contestada

You are doing a measurement to determine the conductance of an artificial membrane. The membrane is selectively permeable to Na . The temperature is 15 degrees C The external concentrate of Na is 500 mM The internal concentration of Na is 70 mM Using a voltage clamp apparatus you clamp the membrane voltage (Vm) at 20 mV At this clamped voltage you measure a current of -71 nA What is the membrane's conductance

Respuesta :

Answer:

What is the membrane's conductance = 2.47miuS

Explanation:

The detailed step and appropriate formula is as shown in the attached file.

Ver imagen olumidechemeng

The membrane's conductance in the given question is;

g_ion = 2.47 * 10⁻⁶ Ω

Conductance calculation

The membrane conductance (g) is gotten from the formula;

g_ion = I_ion/(V_m - E_ion)

Where;

  • I_ion is membrane current
  • V_m is the membrane potential
  • E_ion is the equilibrium potential for the ion or reversal potential.  

Formula for the equilibrium potential is calculated as;

E_ion = (RT/Zf) * In[ion(out)/Ion(in)]

where;

R is gas constant = 8.314 J/K.mol

F is Faraday's constant = 96500 C/mol

T is the temperature in kelvins

z is the ion's charge

[ion]out is the concentration of the ion outside.

[ion]ins is the concentration of the ion inside.  

We are given;

T = 15°C = 288 K

z = +1

[ion]out = 500 mM

[ion]ins = 70 mM

I_ion = -71 nA = -71 * 10⁻⁹ A

Thus;

E_ion = (8.314 * 288/(1 * 96500)) * In(500/70)

E_ion = 48.78 mV = 48.78 * 10⁻³ V

V_m = 20 mV = 20 * 10⁻³ V

g_ion = (-71 * 10⁻⁹)/((20 * 10⁻³) - (48.78 * 10⁻³))

g_ion = 2.47 * 10⁻⁶ Ω

Read more about membrane conductance at; https://brainly.com/question/24943129