Unit 6 Question Bank
Note: TMU slide 45 prints this as a positive question with several individually true options; it is given here as a NOT-question so that it has a single defensible answer.TMU Lecture 6 Slide 45 · Harper's ch.8, p.79
The general shortcut: [S] = Km × f/(1−f), where f is the fraction of Vmax. So 50% → 1 Km, 80% → 4 Km, 90% → 9 Km, 99% → 99 Km. Printed on the TMU deck.TMU Lecture 6 Slide 46 · Harper's ch.8, p.79
Significance:
• It has the dimensions of substrate concentration, not of rate.
• It varies inversely with affinity: a low Km means high affinity for the substrate.
• It is independent of enzyme concentration, and therefore characterises a particular enzyme with a particular substrate. It is affected by the nature of the enzyme and substrate, by temperature and by pH.
• It is determined experimentally as the [S] giving half-maximal velocity, most readily from the negative x-intercept (−1/Km) of a Lineweaver–Burk plot.
• A competitive inhibitor raises the apparent Km; a non-competitive inhibitor leaves it unchanged — which is how the two are distinguished.Harper's ch.8, p.79
It expresses in mathematical terms the relationship between the initial reaction velocity and the substrate concentration, describing the rectangular hyperbola obtained when vi is plotted against [S].
Evaluated under three conditions:
• When [S] ≪ Km, vi is directly proportional to [S].
• When [S] = Km, vi = Vmax/2 — which defines Km.
• When [S] ≫ Km, vi ≈ Vmax and is unaffected by further substrate.Harper's ch.8, p.79
It is reached when the enzyme is saturated — that is, when all the enzyme is present as the ES complex and no free enzyme remains available to bind further substrate. Since only substrate combined with enzyme as ES can be transformed into product, further increases in [S] then produce no increase in rate.TMU Lecture 6 Slide 18 · Harper's ch.8
It acts by decreasing the number of free enzyme molecules available to form ES. Because inhibitor and substrate compete for the same site, their effects are reciprocal: the inhibition is overcome by raising [S].
Kinetics: the apparent Km is INCREASED; Vmax is UNCHANGED. On a double-reciprocal plot the lines converge on the y-axis. Example: malonate inhibiting succinate dehydrogenase; the statins inhibiting HMG-CoA reductase.Harper's ch.8, p.82 · TMU Lecture 6 Slides 32, 43
Because inhibitor binding does not affect substrate binding, Km is unchanged; but the EI complex, while it can still bind substrate, is less efficient at transforming substrate into product. The inhibition cannot be overcome by excess substrate.
Kinetics: Km UNCHANGED; Vmax DECREASED. On a double-reciprocal plot the lines converge on the x-axis.Harper's ch.8, pp.82–83 · TMU Lecture 6 Slides 37–38, 43
Because these covalent changes are relatively stable, an enzyme “poisoned” by an irreversible inhibitor such as a heavy metal atom or an acylating reagent remains inhibited even after the inhibitor is removed from the surrounding medium — the key contrast with reversible inhibitors, from which fully active enzyme is recovered by simple removal.Harper's ch.8, p.83 · TMU Lecture 6 Slide 41
Raising temperature increases both the kinetic energy of molecules — so more exceed the activation-energy barrier — and the frequency of collisions. This holds only over a limited range: continued heating causes the enzyme to denature, with rapid loss of catalytic activity. Every enzyme therefore has a temperature optimum at which activity is maximal.TMU Lecture 6 Slide 14 · Harper's ch.8, p.77
A ratio is used because the benefit of a high kcat can only be realised if Km is sufficiently low — processing speed is worthless if the enzyme cannot capture substrate at physiological concentrations. Enzymes for which formation of the ES complex is rate-limiting are termed diffusion-limited or “catalytically perfect”.Harper's ch.8, p.80
The equation
vi = Vmax[S] / (Km + [S])
where vi is the initial reaction velocity, Vmax the maximal velocity attainable at that enzyme concentration, [S] the substrate concentration, and Km the Michaelis constant. It expresses in mathematical terms the relationship between initial velocity and substrate concentration, and describes the rectangular hyperbola obtained when vi is plotted against [S].
What the curve represents
As [S] rises, vi increases until it reaches Vmax, beyond which further substrate has no effect. The reason is that at any instant only substrate combined with the enzyme as an ES complex can be transformed into product; at Vmax all the enzyme is present as ES complex and no free enzyme remains available.
Significance of the equation — the three conditions
| Condition | The equation reduces to | Meaning |
|---|---|---|
| [S] ≪ Km | vi ≈ (Vmax/Km)[S] | vi is directly proportional to [S] — the linear, first-order region |
| [S] = Km | vi = Vmax/2 | The velocity is half-maximal — this defines Km |
| [S] ≫ Km | vi ≈ Vmax | Velocity is maximal and independent of [S] — the saturated plateau |
Significance of Km
- Definition and dimensions. Km is the substrate concentration giving half-maximal velocity, and therefore has the dimensions of concentration, not of rate.
- An inverse measure of affinity. A low Km indicates high affinity — little substrate is needed to half-saturate the enzyme; a high Km indicates low affinity.
- A constant characterising the enzyme. Km is independent of enzyme concentration (which affects Vmax instead), so it identifies a particular enzyme acting on a particular substrate. It is affected by the nature of the enzyme and substrate, by temperature and by pH.
- Physiological meaning. Two isozymes catalysing the same reaction may have very different Km values, adapting them to different roles — hexokinase has a low Km for glucose and works even at low blood glucose, whereas hepatic glucokinase has a high Km and becomes active only when blood glucose is high.
- Diagnostic of inhibitor mechanism. A competitive inhibitor raises the apparent Km without altering Vmax; a non-competitive inhibitor leaves Km unchanged but lowers Vmax.
