Workbook exercises - Goldman Equation

Question 1

Calculate relative selectivity of an ion channel where solution A is 97 Na+ and 3 K+ EC, and solution B has 0 Na+ and 100 K+ EC. Both IC 100 K+ and 0.1 Na+. VREV NaCl = -70mV, VREV KCl = -4mV

Answer 1

Rearrange Goldman equation to measure the shift of Vrev between the 2 solutions

*Vrev given in the question were measured EXPERIMENTALLY.

• Can see changes through shifts, but can also calculate either side. 2 different circumstances can be plotted and predict what we can see based off of which it shifts towards.
  Nernst potentials can be calculated as we know the values of the conditions. Measure Vrev to see which Nernst potential its closer to to see what its selective for.

Vrev Sol B - Vrev Sol A = Shift in Vrev, therefore:
Shift in Vrev = Goldman B - Goldman A
VREV B (KCl) = 58.2 x log ((0+ 100)/(0.1 + 100))
VREV A (NaCl) = 58.2 x log ((97 + 3)/(0.1+100))
therefore:
-70mV = 58.2 x log ((97 + 3)/(0.1+100))
-4 = 58.2 x log ((0+ 100)/(0.1 + 100))
* This forms a RATIO when rearranged

-4 - -70 = +66mV
Rearranging:
+66 = [58.2 x log ((0+ 100)/(0.1 + 100))]-[58.2 x log ((97 + 3)/(0.1+100))]
*as it is using logirithms, that minus can become a divide
+66 = [58.2 x log ((0+ 100)/(0.1 + 100))]/[58.2 x log ((97 + 3)/(0.1+100))]
+66 = [58.2 x log ((0+ 100)/(0.1 + 100))]x[58.2 x log (((0.1+100/97 + 3)))] (when dividing fractions, keep switch flip
Cancel out top of fraction:
+66 = 58.2 x log 100/3+97
+66/58.2 = log 100/3+97
1.13= log 100/3+97 > to remove log from ratio, take inverse log of 1.13:
13.49 = 100/3+97
13.49(3+97) = 100
40.27PK + 1308.53PNa = 100PK > move over PK so that theyre together
1308.53PNa = 59.53PK
PNa:PK = 59.53/1308.53 = 0.04549379…
Ratio is 0.05 (to 2 decimal places)

Question 2

Calculate the Nernst potential of Na+ at room temperature where EC= 100mM and IC= 5mM

Answer 2

ENa = 58.2 log [Na]o/[Na]i
58.2 log 100/5
=+75.72mV

*can always double check whether calculation is right depending on which way the ions will move > positive will move IN due to electrochemical gradient favouring it.

Question 3

Calculate the Nernst potential of K+ at room temperature where EC= 3 and IC= 100

Answer 3

58.2 x log 3/100 = -88.63mV

*negative nernst potential shows the gradient favours movement outside of the cell.

Question 4

You measure the membrane potential of a cell and get a value of =+50mV. The EC concentration of K is 140mM and Na is 5mM and the IC of K is 15mM and Na is 100 mM. Knowing this, what is the likely identity of channels that are open in the membrane.

Answer 4

*Don’t be tricked by initial values given > they may not be physiological conditions so do not assume its Na as its potential is driven towards ENa.

Do the ions individually when you want to identify:

Na+ channels: 58.2 log (5/100) = -75.72mV (negative as echem gradient drives release of ion)
K+ channels: 58.2 log (140/15) = +56.46mV
* K+ selective, as its Vrev is closer to EK than ENa

Question 5

Using the Nernst equation (equation 1) calculate the Nernst potentials for Na+ and K+ with the 150 NaCl Ringer in the bath and 140 KCl Ringer in the pipette - - The pipette (intracellular) Ringer contained (in mM): 140 KCl, 0.1 NaCl
- The bath (extracellular) Ringer contained (in mM): 150 NaCl, 5 KCl
- All experiments were carried out at a temperature of 20 degrees celcius

Answer 5

Nernst equation: Ex = (RT/zF) log ([X]o/[X]i)
R = CONSTANT, universal gas constant of 8.314J.K-1.mol-1
F= CONSTANT, farradays constant of 96485C.mol-1
T = KELVIN CONVERSION = 20 + 273.15 = 293.15
(8.314x293.15)/96485

ENa = (8.314x293.15)/96485 log (150PNa/0.1PNa) = +184.73mV
EK = (8.314x293.15)/96485 log (5PK/140PK) = -84.17mV

Question 6

Part 2: Current levels were recorded with the NaCl Ringer in the bath and then on substitution of NaCl by KCl. What is the Nernst potential for K+ with the substituted Ringer solution in the bath?
- The pipette (intracellular) Ringer contained (in mM): 140 KCl, 0.1 NaCl
- The bath (extracellular) Ringer contained (in mM): 150 NaCl, 5 KCl

Answer 6

EK = 58.2 x log 155/140
* The 150 EC which is NaCl in the bath was substituted for KCl, therefore, the 150 mM of NaCl is added to the 5 mM of KCl value hence 155 mM of KCl EC
= +2.57mV