Dependence of the rate of a catalysed reaction on substrate concentration?

In my answer to this question: [Master equation for catalyst and substrate binding?](https://chemistry.stackexchange.com/questions/37074/master-equation-for-catalyst-and-substrate-binding). I made the assumption along the following lines: 


If we have a substrate concentration $[S]$ then the rate of binding of any substrate molecule with a given enzyme is proportional to $[S]$. Such that $$\text{rate}=\alpha [S]$$
I have however, read that it is rather a hyperbolic relation. So are there any (and if so what) assumptions required so that we can assume that the rate is proportional to $[S]$ rather then it been a hyperbolic relation?




There is always a maximum convertion rate $V\_{max}$ for every given enzyme-substrate system. In some cases the substrate itself can inhibit the enzyme's activity at high concentrations, because of the formation of some ESS-complexes, which can not dissociate into product and enzyme. 


The frequent correlation is described by the [Michaelis-Menten kinetics](https://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics), which introduces the Michaelis constant $K\_M$ as the substrate concentration at which half of all enzyme is bound to substrate $-$ and thus the reaction is running at a speed of $\frac{V\_{max}}{2}$.[![enter image description here](https://i.stack.imgur.com/Q20EL.png)](https://i.stack.imgur.com/Q20EL.png)


As you can see, the relation is described by a saturation function. But for loose approximations, where $[S] \leq K\_M$ you could say its roughly proportional: $$rate = \dfrac{V\_{max}}{2 \times K\_M} \times [S]$$


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The curve is not hyperbolic. A hyperbolic curve is a result of an equation $f(x)=\dfrac{b}{a}\sqrt{x^2-a^2}$. Which is not the case here.


The enzyme catalysis (Michaelis-Menten model) can be described by the two step reaction.


$$\ce{E + S<=>[k\_f][k\_r] ES ->[k\_{cat}] E + P}$$


In Michaelis-Menten kinetics you make either of the two approximations:


* **Equilibrium approximation**: assuming that first reaction (i.e. enzyme-substrate binding) is in equilibrium. Therefore $k\_r[ES]=k\_f[E][S]$
* **Quasi-steady-state approximation** (QSS): assuming that the concentration of $ES$ complex remains constant over time. Therefore $k\_f[E][S]=(k\_r+k\_{cat})[ES]$


For both these cases you can substitute $[ES]$ with $[E\_0]-[E]$ where $E\_0$ is the total enzyme concentration. You can also represent $[E]$ in terms of $[S]$ using any of the above relationship. Then, you assume that maximum catalytic activity would be when all the available enzyme is complexed with the substrate ($[ES]=[E\_0]$). In that case the maximum rate of catalysis $V\_{max}$ would be $k\_{cat}[E\_0]$. When you put everything together then the rate of catalysis would turn out to be:


$$\frac{d[P]}{dt}=V\_{max}\frac{[S]}{K\_M+[S]}$$


Where:


* $K\_M=\dfrac{k\_r}{k\_f}$ for equilibrium approximation
* $K\_M=\dfrac{k\_r +k\_{cat}}{k\_f}$ for QSS approximation


These kind of curves denote saturation kinetics and can also be seen in case of adsorption. 


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