Question

ssume that life on Mars requires cell potential to be 100mV, and the extracellular concentrations of the three major species are following (mmol/L): Na : 145; K : 4; Cl- : 120. Choose one species and assume that other two permeabilities are zero. Design the cell by calculating the intracellular concentration of the chosen species.

Answer 1

Answer:

Explanation:

Given that:

The cell potential on mars E = + 100 mV

By using Goldman's equation:

$E_m = \dfrac{RT}{zF}In \Big (\dfrac{P_K[K^+]_{out}+P_{Na}[Na^+]_{out}+P_{Cl}[Cl^-]_{out} }{P_K[K^+]_{in}+P_{Na}[Na^+]_{in}+ P_{Cl}[Cl^-]_{in}} \Big )$

Let's take a look at the impermeable cell with respect to two species;

and the two species be Na⁺ and Cl⁻

$E_m = \dfrac{RT}{zF} In \dfrac{[K^+]_{out}}{[K^+]_{in}}$

where;

z = ionic charge on the species = + 1

∴

$100 \times 10^{-3} = \Big (\dfrac{8.314 \times 298}{1\times 96485} \Big) \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}} \Big)$

$100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}} \Big)$

$3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}} \Big)$

$exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\ 53.57 = \dfrac{4}{[K^+]_{in}}$

$[K^+]_{in} = \dfrac{4}{53.57}$

$[K^+]_{in} = 0.0476$

For [Na⁺]:

$100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}} \Big)$

$53.57= \Big ( \dfrac{145}{[Na^+]_{in}} \Big)$

$[Na^+]_{in}= 2.70$

For [Cl⁻]:

$100 \times 10^{-3} = -0.0257 \ \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}} \Big)$

$-3.981 = \ \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}} \Big)$

$0.01867 = \dfrac{120}{[Cl^-]_{in}}$

$[Cl^-]_{in} = \dfrac{120}{0.01867}$

$[Cl^-]_{in} =6427.4$