Q.52 The Kₘ and kcat of an enzyme are 4 mM and 0.1 μM h⁻¹ respectively. In the presence of 1.5 mM inhibitor, the Kₘ and kcat of the enzyme are 6 mM and 0.1 μM h⁻¹, respectively. The value of inhibition constant, Ki (correct to 1 decimal place) is _______ mM.
Km increases from 4 mM to 6 mM with inhibitor present, while kcat remains 0.1 μM h⁻¹, indicating competitive inhibition.
Inhibition Type
Competitive inhibitors bind the enzyme’s active site, competing with substrate. This raises apparent Km (Kmapp) as more substrate is needed for half-maximal velocity, but kcat and Vmax stay unchanged since inhibitor-substrate competition is overcome at high substrate levels.
Ki Calculation
For competitive inhibition, Kmapp = Km (1 + [I]/Ki).
Substitute values: Ki = 1.5 / (6/4 – 1) = 1.5 / 0.5 = 3.0 mM
Correct Ki = 3.0 mM.
SEO Article Content
This CSIR NET enzyme inhibition Ki calculation question tests competitive inhibition identification and formula application. When Km rises from 4 mM to 6 mM while kcat stays 0.1 μM h⁻¹ at 1.5 mM inhibitor, recognize competitive inhibition—Vmax unchanged, Kmapp increased.
Step-by-Step Solution
- Identify type: kcat same → Vmax same; Km ↑ → competitive.
- Formula: Km,app = Km (1 + [I]/Ki)
- Km,app/Km = 6/4 = 1.5
- 1 + 1.5/Ki = 1.5
- 1.5/Ki = 0.5
- Ki = 1.5/0.5 = 3.0 mM
Inhibition Types Table
| Type | Km Effect | kcat/Vmax Effect | Ki Formula |
|---|---|---|---|
| Competitive | ↑ | Unchanged | Ki = [I]/(Km,app/Km – 1) |
| Noncompetitive | Unchanged | ↓ | α = 1 + [I]/Ki on Vmax |
| Uncompetitive | ↓ | ↓ | Binds ES complex |
Master this enzyme inhibition Ki calculation for CSIR NET Life Sciences—practice Lineweaver-Burk plots where competitive lines intersect y-axis.
