Discussion of [S] Values

i.e., the Michaelis-Menten Equation would predict that v = k[S]; that is, the rate is directly proportional to the concentration of one of the reactants (the substrate), so this is a first order reaction. This can be explained by saying that the enzyme is not saturated with substrate and therefore, as we add more substrate we get more ES formed, and more ES breaking down to give P.

At high [S], when [S] > 10

i.e., the Michaelis-Menten Equation would predict that v = k. This means that the rate is independent of the concentration of the substrate. This is a zero order reaction and can be explained by saying that the enzyme is saturated with substrate. Even when more substrate is added to the solution we cannot get any more ES formed and, therefore, the rate does not change. So the Michaelis-Menten Equation does fit the empirical curve at both high and low [S].

Note that when there is more enzyme present the rate is always higher for a given [S]. This means that the will also be higher.

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