11/3/2009Biochem: Enzymes III Enzymes III Andy Howard Introductory Biochemistry 3 November 2008

11/3/2009 Biochem: Enzymes III P. 2 of 43 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.

11/3/2009 Biochem: Enzymes III P. 3 of 43 Mechanism Topics Inhibitors, concluded Types of inhibitors Kinetics of inhibition Pharmaceuticals What makes an inhibitor a useful drug? Mechanisms: Terminology Transition States Stabilization of Transition States

11/3/2009 Biochem: Enzymes III P. 4 of 43 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)

11/3/2009 Biochem: Enzymes III P. 5 of 43 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enzyme from availability Reversible Usually noncovalent (ionic or van der Waals) Several kinds Classifications somewhat superseded by detailed structure-based knowledge of mechanisms, but not entirely

11/3/2009 Biochem: Enzymes III P. 6 of 43 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)

11/3/2009 Biochem: Enzymes III P. 7 of 43 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

11/3/2009 Biochem: Enzymes III P. 8 of 43 Competitive inhibition Put in a lot of substrate: ability of the inhibitor to get in 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 K m

11/3/2009 Biochem: Enzymes III P. 9 of 43 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.

11/3/2009 Biochem: Enzymes III P. 10 of 43 Kinetics of competition Competitive inhibitor hinders binding of substrate but not reaction velocity: Affects the K m of the enzyme, not V max. Which way does it affect it? K m = 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 increases K m

11/3/2009 Biochem: Enzymes III P. 11 of 43 L-B: competitive inhibitor K m goes up so -1/ K m moves toward origin V max unchanged so Y intercept unchanged

11/3/2009 Biochem: Enzymes III P. 12 of 43 Competitive inhibitor: Quantitation of K i Define inhibition constant K i to be the concentration of inhibitor that increases K m by a factor of two. K m,obs = K m (1+[ I c ]/K i ) So [ I c ] that moves K m halfway to the origin is K i. If K i = 100 nM and [ I c ] = 1 µM, then we’ll increase K m,obs elevenfold!

11/3/2009 Biochem: Enzymes III P. 13 of 43 Don’t get lazy! A competitive inhibitor doesn’t automatically double K m The amount by which the inhibitor increases K m is dependent on [I] c If it happens that [I] c = K I, then K m will double, as the equation shows

11/3/2009 Biochem: Enzymes III P. 14 of 43 Noncompetitive inhibition Inhibitor binds distant from active site, so it binds to the enzyme whether the substrate is present or absent. Noncompetitive inhibitor has no influence on how available the binding site for substrate is, so it does not affect K m at all However, it has a profound inhibitory influence on the speed of the reaction, i.e. turnover. So it reduces V max and has no influence on K m. S I

11/3/2009 Biochem: Enzymes III P. 15 of 43 L-B for non-competitives Decrease in V max  1/V max is larger X-intercept unaffected

11/3/2009 Biochem: Enzymes III P. 16 of 43 K i for noncompetitives K i defined as concentration of inhibitor that cuts V max in half V max,obs = V max /(1 + [ I n ]/K i ) In previous figure the “high” concentration of inhibitor is K i If K i = K i ’, this is pure noncompetitive inhibition

11/3/2009 Biochem: Enzymes III P. 17 of 43 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 decreases K m But it interferes with turnover so V max goes down If K m and V max decrease at the same rate, then it’s classical uncompetitive inhibition.

11/3/2009 Biochem: Enzymes III P. 18 of 43 L-B for uncompetitives -1/K m moves away from origin 1/V max moves away from the origin Slope (  K m /V max ) is unchanged

11/3/2009 Biochem: Enzymes III P. 19 of 43 K i for uncompetitives Defined as inhibitor concentration that cuts V max or K m in half Easiest to read from V max value V max,obs = V max /(1+[I] u /K I ) I u labeled “high” is K i in this plot

11/3/2009 Biochem: Enzymes III P. 20 of 43 iClicker quiz, question 1 1. Treatment of enzyme E with compound Y doubles K m and leaves V max unchanged. Compound Y is: (a) an accelerator of the reaction (b) a competitive inhibitor (c) a non-competitive inhibitor (d) an uncompetitive inhibitor

11/3/2009 Biochem: Enzymes III P. 21 of 43 iClicker quiz, question 2 2. Treatment of enzyme E with compound X doubles V max and leaves K m unchanged. Compound X is: (a) an accelerator of the reaction (b) a competitive inhibitor (c) a non-competitive inhibitor (d) an uncompetitive inhibitor

11/3/2009 Biochem: Enzymes III P. 22 of 43 Mixed inhibition Usually involves interference with both binding and catalysis K m goes up, V max 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 K i ≠ K i ’

11/3/2009 Biochem: Enzymes III P. 23 of 43 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.

