Catalysis is Necessary for Life. Chapter 6 Enzymes. Why Study Enzymes? Enzymes are Biological Catalysts

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1 Chapter 6 Enzymes Catalysis is Necessary for Life Definition: a catalyst is a substance that speeds up a chemical reaction, while emerging unchanged at the end Corollary A: a catalyst is never used up, thus a very small amount can facilitate a very large number of reactions Corollary B: a catalyst never affects the equilibrium state of a reaction Dihydrofolate Reductase with NADP + Chapter 6 3 Why Study Enzymes? They catalyze stereo-specific reactions with virtually no undesirable side-products Many disease states are related to altered or absent enzyme activity Many vitamins, drugs, and toxins are enzyme modulators Measurement of activity is diagnostic for many diseases Enzymes are important practical tools in industry, medicine, and agriculture Chapter 6 2 Enzymes are Biological Catalysts Are (almost) always proteins Are extraordinarily efficient ( better than chemical catalysts - see Table 6-5) Are remarkably specific Need mild physiological conditions to function (temperature, salt, ph, etc.) Often require a cofactor (metal ion, prosthetic group, or coenzyme) May be modified for regulatory purposes Chapter 6 4

2 Enzymes are Biological Catalysts Co-factors & Coenzymes/Vitamins An enzyme catalyzes a biochemical reaction where reactants products Holoenzyme is the active form of the enzyme (all cofactors, etc.) Apoenzyme is ONLY the protein component of a holoenzyme Active site is the pocket or cleft in the enzyme where the reaction occurs Substrate is the reactant bound at active site Chapter 6 5 The Secret of Life For a molecule to be stable, it must have a high activation energy for conversion to product (that s what stable means!) Even though equilibrium (thermodynamic) may hugely favor product (remember coal and diamond?) So, to convert stable molecules into other stable molecules (even where equilibrium is unfavorable), requires selective reduction of the activation energy the job of enzymes! Chapter 6 7 Enzymes are Biological Catalysts Enzymes are Sensibly Classified Note that the name of each enzyme class give a hint as to the type of reaction that is being catalyzed Many enzymes are named by adding the suffix ase to the name of their substrate or to a word describing their activity A Note on Free Energy Nomenclature. G o : used by the chemist for reactions under standard conditions of pressure, temperature, and solutes at 1M VERSUS G o : used by the biochemist for reactions under standard conditions of pressure, temperature, and ph 7 (because we can t work at 1M H +!) You should know these 6 classes Chapter 6 6 Chapter 6 8

3 Activation Energy is Needed to Get to the Transition State E +S ES EP E + P Activation energy Enzymes lower the energy of the transition state For S to reach P, it must pass through the high energy transition state the higher the peak, the lower the probability of getting over it. Hence, the slower the rate of conversion Chapter 6 9 Imagine a set Mousetrap Is it stable (in the absence of a mouse)? Is its G o positive, negative, large or small? Negative and Large Is its G positive, negative, large or small? Positive and Large (because it s stable) So, how does a mouse resemble an enzyme? By reducing the activation energy for springing the trap Chapter 6 11 How Does an Enzyme Affect the Rate, but Not the Equilibrium? E +S ES EP E + P Activation energy Enzymes lower the energy of the transition state What does G o tell us? What does it tell us about the rate? What does G tell us? What does it tell us about the equilibrium? Chapter 6 10 An Equation that is REALLY IMPORTANT to Learn There is a direct relationship between net free energy change ( G o ) and the equilibrium constant (K eq ) Which is fixed for any given reaction, because G o refers to standard conditions (ph 7, 25 o C, 1 atm pressure, etc.) G' o = [P] RT ln = RT ln K' S [ ] eq From this can be derived the G actual for a reaction that starts from specific initial conditions: G Actual = G' o + RT ln [P] S [ ] Chapter 6 12

