Question
Regarding the enzymatic reaction a) For an enzyme with K m = 28 * 10 -6 M and V max = 21.4 nMS -1 ,
Regarding the enzymatic reaction
a) For an enzyme with Km = 28 * 10-6 M and Vmax= 21.4 nMS-1, calculate the rate of enzymatic reaction at substrate concentration of 55 * 10-6 M
b) In presence of an inhibitor, Km was increased to 56 * 10-6 M while Vmax was not changed. What type of inhibition occurred? What is the mechanisms of this type of inhibition?
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Introduction to Chemical Engineering Thermodynamics
Authors: J. M. Smith, H. C. Van Ness, M. M. Abbott
7th edition
71247084, 978-0071247085
