (the reaction is complete and substrate is gone), but lowest in the middle when most sites are occupied — it does not simply fall to zero and stay there.
• Thinking the reaction has stopped on the plateau. While substrate is still present, a constant (steady) concentration of enzyme-substrate complexes is maintained — sites are filled as fast as they empty. The reaction is still proceeding, not finished.
• Naming a dependent variable that cannot actually be measured. A dependent variable must be directly measurable, such as volume of CO₂ collected or time taken for a colour change. Vague phrases like "enzyme activity" or "activity of yeast" describe a concept, not something you can read off an instrument.

Exam Tips

• Use precise vocabulary: complementary (not "same" or "matching") for how the substrate fits the active site, and denatured (not "destroyed" or "killed").
• For graph questions, describe the trend and explain it — link each change to active sites, enzyme-substrate complexes and the limiting factor.
• When you read a value off a curve, quote it with its units (for example, state $K_m$ as a concentration).
• When planning an investigation, state a measurable dependent variable (such as volume of gas or time taken) and list the variables you will control to keep it a fair test.
• For "compare" questions on immobilised versus free enzymes, write a single comparative sentence (e.g. "immobilised enzymes are more stable to temperature change than free enzymes"), not two separate descriptions.