Biochemistry I Introduction to Metabolism Bioenergetics: Thermodynamics in Biochemistry ATP 4- + 2 ADP 3- + P i 2- + + Chapter 15 part 2 Dr. Ray 1
Energy flow in biological systems: Energy Transformations in Biochemistry There is an energy cost to create ordered biopolymers: 1. Energy can be used to do make bonds (biosynthesis) 2. Energy can be used to make order out of disorder (proteins, DNA, and RNA with specific sequences), are information rich macromolecules made via templatedirected synthesis Lehninger 5 th Ed, Figure 1.24 2
Glycolysis Stages 1 & 2: Energy Investment & Cleavage C Phosphorylation C 2 Glucose C C 2 1 2 3 C 2 P - C 2 P - G6P - - F6P Stage 3: Energy Recovery xidation coupled to Phosphorylation Isomerization Phosphoylation ATP ADP ATP ADP DG o = -16.7 kj/mol Which steps in glycolysis involve energy utilization, and which involve energy production? Substrate Level Phosphorylation Isomeri C 2 P - - C 2 FBP Retro-Aldol Cleavage P - Dehydra Goal: make two 3-carbon molecules (GAP), each containing a phosphate - C 2 C DAP C 2 P - - C 2 GAP 4 5 P -zation - - -tion - - - C C - C 6 7 C 8 9 C 10 C C P - C P - C C 2 P - C 2 P - C 2 P - - - C 2 C -NAD + NAD - ADP ATP - C 3 + P GAP i 13BPG 3PG 2PG ADP ATP Pyruvate PEP P - Isomeri -zation Goal: generate ATP directly and by forming reduced cofactors (NAD) - Substrate Level Phosphorylation DG o = - 31.4 kj/mol
Bioenergetics Questions Steps 7 and 10 of glycolysis are substrate level phosphorylation reactions catalyzed by kinases, during which ATP is synthesized. 1) Calculate the DG for step 7 of glycolysis: 1,3-BPG + ADP 3-PG + ATP DG o =? Utilize the info given in Text Table 15.1 standard free energies of hydrolysis of phosphorylated compounds 1,3-BPG + 2 3-PG + P i DG o = - 49.4 kj/mol ATP + 2 ADP + P i DG o = - 30.5 kj/mol Solution: set up reactions so appropriate species are reactants or products (change sign of DG o if reverse reaction). Then add reactions & energies: 1,3-BPG + 2 3PG + P i DG o = kj/mol ADP + P i ATP + 2 DG o = kj/mol 1,3-BPG + ADP 3PG + ATP DG o = kj/mol 2) Does oxidation of organic compounds (aldehyde carboxylic acid) produce or require energy? 4
Formation of ATP Using Energy in GAP 1) Can ATP be made directly using compounds with high phosphoryl group transfer potential? Consider glyceraldehyde 3-phosphate, a metabolite formed during the oxidation of glucose via glycolysis. Steps 6 & 7 of Glycolysis: The carbon at C1 is an aldehyde and is NT in its most oxidized state. xidation of the aldehyde to a carboxylic acid will RELEASE energy. This carbon oxidation does not take place directly, rather in 2 steps with 1,3-BPG (1,3-bisphosphoglycerate) generated at first. 2) What functional group is formed at C1 of 1,3-BPG? Step 6 of Glycolysis: The energy released in this step is captured in two ways : 5 (1) NAD (e - carrier) and (2) 1,3-BPG (phosphoryl carrier)
Formation of ATP by Substrate Level Phosphorylation ydrolysis of 1,3-BPG is coupled to the synthesis of ATP: Step 7 of Glycolysis: The oxidation energy of a carbon atom (C1 of glyceraldehyde) is captured in step 6 of glycolysis, and transformed into the phosphoryl transfer potential of ATP in two ways: (1) Cleavage of 1,3-BPG is coupled to direct formation of 1 ATP (step 7) by substrate level phosphorylation (SLP). (2) NAD (electron carrier) formed in glycolysis is later reoxidized in the mitochondria to NAD + via the electron transport chain (ET), resulting in formation of an additional 1.5 ATP molecules by oxidative phosphorylation (P). 1. Why is the phosphate at C1 of 1,3-BPG used to form ATP, rather than that at C3? C1 is a while C3 is a 6
