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Goldman equation

From Wikipedia, the free encyclopedia
Generalization of the Nernst equation for the membrane potential
Not to be confused withGoldman–Hodgkin–Katz flux equation.
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TheGoldman–Hodgkin–Katz voltage equation, sometimes called theGoldman equation, is used in cell membranephysiology to determine theresting potential across a cell's membrane, taking into account all of the ions that arepermeant through that membrane.

The discoverers of this areDavid E. Goldman ofColumbia University, and the Medicine Nobel laureatesAlan Lloyd Hodgkin andBernard Katz.

Equation for monovalent ions

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The GHK voltage equation forM{\displaystyle M} monovalent positiveionic species andA{\displaystyle A} negative:

Em=RTFln(inPMi+[Mi+]out+jmPAj[Aj]ininPMi+[Mi+]in+jmPAj[Aj]out){\displaystyle E_{m}={\frac {RT}{F}}\ln {\left({\frac {\sum _{i}^{n}P_{M_{i}^{+}}[M_{i}^{+}]_{\mathrm {out} }+\sum _{j}^{m}P_{A_{j}^{-}}[A_{j}^{-}]_{\mathrm {in} }}{\sum _{i}^{n}P_{M_{i}^{+}}[M_{i}^{+}]_{\mathrm {in} }+\sum _{j}^{m}P_{A_{j}^{-}}[A_{j}^{-}]_{\mathrm {out} }}}\right)}}

This results in the following if we consider a membrane separating twoKxNa1xCl{\displaystyle \mathrm {K} _{x}\mathrm {Na} _{1-x}\mathrm {Cl} }-solutions:[1][2][3]

Em,KxNa1xCl=RTFln(PNa[Na+]out+PK[K+]out+PCl[Cl]inPNa[Na+]in+PK[K+]in+PCl[Cl]out){\displaystyle E_{m,\mathrm {K} _{x}\mathrm {\text{Na}} _{1-x}\mathrm {Cl} }={\frac {RT}{F}}\ln {\left({\frac {P_{\text{Na}}[{\text{Na}}^{+}]_{\mathrm {out} }+P_{\text{K}}[{\text{K}}^{+}]_{\mathrm {out} }+P_{\text{Cl}}[{\text{Cl}}^{-}]_{\mathrm {in} }}{P_{\text{Na}}[{\text{Na}}^{+}]_{\mathrm {in} }+P_{\text{K}}[{\text{K}}^{+}]_{\mathrm {in} }+P_{\text{Cl}}[{\text{Cl}}^{-}]_{\mathrm {out} }}}\right)}}

It is "Nernst-like" but has a term for each permeant ion:

Em,Na=RTFln(PNa[Na+]outPNa[Na+]in)=RTFln([Na+]out[Na+]in){\displaystyle E_{m,{\text{Na}}}={\frac {RT}{F}}\ln {\left({\frac {P_{\text{Na}}[{\text{Na}}^{+}]_{\mathrm {out} }}{P_{\text{Na}}[{\text{Na}}^{+}]_{\mathrm {in} }}}\right)}={\frac {RT}{F}}\ln {\left({\frac {[{\text{Na}}^{+}]_{\mathrm {out} }}{[{\text{Na}}^{+}]_{\mathrm {in} }}}\right)}}

RTF{\displaystyle {\frac {RT}{F}}} is approximately 26.7 mV at human body temperature (37 °C); when factoring in the change-of-base formula between the natural logarithm, ln, and logarithm with base 10([log10exp(1)]1=ln(10)=2.30258...){\displaystyle ([\log _{10}\exp(1)]^{-1}=\ln(10)=2.30258...)}, it becomes26.7mV2.303=61.5mV{\displaystyle 26.7\,\mathrm {mV} \cdot 2.303=61.5\,\mathrm {mV} }, a value often used in neuroscience.

EX=61.5mVlog([X+]out[X+]in)=61.5mVlog([X]out[X]in){\displaystyle E_{X}=61.5\,\mathrm {mV} \cdot \log {\left({\frac {[X^{+}]_{\mathrm {out} }}{[X^{+}]_{\mathrm {in} }}}\right)}=-61.5\,\mathrm {mV} \cdot \log {\left({\frac {[X^{-}]_{\mathrm {out} }}{[X^{-}]_{\mathrm {in} }}}\right)}}

The ionic charge determines the sign of the membrane potential contribution. During an action potential, although the membrane potential changes about 100mV, the concentrations of ions inside and outside the cell do not change significantly. They are always very close to their respective concentrations when the membrane is at their resting potential.

