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Thermodynamics, Concluded

Thermodynamics, Concluded. Andy Howard Biochemistry, Fall 2014 IIT. What we ’ ll discuss. Thermodynamics Entropy Free energy Equilibrium Protein folding Solving problems. Kinds of thermodynamic properties.

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Thermodynamics, Concluded

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  1. Thermodynamics, Concluded Andy Howard Biochemistry, Fall 2014IIT Biology 401: Thermodynamics

  2. What we’ll discuss • Thermodynamics • Entropy • Free energy • Equilibrium • Protein folding • Solving problems Biology 401: Thermodynamics

  3. Kinds of thermodynamic properties • Extensive properties:Thermodynamic properties that are directly related to the amount (e.g. mass, or # moles) of stuff present (e.g. E, H, S) • Intensive properties: not directly related to mass (e.g. P, T) • E, H, S are state variables;work, heat are not Biology 401: Thermodynamics

  4. Units • Energy unit: Joule (kg m2 s-2) • 1 kJ/mol = 103J/(6.022*1023)= 1.661*10-21 J • 1 cal = 4.184 J:so 1 kcal/mol = 6.948 *10-21 J • 1 eV = 1 e * J/Coulomb =1.602*10-19 C * 1 J/C = 1.602*10-19 J= 96.4 kJ/mol = 23.1 kcal/mol James Prescott Joule Biology 401: Thermodynamics

  5. Typical energies in biochemistry • Go for hydrolysis of high-energy phosphate bond in adenosine triphosphate:33kJ/mol = 7.9kcal/mol = 0.34 eV • Hydrogen bond: 4 kJ/mol=1 kcal/mol • van der Waals force: ~ 1 kJ/mol • See textbook for others Biology 401: Thermodynamics

  6. Entropy • Related to disorder: Boltzmann:S = k ln k=Boltzmann constant = 1.38*10-23 J K-1 • Note that k = R / N0 •  is the number of degrees of freedom in the system • Entropy in 1 mole = N0S = Rln • Number of degrees of freedom can be calculated for simple atoms and molecules Biology 401: Thermodynamics

  7. Components of entropy Liquid propane (as surrogate): Biology 401: Thermodynamics

  8. Real biomolecules • Entropy is mostly translational and rotational, as above • Enthalpy is mostly electronic • Translational entropy = (3/2) R lnMr • So when a molecule dimerizes, the total translational entropy decreases(thereare half as many molecules,but lnMr only goes up by ln 2) • Rigidity decreases entropy Biology 401: Thermodynamics

  9. Entropy in solvation: solute • When molecules go into solution, their entropy increases because theyare freer to move around Biology 401: Thermodynamics

  10. Entropy in solvation: Solvent • Solvent entropy usually decreases because solvent molecules must become more ordered around solute • Overall effect: usually the solute effect wins out, but not always Biology 401: Thermodynamics

  11. Entropy matters a lot! • Most biochemical reactions involve very small ( < 10 kJ/mol) changes in enthalpy • Driving force is often entropic • Increases in solute entropy often is at war with decreases in solvent entropy. • The winner tends to take the prize. Biology 401: Thermodynamics

  12. Apolar molecules in water • Water molecules tend to form ordered structure surrounding apolar molecule • Entropy decreases because they’re so ordered Biology 401: Thermodynamics

  13. Binding to surfaces • Happens a lot in biology, e.g.binding of small molecules to relatively immobile protein surfaces • Bound molecules suffer a decrease in entropy because they’re trapped • Solvent molecules are displaced and liberated from the protein surface Biology 401: Thermodynamics

  14. Free Energy • Gibbs: Free Energy EquationG = H - TS • So if isothermal, G = H - TS • Gibbs showed that a reaction will be spontaneous (proceed to right) if and only if G < 0 Biology 401: Thermodynamics

  15. Standard free energy of formation, Gof • Difference between compound’s free energy & sum of free energy of the elements from which it is composed Biology 401: Thermodynamics

  16. Free energy and equilibrium • Gibbs: Go = -RT ln keq • Rewrite: keq = exp(-Go/RT) • keq is equilibrium constant;formula depends on reaction type • For aA + bB  cC + dD,keq = ([C]c[D]d)/([A]a[B]b) • If all the proportions are equal,keq = ([C][D])/([A][B]) • These values ([C], [D] …) denotes the concentrations at equilibrium Biology 401: Thermodynamics

  17. Spontaneity and free energy • Thus if reaction is just spontaneous, i.e. Go = 0, then keq = 1 • If Go < 0, then keq > 1: Exergonic • If Go > 0, then keq < 1: Endergonic • You may catch me saying “exoergic” and “endoergic” from time to time:these mean the same things. Biology 401: Thermodynamics