Equilibrium Thermodynamics and Kinetics The standard free energy change of reaction (DG o ): is the energy difference between the products and reactants of a rxn is related to the equilibrium constant (K eq ) of a reaction: K For reaction: A + B C + D eq' C D A B o' DG RT log K DG o = D o - TDS o e ' eq DDG DG rxn Physical Chemistry Standard State: DG o Temp = 298 K = 25 o C Concentration of all reactants and products = 1 M (molar) but [ + ] = 1M means p = 0 Biochemical standard state denoted by the symbol DG o [p=7] means [ + ] = 1x10-7 M, the concentration of water is constant = 55.5 M, and concentration of all other species is 1 M 7 Primes ( ) are used to indicate biochemical standard state
Free Energy Changes and Equilibrium Equilibria: A + B C + D The standard free-energy (DG o ) and the equilibrium constant (K eq ) can be calculated from each other: K eq' o' C D A B = DG RT log K e more products less reactants ' eq K eq =10 -DGo /1.36 Each factor of 10 in K eq (10:1 ratio of P to R) means difference of 5.7 kj/mol in DG o R = 1. 987 x 10-3 kcal mol -1 deg -1 T = 298 K (25 o C) Are the following statements TRUE or FALSE? (1) When DG o = 1.0 kj/mol, then K eq = 1 (2) When DG o is negative, then K eq > 1 Ref: Lehninger Biochemistry 8
Bioenergetics Question 1) A mixture of 3-phosphoglycerate (3PG) and 2-phosphoglycerate (2PG) combined in equimolar amounts (1M each), and is incubated at 25 o C until equilibrium is reached. The final mixture contains six times as much 2PG as 3PG. Which one of the following statements is most nearly correct, when applied to the reaction as written? (R=8.315 J/moL. K) 3-phosphoglycerate (3PG) 2-phosphoglycerate (2PG) Start: End: A) DG o is zero B) DG o cannot be calculated from the information given C) DG o is incalculably large and positive D) DG o is + 12.7 kj/mol E) DG o is - 4.44 kj/mol So rxn must have gone to the for to have been formed So DG o must be Remember each factor of 10 in K eq (10:1 ratio of Products to Reactants) 9 means diff 5.7 kj/mol in DG o.
Free Energy Changes in Non-standard Conditions C A + B C + D K DG RT log K eq' D A B DG o informs whether a reaction will proceed in the forward or backward direction at standard conditions (initial 1M concentrations) Does the spontaneity of a reaction depend only on the relative energies of the reactants and products? The DG of a reaction tells us whether a reaction can or cannot occur spontaneously, or whether the system is at equilibrium for ANY starting concentrations! DG = DG o +RT ln Q Q = reaction quotient Ratio of reactant & product concentrations at start of reaction p = 7 [ 2 ] = constant Initial reactant & product concentrations can be ANYTING! o' e ' eq DG gives DRIVING FRCE of reaction. p = 7 [ 2 ] = constant Reactant & product concentrations are ALL 1M at start!
Reactions of Glycolysis (Table 16.1 in text) 1. What is the difference? DG = DG o +RT ln Q Actual DRIVING FRCE is different than standard conditions [since physiological metabolite concentrations are not 1M ] Glycolysis step 1: Cellular concentrations: LARGE SMALL ATP + Glucose Glucose-6-P + ADP DG o = - 16.7 kj/mol ACTUAL CNCENTRATINS IN A CELL CAUSE TE EQUILIBRIUM T SIFT FURTER T TE There is a bigger driving force, so So DG is more than DG o
Concentrations and Free Energy 1) What happens to DG (rxn driving force) when concentrations of reactants and products change? 2) Does DG o = DG? A + B C + D o' DG RT log K e ' eq DG o is another way to express where the equilibrium of a reaction lies, it does NT provide any insights about the spontaneity of a reaction, which is determined by DG. 3) Does the spontaneity of a reaction depend on concentrations of reactants & products (ie, will reaction proceed in forward direction as written without the input of energy)? A reaction will occur spontaneously only if DG < 0 DG = DG o +RT ln Q DG (spontaneity of a reaction its direction of flow in actual cellular conditions) depends on both the chemical nature of reactants (DG o ) and on their relative concentrations. 12
