Calculate vi and the degree of inhibition caused by a competitive inhibitor under the following conditions: (a) [S]=2x10-3 Mand[I]=2x10-3 M (b) [S]=4x10-4 Mand[I]=2x10-3 M (c) [S]=7.5x10-3 Mand[I]=10-5 M Assume that Km = 2 x 10-3 M, Ki = 1.5 x 10-4 M and Vmax = 270 nmoles x liter-1 x min-1.
Problems 14 and 15: some of the exponents are unclear. Here they are:
14. Calculate vi and the degree of inhibition caused by a competitive inhibitor under the following conditions:
(a) [S]=2x10-3 Mand[I]=2x10-3 M
(b) [S]=4x10-4 Mand[I]=2x10-3 M
(c) [S]=7.5x10-3 Mand[I]=10-5 M
Assume that Km = 2 x 10-3 M, Ki = 1.5 x 10-4 M
and Vmax = 270 nmoles x liter-1 x min-1.
The degree of inhibition is the percent of the uninhibited velocity reached in the presence of the inhibitor.
15. (a) What concentration of competitive inhibitor is required to yield 75% inhibition at a substrate concentration of 1.5 x 10-3 M if Km =2.9x10-4 M and Ki =2x10-5 M? (b)Towhatconcentration must the substrate be increased to reestablish the velocity at theoriginal uninhibited value?
Trending now
This is a popular solution!
Step by step
Solved in 4 steps
