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$K_m$ is the Michaelis constant, which is the concentration of substrate needed to achieve a rate of $\frac{V_{max}}{2}$. $V_{max}$ is the maximum number of molecules that can be reacted per second.

enter image description here

My lecturer has said that $K_m$ is the lowest concentration of substrate needed to ensure that the system runs at $V_{max}$.

I do not understand why. To me, I would have assumed that it would need to be at a concentration that would produce $V_{max}$ or something close to it.

I can understand that at a concentration that is too low, then the enzyme will make quick work of the backlog. However, I cannot see why exactly Km should be the point at which this happens. It all feels quite arbitrary to me.

Could someone explain, preferably intuitively, why this is true?

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$K_m$ is the lowest concentration of substrate needed to ensure that the system runs at $V_{max}$?

No. $V_{max}$ is never reached, it is the limit of the rate when you increase the substrate concentration indefinitely. Less dramatically said, when the substrate concentration is about ten times higher than $K_m$, the rate will be about 90% of $V_{max}$ (and so on, so hundred times higher - about 99%).

In practice it is often difficult to get to hundred times $K_m$ because the substrate might not be that soluble or there is substrate inhibition at non-physiologically high substrate concentrations.

Could someone explain, preferably intuitively, why this is true?

As I read in a textbook, imagine a bicycle repair shop with 5 people working in it. When the number of bikes brought in doubles from what it is usual, they will repair more bikes. It does not matter, however, if you park a million or a billion bikes in front of their shop, they will be working at capacity in either case.

If you like math, you can also answer your question by looking at the relationship between substrate concentration and rate described by the Michaelis-Menten equation set up by Leonor Michaelis and Maud Menten:

$$\mathrm{rate} = V_\mathrm{max} \frac{[S]}{K_m + [S]} = V_\mathrm{max} \frac{1}{1 + \frac{K_m}{[S]}}$$

When $[S] \gg K_m$, the fraction approaches 1 and the rate approaches $V_\mathrm{max}$.

My lecturer has said

That is what you heard. If you think the lecturer misspoke or you misheard, it is a good idea to check a textbook or ask here.

The figure

The figure is a conceptual one with two conceptual errors. First, the rate in the figure does exceed $V_\mathrm{max}$.

enter image description here

Secondly, the figure shows a linear increase of the rate for substrate concentrations lower than $K_m$. In the graph showing the real shape of the curve, you can see appreciable curvature at lower substrate concentrations.

enter image description here

For the graph, $V_\mathrm{max}$ was set to 1 and $K_m$ to 0.2. The shape of the curve, however, is the same (if you choose the plotting range in an analogous manner) for any combination of the two parameters.

The problems with the figure are common, as the following gallery shows: enter image description here

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    $\begingroup$ I don't follow, when $[S]<<K_m,\; v_0=V_{max}[S]/K_m$ $\endgroup$ – porphyrin Mar 30 at 12:47
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    $\begingroup$ I assume you meant "When [S] >> Km, the fraction approaches 1 and the rate approaches Vmax"? $\endgroup$ – Andrew Mar 30 at 12:52
  • $\begingroup$ @porphyrin and Andrew: Thanks for the correction. I was looking for the Latex code for >> and << (\gg and \ll ) and did not proof-read. $\endgroup$ – Karsten Theis Mar 30 at 12:56
  • $\begingroup$ Still need to change Km to Vmax at the end of the sentence, right? $\endgroup$ – Andrew Mar 30 at 13:30
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    $\begingroup$ @Andrew I should drink some coffee before saving half-finished answers... thanks again. $\endgroup$ – Karsten Theis Mar 30 at 14:00

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