7. Use the table o f enzyme kinetic data below to answer these three questions:

Question

7. Use the table o f enzyme kinetic data below to answer these three questions: Which enzyme has the most effective at lovw substrate concentration (i.e. binds substrate well?) 0.32 0.40 0.25 0.36 0.21 1500 3400 5500 6200 1800 Which enzyme has the fastest reaction rate at high substrate concentrations? Which enzyme is the most efficient overall in vivo and closest to catalytic perfection? 8. Use the graph below to calculate the kinetic parameters of the enzyme. Be sure to include units. Double Recderecl Plot 1.6 1.4 1.2 1/Vo (uM/min-1 1 0,8 0.6 Control -Inhibitor 150 uM 0.2 -QS-Q4 ,0.3-Q2 "0.1 0 0.1 0.2 0.3 0.4 0.5 1/IS] (mM)-1 Kn= What type of inhibitor was used? Write a chemical binding expression to show how the substrate, enzyme, and inhibitor bind with each other. 9. Suppose you suspect a molecule is an allosteric inhibitor of an enzyme. Describe an experiment to test this hypothesis and what results would you expect to get? (your answer should probably include a graph)

Explanation / Answer

7. most effective at low substrate concentration- E

Km is inversely proportional to binding affinity. Higher the Km, lower the binding affinity. Lower the Km, higher the binding affinity. Enzyme E has the lower Km of 0.21.

Enzyme that has the fastest reaction rate- D

Kcat or turnover number determines how many substrates can be converted to product by one enzyme in one second. Higher the Kcat, higher the reaction rate. Enzyme D has the higher Kcat.

Enzyme that is most efficient- C

Catalytic efficiency is determined by Kcat/Km. Kcat/Km for enzyme C is higher than that of all other enzymes.

8. -1/Km = -0.24

Km = 1/0.24 = 4.16 µM

1/Vmax = 0.5

Vmax = 2 µM/min

-(1+I/Ki)/Km = -0.42

Ki = 203

Type of inhibitor: uncompetitive inhibitor

Expression, 1/Vo = (Km/Vmax) (1/[S]) + (1+I/Ki)/Vmax

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