MathBench > Cellular Processes

Enzyme Kinetics

Km in the crosshairs

So using the Michaelis Menten equation, you can tell me how many bags to put out in order to achieve any possible goop production rate -- 10% of max, or 45%, or 95%... 

Or can you? Specifically, how many bags do I need to put out to get 100% of Vmax?

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How do you get 100% of Vmax?

  Hint... Details...
set up the equation [S] / (Km + [S]) = 1.0
use some algebra [S] = Km + [S]
interpret the results Unless Km is zero, this will NOT work.

Using algebra, we can't find any amount of substrate that will allow the reaction (goop production) to take place at its maximum rate. How about using common sense? Goop production should happen fastest if there's a lot of chocolate around, so let's try making [S] really big:

If [S] is 1000 bags/room, then

[S] / (Km + [S]) = 1000/1010 = 0.99

If [S] is 10000 bags/room, then

[S] / (Km + [S]) = 10000/10010 = 0.999

Hmm, I can get really close to 100%, but I can never actually get there.  Its easy to see the same thing on the graph:

percent of Vmax

The fact that we can never exactly get to Vmax is a bit of a problem when it comes to determining what Vmax is.  In the applet below, you play a researcher trying to determine the Vmax of a new enzyme, yellowcakease.

applet

Have no fear, we will come back to efficient methods for determining Vmax in just a few screens.