pacruzalesrockss
pacruzalesrockss pacruzalesrockss
  • 19-10-2016
  • Biology
contestada

If the total enzyme concentration was 1 nmol/l, how many molecules of substrate can a molecule of enzyme process in each minute?

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Hussain514 Hussain514
  • 25-10-2016
kcat = Vmax/[E] 
(164 μmol L−1 min−1)/(0.001 μmol L−1)
1.64 × 105 min−1
2733 s−1
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