1.1 The table below lists , for a particular cell type , the intracellular ( i ) and extracellular ( e ) concentrations for a number of solutes , X [ in mM ] , including monovalent and divalent ions.

The electrical potential difference across the membrane ( $V_m$$V_m$ ) you measured for this cell type is $-60mV$$-60\ mV$ , cell interior with respect to exterior , at 29°C.

1.A : Enter your estimate of the Nernst potential , $E_{solute}$$E_{solute}$ , for each solute , in mV

• see table

1.B : Considering only the extracellular concentration [X]e , enter your prediction for the intracellular concentration [X]i of each solute, assuming a passive distribution across the membrane ( electrochemical equilibrium according to the measured $V_m$$V_m$ )

• example using $A^{+}$$A^+$

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104 / ( 10**( -0.06 / ( ( 8.314 * ( 29.5 + 274.15 ) ) / ( 1 * 96484 ) ) ) )

1.C : Enter your predictions about the physiological cellular properties concerning these solutes, especially those that are not distributed at electrochemical equilibrium.

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• Use T = 29.25 Celsius

1.2 : Prior to conducting a series of electrophysiological experiments on your favorite cell type, you consider the possible influences on the membrane electrical potential difference ( $V_m$$V_m$ ). You assume that for the mammalian cell that you’re studying , the basal condition has the following properties : $[N{a}^{+}{]}_e=145mM$$[Na^+]_e = 145\ mM$ || $[N{a}^{+}{]}_i=10mM$$[Na^+]_i = 10\ mM$ $[K^{+}{]}_e=4.5mM$$[K^+]_e = 4.5\ mM$ || $[K^{+}{]}_i=150mM$$[K^+]_i = 150\ mM$ $V_m=-64mV$$V_m = -64\ mV$

Before changing the bathing solution so that $[K^{+}{]}_e=25mM$$[K^+]_e = 25\ mM$ , you predict that the measured $V_m$$V_m$ will be altered

1.2.i : Calculate your estimates for the values of $E_{Na}$$E_{Na}$ and $E_K$$E_K$ in the basal condition

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( 60 / 1 ) * math.log10( 145 / 10 )

1.2.ii : From Ohm / Kirchhoff rules for membrane conductances , calculate $\frac{g_{Na}}{g_K}$$\frac{g_{Na}}{g_K}$

• start by calculating each driving force

  $${\text{DF}}_{Na}=(-64mV)-\left(69.68208013409848mV\right)=-133.6820801340985mV$$

  • the driving force is negative , think of inside of the cell as negative

    • sodium is positive charged cation , it will be attracted electrically into the cell

       

    $${\text{DF}}_K=(-64mV)-(-91.37272471682026mV)=27.372724716820258mV$$

  • the driving force is positive , think of inside of the cell as positive

    • potassium is positive charged cation , it will be repelled electrically out of the cell

• We are assuming the cell is in steady-state , no net flow of ions across the membrane. So set all of the currents equal to zero.

• write current equation for each

• set equal to zero

• solve for the ratio of $\frac{g_{Na}}{g_K}$$\frac{g_{Na}}{g_K}$

• multiply both sides by $\frac{1}{g_K}$$\frac{1}{g_K}$

• plug in values

  ( -1 * ( ( -64 ) - ( ( 60 / 1 ) * math.log10( 4.5 / 150 ) ) ) ) / ( ( ( -64 ) - ( ( 60 / 1 ) * math.log10( 145 / 10 ) ) ) )
  

   

• since $0.2$$0.2$ is less than $1$$1$ , the ratio is predominantly influenced by the denominator , potassium conductance

  • if the result of the ratio were greater than $1$$1$ , it would tell us the cell is "conducting" or flowing a lot more sodium ions across the membrane than sodium

1.2.A : Calculate your prediction for $V_m$$V_m$ after increasing the bath $K^{+}$$K^+$ concentration. Calculate the change in $E_K$$E_K$. Draw a time course for $V_m$$V_m$ and $E_K$$E_K$ to illustrate the experiment.

