The K\' concentration inside a nerve cell is found to be 20x higher outside the

Question

The K' concentration inside a nerve cell is found to be 20x higher outside the cell than inside the cell, while the Na concentration is 20x higher inside versus outside If the membrane potential Em (Ein-Eout) is -77.6 mV, identify the ion that must be transported actively versus the ion that is near equilibrium (assuming ideal conditions) A. B. The NA-K'-ATPase (or sodium-potassium pump), moves three sodium ions from inside to outside, and two potassium ions from outside to inside while simultaneously hydrolyzing ATP to ADPPi. If ATP/ADP is 40 and [Pi] is 25mM, estimate the change in potential ( k/mol) for the sodium transport, potassium transport and ATP to ADP + Pi separately (assuming ideal conditions) C. A researcher proposes that perhaps a single ATP hydrolysis is sufficient for two rounds of Na/K' exchange (i.e., move 6 Na" atoms out and 4 K' atoms in). Is the hydrolysis of a single ATP sufficient to account for the active transport of ions for a single cycle? For two? Assume ideal conditions & Justify your answer D. Now imagine we are in non-ideal conditions, and the ionic strength on the inside of the cell is much higher than outside the cell. Considering only the sodium export, will this export be more or less favored when compared to the ideal scenario (Justify your answer)

Explanation / Answer

Ans.A. If we see the membrane potential of the cell of a neuron.There are more potassium ions inside the cells and more sodium ions outside the cells.The cell tries to maintain inside enviroment more negative than outside creating a menbrane potential of -70mV. In the above example it is given that sodium is 20 times higher inside than outside and potassium ion concentration higher outside. So the cell will try to take potaasium inside the cell and throw sodium outside the cells to maintain a equilibrium resting potential.

B.free energy change can directly be associated with change in potential which can be given by

change in potential =std free energy change+ RT ln[ADP][PI]/[ATP]

std free energy change is -30.5kJ/mole

RT is 2.45kJ/mol.

SO change in potential = -30.5+2.45*ln 0.04*0.25/0.04

=-34kJ/mol

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