On this page
Add more substrate to an enzyme reaction and it speeds up. Add more still, and the rate barely changes. This levelling off, called saturation, is the signature of an enzyme-catalysed reaction, and in 1913 Leonor Michaelis and Maud Menten showed how to describe it with a simple equation. More than a century later, the Michaelis–Menten equation is still the starting point for understanding how fast enzymes work, how tightly they bind their substrates and how drugs inhibit them.
The observation: a curve that levels off
Measure the initial rate of an enzyme reaction (v₀) at different substrate concentrations ([S]), keeping the enzyme concentration fixed. The plot of v₀ against [S] has a characteristic shape:
- at low [S], the rate rises almost in proportion to [S] (roughly first order in substrate);
- at high [S], the rate approaches a maximum and hardly changes (roughly zero order in substrate).
The curve is a rectangular hyperbola. The reason is that there are only a limited number of enzyme molecules. Once nearly all of them have substrate bound, adding more substrate can’t make the reaction go faster. The enzymes are working flat out.
(If you’re new to rate orders, see reaction rates and catalysts.)
The model
Michaelis and Menten proposed this scheme:
E + S ⇌ ES → E + P
- The enzyme (E) and substrate (S) bind reversibly to form an enzyme–substrate complex (ES), with rate constants k₁ (binding) and k₋₁ (unbinding).
- ES converts to product (P), releasing free enzyme, with rate constant k₂ (also called k_cat, the turnover number).
Two simplifying assumptions make the maths easy:
- We measure the initial rate, when very little product has formed, so the reverse reaction from P can be ignored.
- The steady-state approximation (introduced by Briggs and Haldane in 1925): shortly after mixing, [ES] stays roughly constant, because it forms as fast as it breaks down.
Deriving the equation
At steady state, the rate of forming ES equals the rate of breaking it down:
k₁[E][S] = (k₋₁ + k₂)[ES]
The free enzyme is whatever isn’t bound: [E] = [E]ₜ − [ES], where [E]ₜ is the total enzyme concentration. Substituting and rearranging:
[ES] = [E]ₜ[S] ⁄ (K_M + [S]), where K_M = (k₋₁ + k₂) ⁄ k₁
The rate of product formation is v₀ = k₂[ES], and the maximum possible rate, when all the enzyme is bound, is V_max = k₂[E]ₜ. So:
v₀ = V_max [S] ⁄ (K_M + [S])
This is the Michaelis–Menten equation.
Checking the limits
- When [S] ≪ K_M: v₀ ≈ (V_max ⁄ K_M)[S]. The rate is proportional to [S]: first order.
- When [S] ≫ K_M: v₀ ≈ V_max. The rate no longer depends on [S]: zero order, saturation.
- When [S] = K_M: v₀ = V_max ⁄ 2.
That last result gives the most useful definition of K_M.
What the parameters mean
V_max: the maximum rate
V_max is the rate when the enzyme is fully saturated with substrate. It depends on how much enzyme is present: double the enzyme and V_max doubles.
k_cat: the turnover number
k_cat = V_max ⁄ [E]ₜ is the number of substrate molecules one enzyme molecule converts per second when saturated. It measures the intrinsic speed of the catalytic step. Values range enormously:
| Enzyme | Approximate k_cat (s⁻¹) |
|---|---|
| Catalase | ~10⁷ |
| Carbonic anhydrase | ~10⁶ |
| Acetylcholinesterase | ~10⁴ |
| Chymotrypsin | ~10² |
| Lysozyme | ~0.5 |
K_M: the Michaelis constant
K_M is the substrate concentration at which the rate is half of V_max. It has units of concentration (for example, mmol L⁻¹ or μmol L⁻¹).
- A low K_M means the enzyme reaches half its maximum rate at a low substrate concentration. It’s often described as having a high affinity for the substrate.
- A high K_M means a lot of substrate is needed.
Strictly, K_M equals the dissociation constant of ES only when k₂ is much smaller than k₋₁; otherwise it includes the catalytic step too. So “K_M measures affinity” is a useful approximation, not a definition.
A biological example: hexokinase (in most tissues) has a K_M for glucose of about 0.1 mmol L⁻¹, well below normal blood glucose (about 5 mmol L⁻¹), so it works near its maximum almost all the time. Glucokinase in the liver has a K_M of about 10 mmol L⁻¹, so its activity rises and falls with blood glucose, letting the liver take up glucose mainly after meals.
k_cat ⁄ K_M: catalytic efficiency
The ratio k_cat ⁄ K_M (the “specificity constant”) measures how efficiently an enzyme works at low substrate concentrations, which is common in cells. Its upper limit is set by how fast enzyme and substrate can meet by diffusion, around 10⁸ to 10⁹ L mol⁻¹ s⁻¹. Enzymes that reach this limit, such as triose phosphate isomerase, are sometimes called “catalytically perfect”: the reaction happens essentially every time the substrate arrives.
