Q.33 For a single substrate enzyme, a reaction is carried out at a substrate concentration four times the
value of Km. The observed initial velocity will be _________ % of Vmax.
Answer: 80%
For [S] = 4Km, v = 80% of Vmax
Michaelis-Menten Calculation
Using the Michaelis-Menten equation:
v = Vmax × [S] / (Km + [S])
Given [S] = 4Km:
v = Vmax × (4Km) / (Km + 4Km)
v = Vmax × 4Km / 5Km
v = Vmax × 4/5
v = 0.8 Vmax = 80% Vmax
v = Vmax × 4Km / 5Km
v = Vmax × 4/5
v = 0.8 Vmax = 80% Vmax
Saturation Levels Reference
| [S]/Km | % of Vmax | Active Sites Occupied |
|---|---|---|
| 0.2 | 17% | Low |
| 4.0 | 80% | High |
| 10.0 | 91% | Near-maximum |
Introduction: Substrate Concentration Four Times Km Vmax
Enzyme kinetics problems calculating initial velocity at specific substrate concentrations relative to Km test Michaelis-Menten equation application. This analysis shows [S] = 4Km yields exactly 80% Vmax, a standard benchmark for biochemical engineering exam preparation.
Common Exam Values Table
| [S] vs Km | % Vmax | Calculation |
|---|---|---|
| 0.5 Km | 33% | 0.5/1.5 = 1/3 |
| 4 Km | 80% | 4/5 |
| 9 Km | 90% | 9/10 |
| 19 Km | 95% | 19/20 |
Physiological Significance
- 4Km represents practical saturating conditions in vivo
- 80% Vmax allows responsive flux control (not fully saturated)
- Enzymes rarely operate at true Vmax in cells (metabolic sensitivity)
