2
$\begingroup$

In my answer to this question: 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?

$\endgroup$
1
$\begingroup$

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, 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

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: $$rate = \dfrac{V_{max}}{2 \times K_M} \times [S]$$

$\endgroup$
  • $\begingroup$ This curve is not called hyperbolic and the model never considers the number of binding sites. So the statement "at which half of the binding sites are taken" is not appropriate. Note that just that mathematical relationship does not talk anything about the mechanism. $\endgroup$ – WYSIWYG Sep 11 '15 at 8:24
  • $\begingroup$ Moreover, Vmax is not a property of the Enzyme-Substrate pair; only kcat is the property of the ES pair. Vmax depends on both the nature of the ES pair and the total enzyme present in the sample. $\endgroup$ – WYSIWYG Sep 11 '15 at 8:34
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.