Study of enzyme catalyzed reactions Rate of reactions Enzyme specificity Mechanism of catalysis There are several good reasons for adding kinetic modeling. Mechanism and dynamics are intimately related, and kinetic modeling reveal new and useful information about biological control systems. Enzymes/ Kinetics Observation during steady state is not informative about the process
Catalyst= Enzyme= biocatalyst e.g. Fermentation of sugar to ethanol by yeast enzymes Conversion of fatty acids to polyketide antibiotics by filamentous fungi Conversion of milk to cheese by microorganisms Conversion of sugar to CO2 by bakers yeast to make leavened bread Catalyst
MOST Enzymes are Proteins Enzyme advantages over chemical catalysis: Enzyme concentration is very small and enzyme is almost always limiting BioCatalyst
S <--------->P 1 2 3 DGo =standard free energy change for chemical reactions DG’o =standard free energy change for biological reactions 298o k (RT); 1 atm, 1M substrate, pH7.0 Active site- pocket in enzyme that binds substrates- most complimentary in structure to transition state Enzyme function
Conformation of proteins and positions of side chains are important for enzyme-substrate interactions and catalysis. Forces involved in protein folding and structure are also involved in catalysis- enzyme-substrate specificity Trypsin hydrolyses proteins by cleaving peptide bond adjacent to Lys/Arg. Aspartate residue in trypsin active site mediates ionic interaction with Lys /Arg and this arranges protein residue at which hydrolysis occurs. To use enzymes in biotechnology, pharmaceutics or drugs that inhibit enzymes in medicine, you NEED TO KNOW KINETIC PARAMETERS OF THE ENZYME REACTION. We may want enzymes that WORK FAST- convert more substrate in a fixed unit of time. To do this optimization we have to perform and analyze the enzyme catalyzed reaction. You can adjust pH, temperature and add co-factors to optimize enzyme activity. You cannot adjust substrate selectivity. Just like chemical reactions, enzyme catalyzed reactions have kinetics and rates Reaction kinetics is Michaelis-Menten kinetics. Rationale for enzyme kinetics
Activation energy Free energy Ground state= starting point (energetically speaking) - - - - -
Energy Barriers [S] Free energy [P] - - - - - -
Sucrose to CO2 and H2O / = DG C12H22O11 + 12O2 <-----> 12CO2 + 11 H2O Reaction has large negative DG’o Therefore Product prevails at equilibrium In your pantry you can store sucrose in the presence of oxygen. It does not spontaneously convert into CO2 and water! ACTIVATION energy of this reaction is HUGE In a cell sucrose is rapidly converted to CO2 and H2O ENZYMES!!! Activation energy barriers are required for ordered life!!! Without barriers molecules would spontaneously interconvert- there would be no regulation
Reaction coordinate Overall std free energy change (S goes to P)
Rate enhancements Enzymatic rate enhancements are 103 to 1017 x
How do ENZYMES carry out catalysis? • ~ • ~ • ~ • - • - • 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!!!
A------------> B------------>C A to B to C The rate constant for B to C is smaller. B to C is slow (rate determining) A is rapidly converted to B but B accumulates because its conversion to C is slow Reaction is in only one direction because DG of B is lower than A and C is lower than B B to C is rate limiting DG conc Rxn coordinate time
A------------> B------------>C The rate constant for A to B is smaller. A to B is slow (rate determining) A is not rapidly converted to B because its conversion to B is slow. B never accumulates because it is rapidly converted to C Reaction is in only one direction because DG of B is lower than A and C is lower than B A to B is rate limiting A to B to C DG conc Rxn coordinate time
A------------> B------------>C <----------- The rate constant for A to B is small. The rate constant for B to A is large. A to B is slow but B to A is very fast and B to C is kinda fast. A is not rapidly converted to B. B never accumulates because it is very rapidly reconverted to A and C accumulation is not very rapid because most of B is reconverted back to A rather than to C Reaction is in two directions because DG of B is higher than A and C is lower than B and A A to B to C DG conc Rxn coordinate time
A------------> B------------>C <----------- The rate constant for A to B is small. The rate constant for B to A is large. A to B is slow but B to A is kinda fast and B to C is very fast. A is not rapidly converted to B. B never accumulates because it is very rapidly converted to C. or reconverted to A. Accumulation of C is rapid because most of B is converted to C rather than reconverted to A Reaction is primarily in one directions because while DG of B is higher than A, C is lower than B and A A to B is rate limiting A to B to C DG conc Rxn coordinate time