4 And you should learn the REALLY IMPORTANT number or K d A factor of 10 difference in equilibrium constant is equivalent to a G o of: 5.7 kj/mole Chapter 6 13 Where does this energy come from? Covalent chemical bonds are rearranged during the reaction, using transient bonds from functional groups on the enzyme With substrate directly (activation) With a group from the substrate (transfer) Binding energy ( G B ), using many weak non-covalent interactions (same as in protein structure) to Trap correct substrates specifically in the active site Stabilize the transition state(s) through which substate(s) may pass on their way to becoming product(s) This is one reason enzymes must be so large Chapter 6 15 The Value of 5.7 kj/mole The Value of 5.7 kj/mole Applies to: A 10-fold difference in K eq (the equilibrium constant) A 10-fold difference in k (the enzyme rate constant) A 10-fold difference in K d (the binding constant) - Binding Lock & Key Hypothesis Lock & Key postulated that substrate(s) fit a perfectly matched active site on the enzyme seems reasonable, yes? Amount of energy contributed by enzyme: 5 x 5.7 = 28.5 kj/mole 11 x 5.7 = 62.7 kj/mole 17 x 5.7 = 96.9 kj/mole Learn this well! Chapter 6 14 Why not? Enzyme here is Stick Breakase Chapter 6 16

5 - Binding Induced Fit Hypothesis Induced Fit postulated that once substrate was in place, the enzyme changed conformation to bind it better and stabilize the transition state Chapter 6 Stick Breakase 17 Specific Mechanisms Enzymes increase reactions rates by several means: Entropy reduction in Enzyme-substrate complex (ES Complex) by orienting the substrates for reaction General acid-base catalysis Covalent catalysis Metal ion catalysis Chapter 6 19 Weak Interactions in ES Complex Multiple weak (non-covalent) interactions between E and S contribute ( G B )to lowering the G for catalysis and also explain substrate specificity. Both E and S are altered in the ES complex. Neither partner is as it was before the combination. The free energy released by the binding of the substrate to the enzyme partially offsets the energy required to reach the top of the energy hill The summation of the unfavorable activation energy (+) and the favorable binding energy (-) yields an overall lower activation energy: G cat = G uncat + G B Chapter 6 18 Entropy Reduction Entropy Reduction occurs when the enzyme places a large restriction on the relative motions of the substrates (usually through binding in the active site) Binding within an enzyme allows for proper orientation of the substrate(s) and an increase in the number of productive collisions Chapter 6 20

6 Many biochemical reactions result in the formation of unstable charged intermediates These intermediates can be stabilized by the transfer of protons to or from the substrate General Acid-Base Catalysis is transfer of protons mediated by molecules other than water General Acid-Base Catalysis A nice range to choose from! R-Group pk a 4.3, , Chapter 6 21 > Metal Ion Catalysis In Metal Ion Catalysis a metal ion participates in the catalysis of the reaction between enzyme and substrates There are several ways this can happen: Ionic interactions for substrate orientation Mediation of redox reactions Metalloenzymes are a class of enzymes that contain one or more metal ions within the holoenzyme Other enzymes may utilize a coenzyme that contains a metal Metal Ions include Fe 2+, Fe 3+, Cu 2+, Zn 2+, Mn 2+, and Co 2+ (Metalloenzymes); and Na +, K +, Mg 2+, and Ca 2+ (Coenzymes) Cytochrome c Oxidase Chapter 6 23 Covalent Catalysis In Covalent Catalysis a transient covalent bond is formed between the enzyme and the substrate This formation results in the alteration of the reaction pathway Catalysis will only occur if this altered pathway has a lower activation energy A number of AAs (D, H, C, K and S) and cofactors can serve as nucleophiles in the formation of covalent bonds with substrates Chapter 6 22 Kinetics A Brief Review Reaction Rate: The change in the concentration of a reactant or a product with time (M/s). Reactant Products aa bb [A] Rate = Rate = t Negative sign for the Reactants Positive sign for the Products [B] t Chapter 6 24

