Discussion
The kinetics of an enzyme in the presence (+I) or absence (-I) of a reversible inhibitor is described in the following graph.
![Double reciprocal plot of 1/V0 against 1/[S] in units of 10^3 per M: the -I and +I lines meet on the 1/V0 axis; the -I line cuts the 1/[S] axis at -4 and the +I line at -2](/uploads/question-images/gate-xl-2012/q44.png)
If concentration of the reversible inhibitor in +I experiment was equal to 3.0 × 10-3 M, then the dissociation constant for the enzyme-inhibitor complex is
Answer: C
The dissociation constant is \(3 \times 10^{-3}\) M.
Both lines meet on the 1/V0 axis, so Vmax is unchanged and the inhibition is competitive. The x intercepts give \(-1/K_{m}\):
\[ \begin{aligned} K_{m} &= \frac{1}{4 \times 10^{3}} = 0.25\ \text{mM} \\ K_{m}^{app} &= \frac{1}{2 \times 10^{3}} = 0.50\ \text{mM} \end{aligned} \]
For competitive inhibition \(K_{m}^{app} = K_{m}(1 + [I]/K_{i})\):
\[ \begin{aligned} 1 + \frac{[I]}{K_{i}} &= 2 \\ K_{i} &= [I] = 3 \times 10^{-3}\ \text{M} \end{aligned} \]
Doubling Km means the inhibitor is present at exactly its Ki. \(1.5 \times 10^{-3}\) M would correspond to a threefold increase and is not offered; the other options do not fit the twofold shift read from the graph.
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