Seven competing ways to recover the Michaelis–Menten equation reveal the alternative approaches to steady state modeling

Abstract

The Michaelis–Menten enzymatic reaction is sufficient to perceive many subtleties of network modeling, including the concentration and time scales separations, the formal equivalence between bulk phase and single-molecule approaches, or the relationships between single-cycle transient probabilities and steady state rates. Seven methods proposed by different authors and yielding the same famous Michaelis–Menten equation, are selected here to illustrate the kinetic and probabilistic use of rate constants and to review basic techniques for handling them. Finally, the general rate of an ordered multistep reaction, of which the Michaelis–Menten reaction is a particular case, is deduced from a Markovian approach.

Appendices

Appendices

1.1 Branch pruning to generate a linear chain

The reduction of disordered cycles to ordered transitions has been addressed for example in [4, 13]. The most general treatment of a linear chain with one input and one output and including a branched module, leads to a Michaelian kinetics with respect to the distribution of the reaction time between the input and output, irrespective of the complexity of the intervening module.

In a simplified case, if the fraction of time spent by the enzyme in the forms $EA$ and $EB$ (Fig. 5), is negligible compared to its total mean cycling time, these alternative forms can be eliminated and the disordered individual rates can be replaced by ordered rates according to

$$\begin{aligned} k_{AB}&= k_{A}\, \frac{k'_{B}}{k'_{B}+k_{-A}} + k_{B}\, \frac{k'_{A}}{k'_{A}+k_{-B}}, \end{aligned}$$

(17a)

$$\begin{aligned} k_{-AB}&= k'_{-B}\, \frac{k_{-A}}{k_{-A} + k'_{B}} + k'_{-A}\, \frac{k_{-B}}{k_{-B}+k'_{A}}. \end{aligned}$$

(17b)

In addition, postulating that the walk is random forces to assume that the transitions $k_{A}$ and $k_{B}$ do not influence each other. For example, if they are the pseudo-first order rates of ligation to two substrates $A$ and $B$, the binding of one substrate has no influence on the binding of the other one. In this case, $k_{A}=k'_{A}$ and $k_{B} = k'_{B}$ and the global rates are

$$\begin{aligned} k_{AB} = \frac{k_{A}k_{B}}{R}, \quad k_{-AB} = \frac{k_{-A}k_{-B}}{R}, \end{aligned}$$

(18a)

with

$$\begin{aligned} R = \frac{(k_{A}+k_{-B})(k_{-A}+k_{B})}{k_{A}+k_{B}+k_{-A}+k_{-B}}. \end{aligned}$$

(18b)

Fig. 5

How to compress lateral alternative states (top scheme) into an unbranched ordered chain (bottom scheme)

1.2 Mean completion time for a general chain

Equation (14) can be rigorously obtained through a continuous time Markovian modeling. In the following formulation, the probabilities of the different enzymatic states are written $P_i(t)$, $i=0,1,\ldots ,n$, and correspond to the amount of final state $i$ at time $t$. Their evolution is determined by a linear differential system,

$$\begin{aligned} {\mathrm{d} \over \mathrm{d}t}\,P_i(t) = \sum _{j=0}^{n} \, \hat{A}_{ij} \, P_j(t), \quad i=0,1,\ldots ,n, \end{aligned}$$

(19)

with the initial condition $P_i(0) = \delta _{i,0}$. The matrix $\hat{A}$ is numerical and contains the transition rates: the coefficient $\hat{A}_{ij}$ is equal to the rate of the transition $j \rightarrow i$. As the only allowed transitions are from $j$ to $j, j\pm 1$, the matrix $\hat{A}$ is tridiagonal; in addition all its column sums are equal to 0, $\sum _{i=0}^n \, \hat{A}_{ij} = 0$ (a consequence of the conservation law $\sum _{i=0}^n P_i(t) = 0$). The last column of $\hat{A}$ is also zero since $n$ is an absorbing state. For this reason, it is sufficient to consider the restriction $A$ of $\hat{A}$ to its first $n$ rows and $n$ columns, $A = (\hat{A}_{ij})_{0 \le i,j \le n-1}$. The differential system can then be written as

$$\begin{aligned} {\mathrm{d} \over \mathrm{d}t} \, \mathbf{P}(t) = A \mathbf{P}(t), \quad \hbox {where} \ \mathbf{P}(t) = (P_i(t))_{0 \le i \le n-1}, \end{aligned}$$

(20)

supplemented by the extra equation ${\mathrm{d} \over \mathrm{d}t} P_n(t) = k^+_{n-1} P_{n-1}(t)$, as well as the initial condition $\mathbf{P}(0) = (1,0,\ldots ,0)^t$.

