Enzymes GOOD

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**ENZYMES** Definitions-- Chemical reactions in cells require specific catalysis. Enzymes are proteins which perform this function. Metabolite acted upon is called the enzyme’s substrate.

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A. Fundamental Properties 1) Enzymes are excellent catalysts, speeding up reactions 108 to 1020 fold. They speed up reactions without being used up. 2) Specificity a) for substrate - ranges from absolute (e.g., aspartase) to relative b) for reaction catalyzed, i.e.,few side-reactions and by-products, etc.) 3) Regulated-- some enzymes can sense metabolic signals.

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B. Enzymes as Molecules 1) Large molecules-- proteins from 12kDa - 1,000kDa or more -- most are much larger than their substrate. 2) Active site -- specific region in enzyme which interacts with its substrate. both binding and catalytic reaction occur here. some residues involved in binding substrate others catalyze reaction

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3) Cofactors for some reactions, the amino acids are not powerful enough for catalysis. some enzymes incorporate additional factors. metal ions as cofactors-- Zn2+, Fe2+, Cu2+, others coenzymes are organic cofactors prosthetic groups are covalently attached

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C. Classification of Enzymes 1) named and classified according to the substrate acted upon and the reaction catalyzed. 2) trivial names-- end in -ase -- urease, hexokinase. 3) named based on a formal systemic catalog (IUB) with six major classifications. (All enzymes should fall into one of these categories and all enzymes therefore have a formal name.)

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Class 1. Oxidoreductases- catalyze redox processes Example: RCH2-OH  RCH=O Class 2. Transferases- transfer chemical groups from one molecule to another or to another part of the same molecule. O O Example: CH3-C-SCoA + XR  CH3-C-XR + HSCoA acetyl CoA acetyl group transferred

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Class 3. Hydrolases- cleave a bond using water to produce two molecules from one. O H2O O example: --CNH-R  --C-OH + H2N-R cleavage of a peptide bond Class 4. Lyases- remove a group from or add a group to double bonds. H-X H X ---C=C---  ---C--C---

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Class 5. Isomerases- interconvert isomeric structures by molecular rearrangements. CH3 CH3 HC-OH HO-CH COOH COOH Class 6. Ligases -- join two separate molecules by the formation of a new chemical bond usually with energy supplied by the cleavage of an ATP. example: O ATP ADP+Pi O -OOC-C-CH3 + CO2 -OOC-C-CH2-COO- pyruvate oxaloacetate enzyme = pyruvate carboxylase

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Enzyme Mechanism Enzymes catalyze difficult reactions by changing the reaction to a series of “easier” steps including nucleophilic attack, general acid-base catalysis, covalent attachment, etc.

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Reaction Rates and the Transition State Enzymes speed up reactions enormously. To understand how they do this, examine the concepts of activation energy & the transition state. In order to react, the molecules involved are distorted, strained or forced to have an unlikely electronic arrangement. That is the molecules must pass through a high energy state.

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This high energy state is called the transition state. The energy required to achieve it is called the activation energy for the reaction. The higher the free energy change for the transition barrier, the slower the reaction rate.

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Enzymes lower energy barrier by forcing the reacting molecules through a different transition state. This transition state involves interactions with the enzyme. Enzyme

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Modes of Enzymatic Enhancement of Rates 1) general acid and general base catalysis-- good proton donors & acceptors positioned just right. 2) covalent catalysis- unstable intermediate 3) metal ion catalysis - electron donor or acceptor 4) electronic effects- “orbital steering” of Koshland, steering aromatic groups

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5) proximity and orientation - close to reactive group and aligned versus random in solution chemistry. 6) conformational strain distortion and induced fit old idea, lock-and-key= substrate fits active site

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induced fit- enzyme changes its conformation to accept the transition state of substrate/product well. Enzyme conformational change works to distort and strain substrate forcing it into transition state. Simultaneous Koshland

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Definition: rate of a reaction For an enzyme acting on its substrate, just as an ordinary chemical reaction, the rate of the reaction depends on the concentration of substrate, S. A reaction leading to formation of product is written: S P Rate = change in [P]/ change in time or rate = v = Δ[P]/Δt

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For a chemical reaction (as contrasted to an enzymatically catalyzed one), the rate is proportional to reactant [S]. A rate constant, k, can be defined: rate = v = Δ[P]/Δt = k [S] rate [S]

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In contrast, found empirically for enzymes: rate depends on [S] but hyperbolic curve & plateaus rate also depends on the enzyme concentration rate [S]

