# Homework Help: Enzyme Kinetics- is my summary correct?

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1. Jan 22, 2015

### Lo.Lee.Ta.

I have been looking at this enzyme kinetics stuff forever and was trying to come up with a summary for things... #=_=

Is this even right...? :/

|---------------|This segment I'm calling "Part 2"
k1 k2
E + S ES E + P
k-1
|---------------|This segment I'm calling "Part 1"

Part 1 and Part 2 are separate reactions, but together, they form the overall enzyme reaction.
Part 1 is dependent on the Km value, Part 2 is dependent on the Kcat value, and the entire enzyme reaction is dependent on the ratio of Kcat to Km (or Kcat/Km).

Part 1:
Since it is dependent on the Km value, and Km = (k-1 + k2)/k1, which means
(ES breakdown)/(ES formation), you have to ask, "Is the ES breakdown higher than the formation?"
If yes, then it means Km, so ES tends to break down (slow).
If no, Km, so ES tends to stay together (fast rxn).

Part 2:
Have to ask, "Is Kcat high or low?"
If high, fast product formation.
If low, slow product formation.

So for the entire enzyme-catalyzed reaction, it seems like there would only be these 4 possibilities:

1. Kcat + Km = slow for ES to form, but once it does, quickly generates product.

2. Kcat + Km = fast for ES to form and quickly generates product.

3. Kcat + Km = slow for ES to form, and once it does, slowly generates product.

4. Kcat + Km = fast for ES to form, but slowly generates product.

...Well, is any of this even right???
Thank you SO much!

Last edited by a moderator: Apr 19, 2017
2. Jan 22, 2015

### Quantum Defect

What book are you using?

The general derivation of Michaelis-Menton kinetics is based upon the steady-state approximation.

In the case of modeling Enzyme kinetics, it says that the change in concentration of the Enzyme-Substrate complex [ES] with time is zero.

The speed with which ES forms is given by k1[E]. The rate of destruction of ES is given by -k-1[ES] - k2[ES] (falling back or going to products)

Km is more like an equilibrium constant. Think about the case without any reaction (k2 very small). If this is the case, then k1[E] = k-1[ES] (definition of equilibrium)
Keq = [ES]/[E] = k1/k-1 (looks almost like 1/Km)

I think the usual way of thinking abou this is similar to the way you described above, except you think about things in terms of how much enzyme is tied up with substrate and the speed with which the enzyme reacts once it has boud substrate.

Last edited by a moderator: Apr 19, 2017
3. Jan 23, 2015

### epenguin

I find your statements tending to the 'not even wrong' I'm afraid.

Formation of ES being fast or slow is irrelevant because in the assumptions of steady-state you are treating a situation where all that is going to be formed already has been. So that leaves the questions how much has been formed? how fast does it give product?

At saturation ( [Substrate] >> Km) practically all the enzyme is in the form ES. The rate is simply kcatEtotal.
The higher kcat the faster the reaction.

Instead if substrate concentration is low such that S << Km your can see your equation simplifies so that v is still proportional to kcat but then also to 1/Km

The first is simple and as before, the second is because the higher the Km the less ES complex there is (at low S!) to be catalysed.
At every stage you should look at formulas but also consider the physical situations they represent, go between one and the other at every
stage.

I am sorry about these unintended cancellations, an artefact of the system. :(

Last edited by a moderator: Apr 19, 2017
4. Jan 23, 2015

### epenguin

Looks like we cannot write S in square brackets!

5. Jan 23, 2015

### Quantum Defect

So that's what is causing all of the strikethroughs?

6. Jan 23, 2015

### epenguin

I think so. We should use x for S and then EX for the complex I guess.

7. Jan 27, 2015