Our purpose is to compute

$$\begin{aligned} \langle T \rangle = \int \limits _0^\infty \mathrm{d}t \, t\,{\mathrm{d}P_n(t) \over \mathrm{d}t} = k^+_{n-1} \int \limits _0^\infty \mathrm{d}t \, t\,P_{n-1}(t). \end{aligned}$$

(21)

Up to the factor $k_{n-1}^+$, this is the last component of the length $n$ vector $\langle \mathbf{T} \rangle = \int _0^\infty \mathrm{d}t \, t\,\mathbf{P}(t)$.

Let us observe that the solution of the differential system is formally given by

$$\begin{aligned} \mathbf{P}(t) = \sum _{k=0}^\infty {t^k \over k!} \, A^k \mathbf{P}(0) = \mathrm{e}^{tA} \, \mathbf{P}(0), \end{aligned}$$

(22)

where, as indicated, the exponential of the matrix $tA$ is defined by its Taylor series. Applying $A^2$ on $\langle \mathbf{T} \rangle$, using $A^2 \mathbf{P}(t) = {\mathrm{d}^2 \over \mathrm{d}t^2} \mathbf{P}(t)$ and integrating by parts, we obtain

$$\begin{aligned} A^2 \langle \mathbf{T} \rangle = \int \limits _0^\infty \mathrm{d}t \, t\,{\mathrm{d}^2 \mathbf{P} (t) \over \mathrm{d}t^2} = \Big [t\,{\mathrm{d} \mathbf{P}(t) \over \mathrm{d}t}\Big ]_0^\infty - \Big [\mathbf{P}(t)\Big ]_0^\infty = \mathbf{P}(0), \end{aligned}$$

(23)

where we have used the fact that $\mathbf{P}(t)$ goes exponentially to zero when $t \rightarrow \infty$. From this we obtain that $\langle \mathbf{T} \rangle = A^{-2} \mathbf{P}(0)$. Taking the last component of this vector equation and using $\mathbf{P}(0) = (1,0,0,\ldots ,0)$, we find that

$$\begin{aligned} \langle T \rangle = k_{n-1}^+ \, (A^{-2})_{n-1,0} = k_{n-1}^+ \, \sum _{i=0}^{n-1} \, A^{-1}_{n-1,i} \, A^{-1}_{i,0} \end{aligned}$$

(24)

is proportional to the scalar product of the last row of $A^{-1}$ and its first column.

Let us first apply this formula to the simplest situation, namely when all transition rates $k^\pm _i$ are equal to 1. In this case, the entries of $A$ are equal to $+1$ on the lower and upper diagonals, namely $A_{i,i\pm 1}=+1$. They are equal to $-2$ on the main diagonal, with the exception of $A_{00}=-1$. One can verify that the inverse of $A$ is given by

$$\begin{aligned} A^{-1}_{ij} = \max (i,j) - n, \quad 0 \le i,j \le n-1. \end{aligned}$$

(25)

We therefore have $A^{-1}_{n-1,i} = -1$ and $A^{-1}_{i,0} = i-n$, and the following result for $\langle T \rangle$,

$$\begin{aligned} \langle T \rangle = \sum _{i=0}^{n-1} \, (n-i) = {n(n+1) \over 2}. \end{aligned}$$

(26)

In the completely general case with arbitrary rates, the matrix $A$ takes the following form

$$\begin{aligned} A = \begin{pmatrix} -k^{+}_{0} &{}\quad k^{-}_{1} &{}\quad 0 &{}\quad \ldots &{}\quad \ldots &{}\quad 0 \\ k^{+}_{0} &{}\quad -(k^{+}_{1}+k^{-}_{1}) &{}\quad k^{-}_{2} &{}\quad \ldots &{}\quad \ldots &{}\quad 0 \\ 0 &{}\quad k^{+}_{1} &{}\quad -(k^{+}_{2}+k^{-}_{2}) &{}\quad \ldots &{}\quad \ldots &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad k^{+}_{2} &{}\quad \ldots &{}\quad \ldots &{}\quad 0 \\ \ldots &{}\quad \ldots &{}\quad \ldots &{}\quad \ldots &{}\quad \ldots &{}\quad k^{-}_{n-1} \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad k^{+}_{n-2} &{}\quad -(k^{+}_{n-1}+k^{-}_{n-1}) \end{pmatrix}.\qquad \end{aligned}$$