7 Kinetics A Brief Review The best was to describe how concentration of reactants affects rates is to use a Rate Law This equation shows the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers. aa + bb cc + dd [A] Rate = = k t X [ A] [ B] x and y are NOT the stoichiometric coefficients. These superscripts are dependent on the Reaction Order! k = the rate constant Chapter 6 25 Y The Enzyme Catalyzed Reaction In enzymatically catalyzed reactions, an enzyme reacts with a substrate in some manner, producing a product: E + S P Before product can be produced, the enzyme-substrate complex (ES) must be formed. This complex only forms when a specific substrate fits into the active site of the enzyme: E + S ES P Chapter 6 27 Kinetics A Brief Review When a reaction is first order the rate is proportional to the concentration of the substrate. S P [ P] Rate = Velocity (v) = = k t [ S] When a reaction is second order the rate is proportional to the concentration of the both substrates. S 1 + S 2 P [ P] 1 Rate = Velocity (v) = = k S 1 S t [ ] [ ] 1 Chapter What happens when you add substrate to an enzyme? Substrate is converted to product, over time How to measure rate when the [S] is always changing? Measure the initial rate (V o ) when the [S] is much greater than the concentration of enzyme With 2X substrate, get 2X product in same time, thus have 2X V o product 2 x V o V o Apparently V o = k[s] time Fixed ample amt of enz [E] Chapter 6 28

8 At higher and higher [S] Reaction Order When the [S] >> [E], every enzyme binds a molecule of substrate and the enzyme is saturated This situation makes the concentration of the substrate essentially constant and the reaction becomes zero order with respect to the substrate Overall the reaction is a pseudo first order reaction: E + S P V = k S max [ ] 0 Chapter 6 29 Chapter 6 31 Does V o = k [S] hold for all values of [S]? Why not what happens if you put in tons of substrate? Output doesn t increase because enzyme is already working as fast as it can saturation! Note different axes Fixed amount of enzyme V o reaches a maximum (V max ) when enzyme becomes saturated Chapter 6 30 Recall the importance of the enzyme-substrate complex in catalysis An aggregate set of weak interactions lowers the activation energy of a reaction Increasing the rate 10-fold for every 5.7 kj/mole that it is lowered The key to understanding the kinetics of enzyme action also lies in focusing on the enzymesubstrate complex, ES, whose rate of formation will depend on the concentrations of both [E] and [S] Chapter 6 32

9 Michaelis-Menten Kinetic Analysis Michaelis-Menten kinetic analysis allows inferences about the mechanism of enzyme action and certain enzyme-substrate / enzyme-product dissociation constants from measuring rates of the catalyzed reaction under varying conditions. Chapter 6 33 E + S Other considerations k 1 k -1 ES Total enzyme present must be the sum of free enzyme and that bound to substrate: E T = E + ES Thus, under steady state conditions where the enzyme is saturated with substrate, the amount of free enzyme is near zero, and E T ES, so that V o = V max = k 2 [ES] = k 2 [E T ] Chapter 6 35 k 2 E + P k -2 Rate Limiting Step The one-substrate reaction E + S k 1 k -1 ES Assumptions: 1. Formation of ES is fast, compared to the 2. Decay of ES to E + P, which is slower (rate limiting) 3. The final constant k -2 is negligible E + P Chapter 6 34 k 2 k -2 With one further assumption That at steady-state, the concentration of the ES intermediate remains constant Which means that reaction rate V o is the rate of product formation, given by k 2 [ES] Very difficult to measure [ES] so need to get at it from the other measurable reaction components ([E T ] and [S]) You should work through the simple algebra of equations 6-10 through 6-20 in the text, so that you understand the derivation Chapter 6 36

10 and a definition of the Michaelis constant (K m ): K m = ( k + k ) 2 k 1 1 Breakdown of ES sec -1 sec -1 M -1 = molarity Formation of ES Chapter 6 37 A Test [S] is much less than K m Condition A [S] << K m So only a small percentage of the overall enzyme is present in a combined form with the substrate (ES) Under these conditions: Vmax V = 0 K [ S] m The initial velocity will depend directly on the [S], making it first order with respect to S Chapter 6 39 A The Michaelis-Menten Equation Derived by Leonor Michaelis and Maud Menten to quantitate the relationship between substrate concentration and reaction rate, immortalized in their eponymous equation: k 2 V = 0 K [ E ][ S] T + [ S] Note: When [S] = K m, then M V V = = Vmax = K + [ S] [ S] Chapter 6 38 m [ S] 1/2 max [ S] max V 0 K + m A Test [S] is much more than K m Condition B [S] >> K m Enzyme is fully saturated and is completely in the ES complex. At this point, the rate is independent of the [S] so the kinetics are zero order. [ S] V V = [S] max = V 0 max B Chapter 6 40

