You have decided to take CHEM 156, and you are doing an enzyme assay in class. Y

Question

You have decided to take CHEM 156, and you are doing an enzyme assay in class. You know that this enzyme obeys Michaelis-Menten kinetics, and you know the Km of the enzyme is 1.0 x 10-4 M.

You measure the initial velocity of the enzyme reaction twice using two different concentrations of substrate and the same concentration of enzyme. You get the following results:

You have decided to take CHEM 156, and you are doing an enzyme assay in class. You know that this enzyme obeys Michaelis-Menten kinetics, and you know the Km of the enzyme is 1.0 x 104 M. You measure the initial velocity of the enzyme reaction twice using two different concentrations of substrate and the same concentration of enzyme. You get the following results: Vo (HM/min) 43 0.2 0.02 43 Explain how two different substrate concentrations can produce the same Vo. Support your explanation by mathematically demonstrating how the Michaelis- Menten equation is consistent/predicts these results

Explanation / Answer

Michaelis-Menten equation is written as :

V = Vm* ( [S] / (Km+[S]) )

Here Vm is the maximum velocity (reaction rate) and Km is the Michaelis constant.

Putting [S] = 0.2 M gives Vm = 43.021 uM/min

Putting [S] = 0.02 M gives Vm = 43.21 uM/min

Thus we see that Vm values are almost the same, but not exactly.

But for an enzyme, Vm value is fixed. This means that keeping the Vm value fixed and putting 2 different [S] values is giving us two Vo values which are not exactly the same, but approximately equal to 43 uM/min.

This is not a big surprise, because in this case we see that [S] is comparitively very larger than Km, so essentially the denominator in the equation [S] + Km approximates [S], and the equation thus reduces to :

V = Vm , which shows reaction rate approaches maximum value and is independant of [S]. Thus we can see that Vo and Vm values are almost the same in this experiment.

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