# calculate km and vmax from the following data

This is ~1000 times slower than the reaction described in the reaction before.

Q2/ Determine the type of inhibition of an enzymatic reaction form the follwoing data collected in the presecnce and absence of the inhibitor [S] [V0] V0 with I present 1 1.3 0.8 2 2.0 1.2 4 2.8 1.7 8 3.6 2.2 12 4.0 2.4

The enzyme solution's activity would therefore not be exactly 100 times less than the undiluted sample, but would be somewhat greater than that value because the proportion of uninhibited enzyme would be greater at the lower concentration. When the concentration of A is 20mM, the reaction velocity is measured as 5µM B produced per minute. Based on some preliminary measurements, you suspect that a sample of enzyme contains an irreversible enzyme inhibitor. Privacy calculating kcat, the concentration of glucose oxidase used in the Seven seperate reactions were examined each containing a different amount of A added. So Vmax = 1/6.0373 = 0.1656 µmol/min/ml The x intercept is -1/km. Question: Calculate The Vmax, Kcat, KM And Catalytic Efficiency For Each Substrate Using Information From The Lineweaver Burk Plots, Then Fill Out The Table Below.