Q.52 Km and Vmax of an enzyme preparation are 5 μM and 30 μM min–1 respectively.
Considering, Ki value of competitive inhibitor is 60 μM, the velocity (V0) of this
enzyme–catalyzed reaction in the presence of 200 μM of substrate and 600 μM of
competitive inhibitor is _______ μM min–1 (rounded off to two decimal places).
Problem Solution
The velocity of the enzyme-catalyzed reaction under competitive inhibition conditions is 23.53 μM min⁻¹.
Given Km = 5 μM, Vmax = 30 μM min⁻¹, Ki = 60 μM, [S] = 200 μM, and [I] = 600 μM, competitive inhibition increases the apparent Km while Vmax remains unchanged.
Formula for initial velocity (V₀):
\[ V_0 = \frac{V_{\max} [S]}{K_m \left(1 + \frac{[I]}{K_i}\right) + [S]} \]
Step-by-Step Calculation
- Calculate apparent Km (Kmapp):
\[ K_m^{app} = 5 \left(1 + \frac{600}{60}\right) = 5 \times 11 = 55 \, \mu M \] - Substitute into velocity equation:
\[ V_0 = \frac{30 \times 200}{55 + 200} = \frac{6000}{255} = 23.53 \, \mu M \, min^{-1} \]
Competitive Inhibition Mechanism
Competitive inhibitors bind reversibly to the enzyme’s active site, competing directly with substrate and reducing effective enzyme availability.
This elevates Kmapp by the factor (1 + [I]/Ki) but Vmax stays constant since high [S] displaces the inhibitor.
Lineweaver-Burk plots show lines intersecting on the y-axis, confirming unchanged Vmax [memory:5].
Why Competitive Inhibition Affects Km Not Vmax
Unlike non-competitive inhibition, competitive types increase apparent Km by substrate competition but allow Vmax recovery at high [S].
CSIR NET questions often use Lineweaver-Burk plots to distinguish: y-axis intercept same, x-intercept shifts right [memory:1].
CSIR NET Exam Tips
- Practice derives Kmapp = Km (1 + [I]/Ki) for quick solves
- Common errors include using wrong inhibition formula or forgetting units—always verify μM consistency
- Memorize: Competitive → ↑Km, Vmax unchanged; Non-competitive → ↓Vmax, Km unchanged [memory:6]
