Question

	Velocity (mM/min)
[S], mM	Uninhibited	Inhibited
1.75	1.94	1.38
2.17	2.26	1.67
3.00	2.85	2.13
5.50	3.55	2.97
10.5	4.39	3.83

(a) Compute for the values of 1/[S] and 1/v for both the uninhibited and inhibited reactions.

	1/Velocity (min/mM)
1/[S], mM-1	Uninhibited	Inhibited

(b) Use MS Excel program to construct the Lineweaver-Burke plots for both reaction systems.Conduct linear regression analysis and compute for the equation of the lines and their respective R² values. Compute for the V_max and K_m values of the uninhibited and inhibited reaction. Determine the mode of inhibition based on the generated graph and the values of the kinetic parameters.

Answer 1

(a):

(b):

Uninhibited Reaction:

Y = 0.596X + 0.167

In Lineweaver Burk plot (Uninhibited):

Y-intercept = 0.167 = 1/Vmax

=> Vmax = 1/0.167 = 5.99 mM/min (Answer)

X-intercept is when Y = 0

=> 0 = 0.596X + 0.167

=> X = - 0.167 / 0.596 = - 0.28020

Also X-intercept = - 1/Km

=> - 0.28020 = - 1/Km

=> Km = 1/0.28020 = 3.57 mM (Answer)

Inhibited Reaction:

Y = 0.965X + 0.161

In Lineweaver Burk plot (Inhibited):

Y-intercept = 0.161 = 1/Vmax

=> Vmax = 1/0.161 = 6.21 mM/min (Answer)

X-intercept is when Y = 0

=> 0 = 0.965X + 0.161

=> X = - 0.161 / 0.965 = - 0.16684

Also X-intercept = - 1/Km

=> - 0.16684 = - 1/Km

=> Km = 1/0.16684 = 5.99 mM (Answer)

Since Km is increased from uninhibited to inhibited, where as Vmax values are nearly same, this is a competitive inhibition.