BCH3033 - CH 6 c.pdf

BCH3033: Biochemistry 1 Chapter 6c 02.23.2024 Donella Beckwith, Ph.D. [email protected] 1 The Constant kcat Is a First-Order Rate Constant k [Et][S] V0 = kcat = Vmax/[Et] cat Km + [S] kcat = turnover number = the number of substrate molecules converted to product in a given unit of time on a single...

BCH3033: Biochemistry 1 Chapter 6c 02.23.2024 Donella Beckwith, Ph.D. [email protected] 1 The Constant kcat Is a First-Order Rate Constant k [Et][S] V0 = kcat = Vmax/[Et] cat Km + [S] kcat = turnover number = the number of substrate molecules converted to product in a given unit of time on a single enzyme molecule when the enzyme is saturated Table 6-7 Turnover Number, kcat, of Some Enzymes Enzyme Substrate kcat (s–1) Catalase H2O2 40,000,000 Carbonic anhydrase HCO3– 400,000 Acetylcholinesterase Acetylcholine 14,000 β-Lactamase Benzylpenicillin 2,000 Fumarase Fumarate 800 RecA protein (an ATPase) ATP 0.5 reciprocal time This would mean that each molecule of enzyme is catalyzing the formation of 400,000 molecules of product every second 2 Comparing Catalytic Mechanisms and Efficiencies k2 + k–1 k1 comparing the ratio kcat/KM for two reactions is the best way to compare catalytic efficiencies or turnover specificity constant = the rate constant for the conversion of E + S to E + P when [S] > [I], the reaction exhibits normal Vmax in the presence of inhibitor, the [S] at which V0 = ½Vmax, the apparent Km, increases by the factor α 15 A Lineweaver-Burk Plot Reveals Competitive Inhibition lines intersect at the y axis the observed KM increases in the presence of an inhibitor Km increases as [I] increases 16 Question Which statement is false for a competitive inhibitor? B. It is often structurally similar to substrate. C. It does not change Vmax. D. It is irreversible. 17 Uncompetitive Inhibition uncompetitive inhibitor: – binds at a site distinct from the substrate active site – unlike a competitive inhibitor, binds only to the ES complex – type of reversible inhibition enzyme ES ESI 18 Uncompetitive Inhibitors Alter the Michaelis-Menten Equation the Michaelis-Menten equation becomes Vmax[S] V0 = Km + α′[S] where [I] α′ = 1 + K′I and (6-32) [ES][I] K′I = [ESI] Km α' = Kmapp 19 Uncompetitive Inhibitors Affect Both the Apparent Km and the Vmax the measured Vmax decreases because at high [S], V0 approaches Vmax/α′ apparent Km decreases because the [S] required to reach ½Vmax decreases by the factor α′ 20 A Lineweaver-Burk Plot Reveals Uncompetitive Inhibition lines are parallel Where α’ = 1 there is no inhibitor Vmax decreases as [I] increases Km decreases as [I] increases 21 Question Which statement is false for an uncompetitive inhibitor? A. Uncompetitive inhibitors bind at a site distinct from the substrate active site. B. Uncompetitive inhibitors bind only to the ES complex. C. Uncompetitive inhibitors are reversible. D. Uncompetitive inhibitors affect the apparent Km, but not the Vmax. 22 Mixed Inhibition mixed inhibitor: – binds at a site distinct from the substrate active site – binds to either E or the ES complex – type of reversible inhibition enzyme ES EI ESI 23 Mixed Inhibitors Alter the MichaelisMenten Equation the Michaelis-Menten equation becomes V0 = Vmax[S] αKm + α′[S] (6-33) 24 Mixed Inhibitors Usually Affect Both the Apparent Km and the Vmax Vmax is affected because the effective [E] on which Vmax depends decreases apparent Km may increase or decrease depending on which enzyme form the inhibitor binds most strongly 25 A Lineweaver-Burk Plot Reveals Mixed Inhibition lines intersect to the left of the y axis Vmax decreases as [I] increases Km increases 26 Question Which statement is true about both mixed inhibitors and uncompetitive inhibitors? A. Both inhibitors bind to the active site of an enzyme. B. Both inhibitors affect the Km, but not the Vmax. C. Both inhibitors can bind to the enzyme-substrate complex. D. Both inhibitors can bind to either the free enzyme or the enzyme-substrate complex. 27 Noncompetitive Inhibition noncompetitive inhibitor: – special case of α = α′ – affects the Vmax but not the Km 28 Effects of Reversible Inhibitors on Apparent Vmax and Apparent Km Table 6-9 Effects of Reversible Inhibitors on Apparent Vmax and Apparent Km Inhibitor type Apparent Vmax Apparent Km None Vmax Km Competitive Vmax αKm Uncompetitive Vmax/α′ Km/α′ Mixed Vmax/α′ αKm/α′ 29 Irreversible Inhibition irreversible inhibitor = bind covalently with or destroy a functional group on an enzyme that is essential for the enzyme’s activity, or form a highly stable noncovalent association DIFP is an irreversible inhibitor of Chymotrypsin Chymotrypsin Diisopropylfluorophosphate 30 Suicide Inactivators suicide inactivator = mechanism-based inactivators = undergo the first few steps of normal enzymatic rxn until it is converted into a compound that combines irreversibly with the enzyme – class of irreversible inhibitors 31 Transition-State Analogs Glycolysis transition-state analogs = stable molecules designed to resemble transition states – type of irreversible inhibitors – bind to an enzyme more tightly than does the substrate in the ES complex Binds 10,000 times better to the enzyme than does the product 32 Question Which type of inhibitor binds to the active site of an enzyme and undergoes the initial steps of the reaction before combining irreversibly with the enzyme? A. mixed inhibitors B. suicide inactivators C. transition-state analogs D. competitive inhibitors 33 The Chymotrypsin Mechanism Involves Acylation and Deacylation of a Ser Residue protease = an enzyme that catalyzes the hydrolytic cleavage of peptide bonds – example = bovine pancreatic chymotrypsin Cleaves at Trp, Tyr, and Phe 34 The Chymotrypsin Mechanism Involves Two Phases acylation phase = the peptide bond is cleaved and an ester linkage is formed between the peptide carbonyl carbon and the enzyme deacylation phase = the ester linkage is hydrolyzed and the nonacylated enzyme is regenerated 35 The pH Dependence of Chymotrypsin-Catalyzed Reactions optimal activity at pH 8 – His57 is unprotonated and Ile16 is protonated the transition just above pH 7 is due to changes in kcat – results from protonation of His57 the transition above pH 8.5 is due to changes in 1/Km – results from the ionization of the α-amino group of Ile16 36 The Chymotrypsin Reaction serine proteases = proteases with a Ser residue that acts as the nucleophile – acylation phase nucleophile is the oxygen of Ser195 catalytic triad = hydrogen bonding network – in chymotrypsin, Ser195 is linked to His57 and Asp102 37 Question Which three amino acids form the catalytic triad in chymotrypsin? A. Ser, Asp, Lys B. Glu, Ser, Asp C. Ser, Asp, His D. Phe, Tyr, Ala 38