Uncompetitive Inhibition: Why Km and Vmax Both Fall

An uncompetitive inhibitor binds only the enzyme-substrate (ES) complex, never the free enzyme. It locks substrate into a complex that cannot make product, so Vmax decreases. It also pulls the E + S equilibrium toward ES, so the enzyme looks like it binds substrate more tightly and apparent Km decreases. On a Lineweaver-Burk plot the lines are parallel.

Effect on the constants

Both constants shrink by the same factor, alpha' = 1 + [I]/Ki'.

• Vmax(app) = Vmax / alpha'
• Km(app) = Km / alpha'
• Vmax/Km is unchanged, so the slope of the double-reciprocal line is unchanged. That is why the lines are parallel.

Why more substrate makes it worse

Inhibitor needs ES to bind. More substrate means more ES, so more inhibitor binds. This is the opposite of competitive inhibition, where more substrate overcomes the block.

Side by side

Worked example

Without inhibitor: Vmax = 100 umol/min, Km = 20 uM. With an inhibitor that gives alpha' = 2:

• Vmax(app) = 100 / 2 = 50 umol/min
• Km(app) = 20 / 2 = 10 uM
• 1/Vmax goes from 0.010 to 0.020 (y-intercept doubles). -1/Km goes from -0.05 to -0.10 (x-intercept doubles in magnitude).
• Slope Km/Vmax = 0.20 before and after.

Common traps

• Saying Km increases. That is competitive. Uncompetitive lowers Km.
• Expecting the lines to cross. Same slope means parallel.
• Thinking substrate rescues the enzyme. It does the reverse.

One-minute recap

• Binds ES only.
• Vmax down, Km down, slope unchanged.
• Parallel Lineweaver-Burk lines.