• Synaptic vesicles of cholinergic neurons are loaded with the neurotransmitter acetylcholine ( $AC{h}^{+}$$ACh^+$ )

• The membrane of these synaptic vesicles contains an electrogenic $H^{+}-\text{ATPase}$$H^+-\text{ATPase}$ that pumps $H^{+}$$H^+$ into the vesicle , together with $AC{h}^{+}/H^{+}$$ACh^+/H^+$-exchangers ( vAChT ) and $C{l}^{–}/H^{+}$$Cl^–/H^+$-exchangers ( CLC )

• The coupling stoichiometry of the vAChT is one ACh+ exchanged for $2H^{+}$$2 H^+$ , and of the CLC is $2Cl–$$2\ Cl–$ exchanged for $1H+$$1\ H+$ , with the $H^{+}-\text{ATPase}$$H^+-\text{ATPase}$ hydrolyzing $1\text{ATP}$$1\ \text{ATP}$ for $2H^{+}$$2 H^+$ pumped

• These vesicles exhibit a membrane potential $V_{\text{sv}}=80mV$$V_{\text{sv}} = 80\ mV$ , inside positive , and a pH more acidic than cytoplasm by $1.5$$1.5$ units

• Assume that the cytosol of the pre-synaptic terminal has the following properties :

  • $[{\text{ACh}}^{+}{]}_{\text{intracellular}}=1\text{mM}$$[\text{ACh}^+]_{\text{intracellular}} = 1\ \text{mM}$

  • $[{\text{H}}^{+}{]}_{\text{intracellular}}=1\mu \text{M}$$[\text{H}^+]_{\text{intracellular}} = 1\ \mu \text{M}$

  • $[{\text{Cl}}^{-}{]}_{\text{intracellular}}=10\text{mM}$$[\text{Cl}^-]_{\text{intracellular}} = 10\ \text{mM}$

• You predict that the vesicular transporters constitute a secondarily active system maintaining a high vesicular $[{\text{ACh}}^{+}]$$[\ \text{ACh}^+\ ]$

1. Draw a diagram of the synaptic vesicle that includes the transporters and indicates the relative rates for each that will achieve a steady-state $V_{\text{sv}}$$V_{\text{sv}}$ and ${\text{pH}}_{\text{sv}}$$\text{pH}_{\text{sv}}$

   

   • in order to know what "steady-state" looks like , we have to balance the stoichiometry numbers for all of the inputs and outputs

   • we don't need to worry about balancing ACh , because we want and expect it to accumulate in the synaptic vesicle

   • and since in our simplified model we don't have a way to export chloride , we want to then minimize the number of cycles that transporter

   • so lets just balance hydrogen :

     • at steady-state , we have more leaving ( 3 ) the cell than entering ( 2 )

     • if we run the ATPase $1.5$$1.5$ times instead of just once :

       $$1.5\ast 2=3H^{+}$$

     • so then inputs = 3 , and outputs = 3

     • but we can't run the ATPase half-way , its all or none.

     • double $1.5$$1.5$ to get the next whole number $3$$3$

     $$3\ast 2=6H^{+}$$

     • but now total hydrogen inputs = 6 , and total hydrogen outputs = 3

     • so make them also run an additional cycle

   

   • ok , but the charges are messed up and its not electrically neutral

   $$\text{Total Charge}=(+6\text{ from }{H}^{+})+(+2\text{ from }AC{h}^{+})+(-4\text{ from }C{l}^{-})$$
   $$\text{Total Charge}=+6+2-4=+4$$

   

   • now $V_{\text{sv}}$$V_{\text{sv}}$ and ${\text{pH}}_{\text{sv}}$$\text{pH}_{\text{sv}}$ are balanced

2. You decide to calculate how high the vesicular $[{\text{ACh}}^{+}]$$[\ \text{ACh}^+\ ]$ can reach with the available electrochemical gradients , $E_{\text{ACh}}$$E_{\text{ACh}}$

   $$E_{\text{ACh}}=\frac{\frac{8.314Joules}{1Kelvin\cdot mol}\ast (37.0+{274.15}^{\circ }K)}{+1\ast \frac{96484Joules}{Volts\cdot mol}}\ast ln\left(\frac{[{\text{ACh}}^{+}{]}_{\text{vesicle}}}{[{\text{ACh}}^{+}{]}_{\text{cytosol}}}\right)$$
   $$E_{\text{ACh}}=\frac{\frac{8.314Joules}{1Kelvin\cdot mol}\ast (37.0+{274.15}^{\circ }K)}{+1\ast \frac{96484Joules}{Volts\cdot mol}}\ast ln\left(\frac{[{\text{ACh}}^{+}{]}_{\text{vesicle}}}{1mM}\right)$$

