Thermodynamics of glycolysis. Chemistry 433. The role of enzymes. Phosphorylation of glucose. Question. Question

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1 hemistry 433 Lecture 15 pplications of Free Energy N State University hermodynamics of glycolysis Reaction kj/mol D-glucose P D-glucose-6-phosphate DP Δ o = D-glucose-6-phosphate D-fructose-6-phosphate Δ o = 1.7 D-fructose- 6-diphosphate P D-fructose-1,6-diphosphate DP Δ o = D-fructose-1,6-diphosphate glyceraldehyde-3-phosphate dihydroxyacetone phosphate Δ o = 23.8 dihydroxyacetone phosphate glyceraldehyde-3-phosphate Δ o = glyceraldehyde-3-phosphate phosphate ND 1,3-diphosphoglycerate NDH H Δ o = 6.3 1,3-diphosphoglycerate DP 3-phosphoglycerate P Δ o = phosphoglycerate 2-phosphoglycerate Δ o = phosphoglycerate 2-phosphoenolpyruvate H 2 O Δ o = phosphoenolpyruvate DP pyruvate P Δ o = pyruvate NDH H lactate ND Δ o = pyruvate acetaldehyde O 2 Δ o = acetaldehyde NDH H ethanol ND Δ o = Phosphorylation of glucose D-glucose P D-glucose-6-phosphate DP Δ o = he reaction can be decomposed into two reactions. D-glucose phosphate D-glucose-6-phosphate H 2 O Δ o = 14.3 P H 2 O DP phosphate Δ o = he sum of the two reactions results in an overall negative free energy change under standard conditions. In this manner the strongly spontaneous hydrolysis of P is coupled to the otherwise unspontaneous glucose phosphorylation. his reaction is typical of the role played by P in the cell. Note that the values for Δ o assume a concentration of 1 M. learly, 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. he role of enzymes ll of the reactions in the glycolytic pathway are catalyzed by enzymes. For example, the reaction considered on the previous slide is catalyzed by hexokinase. he role of the enzyme is to speed up the reaction, but the enzyme does not change thermodynamics of the process. he role of enzymes is the same as that of any catalyst. atalysts 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 Δ o for certain steps is positive. For example, D-glucose-6-phosphate D-fructose-6-phosphate Δ o = 1.7 is catalyzed by phosphoglucose isomerase. he equilibrium constant for this process is K = exp{-δ o /R} = exp{-1700/8.31/310} ~ 0.5 he concentration of D-fructose-6-phosphate at equilibrium will be less than that of D-glucose-6-phosphate. Question iven that K ~ 0.5 for the reaction D-glucose-6-phosphate D-fructose-6-phosphate alculate the concentration of D-fructose-6-phosphate at equilibrium under standard conditions (i.e. for initial concentrations of 1 M).. 1 M Question iven that K ~ 0.5 for the reaction D-glucose-6-phosphate D-fructose-6-phosphate alculate the concentration of D-fructose-6-phosphate at equilibrium under standard conditions (i.e. for initial concentrations of 1 M).. 1 M B. 0.5 M M D M B. 0.5 M M D M K = 1x, K(1 x) =1x, K 1=(1K)x 1 x x = K 1 1K = = [D gluc 6 phos] =1 x =1.33 [D fruc 6 phos] =1x =0.67 1

