Enzyme kinetics

Concept

Enzyme kinetics

T-094Home BU-306Threads structure · energy

Statement

The Michaelis-Menten model of catalysis.

Why it matters

protein-folding explains how a polypeptide reaches the specific three-dimensional shape that gives an enzyme its active site, but shape alone does not say how fast a reaction actually proceeds once substrate is present; Michaelis-Menten kinetics is the quantitative model that connects an enzyme's structure to its measurable catalytic performance, and is the standard language in which catalytic efficiency is discussed and compared across enzymes. allosteric-regulation and enzyme inhibition (both routinely analysed against this same kinetic framework) are essentially deviations from, or perturbations of, the Michaelis-Menten baseline, which is exactly why the baseline model must be established first.

It matters practically as well: \(K_M\) and \(k_{cat}\) are the two numbers used throughout biochemistry and pharmacology to characterise an enzyme's substrate affinity and turnover speed, and drug design against enzyme targets is routinely expressed directly in these kinetic terms.

Hypotheses

Substrate concentration \([S]\) is much greater than enzyme concentration \([E]\), so that a steady-state concentration of enzyme-substrate complex is reached and maintained.Under this condition the rate of ES complex formation equals its rate of breakdown almost immediately relative to the timescale of the overall reaction (the steady-state approximation), which is what makes the algebra of Step 2 tractable; without it, \([ES]\) itself changes on the same timescale as the reaction being studied and no simple closed-form rate law exists. The reaction is measured at initial velocity, before significant product has accumulated.This avoids product inhibition and the reverse reaction complicating the rate law — the entire derivation implicitly assumes the reverse reaction (\(E+P\to ES\)) is negligible, which is only true early in the reaction, before appreciable product \(P\) has built up. The classical derivation additionally assumes a single substrate and a single catalytic step; enzymes with multiple substrates, multiple intermediate complexes, or cooperative subunit interactions (allosteric-regulation) require extended kinetic treatments that reduce to the simple hyperbolic form only in special limiting cases.

Proof

E+S \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} ES \xrightarrow{k_2} E+P

Enzyme and substrate reversibly form a complex \(ES\), which then breaks down irreversibly (on the timescale of interest, Hypotheses) to release product and regenerate free enzyme. A

\frac{d[ES]}{dt}=k_1[E][S]-k_{-1}[ES]-k_2[ES]=0\ \ (\text{steady state})

Applying the steady-state approximation (Hypotheses) to \([ES]\), its formation rate must equal its breakdown rate; setting the net rate of change to zero is the key simplifying step that converts a coupled differential system into solvable algebra. B

K_M \equiv \frac{k_{-1}+k_2}{k_1} = \frac{[E][S]}{[ES]}

Rearranging Step 2 and substituting \([E]=[E]_{total}-[ES]\) defines the Michaelis constant \(K_M\), a composite of all three rate constants, as the substrate concentration at which half the enzyme population is bound as \(ES\) (shown explicitly in Step 4). B

v_0 = \frac{V_{max}[S]}{K_M+[S]},\qquad V_{max}=k_2[E]_{total}

Since initial velocity \(v_0=k_2[ES]\) (product release, Step 1) and \(V_{max}\) is the rate when every enzyme molecule is saturated as \(ES\) (i.e. \([ES]=[E]_{total}\)), solving Step 3 for \([ES]\) and substituting gives the Michaelis-Menten equation directly; setting \([S]=K_M\) gives \(v_0=V_{max}/2\), confirming \(K_M\)'s interpretation as a half-saturation constant. A

k_{cat}\equiv k_2,\qquad \frac{k_{cat}}{K_M}\ \text{is the catalytic efficiency}

\(k_{cat}\) (the turnover number, equal to \(k_2\) in this simple mechanism) gives the maximum number of substrate molecules converted per enzyme molecule per second; the ratio \(k_{cat}/K_M\) combines turnover speed and substrate affinity into a single figure of merit used to compare different enzymes, or the same enzyme against different substrates. A

Result

v_0 = \dfrac{V_{max}[S]}{K_M+[S]}

Reading. Reaction velocity rises with substrate concentration but saturates toward \(V_{max}\) as the enzyme's finite population of active sites becomes fully occupied; \(K_M\) sets the substrate concentration at which the enzyme is running at half its maximum rate, a rough inverse measure of substrate affinity.

