06 Thermodynamics Answ PDF

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Critical Thinking Problems 1. List the three laws of thermodynamics and comments on how they relate to biological systems. First Law: The total energy of an isolated system is conserved. Organisms are open systems but we can define the universe as the closed system. We take energy from the sun, which is conserved by plants in the form of glucose. We can us the glucose to release the stored energy to drive other reactions and produced heat. No energy is lost. Second Law: Systems tend toward disorder and randomness. The second law states that entropy of the universe increases during all chemical and physical processes. Living organisms preserve their internal order, -ΔS, by taking from the surroundings free energy in the form of nutrients or sunlight and returning to their surroundings an equal amount of energy as heat and entropy, +ΔS. Third Law: At T = 0 K entropy is exactly zero. Every compound and system have a certain energy. It is too complex to calculate the exact values for biological compounds. Therefore, when looking at biological processes relative values are used.

2. Draw the reaction coordinate diagram for an endergonic reaction. Indicate in the figure the activation energies for the forward and reverse reactions. Indicate the free energy change.

3. Identify the criteria for a system to be at equilibrium. The criteria for a system to be at equilibrium is when no energy flow occurs to any direction within the system or out of the system. For a reaction, ΔG´ (or ΔG) = 0.

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4. What is the difference between free energy change (ΔG°) and standard free energy (ΔG'°)? Which is a more accurate representation of chemical reactions in a biological cell? The standard free-energy change, ΔG°, is a constant determined under so-called standard conditions: 298 K = 25°C, reactants and products are initially present at 1 M concentrations, and gases are present at partial pressures of 101.3 kilopascals (kPa), or 1 atm. In biology we use a modified standard state, designated with prime (‘) symbols: ΔG’°. This is because concentrations of 1 M for some compounds are not biologically relevant. For example since the pH = 7 the proton concentration is set at [H +] = 10-7 M. By convention, when H2O, H+, and/or Mg2+ are reactants or products, their concentrations ([H2O] = 55.5 M [Mg2+] = 1 mM) are not included in the equations, but are instead incorporated into the constants K’eq and ΔG’°. 5. Consider the reaction A + B ⇄ C + D. If the equilibrium constant for this reaction is a large number (say, 10,000), what do we know about the standard free-energy change (ΔG'°) for the reaction? Describe the relationship between Keq' and ΔG'°. ΔG'° = –RT ln Keq'. If Keq' is a large (positive) number, the term –RT ln Keq' (and therefore ΔG'°) has a relatively large, negative value. 6. Does the actual concentration of metabolites affect ΔG´? ΔG°?

∆ G = ∆ G° +RT ln

[C ] z [ D]r x y [ A] [ B ]

By definition ΔG° (or ΔG’°) does not change since it is a constant determined at set concentrations and conditions. Only the ΔG´ (or ΔG) is affected by the concentration of metabolites.

7. Explain in quantitative terms the circumstances under which the following reaction can proceed. Citrate

⇄

Isocitrate

ΔG'° = +13.3 kJ/mol

A reaction for which ΔG'° is positive can proceed under conditions in which ΔG is negative. From the relationship

∆ G = ∆ G° + RT ln

[ product ] [ reactant ]

it is clear that if the concentration of product is kept very low (by its subsequent metabolic removal, for instance), the logarithmic term becomes negative and ΔG can then have a negative value.

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8. The standard free energy change (ΔG'°) for ATP hydrolysis is –30.5 kJ/mol. ATP, ADP, and Pi are mixed together at initial concentrations of 1 M of each, then left alone until the reaction: ADP + Pi ⇄ ATP has come to equilibrium. For each species (i.e., ATP, ADP, and P i) indicate whether the concentration will be equal to 1 M, less than 1 M, or greater than 1 M. At equilibrium, ATP < 1 M; ADP > 1 M; Pi > 1 M.

9. Metabolically far from equilibrium reactions are said to be "metabolically irreversible.” What do you think that means? Be sure to consider free energy (magnitude and sign) and reactant/product concentrations in your answer. The ΔG for the reaction is either highly positive or highly negative. This is mainly due to ΔG° being highly positive or highly negative. The actual concentrations of reactants and products have very little effect on ΔG itself. Only very large changes in concentration or reaction conditions could affect the process but these conditions will not occur in a living cell.

10. The figure below shows the relative changes in energy in the conversion of glucose into lactate using the different steps of glycolysis and fermentation. The energy change in each step is included in the figure. The values are given relatively toward that of glucose, which is set at zero. In the figure, circle the reactions that are near equilibrium. Justify your choices. The reactions that are near equilibrium are the reactions that have a very small ΔG´, either positive or negative. In such a case, the equilibrium is not pushed towards either the products or reactants (making the other non-existent) but both are present under the equilibrium conditions.

11. Using the same figure, box the reactions that are far from equilibrium. Justify your choice. In this case, there are three reactions with a highly negative ΔG´. The reaction will proceed for 100% towards the formation of products. In the first two of these the coupling with a high energy compound makes the reactions endergonic. The conversion of PEP in pyruvate is highly endergonic, enough to generate ATP.

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12. Answer the following TRUE or FALSE. a) b) c) d) e)

___F__ Breaking any bond is always exothermic. ___T__ Breaking any bond is always endothermic. ___F__ Whether breaking a bond is endothermic or exothermic depends on the bond. ___T__ Any reaction that involves ATP as a reactant will have a large negative ΔG. ___F__ Metabolic reactions with large, negative ΔG are said to be reversible.

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Given below are the balanced chemical equation for the hydrolysis of ATP and the structure of ATP. Refer to these to answer the following questions.

