Question

1. a) If one were to decrease the enzyme concentration by half, how would this affect the V_maxvalue? Explain why this is so. What assumption are we making in regards to the relative amounts of substrate to enzyme?

b) Is the turnover rate (k_cat value) of an enzyme expected to change if the enzyme concentration was decreased by half? Briefly explain.

c) Is the turnover rate (k_cat value) of an enzyme expected to change in the presence of a competitive inhibitor? Briefly explain.

Please answer the questions listed and explain. Thank you!

Answer 1

V 7. If one were to decrease the enzyne concentration! y half, the Umax is decreased by half. max is directly proportional to the total enzyme Concentration, E and the catalytic Constant of the enzyme, keat, which describes the frequency at which the enzyme-substrate complex is convested to product. " When the enzyme concentration is small, Vmax is much smaller, The reaction rate still increase with increasing Substrate concentration, but lovel off at a much lowes rate. By increasing the enzyme concentration, the maximum reaction date greatly increases. by so we know that turnover rate keat =Vmax/[E]; ther a decrease in the Vmax would have to be accompanied by a decrease in the enzyme concentra -tion for keat to remain the same. Therefor, Keat. will rerrain some aftes changing the enzyme concent gation by half c) In case of competive to inhibitors. Etsikon applies) -> ES I feat applies) - E+P. 7 In this scenario keat our inhibitor does nothing to the enzyme substrate complex and its ability to turnover, instead it just raise the km I concentration of bound substrate to achieve Imax). by Lo, The turnover ratel keat values of an enzyme o does not change in the presence of an competitive inhibitore