11 A Test [S] is equal to K m Condition C [S] = K m So only a small percentage of the overall enzyme is present in a combined form with the substrate (ES) Under these conditions: 1 V = K The Lineweaver-Burk equation m V V 0 max max y = m x + b 1 [S] + 1 [ S] V 1 V = [S] + [S] 2 max = V 0 max C This equation is very useful for quantitation! Chapter 6 41 Chapter 6 43 Problem with Data When running a kinetics experiment, you cannot always read V max from the hyperbolic plot. Chapter 6 42 So, what is the K m really about? Remember pk a? Remember K d? Same idea Although it is formally a ratio of rate constants, it can be though of as. The [S] at which half of the enzyme is in [ES], and half is free [E], so rate is half-maximal Free Enzyme K m Bound Enzyme Chapter 6 44

12 Why Do the K m s of Enzymes Vary So Much? Comparison of Turnover Numbers What should the K m be matched to in vivo? What do we REALLY care about? Chapter 6 45 Look at the difference from highest to lowest (10 9!) So k cat will give us the best way to compare enzymes right?? Chapter 6 47 The Turnover Number (k cat ) Dividing V max by the enzyme concentration [E T ] [ES] tells the amount of product made when the enzyme is working full blast, giving the catalytic rate constant, k cat (aka. the turnover number) V k = cat [ E ] In other words, k cat tells us how many molecules of product are made per enzyme molecule per second Think of k cat as the molecular rate of the reaction Chapter 6 46 max T What about the catalytic efficiency? Recall that at saturation the turnover number times the number of enzyme molecules present gives: k cat [E T ]= V max Which can substitute in the M-M Equation, yielding: k cat V = 0 K [ E ][ S] T + [ S] k E S cat T Thus when [S] << K m : V = 0 K m k cat /K m is termed the Specificity Constant (a 2 nd order rate constant!) Chapter 6 48 m [ ][ ]

13 K m, k cat and Catalytic Efficiency K M provides a measure of the affinity of the enzyme for its substrate: The higher is K M, the higher is the concentration of the substrate required to reach ½ V max That is, at a lower affinity, a higher concentration of substrate is required to reach ½ V max Turnover number or catalytic constant (k cat ): Maximum # molecules of substrate converted per unit time when the enzyme is saturated Turnover # (per second) of 1 to 1000 is common Different enzymes may be compared in their catalytic efficiencies: the ratio k cat /K m combines information about product/sec produced and how well substrate is captured Because this depends on encounters between E and S can get together, there is a diffusion limit to its value, around 10 8 to 10 9 M -1 s -1 In this range lies catalytic perfection How Fast, How Far, How Well? You should now appreciate several important parameters that describe an enzymatic reaction: K m k cat V o V max - the initial rate of the reaction - the maximum rate (saturated) - substrate concentration to get 1/2 V max - the turnover number of the enzyme Chapter 6 49 Chapter 6 51 Comparison of Catalytic Efficiencies Factors that Affect Enzyme Activity What do you think are the three factors that affect an enzyme s activity? It doesn t help to have a low K m, if an enzyme isn t very fast Temperature ph Inhibitors It doesn t help to be really fast, if an enzyme has a very high K m Chapter 6 50 Chapter 6 52

