54. The Ki of a novel competitive inhibitor designed against an enzyme is
2.5 µM. The enzyme was assayed in the absence or presence of the inhibitor
(5 µM) under identical conditions. The Km in the presence of the
inhibitor was found to be 30 µM. The Km in the absence of the
inhibitor is __________ µM.
For competitive inhibition, the apparent Km increases based on inhibitor concentration and Ki. Here, with Ki at 2.5 µM and inhibitor at 5 µM, the Km in the presence is 30 µM, yielding Km = 10 µM without inhibitor.
Competitive Inhibition Basics
Competitive inhibitors bind the enzyme’s active site, competing with substrate. This raises the apparent Km to Km(app) = Km × (1 + [I]/Ki), where Vmax stays unchanged.
Km without inhibitor solves from rearranging: Km = Km(app) / (1 + [I]/Ki).
Step-by-Step Calculation
Given Ki = 2.5 µM, [I] = 5 µM, Km(app) = 30 µM.
- Compute α = 1 + [I]/Ki = 1 + 5/2.5 = 1 + 2 = 3
- Km = 30 / 3 = 10 µM
Verification: Km(app) = 10 × 3 = 30 µM matches data perfectly.
Options Analysis
Options from source imply pairs like 90/10 or 10/30 µM. Here’s the complete breakdown:
| Option Pair | Calculation Check | Correct? |
|---|---|---|
| 80 or 30 | 30 / (1+5/2.5) = 10 ≠ 80 or 30 | No |
| 100 or 40 | Yields ~33 µM apparent, mismatch | No |
| 90 or 10 | 10 × 3 = 30 µM ✓ | Yes (Km = 10 µM) |
| 60 or 100 | 100×3=300 ≠30; 60×3=180 mismatch | No |
Correct Answer: Km = 10 µM (option C aligns if listing apparent/absent variants).
