Bioenergetics Lecture 1 PDF

MDSC1402 – Digestion & Metabolism Bioenergetics Dr. J. Foster Biochemistry Unit, Dept. Preclinical Sciences Faculty of Medical Sciences, U.W.I. What is meant by the term “bioenergetics”? bio = biology energetics = energy flow What is the significance of “bioenergetics” to us? Study of how such energ...

MDSC1402 – Digestion & Metabolism Bioenergetics Dr. J. Foster Biochemistry Unit, Dept. Preclinical Sciences Faculty of Medical Sciences, U.W.I. What is meant by the term “bioenergetics”? bio = biology energetics = energy flow What is the significance of “bioenergetics” to us? Study of how such energy flows, transforms and harnessed Processes by which the body meets energy demands e.g. digestion & other metabolism Bioenergetics Transformation and flow of energy within biological systems, and their environment Concerned with the initial and final energy states of reactants, not the mechanism or kinetics i.e. Biochemical thermodynamics Bioenergetics Thermodynamics - laws & principles describing the flow and interchanges of heat, energy, & matter in systems Concepts applicable to biological systems System – part of the universe we are concerned with Surroundings – everything else! Bioenergetics Three types of systems Isolated - cannot exchange matter or energy with its surroundings Closed - may exchange energy, but not matter, with the surroundings Open - may exchange matter, energy, or both with the surroundings Laws of Thermodynamics 1st Law: the total energy of a system (including surroundings) remains constant energy cannot be gained or lost it can be transferred from part to part It can be converted from one form to another 2nd law: total entropy of a system must increase for a process to occur spontaneously Laws of Thermodynamics What determines how energy flows and whether reactions occur? Gibbs free energy (G) Laws of Thermodynamics Gibbs free energy energy available to reactants/products in rxn determines the feasibility of reactions i.e. direction & extent (predictive) Two forms of G used in chemical reactions ΔG, the change in G of rxn ΔG°, the standard ΔG (reactants/products @ 1mol/L) ΔG° useful only under standard conditions Free Energy Given a reaction where A ⇆ B, if ΔG is negative – rxn is exergonic*, energy is lost from system, spontaneous from A → B is positive – rxn is endergonic*, energy is required by system from surroundings for rxn to occur equals zero – rxn is at equilibrium; no direction favoured also ΔG A→B = – ΔG B→A Spontaneous reactions move towards equilibrium * differ from exothermic/endothermic which relate to only heat Garrett, Grisham. Biochemistry, 2nd ed © 2000 Free Energy ΔG is determined by two factors Enthalpy (ΔH) – change in heat of reactants and products of a rxn (e.g. chemical bonds) Entropy (ΔS) – change in randomness/disorder of reactants & products Neither ΔH or ΔS can predict rxn feasibility alone °K J/mol = 273 + °C ΔG= ΔH – TΔS ΔG= ΔH – TΔS J/mol J/mol/K as Δ S increases, ΔG becomes more -ve Free Energy ΔG can also be defined for ideal gas reactions: ΔG = ΔG° + RT ln [B]/[A] At constant P (pressure) & T (absolute) – thermal equilibrium R = gas constant (8.315 J/mol/K) In [B] [A] = natural logarithm = concentration of product = concentration of reactant Note that ΔG and ΔG° can have different signs Free Energy Under standard conditions [A]=[B]= 1 mol/L ΔG = ΔG° + RT ln [B]/[A] ΔG = ΔG° + RT ln 1 ΔG = ΔG° (ln 1 = 0) ΔG° is predictive only under standard conditions ΔG & ΔG° can differ greatly depending on [A], [B] Free Energy At equilibrium (steady-state) [B] / [A] = constant = Keq ΔG = ΔG° + RT ln([B]/[A]) becomes ΔG = ΔG° + RT lnKeq Thus, At equilibrium ΔG = 0 0 = ΔG° + RT lnKeq ΔG° = – RT lnKeq Bioenergetics of pathways Biochemical pathway - series of rxns each with characteristic ΔG Thermodynamically for a pathway ΔGpathway can be considered additive feasibility depends on sum of individual ΔG’s As long as the sum of ΔG is –ve the pathway is feasible A rxn can still occur even if ΔG is +ve, if it is kinetically favoured… how?? Enzymes They reduce the activation energy needed for a rxn Bioenergetics