14 Factors that Affect Enzyme Activity Temperature: In general, a 10 C increase in T doubles the rate. There is an upper T limit for each enzyme where it will denature (its T m ) ph (of [H + ]) Each enzyme has its own range of ph in which it will work. The maximum is its ph optimum Chapter 6 53 Types of Enzyme Inhibition There are two basic types of inhibition: Reversible or Irreversible Reversible inhibition Inhibitors that can reversibly bind and dissociate from enzyme Activity of enzyme recovers when inhibitor diluted out Usually non-covalent interaction Competitive Uncompetitive Mixed (non-competitive) Irreversible inhibition Inactivators that irreversibly associate with enzyme Activity of enzyme does not recover with dilution Usually covalent interaction Examples include suicide inhibitors (mechanism-based) Chapter 6 55 Types of Enzyme Inhibition Enzyme Inhibitors (I) are molecules that interfere with catalysis by slowing or halting enzymatic reactions Inhibitors may affect the K m and/or V max Inhibition can be reversible or irreversible The Inhibition Constant (K I ; analogous to K m ) is the concentration of inhibitor at which it occupies half of the sites on the enzyme It is conceptually equivalent to the binding constant of the enzyme, or enzyme-substrate complex, for the inhibitor The type of (reversible) inhibition can often be discerned from double-reciprocal plots where [I] and [S] have both been varied Chapter 6 54 Reversible Inhibition: Competitive K app M = αk M ; K app M > K M α = 1 + [I]/K I V app max = V max Inhibitor binds to the same site as substrate, but can be competed away by adding more substrate, so V max is not affected Apparent K m will increase with Inhibitor concentration Why do you think K m is affected? Chapter 6 56

15 A Substrate and Its Competitive Inhibitor Reversible Inhibition: Non-competitive Inhibitor binds at a non-overlapping site, blocking either E or ES; K i is fixed, and if it is the same for both (α = α ) V max is decreased, but K m does not change as though a certain amount of E was simply out of the picture K app M = αk M / α ; K app M > K M V app max = V max / α ; V app max < V max α = 1 + [I]/K I α = 1 + [I]/K I If α = α Chapter 6 57 Why do you think V max is affected? Chapter 6 59 Reversible Inhibition: Uncompetitive K app M = αk M ; K app M < K M Inhibitor binds only to ES complex so increased [S] cannot overcome it, and since [I] removes some fraction of ES from participation: V max is decreased K m is proportionally decreased Let s look at the detailed reaction mechanism for chymotrypsin, a protease: It hydrolyzes peptide bonds immediately C-terminal to aromatic amino acids (W, F and Y) With a rate enhancement of > 10 9 And illustrates several principles: Transition state stabilization Acid-base catalysis Covalent catalysis A two-phase reaction V app max = V max / α; V app max < V max Chapter 6 58 Leu-Ala-Tyr-Ile-Asp Chapter 6 60

16 Chymotrypsin The Active Site In the active site of chymotrypsin, know the chemical roles of: Chymotrypsin Mechanism Step 2 Histidine 57 Aspartate 102 Serine 195 Glycine 193 PDB: 7GCH Chapter 6 61 Chapter 6 63 Chymotrypsin Mechanism Step 1 3 Chymotrypsin Mechanism Steps 3 and 4 4 Chapter 6 62 Chapter 6 64

17 Chymotrypsin Mechanism Step 5 Chymotrypsin Mechanism Step 7 Chapter 6 65 Chapter 6 67 Chymotrypsin Mechanism Step 6 Effects of Substrate Structure on the Cleavage Parameters of Chymotrypsin Glycine Alanine Chapter 6 66 What does this mean? Chapter 6 68

18 Chymotrypsin Inhibition Normally, the acyl-enzyme intermediate is short-lived But what if a substrate is such that it is not? Hexokinase Demonstrates Induced Fit H 2 O In this case, if the next step (base catalysis via an incoming water molecule) cannot occur, the reaction becomes stuck, and active enzyme cannot be regenerated, leading to Chapter 6 69 Hexokinase undergoes a dramatic conformational change upon binding glucose. Two lobes of the enzyme come together to surround glucose and exclude water from the active site. The ATP binding site is formed after glucose binds to the enzyme Chapter 6 71 Suicide Inhibition Irreversible! Diisopropylfluorophosphate (DIFP) reacts covalently with the active-site serine of chymotrypsin Every molecule that reacts is inactivated permanently Chymotrypsin Chapter 6 70 But the enzyme can be fooled! Xylose can also cause a similar conformational change Though it does not get phosphorylated, ATP hydrolysis is stimulated With phosphoryl group transfer to water! H-O-H Chapter 6 72

