Problem 3 You measure the following extracellular and intracellular concentratio
ID: 3477695 • Letter: P
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Problem 3 You measure the following extracellular and intracellular concentrations of ions (in mM) in a giant squid neuron at 25°C. [Na] = 510 [Na] = 74 [K]. = 22 (K] =436 [CI, = 482 [CI] = 83 You also measure the conductance (g) of K* to be 127nS, of Nat to be 32ns, and of Cl to be 89ns. (a) Calculate Vrest of the neuron. (b) On graph paper (or using your favorite graphing program, plot the I-V relationship for IK, Ina, ICIand total. Be sure to indicate points of current carried by each ion at the following voltages: a. -35mV b. Omv C. +90mV d. EK e. ENA f. Eci (c) What does the slope of these lines represent? What is the significance of the line(s) crossing the x-axis?Explanation / Answer
For cells in resting state ,the relativily static membrane potential is known as resting membrane potential (Vrest).
The resting membrane potential is at equilibrium since it relies on the constant expenditure of energy for its maintenance. It is dominated by ionic species having greatest conductance and for most cells that is potassium ion (K+). As potassium is the ion with most negative equilibrium potential the cells resting potential is also negative.
The voltage I.e Vrest is caused by differences in concentration of ions inside and outside the cell.The difference in concentration of potassium ions inside and outside the cell dominates the resting potential. The negative resting membrane potential is created and maintained by increasing the cations outside the cell.
The Vrest or membrane potential can be obtained using Ionic Ohms Law and Nernst equation and conductance of the ions.
Ionic Ohms law :Em= (gKEK + gNaENa + gClECl)/(gK +gNa +gCl)
where Em= membrane potential
g= conductance thus gK is cconductance for potassium and similarly gNa and gCl.
EK= equilibrium potential for potassium ion and similarly ENa for sodium and ECl for chloride ion.
The equilibrium potential can bE derived using Nernst equation
Eion = 2.303 RT/ZF × log [ion concentration] outside cell/[ion concentration] inside cell.
where
Eion= equilibrium potential of ion
R= gas constant= 8.314 Jmol-1K-1
T= temperature in Kelvin.
Z= Ion charge
F= Faraday's constant= 9.649×104 JV-1mol-1
Therefore Temp= 25degree Celsius = 298Kelvin
2.303RT/F = 59
Therefore EK= 59/1× log [22]/[436]= -76.52mV.
ENa= 59/1 × log [510]/[74]= 49.46 mV.
ECl = -59/1 × log [482]/[83]= -45.07 mV.
Em= (gKEK + gNaENa + gClECl)/(gK +gNa +gCl)
Therefore
Em =(127×-76.52 +32×49.46 + 89×-45.07)/(127+32+89)=-61.74 mV.
Therefore Vrest = Em= -61.74mV.
Itotal =IK + INa+ ICl
Iion=gion × (Eion- Em) as per Ionic Ohms law.
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