Lecture 10 Enzymes: Introduction

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1 Lecture 10 Enzymes: Introduction Reading: Berg, Tymoczko & Stryer, 6th ed., Chapter 8, pp (These pages in textbook are very important -- concepts of thermodynamics are fundamental to all of biochemistry.) Thermodynamics practice problems (same as for Lecture 2): (also linked in lecture notes directory) Enzymes introduction sample problems: (also linked in lecture notes directory) Key Concepts Enzymes are biological catalysts, very powerful and very specific. Enzymes increase rates of (bio)chemical reactions but have no effect on K eq (and no effect on overall ΔG) of the reaction. Some enzymes need cofactors (inorganic ions or organic/metalloorganic coenzymes, derived from vitamins) for their catalytic activities. Different cofactors are useful for different kinds of chemical reactions, including transfers of specific kinds of groups or transfers of electrons. Kinetics: the study of reaction rates. Rates depend on rate constants. Rate constants depend inversely and exponentially on Arrhenius activation energy, ΔG, the difference in free energy between free energy of transition state and free energy of reactant(s). Rate constants are increased by catalysts (enzymes), because enzymes decrease ΔG. Enzymes lower ΔG by affecting either ΔH or ΔS (or both). One way enzymes reduce ΔG is by tight binding (noncovalent) of the transition state. Enzymes generally change the pathways by which reactions occur. Rate enhancement (factor by which enzyme increases the rate of a reaction) is determined by ΔΔG, the decrease in ΔG brought about by enzyme compared with uncatalyzed reaction's ΔG. LEC 10, Enzymes - Introduction 1

2 Learning Objectives (See also posted Thermo and Enzymes-Introduction sample problems.) Terminology: rate enhancement, cofactor, coenzyme, apoenzyme, holoenzyme, prosthetic group, catalyst, activation energy, transition state. (Review: equilibrium constant, mass action ratio for a reaction, biochemical standard conditions, standard free energy change, actual free energy change). Describe the general properties of enzymes as catalysts that are especially important for their roles as biological catalysts. Explain the effect of a catalyst on the rate of a reaction, and on the equilibrium constant of a reaction. Define "standard free energy change" and give the symbol for that parameter. Write the mathematical expression relating ΔG ' to K eq ', and be able to interconvert ΔG ' and K eq '. Calculate the actual free energy change (ΔG'), given the starting concentrations of appropriate chemical species and either ΔG ' or K eq '. Describe the relation between ΔG' and the rate of a reaction; using ΔG', predict reaction direction. Learning Objectives, continued Express the velocity of a simple reaction in terms of the rate constant and the concentration of the reactant. Express the equilibrium constant of a reaction in terms of the equilibrium mass action ratio. Express the equilibrium constant of a reaction in terms of the rate constants for the forward and reverse directions. (Note that equilibrium constants are symbolized with upper case K and rate constants with lower case k.) If an enzyme increases the rate constant for the forward reaction by a factor of 10 8, by what factor does it increase the rate constant for the back reaction? What is the rate enhancement brought about by the catalyst for that reaction? Draw the free energy diagram of a hypothetical reaction and show how a catalyst may increase the rate of the reaction, pointing out on the diagram ΔG for the overall reaction, ΔG uncat, and ΔG cat. Indicate (and name) the quantity on a free energy diagram (HINT: it's a specific kind of ΔG) that determines the magnitude of the rate constant for the reaction at a given temperature. You don't have to memorize the equation relating this quantity to k. What reaction parameter (kinetic parameter) do enzymes affect in order to increase the rate? LEC 10, Enzymes - Introduction 2

3 Enzymes: Introduction Enzymes are proteins. (ribozymes: catalytic RNA molecules) biological catalysts not chemically altered in reaction do not change equilibrium constant (K eq ) for reaction increase rate of reaction by providing a pathway of lower activation energy to get from reactants to products operate under physiological conditions (moderate temps., around neutral ph, low conc. in aqueous environment) work by forming complexes with their substrates (binding), thus providing unique microenvironment for reaction to proceed, the active site VERY HIGH SPECIFICITY for both reaction catalyzed and substrate used VERY HIGH CATALYTIC EFFICIENCY ACTIVITIES of some enzymes REGULATED CATALYTIC POWER OF ENZYMES RATE ENHANCEMENT = catalyzed rate constant/uncatalyzed rate constant = factor by which catalyst increases rate of reaction Examples: Berg et al., Table 8-1 LEC 10, Enzymes - Introduction 3

