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Biochemistry Unit 5: Enzyme Inhibition and Regulation

Cells control enzymes, and drugs block them. This unit covers the kinetic signatures of reversible inhibitors, irreversible inhibitors, and the main ways cells regulate enzymes: allostery, covalent modification, cleavage and feedback.

Covers: Competitive inhibition, Uncompetitive inhibition, Noncompetitive and mixed inhibition, Irreversible and suicide inhibitors, Allosteric regulation, Covalent modification, Zymogens and feedback control

Study guide

Reversible inhibitors bind noncovalently and can leave. A competitive inhibitor binds the free enzyme's active site, often because it looks like the substrate. It raises the apparent Km by a factor α = 1 + [I]/Ki but leaves Vmax unchanged, because enough substrate can outcompete it. On a Lineweaver-Burk plot the lines cross on the y-axis. Methotrexate, a folate analog, and ethanol used against methanol poisoning are classic examples. To find Ki from data, compute α = Km,app/Km and solve for Ki.

An uncompetitive inhibitor binds only the ES complex. It lowers both Vmax and Km by the same factor α′ = 1 + [I]/Ki′, so Lineweaver-Burk lines are parallel. A mixed inhibitor binds both E and ES, but with different affinities, so both Km and Vmax change; if it prefers free E, Km rises, and if it prefers ES, Km falls. The special case of equal binding is pure noncompetitive inhibition: Vmax falls by α and Km stays the same, and lines cross on the x-axis. Substrate cannot overcome uncompetitive, mixed or noncompetitive inhibition.

Irreversible inhibitors form covalent bonds or bind so tightly they never leave. Aspirin acetylates a serine in cyclooxygenase, and nerve agents like sarin modify the active-site serine of acetylcholinesterase. Suicide (mechanism-based) inhibitors are unreactive until the enzyme begins its normal reaction on them, then trap the enzyme; penicillin inactivates bacterial transpeptidase this way. Because the enzyme is removed from action, adding substrate does not help, and recovery requires new enzyme synthesis. Kinetically, an irreversible inhibitor simply lowers the amount of working enzyme, so Vmax falls while the Km of the remaining active molecules is unchanged.

Cells regulate enzymes in several ways. Allosteric enzymes have regulatory sites separate from the active site and usually several subunits, giving sigmoidal kinetics. Activators stabilize the high-activity R state and inhibitors stabilize the T state. Feedback inhibition, in which an end product inhibits the committed step, is common: CTP inhibits aspartate transcarbamoylase, while ATP activates it. PFK-1 in glycolysis is inhibited by ATP and citrate and activated by AMP and fructose 2,6-bisphosphate, so it tracks the energy state. On graphs, remember that allosteric enzymes do not follow Michaelis-Menten kinetics, so the term K0.5 is used for the substrate level giving half-maximal rate.

Covalent modification, most often phosphorylation by kinases and removal by phosphatases, switches enzymes on or off in seconds. Hormones such as glucagon and epinephrine work through cAMP and protein kinase A, phosphorylating glycogen phosphorylase (activating it) and glycogen synthase (inactivating it). Proteolytic activation is irreversible: zymogens such as trypsinogen, chymotrypsinogen and pepsinogen, and blood-clotting factors, are made inactive and cleaved when and where they are needed. Cleavage cascades amplify signals, so a small trigger activates many enzyme molecules.

Key ideas

Formulas and rules

α = 1 + [I]/Ki (binding to E); α′ = 1 + [I]/Ki′ (binding to ES)
Competitive: Km,app = αKm; Vmax,app = Vmax
Uncompetitive: Km,app = Km/α′; Vmax,app = Vmax/α′
Pure noncompetitive: Km unchanged; Vmax,app = Vmax/α
Mixed: v = Vmax[S]/(αKm + α′[S])
Competitive IC50 = Ki(1 + [S]/Km)

Common mistakes

Before the test, make sure you can

Words to know

Competitive inhibitor
Binds the free enzyme's active site; raises Km, Vmax unchanged.
Uncompetitive inhibitor
Binds only ES; lowers Km and Vmax by the same factor.
Mixed inhibitor
Binds E and ES with different affinities; changes both Km and Vmax.
Pure noncompetitive inhibitor
Binds E and ES equally; lowers Vmax, Km unchanged.
Ki
The dissociation constant of the enzyme-inhibitor complex; smaller means tighter binding.
IC50
The inhibitor concentration that reduces activity by half under given assay conditions.
Irreversible inhibitor
Covalently or permanently inactivates an enzyme, such as aspirin or sarin.
Suicide inhibitor
A mechanism-based inhibitor activated by the enzyme's own catalysis, such as penicillin.
Allosteric site
A regulatory site distinct from the active site where effectors bind.
Allosteric activator
An effector that stabilizes the R state and shifts the curve left.
Feedback inhibition
Inhibition of an early pathway enzyme by the pathway's end product.
Committed step
The first irreversible step unique to a pathway, usually the regulated one.
Protein kinase
An enzyme that transfers phosphate from ATP to a protein's Ser, Thr or Tyr.
Phosphatase
An enzyme that removes phosphate groups from proteins.
Zymogen
An inactive enzyme precursor activated by proteolytic cleavage.
Enteropeptidase
An intestinal protease that activates trypsinogen to trypsin.
Aspartate transcarbamoylase
An allosteric enzyme of pyrimidine synthesis inhibited by CTP and activated by ATP.

Sample practice questions

30 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

A competitive inhibitor binds:

  1. The active site of the free enzyme
  2. The substrate rather than the enzyme
  3. Only the enzyme-substrate complex
  4. A site that blocks product release only
Show the answer and steps

Answer: A) The active site of the free enzyme

  1. A competitive inhibitor resembles the substrate and competes for the same site.
  2. It binds free E, not ES.
  3. Enough substrate can outcompete it, so Vmax is unchanged.
Question 2 (level 1)

How does a competitive inhibitor change the apparent Km and Vmax?

  1. Km and Vmax both increase
  2. Km decreases; Vmax decreases
  3. Km is unchanged; Vmax decreases
  4. Km increases; Vmax is unchanged
Show the answer and steps

Answer: D) Km increases; Vmax is unchanged

  1. Km,app = αKm with α = 1 + [I]/Ki > 1, so Km rises.
  2. At very high [S] the substrate wins, so Vmax is the same.
  3. On a Lineweaver-Burk plot, lines meet on the y-axis.
Question 3 (level 3)

Km rises from 4.0 to 12 µM when 20 µM of a competitive inhibitor is added. What is Ki?

  1. 6.7 µM
  2. 20 µM
  3. 10 µM
  4. 3.0 µM
Show the answer and steps

Answer: C) 10 µM

  1. α = Km,app/Km = 12/4.0 = 3.0.
  2. α = 1 + [I]/Ki, so [I]/Ki = 2.0 and Ki = 20/2.0 = 10 µM.
  3. 6.7 µM uses α = 3 directly as [I]/Ki.