Introduction to Kinetics III (Biocatalysis)

Turnover rate and kinetic optimum

The turnover number (catalytic constant) $k_{\text{cat}}$ of an enzyme indicates how many substrate molecules are converted into the reaction product per unit of time when the enzyme is completely saturated and is used in ${\text{s}}^{-1}$ expressed. It therefore corresponds to the kinetic constant $k_2$ and is about the maximum speed $v_{\text{Max}}$ and the total concentration of the enzyme (total concentration of active centers) ${[\mathrm{E.}]}_0$ calculated.

$$\begin{array}{ccccc}\hfill \text{S.}+\text{E.}& \underset{k_{-1}}{\overset{k_2}{\rightleftharpoons }}& \hfill \text{E.}\text{S.}& \stackrel{k_2}{\to }& \text{P.}+\text{E.}\hfill \end{array}$$

$$\begin{array}{l}v=\frac{v_{\text{Max}}\cdot [\mathrm{S.}]}{K_m+[\mathrm{S.}]}=k_2\frac{{[\mathrm{E.}]}_0\cdot [\mathrm{S.}]}{K_m+[\mathrm{S.}]}\\ \text{if}[\mathrm{S.}]>>K_m\text{then}v=v_{\text{Max}}=k_2{[\mathrm{E.}]}_0\text{or}{k}_2=\frac{v_{\text{Max}}}{{[\mathrm{E.}]}_0}=\frac{{\left(\frac{d[\mathrm{P.}]}{dt}\right)}_{\text{Max}}}{{[\mathrm{E.}]}_0}=k_{\text{cat}}\\ \text{if}[\mathrm{S.}]<<K_m\text{then}v=\frac{v_{\text{Max}}\cdot [\mathrm{S.}]}{K_m}=k_2\frac{{[\mathrm{E.}]}_0\cdot [\mathrm{S.}]}{K_m}\end{array}$$

At low substrate concentrations, i.e. with very incomplete saturation of the enzyme, the reaction rate is beyond ${\mathit{k}}_2$ also from the substrate concentration $[\mathrm{S.}]$ and the Michaelis-Menten constant$K_m$ addicted.

The catalytic efficiency $k_{\text{cat}}/K_m=k_2/K_m$ cannot become larger than the diffusion-controlled encounter between enzyme and substrate. These physical limit values ​​are between${10}^8$ and ${10}^9\text{L.}{\text{mol}}^{-1}{\text{s}}^{-1}$

This also applies to complex, multi-stage reactions. Here is the number of changes $k_{\text{cat}}$ how many substrate molecules are converted into the corresponding product by a single enzyme molecule in a given unit of time when the enzyme is saturated with substrate. This corresponds to the maximum rate of catalysis or the kinetic optimum. Again, the reaction rate is best determined by the catalytic efficiency $k_{\text{cat}}/K_m$ described. The factor $k_{\text{cat}}/K_m$ therefore represents the best parameter for comparing catalytic efficiencies.

Use

$K_m$ and $k_{\text{cat}}$ are therefore suitable for comparing different enzymes, regardless of whether the reactions are complex or simple. When comparing enzyme activities, both $K_m$ as well as $k_{\text{cat}}$ to be determined. Two enzymes can have the same turnover number even though they catalyze different reactions. Due to the speed with which enzyme E and substrate S diffuse towards one another in aqueous solution, there is an upper limit on the catalytic efficiency $k_{\text{cat}}/K_m$who at${10}^8$ and ${10}^9\text{L.}{\text{mol}}^{-1}{\text{s}}^{-1}$the ${\mathit{k}}_{\mathit{cat}}\text{/}{K}_m$- Values ​​for most enzymes are close to this range. These enzymes have thus reached their kinetic optimum. Their rate of reaction is limited only by the rate at which the enzyme and the substrate meet in the solution. The reaction speed can only be increased by reducing this diffusion time, for example by bringing the enzyme and substrate closer together. This approximation is ensured, for example, by multi-enzyme complexes in which the product of one enzyme serves as a substrate for the next enzyme and is passed on as if on a conveyor belt. The space in which the enzyme and substrate meet is thus limited and thereby enables the diffusion time to be reduced and the rate of reaction to be increased.

Tab. 1

Change numbers and $k_{\text{cat}}/K_m$Values ​​of some enzymes

Video: Steady states and the Michaelis Menten equation. Biomolecules. MCAT. Khan Academy (May 2022).