Calculating the first term

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UsingR8.3 JKmol{\displaystyle R\approx {\frac {8.3\ \mathrm {J} }{\mathrm {K} \cdot \mathrm {mol} }}},F9.6×104 JmolV{\displaystyle F\approx {\frac {9.6\times 10^{4}\ \mathrm {J} }{\mathrm {mol} \cdot \mathrm {V} }}}, (assuming body temperature)T=37 C=310 K{\displaystyle T=37\ ^{\circ }\mathrm {C} =310\ \mathrm {K} } and the fact that one volt is equal to one joule of energy per coulomb of charge, the equation

EX=RTzFlnXoXi{\displaystyle E_{X}={\frac {RT}{zF}}\ln {\frac {X_{o}}{X_{i}}}}

can be reduced to

EX0.0267 VzlnXoXi=26.7 mVzlnXoXi61.5 mVzlogXoXi since ln102.303{\displaystyle {\begin{aligned}E_{X}&\approx {\frac {0.0267\ \mathrm {V} }{z}}\ln {\frac {X_{o}}{X_{i}}}\\&={\frac {26.7\ \mathrm {mV} }{z}}\ln {\frac {X_{o}}{X_{i}}}\\&\approx {\frac {61.5\ \mathrm {mV} }{z}}\log {\frac {X_{o}}{X_{i}}}&{\text{ since }}\ln 10\approx 2.303\end{aligned}}}

which is theNernst equation.

Derivation

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Goldman's equation seeks to determine thevoltageEm across a membrane.[5] ACartesian coordinate system is used to describe the system, with thez direction being perpendicular to the membrane. Assuming that the system is symmetrical in thex andy directions (around and along the axon, respectively), only thez direction need be considered; thus, the voltageEm is theintegral of thez component of theelectric field across the membrane.

According to Goldman's model, only two factors influence the motion of ions across a permeable membrane: the average electric field and the difference in ionicconcentration from one side of the membrane to the other. The electric field is assumed to be constant across the membrane, so that it can be set equal toEm/L, whereL is the thickness of the membrane. For a given ion denoted A with valencenA, itsfluxjA—in other words, the number of ions crossing per time and per area of the membrane—is given by the formula

jA=DA(d[A]dznAFRTEmL[A]){\displaystyle j_{\mathrm {A} }=-D_{\mathrm {A} }\left({\frac {d\left[\mathrm {A} \right]}{dz}}-{\frac {n_{\mathrm {A} }F}{RT}}{\frac {E_{m}}{L}}\left[\mathrm {A} \right]\right)}

The first term corresponds toFick's law of diffusion, which gives the flux due todiffusion down theconcentration gradient, i.e., from high to low concentration. The constantDA is thediffusion constant of the ion A. The second term reflects theflux due to the electric field, which increases linearly with the electric field; Formally, it is [A] multiplied by the drift velocity of the ions, with thedrift velocity expressed using theStokes–Einstein relation applied toelectrophoretic mobility. The constants here are thechargevalencenA of the ion A (e.g., +1 for K+, +2 for Ca2+ and −1 for Cl), thetemperatureT (inkelvins), the molargas constantR, and thefaradayF, which is the total charge of a mole ofelectrons.

This is a first-orderODE of the formy' = ay + b, withy = [A] andy' = d[A]/dz; integrating both sides fromz=0 toz=L with the boundary conditions [A](0) = [A]in and [A](L) = [A]out, one gets the solution

jA=μnAPA[A]out[A]inenAμ1enAμ{\displaystyle j_{\mathrm {A} }=\mu n_{\mathrm {A} }P_{\mathrm {A} }{\frac {\left[\mathrm {A} \right]_{\mathrm {out} }-\left[\mathrm {A} \right]_{\mathrm {in} }e^{n_{\mathrm {A} }\mu }}{1-e^{n_{\mathrm {A} }\mu }}}}

where μ is a dimensionless number

μ=FEmRT{\displaystyle \mu ={\frac {FE_{m}}{RT}}}

andPA is the ionic permeability, defined here as

PA=DAL{\displaystyle P_{\mathrm {A} }={\frac {D_{\mathrm {A} }}{L}}}

Theelectric currentdensityJA equals the chargeqA of the ion multiplied by the fluxjA

JA=qAjA{\displaystyle J_{A}=q_{\mathrm {A} }j_{\mathrm {A} }}

Current density has units of (Amperes/m2). Molar flux has units of (mol/(s m2)). Thus, to get current density from molar flux one needs to multiply by Faraday's constant F (Coulombs/mol). F will then cancel from the equation below. Since the valence has already been accounted for above, the charge qA of each ion in the equation above, therefore, should be interpreted as +1 or -1 depending on the polarity of the ion.

