Changes in the extracellular potassium concentration can significantly affect excitable cells, particularly those of the heart. Hypokalemia, which refers to a decrease in extracellular potassium, causes the membrane potential (Vm) of cardiac myocytes to hyperpolarize. This makes initiating an action potential more difficult because a larger depolarizing current is required to reach the threshold. Severe hypokalemia can lead to cardiac arrhythmias and eventually cause the heart to stop contracting (asystole). On the other hand, hyperkalemia, an increase in extracellular potassium, can also be detrimental to cardiac function. In hyperkalemia, the Vm becomes depolarized, making it easier to initiate an action potential. However, once the action potential fires, the channels become inactivated and are unable to initiate another action potential until they are reactivated by the normal repolarization of the Vm. Since the Vm remains depolarized in hyperkalemia, it can lead to cardiac arrhythmias and loss of cardiac muscle contraction.

Eq. 2.7 defines the limits for the membrane potential. In the depicted example in Fig. 2.4, it is evident that the Vm cannot be more negative than the potassium equilibrium potential (EK+; -90.8 mV) if the membrane is only conductive to potassium. Conversely, the Vm cannot be more positive than the sodium equilibrium potential (ENa+; 66.6 mV) if the membrane is conductive solely to sodium. The dependence of the Vm on the membrane's ion conductance is the foundation for generating action potentials in excitable cells (Fig. 2.5). As mentioned before, in all excitable cells, the resting membrane is predominantly conductive to potassium, resulting in Vm being close to EK+. When an action potential is initiated, sodium channels open, and the membrane becomes predominantly conductive to sodium. Consequently, Vm approaches ENa+. The generation of action potentials is discussed in greater detail in the section on homeostasis.

Establishing the Vm requires charge separation across the plasma membrane. However, the number of ions that need to move across the membrane is only a tiny fraction of the total number of ions in the cell. For instance, let's consider a spherical cell with a diameter of 20 µm and a Vm of -80 mV. Additionally, assume that this Vm of -80 mV is a result of potassium diffusing out of the cell, with an intracellular potassium concentration of 120 mmol/L. To calculate the amount of potassium that must diffuse out of the cell to establish a Vm of -80 mV, the following steps are taken:

First, the charge separation across the membrane needs to be calculated. Since the plasma membrane behaves electrically like a capacitor, with a capacitance (C) of approximately 1 µF/cm2, the charge (Q) can be expressed in units of coulombs. If the surface area of the cell is 4πr^2 or 1.26 × 10^-5 cm^2, the cell's capacitance is calculated as follows: 1 × 10^-6 F/cm × 1.26 × 10^-5 cm^2 = 1.26 × 10^-11 F.

Hence, the charge separation across the membrane is calculated as follows: 1.26 × 10^-11 coulombs.
Changesintheextracellular[K+]canhaveimportanteffectsonexcitablecells,especiallythoseoftheheart.Adecreaseinextracellular[K+](hypokalemia) hyperpolarizestheVmofcardiacmyocytesand,insodoing,makesinitiatinganactionpotentialmoredifficult,becausealargerdepolarizingcurrentisneededtoreachthreshold(see ).Ifsevere,hypokalemiacanleadtocardiacarrhythmias,andeventuallytheheartcanstopcontracting(asystole). Anincreaseintheextracellular[K+](hyperkalemia) canbeequallydeleterioustocardiacfunction.Withhyperkalemia,theVmisdepolarized,anditiseasiertoinitiateanactionpotential.However,oncetheactionpotentialfiresthechannelsbecomeinactivated,andareunabletoinitiateanotheractionpotential,untiltheyarereactivatedbynormalrepolarizationoftheVm.BecausetheVmisdepolarizedinhyperkalemia,thechannelsstayinaninactivatedstate.ThusdepolarizationoftheVmwithhyperkalemiacanleadtocardiacarrhythmiasandlossofcardiacmusclecontraction.
Eq. 2.7 also defines the limits for the membrane potential. In the example depicted in
Fig. 2.4, it is apparent that the Vm cannot be more negative than EK+ (−90.8 mV), as would be the case if the membrane were only conductive to K+ . Conversely, the Vm could not be more positive than ENa+ (66.6 mV); such a condition would be met if the membrane were conductive only to Na+ . The dependence of the Vm on the conductance of the membrane to specific ions is the basis by which action potentials in excitable cells are generated (
Fig. 2.5). As noted previously, in all excitable cells, the
CHAPTER 2 Homeostasis: Volume and Composition of Body Fluid Compartments membrane at rest is conductive predominantly to K+ , and thus Vm is near EK+. When an action potential is initiated, Na+-channels open and the membrane is now conductive predominantly to Na+ . As a result, Vm now approaches ENa+. The generation of action potentials is discussed in more detail in
TheestablishmentoftheVmrequirestheseparationofchargeacrosstheplasmamembrane.However,thenumberofionsthatmustmoveacrossthemembraneisatinyfractionofthetotalnumberofionsinthecell.Forexample,considerasphericalcellwithadiameterof20µmandaVmof−80mV.Furthermore,assumethatthisVmof−80mVistheresultofthediffusionofK+ outofthecellandthattheintracellular[K+]is120mmol/L.TheamountofK+ thatwouldhavetodiffuseoutofthecelltoestablishtheVmof−80mVisthencalculatedasfollows.
Firstthechargeseparationacrossthemembraneneedstobecalculated.Thisisdonewiththeknowledgethattheplasmamembranebehaveselectricallylikeacapacitor,thecapacitance(C)ofwhichisapproximately1µF/cm2,and whereQ= chargeandisexpressedinunitsofcoulombs.Ifthesurfaceareaofthecellis4πr2or1.26× 10−5cm2,thecapacitanceofthecelliscalculatedasfollows: 1× 10 F/cm × . × 10 cm = . × 10 126 126 F.
Thusthechargeseparationacrossthemembraneiscalculatedasfollows: 1 01× 10−12 coulombs.