BCH3033 Biochemistry 1 Chapter 6b PDF

BCH3033: Biochemistry 1 Chapter 6b 02.21.2024 Donella Beckwith, Ph.D. [email protected] 1 Enzyme Models Let’s introduce 3 important concepts before talking about induced fit 2 Enzyme Specificity specificity = ability to discriminate between a substrate and a competing molecule maximize weak interact...

BCH3033: Biochemistry 1 Chapter 6b 02.21.2024 Donella Beckwith, Ph.D. [email protected] 1 Enzyme Models Let’s introduce 3 important concepts before talking about induced fit 2 Enzyme Specificity specificity = ability to discriminate between a substrate and a competing molecule maximize weak interactions Substrate must be distorted slightly in order to bind to the enzyme, causing an unfavorable free-energy change but binding energy compensates thermodynamically for this 3 Rate Enhancement by Entropy Reduction entropy reduction = large restriction in the relative motions of two substrates that are to react ester anhydride carboxylate binding energy constrains substrates in the proper orientation to reaction What are the rate constants? What happens to entropy? 4 Desolvation of the Substrate desolvation = replacement of the solvation shell of structured water around the substrate with weak bonds between substrate and enzyme – replaces most or all hydrogen bonds between the substrate and H2O 5 Question What is the primary source of the energy enzymes used to reduce activation energies? A. desolvation of the substrate B. kinetic energy from the transition state C. noncovalent enzyme-substrate interaction D. covalent enzyme-product interaction 6 The Induced Fix Mechanism induced fit = mechanism by which the enzyme undergoes a conformational change when the substrate binds, induced by multiple weak interactions with the substrate – enhances catalytic properties – the active site of an enzyme gets continually reformed based on the interactions that it establishes with the substrate Example: Hexokinase: transfers phosphate from ATP to glucose (forms glucose 6-phosphate in glucose metabolism) glucose + adenosine triphosphate 7 Induced Fix: Hexokinase ATP + glucose 1) Initially, the enzyme has a conformation that attracts its substrate 2) Only the correct catalyst can induce interaction 3) Conformational changes MAY occur as the substrate is bound 4) Catalysis occurs 5) Lastly, the reaction products will move away from the enzyme and the active site returns to the initial shape 8 Question How does the induced fit mechanism of enzyme catalysis work? A. The enzyme assumes a conformation identical to the substrate. B. The enzyme undergoes a conformational change to maximize weak interactions to the substrate. C. The substrate binds the active site of the enzyme. D. The enzyme undergoes entropy reduction to accommodate substrate. 9 Question How does the concept of “induced fit” support the current theory of substrate-enzyme interaction? A. Enzyme structure is rigid, allowing for competition for the active site and easy ejection of the final product. B. Enzyme conformational changes demonstrate feedback regulation of the enzyme and the ability to “turn off” activity. C. Having a rigid structure allows the enzyme to bind the substrate and stabilize the ES complex. D. Enzyme conformation changes allow additional stabilizing interactions with the transition state and enhance the activity. 10 Question Which statement about enzymes is false? A. They lower the activation energy of a reaction. B. They alter the overall thermodynamics of a reaction. C. A substantial amount of their catalytic power results from binding of the substrate(s) through weak interactions. D. They increase the rates of reactions by 105-fold to 1017-fold. 11 Covalent Interactions and Metal Ions Contribute to Catalysis catalytic functional groups aid in the cleavage and formation of bonds by a variety of mechanisms: – general acid-base catalysis – covalent catalysis – metal ion catalysis Really pay attention to the next 2 slides 12 General Acid-Base Catalysis protons are transferred between an enzyme