Thermodynamics and Kinetics

Transcription

1 Thermodynamics and Kinetics Lecture 12 Free Energy Applications NC State University

2 Isolated system requires DS > 0 DS sys > 0 Isolated system: Entropy increases for any spontaneous process

3 System and surroundings both play in role in the entropy DS sys + DS surr > 0 DS surr > 0 DS sys < 0 or DS sys > 0 If the system is not isolated, then the entropy can have either sign for the System as long as the surroundings can compensate. Closed system in contact with surroundings: Entropy increases for any spontaneous process

4 We derive one function that includes both sys and surr DS sys + DH surr /T > 0 DH surr /T > 0 DS sys < 0 or DS sys > 0 However, if we now turn to a case where the entropy change in the surroundings compensates for a change in the system, we can include the heat flow from surr to sys.

5 We derive one function that includes both sys and surr DS sys - DH sys /T > 0 -DH sys /T > 0 DS sys < 0 or DS sys > 0 The heat flow to or from the system must opposite to that of the surroundings. Hence, there must be a minus sign In front of the enthalpy term.

6 We derive one function that includes both sys and surr DH sys - TDS sys < 0 -DH sys /T > 0 DS sys < 0 or DS sys > 0 Here we have expressed the total change of system and Surroundings in terms of quantities that relate to the System. Hence, DH sys -TDS sys constitutes a new state function

7 We derive one function that includes both sys and surr DG sys < 0 -DH sys /T > 0 DS sys < 0 or DS sys > 0 Note the sign change when we multiplied by T. Here we have written the new state function explicitly as the free energy change, DG, which is defined for the system.

8 A Definition of Gibbs Free Energy Reactions will proceed in the direction written if DG <0. G decreases as shown in the figure below until G is constant. At equilibrium, DG = 0. DG = DG o + RT ln Q where Q is the reaction quotient. Q changes as the Reaction proceeds. Finally, at equilibrium Q = K. Therefore, DG o = -RT ln K time

9 Alternative form for free energy expression Since DG = DG o + RT ln Q and DG o = -RT ln K we can also write DG = -RT ln K + RT ln Q which can be rearranged to DG = RT ln (Q/K)

10 Van t Hoff plot for drug binding A practical example of the application of the van t Hoff equation can be found in drug binding. The equilibrium constant for drug binding to an active site can be measured by fluorescence, NMR, etc. at various temperatures. Then one may plot ln K vs. 1/T and fit the result to a line. In most cases the binding will be exothermic to that DH o < 0 and then slope of the line will be positive rather than negative as shown in the previous slide.

11 Van t Hoff equation The fundamental relation that leads to the temperature dependence of the equilibrium constant is: If there are more than two data points we can plot the data as ln(k) vs. 1/T.

12 Van t Hoff plot for drug binding Slope = -DH o /R Note: DH o < 0 In this example, the slope is positive because the enthalpy of binding is negative (i.e. binding is exothermic).

13 Example: preventing inflammation by binding to prostaglandin synthase 2 COX-2 crystal structure DH o and DS o can be measured using ln(k) as a function 1/T.

14 Example: preventing depression serotonin transport inhibitors DH o and DS o can be measured using ln(k) as a function 1/T. This article shows specific differences in the enthalpy of binding of drugs based on analysis of so-called van t Hoff plots (i.e. plots of ln(k) vs. 1/T). Kinetic and Thermodynamic Assessment of Binding of Serotonin Transporter Inhibitors. J Pharm Exp Tech (2008) vol. 327, pp

15 Basic conclusion: fluvoxamine binds exclusively based on entropic driving force Entropically driven binding is relatively rare. Usually the entropy of binding is unfavorable since a flexible drug molecule (large W) will be forced to adopt a fixed conformation (small W or W = 1) upon binding to a protein. Kinetic and Thermodynamic Assessment of Binding of Serotonin Transporter Inhibitors. J Pharm Exp Tech (2008) vol. 327, pp

