QUESTIONS: alpha-helix "signals" in proteins

Ken Prehoda kenp at nmrfam.wisc.edu
Fri Jul 8 17:57:40 EST 1994

In article <CHERRY.94Jul8152635 at watneys.med.utah.edu>
Joshua Cherry, cherry at watneys.med.utah.edu writes:
>   That's my point - they're kinetically inaccessible (in most cases) if
>   they are there at all.  So of what importance are they?  
>That's exactly what kinetic control means--that the distribution of
>products of a reaction (in this case folding) is not in accord with
>the equilibrium distribution due to kinetic considerations.  Part of 
>the importance of such states, if they do exist, is that 
>they would imply that thermodynamic stability is no guaranty that a 

You missed my point.  What I am saying is that if there is some
state that has a free energy lower than our observed state, it is
irrelevant, if it cannot be reached.  In other words, although
having a lower free energy state implies that the system is not
at equilibrium, for all practical purposes, the system is at
equilibrium if it takes the age of the universe to reach that 
global minimum.

>I remember something about ts mutants which don't fold at the
>restrictive temperature, but, if folded at the permissive temperature,
>remain stable at the restrictive temperature.  If such mutations 
>are not rare aberrations, but common ways of breaking a refined
>function, then the kinetics of folding are quite important.

You are talking about mutations that change the stability of the
protein.  What does this have to do with the folding pathway?

>>   To put it
>>   another way, do you believe it is valid to use equilibrium constants
>>   when studying protein systems?
>>Certainly in some contexts.  It must be true that the native state
>is (under native conditions) favored over the unfolded state
>(otherwise we'd never call it the folded state, it would just be
>one of the substates of unfolded protein).  However, if Simon
>is right, then application of equilibrium constants to determine
>which state is thenative state would sometimes lead you to false
>You would conclude that the (hypothetical) more stable, kinetically
>inaccessible state was the native one.

First, the equilibrium constant is not used to determine which state
is the native state.  This must be know before hand (spectroscopically,
for example).

Second, how would you conclude that an inaccessible state is the native
state, if by definition the inaccessible state can't be reached?
Sounds pretty strange...

>Josh Cherry
>cherry at watneys.med.utah.edu

-Ken Prehoda
kenp at nmrfam.wisc.edu

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