Herbert Sauro wrote:>>>>>>>>>>>
I don't have easy access to University libraries etc because if I did I
could probably look it up but my question concerns membrane transport
kinetics. In a mixed solution containing two components, A and B and an
enzyme which can convert A + B to C, the rate at which the enzyme
complex EAB forms is proportional to E*A*B - I hope I've got that bit
right. I presume this comes from considerations of collision
probabilities in a perfectly mixed compartment. What I want to know is
what would happen if the enzyme were stuck on a membrane, i.e confined
to a 2D surface but with the substrates, A + B still mixed in the bulk
solution. What would be the rate of formation of EAB be proportional
too? Would it be the same (i.e E*A*B) or different because one of the
colliding particles (E) can no longer freely diffuse in 3D?<<<<<<<<
How about the equilibrium of the reaction A + B <-> C on adding the
catalyst? The forward reaction was diffusion-dependent and now it is in
a sense "more" diffusion-dependent. The back reaction was NOT diffusion-
dependent but now it is. Can we expect that the rate constants change
proportionally so that their ratio K = Kf/Kb remains the same?
Pentcho
