The data to the right was collected for a certain enzyme-catalyzed DSI (unol. L-

Question

The data to the right was collected for a certain enzyme-catalyzed DSI (unol. L-ty Rumol. L-1. min 1) reaction. Use these data to determine the values of Rmax (the 22 5.0 maximum rate of the enzyme, often called the maximum velocity, 39 10.0 Vmax) and KM (the Michaelis-Menten constant of the enzyme). 65 Answer the three parts below. 20.0 50.0 102 a) If you were to plot this data to graphically determine max and KM R 120 100.0 using a Lineweaver-Burk plot, what would you plot? If you would he original data, enter the original value in the corresponding 200.0 blank 135 [sl (Amol. L) Graph as R (Amol min Graph as Number Number 5.00 0.2 0.045 22 Number Number 20.0 0.05 0.015 65

Explanation / Answer

c)

Rmax --> max rate, is essentially given by the linewaver burk plot as follows:

y-intercept --> 1/Vmax

y.intercept = 0.00609

so

Vmax = 1/(y-intercept) = 1/(0.00609) = 164.203 micromol / L min

Vmax = 164.203 micromol / L min

b)

Km is given by linewaver burk plot as follows:

slope = Km/Vmax

slope = 0.196

Vmax = 164.203

then

Km = slope*Vmax = (0.196)(164.203 ) = 32.1837 micromol/L

Km =  32.1837 micromol/L

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