Answer key: Enzyme Kinetics Activity
For teachers. This page is hidden from search engines and not linked from student pages, but it is not password-protected, so please don’t share the link with students. Student activity: 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.
Completed results table
| Setting | Reaction rate V | Apparent Km | Apparent Vmax |
|---|---|---|---|
| [S] = Km (half of Vmax) | 50 µmol/min | 2 mM | 100 µmol/min |
| Saturating: [S] = 10 × Km | 90.909 µmol/min | 2 mM | 100 µmol/min |
| Competitive: [I] = Ki | 33.333 µmol/min | 4 mM | 100 µmol/min |
| Competitive, outcompeted by 10 × Km substrate | 83.333 µmol/min | 4 mM | 100 µmol/min |
| Noncompetitive: [I] = Ki | 25 µmol/min | 2 mM | 50 µmol/min |
Answers
When [S] = Km, the rate was exactly half of Vmax. Use V = Vmax[S] ÷ (Km + [S]) to show why.
With [S] = Km, V = Vmax × Km ÷ (Km + Km) = Vmax ÷ 2. Here 100 × 2 ÷ (2 + 2) = 50 µmol/min.
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?
At high [S] almost every active site is already occupied, so the enzyme is close to saturation and the rate levels off toward Vmax = 100 µmol/min: V = 100 × 20 ÷ (2 + 20) = 90.9 µmol/min.
How did the competitive inhibitor change the apparent Km and Vmax, and why?
The apparent Km doubled from 2 to 4 mM, but Vmax stayed at 100 µmol/min. A competitive inhibitor competes with the substrate for the active site, so more substrate is needed to reach half speed, but enough substrate can still outcompete it.
With the competitive inhibitor present, what happened to the rate when [S] rose to 20 mM? What does this show?
The rate rose from 33.3 to 83.3 µmol/min, close to the uninhibited 90.9. Flooding the enzyme with substrate overcomes a competitive inhibitor.
How did the noncompetitive inhibitor change Km and Vmax? Why can’t adding more substrate overcome it?
Vmax halved from 100 to 50 µmol/min while Km stayed at 2 mM, so the rate at [S] = Km fell to 25 µmol/min. The inhibitor binds elsewhere on the enzyme, not at the active site, and stops it working whether or not substrate is bound, so more substrate cannot reverse 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?
V ÷ Vmax = [S] ÷ (Km + [S]) = 5 ÷ (0.05 + 5) = 0.990, so about 99% of Vmax. Glucose is at 100 times Km, so the enzyme is almost saturated.