Prof. R. Shanthini 09 Nov 2012 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 CP504 – ppt_Set 02

Prof. R. Shanthini 09 Nov 2012 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 09 Nov 2012 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. Enzymes does not get consumed in the reaction that it catalyses. What is an Enzyme?

Prof. R. Shanthini 09 Nov 2012 How does an Enzyme help? Enzymes speed up reactions enormously. To understand how they do this, examine the concepts of activation energy & the transition state. In order to react, the molecules involved are distorted, strained or forced to have an unlikely electronic arrangement. That is the molecules must pass through a high energy state. transition state This high energy state is called the transition state. activation energy The energy required to achieve it is called the activation energy for the reaction.

Prof. R. Shanthini 09 Nov 2012 How does an Enzyme help? The higher the free energy change for the transition barrier, the slower the reaction rate.

Prof. R. Shanthini 09 Nov 2012 How does an Enzyme help? Enzymes lower energy barrier by forcing the reacting molecules through a different transition state. This transition state involves interactions with the enzyme. Enzyme

Prof. R. Shanthini 09 Nov 2012 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 Enzyme classification

Prof. R. Shanthini 09 Nov 2012 Examples of Enzyme Catalysed Reactions CO 2 + H 2 O H 2 CO 3 Carbonic anhydrase 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 CO 2 ) to the lungs (low CO 2 ). One molecule of carbonic anhydrase can process millions of molecules of CO 2 per second. Example 1: Examples of enzyme catalyzed reactions

Prof. R. Shanthini 09 Nov H 2 O 2 2H 2 O + O 2 Catalase 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. Example 2: Examples of enzyme catalyzed reactions

Prof. R. Shanthini 09 Nov 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 Industrial use of catalase

Prof. R. Shanthini 09 Nov 2012 See the hand out on the same topic Examples of Industrial Enzymes

Prof. R. Shanthini 09 Nov 2012 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 More on enzymes

Prof. R. Shanthini 09 Nov 2012 Source:

Prof. R. Shanthini 09 Nov 2012 Source: E + S ES

Prof. R. Shanthini 09 Nov 2012 Source: Lock & Key Theory Of Enzyme Specificity (postulated in 1894 by Emil Fischer) E + S ESE + P

Prof. R. Shanthini 09 Nov 2012

Active Site Of Enzyme Blocked By Poison Molecule Source:

Prof. R. Shanthini 09 Nov 2012 Induced Fit Model (postulated in 1958 by Daniel Koshland ) Source: 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. E + S ESE + P

Prof. R. Shanthini 09 Nov 2012 E + S ESE + P k1k1 k2k2 k3k3 which is equivalent to S P [E] S for substrate (reactant) E for enzyme ES for enzyme-substrate complex P for product Simple Enzyme Kinetics

Prof. R. Shanthini 09 Nov 2012 Michaelis-Menten approach to the rate equation: Assumptions: 1.Product releasing step is slower and it determines the reaction rate 2. ES forming reaction is at equilibrium 3. Conservation of mass (C E0 = C E + C ES ) E + S ESE + P k1k1 k2k2 k3k3 Initial concentration of E Concentration of E at time t Concentration of ES at time t

Prof. R. Shanthini 09 Nov 2012 Michaelis-Menten approach to the rate equation: E + S ESE + P k1k1 k2k2 k3k3 r P = - r S = k 3 C ES Product formation (= substrate utilization) rate: k 1 C E C S = k 2 C ES Since ES forming reaction is at equilibrium, we get (1) (2)

Prof. R. Shanthini 09 Nov 2012 Michaelis-Menten approach to the rate equation: E + S ESE + P k1k1 k2k2 k3k3 k 1 (C E0 – C ES ) C S = k 2 C ES Using C E0 = C E + C ES in (2) to eliminate C E, we get which is rearranged to give C E0 C S C ES = k 2 /k 1 + C S (3)

Prof. R. Shanthini 09 Nov 2012 Michaelis-Menten approach to the rate equation: E + S ESE + P k1k1 k2k2 k3k3 k 3 C E0 C S r P = r max C S = K M + C S (4) Using (3) in (1), we get k 2 /k 1 + C S - r S = where r max = k 3 C E0 and K M = k 2 / k 1 (6) (5)

Prof. R. Shanthini 09 Nov 2012 E + S ESE + P k1k1 k2k2 k3k3 Substrate binding step Other terminology used Catalytic step k 3 = k cat r max = k 3 C E0 = k cat C E0 K M = k 2 / k 1 (6) (5a)

Prof. R. Shanthini 09 Nov 2012 Assumptions: 1. Steady-state of the intermediate complex ES 2. Conservation of mass (C E0 = C E + C ES ) E + S ESE + P k1k1 k2k2 k3k3 Initial concentration of E Concentration of E at time t Concentration of ES at time t Briggs-Haldane approach to the rate equation:

Prof. R. Shanthini 09 Nov 2012 E + S ESE + P k1k1 k2k2 k3k3 Briggs-Haldane approach to the rate equation: r P = k 3 C ES Product formation rate: (7) r s = - k 1 C E C S + k 2 C ES Substrate utilization rate: (8) k 1 C E C S = k 2 C ES + k 3 C ES Since steady-state of the intermediate complex ES is assumed, we get (9)

Prof. R. Shanthini 09 Nov 2012 E + S ESE + P k1k1 k2k2 k3k3 Briggs-Haldane approach to the rate equation: r P = - r S = k 3 C ES Combining (7), (8) and (9), we get (10) k 1 (C E0 - C ES )C S = (k 2 + k 3 )C ES Using C E0 = C E + C ES in (9) to eliminate C E, we get which is rearranged to give C E0 C S C ES = (k 2 +k 3 )/k 1 + C S (11)

Prof. R. Shanthini 09 Nov 2012 E + S ESE + P k1k1 k2k2 k3k3 Briggs-Haldane approach to the rate equation: where r max = k 3 C E0 and K M = (k 2 + k 3 ) / k 1 (13) Combining (10) and (11), we get k 3 C E0 C S - r S = (k 2 +k 3 )/k 1 +C S r max C S = K M + C S (5) (12)r P = When k 3 << k 2 (i.e. product forming step is slow), K M = k 2 / k 1 (6)

Prof. R. Shanthini 09 Nov 2012 where r max = k 3 C E0 = k cat C E0 and K M = f(rate constants) - r S r max C S = K M + C S r P = Simple Enzyme Kinetics (in summary) S P [E] r max is proportional to the initial concentration of the enzyme K M is a constant