Principles Of Bioenergetics PDF

PRINCIPLES OF BIOENERGETICS Professor Özlem DALMIZRAK, PhD Department of Medical Biochemistry Faculty of Medicine Near East University Nicosia, TRNC Prepared by Professor Nazmi Ozer  PRINCIPLES OF BIOENERGETICS Living cells and organism...

PRINCIPLES OF BIOENERGETICS Professor Özlem DALMIZRAK, PhD Department of Medical Biochemistry Faculty of Medicine Near East University Nicosia, TRNC Prepared by Professor Nazmi Ozer  PRINCIPLES OF BIOENERGETICS Living cells and organisms must perform work to stay alive, to grow, and to reproduce. The ability to harness energy and to channel it into biological work is a fundamental property of all living organisms; it must have been acquired very early in cellular evolution. Modern organisms carry out a remarkable variety of energy transductions, conversions of one form of energy to another. They use the chemical energy in fuels to bring about the synthesis of complex, highly ordered macromolecules from simple precursors. They also convert the chemical energy of fuels into concentration gradients and electrical gradients, into motion and heat, and, in a few organisms such as fireflies and some deep-sea fish, into light. Photosynthetic organisms transduce light energy into all these other forms of energy.  PRINCIPLES OF BIOENERGETICS This lecture considers; The nature of biological energy transformations Thermodynamics of biological reactions There are two important questions in Biochemistry : i. How do the living organisms extract energy from their surroundings? ii. How do they use this energy for (performing biosynthesis, transport and motility) survival?  PRINCIPLES OF BIOENERGETICS Bioenergetics is the quantitative study of energy transductions—changes of one form of energy into another— that occur in living cells, and of the nature and function of the chemical processes underlying these transductions. Biological energy transformations obey the laws of Thermodynamics  PRINCIPLES OF BIOENERGETICS The first law is the principle of the conservation of energy: «for any physical or chemical change, the total amount of energy in the universe remains constant; energy may change form or it may be transported from one region to another, but it cannot be created or destroyed.» EUniverse = ESystem + ESurrroundings Energy change is independent of the path of transformation; It is just the difference Energy E* of energies at initial and final states of the A system as shown: B E ΔE = EB – EA  The second law of thermodynamics, which can be stated in several forms, says that: «the universe always tends toward increasing disorder: in all natural processes, the entropy of the universe» The second law of thermodynamics states that “for a spontaneous reaction” the sum of the entropy differences of the system and the surroundings is greater than zero: Ssystem + Ssurrounding > 0 e.g. The entropy of this sentence is smaller than the letters forming this sentence.  PRINCIPLES OF BIOENERGETICS The reacting system is the collection of matter that is undergoing a particular chemical or physical process; it may be an organism, a cell, or two reacting compounds. The reacting system and its surroundings together constitute the universe. In the laboratory, some chemical or physical processes can be carried out in isolated or closed systems, in which no material or energy is exchanged with the surroundings. Living cells and organisms, however, are open systems, exchanging both material and energy with their surroundings; living systems are never