   

   $$\mu ={\mu }_0+(z\ast F\ast V)+(R\ast T\ast ln\left(\frac{[X{]}_{\text{inside}}}{[X{]}_{\text{outside}}}\right))$$

   • so we were given a pH difference instead of concentrations

     $$\frac{[H^{+}{]}_{\text{inside}}}{[H^{+}{]}_{\text{outside}}}={10}^{-\mathrm{\Delta }pH}$$

   • and we also need the potential across the membrane ( $\mathrm{\Delta }\mu$$\Delta \mu$ ) , instead of just for one ion ( $\mu$$\mu$ )

     $$\mathrm{\Delta }{\mu }_X={\mu }_{\text{inside}}-{\mu }_{\text{outside}}$$
     $$\mathrm{\Delta }\mu H=(z\ast F\ast \left((V_{\text{inside}}-\left(V_{\text{outside}}\right))\right))+(R\ast T\ast ln\left({10}^{-\mathrm{\Delta }pH}\right))$$

   • we can simplify the natural log to :

     $$\mathrm{\Delta }\mu H=(z\ast F\ast \left((V_{\text{inside}}-\left(V_{\text{outside}}\right))\right))+(R\ast T\ast \mathrm{\Delta }pH\ast ln\left(10\right))$$

   • ok , but $V_{\text{sv}}$$V_{\text{sv}}$ is defined as :

   $$V_{\text{sv}}=\left(V_{\text{vesicle}}\right)-\left(V_{\text{cytosol}}\right)$$

   • so , we have to flip the order of $\mathrm{\Delta }V$$\Delta V$ in the equation :

   $$\mathrm{\Delta }\mu H=(z\ast F\ast \left((V_{\text{outside}}-\left(V_{\text{inside}}\right))\right))+(R\ast T\ast \mathrm{\Delta }pH\ast ln\left(10\right))$$

   • being explicit :

     • $V_{\text{outside}}=V_{\text{cytosol}}$$V_{\text{outside}} = V_{\text{cytosol}}$

     • $V_{\text{inside}}=V_{\text{vesicle}}$$V_{\text{inside}} = V_{\text{vesicle}}$

     $$\mathrm{\Delta }\mu H=(z\ast F\ast \left((V_{\text{cytosol}}-\left(V_{\text{vesicle}}\right))\right))+(R\ast T\ast \mathrm{\Delta }pH\ast ln\left(10\right))$$

   • finally solving for $\mathrm{\Delta }\mu H$$\Delta \mu H$ :

   $$\mathrm{\Delta }\mu H=(+1\ast \frac{96484Joules}{Volts\cdot mol}\ast (-80.0\ast {10}^{-3}V))+(\frac{8.314Joules}{1Kelvin\cdot mol}\ast (37.0+{274.15}^{\circ }K)\ast \left(1.5\right)\ast ln\left(10\right))$$
   $$=\frac{1216.119864864849291806203624\text{Joules}}{\text{mol}}$$

   xxxxxxxxxx
   import math
   from decimal import Decimal as D
   ( D('1') * D('96484') * D(-80.0 * 10**-3) ) + ( D('8.314') * D('311.15') * D('1.5') * D(math.log(10)) )
   

   • but now we need $\mathrm{\Delta }{\mu }_{\text{ACh}}$$\Delta \mu_{\text{ACh}}$

     • and its coupled to hydrogen in the vAChT transporter at 1 : 2 ratio

     $$\mathrm{\Delta }{\mu }_{\text{ACh}}=2\ast \mathrm{\Delta }{\mu }_H=\frac{2432.239729729698583612407248\text{Joules}}{mol}$$

   • now we can plug this into the nersnt equation , and solve for the vesicular concentration of ACh :

   $$\mathrm{\Delta }{\mu }_{\text{ACh}}=R\ast T\ast ln\left(\frac{[{\text{ACh}}^{+}{]}_{\text{vesicle}}}{[{\text{ACh}}^{+}{]}_{\text{cytosol}}}\right)$$
   $$\frac{2432.239729729698583612407248\text{Joules}}{mol}=\frac{8.314Joules}{1Kelvin\cdot mol}\ast (37.0+{274.15}^{\circ }K)\ast ln\left(\frac{[{\text{ACh}}^{+}{]}_{\text{vesicle}}}{1mM}\right)$$
   $$\frac{\frac{2432.239729729698583612407248\text{Joules}}{mol}}{\frac{8.314Joules}{1Kelvin\cdot mol}\ast (37.0+{274.15}^{\circ }K)}=ln\left(\frac{[{\text{ACh}}^{+}{]}_{\text{vesicle}}}{1mM}\right)$$
   $$e^{\left(\frac{\frac{2432.239729729698583612407248\text{Joules}}{mol}}{\frac{8.314Joules}{1Kelvin\cdot mol}\ast (37.0+{274.15}^{\circ }K)}\right)}=e^{\left(ln\left(\frac{[{\text{ACh}}^{+}{]}_{\text{vesicle}}}{1mM}\right)\right)}$$
   $$e^{\left(\frac{\frac{2432.239729729698583612407248\text{Joules}}{mol}}{\frac{8.314Joules}{1Kelvin\cdot mol}\ast (37.0+{274.15}^{\circ }K)}\right)}=\frac{[{\text{ACh}}^{+}{]}_{\text{vesicle}}}{1mM}$$
   $$e^{\left(\frac{\frac{2432.239729729698583612407248\text{Joules}}{mol}}{\frac{8.314Joules}{1Kelvin\cdot mol}\ast (37.0+{274.15}^{\circ }K)}\right)}\ast \left(1.0mM\right)=[{\text{ACh}}^{+}{]}_{\text{vesicle}}$$
   $$[{\text{ACh}}^{+}{]}_{\text{vesicle}}=2.560528421056074965246751589mM$$

