I have encountered comparisons of the Michaelis-Menten constant ($K_m$) a few times. Generally speaking if the $K_m$ of an enzyme is higher, then its affinity to its substrate is lower. How does this make sense?

Maybe the maximum velocity ($V_{\mathrm{max}}$) of higher $K_m$ enzymes is higher? Then of course, $K_m$ can have a higher value. Because $K_m$ is the substrate concentration at half of the $V_{\mathrm{max}}$. But I think we cannot determine affinity with $K_m$.

  • $\begingroup$ I think you need to be very, very careful about equating Km values with affinities. It is the binding constant that is a true measure of affinity. Biochemists talk about dissociation constants (Kd), and the lower the Kd value the lower the affinity. Chemists speak of stability constants, which are association constants: The higher the stability constant the higher the affinity. $\endgroup$ – user1136 Dec 2 '18 at 12:56
  • $\begingroup$ Consider the simplest case: E +A = EA -> E + P . The Kd is k-1/k1 (a first-order rate constant divided by a second order rate constant). but Km is (k-1 + k2)/k1 (the sum of the first-order rate constants divided by the SO rate constant. Now consider two enzymes where k1 and k-1 are equal for a given substrate. The affinities are equal. But consider that enz 1 has k2 of 1 per sec and that enz 2 has k2 of 10000 per sec. The Km for enz 2 ((k-1+k2)/k1) will be much larger! (note k2 is the rate constant governing EA -> E+P). $\endgroup$ – user1136 Dec 2 '18 at 13:02

Since the Michaelis-Menton constant Km is the concentration of substrate at 0.5Vmax, it is an inverse measure of its substrate affinity, because a lower Km indicates that less substrate is needed to reach a certain reaction speed. Hence, a low Km means a high substrate affinity.

Your statement

"Maybe the maximum velocity (Vmax) of higher-Km enzymes is higher? Then of course Km can have a higher value."

is incorrect. Km characterizes how steep reaction speed increases with substrate availability; it does not determine maximum speed.

Lastly, to address your title question, comparing affinities can make a lot of sense. For example, consider the case where enzymes catalyzing similar reactions in different species of organisms are compared. Very low Km means optimal use of small substrate levels, while a high Vmax shows optimized reaction speeds. This in turn may tell you something about optimal habitats and evolutionary pressure.

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  • $\begingroup$ Am mobile, will add ref later: ucl.ac.uk/~ucbcdab/enzass/substrate.htm $\endgroup$ – AliceD Jan 25 '15 at 13:39
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    $\begingroup$ To be spesific, here is what my professor says: "Glucokinase has much higher (Km) value for glucose than hexokinase-1. The Km value of glucokinase is 10(micromolar), whereas the (Km) value for hexokinase-1 is 0,1(micromolar). Therefore glucose affinity of glucokinase is low.". Now, lets say glucokinase has 100 times higher velocity than hexokinase, then affinities of these two become about the same.. Because at the Km concentration of hexokinase, they produce nearly same amount of product. (I know that they don't in real because the graphic is not lineer, but.. could I explain myself?) $\endgroup$ – imcakmak Jan 25 '15 at 14:20
  • $\begingroup$ Affinity to substrate resembles the desire of reaction, which means yield of product? Why should we forget about V and speed then? $\endgroup$ – imcakmak Jan 25 '15 at 14:38
  • $\begingroup$ No, affinity reflects bonding. Km is measured using Vmax, but the absolute value of Vmax is totally irrelevant for Km. It merely reflects 0.5vmax, whatever the absolute value of Vmax. You are coupling Km and Vmax, while in reality Km says nothing about speed or product. It says something about substrate. Period. $\endgroup$ – AliceD Jan 25 '15 at 14:47
  • $\begingroup$ I consider the following a statement of fact: for a diffusion controlled enzyme (kcat/Km at the upper limit allowed by diffusion), the higher the Vmax, the higher the Km. $\endgroup$ – user1136 Dec 2 '18 at 13:23

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