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The catalytic efficiency of an enzyme is given by $k_{cat}/k_M$ where $k_{cat}$ is the turnover number, or the number of molecules that can be produced per second per active site of an enzyme.

$K_{M}$ is a measure of the affinity of the enzyme with the substrate, or the likelihood of binding.

Why bother dividing the $k_{cat}$ by $K_{M}$? Isn't the affinity of the enzyme already encoded into the quantity of $k_{cat}$? How could you be an enzyme that has low affinity, but still have a huge turnover? To me this doesn't seem possible, and thus it is redundant to divide by $K_{M}$. Likewise, could there be a situation where $k_{cat}$ is low, but $K_{M}$ is high?

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The turnover number or catalytic constant $ k_{\mathrm{cat}}$ in the Michaelis-Menten model is the rate constant for the productive dissociation of intermediate $\ce{ES}$:

$$\nu = k_{\mathrm{cat}}[\ce{ES}]$$

The constant $k_{\mathrm{cat}}$ says how much product forms from intermediate but does not say how much intermediate forms in the first place. It is assumed that there is a rapid equilibrium between enzyme, substrate and intermediate: $$\ce{E + S <=>[fast] ES}$$

that can be described by either an association or a dissociation constant for the equilibrium between intermediate, apo enzyme and substrate:

$$K_\mathrm{a}=K_\mathrm{d}^{-1}=K_\mathrm{M}^{-1}=\frac{[\ce{ES}]}{[\ce{E}][\ce{S}]}$$

At low substrate concentration $[\ce{E}]\approx [\ce{E}]_0$ and

$$[\ce{ES}] \approx \frac{[\ce{E}]_0[\ce{S}]}{K_\mathrm{M}}$$

so that

$$\nu \approx k_{\mathrm{cat}}K_\mathrm{a} [\ce{E}]_0[\ce{S}] = \frac{k_{\mathrm{cat}}}{K_\mathrm{M}} [\ce{E}]_0[\ce{S}]$$

In summary, the reason the ratio might look strange is because $K_\mathrm{M}$ is a dissociation constant, and $k_{\mathrm{cat}}/K_\mathrm{M}$ is the rate constant for low substrate concentration.

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  • $\begingroup$ I think when you wrote “early in the reaction” you meant “at low substrate concentration”? Otherwise it doesn’t really make sense. $\endgroup$ – Andrew Apr 19 at 13:51
  • $\begingroup$ Yes, you are totally right, thank you. $\endgroup$ – Buck Thorn Apr 19 at 13:54
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The most basic kinetic scheme for enzymes is represented as

$$\ce{E + S <=>[K_m] ES ->[k_{cat}] E + P}$$

As should be clear, the $k_{cat}$ is the rate constant for the reaction that occurs after substrate is bound to the enzyme. The resulting rate (kcat[E]tot) is only achieved when every molecule of enzyme essentially always is in the act of converting substrate to product. That is, every time a product molecule is released, a substrate molecule immediately binds. Even if the substrate molecule dissociates before reacting, another immediately takes its place. We describe this situation as the enzyme being "saturated" with substrate.

The Km is a measure of how tightly the substrate binds to the enzyme, approximately equal to the equilibrium constant for the dissociation of the substrate from the enzyme. If the substrate is at a low concentration relative to this Km value, then many of the enzyme molecules will not have substrate molecules bound to them and will be unproductive as a result. The overall rate will be substantially below the maximum kcat[E]tot.

All of this is captured in the basic Michaelis-Menten kinetics equation:

$$Rate = \frac{k_{cat}[E]_{tot}[S]}{K_m + [S]}$$

You can see that the impact of $K_m$ on the rate increases substantially as [S] decreases relative to $K_m$.

Since most substrates exist physiologically at concentrations below what is required for maximum rate, it is the combination of $k_{cat}$ and $K_m$ relative to [S] that determines the in vivo rate of reaction (along with the amount of enzyme of course).

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  • $\begingroup$ how did you get kcat[E]tot? Is that just by assuming [ES]=[E]tot? $\endgroup$ – John Hon Apr 19 at 13:59
  • $\begingroup$ But besides that, I think I get it. Km and Kcat are controlling different aspects of the reaction. Km is for the part 1 of the reaction and Kcat is for the second part. Is this a correct understanding? $\endgroup$ – John Hon Apr 19 at 14:00
  • $\begingroup$ Yes, when the enzyme is saturated, [ES] is effectively equal to [E]tot. $\endgroup$ – Andrew Apr 19 at 16:57
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[OP] The catalytic efficiency of an enzyme is given by $k_\mathrm{cat}/K_\mathrm{M}$ where $k_\mathrm{cat}$ is the turnover number, or the number of molecules that can be produced per second per active site of an enzyme.

The last part is not quite accurate. $k_\mathrm{cat}$ is the rate of the reaction under saturating conditions divided by the enzyme concentration. The dimensions are one divided by time (a first-order rate constant).

[OP] $K_\mathrm{M}$ is a measure of the affinity of the enzyme with the substrate, or the likelihood of binding.

The Michaelis-Menten constant $K_\mathrm{M}$ has a rigorous definition based on rate constants, and its dimensions are the same as those of a concentration. If you interpret $K_\mathrm{M}$ as the affinitiy of the enzyme to the substrate, you have to know that higher values of $K_\mathrm{M}$ correspond to lower degree of binding. The likelihood of binding strongly depends on the concentration of substrate. For discussion of the catalytic efficiency, we are interested in substrate concentrations lower than $K_\mathrm{M}$.

Why bother dividing the $k_\mathrm{cat}$ by $K_\mathrm{M}$? Isn't the affinity of the enzyme already encoded into the quantity of $k_\mathrm{cat}$? How could you be an enzyme that has low affinity, but still have a huge turnover? To me this doesn't seem possible, and thus it is redundant to divide by $K_\mathrm{M}$. Likewise, could there be a situation where $k_\mathrm{cat}$ is low, but $K_\mathrm{M}$ is high?

Here are three examples showing rate vs. substrate concentration. Let's say the red curve shows kinetics of a given enzyme. If we compare the red enzyme to one (in green) that has the same $K_\mathrm{M}$ but a $k_\mathrm{cat}$ smaller by a factor of two, the rate is half the "red" rate at all concentrations. On the other hand, the blue enzyme has the same $k_\mathrm{cat}$ as the red enzyme, but twice the $K_\mathrm{M}$ (remember, this means it is hard to get the substrate to bind). At high substrate concentration, red and blue show the same miaximal rate, but at low concentrations, the "red reaction" is twice as fast as the "blue reaction".

In fact, the green and the blue enzyme show identical behavior at low substrate concentrations because $\frac{k_\mathrm{cat}}{K_\mathrm{M}}$ for the two enzymes is the same. That is the idea of catalytic efficiency.

enter image description here

image source: https://www.desmos.com/calculator/ue4eeg4xyh

Try it yourself

Here is a tool where you can change $K_\mathrm{M}$ (abbreviated as "K" in the tool) and $k_\mathrm{cat}$ (via $v_\mathrm{max}$ abbreviated as "v") to explore how the graph of rate (abbreviated as "y") vs. substrate concentration (abbreviated as "x") changes while varying those two parameters: https://www.desmos.com/calculator/egmxfzxct8

enter image description here

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