Enzyme Reactions - Introduction to Biochemistry Laboratory | CHEM 603, Lab Reports of Biochemistry

Download Enzyme Reactions - Introduction to Biochemistry Laboratory | CHEM 603 and more Lab Reports Biochemistry in PDF only on Docsity! Chemistry 603 Introduction to Biochemistry Laboratory LO Fe(II) Fe(II) OH OO O HPP HPP bidentate association Fe(III) OH OO O O2 O O nucleophilic attack Fe(IV) OH OO O O O decarboxylation & heterolytic cleavage Fe(IV) OH O O CO2 O electrophilic attack Fe(II) OH O O O substituent migration Fe(II) OH O O H O tautomerization Fe(II) OH O O OH dissociation HG HG Michaelis – Menten Equation This saturation phenomenon was first accounted for by M&M is 1913. k1 E + S <-> ES k-1 based on the formation of an enzyme substrate complex. Where association and dissociation processes defined a pre-equilibrium before catalysis of the reaction. For simple reactions the equilibrium constant approximates the association constant Ks =[ES]/[E][S] = k1/k-1 Units = M-1 s-1/ s-1 = M-1 The dissociation constant is the inverse of this number k-1 [ES] =[E][S] k1 Forward (second order) Reverse (first order) [ES]/[E][S] = k1 /k-1 so when product is formed we have E + S <-> ES <-> E + P The Steady State Assumption. The interpretation of MM was refined by Briggs and Haldane. B&H said that a given level of ES is formed rapidly (ie equilibrium level is rapidly reached.) They said that ES can decay to either E + S or E + P The steady state assumption states d[ES]/dt = 0 The Initial Velocity Assumption (Vo) Many enzymes catalyze reversible reactions. ie They accelerate the reaction in either direction. It is simpler to assume that there is no back reaction. This assumption would only be reasonable however when the concentration of products is essentially zero. The idea is that if we observe only the initial velocity we will have the best estimate of the forward velocity. Rearranging [ES] = [Et] [S] Km + [S] Now the rate of product formation is v = d[P]/dt and for this reaction v = k2[ES] Substituting for ES into our expression gives v = k2[Et] [S] Km + [S] k2 [Et] has special meaning When [S] is high essentially all E is in the ES form ie [Et] = [ES] Thus the enzyme is saturated with S. At this point v = Vmax Vmax = k2[Et] Substituting this into the expression above v =Vmax [S] Km + [S] This equation says that the rate of an enzyme catalyzed reaction at any time is a function of the two constants Vmax and Km. When [S] = Km, v = Vmax/2 So Km is the concentration of S at which v is half maximal. HOHETUIIUOT) uonenuacu0Z