Factors that affect enzyme action

Biology AS · Enzymes. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

The rate of an enzyme reaction depends on how often substrate molecules collide with free active sites and on whether the active site keeps its shape.

Temperature and pH act on shape. Beyond the optimum, bonds holding the tertiary structure break, the active site changes shape and the enzyme is denatured. Enzyme and substrate concentration act on the number of collisions. When every active site is occupied the rate cannot rise further: this maximum rate is V_max.

Inhibitors lower the rate. A competitive inhibitor competes with the substrate for the active site. A non-competitive inhibitor binds somewhere else and changes the shape of the active site.

Rules to remember

Temperature: up to the optimum, more kinetic energy gives more successful collisions; above it, hydrogen and ionic bonds break and the enzyme is denatured
pH: away from the optimum, the charges on R groups change and ionic bonds break; buffer solutions keep pH constant
With substrate in excess, rate is directly proportional to enzyme concentration
Michaelis-Menten constant (K_m) = the substrate concentration that gives half of V_max; a low K_m means a high affinity for the substrate
Competitive inhibitor: similar shape to the substrate, binds to the active site; more substrate overcomes it; V_max unchanged, K_m increased
Non-competitive inhibitor: binds away from the active site; more substrate does not overcome it; V_max lowered, K_m unchanged
Immobilised enzymes (in alginate beads): reusable, product is free of enzyme, more stable to heat and pH changes; the rate may be slower because substrate must diffuse into the beads

Common mistake

Reading K_m as half of the highest substrate concentration. First find V_max, halve the rate, then read the substrate concentration that gives that rate.

Worked example

An enzyme has a V_max of 60 µmol min⁻¹. The rate is 30 µmol min⁻¹ at a substrate concentration of 3 mmol dm⁻³. With an inhibitor, V_max is still 60 but a rate of 30 is only reached at 9 mmol dm⁻³. What is K_m in each case, and what type of inhibitor is it?

Half of V_max = 60 ÷ 2 = 30 µmol min⁻¹.
Without inhibitor this rate occurs at 3 mmol dm⁻³, so K_m = 3 mmol dm⁻³.
With inhibitor it occurs at 9 mmol dm⁻³, so K_m = 9 mmol dm⁻³.
V_max is unchanged and K_m has increased, which is the pattern for a competitive inhibitor.

Answer: K_m is 3 mmol dm⁻³ without and 9 mmol dm⁻³ with the inhibitor; it is a competitive inhibitor.

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