A) turn in two plots: 1) vo versus [S]; and 2) 1/ vo versus 1/[S]. Be sure to use proper units for these plots: vo = amount of pNP formed/unit time; and [pNPP] = mM. (Important to get the total volume of the reaction mixture correct including the addition of enzyme to the final mixture for calculating substrate concentrations.)
B) From the second plot (ie the double reciprocal plot or 1/ vo versus 1/[S] ), calculate the Km and Vmax for the uninhibited reaction and the apparent Km (ie Km’) and apparent Vmax (ie Vmax’) from inhibited reactions. By apparent Km or Km’ and apparent Vmax or Vmax’, I mean the values obtained from the double reciprocal plot for the two inhibited reactions. They are apparent values because an enzyme can only have one Km and Vmax under a given set of conditions of substrate concentrations and enzyme concentration. So in the inhibited reactions, the values you get from the graph for Km and Vmax are only appearing to be those values because the inhibitor is present.
C) For each inhibitor, calculate the Ki for inhibitor binding to the enzyme using the equations shown in the table below. Ki values are in units of concentration like the Km, so you need to use the final concentration of the inhibitor in the calculations - [Pi] = 2 mM and [F-] = 0.2 mM. Show how you arrive at these inhibitor concentrations. Finally, compare the Ki values for the two inhibitors and say which one is the more effective inhibitor (ie which one has smaller Ki).
Equations for calculation of Ki constants for the inhibitors:
| Type of Inhibitor | Apparent Km (ie. Km') | Apparent Vmax (ie. Vmax') |
| Competitive | Km(1+[I]/Ki) | Vmax |
| Non-Competitive | Km | Vmax / (1 + [I]/Ki) |
Rearrange the equations to find the Ki values.
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©Wilbur H. Campbell, 1996, 1998; wcampbel@mtu.edu