1. Calculate how many free $H^{+}$$H^+$ ions are inside a mitochondrion

   • You start by assuming that :

     • a mitochondrion is a sphere with a diameter of 1 µm

     • a mitochondrial matrix pH of 8.0 , then 7.4 , and then 7.0

   • 1 cm3 = 1 milli-Liter = 10-3 Liter

   • 1 µm3 = 1 femto-Liter = 10-15 Liter

   • 1 m3 = 1,000 Liters

   $$pH=-log([H^{+}])$$
   $$V=\frac{\pi \ast d^3}{6}$$
   $$V=\frac{\pi \ast {(1\ast {10}^{-6}m)}^3}{6}=5.2360\ast {10}^{-19}{m}^3$$
   $$L=5.2360\ast {10}^{-19}{m}^3\ast {10}^3=5.2360\ast {10}^{-16}Liters$$

   ---

   $$\begin{array}{rl}pH=8& =-1o{g}_{10}([1\ast {10}^{-8}])\\ [H^{+}]& =\frac{1\ast {10}^{-8}mol{H}^{+}}{1Liter}\end{array}$$
   $$\frac{1\ast {10}^{-8}mol{H}^{+}}{1Liter}\ast 5.2360\ast {10}^{-16}Liters=5.2360\ast {10}^{-24}mol{H}^{+}$$
   $$5.2360\ast {10}^{-24}mol{H}^{+}\ast \frac{6.02214076\ast {10}^{23}\text{particles}}{\text{mol}}=3.1532H^{+}\text{ions}$$

   ---

   $$\begin{array}{rl}pH=7.4& =-1o{g}_{10}([1\ast {10}^{-7.4}])\\ [H^{+}]& =\frac{1\ast {10}^{-7.4}mol{H}^{+}}{1Liter}\end{array}$$
   $$\frac{1\ast {10}^{-7.4}mol{H}^{+}}{1Liter}\ast 5.2360\ast {10}^{-16}Liters=2.0845\ast {10}^{-23}mol{H}^{+}$$
   $$2.0845\ast {10}^{-23}mol{H}^{+}\ast \frac{6.02214076\ast {10}^{23}\text{particles}}{\text{mol}}=12.553H^{+}\text{ions}$$

   ---

   $$\begin{array}{rl}pH=7.0& =-1o{g}_{10}([1\ast {10}^{-7.0}])\\ [H^{+}]& =\frac{1\ast {10}^{-7.0}mol{H}^{+}}{1Liter}\end{array}$$
   $$\frac{1\ast {10}^{-7.0}mol{H}^{+}}{1Liter}\ast 5.2360\ast {10}^{-16}Liters=5.2360\ast {10}^{-23}mol{H}^{+}$$
   $$5.2360\ast {10}^{-23}mol{H}^{+}\ast \frac{6.02214076\ast {10}^{23}\text{particles}}{\text{mol}}=31.532H^{+}\text{ions}$$

   • Explain how this can occur

     • in the definition of pH

       • more positive pH = less available protons

       • less positive pH = more available protons

• You have measured the recovery of intracellular pH in response to acidification induced after an NH4Cl pre-pulse in cells under two conditions.

• The two cellular conditions had differences in Vm , normal and depolarized , which yielded the following data :

• The data was gathered from large spherical cells ( $120\mu m$$120\ \mu m$ diameter )

• Amiloride = blocks the $N{a}^{+}/H^{+}$$Na^+/H^+$ exchanger from functioning

• All of the intracellular pH recovery in depolarized cells was inhibited by amiloride

2. Calculate the net fluxes of $H^{+}$$H^+$ and $N{a}^{+}$$Na^+$ through the $N{a}^{+}/H^{+}$$Na^+/H^+$ exchanger in the normal cells and in the depolarized cells

   $$1m=100cm$$
   $$1m^3=(100{)}^{3}cm$$
   $$1m^3={(1\ast {10}^2)}^{3}cm$$
   $$1m^3=1\ast {10}^{6}c{m}^3$$
   $$1m^2=1\ast {10}^{4}c{m}^2$$

