07Fexam5key - Name_KEY EID CH 339K Exam 5 13 Nov 2007 100...

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Name_____________ KEY ____________________________ EID__________________________________________ 1 CH 339K Exam 5 13 Nov 2007 100 points total 10 pts. 1 . Free energy changes under intracellular conditions differ markedly from those determined under standard conditions. G°’ for ATP hydrolysis to ADP + P i is -30.5 kJ/mol. Calculate G for ATP hydrolysis in a cell at 37°C that contains ATP at 3 mM, ADP at 2 mM, and P i at 1 mM. " G = " G ° ' + RTln [ ADP ][ P i ] [ ATP ] " G = # 30.5 kJ / mol + (8.315 J / mol ° K )(310 ° K )ln (2x10 # 3 M )(1x10 # 3 M ) 3x10 # 3 M " G = # 30.5 kJ / mol # 18.9 kJ / mol = # 49.4 kJ / mol 10 pts. 2 . a . Calculate the average total negative charge on the phosphate groups of ATP at pH = 7.0, given that for three of the ionizable –OH groups, pK dn = 2.5, and for the fourth, pK dn = 6.5. At pH = 7, the 3 phosphate oxygens with pK dn = 2.5 will be fully ionized = -3 charge. For the 4 th oxygen: 7.0 = 6.5 + log [ " O " ] [ " OH ] # 0.5 = log [ " O " ] [ " OH ] # [ " O " ] [ " OH ] = 10 0.5 = 3.2 [ " O " ] = 3.2[ " OH ]; [ " OH ] + [ " O " ] = 1 # [ " OH ] + 3.2[ " OH ] = 1 4.2[ " OH ] = 1 [ " OH ] = 0.24 # [ " O " ] = 0.76 So the average total negative charge on the phosphate groups of ATP = -3.8 at pH = 7.0
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Name_____________ KEY ____________________________ EID__________________________________________ 2 5 pts. b . Make the same calculation for the average total negative charge on the phosphate groups of ADP at pH = 7.0, given that for two of the ionizable –OH groups, pK dn = 2.5, and for the third, pK dn = 7.2. At pH = 7, the 2 phosphate oxygens with pK
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07Fexam5key - Name_KEY EID CH 339K Exam 5 13 Nov 2007 100...

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