16 Thermodynamics of glycolysis Reaction kj/mol D-glucose + ATP D-glucose-6-phosphate + ADP DG o = D-glucose-6-phosphate D-fructose-6-phosphate DG o = +1.7 D-fructose- 6-diphosphate + ATP D-fructose-1,6-diphosphate + ADP DG o = D-fructose-1,6-diphosphate glyceraldehyde-3-phosphate + dihydroxyacetone phosphate DG o = dihydroxyacetone phosphate glyceraldehyde-3-phosphate DG o = glyceraldehyde-3-phosphate + phosphate + NAD+ 1,3-diphosphoglycerate + NADH + H+ DG o = ,3-diphosphoglycerate + ADP 3-phosphoglycerate + ATP DG o = phosphoglycerate 2-phosphoglycerate DG o = phosphoglycerate 2-phosphoenolpyruvate + H 2 O DG o = phosphoenolpyruvate + ADP pyruvate + ATP DG o = pyruvate + NADH + H+ lactate + NAD+ DG o = pyruvate acetaldehyde + CO 2 DG o = acetaldehyde + NADH + H+ ethanol + NAD+ DG o = -23.7

17 Phosphorylation of glucose D-glucose + ATP D-glucose-6-phosphate + ADP DG o = The reaction can be decomposed into two reactions. D-glucose + phosphate D-glucose-6-phosphate + H 2 O DG o = ATP + H 2 O ADP + phosphate DG o = The sum of the two reactions results in an overall negative free energy change under standard conditions. In this manner the strongly spontaneous hydrolysis of ATP is coupled to the otherwise unspontaneous glucose phosphorylation. This reaction is typical of the role played by ATP in the cell. Note that the values for DG o assume a concentration of 1 M. Clearly, the concentrations in the cell are often quite different from the standard state and this will have profound consequences for the direction of spontaneous change.

18 Role of enzymes as catalysts All of the reactions in the glycolytic pathway are catalyzed by enzymes. For example, the phosphorylation of glucose is Catalyzed by hexokinase. The role of the enzyme is to speed up the reaction, but the enzyme does not change thermodynamics of the process. The role of enzymes is the same as that of any catalyst. Catalysts affect the kinetics of the reaction, but not the thermodynamics. We will consider the role of catalysts in the section on kinetics.

19 Equilibrium can be important even when DG o > 0 Notice that DG o for certain steps is positive. For example, D-glucose-6-phosphate D-fructose-6-phosphate DG o = +1.7 is catalyzed by phosphoglucose isomerase. The equilibrium constant for this process is K = exp{-dg o /RT} = exp{-1700/8.31/310} ~ 0.5 The concentration of D-fructose-6-phosphate at equilibrium will be less than that of D-glucose-6-phosphate.

20 Question Given that K ~ 0.5 for the reaction D-glucose-6-phosphate D-fructose-6-phosphate Calculate the concentration of D-fructose-6-phosphate at equilibrium under standard conditions (i.e. for initial concentrations of 1 M). A. 1 M B. 0.5 M C M D M

21 Question Given that K ~ 0.5 for the reaction D-glucose-6-phosphate D-fructose-6-phosphate Calculate the concentration of D-fructose-6-phosphate at equilibrium under standard conditions (i.e. for initial concentrations of 1 M). A. 1 M B. 0.5 M C M K = 1 + x, K(1 x) = 1 + x, K 1 = (1 + K)x 1 x x = K K = = [D gluc 6 phos] = 1 x = 1.33 [D fruc 6 phos] = 1 + x = 0.67 D M

22 Intracellular conditions are not equilibrium conditions If the subsequent step in a series of reactions is spontaneous this will tend to deplete the product for the previous reaction. Thus, more of the product will form by L Chateliers principle. We can observe this quantitatively by considering the value of Q, the reaction quotient. Since, DG = DG o + RT ln Q the value of DG may not be zero. In other words, the coupled Series of reactions in the cell are not at equilibrium. Rather, they proceed under steady state conditions where the concentrations are not at 1 M, but are poised so that the overall of effect on a series of reactions is to produce a net spontaneous change.

