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Catalase (MW250,000Da) has one of the highest turnover numbers we know of. It catalyzes the decomposition...

  1. Catalase (MW250,000Da) has one of the highest turnover numbers we know of.

It catalyzes the decomposition of hydrogen peroxide as shown in the reaction

                                    (2H2O2à2H2O + O2)

If 15.0 mg of pure catalasecatalyzes the decomposition of 0.40 g of H2O2in 1 min at 37 °C at its Vmax, what is the turnover number (kcat) of catalase(in units of sec-1)?

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Answer #1

Vmax = 0.4 gram in 1 minute = 0.4 gram in 60 seconds = 0.4/60 = 0.00667 gram per second

Turnover number = Kcat = Vmax/Et

Et is enzyme concentration = 15 mg = 15000 g

Kcat = 15000/0.00667 = 2,248,875.56 per second

It means around 2,248,875 molecules of substrate are converted to product.

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