There is such a current associated with every type of ion that can cross the membrane; this is because each type of ion would require a distinct membrane potential to balance diffusion, but there can only be one membrane potential. By assumption, at the Goldman voltageEm, the total current density is zero

Jtot=AJA=0{\displaystyle J_{tot}=\sum _{A}J_{A}=0}

(Although the current for each ion type considered here is nonzero, there are other pumps in the membrane, e.g.Na+/K+-ATPase, not considered here which serve to balance each individual ion's current, so that the ion concentrations on either side of the membrane do not change over time in equilibrium.) If all the ions are monovalent—that is, if all thenA equal either +1 or -1—this equation can be written

wveμ=0{\displaystyle w-ve^{\mu }=0}

whose solution is the Goldman equation

FEmRT=μ=lnwv{\displaystyle {\frac {FE_{m}}{RT}}=\mu =\ln {\frac {w}{v}}}

where

w=cations CPC[C+]out+anions APA[A]in{\displaystyle w=\sum _{\mathrm {cations\ C} }P_{\mathrm {C} }\left[\mathrm {C} ^{+}\right]_{\mathrm {out} }+\sum _{\mathrm {anions\ A} }P_{\mathrm {A} }\left[\mathrm {A} ^{-}\right]_{\mathrm {in} }}
v=cations CPC[C+]in+anions APA[A]out{\displaystyle v=\sum _{\mathrm {cations\ C} }P_{\mathrm {C} }\left[\mathrm {C} ^{+}\right]_{\mathrm {in} }+\sum _{\mathrm {anions\ A} }P_{\mathrm {A} }\left[\mathrm {A} ^{-}\right]_{\mathrm {out} }}

If divalent ions such ascalcium are considered, terms such ase appear, which is thesquare ofeμ; in this case, the formula for the Goldman equation can be solved using thequadratic formula.

See also

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References

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  1. ^Enderle, John (2005-01-01), Enderle, John D.; Blanchard, Susan M.; Bronzino, Joseph D. (eds.),"Bioelectric Phenomena",Introduction to Biomedical Engineering (Second Edition), Biomedical Engineering, Boston: Academic Press, pp. 627–691,doi:10.1016/b978-0-12-238662-6.50013-6,ISBN 978-0-12-238662-6, retrieved2020-10-23
  2. ^Reuss, Luis (2008-01-01), Alpern, Robert J.; Hebert, Steven C. (eds.),"Chapter 2 – Mechanisms of Ion Transport Across Cell Membranes and Epithelia",Seldin and Giebisch's The Kidney (Fourth Edition), San Diego: Academic Press, pp. 35–56,doi:10.1016/b978-012088488-9.50005-x,ISBN 978-0-12-088488-9, retrieved2020-10-23
  3. ^Enderle, John D. (2012-01-01), Enderle, John D.; Bronzino, Joseph D. (eds.),"Chapter 12 – Bioelectric Phenomena",Introduction to Biomedical Engineering (Third Edition), Biomedical Engineering, Boston: Academic Press, pp. 747–815,doi:10.1016/b978-0-12-374979-6.00012-5,ISBN 978-0-12-374979-6, retrieved2020-10-23
  4. ^abcdBhadra, Narendra (2015-01-01), Kilgore, Kevin (ed.),"2 - Physiological principles of electrical stimulation",Implantable Neuroprostheses for Restoring Function, Woodhead Publishing Series in Biomaterials, Woodhead Publishing, pp. 13–43,doi:10.1016/b978-1-78242-101-6.00002-1,ISBN 978-1-78242-101-6, retrieved2020-10-23
  5. ^Junge D (1981).Nerve and Muscle Excitation (2nd ed.). Sunderland, Massachusetts: Sinauer Associates. pp. 33–37.ISBN 0-87893-410-3.

External links

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