BCH3034 - Chapter 3 - Thermodynamics of Biological Systems PDF

Chapter 3 Thermodynamics of Biological Systems  Outline Basic Thermodynamic Concepts State functions (E,H,S,G) Physical Significance of Thermodynamic Properties pH and the Standard State The Effect of Concentration Coupled Processes High-Energy Biomolecules ...

Chapter 3 Thermodynamics of Biological Systems  Outline Basic Thermodynamic Concepts State functions (E,H,S,G) Physical Significance of Thermodynamic Properties pH and the Standard State The Effect of Concentration Coupled Processes High-Energy Biomolecules  Outline cont’d ATP – Intermediate Energy-Shuttle Molecule The Daily Human Requirement for ATP Electromotive force  Basic Concepts The system: the portion of the universe with which we are concerned The surroundings: everything else Isolated system cannot exchange matter or energy Closed system can exchange energy Open system can exchange either or both  The First Law The total energy of an isolated system is conserved. E (or U) is the internal energy - a function that keeps track of heat transfer and work expenditure in the system E2 - E1 = E = Q + w Q >0 is heat absorbed BY the system W >0 is work done ON the system E is heat exchanged at constant volume (q v) E = Q + w = Q + (-PV) = Q (V=0) E is independent of path  State Functions A variable whose value depends on the initial and final state, but not how it got from one state to the other Energy (E=q+w) is a state function Temperature is a state function Work (w) and heat (q) are not state functions, but q+w is E and H are state function whose values are those of the heat exchanged at constant V or P, respectively.  Enthalpy A better function for constant pressure H = E + PV If P is constant, H = Q H is the heat absorbed at constant P Volume is approx. constant for biochemical reactions (in solution) So H is approx. same as E H = E + PV H = E + (PV) =Q + W + PV + VP = Q + (-PV) + PV + VP = Q + VP = Q ( when P = 0)  Hess Law The variation of enthalpy in going from one state to another state is independent of the pathway The enthalpy released or absorbed for any chemical change is the same regardless of the pathway The energy to make CO2 from C and O2, is the same as the energy to make CO from C and O2 plus the energy to make CO2 from CO and O2. H30=H10 + H20  Example Heat of formations are useful to illustrate the law. CH3-OH + 1/2O2 ----- HCO2H + H2 H30 can be written as decomposition followed by recombination CH3-OH ----- 1/2O2(g) + C(s) + 2H2(g) H10f HCO2H ----- O2(g) + C(s) + H2(g) H20f If we turn the second equation around, the elements cancel each other and we have the first equation. The H30 = H10f - H20f  Van’t Hoff Law ΔHo = -R d(lnKeq)/d(1/T) Where ΔHo is standard state van’t Hoff plot is 1/T vs R(lnKeq)  The Second Law Systems tend to proceed from ordered to disordered states The entropy change for (system + surroundings = universe) is unchanged in reversible processes and positive for irreversible processes All processes proceed toward equilibrium - i.e., minimum potential energy  Entropy A measure of disorder An ordered state is low entropy A disordered state is high entropy dSreversible = dq/T (ideal process, not real) Spontaneous process, Suniverse >0 Reversible process, Suniverse = 0 The universe always tends toward increasing disorder  More order (energetically costly) → more information Your house/room does not spontaneously become organized Work must be put in to clean up the room to decrease entropy More order: more information Less order “A theory is the more impressive the greater is the simplicity of its premise, the more different are the kinds of things it relates and the more extended is its range of application.”  