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Does an enzyme catalyzed reaction going at Vmax implies that the enzyme is nearly saturated?(no inhibitor present)
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Yes it does. You can also see it from math. We define Vmax within the MM-framework as: \[\Large V_{\max}=k_{cat} \times [E]_T\] In words this mean, that the highest rate is archived when all enzymes (E_T) is bound as enzyme-substrate complex.
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