![Q.40 The kinetic data for a single substrate enzyme is shown below. The concentration of inhibitor [I] used in the reaction was equal to the Ki of the inhibitor. The Km value of an uninhibited reaction is 2 × 10−5 M. In the presence of the inhibitor, the observed Km value is ________ × 10−5 M.](https://www.letstalkacademy.com/wp-content/uploads/2026/01/slide_40-5.jpg)
Q.40 The kinetic data for a single substrate enzyme is shown below.
The concentration of inhibitor [I] used in the reaction was equal to the
Ki of the inhibitor.
The Km value of an uninhibited reaction is
2 × 10−5 M.
In the presence of the inhibitor, the observed Km value is
________ × 10−5 M.
Given Data
- Type of plot: Lineweaver–Burk (1/v vs 1/[S])
- Inhibitor concentration: [I] = Ki
- Uninhibited Km = 2 × 10−5 M
Step 1: Identify the Type of Inhibition
From the Lineweaver–Burk plot:
- All lines intersect at the same Y-axis point
- Y-intercept (1/Vmax) remains unchanged
- Slope increases in the presence of inhibitor
These features indicate competitive inhibition.
Key Concept: Competitive Inhibition
In competitive inhibition, the apparent Michaelis constant is given by:
Kmapp = Km (1 + [I]/Ki)
Step 2: Substitute Given Values
Since [I] = Ki:
[I]/Ki = 1
Therefore:
Kmapp = Km (1 + 1) = 2Km
Step 3: Calculate the New Km
Kmapp = 2 × (2 × 10−5)
Kmapp = 4 × 10−5 M
Final Answer
The observed Km value in the presence of inhibitor is
4 × 10−5 M.
Conclusion
The Lineweaver–Burk plot clearly indicates competitive inhibition, where the
inhibitor increases the apparent Km without affecting Vmax.
When the inhibitor concentration equals Ki, the Km value
doubles. Hence, the correct observed Km is 4 × 10−5 M.
