Factors affecting enzyme activity – Enzymes (Senior 4 Biology) | Ganzaa

Enzymes

Lessons

Senior 4 · Biology

Enzymes

4 lessons

Lesson 3 of 4

Factors affecting enzyme activity

25 min

An enzyme that loves boiling water

In 1976 scientists isolated a DNA polymerase from Thermus aquaticus, a bacterium that lives in hot springs. This enzyme, Taq polymerase, works fastest at 75–80 °C and survives being heated to 95 °C. At 37 °C, where your own enzymes work best, it adds only about 1.5 nucleotides per second; at 70 °C, about 60. Most of your enzymes would be destroyed long before 70 °C. Every enzyme has conditions in which it works best, and this lesson shows how temperature, pH, the concentrations of substrate and enzyme, and inhibitors change the rate.

Definition

Optimum: the temperature or pH at which an enzyme works fastest.

Temperature coefficient, $Q_{10}$ : the factor by which the rate increases for a 10 °C rise: $Q_{10} = \frac{\text{rate at } (T + 10)\,{}^\circ\mathrm{C}}{\text{rate at } T\,{}^\circ\mathrm{C}}$ .

$V_{max}$ : the maximum rate, reached when every active site is occupied (the enzyme is saturated). Michaelis constant, $K_{m}$ : the substrate concentration at which the rate is half of $V_{max}$ ; a low $K_{m}$ means the enzyme binds its substrate strongly.

Limiting factor: the factor in shortest supply, which sets the rate.

Inhibitor: a substance that reduces the rate of an enzyme reaction. Competitive inhibitors bind to the active site; non-competitive inhibitors bind elsewhere and change the shape of the active site. Inhibition is reversible if the inhibitor can leave, irreversible if it binds permanently.

The four main factors

Figure 1. How four factors affect the rate of an enzyme reaction (general shapes).

Factor	Effect	Explanation
Temperature	Rate rises to an optimum (about 37–40 °C for most human enzymes), then falls steeply	Below the optimum, molecules gain kinetic energy, so enzyme–substrate collisions are more frequent and more of them succeed; $Q_{10}$ is about 2. Above it, vibrations break hydrogen and ionic bonds, the tertiary structure and active site change shape, and the enzyme is denatured; this is usually permanent. At low temperatures the enzyme is only inactive, and works again when warmed.
pH	Each enzyme has an optimum pH: pepsin (stomach) about 1.5–2.5; trypsin (small intestine) about 8	$\mathrm{H^+}$ and $\mathrm{OH^-}$ ions change the charges on R groups, breaking ionic and hydrogen bonds, so the active site loses its shape and charge pattern and the substrate binds less well. A small change is reversible: pepsin is inactive at pH 7 but works again if acidified. Extreme pH denatures permanently.
Substrate concentration	Rate rises, then levels off at $V_{max}$	At low concentration, substrate is the limiting factor: more substrate, more collisions. At high concentration all active sites are occupied; the enzyme concentration now limits the rate.
Enzyme concentration	Rate is proportional to enzyme concentration, while substrate is in excess	More active sites are available. If the substrate runs short, it becomes limiting and the line levels off.

Worked example 1: temperature coefficient

An enzyme produces product at $0.40\,\mu\mathrm{mol\,min^{-1}}$ at 25 °C and $0.82\,\mu\mathrm{mol\,min^{-1}}$ at 35 °C (illustrative). (a) Calculate $Q_{10}$ . (b) Taq polymerase adds 1.5 nucleotides per second at 37 °C and 60 per second at 70 °C. How many times faster is it at 70 °C?

(a) $Q_{10} = \frac{0.82}{0.40} = 2.05 \approx 2.1$ : the rate roughly doubles for a 10 °C rise, typical below the optimum.
(b) $\frac{60}{1.5} = 40$ times faster. Taq's optimum is far above 37 °C because the bonds holding its shape are unusually stable.

Worked example 2: finding $V_{max}$ and $K_{m}$

Initial rates of an enzyme reaction (illustrative):

Substrate / $\mathrm{mmol\,dm^{-3}}$	1	2	4	8	16	32
Rate / $\mu\mathrm{mol\,min^{-1}}$	16.7	25.0	33.3	40.0	44.4	47.1

(a) Estimate $V_{max}$ . (b) Find $K_{m}$ . (c) What limits the rate at 32 $\mathrm{mmol\,dm^{-3}}$ ?

(a) The rate levels off, approaching about $50\,\mu\mathrm{mol\,min^{-1}}$ : $V_{max} \approx 50$ .
(b) Half of $V_{max}$ is 25.0, reached at a substrate concentration of 2: $K_m = 2\,\mathrm{mmol\,dm^{-3}}$ .
(c) Nearly all active sites are occupied, so the enzyme concentration (the number of active sites) limits the rate, not the substrate.

Inhibitors

Figure 2. Competitive and non-competitive inhibition.

Inhibitor	Type	Enzyme affected and effect
Malonate	Competitive, reversible	Similar in shape to succinate; blocks succinate dehydrogenase in respiration
Cyanide	Non-competitive	Blocks cytochrome oxidase, the last enzyme of aerobic respiration, so cells cannot make enough ATP: a deadly poison
Heavy-metal ions (lead, mercury)	Non-competitive, often irreversible	Bind to sulfur-containing groups and distort many enzymes, which is part of what makes these metals poisonous
Penicillin	Irreversible	Binds permanently to a bacterial enzyme that builds cell walls, so growing bacteria burst

You can tell the two kinds of reversible inhibitor apart with an experiment: increase the substrate concentration. A competitive inhibitor is outcompeted, so the rate climbs back towards the normal $V_{max}$ . A non-competitive inhibitor still holds some enzyme molecules out of action, so $V_{max}$ stays low (Figure 1C).