• new nernst :

  $$E_K=\frac{60}{+1}\ast lo{g}_{10}\left(\frac{25}{150}\right)=-46.68907502301862mV$$

• "change in $E_K$$E_K$" ( final - initial ) :

• new driving force :

• write balanced equation :

• we use the same conductance ration as before , and solve for one of them , so we can simplify the equation :

• replace $g_{Na}$$g_{Na}$ in steady-state equation *** :

• divide both sides of equation by $g_k$$g_k$

• plug numbers in , solve for $V_m$$V_m$

• *** also , instead of replacing $gNa$$g{Na}$ in steady-state equation ,

  • we also could find $g_{total}$$g_{total}$

    • aka , we let the numerator be the answer $\approx 0.2047$$\approx 0.2047$

    • which means , the denominator , $g_K$$g_K$ must be $1.0$$1.0$

    • therefore , $g_{total}=1.0+0.2047=1.2047$$g_{total} = 1.0 + 0.2047 = 1.2047$

  • now we just use the full parallel conductance equation :

1.2.B : List assumptions that might complicate your estimate.

• steady-state with no net ionic current

• conductances are constant

• ignores other ions ( calcium , chloride )

1.2.C : You apply an experimental stimulus to the cell that quickly depolarizes $V_m$$V_m$ to a new steady-state value of -45mV ( assume the basal ion concentrations )

Calculate your estimate for gNa / gK during this period of depolarization

If the stimulus only altered $N{a}^{+}$$Na^+$ conductance ( $g_{Na}$$g_{Na}$ ) , calculate the change in $g_{Na}$$g_{Na}$

If the stimulus only altered $K^{+}$$K^+$ conductance ( $g_K$$g_K$ ) , calculate the change in $g_K$$g_K$

1.2.D : After you add a hormone to the bathing solution , $V_m$$V_m$ slowly hyperpolarizes from $-64mV$$-64\ mV$ ( original basal condition ) to $-87mV$$-87\ mV$ , over the next 15 min. Because of your concerns that the cellular condition may have changed considerably over this time course , you test for a change in the relative $g_K$$g_K$ by increasing $[K^{+}{]}_e$$[K^+]_e$ to $10mM$$10\ mM$. The resulting change in $V_m$$V_m$ was $+6mV$$+6\ mV$ , allowing you to estimate $\frac{g_K}{g_{total}}$$\frac{g_K}{g_{total}}$

• assuming conductances stay constant after $[K^{+}{]}_e$$[\ K^+\ ]_e$ changes ,

  • then $\mathrm{\Delta }{V}_m$$\Delta V_m$ should scale with the amount of potassium conductance

  • let $g_{rest}$$g_{rest}$ = all of the other conductances combined , $(g_{N{a}^{+}},g_{C{l}^{-}},g_{C{a}^{}})$$\left(\ g_{Na^+}\ ,\ g_{Cl^-}\ ,\ g_{Ca^{}} \ \right)$

• if only $[K^{+}{]}_e$$[\ K^+\ ]_e$ , then $E_K$$E_K$ changes , but $E_{Na}$$E_{Na}$ doesn't really change much

• therefore ,

1.3 : Changing the membrane electrical potential difference ( $V_m$$V_m$ ) requires separation of charges between the inside and outside of the cell. You predict that generating $V_m=-60mV$$V_m = -60\ mV$ for a large cell will require more charge separation than for a smaller cell. You also are curious about the amount of charge separation needed compared with the total number of charges inside of each cell.

1.3.A : You decide to calculate how many charges you have to separate across the cell membrane to achieve $V_m=-60mV$$V_m = -60\ mV$ for a $10\mu m$$10\ \mu m$ diameter cell. Also for comparison , you decide to calculate how many charges likely are inside of this cell. ( Assume a $150mM$$150\ mM$ KCl solution inside the cell and a $150mM$$150\ mM$ NaCl solution outside the cell.

• compute surface area assuming sphere shape :

  $$\text{Surface Area}\left(A\right)=4\ast \pi \ast r^2$$
  $$A=4\ast \pi \ast {\left(5\mu m\right)}^2$$
  $$A=3.1415926535897923\ast {10}^{-10}{\text{meter}}^2$$

• capacitance units is just Farads , but for some "specific membrane capacitance" $\left(C_m\right)$$\left(\ C_m \ \right)$ of the lipid-bilayer material itself , we will find "capacitance per area" :

  $$C=\frac{Farad}{mete{r}^2}$$

• we are going to assume $C_m=\frac{1\mu F}{c{m}^2}$$C_m = \frac{1\ \mu F}{cm^2}$ :

  $$A=3.1415926535897923\ast {10}^{-10}{\text{meter}}^2\ast \frac{{10}^{4}c{m}^2}{1m^2}=3.1415926535897925\ast {10}^{-6}c{m}^2$$

