G = (G C+ G D ) (G A + G B ) = ΣG P ΣG R Η = (H C+ H D ) (H A + H B ) = ΣΗ P ΣΗ S = (S C+ S D ) (S A + S B ) = ΣS P ΣS R BACKGROUND

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1 BACKGRUND BACKGRUND Α + Β = C + D G = Η Τ Τ S G = (G C+ G D ) (G A + G B ) = ΣG P ΣG R Η = (H C+ H D ) (H A + H B ) = ΣΗ P ΣΗ R S = (S C+ S D ) (S A + S B ) = ΣS P ΣS R 1

2 Standard versus Physiological Conditions Quantity Standard Physiological Conditions Conditions Solution 1.0 m 1.0 M Gas 1.0 atm 1.0 atm [H + ] 1.0 M 1.0 x 10-7 M ph 0.0 ph 7.0 T 298K (25 o C) 298K (25 o C) H M 1.0 M R = J/mol/K ( cal/mol/k) SIGN F G G = ΣG P ΣG R G = - 2

3 SIGN F H H = ΣH P ΣH R H = - SIGN F S S = ΣS P ΣS R S = + P T S - S = - T S + 3

4 TW TYPES F CUPLED REACTINS Hard! Easy! INTERMEDIATE CUPLED REACTINS A,B,C,D, etc are intermediates in metabolic pathways Glucose + ATP glucose-6-pi + ADP K 1 glucose-6-pi º fructose-6-pi K 2 X X X X fructose-6-pi + ATP fructose-1,6-bisphosphate (FBP) K 3 VERALL: Glucose + 2 ATP FBP + 2ADP K 1 K 2 K 3 4

5 ENERGY CUPLED REACTINS Note the following: 1. Endergonic reaction as written is not occurring! 2. Exergonic reaction as written is not occurring! 3. VERALL REACTIN AS WRITTEN IS CCURRING in the active site of an enzyme 4. THE ENZYME IS THE CUPLING AGENT! ENERGY CUPLED REACTINS Note the following: Without the enzyme, the energy provided by ATP would be lost as heat and entropy as the ATP is hydrolyzed!!! 5

6 Maximum Energy Available: t = 0 t = 0 t = long times Difference between G ' and G 0' G 0' determined in the laboratory Don t memorize! Use conditions of reaction to find the sign and magnitude of this correction term 6

7 What is G 0' when G = 0? AT CHEMICAL EQUILIBRIUM Using spectroscopic methods, the amount of reactants and products can be determined and a value of K eq can be calculated. and log K eq = - G 0' / R. T 7

8 SIGN F G 0' AND MAGNITUDE F K eq EXTENT F REACTIN Special case where G 0' = 0 so K eq = 1.0? Does not mean that G ' has to be zero nor that rxn is at chemical equilibrium 8

9 FIRST NTE SECND NTE 9

10 THIRD NTE 1 - Problem illustrating the difference between G ' and G 0' 2PG º PEP + H 2 K eq = M H C 2 - C CH 2 H P - - C 2 - C CH 2 P H 2 2-phosphoglycerate 2PG What is G 0'? phosphoenol pyruvate PEP R = J/mol/K T = 298 K What is G 0' = R. T. log K eq = R. T. ( ) = J/mol (1840 J/mol if answer) Carry extra digit if used in another calculation Carry extra digit 10

11 2 - Problem illustrating the difference between G ' and G 0' 2PG º PEP + H 2 K eq = M Given: 5.00 mm 1.00 mm What are the 2PG and PEP concentrations at equilibrium and what is G' (+ or -)? Is there net synthesis of product? 3 - Problem illustrating the difference between G ' and G 0' 2PG º PEP + H 2 K eq = M Given: 5.00 mm 1.00 mm let X = new PEP at equilibrium At Eq: 5.00 mm X 1.00 mm + X [PEP][H 2 ] K eq = = M [2PG] [1 mm + X][1.0 M] K eq = = M [5 mm - X] 11

12 4 - Problem illustrating the difference between G ' and G 0' 2PG º PEP + H 2 K eq = M Given: 5.00 mm 1.00 mm let X = new PEP at equilibrium At Eq: 5.00 mm X 1.00 mm + X X = mm positive solution indicates net synthesis of product What must be sign of G'? _ 5 - Problem illustrating the difference between G ' and G 0' 2PG º PEP + H 2 K eq = M Given: 5.00 mm 1.00 mm Calculate the value of G'! Use initial values since G' = 0 at equilibrium! G ' = G 0' R T log {[PEP][H 2 ]/[2PG]} G ' = G 0' *8.3145J/mol/K* 298K ( ) = J/mol + (-3988 J/mol) = J/mol Use extra sig fig as previously calc d G' = - Correct sig fig 12

13 HIGH ENERGY PHSHPATE CMPUNDS J = 1 cal HIGH ENERGY PHSHPATE CMPUNDS [kj/mol] [kcal/mol] 13

14 Why is G 0' for ATP hydrolysis so large and negative? [kj/mol] [kcal/mol] Why is G 0' for ATP hydrolysis so large and negative? [kj/mol] [kcal/mol] ATP -4 + H 2 ADP -3 + Pi -2 + H + HATP 3- º ATP 4- + H + pk 1 ' = 6.95 HADP 2- º ADP 3- + H + pk 2 ' = 6.88 Compare to: 6.75 H 2 P 4- º HP H + pk a ' =

15 Why is G 0' for ATP hydrolysis so large and negative? ATP -4 + H 2 ADP -3 + Pi -2 + H + Fig 14-1 Summary of why G 0' for ATP hydrolysis so large and negative ATP -4 + H 2 ADP -3 + Pi -2 + H + A-R P - P - P - - H + H A-R P P - + H P H Higher charge repulsion Lower charge repulsion because of separation of charge H charge repulsion H = - H solvation Fewer H 2 (24) More H 2 (26) H = - S 24 frozen H 2 26 frozen H 2 S = - S ne molecule Two molecules + H + S = + S Fewer ways to distribute negative charge More ways to distribute negative charge S = + + S contributes to G (-T S); - S contributes to + G ( -(T)(- S) = +) 15

16 Explain the ph dependence of G 0 for ATP ATP -4 + H 2 ADP -3 + Pi -2 + H + kcal/mol Alberty, R. A. (1969) Standard Gibbs free energy. J. Biol. Chem. 244, Explain the Mg 2+ dependence of G 0 for ATP A-R P - P - P - - Mg 2+ kcal/mol Alberty, R. A. (1969) Standard Gibbs free energy. J. Biol. Chem. 244,