# Homework Help: Competitive kinetics

1. Mar 23, 2014

Rate is being defined as d[P]/dt, and d[P]/dt = k3[ES]. So the main challenge is expressing [ES]. I applied the steady-state approximation to find [ES] = [E][noparse][/noparse]k1+k-2[ES2] / (k3+k-1+k2[noparse][/noparse]).

The other equation I've used is [E]0=[E]+[ES]+[ES2], the E mass balance.

But I can't seem to eliminate all of the extra variables doing that alone. What should I do?

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2. Mar 23, 2014

### epenguin

Whenever you have a 'dead end' branch like that, what's on it has (in the steady state) an equilibrium ratio with what it's connected to, in this case

[ES2] = (k2/k-2)[ES]

Last edited: Mar 23, 2014
3. Mar 23, 2014

Ok, thanks, but why? How can you derive that this is the case? That sounds like a pre-equilibrium approximation, but all we're told about in the question is to apply the steady-state approximation ... so given that ES is the intermediate and must have constant concentration, how can we derive the expression you gave for [ES2] in terms of [ES]?

4. Mar 23, 2014

### epenguin

Just look at ES2 - what is d[ES2]/dt ?

5. Mar 23, 2014

d[ES2]/dt = (k2/k-2)[ES]. But we need to find [ES2], not its rate of production.

6. Mar 23, 2014

### epenguin

That's quite wrong, try again.

Last edited: Mar 23, 2014
7. Mar 23, 2014

Ah now I see what you're getting at. d[ES2]/dt = k2[ES] - k-2[ES2]; if d[ES2]/dt ≈ 0 then [ES2] = (k2/k-2)[ES].

But why did you apply the steady-state approximation to ES2 - or rather, how did you know to? By definition SSA is meant to apply to "intermediates", and ES2 is not an intermediate but rather a product ...

8. Mar 24, 2014

### epenguin

That's right - it is after all one of the most elementary things in kinetics.

correction
I apply it to ES2 because I have to apply it to all the forms of the enzyme in order to get their concentration in terms of the concentration of one of them, e.g. of [E]. You have not written it here but you must surely have written

Rate = k3[ES]

and to get [ES] use [ES]/[Etotal] = [ES]/([E] + [ES] + [ES2])

and then you have to get all the forms of E in terms of one of them, say [E] which you will then be able to remove from the equation so as to get [ES] as a fraction of [Etotal].

Last edited: Mar 24, 2014
9. Mar 24, 2014

Ok that's fine, but how was I supposed to know that this would be a valid approach? All the question says is steady-state approximation, but that normally applies only to intermediates, not to every form of the enzyme ...

If this sort of thing is commonly done, can you refer me to a book where I can find more treatments "off the back" of steady-state which are not simply straightforward applications of the steady-state approximation but rather, like this case, require me to apply a constant-concentration assumption to species which are not merely intermediates?

Also, is there a general message to take from this lesson? Such as: "to express rate in terms of the concentration of a certain initial concentration via a mass balance, you must apply the steady-state approximation to all species within the mass balance for that element - and can then find rate in terms of the initial concentration of that element, and potentially any of the time-dependent concentrations of species not in the mass balance for that element" - in other words, in order to ensure that [E], [ES] and ES2 do not show up in the final rate equation we apply the steady-state approximation to each of them (and would do the same to any other species we do not want to show up in the final approximate rate equation). Hence in this case we must apply the steady-state approximation for E, ES and ES2 to get rate in terms of [E]0 and, to be precise, in terms of anything but [E], [ES] and ES2.

Last edited: Mar 24, 2014
10. Mar 24, 2014

### Ygggdrasil

ES2 is an intermediate. Products are chemical species that will accumulate as the reaction proceeds. Because ES2 --> ES + S proceeds at a reasonable rate, ES2 will not accumulate. P is a product because E + P --> ES occurs at a negligible rate under the conditions studied. If ES2 dissociation did not occur reasonably quickly, you are correct and it would not be correct to apply the steady-state assumption to the concentration on ES2.

11. Mar 24, 2014

### epenguin

if you have several species all interconnected by chemical reactions then if you think about it (preferably with a particular example in mind) if they are not all in a steady state none of them can be.

See above. And have you not been through yet competitive inhibition by and inhibitor I which has a scheme similar to that in your attachment but with a form EI instead of ES2?

I do not encourage you to think the way to understand a single point that can be explained in less than a page is to wade through a whole book. Where is your item quoted in #1 from? - it looks quite conventional. Then you can find treatments in general biochemistry books, then in an enzymology book like Fersht (from memory, I don't have any books by me at this moment). Probably the most used book at the moment on enzyme kinetics is that by Cornish-Bowden (but wider than just steady-state). There is a summary here: http://www.inf.ed.ac.uk/teaching/courses/csb/CSB_lecture_enzyme_kinetics.pdf Best of all is to work things out yourself with the books ore for hints or indications.

A bit of a mouthful!

I'd say take the case where there is just one enzyme species, ES, that decomposes into product.(Easy to extend to other cases). Then we'll have, as already, something like
Rate = kcat[ES]
Then get what fraction [ES] is of all the enzyme species present, which total is constant. All means all, why not? If anything is converted into something which you call 'not an intermediate' that affects the reaction rate - there is less enzyme in the ES form.

Finally for this dead end ES2 you were stumbling with, if you have anything like

A ⇔ B then after sufficient time you end up with [A] = (kBA/kAB) (with obvious notation).

This will still be true if B is in equilibrium with something else.
It will still be true of A is in equilibrium with something else.
And it will still be true if B is not in equilibrium with something else but in a steady state as long as A is a dead end - or indeed on a chain of reactions just leading to a dead end.
When there is also another process Z ← A as well as the above it gets more complicated.