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LI NI U Vit Nam v trn th gii L sinh (biophysics) l mn hc c s c dy trong cc trng i hc Khoa hc, ai hc Y - Dc, i hc Nng nghip, i hc S phm, i hc Thu sn v.v... Hiu c L sinh, cng vi mt s mn khoa hc c bn khc s hiu c cc nguyn l ca cc qu trnh sinh hc, c bit l c ch ho l v bn cht vt l ca cc hin tng sng. vit c gio trnh L sinh va bao hm nhng kin thc c bn va cp nht nhng thng tin mi, tht l mt cng vic ht sc kh khn. Nhng vi mong mun phc v kp thi kin thc c bn v L sinh trong khun kh ca d n Gio dc i hc mc B, chng ti mnh dn vit gio trnh ny. Gio trnh L sinh khng ch phc v cho cc sinh vin ca cc trng thuc i hc Hu m cn l ti liu cn thit cho nhng ai quan tm n chuyn ngnh L sinh. Phn cng ni dung: - M u, chng 1, 2, 3, 7, 8, 9 do Tin s on Suy Ngh bin son. - Chng 4, 5, 6 do Tin s L Vn Trng bin son. vit c gio trnh ny, chng ti xin chn thnh cm n Ban d n Gio dc i hc ca i hc Hu, GS. TS. Nguyn Th Kim Ngn v GS. TSKH. L Don Din c v ng gp nhng kin qu bu cho gio trnh ny. gio trnh ngy cng hon thin, chng ti mong nhn c s gp ca ngi c gn xa. Ch bin Tin s: on Suy Ngh

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M U L SINH, S HNH THNH V PHT TRINS p dng kin thc vt l vo nghin cu sinh hc c thc hin vo cui th k XVIII. Nm 1780 hai nh khoa hc Php l Lavoadie v Laplace tin hnh th nghim kho st tnh ng n ca nh lut I nhit ng hc khi p dng vo h thng sng. Nm 1791, Galvani, gio s gii phu trng i hc Bolon (Italia) cng b kt qu nghin cu trong quyn sch "Bn v cc lc in ng vt trong co c", khng nh c tn ti dng in sinh vt. Nm 1859, Raymond pht hin phn trc v phn sau cu mt ng vt c xng sng tn ti mt hiu in th v o c gi tr t 10 n 38mV, gi l in th tnh (hay in th ngh ngi). Nm 1865, Holgreen pht hin c gi tr hiu in th gia phn trc v phn sau cu mt ng vt c xng sng s tng ln khi mt c chiu sng. Sau ny cc nh khoa hc xc nh, chnh l in th hot ng (hay in th hng phn). Nm 1875, Calton khng nh khi mt c chiu sng, khng nhng in cu mt tng ln nh Holgreen pht hin m in vng th gic trn bn cu i no cng tng ln. Sau ny cc nh khoa hc xc nh chnh l dng in hng phn xut hin khi mt c chiu sng, lan truyn theo dy thn kinh th gic ti vng th gic trn bn cu i no, dn ti hiu ng sinh hc l cm nhn c nh sng. Nm 1922, Erlanger v Gasser dng dao ng k m cc o dng in hng phn xut hin trong dy thn kinh. Nm 1922,Vin L sinh Lin X c c thnh lp. Nm 1929, Berger ghi c in no ca ng vt. Lch s hnh thnh L sinh c Taruxop, gio s trng ai hc tng hp Lomonoxop khng nh: "L sinh c xem nh l mt khoa hc bt u c hnh thnh t th k XIX". Th k XX l th k pht trin mnh m nhng nghin cu khoa hc v L sinh trong cc lnh vc: Nhit ng hc, ng hc ca cc qu trnh sinh vt, vn chuyn cht qua mng t bo, quang sinh hc v phng x sinh hc v.v... Thi k u L Sinh c xc nh nh l mt ngnh khoa hc nghin cu cc hin tng vt l trong h thng sng. Sau L sinh c xc nh nh l mt ngnh khoa hc nghin cu cc c ch vt l, c bit l c ch ho l ca cc qu trnh xy ra trong h thng sng mc phn t, t bo, m v c th.

