enzymes and activation energy?

Dr. Ron rxw13 at po.cwru.edu
Thu Oct 5 13:28:52 EST 1995

In article <t.chappell-2709952206060001 at t-chappell.mcbl.ucl.ac.uk> t.chappell at ucl.ac.uk (tom chappell) writes:
>In article <rxw13.277.04C9D10D at po.cwru.edu>, rxw13 at po.cwru.edu (Dr. Ron) 
>> In article <447hur$85s at rover.ucs.ualberta.ca>
>msingh at gpu2.srv.ualberta.ca (Madhu Singh) writes:
>> >Paul. S. Brookes. (psb at mole.bio.cam.ac.uk) wrote:
>> >: rray at ix.netcom.com (ROMAN RAY ) wrote:-
>> >: I thought enzymes catalysed by stabilising the transition state of a
>> >: reaction (presumably by transiently bonding/interacting with it)
>> >I think the transition states still need changes in chemical bonding.  
>> Enzymes often change the transition state to one requiring lower activation, 
>> e.g. in group transfer reactions, first transfering the group to the enzyme, 
>> thence to the acceptor.

>That is a really tough statement to interpret without being able to draw
>the free energy diagrams.

>I wouldn't consider a group transfer reaction to be changing a transition
>state, but using a different free energy pathway with 2 new transitions
>states on either side of a high energy intermediate. Straight organic
>chemistry--a classical SN2 reaction has an energy profile with a single
>hump in going from A to B. The height of that hump is the delta G of
>activation. A catalyst (enzyme) provides an energy profile with a lower
>delta G of activiation. The catalyst might provide an energy profile that
>still has a single hump, in which case one can consider the transition
>state to be stablized (delta G is less to get from A to transition state).
>The catalyst might, however, provide an energy profile that has 2 humps,
>by generating a finite lived high energy intermediate. In the classical
>SN2 reaction, such an intermediate might be a carbonium ion. A group
>transfer reaction would definitely be an example of the latter and not the
>former mechanism of catalysis.

You are obviously correct - -  changing the pathway is precisely what I meant. 
Forgive my semantic sloppiness, but I was answering the question at the level 
it was asked. If all you know about a "reaction" is the composition of 
reactants and products, then everything that occurs in the black box that 
connects the two can (sloppily) be considered "transition state".  Changing 
the path of the reaction lowers the free energy of that "transition". 

Dr. Ron (picky, picky, picky)

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