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Reaction Kinetics and Mass Action · Tutorial 40 of 799

The Enzyme-Substrate Reaction Mechanism

Learn to translate the elementary enzyme-substrate mechanism into concentration ODEs and verify its enzyme and substrate balances.

Beginner 9 min read

What You'll Learn

  • Write the elementary steps for enzyme binding, dissociation, and catalytic product formation.
  • Define free enzyme, enzyme-substrate complex, and total enzyme concentration.
  • Derive each concentration ODE from its reaction flux and stoichiometry.
  • Check enzyme and substrate-equivalent conservation in a closed system.
  • Diagnose sign, unit, and boundary errors in the mechanism equations.

From Binding Steps to an Enzyme Mechanism

An enzyme can bind a substrate, release it again, or convert it to product while returning to its free form. A useful starting point is to represent these events as separate elementary reactions. The resulting equations make clear which species change in each step, and whether the enzyme is free or bound at a given time.

Consider the minimal mechanism in a well-mixed compartment of constant volume:

$$ E+S \xrightleftharpoons[k_{\mathrm{off}}]{k_{\mathrm{on}}} ES \xrightarrow{k_{\mathrm{cat}}} E+P $$

Here \(E\) is free enzyme, \(S\) is free substrate, \(ES\) is the enzyme-substrate complex, and \(P\) is product. The forward binding step joins one enzyme molecule and one substrate molecule. Dissociation separates the complex into free enzyme and free substrate. Catalysis converts the bound substrate to product and returns the enzyme to its free form.

Definition: The free enzyme concentration is \([E]\), the concentration of enzyme not bound to substrate. The bound enzyme concentration is \([ES]\), the concentration of enzyme in the complex. In this tutorial, \(E\) and \(ES\) denote these concentrations directly. The total enzyme concentration is \(E_T=E+ES\), provided the enzyme has no other forms and is neither created nor removed from the compartment.

The rate constants are nonnegative. For concentration-based mass-action kinetics, \(k_{\mathrm{on}}\) has units of concentration\(^{-1}\) time\(^{-1}\), while \(k_{\mathrm{off}}\) and \(k_{\mathrm{cat}}\) have units of time\(^{-1}\). Thus each reaction flux has units of concentration per time. This minimal model assumes a single substrate-binding site, one-to-one binding, no enzyme loss or synthesis, no transport across the compartment boundary, and no product reaction or reverse catalytic step.

Mass-Action Fluxes and ODEs

The mass-action flux of a reaction is its rate of progress in the forward direction. For the three steps here, the fluxes are \(v_{\mathrm{on}}=k_{\mathrm{on}}ES\), \(v_{\mathrm{off}}=k_{\mathrm{off}}ES\), and \(v_{\mathrm{cat}}=k_{\mathrm{cat}}ES\). Binding depends on both free reactant concentrations; dissociation and catalysis each depend on the concentration of complex.

Each step changes the species according to its stoichiometry. Binding removes one \(E\) and one \(S\) and forms one \(ES\). Dissociation does the reverse. Catalysis removes one \(ES\) and forms one \(E\) and one \(P\). Adding the contributions from all three steps gives:

$$ \begin{aligned} \frac{dE}{dt} &=-k_{\mathrm{on}}ES+k_{\mathrm{off}}ES+k_{\mathrm{cat}}ES,\\ \frac{dS}{dt} &=-k_{\mathrm{on}}ES+k_{\mathrm{off}}ES,\\ \frac{d(ES)}{dt} &=k_{\mathrm{on}}ES-k_{\mathrm{off}}ES-k_{\mathrm{cat}}ES,\\ \frac{dP}{dt} &=k_{\mathrm{cat}}ES. \end{aligned} $$

In these equations, products such as \(ES\) mean the product of the concentrations \(E\) and \(S\); the complex concentration is written \(ES\) as a single species. To avoid confusing the two, the binding term can equivalently be written \(k_{\mathrm{on}}[E][S]\) and the complex terms \(k_{\mathrm{off}}[ES]\) and \(k_{\mathrm{cat}}[ES]\). Below, the symbols \(E,S,ES,P\) continue to denote concentrations.