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Michaelis-Menten Model or Interpretation E + S  E●S  E + P where E●S is enzyme-substrate complex, i.e., an intermediate complex. rate stops increasing or plateaus because the complex E●S becomes filled at high [S]

I. Enzyme Kinetics : 

I. Enzyme Kinetics Expression for enzyme catalyzed reaction:

Michaelis-Menton Equation : 

Michaelis-Menton Equation Relates rate and [S] Rate = V0 = k2[ES] V0 = initial velocity/rate Neglects complications from inhibition, deactivation, etc. What is the [ES]? To determine, look at rate of formation and disappearance of [ES]

[ES] : 

[ES] Rate of formation = k1[E][S] Rate of breakdown = k-1[ES] + k2[ES] Assume steady state As ES is produced, it reacts [ES] remains constant Rate formation = rate breakdown So, k1[E][S] = k-1[ES] + k2[ES]

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k1[E][S] = k-1[ES] + k2[ES] k1[E][S] = (k-1 + k2)[ES] Rearrange [E][S] / [ES] = (k-1 + k2) / k1 Where (k-1 + k2) / k1 = KM (Michaelis constant) Define total enzyme concentration [E]T = [E] + [ES] Substitute for [E] ([E]T – [ES])[S] / [ES] = KM Solve for [ES] [ES] = [E]T[S] / KM + [S]

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[ES] = [E]T[S] / KM + [S] Since V0 = k2[ES] V0 = k2[E]T[S] / KM + [S] Define maximum velocity Vmax Occurs at high [S] Enzyme is all in ES form [ES] = [E]T Vmax = k2[E]T Therefore V0 = Vmax[S] / KM + [S]

Michaelis-Menton Equation : 

Michaelis-Menton Equation Rate increase with [S] Rate levels off as approach Vmax More S than active sites in E Adding S has no effect At V0 = ½ Vmax [S] = KM V0 = Vmax[S] / KM + [S]

KM : 

KM Unique for each E-S pair Magnitude varies Depends on E and S Function of Temperature Rate increases with T Optimum T ~ 37°C for most enzymes Function of pH Change catalytic activity Most enzymes active in narrow pH range close to pH of environment of enzyme

Catalytic Constant kcat : 

Catalytic Constant kcat kcat = Vmax / [E]T Turnover number Number of reaction processes each active site catalyzes per unit time Measure of how quickly an enzyme can catalyze a specific reaction For M-M systems kcat = k2 Rate constant of RDS

kcat / kM : 

Rate constant of rxn E + S → E + P Specificity constant Gauge of catalytic efficiency Catalytic perfection ~ 108 ─ 109 M-1 s-1 kcat / kM

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A plot of v = Vmax[S] [S] + Km is not accurate enough to derive good Km & Vmax. Computer analysis is done. Reciprocal Plot A double reciprocal plot or Lineweaver-Burk plot is linear and more eye-appealing for presentation. mathematically = “linear transformation”

Line weaver-Burk Plot : 

Line weaver-Burk Plot Another method to determine Vmax and KM Reciprocal of M-M equation Linear equation

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Competitive Inhibition presented as double reciprocal plot Model: E + S  E●S  E + P + I  E●I I resembles S I binds at active site reversibly E●I cannot bind to S so no reaction

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No I Competitive Inhibition +I +more I Vmax Km In competitive inhibition, can always add enough [S] to overcome inhibition.  same Vmax

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1/[S] 1/v 1/Vmax Competitive inhibition Double reciprocal plot Same 1/v intercept, same Vmax Different slopes, competitive Inhibition changes apparent Km Note: inhibition line always above no inhibition. + more I +I No I

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Molecular interpretation for competitive inhibition competitive inhibitor binds to same site as the substrate (competes). its structure usually resembles substrate or product. While the inhibitor is bound, the enzyme cannot bind substrate and no reaction possible.

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Noncompetitive Inhibition E + S  E●S E + P + + I I   E●I  E●S●l inhibitor substrate E●I and E●S●I not productive, depletes E and E●S

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No I + I 1/[S] 1/v Noncompetitive Inhibition different slopes, different 1/v intercepts.

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Molecular Interpretation: Inhibitor binds the enzyme somewhere different from where the substrate binds. So the inhibitor does not care whether substrate is bound or not. Inhibitor changes the conformation of the enzyme at the active site so no reaction is possible with inhibitor bound. E●I and E●S●I not productive

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Irreversible Inhibition reactive compounds combine covalently to enzyme so as to permanently inactivate it (previous examples are all reversible) almost all are very toxic most bind to a functional group in active site of enzyme to block that site

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