Unfortunately, the articles in which these questions were first
brought up have been expired on our system. I'll therefore have to wing
it a bit.
Someone (whose name escapes me... sorry) said that the reaction of
glucose to form ethanol and glycerol in vivo went (within limits of
detection) to completion. He believed that this meant that there was a
"forward" enzyme (or set of enzymes) for this reaction, distinct from
the reverse-direction enzymes. I pointed out that there is no physical
mechanism that would account for this, unless it weren't a simple
enzyme-catalyzed reaction at all. I will now show that the reaction is
in fact quite likely to go to completion (or just next to it) under ordinary
circumstances so that there is no need to assume that anything odd is
going on.
The stoichiometry provided was
glucose + 0.29 H2 -> 1.71 (ethanol + CO2(aq)) + 0.29 glycerol.
Let's call this R1. It matters at least a little what the precise
reactants and products are but let's study this stoichiometry for the
sake of argument. Delta G^0 for this reaction is
Delta G^0_{R1} = -185.63 kJ/mol
and thus the equilibrium constant is
Keq_{R1} = 3.45e32.
The dissolved CO2 is in equilibrium with atmospheric gas. Let's assume
that the gas mixture over the solution has a similar composition to the
atmosphere, i.e. 5% CO2, total pressure 1 atm. For the reaction
CO2(g) -> CO2(aq) (R2)
we have
Delta G^0_{R2} = 8.38 kJ/mol
so that
Keq_{R2} = 0.0340
This gives
[CO2] = 0.0017 M.
Going back to R1, we find (assuming that everything is ideal)
[ethanol]^1.71 [glycerol]^0.29
------------------------------ = 1.9e37
[glucose]
For reasonable values of [ethanol] and [glycerol], [glucose] has to be
quite small. This conclusion is unaltered by changes of even a few
orders of magnitude in the CO2 pressure. The only thing that might make
a material difference would be the detailed inventory of reactants and
products of the reaction.
Let's get back to the question which started this. Can an enzyme
affect the equilibrium constant of a reaction? No, at least not if
we're talking about the usual sort of catalytic body which is
regenerated. The equilibrium constant is fixed by the free energy
difference of the reactants and products. The way you get from here to
there doesn't matter at all.
What does this mean for enzymes? Every enzyme (again understood to
be the usual, run-of-the-mill catalyst) must catalyze a reaction in both
directions in such a way as to leave the equilibrium constant
undisturbed. An enzyme which acted otherwise would violate the second
law and could be used as the basis of a perpetual motion machine. (It
might take some imagination to design such a machine, but I know of at
least one gedanken experiment in which a uni-directional catalyst plays
the part of a Maxwell's demon. Hint: Reactions that generate gases are ideal
candidates for this sort of thing.)
I have been careful to exclude enzymes which are not classical
catalysts. Many enzymes use ATP to push a reaction in one direction or
another. In these cases though, we must deplete the cell's store of ATP
(or equivalently have access to an external source of free energy with
which to maintain the ATP level) and thus the overall reaction has
changed: ATP has been added as a reactant and ADP and phosphate as
products. Of course, in a sense, that's cheating. I can do the same
thing with inorganic reactions by adding appropriate reagents to engage
in side reactions. As long as the reactions are coupled by a common
species, I can have an almost arbitrary effect on the partial
mass-action ratios.
I hope that I'm not being too pedantic. I'm sure that most of the
members of this audience (perhaps even the original poster) know all
about this. Perhaps this exposition will bring out the real question,
if I haven't addressed it.
Marc R. Roussel
mroussel at alchemy.chem.utoronto.ca
