1. You decide to measure the Vm of the membrane below to verify your calculation
ID: 142514 • Letter: 1
Question
1. You decide to measure the Vm of the membrane below to verify your calculation. You place an axon in a dish containing a solution that matches the table above. Next you put your reference electrode and the recording electrode into the solution. What do you read on the volt meter?
2. Now you puncture the axon membrane with the recording electrode and the voltmeter reads -62.1 mV. Is this different from your calculated Vm? Explain why it is different.
150
Ion ICF (mM) ECF (mM) Permeability (p) K+150
15 110 Na+ 25 200 1.5 Cl- 10 150 15Explanation / Answer
1 . The membrane potential Vm can be calculated through the following formula .
Vm =58 log (Pk[K]out + PNa[Na]out + PCl[Cl]in)/ (Pk[K]in + PNa[Na]in + PCl[Cl]out)
Where Pk ,PNa , PCl are the permeability of the ions .
Therefore Vm = 58log[110(15) +1.5(200) + 15(10)]/
[110(150) + 1.5(25) + 15(150)]
= 58log (2100/18787.5)= 58log(0.111)
= 58 x -0.95 = -55.3 mV .
The voltage recorded will be negative will be negative -
2. The membrane potential will be different from what we have calculated . The calculated Vm value is -55.3 mV .
The recorded value is -62.1 mV which is much more negative than the outer solution .
The inside of the cell does not only contains chloride ions (10mM ) which contributes to the negative charge , but also contains negatively charged proteins , which also contributes to the negative charge inside the cell.
We calculated the Vm with respect to only sodium potassium and chloride , but didn't mention the negatively charged protein molecules . That's why the observed and the calculated values are different .
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