Question

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: Create a Michaelis-Menten plot (inhibited and uninhibited should be on the same plot! You MUST use Excel and follow the provided instructions "Non-Linear Regression Fitting of Kinetics Data". Calculate the V_max in the presence and absence of the inhibitor. Calculate the K_m in the presence and absence of the inhibitor. What type of inhibitor is this? How do you know (clearly explain)?

Answer 1

[S] mM 1/ [S] mm 1 Rate or V (absence of inhibitor) mm/min 33 Rate or V (presence of inhibitor) mm/min 17 1/rate or 1/V (absence of inhibitor) mMmin 0.0303 0.02 0.01408 0.0120 0.0109 0.0001 0.0002 0.0005 0.001 0.002 10000 5000 2000 1000 500 29 1/rate or 1/v (presence of inhibitor) mM-min 0.0589 0.0345 0.02 0.0175 0.0125 50 91

y= 5E-06x + 0.011 1/rate or 1/ mM'min y = 2 E-06x + 0.01 2000 4000 8000 10000 12000 5000 1/ [S] mM1 Red line: without inhibitor Greenline: with inhibitor

Without inhibitor

Equation is   y = 2E-06x + 0.01

Comparing it with line weaver burk equation i.e

1/Vo = Km / Vmax( [S] + 1/Vmax

Comparing slope with                            Km/Vmax = 2 x 10-6                                                                            -i)

And intercept with                                   1/Vmax = 0.01                                                                               -ii)

Multiply equation i) with reciprocal of equation ii)

We get                                   Km =    2 x 10-6/0.01      =   200 x 10-6 mM = 0.2 x 10-3

                           And by ii)     Vmax = 100 mM/min

With inhibitor

Equation is y = 5E-06x + 0.011

Comparing it with line weaver burk equation i.e

1/Vo = K ‘m / Vmax( [S] + 1/Vmax

where k ‘m = km (1 + KI[In])       i.e. Michaelis-Menten constant in presence of inhibitor.

Comparing slope with                            K ‘m/Vmax = 5 x 10-6                                                                            -i)

And intercept with                                   1/Vmax = 0.011                                                                               -ii)

Multiply equation i) with reciprocal of equation ii)

We get                                   K ‘m =    5 x 10-6/0.011      =   454.55 x 10-6 mM = 0.45455 x 10-3 mM

                           And by ii)     Vmax = 90.91 mM/min

From graph it is clear that both the lines have almost same y intercept ( i.e 0.01 and 0.011) hence the inhibition by inhibitor is competitive.