Rate constant and Michaelis Menten

Question

Please consider the following figure of a concentration over time (msec) diagram with variables for an enzyme, a substrate bound enzyme and a product.

Recall that $\ce{E + S \leftrightharpoons ES \rightharpoonup E + P}$ and also recall that the velocity constant of the respective reactions are noted $k_1 , k_2, k_3$.

What I would like to know is:

On the plot provided, is the slope of $[ES]$ equal to $k_2$ and is the slope of $[E]$ equal to $k_1 + k_3$ thereby giving some insight/explanation for $k_m$ which is used to determine the affinity of an enzyme?

Any more details on Michaelis Menten is greatly appreciated but please note that my level of understanding is limited on the topic.

Answer 1

Given $\ce{E + S <=>ES -> E + P}$

$\frac{d[\ce{ES}]}{dt}=-\frac{d[\ce{E}]}{dt}= k_1[\ce{E}][\ce{S}]-k_2[\ce{ES}] -k_3[\ce{ES}] $

is the slope of $[ES]$ equal to $k_2$ ?

no

is the slope of $[E]$ equal to $k_1 + k_3$

no

Answer 2

On the plot provided, is the slope of [ES] equal to k2 and is the slope of [E] equal to k1+k3 thereby giving some insight/explanation for km which is used to determine the affinity of an enzyme?

As DavePhD said, no to both counts. First note that in your system, $\ce{E_0=E +ES}$, where $\ce{E_0}$ is the total amount of enzyme added, and $\ce{E_0}$ is conserved -- enzymes, as catalysts, are created or destroyed in the reaction. This means $\frac{d\ce{E_0}}{dt}=0$, which implies that $\frac{d\ce{E}}{dt}=-\frac{d\ce{ES}}{dt}$. So the slope of [ES] is negative the slope of [E].

Second, as you study more about Michaelis Menten kinetics, you will come to appreciate that this data is (mostly) for the unsteady-state regime, also often called the presteady-state regime.