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Nerve signals in the body occur when a small voltage, called an action potential, is applied across the membrane of a cell. When this action potential is applied across a region of the cell membrane called an ion channel, current in the form of moving potassium ions will be established across the cell. If, during an action potential, a single ion channel with a resistance of 1.8 GΩ, is opened for 0.6 ms, approximately 90000 singly-ionized potassium ions travel through the channel during this time. What was the voltage of the action potential? You can approximate the ion channel to be Ohmic.

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cjrodriguez89

Answer:

V = 43.2 mV

Explanation:

  • If we can approximate the ion channel to be Ohmic, this means that  we can apply Ohm's Law to get the voltage of the action potential, as follows:

       [tex]V = I*R (1)[/tex]

  • The definition of electric current, is the rate of change of the charge, i.e., the amount of total charge crossing a given area, perpendicular to the flow of charge carriers, per unit time.
  • If we know that approximately 90000 single ionized potassium ions travel through the channel during 0.6 mseg, we can find the value of the current.
  • The charge represented by 90000 single ionized potassium ions, is the same as 90000 positive elementary charges, equal to 1.6*10⁻¹⁹ C each (as the potassium single ionized carries a +1 e charge).
  • So, total charge is just the product of 90000 by 1.6*10⁻¹⁹ C:

       [tex]Q = 90000* 1.6e-19C = 1.44e-14 C[/tex]

        ⇒ [tex]I =\frac{Q}{t} = \frac{1.44e-14C}{0.6e-3s} = 24e-12 A = 24 pA[/tex]

  • Replacing the values of I and R in (1), we get:
  • [tex]V = I*R = 24e-12A * 1.8e9 \Omega = 43.2e-3 V = 43.2 mV[/tex]

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