The Lineweaver–Burk plot
Estimating V_max from a curve that only approaches it is awkward. Taking reciprocals of both sides of the Michaelis–Menten equation gives a straight line:
1 ⁄ v₀ = (K_M ⁄ V_max)(1 ⁄ [S]) + 1 ⁄ V_max
Plotting 1 ⁄ v₀ against 1 ⁄ [S] (the Lineweaver–Burk, or double-reciprocal, plot) gives:
- y-intercept = 1 ⁄ V_max
- x-intercept = −1 ⁄ K_M
- gradient = K_M ⁄ V_max
The plot is very useful for seeing how inhibitors change K_M and V_max (see competitive vs non-competitive inhibition). Its drawback is that it gives too much weight to measurements at low [S], which are usually the least accurate, so modern analyses fit the curve directly with computer software. (For fitting data to straight lines in general, see graphs in chemistry.)
Worked example
An enzyme at a total concentration of 2.0 × 10⁻⁹ mol L⁻¹ gives these initial rates:
| [S] (mmol L⁻¹) | v₀ (μmol L⁻¹ s⁻¹) |
|---|---|
| 0.5 | 1.00 |
| 1.0 | 1.67 |
| 2.0 | 2.50 |
| 4.0 | 3.33 |
| 8.0 | 4.00 |
Step 1: take reciprocals.
| 1 ⁄ [S] (L mmol⁻¹) | 1 ⁄ v₀ (L s μmol⁻¹) |
|---|---|
| 2.00 | 1.000 |
| 1.00 | 0.600 |
| 0.50 | 0.400 |
| 0.25 | 0.300 |
| 0.125 | 0.250 |
Step 2: find the line. The points lie on a straight line. Between 1 ⁄ [S] = 2.00 and 0.125, 1 ⁄ v₀ changes from 1.000 to 0.250, so:
gradient = (1.000 − 0.250) ⁄ (2.00 − 0.125) = 0.750 ⁄ 1.875 = 0.400
y-intercept: 0.250 − 0.400 × 0.125 = 0.200
Step 3: read off the constants.
V_max = 1 ⁄ 0.200 = 5.0 μmol L⁻¹ s⁻¹ K_M = gradient × V_max = 0.400 × 5.0 = 2.0 mmol L⁻¹
Check: at [S] = 2.0 mmol L⁻¹ = K_M, v₀ should be V_max ⁄ 2 = 2.5 μmol L⁻¹ s⁻¹. It is. ✓
Step 4: turnover number.
k_cat = V_max ⁄ [E]ₜ = (5.0 × 10⁻⁶ mol L⁻¹ s⁻¹) ⁄ (2.0 × 10⁻⁹ mol L⁻¹) = 2.5 × 10³ s⁻¹
Each enzyme molecule converts 2,500 substrate molecules per second when saturated.
Step 5: catalytic efficiency.
k_cat ⁄ K_M = 2.5 × 10³ s⁻¹ ⁄ 2.0 × 10⁻³ mol L⁻¹ = 1.25 × 10⁶ L mol⁻¹ s⁻¹
Limitations of the model
- It describes single-substrate reactions measured at initial rates. Many real enzymes use two substrates, though the same ideas can be extended.
- Allosteric enzymes, which have several interacting active sites, give sigmoidal (S-shaped) curves rather than hyperbolas, and don’t follow the simple equation.
- Conditions such as pH, temperature and ionic strength must be held constant (see how temperature and pH affect enzymes).
Key takeaways
- Enzyme rates saturate at high substrate concentration because the enzyme molecules are all occupied.
- The Michaelis–Menten equation, v₀ = V_max[S] ⁄ (K_M + [S]), follows from the E + S ⇌ ES → E + P scheme and the steady-state assumption.
- V_max is the saturated rate; k_cat is the turnover number; K_M is the [S] giving half V_max.
- k_cat ⁄ K_M measures catalytic efficiency, capped by diffusion at about 10⁸ to 10⁹ L mol⁻¹ s⁻¹.
- The Lineweaver–Burk plot turns the hyperbola into a straight line: intercepts give 1 ⁄ V_max and −1 ⁄ K_M. For the basics, see enzymes: how they work.
Advertisement