  18. Protein folding and thermodynamics • An oversimplified model of a protein would suggest that it can exist in two possible states: • Completely unfolded and free to wobble • Completely fixed in conformation • There are solvent (H2O) molecules present in both cases • What does thermodynamics tell us? • We’ll describe this in terms of the reaction (Protein + Solvent)unfolded  (Protein + Solvent)folded Biology 401: Thermodynamics

  19. Protein folding:enthalpy of the protein • Fully folded form has many hydrogen bonds, ion pairs (“salt bridges”), and van der Waals interactions • These are absent or fleeting in the unfolded form • So in general for the reactionPunfolded  Pfoldedthe Hp < 0 for the protein itself Biology 401: Thermodynamics

  20. Protein folding:enthalpy of the solvent • Primary enthalpic contribution involves hydrogen bonds among water molecules and H-bonds between water molecules and protein molecules • These will balance out between folded and unfolded states. • So Hs ~ 0 for the reaction Biology 401: Thermodynamics

  21. Protein folding:entropy of the protein • In the unfolded state, the protein can adopt many slightly different conformations • In the folded state, it has only one or a very few feasible conformations • Therefore it is more ordered in the folded state. • Thus Sp < 0 and -TSp > 0 Biology 401: Thermodynamics

  22. Protein folding:entropy of the solvent • In the unfolded state, many water molecules will be attracted to and held in place near the polar parts of the protein • Many of those water molecules will be released into the bulk solvent when the protein folds up • Therefore those water molecules become more disordered when the protein folds up. • Thus Ss > 0 and -TSs < 0. Biology 401: Thermodynamics

  23. Put this all together • For the overall reaction • (Protein + Solvent)unfolded  (Protein + Solvent)folded • We determine whether the reaction is spontaneous (favorable) by calculating G = H - TS • So that’sGp+s = Hp + Hs - T(Sp + Ss) Biology 401: Thermodynamics

  24. … and what do we get? • We said Hp < 0, Hs ~ 0,Sp < 0, Ss > 0 • Therefore in Gp+s we have: • One term (Hp) that is negative • One term (Hs) that is close to zero • One term (-TSp) that is positive • One term (-TSs) that is negative • So it’s not obvious what the total should be! Biology 401: Thermodynamics

  25. Reality I: folded at room temp • In practiceSp is a bit more negative thanSs is positive, so the overall term-T(Sp+ Ss) is slightly positive. • At room temperature the enthalpic stabilization forces (Hp) are slightly more important than the entropic destabilization, so Gp+s is slightly negative, i.e. the protein will be folded Biology 401: Thermodynamics

  26. Reality II: unfolded at higher T • However, at a slightly higher temperature, the entropic destabilization -T(Sp+ Ss) becomes more important (more positive) as T gets bigger, so the overall G becomes positive, and the reaction becomes unfavorable; so the protein unfolds. Biology 401: Thermodynamics

  27. What’s the melting temperature? • That depends on the protein • Proteins derived from thermophilic organisms have more stabilizing interactions in them (both enthalpic and entropic) so Tm ~ 70ºC for those • Proteins from ordinary (mesophilic) organisms generally have Tm ~ 50ºC Biology 401: Thermodynamics

  28. Free energy as a source of work • Change in free energy indicates that the reaction could be used to perform useful work • If Go < 0, the reaction can do work • If Go > 0, we need to do work to make the reaction occur Biology 401: Thermodynamics

  29. What kind of work? • Movement (flagella, muscles) • Chemical work: • Transport molecules against concentration gradients • Transport ions against potential gradients • To drive otherwise endergonic reactions • by direct coupling of reactions • by depletion of products Biology 401: Thermodynamics

  30. Coupled reactions • Often a single enzyme catalyzes 2 reactions, shoving them together:reaction 1, A  B: Go1 < 0 reaction 2, C D: Go2 > 0 • Coupled reaction:A + C  B + D: GoC = Go1 + Go2 • If GoC < 0,then reaction 1 is driving reaction 2! Biology 401: Thermodynamics

  31. How else can we win? • Concentration of product may play a role • As we’ll discuss in a moment, the actual free energy depends on Go and on concentration of products and reactants • So if the first reaction withdraws product of reaction B away,that drives the equilibrium of reaction 2 to the right Biology 401: Thermodynamics

