Introduction to enzymes, enzyme catalyzed reactions and CP504 – Lecture 3 Enzyme kinetics and associated reactor design: Introduction to enzymes, enzyme catalyzed reactions and simple enzyme kinetics learn about enzymes learn about enzyme catalyzed reactions study the kinetics of simple enzyme catalyzed reactions Prof. R. Shanthini 23 Sept 2011

What is an Enzyme? Enzymes are mostly proteins, and hence they consists of amino acids. Enzymes are present in all living cells, where they help converting nutrients into energy and fresh cell material. Enzymes breakdown of food materials into simpler compounds. Examples: - pepsin, trypsin and peptidases break down proteins into amino acids - lipases split fats into glycerol and fatty acids - amylases break down starch into simple sugars Prof. R. Shanthini 23 Sept 2011

What is an Enzyme? Enzymes are very efficient (biological) catalysts. Enzyme catalytic function is very specific and effective. Enzymes bind temporarily to one or more of the reactants of the reaction they catalyze. By that means, they lower the amount of activation energy needed and thus speed up the reaction. Enzymes does not get consumed in the reaction that it catalyses. Prof. R. Shanthini 23 Sept 2011

Enzyme classification Oxidoreductase: transfer oxygen atoms or electron Transferase: transfer a group (amine, phosphate, aldehyde, oxo, sulphur, etc) Hydrolase: hydrolysis Lyase: transfer non-hydrolytic group from substrate Isomerase: isomerazion reactions Ligase: bonds synthesis, using energy from ATPs Prof. R. Shanthini 23 Sept 2011

Examples of enzyme catalyzed reactions Examples of Enzyme Catalysed Reactions Example 1: Carbonic anhydrase CO2+ H2O H2CO3 Carbonic anhydrase is found in red blood cells. It catalyzes the above reaction enabling red blood cells to transport carbon dioxide from the tissues (high CO2) to the lungs (low CO2). One molecule of carbonic anhydrase can process millions of molecules of CO2 per second. Prof. R. Shanthini 23 Sept 2011

Examples of enzyme catalyzed reactions Catalase 2H2O2 2H2O + O2 Catalase is found abundantly in the liver and in the red blood cells. One molecule of catalase can breakdown millions of molecules of hydrogen peroxide per second. Hydrogen peroxide is a by-product of many normal metabolic processes. It is a powerful oxidizing agent and is potentially damaging to cells which must be quickly converted into less dangerous substances. Prof. R. Shanthini 23 Sept 2011

Industrial use of catalase - in the food industry for removing hydrogen peroxide from milk prior to cheese production - in food-wrappers to prevent food from oxidizing - in the textile industry to remove hydrogen peroxide from fabrics to make sure the material is peroxide-free - to decompose the hydrogen peroxide which is used (in some cases) to disinfect the contact lens Prof. R. Shanthini 23 Sept 2011

Examples of Industrial Enzymes See the hand out on the same topic Prof. R. Shanthini 23 Sept 2011

More on enzymes Enzymes are very specific. Absolute specificity - the enzyme will catalyze only one reaction Group specificity - the enzyme will act only on molecules that have specific functional groups, such as amino, phosphate or methyl groups Linkage specificity - the enzyme will act on a particular type of chemical bond regardless of the rest of the molecular structure Stereochemical specificity - the enzyme will act on a particular steric or optical isomer Prof. R. Shanthini 23 Sept 2011

Source: http://waynesword.palomar.edu/molecu1.htm Prof. R. Shanthini 23 Sept 2011 Source: http://waynesword.palomar.edu/molecu1.htm

E + S ES Source: http://waynesword.palomar.edu/molecu1.htm Prof. R. Shanthini 23 Sept 2011 Source: http://waynesword.palomar.edu/molecu1.htm

Lock & Key Theory Of Enzyme Specificity (postulated in 1894 by Emil Fischer) E + S ES E + P Prof. R. Shanthini 23 Sept 2011 Source: http://waynesword.palomar.edu/molecu1.htm

