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Km is [S] at 1/2 Vmax It is a constant for a given enzyme at a particular temp and pressure Km is unique to each Enzyme and Substrate. It describes properties of enzyme-substrate interactions. Dependent on temp, pH etc. Independent of enzyme conc .
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Km is [S] at 1/2 Vmax It is a constant for a given enzyme at a particular temp and pressure Km is unique to each Enzyme and Substrate. It describes properties of enzyme-substrate interactions. Dependent on temp, pH etc. Independent of enzyme conc. It is an ESTIMATE of equilibrium constant for substrate binding to enzyme Small Km= tight binding, large Km=weak binding It is a measure of substrate concentration required for effective catalysis Vmax is THEORETICAL MAXIMAL VELOCITY Vmax is constant for a given enzyme. 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) To reach Vmax, ALL enzyme molecules have to be bound by substrate Kcat is a measure of catalytic activity- direct measure of production of product under saturating conditions. Kcat is turnover number- number of substrate molecules converted to product per enzyme molecule per unit time Catalytic efficiency = kcat/km Allows comparison of effectiveness of an enzyme for different substrates Km Vmax Kcat
Enzyme Km examples Hexokinase prefers glucose as a substrate over ATP Km values of enzymes range from 10-1M to 10-7M for their substrates. It also varies depending on substrate, pH, temp, ionic strength etc.
Kcat Catalase is very efficient-it generates 40 million molecules of product per second. Fumarase is not efficient-it generates only 800 molecules/per second 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 • Kcat is turnover number for the enzyme • number of substrate molecules converted into product per unit time by that enzyme
Kcat/Km The catalytic constant, kcat is kcat = Vmax/[ET] for simple reactions, kcat = k2 When [S] << Km, very little ES is formed, such that [E] is almost equal to [ET]. In this case, vo = k2[ET][S]/Km + [S] becomes = (k2/Km)[ET][S] = (kcat/Km)[ET][S] • Rate constant of rxn E + S ---> E + P • Specificity constant • Gauge of catalytic efficiency • Catalytic perfection ~ 108-109 M-1 s-1 (close to diffusion)
Kcat/Km • Rate constant of rxn E + S ---> E + P • Specificity constant • Gauge of catalytic efficiency • Catalytic perfection ~ 108-109 M-1 s-1 (close to diffusion)
How do ENZYMES carry out catalysis? • Entorpy reduction- holds substrates in proper position • Bringing two reactants in close proximity (reduce entropy & increase effective reactant concentrat) • Substrate is desatbilized when bound to enzyme favoring reaction-(change of solvent, charge-charge interactions strain on chemical bonds). • Desolvation of substrate- H bonds with water are replaced by H bonds with active site • Enzymes form a covalent bond with substrate which stabilizes ES complex (Transition state is stabilized) • Enzyme also interacts non-covalently via MANY weak interactions • Bond formation also provides selectivity and specificity • (H bonds- substrates that lack appropriate groups cannot form H bonds and will be poor substrates) • (Multiple weak interactions between enzyme and substrate) • Free energy released by forming bonds is used to activate substrate (decrease energy barrier/lower activation energy of reaction) • Induced fit-binding contributes to conformation change in enzyme • Whats the Bill? • 5.7 kJ/mol is needed to achieve a 10x increase in rate of a reaction • Typical weak interactions are 4-30 kJ/mol • Typical binding event yields 60-100 kJ/mol • MORE THAN ENOUGH ENERGY!!!
Breaking a stick Imagine you have to break a stick. You hold the two ends of the stick together and apply force. The stick bends and finally breaks. You are the catalyst. The force you are applying helps overcome the barrier.
A stickase with a pocket complementary in structure to the stick (the substrate) stabilizes the substrate. Bending is impeded by the attraction between stick and stickase.
An enzyme with a pocket complementary to the reaction transition state helps to destabilize the stick, contributing to catalysis of the reaction. The binding energy of the interactions between stickase and stick compensates for the energy required to bend the stick.
Role of binding energy in catalysis. The system must acquire energy equivalent to the amount by which DG‡ is lowered. Much of this energy comes from binding energy (DGB) contributed by formation of weak noncovalent interactions between substrate and enzyme in the transition state.
Lock/Key- Complementary shape The enzyme dihydrofolate reductase with its substrate NADP+ NADP+ binds to a pocket that is complementary to it in shape and ionic properties, an illustration of "lock and key" hypothesis of enzyme action. In reality, the complementarity between protein and ligand (in this case substrate) is rarely perfect,
Induced Fit Hexokinase has a U-shaped structure (PDB ID 2YHX). The ends pinch toward each other in a conformational change induced by binding of D-glucose (red).
