Question

The kinetics of an enzyme were studied in the absence and presence of two different concentrations of an inhibitor. Determine the K_m and V_max of the uninhibited enzyme and identify which class of inhibitor is present using the data presented below. Use this data to determine the K_I, the binding constant of the inhibitor to the enzyme. Note that this question will require you to use a computer-graphing program such as EXCEL.  

[S] mmol/L	vo (mmol/L)min-1 (no inhibitor)	vo (mmol/L)min-1 (3 mM inhibitor)	vo (mmol/L)min-1 (5 mM inhibitor)
1.25	1.72	1.25	1.01
1.67	2.04	1.54	1.26
2.50	2.63	2.00	1.72
5.00	3.33	2.86	2.56
10.00	4.17	3.70	3.49

Answer 1

for determining the KI, Lineweaver- burk plot need to be generated. The plot is

1/V= (KM/Vmax)*1/S + 1/Vmax

so a plot of 1/V vs 1/S is straight line with intercept of 1/Vmax and slope of KM/Vmax

for the case of inhibition, Vmax becomes Vmaxapp and KM becomes Kmapp.

Both plots of 1/V vs 1/S is generated and they are shown below

from the plot of 1/V vs 1/S for no inhibitor, 1/Vmax = intercept= 0.1861, Vmax= 1/0.1861= 5.37 mmol/L.min, Slope is KM/Vmax= 0.4999, Km = 0.4999*5.37 =2.68 mmol/L

for the case of inhibitior, 1/Vmaxapp = 1/0.1861, Vmax= 1/0.1861= 5.37 mmol/L.min  

slope is Kmapp/Vmaxapp = 1.0062. Kmapp = 1.0062*5.37 =5.4 mmoles/L.

Since Vmax is same as Vmax app. , the inhibition is competitive where the enzyme and inhibitor competes for binding to the substrate.

KMapp = KM*(1+I/KI)

KMapp/KM= 1+I/KI

5.4/2.68-1 = I/KI

KI= I/1.0149 = 5mM/1.0149= 4.93