11/3/2009 Biochem: Enzymes III P. 24 of 43 Characteristics of Pharmaceutical Inhibitors Usually competitive, i.e. they raise K m without affecting V max Some are mixed, i.e. K m up, V max down Iterative design work will decrease K i from millimolar down to nanomolar Sometimes design work is purely blind HTS; other times, it’s structure-based

11/3/2009 Biochem: Enzymes III P. 25 of 43 Amprenavir Competitive inhibitor of HIV protease, K i = 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

11/3/2009 Biochem: Enzymes III P. 26 of 43 When is a good inhibitor a good drug? It needs to be bioavailable and nontoxic Beautiful 20nM inhibitor is often neither Modest sacrifices of K i in improving bioavailability and non-toxicity are okay if K i is low enough when you start sacrificing

11/3/2009 Biochem: Enzymes III P. 27 of 43 How do we lessen toxicity and improve bioavailability? Increase solubility… that often increases K i 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

11/3/2009 Biochem: Enzymes III P. 28 of 43 Drug-design timeline 2 years of research, 8 years of trials log K i Time, Yrs Cost/yr, 10 6 $ Improving affinity Toxicity and bioavailability Research Clinical Trials Preliminary toxicity testing Stage I clinical trials Stage II clinical trials

11/3/2009 Biochem: Enzymes III P. 29 of 43 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.

11/3/2009 Biochem: Enzymes III P. 30 of 43 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.

11/3/2009 Biochem: Enzymes III P. 31 of 43 Overcoming the barrier Simple system: single high-energy transition state intermediate between reactants, products Free Energy Reaction Coordinate R P G‡G‡

11/3/2009 Biochem: Enzymes III P. 32 of 43 Intermediates Often there is a quasi-stable intermediate state midway between reactants & products; transition states on either side Free Energy R P T1 T2 I Reaction Coordinate

11/3/2009 Biochem: Enzymes III P. 33 of 43 Activation energy & temperature It’s intuitively sensible that higher temperatures would make it easier to overcome an activation barrier Rate k(T) = Q 0 exp(-  G ‡ /RT)  G ‡ = activation energy or Arrhenius energy This provides tool for measuring  G ‡ Svante Arrhenius

11/3/2009 Biochem: Enzymes III P. 34 of 43 Determining  G ‡ Remember k(T) = Q 0 exp(-  G ‡ /RT) ln k = lnQ 0 -  G ‡ /RT Measure reaction rate as function of temperature Plot ln k vs 1/T; slope will be -  G ‡ /R ln k 1/T, K -1 uncatalyzed catalyzed

11/3/2009 Biochem: Enzymes III P. 35 of 43 How enzymes alter  G ‡ Enzymes reduce  G ‡ 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).

11/3/2009 Biochem: Enzymes III P. 36 of 43 Dissociation constants for ES and EX* Dissociation constant for ES: K s = [E][S]/[ES] Dissociation constant for EX ‡ : K T = [E][X ‡ ]/[EX ‡ ] Transition state theory says the ratio of reaction rates is related to the ratio of these: k e /k u = K s / K T

11/3/2009 Biochem: Enzymes III P. 37 of 43 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

11/3/2009 Biochem: Enzymes III P. 38 of 43 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 ~ times more avidly than substrates 1,6-hydrate of purine nucleoside binds to adenosine deaminase with K I = 3* M

11/3/2009 Biochem: Enzymes III P. 39 of 43  G ‡ 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.

11/3/2009 Biochem: Enzymes III P. 40 of 43 Enthalpy and transition states Often an enthalpic component to the reduction in  G ‡ 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.

11/3/2009 Biochem: Enzymes III P. 41 of 43 Two ways to change  G ‡ Reactants bound by enzyme are properly positioned Get into transition- state geometry more readily Transition state is stabilized E ABAB E ABAB A+B A-B

11/3/2009 Biochem: Enzymes III P. 42 of 43 Binding modes: proximity 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. William Jencks

11/3/2009 Biochem: Enzymes III P. 43 of 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