23 Sample Problem in Metabolism The enzyme aldolase catalyzes the conversion of fructose 1,6-diphosphate (FDP) to dihydroxyactone phosphate (DHAP) and glyceraldehyde-3-phosphoate (GAP). Under physiological conditions the concentrations of these species in red blood cells (erythrocytes) are [FDP] = 35 mm, [DHAP] = 130 mm and [GAP] = 15 mm. Will the conversion occur spontaneously under these conditions? Solution: The standard free energy change for the reaction is FDP DHAP + GAP DG o = kj and Q = [DHAP][GAP]/[FDP] = (130 x 10-6 )(15 x 10-6 )/ (35 x 10-6 ) = 5.57 x 10-5 DG = DG o + RT ln Q = J/mol + (8.31 J/mol-K)(310 K)ln(5.57 x 10-5 ) = J/mol or kj/mol The reaction will occur spontaneously under the conditions of the cell.

24 Thermodynamics of DNA hybridization A combination of spectroscopy and calorimetry was used to determine the free energies of melting of short oligonucleotides. Based on these measurements the free energy of a helix can be determined based on10 sets of nearest-neighbor pairs shown on the next slide. In addition to these values we need to know the free energy of the initiation (i.e. the first base pair). The overall free energy is then calculated from: DG o = DG o (initiation) + S DG o (nearest neighbors)

25 A 5 A 3 T T T 5 A 3 T A A 5 T 3 A T C 5 A 3 T G A 5 C 3 G T 5 AA 3 TT 5 AT 3 TA 5 TA 3 AT 5 AC 3 TG 5 CA 3 GT DG o DH o DS o Breslauer et al. PNAS (1986)

26 G 5 A 3 T C A 5 G 3 C T G 5 C 3 G C C 5 G 3 C G C 5 C 3 G G 5 AG 3 TC 5 GA 3 CT 5 CG 3 GC 5 GC 3 CG 5 CA 3 GT DG o DH o DS o Breslauer et al. PNAS (1986)

27 Sample problem Determine the melt temperature for the oligonucleotide 5 -ATAGCA TATCGT-5 5 -ATAGCA-3 3 -TATCGT-5 Solution: DG o = DG o (initiation) + S DG o (nearest neighbors) = DG o GC {initial) + DG o AT TA + TA DGo + DG o AG AT TC + GC DGo CG + CA DGo GT = = kj Initiation can be treated as a single value required to start the hybridization. It is unfavorable, but then all other interactions are favorable.

28 Sample problem Determine the melt temperature for the oligonucleotide 5 -ATAGCA TATCGT-5 5 -ATAGCA-3 3 -TATCGT-5 Solution (cont;d): DG o = kj Notice that the the free energy of initiation is positive. Initiation is unfavorable because of the entropy that must be overcome to bring the chains together. To calculate the melt temperature we need the enthalpy of reaction as well. DH o = DH o AT TA + DHo TA AT + DH o AG TC + GC DHo CG + CA DHo GT = = kj

29 Sample problem Given that DG o = kj/mol and DH o = kj/mol for the hexamer, determine the melt temperature. A. 42 o C B. 48 o C C. 52 o C D. 58 o C

30 Sample problem Given that DG o = kj/mol and DH o = kj/mol for the hexamer, determine the melt temperature. A. 42 o C B. 48 o C C. 52 o C D. 58 o C DS o = (DH o - DG o )/T = ( )/298 x1000 = J/mol-K The melt temperature occurs when DG o = 0. T = DH o /DS o = -164,400/(-495.3) = 331 K = 58 o C

31 DNA hyperchromicity

32 DNA hyperchromicity Observed in reverse by fluorescence

33 Folding and hybridization DNA hybridization is similar to protein folding in that they both have a melt temperature. The melt temperature refers to the temperature at which proteins unfold and DNA becomes single-stranded. We next consider the interactions that stabilize folded proteins.

