Reducing Fast Binding to an Effective Rate

Why Fast Binding Changes the Rate Law

A reaction can contain a fast reversible binding step and a slower step that removes or converts the bound complex. When binding relaxes much faster than conversion, the complex concentration need not be tracked as an independent, rapidly changing state. Instead, it can be approximated from the binding equilibrium, and the slower process can be expressed as an effective rate in terms of the remaining reactant and the available binding partner.

The reduced rate depends on what is held fixed. A specified free concentration of binding partner is not the same as a specified total concentration. The first case can have an externally maintained reservoir; the second describes a finite pool that is shared between free and bound forms. Keeping that distinction explicit is essential to deriving a valid rate law.

Definition: Consider \(A+B \rightleftharpoons C\), with forward rate constant \(k_{\mathrm{on}}\) and reverse rate constant \(k_{\mathrm{off}}\). Suppose \(C\) is converted at rate \(k_{\mathrm{cat}}C\), with the conversion returning \(B\): \(C \rightarrow P+B\). The dissociation constant is \(K_d=k_{\mathrm{off}}/k_{\mathrm{on}}\). Under rapid equilibrium, the binding relation is \(C=AB/K_d\); the conversion rate is then approximated by substituting this equilibrium complex concentration into \(k_{\mathrm{cat}}C\).

Concentrations are assumed to be well mixed and measured in a constant volume. The reaction is one-to-one, the binding step follows mass action, and no other reactions change the species during the reduction. Returning \(B\) in the conversion step makes its total concentration \(B_T=B+C\) conserved in a closed system. The amount of unconverted \(A\), including bound \(A\), is \(A_T=A+C\); it decreases when \(C\) converts to product.

From Binding Equilibrium to an Effective Rate

Before making the approximation, the mass-action equation for the complex is

\frac{dC}{dt}=k_{\mathrm{on}}AB-k_{\mathrm{off}}C-k_{\mathrm{cat}}C.

The rapid-equilibrium approximation applies when binding relaxes on a much shorter time scale than conversion. To leading order, the binding association and dissociation fluxes balance, \(k_{\mathrm{on}}AB\approx k_{\mathrm{off}}C\), so \(C\approx AB/K_d\). This is not the exact condition \(dC/dt=0\) for the full equation when conversion is occurring: the small departure from binding equilibrium supplies the conversion flux. The approximation treats that departure as negligible in calculating the leading-order complex concentration.

There are two common ways to specify the binding partner. If its free concentration is held at \(B_f\), then

C=\frac{AB_f}{K_d}.

This relation is linear in free \(A\). If \(B_f\) is maintained by a reservoir, the amount of \(B\) bound in \(C\) does not have to be less than the specified free concentration: the reservoir supplies or absorbs \(B\) as needed.

If instead the system is closed and the total amount of binding partner is fixed at \(B_T\), then \(B=B_T-C\). Substitution into the equilibrium relation gives

C=\frac{A(B_T-C)}{K_d} \qquad\Longrightarrow\qquad C=\frac{B_TA}{K_d+A}.

This saturating expression is in terms of free \(A\). It cannot be obtained by treating a fixed free \(B_f\) as though it were a fixed total \(B_T\).

Worked Example: Fixed Free Partner Versus Fixed Total Partner

Suppose \(K_d=0.15\ \mu\mathrm{M}\), and free \(A=0.30\ \mu\mathrm{M}\). Compare the equilibrium complex concentration when free \(B\) is maintained at \(B_f=0.60\ \mu\mathrm{M}\) with the result when a closed system has total \(B_T=0.60\ \mu\mathrm{M}\).

Fixed free \(B\). Use \(C=AB_f/K_d\):

C=\frac{(0.30\ \mu\mathrm{M})(0.60\ \mu\mathrm{M})}{0.15\ \mu\mathrm{M}} =1.20\ \mu\mathrm{M}.

This is consistent with free \(B\) maintained by a reservoir. In this case, \(0.60\ \mu\mathrm{M}\) is the free concentration, not the total available amount of \(B\).

Fixed total \(B\). Use \(C=B_TA/(K_d+A)\):

C=\frac{(0.60\ \mu\mathrm{M})(0.30\ \mu\mathrm{M})} {0.15\ \mu\mathrm{M}+0.30\ \mu\mathrm{M}} =0.400\ \mu\mathrm{M}.

Here \(C\) is less than \(B_T=0.60\ \mu\mathrm{M}\), as required because the complex contains \(B\) drawn from that finite total pool. The two answers differ because the stated conditions describe different systems.