2 Intracellular conditions are not equilibrium conditions If the subsequent step in a series of reactions is highly spontaneous this will tend to deplete the product for the previous reaction. hus, more of the product will tend to be formed by L hateliers principle. We can observe this quantitatively by considering the value of Q, the reaction quotient. Since, Δ = Δ o R ln Q the value of Δ 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 he enzyme aldolase catalyzes the conversion of fructose 1,6-diphosphate (FDP) to dihydroxyactone phosphate (DHP) and glyceraldehyde-3-phosphoate (P). Under physiological conditions the concentrations of these species in red blood cells (erythrocytes) are [FDP] = 35 μm, [DHP] = 130 μm and [P] = 15 μm. Will the conversion occur spontaneously under these conditions? Solution: he standard free energy change for the reaction is FDP DHP P Δ o = 23.8 kj and Q = [DHP][P]/[FDP] = (130 x 10-6 )(15 x 10-6 )/ (35 x 10-6 ) = 5.57 x 10-5 Δ = Δ o R ln Q = J/mol (8.31 J/mol-K)(310 K)ln(5.57 x 10-5 ) = J/mol or kj/mol he reaction will occur spontaneously under the conditions of the cell. Protein folding example: wo state model U Unfolded k f k u K = [F]/[U] F Folded K = ff/(1-ff) Fraction folded ff Fraction unfolded 1-ff hermodynamic model for two-state equilibria K=ff/(1-ff) ff = K/(K1) K = e -Δo /R ff = 1/(1 e Δo /R ) ff = 1/(1 e ΔHo /R e -ΔSo /R ) he temperature at which the protein is 50% folded or DN is 50% hybridized can be defined as m the melt temperature. t m, Δ o = 0 or m = ΔH o /ΔS o. Equilibrium melt curves Proteins or DN Van t Hoff plots o o Slope = -ΔH o /R Mostly folded Mostly hybridized d m Mostly unfolded Mostly ssdn In this case: m = 300 K = ΔH o /ΔS o he standard method for obtaining the reaction enthalpy is a plot of ln K vs. 1/ 2

3 hermodynamics of DN hybridization 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). he overall free energy is then calculated from: Δ o = Δ o (initiation) Σ Δ o (nearest neighbors) Δ o ΔH o ΔS o Breslauer et al. PNS (1986) Δ o ΔH o ΔS o Breslauer et al. PNS (1986) Determine the melt temperature for the oligonucleotide Solution: Δ o = Δ o (initiation) Σ Δ o (nearest neighbors) = Δ o init Δ o Δo Δo Δo Δo = = kj Note that we use initiation if there is a single base pair. Only use initiation of the strands are all and. Determine the melt temperature for the oligonucleotide iven that Δ o = kj/mol and ΔH o = kj/mol for the hexamer, determine the melt temperature.. 42 o B. 48 o. 52 o D. 58 o Solution (cont;d): Δ 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. o calculate the melt temperature we need the enthalpy of reaction as well. ΔH o = ΔH o ΔHo ΔHo ΔHo ΔHo = = kj 3

4 iven that Δ o = kj/mol and ΔH o = kj/mol for the hexamer, determine the melt temperature.. 42 o B. 48 o. 52 o D. 58 o ΔS o = (ΔH o - Δ o )/ = ( )/298 x1000 = J/mol-K he melt temperature occurs when Δ o = 0. = ΔH o /ΔS o = -164,400/(-495.3) = 331 K = 58 o 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? onformational Entropy Dipole-Dipole Interactions Main hain Electrostatic Interactions oulomb s Law: V = q 1 q 2 /εr Example of a hydrogen bond -N-H.. O=- Example of a Salt Bridge Main hain Dipoles often line up in this manner. Example: α-helix Lysine lutamate Hydrogen bonding in water Hydrophobic interactions 4

5 ontributions to Δ - 0 ΔH Internal Interactions -ΔS onformational Entropy he thermal unfolding of a serine protease (SP) from a thermophilic bacterium was studied and gave the following hermodynamic data: Δ o = kj/mol and ΔH o = kj/mol for the process SP (random coil) SP (native) Determine the melt temperature. t Net: -ΔS Hydrophobic Effect Δ Folding. 82 o B. 92 o. 102 o D. 112 o he thermal unfolding of a serine protease (SP) from a thermophilic bacterium was studied and gave the following hermodynamic data: Δ o = kj/mol and ΔH o = kj/mol for the process SP (random coil) SP (native) Determine the melt temperature. t. 82 o B. 92 o. 102 o D. 112 o ΔS o = (ΔH o - Δ o )/ = ( )/298 x1000 = -200 J/mol-K he melt temperature occurs when Δ o = 0. = ΔH o /ΔS o = -75,000/(-200) = 375 K = 102 o! 5