MathBench > Cellular Processes

Enzyme Kinetics

Using (and abusing) Km

tea leavesWe need to make this concrete: we'll investigate how the production of goop piles per minute depends on the concentration of chocolate chip bags -- that is, the number of bags per cubic meter.  Furthermore, let's just assume for a moment that we know, by divine inspiration or tealeaves or whatever method you choose, that Km = 5. (Later, we'll go through how to get this number in excruciating detail -- you may end up preferring tea leaves).

First, an easy problem:

  If [S] = 2.5 bags per cubic meter, then what percentage of Vmax is obtained?

(To make this problem interactive, turn on javascript!)

I think I have the answer: 5/(2.5+5) = 5/7.5 = 0.67 = 67%

Now for a slightly more difficult problem:

Turn on javascript to make this table interactive.

At what [S] is 80% of Vmax obtained?

  Hint... Details...
set up the equation [S] / (Km + [S]) = 0.8
use some algebra [S] = 0.8 (Km + [S])  
[S] = 0.8 Km + 0.8 [S]
0.2 [S] = 0.8 Km
[S] = 0.8/0.2 Km
[S] = 4 Km
interpret the answer
(remember Km = 5) 		in order to get a goop production rate of 80% of maximum, you would need to put out 20 bags

Here is an applet which will allow you to practice this calculation.  If you feel confident, just solve the first problem.  Otherwise, keep going until you do feel confident.  Each time you hit "new problem", a new set of parameters will appear.

   
value of Km  
% of Vmax desired %  
[S] needed bags/room