>From: Herbert M Sauro <HSauro at fssc.demon.co.uk>
>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.
That's assuming that all three must collide at once to form the complex,
which I don't think is typically the case. What you should consider is
the sequence of elementary steps:
A + E -> A*E or B + E -> B*E
then
A*E + B -> A*B*E or B*E + A -> A*B*E
This is because the enzyme is capable of binding one substrate and
holding onto it while waiting for the second. In some cases (I think)
the order of binding is specific, while in others you can bind either
A or B first. With certain simplifying assumptions you may be able to
show that this is approximated by E*A*B, but I'm not sure of that.
>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. ...
Here the concentration of E becomes defined as per unit surface area
rather than per unit volume, but the form of the rate equation remains
the same. This is because it still matters how much enzyme is on the
membrane: if you have 100 molecules/mm^2 you would expect a reaction
rate 10 times faster than if you have 10 molecules/mm^2, given the same
total surface area of membrane and the same concentration of A and B in
the adjacent medium.
-Paul Schlosser
