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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?

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    $\begingroup$ 1/2 is the least arbitrary choice of a number between 0 and 1. To me, that's good enough. $\endgroup$ – Ivan Neretin Jun 6 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 at 7:38
  • $\begingroup$ If you need a formal definition, then yeah, this one will do. $\endgroup$ – Ivan Neretin Jun 6 at 7:39
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    $\begingroup$ 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 at 13:00
  • $\begingroup$ @Andrew I think that's an answer. $\endgroup$ – Buck Thorn Jun 6 at 17:07
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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

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

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

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 $[\ce{S}]_o$. 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

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

Combining everything, we get

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

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

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

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 $=\frac{V_{max}[\ce{S}]_o}{K_m+[\ce{S}]_o}$.

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 $K_d$.

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  • $\begingroup$ Now i am satisfied, thank you very much $\endgroup$ – Paul Jun 6 at 21:10

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