A useful way to read the free-enzyme equation is by reaction rather than by sign alone. Binding removes free enzyme, dissociation returns free enzyme, and catalysis also returns free enzyme because it consumes the complex and produces \(E+P\). The free enzyme generally changes as complex forms and dissociates, and it is produced when the complex is consumed catalytically; it is the sum of free and bound enzyme that stays fixed under this model.

Key result: For this closed, minimal mechanism, \(E_T=E+ES\) is constant. The substrate-containing pool \(S+ES+P\) is also constant. These balances follow from the reaction stoichiometry; they do not require equilibrium or any assumption that concentrations change slowly.

Why the Balances Hold

The conservation statements can be checked directly from the ODEs. Adding the free-enzyme and complex equations cancels the binding, dissociation, and catalytic terms:

$$ \frac{d(E+ES)}{dt} =(-k_{\mathrm{on}}ES+k_{\mathrm{off}}ES+k_{\mathrm{cat}}ES) +(k_{\mathrm{on}}ES-k_{\mathrm{off}}ES-k_{\mathrm{cat}}ES)=0. $$

Therefore \(E+ES\) is constant in time. In the same way, adding the substrate, complex, and product equations gives:

$$ \frac{d(S+ES+P)}{dt} =(-k_{\mathrm{on}}ES+k_{\mathrm{off}}ES) +(k_{\mathrm{on}}ES-k_{\mathrm{off}}ES-k_{\mathrm{cat}}ES) +k_{\mathrm{cat}}ES=0. $$

So if initial concentrations are \(E(0), S(0), ES(0), P(0)\), then \(E_T=E(0)+ES(0)\) and \(S+ES+P=S(0)+ES(0)+P(0)\) at all later times. The second balance counts substrate-derived material in free substrate, complex, and product. It presumes that the product contains the substrate moiety represented by this mechanism and that no material enters or leaves the modeled system.

The same result follows by writing the stoichiometric change vectors in the species order \((E,S,ES,P)\). Binding changes the vector by \((-1,-1,+1,0)\), dissociation by \((+1,+1,-1,0)\), and catalysis by \((+1,0,-1,+1)\). The enzyme-counting weights \((1,0,1,0)\) and substrate-equivalent weights \((0,1,1,1)\) each have zero dot product with every change vector. Thus every reaction preserves both totals, and any sum of their reaction fluxes preserves them as well.

Worked Example: Calculate the Initial Concentration Rates

Suppose a hypothetical system has \(E=0.20\ \mu\mathrm{M}\), \(S=10.0\ \mu\mathrm{M}\), \(ES=0.050\ \mu\mathrm{M}\), and \(P=0.10\ \mu\mathrm{M}\) at the instant of interest. Let \(k_{\mathrm{on}}=0.040\ \mu\mathrm{M}^{-1}\mathrm{min}^{-1}\), \(k_{\mathrm{off}}=0.30\ \mathrm{min}^{-1}\), and \(k_{\mathrm{cat}}=0.80\ \mathrm{min}^{-1}\). Find the instantaneous rates of change.

The binding flux is \(v_{\mathrm{on}}=k_{\mathrm{on}}ES=(0.040)(0.20)(10.0)=0.0800\ \mu\mathrm{M}/\mathrm{min}\). The dissociation flux is \(v_{\mathrm{off}}=k_{\mathrm{off}}ES=(0.30)(0.050)=0.0150\ \mu\mathrm{M}/\mathrm{min}\), and the catalytic flux is \(v_{\mathrm{cat}}=k_{\mathrm{cat}}ES=(0.80)(0.050)=0.0400\ \mu\mathrm{M}/\mathrm{min}\).

Substituting the fluxes into the ODEs:

$$ \begin{aligned} \frac{dE}{dt}&=-0.0800+0.0150+0.0400=-0.0250\ \mu\mathrm{M}/\mathrm{min},\\ \frac{dS}{dt}&=-0.0800+0.0150=-0.0650\ \mu\mathrm{M}/\mathrm{min},\\ \frac{d(ES)}{dt}&=0.0800-0.0150-0.0400=0.0250\ \mu\mathrm{M}/\mathrm{min},\\ \frac{dP}{dt}&=0.0400\ \mu\mathrm{M}/\mathrm{min}. \end{aligned} $$

The free enzyme is decreasing at this instant because the binding flux exceeds the two fluxes that return enzyme to its free form. The enzyme balance checks: \(-0.0250+0.0250=0\). The substrate-equivalent balance also checks: \(-0.0650+0.0250+0.0400=0\).

