Extension and Justification of Quasi-Steady-State Approximation for Reversible Bimolecular Binding

Abstract

The quasi-steady-state approximation (QSSA) is commonly applied in chemical kinetics without rigorous justification. We provide details of such a justification in the ubiquitous case of reversible two-step bimolecular binding in which molecules as an intermediate step of the reaction form a transient complex. First, we justify QSSA in the regime that agrees with the results in the literature and is characterized by $\max \{R_0, L_0\} \ll K_m$. Here, $R_0$ and $L_0$ are the initial concentrations of reacting receptor and ligand, respectively, and $K_m$ is the Michaelis constant. We also validate QSSA under an alternative condition that can be viewed as partially irreversible binding, and it does not require a tight bound on $R_0$ and $L_0$ but rather requires $k_2 + k_{-2} \ll k_{-1}$. Here, $k_{-1}$ is the rate constant of decomposition of the transient complex to the ligand and the receptor, and $k_2$ and $k_{-2}$ are the forward and the reverse rate constants of transformation of the complex to the product, respectively. Furthermore, we provide arguments that QSSA can also be accurate in a regime when $\max \{R_0, L_0\} \approx K_m$ and $k_2 + k_{-2} \approx k_{-1}$ if $|R_0 - L_0| \ll K_m$. The derived conditions may be of practical use as they provide weaker requirements for the validity of QSSA compared to the existing results.

Appendix

1.1 Monotonicity

First we prove that $\hbox {d}L/\hbox {d}t \le 0$ for all $t\ge 0$ for the solution of (12)–(13) subject to the initial condition $(L(0), C(0)) = (L_0, 0)$. The same property then holds for the solution of (6)–(9) subject to (10). Let us assume the contrary, i.e., there is a trajectory (L(t), C(t)) starting at $(L_0, 0)$ such that $dL / \hbox {d}t(t_1) > 0$ for some $t_1 > 0$. Since the vector field $(\hbox {d}L/\hbox {d}t, \hbox {d}C/\hbox {d}t)$ points downwards on the C–nullcline for $L > L^{*}$, the trajectory (L(t), C(t)) can only cross the nullcline for $L \in (0,L^{*})$. Therefore, there exist $t_2 \in (0, t_1)$ such that the trajectory (L(t), C(t)) crosses the line segment $L = L^{*}, C \in [0,C^{*})$ in the phase plane for the first time, i.e., $L(t) > L^{*}$ for all $t \in [0, t_2)$. On that segment, $L(t_2) = L^{*}$ and $C(t_2) < C^{*}$, and thus by (11) also $P(t_2) > P^{*}$. Since $P(0) = 0$ by continuity, there must exist $t_3 \in (0, t_2)$ such that $P(t_3) = P^{*}$ and $\hbox {d}P / \hbox {d}t (t_3) \ge 0$. But (9) implies $0 \le \hbox {d}P / \hbox {d}t (t_3) = k_2 C(t_3) - k_{-2} P^{*}$, i.e., $C(t_3) \ge k_{-2}/k_2 P^{*} = C^{*}$. Therefore, $P(t_3) = P^{*}, C(t_3) \ge C^{*}$, and hence $L(t_3) \le L^{*}$, yielding a contradiction because $L(t) > L^{*}$ for all $t \in [0,t_2)$. Therefore, $dL/dt \le 0$ for all $t >0$.

Furthermore, we prove $\hbox {d}P/\hbox {d}t \ge 0$ for all $t\ge 0$ for the solution satisfying (6)–(9) subject to (10) by using an argument similar to Darvey and Matlak (1967). Using (6)–(9), the initial conditions, and continuity of the solution, we derive

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t} (k_2 C - k_{-2} P) = k_2 k_1 L R - k_2 (k_{-1} + k_2 + k_{-2}) C + k_{-2}(k_2 + k_{-2}) P \ge 0\nonumber \\ \end{aligned}$$

(62)

for all small enough positive t. The stoichiometry restrictions (11) allow us to reduce the system (6)–(9) to the two-dimensional dynamical system in variables (L, P) of the form

$$\begin{aligned} \frac{\hbox {d}L}{\hbox {d}t}= & {} -k_1 L(L + R_0 - L_0) + k_{-1} (L_0 - L - P), \\ \frac{\hbox {d}P}{\hbox {d}t}= & {} k_2 (L_0 - L) - (k_2 + k_{-2}) P. \end{aligned}$$

