Kinetic vs. Thermo Control for Protein Folding

Ken Prehoda kenp at nmrfam.wisc.edu
Fri Jul 8 08:55:09 EST 1994

In article <2vivmp$qvh at lyra.csx.cam.ac.uk>
Simon Brocklehurst, smb18 at mole.bio.cam.ac.uk writes:
>>You deleted my question: again I ask you, do you believe it is valid
>>to use equilibrium constants when studying protein systems???  If so,
>>I don't see how you can argue kinetic control.
>   I still think we might be talking at cross purposes! But here
>   OK, start let's start with 100% of our protein unfolded and quickly
>change the conditions so that the protein is allowed to start folding
>straight away.
>   At the point where we "start" the folding, the system is not at 
>equilibrium, is it?   If you were studying the early stages of
>folding in this way, why would equilibrium be relevant.
>If the system proceeds to an equilibrium between unfolded and folded 
>states, then by all means use an equilibrium constant in that context - 
>but only when it's got there.

Since you think we are talking cross purposes let's review the question:
Is the native state of a protein determined by kinetics or thermodyamics?

If you answer kinetics, then the folded state is not the equilibrium

However, I argue that even if this was the case, there is still 
thermodynamic control between the native and unfolded state.  You
would have to argue against using equilibrium constants to
study protein systems if this weren't the case.  Good luck here.

Anyways, you have pretty much conceded my point.  If you believe
you can use an equilibrium constant between unfolded and folded
states then  it is undeniable that there is path independence
between these two states.

>>Of course there is some "transition state" involved in the folding
>>process.  However, this determines the _rate_ of folding, not the final
>>structure.  As such, the structure of the transition state only affects
>>the rate of folding.
>  If you can only to proceed along the "folding coordinate" to the native
>state, via the transition state, then the structure of the native state 
>has been influenced by the structure of the transition state.  My point
>is that if the system goes through a particular transition state, then
>this determines what native state you see.
>  So you could imagine that proteins could fold unproductively if
>they go through the wrong transition state.
>  This does seem to happen in real life.  Ever tried to overexpress a 
>protein in a system that expresses really quickly?  Often the
>protein doesn't fold to it's usual native state.  It's gone along
>some "incorrect" folding pathway from which there's no return.  Express 
>the protein more slowly and the protein folds properly.

OK, now I see your confusion.  In the case you discuss, there is 
kinetic control _because there is a different folded state_ but the
kinetic control is only for this alternate state, which you
obviously don't care about.  While it is in that different folded
state, presumaby it is metastable, meaning that for kinetic reasons
it cannot get to the normal folded state (and therefore the normal
folded state is at a lower (global?) free energy minimum).

However, this does not mean the the normal folded state is determined
kinetically.  I repeat, between the normal folded state and the unfolded
state, there is path *independence*.  The question that we have
been trying to answer, I repeat, is kinetic vs. thermodynamic control
for the folding of the native state.  As I have said before, these
alternate states are besides the point.

>> It is undeniable that equilibrium (for all
>>practical purposes), and therefore thermodynamic control is going
>>on here.  Otherwise, we wouldn't be using Keq's to describe the system.
>   What about this model for protein folding?
>      Unfolded    --->  Intermediate -----> Native 
>                 <---
>    There's no equilibrium between the Unfolded and native states here.

I think you're grasping at straws here.

You can dream up mystical cases to argue your point, but the 
overwhelming evidence shows that there are no stable intermediates in
protein folding, in most cases.  And again, your mystical case is
unrealistic:  there is not a single system I know of where there
is irreversibility between a folding intermediate (in the cases
where they exist) and the folded state itself.

Let's restrict ourself to real world examples - what do you say?  And
furthermore to the majority of proteins.  In my original reply I stated
that some proteins may show kinetic control (although this is even

And anyway, equilibrium constants are used extensively in studying
proteins. For all intensive purposes, there is an equilibrium between
the unfolded and folded states - meaning path independence.  And this
means that no matter how you *imagine* the folding pathway, it doesn't
matter as far as the folded state is concerned.

> |
> |  ,_ o     Simon M. Brocklehurst,
> | /  //\,   Oxford Centre for Molecular Sciences,
> |   \>> |   Department of Biochemistry, University of Oxford,
> |    \\,    Oxford, UK.
> |           E-mail: smb at bioch.ox.ac.uk

-Ken Prehoda
kenp at nmrfam.wisc.edu

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