Quantitative relationships between rate constants to calculate Km, kinetic efficiency (kcat/Km) and Vmax –VI
The enzyme catalase catalyzes the decomposition of hydrogen peroxide:
2 H2O2 ⇌ 2 H2O + O2
The turnover number (kcat) for catalase is 40,000,000 sec-1. The Km of catalase for its substrate H2O2 is 0.11 M.
a. In an experiment using 3 nanomole/mL of catalase, what is Vmax?
b. What is v when [H2O2] is 0.75 M?
c. What is the catalytic efficiency of catalase?
d. Does catalase approach “catalytic perfection”?
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