Question

assume 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 probabilities are zero. Design the cell by calculating the intracellular concentration of the chosen species.

Answer 1

Step-by-step explanation:

From the information given:

The cell potential on mars E = + 100 mV

By using Goldman's equation:

`E_m = (RT)/(zF)In \Big ((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 = (RT)/(zF) In ([K^+]_(out))/([K^+]_(in))`

where;

z = ionic charge on the species = + 1

F = faraday constant

∴

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

`100 * 10^(-3) = 0.0257 \Big ( (4)/([K^+]_(in)) \Big)`

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

`exp ( 3.981) = (4)/([K^+]_(in)) \\ \\ 53.57 = (4)/([K^+]_(in))`

`[K^+]_(in) = (4)/(53.57)`

`[K^+]_(in) = 0.0476`

For [Cl⁻]:

`100 * 10^(-3) = -0.0257 \ \mathtt{In} \Big ( (120)/([Cl^-]_(in)) \Big)`

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

`0.01867 = (120)/([Cl^-]_(in))`

`[Cl^-]_(in) = (120)/(0.01867)`

`[Cl^-]_(in) =6427.4`

For [Na⁺]:

`100 * 10^(-3) = 0.0257 \Big ( (145)/([Na^+]_(in)) \Big)`

`53.57= \Big ( (145)/([Na^+]_(in)) \Big)`

`[Na^+]_(in)= 2.70`