Final Exam Solutions, 1998

Transcription

1 Fnal Exa Solutons, 1998 roble 1 art a: Equlbru eans that the therodynac potental of a consttuent s the sae everywhere n a syste. An exaple s the Nernst potental. If the potental across a ebrane equals an on s Nernst potental, then that on s at equlbru across the ebrane ( V RT zf ln C C ). out n Steady state eans that all varables are constant n te. The concentratons of ons nsde a restng cell, for exaple, are constant n te. Ths occurs because the su of all actve and passve transports through the ebrane are zero for each on. art b: At the nodes, there ust be an equaton for balance of current through the ebrane capactance, the on channels present n the node, and the currents flowng nto the nodes fro the adjacent nternodes. For exaple C dv dt ( Vt, )( V E ) + Ileft cable + I all on channels rght cable where I left cable and I rght cable are the boundary condton currents for the cable equaton at the adjacent nternodes (see below). In addton, at the nodes, the on channel conductances ust be odeled by approprate Hodgkn-Huxley equatons. For the nternodes, there s a cable equaton whose boundary condtons are the currents flowng fro the nternodes nto the nodes (I left cable and I rght cable above). V χ V + V t V I V rght cable and I χ χ χ χ left cable wth whatever te boundary condton s approprate (say all voltages are at te ). There wll be one such equaton for each node and nternode n the proble. At the ends of the axon there wll be soe approprate boundary condton. For exaple, at the left end, I left cable would be a current njected nto the frst node to ntate the acton potental. At the rght end, the cable ght just end wth a node to odel an axon ternal. The equatons are slar for the cell n the proble. For exaple, we could assue that the dendrtes can be reduced to equvalent cylnders; they would then be odeled by the cable equaton wth zero current boundary condtons at ther dstal ends

2 and wth the current nto the soa at the other end. The soa could be odeled as a sngle ebrane patch wth a capactance and on channels. Agan the current njected nto the soa would be the su of the boundary-condton currents for the dendrtes. The axon would be odeled as a sequence of ebrane patches contanng a capactance and on channels, coupled by resstors. art c: The soclnes are gven by: W Vso ( V) Iext ( V)( V E) ( V E) Na( V ENa) ( V E ) W ( V) w ( V) Wso where the ter Na ( V ENa) s the added sodu leak channel. Clearly there s no dfference between the soclnes wth a partcular value of I ext and the soclnes wth an equal value of Na ( V ENa) n the.c. steady state. Notce that Na ( V ENa) s negatve, because V<E Na, so actvatng ths sodu leak channel has the sae effect as njectng a postve current I ext. Thus any equlbru pont and soclne whch can be acheved by njectng a current can be acheved by adjustng the sodu leak (up to the sodu equlbru potental). However, the egenvalues at a gven equlbru pont wll not necessarly be the sae because the leak ter enters the Jacoban whereas the external current does not. To see ths, consder the FV / V ter n the Jacoban. I ext has no effect on ths ter, whereas the sodu leak adds Na / C to ths ter. The other ters n the Jacoban are not changed. Nevertheless, at a gven equlbru pont, the egenvalues wll, n general, be dfferent. art d: Wth only the transporter n the ER ebrane, there wll be a ebrane potental, postve nsde the ER, because of the charge carred by calcu nto the ER. The steady state concentraton dfference wll be gven by the concentraton and potental dfference necessary to stop the pup, whch occurs when the free energy ganed by the pup by hydrolyzng AT equals the free energy dfference of calcu nsde relatve to outsde the ER. It doesn t atter how any pup olecules there are. (Note that ths s a totally unrealstc stuaton because n the absence of any leak, the ebrane potental across the ER ebrane wll rapdly buld up to a hgh enough value to stop the pup.) In the presence of a leak, there wll stll be a potental dfference, postve nsde the ER, agan due to the charge carred by calcu nto the ER. The steady state occurs when the flux of calcu through the leak channels equals the actve flux of calcu due to the transporter. In ths case the nuber of transporter olecules atters because the ore olecules there are, the larger the calcu flux, at a gven electrochecal gradent. roble art a: Because the electrotonc lengths are the sae,

