Lecture Notes: Enzymes PDF
Document Details
Uploaded by FriendlyTrust
University of KwaZulu-Natal - Westville
Dr S Naidoo
Tags
Summary
These lecture notes provide an introduction to enzymes, covering topics such as enzyme catalysis, enzyme kinetics, and enzyme inhibition. The notes also discuss various models for enzyme action, such as lock-and-key and induced fit.
Full Transcript
PHRM246 BIOCHEMISTRY INTRODUCTION TO ENZYMES PHRM 246/Dr S Naidoo/Pharm Sciences/Health Sciences/UKZN Proteins Enzymes as Biological Catalysts Increase reaction rates by over 1,000,000-fold Two fundamental properties Increase the reaction rate with no a...
PHRM246 BIOCHEMISTRY INTRODUCTION TO ENZYMES PHRM 246/Dr S Naidoo/Pharm Sciences/Health Sciences/UKZN Proteins Enzymes as Biological Catalysts Increase reaction rates by over 1,000,000-fold Two fundamental properties Increase the reaction rate with no alteration of the enzyme Increase the reaction rate without altering the equilibrium Reduce the activation energy Proteins Enzymes as Biological Catalysts The substrate binds to a specific region called the active site Enzyme-substrate cycle Proteins Enzymes as Biological Catalysts Two popular models provide an aid to understanding the mechanisms of enzyme action: Lock-and-key Induced fit Proteins Regulation of Enzyme activity Feedback inhibition – an example of allosteric regulation Topic 2-3 6 Stickase Theoretical Model Substrate Transition state Product X If enzyme just binds substrate then there will be no further reaction Enzyme not only recognizes substrate, but also induces the formation of transition state Adapted from Nelson & Cox (2000) Lehninger Principles of Biochemistry (3e) p.252 The Nature of Enzyme Catalysis Enzyme provides a catalytic surface This surface stabilizes transition state Transformed transition state to product B A A B Catalytic surface Juang RH (2004) BCbasics Enzyme Stabilizes Transition State Energy change ST Energy required (no catalysis) Energy decreases (under catalysis) EST S ES EP P Reaction direction T = Transition state What’s the difference? Adapted from Alberts et al (2002) Molecular Biology of the Cell (4e) p.166 Active Site Is a Deep Buried Pocket Why energy required to reach transition state is lower in the active site? + (1) Stabilizes transition (2) Expels water CoE (1) (2) (3) Reactive groups (4) - (4) Coenzyme helps (3) Juang RH (2004) BCbasics Enzyme Active Site Is Deeper than Ab Binding Instead, active site on enzyme Ag binding site also recognizes substrate, but Ab binds to Ag complementarily actually complementarily fits the no further reaction occurs. transition state and stabilized it. X Adapted from Nelson & Cox (2000) Lehninger Principles of Biochemistry (3e) p.252 Active Site Avoids the Influence of Water + - Preventing the influence of water sustains the formation of stable ionic bonds Adapted from Alberts et al (2002) Molecular Biology of the Cell (4e) p.115 Enzyme Kinetics Enzyme activity K L Score J J 0 1 2 Exam Chapters 3 4 0 1 2 3 Substrate concentration 4 Increase the substrate concentration, observe the change of enzyme activity This represents the amount of enzyme activity in relation to amount of substrate available; observe the same trend where the concentration of the substrate increases, and at some point the reaction plateaus off and there is no increase in enzyme activity but just a maintenance