# Master equation for catalyst and substrate binding?

I recently asked this on Biology.SE, and need to start with the master equation for the reaction: $$\ce{E + S -> ES}$$ Where $\ce{E}$ is our enzyme and $\ce{S}$ the substrate to get a probability density function for the binding time of a substrate with one specific enzyme to be: $$f{\left(t\right)}=k\left[\ce{S}\right]\mathrm e^{-k\left[\ce{S}\right]t}$$ I have read up on master equations, but really have no idea where to start in this derivation, any help would be appreciated?

• It is not clear what system you want to discuss and what is the question. – Greg Sep 9 '15 at 15:40

The way to approach this is not how you have mentioned in your answer.

The representation of the master equation is correct and the generalized form of the chemical master equation (CME) [1] would be:

$$\frac{\partial P(x,t)}{\partial t}=\displaystyle\sum _{\mu=1}^M a_\mu(x-v_\mu)\ P(x-v_\mu,t) - a_\mu(x)\ P(x,t)$$

Where:

• $x$ denotes the state of the chemical system (number of molecules of each species)
• $\mu$ denotes a type of reaction
• $a_\mu$ denotes the probability of happening of reaction-$\mu$ in a small interval (also called reaction propensity)
• $v_\mu$ denotes the change in the state due to the reaction-$\mu$
• $M$ denotes the total number of possible reactions

As far as I can understand (also from your question on Biology.SE), you are interested in knowing the distribution of the time interval when no reaction happens. A chemical reaction can be modelled as a Poisson process which is based on 3 main postulates:

1. $\lim\limits_{h\to0+}\dfrac{P(N_h=1)}h=\lambda$

i.e. the probability of occurrence of one event in a very small interval of time is equal to the macroscopic rate or intensity ($\lambda\,$).

2. $P(N_h\geqslant2)=o(h)$

i.e. the probability of occurrence of more than one events in an infinitesimal interval is essentially zero.

3. Consecutive events are independent of each other.

A chemical reaction seems to follow all these postulates. The detailed derivation of the CME can be found in [1]; I am trying to present a short derivation based on Poisson postulates.

The probability that system is in state $x$ at time $t+\Delta t$ would be the probability that system is in state $x-v_\mu$ at time $t$ times the probability that reaction $\mu$ happens in the interval $\Delta t$, or the probability that system is in state $x$ at time $t$ times probability that no reaction happens in the interval $\Delta t$.

$$P(x,t+\Delta t)=\sum_\mu P(x-v_\mu,t)\ P(\mu) + P(x,t)\left(1-\sum_\mu P(\mu) \right)$$

when you rearrange this equation, divide by $\Delta t$ and take the limit of $\Delta t \to 0$, you will end up with the CME. Note that, as per the first postulate: $$\lim\limits_{\Delta t\to 0}\frac{P(\mu)}{\Delta t}=a_\mu$$

Now, the probability that $k$ reactions would have happened in a time interval $\tau$ follows the Poisson distribution (see [2] for the derivation).

$$P(N=k)=\frac{(\lambda\tau)^k e^{-\lambda \tau}}{k!}$$

This is a well known property of the Poisson distribution that the time interval between reactions will follow exponential distribution. The probability that no reaction has happened in time interval $\tau$ would be:

$$P(N=0)=\frac{(\lambda\tau)^0 e^{-\lambda \tau}}{0!}=e^{-\lambda\tau}$$

in this case, $\lambda=\displaystyle\sum _{\mu=1}^M a_\mu$ i.e. probability of any reaction.

Conversely, the probability that some reaction would happen in time $\tau$ would be $P(r\leq \tau) = 1-e^{-\lambda\tau}$. Where $r$ denotes the first reaction time. This is the cumulative distribution function for the exponential distribution.

In other words, the reaction time between intervals would follow an exponential distribution such that the probability that no reaction would happen will exponentially reduce with the size of the interval $\tau$.

In your specific case there are two reactions — association and dissociation, the propensities of whose will be:

$$a_f=\frac{k_f}{V}n_E.n_S\\[1em] a_r=k_r n_{ES}=k_r(n_{E_0}-n_E)$$

Where $k_f$ is the rate constant of association, $k_r$ is the rate constant for dissociation (as per mass action kinetics), $n_i$ denotes the number of molecules of species-$i$ and $V$ is the volume (of the reaction vessel). See [3] for details. Note that the amounts of the species change with time, and so will the reaction propensities. You can calculate the exact reaction time distribution when the system is at the deterministic steady state (using the steady state concentrations).

## References:

1. Gillespie, Daniel T. "A rigorous derivation of the chemical master equation". Physica A: Statistical Mechanics and its Applications, 188 (1992): 404-425.
2. Hogg, Robert V., and Allen T. Craig. Introduction to mathematical statistics. New York: Macmillan, 1978.
3. Gillespie, Daniel T. "A general method for numerically simulating the stochastic time evolution of coupled chemical reactions." Journal of computational physics 22.4 (1976): 403-434.
• You can use \displaystyle to force the "normal" behaviour in inline math, hence the formula will be bigger and you don't need explicit \limits - useful for any lists. – Martin - マーチン Sep 11 '15 at 12:15

(Note I am the OP)

Master equation

The master equation can be written as: $$\frac{\mathrm d p_n{\left(t\right)}}{\mathrm dt}=\sum_{n'}\left\{W_{nn'}p_{n'}{\left(t\right)}-W_{n'n}p_n{\left(t\right)}\right\}$$ Where $p_n$ denotes the probability that the system is in state $n$ at time $t$. And $W_{nn'}\Delta t$ denotes the probability of the system changing from state $n'$ to the state $n$ in the time interval $\Delta t$.

Specific case

Let us say we have $n=1$ free enzyme, then: $$\frac{\mathrm dp_0{\left(t\right)}}{\mathrm dt}=W_{01}p_1{\left(t\right)}$$ So what is $W_{01}$? This is the probability that the one enzyme is used up, let us call this $\mu$. $$\frac{\mathrm dp_0{\left(t\right)}}{\mathrm dt}=\mu p_1{\left(t\right)}$$ The average number of 'free enzymes' is given by: $$\left\langle n \right\rangle=p_1{\left(t\right)}$$ Differentiating this gives: $$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=\frac{\mathrm dp_1{\left(t\right)}}{\mathrm dt}$$ But in our case, $$\frac{\mathrm dp_1{\left(t\right)}}{\mathrm dt}=-\mu p_1{\left(t\right)}$$ Thus $$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=-\mu p_1{\left(t\right)}$$ $$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=-\mu\left\langle n \right\rangle$$ From here with the assumption that $\mu\propto [\ce{S}]$ you can follow through my derivation on Biology.SE

The reason for the assumption $\mu\propto [\ce{S}]$ can be explained by the fact that $\mu$ is related to the rate of the reaction, which increases (approximately) linearly with substrate concentration. If this assumption is not valid, please can you comment.

Sources

1.'Stochastic processes in physics and chemistry' by N.G. Van Kampen chapter V