4. (6 points) The initial rates at various substrate concentrations for an enzyme-catalyzed reaction are as follows:
[S] (M) Vo /10^-6 (M min^-1)
2.5x10^-5 38.0
4.0x10^-5 53.4
6.0x10^-5 68.6
8.0x10^-5 80.0
16.0x10^-5 106.8
20.0x10^-5 114.0
Hint: You will need the Excel software to make a plot. Use the double reciprocal plot to determine KM and Vmax parameters. Show all your work.
Make a plot of the initial rate as a function of substrate concentration as well as a double reciprocal plot. Does this reaction follow Michaelis-Menten kinetics?
Yes, this reaction follows Michaelis-Menten kinetics. At higher concentrations of substrate, the rate doesn't depend on substrate concentration. Linear plot.
Calculate the value of Vmax.
V = Vmax * [S] / (Km + [S])
1/V = (Km + [S]) / (Vmax * [S]) = Km / Vmax * [S] + 1 / Vmax
The intercept is at 1/Vmax.
The slope = Km / Vmax = Km * intercept
Vmax = 1 / 0.00623 = 160.51
Vmax = 160.51 M min^-1
Calculate the KM value of the reaction.
Km = slope / intercept
5.01 * 10^-7 / 0.00626 = 8.04 * 10^-5
Calculate the initial rates at [S] = 5.00 x 10^-5 M and [S] = 3.00 x 10^-1 M.
V = Vmax * [S] / (Km + [S])
Vmax = 160.51 M min and Km = 8.04 * 10^-5
At [S] = 5.00 * 10^-5
V = 160.51 * 5.00 * 10^-5 / (8.04 * 10^-5 + 5.00 * 10^-5) = 61.55
When [S] = 3.00 * 10^-1
V = 160.51 * 3.00 * 10^-1 / (8.04 * 10^-5 + 3.00 * 10^-1) = 61.55
Rate = 160.47 min^-1
What is the total amount of product formed during the first 3 min at [S] = 7.2 x 10^-5 M?
How would an increase in the enzyme concentration by a factor of 3 affect each of the following quantities: KM, Vmax, and Vo (at [S] = 5.00 x 10^-5 M)?
You added a competitive inhibitor [I] = 4.8 x 10^-4 M with KI = 1.7 x 10^-5 M. Determine the apparent Vmax and KM in the presence of the inhibitor.