Answer: (a) 0 no free enzyme left
(b) 10^-9M
Explanation:
ANSWER:
Given that
Kcat = 10 sec-1
Km = 10^-6 M
[S] = 10^-3 M
[Enzyme] = 10^-9 M
The reaction follows the following path-
Enzyme (E) + Substrate (S) <=> ES complex -> E + Product (P)
According to the improved model of Michaelis-Menten kinetics, upon addition of substrate and the enzyme, instead of dynamic equilibrium, a steady state is reached. The time taken is very less, almost instantaneously (since Kcat is much higher than the concentrations we are dealing with (10 per second! Whereas we are dealing with concentrations as low as 10-9).
In this steady state, the Enzyme and substrate instead of existing individually, exist as an Enzyme-Substrate complex, or ES complex.
Physically, Km is a measure of how well substrate complexes with an enzyme, i.e. It's binding affinity.
You can imagine this as if 1 unit of the substrate can bind to "Km" units of Enzyme. For the give conditions, 1 M of the substrate requires 10^-6 M enzyme for complete binding. So, 10^-3 M of the substrate will require 10^-3 x 10^-6 = 10^-9 M of the enzyme, which is the exact amount of enzyme added to the reaction mixture.
So it is safe to assume that when the steady state is reached, all of the enzyme is bound to the available substrate producing the ES complex with the concentration equal to the limiting reactant, i.e. the enzyme = 10^-9 M
Hence, there will be no free enzyme left after the short duration of the reaction. And the concentration of the ES complex will be 10^-9 M