The P-nullcline is the line connecting the initial condition $(L_0, 0)$ and the point $(0, k_2L_0/(k_2 + k_{-2}))$ in the (L, P) phase plane. Furthermore, the L-nullcline is given by the parabola $P = L_0 - L - k_1L(L+R_0 - L_0)/k_{-1}$ that contains points $(0, L_0), (L_0,-k_1L_0R_0/k_{-1})$ and that is decreasing for all $L \ge 0$. Therefore, there is a unique equilibrium of the system in the positive quadrant. The condition (62) and the fact that $dP /dt = 0$ at $t=0$ imply that the solution (L(t), P(t)) lies under the P-nullcline for small positive t. Since the solution (L(t), P(t)) cannot cross neither of the nullclines, it remains wedged in between them for all positive t. That immediately implies $\hbox {d}P / \hbox {d}t \ge 0$ for all $t \ge 0$.

1.2 Estimates

Estimate of $t_{L}$. The numerator in the expression (16) for $t_L$ is equal to the range of values of L given by $L_\mathrm{max}-L_\mathrm{min}$. It can be estimated by $L_0$ as $L + C + P = L_0$, and both P and C are nonnegative. Since $\hbox {d}L / \hbox {d}t \le 0$ for all t, a sharper estimate $L_\mathrm{max} - L_\mathrm{min} = L_0 - L^{*}$, where $L^{*}$ is the equilibrium concentration of L, can be used as well. On the other hand, the denominator in (16) that is equal to the maximal value of $\hbox {d}L / \hbox {d}t$ can be estimated in the following way. By (12) at the point of its maximum, the function

$$\begin{aligned} U(t) = -k_1 L(t) (L(t) + R_0 - L_0) + k_{-1}C(t) \end{aligned}$$

has its local extreme, i.e., $\hbox {d}U/ \hbox {d}t = 0$. Hence,

$$\begin{aligned} k_{-1} \frac{\hbox {d}C}{\hbox {d}t} = k_1 \frac{\hbox {d}L}{\hbox {d}t} (2L + R_0 - L_0). \end{aligned}$$

Using (12)–(13), it reduces to an expression for $C = C(L)$ that is valid at the point of the local maximum:

$$\begin{aligned} C = \frac{\frac{k_{-1}k_{-2}}{k_1} (L_0 - L) + L(L+R_0-L_0)\left[ k_{-1} + k_1(2L + R_0 - L_0)\right] }{k_{-1}(2 L + R_0 - L_0 + K_m)}. \end{aligned}$$

The value of $\hbox {d}L/\hbox {d}t$ at its eventual point of local maximum is then

$$\begin{aligned} \frac{\hbox {d}L}{\hbox {d}t} \Big |_{\mathrm{max}} = \frac{-(k_2 + k_{-2})L(L+R_0 - L_0) + \frac{k_{-1}k_{-2}}{k_1}(L_0 - L)}{2L + R_0 - L_0 + K_m}\, . \end{aligned}$$

(63)

If we consider the right-hand side of (63) as a function of L, it is easy to see that it is a decreasing function on its natural domain $L \in [0, L_0]$, and thus, it reaches its maximum absolute value at one of its end points. Its value at $L = 0$ is positive and at $L = L_0$ negative. Since we have proved that $\hbox {d}L / \hbox {d}t \le 0$ for all $t \ge 0$, it is enough to consider the value at $L = L_0$ given by

$$\begin{aligned} \Big |\frac{\hbox {d}L}{\hbox {d}t}\Big |_{\mathrm{max}, L = L_0} = (k_2 + k_{-2}) \frac{L_0R_0}{R_0 + L_0 + K_m}. \end{aligned}$$

The maximum possible value of |dL / dt| is then bounded by the maximum of the values of U(t) at $t = 0 (L = L_0$ and $C = 0)$, at $t = \infty$, i.e., at the system equilibrium for which $U(t) = 0$ (this follows from the analysis performed in Sect. 2.5) and at its eventual local extreme. Since at $t = 0, |U(0)|= k_1 L_0R_0$ and $k_1 K_m \ge k_2 + k_{-2}$, it easy to see that

$$\begin{aligned} \Big |\frac{\hbox {d}L}{\hbox {d}t}\Big |_{\mathrm{max}} \le \max \Big \{k_1 L_0R_0, \frac{(k_2 + k_{-2})L_0R_0}{R_0 + L_0 + K_m}\Big \} = k_1 L_0R_0, \end{aligned}$$

that leads to (17).

