1. LECTURE 4: Principles of Enzyme Catalysis Reading: Berg, Tymoczko & Stryer: Chapter 8 ENZYME An ENZYME is a biomolecular catalyst that accelerates the rate of a specific reaction Enzymes DO NOT make a chemical reaction more energetically favorable; They only ACCELERATE the rate of the reaction towards its energetic equilibrium Enzymes work by stabilizing chemical transition states, the high-energy intermediates that normally act as a barrier to spontaneous reaction Most enzymes are FOLDED PROTEINS : proteins have the ability to fold into scaffolds with binding surfaces for substrates that position the substrates for chemical reaction A few enzymes are RNA molecules! RNAs also have ability to adopt tertiary structures. Some RNAs (called RIBOZYMES) act as enzymes catalyzing their own site-specific cleavage or that of other RNA molecules As catalysts, enzymes are NOT CONSUMED during reactions. S + E -----> ES -------> P + E In some cases, enyzmes are chemically modified during catalysis, but return to their original form after reaction cycle to allow further catalysis of substrate S + E -----> E*S -------> P + E

2. Proteases Are Examples of Enzymes That Catalyze An Energetically Favored Process Fig 8.0new (peptide hydryolys) Peptide hydrolysis is an energetically favorable process, but normally occurs very slowly. PROTEASES are enzymes that catalyze peptide hydrolysis. Some proteases are rather NONSELECTIVE (e.g., papain) Other proteases are VERY SELECTIVE (e.g., trypsin, thrombin, fibrin) Fig 8.1new

3. Some Enzymes Employ Cofactors Some enzymes use cofactors as part of the active site in enzymatic catalysis APOENZYME + COFACTOR --------> HOLOENZYME Many cofactors cannot be synthesized by humans, and must be obtained through diet as vitamins and minerals Tab8.2new

4. Free Energy of Biochemical Reactions For reaction A + B C + D  G is the differential in free energy between the products vs. reactants If  G < 0, reaction is energetically favorable I.e., reactants will convert to products as system moves to equilibrium If  G = 0, reaction is already at equilibrium I.e., there will be no NET conversion of reactants to products If  G > 0, reaction is disfavored I.e., products will convert to reactants as system moves to equilibrium; the reverse reaction is favored Free energy  G is usually expressed in units kcal / mol or kcal x mol -1

5. Standard Free Energy is Related to Equilibrium Constant For reaction A + B C + D  G =  G o + RT ln [C] [D] [A] [B]  G is free energy change when reactant and product concentrations are [A],[B],[C],[D]  G o is free energy change when reactant and product concentrations are each 1M WHAT DOES THIS MEAN?  G o is measure of whether reactants or products are favored if all components are at same concentration The actual concentration of reactants and products impacts on  G; even if  G o is unfavorable, high ratio of reactants to products can give favorable  G  G o can be related to the equilibrium constant, K eq, of the reaction K eq = [C] eq [D] eq [A] eq [B] eq Since  G o = - RT lnK eq = - 2.3RT log 10 K eq

6. Tab8.3new

7. Enzymes Accelerate Rate Constant Without Altering Equilibrium Constant SP Without enzyme With enzyme SP K eq = k f / k r = Fig8.2new

8. Enzymes Stabilize Reaction Transition State(s) Fig8.3new

9. Properties of Enzyme Active Sites Active site consists of atoms on residue side chains that are brought together by the fold Most of the enzyme structure is a scaffold to precisely position active site residues Active site uses range of noncovalent bonding mechanisms to bind substrate By binding multiple substrates in a favorable interspatial relationship and/or by altering charge distribution (resonance) within substrates,  G transition is much smaller than would be spontaneously Fig8.7new Fig8.8new

10. The Michaelis-Menton Model of Enzyme Function E + S ES E + P k2k2 k-1k-1 k1k1 k-2k-2 At time=0, if [P]=0, then E + S ES E + P k-1k-1 k1k1 Vo = k cat [ES] [ES] determined by [E], [S], and rate constants THESE EQUATIONS CAN BE SOLVED TO EXPRESS THE REACTION RATE AS A FUNCTION OF THE SUBSTRATE CONCENTRATION [S] AND TWO INHERENT PROPERTIES OF THE ENZYME: K M AND k cat V o = V MAX [S] [S] + K M Michaelis-Menton Equation k2k2 k cat = where V MAX = k cat [E] K M = k -1 / k 1 = [E][S] / [ES] and

11. The Michaelis-Menton Equation: Meaning Behind The Terms Fig8.12new V o = V MAX [S] [S] + K M No matter how large the substrate concentration, reaction rate can never exceed V MAX V MAX reflects the TURNOVER RATE of substrate molecules through the enzyme (k cat ) and the enzyme concentration K M is the substrate concentration at which reaction rate is HALF MAXIMAL K M reflects the BINDING AFFINITY of the enzyme for the substrate; The higher the affinity, the smaller is K m By performing experiments to calculate V o at different substrate concentrations, properties V MAX and K m can be calculated If the enzyme concentration is known, V MAX can be used to calculate k cat

12. Lineweaver-Burk Plot Facilitates Calculation of K M and V MAX V o = V MAX [S] [S] + K M By inverting equation, get: VoVo 11 V MAX =+ K M V MAX () 1 [S] Fig8.13new Tab8.5new

13. Many Enzymatic Reactions Proceed Through Fixed Sequential Steps PyrLac.p223new TransAm.p224new Reaction Intermediate May Utilize Covalently Modified Enzyme or Cofactor

14. Enzyme Inhibition Many small molecules can bind to enzymes and inhibit them. Inhibitors can be described as REVERSIBLE or IRREVERSIBLE. Inhibitors may be naturally occuring within the homologous organism or in a heterologous organism Other inhibitors are synthetic and have been developed as pharmaceuticals for research and clinical applications COMPETITIVE INHIBITORS act by occupying the enzyme active site in place of the substrate NONCOMPETITIVE INHIBITORS bind away from the active site, but their binding exerts allosteric effects that prevents bound substrate conversion to product Fig8.15Anew Fig8.15BnewFig8.15Dnew

15. Kinetics of Competitive Inhibition Fig8.17new V o = V MAX [S] [S] + K M ( KiKi [ I ] 1+ ) K i reflects the AFFINITY OF INHIBITOR for the enzyme Inhibitor in effect raises the apparent K M term The potency of inhibitor determined by K i Therefore, the amount of substrate needed for half-maximum rate is increased THERE IS NO EFFECT ON V MAX I.e., a competitive inhibitor can be overcome by sufficiently high substrate concentration

16. Kinetics of Noncompetitive Inhibition Fig8.19new ( KiKi [ I ] 1+ ) V o = [S] [S] + K M V MAX x As before, K i reflects the AFFINITY OF INHIBITOR for the enzyme Inhibitor reduces V MAX of the enzyme The potency of inhibitor determined by K i The inhibitor does not affect K M, i.e., the binding of substrate to enzyme