(27)

Because the column sums of $A$ are all zero except the last one which is $-k^+_{n-1}$, the last row of $A^{-1}$ is constant and equal to $A^{-1}_{n-1,i} = -{1 \over k^+_{n-1}}$. Beside, the first column of $A^{-1}$ is somewhat more complicated and given by

$$\begin{aligned} A^{-1}_{i,0} = -{1 \over k^+_i} \; \sum _{\ell =0}^{n-1-i} \, {k^{-}_{i+1} \ldots k^{-}_{i+\ell } \over k^{+}_{i+1} \ldots k^{+}_{i+\ell }}, \quad i=0,1,\ldots ,n-1, \end{aligned}$$

(28)

where, by convention, the term $\ell =0$ is set to 1. We then obtain

$$\begin{aligned} \langle T \rangle = \sum _{i=0}^{n-1} \, {1 \over k^{+}_i} \; \sum _{\ell =0}^{n-1-i} \, {k^{-}_{i+1} \ldots k^{-}_{i+\ell } \over k^{+}_{i+1} \ldots k^{+}_{i+\ell }}, \end{aligned}$$

(29)

which is equivalent to (14) upon the interchange of the two summations.

1.3 Testing another example: the 3-step enzymatic reaction

To validate the above recipes, let us apply them to a slightly more complicated (3-step) enzymatic reaction shown in Fig. 6.

Fig. 6

Extended MM reaction where the catalytic reaction is reversible

The 3-step reaction is more realistic than the traditional 2-step MM reaction of Fig. 2. Indeed, the last transition of the MM scheme (rate $k_{c}$ in Fig. 2) mixes two elementary reactions: the catalysis sensu-stricto and the dissociation of the product from the enzyme. From a chemical viewpoint, the catalytic reaction has no obvious reason to be micro-irreversible. Instead, irreversibility can be understood as the consequence of the very low concentration of the product in the medium preventing it to rebind to the enzyme. This is true in vitro when measuring initial reaction rates because no product molecules are added in the reaction mixture. This is also generally true in vivo because the products are immediately removed by sequestration or by subsequent reactions of which they are the substrates, so that their steady state concentration remains negligible in the cell. Whatever the method used, the rate corresponding to the scheme of Fig. 6 is the following function of the substrate concentration

$$\begin{aligned} k = \frac{k_{c}k_{p}k_{a}[S]}{k_{p}k_{c}+k_{p}k_{d}+ k_{d}k_{-c}+(k_{p}+k_{c}+k_{-c})k_{a}[S]}. \end{aligned}$$

(30)

The most direct methods to obtain this result are listed below.

1.3.1 The King and Altman method

The first order enzyme cycle corresponding to Fig. 6 is represented in Fig. 7.

Fig. 7

The first order cycle of the reaction of Fig. 6, whose drawing is a prerequisite for using the King and Altman procedure

The turnover rate of a single-enzyme corresponds to the reaction rate and is

$$\begin{aligned} k = k_{p}P_{EP}, \end{aligned}$$

(31)

where $P_{EP}$ is given by the graphical method of King and Altman.

$P_{EP}$ can be calculated by hand by replacing every graph by the product of rates corresponding to the arrows, but the result can be more safely obtained using an algorithm such as the KAPattern algorithm of [19], very useful for more complex networks.

1.3.2 The sum of direct conversion times

Here the rate reads

$$\begin{aligned} \frac{1}{k} = \frac{1}{k_{p}} + \frac{1}{k_{c}\frac{k_{p}}{k_{-c}+k_{p}}} + \frac{1}{k_{a}[S]\frac{k_{c}\frac{k_{p}}{k_{-c}+k_{p}}}{k_{d}+ k_{c}\frac{k_{p}}{k_{-c}+k_{p}}}}. \end{aligned}$$

(32)

1.3.3 Markovian modeling of a random walk

We obtain by this method

$$\begin{aligned} \frac{1}{k} = \frac{1}{k_{a}[S]} + \frac{1}{k_{c}}+ \frac{1}{k_{p}} + \frac{k_{d}}{k_{a}[S] k_{c}} + \frac{k_{-c}}{k_{c}k_{p}} + \frac{k_{d}k_{-c}}{k_{a}[S]k_{c}k_{p}}. \end{aligned}$$

(33)