3 3 l l l3 l3 and λ R R a λ3 R R a 3 so l C a l3 l or C a a 3 l3 a 3 where C R R. art b: The requreents of the equvalent cylnder theore are et except for the requreent +, whch eans that / 3 / 3 3 / a a + a art c: The potental V N can be coputed usng the transfer pedance rule across the cable of the branch nto whch the current was njected: V I N nj IN Inj ( Yn + Yother ) ( + Yn + Yother ) cosh q + + q snh q q where the load adttance at the rght-hand end of the branch s (closed-end boundary condton) and the load adttance at the branch pont s Y n +Y other. Y n s the nput adttance of the parent branch and Y other s the nput adttance of the other branch (.e. the one nto whch current s not njected). Because of the closed-end boundary condton, Cobnng these equatons, Y q tanh q other other V N Inj Y + qtanh q cosh q qsnh q ( ) + n other Inj Y cosh q + qsnh q + qsnh q n other Inj Y cosh q + ( q+ q) snh q n other where use has been ade of the fact that tanh x snh x / cosh x. The last equaton above s the sae regardless of whch branch the current s njected nto. Thus, V N s the sae n the two cases.

4 4 Note that ths s one of the conclusons of the equvalent cylnder theore, but s true n the absence of one of the assuptons of that theore. Ths s a specal case, whch s only true wth the currents at the ends of the chld cylnders. art d: The easest way to see that the potentals n the chld branches are not the sae s to consder the potental at the njecton ste, whch s Inj II where II s the nput pedance at that pont. The nput adttance s gven by the second of the three rules developed n class: ( 1 Y q Y n + Y other ) q + tanh q II II 1 + tanh q ( Y + Y ) q Where the sybols are defned as n part c. Replacng Y other wth ts value and rearrangng n other ( Yn + q q q q YII q other tanh ) + tanh 1 + tanh q ( Y + q tanh q ) q n other Yn + otherqtanh q + qtanh q 1 + tanh q ( Y + q tanh q ) q n other ( ) Yn + otherq+ q tanh q 1 + tanh q ( Y + q tanh q ) q n other Consder the last equaton above. The nuerator s the sae regardless of whch branch the current s njected nto. However, the second ter n the denonator s dfferent f the s of the two chld branches are dfferent. Thus the nput pedances of the two current njecton ponts are dfferent. Because the nput pedances are dfferent, the potental at the pont of current njecton s also dfferent and so the potental everywhere n the chld cylnders wll be dfferent. Inspecton of the equatons above shows that the cylnder wth the larger value wll have the saller potental. roble 3 art a: Usng the usual defntons and the paraeters n the proble:

5 5 4 4 R a 1 x Ωc 51 x c λ. 36 c 36 µ R 1 Ωc length λ 1 π π ( ) µ R R a 3 / 4 3 / 4 5x1 c. 35 S 1 x Ωc 1 Ωc apcal end πa π ( 51 x c) R. 4 Ωc,apcal µ S Note that apcal end. art b: Ths proble can be done n several ways. Below s the ost drect. The cable equaton and boundary condtons for the.c. steady state ( V t ) s dv dχ V dv I dv and V χ V dχ dχ χ χ end χ where end s the conductance of the apcal end of the cell, equal to. The soluton to the cable equaton s At χ: and at χ: V( χ) Asnh χ+ Bcosh χ I ( χ) Acosh χ+ Bsnh χ [ ] I I( ) A I A

