Energetics of metabolism

Transcription

1 Energetics of metabolism Dr. Bódis Emőke October 7, 2015 JJ9 Why do we study difficult thermodynamics? The laws and principles of the thermodynamics describe the characteristics of matter- and energy flow of the different biochemical systems. - From these we can determine the direction of a spontaneous reaction. The 0. law of the thermodynamics (equilibrium and stability) If system A is in thermal equilibrium with system C, and system B is in thermal equilibrium with system C, then system A is in thermal equilibrium with system B. Two systems are said to be in thermal equilibrium if, when they are placed in ``thermal contact'' (basically, contact that permits the exchange of energy between them), their state variables do not change. 1

2 The I. law of the thermodynamic (conservation of energy) ΔU = U 2 -U 1 = Q + W Energy can be converted from one form to another with the interaction of heat, work and internal energy. Energy can be changed from one form to another, but it cannot be created or destroyed. BUT! In biological systems instead of internal energy it is more useful to apply ENTHALPY. H = U + pv ΔH = ΔU + pδv = Q + W + pδv = Q - pδv + pδv = Q ΔH: Enthalpy change is the amount of heat content used or released in a system at constant pressure. (ΔU: Internal energy change is the amount of heat content used or released in a system at constant volume. ) U H pv The II. Law of the thermodynamics (The direction of the processes) Systems tend to proceed from ordered (low-entropy or low-probability) states to disordered (high-entropy or high-probability) states. ΔQ = T ΔS Heat transferred in a process change the entropy of the system. Entropy represents energy dispersion: large number of molecular motions is relatable to quantized states (microstate) S = k ln W = k ln (W 2 -W 1 ) 2

3 A termodinamika III. főtétele Tö ke letes krista lyos, tiszta anyagok entrópia ja (vegyu leteke is!) T = 0 K-en nulla. Következme nye: T= 0 K-en ismerju k az entrópia abszolu t e rte ke t DE! Biológiai folyamatokban az entrópia va ltoza s hasznosabb, mint az abszolu t entrópia. Pl: ΔG = ΔH TΔS A III. főte tel statisztikus magyara zata: S = k ln W egyenletből T = 0 K hőme rse kleten a töke letesen krista lyos anyagok re szecske i mind a leheto legalacsonyabb energia ju a llapotban vannak W = 1 e s S = 0 A legalacsonyabb me rt hőme rse klet a terme szetben 1 K (Boomerang csillagködben, 5000 fe nye vnyire tőlu nk, a Kentaur csillagke pben) Question: Will the reaction proceed in the direction written? 1. The thermodynamic condition: Gibbs free energy must decrease. Gibbs free energy: measures the "usefulness" or process-initiating work obtainable from a thermodynamic system at constant temperature and pressure. It is usefull to apply for biological systems. 3

4 2. The reaction kinetic condition: the reactants must decrease the activation energy 4

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6 The catabolic ways are convergent, the anabolic ways are divergent. 6

7 First period of the cellular respiration: Glycolysis From 1 glucose molecule aroses 2 pyruvate molecules. 7

8 Pyruvate, a versatile metabolite The final product of glycolysis Can be used in several ways Oxygen is present (aerobic condition): pyruvate is oxidized and will be the acetyl group of acetyl-coa and will further metabolized in the tricarboxylic acid cycle Oxygen is absent (anaerobic condition): - lactic acid fermentation - alcoholic fermentation Glikolízis, tejsavas fermenta ció C 6 H 12 O6 2 H 3 C - CHOH COO - +2 H + ΔG = - 183,6 kj/mol Emellett netto 2 ATP is keletkezik 2 ADP + 2 P i 2 ATP + 2 H 2 O ΔG = 61 kj/mol ΔG = - 183, = - 122,6 kj/mol A szabad energia va ltoza s bőven fedezi a 2 ATP szinte zise t: (61 / 183,6) * 100% = 33,2 % (A felszabaduló energia 33,2 %-a fordítódik a 2 ATP szinte zise re.) 8

9 Why does Glycolysis occur? Are the thermodynamic and kinetic conditions necessary for the reactions met? The first phase of Glycolysis The first phase of Glycolysis: First energy priming reaction Kinase: enzyme responsible for transferring the γ-phosphate of ATP to molecules Hexokinase: phosphorylate hexose sugars (C-6) The first and third steps of the Glycolysis pathway will be primed with energy from hydrolysis of 2 ATP so that at the end of Glycolysis 4 ATP will be formed (net 2 ATP) ATP cleavage: ΔG = -30,5 kj/mol Phosphorylation: ΔG = 13,8 kj/mol Firs step: ΔG = -30,5 + 13,8 = - 16,7 kj/mol Water pump has to be primed with small amount of water to deliver more water out. 9

10 Phosphorylation of glucose to glucose-6-phosphate Hexokinase (enzyme) Mg 2+ - ATP Glucose - phosphorylation: glucose glucose-6- phosphate - Conformational change in the presence of Mg 2+ - ATP and glucose (open-closed) - Regulation: The enzyme function is inhibited allosterically in the presence of high concentration of glucose-6-phosphate glucose-6-phosphate is a the brancs point of several metabolic pathways. Glucokinase: hexokinase in liver and pancreas 10

11 The first phase of Glycolysis: glucose to glucose-6-phosphate The reaction occur, because 1. The thermodynamic condition meets (ΔG = - 16,7 kj/mol) 2. The kinetic condition meets, because the presence of hexokinase decrease the activation energy of the step. Gibbs free energy change during the whole process of Glycolysis 11

12 Third step of Glycolysis Third step of Glycolysis: Second energy priming reaction Catalysator: Phosphofructokinase ATP hydrolysis: ΔG = -30,5 kj/mol 2. Phosphorylation: ΔG = 11,7 kj/mol Third step: ΔG = -30,5 + 13,8 = - 18,8 kj/mol Regulation of the protein: - 2 ATP binding sites - Low concentration of ATP: ATP binds to the high affinity binding site (ATP at active site) - High concentration of ATP: : ATP binds also to the low affinity binding site (ATP at regulatory site) glycolysis turns off 12

13 Fourth step of Glycolysis Fourth step of Glycolysis: concentration-dependent reaction Fructose-Bisphosphate aldolase: claves between C3-C4 carbons yield two triose phosphates Concentration-dependent reaction: - in vitro: the reaction does not proceed effectively from left to right (ΔG positive: 23,97 kj/mol) - in vivo: efficient (ΔG negative: -0,23 kj/mol 13

14 Seventh step of Glycolysis Seventh step of Glycolysis: Synthesis of the first two ATPs - Phosphoglycerate kinase: phophorylation of ADP (substrate-level phophorylation ) - Two ATP were consumed in the first phase of Glycolysis, the Phosphoglycerate kinase reaction pays off the ATP debt created by the priming reactions - In living cells: ΔG : -0,1 kj/mol ADP Mg 2+ phosphate 14

15 Tenth (last) step of Glycolysis Tenth (last) step of Glycolysis: Synthesis of the second two ATPs Catalysator : Pyruvate kinase Presence of Mg 2+ e s K + are necessary 1. step: phosphorylation of ADP 2. step: enol-ketone conversion Highly favorable and spontaneous conversion of the enol to the more stable keto form following the phosphorylation group transfer step. ΔG = -31,7 kj/mol Gibbs free energy pays the production of ATP. ΔG = 30,5 kj/mol 15

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17 Thank you! 17