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3 Bio 98 - Lecture 13 Energetics

4 Thermodynamics

5 The 2 nd Law Unsustainable “All natural and technological processes proceed in such a way that the availability of the remaining energy decreases. In all energy exchanges, if no energy enters or leaves an isolated system, the entropy of that system increases. Energy continuously flows from being concentrated, to becoming dispersed, spread out, wasted, and useless. New energy cannot be created and high-grade energy is being destroyed. An economy based on endless growth is unsustainable.”

6 Overview 1.What is the relationship between the free energy change (  G) of a reaction and its equilibrium constant? 2.How can compounds like ATP be used to drive metabolic reactions that are unfavorable? 3.What is it about the structure of ATP and related compounds that make them so “full of energy”?


8 Simple Thermodynamics aA + bB cC + dD If a process is spontaneous then the reaction moving to the right (products) is favored and K eq > 1 or G reactants > G productsprocess is spontaneous

9 Glucose-6-phosphate G6P Fructose-6-phosphate F6P Example: an important reaction in biology is the inter-conversion of various sugars. The equilibrium shown below is the inter-conversion of glucose-6-phosphate (G6P) and fructose-6- phosphate (F6P):

10 Suppose you dissolve 1 mole of G6P and 0.3 moles of F6P into a liter of water. What is the concentration of each at equilibrium? The sum is 1.3 moles in a liter or 1.3 M, so [F6P]+[G6P] = 1.3 M, thus [G6P] = 1.3 M – [F6P] Solve for [F6P] = 0.44 M. And the concentration of G6P = [G6P] = M = 0.86 M.

11 Free Energy A measure of the spontaneity of a reaction is called the Gibbs free energy (G). Gibbs free energy The reactants and products have a free energy. We can only observe the change (  ) in free energy as we go from reactants to products:  Gº reaction = G products - G reactants For a spontaneous process G reactants > G products, sospontaneous process  Gº reaction < 0

12 1. Enzymes do not alter the equilibrium or  G. 2. They accelerate reaction rates by decreasing  G ‡. 3. They accomplish this by stabilizing the transition state(s). Enzyme-catalyzed reaction: G (free energy) Reaction coordinate E+P ES ‡ E+S G‡G‡ GG S ‡ SP E

13  G° is called the standard Gibbs free energy change for a given reaction under standard conditions of pressure (1 atm), temperature (25 °C or 298 K) and [H + ] (pH 7.0 or 0.1  M). Under standard conditions the concentration of every reactant and product is 1 M (and water is 55 M!).Gibbs free energy It basically tells us whether, at equilibrium, a reaction lies to the right or left. However, inside living systems reactions are seldom at equilibrium, so a more useful quantity is  G under non-equilibrium conditions. Free Energy

14 1. Gibbs free energy (  G) - free energy needed to convert reactants to products under a defined set of non- equilibrium conditions. Which direction is favored? aA + bB cC + dD [C] c [D] d  G = ∆G° + RT ln –––––––– = ∆G° + RT ln Q [A] a [B] b ∆G°: Standard free energy of a reaction. Depends on free energies inherent in the structures of the chemicals. RT ln Q : accounts for differences in the concentrations of reactants and products (law of mass action); R is the gas constant (8.31 J K -1 mol -1 ), T is the temperature in of mass actiongas constantKelvin ln(x) is the logarithm base e not base 10!e

15 [C] c [D] d ∆G = ∆G° + RT ln –––––––– = ∆G° + RT ln Q [A] a [B] b (3) Q is a very large number (eg. 1x10 6 ): [products] >> [reactants] Will make ∆G more positive/unfavorable with respect to ∆G°. Follows law of mass action, ie. high [products] will push balance to left. Consider special cases to “test” this equation (1) [A], [B], [C], [D] = 1 M Then ∆G = ∆G° (by definition!) (2) Q is a very small number (eg. 1x10 -6 ): [reactants] >> [products] Will make ∆G more negative/favorable with respect to ∆G°. Follows law of mass action, ie. low [products] will pull balance to right. Which direction is favored? aA + bB cC + dD

16 The Equilibrium State ∆G = ∆G° + RT ln K eq At equilibrium, neither direction of the reaction is favored, therefore ∆G = 0 (by definition). With ∆G° + RT ln K eq = 0 ln K eq = -∆G°/RT K eq = e -∆G°/RT

17 K eq = e [-∆G°/RT] = e (-14,000 J/mol)/(8.314 J/K/mol x 298 K) = e = 3.5 x = –––––––––– The equilibrium lies far to the left; thus the reaction is not favored. G6P has a higher inherent free energy than do glucose and P i together. Example Glucose + HPO 4 -- Glu-6-phosphate ∆G° = (Glu) (P i ) (G6P) +14 kJ/mol [G6P] [Glu][HPO 4 -- ]

18 By coupling (1) and and (2), an enzyme can “tap” the energy of a highly favorable reaction to drive an otherwise unfavorable reaction! (1) A B ∆G° = +14 kJ/mol (unfavorable) (2) C D ∆G° = -30 kJ/mol (favorable) A + C B + D ∆G° = -16 kJ/mol (favorable) Enz Energy Coupling - an important role for enzymes and for high-energy compounds like ATP.

