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Membrane Transport. Δ G of Transport. Using a concentration gradient only: Assume that “A” is 0.05 mM extracellular and that a 15 mM intracellular concentration is desired at 37 o C. Calculate the free energy of movement of 1 mole of “A” from outside to inside the cell.
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ΔG of Transport Using a concentration gradient only: Assume that “A” is 0.05 mM extracellular and that a 15 mM intracellular concentration is desired at 37oC. Calculate the free energy of movement of 1 mole of “A” from outside to inside the cell.
ΔG of Transport For movement outside to inside: G = RT ln Ci/Co = 8.314(310) ln 15 x 10-3/0.05 x 10-3 = 2577 (ln 300) = 2577 (5.7) = 14700 J/mol or 14.7 kJ/mol G is (+) so this amount of energy must be provided in order to move 1 mol of “A”.
ΔG of ATP Hydrolysis Assume that ATP is available to provide energy for this transport. For ATP hydrolysis : ATP ADP + Pi Go' = -30.5 kJ/mol Cellular concentrations: ATP = 2.5 mM; ADP = 1.5 mM and Pi = 0.5 mM Calculate the free energy available from ATP hydrolysis at these concentrations.
ΔG of ATP Hydrolysis The overallG for ATP hydrolysis: G = Go' + RT ln ([ADP][Pi]/[ATP]) G = -30500 + 2577ln ([1.5x10-3][0.5x10-3]/[2.5x10-3 ]) G = -30500 + 2577 ln (3 x 10-4) G = -30500 + 2577 (-8.111) = -30500 - 20900 G = -51.4 kJ/mol There is sufficient energy from 1 mol of ATP hydrolysis to move 3 mols of “A”.
ΔG from Membrane Potential Assume that a membrane potential can be used to drive this transport. Calculate the membrane potential that would be needed to move 1 mol of “A” across the membrane using a unit charge. G = zF Y 14700 = 1(96480) YY = 14700/96480 = 0.152 V = 152 mV This is greater than the normal cell potential.
ΔG from Membrane Potential Normal membrane potentials range from ~ 60 mV to 100 mV. Assume a membrane potential of 60 mV is available. Calculate the intracellular concentration of “A” could be reached if driven by a cell potential of 60 mV.
ΔG from Membrane Potential Intracellular concentration of “A” attained from a cell potential of 60 mV: zF Y = RT ln Ci /Co 1(96480)(0.06) = 2577 (ln Ci /0.05 x 10-3) 5789 / 2577 = 2.246 = ln Ci /0.05 x 10-3 9.454 = Ci /0.05 x 10-3 and Ci = 4.72 x 10-4 M Therefore, a normal cell potential difference of 60 mV could attain only 0.472 mM “A” inside the cell which is less that the initially desired concentration of 15 mM.