34 Dipole-Dipole Interactions Dipoles often line up in this manner. Example: a-helix

35 Main Chain Electrostatic Interactions Coulomb s Law: V = q 1 q 2 /er Example of a hydrogen bond -N-H.. O=C- Example of a Salt Bridge Main Chain Lysine Glutamate

36 Hydrogen bonding in water

37 Hydrophobic interactions

38 Protein Folding Non-covalent forces in proteins What holds them together? Hydrogen bonds Salt-bridges Dipole-dipole interactions Hydrophobic effect Van der Waals forces What pulls them apart? Conformational Entropy

39 Contributions to DG TDS DH Internal Interactions Conformational Entropy -TDS Hydrophobic Effect Net: DG Folding

40 Protein folding example: Two state model U k f F Unfolded k u Folded K = [F]/[U] K = ff/(1-ff) Fraction folded ff Fraction unfolded 1-ff

41 Thermodynamic model ff = K/(1+K) K = e -DGo /RT ff = 1/(1 + e DGo /RT ) ff = 1/(1 + e DHo /RT e -DSo /R ) The temperature at which the protein is 50% folded can be defined as T m the melt temperature. At T m, DG o = 0 or T m = DH o /DS o.

42 Equilibrium melt curves Mostly folded o o Mostly unfolded T m In this case: T m = 300 K = DH o /DS o

43 Absorbance Delta Absorbance Thermal melt data 90x Temp ( C) Temp ( C) Temp ( C) x Wavenumber (cm -1 ) Wavenumber (cm -1 ) 1720 Infrared absorption spectra in the amide I region of the peptide. This is a probe of degree of folding.

44 The iron in heme is the binding site for oxygen and peroxide Heme is iron protoporphyrin IX. Functional aspects in Mb N O // O N 1. Discrimination against CO binding. N Fe N 2. O 2 is the physiologically relevant ligand, but it can oxidize iron (autooxidation). O O - O O -

45 Ligand binding in myoglobin Ligand binding to myoglobin is described as a chemical equilibrium between a bound state MbCO, a dissociated state Mb:CO and a solvent state Mb + CO. Mb + CO Mb:CO MbCO If we ignore the intermediate state the binding can be described by An overall equilibrium process. Mb + CO with equilibrium constant K. We have: The fraction bound is: f = K = [MbCO] [Mb][CO] MbCO [MbCO] [Mb] + [MbCO] We use CO in Many studies. The same equations Hold for O 2.

46 Ligand binding curve The fraction bound f can be related to the binding constant K: K = [MbCO] [Mb][CO] [MbCO] f = [Mb]+[MbCO] = K[Mb][CO] [Mb] + K[Mb][CO] = K[CO] 1 + K[CO] This type of binding curve is plotted below.

47 Cooperativity: hemoglobin If there are multiple binding sites in a macromolecule there can be two possible cases. First, they can be completely non-interacting and independent. In that case they behave exactly as a collection of monomers. On the other hand, the sites can interact. In that case we can speak of cooperativity or anti-cooperativity of the binding sites. We can use hemoglobin as an example.

48 Multiple equilibria There are four hemes in hemoglobin. Thus, there are four binding sites. The equilbrium constants are: Hb + O 2 Hb(O 2 ) Hb(O 2 ) + O 2 Hb(O 2 ) 2 K 1 = [Hb(O 2)] [Hb][O 2 ] K 2 = [Hb(O 2) 2 ] [Hb(O 2 )][O 2 ] Hb(O 2 ) 2 + O 2 Hb(O 2 ) 3 Hb(O 2 ) 3 + O 2 Hb(O 2 ) 4 K 3 = K 4 = [Hb(O 2 ) 3 ] [Hb(O 2 ) 2 ][O 2 ] [Hb(O 2 ) 4 ] [Hb(O 2 ) 3 ][O 2 ]