• 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

3. Calculate how many charges you have to separate across the cell membrane to achieve $Vm=-60mV$$Vm = -60\ mV$ for a $10\mu m$$10\ \mu m$ diameter cell

   $$Q=C\ast V_m$$
   $$\text{\# of elementary charges}=\frac{Q}{e}$$

   • $e\approx 1.602\ast {10}^{-19}C$$e \approx 1.602 * 10^{-19}\ C$

   • Membrane Specific Capacitance = $\frac{1\mu F}{c{m}^2}$$\frac{1\ \mu F}{cm^2}$

   • Surface Area :

     $$A=\pi \ast {(5\ast {10}^{-6}meters)}^2=7.84540\ast {10}^{-11}meter{s}^2$$

   • $1m^2={10}^{4}c{m}^2$$1\ m^2 = 10^{4}\ cm^2$

   $$A=7.84540\ast {10}^{-11}meter{s}^2\ast \frac{{10}^{4}c{m}^2}{m^2}=7.8540\ast {10}^{-7}{\text{cm}}^2$$

   • back to capacitance :

   $$C=\left(\frac{1\ast {10}^{-6}F}{c{m}^2}\right)\ast (7.8540\ast {10}^{-7}{\text{cm}}^2)=7.8540\ast {10}^{-13}F$$

   • total charge :

   $$Q=(7.8540\ast {10}^{-13}F)\ast (60\ast {10}^{-3}V)=4.7124\ast {10}^{-14}C$$

   • and finally , converting to number of elementary charges :

   $$\text{\# of elementary charges}=\frac{4.7124\ast {10}^{-14}C}{1.602\ast {10}^{-19}C}=2.9412\ast {10}^5\text{charges}$$

4. Calculate how many charges likely are inside of this cell. Assume a $150mM$$150\ mM$ KCl solution inside of cell and a $150mM$$150\ mM$ NaCl solution outside of cell. ( $1c{m}^3$$1 cm^3$ = 1 milli-Liter = ${10}^{-3}$$10^{-3}$ Liter )

   • volume of a sphere :

   $$V=\frac{4}{3}\ast \pi \ast {(5\ast {10}^{-6}m)}^3=5.2360\ast {10}^{-16}{m}^3$$
   $$V=5.2360\ast {10}^{-16}{m}^3\ast \frac{1000L}{1m^3}=5.2360\ast {10}^{-13}L$$

   • convert concentration of $KCl$$KCl$ to moles :

   $$\text{moles of KCl}=\frac{0.150mol}{L}\ast 5.2360\ast {10}^{-13}L=7.8540\ast {10}^{-14}mol$$

   • now into total number of particles :

     $$7.8540\ast {10}^{-14}mol\ast \frac{6.022\ast {10}^{23}\text{particles}}{\text{mol}}=4.7298\ast {10}^{10}\text{particles}$$

   • but KCl breaks apart into $K^{+}$$K^+$ and $C{l}^{-}$$Cl^-$ , so van't Hoff factor = 2

   $$2\ast 4.7298\ast {10}^{10}\text{particles}=9.4596\ast {10}^{10}\text{particles}$$

5. For synaptic vesicles $100nm$$100\ nm$ and $40nm$$40\ nm$ in diameter , you calculate how many charges have to be separated across the membrane to achieve $V_{\text{sv}}=+60mV$$V_{\text{sv}} = +60\ mV$

   • surface area of the $100nm$$100\ nm$ sphere :

     $$SA=4\ast \pi \ast {(50\ast {10}^{-9}m)}^2=3.1416\ast {10}^{-14}{m}^2$$
     $$SA=3.1416\ast {10}^{-14}{m}^2\ast \frac{{10}^{4}c{m}^2}{m^2}=3.1416\ast {10}^{-10}c{m}^2$$

