BCH244 A3 Eng (1) Biochemistry 244 Past Paper PDF

Biochemistry 244 – A3 Theme A: The Energy of Life The study of thermodynamic forces for biological reactions Strategies that living organisms use to permit reactions with positive Gibbs energy changes to proceed spontaneously Stu...

Biochemistry 244 – A3 Theme A: The Energy of Life The study of thermodynamic forces for biological reactions Strategies that living organisms use to permit reactions with positive Gibbs energy changes to proceed spontaneously Study material: Appling: pp 89 – 92 Additional notes © Department of Biochemistry, Stellenbosch University J Willard Gibbs  Biochemistry 244 – A2 Key concepts The Gibbs energy change (∆G) of a reaction gives an indication of the direction of the driving force of the reaction and how far the reaction is from equilibrium. The Gibbs energy change (∆G) depends on the equilibrium constant (Keq) and the prevailing ratio between the product and reactant concentrations (the mass-action ratio, G). There is a direct relationship between the standard Gibbs energy change (∆G°) and the equilibrium constant (Keq) of a reaction.  Biochemie 244 – A2 Sleutelkonsepte Die Gibbs-energie verandering (∆G) van 'n reaksie dui aan na watter kant die dryfkrag vir 'n reaksie is en hoe ver van ewewig die reaksie is. Die Gibbs-energie verandering van 'n reaksie (∆G) hang af van die ewewigskonstante (Keq) en die heersende konsentrasies van reaktante en produkte (massawerkingsverhouding , 𝚪). Daar is 'n direkte verband tussen die standaard Gibbs- energie verandering (∆G°) en die ewewigskonstante (Keq) van 'n reaksie.  Biochemistry 244 – A3 A⇋B DGº’ = +10 kJ/mol 1. The reaction is endergonic, in other words ∆G°' > 0. 2. If 1M of A and 1M of B (standard conditions) are added together the reaction will proceed to the left until equilibrium is reached. 3. The Keq will therefore be < 1 (0.0177 to be specific). 4. At equilibrium there will therefore be much more of A (reagent) present than B (product). To be specific, the [B]/[A] ratio at equilibrium is 1/56.6 = 0.0177  Biochemistry 244 – A3 Keq = e–DGº / RT = e–10000 / (8.314 x 298) = 0.01766  Biochemistry 244 – A3 Keq = e–DGº / RT = e–10000 / 8.314 x 298 = 0.01766 Keq = [products]eq / [reagents]eq = [B]eq / [A]eq = 0.0177 = 0.0177 / 1 = 1 / 56.6 [B]eq / [A]eq = 1 / 56.6 Total of B and A = 57.6 f(B)eq = 1 / 57.6 f(A)eq = 56.6 / 57.6 = 0.017 = 0.98  Strategies that living organisms use to enable reactions with positive Gibbs energy changes to proceed spontaneously 1. Mass action (Maintaining 𝚪 < Keq) 2. Chemical coupling to an exergonic reaction Biochemistry 244 – A3  Biochemistry 244 – A3 A ⇋ B DGº’ = +10 kJ/mol Mass action Increase the [substrate] or reduce the [product] DG’ = DGº’ + RT ln [B] / [A] Ø At standard conditions: ∆G’ = ∆Gº’ because 𝚪=([B]/[A]) = 1 Ø For ∆G’ < 0 (exergonic): ∆G’ = 10000 + RT ln [B] / [A] < 0 RT ln [B] / [A] < -10000 Therefore at 25 °C: 8.314 x 298 ln [B] / [A] < -10000 ln [B] / [A] < -4.036 [B] / [A] < e-4.036 [B] / [A] < 0.0177 ( 𝚪 < Keq value) [A] / [B] > 56.6  Biochemistry 244 – A3 A⇋B DGº’ = +10 kJ/mol Mass action increase the [substrate] or decrease the [product] [A] / [B] > 56.6 e.g. [A] / [B] = 60 ∆G’ = ∆Gº’ + RT ln [B] / [A] ∆G’ = 10000 + RT ln 1 / 60 Therefore at 25 °C: = 10000 + 8.314 x 298 ln 0.0167 = 10000 + -10139.08 = -139.08 J/mol  Biochemistry 244 – A3 glucose-6-phophate ⇋ fructose-6-phosphate ∆Gº’ = +1.7 kJ/mol Mass action At standard conditions (∆Gº’): Ø fG6P = 0.5 (1 / 2 = 1/(1+1)) Ø fF6P = 0.5 (1 / 2) Ø ∆Gº’ = +1.7 kJ/mol At equilibrium (Keq): Ø ∆G’ = 0 kJ/mol Ø fG6P = 0.665 Ø fF6P = 0.335 ∆G’ < 0: Ø fG6P > 0.665 Ø fF6P < 0.335 Ø 𝚪 < Keq value  Biochemistry 244 – A3 A⇋B ∆Gº’ = +10 kJ/mol Chemical coupling to an exergonic reaction Therefore, coupling to a reaction with a greater negative ∆Gº’ A⇋B ∆Gº’ = +10 kJ/mol C⇋D ∆Gº’ = -31 kJ/mol A+C⇋B+D A⇋B ∆Gº’ = +10 kJ/mol C⇋D ∆Gº’ = -31 kJ/mol A+C⇋B+D ∆Gº’ = -21 kJ/mol  Biochemistry 244 – A3 Examples of chemical coupling Ø Couples to a reaction with a greater negative ∆Gº’ Ø Often ATP hydrolysis Page 567  Biochemistry 244 – A3 Examples of chemical coupling ATP synthesis is endergonic but by coupling it becomes exergonic Page 567  Biochemie 244 – A3 Sleutelkonsep Lewende organismes gebruik twee strategieë, massawerking en chemiese koppeling, om reaksies met positiewe Gibbs-energie waardes wel spontaan te laat verloop Doelwitte Doelwit 1: Massawerking, Gekoppelde reaksies; 10. Die twee strategieë wat lewende organismes gebruik om reaksies met positiewe ∆G-waardes wel spontaan te laat verloop: (a) kan noem; (b) kan verduidelik aan die hand van 'n voorbeeld; (c) kan identifiseer in 'n metaboliese pad. 11. Gegee die ∆G- of ∆G°-waardes vir individuele reaksies, kan bereken wat die ∆G- of ∆G°-waardes is vir die netto gekoppelde reaksie.  Key concept Biochemistry 244 – A3 Living organisms use two strategies, mass action and chemical coupling, to enable reactions with positive Gibbs energy values to take place spontaneously. Objectives Objective 1: mass action coupled reactions; 10. There are two strategies that living organisms employ to enable reactions with positive ∆G-values to take place spontaneously: (a) name them; (b) explain them using an example; (c) identify them in a metabolic pathway. 11. If given the values of either ∆G or ∆G° for individual reactions, calculate their values for the net, coupled reaction.  NEXT LECTURE: Biochemistry 244 – A4 Theme A: The Energy of Life The study of thermodynamic forces for biological reactions Activated metabolites Study material: (ATP, acyl-CoA, Appling: pp. 92 – 95 Appling: pp. 382 – 390, NAD(P)H, FADH2 and 460-463 Additional notes FMNH2)