Rates in Terms of Total Unconverted Reactant

For many applications the quantity of interest is \(A_T=A+C\), not free \(A\) alone. Because conversion removes \(A\) from the unconverted pool while returning \(B\), its balance is \(dA_T/dt=-k_{\mathrm{cat}}C\). The reduction therefore requires expressing the equilibrium complex in terms of \(A_T\) and the appropriate description of \(B\).

When free \(B=B_f\) is maintained, combine \(C=AB_f/K_d\) with \(A_T=A+C\). This gives \(A_T=A(1+B_f/K_d)\), and hence

C=\frac{B_f}{K_d+B_f}A_T, \qquad -\frac{dA_T}{dt} =k_{\mathrm{cat}}\frac{B_f}{K_d+B_f}A_T.

The effective first-order rate constant for loss of total unconverted \(A\) is \(k_{\mathrm{cat}}B_f/(K_d+B_f)\), with units of inverse time. The fraction \(B_f/(K_d+B_f)\) is the fraction of \(A_T\) bound under this equilibrium relation.

For a closed system with fixed total \(B_T\), both totals include complex: \(A_T=A+C\) and \(B_T=B+C\). Substituting \(A=A_T-C\) and \(B=B_T-C\) into \(K_dC=AB\) yields

$$ C^2-(A_T+B_T+K_d)C+A_TB_T=0. $$

The physically meaningful root is the smaller one, because \(C\) cannot exceed either total concentration:

C=\frac{A_T+B_T+K_d- \sqrt{(A_T+B_T+K_d)^2-4A_TB_T}}{2}.

The reduced rate is \( -dA_T/dt=k_{\mathrm{cat}}C\), with this expression for \(C\). Unlike the buffered-free-partner result, it is generally not a simple first-order rate in \(A_T\): the fraction of \(B_T\) bound can change as \(A_T\) changes.

Worked Example: Effective First-Order Rate With Buffered Free Partner

Suppose \(K_d=0.15\ \mu\mathrm{M}\), \(B_f=0.60\ \mu\mathrm{M}\), \(A_T=2.00\ \mu\mathrm{M}\), and \(k_{\mathrm{cat}}=0.10\ \mathrm{min}^{-1}\). Assume free \(B\) is maintained at the stated concentration and binding is fast relative to conversion. Find the complex concentration and the initial rate of loss of \(A_T\).

The bound fraction is

\frac{B_f}{K_d+B_f} =\frac{0.60}{0.15+0.60} =0.800.

Thus \(C=0.800(2.00\ \mu\mathrm{M})=1.60\ \mu\mathrm{M}\). The corresponding rate is

-\frac{dA_T}{dt} =k_{\mathrm{cat}}C =(0.10\ \mathrm{min}^{-1})(1.60\ \mu\mathrm{M}) =0.160\ \mu\mathrm{M}\,\mathrm{min}^{-1}.

Equivalently, the effective first-order rate constant is \(0.10(0.800)=0.0800\ \mathrm{min}^{-1}\), and multiplying it by \(A_T=2.00\ \mu\mathrm{M}\) gives the same rate. The calculated complex concentration can exceed \(B_f\) because \(B_f\) is maintained by a reservoir, not specified as the total amount of \(B\).

Worked Example: Effective Rate With a Finite Total Partner Pool

Now suppose \(K_d=0.15\ \mu\mathrm{M}\), \(A_T=2.00\ \mu\mathrm{M}\), \(B_T=0.600\ \mu\mathrm{M}\), and \(k_{\mathrm{cat}}=0.10\ \mathrm{min}^{-1}\). Assume a closed system with \(C\rightarrow P+B\), so \(B_T\) stays fixed. Calculate the equilibrium complex and the initial conversion rate.

The quadratic is

$$ C^2-(2.00+0.600+0.150)C+(2.00)(0.600)=0, $$

where concentrations in the quadratic are in \(\mu\mathrm{M}\). Its smaller root is

C=\frac{2.75-\sqrt{(2.75)^2-4(2.00)(0.600)}}{2} =\frac{2.75-\sqrt{2.7625}}{2} \approx 0.54396\ \mu\mathrm{M}.

The larger root exceeds the available \(B_T\), so it is not physically possible. The smaller root is below both \(A_T\) and \(B_T\). The initial rate is therefore

-\frac{dA_T}{dt} =(0.10\ \mathrm{min}^{-1})(0.54396\ \mu\mathrm{M}) =0.054396\ \mu\mathrm{M}\,\mathrm{min}^{-1}.

This rate differs from the buffered-\(B\) example because the finite \(B_T\) pool limits how much complex can form. The total \(B_T\) cannot be substituted for a maintained free \(B_f\).

