1 / 9

Membrane Transport

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.

demetria-lott
Télécharger la présentation

Membrane Transport

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Membrane Transport

  2. 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.

  3. 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”.

  4. 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.

  5. ΔG of ATP Hydrolysis The overallG 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”.

  6. 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) YY = 14700/96480 = 0.152 V = 152 mV This is greater than the normal cell potential.

  7. 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.

  8. 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.

  9. End of Membrane Transport

More Related