4 SPECIFICITY OF ENZYMES Enzymes very specific for substrate acted upon for reaction catalyzed Example: Proteases are a whole class of enzymes that all catalyze hydrolysis of peptide bonds: Substrate Specificity -- proteases as an example Substrate specificity (e.g., of proteases) due to precise interaction of enzyme with substrate result of 3-D structure of enzyme active site where substrate has to bind and be properly oriented for catalysis to occur (A) Trypsin catalyzes hydrolysis of peptide bonds on carboxyl side of Lys and Arg residues (digestive function in small intestine, cleaves just about any protein it encounters after (eventually) every Lys and Arg) (B) Thrombin (involved in blood clotting cascade) catalyzes hydrolysis of peptide bonds between Arg and Gly residues in specific sequences in specific protein substrates (activated only where blood needs to clot, works only on very specific target protein) Berg et al., Fig. 8-1 LEC 10, Enzymes - Introduction 4

5 Enzyme Specificity, continued substrate specificity of proteases -- another example, chymotrypsin: cleaves on carboxyl side of aromatic and hydrophobic amino acid residues evolutionarily related to trypsin Genes for trypsin and chymotrypsin are homologous. Ancestral gene duplicated and sequences diverged through evolution. Substrate specificities for site of cleavage diverged, but catalytic mechanism and overall tertiary structure was conserved. Specificity of reaction catalyzed: Many proteases also catalyze hydrolysis of carboxylic ester bonds: Cofactors Some enzymes need cofactors for their activity. COFACTORS: small organic or metalloorganic molecules (coenzymes) or metal ions Cofactors can bind tightly or weakly to enzymes. (Equilibrium below can lie far to left, weak binding, or far to right, tight binding.) Prosthetic groups (e.g. heme in hemoglobin): tightly bound cofactors (either coenzymes or metals) remain associated with their enzymes even between reaction cycles. Weakly bound coenzymes (which are NOT prosthetic groups) can associate and dissociate from enzymes between reaction cycles, behaving like substrates sometimes referred to as "cosubstrates" LEC 10, Enzymes - Introduction 5

6 Common Coenzymes and Reactions They Mediate See also Berg et al., Table 8.2 (for reference, not for memorization) Coenzyme (precursor/vitamin) Reaction Mediated (Group Transferred) Biotin Carboxylation (CO 2 ) Cobalamin (B 12 ) Coenzyme A (pantothenate) Flavin coenzymes (B 2 ) Lipoic acid Nicotinamide coenzymes (niacin) Pyridoxal phosphate (B 6 ) Tetrahydrofolate (folic acid) Thiamine pyrophosphate (B 1 ) Uridine diphosphate [UDP] Alkylation (methyl group), intramolecular rearrangements, and ribonucleotide reduction Acyl transfer (R C=O group) Oxidation-reduction (hydrogen atoms) (1 or 2 e transfer) Acyl group transfer Oxidation reduction (hydride ions H:, 2 e transfers) Amino group transfer (and many other reactions) One-carbon transfer Aldehyde transfer Sugar transfer (hexose units) Review Biological Thermodynamics (Thermodynamics will not be covered again in class. Go back over notes for Lecture 2, and Berg, Tymoczko & Stryer, 6th ed. Chapter 1, pp and Chapter 8, pp ) ΔG = ΔH TΔS ΔG = G products (final state) G reactants (initial state) If ΔG < 0, reaction proceeds left to right (as written) reaction is exergonic, free energy decreases in going from reactants to products) If ΔG = 0, reaction is at equilibrium (no net reaction occurs in either direction) If ΔG > 0, reaction proceeds right to left (reverse direction from what's written) Forward reaction would be endergonic, requiring INPUT of free energy to go left to right. ΔG forward reaction = ΔG reverse reaction Just change sign of ΔG for reversing direction. To get a reaction for which ΔG > 0 (a reaction with a positive ΔG) to go forward, couple it with a reaction for which ΔG < 0 (negative ΔG) to "drive" the process. (Exergonic reaction provides free energy "input" to drive endergonic reaction.) LEC 10, Enzymes - Introduction 6