Estimate of $\triangle L / L_0$. Next, we estimate the relative depletion of the ligand during the initial short timescale $t \in [0, t_C]$, where $t_C$ is given by (15). The formula for $\triangle L = L_0 - L(t_C)$ follows from (12):

$$\begin{aligned} \triangle L = \left| -\int _0^{t_C} \frac{\hbox {d}L}{\hbox {d}t} \, \hbox {d}t \right| = \left| -k_{-1}\int _0^{t_C} C \, \hbox {d}t + k_1 \int _0^{t_C} L (L + R_0- L_0)\, \hbox {d}t \right| . \end{aligned}$$

(64)

Here, both integrals in (64) are positive. Next, we estimate the first integral in (64). Note that $\hbox {d} / \hbox {d}L (\hbox {d}C/\hbox {d}t) > 0$; therefore, the production of C is increasing with concentration of L. Since $L \le L_0$ and at the concentration $L = L_0$ the evolution of C is governed by (14), it follows that

$$\begin{aligned} C(t) \le \frac{L_0R_0}{K_m} \left( 1 - \hbox {e}^{-(k_{-1}+k_2 + k_{-2})t} \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} k_{-1}\int _0^{t_C} C \, \hbox {d}t \le k_{-1}\frac{L_0R_0}{K_m} \int _0^{t_C} 1 - \hbox {e}^{-(k_{-1}+k_2 + k_{-2})t} \hbox {d}t = \frac{L_0R_0}{K_m}\frac{1-\kappa }{\hbox {e}}. \end{aligned}$$

The second integral in (64) can be estimated by the trapezoid rule using the fact that $L(t_C) = L_0 - \triangle L$:

$$\begin{aligned} k_1 \int _0^{t_C} L(L + R_0 - L_0) \hbox {d}t\approx & {} k_1 t_C \frac{1}{2} \left( L_0 R_0 + (L_0 - \triangle L)(R_0 - \triangle L) \right) \\\approx & {} \frac{L_0R_0}{K_m} - \frac{\triangle L}{K_m} \frac{L_0 + R_0}{2}, \end{aligned}$$

where we neglected the quadratic terms $O(\triangle ^2 L)$. The quantity $\triangle L$ in (64) can be then estimated as

$$\begin{aligned} \triangle L \approx -\frac{L_0 R_0}{K_m} \frac{1-\kappa }{\hbox {e}} + \frac{L_0R_0}{K_m} - \triangle L \frac{L_0 + R_0}{2K_m}, \end{aligned}$$

that can be rewritten as

$$\begin{aligned} \triangle L \left( 1 +\frac{L_0 + R_0}{2K_m}\right) \approx \frac{L_0 R_0}{K_m} \left( 1 - \frac{1-\kappa }{e}\right) . \end{aligned}$$

Note that $\kappa \in [0,1]$, i.e., $1 - (1-\kappa ) / e= O(1)$. Hence,

$$\begin{aligned} |\triangle L| \approx O\left( \frac{L_0R_0}{2K_m + L_0+R_0}\right) . \end{aligned}$$

(65)

The reactant stationary assumption $\triangle L/L_0 \ll 1$ can be then expressed as

$$\begin{aligned} \frac{R_0}{2K_m + L_0+R_0} \ll 1 \end{aligned}$$

that is for $L_0 \le R_0$ equivalent to $R_0 / K_m \ll 1$. However, note that for any $R_0 - L_0 < 2K_m$, one has

$$\begin{aligned} \frac{R_0}{2K_m + L_0+R_0} = \frac{R_0}{2K_m + 2R_0 + (L_0 - R_0)} < \frac{1}{2}. \end{aligned}$$

For values of $\kappa$ close to $0^+$, the neglected factor $(1-\kappa ) /\hbox {e}$ in (65) is smaller than 2 / 3, i.e., $\triangle L / L_0 < 1/3$, regardless of the value of $R_0/K_m$. For moderate values of $R_0/K_m \approx \frac{1}{2}$ and $R_0 = L_0$, the ratio $\triangle L / L_0$ can even reach the value 0.1.