of reactivity. Essential of Enzyme Kinetics Steady State Theory E + S E S E +P In steady state, the production and consumption of the transition state proceed at the same rate. So the concentration of transition state keeps a constant. Juang RH (2004) BCbasics Constant ES Concentration at Steady State Concentration S P E ES Reaction Time constant enzyme-substrate combination at transition. Consider the area under the line = steady state condition where there is a constant concentration of enzyme-substrate complex over the duration of the reaction. As substrate is used up, there is an increase in product formation There is initial linear increase of E-S at transitional state which then achieves a constant conc. At same time, there is a linear decrease of enzyme in this reaction, which then levels out for the duration of reaction. Enzyme Inhibition (Mechanism) I Competitive I Non-competitive I Uncompetitive Substrate E X Cartoon Guide S S E I S E I S I I Compete for S I Inhibitor active site Different site E + S← → ES → E + P E + S←→ ES → E + P E + S← → ES → E + P Equation and Description + + + + I I I I ↓↑ ↓↑ ↓↑ ↓↑ EI EI + S →EIS EIS [I] binds to free [E] only, [I] binds to free [E] or [ES] [I] binds to [ES] complex and competes with [S]; complex; Increasing [S] can only, increasing [S] favors increasing [S] overcomes not overcome [I] inhibition. the inhibition by [I]. Inhibition by [I]. Juang RH (2004) BCbasics Competitive Inhibition Product Substrate Competitive Inhibitor Succinate Glutarate Malonate Oxalate C-OO- C-OO- C-OO- C-OO- C-OO- C-H H-C-H H-C-H H-C-H C-OO- C-H H-C-H H-C-H C-OO- C-OO- C-OO- H-C-H Inhibitor structures are similar to substrate C-OO- Succinate Dehydrogenase Adapted from Kleinsmith & Kish (1995) Principles of Cell and Molecular Biology (2e) p.49 MICHAELIS-MENTON REACTION KINETICS In an enzyme-catalysis reaction: [substrate] is many orders of magnitude larger than [enzyme] Various states of complexation before product formation S+E ES ES* EP E+P V1 = {Vmax [S]} {Km + [S]} This equation describes the rate of the reaction (V) to the substrate concentration related to the two constants Vmax and Km (Michaelis-Menton constant) Means that 1 enzyme will catalyse the reaction of many substrate molecules. This equation demonstrates that substrate concentration at which the reaction is half its max rate, so that at this reaction rate, substrate concentration is equal to Km Another way of saying that the substrate concentration is half that required to support max reaction rate. Km is unique to each Enzyme and Substrate. It describes properties of enzyme-substrate interactions Km Independent of enzyme conc. Dependent on temp, pH etc. Vmax is maximal velocity POSSIBLE. It is directly dependent on enzyme conc. It is attained when all of the enzyme binds the substrate. (Since these are equilibrium reactions enzymes tend towards Vmax at high substrate conc but Vmax is never achieved. So it is difficult to measure). When an enzyme is operating at Vmax, all enzyme is bound to substrate and adding more substrate will not change rate of reaction (enzyme is saturated). Invertase (IT) IT Sucrose Glucose + Fructose HOCH2 HOCH2 HOCH2 HOCH2 HOCH2 HOCH2 6 O 6 1 O O 5 O 5 2 1 2 4 1 3 OH 2 OH 4 OH HO 3 b b H2 O CHO 1 H2C-OH HOCH2 H-C-OH 2 C=O Juang RH (2004) BCbasics HOCH2 HOCH2 O O HO-C-H 3 HO-C-H H-C-OH 4 H-C-OH O H-C-OH 5 H-C-OH H2-C-OH 6 