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Bc sang th k XXI, hng lot vn ang c t ra cho cc nh L sinh cn phi nghin cu. l nng lng sinh hc, s chuyn ho nng lng v s dng nng lng ca h thng sng? Bn cht v c ch hnh thnh in th sinh vt? Hin tng phn cc trong h thng sng xy ra nh th no v c g khc so vi h vt l ? Bn cht ca qu trnh hng phn l vn cn phi tip tc nghin cu. Cc ch s c trng v vt l v ho l i vi t bo, m, c quan, c th c mi lin quan nh th no trong h thng tin ho ? Vn t iu chnh cc qu trnh sinh hc ca c th sng trc nhng thay i ca yu t mi trng cng ang c cc nh L sinh quan tm nghin cu. Sinh hc phng x hin ang thu ht nhiu nh khoa hc i su nghin cu nhm phc v cho cng tc chn ging mi, bo qun lng thc, thc phm, cng cuc chinh phc v tr, s dng nng lng ht nhn v mc ch ho bnh v khng loi tr kh nng c cuc chy ua v trang trong vic nm gi "n ht nhn u tin" vi tham vng b quyn th gii ?

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Chng 1

NHIT NG HC H SINH VTI. Nhit ng hc h sinh vt v hng nghin cuNhit ng hc h sinh vt l lnh vc nghin cu hiu ng nng lng, s chuyn ho gia cc dng nng lng, kh nng tin trin, chiu hng v gii hn t din bin ca cc qu trnh xy ra trong h thng sng. C th sng trong qu trnh sinh trng v pht trin u c s dng nng lng v vy nhit ng hc h sinh vt l lnh vc cn c nghin cu. i tng nghin cu ca nhit ng hc h sinh vt l c th sng, l mt h m do lun xy ra s trao i vt cht v nng lng vi mi trng xung quanh, c kh nng t iu chnh, t sinh sn... nn khc vi h vt l nh cht rn, cht lng hay cht kh... Hin nay nhit ng hc h sinh vt c cc hng nghin cu ch yu sau: - Nghin cu s chuyn bin nng lng mc phn t, t bo, m, c quan hay ton b c th khi trng thi sinh l bnh thng v trng thi ang hot ng. Xc nh hiu sut s dng nng lng ca cc qu trnh sinh vt v nng lng lin kt trong cc lin kt ca cc cao phn t sinh hc. - Nghin cu tnh cht nhit ng ca cc qu trnh din ra trong c th sng nh qu trnh khuych tn, thm thu, vn chuyn tch cc... - Nghin cu c ch tc ng ca s thay i cc yu t mi trng ln qu trnh chuyn ho nng lng v s trao i nng lng gia c th sng vi mi trng.

II. Mt s khi nim v i lng c bn- H: H l mt vt th hay mt nhm vt th c dng lm i tng nghin cu. V d khi chn c th nghin cu th c th l mt h cn khi chn qun th nghin cu th qun th l mt h. - H c lp: L h khng c s trao i vt cht v nng lng gia h vi mi trng xung quanh. Trn thc t kh xc nh c mt h c lp hon ton nhng qui m th nghim cc nh khoa hc c th thit k c h c lp nh bom nhit lng dng nghin cu hiu ng nhit ca cc phn ng oxy ha. - H kn: L h khng trao i vt cht vi mi trng xung quanh nhng c trao i nng lng vi mi trng xung quanh.

