asked 203k views
2 votes
If the total enzyme concentration was 1 nmol/l, how many molecules of substrate can a molecule of enzyme process in each minute?

asked
User Zion
by
7.4k points

1 Answer

6 votes
kcat = Vmax/[E]
(164 μmol L−1 min−1)/(0.001 μmol L−1)
1.64 × 105 min−1
2733 s−1
answered
User Arnav Aggarwal
by
7.8k points
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