Question

Neuron cells generate electrical signals by concentration gradients across membranes. Assuming a potassium ion concentration of 0.00300 M inside the cell, and a concentration of 0.115 M outside the cell, what is the electrical potential across the cell membrane? Body temperature is 300 K. The sign identifies the change in the electrical potential across the membrane and which way the ions flow (answer in mV) Answer 04684 Check

Answer 1

Here we have to calculate the membrane potentical. Now here the potential lis generated due to the movement of K+ ion inside and outside cell. The given concentration of K+ inside = 0.00300 M i.e 0.00300 mol of K+ per Liter is present inside the cell. Similarly, K+ outside = 0.115 M i.e 0.115 mol of K+ per Liter is present inside the cell.

Thus the number of moles of K+ outside is smore than K+ inside. Thus the flow of K+ ion will be from outside to inside.

Now the Potential can be calculated by using the Nernst Equation-

Potential(E) = (RT/zF) log([X]_out/[X]_in)

where-

• R is the universal gas constant (8.314 J.K^-1.mol^-1).
• T is the temperature in Kelvin (°K = °C + 273.15).= 300K
• z is the ionic charge for an ion. For example, z is +1 for K⁺
• F is the Faraday's constant (96485 C.mol^-1).
• [X]_out is the concentration of the ion outside of the species. For example the molarity outside of a neuron.
• [X]_in is the concentration of the ion inside of the species. For example, the molarity inside of a neuron.

Now putting the values, the potential will be-

= (RT/zF) log([X]_out/[X]_in)

= (8.314 J.K^-1.mol^-1 * 300K/ (+1)(96485 C.mol^-1) ) * log(0.115 / 0.00300)

= [(8.314 * 300)/ 96485] * 1.71 (J.K^-1.mol^-1 * K / C.mol^-1)

= 0.044 J/C = 0.044 V = 44mV