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- H m: L h c trao i c vt cht v nng lng vi mi trng xung quanh. V d: c th sng l mt h m. - Tham s trng thi: L cc i lng c trng cho trng thi ca mt h, v d nh nhit , p sut, th tch, ni nng, entropi... - Trng thi cn bng: L trng thi trong cc tham s trng thi t mt gi tr nht nh v khng i theo thi gian. - Qu trnh cn bng: L qu trnh trong cc tham s trng thi thay i vi tc chm ti mc sao cho ti mi thi im c th xem nh trng thi ca h l trng thi cn bng. - Qu trnh ng nhit, ng p, ng tch l qu trnh din ra trong nhit , p sut v th tch lun khng i trong sut qu trnh din ra. - Qu trnh thun nghch: L qu trnh bin i m khi tr v trng thi ban u khng km theo bt c mt s bin i no ca mi trng xung quanh. - Qu trnh bt thun nghch: L qu trnh bin i m khi tr v trng thi ban u lm thay i mi trng xung quanh. - Hm trng thi: Mt i lng c xem l mt hm trng thi, c trng cho trng thi ca h, khi s bin thin gi tr ca n trong bt c qu trnh no cng ch ph thuc vo gi tr u v gi tr cui m khng ph thuc vo con ng chuyn bin. Ni nng (U), nng lng t do (F), th nhit ng (Z hay G), entanpi (H), entropi (S) l nhng hm trng thi. - Nng lng: Nng lng l i lng c th o c, c th bin i mt cch nh lng lun theo cng mt t l thnh nhit lng. Nng lng phn nh kh nng sinh cng ca mt h. n v dng o nng lng l Calo (Cal) hay Joule (J). - Cng v nhit: l hai hnh thc truyn nng lng t h ny sang h khc. Nu nh s truyn nng lng t h ny sang h khc gn lin vi s di chuyn v tr ca h th s truyn c thc hin di dng cng. V d khi chy 100 mt th nng lng tiu tn c dng vo thc hin cng di chuyn v tr. Nu s truyn nng lng t h ny sang h khc lm tng tc chuyn ng ca phn t h nhn nng lng th s truyn c thc hin di dng nhit. Cng v nhit l hm s ca qu trnh v chng u ph thuc vo cch chuyn bin. - Ni nng: Ni nng ca mt vt th bao gm ng nng ca cc phn t chuyn ng v th nng tng tc do s ht v y ln nhau gia cc

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phn t cng vi nng lng ca ht nhn nguyn t v nng lng ca cc in t. III. nh lut I nhit ng hc v nhng h qu ca n nh lut I nhit ng hc c hnh thnh qua cc cng trnh nghin cu ca cc tc gi nh M. V. Lomonoxob (1744), G. I. Heccer (1836), R. Majo (1842), Helmholtz (1849), Joule (1877)... nh lut I nhit ng hc c pht biu nh sau: "Trong mt qu trnh nu nng lng dng ny bin i th nng lng dng khc s xut hin vi lng hon ton tng ng vi gi tr ca nng lng dng ban u". nh lut I nhit ng hc bao gm hai phn: - Phn nh tnh khng nh nng lng khng mt i m n ch chuyn t dng ny sang dng khc. - Phn nh lng khng nh gi tr nng lng vn c bo ton (tc gi nguyn gi tr khi qui i thnh nhit lng) khi chuyn t dng nng lng ny sang dng nng lng khc. Gi tr nng lng ch c bo ton khi qu trnh xy ra l qu trnh thun nghch v hiu sut ca qu trnh t 100%. i vi qu trnh bt thun nghch, hiu sut ca qu trnh nh hn 100% th ngoi phn nng lng truyn cho h phi cng thm phn nng lng to ra mi trng xung quanh. Biu thc ton hc ca nh lut I nhit ng hc: Mt h c lp trng thi ban u c ni nng U1, nu cung cp cho h mt nhit lng Q th mt phn nhit lng h s dng thc hin cng A, phn cn li lm thay i trng thi ca h t trng thi ban u c ni nng U1 sang trng thi mi c ni nng U2 (U2>U1). T nhn xt trn ta c biu thc: Q = U + A (1.1) Trong U = U2 - U1 Cng thc (1.1) c th vit di dng: (1.2) U = U1 - U1 = Q - A i vi qu trnh bin i v cng nh, phng trnh (1.2) c th vit di dng: dU = Q - A (1.3) dU: Ch s bin i ni nng, l hm s trng thi Q v A: Ch s bin i nhit v cng, l hm s ca qu trnh. T biu thc (1.2), nh lut I nhit ng hc c th pht biu nh sau:

