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31 March, 15:57

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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  1. 31 March, 17:36
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    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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