At what substrate concentration would an enzyme with a Km of 0.005M operate at one quarter of its Vmax? What would be the Km for an enzyme if it is operating at 90% of its Vmax when the substrate concentration is 0.01M

The following data was obtained during kinetic analysis of an enzyme with and without an inhibitor. Substrate concentration (mM) Reaction rate without inhibitor (µM/s) Reaction rate with inhibitor (µM/s) 10 28 12 20 50 23 40 83 42 60 107 58 100 139 83 200 179 125 300 197 150 400 209 167 560 227 197 How do you calculate the KM for the enzyme in the absence of an inhibitor. And how do you calculate kcat with the given enzymatic concentration of 5 µM.

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The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. Vmax [S] Km + [S] where v is the velocity, or rate, Vmax is the maximum velocity. K is the Michaelis-Menten constant, and [S] is the substrate concentration. A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (ro) at different substrate concentrations ([S]). First, move the line labeled Vmax to a position that represents the maximum velocity of the enzyme. Next, move the line labeled 1/2 Vmax to its correct position. Then, move the line labeled Km to its correct position. Estimate the values for Vmax and Km- Vmax= µM/min v (µM/min) 300 275 250 225 200 175 150 125 100 75 50 Km = 25 K 0 10 20 30 V max 40 50 [S] (M) 1/2 V max Michaelis-Menten curve 60 70 80 90 100 HM