Enzyme Kinetics Activity
Goal: Use the Michaelis–Menten equation to find reaction rates from Vmax, Km and substrate concentration, and compare how competitive and noncompetitive inhibitors change an enzyme.
What to do
- On the No inhibitor tab, press “[S] = Km (half of Vmax)” and then “Saturating: [S] = 10 × Km”. Record V, the apparent Km and the apparent Vmax.
- Switch to Competitive inhibitor and press “Competitive: [I] = Ki”, then “Competitive, outcompeted by 10 × Km substrate”. Record the same readouts.
- Switch to Noncompetitive inhibitor and press “Noncompetitive: [I] = Ki”. Record the same readouts.
- For the challenge, go back to No inhibitor and press “Hexokinase (Km ≈ 0.05 mM) at 5 mM glucose”.
Record your results
| Setting | Reaction rate V | Apparent Km | Apparent Vmax |
|---|---|---|---|
| [S] = Km (half of Vmax) | |||
| Saturating: [S] = 10 × Km | |||
| Competitive: [I] = Ki | |||
| Competitive, outcompeted by 10 × Km substrate | |||
| Noncompetitive: [I] = Ki |
Questions
When [S] = Km, the rate was exactly half of Vmax. Use V = Vmax[S] ÷ (Km + [S]) to show why.
Raising [S] from 2 mM to 20 mM (10 times as much) only raised the rate from 50 to about 90.9 µmol/min. Why not 10 times faster?
How did the competitive inhibitor change the apparent Km and Vmax, and why?
With the competitive inhibitor present, what happened to the rate when [S] rose to 20 mM? What does this show?
How did the noncompetitive inhibitor change Km and Vmax? Why can’t adding more substrate overcome it?
Challenge: hexokinase has a Km of about 0.05 mM, and cells contain about 5 mM glucose. At what percentage of Vmax does it run?
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