Why do we say there is an overall negative charge on the intracellular side of the plasma membrane at rest, and an overall positive charge on the extracellular side when both potassium and sodium are positively charged ions, and they are in relatively equal amounts on either side? It seems to me that both sides should be positively charged, and I can't get my head around it.
$\begingroup$
$\endgroup$
1
-
$\begingroup$ great question... with an equally great answer... where have you looked for this answer? $\endgroup$– Vance L AlbaughAug 12, 2017 at 18:52
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
The resting membrane potential is the voltage at which there is no net flow of ions across the membrane.
Every ion will tend to flow down it's concentration gradient (this is just a basic principle of physics/chemistry). So, without considering charge, for a typical cell, potassium will tend to flow out, sodium will tend to flow in, and other ions will also behave according to their concentration gradients.
The other crucial part of the equation is the relative permeability of the membrane to each ion. This comes mainly from the presence of ion channels. At rest, a typical cell has more permeability to potassium than sodium.
That means that if the driving force for sodium and potassium is the same (it's not identical at 0 mV but fairly close), more potassium will move than sodium.
If the definition of resting potential is the potential where net flow of charge is zero, and considering only sodium and potassium, then rest has to be where sodium flowing in equals potassium flowing out. This situation occurs when the inside of the cell is negatively charged compared to the outside: the membrane voltage "holds" potassium in and "pulls" some sodium in, such that their currents are equal and opposite despite the higher permeability to potassium.
Note that the voltages we are talking about in neurons, on the order of 10s of mV, are very small, and electrical forces are very strong. The concentrations of positive and negative ions are almost exactly identical on both sides of the membrane: only a few ions actually have to move to make a (biologically) large membrane voltage. What is actually important is the permeabilities: that's what makes action potentials and other forms of neuronal signaling feasible.
answered Aug 12, 2017 at 14:28
$\begingroup$
$\endgroup$
I had the same question and wanted to re state the answer with more steps. This might be helpful as I had no idea what was going on and this will explain some common misunderstandings that I and probably most people have with this.
At resting potential we have Na+ in the exterior and K+ and organic acids in the ctyosol. The membrane potential relies on electrochemistry to propagate a signal from dendrite to distal synpase. To propagate neuron signals it would not be ideal to transport these anions across the membrane given their size and utility inside the cell. It would be slow and require intensive transporters in the membrane. We can however use cations which are relatively abundant, small and with relatively low thermodynamic cost.
The -70mV resting potential describes the relative charge state across the membrane. It describes that the cytoplasm is relatively more negative compared to the extracellular fluid. Since we have organic acids giving the cytosol a slight negative character, we can use the thermodynamics of diffusion to our advantage. This is where our cations come in to use. At rest we balance the concentrations of cation movement across the membrane via the sodium potassium ATPase which maintains the gradients of Na (high on extracellular side) and K (high on cytosol side). We only need the balance the movement of these cations to create the polarised character on either side of the membrane given the organic acids.
When a dendrite receives excitation signals Na+ transporter proteins open and sodium rapidly enters the cytosol across the gradient. The potential falls as the cytosol becomes more positive and we have a change in the potential (action potential).
This doesn't explain why K+ is needed here. Propagation of the signal along the neuron requires the cytosol beside the site of action potential to change potential and become more positive. Diffusion takes time. If we reverse the Na pumps and immediately removed all Na+ from the cytosol this wouldn't create the necessary change in potential along the neuron because the Na+ is being pulled into the extracellular space rather than given the time to move along the cytosol.
K+ plays the role of repolarising instead. While Na+ allows the signal to propagate, K+ is moved out of the cytosol into the extracellular space to move the potential back to -70mV. When the signal is sufficiently distal, Na+ can then be moved to the extracellular space and the resting potential can be restored.