H2-C-OH An Example for Enzyme Kinetics (Invertase) 1) Use predefined amount of Enzyme →E 2) Add substrate in various concentrations → S (x) 3) Measure Product in fixed Time (P/t)→ vo (y) 4) (x, y) plot get hyperbolic curve, estimate → Vmax 5) When y = 1/2 Vmax calculate x ([S]) → Km Vmax 1 vo vo 1/2 Juang RH (2004) BCbasics -1 Km 1 Vmax Double reciprocal 1/S Km Direct plot S Reciprocal It is not easy to extrapolate a hyperbola to its limiting value The Michaelis-Menten equation can be recast into a linear form To obtain parameters of interest Reciprocal form of equation 1 = Km 1 + 1 V Vmax S Vmax Y= m x + c The y-intercept gives the Vmax value and the slope gives Km/Vmax Vmax is determined by the point where the line crosses the 1/Vi = 0 axis (so the [S] is infinite). Km equals Vmax times the slope of line. This is easily determined from the intercept on the X axis. A Real Example for Enzyme Kinetics (sucrase/invertase reaction) Substrate Product Velocity Double reciprocal no [S] Absorbance v (mmole/min) 1/S 1/v 1 0.25 0.21 → 0.42 4 2.08 2 0.50 0.36 → 0.72 2 1.56 Data 3 1.0 0.40 → 0.80 1 1.35 4 2.0 0.46 → 0.92 0.5 1.16 (1) The product was measured by spectroscopy at 600 nm for 0.05 per µmole (2) Reaction time was 10 min 2.0 Double reciprocal 1.0 Direct plot v 1/v 1.0 1.0 uang RH (2004) BCbasics 0.5 -3.8 0 0 0 1 2 -4 -2 0 2 4 [S] 1/[S] Km: Affinity with Substrate Vmax [S] Vmax vo = If vo = Km + [S] 2 When using different substrate Vmax Vmax [S] Vmax = S2 2 Km + [S] S1 S3 1/2 Km + [S] = 2[S] Juang RH (2004) BCbasics S1 S2 S3 Km = [S] Km Affinity changes Turn Over Number, kcat k1 k3 E+S k2 ES (vo) E + P When substrate excess, k3 = kcat, turn over number (t.o.n) When [substrate] is low Vmax [S] k3 [E][S] k3 vo = = = [E][S] Km + [S] Km + [S] Km Start from M-M eq. Omit the [substrate] Substrate specificity Describes the maximum number of molecules of substrate that an enzyme can convert to product per catalytic site per unit of time Turn Over Numbers of Enzymes Enzymes Substrate kcat (s-1) Catalase H2O2 40,000,000 Carbonic anhydrase HCO3- 400,000 Acetylcholinesterase Acetylcholine 140,000 b-Lactamase Benzylpenicillin 2,000 Fumarase Fumarate 800 RecA protein (ATPase) ATP 0.4 The number of product transformed from substrate by one enzyme molecule in one second Adapted from Nelson & Cox (2000) Lehninger Principles of Biochemistry (3e) p.263 Chymotrypsin Has Distinct kcat /Km to Different Substrates O R O = = – – H3C–C–N–C–C–O–CH3 – H H R= kcat / Km Glycine –H 1.3 ╳ 10-1 Norvaline –CH2–CH2–CH3 3.6 ╳ 102 Norleucine –CH2–CH2–CH2–CH3 3.0 ╳ 103 Phenylalanine –CH2– 1.0 ╳ 105 (M-1 s-1) Enzyme Inhibition (Mechanism) I Competitive I Non-competitive I Uncompetitive Substrate E X Cartoon Guide S S E I S E I S I I Compete for S I Inhibitor active site Different site E + S← → ES → E + P E + S←→ ES → E + P E + S← → ES → E + P Equation and Description + + + + I I I I ↓↑ ↓↑ ↓↑ ↓↑ EI EI + S →EIS EIS [I] binds to free [E] only, [I] binds to free [E] or [ES] [I] binds to [ES] complex and competes with [S]; complex; Increasing [S] can only, increasing [S] favors increasing [S] overcomes not overcome [I] inhibition. the inhibition by [I]. Inhibition by [I]. Juang RH (2004) BCbasics Enzyme Inhibition (Plots) I Competitive I Non-competitive I Uncompetitive Vmax Vmax Vmax vo vo Direct Plots Vmax’ Vmax’ I I I Km Km’ [S], mM Km = Km’ [S], mM Km’ Km [S], mM Vmax unchanged Vmax