49 Combined equilibrium constant There are four hemes in hemoglobin. Thus, there are four binding sites. The equilbrium constants are: [Hb(O 2 )] = K 1 [Hb][O 2 ] [Hb(O 2 ) 2 ] = K 2 [Hb(O 2 )][O 2 ] = K 2 K 1 [Hb][O 2 ] 2 [Hb(O 2 ) 3 ] = K 3 [Hb(O 2 ) 2 ][O 2 ] = K 3 K 2 K 1 [Hb][O 2 ] 3 [Hb(O 2 ) 4 ] = K 4 [Hb(O 2 ) 3 ][O 2 ] = K 4 K 3 K 2 K 1 [Hb][O 2 ] 4 Define K = K 4 K 3 K 2 K 1 which is the equilbriumconstant for the overall reaction: Hb + 4O 2 Hb(O 2 ) 4 K = [Hb(O 2) 4 ] [Hb][O 2 ] 4

50 Cooperative binding isotherm The combined equilibrium treats the binding to hemoglobin as an all-or-nothing process. That is, either there are no O2 ligands or there are four O2 ligands. The is the extreme of cooperative binding. In this case the fraction bound is: f = [Hb(O 2 ) 4 ] [Hb] + [Hb(O 2 ) 4 ] = K [Hb][O 2 ] 4 [Hb] + K [Hb][O 2 ] = K [O 2] K [O 2 ] 4 The isotherm for cooperative binding is shown in the red curve below:

51 The Hill Coefficient In reality the cooperativity is not complete. In other words the binding of ligands does not occur all at once. The binding is sequential such that when then first oxygen (or CO) binds it increases the binding affinity of the second one etc. One way to express the fact that the cooperativity is less than perfect is to use n as a parameter. For hemoglobin n = 1 to 4 depending on the degree of communication between the four subunits. f = K [O 2] n 1 + K [O 2 ] n Experimentally the degree of ligation (O 2 or CO binding) can be followed using UV-vis spectroscopy. The effects of mutations on the local Environment of the bound ligand can be probed using infrared spectroscopy of bound CO.

52 Monod-Wyman-Changeux Theory The original idea of the MWC theory was to describe protein cooperativity in terms of two states R = relaxed state T = tense state In the relaxed state there was little interaction between the protein subunits. A binding event would cause the structure to change and the protein subunits would begin to interact strongly. This would give rise to the tense state (T). In hemoglobin the R state is the oxy form. When oxygen dissociates the heme iron moves out of the heme plane. This occurs because the iron changes from a low spin to a high spin metal. The change in iron position permits the F-helix to move into contact with the CD loop region of the neighboring subunit. T R

53 Structural changes in Hb The change in intersubunit contacts results in a rotation of the a/b subunits so that the overall structure of the tetrameric protein is changed in the R to T transition.

54 The origin of hemoglobin cooperativity The key amino acid residues that change their hydrogen bonding are shown below. The intersubunit contacts are altered by the binding of oxygen to the heme iron because of the change in the position of the heme iron. In-plane iron locks in the R state (pulls subunits apart) Out-of-plane iron iron gives the T state (allows subunits to interact)

55 The Bohr effect Acid Conditions Help Release O 2 in the Tissues Hemoglobin must bind oxygen tightly in the lungs Some unloading in tissue results from the low po 2 Active tissues make lots of acid (carbonic and lactic), which helps to unload oxygen from the hemoglobin Bohr effect is the increased unloading of O 2 at low ph Figure The black curve is at ph 7.4, The red curve is at ph 7.1 The blue curve is at ph 6.8

56 The physiology of Hb and Mb Oxygen Uptake in the Lungs is Increased About 70 times by Hemoglobin in the Red Cells A small amount of oxygen dissolves directly in the serum, but 98.5% of the oxygen is carried by Hb Hb is found within the red blood cells Hb content of blood is ~15 gm/100 ml Each Hemoglobin Can Bind Four O 2 Molecules (100% Saturation) Hemoglobin is a protein molecule with (2 a and 2 b) When hemoglobin is saturated with oxygen it has a bright red color; as it loses oxygen it becomes bluish (cyanosis)