   ---

   $$\text{Surface Area}=\pi \ast d^2$$
   $$SA=\pi \ast {(120\ast {10}^{-6}m)}^2=4.5239\ast {10}^{-8}{m}^2\ast \frac{1\ast {10}^{4}cm}{m^2}=4.5239\ast {10}^{-4}c{m}^2$$

   ---

   $$\text{Volume}=\frac{\pi \ast d^3}{6}$$
   $$V=\frac{\pi \ast {(120\ast {10}^{-6}m)}^3}{6}=9.0478\ast {10}^{-13}{m}^3\ast \frac{1\ast {10}^{6}c{m}^3}{m^3}=9.0478\ast {10}^{-7}c{m}^3$$

   ---

   $$H^{+}\text{flux}=\frac{\beta \ast \mathrm{\Delta }pH}{\mathrm{\Delta }\text{time}}$$
   $$H^{+}\text{flux}=\frac{\beta \ast \frac{\mathrm{\Delta }pH}{\mathrm{\Delta }\text{time}}\ast Volume}{\text{Surface Area}}$$
   $$H^{+}\text{flux}=\frac{\frac{25\ast {10}^{-6}mol}{c{m}^3\cdot \text{pH unit}}\ast \frac{100\ast {10}^6\text{pH unit}}{\text{second}}\ast 9.0478\ast {10}^{-7}c{m}^3}{4.5239\ast {10}^{-4}c{m}^2}=\frac{5\ast {10}^{-12}mol}{c{m}^2\cdot \text{second}}$$

   ---

   $$H^{+}\text{flux}=\frac{\frac{50\ast {10}^{-6}mol}{c{m}^3\cdot \text{pH unit}}\ast \frac{100\ast {10}^6\text{pH unit}}{\text{second}}\ast 9.0478\ast {10}^{-7}c{m}^3}{4.5239\ast {10}^{-4}c{m}^2}=\frac{1\ast {10}^{-11}mol}{c{m}^2\cdot \text{second}}$$

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   ((50E-6)*(1E-4)*(((pi*((120E-6)^3))/6)*1E6))/((pi*((120E-6)^2))*1E4)
   

• The cells were then exposed to a weak acid that doubles the intrinsic buffering power $\left({\beta }_{int}\right)$$\left(\ \beta_{int} \ \right)$

3. Calculate the rate of recovery of $p{H}_i$$pH_i$ in the depolarized cell

   $$\frac{\mathrm{\Delta }pH}{\mathrm{\Delta }t}=\frac{H^{+}\text{flux}\ast SA}{\beta \ast V}$$
   $$p{H}_i=\frac{\frac{5\ast {10}^{-12}mol}{c{m}^2\cdot \text{second}}\ast 4.5239\ast {10}^{-4}c{m}^2}{\frac{50\ast {10}^{-6}mol}{c{m}^3\cdot \text{pH unit}}\ast 9.0478\ast {10}^{-7}c{m}^3}=\frac{5\ast {10}^{-5}\text{pH unit}}{\text{second}}$$

4. Calculate the influx of $N{a}^{+}$$Na^+$ into the depolarized cell

   • 1 : 1 , so $\frac{5\ast {10}^{-12}mol}{c{m}^2\cdot \text{second}}$$\frac{5 * 10^{-12}\ mol}{cm^2 \cdot \text{second}}$

5. Explain what the measured fluxes imply about the properties of the $N{a}^{+}/H^{+}$$Na^+/H^+$ exchanger

   • At RMP ( -90 mV ) , there is no flux

   • But it is flowing at depolarized potentials ( -20 mV )

6. Estimate the driving force for this exchanger. ( Try plotting driving force versus $V_m$$V_m$ )

   $$\mathrm{\Delta }\mu =R\ast T\ast ln\left(\frac{[Na{]}_i\ast [H{]}_o}{[Na{]}_o\ast [H{]}_i}\right)$$
   $$\mathrm{\Delta }\mu =\frac{8.31446261815324J}{mol\cdot {}^{\circ }K}\ast (273.15+37)\ast ln\left(\frac{(10\ast {10}^{-3})\ast \left({10}^{-7.4}\right)}{(145\ast {10}^{-3})\ast \left({10}^{-7.0}\right)}\right)=\frac{-1.5209\ast {10}^{4}J}{mol}$$