Energy dispersion Entropy can be defined as S = k ln W And ΔS = k ln Wfinal – k ln Winitial Where Wfinal and Winitial are the final and initial number of microstates of a system, and k is Boltzmann’s constant. Viewed in this way, entropy represents energy dispersion – the dispersion of energy among a large number of molecular motions relatable to quantized states (microstates). The definition of entropy above is engraved on the tombstone of Ludwig Boltzmann in Vienna, Austria  The Third Law The entropy of any crystalline, perfectly ordered substance must approach zero as the temperature approaches 0 K At T = 0 K, entropy is exactly zero Heat Capacity: the amount of heat 1mole of the substance can store as the temperature of that substance is raised by 1 degree For a constant pressure process: Cp = dH/dT  Free Energy Hypothetical quantity - allows chemists to asses whether reactions will occur G = H - TS For any process at constant P and T: G = H - TS If G = 0, reaction is at equilibrium If G < 0, reaction proceeds as written For a chemical reaction, aA + bB = cC + dD G0 = G0f (cC + dD) - G0f (aA + bB) Where G0 is free energy at a reference point ( standard conditions) and G0f formation free energies at same ref. point. Standard free energy = constant Unchanged for a reaction G at non-standard chemical conditions is given by G = G0 + RT Ln Q Where Q is [C ]c[ D ]d / [A ]a[ B ]b. For Q=Keq., G = 0 and G0 = - RT Ln Keq.  Keq and Go Keq DGo (kJ/mol) 104 – 23 102 – 11 100 = 1 0 10–2 + 11 10–4 + 23 What is the Effect of pH on Standard State Free Energies? A standard state of 1 M for H+ is awkward for biochemical reactions. It makes more sense to adopt a modified standard state – i.e., 1 M for all constituents except protons, for which the standard state is pH 7. This standard state is denoted with a superscript “°´” For reactions in which H+ is produced, we have ΔG°´= ΔG° + RT ln [H+] And for reactions in which H+ is consumed, we have ΔG°´ = ΔG° - RT ln [H+] For rx that produces H+, under chemical conditions, A = B + H+, then G0 =-RTLnKeq =-RTLn( [B] [H+])/ [A] Under biochemical conditions, [H+] = 10-7M ≡ 1 and the free energy is G0’ G0’ = =-RTLn( [B] / [A]). Thus, G0’ is part of the G0 expression (in yellow) G0 = -RTLn( [B] [H+])/ [A] = G0 = -RTLn( [B] / [A] ) – RTLn [H+] G0’ = G0 + RT Ln [H+] (where [H+] = 10-7 M)  G versus Go’ How can we calculate the free energy change for rxns not at standard state? Consider a reaction: A + B C + D Then: G = Go’ + RT ln ([C][D]/[A][B]) (If no water or protons are involved) Which the same as per chem. standard conditions Go’ = Go  ΔG° Can Be Temperature Dependent Δ G = Δ H - TΔS + + + +/- - + +/- + - - - -  ΔS° Can Be Temperature Dependent S and ΔS increases with temperature  Energy Transfer A Crucial Biological Need Energy acquired from sunlight or food must be used to drive endergonic (energy-requiring) processes in the organism Two classes of biomolecules do this: – Reduced coenzymes (NADH, FADH2) – High-energy phosphate compounds - free energy of hydrolysis larger than -25 kJ/mol)  High-Energy Biomolecules Study Table 3.3! Note what's high - PEP and 1,3-BPG Note what's low - sugar phosphates, etc. Note what's in between - ATP! The ADP/ATP pair requires a higher energy than that of the ATP molecule to activate the ADP Note difference (Figure 3.6) between overall free energy change - noted in Table 3.3 - and the energy of activation for phosphoryl-group transfer! ATP is kinetically stable, but Thermodynamically unstable  ATP An Intermediate Energy Shuttle Device PEP and 1,3-BPG are created in the course of glucose breakdown Their energy (and phosphates) are transferred to ADP to form ATP But ATP is only a transient energy carrier - it quickly passes its energy to a host of energy-requiring processes  Phosphoric Acid Anhydrides Why ATP does what it does! ADP and ATP are examples of phosphoric acid anhydrides Note the similarity to acyl anhydrides Large negative free energy change on hydrolysis is due to: – electrostatic repulsion – stabilization