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(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

2. http://www.jpoffline.com/physics_docs/y4s7/advstatmech_ln.pdf

(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

2. http://www.jpoffline.com/physics_docs/y4s7/advstatmech_ln.pdf

(Note I am the OP)

Master equation

The master equation can be written as: $$\frac{d p_n(t)}{dt}=\sum_{n'}\{W_{nn'}p_{n'}(t)-W_{n'n}p_n(t)\}$$ 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: $$ \frac{dp_0(t)}{dt}=W_{01}p_1(t)$$ So what is $W_{01}$? This is the probability that the one enzyme is used up, let us call this $\mu$. $$\frac{dp_0(t)}{dt}=\mu p_1(t)$$ The average number of 'free enzymes' is given by: $$<n>=p_1(t)$$ Differentiating this gives: $$\frac{d<n>}{dt}=\frac{dp_1(t)}{dt}$$ But in our case, $$ \frac{dp_1(t)}{dt}=-\mu p_1(t)$$ Thus $$\frac{d<n>}{dt}=-\mu p_1(t)$$ $$\frac{d<n>}{dt}=-\mu<n>$$ From here with the assumption that $\mu\propto [S]$ you can follow through my derivation on Biology.SE

The reason for the assumption $\mu\propto [S]$ 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

2. http://www.jpoffline.com/physics_docs/y4s7/advstatmech_ln.pdf

(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

2. http://www.jpoffline.com/physics_docs/y4s7/advstatmech_ln.pdf

(Note I am the OP)

Master equation

The master equation can be written as: $$\frac{d p_n(t)}{dt}=\sum_{n'}\{W_{nn'}p_{n'}(t)-W_{n'n}p_n(t)\}$$ 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: $$ \frac{dp_0(t)}{dt}=W_{01}p_1(t)$$ So what is $W_{01}$ this is the probability that the one enzyme is used up, let us call this $\mu$. $$\frac{dp_0(t)}{dt}=\mu p_1(t)$$ The average number of 'free enzymes' is given by: $$<n>=p_1(t)$$ Differentiating this gives: $$\frac{d<n>}{dt}=\frac{dp_1(t)}{dt}$$ But in our case, $$ \frac{dp_1(t)}{dt}=-\mu p_1(t)$$ Thus $$\frac{d<n>}{dt}=-\mu p_1(t)$$ $$\frac{d<n>}{dt}=-\mu<n>$$ From here with the assumption that $\mu\propto [S]$ you can follow through my derivation on Biology.SE

Sources

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

2. http://www.jpoffline.com/physics_docs/y4s7/advstatmech_ln.pdf

(Note I am the OP)

Master equation

The master equation can be written as: $$\frac{d p_n(t)}{dt}=\sum_{n'}\{W_{nn'}p_{n'}(t)-W_{n'n}p_n(t)\}$$ 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: $$ \frac{dp_0(t)}{dt}=W_{01}p_1(t)$$ So what is $W_{01}$? This is the probability that the one enzyme is used up, let us call this $\mu$. $$\frac{dp_0(t)}{dt}=\mu p_1(t)$$ The average number of 'free enzymes' is given by: $$<n>=p_1(t)$$ Differentiating this gives: $$\frac{d<n>}{dt}=\frac{dp_1(t)}{dt}$$ But in our case, $$ \frac{dp_1(t)}{dt}=-\mu p_1(t)$$ Thus $$\frac{d<n>}{dt}=-\mu p_1(t)$$ $$\frac{d<n>}{dt}=-\mu<n>$$ From here with the assumption that $\mu\propto [S]$ you can follow through my derivation on Biology.SE

The reason for the assumption $\mu\propto [S]$ 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

2. http://www.jpoffline.com/physics_docs/y4s7/advstatmech_ln.pdf