Kinetic vs. Thermo Control for Protein Folding

Robert Solomon Bioc rgs at mole.bio.cam.ac.uk
Mon Jul 11 07:43:01 EST 1994


In article <2vjlrt$b2o at news.doit.wisc.edu> Ken Prehoda <kenp at nmrfam.wisc.edu> writes:
>In article <2vivmp$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
>>goes.
>>
>>   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
>state.

au contraire - this seems to me a classic example of a system out
of equilibrium moving to regain that equilibrium.  The means by which it
arrives at that equilibrium (the means which you argue is irrelevant)
is the field of study for scores of scientists, who believe (rightly,
IMHO) that they are studying the folding pathway (I'll allow the plural
there - i.e. pathways, although I believe there will be few for any one
protein starting out from a particular set of conditions).

>
>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.
>

OK, I'll bite here - Your argument for path independence, which you keep
coming back to, seems to me a little simplistic - are you arguing that
a folding pathway, per se, is irrelevant to the energy difference between
unfolded (sic) and folded forms of a protein, or that there exists no
folding pathway?  If the latter, then you seem awfully out of step with
experimentalists in the field (and even many theorists).  And in the
case above, of wrongly folded versus correctly folded forms, you can not
seriously argue that the incorrectly folded form obeys kinetic control
while the correctly folded (love these perjorative terms) form is under
thermodynamic control.  Seems like having cake and eating same.

>>> 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
>unlikely).
>
>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

Rob Solomon
<.sig deleted due to loss of sense of humour>




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