Question

1. The kinetics of an enzyme was examined at various substrate concentrations in both the presence and absence of 3 mM inhibitor Z. The initial velocity data obtained are shown below: [S] (mmoles liter) v (mmoles"litermin) no inhibitor inhibitor Z 1.25 1.67 2.50 5.00 10.0 1.72 2.04 2.63 3.33 4.17 0.98 1.17 1.47 1.96 2.38 (4 pts) Estimat e Vmax and Kw in the presence and absence of inhibitor using the Michaelis Menton curve-fitting program on Kaleidagraph (see lab manual). Make sure to include units. Kaleidagraph plots must be included to receive full credit. no inhibitor KM inhibitor Z KM Vm b) (1 pt) What type of inhibitor is inhibitor Z? Explain. c). (1 pts) What is the kcat in the presence and absence of inhibitor if the total enzyme concentration used in each assay was 1 nmole liter d) (1 pts) What is the efficiency of the enzyme in the absence and presence of the inhibitor? 2. (3 pts) What is the dissociation constant for the inhibitor?

Answer 1

Answer 1) a

no inhibitor K_m=2.5 mmoles/L V_max= 2.97 mmoles/L/min

with inhibitor Z K_m=2.5 mmoles/L V_max= 5.13 mmoles/L/min

1 b) The Inhibitor Z is non-competitive type of inhibitor as the Km value is similar to that without inhibitor, while the Vmax has increased when compared to without inhibitor.

1 c)The Kcat value without inhibitor

no inhibitor K_cat=V_max/E_T = 2.97/(1*10^-6) = 2.97*10⁶ min^-1 ​

with inhibitor Z K_cat==V_max/E_T=​5.13/(1*10^-6)​ = 5.13*10⁶ min^-1

1d) Efficiency of enzyme

absence of inhibitor Kcat/Km=2.97*10⁶ /2.5 = 1.19*10⁶ (mmoles/L)^-1  min^-1   

Presence of inhibitor Z Kcat/Km=5.13*10⁶ /2.5 = 2.05*10⁶ (mmoles/L)^-1  min^-1