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"S bin thin ni nng ca h bng nhit lng do h nhn c tr i cng do h thc hin". T nh lut I nhit ng hc dn n cc h qu sau y: - Nu h bin i theo mt chu trnh kn (c trng thi u v trng thi cui trng nhau) th ni nng ca h s khng thay i (U2 = U1U = 0). - Khi cung cp cho h mt nhit lng, nu h khng thc hin cng th ton b nhit lng m h nhn c s lm tng ni nng ca h. Theo (1.2) U = U2 - U1 = Q - A, nu A = 0 U2 - U1 = Q. H nhn nhit nn Q > 0 U2 - U1 = Q > 0 U2 > U1. - Khi khng cung cp nhit lng cho h m h mun thc hin cng th ch c cch l lm gim ni nng ca h. Theo (1.2) U = U2 - U1 = Q - A, nu Q = 0 U2 - U1 = -A A = U1 - U2. H mun thc hin cng, tc A > 0 U1 - U2 > 0 U1 > U2. Sau khi thc hin cng (tc A > 0), ni nng ca h gim t U1 xung U2 nh hn. - H thc hin theo chu trnh kn, nu khng cung cp nhit lng cho h th h s khng c kh nng sinh cng. Theo (1.2) U = Q - A, nu h thc hin theo chu trnh kn, theo h qu 1 th U = 0 Q - A = 0 Q = A Do vy, nu Q = 0, tc khng cung cp nhit lng cho h th h cng khng c kh nng sinh cng, tc A = 0. H qu ny, c th pht biu di dng: "Khng th ch to c ng c vnh cu loi mt, l loi ng c khng cn cung cp nng lng nhng vn c kh nng sinh cng". IV. nh lut Heccer c Heccer tm ra nm 1836, sau ny c cc nh khoa hc xp thuc vo h qu ca nh lut I nhit ng hc. nh lut Heccer pht biu nh sau: "Hiu ng nhit ca cc phn ng ho hc ch ph thuc vo dng v trng thi ca cht u v cht cui m khng ph thuc vo cch chuyn bin". V d: Phn ng to kh CO2 t than nguyn cht l cacbon (C) c th tin hnh theo 2 cch sau: Cch 1: t trc tip than nguyn cht thnh kh CO2 s gii phng nhit lng l Q1. Phn ng xy ra: C + O2 CO2 + Q1 Cch 2: Chuyn than nguyn cht thnh CO theo phn ng:

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C+

1 O2 CO + Q2 2

T CO chuyn tip thnh CO2 theo phn ng: CO +

1 O2 CO2 + Q3 2Q1

S minh ho: C Q2 CO CO2 Q3

Theo nh lut Heccer, cht u tham gia phn ng (C) v sn phm ca phn ng (CO2) ging nhau nn c hiu ng nhit ging nhau: Q1 = Q2 + Q3 Trong thc nghim, hiu ng nhit ca qu trnh t than thnh CO khng th o trc tip c v khi than chy khng bao gi ch cho CO m cn cho mt t CO2. Nhng thc nghim li o trc tip c Q1 = 97 KCal/M v Q3 = 68 KCal/M. T d dng suy ra gi tr Q2 = Q1 - Q3 = 97 KCal - 68 KCal Q2 = 29 KCal/M. nh lut Heccer c ngha rt quan trng i vi h sinh vt. Trong h sinh vt din ra nhiu phn ng phc tp, cho n nay vn cn nhiu phn ng trung gian cha c th o trc tip c hiu ng nhit. Da vo nh lut Heccer c th gii quyt c kh khn ny. V. nh lut I nhit ng hc p dng vo h sinh vt Ngi u tin tin hnh th nghim chng minh tnh ng n ca nh lut I nhit ng hc khi p dng vo h thng sng l hai nh khoa hc Php Lavoisier v Laplace vo nm 1780. i tng th nghim l chut khoang. Th nghim cch ly c th khi mi trng bn ngoi bng

