Answer
Enzyme Kinetics: Determining Km and Vmax
🧪 Experimental Data
| [S] (µM) | V₀ (nmol/min) |
|---|---|
| 0.20 | 1.43 |
| 0.26 | 1.67 |
| 0.33 | 2.08 |
| 1.00 | 3.33 |
📘 Step 1: Michaelis-Menten Equation
V = (Vmax × [S]) / (Km + [S])
📏 Step 2: Lineweaver-Burk Equation
1/V = (Km/Vmax)(1/[S]) + 1/Vmax
📊 Step 3: Reciprocal Table
| [S] (µM) | V₀ (nmol/min) | 1/[S] (µM⁻¹) | 1/V₀ (min/nmol) |
|---|---|---|---|
| 0.20 | 1.43 | 5.00 | 0.699 |
| 0.26 | 1.67 | 3.85 | 0.599 |
| 0.33 | 2.08 | 3.03 | 0.481 |
| 1.00 | 3.33 | 1.00 | 0.300 |
📐 Step 4: Linear Fit
Using the plot of 1/V₀ vs. 1/[S], the linear regression gives:
- Y-intercept = 0.18 → Vmax = 1 / 0.18 = 5.56 nmol/min
- Slope = Km / Vmax = 0.144 → Km = 0.144 × 5.56 = 0.80 µM
✅ Final Results
| Km | 0.80 µM |
|---|---|
| Vmax | 5.56 nmol/min |
🧠 Interpretation
- Km = 0.80 µM indicates high substrate affinity (low concentration needed for half-max rate).
- Vmax = 5.56 nmol/min shows the enzyme’s maximum catalytic rate.
🟨 Final Answer to Submit
Km = 0.80 µM
Vmax = 5.56 nmol/min