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SI session Sept. 18 Ch. 5 and 6 answers

SI session Sept. 18 Ch. 5 and 6 answers - Kaiser Imam BIOL...

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Kaiser Imam September 18, 2007 BIOL 215 SI Session Chapters 5 and 6 Chapter 5: Bioenergetics 1. Write the (P/R) ratio for the following process: 2A + 3B 4C + 5D The (P/R) ratio would be [C] 4 [D] 5 / [A] 2 [B] 3 2. Write an equation to find Δ G’ . Δ G’ = -RT ln(K eq ) + RT ln(([C] c cell )([D] d cell )/([A] a cell )([B] b cell )) Δ G’ = -RT ln(K eq ) + RT ln([P]/[R]) 3. Write an equation to find Δ G°’ . Δ G°’ = -RT ln(K eq ) + RT ln(1) Δ G°’ = -RT ln(K eq ) 4. Write an equation to relate Δ G’ and Δ G°’ . Δ G’ = Δ G°’ + RT ln([P]/[R]) 5. Fill in the following table: Sign of Δ G°’ Spontaneous or Non-spontaneous (P/R) cell ? K eq Direction of Reaction Can work be done? < 0 Spontaneous < Reactants to products Yes; free energy released > 0 Non-spontaneous > Products to reactants Yes if energy is applied = 0 Equilibrium = Equilibrium No; reaction at equilibrium 1
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6. Using information from the table below determine if each of the following reactions is thermodynamically favorable. Reaction Δ H -TΔ S C 6 H 12 O 6 → 2C 2 H 5 OH + 2CO 2 -82 kJ/mol -136 kJ/mol 2CO 2 + 3H 2 O → C 2 H 5 OH + 3O 2 +1367 kJ/mol -41 kJ/mol N 2 O 5 → 2NO 2 + 1/2O 2 +110 kJ/mol -140 kJ/mol Use the equation Δ G = Δ H – TΔ S when finding the free energy changes for each of these reactions. The first reaction has a Δ G of -218 kJ/mol and is spontaneous as written. The second reaction has a Δ G of +1326 kJ/mol and is nonspontaneous. Finally, the third reaction has a Δ G of -30 kJ/mol and is spontaneous. 7. The interconversion of dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3- phosphate (G3P) is part of the glycolytic pathway for producing energy for the cell. DHAP ↔ G3P The value for Δ G°’ for this reaction is +1.8 kcal/mol at 25 °C. In the glycolytic pathway, this reaction goes to the right, converting DHAP to G3P. a) In which direction does equilibrium lie? What is the equilibrium constant at 25 °C? The Δ G°’ for this reaction is positive so the K eq value is less than one. Δ G°’ = -(1.987 cal/mol-K)(298 K) ln K eq = +1800 cal/mol. 1800/-592 = -3.04, so K eq = 0.048. Therefore, this equilibrium lies far to the left. b) In which direction does this reaction proceed under standard conditions? What is the Δ G’ for the reaction in that direction? Because the Δ G°’ value is the Δ G’ value under standard conditions, the reaction will tend to proceed to the left. The Δ G’ value in that condition is +1.8 kcal/mol. c) In the glycolytic pathway, this reaction is driven to the right because G3P is consumed by the next reaction in the sequence, thereby maintaining a low G3P concentration. What will Δ G’ be (at 25 °C) if the concentration of G3P is maintained at 1% of the DHAP concentration (i.e., if [G3P]/[DHAP] = 0.01)? Δ G’ = Δ G°’ + RT ln ([G3P]/[DHAP]) = +1800 + (1.987 cal/mol-K)(298) ln(0.01) = +1800 + 592(-4.605) = 1800 – 2726 = -926 cal/mol = -0.93 kcal/mol 2
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Chapter 6: Enzymes 8. Suppose that the standard free energy change for an uncatalyzed reaction (S ↔ P) is +60 kJ/mol, and a specific enzyme enhances the rate of the reaction by 10 7 -fold. Under standard conditions in the presence of the enzyme, in which direction will the reaction go?
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