decreased Km increased Km unchanged Both Vmax & Km decreased Double Reciprocal 1/vo I 1/vo I 1/vo I Two parallel Intersect lines at Y axis 1/ Vmax Intersect 1/ Vmax 1/ Vmax at X axis 1/Km 1/[S] 1/Km 1/[S] 1/Km 1/[S] Juang RH (2004) BCbasics Sulfa Drug Is Competitive Inhibitor Domagk (1939) Para-aminobenzoic acid (PABA) Bacteria needs PABA for H2N- -COOH the biosynthesis of folic acid Folic Tetrahydro- Precursor acid folic acid Sulfa drugs has similar H2N- -SONH2 structure with PABA, and inhibit bacteria growth. Sulfanilamide Sulfa drug (anti-inflammation) Adapted from Bohinski (1987) Modern Concepts in Biochemistry (5e) p.197 Enzyme Inhibitors Are Extensively Used Sulfa drug (anti-inflammation) Pseudo substrate competitive inhibitor Protease inhibitor Alzheimer's disease Plaques in brain contains protein inhibitor HIV protease is critical to life cycle of HIV HIV protease (homodimer): subunit 1 subunit 2 ↑inhibitor is used to treat AIDS As As Symmetry p p → Human aspartyl protease: domain 1 domain 2 Not symmetry (monodimer) As As p p Juang RH (2004) BCbasics Chymotrypsin Catalytic Mechanism A1 Catalytic Triad His57 Asp102 Ser195 H H O [HOOC] C N C C [NH2] N C H C N C C O Check substrate specificity Chymotrypsin Catalytic Mechanism A2 His57 Asp102 Ser195 H H O C C [NH2] C N [HOOC] C N C N C H C O First Transition State Chymotrypsin Catalytic Mechanism A3 H O H C C C [NH2] C N C N [HOOC] C N C H O Acyl-Enzyme Intermediate Chymotrypsin Catalytic Mechanism D1 H O H O C C H C [NH2] C N C O C N-H [HOOC] N C H Acyl-Enzyme Water Intermediate Chymotrypsin Catalytic Mechanism D2 H O H C C O C [NH2] H C N C O Second Transition State Chymotrypsin Catalytic Mechanism D3 H H O O C C [NH2] H C C N C O Deacylation Chymotrypsin Is Activated by Proteolysis Chymotrypsinogen (inactive) 245 Cleavage by Trypsin p-Chymotrypsin (active) R15-I16 a-Chymotrypsinogen (active) S14-R15 T147-N148 L13 I16 Y146 A149 Disulfide bonds Charge Relay in Active Site H O C C–O - = H–N N H–O–CH2 Ser 195 C C Adapted from Alberts et al (2002) Molecular Biology of the Cell (4e) p.158 Asp 102 H CH2 His 57 H Active Ser O C = C–O–H N N–H - O–CH Ser 2 195 C C Asp 102 H CH2 His 57 pH Influences Chymotrypsin Activity Relative Activity 5 6 7 8 9 10 11 pH Adapted from Dressler & Potter (1991) Discovering Enzymes, p.162 Imidazole on Histidine Is Affected by pH Adapted from Dressler & Potter (2000) Discovering Enzymes, p.163 pH 6 H H pH 7 C + C H–N N–H H–N N H+ C C C C H H H O Inactive C + C–O - = H–N N–H H–O–CH2 Ser 195 C C-H Asp 102 Change in pH changes CH2 ionic charge on amino acids in the enzyme binding site and affects binding properties of the His 57 active site Adapted from Alberts et al (2002) Molecular Biology of the Cell (4e) p.158 Regulation of Enzyme Activity Inhibitor Proteolysis or proteolysis o I I x x o S I I S inhibitor Feedback regulation Phosphorylation o R x x P o S R S P (+) regulator effector phosphorylation Signal transduction Juang RH (2004) BCbasics A or A o + x Regulatory cAMP or S (-) subunit calmodulin Mechanism and Example of Allosteric Effect Kinetics Models Co-operation R = Relax Allosteric site (active) vo R Homotropic (+) S S (+) S R Concerted Allosteric site [S] R (+) A vo T Heterotropic (+) S S (+) R Sequential X [S] T T = Tense I vo T Heterotropic (inactive) (-) X (-) X (-) X = inhibitor T Concerted S = substrate I = inhibitor [S] Juang RH (2004) BCbasics