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I'm trying to understand enzyme kinetics, the formula for Km and Kcat make sense to me.

Km , the substrate concentration at which the reaction rate is half of Vmax

Kcat, used to describe the limiting rate of any enzyme-catalyzed reaction at saturation. Most of the time Kcat just equals K2 (NOT the case when there are more reaction steps)

I can find information about the calculation of the specificity constant (Kcat / Km) and what it means:

specificity constant,the rate constant for the conversion of E + S to E + P

However I can't find any detailed explanation WHY you should devide the Kcat by Km. Secondly Km is HALF of Vmax (so actullay at half saturation) but Kcat is at full saturation, deviding these two numbers makes no sense to me. So to summerize I'm asking : What is the "meaning" of the division of these two consants?

definitions are derived/edited from Lehninger principles of biochemistry

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Just as $k_{cat}$ represents the rate of reaction at saturating substrate concentration, $k_{cat} / K_m$ represents the rate of the reaction at negligible substrate concentration.

If we take a look at the standard one substrate/one product Michaelis–Menten kinetics rate equation:

$$v = \frac{k_{cat}[E][S]}{K_m + [S]}$$

We can imagine what happens when $[S] \to 0$, we see that when $[S] \ll K_m$, the denomiator can be reduced to $K_m$, and thus the rate equation becomes

$$v = \frac{k_{cat}}{K_m}[E][S]$$

Or in other words, $k_{cat} / K_m$ is the (pseudo-)second order rate constant between the enzyme and the substrate, when $[S] \ll K_m$.

In fact, when you get into more complex rate equations (like inhibitors, pH effects, and kinetic isotope effects) there's a decent argument (one often made by W.W. Cleland) to be made that the two key constants for enzymatic reactions are $k_{cat}$ and $k_{cat} / K_m$, and it's $K_m$ that should be thought of as the "derived" constant. (The fact that we write the rate equation in terms of $k_{cat}$ and $K_m$ is a historical accident - we could have just as easily had $ v = \frac{k_{cat}\Phi [E][S]}{K_m + \Phi [S]}$, where I've arbitrarily chosen $\Phi$ as the symbol for $k_{cat}/K_m$.)

This still leaves the issue of why $k_{cat} / K_m$ is often referred to as the "specificity constant" of the enzyme. The reason for this is that if you have a single enzyme in the presence of two different substrates, you have a competitive inhibition setup. The math is a little dense (see here or here for examples), but the end result is that the ratio of reaction rates for the two substrates is related to the ratio of their respective $k_{cat} / K_m$'s:

$$\frac{v_a}{v_b} = \frac{k_{cat,A} / K_{m,A}}{k_{cat,B} / K_{m,B}}\frac{[A]}{[B]}$$

This actually makes intuitive sense, with the right mindset - the only stage where the two substrates are competing (where you make the decision to do reaction A versus reaction B) is when they're binding to free enzyme. "The rate when you only care about free enzyme" is equivalent to the negligible substrate case. With negligible substrate, all you have is free enzyme - there isn't enough substrate to have appreciable amounts of substrate-bound form, and the rate of enzyme-substrate encounter is much lower than the rate of product formation, meaning that in the steady state you don't have appreciable amounts of product-bound form, either. Thus the $k_{cat} / K_m$ for a particular substrate is representing how good the free enzyme is at performing that reaction.

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  • $\begingroup$ Thankyou nice explanation! However there is one thing that's not quite clear for me because your example with two substrates for the SAME enzyme makes sense now (negligible [S] --> [S] <-- Km etc.) However Kcat/Km is also used to compare between different enzymes, doing that the argument about low [S] is not valid (unless they are both using the same substrate offcourse) or am I missing something? @R.M. $\endgroup$ – KingBoomie Nov 19 '16 at 8:01
  • $\begingroup$ @RickBeeloo Do you have a particular example of the sort of comparison of $k_{cat}/K_m$ between different enzymes that you're talking about? Because you can certainly compare different enzymes via $k_{cat}/K_m$ -- just like you can compare different enzymes based on $k_{cat}$ or $K_m$. Comparing based on $k_{cat}/K_m$ gives you information about rates in the low (respective) substrate regime, just as comparing $k_{cat}$ gives information about rates in the high (respective) substrate regime. - I'm not sure if you're talking about this sort of comparison, or something else. $\endgroup$ – R.M. Nov 19 '16 at 15:54
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This ratio is more often called the "catalytic efficiency" and it's a bad measure. In many cases, enzymes are only better when they have a higher kcat, because the substrate concentration is always way above the Km. Using the ratio when this is obviously the case is just misleading imo, but this happens way too often.

In some cases it could be a useful measure to compare to enzymes. For example if you're unable to measure at higher substrate concentrations. You'd just have the initial slope of the [S] vs vmax graph. At low concentrations this slope approaches kcat/Km.

The kcat/Km ratio doesn't represent something real, it's just a lazy way to compare enzymes, combining two numbers that each represent something different into one number that's only looking good because it's one number.

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  • $\begingroup$ "There is an upper limit to Kcat / Km, imposed by the rate at which E and S can diffuse together in an aqueous solution. This diffusion controlled limit is 10^8 to 10^9 , enzymes with a Kcat / Km near this range are said to have achieved catalytic perfection"(Lehninger) This would be a strange assumption if the value "doesn't represent something real' ? @VonBeche $\endgroup$ – KingBoomie Nov 18 '16 at 12:36
  • $\begingroup$ I don't think that value is based directly on kcat / Km. It can be described in this way because Km = (k-1 + k2) / k1. But someone else might be able to help you more here, I'm not that at home in the maths behind this. $\endgroup$ – VonBeche Nov 18 '16 at 14:26

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