19 ADP vs. ATP ADP, net charge: -3ATP, net charge: -4

20 Mitochondria and ATP Movie: Powering the Cell

21 Example of energy coupling: hexokinase reaction Reaction ∆G° (kJ/mol) Glu + P i G6P +14 (unfavored) ATP ADP + P i -31 (favored) hexokinase Glu + ATP G6P + ADP -17 (favored) Using K eq = e [-∆G°/RT] with R = J/K/mol and T = 298 K we obtain K eq = 955

22 O O O Ad-Rib-O-P-O-P-O + O-P-O + H + O O OH ADP (P i ) ∆G° = -31 kJ/mol 2. ATP hydrolysis is highly exergonicexergonic O O O Ad-Rib-O-P-O-P-O-P-O + H 2 O O O O ATP

23 3. Proton release [H + ] is kept very low in the cell (~10 -7 M or 0.1  M) This favors reactions that release protons (mass action effect) Why is ATP hydrolysis so highly exergonic?exergonic 2. Resonance forms - more with products 4. Hydration stabilization - more with products 1. Electrostatic repulsion - less with products Similar considerations explain why other phosphate ester compounds also store energy; for example phosphoenolpyruvate, phosphocreatine, etc.

24 Why is ATP hydrolysis so highly exergonic?ATP hydrolysis exergonic 2. Resonance forms - more with products 1. Electrostatic repulsion - less with products -

25 ATP turnover Each of the approximately one hundred trillion human cells contains about one billion ATP molecules, or about ATP molecules in the body. For each of these ATP “the terminal phosphate is added and removed 3 times each minute” (Kornberg, 1989). This amount is sufficient for only a few minutes and must be rapidly recycled. The total human body content of ATP is only about 50 grams. The ultimate source of energy for generating ATP is food; ATP is simply the carrier or storage unit of energy. The average daily intake of 2,500 food calories translates into a turnover of a whopping 180 kg (400 lbs) of ATP (Kornberg, 1989).

26 PhosphocreatinePhosphocreatine is an “ATP buffer” in skeletal muscle ATPADP creatine kinase phosphocreatinecreatine  G° = kJ/mol Mg ++

27 Enthalpy and Entropy  G =  H - T  S  H or enthalpy is the heat given off in a reaction.enthalpy  H is defined as H products – H reactants. When heat is given off then  H < 0. This means a given reaction is enthalpically favorable. Suppose we have a reaction going on in a beaker. If the beaker heats up then the reaction is giving off heat so  H < 0.

28  S or entropy is a measure of disorder in a system.entropy Similar to  H,  S is defined as S products - S reactants. A spontaneous or favorable process tends toward higher disorder so  S > 0 is favorable. Enthalpy and Entropy  G =  H - T  S

29 In this container the gas molecules are confined to a small space. The order in the system is relatively high, so the entropy, S, is small. In this container the same number of gas molecules have more space. Each molecule has more “freedom” and the system is more disordered. Here the entropy, S, is larger. Enthalpy and Entropy  G =  H - T  S

30 Consider what happens to the thermodynamic quantities in a reaction between 2 molecules to form 1 molecule: A + B AB The formation of a chemical bond is usually a heat-generating process so  H < 0. However, in going from 2 molecules to 1 molecule, the system will become more ordered. Therefore,  S < 0 (decrease in entropy). Whether or not a reaction is favorable depends on the balance between enthalpy and entropy. It is typical for enthalpy and entropy to balance or compensate for one another. Enthalpy and Entropy  G =  H - T  S

31 A classic example is the reaction between oxygen and hydrogen to form water O H 2 2 H 2 O Take a balloon filled with hydrogen gas. Touch a lit match to the balloon. What happens? A big boom and lots of heat given off. Is  H favorable or unfavorable?? Add solid urea to a beaker of water. The beaker gets very cold. Is  H favorable or unfavorable???

32 ANOTHER EXAMPLE N 2 O 5 (solid) 2 NO 2 (gas) + ½ O 2 (gas)NO 2  G 0 = -9.5 kcal/mol  H 0 = kcal/mol T  S 0 = kcal/mol at room temperature (293 K) So this reaction is entropically favorable even though the change in enthalpy is unfavorable.  G =  H - T  S

33 First one to send an with all correct answers with “98challenge“ in the subject line to wins $10! Double-or-nothing: You can double your winnings by playing me (and winning) best-of-3 in racquetball at the The 2 nd Bio98C/F $10 Challenge

34 Bio97 Bio98 Bio99 BIOCHEMISTRY

35 THE END Good luck with the midterm! My lab is looking for Bio199 students (GPA of 3.5 or better, A- or better in 97, 90% or better in 98 midterm, hours per week / 3-4 units per quarter, 2 year commitment and participation in Excellence in Research. Please after the midterm with “199” in the subject line if you’re interested.)Excellence in Research

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