Enz + Sub <-------> Enz-Sub <-------> Enz + Prod [S] [P] Enzyme catalyzed Steady State Conc During steady state formation of ES equals its breakdown k1[E][S]=k-1[ES]+k2[ES] Km= k-1+k2/k1 [ES] [E] Time Pre steady state Steady state
Rate of a Reaction What is the rate of a chemical reaction S---->P - - - - - A+B----.>P is a bimolecular reaction (there are two reactants) and this is a second order reaction - Sometimes second order reactions appear as first order. E.g. If conc of B is very large and does not change much then reaction rate is entirely dependent on conc of A only
Rate of catalysis k1 ES E+S k-1 Vo= For a fixed amount of enzyme, The rate of catalysis
All measurements are done at very high substrate conc and very low product conc so the reverse reaction is rare. We can simplify the above reaction scheme as
k1k2 A -----> B ------> C <----- k-1 If k-1 is greater than k2 (B reforms A faster than it forms C) then there will be a rate-determining pre-equilibrium and overall rate of formation of C will depend on ALL THREE RATE CONSTANTS ((((d[C]/dt= [k1][k2]/[k-1][A])))) If k2 is vastly greater than k-1 (B forms C faster than it reforms A) then we can effectively ignore the back reaction (B to A) and the only question then is whether k1 or k2 is rate limiting. Rate constants k1 k2 ES E+S E+P k-1 Binding of substrate to enzyme is reversible Product release from enzyme is instantaneous k1 is rate constant for formation of ES k-1 is rate constant for conversion of ES to E+S k2 is rate constant for formation of product = kcat
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] • 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] [ES]
[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] • Rate of reaction 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 = initial velocity/rate
The classic Michaelis-Menten rate equation The first order rate constant kcat is k2 Vmax is product of kcat and enzyme conc Km is mixture of rate constants that describe formation and dissociation of enzyme-substrate complexes Km gives a sense of the affinity of enzyme for substrate When [substrate] is >>> Km, the reaction kinetics are equal to max rate (Vmax). When [substrate] is <<< Km, the reaction rate is kcat/Km All these analyzes are done at saturation kinetics Kcat and Km are PRIMARY INDICATORS of how well an enzyme will react with a particular substrate. High kcat= fast reactions. low kcat= slow reactions High Km = low affinity of enzyme for substrate. Low Km = high affinity of enzyme for substrate Michaelis-Menten
- • - • V0 = • - • What is the [ES]? Michaelis-Menton Equation
[S]4 [S]3 Product (mol/lit) [S]2 Vo(4) Vo(3) Vo(2) [S]1 Vo(1) Time - - - - ---- ----- What do we measure? A+B<--->AB At low [S], Km >> [S] At high [S], [S] >> Km
Velocity measured at very beginning of reaction when very little product is made Initial velocity is measured at saturation kinetics- at high [S], enzyme is saturated with respect to substrate Vmax Initial Velocity Vo [Product] [Substrate] Time
Vmax and Km [S] V0 = Vmax [S]+Km k-1 + k2 Km= k1 When V0=Vmax, Km= [S] Km is unique to each Enzyme and Substrate. It describes properties of enzyme-substrate interactions 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). (adding more enzyme will change the reaction).
Vmax vo [Substrate] 1/vo 1/Vmax 1/[S] -1/Km Measuring Km and Vmax You can use a curve fitting algorithm to determine Km and Vmax from a V vs [S] plot (need a computer) Reaction rates are initial rates determined when the substrate is in vast excess and isn’t changing much. Alternatively you can convert the curve to a straight line via a double reciprocal plot (1/Vmax and 1/[S])
Reciprocal It is not easy to extrapolate a hyperbola to its limiting value (computers can do this) 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 + b The y-intercept gives the Vmaxvalue 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.
1/Vo It is difficult to accurately measure Vmax [S] V0 = Vmax [S]+Km Reciprocal 1 Km 1 S 1 Vma + = V0 Vmax Vmax = k2[Et] k2 is also called Kcat Et is conc of active sites in enzyme Km values of enzymes range from 10-1M to 10-7M for their substrates. It varies depending on substrate, pH, temp, ionic strength etc. Kcat is turnover number for the enzyme-number of substrate molecules converted into product per unit time by that enzyme
Km is [S] at 1/2 Vmax It is a constant for a given enzyme at a particular temp and pressure 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 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 and Vmax
Enzyme Km examples Hexokinase prefers glucose as a substrate over ATP
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/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 an amount of 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).