at equilibrium with their surroundings, and the constant transactions between system and surroundings explain how organisms can create order within themselves while operating within the second law of thermodynamics. Organisms convert raw materials into living matter at expense of energy extracted from their environment There are two groups of organisms: - Chemotrophs: Use fuel molecules (carbohydrates, fats, proteins) to obtain chemically useful energy - Phototrophs: convert radian energy (photon) of sunlight into chemically useful energy Each class of organisms is dependent on the other. Organic Molecules O2 Radian Energy Chemical Energy Energy Phototrophs Chemototrophs CO2 H2O NO3-  Free energy is one of the most useful form of energy in Biochemistry - 1 Josiah Willard Gibbs, in 1878, obtained a new equation by combining the first and second laws of thermodynamics. This new equation was able to demonstrate whether a reaction is spontaneous (Exergonic : can occur by itself) or not (Endergonic : can not occur by itself, energy must be added). In this equation a term, free energy, G, can be easily measured. G = H – T S and H = E – T V; In biological systems volume change is negligible so, in the above equation T V  0 and the free energy equation becomes as shown below; G  E – T S G, Free Energy change; H, Enthalpy change; T, Temperature in Kelvin degrees; S, Entropy change; E, Internal energy change; V, Volume change.  Gibbs free energy, G, expresses the amount of energy capable of doing work during a reaction at constant temperature and pressure. Enthalpy, H, is the heat content of the reacting system. It reflects the number and kinds of chemical bonds in the reactants and products. When a chemical reaction releases heat, it is said to be exothermic; the heat content of the products is less than that of the reactants, and the change in enthalpy, ΔH, has, by convention, a negative value. Reacting systems that take up heat from their surroundings are endothermic and have positive values of ΔH. Entropy, S, is a quantitative expression for the randomness or disorder in a system.  Free energy is one of the most useful form of energy in Biochemistry - 2 Let us analyze this equation; G  E – TS This equation contains spontaneity term, S, internal energy change term, E, and easily measurable free energy term, G. E is constant for a system so, the spontaneity will be determined by S terms. If, G < 0 then; Reaction is spotaneous. If, G > 0 then; Reaction is not spotaneous. If, G = 0 then; Reaction is at equilibrium. Free energy is one of the most useful form of energy in Biochemistry - 3 GA G GB  G, like E, is independent of the path of transformation Enzymes (Glycolysis etc.) Glucose + O2 CO2 + H2O G = -686 Kcal/mol Burning (in oven) A G* B G Standard free-energy change is directly related to the equilibrium constant- 1 1. aA + bB  cC + dD A, B = Reactants; C, D = Products [C]c [D]d 2. G = G°+ RT Ln [A]a [B]b Under standard conditions (298 K = 25oC), when reactants and products are initially present at 1 M concentrations or, for gases, at partial pressures of 101.3 kilopascals (kPa), or 1 atm, the force driving the system toward equilibrium is defined as the standard free-energy change, ΔG°. G, Free energy change; G°, Standard free energy change Standard free-energy change is directly related to the equilibrium constant- 2 At equilibrium, G = 0; If we substitute in the following equation 0 = G° + RT Ln {[C]c [D]d / [A]a [B]b} and, G° = - RT Ln {[C]c [D]d / [A]a [B]b} and, Ke = {[C]c [D]d / [A]a [B]b} G° = - RT Ln* Ke ; If we subtitute the values of constants and reorganize the above equation we will obtain the below equation; Ke = 10- G°/1.36 is obtained. *Ln = 2.303 Log ; R = 1.987 x 10-3 Kcal / mol – deg ; T = 273 + 25 = 298 oK  Some Definitions And Assumptions i. Chemists derived all thermodynamical terms assuming pH as 0.0. In biological systems, pH is not 0.0 but 7.0 so, to differentiate biochemical thermodynamic terms from chemical thermodynamic terms we use a prime (′) for biochemical terms; Such as, G′, G′o and Ke′ ii. The energy unit is in calories (cal): It is equal the amount of heat required to raise the temperature of one gram of water from 14.5°C to 15.5°C. Thousand calories is one kilocalorie (Kcal) and one Kcal is equal to 4.184 KJ.  G′o & Ke′ are inversely related K'e G′o (Kcal/mol) Inverse relation between G′o & Ke' 10-4 + 5.44 10-2 G°' + 2.72 10-1 1.36 1 0 Slope = -1.36 10+1 - 1.36 LogKe' 10+2 - 2.72 10+4 - 5.44 Demonstration of the interrelation of G′ with G′o - 1 Example - 1: Triose isomerase Dihydroxyacetone phosphate (DHAP)  Glyceraldehyde-3-Phosphate (GAL-3P) At equilibrium, pH = 7 and T = 298 K (25°C) Under these conditions; Ke′ = 0.0475 (= [GAL-3-P] / [DHAP]), and G′o = -RT Ln*Ke’ G′o = - 2.303* RTLog Ke′; -2.303 x 1.987 x 10-3 x 298 x Log(0.0475) G′o = 1.8 Kcal / mole is obtained. *Ln = 2.303 Log Demonstration of the interrelation of G′ with G′o - 2 Example - 2: If, [DHAP] and [GAL-3-P] were added at concentrations of 2 x 10-4 and 3 x 10-6 M, respectively; i. What is the value for free energy change G′? ii. Is the reaction spontaneous? G′ = 1.8 + 2.303 x 1.987 x 10-3 x 298 x Log (3 x 10-6 / 2 x 10-4) G′ = 1.8 – 2.5 and G′ = - 0.7 Kcal / mole and G′ < 0. So, reaction is spontaneous. Although G′o has very high positive value, under these conditions (at these [GAL-3-P] and [DHAP]), the reaction is spontaneous. Reactions can be coupled by means of free energy - 1 A thermodynamically unfavorable reaction becomes favorable when it is coupled to a favorable reaction Reactions can be coupled by means of free energy - 2 1. A non-spontaneous reaction may driven by a spontaneous reaction 2. Free energy changes of coupled reactions are additive and substractive. (Free energies are addible and substractable quantities) 3. Some enzyme catalyzed reactions are interpretable as two coupled half – reactions, one spontaneous and the other non-spontaneous. 4. At the active site of enzymes, the coupled reaction is kinetically facilitated, while individual reactions are prevented. Reactions can be coupled by means of free energy - 3 Example 1. In the reaction catalyzed by the Glycolysis enzyme Hexokinase, the half-reactions are: i. ATP + H2O ADP + Pi Go' = - 7.3Kcal/mol ii. Pi + glucose  glucose-6-P + H2O Go' = +3.3Kcal/mol ATP + glucose  ADP + glucose-6-P Go' = - 4 Kcal/mol The structure of the enzyme active site, from which H2O is excluded, prevents the individual hydrolytic reactions, while favoring the coupled reaction. Reactions can be coupled by means of free energy - 4 Example 2. Two separate reactions, one enzymatic and the other non- enzymatic may be coupled by a common intermediate. A typical example involving PPi hydrolysis is shown below: Reaction 1; Enzymatic PEP + ADP  Enolpyruvate + ATP Go' = + 0.0 Kcal/mol Reaction 2; Non-enzymatic Enolpyruvate + H2O Pyruvate Go' = – 7.5 Kcal/mol Overall spontaneous reaction: PEP + ADP Pyruvate + ATP Go' = – 7.5 Kcal/mol In this reaction; enolpyruvate, like pyrophosphate in several reactions that needs a driving force, makes the