Worked Example: Use Total Enzyme to Set the Free Concentration

In a second hypothetical system, total enzyme is \(E_T=0.500\ \mu\mathrm{M}\), and \(ES=0.120\ \mu\mathrm{M}\). The free substrate is \(S=2.00\ \mu\mathrm{M}\). Let \(k_{\mathrm{on}}=0.100\ \mu\mathrm{M}^{-1}\mathrm{min}^{-1}\), \(k_{\mathrm{off}}=0.200\ \mathrm{min}^{-1}\), and \(k_{\mathrm{cat}}=0.500\ \mathrm{min}^{-1}\). First find \(E\), then calculate the concentration rates.

By definition, \(E=E_T-ES=0.500-0.120=0.380\ \mu\mathrm{M}\). The three fluxes are \(v_{\mathrm{on}}=(0.100)(0.380)(2.00)=0.0760\ \mu\mathrm{M}/\mathrm{min}\), \(v_{\mathrm{off}}=(0.200)(0.120)=0.0240\ \mu\mathrm{M}/\mathrm{min}\), and \(v_{\mathrm{cat}}=(0.500)(0.120)=0.0600\ \mu\mathrm{M}/\mathrm{min}\).

Therefore:

$$ \begin{aligned} \frac{dE}{dt}&=-0.0760+0.0240+0.0600=0.00800\ \mu\mathrm{M}/\mathrm{min},\\ \frac{dS}{dt}&=-0.0760+0.0240=-0.0520\ \mu\mathrm{M}/\mathrm{min},\\ \frac{d(ES)}{dt}&=0.0760-0.0240-0.0600=-0.00800\ \mu\mathrm{M}/\mathrm{min},\\ \frac{dP}{dt}&=0.0600\ \mu\mathrm{M}/\mathrm{min}. \end{aligned} $$

Here free enzyme increases because the complex is being consumed overall. In particular, catalysis contributes positively to \(dE/dt\): it removes \(ES\) and produces free \(E\). The rates satisfy \(d(E+ES)/dt=0\), as required.

Worked Example: Check Rates and Units in a Second State

For a third hypothetical state, let \(E=0.300\ \mu\mathrm{M}\), \(S=5.00\ \mu\mathrm{M}\), and \(ES=0.100\ \mu\mathrm{M}\). Take \(k_{\mathrm{on}}=0.0200\ \mu\mathrm{M}^{-1}\mathrm{min}^{-1}\), \(k_{\mathrm{off}}=0.100\ \mathrm{min}^{-1}\), and \(k_{\mathrm{cat}}=0.400\ \mathrm{min}^{-1}\). The binding flux is \((0.0200)(0.300)(5.00)=0.0300\ \mu\mathrm{M}/\mathrm{min}\), dissociation is \((0.100)(0.100)=0.0100\ \mu\mathrm{M}/\mathrm{min}\), and catalysis is \((0.400)(0.100)=0.0400\ \mu\mathrm{M}/\mathrm{min}\). Each product has units of concentration per time.

It follows that \(dE/dt=-0.0300+0.0100+0.0400=0.0200\ \mu\mathrm{M}/\mathrm{min}\), \(dS/dt=-0.0300+0.0100=-0.0200\ \mu\mathrm{M}/\mathrm{min}\), \(d(ES)/dt=0.0300-0.0100-0.0400=-0.0200\ \mu\mathrm{M}/\mathrm{min}\), and \(dP/dt=0.0400\ \mu\mathrm{M}/\mathrm{min}\). The enzyme-balance derivative is \(0.0200-0.0200=0\), and the substrate-equivalent derivative is \(-0.0200-0.0200+0.0400=0\).

Assumptions, Diagnostics, and Scope

These ODEs apply when the stated elementary steps are an adequate description of the system and concentrations are well mixed in a fixed volume. Concentration changes are attributed only to binding, dissociation, and catalysis. If enzyme is synthesized or degraded, if substrate or product is transported, or if product can rebind or react further, additional species or fluxes are needed and the corresponding totals need not remain constant.

