Question

5) (14 marks) The following kinetic data were obtained for an enzyme in the absence of inhibitor (1), and in the presence of an inhibitor at 5 mM concentration (2). Assume[ET] is the same in each experiment. [S] (MM) (1) v(umol/mL sec) 12 (2) v(umol/mL sec) 4.3 1 8 2 4 20 29 14 21 8 35 12 40 26 a. Using a graphing program (excel or sigmaplot) construct a lineweaver burke plot representing the uninhibited reaction and the inhibited reaction on the same plot (to submit your graphs please paste into either a word or a pdf document along with your solutions) b. What is the equation for each of the lines? (on graph) c. Determine Vmax and Km for the enzyme. d. Determine the type of inhibition and the Kl for the inhibitor.

Answer 1

Vmax is the maximum rate of reaction or the maximum velocity attained by an enzyme. At Vmax, the enzyme is totally saturated with the substrate molecules; and any further increase in the substrate concentration will not cause any increase in the rate of reaction.

K_mis the substrate concentration at which the velocity of the reaction is equal to the half of the maximum velocity (Vmax). Lower the Km of an enzyme, higher is its affinity towards the substrate and vice versa. It is also called Michaelis constant.

The reciprocal values of the substrate concentration [S] and the initial velocity vo or v are calculated for the preparation of the Lineweaver Burk Plot, which can be obtained by taking the reciprocal of Michelis Menten Equation.

vo= Vmax. [S]/ Km +[S]

Taking reciprocal of the equation

1/vo= (Km/Vmax).1/[S]+ 1/Vmax

On comparing the Lineweaver Burk equation with the equation of a straight line, we can observe the following:

1/vo= (Km/Vmax).1/[S]+ 1/Vmax

y = m. x + c

y=mx+c

y=1/vo

m= Km/Vmax

x= 1/[S]

c=1/Vmax

So, y-axis: 1/Vo, and x-axis: 1/[S] ;  x-intercept= -1/Km, and y-intercept= 1/Vmax

a.

[S] (mM)	v (no inhibition) (uM/mL sec)	v (inhibition) (uM/mL sec)	1/[S] (1/mM)	1/v (no inhibition) (1/(uM/mL sec))	1/v (inhibition) (1/(uM/mL sec))
1	12	4.3	1	0.083333	0.232558
2	20	8	0.5	0.05	0.125
4	29	14	0.25	0.034483	0.071429
8	35	21	0.125	0.028571	0.047619
12	40	26	0.083333	0.025	0.038462

0.25 0.2 1/ VUM / mL sec 0.15 0.1 0.05 -1/v(no inhibition) uM / mL sec -1/v (inhibitor at conc.Sum / mL sec) -1/Km 1/Vmax € -1.5 -1 -0.5 1 1.5 0.05 0.5 1 1/[S] MM -1/Km -0.1

b.

y=1/vo

m (slope) = Km/Vmax

x= 1/[S]

c=1/Vmax

Therefore, y-axis: 1/Vo, and x-axis: 1/[S] ;

x-intercept= -1/Km, and y-intercept= 1/Vmax

c.

1/Km (1/mM)	1/Km' (1/mM)	1/Vmax (1/(uM/mL sec))	Km (mM)	Km' (mM)	Vmax (uM/mL sec)
0.3	0.1	0.02	3.33	10	50

d. Since the Km of the enzyme is increasing in the presence of the inhibitor, while Vmax is constant, it means that the type of inhibition is competitive.

Km' or the apparent Km (Km of tyhte enzyme in presence of the inhibitor) cn also be written as = αKm.

Therefore: αKm= 10

α= 10/3.33

α=3

Ki is the inhibitor constant and is determined by the following formula:

α= 1+ [I]/Ki

α =3

[I]= 5 mM

3=1+ 5/Ki

3Ki= Ki+5

2Ki= 5

Ki= 2.5 mM