ENZYME_KINETICS

ENZYME_KINETICS - 1 Enzyme Kinetics In this exercise we...

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1 Enzyme Kinetics In this exercise we will look at the catalytic behavior of enzymes. You will use Excel to answer the questions in the exercise section. At the end of this session, you must hand in answers to all the questions, along with print outs of any plots you created. Background Enzymes are the catalysts of biological systems and are extremely efficient and specific as catalysts. In fact, typically, an enzyme accelerates the rate of a reaction by factors of at least a million compared to the rate of the same reaction in the absence of the enzyme. Most biological reactions do not occur at perceptible rates in the absence of enzymes. One of the simplest biological reactions catalyzed by an enzyme is the hydration of CO 2 . The catalyst in this reaction is carbonic anhydrase. This reaction is part of the respiration cycle which expels CO 2 from the body. Carbonic anhydrase is a highly efficient enzyme – each enzyme molecule can catalyze the hydration of 10 5 CO 2 molecules per second. Enzymes are highly specific. Typically a particular enzyme catalyzes only a single chemical reaction or a set of closely related chemical reactions. As is true of any catalyst, enzymes do not alter the equilibrium point of the reaction. This means that the enzyme accelerates the forward and reverse reaction by precisely the same factor. For example, consider the interconversion of A and B. A B ( 1 ) Suppose that in the absence of the enzyme the forward rate constant (k f ) is 10 -4 s -1 and the reverse rate constant (k r ) is 10 -6 s -1 . The equilibrium constant (K eq ) is given by the ratio of the two rate constants. K eq = [B] [A] = k f k r = 10 4 10 6 = 100 (2) The equilibrium concentration of B is 100 times that of A whether or not an enzyme is present.
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2 second. The enzyme lowers the height of the energy barrier to the reaction thereby increasing the rate of the reaction, but since the rate of both the forward and reverse reaction are affected by the same amount, the equilibrium constant is not affected by the presence of the enzyme. the same amount (see Figure 1) Figure 1 where, E Af is the activation energy for the forward reaction (A Æ B) in the absence of a catalyst and E’ Af is the activation energy for the forward reaction (A Æ B) in the presence of a catalyst, and G o is the change in free energy for the reaction. The equilibrium constant is related to G o as follows: K eq = e −∆ G o /RT Since G o is the same for the catalyzed and uncatalyzed reaction, K eq is the same for both reactions. One reason for the efficiency and specificity of an enzyme is the way the enzyme interacts with
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This note was uploaded on 11/21/2011 for the course CHEE 370 taught by Professor Tufenkji during the Fall '07 term at McGill.

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ENZYME_KINETICS - 1 Enzyme Kinetics In this exercise we...

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