Factors Affecting Enzyme Activity

Enzymes are remarkably effective, but they are also fussy: change their conditions and they will speed up, slow down, or stop altogether. This lesson, part of Topic B1 of OCR Gateway Combined Science, works through the three factors you need to know — temperature, pH and substrate concentration — and the crucial idea of denaturation. It then covers the required practical on how pH affects the enzyme amylase, and how to calculate a rate of reaction. It builds directly on the lock-and-key model from the last lesson, because every effect here comes back to the shape of the active site.

By the end of this lesson you should be able to describe and explain how temperature, pH and substrate concentration affect enzyme activity, explain denaturation, calculate a rate, and describe the amylase pH practical.

This lesson builds AO1 (understanding of how each factor affects enzyme activity and what denaturation is), AO2 (applying it in the rate calculation and the required practical) and AO3 (analysing and interpreting the rate-against-factor graphs).

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A Quick Recap: Why Shape Matters

From the lock-and-key model, an enzyme only works when the substrate fits its active site. So anything that changes the shape of the active site changes how well the enzyme works. Both high temperature and the wrong pH can change that shape permanently — this is denaturation.

Exam Tip: A frequent misconception: "denatured" does not mean "killed" (enzymes are not alive), and it is not the same as "used up". Denatured means the enzyme's shape — especially its active site — has been permanently changed, so the substrate no longer fits.

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Factor 1: Temperature

Temperature has two opposing effects, which is why an enzyme has an optimum temperature where it works fastest.

• As the temperature rises towards the optimum, the enzyme and substrate molecules gain kinetic energy and move faster, so they collide more often and with more energy. More successful collisions means a faster rate.
• Above the optimum, the heat makes the enzyme vibrate so much that the bonds holding its shape break. The active site changes shape, so the substrate no longer fits — the enzyme is denatured and the rate falls sharply, eventually to zero.

For many human enzymes the optimum is around 37 °C (body temperature). The temperature graph is therefore a rise to a peak, followed by a steep fall.

Rate of an enzyme reaction against temperature (typical data):

Temperature (°C)	Relative rate
10	2
20	4
30	7
37	10 (optimum)
45	5
55	1 (denatured)

Notice the rate rises to a peak at the optimum and then drops away as the enzyme denatures — it is not symmetrical, because the fall is caused by permanent denaturation, not simply by fewer collisions.

Exam Tip: The rise and the fall have different causes. The rise is due to more frequent, higher-energy collisions; the fall is due to denaturation (the active site changing shape). State both causes separately for full marks.

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Factor 2: pH

Each enzyme also has an optimum pH at which it works fastest. Move the pH too far from the optimum in either direction and the enzyme denatures: the change in acidity or alkalinity alters the bonds that hold the active site's shape, so the substrate no longer fits.

Different enzymes have different optimum pH values to suit where they work in the body:

Enzyme	Where it works	Approximate optimum pH
Amylase	Mouth (saliva) and small intestine	About pH 7 (neutral)
Pepsin (a protease)	Stomach	About pH 2 (very acidic)
Lipase	Small intestine	Slightly alkaline (about pH 8)

Pepsin's low optimum suits the acidic stomach, while amylase prefers the near-neutral mouth — a neat link between an enzyme's properties and where it does its job.

Exam Tip: Either side of the optimum pH the rate falls because the enzyme is denatured — the active site changes shape. Do not write that the enzyme is "used up" or that "the substrate denatures".

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Factor 3: Substrate Concentration

Increasing the substrate concentration generally speeds the reaction up — but only up to a point.

• At low substrate concentration, there are few substrate molecules, so collisions with active sites are rare. Adding more substrate gives more frequent collisions and a faster rate.
• At high substrate concentration, all the active sites are already working as fast as they can — the enzymes are saturated. Adding still more substrate makes no further difference, so the rate levels off (plateaus).

The graph therefore rises and then flattens. (Note: this plateau is not denaturation — the enzymes are simply all occupied.)

A helpful way to picture this is to imagine a fixed number of enzyme molecules as a set of checkout tills in a shop. When only a few customers (substrate molecules) arrive, adding more customers means more tills are used and the shop serves people faster. But once every till is busy all the time, extra customers simply have to queue — the tills cannot work any faster, so the rate of serving stays the same however many more people arrive. In the same way, once every active site is occupied, adding more substrate cannot speed the reaction up. The only way to raise the plateau would be to add more enzyme (open more tills). This is why the amount of enzyme, as well as the amount of substrate, matters, and why it is important to keep the enzyme concentration constant when you are investigating the effect of substrate concentration.

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Calculating Rate of Reaction

You are expected to calculate a rate from experimental data. Rate is how much something changes per unit of time:

$\text{rate} = \frac{\text{change in quantity}}{\text{time taken}}$rate=time takenchange in quantity​

Worked Example 1 — Rate of product formed

In an enzyme experiment, $24 \text{ cm}^3$24 cm3 of oxygen is produced in $40 \text{ s}$40 s. Calculate the mean rate of reaction.

$\text{rate} = \frac{\text{volume of gas}}{\text{time}} = \frac{24 \text{ cm}^3}{40 \text{ s}} = 0.6 \text{ cm}^3/\text{s}$rate=timevolume of gas​=40 s24 cm3​=0.6 cm3/s

Answer: $0.6 \text{ cm}^3/\text{s}$0.6 cm3/s (cubic centimetres per second).

Common error: dividing time by volume instead of volume by time. Rate is always the quantity per unit of time, so time goes on the bottom.

Worked Example 2 — Rate from a time-to-complete

A starch solution is fully broken down by amylase in $50 \text{ s}$50 s. A common way to express the rate is $\dfrac{1}{\text{time}}$time1​. Calculate this rate.

$\text{rate} = \frac{1}{\text{time}} = \frac{1}{50 \text{ s}} = 0.02 \text{ s}^{-1}$rate=time1​=50 s1​=0.02 s−1

Answer: $0.02 \text{ s}^{-1}$0.02 s−1 (per second). A larger value of $\frac{1}{\text{time}}$time1​ means a faster reaction, because the reaction finished in a shorter time.

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Required Practical: Effect of pH on Amylase

A core practical of this topic investigates how pH affects the breakdown of starch by amylase. The clever part is using iodine solution as an indicator: iodine turns blue-black when starch is present, but stays orange/brown once the starch has been digested. So the time taken for the blue-black colour to disappear tells you how fast the amylase has worked.

flowchart TD
  A["Set up buffer solutions at a range of pH values<br/>(e.g. pH 4, 5, 6, 7, 8)"] --> B["Place drops of iodine in the wells of a spotting tile"]
  B --> C["Mix starch, amylase and one buffer in a test tube;<br/>start the timer"]
  C --> D["Every 10 s, take a drop of the mixture<br/>and add it to a fresh well of iodine"]
  D --> E["Note the time when iodine no longer turns blue-black<br/>(starch fully broken down)"]
  E --> F["Repeat for each pH; control variables<br/>(temperature, volumes, concentrations)"]

Method points the exam rewards: