Enzymes III

1. Enzymes III Andy HowardIntroductory Biochemistry 3 November 2008 Biochem: Enzymes III

2. How do enzymes reduce activation energies? • We want to understand what is really happening chemically when an enzyme does its job. • We’d also like to know how biochemists probe these systems. Biochem: Enzymes III

3. Inhibitors, concluded Types of inhibitors Kinetics of inhibition Pharmaceuticals What makes an inhibitor a useful drug? Mechanisms:Terminology Transition States Stabilization of Transition States Mechanism Topics Biochem: Enzymes III

4. Distinctions we can make • Inhibitors can be reversible or irreversible • Where do they bind? • At the enzyme’s active site • At a site distant from the active site. • To what do they bind? • To the unliganded enzyme E • To the enzyme-intermediate complex or the enzyme-substrate complex (ES) • To both (E or ES) Biochem: Enzymes III

5. Types of inhibitors • Irreversible • Inhibitor binds without possibility of release • Usually covalent • Each inhibition event effectively removes a molecule of enzyme from availability • Reversible • Usually noncovalent (ionic or van der Waals) • Several kinds • Classifications somewhat superseded by detailed structure-based knowledge of mechanisms, but not entirely Biochem: Enzymes III

6. Types of reversible inhibition • Competitive • Inhibitor binds at active site of unliganded enzyme • Prevents binding of substrate • Noncompetitive • Inhibitor binds distant from active site (E or ES) • Interferes with turnover • Uncompetitive (rare?) • Inhibitor binds only to ES complex • Removes ES, interferes with turnover • Mixed(usually Competitive + Noncompetitive) Biochem: Enzymes III

7. How to tell them apart • Reversible vs irreversible • dialyze an enzyme-inhibitor complex against a buffer free of inhibitor • if turnover or binding still suffers, it’s irreversible • Competitive vs. other reversible: • Structural studies if feasible • Kinetics Biochem: Enzymes III

8. Competitive inhibition • Put in a lot of substrate:ability of the inhibitor to getin the way of the binding is hindered:out-competed by sheer #s of substrate molecules. • This kind of inhibition manifests itself as interference with binding, i.e. with an increase of Km Biochem: Enzymes III

9. Competitive inhibitors don’t affect turnover • If the substrates manages to bind even though there is inhibitor present, then it can be turned over just as quickly as if the inhibitor is absent; so the inhibitor influences binding but not turnover. Biochem: Enzymes III

10. Kinetics of competition • Competitive inhibitor hinders binding of substrate but not reaction velocity: • Affects the Km of the enzyme, not Vmax. • Which way does it affect it? • Km = amount of substrate that needs to be present to run the reaction velocity up to half its saturation velocity. • Competitive inhibitor requires us to shove more substrate into the reaction in order to achieve that half-maximal velocity. • So: competitive inhibitor increasesKm Biochem: Enzymes III

11. L-B: competitive inhibitor • Km goes up so -1/ Km moves toward origin • Vmax unchanged so Y intercept unchanged Biochem: Enzymes III

12. Competitive inhibitor:Quantitation of Ki • Define inhibition constant Ki to be the concentration of inhibitor that increases Km by a factor of two. • Km,obs = Km(1+[Ic]/Ki) • So [Ic] that moves Km halfway to the origin is Ki. • If Ki = 100 nM and [Ic] = 1 µM, then we’ll increase Km,obs elevenfold! Biochem: Enzymes III

13. Don’t get lazy! • A competitive inhibitor doesn’t automatically double Km • The amount by which the inhibitor increases Km is dependent on [I]c • If it happens that [I]c = KI, then Km will double, as the equation shows Biochem: Enzymes III

14. Noncompetitive inhibition S I • Inhibitor binds distant fromactive site, so it binds to theenzyme whether the substrateis present or absent. • Noncompetitive inhibitor has no influence on how available the binding site for substrate is, so it does not affect Km at all • However, it has a profound inhibitory influence on the speed of the reaction, i.e. turnover. So it reducesVmax and has no influence on Km. Biochem: Enzymes III

15. L-B for non-competitives • Decrease in Vmax 1/Vmax is larger • X-intercept unaffected Biochem: Enzymes III

16. Ki for noncompetitives • Ki defined as concentration of inhibitor that cuts Vmax in half • Vmax,obs =Vmax/(1 + [In]/Ki) • In previous figure the “high” concentration of inhibitor is Ki • If Ki = Ki’, this is pure noncompetitive inhibition Biochem: Enzymes III

17. Uncompetitive inhibition • Inhibitor binds only if ES has already formed • It creates a ternary ESI complex • This removes ES, so by LeChatlier’s Principle it actually drives the original reaction (E + S  ES) to the right; so it decreasesKm • But it interferes with turnover so Vmax goes down • If Km and Vmax decrease at the same rate, then it’s classical uncompetitive inhibition. Biochem: Enzymes III

18. L-B for uncompetitives • -1/Km moves away from origin • 1/Vmax moves away from the origin • Slope (Km/Vmax) is unchanged Biochem: Enzymes III

19. Ki for uncompetitives • Defined as inhibitor concentration that cuts Vmax or Km in half • Easiest to read from Vmax value • Vmax,obs = Vmax/(1+[I]u/KI) • Iu labeled “high” is Ki in this plot Biochem: Enzymes III

20. iClicker quiz, question 1 1. Treatment of enzyme E with compound Y doubles Km and leaves Vmax unchanged. Compound Y is: • (a) an accelerator of the reaction • (b) a competitive inhibitor • (c) a non-competitive inhibitor • (d) an uncompetitive inhibitor Biochem: Enzymes III

21. iClicker quiz, question 2 • 2. Treatment of enzyme E with compound X doubles Vmax and leaves Km unchanged. Compound X is: • (a) an accelerator of the reaction • (b) a competitive inhibitor • (c) a non-competitive inhibitor • (d) an uncompetitive inhibitor Biochem: Enzymes III

22. Mixed inhibition • Usually involves interference with both binding and catalysis • Km goes up, Vmax goes down • Easy to imagine the mechanism: • Binding of inhibitor alters the active-site configuration to interfere with binding, but it also alters turnover • Same picture as with pure noncompetitive inhibition, but with Ki ≠ Ki’ Biochem: Enzymes III

23. Most pharmaceuticals are enzyme inhibitors • Some are inhibitors of enzymes that are necessary for functioning of pathogens • Others are inhibitors of some protein whose inappropriate expression in a human causes a disease. • Others are targeted at enzymes that are produced more energetically by tumors than they are by normal tissues. Biochem: Enzymes III

24. Characteristics of Pharmaceutical Inhibitors • Usually competitive, i.e. they raise Km without affecting Vmax • Some are mixed, i.e. Km up, Vmax down • Iterative design work will decrease Kifrom millimolar down to nanomolar • Sometimes design work is purely blind HTS; other times, it’s structure-based Biochem: Enzymes III

25. Amprenavir • Competitive inhibitor of HIV protease,Ki = 0.6 nM for HIV-1 • No longer sold: mutual interference with rifabutin, which is an antibiotic used against a common HIV secondary bacterial infection, Mycobacterium avium Biochem: Enzymes III

26. When is a good inhibitor a good drug? • It needs to be bioavailable and nontoxic • Beautiful 20nM inhibitor is often neither • Modest sacrifices of Ki in improving bioavailability and non-toxicity are okay if Ki is low enough when you start sacrificing Biochem: Enzymes III

27. How do we lessen toxicity and improve bioavailability? • Increase solubility…that often increases Ki because the van der Waals interactions diminish • Solubility makes it easier to get the compound to travel through the bloodstream • Toxicity is often associated with fat storage, which is more likely with insoluble compounds Biochem: Enzymes III

28. Drug-design timeline 100 -3 • 2 years of research, 8 years of trials Improving affinity Toxicity and bioavailability Stage II clinical trials Cost/yr, 106 $ Stage I clinical trials Preliminary toxicity testing log Ki -8 10 Research Clinical Trials 0 2 Time, Yrs 10 Biochem: Enzymes III

