Chemistry 433 Lecture 15 Applications of Free Energy NC State University
Thermodynamics of glycolysis Reaction kj/mol D-glucose + ATP D-glucose-6-phosphate + ADP ΔG o = -16.7 D-glucose-6-phosphate p D-fructose-6-phosphate p ΔG o = +1.7 D-fructose- 6-diphosphate + ATP D-fructose-1,6-diphosphate + ADP ΔG o = -14.2 D-fructose-1,6-diphosphate glyceraldehyde-3-phosphate + dihydroxyacetone phosphate ΔG o = +23.8 dihydroxyacetone phosphate glyceraldehyde-3-phosphate ΔG o = + 7.5 glyceraldehyde-3-phosphate + phosphate + NAD+ 13di 1,3-diphosphoglycerate h + NADH + H+ ΔG o = + 63 6.3 1,3-diphosphoglycerate + ADP 3-phosphoglycerate + ATP ΔG o = -18.8 3-phosphoglycerate 2-phosphoglycerate ΔG o = +4.6 2-phosphoglycerate 2-phosphoenolpyruvate + H 2 O ΔG o = +1.7 2- phosphoenolpyruvate + ADP pyruvate + ATP ΔG o = -31.4 pyruvate + NADH + H+ lactate + NAD+ ΔG o = -25.1 pyruvate acetaldehyde + CO o 2 ΔG =-19.8 acetaldehyde + NADH + H+ ethanol + NAD+ ΔG o = -23.7
Phosphorylation of glucose D-glucose + ATP D-glucose-6-phosphate + ADP ΔG o = -16.7 The reaction can be decomposed into two reactions. D-glucose + phosphate D-glucose-6-phosphate + H 2 O ΔG o = +14.3 ATP + H 2 O ADP + phosphate ΔG o = -31.0 The sum of the two reactions results in an overall negative free energy change under standard d 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 ΔG 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.
The role of enzymes All of the reactions in the glycolytic pathway are catalyzed by enzymes. For example, the reaction considered on the previous slide 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 second half of the course. Notice that ΔG o for certain steps is positive. For example, D-glucose-6-phosphate h D-fructose-6-phosphate h t ΔG o = +1.7 17 is catalyzed by phosphoglucose isomerase. The equilibrium constant for this process is K = exp{-δg 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.
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. 0.667 M D. 0.15 M
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 K = 1+x, K(1 ( x) ) =1+x, K 1=(1 +K)x 1 x C. 0.667 M x = K 1 1+K = 0.5 1.5 = 0.333 D. 0.15 M [D gluc 6 phos] =1 x =1.33 [D fruc 6 phos] =1+x =0.67
Intracellular conditions are not equilibrium conditions If the subsequent step in a series of reactions is highly hl spontaneous this will tend to deplete the product for the previous reaction. Thus, more of the product will tend to be formed by L Chateliers principle. We can observe this quantitatively by considering the value of Q, the reaction quotient. Since, ΔG = ΔG o + RT ln Q the value of ΔG may not be zero. In other words, the coupled series of reactions in the cell are not at equilibrium but 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.
Sample Problem in Metabolism The enzyme aldolase catalyzes the conversion of fructose 1,6-diphosphate (FDP) to dihydroxyactone phosphate (DHAP) and glyceraldehyde-3-phosphoate gy y p (GAP). Under physiological conditions the concentrations of these species in red blood cells (erythrocytes) are [FDP] = 35 μm, [DHAP] = 130 μm and [GAP] = 15 μm. Will the conversion occur spontaneously under these conditions? Solution: The standard free energy change for the reaction is FDP DHAP + GAP ΔG o = +23.8 kj and Q = [DHAP][GAP]/[FDP] = (130 x 10-6 )(15 x 10-6 )/ (35 x 10-6 ) = 5.57 x 10-5 ΔG = ΔG o + RT ln Q = 23800 J/mol + (8.31 J/mol-K)(310 K)ln(5.57 x 10-5 ) = -1434 J/mol or -1.43 kj/mol The reaction will occur spontaneously under the conditions of the cell.
Protein folding example: Two state t model k f U F k u Unfolded Folded K = [F]/[U] K = ff/(1-ff) Fraction folded ff Fraction unfolded 1-ff
Thermodynamic model for two-state equilibria K=ff/(1-ff) ff) ff = K/(K+1) K = e -ΔGo /RT ff = 1/(1 + e ΔGo /RT ) ff = 1/(1 + e ΔHo /RT e -ΔSo /R ) The temperature at which the protein is 50% folded or DNA is 50% hybridized can be defined as T m the melt temperature. At T m, ΔG o = 0 or T m = ΔH o /ΔS o.
