If the cell above is equally permeable to $N{a}^{+}$$Na^+$ , $K^{+}$$K^+$ , and $C{l}^{-}$$Cl^-$ $(g_K=g_{Na}=g_{Cl}=\frac{1}{3})$$\left(\ g_K = g_{Na} = g_{Cl} = \frac{1}{3} \ \right)$ , the resting membrane potential will be closest to which of the following values?

What direction does $N{a}^{+}$$Na^+$ move, producing what direction of current flow, and how will this impact the $V_m$$V_m$ ?

• Driving Force = $(-22.9856mV)-\left(61.537mV\right)=-84.5226mV$$\left(\ -22.9856\ mV \ \right) - \left(\ 61.537\ mV \ \right) = -84.5226\ mV$

  • think of this as the inside of the cell being negative

    • cations will want to go inside

    • anions will want to go outside

• $N{a}^{+}$$Na^+$ enters the cell

• depolarizes the membrane

• current flow = inward

What direction does $K^{+}$$K^+$ move , producing what direction of current flow , and how will this impact the $V_m$$V_m$ ?

• Driving Force = $(-22.9856mV)-(-66.410mV)=+43.42439mV$$\left(\ -22.9856\ mV \ \right) - \left(\ -66.410\ mV \ \right) = +43.42439\ mV$

  • think of this as the inside of the cell being positive

    • cations will want to go outside

    • anions will want to go inside

• $K^{+}$$K^+$ leaves the cell

• hyperpolarizes the membrane

• current flow = outward

What direction does $C{l}^{-}$$Cl^-$ move, producing what direction of current flow, and how will this impact the $V_m$$V_m$ ?

• Driving Force = $(-22.9856mV)-(-64.084mV)=+41.0984mV$$\left(\ -22.9856\ mV \ \right) - \left(\ −64.084\ mV \ \right) = +41.0984\ mV$

  • think of this as the inside of the cell being positive

    • cations will want to go outside

    • anions will want to go inside

• $C{l}^{-}$$Cl^-$ enters the cell

• hyperpolarizes the membrane

• current flow = the movement of negative charges $\left(C{l}^{-}\right)$$\left(\ Cl^- \ \right)$ into the cell is equivalent to an outward current

  • since negative ions moving in is the same as positive ions moving out

If $K^{+}$$K^+$ conductance increases such that $g_K=0.6$$g_K=0.6$ , and $g_{Na}=g_{Cl}=0.2$$g_{Na} = g_{Cl} = 0.2$ , the membrane potential will be closest to which of the following values?

• Answer = B = closest to $-40mV$$-40\ mV$

At this new membrane potential, what direction does $C{l}^{-}$$Cl^-$ move, producing what direction of current flow, and how will this impact the $V_m$$V_m$ ?

• Driving Force = $(-40.35539mV)-(-64.084mV)=+23.72861mV$$\left(\ -40.35539\ mV \ \right) - \left(\ −64.084\ mV \ \right) = + 23.72861\ mV$

  • think of this as the inside of the cell being positive

    • cations will want to go outside

    • anions will want to go inside

• $C{l}^{-}$$Cl^-$ enters the cell

• hyperpolarizes the membrane

• current flow = the movement of negative charges $\left(C{l}^{-}\right)$$\left(\ Cl^- \ \right)$ into the cell is equivalent to an outward current

  • since negative ions moving in is the same as positive ions moving out

If the serum $K^{+}$$K^+$ levels are dropped to 5 meq/L (mM) , with the relative ionic conductances as listed in question 5 , will the membrane potential hyperpolarize or depolarize, and closest to which of the following values?

Separate Problem :

• Part A : A cell is permeable to $K^{+}$$K^+$ , $N{a}^{+}$$Na^+$ , and $C{l}^{-}$$Cl^-$ , with resting relative conductances of $0.7$$0.7$ , $0.1$$0.1$ , and $0.2$$0.2$ , respectively , and $E_K=-100mV$$E_K = -100\ mV$ , $E_{Na}=+67mV$$E_{Na} = +67\ mV$ and $E_{Cl}=-60mV$$E_{Cl} = -60\ mV$. What is the approximate resting membrane potential?

  $$E_m=\frac{(0.7\ast -100.0)+(0.1\ast 67.0)+(0.2\ast -60.0)}{0.7+0.1+0.2}=-75.3mV$$

• Part B : Then, a channel for $N{a}^{+}$$Na^+$ opens and increases the relative permeability for $N{a}^{+}$$Na^+$ to $0.9$$0.9$ , reducing relative permeability of $K^{+}$$K^+$ to $0.07$$0.07$ and $C{l}^{-}$$Cl^-$ to $0.03$$0.03$ . What is the expected membrane potential after this occurs , and what direction will each ion move , if at all?

• Resting membrane potential before the change: $-75.3mV$$-75.3\ mV$

• Membrane potential after $N{a}^{+}$$Na^+$ permeability increases: $+51.5mV$$+51.5\ mV$

• Ion movements:

  • $N{a}^{+}$$Na^+$ :

    • Driving Force = $(+51.5mV)-\left(67.0mV\right)=-15.5mV$$\left(\ +51.5\ mV \ \right) - \left(\ 67.0\ mV \ \right) = -15.5\ mV$

      • think of this as the inside of the cell being negative

        • cations will want to go inside

        • anions will want to go outside

    • $N{a}^{+}$$Na^+$ enters the cell

    • depolarizes the membrane

    • current flow = inward

  • $K^{+}$$K^+$ :

    • Driving Force = $(+51.5mV)-\left(100.0mV\right)=-48.5mV$$\left(\ +51.5\ mV \ \right) - \left(\ 100.0\ mV \ \right) = -48.5\ mV$

      • think of this as the inside of the cell being negative

        • cations will want to go inside

        • anions will want to go outside

    • $K^{+}$$K^+$ enters the cell

    • depolarizes the membrane

    • current flow = inward

  • $C{l}^{-}$$Cl^-$ :

    • Driving Force = $(+51.5mV)-(-60.0mV)=+111.5mV$$\left(\ +51.5\ mV \ \right) - \left(\ −60.0\ mV \ \right) = + 111.5\ mV$

      • think of this as the inside of the cell being positive

        • cations will want to go outside

        • anions will want to go inside

    • $C{l}^{-}$$Cl^-$ enters the cell

    • hyperpolarizes the membrane

    • current flow = the movement of negative charges $\left(C{l}^{-}\right)$$\left(\ Cl^- \ \right)$ into the cell is equivalent to an outward current

      • since negative ions moving in is the same as positive ions moving out