Free Energy Changes and Spontaneity A + B C + D DG = D - TDS K C D o' eq' DG RT log K K eq =10 -DGo /1.36 A B 1. Biochemical Standard State ( DG o ) has p = 7, [ 2 ] = 55.5 M, and concentration of all other species is 1 M. DG o informs on the chemical nature of reactants and products: their tendency to react and the relative stability of reactants & products If reaction starts with 1M of all species, K eq and DG o inform on what concentrations will occur once the reaction has reached equilibrium. 2. The DG of a reaction is independent of the path of the transformation. 3. The DG of a reaction tells us whether a reaction can or cannot occur spontaneously, or whether the system is at equilibrium at given conditions (ANY CNCENTRATINS)! DG < 0 reaction occur spontaneously from A, B to C, (forward reaction as written occurs) DG > 0 reaction occur spontaneously from C, D to A, B (backwards reaction occurs, but forward does not) DG = 0 reaction is at equilibrium (no change in [ ] of all) e ' eq 13
ATP ydrolysis Energy Depends on Concentrations 1) ow is the energy stored in ATP released? A large amount of free energy is liberated when ATP is. ATP + 2 ADP + P i DG o = - 30.5 kj/mol 2) Under typical cellular concentrations of ATP, ADP & P i is the actual DG for this hydrolysis reaction as given above or different? WY? DG is approximately - 50 kj mol -1 (- 12 kcal mol -1 ) with a stronger driving force than at standard conditions, because it depends on BT: (1) the standard free energy change and DG = DG o + RT ln Q (2) the concentrations of all species present Cells usually maintain a [ATP]/[ADP][P i ] ratio of 500:1 (known as the cell s energy charge ). DG = Non-standard conditions: reactants & products are initially present at ANY concentrations DG o = Biochemical standard conditions: all species initially present at 1 M concentration, except [ + ] = 1 x 10-7 M (at p = 7) Q = [products]/[reactants], reaction quotient for actual cellular concentrations (not equilibrium concentrations)
Stryer Textbook Problem 21 The concentrations of ATP, ADP, and P i differ with cell type. Consequently, the release of free energy with the hydrolysis of ATP will vary with cell type. Using the following table, calculate the DG for the hydrolysis of ATP in liver, muscle, and brain cells. In which cell type is the free energy of ATP hydrolysis most negative? [ADP][Pi] ATP (mm) ADP (mm) Pi (mm) ln [ATP] DG Liver 3.5 1.8 5.0-5.963 Muscle 8.0 0.9 8.0-7.013 Brain 2.6 0.7 2.7-7.227 ATP + 2 ADP + P i DG o = - 30.5 kj/mol DG = DG o + RT ln 15 T = 37 o C
Metabolic Pathway Integration There is coordination between glycolysis and gluconeogenesis during a sprint, to ensure the energy needs of all types of cells are met.
Regulation of Metabolism via Cell s Energy Charge There is 10 8 times as much product B formed when the conversion of A to B is coupled with the hydrolysis of ATP. Energy Charge = [ATP] + ½ [ADP] [ATP] + [ADP] + [AMP] 0 = all AMP 1 = all ATP igh energy charge: inhibits ATP generating pathways stimulates ATP utilizing pathways Energy charge, like p of cell, is buffered between 0.80 0.95 Phosphorylation potential is: [ATP]/[ADP][P i ] cells usually maintain 500:1 ratio depends on [P i ], so directly related to free energy storage available from ATP
ow Far Does Coupling a Reaction to ATP ydrolysis Shift the Equilibrium Constant of the Reaction? FYI details Consider a chemical reaction that is thermodynamically unfavorable without an input of free energy, a situation common to many biosynthetic (anabolic) reactions, and some catabolic ones (ex: steps 1 & 3 of glycolysis). Suppose that the standard free energy of the conversion of compound A into compound B is: So rxn A B will NT occur since K eq << 1 K eq = [B] eq = 1.15 x 10-3 = 1 [A] eq 1150 owever, A can be converted into B under these conditions IF the reaction is coupled to the hydrolysis of ATP. The new overall reaction is: Now, rxn A B will occur since Keq >> 1 K eq = 2.67 x 10 2 = [products] eq = 267 [reactants] eq 1 1) Why are these two K eq so different? 2 nd reaction is coupled to ATP hydrolysis 2) What is the expression for the equilibrium constant of the 2 nd reaction?
