Rate constant and Michaelis Menten

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


![enter image description here](https://i.stack.imgur.com/fTcYo.png)


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.




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


 4


> 
> 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 *un*steady-state regime, also often called the *pre*steady-state regime.


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