Worked example 3: which kind of inhibitor?

The enzyme of worked example 2 was tested with inhibitor X and, separately, inhibitor Y (illustrative rates in $\mu\mathrm{mol\,min^{-1}}$ ):

Substrate / $\mathrm{mmol\,dm^{-3}}$	2	8	32
No inhibitor	25.0	40.0	47.1
With X	10.0	25.0	40.0
With Y	12.5	20.0	23.5

X: at low substrate it cuts the rate by $\frac{25.0 - 10.0}{25.0} = 60\,\%$ , but at 32 only by $\frac{47.1 - 40.0}{47.1} = 15\,\%$ . Extra substrate overcomes it: competitive.
Y: the rate is halved at every concentration ( $\frac{12.5}{25.0} = \frac{20.0}{40.0} = 0.5$ ), so $V_{max}$ is halved and extra substrate does not help: non-competitive.

FuseSchool: "How Enzymes Denature". Watch for why the rate rises with temperature up to the optimum, and what happens to the active site above it or at the wrong pH.

Warning

Common mistakes

Writing that enzymes are "killed" by heat. Enzymes are not alive: they are denatured.
Saying low temperature denatures enzymes. Cold only slows them; they work again when warmed (that is why food keeps longer in a fridge).
Explaining the plateau of a substrate graph as "the enzyme is used up". The enzyme is saturated: all its active sites are occupied at any moment.
Saying a competitive inhibitor "has the same shape" as the substrate. It has a similar shape, enough to bind in the active site but not to react.

Tip

Exam technique for graphs. Describe first (quote numbers from both axes, name the optimum or where the curve levels off), then explain using kinetic energy, collisions, enzyme–substrate complexes, active sites and denaturation. For a plateau, always say which factor is now limiting.

✦

Practice: factors affecting enzyme activity

Explain why the rate of an enzyme reaction increases from 20 °C to 35 °C.
An enzyme reaction has a rate of 12 units at 20 °C and 25 units at 30 °C. Calculate $Q_{10}$ .
Explain why pepsin works in the stomach but stops working in the small intestine.
Using worked example 2, state the rate when the substrate concentration is 4 $\mathrm{mmol\,dm^{-3}}$ , and what fraction of $V_{max}$ this is.
Explain why adding more enzyme increases the rate only while there is plenty of substrate.
Describe an experiment to find out whether a new inhibitor is competitive or non-competitive, and state the results you would expect for each.
A pupil boils an amylase solution, cools it and adds it to starch. After an hour the starch is still there. She concludes that "boiling just slowed the enzyme down". Evaluate her conclusion.

Show answersHide answers

Molecules gain kinetic energy and move faster, so enzyme and substrate collide more often and more collisions have enough energy to react; more enzyme–substrate complexes form per second.
$Q_{10} = \frac{25}{12} = 2.1$ (2 s.f.).
Pepsin's optimum pH is about 1.5–2.5, as in the stomach. In the alkaline small intestine (about pH 8), the charges on its R groups change, the active site changes shape and the substrate no longer binds well, so it stops working.
33.3 $\mu\mathrm{mol\,min^{-1}}$ , which is $\frac{33.3}{50} = \frac{2}{3}$ of $V_{max}$ .
Each extra enzyme molecule adds active sites, so more complexes form per second while substrate is in excess. When the substrate runs short, there are not enough substrate molecules for the extra active sites; substrate becomes the limiting factor.
Measure the initial rate at several substrate concentrations, with and without the inhibitor, keeping enzyme concentration, temperature and pH constant. If competitive, the inhibited rate approaches the normal $V_{max}$ at high substrate concentrations; if non-competitive, $V_{max}$ stays lower however much substrate is added.
Her conclusion is wrong. Boiling denatures amylase: its hydrogen and ionic bonds break and the active site loses its shape permanently, so it does not recover on cooling. If it had been slowed only, some starch would have been digested in an hour. A fair test would compare it with unboiled amylase.

Quick check

On a graph of rate against substrate concentration, the curve levels off at high concentration. Why?

Key points

Temperature: rate rises ( $Q_{10} \approx 2$ ) to an optimum, then falls as the enzyme denatures. pH: each enzyme has an optimum; extreme pH denatures.
Substrate concentration: the rate rises to $V_{max}$ as the active sites become saturated; $K_{m}$ is the substrate concentration at $\frac{1}{2}V_{max}$ . Enzyme concentration: rate proportional while substrate is in excess.
Competitive inhibitors occupy the active site and are overcome by more substrate; non-competitive inhibitors bind elsewhere and lower $V_{max}$ .

Sources and credits

2.3.6: Enzymes, from Biology 2e by OpenStax (Mary Ann Clark, Matthew Douglas, Jung Choi), via LibreTexts Biology, licensed under CC BY 4.0. Used for the effects of temperature and pH and for competitive and non-competitive inhibition. Adapted by Ganzaa.
"Taq polymerase" and "Pepsin", Wikipedia contributors, licensed under CC BY-SA 4.0. Used for the optimum temperature and rates of Taq polymerase and the optimum pH of pepsin. Adapted by Ganzaa.
Figures 1 and 2, the tables, worked examples (all data except Taq's rates are illustrative), practice and quick check: written by Ganzaa.
Photo: Jim Peaco, National Park Service, public domain, via Wikimedia Commons.
Video: "How Enzymes Denature | Cells | Biology | FuseSchool", FuseSchool - Global Education, embedded from YouTube.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1An enzyme that loves boiling water

2The four main factors

3Inhibitors

An enzyme that loves boiling water

The four main factors

Inhibitors