ATP ydrolysis Shifts the Equilibria of Coupled Reaction by a Factor of 10 8 FYI details 2) What is the expression for the equilibrium constant of the 2 nd reaction? At equilibrium, the ratio of [B] to [A] for the coupled reaction is given by: A B The ATP-generating system of cells maintains the [ATP]/[ADP][P i ] ratio (the energy charge at a high level, typically of the order of 500 M -1 ). For this ratio: = 134,000 1 Coupling the hydrolysis of ATP with the conversion of A into B, changes the equilibrium ratio of B to A by a factor of about 10 8! There is 10 8 times as much product B formed when the conversion of A to B is coupled with the hydrolysis of ATP. Cells maintain a high level of ATP by using oxidizable substrates or light as sources of free energy. Know this 19
Bioenergetics Questions 1) When one mole of maltose is degraded to pyruvate through the actions of an a-glucosidase enzyme and the glycolytic pathway, the net production of ATP is: A) 4 mol B) 3 mol C) 2 mol D) 1 mol E) 0 mol 2) In glycolysis, fructose-1,6-bisphosphate is converted by aldolase into two products, with a standard free-energy change DG o of + 23.8 kj/mol. Under what conditions encountered in a normal cell will the free-energy change DG be negative, enabling the reaction to proceed spontaneously to the right? A) under standard conditions, enough energy is released to drive the reaction to right B) the reaction will not go to the right spontaneously under any conditions because the DG o is positive. C) the reaction will spontaneously proceed to the right if initially there is a high concentration of products relative to the concentration of fructose-1,6-bisphosphate. D) the reaction will spontaneously proceed to the right if initially there is a high concentration of fructose-1,6-bisphosphate relative to the concentration of the products. E) the driving force of this reaction is supplied by a coupled exergonic reaction, the hydrolysis of ATP.
Bioenergetics Questions 1) If the DG of the reaction A B is - 2.0 kj/mol, which one(s) of the following statements are TRUE? (a) The reaction will proceed spontaneously from left to right at experimental conditions. (b) The reaction will proceed spontaneously from right to left at standard conditions. (c) The equilibrium constant favors the formation of B over the formation of A. (d) The equilibrium constant could be calculated if the initial concentrations of A and B were known. (e) The value of DG o is also negative. DG o = - RT ln K eq Spontaneity depends on DG not DG o! 21 DG = DG o + RT ln Q
verview of Carbohydrate Catabolism makes 26 ATP xidative Phosphorylation C C 2 GLYCLYSIS 2 x pyruvate (3C) 2 x C 2 C 2 - C 3 C 2 CoAS 4C oxaloacetate D-glucose (6C) Total output for NE molecule of glucose - C Make 2 ATP C 3 Pyruvate 2 NAD CoAS 2 NAD PDC C 3 Acetyl CoA Substrate Level Phosphorylation 2 C 2 C 3 CITRIC ACID CYCLE 2 x Acetyl CoA (2C) 4 x C 2 (1C) 6C citrate 6 NAD 2 FAD 2 ELECTRN TRANSPRT CAIN Result for one glucose: Make 30 ATP PDC = Pyruvate Dehydrogenase Complex 2 GTP
Bioenergetics Question 1) A mixture of 3-phosphoglycerate (3PG) and 2-phosphoglycerate (2PG) combined in equimolar amounts (1M each), and is incubated at 25 o C until equilibrium is reached. The final mixture contains six times as much 2PG as 3PG. Which one of the following statements is most nearly correct, when applied to the reaction as written? (R=8.315 J/moL. K) 3-phosphoglycerate (3PG) 2-phosphoglycerate (2PG) If 7x = 2 M then x = 2 M 7 Start: 1M 1M = 2M total End: 2 M 12 M = 14 1x 6x M = 2M total 7 7 7 A) DG o So rxn must have gone to the right (forward) is zero for more product 2PG to have been formed B) DG o cannot be calculated from the information given C) DG o is incalculably large and positive D) DG o is + 12.7 kj/mol So DG o must be negative E) DG o is - 4.44 kj/mol Ans: E Remember each factor of 10 in K eq (10:1 ratio of Products to Reactants) means diff 5.7 kj/mol in DG o. ere K eq = 6/1, so DG o < 5.7 kj/mol.
Bioenergetics Questions 3) If the DG of the reaction A B is - 2.0 kj/mol, which one(s) of the following statement(s) is TRUE? (a) The reaction will proceed spontaneously from left to right at experimental conditions. (b) The reaction will proceed spontaneously from right to left at standard conditions. (c) The equilibrium constant favors the formation of B over the formation of A. (d) The equilibrium constant could be calculated if the initial concentrations of A and B were known. Ans: A, D (e) The value of DG o is also negative. DG o = RT ln K eq (two correct answers) Spontaneity depends on DG not DG o! 24 - - or + if know Q and DG DG = DG o + RT ln Q standard can calc DG o actual concs