1. Enzyme amount route: 12 µmol/min Vmax with 0.02 µmol enzyme
Biochemistry studentBackground
A saturated assay measures Vmax as 12 µmol/min and contains 0.02 µmol of active enzyme.
Problem
The student needs the turnover number in both minutes and seconds.
How to use
Select the amount mode and enter Vmax = 12 µmol/min and enzyme amount = 0.02 µmol.
{"mode":"amount","vmaxUmolMin":12,"enzymeUmol":0.02,"decimalPlaces":4}Outcome
The calculator returns kcat ≈ 600 min⁻¹ = 10 s⁻¹, classified in the moderate range.