1.1 The table below lists , for a particular cell type , the intracellular ( i ) and extracellular ( e ) concentrations for a number of solutes , X [ in mM ] , including monovalent and divalent ions.

The electrical potential difference across the membrane ( Vm ) you measured for this cell type is 60 mV , cell interior with respect to exterior , at 29°C.

1.A : Enter your estimate of the Nernst potential , Esolute , for each solute , in mV

1.B : Considering only the extracellular concentration [X]e , enter your prediction for the intracellular concentration [X]i of each solute, assuming a passive distribution across the membrane ( electrochemical equilibrium according to the measured Vm )

0.06 V=8.314 Joules1 Kelvinmol( 29.25+274.15K )+196484 JoulesVoltsmolln( 104 mM[X]i )
0.06 V8.314 Joules1 Kelvinmol( 29.25+274.15K )+196484 JoulesVoltsmol=ln( 104 mM[X]i )
2.293101322253533=ln( 104 mM[X]i )
e2.293101322253533=eln( 104 mM[X]i )
0.10095288841966855=104 mM[X]i
[X]i=104 mM0.10095288841966855=1030.1835007203001 mM

1.C : Enter your predictions about the physiological cellular properties concerning these solutes, especially those that are not distributed at electrochemical equilibrium.

Solute X[X]e[X]i[X]ipredictedEsolutePredictions
A+10471030.18470.316 mV 
E+81101571.055-68.298 mV 
G+20.011385.658-60.0 mV 
J1050.0509-18.062 mV 
L100100.5092-60.0 mV 
M220.010180 mV 
Q44 0 mV 
R13 0 mV 
Ef=( 60z )log10( [X]extracellular[X]intracellular )

1.2 : Prior to conducting a series of electrophysiological experiments on your favorite cell type, you consider the possible influences on the membrane electrical potential difference ( Vm ). You assume that for the mammalian cell that you’re studying , the basal condition has the following properties : [Na+]e=145 mM || [Na+]i=10 mM [K+]e=4.5 mM || [K+]i=150 mM Vm=64 mV

Before changing the bathing solution so that [K+]e=25 mM , you predict that the measured Vm will be altered

1.2.i : Calculate your estimates for the values of ENa and EK in the basal condition

ENa=60+1log10( 14510 )=69.68208013409848 mV
EK=60+1log10( 4.5150 )=91.37272471682026 mV

1.2.ii : From Ohm / Kirchhoff rules for membrane conductances , calculate gNagK

Driving Force=( Vm )( Eion )
Current ( I )=ConductanceDriving Force
V=IR
R=1g
V=Ig
I=gV
INa+IK=0
INa=gNaDFNa
IK=gKDFNa
[ gNaDFNa ]+[ gKDFK ]=0
[ gNaDFNa ]=[ gKDFK ]
gNa=[ gKDFK ]DFNa
gNagK=[ DFK ]DFNa
gNagK=[ 27.372724716820258 mV ]133.6820801340985 mV=0.20475986526662562

1.2.A : Calculate your prediction for Vm after increasing the bath K+ concentration. Calculate the change in EK. Draw a time course for Vm and EK to illustrate the experiment.

( 46.68907502301862 mV )( 91.37272471682026 mV )=44.68364969380164 mV
DFK=( 64 mV )( 46.68907502301862 mV )=17.310924976981383 mV
[ gNaDFNa ]+[ gKDFK ]=0
[ gNa( ( Vm )( ENa ) ) ]+[ gK( ( Vm )( EK ) ) ]=0
gNagK=[ DFK ]DFNa=0.20475986526662562
gNa=gk0.20475986526662562
[ ( gk0.20475986526662562 )( ( Vm )( ENa ) ) ]+[ gK( ( Vm )( EK ) ) ]=0
[ ( 0.20475986526662562 )( ( Vm )( ENa ) ) ]+[ ( ( Vm )( EK ) ) ]=0
[ ( 0.20475986526662562 )( ( Vm )( 69.68208013409848 mV ) ) ]+[ ( ( Vm )( 46.68907502301862 ) ) ]=0
( 0.20475986526662562 )( ( Vm )( 69.68208013409848 mV ) ) + ( ( Vm )( 46.68907502301862 ) )=0
( 0.20475986526662562Vm )14.268093339756216+Vm+46.68907502301862=0
1.20475986526662562Vm+32.4209816832624=0
1.20475986526662562Vm=32.4209816832624
Vm=32.42098168326241.20475986526662562=26.91074181499838
Vm=( 0.20471.204769.68 mV )+( 1.01.204746.689 mV )27.09 mV

1.2.B : List assumptions that might complicate your estimate.