7 Review thermodynamics (continued) Any free energy change, from any starting conditions, can be described by the important equation: is the actual mass action ratio, NOT K eq (unless actual m.a. ratio happens to be the equilibrium ratio, in which case ΔG = 0). ACTUAL free energy change ΔG' for any reaction or process depends on 3 things: 1. standard free energy change for that reaction (ΔG ', a "reference telling where equilibrium lies) 2. actual starting concentrations of reactants and products (mass action ratio) 3. temperature CHEMICAL KINETICS (review from gen chem) For the reaction k = rate constant NOTE: Rate constants are lower case k's. Equilibrium constants are upper case K's. Velocity (rate) of forward reaction = v F = k F [S] eq Velocity (rate) of reverse reaction = v R = k R [P] eq At equilibrium, v F = v R, so k F [S] eq = k R [P] eq. The equilibrium constant is NOTE: Enzymes do NOT alter K eq. As catalysts, enzymes DO increase rate constants and thus increase rates of reactions. COROLLARY: An enzyme that increases k F by a factor of must also increase k R by a factor of There are a few Enzymes-Introd sample problems posted online. LEC 10, Enzymes - Introduction 7

8 TRANSITION STATE THEORY transition state: an activated complex at the highest free energy point on the reaction coordinate a PEAK on the free energy diagram not isolatable as structures (lifetimes ~10 13 sec) -- they re "in transition", sort of with bonds half-made, half-broken. Chemical example: an S N 2 reaction, attack of a thiolate anion on iodoacetate: transition state (in brackets)( ): a trigonal bipyramid, with 3 covalent bonds + 2 more "half" bonds: FREE ENERGY DIAGRAM FOR THE REACTION S P free energy G vs. progress of reaction (i.e., the "reaction coordinate") Enzymes decrease activation energy (ΔG ) for reactions they catalyze. ΔG = overall difference in free energy between final (P) and starting (S), not affected by enzyme. RATE of reaction IS affected by enzyme. RATE depends on ΔG, the Arrhenius activation energy (i.e., the free energy of activation for the reaction). Berg et al., Fig. 8.3 LEC 10, Enzymes - Introduction 8

9 Dependence of rate constant on ΔG, the activation energy Rate constant (k) depends on ΔG, the Arrhenius activation energy (i.e., the free energy of activation for the reaction) ΔG = G G S = difference in free energy between transition state and starting state (S in this case), the "barrier" over which the reaction must go in order to proceed. ΔG has POSITIVE values (ΔG > 0) -- it's a free energy BARRIER. k is rate constant for the reaction. κ is Boltzmann s constant and h is Planck s constant. NOTE: Rate constant k is inversely and exponentially dependent on the activation energy, ΔG. Velocity of the reaction: (rate constant k is what s inside large brackets.) How could you increase the reaction rate of S P? How could you increase the reaction rate of S P? Rate of S P = velocity = k [S] to increase rate, 1) increase concentration of a reactant [S], or 2) increase the rate constant HOW? a) increase temperature, or b) decrease ΔG (catalyst) Enzymes increase reaction rates by decreasing ΔG and thus increasing k. step1 step2 step3 How do enzymes increase k, i.e., how do they decrease ΔG? Nelson & Cox, Lehninger Principles of Biochemistry, 4th ed., Fig. 6-3 LEC 10, Enzymes - Introduction 9

10 How do enzymes increase k (decrease ΔG )? by changing the pathway of reaction, and by tightly binding transition state(s). New pathways : often multiple steps in an enzyme-catalyzed reaction. Intermediates are troughs between steps on free energy diagram (e.g., ES and EP). Each step has a transition state (peak), so each step has its own ΔG. step1 step2 step3 Slowest step in pathway (the "rate-limiting step") = step with highest ΔG. NOTE: Even the highest ΔG (step #2 in figure above, ES < == > EP) for a catalyzed reaction is less than the ΔG for an uncatalyzed reaction. K eq is not affected by the catalyst, and ΔG' is not affected by the catalyst. LEC 10, Enzymes - Introduction 10

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