When the Reduction Is Appropriate

Rapid equilibrium is a time-scale approximation, not a consequence of binding being reversible by itself. A useful check is to compare the binding relaxation rate with the conversion rate. For fixed free \(B_f\), the binding relaxation rate is approximately \(k_{\mathrm{on}}B_f+k_{\mathrm{off}}\); it should be large compared with \(k_{\mathrm{cat}}\). For a finite pool, the local binding relaxation rate around equilibrium is approximately \(k_{\mathrm{on}}(A+B)+k_{\mathrm{off}}\). The comparison should hold over the concentration range of interest, not just at the initial condition.

The approximation is most credible after any short initial binding transient has passed. If the initial concentrations are far from equilibrium, the full model may show a fast adjustment before the slower effective rate describes the subsequent behavior. If binding and conversion occur on comparable time scales, setting the binding relation to equilibrium can misstate both the transient and the rate.

A common alternative is a quasi-steady-state reduction, which sets the net change in \(C\) approximately to zero while including conversion in the complex balance. That condition balances association against both dissociation and conversion; rapid equilibrium instead balances association and dissociation to leading order. The earlier material on rapid equilibrium versus quasi-steady state develops this distinction. When neither reduction is adequately justified, retain the full mass-action differential equations and model the complex explicitly.

The returning-\(B\) assumption also matters. If \(C\) is cleared together with \(B\), then \(B_T\) is not conserved, and the fixed-total-\(B\) reduction above does not apply unchanged. One must specify the products and balances of that reaction before deriving a reduced rate. Likewise, additional binding partners, competing reactions, or changing volume can invalidate the simple expressions unless incorporated explicitly.

Common Mistakes and Practical Checks

Confusing free and total partner. For fixed free \(B_f\), the relation is \(C=AB_f/K_d\). For fixed total \(B_T\), the relation in terms of free \(A\) is \(C=B_TA/(K_d+A)\).
Rejecting a complex concentration because it exceeds a free concentration. A maintained free concentration does not set a bound on total \(B\). The physical constraint \(C\leq B_T\) applies when \(B_T\) is the finite total pool.
Applying the equilibrium formula without checking time scales. A fast-looking binding step is not enough; binding relaxation must be fast compared with conversion throughout the relevant range.
Using a free-species rate as though it were a total-species rate. If the outcome is loss of \(A_T\), derive \(C\) in terms of \(A_T\) before writing the effective rate.

A practical diagnostic is to compare the reduced model with the full mass-action model at representative parameter values. Large discrepancies, especially during the time interval used for interpretation, suggest that binding is not sufficiently fast or that the assumed fixed-free or fixed-total condition is wrong.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

For fixed free \(B_f=0.20\ \mu\mathrm{M}\), \(A=0.50\ \mu\mathrm{M}\), and \(K_d=0.10\ \mu\mathrm{M}\), what is \(C\)?
For fixed total \(B_T=0.30\ \mu\mathrm{M}\), free \(A=0.20\ \mu\mathrm{M}\), and \(K_d=0.10\ \mu\mathrm{M}\), calculate \(C\).
With maintained \(B_f=0.30\ \mu\mathrm{M}\), \(K_d=0.10\ \mu\mathrm{M}\), \(A_T=1.50\ \mu\mathrm{M}\), and \(k_{\mathrm{cat}}=0.20\ \mathrm{min}^{-1}\), find the effective first-order constant and initial rate.
Why is \(C=1.20\ \mu\mathrm{M}\) not inconsistent with a maintained free \(B_f=0.60\ \mu\mathrm{M}\)? What would change if \(0.60\ \mu\mathrm{M}\) instead described total \(B_T\)?

Solutions. (1) \(C=AB_f/K_d=(0.50)(0.20)/0.10=1.00\ \mu\mathrm{M}\). (2) \(C=B_TA/(K_d+A)=(0.30)(0.20)/(0.10+0.20)=0.200\ \mu\mathrm{M}\), which is below \(B_T\). (3) The effective constant is \(k_{\mathrm{cat}}B_f/(K_d+B_f)=0.20(0.30/0.40)=0.150\ \mathrm{min}^{-1}\). The rate is \(0.150(1.50)=0.225\ \mu\mathrm{M}\,\mathrm{min}^{-1}\). (4) A reservoir can maintain the stated free concentration while supplying additional \(B\) into complex, so the bound amount need not be below \(B_f\). If the value is total \(B_T\), use \(C=B_TA/(K_d+A)\), which constrains \(C\leq B_T\).