Question

Assume Km = 2 x 10-3; Ki = 1.5 x 10-4 and Vmax = 270 nmoles/l/min. Calculate vi and the degree (%) of inhibition caused by a competitive inhibitor under the following conditions: a) [S] = 2 x 10-3 M an [I] = 2 x 10-3 M b) [S] = 4 x 10-4 M an [I] = 2 x 10-3 M c) [S] = 7.5 x 10-3 M an [I] = 1 x 10-5 M

Answer 1

for competitive inhibition,

V= VmaxS/(KMapp+S)

KMapp= KM*(1+I/Ki)

Kmapp = 2*10-3*(1+2*10^-3/(1.5*10^-4)= 0.029M

V= 270*10^-9 moles/min* 2*10^-3/(0.029+2*10^-3)= 1.74*10-8 moles/min= 17.4*10^-9 moles/min

In the absence of an enzyme, V= VmaxS/(KM+S)= 270*10^-9* 2*10^-3/(2*10^-3+ 2*10^-3)=1.35*10^-7 moles/min

Hence % inhibition= 100* V in presence of inhibitor/V in absence of inhibitor= 100*17.4*10^-9/ (1.35*10^-7)= 12.89%

b)

Assume Km = 2 x 10-3; Ki = 1.5 x 10-4 and Vmax = 270 nmoles/l/min

Kmapp = 2*10-³*(1+2*10^-3/1.5*10^-4)= 0.029, V= 270*10-9* 4*10^-4/ (0.029+ 4*10^-4) in the presence of inghibitor=3.674*10^-9, in the absence of inhibitor, V= 270*10^-9*4*10^-4/ (4*10^-4+ 2*10^-3)=4.5*10-8 moles/min

% inhibition = 100* 3.674*10^-9/ 4.5*10^-8)= 8.164%

3. in the presence of inhibitor, KMapp = 2*10^-3*(1+1*10^-5/ 1.5*10^-4)= 0.002133M, V= 270*10^-9* 7.5*10^-3/(7.5*10^-3+ 0.002133)= 2.10*10^-7   in the absence of inhibitor, V= 270*10^-9* 7.5*10^-3/ (7.5*10^-3+ 2*10^-3)= 2.13*10-7 moles/min

% inhibition = 100*2.10*10^-7/ 2.13*10^-7 = 98.59%

ssume Km = 2 x 10-3; Ki = 1.5 x 10-4 and Vmax = 270 nmoles/l/min. Calculate vi and the degree (%) of inhibition caused by a competitive inhibitor under the following conditions: a) [S] = 2 x 10-3 M an [I] = 2 x 10-3