($v_0$ = initial velocity, $[S]$ = free substrate concentration)

This question asks to identify the correct Eadie-Hofstee plot for an enzyme exhibiting Michaelis-Menten kinetics in the presence of an uncompetitive inhibitor.
The Eadie-Hofstee plot graphs the initial velocity ($v_0$) on the y-axis against the ratio $v_0/[S]$ (where $[S]$ is substrate concentration) on the x-axis. The equation is represented as: $v_0 = V_{max} - K_m \frac{v_0}{[S]}$ This equation is in the form $y = c + mx$, where:
An uncompetitive inhibitor binds exclusively to the enzyme-substrate (ES) complex. This type of inhibition affects the enzyme kinetics by:
Based on the Eadie-Hofstee equation and the effects of uncompetitive inhibition:
Therefore, the line representing the inhibited enzyme (+I, solid line) should be positioned below the uninhibited enzyme line (-I, dotted line) and have a less steep slope.
We need to find the plot that shows a decreased y-intercept and a less steep slope.
Plot 4 (Option D) best depicts the decrease in both $V_{max}$ (y-intercept) and $K_m$ (slope) characteristic of uncompetitive inhibition on an Eadie-Hofstee plot.
The kinetics of an enzyme in the presence (+I) or absence (-I) of a reversible inhibitor is described in the following graph.

If concentration of the reversible inhibitor in +I experiment was equal to $3.0 \times 10^{-3}$ M, then the dissociation constant for the enzyme-inhibitor complex is