Determining Km and Vmax in practice
Direct measurement of Vmax requires impractically high substrate concentrations — 99% of Vmax needs 99 Km. Inverting the equation gives the linear Lineweaver–Burk (double-reciprocal) form:
1/vi = (Km/Vmax)(1/[S]) + 1/Vmax
A plot of 1/vi against 1/[S] gives a straight line with y-intercept 1/Vmax, slope Km/Vmax and x-intercept −1/Km, allowing both constants to be extrapolated from data obtained below saturation.
A limitation worth noting
Neither the Michaelis–Menten expression nor its double-reciprocal plots can be applied to enzymes showing positive cooperativity, whose vi versus [S] plot is sigmoidal; these are analysed using the Hill equation instead.
The comparison
| Competitive inhibitor | Non-competitive inhibitor | |
|---|---|---|
| Structure of inhibitor | Resembles the substrate — a substrate analogue | Little or no resemblance to the substrate |
| Binding site on enzyme | The active site — specifically the substrate-binding portion, blocking access | A site distinct from the substrate-binding site; binds either E or ES |
| Effect of increasing [S] | Overcomes the inhibition | Cannot overcome the inhibition |
| Km | Increased (apparent Km, K′m) | Unchanged |
| Vmax | Unchanged | Decreased |
| Lineweaver–Burk lines converge on | the y-axis | the x-axis |
Why competitive inhibition behaves as it does
Inhibitor and substrate compete for the same site, so they exert reciprocal effects on the concentrations of the EI and ES complexes: forming ES removes free enzyme available to bind inhibitor, so raising [S] decreases [EI] and raises the velocity. In effect the inhibitor works by decreasing the number of free enzyme molecules available to form ES. Because sufficient substrate can always outcompete it, Vmax is still attainable — more substrate is simply required to reach it, which is precisely what a raised apparent Km signifies.
Classic example. Succinate dehydrogenase removes one hydrogen atom from each of the two methylene carbons of succinate. Its structural analogue malonate (⁻OOC–CH₂–COO⁻) binds the same active site to form an EI complex, but since it possesses only one methylene carbon it cannot undergo dehydrogenation. The statins, competitive inhibitors of HMG-CoA reductase, are the clinical example.
Why non-competitive inhibition behaves as it does
Because the inhibitor binds at a site distinct from the active site, its binding does not affect the binding of substrate — hence Km is unchanged. However, although the EI complex can still bind substrate, its efficiency at transforming substrate into product is decreased, and this is reflected in a reduced Vmax. Since the difficulty was never one of access to the site, excess substrate cannot overcome the inhibition.
Measuring potency
Inhibitors of the same enzyme are compared by Ki, the equilibrium dissociation constant of the EI complex: the lower the Ki, the more effective the inhibitor. A less rigorous alternative is IC₅₀, the concentration producing 50% inhibition, whose numerical value varies with the conditions of measurement.
Contrast with irreversible inhibition
Both of the above form dissociable, dynamic complexes, so fully active enzyme is recovered by removing the inhibitor from the medium. Irreversible inhibitors instead chemically modify the enzyme — making or breaking covalent bonds with residues essential for substrate binding, catalysis or conformation — so an enzyme “poisoned” by a heavy metal or acylating reagent remains inhibited even after the inhibitor is removed.
Framing
Enzyme kinetics is the quantitative measurement of reaction rates and the systematic study of the factors affecting them. Four principal factors are considered: substrate concentration, enzyme concentration, temperature and pH — together with the presence of inhibitors.
1 · Substrate concentration
As [S] increases, vi rises along a rectangular hyperbola towards Vmax, described by the Michaelis–Menten equation, vi = Vmax[S]/(Km + [S]). Three regions:
- When [S] ≪ Km, velocity is directly proportional to [S].
- When [S] = Km, velocity is half-maximal.
- When [S] ≫ Km, velocity is maximal and independent of [S], because all the enzyme exists as ES complex and no free enzyme remains available.
2 · Enzyme concentration
Under initial rate conditions with a large molar excess of substrate over enzyme (10³–10⁷), vi is directly proportional to the concentration of enzyme. This is the basis of clinical enzyme assay: measuring the initial velocity permits the quantity of enzyme in a biological sample to be estimated. Note that enzyme concentration alters Vmax but not Km.
3 · Temperature
Raising the temperature increases the kinetic energy of molecules, so a greater number exceed the activation-energy barrier, and increases their motion and hence the frequency of collisions. More frequent and more energetic collisions accelerate the reaction. The Q₁₀, or temperature coefficient, is the factor by which the rate increases for a 10 °C rise.
This holds only over a limited range. With continued heating the enzyme denatures, with rapid loss of catalytic activity; every enzyme therefore possesses a temperature optimum at which activity is maximal, with a steep decline beyond it.
4 · pH
Activity is maximal over a narrow pH range because both enzyme and substrate must bear the appropriate ionic charge. Consider a negatively charged enzyme (EH⁻) binding a positively charged substrate (SH⁺): as pH rises the proportion of EH⁻ increases while the proportion of SH⁺ falls, and only where the two overlap are both correctly charged. Extremes of pH additionally denature the protein.
5 · Inhibitors
Competitive inhibitors resemble the substrate and bind the active site, raising the apparent Km without altering Vmax; their effect is overcome by raising [S]. Non-competitive inhibitors bind elsewhere, leaving Km unchanged but decreasing Vmax, and cannot be overcome by excess substrate. Irreversible inhibitors chemically modify the enzyme and its activity is not restored by removing them.
What enzymes do NOT do
Whatever the conditions, an enzyme increases the rate by lowering the activation energy. It does not alter ΔG or the position of equilibrium, and therefore cannot force a reaction that is energetically unfavourable.