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    Problems in electrophysiology

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    Part 1. Complete problem 1 from the Problems in electrophysiology 2006 handout. You can find it in the electrophysiology handouts document in the course documents folder. Use the "RT/F ln" form for the GHK equation as was shown in class but multiply RT/F by 1000 to convert volts to millivolts. (Remember that the Nernst equation is the statement of electrochemical equilibrium for a permeable ion.)

    The Excel program uses "60 log10" so if you use the excel sheet fix the equation first. Note that the units of RT/F will be volts. The "60" in the other form is 1000 x 2.303 x RT/F (the temperature was not 20C). The "2.303" converts log10 to ln; the factor "1000" converts volts to millivolts.

    Part 2. Complete problem 2, parts a through d, from the same handout. Q = CV for a parallel plate capacitor. The same equation can be used for a cell membrane because the membrane is so thin that curvature doesn't affect the results. With axon radius given and by assuming that the total membrane surface area is 1 cm2, the length and volume of the piece of axon can be computed. Then you can use those values to find out how much K is in the piece of axon.

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    Solution Preview

    Please see the attachment for detailed calculations.

    1a. Use the GHK equation for Vm to determine the membrane potential. Remember that [Cli], the concentration of Cl inside the cell, appears on top because of the negative charge. (In the Nernst equation the negative charge on Cl results in a valence Z = -1).

    1b. Apply the Nernst equation to determine the individual potentials. Ions are in electrochemical equilibrium only if their equilibrium potential less the membrane potential is zero. If the equilibrium potential less the membrane ...

    Solution Summary

    The solution determines how much K is in the piece of axon.