6 6 I ( ) V( ) I I cosh + Bsnh snh Bcosh + [ ] [ ] B I snh + cosh snh + cosh I so that I V( χ) [ cosh χ snh χ] I e χ Ths sae result could have been gotten by argung sply that, because the apcal ternal adttance s equal to, the cylnder behaves lke a se-nfnte one, so the ebrane potental s equal to V()e -χ, V()I / n, and n. art c: Begnnng wth Eqn. 7 of the appendx C x 1 + B C + C t roceedng n the sae way as for the lnear cable equaton, dvde through by C x ( ) 1 + B C + C t Now, as for the electrcal cable equaton, the length and te constants are λ 1 B and τ + Note that these have the correct unts, of dstance and te. / c s / s B ole / c 1+ ole / c λ c and τ / s s The equvalent of can be gotten by analogy wth the electrcal cable equaton where 1/r λ by consderng what the equvalent of 1/r λ should be. r s the rato between the ebane potental gradent ( V x, gnorng V e ) and axal current. The equvalent n the calcu cable equaton s 1/, the rato beteen the gradent of calcu concentraton and calcu flux (Eqn. 6 n the appendx). Thus the calcu equvalent of s

7 7 Ths can also be derved fro the results n part d below as the rato between the calcu flux njected nto the basal ebrane of the cell I /F and C(). Note that alternate defntons of can be wrtten, dependng on what s assued to be the analog of electrcal current flow n the calcu syste. Above t was assued that calcu flux J s the approprate analog, but electrcal current FJ could also be used, gvng a slghtly dfferent value. art d: Assung the.c. steady state, the calcu equaton becoes dc C dx (*) The boundary condton at x s that the flux of calcu down the cylnder (J n Eqn. 6 of the appendx) s equal to the calcu carred nto the cell by I. Thus J ( ) dc dx x I F (**) where F s the Faraday, to convert I fro electrcal (aps) to checal (oles/s) unts. Ths provdes one boundary condton. The second boundary condton could be soe condton on calcu flux at the apcal end of the cell. However, the for of Eqn (*) above s the sae as the electrcal cable equaton before the spatal varable s nondensonalzed. The calcu space constant s / and ths dstance s a easure of how far dsturbances of calcu wll spread. ven that / 5 µ, t s safe to conclude that calcu wll not spread far fro the base of the cell, snce the length of the cell s uch larger, 36 µ. In other words, t s safe to assue that the cell s senfnte n length, so the second boundary condton s that C s fnte as x, a regularty condton. The soluton to the calcu cable equaton s x/ / x/ / C( x) Ae + Be The regularty condton requres that B and the boundary condton at x (Eqn. (**))gves dc A dx e / / I I A / F F x so that the calcu concentraton s

8 8 Cx ( ) I F x/ / e Note, f you ade a dfferent assupton about the second boundary condton, such as calcu flux at the apcal pole equals zero, then you would have gotten a ore coplex equaton, whch would reduce to the equaton above, gven that the length of the cylnder s large copared to. art e: For the electrcal cable theory, begn wth the non-lnear cable equaton: 1 V r x c V I xtv t + (,, ) where I s the onc ebrane current/length and the external resstance r e has been gnored. In dervng the cable equaton used n part b above, t was assued that I g V, the usual assupton n the lnear cable theory. However, n ths case, there s an addtonal current due to calcu pups n the lateral ebrane, so that I ( x) g V( x) + I ( x) g V( x) +FC( x) where FC s the odel for calcu pup current fro the dervaton of the calcu cable equaton (Eqn. 5). The electrcal cable equaton now becoes 1 V r x c V gv FC t + + or, n non-densonal ters V χ V + + F V C T g The calcu equaton ust be odfed to nclude the effects of the potental gradent n the cylnder on the calcu flux. The relevant equaton s Eqn. 6 n the appendx, whch contans only a dffuson ter. The Nernst-lanck equaton for calcu s J u RT dc dx u z FC dv dx The frst ter on the rght s the sae as Eqn. 6 of the appendx, snce u RT. The second ter s flux due to the electrcal potental gradent. Note that ebrane potental V has been equated to ntracellular potental V here, gnorng V e as usual. Cobnng ths equaton wth Eqns. 4 and 5 n the appendx gves

9 9 C z F x RT x C + V + x 1 B C + C t where the second ter on the left hand sde s the effect of electrcal potental gradent on the calcu flux.