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# Class Note for BIOC 460 at UA

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This 3 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 17 views.

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Date Created: 02/06/15
Bioc 460 Summer 2009 Bioenergetics Practice Questions Note In thermodynamic calculations temperature is always given in Kelvins K where K C 273 The gas constant R converts temperature to energy R 83 Jmol K 199 calmol K For this course we will use joules J for our units of energy 1 J 0239 cal or 1 cal 4184 J 1 For the dissociation of a ligand L from a membrane receptor R 2For the hydrolysis of ATP LRlt1gtLR The equilibrium constant for dissociation is K 2 X 10396 M lntuitively how much dissociation do you expect a little or a lot If L 1 x 10395 M R 5 x107 M calculate LR For the dissociation reaction at equilibrium calculate AG at 25 C as written Which reaction is favored dissociation or association ATP ltgt ADP Pi lfthe eguilibrium concentrations of ATP lx lO397 M ADP 9165 M and Pi 01 M what are the equilibrium constant and AG for the hydrolysis of ATP at 37 C 3 In a typical cell at 37 C the concentration of ATP 8 X 10393 M ADP 1 x103 M and Pi 8 x10393 M What is the actual free energy change AG39 for ATP hydrolysis under these conditions 4 The two reactions below are coupled A B ltgt c AG 391 15 kJmol C D ltigt E AG IZ 35 kJmol Calculate the total AGd for the coupled reactions Bioc 460 Summer 2009 2 Bioenergetics Practice Questions Answers 1For the dissociation of a ligand L from a membrane receptor R LR ltgt L R o The equilibrium constant for dissociation is K 2 X 10396 M lntuitively how much dissociation do you expect a little or a lot o lfL 1 x105 M R 5 x107 M calculate LR o For the dissociation reaction at equilibrium calculate AG at 25 C as written 0 Which reaction is favored dissociation or association Since K is such a small number you would expect to get very little dissociation Kd LRLR 1 x 105 M 5 x 107 MLR 2 x 106M LR 5x 1072 M22 x105 M 25x 105 M The system is at equilibrium so AGO RTnKeq 8315X103 kJ K mol298K ln 2 x 106M 32 52 kJmo Since AGO gtgt 0 the dissociation reaction is not favorable but we knew that because of the value for Kdl 2For the hydrolysis of ATP ATP ltgt ADP P lfthe eguilibrium concentrations of ATP lx lO397 M ADP 9165 M and P 01 M what are the equilibrium constant and AG for the hydrolysis of ATP at 37 C Bioc 460 Summer 2009 3 Use the data to calculate Keq39 Keq39 ADPMPl 0165Ml01Ml ATP 1x107 M 165x105 M I Use the value of Keq39 to calculate AGO AGO RTnKeq39 8315x103 kJ K mol310 Kln165X105 AGO 257812 310kJ mol 3ln a typical cell at 37 C the concentration of ATP 8 X 10393 M ADP 1 x103 M and Pi 8 x10393 M What is the actual free energy change AG39 for ATP hydrolysis under these conditions The KEYs to recoqnize that this system is NOT at equilibrium AG39 AG039 RTlnLADPlLPil A TP AG39 310kJmol 8315x10 3 kJmolIo 310 K ln Mlel 01M 8x10 3 M AG39 310kJmol 1 78 kJmol 49 kJmol 4 The conversion ofA to E is coupled by the two reactions below A B ltigt C AG 391 15 kJmol c D ltgt E AG 392 35 kJmol Calculate the total AGd for the coupled reactions Aco m AGO AGO 2 15 kJmol 35 kJmol 20 kJmo Thus coupling these two reactions together permits the thermodynamicaly favorable conversion ofA to E

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