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?
Homework Answers
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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