and a substrate or intermediate specific acid-base catalysis = uses only the H+ (H3O+) or OH– ions present in water general acid-base catalysis = mediated by weak acids or bases other than water 13 Amino Acids in General Acid-Base Catalysis found in the vast majority of enzymes Proton donors (acids) Proton acceptors (bases) These amino acids are found in enzyme active sites to assist in the catalytic process as proton donors/acceptors 14 Question Which amino acid might be expected to have the LEAST effect on the function of an enzyme if it replaces a Glu residue in the enzyme? A. Gly B. Lys C. His D. Asp 15 Covalent Catalysis covalent catalysis = transient covalent bond forms between the enzyme and the substrate – catalysis only results when the new pathway has a lower activation energy than the uncatalyzed pathway – all new steps must be faster than the uncatalyzed reaction hydrolysis of a bond: hydrolysis of a bond between A and B in the presence of a covalent catalyst: H2O A ⎯ B ⎯⎯⎯ →A + B H2O A ⎯ B + X: ⎯⎯ → A ⎯ X + B ⎯⎯⎯ → A + X: + B 16 Metal Ion Catalysis metals: – help orient the substrate for reaction – stabilize charged reaction transition states – mediate oxidation-reduction reactions by reversible changes in the metal ion’s oxidation state nearly 1/3 of all known enzymes require 1+ metal ions for catalytic activity Carbonic anhydrase (metalloenzyme) 17 Question Which statement is NOT associated with covalent catalysis by enzymes? A. When the reaction is complete, the enzyme returns to its original state. B. It never involves coenzymes. C. A transient covalent bond is formed between the enzyme and the substrate. D. A new pathway from substrate(s) to product(s) is formed that is faster than the uncatalyzed reaction. 18 Enzyme Kinetics enzyme kinetics = the discipline focused on determining the rate of a reaction and how it changes in response to changes in experimental parameters 19 Substrate Concentration Affects the Rate of Enzyme-Catalyzed Reactions = initial transient period during which ES builds up to a constant level = period during which [ES] and other intermediates remain constant steady-state kinetics = the traditional analysis of reaction rates 20 Initial Velocities of EnzymeCatalyzed Reactions initial rate (initial velocity), V0 = tangent to each curve taken at time = 0 – reflects a at the beginning of the reaction, [S] is regarded as constant [enzyme] present in nanomolar quantities compared to [S] that is 5-6 orders of magnitude higher rate of rxn declines as substrate converted to product 21 Question After you mix a substrate and enzyme together, there is an initial transient period, the pre–steady state. What happens during the pre–steady state? A. [Et] decreases until it reaches a constant level. B. [S] decreases until it reaches a constant level. C. [P] increases until it reaches a constant level. D. [ES] increases until it reaches a constant level. 22 General Theory of Enzyme Action Proposed by Michaelis and Menten step 1 – enzyme and substrate combine to form a complex in a reversible, relatively fast step: (6-7) step 2 – the complex breaks down to yield the free enzyme and the reaction product in a slower step: (6-8) because the second step limits the overall reaction rate, the overall rate is proportional to [ES] 23 The Saturation Effect Vmax (maximum velocity) is observed when virtually all the enzyme is present as the ES complex – further increases in [S] have no effect on rate – responsible for the plateau observed Increase in V0 is vanishing [S] very high 24 Question Which statement is false about the saturation effect? A. The maximum initial rate of the catalyzed reaction is observed. B. Virtually all of the enzyme is present as the ES complex. C. It is responsible for the linear increase of V0 with an increase in [S]. D. Further increases in [S] have no effect on rate. 25 The Relationship between Substrate Concentration and Reaction Rate Can Be Expressed with the Michaelis-Menten Equation the curve expressing the relationship between [S] and V0 can be