1.2.C : You apply an experimental stimulus to the cell that quickly depolarizes Vm to a new steady-state value of -45mV ( assume the basal ion concentrations )

Calculate your estimate for gNa / gK during this period of depolarization

gNagK=[( 45 mV )( 91.37 mV )]( 45 mV )( 69.68 mV )=0.404

If the stimulus only altered Na+ conductance ( gNa ) , calculate the change in gNa

gNa=gK0.404
ΔgNa=0.4040.204=0.2

If the stimulus only altered K+ conductance ( gK ) , calculate the change in gK

ΔgK=10.40410.204=2.43

1.2.D : After you add a hormone to the bathing solution , Vm slowly hyperpolarizes from 64 mV ( original basal condition ) to 87 mV , over the next 15 min. Because of your concerns that the cellular condition may have changed considerably over this time course , you test for a change in the relative gK by increasing [K+]e to 10 mM. The resulting change in Vm was +6 mV , allowing you to estimate gKgtotal

Hyperpolarization Vm=( 87 mV )( 64 mV )=23 mV=New Resting Membrane Potential
ΔVm=( X mV )( 23 mV )
+6 mV=( X mV )( 23 mV )
Final Vm=17 mV

 

EK=60+1log10( 4.5150 )=91.37272471682026 mV
EK=60+1log10( 10150 )=70.56547554334088 mV

 

ΔEK=( 70.56547554334088 mV )( 91.37272471682026 mV )=20.807249173479377 mV
Vm=( gKEK )+( gNaENa )gtotal
Δ Vm( gKgtotal )ΔEK
gKgtotal=ΔVmΔEK=+6 mV+20.807249173479377 mV=0.28836103946155045

1.3 : Changing the membrane electrical potential difference ( Vm ) requires separation of charges between the inside and outside of the cell. You predict that generating Vm=60 mV for a large cell will require more charge separation than for a smaller cell. You also are curious about the amount of charge separation needed compared with the total number of charges inside of each cell.

1.3.A : You decide to calculate how many charges you have to separate across the cell membrane to achieve Vm=60 mV for a 10 μm diameter cell. Also for comparison , you decide to calculate how many charges likely are inside of this cell. ( Assume a 150 mM KCl solution inside the cell and a 150 mM NaCl solution outside the cell.

1 cm3=1 milli Liter=1103=1 μm3=1 femto Liter=11015 Liter
C=ACm
C=( 3.14106 cm2 )( 1106 Fcm2 )=3.14159265358979231012 F
Charge ( Q )=( 3.14159265358979231012 Farads )( 0.06 Volts )=1.88495559215387531013 Coulombs
1elementary charge ( e )=11.6021766341019 Coulombs=6.2415090744607621018 elementary charges
1.88495559215387531013 Coulombs1.6021766341019 Coulombselementary charge=1176496.7433383972 ions
V=43πr3
V=43π( 5 μm )3=5.2359877559829861016 meters3
5.2359877559829861016 meters31000 L1 m3=5.2359877559829871013 L
0.15 mol KCl1 Liter5.2359877559829871013 L=7.853981633974481014 mol KCl
7.853981633974481014 mol K+ from KCl6.022140761023 ionsmol=4.72977829262491151010 ions

1.3.B : You decide to calculate how many charges you have to separate across the cell membrane to achieve Vm=60 mV for a Xenopus laevis oocyte 1.0 mm in diameter. Also , you decide to calculate how many charges likely are inside of this oocyte.

1.884955592153876109 Coulombs1.6021766341019 Coulombsion=11764967433.383976 ions
V=43π( 0.5103 m )3=5.2359877559829891010 meters3
5.2359877559829891010 meters31000 L1 m3=5.235987755982988107 L
7.853981633974482108 mol K6.022140761023 ionsmol=4.7297782926249131016 ions

R1 : Consider two compartments of equal volume separated by a membrane ( shown below )

image-20250122141541685

Condition A : The membrane is equally permeable to both K+ and Cl-.

Condition B : The permeability of the membrane to K+ is made greater than that to Cl-

Condition C : The membrane is made impermeant to Cl-, but remains permeable to K+.


R2 : Consider two compartments of equal volume separated by a membrane (shown below).

image-20250122141731226

Evaluate the following issues for each permeability condition of the membrane :

  1. Comparing side A with side B, predict the orientation of the electrical potential difference (ePD) across the membrane during the initial moments of flow (before noticeable changes in chemical composition have occurred)

  2. Predict the magnitude of the ePD across the membrane during the initial moments of flow.

  3. Predict the ePD and chemical composition of compartments A and B at electrochemical equilibrium.

Condition D : The membrane is permeable to K+, but not to Na+ or anions.

Condition E : The membrane becomes permeable only to Na+

Condition F : The membrane becomes permeable to both Na+ and K+ , with Na+ permeability greater than K+ permeability ( kpNa>kpK )