462a_exam1_2013_A_key

# Therefore g 303 kjmol g rt ln keq ln keq

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Unformatted text preview: 1) Oxaloacetate + acetyl ­CoA + H2O ⟷ citrate + CoA ΔG'° = –32.2 kJ/mol Reaction 2) Oxaloacetate + acetate ⟷ citrate ΔG'° = –1.9 kJ/mol Using this information, calculate the equilibrium constant, Keq’, for the hydrolysis of acetyl ­CoA at T = 298 K. The reaction is the sum of reaction 1 and the reverse of reaction 2. Therefore, Δ G'° = –30.3 kJ/mol. Δ G'° = –RT ln Keq' ln Keq' = –Δ G'°/RT = 30.3 kJ/mol / [(8.314 J/mol·K)(298 K)] ln Keq' = 12.22 Keq' = 2.04 x 105 (can have M units on Keq') 2. [12 pts] The free energy change under standard conditions for Citrate → Isocitrate is ΔG'° = +13.3 kJ/mol at 37 °C. Assume that, in the cell, the concentration of citrate is constant at 1 mM. What range of concentration of isocitrate will allow this reaction to proceed spontaneously in the cell? Δ G’ needs to be negative. Δ G’ = Δ G’° + RT ln [Iso]/[Cit] must be < 0 so ln[Iso]/[Cit] < –Δ G’°/RT = –13.3x103/(8.314x310) = –5.159 [Iso]/[Cit] < exp(–5.160) = 0.00574 [Iso] < 0.00574 mM = 5.74 μM /24 Name: ______________ KEY_______________________________________ 3 3. [6 pts] A reaction has ΔH =  ­40 kJ/mol and ΔS = +0.4 kJ/mol•K. Over what temperature range is the reaction spontaneous? All temperatures. It is favorable both for Δ H and Δ S 4. [8 pts]. An α ­helix within a protein has the sequence Leu1 ­Ala2 ­Glu3 ­Ser4 ­Leu5 ­Lys6. 4a) [4 pts] If Glu3 were mutated to Lys, and Lys6 were mutated to Glu, switching their positions in the sequence, how would you expect the helix stability to be affected? (circle one and explain your choice): more stable less...
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## This test prep was uploaded on 02/25/2014 for the course BIOC 462A taught by Professor Ziegler,baldwin during the Spring '08 term at University of Arizona- Tucson.

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