Scope. Valid for a single-substrate, single-intermediate enzyme under steady-state, initial-velocity conditions (Hypotheses); does not describe cooperative (sigmoidal, non-hyperbolic) kinetics shown by allosteric multi-subunit enzymes.

Corollaries & converses

Competitive inhibition raises the apparent \(K_M\) without changing \(V_{max}\) (the inhibitor and substrate compete for the same site, but enough substrate can always outcompete the inhibitor), while noncompetitive inhibition lowers \(V_{max}\) without changing \(K_M\) — both are standard diagnostic deviations from the Result, analysed by exactly which parameter shifts.
allosteric-regulation produces a sigmoidal, not hyperbolic, velocity-versus-substrate curve, because cooperative subunit interactions violate Step 1's single-site, single-step assumption — the departure from the Michaelis-Menten hyperbola is itself the standard signature used to detect cooperativity.
Converse: a measured \(v_0\) versus \([S]\) plot that fits a hyperbola well, using the double-reciprocal (Lineweaver-Burk) linear transform of the Result to extract \(K_M\) and \(V_{max}\) from a straight line, is itself evidence the enzyme is behaving as a simple, non-cooperative Michaelis-Menten catalyst under the conditions tested.

Fails without

Drop the steady-state assumption (Hypotheses): at very early or very late timepoints, or if \([E]\) is not much smaller than \([S]\), \([ES]\) is itself still changing rather than constant, and Step 2's key simplification is invalid — the resulting instantaneous rate no longer follows the simple hyperbolic law of Step 4, and full time-course kinetics (a harder problem) must be solved instead.
Measure well past initial velocity, after significant product has accumulated: the reverse reaction \(E+P\to ES\) (assumed negligible, Hypotheses) becomes non-negligible, and product may itself competitively inhibit the enzyme; the observed rate falls below what the Result predicts purely from \([S]\), not because the kinetic mechanism has changed but because a violated assumption is distorting the measurement.

Common errors

Interpreting \(K_M\) as literally the dissociation constant of the \(ES\) complex; it equals \((k_{-1}+k_2)/k_1\) (Step 3), which reduces to the true dissociation constant \(k_{-1}/k_1\) only in the special case \(k_2\ll k_{-1}\), not in general.
Confusing \(K_M\) (a concentration, with units of molarity) with \(k_{cat}\) (a rate constant, with units of inverse time) — the two describe different physical quantities and are combined, not interchanged, in the efficiency ratio of Step 5.
Assuming a lower \(K_M\) always means a "better" enzyme; a low \(K_M\) indicates high apparent affinity for substrate at low concentration, but says nothing about \(k_{cat}\) itself, so overall catalytic efficiency requires both parameters together (Step 5).
Extrapolating the hyperbolic Result to enzymes with multiple interacting subunits without checking for the sigmoidal deviation characteristic of cooperativity (Corollaries).

Discussion

Victor Henri first proposed the underlying rate equation around 1903; Leonor Michaelis and Maud Menten put it on a firm, quantitatively verified experimental footing in 1913, and the steady-state derivation in the form given here (Step 2) is due to George Briggs and John Haldane in 1925, refining the original quasi-equilibrium assumption into the more general steady-state approximation still used today.

The double-reciprocal (Lineweaver-Burk) linearisation, \(1/v_0 = (K_M/V_{max})(1/[S]) + 1/V_{max}\), was historically important for extracting \(K_M\) and \(V_{max}\) from limited data by hand using linear regression, but it distorts experimental error disproportionately at low \([S]\) (where \(1/[S]\) is large); nonlinear regression directly on the hyperbolic Result is now standard wherever computation is available.

