Enzyme Kinetic Sample Problems #1
Given the reaction
E + S ES E + P
where k1 = 1 x 107 M-1 sec-1 k-1 = 1 x 102 sec-1, and kp = 3 x 102 sec-1
a) Calculate Ks
b) Calculate Km
k-1 1 x 102 sec-1
Ks = k1 = 1 x 107 M-1 sec-1 = 1 x 10-5 M
k-1 + kp (1 x 102 sec-1) + (3 x 102 sec-1) Km = k1 = 1 x 107 M-1 sec-1 = 4 x 10-5 M
2) An enzyme was assayed at an initial substrate concentration of 2 x 10-5 M. In 6 minutes half of the substrate had been consumed. Km for this enzyme’s substrate is 5 x 10 -3 M.
a) Calculate k
b) Calculate Vmax
Calculate the concentration of product produced in 15 minutes.
One can by inspection come to the conclusion that this reaction is first-order. This can be seen as follows:
[2 x 10-5]
Is Km < 0.01 ? [5 x 10 -3] = .004 Yes, first order
Since this is a first order reaction we can use the following equation to determine the rate constant, k.
k = t1/2 k = 6 min k = 6 min k = 0.115 min-1
k = Vmax / Km
Vmax = k x Km
Vmax = 0.115 min-1 x 5 x 10 -3 M = 0.575 x 10 -3 M min-1
Vmax = 0.575 x 10 -3 mole liter-1 min-1 = 575 moles liter-1 min-1
We know the initial concentration, [S] o, and we have calculated the rate constant, k. We need to determine an arbitrary [S] at 15 minutes after the initial concentration for a first order reaction.
2.303 log [S] =kt
[2 x 10-5]
2.303 log [S] = 0.115 min-1 x 15 min.
2.303 log [2 x 10-5] - 2.303 log [S] = 1.725
2.303 log [S] = 2.303 log [2 x 10-5] -1.725
2.303 log [S] = (2.303 (-4.699)) -1.725
2.303 log [S] = (-10.82) -1.725
log [S] = ((-10.82) -1.725) / 2.303
log [S] = (-12.547) / 2.303
log [S] = - 5.47
[S] = 3.56 x 10 -6 M at 15 minutes of reaction
We know that So = 2 x 10-5 M, the initial concentration
[P] = [S]o – [S] = 2 x 10-5 M - 3.56 x 10 -6 M = 1.644 x 10-5 M
An enzyme catalyzes the reaction S P. The following data has been obtained. Plot the data to determine Km and V max using both a Lineweaver-Burke plot and also an Eadie-Hofstee plot.
Manipulate the given data on the left into forms that can be plotted:
Sample Kinetics Problems 2004
Data for Problem #
M -1 min
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