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Suppose that the concentration of Na outside the cell is 100 and inside the cell is 100 mmol/liter. The Nernst equation at 20oC is: You set the membrane voltage at 58 millivolts using a voltage clamp, and measure membrane current. If Na is the only ion crossing the membrane, you would expect to see:

User Bryan Edds
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2 Answers

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Final answer:

With equal Na+ concentrations inside and outside the cell and the membrane voltage set to the Nernst potential for Na+, little to no Na+ movement would occur since this represents an equilibrium state between concentration and electrical gradients.

Step-by-step explanation:

If the concentration of Na+ outside and inside the cell are both 100 mmol/liter and you set a membrane voltage of 58 millivolts using a voltage clamp, measuring membrane current with Na+ being the only ion crossing the membrane, you would expect to see little to no movement of Na+ ions. This is because the membrane voltage equals the Nernst potential for Na+, which is the voltage at which there is no net flow of ions across the membrane due to equilibrium between the concentration gradient and the electrical gradient.

Typically, the resting potential inside a neuron cell membrane is about -70 mV due to differing ion concentrations on either side, creating an electrical charge difference. However, when Na+ channels open, positively charged sodium ions rush into the cell, reducing this charge difference in a process called depolarization. If the concentration of Na+ was higher outside the cell, the ions would move into the cell following the concentration gradient until the electrical charge inside the cell increases, potentially reaching as high as +30 mV, and causing further depolarization.

User Pierre Gayvallet
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3 votes

Answer:

0 current

Step-by-step explanation:

Given that:

The concentration of Na inside the cell is 100 mmol/liter

The concentration of Na outside the cell is 100 mmol/liter

where;

z(valence of the ionic species for Na⁺ ) = +1, and the membrane voltage is 58 millivolt

Therefore;


E_(ion) = (58)/(z) * log _(10) \bigg (([ion]_(out))/([ion]_(in))\bigg)


E_(ion) = (58)/(1) * log _(10) \bigg ((100)/(100)\bigg)


E_(ion) =58 * log _(10) \bigg (1\bigg)


E_(ion) = 58 * 0


\mathbf{E_(ion) =0}

User Mehrdad Moradi
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