  32. Adenosine Triphosphate • ATP readily available in cells • Derived from catabolic reactions • Contains two high-energy phosphate bonds that can be hydrolyzed to release energy: O O- || |(AMP)-O~P-O~P-O- | || O- O Biology 401: Thermodynamics

  33. Hydrolysis of ATP • Hydrolysis at the rightmost high-energy bond:ATP + H2O  ADP + PiGo = -33kJ/mol • Hydrolysis of middle bond:ATP + H2O  AMP + PPiGo = -40kJ/mol • BUT PPi + H2O 2 Pi,Go = -31 kJ/mol • So, appropriately coupled, we get roughly twice as much! Biology 401: Thermodynamics

  34. ATP as energy currency • Any time we wish to drive a reaction that has Go < +30 kJ/mol, we can couple it to ATP hydrolysis and come out ahead • If the reaction we want hasGo < +60 kJ/mol, we can couple it toATP  AMP and come out ahead • So ATP is a convenient source of energy — an energy currency for the cell Biology 401: Thermodynamics

  35. Coin analogy • Think of store of ATPas a roll of quarters • Vendors don’t give change • Use one quarter for some reactions,two for others • Inefficient for buying $0.35 items Biology 401: Thermodynamics

  36. Other high-energy compounds • Creatine phosphate: ~ $0.40 • Phosphoenolpyruvate: ~ $0.45 • So for some reactions, they’re more efficient than ATP Biology 401: Thermodynamics

  37. Why not use those always? • There’s no such thing as a free lunch! • In order to store a compound, you have to create it in the first place • So an intermediate-energy currency is the most appropriate Biology 401: Thermodynamics

  38. Dependence on Concentration • Actual G of a reaction is related to the concentrations / activities of products and reactants:G = Go + RT ln [products]/[reactants] • If all products and reactants are at 1M, then the second term drops away; that’s why we describe Go as the standard free energy Biology 401: Thermodynamics

  39. Is [A] = [B] = 1M… realistic? • No: often [A] may not even be soluble at as high a concentration as that • But it doesn’t matter;as long as we can define the concentrations, we can correct for them • Often we can rig it so[products]/[reactants] = 1even if all the concentrations are small • Typically [ATP]/[ADP] > 1 so ATP coupling helps even more than 33 kJ/mol! Biology 401: Thermodynamics

  40. How does this matter? • Often coupled reactions involve withdrawal of a product from availability • If that happens,[product] / [reactant]shrinks, the second term becomes negative,and G < 0 even if Go > 0 Biology 401: Thermodynamics

  41. Example: glycolysis • During 402 we’ll spend at least one lecture looking at glycolysis, one of the fundamental pathways • Some of the glycolytic reactions haveGo’ or Go > 0 • But all have G values that are negative or zero because of this concentration effect Biology 401: Thermodynamics

  42. How to solve energy problems involving coupled equations • General principles: • If two equations are added, their energetics add • An item that appears on the left and right side of the combined equation can be cancelled • Reversing a reaction reverses the sign of G. Biology 401: Thermodynamics

  43. A bit more detail • Suppose we couple two equations:A + B  C + D, DGo’ = xC + F  B + G, DGo’ = y • The result is:A + B + C + F  B + C + D + GorA + F  D + G, DGo’ = x + y • … since B & C appear on both sides Biology 401: Thermodynamics

  44. Slightly more complex… • Suppose we couple two equations:A + B  C + D, DGo’ = xH + A  J + C, DGo’ = z • Reverse the second equation:J + C  A + H, DGo’ = -z • Add this to 1st eqn. & simplify:B + J  D + H, DGo’ = x - z • … since A & C appear on both sides Biology 401: Thermodynamics

  45. What do we mean by hydrolysis? • It simply means a reaction with water • Typically involves cleaving a bond: • U + H2O  V + Wis described as hydrolysis of Uto yield V and W Biology 401: Thermodynamics

  46. iClicker quiz! • A reaction with ΔGo < 0 is described as • (a) exothermic • (b) endothermic • (c) endergonic • (d) exergonic • (e) none of the above. Biology 401: Thermodynamics

  47. iClicker quiz, question 2 2. If the reaction A + B  C + D has ΔGo = -8 kJ mol-1,then the reaction D + C  B + A will have a ΔGo equal to • (a) -8 kJ mol-1 • (b) 0 kJ mol-1 • (c) 8 kJ mol-1 • (d) undeterminable from the data given. Biology 401: Thermodynamics

  48. iClicker quiz, question 3 3. A reaction is described as a hydrolysis reaction if • (a) water is a reactant • (b) water is a product • (c) water is both a reactant and a product • (d) none of the above. Biology 401: Thermodynamics

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