BB lecture 10-17 enzymes

BB lecture 10-17 enzymes - Chapter 8 (pp.-157) Enzymes...

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Chapter 8 (pp.-157) – Enzymes Learning objectives: •From an equilibrium constant (K eq ), be able to determine what it implies about the Gibbs free energy of a reaction. •Describe the structural and functional characteristics of an enzyme •Explain how the rate of an enzyme-catalyzed reaction is influenced by concentration of substrate, concentration of enzyme, temperature, pH… •Explain what is meant by the induced-fit model of enzyme function. •Know the difference between a competitive and a non-competitive inhibitor of an enzyme •Learn how enzyme activity may be regulated via allosteric configurations and feedback mechanisms
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Chemical equilibrium is the balance between forward and reverse reactions. At equilibrium the concentration of reactants and products remain constant as measured by K eq ( equilibrium constant ). K eq is known as the ratio of the concentration of products over the concentration of reactants at equilibrium. For example: A + B C + D [ means reversible rx ] [reactants] [products] K eq = [C] x [D] [A] x [B] The K eq can let you know which way the reaction is going. - If K eq is greater than 1………. favors products - If K eq is less than 1 …………favors reactants The higher the K eq , the faster the rx happens A rx with higher Keq can be expected to have a -^G
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Practice Start out with 10 moles of glucose in a flask. Glucose changes to fructose slowly. When the reaction has reached equilibrium we have 7 moles glucose and 3 moles fructose. We can express this equilibrium in terms of an equilibrium constant K eq . [ ] means concentration in moles/liter [Product ] K eq = [Reactant] = ? [fructose] K eq = [glucose] 3 moles/liter K eq = 7 moles/liter K eq = 0.43 or 4.3 x 10 -1
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Note: Coefficients become exponents Consider this example: N 2 + O 2 2 NO [N 2 ] = .95 moles/liter [O 2 ] = .95 moles/liter
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BB lecture 10-17 enzymes - Chapter 8 (pp.-157) Enzymes...

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