So the michaelis constant is defined as the substrate concentration at half maximal „reaction-speed“. I was wondering why 1/2 ? I guess for higher values you need more substrate, which could be a problem. But would 1/3 also work ? Or does 1/2 optimize some other parameter?

  • 2
    $\begingroup$ 1/2 is the least arbitrary choice of a number between 0 and 1. To me, that's good enough. $\endgroup$ Jun 6 '19 at 7:34
  • $\begingroup$ The „least arbitrary choice“ is defined in which way? Smalles sum of denominator and nominator ? Seems kinda odd to me, that there is no further considerations defining such a important constant $\endgroup$
    – Paul
    Jun 6 '19 at 7:38
  • $\begingroup$ If you need a formal definition, then yeah, this one will do. $\endgroup$ Jun 6 '19 at 7:39
  • 2
    $\begingroup$ in the derivation of the M-M equation using rate constants (lower case k's), the k's are grouped in such a way as to make the entire expression most simple, in particular to isolate [S]. When you then define Km as the other term in the denominator, the coincidental result of that denominator (Km + [S]) is that when [S]=Km, we can replace Km with [S] and get rate = Vmax/2. This definition of Km is also conceptually useful because as $k_{-1}/k_2$ goes to infinity, Km becomes equal to Kd for the ES complex in the simple case. $\endgroup$
    – Andrew
    Jun 6 '19 at 13:00
  • $\begingroup$ @Andrew I think that's an answer. $\endgroup$
    – Buck Thorn
    Jun 6 '19 at 17:07

To see why the $K_m$ is equivalent to the substrate concentration at which the rate is equal to $V_{max}/2$, we need to look at the derivation of the Michaelis-Menten, which is usually included in detail in any introductory biochemistry book. Here is a brief version.

We will consider an enzyme-catalyzed reaction described by the following scheme:

$$\ce{E + S <=>[k_1][k_{-1}] ES ->[k_2] E + P},$$

where "E" is enzyme, "S" is substrate and "P" is product.

To get an expression for the rate, we start by noting that the rate of formation of product is $k_2[\ce{ES}]$, so we need to find an expression for $[\ce{ES}]$ using known quantities. Next, we make a steady-state approximation, setting $\frac{d[\ce{ES}]}{dt}=0$, so we have

$\displaystyle k_1[\ce{E}][\ce{S}]-(k_{-1}+k_2)[\ce{ES}]=0$

$\displaystyle \implies [\ce{ES}]=\frac{k_1}{k_{-1}+k_2}[\ce{E}][\ce{S}].$

To finish, we note that when the reaction has not consumed very much substrate, we can approximate $[\ce{S}]$ as equal to the starting substrate concentration $[\ce{S}]_0$. We also know that the total amount of enzyme in the reaction must be equal to the sum of the concentrations of E and ES, ie $[\ce{E}]+[\ce{ES}]=[\ce{E}]_{Total}$. We already have an equation for $[\ce{ES}]$ in terms of $[\ce{E}]$, so with some algebra, we can find that

$\displaystyle [\ce{ES}]=\frac{k_1[\ce{E}]_{Tot}[\ce{S}]}{k_{-1}+k_2+k_1[\ce{S}]}$.

Combining everything, we get

rate of formation of $\displaystyle \mathrm{P} =k_2[\ce{ES}]=\frac{k_2k_1[\ce{E}]_{Tot}[\ce{S}]_0}{k_{-1}+k_2+k_1[\ce{S}]_0}$.

In order to isolate $[\ce{S}]_0$ in the denominator, we divide the numerator and denominator by $k_1$:

rate $\displaystyle =\frac{k_2[\ce{E}]_{Tot}[\ce{S}]_0}{\frac{k_{-1}+k_2}{k_1}+[\ce{S}]_0}$,

and our final simplification is to define the unwieldy $\frac{k_{-1}+k_2}{k_1}$ as $K_m$ and $k_2[\ce{E}]_{Tot}$ as $V_{max}$:

rate $\displaystyle =\frac{V_{max}[\ce{S}]_0}{K_m+[\ce{S}]_0}$.

Getting back to your original question: Note that we never explicitly defined $K_m$ to be the substrate concentration at half-maximal rate. It simply resulted from our choice to isolate the substrate term in the denominator. However, that was not a completely arbitrary choice.

If we look at $K_m=\frac{k_{-1}+k_2}{k_1}$, we can see that if $k_{-1}$ is large and $k_2$ is small, $K_m$ approaches $\frac{k_{-1}}{k_1}$, which is the equilibrium constant for dissociation of the substrate from ES. For that reason, $K_m$ is often treated as an apparent measure of the affinity of the enzyme for the substrate, and that is an additional reason why we choose to define $K_m$ this way. Always remember that it is not actually the dissociation constant, and in some reactions will be quite far off from the actual $K_d$.

  • $\begingroup$ Now i am satisfied, thank you very much $\endgroup$
    – Paul
    Jun 6 '19 at 21:10
  • $\begingroup$ "quite far off from the actual Kd": You can't have a Km and a Kd at the same time. The Kd is for a molecule that binds but does not react (i.e. inhibitor bound to enzyme, or substrate bound to inactivated enzyme) while the Km is for a molecule that binds and reacts. So there is no "actual Kd", it is a though experiment. $\endgroup$ Apr 20 '21 at 13:33
  • $\begingroup$ @KarstenTheis - I disagree. A substrate molecule can react or it can fall back off. There is no reason why we cannot limit our focus to the elementary step of reversible binding, just as we often break down kinetic schemes into elementary steps. In the most basic M-M scheme, we have $\ce{E + S <=>[k_1][k_{-1}] ES ->[k_2] E + P}$. The $K_d$ is given by $\frac{k_{-1}}{k_1}$, while $K_m$ is $\frac{k_{-1} + k_2}{k_1}$. $\endgroup$
    – Andrew
    Apr 20 '21 at 13:51
  • $\begingroup$ I'm just saying that you can't measure Kd if you can measure Km. In a thought experiment, you can disregard k2, but in reality, the enzyme substrate complex has two fates, not one. You are right that in cases where k(-1) is much larger than k2, Km approaches what Kd would be if k2 were zero. If you simply define Kd as the ratio of k(-1) and k1 (and not as a measurable quantity), then you are absolutely right and the adjective "actual" is appropriate. $\endgroup$ Apr 20 '21 at 13:59
  • 1
    $\begingroup$ @KarstenTheis - I've never heard the claim that equilibrium constants must be easily measurable in order to be real. Furthermore, even if one uses applies that constraint, it is incorrect to say categorically that "you can't measure Kd if you can measure Km". It simply requires more and different data so that you can accurately fit to a model that separates k1, k-1 and k2. Pre-steady state kinetics are often used for this. Here's a good review of the subject: molbiolcell.org/doi/full/10.1091/mbc.e13-01-0030. $\endgroup$
    – Andrew
    Apr 20 '21 at 17:53

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.