ATP4- + H2O

⟶ ADP3- + HPO42- + H+

13. The standard free energy change for the hydrolysis of ATP is large and negative. Provide the 5 reasons why hydrolysis of ATP into ADP in a biological cell results in the release of energy. 1) The hydrolytic cleavage of the gamma-phosphate anhydride bond relieves electrostatic repulsion in ATP. 2) The phosphate formed in stabilized by several resonance forms that are not possible in ATP. 3) The ADP product immediately ionizes, releasing H+ in a medium with low hydrogen ion concentration at a pH of 7. 4) ATP has a small solvation energy compared to the solvation energies of ADP, Pi, and H+. Thus the products of hydrolysis are stabilized more by solvation than the reactant ATP. 5) Cellular conditions (see below) See page 12-13

14. Is the standard free energy change for the hydrolysis of ATP the same as the actual free energy change that would occur in a cell? Why or why not? No, the amount of ATP, ADP, and Pi are not present at 1 M concentrations.

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15. Cellular ATP concentrations are maintained well above equilibrium concentrations in cells. a) Calculate G when ATP, ADP, and Pi are 22.5, 0.25, 1.65 mM, respectively. Assume standard temperature and pH. b) Explain how ATP concentration relates to actual free energy changes of reactions that involve ATP and why it is important to keep ATP levels high. a)

∆ G = ∆ G' ° +RTln

[ ADP ] [ Pi ] [ ATP ]

[ 0.25 ∙ 10−3] [1.65 ∙ 10−3 ] kJ J kJ ( 298 K ) ln ∆ G=−30.5 =−53 + 8.315 −3 mol ∙ K mol mol [2.25 ∙10 ]

(

)

b) The main point is that at lower concentrations ATP loses its potency, 30.5 kJ/mol vs. 53 kJ/mol. The cell needs to keep the ATP levels high because it needs it for reactions but it is also needed to keep the ATP potent enough. This is a big waste since the cell needs much more ATP that it will be able to use. (This all goes back to thermodynamics: If cells were in an isolated system, its chemical reactions would reach equilibrium. If cells reached equilibrium the cell would die because there is no free energy left to perform the work needed to keep it alive. Cells stay out of equilibrium by manipulating concentrations of reactants and products to keep their metabolic reactions running in the correct direction. Providing a high concentration of a reactant (ATP in this case) can “push” the chemical reaction in the direction of the products. Intracellular concentrations of ATP are higher than those of ADP. These differences between intracellular concentrations and those of the standard state favor ATP hydrolysis.)

16. Explain the difference between kinetic and thermodynamic stability. How do they apply to ATP. Thermodynamic stability describes compounds with low bond energy while kinetic stability describes compound that are relatively inert due to high activation energies associated with its reaction. ATP is kinetically stable (i.e. it has a large half-life in a neutral solution of water) but is thermodynamically unstable due to the high-energy phosphodiester bond and will quickly reaction to form ADP if the activation energy is overcome. In cells, enzymes catalyze ATP-coupled reactions to lower the activation energy.

17. Explain what is meant by the statement: “Standard free-energy changes are additive.” Give an example of the usefulness of this additive property in understanding how cells carry out thermodynamically unfavorable chemical reactions.

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When two chemical reactions can be summed to give a third reaction, the standard freeenergy change for the third reaction is the arithmetic sum of the standard free-energy changes of the other two reactions. 18. Consider the following mini-pathway shown at right. The pathway starts with substrate A and ends with substrate D. Substrate B is converted into C in a far from equilibrium reaction that involves more than just B and C although those are the only substances shown in the diagram. Conversion of C back into B requires a completely different pathway with different substances involved although they are not shown. The reaction B to C is catalyzed by enzyme E1 and the reaction from C to B is catalyzed by enzyme E2. For both pathways to happen the conversion of A into D has to be exergonic. The same is true for the conversion of D back into A. Thermodynamically that would not make any sense but there is an essential difference in how you get from A to D and back to A. Of course, the involvement of different enzymes is the key to this. To provide a real example, below are the reactions for the phosphorylation of glucose to form glucose-6-phosphate (G-6-P) and the dephosphorylation of G-6-P to regenerate glucose, which are reactions of glycolysis and gluconeogenesis, respectively. Examine both and explain how it is possible that both of these reactions have negative ΔG values.

hexokinase glucose-6-phosphate + ADP G°’ = -16 kJ/mol →

Glycolysis

Glucose + ATP

Gluconeogenesis

Glucose-6-phosphate + H2O

G 6 P phosphatase glucose + PO43→

G°’ =

-12 kJ/mol

Both reactions have glucose and glucose-6-phosphate in common. The enzymes are different. The conversion of glucose into glucose-6-phosphate is actually endergonic without the coupling of the reaction to ATP hydrolysis which puts an additional -30 kJ/mol into the reaction. To make this reaction go in the opposite direction in gluconeogenesis the coupling with ATP is removed.

19. Why are the above reactions metabolically irreversible? Would these far from equilibrium reactions be irreversible outside the context of a metabolic pathway? How could you test your assertion? The reactions have a highly negative ΔG´, and will be irreversible in and outside the context of a reaction pathway. The reaction could be performed in a reaction tube where only the relevant components are present.

20. In general, when ATP hydrolysis is coupled to an energy-requiring reaction, the actual reaction often consists of the transfer of a phosphate group from ATP to another substrate, rather than an actual hydrolysis of the ATP. Explain.

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Hydrolysis of the ATP would result in the loss of most of the free energy as heat. In a transfer reaction, the gamma (third) phosphate of ATP is transferred to the substrate to produce a high-energy phosphorylated intermediate, which can then form the product in an exergonic reaction.

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