of products by ionization and resonance – entropy factors: more particles, ADP , P electrostatic repulsion Increases the stored energy Hydrolysis of Phosphoric Anhydrides is Highly Favorable  Phosphoric-Carboxylic Anhydrides These mixed anhydrides - also called acyl phosphates - are very energy-rich Acetyl-phosphate: G°´ = -43.3 kJ/mol 1,3-BPG: G°´ = -49.6 kJ/mol Bond strain, electrostatics, and resonance are responsible  Enol Phosphates Phosphoenolpyruvate (PEP) has the largest free energy of hydrolysis of any biomolecule Formed by dehydration of 2-phospho- glycerate Hydrolysis of PEP yields the enol form of pyruvate - and tautomerization to the keto form is very favorable Formation and hydrolysis of PEP 62.2 KJ/ mol Phosphate hydrolysis And so does strain relief releases energy 62.2 KJ/ mol  Ionization States of ATP ATP has four dissociable protons pKa values range from 0-1 to 6.95 Free energy of hydrolysis of ATP is relatively constant from pH 1 to 6, but rises steeply at high pH Since most biological reactions occur near pH 7, this variation is usually of little consequence Metal Ions Affect the Free Energy of Hydrolysis of ATP No Mg2+ Mg2+ With Mg2+  The Effect of Concentration Free energy changes are concentration dependent We will use the value of -30.5 kJ/mol for the standard free energy of hydrolysis of ATP But at non-standard-state conditions (in a cell, for example), the G is different! Equation 3.12 is crucial - be sure you can use it properly In typical cells, the free energy change for ATP hydrolysis is normally -50 kJ/mol For the reaction ATP = ADP + Pi, ΔG = ΔGo’ + RT Ln ([ADP][Pi]/[ATP]) Thus, the free energy of ATP hydrolysis (or coupling) changes with concentration  Energy Coupling Many reactions of cells and organisms run against their thermodynamic potential – that is, in the direction of positive ΔG Examples – synthesis of ATP, creation of ion gradients These processes are driven in the thermodynamically unfavorable direction via coupling with highly favorable processes  Energy Coupling Consider A ↔ B , G0’ = +13.8 kJ/mol and G0’ = - RT Ln Keq., thus +13,800 = - (8.31J/K.mol)298K LnKeq. Or Ln Keq = -5.57 and Keq = 0.0038 = [B]/[A] and On the other hand, we know that ATP+ H2O ↔ ADP + Pi has a G0’ = -30.5 kJ/mol This is a strongly negative G0’  Energy Coupling (contd.) Now consider A + ATP + H2O ↔ B + ADP + Pi [ADP]=8x10-3M, [Pi]=10-3M, [ATP]=8x10-3M For the latter, the G0’ is given by the addition of both G0’s of the separate reactions A↔ B G0’ = +13.8 kJ/mol ATP+ H2O ↔ ADP + Pi G0’ = -30.5 kJ/mol __________________________________________ A + ATP + H2O ↔ B + ADP + Pi G0’ = -16.7 kJ Plugging this G0’ in the previous equations, we get a Keq = [B][ADP][Pi]/[A][ATP] = 850 and a [B]/[A] = 850,000 or a 850,000/0.0038=2.2 x108 INCREASE!  Energy Coupling (Contd.) Note that the previous reactions must have a common species to be coupled. In this case it could be: A + P -----→ A-P G = +13.8 kJ/mol 0’ ATP -----→ ADP + P G = -30.5 kJ/mol 0’ Where P is the common species, and A-P would be the precursor to the B species from the previous example.  Daily Human Requirement for ATP The average adult human consumes approximately 11,700 kJ of food energy per day Assuming thermodynamic efficiency of 50%, about 5860 kJ of this energy ends up in form of ATP Assuming 50 kJ of energy required to synthesize one mole of ATP, the body must cycle through 5860/50 or 117 moles of ATP per day This is equivalent to 65 kg of ATP per day The typical adult human body contains 50 g of ATP/ADP Thus each ATP molecule must be recycled nearly 1300 times per day Concentration gradients and Free Energy A difference in proton (or any other particle) concentration on two sides of semi permeable membrane has a difference on free energy given by: G = -RT Ln [H+] high / [H+] low Plus an electric component, ZFΔΨ This is called the ELECTROMOTIVE FORCE