19 A Key to Understanding Enzyme Regulation Lies in Such Conformational Changes As with both hemoglobin and hexokinase, changes in structure can have profound changes on the function or activity of a protein molecule Conformation changes can occur in response to interactions with other molecules These other molecules may be: Substrate, substrate analogs, or pathway end-products Allosteric modulators Covalently added groups Regulatory subunits or accessory proteins Proteolytic enzymes Chapter 6 73 Enzyme Regulation Some Classes of Enzyme Regulation: Feedback-inhibited enzymes: excessive buildup of an end product in a pathway inhibits the activity of an early enzyme in the pathway Covalently modified enzymes: reversible attach-ment of small molecules alter their activity (by phosphorylation, methylation, ribosylation, etc.) Allosteric enzymes: incur conformational changes via reversible, non-covalent binding of small metabolites or co-factors, or regulatory subunits Zymogens and proenzymes: become activated by proteolytic cleavage of inactive precursor molecules Note: these classes are not mutually exclusive! Chapter 6 75 Why Regulate Enzyme Activity? Metabolic pathways rarely stand alone, and usually intersect with numerous other pathways In enzyme pathways of several steps there are often one (or more) key enzymes that are ratelimiting in controlling the flux through the pathway They can respond to a variety of exogenous signals, which in turn can increase or decrease their catalytic activity, Thus integrating expression of the pathway with other metabolic needs of the cell Chapter 6 74 Enzyme Regulation Allosteric Activation or Inhibition Allosteric modulators may bind directly to an enzyme (e.g. threonine dehydratase), or indirectly to a regulatory subunit Which in turn activates (or inhibits) the catalytic subunit In some cases, allosteric effects are mediated by covalent modification of the enzyme Chapter 6 76

20 Enzyme Regulation Allosteric Enzymes Diverge from Michaelis-Menten Kinetics A sigmoid rather than hyperbolic curve for a plot of Positive cooperativity V o vs. [S] suggests allostery, or cooperative interactions among subunits Modulating activators and/or inhibitors can cause major shifts in these curves Negative cooperativity No mechanism is implied shifts may be in response to noncovalent binding or to covalent attachment of a modifying group Normal Chapter 6 77 Enzyme Regulation Enzymes can be reversibly covalently modified to change their characteristics Chapter 6 79 Enzyme Regulation Feedback Inhibition Feedback inhibition occurs when a late or ultimate product of a multi-step pathway inhibits an early enzyme in the pathway (usually at a rate-limiting step). Here, inhibition by isoleucine is: Allosteric (at a site on E 1 remote from the active site) Specific, in that (1) Only isoleucine inhibits E 1 none of the other intermediates do, and (2) Isoleucine inhibits only E 1 not any of the other enzymes (E 2 E 5 ) Chapter 6 78 Enzyme Regulation Reversible Covalent Modification of Glycogen Phosphorylase Interconversion between a and b forms of the enzyme adds or removes phosphate from a serine residue to control its activity The catalytic activity correlates with differences in their secondary, tertiary, and quaternary structures The kinase and phosphatase that carry out these modifications respond in turn to the metabolic demands of the cell as part of a complex cascade pathway that mobilizes the energy stored in glycogen as it is needed. Chapter 6 80

21 Enzyme Regulation Cleaving a Precursor Can Change Its Shape and Activity, with Preprohormones Insulin begins life as a single poly-peptide chain that is inactive Its signal sequence targets it to secretory vesicles, where 3 disulfides pin it into its near-final (but still inactive) shape Upon secretion, proteolytic removal of the C-peptide converts it into active hormone Chapter 6 81 Enzyme Regulation Zymogens Chymotrypsinogen One Polypeptide to Three! Zymogens, also called proenzymes, are inactive precursors of enzymes Cleavage in the correct cellular compartment causes conformational changes that expose the active site Such changes are irreversible Chapter 6 82

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