above reaction irreversible. Reactions can be coupled by means of free energy - 5 The equilibrium constant, Ke of a reaction coupled to ATP hydrolysis increased by a factor of 108 1. A  B G°'= +4.0 Kcal/mol 2. ATP + H2O  ADP + Pi + H+ G°'= -7.3 Kcal/mol 3. A + ATP + H2O  B + ADP + Pi + H+ G°'= -3.3 Kcal/mol From rxs 1. G°' = 4 = -2.303 x 1.987 x 10-3 x 298 x Log Ke'; and, Ke' = [B]e/[A]e = 1.15 x 10-3 was calculated for uncoupled reaction. For the coupled reaction (3. reaction) if Ke'' could be calculated; -3.3 = -2.303 x 1.987 x 10-3 x 298 x Log Ke'' is obtained. Reactions can be coupled by means of free energy - 6 The equilibrium constant, Ke of a reaction coupled to ATP hydrolysis increased by a factor of 108 Then, Ke''=[B]e/[A]e x {[ADP]e[Pi]e/[ATP]e} = 2.67x102 is calculated. In a metabolically active cell: { [ATP]e / [ADP]e [Pi]e} = 500. Calculate the new [B]e/[A]e ratio; New, [B]e/[A]e = 2.67 x102 x 500 = 1.34 x 105. Now, divide new (coupled) [B]e/[A]e by old (uncoupled) [B]e/[A]e {[B]/[A]}c / {[B]/[A]}uc = 1.34 x 105 / 1.15 x10-3 = 108 In a coupled rxs. if n moles of ATP is hydrolyzed, Ke' will increase by a factor 108n. Organisms obtain their energy from redox reactions - 1 Ared  Aox + 2e- Box + 2e-  Bred Organisms obtain their energy from redox reactions - 2 An electron transfer reaction: Aox + Bred  Ared + Box Aox is the oxidized form of A (Oxidizing agent) Bred is the reduced form of B (Reducing Agent) For such an electron transfer, one may consider two half-cell reactions: i. Aox + ne-  Ared ii. Box + ne-  Bred Aox + Bred  Ared + Box As a general rule, even a molecule is oxidized in the reaction, the reduced form is written to the right. Organisms obtain their energy from redox reactions - 3 i. Aox + ne-  Ared ii. Box + ne-  Bred For each half reaction: E' = Eo' – RT/nF (Ln [reduced]/[oxidized]) e.g., for the first half reaction: E' = Eo' – RT/nF (Ln [Ared]/[Aox]) E = voltage, R = gas const., F = Faraday, n = number of e- transferred Organisms obtain their energy from redox reactions - 4 When [Ared] = [Aox], E' = Eo' – RT/nF (Ln [Ared]/[Aox]) and E' = Eo' - 2.303 RT / nF (Log 1) and E' = Eo’ Eo' is called as the mid-point potential, or standard redox potential. Standard redox potential, Eo' , is the potential at which, for the A half reaction; [A, oxidized form] = [A, reduced form] Organisms obtain their energy from redox reactions - 5 For an electron transfer:  Eo' = Eo' (reduced half-cell) – Eo' (oxidized half-cell) = Eo' (electron acceptor) – Eo' (electron donor)  Eo' = substracting Eo' of half-cell undergoing oxidation from Eo' of half- cell undergoing reduction Go′ = – nF Eo' An electron transfer reaction is spontaneous (negative G′) if Eo' of the donor is more negative than Eo' of the acceptor, i.e., when there is a positive  Eo' Organisms obtain their energy from redox reactions - 6 Similarly, G′ = - nF E′ The equation given below is called as Nernst equation and used in free energy calculations. It also shows that a redox reaction is spontaneous or not. E′ =  Eo' – (RT /nF) Ln {[Ared][Box] / [Aox][Bred]} If; E′ > 0; then the reaction is spontaneous if E′ < 0; then the reaction is not spontaneous. Organisms obtain their energy from redox reactions - 7 Let’s assume that two electron is transferred from NADPH to GSSG : a. GSSG + 2H+ + 2e-  2 Glutathione (2GSH) E°' = -0.230 V b. NADP+ + 2H+ + 2e-  NADPH + H+ E°' = -0.324 V Here the oxidized reaction is reaction b so, the half potential (Eo' ) of reaction b must be substracted from the half-cell potential (Eo' ) of reaction