Solved by a verified expert: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