• capacitance is also $C=\frac{\text{charge ( q )}}{V_m}$$C = \frac{\text{charge ( q )}}{V_m}$ :

• a Coulomb is also the total charge of around $6.242\ast {10}^{18}$$6.242 * 10^{18}$ elementary charges :

• so that was about $1.17\ast {10}^6$$1.17 * 10^{6}$ ions that were used to create the $-60mV$$-60\ mV$

• now to calculate cell volume ( assuming sphere ) :

• convert ${\text{meters}}^3$$\text{meters}^3$ into just $\text{Liters}$$\text{Liters}$ :

• convert to moles of potassium ions :

• convert to total ions :

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  ( ( ( 4.0 / 3.0 ) * math.pi * ( ( 5.0 * 10**-6 )**3 ) ) * 1000 ) * ( 0.15 ) * ( 6.02214076 * 10**23 )
  

1.3.B : You decide to calculate how many charges you have to separate across the cell membrane to achieve $V_m=-60mV$$V_m = -60\ mV$ for a Xenopus laevis oocyte $1.0mm$$1.0\ mm$ in diameter. Also , you decide to calculate how many charges likely are inside of this oocyte.

• surface area :

  $$A=4\ast \pi \ast {\left(0.5mm\right)}^2=3.141592653589793m{m}^2$$
  $$A=3.141592653589793m{m}^2\ast \frac{1\ast {10}^{-2}c{m}^2}{m{m}^2}=0.031415926535897934c{m}^2$$

• capacitance , again assuming $C_m=\frac{1\mu F}{c{m}^2}$$C_m = \frac{1\ \mu F}{cm^2}$ :

  $$C=\left(0.031415926535897934c{m}^2\right)\ast \left(\frac{1\ast {10}^{-6}F}{c{m}^2}\right)=3.141592653589793\ast {10}^{-8}F$$

• charge needed to establish $-60mV$$-60\ mV$ resting membrane potential :

  $$Q=(3.141592653589793\ast {10}^{-8}F)\ast (60\ast {10}^{-3}V)=1.884955592153876\ast {10}^{-9}\text{Coulombs}$$

• converting to elementary charges :

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  ( ( 4.0 * math.pi * ( 0.5 )**2 ) * ( 10**-2 ) * ( 10**-6 ) * ( 60 * 10**-3 ) ) / ( 1.602176634 * 10**-19 )
  

• now the volume :

• converting to Liters :

• converting to moles using same concentration as before :

  $$5.235987755982988\ast {10}^{-7}L\ast \frac{0.15molKCl}{1Liter}=7.853981633974482\ast {10}^{-8}molK$$

• converting to total ions :

R1 : Consider two compartments of equal volume separated by a membrane ( shown below )

• Evaluate the following issues for each permeability condition of the membrane :

  1. Comparing side A with side B, predict the orientation of the electrical potential difference (ePD) across the membrane during the initial moments of flow (before noticeable changes in chemical composition have occurred)

  2. Predict the magnitude of the ePD across the membrane during the initial moments of flow.

  3. Predict the ePD and chemical composition of compartments A and B at electrochemical equilibrium.

Condition A : The membrane is equally permeable to both K+ and Cl-.

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Condition B : The permeability of the membrane to K+ is made greater than that to Cl-

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Condition C : The membrane is made impermeant to Cl-, but remains permeable to K+.

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R2 : Consider two compartments of equal volume separated by a membrane (shown below).

Evaluate the following issues for each permeability condition of the membrane :

1. Comparing side A with side B, predict the orientation of the electrical potential difference (ePD) across the membrane during the initial moments of flow (before noticeable changes in chemical composition have occurred)

2. Predict the magnitude of the ePD across the membrane during the initial moments of flow.

3. Predict the ePD and chemical composition of compartments A and B at electrochemical equilibrium.

Condition D : The membrane is permeable to K+, but not to Na+ or anions.

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Condition E : The membrane becomes permeable only to Na+

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• Also, compare these outcomes with those for condition D.

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Condition F : The membrane becomes permeable to both Na+ and K+ , with Na+ permeability greater than K+ permeability $(k_p^{Na}>k_p^K)$$\left(\ k^{Na}_{p} > k^{K}_{p} \ \right)$

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• Also, indicate the factors that the magnitude of the ePD depend on.

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