That is equal to +0.094 volt. c. GSSG + NADPH + H+  2 GSH + NADP+ E°'=+0.094V Go’ = - nF Eo' = – 2 (96 494)(+0.094); and Go’ = - 18.14 kJ/mol = - 4.33 Kcal/mole is calculated. Although G’ determines the spontaneity of a reaction the large negative value for Go’ shows the tendency of the reaction for spontaneity. Organisms obtain their energy from redox reactions - 8 If two half-cells were combined by a metal wire Eo' can directly be read from voltmeter in volts. The potential of an half-cell could be measured by comparing it to a standard half-cell. Standard half-cell is a H2 half-cell because the standard half-cell potential of H2 (Eo' ) is equal to zero at pH 0.0. 2H+ + 2e-  H2(gas) E°′ = 0.00 volt A complete electrochemical- cell is shown on left. Organisms obtain their energy from redox reactions - 9 In biological systems pH is not zero; is 7 so, the standard half-cell potential of H2 at pH 7, is not zero but -0.42 volt. The half-cells with smaller half-cell potential than H2 half-cell will reduce H2 half-cell (they themselves will be oxidized) whereas half-cells with greater half-cell potential than H2 half-cell will oxidize H2 half-cell (they themselves will be reduced). A half-cell with more negative value will reduce a half-cell with less negative value and a half-cell with less positive value will reduce a half-cell with more positive value.  Standard reduction potentials of some biologically important reactions (25°C, pH 7) Half-reaction E°′ (Volt) 1/2O2 + 2H+ +2e-  H2 O 0.816 Fe3+ + e-  Fe2+ 0.771 Cytochrome a3 (Fe3+) + e-  Cyt a3 (Fe2+) 0.360 O2 + 2H+ + 2e-  H2 O 2 0.295 Cytochrome c1 (Fe ) + e  Cyt c1 (Fe2+) 3+ - 0.220 Fumarate2- + 2H+ + 2e-  Succinate2- 0.031 2H+ + 2e-  H2(gas) 0.000 Pyruvate- + 2H+ + 2e-  Lactate- -0.031 FAD + 2H+ + 2e-  FADH2 -0.219 Glutathione (GSH)+2H + 2e + -  GSSG -0.230 NAD + 2H + 2e + + -  NADH + H+ -0.320 NADP+ + 2H+ + 2e-  NADPH + H+ -0.324 2H+ + 2e-  H2(gas) -0.414 Ferrodoxin (Fe3+) + e-  Ferdx (Fe2+) -0.432 Hydrogen half-cell, red at pH = 0 and violet at pH = 7 Problems Question 1. What is the direction of each of the following reaction when the reactants are initially present at equimolar amounts? Use the data given on Slide 54. Q1. ATP + Creatine  Creatine phosphate + ADP Solution : Calculate first ΔGo’, then ΔG’ ATP + Cr  ADP + CrP ΔGo’ = ? ATP + H2O ► ADP + Pi -7.3 Kcal / mole Cr + Pi ► CrP +10.3 Kcal / mole __________________________________________________ ATP + Cr ► CrP + ADP 3.0 Kcal / mole ΔG’ = ΔGo’ + (1.987)(10-3)(298) Ln / ΔG’ = 3.0 + 0 Answer : To the left (ATP synthesis)  Problems Question 2. Consider the following reaction: ATP + Pyruvate  Phosphoenolpyruvate + ADP (PEP synthesis) 2.a. Calculate G°′ and K’e at 25C for this reaction, by using the data given on Slide 54. 2.b. What is the equilibrium ratio of pyruvate to PEP if the ratio of ATP to ADP is 10? Solution : Calculate first ΔGo’ then K'eq Pyr + Pi ► PEP + H2O +14.8 Kcal / mole ATP + H2O ► ADP + Pi -7.3 Kcal / mole --------------------------------------------------------------------------------------------------------------- ATP + Pyr ► PEP + ADP +7.5 Kcal / mole Answer : G°′ = 7.5 = 2.303 x 298 x 1.987 x 10-3Log K'eq; 7.5 = 1.36 Log K'eq; G°′ = 7.5 Kcal / mole K′eq = 3.25 x 105 K'eq = [PEP][ADP]/[ATP][Pyr]; 3.25 x 105 = [PEP](1)/(10)[Pyr] = 3.25 x 106 = [PEP]/[Pyr] [Pyr]/[PEP] = 1 / 3.25 x 106 ; [Pyr]/[PEP] = 3.07 x 10-7 Problems Question 3. What is the direction of each of the following reactions when the reactants are initially present at equimolar amounts? Use the