expressed by the Michaelis-Menten equation: V0 = Vmax [S] (6-9) Km + [S] where V0 is the initial velocity, Vmax is the maximum velocity, [S] is the initial substrate concentration, and Km is a constant called the Michaelis constant the Michaelis-Menten equation is the rate equation for a one-substrate enzyme-catalyzed reaction V0, Vmax, [S], and Km are readily measured experimentally 26 Deriving the Michaelis-Menten Equation the overall reaction for the formation and breakdown of ES simplifies to (6-10) k-2 can be ignored because there is so little [P] early in the rxn V0 is determined by the breakdown of ES to form product, which is determined by [ES]: V0 = k2[ES] (6-11) 27 Deriving the Michaelis-Menten Equation, [Et] [ES] is not easily measured experimentally [Et] = the total enzyme concentration (the sum of free and substrate-bound enzyme) free or unbound enzyme [E] = [Et] – [ES] because [S] >> [Et], the amount of substrate bound by the enzyme at any given time is negligible compared with the total [S] [enzyme] is present in nanomolar quantities compared to [S] that is 5-6 orders of magnitude higher 28 Question For a specific enzyme/substrate system, which condition will ALWAYS result in an increase in V0 (assuming nothing else changes)? Vmax [S] V0 = Km + [S] A. a decrease in Km and an increase in Vmax B. an increase in pH C. an increase in [S] D. a decrease in [Et] 29 Deriving the Michaelis-Menten Equation steady-state assumption = the rate of formation of ES is equal to the rate of its breakdown the Michaelis constant, Km = (k–1 + k2)/ k1 expressing V0 in terms of [ES]: V0 = k2[ES] (6-11) k2[Et][S] V0 = Km + [S] (6-20) given Vmax occurs when the enzyme is saturated (when [ES] = [Et]), Vmax = k2[Et]: V0 = Vmax[S] Km + [S] (6-9) 30 Km = [S], When V0 = ½Vmax when V0 = ½Vmax: Vmax Vmax[S] = 2 Km + [S] (6-21) 1 [S] = 2 Km + [S] (6-22) solving for Km: Km = [S] when V0 = ½Vmax (6-23) 31 Dependence of V0 on [S] [S] very low Km >> [S] Shape of curve defined by Vmax/Km [S] very high [S] >> Km Shape of curve defined by Vmax 32 Question In deriving the Michaelis-Menten equation, what is the steadystate assumption? A. [P] is negligible and k–2 can be ignored. B. The rate of formation of ES is equal to the rate of its breakdown. C. The amount of substrate bound by the enzyme at any given time is negligible compared with the total [S]. D. The reaction is a one-substrate enzyme-catalyzed reaction. 33 Question What is the definition of Km, the Michaelis constant? A. the rate at which enzyme is turning substrate into product at ½Vmax B. the V0 at ½ maximal [S] C. the number of molecules of substrate converted to product per second D. the concentration of substrate at which the enzyme is operating at half its maximal velocity 34 Michaelis-Menten Kinetics Can Be Analyzed Quantitatively an algebraic transformation of the Michaelis-Menten equation converts the hyperbolic curve into a linear form: Vmax[S] V0 = Km + [S] 1 Km + [S] = V0 Vmax[S] (6-24) 35 Deriving the Lineweaver-Burk Equation simplifying this equation gives the Lineweaver-Burk equation: 1 Km + [S] = V0 Vmax[S] (6-24) 1 Km [S] = + V0 Vmax[S] Vmax[S] (6-25) 1 Km 1 = + V0 Vmax[S] Vmax (6-26) 36 A Double-Reciprocal, or LineweaverBurk, Plot for enzymes obeying the Michaelis-Menten relationship, a plot of 1/V0 versus 1/[S] yields a straight line 1 Km 1 = + V0 Vmax[S] Vmax 37 Question 25 The Km on a Lineweaver-Burk plot is: A. equal to the y intercept. B. equal to the slope of the line. C. the reciprocal of the y intercept. D. the negative reciprocal of the x intercept. 38 Interpreting Km and Vmax Km can vary for different substrates of the same enzyme Table 6-6 Km for Some Enzymes and Substrates Enzyme Substrate Km (mM) Hexokinase (brain) ATP D-Glucose D-Fructose 0.4 0.05 1.5 Carbonic anhydrase HCO3– 26 Chymotrypsin Glycyltyrosinylglycine N-Benzoyltyrosinamide 108 2.5 β-Galactosidase D-Lactose 4.0 Threonine dehydratase L-Threonine 5.0 39 The Dissociation Constant, Kd for reactions with two steps: k2 + k–1 Km = k1 (6-27) when k2 is rate-limiting, k2