Enzymes Fall 2007 Lecture 2 Downloaded from

Enzymes High molecular weight proteins 15,000 < MW <4,000,000 Catalysts Nomenclature - end with “ase” Holoenzyme - enzyme containing a non-protein group like a metal –(apoenzyme + cofactor) –(protein + metal) Isoenzymes - catalyze same rxn Classified based on type of rxn catalyzed (Table 3.1) Downloaded from

Examples (enzyme - use - source) Trypsin - anti-inflammatory, meat tenderizers - animal pancreas amylase - syrup, glucose production - Bacillus subtilis protease - detergents, silver recovery - B. subilis invertase - confectionaries- Sacharomyces cerevisiae cellulase - breaks down cellulose - bacteria/yeast/mold penicillinase - remove penicillin from allergic individuals - bacteria Downloaded from

Enzyme Function Lower activation energy of a reaction by binding to the substrate and forming a substrate-enzyme complex Interaction is due to van der Waals and H-bonding at the active site Interaction is complex - studied via Raman Spectroscopy and X-ray Enzyme does not affect equilibrium constant or free energy change Downloaded from

Lowers the activation energy of a reaction - highly specific Downloaded from

Lock and Key Mechanism Specific binding site Downloaded from

Enzyme Kinetics S P r = v = dP/dt = k 1 (S) E S + E P + E r = v = dP/dt = k 1 (S)(E) First order Zero order Downloaded from

Enzyme Kinetics Michaelis Menten Approach (Henri) S + E ES > E + P Based on experimental data k1k1 k2k2 k3k3 Downloaded from

Rapid Equilibrium MM k 1 (E)(S) = k 2 (ES) or(1) Rate Equation(2) substituting for (ES) from Eq 1 into Eq 2 Downloaded from

Using the total enzyme balance E 0 = E + ES or substituting into the rate equation Downloaded from

we obtain K M = k 2 /k 1 here is a dissociation constant, it characterizes the interaction of an enzyme with a given substrate S= K M when v = ½ v max v max = k 3 E 0 - maximum reaction rate, proportional to the initial enzyme concentration Downloaded from

V max and K M V max KMKM Figure 3.3 in book Downloaded from

In many cases the assumption of rapid equilibrium is not valid although the enzyme-substrate reaction still shows saturation type kinetics. In Class Exercise S + E ES E + P Assume rapid equilibrium and determine the rate expression for product formation k2k2 k1k1 k4k4 k3k3 Downloaded from

Quasi Steady State Approach Briggs and Haldane – another mathematical approach to the observed experimental MM eqn S + E ES > E + P Assume that the change in (ES) with time is very small compared with to changes in S or P Rate Equation of (ES) k2k2 k1k1 k3k3 Downloaded from

Assuming quasi steady state d(ES)/dt = 0 Solve equation for (ES) Rate Equation for P is substituting for (ES) Using the total enzyme balance E 0 = E + ES Hint: Solve SS eqn for whatever quantity set equal to zero Downloaded from

Or Substituting into the rate equation for E Rearranging results in the following: Hint: Do not multiply through k values Downloaded from

The equation has the same form as the MM eqn Where v max = k 3 E 0 and OrFigure 3.4 In Class Exercise S + E ES E + P Assume rapid quasi steady state and determine the rate expression for product formation k1k1 k2k2 k4k4 k3k3 Downloaded from