29. Atomic-Level Mechanisms • We want to understand atomic-level events during an enzymatically catalyzed reaction. • Sometimes we want to find a way to inhibit an enzyme • in other cases we're looking for more fundamental knowledge, viz. the ways that biological organisms employ chemistry and how enzymes make that chemistry possible. Biochem: Enzymes III

30. How we study mechanisms • There are a variety of experimental tools available for understanding mechanisms, including isotopic labeling of substrates, structural methods, and spectroscopic kinetic techniques. Biochem: Enzymes III

31. Overcoming the barrier • Simple system:single high-energy transition state intermediate between reactants, products Free Energy G‡ R P Reaction Coordinate Biochem: Enzymes III

32. Intermediates • Often there is a quasi-stable intermediate state midway between reactants & products; transition states on either side T2 T1 Free Energy I R P Reaction Coordinate Biochem: Enzymes III

33. Activation energy & temperature • It’s intuitively sensible that higher temperatures would make it easier to overcome an activation barrier • Rate k(T) = Q0exp(-G‡/RT) • G‡ = activation energy or Arrhenius energy • This provides tool for measuring G‡ Svante Arrhenius Biochem: Enzymes III

34. Determining G‡ • Rememberk(T) = Q0exp(-G‡/RT) • ln k = lnQ0 - G‡/RT • Measure reaction rate as function of temperature • Plot ln k vs 1/T; slope will be -G‡/R catalyzed ln k uncatalyzed 1/T, K-1 Biochem: Enzymes III

35. How enzymes alter G‡ • Enzymes reduce DG‡ by allowing the binding of the transition state into the active site • Binding of the transition state needs to be tighter than the binding of either the reactants or the products. • In fact, the enzyme must stabilize the transition-state complex EX‡more than it stabilizes the substrate complex ES (see section 14.2). Biochem: Enzymes III

36. Dissociation constants for ES and EX* • Dissociation constant for ES:Ks = [E][S]/[ES] • Dissociation constant for EX‡:KT = [E][X‡]/[EX‡] • Transition state theory says the ratio of reaction rates is related to the ratio of these: • ke/ku = Ks / KT Biochem: Enzymes III

37. What makes EX‡ more stable than ES? • Intrinsic (enthalpic) binding energy of ES makes it a lower-energy species than E+S; but we want EX* to be lower. • ES loses entropy relative to E + S • ES is sometimes strained, distorted, or desolvated relative to E+S • So if EX‡ is less strained and has more entropy, we win • See section 14.3 Biochem: Enzymes III

38. How tight is the binding? • Section 14.4 gives some examples • Transition-state analogs are stable molecules that are geometrically and electrostatically similar to transition states • Sometimes the analogs bind ~ 160 - 40000 times more avidly than substrates • 1,6-hydrate of purine nucleoside binds to adenosine deaminase with KI = 3*10-13M Biochem: Enzymes III

39. DG‡ and Entropy • Effect is partly entropic: • When a substrate binds,it loses a lot of entropy. • Thus the entropic disadvantage of (say) a bimolecular reaction is soaked up in the process of binding the first of the two substrates into the enzyme's active site. Biochem: Enzymes III

40. Enthalpy and transition states • Often an enthalpic component to the reduction in DG‡ as well • Ionic or hydrophobic interactions between the enzyme's active site residues and the components of the transition state make that transition state more stable. Biochem: Enzymes III

41. Reactants bound by enzyme are properly positioned Get into transition-state geometry more readily Transition state is stabilized Two ways to change G‡ AB AB E E A+B A+B A-B A-B Biochem: Enzymes III

42. Binding modes: proximity William Jencks • We describe enzymatic mechanisms in terms of the binding modes of the substrates (or, more properly, the transition-state species) to the enzyme. • One of these involves the proximity effect, in which two (or more) substrates are directed down potential-energy gradients to positions where they are close to one another. Thus the enzyme is able to defeat the entropic difficulty of bringing substrates together. Biochem: Enzymes III

43. Near-Attack Conformations • Substrate is preorganized in the active site such that the reacting atoms are in van der Waals contact and at an angle resembling the bond to be formed in the transition state. • The NAC would form anyway (0.0001% of the time?) but with the help of the enzyme, it forms 1-70% of the time Biochem: Enzymes III