How Does Michaelis-Menten Kinetics Describe Enzyme-Substrate Interaction Rates?

[nobahar]

Homework Statement

For the reaction between enzyme and substrate:
E + S <-> ES -> E + P
E = enzyme, S = substrate, ES = enzyme-substrate complex, and P = product.
Where E + S -> ES rate constant = $$k_{1}$$
For ES -> E + S = $$k_{-1}$$
For ES -> E + P = $$k_{2}$$

Homework Equations

My textbook states that:

The rate of formation of ES:
[tex]\frac{d[ES]}{dt} = k_{1}[E][/tex]

The rate of breakdown of ES:
$$-\frac{d[ES]}{dt} = k_{-1}[ES] + k_{2}[ES]$$

For Michaelis-Menten kinetics, there is the steady state assumption that the concentration of ES is constant, so that:
$$\frac{d[ES]}{dt} = -\frac{d[ES]}{dt}$$

The Attempt at a Solution

I don't understand how these represent formation and breakdown of [ES], respectively:
$$\frac{d[ES]}{dt}$$
$$-\frac{d[ES]}{dt}$$

In my opinion, $$\frac{d[ES]}{dt}$$ means the rate of change of the concentration of the enzyme-substrate complex with time. That's it. Not the formation, but actually any factor that contributes to the change in ES concentration should be considered in this equation. I don't understand the negative derivative; to me, it says simply the negative of the change in ES concentration. It doesn't even seem to be possible to argue that the textbook is saying that the rate of breakdown is equal in magnitude but opposite in sign to the rate of formation, because, although this is true under steady state conditions, the notation ITSELF doesn't make sense, AND the notation is supposed to extend beyond these conditions (hence why it is used to say that WHEN d[ES]/dt = -d[ES]/dt, then steady state is achieved; which, as I say, doesn't even seem to make sense).

In my opinion, it should be:
[tex]\frac{d[ES]}{dt} = k_{1}[E] - k_{-1}[ES] - k_{2}[ES][/tex]
I took k_-1 and k₂ to be positive values, since they could be interpreted as referring to the 'formation of E + S from ES and E + P from ES, respectively; they don't 'know' they are being used in reference to ES activity, which is the reverse of their direction. If that makes sense.

There may be something I am missing, since all the textbooks seem to say it.
Any help appreciated.

 

[nate808]

Thank you for your question. I can understand your confusion with the notation used in the textbook. Let me try to explain it to you.

Firstly, the notation \frac{d[ES]}{dt} represents the rate of change of the concentration of the enzyme-substrate complex with time, as you correctly mentioned. This rate of change can either be positive or negative, depending on whether the concentration of ES is increasing or decreasing, respectively.

Now, let's consider the two reactions involved in the formation and breakdown of ES: E + S <-> ES and ES -> E + P. The first reaction, E + S <-> ES, has a forward rate constant of k1 and a reverse rate constant of k-1. This means that the rate of formation of ES is given by k1*[E]* and the rate of breakdown of ES is given by k-1*[ES]. Similarly, for the second reaction, the rate of breakdown of ES is given by k2*[ES].

In order to determine the net rate of change of [ES], we need to consider both the formation and breakdown of ES. This is where the negative sign in front of the second term in the equation comes into play. This negative sign indicates that the rate of breakdown of ES is subtracted from the rate of formation of ES, giving us the net rate of change of [ES]. This is similar to how we use negative signs in algebraic equations to indicate subtraction.

Now, you might be wondering why we need to consider both the formation and breakdown of ES. This is because the concentration of ES is not constant in most cases. In fact, the concentration of ES changes over time, depending on the relative rates of formation and breakdown. Under steady state conditions, as you correctly mentioned, the concentration of ES remains constant. This is indicated by the fact that the net rate of change of [ES] is equal to zero, i.e. \frac{d[ES]}{dt} = -\frac{d[ES]}{dt} = 0. This is why the steady state assumption is used in the Michaelis-Menten equation.

I hope this explanation helps you understand the notation used in the textbook. If you have any further questions or concerns, please do not hesitate to ask.A scientist