Prof. R. Shanthini 23 Sept 2011

Active Site Of Enzyme Blocked By Poison Molecule Prof. R. Shanthini 23 Sept 2011 Source: http://waynesword.palomar.edu/molecu1.htm

Induced Fit Model E + S ES E + P (postulated in 1958 by Daniel Koshland ) E + S ES E + P Binding of the first substrate induces a conformational shift that helps binding of the second substrate with far lower energy than otherwise required. When catalysis is complete, the product is released, and the enzyme returns to its uninduced state. Prof. R. Shanthini 23 Sept 2011 Source: http://www.mun.ca/biology/scarr/Induced-Fit_Model.html

Simple Enzyme Kinetics E + S ES E + P k2 which is equivalent to [E] S P S for substrate (reactant) E for enzyme ES for enzyme-substrate complex P for product Prof. R. Shanthini 23 Sept 2011

Michaelis-Menten approach to the rate equation: k1 k3 E + S ES E + P k2 Assumptions: Product releasing step is slower and it determines the reaction rate 2. ES forming reaction is at equilibrium 3. Conservation of mass (CE0 = CE + CES) Initial concentration of E Concentration of E at time t Concentration of ES at time t Prof. R. Shanthini 23 Sept 2011

Michaelis-Menten approach to the rate equation: k1 k3 E + S ES E + P k2 Product formation (= substrate utilization) rate: rP = - rS = k3 CES (1) Since ES forming reaction is at equilibrium, we get k1 CE CS = k2 CES (2) Prof. R. Shanthini 23 Sept 2011

Michaelis-Menten approach to the rate equation: k1 k3 E + S ES E + P k2 Using CE0 = CE + CES in (2) to eliminate CE, we get k1 (CE0 – CES) CS = k2 CES which is rearranged to give CE0CS CES = (3) k2/k1 + CS Prof. R. Shanthini 23 Sept 2011

Michaelis-Menten approach to the rate equation: k1 k3 E + S ES E + P k2 Using (3) in (1), we get k3CE0CS rmaxCS rP = - rS = = (4) k2/k1 + CS KM + CS where rmax = k3CE0 and KM = k2 / k1 (5) (6) Prof. R. Shanthini 23 Sept 2011

Briggs-Haldane approach to the rate equation: k1 k3 E + S ES E + P k2 Assumptions: 1. Steady-state of the intermediate complex ES 2. Conservation of mass (CE0 = CE + CES) Initial concentration of E Concentration of E at time t Concentration of ES at time t Prof. R. Shanthini 23 Sept 2011

Briggs-Haldane approach to the rate equation: k1 k3 E + S ES E + P k2 Product formation rate: rP = k3 CES (7) Substrate utilization rate: rs = - k1 CECS + k2 CES (8) Since steady-state of the intermediate complex ES is assumed, we get k1 CECS = k2 CES + k3 CES (9) Prof. R. Shanthini 23 Sept 2011

Briggs-Haldane approach to the rate equation: k1 k3 E + S ES E + P k2 Combining (7), (8) and (9), we get rP = - rS = k3 CES (10) Using CE0 = CE + CES in (9) to eliminate CE, we get k1 (CE0 - CES)CS = (k2 + k3)CES which is rearranged to give CE0CS CES = (11) (k2+k3)/k1 + CS Prof. R. Shanthini 23 Sept 2011

Briggs-Haldane approach to the rate equation: k1 k3 E + S ES E + P k2 Combining (10) and (11), we get k3CE0CS rmaxCS rP = - rS = = (12) KM + CS (k2+k3)/k1 +CS where rmax = k3CE0 and KM = (k2 + k3) / k1 (5) (13) When k3 << k2 (i.e. product forming step is slow), KM = k2 / k1 (6) Prof. R. Shanthini 23 Sept 2011

Simple Enzyme Kinetics (in summary) rmaxCS rP = - rS = KM + CS where rmax = k3CE0 and KM = f(rate constants) rmax is proportional to the initial concentration of the enzyme KM is a constant Prof. R. Shanthini 23 Sept 2011