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An enzyme has a molecular weight of 25,000 Da. If Vmax = 436 umol/min and you...

An enzyme has a molecular weight of 25,000 Da. If Vmax = 436 umol/min and you use 197 uL of a 1mg/mL solution of enzyme, what is the turnover number?

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Answer #1

1mg/mL = 1ug//uL

Et = mass / molar mass = 197 / 25000 = 0.00788 umol

Vmax = 436 umol/min
Vmax = Kcat x Et
and kcat = Vmax / Et = turnover number = 436 / 0.00788    = 55330 umol/min/umol

        = 922 umol/sec/umol

      turnover number = 922 sec-1

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