The kinetics analyses of an enzyme are measured as a function of substrate concentration in the presence and in the absence of 2mM inhibitor [S] (microM): 3, 5, 10, 30, 90

Velocity (micromol/min) No Inhibitor: 10.4, 14.5, 22.5, 33.8, 40.5
Velocity (micromol/min) With Inhibitor: 4.1, 6.4, 11.3, 22.6, 33.8

Required:
What are the values of Vmax and Km in the absence of inhibitor? In its presence?

take the reciprocal of the values as 1/S and 1/V and plot the values on lineweaver-burk plot.

so the re-arranged data is as following

1/S               1/V (no inhibitor)       1/V (with inhibitor)

0.333           0.096                       0.244

0.2              0.068                         0.156

0.1               0.044                         0.088

0.033          0.029                        0.044

0.011            0.024                         0.029

and the plotted lineweaver-burk plot with the equation of straight line would be

where, first line represents the velocity of reaction with no inhibitor and second line represent the velocity in presence of inhibitor. X-axis has the substrate concentration as 1/S and y-axis has the velocity as 1/V.

as per the plot, equation of line on the lineweaver-burk plot, Y = mx + c, where c = 1/Vmax and m = Km/Vmax.

so solve for the given equations.

First, equation of line in the absence of inhibitor is Y = 0.23x + 0.021

so Vmax = 1/0.021 = 47.6 micromol/min

and Km/Vmax = 0.23, or Km = 0.23 x 47.6

Km = 10.95

Km do not have a unit, so 10.95 x 10-6 or as per your values round it up to 1 decimal point as 1.1 x 10-5

so value of Vmax = 47.6micromol/min and Km = 1.1 x 10-5 in the absence of inhibitor.

similarly, equation of line in the presence of inhibitor is given as Y = 0.67x + 0.021

so Vmax = 1/0.021 = 47.6 micromol/min

and Km/Vmax = 0.67

or Km = 0.67 x Vmax = 0.67 x 47.6 = 31.89

Km do not have a unit, so 31.89 x 10-6 or 3.1 x 10-5

so value of Vmax = 47.6micromol/min and Km = 3.1 x 10-5 in the absence of inhibitor.

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