of pathways Biological systems have rxns with +ve ΔG’s How do they overcome these +ve ΔG’s? Exergonic reactions are usually coupled with endergonic ones Such coupling of rxns involves using a common intermediate – an energy coupler E.g. First step of Glycolysis The individual half-reactions in aqueous solution: ATP + H2O ↔ ADP + Pi ΔGo' = −31 kJ/mol (exo) Pi + glucose ↔ glucose-6-P + H2O ΔGo' = +14 kJ/mol (end) Hexokinase catalyses this rxn (active site excludes H2O, promotes coupled over individual rxns) ATP + glucose ↔ ADP + glucose-6-P ΔGo'= −17 kJ/mol Adenosine triphosphate (ATP) is the coupler for the reaction Garrett, Grisham. Biochemistry, 2nd ed © 2000 ATP as an energy coupler The energy currency of cells – universal coupler Intermediate in the rank of high-energy phosphates in the body Allows it to accept and donate energy in many rxns ATP as an energy coupler Terminal phosphate bonds are “high energy” (~) release a large amount of energy on hydrolysis/breaking These phosphate bonds allow ATP to release energy for some processes when hydrolysed to ADP + Pi ADP to store energy from other processes as chemical potential energy in the form of ATP Thermodynamics vs Kinetics of ATP High activation energy barrier for its bonds means very slow hydrolysis of “high energy” bonds w/o enzyme catalyst provide kinetic stability - essential to ATP’s role (and other compounds with ~ bonds) Why is this kinetic stability for ATP hydrolysis a good thing?? Rapid hydrolysis (due to low barriers) would hinder ATP’s role in metabolism enzymes lower these barriers, coupling the rxn with other useful ones prevents free energy released from ATP hydrolysis being wasted Harper’s Biochemistry 26th ed, Appleton and Lange, USA ▪ Anabolic – complex molecules from simple ones (endergonic) ▪ Catabolic – simple molecules from complex ones (exergonic) ▪ ATP bridges the 2 types of metabolism Lehninger, Biochemistry, 4nd edition © 2005 Redox Reactions Redox (reduction-oxidation) rxns are inherently coupled & involve both donating e- (oxidation) accepting e- (reduction) The two halves of a redox rxn are considered separately: Fe2+ + Cu2+ ↔ Fe3+ + Cu+ can be rewritten in half-reactions as 1. Fe2+ → Fe3+ + e2. Cu2++ e- → Cu+ Redox Reactions Free energy is also transferred when these electrons move The transfer of e- can be measured as reduction potential (E) E - the tendency of a chemical species to acquire electrons and thereby be reduced Redox Reactions For a rxn, E can be used to calculate G: ΔG = -nFΔE likewise, ΔG o = -nFΔE o n = number of electrons transferred, F = Faraday’s constant (96,480 J/V/mol; charge of 1e-) E = reduction potential E o = std reduction potential +ve E favours a –ve G (forward rxn) –ve E favours a +ve G (backward rxn) https://chem.libretexts.org/ Redox Reactions Note: the Eo can also be calculated using: Ecell = Ecathode - Eanode Going back to this reaction: Fe2+ + Cu2+ ↔ Fe3+ + Cu+ can be rewritten in half-reactions as 1. Fe2+ + e- → Fe3+ cathode 2. Cu2++ e- → Cu+ anode The cathode donates e- and the anode accepts Redox Reactions Note: the Eo can also be calculated using: Ecell = Ecathode - Eanode Going back to this reaction: Fe2+ + Cu2+ ↔ Fe3+ + Cu+ standard reduction potentials are 1. Fe3+ + e- → Fe2+ Eo = +0.77v 2. Cu2++ e- → Cu+ Eo = +0.16v Thus Eocell = +0.77-0.16 = +0.61v Redox Reactions Note: the Eo can also be calculated using: Ecell = Ecathode - Eanode Going back to this reaction: Fe2+ + Cu2+ ↔ Fe3+ + Cu+ ΔG o = -nFΔE o = -(1)(96,480)(0.61) = -(1)(96,480)(0.61) = -58,852 J Quick Quiz ❖ A simple pathway has 2 steps: A→B → C , where ΔG° for A → B = -53.4 kJ/mol, and Keq is 2.8x103 ΔG° for B → C = +28.2 kJ/mol and Keq is 1.5x103 What is ΔG° for A → C? ΔG° = (-53.4) + (28.2) = -25.2 kJ/mol Select an eqn for ΔG for A → B, taking into account the concentrations of reactant & product: ΔG= ΔG° + RTlnKeq Quick Quiz ❖ Calculate the equilibrium constant for the hydrolysis of Compound A to components B & C, if the ΔG° is -3.496 KJ/mol at 25oC. T in oK = oC + 273; R = 8.315 J/K mol Go = -RTlnKeq lnKeq = G/-RT Keq = e(G/-RT) = e(-3,496/(-8.315x298)) = e(3,496 /2478) = e(1.41) = 4.1