Several checks can reveal an incorrectly written model:

  • Check the reaction signs. Binding must decrease free \(E\) and \(S\) while increasing \(ES\). Catalysis must decrease \(ES\) while increasing both \(E\) and \(P\).
  • Check the enzyme balance. If the equations do not give \(d(E+ES)/dt=0\), a term may be missing or have the wrong sign for this mechanism.
  • Check units. \(k_{\mathrm{on}}ES\), \(k_{\mathrm{off}}ES\), and \(k_{\mathrm{cat}}ES\) must all have units of concentration per time. A bimolecular association constant used with concentration variables must carry inverse-concentration units.
  • Check the modeled species. \(E_T=E+ES\) is valid only if these are the only enzyme forms represented. A second bound state would need to be included in the total.
  • Check boundaries. With nonnegative rate constants and concentrations, a species at zero cannot have a negative derivative under these equations. For example, at \(E=0\), \(dE/dt=(k_{\mathrm{off}}+k_{\mathrm{cat}})ES\geq0\); at \(ES=0\), \(d(ES)/dt=k_{\mathrm{on}}ES\geq0\). A negative derivative at such a boundary suggests a sign or implementation error.

This explicit mechanism is more informative than a single empirical expression for product formation because it tracks free enzyme and the bound complex separately. That detail is important when complex formation, dissociation, and catalysis each matter to the model. The tradeoff is that the mechanism requires separate state variables and rate constants, and the assumed steps must be appropriate for the system. The equations above specify the mechanism itself; they do not assume that the complex is at equilibrium or that its concentration changes slowly.

Common pitfall: It is wrong to say that catalysis consumes free enzyme because the catalytic rate is proportional to the amount of enzyme. In this mechanism, catalysis consumes the complex \(ES\) and produces free enzyme \(E\) plus product \(P\). The catalytic contribution is therefore \(-k_{\mathrm{cat}}ES\) in \(d(ES)/dt\), \(+k_{\mathrm{cat}}ES\) in \(dE/dt\), and \(+k_{\mathrm{cat}}ES\) in \(dP/dt\). Free enzyme can still decrease overall if binding exceeds its return through dissociation and catalysis.

Check Your Understanding

Use the mechanism and balances from this tutorial. The solutions follow each question so that the sign and conservation checks can be compared directly.

  1. If \(E=0.10\ \mu\mathrm{M}\), \(S=3.0\ \mu\mathrm{M}\), \(ES=0.020\ \mu\mathrm{M}\), \(k_{\mathrm{on}}=0.050\ \mu\mathrm{M}^{-1}\mathrm{min}^{-1}\), \(k_{\mathrm{off}}=0.10\ \mathrm{min}^{-1}\), and \(k_{\mathrm{cat}}=0.20\ \mathrm{min}^{-1}\), calculate the three reaction fluxes.
  2. For the same state and parameters, calculate \(dE/dt\), \(dS/dt\), \(d(ES)/dt\), and \(dP/dt\).
  3. Why does \(E_T=E+ES\) remain constant even though catalysis decreases the complex concentration?
  4. Which additional enzyme state must be included in the total if the model also represents an enzyme bound to an inhibitor?

Solutions. For question 1, \(v_{\mathrm{on}}=(0.050)(0.10)(3.0)=0.0150\ \mu\mathrm{M}/\mathrm{min}\), \(v_{\mathrm{off}}=(0.10)(0.020)=0.00200\ \mu\mathrm{M}/\mathrm{min}\), and \(v_{\mathrm{cat}}=(0.20)(0.020)=0.00400\ \mu\mathrm{M}/\mathrm{min}\). For question 2, \(dE/dt=-0.0150+0.00200+0.00400=-0.00900\ \mu\mathrm{M}/\mathrm{min}\), \(dS/dt=-0.0150+0.00200=-0.0130\ \mu\mathrm{M}/\mathrm{min}\), \(d(ES)/dt=0.0150-0.00200-0.00400=0.00900\ \mu\mathrm{M}/\mathrm{min}\), and \(dP/dt=0.00400\ \mu\mathrm{M}/\mathrm{min}\). The enzyme rates sum to zero, and the substrate-equivalent rates sum to \(-0.0130+0.00900+0.00400=0\). For question 3, catalysis removes one enzyme-containing complex but produces one free enzyme, so it changes the partition between \(ES\) and \(E\), not their sum. For question 4, include the inhibitor-bound enzyme concentration as another enzyme-containing state in the total.

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