Flashcards in membrane potential Deck (17):
1. Compare qualitatively the relative strengths of electric and osmotic forces.
Electric forces are much, much stronger than osmotic forces – meaning that relatively few excess ions are needed to counter large concentration differences
2. Describe the two forces acting on an ion moving across a membrane.
The movement of ions across a membrane is governed by the concentration difference (chemical gradient) and the electrical potential difference across the membrane (the membrane potential – produced by an imbalance in the number of cations and anions inside a cell). Combined, these two forces make an electrochemical gradient.
3. Define equilibrium potential.
The equilibrium potential relates the concentration gradient to the electrical force and it is the electrical potential difference across the membrane that must exist if the ion is to be at equilibrium at the given concentrations. It is specific to ONE ion!
If the membrane potential is equal to the equilibrium potential – the ion is said to be distributed at its electrochemical equilibrium. If this is not the case (which it most often is not!), one of two things must be true – either the membrane is impermeable to that ion or the ion is being pumped across the membrane because it is not distributed at equilibrium.
4. Understand the difference between an equilibrium potential and a recorded membrane potential.
See above – additionally, the membrane potential is a measure of the real difference in voltage between the internal environment of the cell and the ECF, while the equilibrium potential is what the voltage difference would be if a given ion was in equilibrium given a particular ICF:ECF ratio.
5. Know that each and every ion species has its own, independent equilibrium potential.
This is important to note because equilibrium potential can be calculated for each individual ion, while the cell will have only one membrane potential – comparison of these values will indicate whether or not a pump must exist for the ion in question and which direction it must be pumping.
6. In a cell at rest (steady state), given an ion’s concentration inside and out, the membrane potential, and knowledge that the membrane is permeable to the ion in question – be able to determine whether or not a pump for that ion must exist and if so, in which direction it pumps the ion.
Look at [ ]i vs [ ]o – determine which way ion would move.
Look at charge of ion vs membrane potential – determine which way ion would move.
If these are the same direction, there must be a pump moving ions against the gradients.
If these are different directions – must compare the equilibrium potential to the membrane potential:
-If they are equal – there is no pump required.
-If they are different – a pump exists and to determine the direction of the pump - must consider which direction of ion movement would make Vm closer to E… pump pushes the ions in the opposite direction!
1. Understand that the number of excess anions in a typical cell is small compared to the total number of anions.
For example – in a typical cell with resting membrane potential of -80mV, for every 100,000 cations, there are approximately 100,001 anions. This gives an indication of just how strong the electric force is.
2. Know that bulk solutions are always electrically neutral.
This is the Principle of Electrical Neutrality
2. Know that bulk solutions are always electrically neutral.
This is the Principle of Electrical Neutrality - [+]o=[-]o and [+]i=[-]i Though this seems to contradict the idea that membrane potentials arise from an excess of one type of ion in the cells – the key is to note that the electric force is so powerful that the excess number of anions is very, very small compared to the total number of ions present in the cell.
3. Understand how an artificial cell can be in a state of equilibrium even though the concentrations of ions are not the same inside and out.
The specific ions do not have to be equal inside and out – the cell simply must follow the Donnan rule, that is the products of the ion concentrations inside the cell must equal the product of the ion concentrations outside the cell.
4. Be able to apply osmotic balance, charge neutrality, and the Nernst equation to calculate ion concentrations and the membrane potential of an artificial cell.
-osmotic balance: mosM inside and out will be equal
-charge neutrality: can assume an equal number of cations and anions inside the cell and also outside the cell (note this means in each environment +=- NOT that inside is equal to outside)
-Nernst: E=-60*log([ ]o/[ ]i)
-if in a steady state with no pumps acting on cell – Vm = E
5. Know what ions are moved (and how many and in what direction) by one cycle of the sodium pump.
With each cycle of the Na+ pump – 3 sodium ions are pumped out of the cell (from ICF to ECF) and 2 potassium ions are pumped into the cell (from ECF to ICF). The action of this pump requires ATP!
Know the difference between equilibrium and steady state.
Real cells are often in a steady state, which means that, like the model cell at equilibrium, the ion concentrations aren't changing over time, but unlike the model cell, a constant input of energy is needed in the real cell (in the form of ATP, to drive the Na/K pump).
Understand that membrane potential depends on relative, not absolute,
permeabilities to ions.
Understand that the primary short term determinant of membrane potential is not
the Na/K pump, but relative membrane permeabilities to the different ions.
Define Driving Force on an ion.
The driving force at any instant is the difference between Vm and Eion.
Understand why, in neurons and other excitable cells, membrane potential is
sensitive to small changes in [K+]o, but not [Na+]o.
Because the membrane potential of the cell is nearly at the E of K+ anyway, the loss of extracellular Na+ (which brings the E of Na+ closer to 0) will make the Vm slightly hyperpolarized (closer to -80 mV) but have not much other effect.
Its because the cell is WAY more permeable to K+ than to Na+
Sensitivity to [K+]o is higher because its starting concentration outside the cell are much smaller than Na+, and thus is much more sensitive to small changes in concentration.