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Specialty Chemistry Forums => Biochemistry and Chemical Biology Forum => Topic started by: joyb on April 22, 2013, 12:30:53 PM

Title: How does rate vary with time for an enzyme?
Post by: joyb on April 22, 2013, 12:30:53 PM

The question says: "Sketch rate vs. time and substrate concentration vs. time for an enzyme catalysed reaction that follows Michaelis-Menten kinetics when the initial substrate concentration is approximately equal to K_m."

I'm not sure how to draw either of these. So I guess the initial rate will equal V_max/2; does this mean that at this point the reaction is zero order with respect to the substrate? And then as the substrate concentration falls will substrate concentration tend to zero or is it more likely to reach equilibrium? (Otherwise can I assume as substrate concentration tends to zero v=V_max[ S ]/K_m so the reaction goes from zero order to 1st order?) Am I on the right track?

Title: Re: How does rate vary with time for an enzyme?
Post by: Babcock_Hall on April 22, 2013, 04:57:00 PM

At the point where [ S] = K_m, the rate is neither zero order nor first order, as one with better math skills than I have can probably derive from the full Michaelis-Menten equation.  The reaction is approximately zero order only when [ S] >> K_m.  I would probably assume that the reaction proceeds until substrate is completely consumed, for the sake of simplicity.