Question

a) The Km of an enzyme of an enzyme-catalyzed reaction is 7.5 µM. What substrate concentration will be required to obtain 65% of Vmax for this enzyme? (same enzyme was used in part a and b)

b) Calculate the Ki for a competitive inhibitor whose concentration is 7 x10-6 M. The Km in the presence of inhibitor was found to be 1.2x10-5 M. 

Answer 1

V = Vmax[S]/[KM+S]

where V= 0.65Vmax [S] =7.5*10-6M

0.65Vmax= Vmax[S]/(KM+S)

0.65 = S/ (KM+S)

S= 0.65KM+0.65S

0.35S= 0.65KM

KM= S(0.35/0.65)=0.54*7.5uM=4.05uM

2. V= VmaxS/(K_app+S)

Vmax is not going to be influenced by competitive inhibitor. Only KM changes

0.65Vmax= Vmax*7*10-6/ (Kapp+S)

0.65= 7.5*10-6/ (Kapp+7.5^10-6)

0.65*(Kapp+7.5*10^-6)= 7.5*10^-6

0.65Kapp +4.875*10^-6 = 7.5*10^-6

0.65Kapp = (7.5-4.875)*10^-6

Kapp = 4.04*10^-6

KM app =(1+I/KI)

4.04*10^-6 =1+7*10^-6/ Ki

7*10^-6/Ki= 1-4.04*10^-6 =0.999999

Ki = 7*10^-6/ 0.99999 = 7.00003*10^-6

Kapp = 3.77*10^-6