Common misconception: that \(V_{max}\) is a fixed, universal property of an enzyme. It is proportional to total enzyme concentration \([E]_{total}\) (Step 4) and therefore depends on how much enzyme is present in a given assay; \(k_{cat}\), the per-molecule turnover number, is the concentration-independent intrinsic property, not \(V_{max}\) itself.

Worked examples

K_M = 2\times10^{-3}\ \text{M},\quad V_{max}=10\ \mu\text{M/s},\quad [S]=8\times10^{-3}\ \text{M}

Substituting directly into the Result: \(v_0 = \dfrac{10\times8\times10^{-3}}{2\times10^{-3}+8\times10^{-3}} = \dfrac{10\times8}{10} = 8\ \mu\text{M/s}\), i.e. 80% of \(V_{max}\), consistent with \([S]\) being four times \(K_M\). A

v_0 = 8\ \mu\text{M/s}\ \ (80\%\ \text{of}\ V_{max})

Reading. Once \([S]\) substantially exceeds \(K_M\), further increases in substrate concentration yield rapidly diminishing returns in rate, since the enzyme population is already mostly saturated.

Scope. The same calculation at \([S]=K_M\) would give exactly \(v_0=V_{max}/2=5\ \mu\text{M/s}\), the defining property of \(K_M\) from Step 4.

Problems

An enzyme has \(K_M=5\times10^{-4}\ \text{M}\) and \(V_{max}=20\ \mu\text{M/s}\). Find \(v_0\) at \([S]=5\times10^{-4}\ \text{M}\) and at \([S]=5\times10^{-2}\ \text{M}\).
Solution
At \([S]=K_M\): \(v_0=V_{max}/2=10\ \mu\text{M/s}\) directly (Step 4's half-saturation property). At \([S]=5\times10^{-2}\ \text{M}\) (100× \(K_M\)): \(v_0=\dfrac{20\times5\times10^{-2}}{5\times10^{-4}+5\times10^{-2}}\approx\dfrac{1.0}{0.0505}\approx19.8\ \mu\text{M/s}\), close to \(V_{max}\) since \([S]\gg K_M\).
Two enzymes act on the same substrate: Enzyme A has \(K_M=1\times10^{-5}\ \text{M}\), \(k_{cat}=10\ \text{s}^{-1}\); Enzyme B has \(K_M=1\times10^{-3}\ \text{M}\), \(k_{cat}=1000\ \text{s}^{-1}\). Using \(k_{cat}/K_M\) (Step 5), determine which is the more catalytically efficient enzyme at low substrate concentration.
Solution
Enzyme A: \(k_{cat}/K_M = 10/10^{-5}=10^{6}\ \text{M}^{-1}\text{s}^{-1}\). Enzyme B: \(k_{cat}/K_M=1000/10^{-3}=10^{6}\ \text{M}^{-1}\text{s}^{-1}\). The two enzymes have identical catalytic efficiency despite very different individual \(K_M\) and \(k_{cat}\) values — a direct illustration of why efficiency comparisons must use the combined ratio (Common errors), not either parameter alone.
A researcher measures reaction rate well after 50% of substrate has already been converted to product, rather than at true initial velocity, and obtains an apparent \(K_M\) noticeably higher than the enzyme's true value. Explain this discrepancy using Fails without.
Solution
Measuring after substantial product accumulation violates the initial-velocity assumption (Hypotheses); the reverse reaction and/or product inhibition (Fails without, second bullet) both act to slow the observed rate below what the true forward-only kinetics would predict at the nominal \([S]\), which is measured (bookkeeping) rather than the actual remaining substrate concentration during the assay interval. This lowered observed rate, misfit to the simple hyperbolic Result, yields an inflated apparent \(K_M\).