Question

Use the following table to answer the questions

Ion	Extracellular Concentration (mM)	Intracellular Concentration (mM)
Na+	440	50
K+	20	400
Cl-	560	52
Ca++	10	1

1. Calculate the equilibrium potential for Na+, K+, Ca++, and Cl- ions? (4 points)
2. If the resting membrane potential is -65, give the direction of the flow of Na+, Ca++, K+, and Cl- ion. (4 points)
3. What will happen to the Cl- equilibrium potential, if one changes the extracellular chloride concentration from 560 mM to 100 mM? Give the direction of Cl- flow. (2 points)

Answer 1

A.It can be calculated using the Nernst equation:

Eeq Na+ = RT/zF ln [Na+]o/ [Na+]i

Eeq,Na+ is the equilibrium potential for Sodium, measured in volts

R is the universal gas constant, equal to 8.314 joules·K−1·mol−1

T is the absolute temperature, measured in kelvins (= K = degrees Celsius + 273.15)

For Normal body temperature, 37°C

T= 310.15K

z is the number of elementary charges of the ion in question involved in the reaction

F is the Faraday constant, equal to 96,485 coulombs·mol−1 or J·V−1·mol−1

[Na+]o is the extracellular concentration of sodium, measured in mol·m−3 or mmol·l−1

[Na+]i is likewise the intracellular concentration of sodium.

Putting concentration the values from the given table in the above equation with the constant terms

We have Eeq,Na+ = (8.314×310.15)/(1×96485) • ln (440/50)

                              = 0.05812 V

                              = 58.12 mV

Just alike Na+,

Equilibrium potential of K+ is given by,

Eeq,K+ = (8.314×310.15)/(1×96485) • ln (20/400)

            = -0.08006 V

            = -80.06 mV

Equilibrium potential of Cl- is given by,

Eeq,Cl- = (8.314×310.15)/(-1×96485) • ln (560/52)

             = - 0.06352 V

             = -63.52mV

Equilibrium potential of calcium is given by,

Eeq,Ca++ = (8.314×310.15)/(2×96485) • ln (10/1)

                 = 0.03077 V

                 = 30.77 mV

B. At the presence of Na+ ions, -ve value of resting potential decreases hence depolarization occurred. So positively charged Na+ will enter to the membrane . Hence the direction of sodium ions are towards the membrane.

For positively charged potassium K+ ions, negative value of resting potential increases hence hyperpolarization occurs. So direction of K+ ions will be outward of cell membrane .

For negatively charged Cl- chlorine ions negative value of resting potential increases and hyperpolarization occurs but the direction of the Cl- ions are into the/ towards the cell membrane.

For Ca2+ ions, negative values of resting potential decreases hence depolarisation occurs. So positively charged Ca++ ions will enter to the cell membrane or the direction will be towards the cell membrane.

C.

if one changes the extracellular chloride concentration from 560 mM to 100 mM then,

Eeq,Cl- = (8.314×310.15)/(-1×96485) • ln (100/52)

= - 0.01747 V

= - 17.47 mV

The direction of Cl- ions will be into the cell membrane or towards the cell membrane.