# 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, 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 $$Q$$ 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 $$K_\mathrm e$$.

#### 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 $$v$$?

• Here's a good description: bio-physics.at/wiki/… Also, I don't think you mean the Nernst equation. That is only applicable to redox reactions. Commented May 30, 2023 at 12:46
• Then, I have a follow up question to this. In the link's equation, for simplicity let's assume $V_{max}^f = V_{max}^b = V_{max}$ and $K_{M,1} = K_{M,2} = K_M$, we get $v = V_{max} \frac{S - P}{K_M + S + P}$. So that means even if $S \rightleftharpoons P | -100kJ$, the enzyme would stop ($v = 0$) once $P = S$? My understanding was always that the reaction would continue until $K_e = Q$ Commented May 30, 2023 at 13:52
• Or in other words: Say $K_{M,2} >> K_{M,1}$. So, even though a lot of $P$ might exist, it will not have a big effect and $v \approx V_{max}^f$. Then, would $v \approx V_{max}^f$ until the last molecule $S$ was converted to $P$ that sets $Q = K_e$, then suddenly $v = 0$? I would have imagined that there is a kind of slow-down in $v$ as $Q \rightarrow K_e$ Commented May 30, 2023 at 14:06
• If P is present, you are no longer under Vmax conditions, since P and S are competing for the enzyme. Also, if the Vmax and Km values are the same for forward and reverse reaction, then Ke necessarily equals 1, so your first hypothetical is not possible. Commented May 30, 2023 at 14:29
• The kinetics and thermodynamics are related by the fact that equilibrium means that the conditions are such that the forward and reverse reaction are occurring at identical rates, thus no net reaction in either direction is observed. If the Km's and Vmax's are the same, then the rates will be the same when S=P Commented May 30, 2023 at 17:15

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 $$$$r_\mathrm{P} = \frac{\mathrm{d}[P]}{\mathrm{d}t} = \frac{k_2[E_\mathrm{t}] [S]}{[S] + K_\mathrm{M}} = \frac{r_\mathrm{P,max} [S]}{[S] + K_\mathrm{M}} \tag{1}$$$$ where $$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 $$Q<. What happens if $$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.

• Is this the Briggs-Haldane rate law? I find it mentioned a lot on google but no explicit formulas. Commented May 30, 2023 at 18:23
• @mRcSchwering No. That web has in fact the mechanism I proposed to you. The one I am referring to is $\ce{E + S <--> ES <--> PE <--> P + E}$, where a product-enzyme complex is added. Commented May 30, 2023 at 18:27
• Ah, yes now I see it. I think I will take your mechanism. It was mentioned in comments earlier as well. Does it have a name? Commented May 30, 2023 at 18:28
• @mRcSchwering Yes, my name. Joking! Just say that is the classic M-M mechanism, but you allow the possibility of reversibility on the second step. This means that the presence of $P$ and $E$ will hinder the rate of reaction for the formation of your desired product $P$. Commented May 30, 2023 at 18:32

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.