Equilibrium melt curves Proteins or DNA o o Mostly folded Mostly hybridized T m Mostly unfolded Mostly ssdna In this case: T m = 300 K = ΔH o /ΔS o
Van t Hoff plots Slope = -ΔH o /R The standard method for obtaining the The standard method for obtaining the reaction enthalpy is a plot of ln K vs. 1/T
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 d 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: ΔG o = ΔG o (initiation) + Σ ΔG o (nearest neighbors)
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 ΔG o ΔH o ΔS o 7.9 38.1 101.3 6.3 36.0 99.7 3.8 25.1 71.5 5.4 27.2 73.2 7.9 24.3 55.0 Breslauer et al. PNAS 93 3746 (1986)
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 ΔG o ΔH o ΔS o 6.7 32.6 86.9 6.7 23.4 56.0 15.1 49.8 116.4 13.0 46.4 112.1 13.0 46.0 110.7 Breslauer et al. PNAS 93 3746 (1986)
Sample problem Determine the melt temperature for the oligonucleotide 5 -ATAGCA-3 + 3 -TATCGT-5 5 -ATAGCA-3 3 -TATCGT-5 Solution: ΔG o = ΔG o (initiation) + Σ ΔG o (nearest neighbors) = ΔG o GC init + ΔG o AT + TA ΔGo + ΔG o AG + GC ΔGo + CA TA AT TC CG ΔGo GT = 20.9-6.3-3.8-6.7-13.0-6.9 = -16.8 kj Note that we use GC initiation if there is a single GC base pair. Only use AT initiation of the strands are all A and T.
Sample problem Determine the melt temperature for the oligonucleotide 5 -ATAGCA-3 + 3 -TATCGT-5 5 -ATAGCA-3 3 -TATCGT-5 Solution (cont;d): ΔG o = -16.8 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. ΔH o = ΔH o AT + TA ΔHo + ΔH o AG + GC ΔHo + CA TA AT TC CG ΔHo GT = - 36.0-25.1-32.6-46.4-24.3 = -164.4 kj
Sample problem Given that ΔG o = -16.8 kj/mol and ΔH o = -164.4 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
Sample problem Given that ΔG o = -16.8 kj/mol and ΔH o = -164.4 kj/mol for the hexamer, determine the melt temperature. A. 42 o C B. 48 o C C. 52 o ΔS C o = (ΔH o - ΔG o )/T = (-164.4+16.8)/298 x1000 D. 58 o = -495.3 J/mol-K C The melt temperature occurs when ΔG o = 0. T = ΔH o /ΔS o = -164,400/(-495.3) = 331 K = 58 o C
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
Dipole-Dipole p Interactions Dipoles often line up in this manner. Example: α-helixh
Main Chain Electrostatic Interactions Coulomb s Law: V = q 1 q 2 /εr Example of a hydrogen bond -N-H.. O=C- Example of a Salt Bridge Main Chain Lysine Glutamate
Hydrogen bonding in water
Hydrophobic interactions
Contributions to ΔG - 0 + -TΔS ΔH Internal Interactions Conformational Entropy -TΔS Hydrophobic Effect Net: ΔG Folding
Sample problem The thermal unfolding of a serine protease (SP) from a thermophilic bacterium was studied and gave the following Thermodynamic data: ΔG o = -15.4 kj/mol and ΔH o = -75.0 kj/mol for the process SP (random coil) SP (native) Determine the melt temperature. A. 82 o C B. 92 o C C. 102 o C D. 112 o C
Sample problem The thermal unfolding of a serine protease (SP) from a thermophilic bacterium was studied and gave the following Thermodynamic data: ΔG o = -15.4 kj/mol and ΔH o = -75.0 kj/mol for the process SP (random coil) SP (native) Determine the melt temperature. A. 82 o C B. 92 o C C. 102 o C D. 112 o C ΔS o = (ΔH o - ΔG o )/T = (-75.0+15.4)/298 x1000 = -200 J/mol-K The melt temperature occurs when ΔGG o = 0. T = ΔH o /ΔS o = -75,000/(-200) = 375 K = 102 o C!