The initial rate for an enzyme-catalyzed reaction has been determined at several substrate concentrations. Data are as follows: [S] µmol/L νί(μmol/L) min1] 22 10 39 20 65 50 102 100 120 200 135 If the total enzyme concentration was 1nmol/L, how many molecules of substrate can a molecule of enzyme process in each minute? A. 175000 В. 325000 C. 900000 D. 10000 E. 375000 F. 166000 G. 3750 H. 1000000 I. 12500 5.

The following data was obtained during kinetic analysis of an enzyme with and without an inhibitor. Substrate concentration (mM) Reaction rate without inhibitor (µM/s) Reaction rate with inhibitor (µM/s) 10 28 12 20 50 23 40 83 42 60 107 58 100 139 83 200 179 125 300 197 150 400 209 167 560 227 197 How do you calculate the KM for the enzyme in the absence of an inhibitor. And how do you calculate kcat with the given enzymatic concentration of 5 µM.

Below is kinetic data obtained for an enzyme-catalyzed reaction. The enzyme concentration is fixed at 100 nM. Using a Lineweaver-Burke plot, calculate the kcat value for this reaction. Report your answer to three significant figures in units of 1/sec.

From the chart below, comment on the temperature dependence of this enzyme catalyzed reaction.