In <9401120304.AA04594 at net.bio.net>
mroussel at alchemy.chem.utoronto.ca (Marc Roussel) writes:
>This is only possible if there is a reactant flux. A closed chemical
>system (with or without enzyme) will always relax to a state of
>thermodynamic equilibrium. Enzymes cannot alter the equilibrium state.
Yes, but on what time scale? A mixture of H2 and O2 is not at thermo.
equilibrium, but sans spark the relaxation to equilibrium is not of
practical significance. I expect that the same would hold for the
products of many enzymatic reactions. For example I performed a series of
experiments with yeast in which the the yeast was harvested and placed in
an *anaerobic* medium containing glucose, but no nitrogen source. The
glucose was completely consumed (to the limit of detection) being converted
primarily to ethanol and glycerol. I didn't do the calculations, but this
state is probably far from thermodynamic equilibrium. Moreover, if I
were to take some of the "spent" broth (or any mixture of ethanol and
glycerol, keeping it sterile at 4C) it would probably take a very long time
to reach equilibrium.
:-)
So, presumably, the "Keq" that one would want to model a reversible step
(one in which the enzymes for both the forward and reverse reactions are
present at signficant activities) in a metabolic pathway is not
the true (thermodynamic) Keq, but the ratio of concentrations that exist
at steady state. Unfortunately, as Marc pointed out in a previous message,
values for the steady state ratios are often not available. If one could
measure the metabolite levels at one steady state (ie., before changing the
activities of *other* enzymes in the pathway) then the measured ratio could
be assumed to be maintained after such a change.
Paul
schlosser at beta.ciit.org
