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Principles of Bioenergetics
Lehninger Ch. 13 BIO 322 Recitation 1 / Spring 2013
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Two Fundamental Laws of Thermodynamics
For any physical or chemical change, the total amount of energy in the universe remains constant – Conservation of energy In all natural processes, the entropy of universe increases – Tends towards increasing disorder In living organisms - Molecules are highly organized, maintain order Controversial with 2nd ? Order in themselves & exchange energy with surrounding
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Bio systems (Ex. Cells) – constant temp. & pressure
When ΔG is negative – spontaneous system Cellular order is maintained by taking free energy from surroundings as nutrients and sun light, and by returning energy back to surroundings in the form of heat and entropy.
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∆G’° & K’eq Concentration of reactants and products at equilibrium. (Keq) a, b, c, d - # of molecules [A], [B], [C], [D] – molar concentrations When a rxn is not at eq., the magnitude of the tendency to move forward to eq – ΔG Under std conditions (298 K = 25°C), when reactants and products are at 1 M or for gases 1 atm, the magnitude of tendency to move toward to eq, is called standard free energy change - ΔG° (pH=0) ΔG’° (pH=7) Small changes in ΔG’° - large changes in Keq due to exponential relationship.
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ΔG vs. ΔG’° ΔG’° (standard free energy change) – characteristic, unchanging value for each chemical reaction, represents a constant at which initial conc. of each component at 1 M, pH=7, 25°C, 1 atm ΔG (actual free energy change) – function of reactant and product conc, which will not necessarily match the standard conditions as stated above. ΔG negative – spontaneous, becomes less negative as reaction proceeds, is zero at eq., indicating no more work can be done by the rxn. The criterion for spontaneity of rxn is ΔG, not ΔG’°
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Biological Oxidation – Reduction Rxns
Two chemical species – different affinities to electrons results in spontaneous electron flow, driven by the force proportional to the difference in electron affinity – ELECTROMOTIVE FORCE (emf) Living cells use glucose as the source of electrons. Oxidation of glucose via enzymes- flow of electrons to oxygen. (exergonic because oxygen has higher affinity for electrons than the other electron carrier intermediates) – this process results in a emf for biological processes. Ex: Membrane bound enzymes couple electron flow to the production of TM pH difference – also called proton motive force- ATP sythase use this proton motive force for ATP synthesis. (E-Coli flagella movement) Electron donating molecule (oxidized) – reducing agent, reductant Electron accepting molecule (reduced) – oxidizing agent, oxidant
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Biological Oxidation – Reduction Rxns
H<C<S<N<O – order of increasing electronegativity In biological systems, oxidation (loss of electrons coincident with loss of hydrogen) ~ dehydrogenation – oxidation rxns are catalyzed by dehydrogenases. The more reduced compunds are richer in hydrogen than in oxygen, whereas the more oxidized compounds have more oxygen and less hydrogen.
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Biological Oxidation – Reduction Rxns
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Biological Oxidation – Reduction Rxns
Two conjugate redox pairs are in solution – electrons transfer from electron donor of one pair to electron acceptor of the other pair. This tendency for such a rxn is dependent on the relative affinity of electron acceptor of each redox pair for electrons. Standard reduction potential, E°, is a measure in volts of this affinity, can be determined as in Figure Standard of reference (pH=0) The cell that gains electrons (stronger tendency to acquire electrons) is assigned positive E° value, by convention. E’° - at pH=7, conjugate redox pair at 1 M, connected to standard hydrogen electrode.
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Biological Oxidation – Reduction Rxns
Electrons tend to flow to the half-cell with the more positive E and the strength of this tendecy is proportional to the difference in reduction potentials, ΔE . The energy made available by this spontaneous electron flow is proportional to ΔE. n is the number of electrons transferred. See Table 13-7 for stardard reduction potential of half rxns. ∆E'° = E'°(acceptor) – E'°(donor)
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Example. 1 In glycolysis, the enzyme pyruvate kinase catalyzes this reaction: Phosphoenolpyruvate + ADP pyruvate + ATP Given the information below, show how you would calculate the equilibrium constant for this reaction. (R = J/mol·K; T = 298 K) Reaction 1) ATP ADP + Pi ∆G'° = –30.5 kJ/mol Reaction 2) phosphoenolpyruvate pyruvate + Pi ∆G'° = –61.9 kJ/mol Ans: The reaction is the sum of reaction 2 and the reverse of reaction 1. Therefore, ∆G'° = –31.4 kJ/mol. ∆G'° = –RT ln Keq‘ ln Keq' = –∆G'°/RT = 31.4 kJ/mol / [(8.315 J/mol·K)(298 K)] ln Keq' = Keq' = 3.19 x 105
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Example. 2 Lactate dehydrogenase catalyzes the reversible reaction:
Pyruvate + NADH + H Lactate + NAD+ Given the following facts (a) tell in which direction the reaction will tend to go if NAD+, NADH, pyruvate, and lactate were mixed, all at 1 M concentrations, in the presence of lactate dehydrogenase at pH 7; (b) calculate ∆G'° for this reaction. Show your work. NAD+ + H+ + 2e– NADH E'° = –0.32 V pyruvate + 2H+ + 2e– lactate E'° = –0.19 V The Faraday constant, , is kJ/V·mol. Ans: ∆E'° = E'°(acceptor) – E'°(donor) = –0.19 V – (–0.32 V) = V ∆G'° = –n ∆E'° = (–2)(96.48 kJ/V·mol)(0.13 V) ∆G'° = –25.1 kJ/mol
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