Removal of Non-Competitive Inhibitor Restores V_max

Steps:

• Non-competitive inhibitors bind to an allosteric site, reducing effective enzyme concentration and thus V_max, without altering substrate binding.
• K_m remains unchanged because the active site affinity for substrate is unaffected.
• Removing the reversible inhibitor eliminates this binding, restoring full enzyme activity.
• Therefore, V_max increases to its uninhibited level, while K_m stays the same.

Why B is correct:

• Non-competitive inhibition decreases V_max (via the Michaelis-Menten equation, V_max = k_cat [E_total]) but leaves K_m unchanged; removal reverses the V_max reduction.

Why the others are wrong:

• A: K_m does not decrease, as non-competitive inhibition does not affect substrate affinity.
• C: V_max increases upon inhibitor removal, restoring maximum velocity.
• D: V_max increases, not decreases, when inhibition is lifted.

Final answer: B