How does enzyme velocity decrease as reaction quotient approaches equilibrium constant?

I created a simulation which approximates molecule concentrations in a cell. One part of it are enzymes that can catalyze pre-defined reactions, *e.g.* $\ce{A <=> B}$.


I based the kinetics for these reactions on [Michaelis-Menten kinetics](https://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics), so every enzyme has a $V\_\text{max}$ and $K\_\mathrm M$, and has velocity $v=V\_\text{max}\frac x{K\_\mathrm M+x}$ at a given time point, where $x$ is the concentration of substrates.


I use the [Reaction Quotient](https://en.wikipedia.org/wiki/Reaction_quotient) $Q$ at any point in time to determine in which direction a reaction would proceed. *E.g.* would (net/overall) A be converted to B or vice versa. Each reaction is defined with a reaction energy which is used to calculate [Equilibrium constant](https://en.wikipedia.org/wiki/Equilibrium_constant) $K\_\mathrm e$.


#### Here is my question


My understanding is that Miachelis-Menten kinetics only describe the case in which $Q\ll K\_\mathrm e$. What happens if $Q\to K\_\mathrm e$?
Intuitively I would assume that the overall reaction velocity slows down as $Q$ approaches $K\_\mathrm e$.
Is there a term that I can add to the Michaelis-Menten equation which would describe this decreasing $v$?




The Michaelis-Menten mechanism is the following:
\begin{align}
 \ce{E + S &<-->[$k\_1$][$k\_{-1}$] ES} \tag{R1} \\
 \ce{ES &->[$k\_2$] P + E} \tag{R2}
\end{align}
If you apply: (1) steady-state approximation for the enzyme-substrate complex $\ce{ES}$, and (2) enzyme concentration balance $[E\_\mathrm{t}] = [E] + [ES]$, then you arrive at the rate law or the Michaelis-Menten equation
\begin{equation}
 r\_\mathrm{P} = \frac{\mathrm{d}[P]}{\mathrm{d}t} =
 \frac{k\_2[E\_\mathrm{t}] [S]}{[S] + K\_\mathrm{M}} =
 \frac{r\_\mathrm{P,max} [S]}{[S] + K\_\mathrm{M}} \tag{1}
\end{equation}
where $K\_\mathrm{M} = (k\_2 + k\_{-1})/k\_1$ and $r\_\mathrm{P,max}:=k\_2[E\_\mathrm{t}]$, which is a constant.



> 
> [OP] I based the kinetics for these reactions on Michaelis-Menten kinetics, so every enzyme has a $V\_{max}$ and $K\_M$, and has velocity $v=V\_{max}x /(K\_M+x)$
> 
> 
> 


The **product** $P$ has a maximum rate of $r\_\mathrm{P,max}$, not the enzyme. This value is obtained when the substrate is so abundant that $[S] \gg K\_\mathrm{M}$.



> 
> [OP] I use the Reaction Quotient $Q$ at any point in time to determine in which direction a reaction would proceed. E.g. would (net/overall) $A$ be converted to $B$ or vice versa. Each reaction is defined with a reaction energy which is used to calculate Equilibrium constant $K\_e$.
> 
> 
> 


You have as many reaction quotients $Q$ as reversible chemical reactions. You can do this for $\text{R1}$ because this step is reversible. However, you cannot do this for $\text{R2}$, because for the M-M mechanism this step is **irreversible**. Thus, once any quantity of $\ce{ES}$ complex has been formed, $\text{R2}$ will proceed, independently if you had a lot of $\pu{P}$ at $t=0$.



> 
> My understanding is that Miachelis-Menten kinetics only describe the case in which $Q<<K\_e$. What happens if $Q→K\_e$?
> 
> 
> 


A reaction quotient is not defined for irreversible reactions.



> 
> Intuitively I would assume that the overall reaction velocity slows down as $Q$ approaches $K\_e$. Is there a term that I can add to the Michaelis-Menten equation which would describe this decreasing $v$?
> 
> 
> 


I propose to modify the reaction mechanism in the second step
\begin{align}
 \ce{ES &<-->[$k\_2$][$k\_{-2}$]P + E} \tag{new R2}
\end{align}
However, the math has to be carried again, because the old $\text{R2}$ had a rate of $k\_2[ES]$, and now we have
\begin{align}
 r\_\mathrm{P} = \frac{\mathrm{d}[P]}{\mathrm{d}t} =
 k\_2[ES] - k\_{-2} [P][E] \tag{2}
\end{align}
This adds an additional rate constant $k\_{-2}$ to your simulation. Another option is to consider the Briggs-Haldane rate law, where every step is reversible. If you need help with the math just tell in the comments.


 7

You can use the formula you have for the rate of the forward reaction. Once you have product, you need a second set of kinetic parameters for the reverse reaction. As you reach equilibrium, one rate increases and the other decreases until they are equal. So you can’t say the reaction slows down. Instead, the net reaction tends to zero while the forward and reverse rate remain non-zero and equal.


 0