data given on Slide 54. a.ATP + Glycerol  Glycerol-3-phosphate + ADP b.ATP + Glucose  Glucose-6-phosphate + ADP Question 4. Calculate G°′ for the isomerisation of Glucose-6-Phosphate to Glucose-1-phosphate. What is the equilibrium ratio of G-6-P to G-1-P at 25 C? Use the data given on Slide 54. Question 5. The formation of acetyl-CoA from acetate is an ATP-driven reaction:Consider the following reaction: Acetate + ATP + CoA  Acetyl-CoA + AMP + PPi a. Calculate G°′ for this reaction, by using the data given on Slide 54. b. The PPi formed in the prceeding reaction is rapidly hydrolyzed in vivo because of ubiquity of inorganic pyrophosphatase. The G°′ for the hydrolysis of PPi is -4.6 Kcal mol-1. Calculate G°′ for the overall reaction. What effect does the hydrolysis of PPi have on the formation of acetyl- CoA?.  Problemler Solution of Question 5. : Activation of acetate. Formation of acetyl-CoA is carried out by ATP : Consider the below reaction : Asetat + ATP + KoA  Asetil-KoA + AMP + PPi a. Calculate the G°′ value using the data on slide 54. ATP ------------------------------- AMP + PPi ΔGo’ = - 10.9 Kkal/mol Acetate + CoA --------------------Acetyl-CoA ΔGo’ = + 7.5 Kkal/mol ATP + Acetate + CoA -------------AMP + PPi + Acetyl-CoA ΔGo’ = - 3.4 Kkal/mol b. The PPi released to the medium immediately hydrolyzed because cells are very in inorganic pyrophosphotase. The G°′ value of PPi hydrolysis is -4.6 Kcal/mole. Calculate the G°′ value for whole reactin (coupled reaction). What will be the effect PPi hydrolysis on acetyl-CoA formation ? ATP ------------------------------- AMP + PPi ΔGo’ = - 10.9 Kcal/mole Acetate + CoA -----------------Acetyl-CoA ΔGo’ = + 7.5 Kcal/mole PPi ------------------------------- 2Pi ΔGo’ = - 4.6 Kcal/mole Acetate + CoA -------------AMP + PPi + Acetyl-CoA ΔGo’ = - 8.0 Kcal/mole The hydrolysis of PPi’ will shift the equilibrium constant to the right.  Standard Reduction Potentials Of Some Biologically Important Reactions (25°C, pH 7) Half-reaction E°′ (Volt) 1/2O2 + 2H+ +2e-  H2O 0.816 Fe3+ + e-  Fe2+ 0.771 Cytochrome a3 (Fe3+) + e-  Cyt a3 (Fe2+) 0.360 O2 + 2H+ + 2e-  H2O2 0.295 Cytochrome c1 (Fe3+) + e-  Cyt c1 (Fe2+) 0.220 Fumarate2- + 2H+ + 2e-  Succinate2- 0.031 2H+ + 2e-  H2(gas) 0.000 Pyruvate- + 2H+ + 2e-  Lactate- -0.031 FAD + 2H+ + 2e-  FADH2 -0.219 Glutathione (GSH)+2H+ + 2e-  GSSG -0.230 NAD+ + 2H+ + 2e-  NADH + H+ -0.320 NADP+ + 2H+ + 2e-  NADPH + H+ -0.324 2H+ + 2e-  H2(gas) -0.414 Ferrodoxin (Fe3+) + e-  Ferdx (Fe2+) -0.432 Hydrogen half-cell, red at pH = 0 and violet at pH = 7 Question 1. Regeneration of NAD+ in Glycolysis 1. NAD+ +2H+ + 2e-  NADH + H+ E = -0.320 Volt 2. Pyruvate + 2H+ + 2e-  Lactate- E = -0.031 Volt When two reactions above are coupled; i. What will be E ? ii. What will be G? Question 2. Electron transport and ATP synthesis (Oxidative phosphorylation) When two reactions above are coupled; 1/2O2 + 2H+ +2e-  H 2O E = 0.816 NAD+ + 2H+ + 2e-  NADH + H+ E = -0.320 i. What will be E ? ii. What will be G? iii. The released enegry is sufficent for the synthesis of how many moles of ATP (theoretically) from ADP and Pi? Assume that for ATP synthesis 7.5 Kcal/mole. References: 1. Berg, J.M., Tymoczko, J.L., and Stryer, L., Biochemistry, Fifth Edition, 2002 2. Nelson, D.L., and Cox, M.M., Lehninger’s Principles of Biochemistry, Third Edition, 2000. 3. Mathews, C.K. And Van Holde, K.E., Biochemistry, Second Edition,1996. 4. Wood, W.B., Wilson, J.H., Benbow, R.M., and Hood, L.E., Biochemistry, A Problems Approach, Second Edition, 1981. THANK YOU! 43