Question

1. You measure the initial rate of an enzyme reaction as a function of substrate concentration in the presence and absence of an inhibitor. The following data was obtained: V. (-) Inhibitor (mm/min) (+) Inhibitor (mM/min) 17 [S] (MM) 0.0001 0.0002 0.0005 0.001 0.002 Please submit calculations and graph for full credit! Note: You are required to use Excel to generate the Lineweaver-Burk plot (a) (10 points) Create a Lineweaver-Burk plot inhibited and uninhibited should be on the same plot!) (b) (15 points) Calculate the Vmax in the presence and absence of the inhibitor. Show your calculation (e) (15 points) Calculate the Km in the presence and absence of the inhibitor. Show your calculation (d) (10 points) What type of inhibitor is this? How do you know (clearly explain)?

need B C and D done please please please help!!!

Answer 1

0.08 • + 1/[(-) 1/[V(+) 0.06 1/[H](min/mM) 0.00+ 0.00 TTTTTTTT 1/[S] | 10000 - 15000

Slope =Km/Vmax, The y-intercept is 1/Vmax

Thus the slope for V(-) is 2.032 e^-006. So 1/slope is 492168. The y-intercept is 0.009946

The slope for V(+) is 4.689 e^-006. So 1/slope is 205358. The y-intercept is 0.01010

putting the values in the equation,

we get Vmax for without inhibitor and with an inhibitor to be 100.54 mM/min and 99.009 mM/min respectively

and by substituting the Vmax values in the slope equation we get the Km for the without inhibitor and with an inhibitor to be 0.000204 and 0.000481 respectively

This is competitive inhibition as this does not have an effect on Vmax when the substrate concentration increases. The Vmax is conserved.