   • capacitance :

     $$C=\frac{1\ast {10}^{-6}F}{c{m}^2}\ast 3.1416\ast {10}^{-10}c{m}^2=3.1416\ast {10}^{-16}F$$

   • charge :

     $$Q=3.1416\ast {10}^{-16}F\ast (60\ast {10}^{-3}V)=1.8850\ast {10}^{-17}C$$

   • number of charges :

     $$\frac{1.8850\ast {10}^{-17}C}{1.602\ast {10}^{-19}C}=117.65\text{charges}$$

   ---

   • now the same thing for $40nm$$40\ nm$ vesicle

   $$SA=4\ast \pi \ast {(20\ast {10}^{-9}m)}^2=5.0265\ast {10}^{-15}{m}^2$$
   $$SA=5.0265\ast {10}^{-15}{m}^2\ast \frac{{10}^{4}c{m}^2}{m^2}=5.0265\ast {10}^{-11}c{m}^2$$

   • capacitance :

     $$C=\frac{1\ast {10}^{-6}F}{c{m}^2}\ast 5.0265\ast {10}^{-11}c{m}^2=5.0265\ast {10}^{-17}F$$

   • charge :

     $$Q=5.0265\ast {10}^{-17}F\ast (60\ast {10}^{-3}V)=3.0159\ast {10}^{-18}C$$

   • number of charges :

     $$\frac{3.0159\ast {10}^{-18}C}{1.602\ast {10}^{-19}C}=18.824\text{charges}$$

6. Calculate how many charges likely are inside of these synaptic vesicles

   • $100nm$$100\ nm$ :

     • volume :

       $$V=\frac{4}{3}\ast \pi \ast {(50\ast {10}^{-9}m)}^3=5.2360\ast {10}^{-22}{m}^3$$
       $$V=5.2360\ast {10}^{-22}{m}^3\ast \frac{1000L}{1m^3}=5.2360\ast {10}^{-19}L$$

     • moles :

     $$\text{moles of KCl}=\frac{0.150mol}{L}\ast 5.2360\ast {10}^{-19}L=7.8540\ast {10}^{-20}mol$$

     • total particles :

       $$7.8540\ast {10}^{-20}mol\ast \frac{6.022\ast {10}^{23}\text{particles}}{\text{mol}}=4.7298\ast {10}^4\text{particles}$$

     • times 2 for van't Hoff factor :

       $$4.7298\ast {10}^4\text{particles}\ast 2=9.4596\ast {10}^4\text{particles}$$

   • $40nm$$40\ nm$ :

     • volume :

       $$V=\frac{4}{3}\ast \pi \ast {(20\ast {10}^{-9}m)}^3=3.3510\ast {10}^{-23}{m}^3$$
       $$V=3.3510\ast {10}^{-23}{m}^3\ast \frac{1000L}{1m^3}=3.3510\ast {10}^{-20}L$$

     • moles :

       $$\text{moles of KCl}=\frac{0.150mol}{L}\ast 3.3510\ast {10}^{-20}L=5.0265\ast {10}^{-21}mol$$

     • total particles :

       $$5.0265\ast {10}^{-21}mol\ast \frac{6.022\ast {10}^{23}\text{particles}}{\text{mol}}=3027.1\text{particles}$$

     • times 2 for van't Hoff factor :

       $$3027.1\text{particles}\ast 2=6054.1\text{particles}$$

Review Questions

• Consider a system with two compartments separated by a rigid membrane that is freely permeable to water but impermeant to sucrose.

• Both compartments contain pure water and pressure is applied to the piston establishing a pressure difference , $P_B>P_A$$P_B > P_A$ , across the membrane

1. Describe the relation between volume flow across the membrane and the hydrostatic pressure difference , $P_B-P_A$$P_B - P_A$

• No force is applied to the piston and 100mM sucrose is placed in compartment B.

2. In what direction will the meniscus ( in compartment A ) move?

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3. What is the driving force for this volume flow?

   • asdf

4. Adding NaCl ( also impermeant ) to what compartment could prevent volume displacement?

   • asdf

5. What concentration of NaCl would have to be added to prevent volume displacement?

   • asdf

6. What concentration of MgCl2 would prevent volume displacement?

   • asdf

7. What pressure ( and orientation ) must be applied to the piston to prevent volume flow?

   • asdf

• Consider the membrane permeable to glycerol , but less permeable to glycerol than to water

• No force is applied to the piston and $100mM$$100\ mM$ glycerol is placed in compartment B

8. How will the meniscus in the capillary of A move?

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9. How will the rate of movement compare with that observed when $100mM$$100\ mM$ sucrose was in compartment B?

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10. What pressure ( and orientation ) must be applied to the piston to prevent volume flow?