Substrate specificity The specific attachment of a prochiral center (C) to an enzyme binding site permits enzyme to differentiate between prochiral grps
Catalysis • • • • •
Catalysis •Acid-Base Catalysis- donate or accept protons/electrons from and to substrate •Covalent Catalysis-transient covalent link between substrate and enzyme side chain •Metal-Ion Catalysis-Metal in active site donate or accept protons with substrate •Proximity & Orientation Effects (reduction in entropy-two mol brought together and oriented in specific manner) •Transition State Preferential Binding
R Groups The active sites of enzymes contain amino acid R groups. Active site is lined with hydrophobic residues Polar amino acid residues in active site are ionizable and participate in the reaction. Anion/cation of some amino acids are involved in catalysis Lysozyme: Cleaves glycosidic bonds in carbohydrates
Covalent Catalysis All or part of a substrate is transiently covalently bound to the enzyme to form a reactive intermediate Group X can be transferred from A-X to B in two steps via the covalent ES complex -EX A-X+ E <-----> X-E + A X-E + B <-----> B-X+ E
Specific acid catalysis: - A proton is transferred to the substrate in a rapid preequilibrium; Subsequently, the protonated substrate reacts further to form the product Specific acid–base catalysis means specifically, –OH or H+ accelerate the reaction. The reaction rate is dependent on pH only. The rate is only dependent on the pH, not on [HA] General acid catalysis: - Proton transfer occurs in a slow, rate determining step; Subsequently, the protonated substrate rapidly reacts to give the product. the reaction rate is dependent on all acids/bases present, dependent on the buffer concentration, at constant pH. Two mechanisms for acid catalysis
Enzymes often use general acid or base catalysis: • They work at neutral pH, so low [H+] and [OH-] • High localized concentration of general acid/base • Correct orientation of the acidic/basic group around the substrate • Optimum catalysis at pH around pKa • General acid-base catalysis involves a molecule besides water that acts as a proton donor or acceptor during the enzymatic reaction. It facilitates a reaction by stabilizing charges in the transition state through the use of an acid or base, which donates protons or accepts them, respectively. • Nucleophilic and electrophilic groups are activated as a result of the proton addition or removal and causes the reaction to proceed. • Side chains of various amino acids act as general acids or general basis • Amino acid residues such as His often have a pKa that is close to neutral pH and are therefore able to act as a general acid or base catalysts :B :X- X H B+ H General acid/base catalysis by enzymes
Catalysts Without a catalyst the intermediate converts back to the reactants and does not proceed forward (high barrier). Donation of a proton by water or an acid helps the process move forward. The active sites of enzymes contain amino acid R groups, that participate in the catalytic process as proton donors or proton acceptors.
Proton donor/acceptor (Nucleophile/electrophile) R +C O R’ Asp and Glu are negatively charged at pH7.0 and their side chains are acidic. These side chains ACCEPT protons which neutralize the charge. Lys, Arg, His are positively charged at pH 7.0 and their side chains are basic. These side chains DONATE protons to neutralize their charge. Asp/Glu COO- + H+ <-----------> COOH Lys/Arg NH3+ <----------> NH2 + H+ Nucleophiles R-OH <---> R-O: + H+ (hydroxyl) R-SH <---> R-S: + H+ (sulphydryl) R-NH3 <---> R-NH2: + H+ (amino) Electrophiles H+ Proton M+ Metal ion Carbonyl Nucleophiles-groups rich in and capable of donating electron (attracted to nucleus) Electrophile- group deficient in electron (attracted to electron) Reactions are promoted by proton donors (general acids) or proton acceptors (general bases). The active sites of some enzymes contain side groups, that can participate in the catalytic process as proton donors or proton acceptors.
Acid base catalysis RNaseA cleavage of RNA
Metal Ion catalysis • Roles of metals in catalysis: • As “super acid”: comparable to H+ but stronger • As template: metal ions are able to coordinate to more than 2 ligands and can thereby bring molecules together • As redox catalyst: many metal ions can accept or donate electrons by changing their redox state
E E P S S E + S P E E E ES EP E + P Enzyme Inhibition • What happens if a reactant does not leave the active site? • Enzyme is blocked (inhibited) from further interactions • Why inhibit? • To control [S] or [P] • Increase [S] unreacted • Decrease [P] formed
Enzyme Inhibition Many molecules inhibit enzymes
Enzyme Inhibition • Reversible • Competitive • Uncompetitive • Mixed • Noncompetitive • Irreversible • Suicide inactivators Many molecules inhibit enzymes Reversible Competes with substrate Does not compete with substrate Irreversible (covalently bound to enzyme) To figure out what kind of inhibitor we have, conduct two sets of rate experiments: ([E] constant in each case) • [S] constant, test effect of increasing [I] on Vo • [I] constant, vary [S] • Plot the results as 1/Vo vs. 1/[S]
Competitive Inhibitor • Most common • Inhibitor competes with natural substrate for binding to active site • Inhibitor similar in structure to natural substrate and binds active site of enzyme (reducing effective enzyme conc) • Binds more strongly • May or may not react • If reacts, does so very slowly • Gives info about active site through comparison of structures
I and S compete for E • Increasing [I] • Increases [EI] • Reduces [E] available for substrate binding • Need to keep [I] high to ensure inhibition • Dosage • S overcome inhibitor effects; saturate • As [I] increases, KM increases • More S required to reach ½ Vmax • Rate does not increase rapidly with [S] due to inhibition but the rate reaches the same maximal rate, just at a higher substrate conc
Reversible Inhibition (competitive) +Inh Vmax 1/v -Inh -Inh +inh vo 1/Vmax 1/2 Vmax 1/[S] -1/Km -1/Km (app) [Substrate] Km Km (app) a • Inhibitor competes with substrates for binding to active site • Inhibitor is similar in structure to substrate • binds more strongly • reacts more slowly • Increasing [I] increases [EI] and reduces [E] that is available for substrate binding • Need to constantly keep [I] high for effective inhibition (cannot be metabolized away in body) • Slope is larger • Intercept does not change (Vmax is the same) • KM is larger
Uncompetitive Inhibitor Binds only to ES complex but not free enzyme Binds at location other than active site Does not look like substrate. Binding of inhibitor distorts active site thus preventing substrate binding and catalysis Cannot be competed away by increasing conc of substrate (Vmax is affected by [I]) Increasing [I] lowers Vmax and lowers Km.
Increasing [I] • Lowers Vmax (y-intercept increases) • Lowers KM (x-intercept decreases) • Ratio of KM/Vmax is the same (slope)