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cch nui chut trong nhit lng k nhit 0oC. Dng mt lng thc n xc nh trc nui chut th nghim. Trong c th chut s din ra cc phn ng phn hu thc n ti sn phm cui cng l kh CO2 v H2O, ng thi gii phng ra nhit lng Q1. Nu coi iu kin 0oC, chut ng yn, khng thc hin cng m ch s dng nhit lng gii phng ra do oxy ho thc n cung cp nhit lng cho c th v ta nhit ra mi trng, qua nhit k o c s tng nhit , theo cng thc s tnh c nhit lng Q1. ng thi ly mt lng thc n tng ng vi lng thc n cho chut n trc khi th nghim em t chy trong bom nhit lng k cng ti kh CO2 v H2O, gii phng ra nhit lng Q2. So snh hai kt qu th nghim thy gi tr Q1 tng ng vi Q2. iu ny chng t nhit lng gii phng ra t cc phn ng ho sinh din ra trong c th sng hon ton tng ng vi nhit lng gii phng ra t cc phn ng xy ho din ra ngoi c th sng. Ni cch khc, hiu ng nhit ca qu trnh xy ho cht din ra trong c th sng v hiu ng nhit ca qu trnh xy ho cht din ra ngoi c th sng l hon ton tng ng. tng chnh xc ca th nghim, sau ny c nhiu m hnh th nghim ca nhiu nh nghin cu c tin hnh nhng ng ch nht l ca Atwater v Rosa vo nm 1904. i tng th nghim l ngi v thi gian th nghim l mt ngy m (24 gi). Trong thi gian th nghim, cho ngi tiu th mt lng thc n nht nh, thng qua o lng kh xy ht vo (hay kh CO2 th ra), nhit thi ra t phn v nc tiu... s tnh c hiu ng nhit ca cc phn ng phn hu thc n din ra c th ngi trong 24 gi. ng thi t lng thc n tng ng vi lng thc n m ngi tiu th trong bom nhit lng k s o c nhit lng to ra. Kt qu th nghim: Hiu ng nhit ca cc phn ng din ra c th ngi trong 24 gi: Nhit lng to ra xung quanh : 1374 KCal Nhit lng to ra do th ra : 181 KCal Nhit lng to ra do bc hi qua da: 227 KCal Nhit do kh thi ra : 43 KCal Nhit to ra t phn v nc tiu : 23 KCal Hiu nh (do sai s) : 31 KCal Tng cng nhit lng thi ra : 1879 KCal Nhit lng do thc n cung cp:

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56,8 gam Protein : 237 KCal 79,9 gam Gluxit : 335 KCal 140,0 gam Lipit : 1307 KCal Tng cng : 1879 KCal Lu : Khi xy ho 1 gam Protein trong bom nhit lng k ti kh CO2 v H2O, gii phng ra 5,4 KCal cn trong c th sng phn gii 1 gam Protein ti ur ch gii phng khong 4,2 KCal. Khi oxy ho hon ton 1 gam Gluxit, gii phng khong 4,2 KCal cn xy ho hon ton 1 gam Lipit gii phng t 9,3 n 9,5 KCal. Kt qu th nghim ca Atwater v Rosa khng nh nng lng cha trong thc n sau khi c th tiu th chuyn thnh nng lng gii phng thng qua qu trnh phn gii bi cc phn ng ho sinh din ra trong c th sng. Nng lng cha trong thc n v nng lng gii phng ra sau khi c th phn gii thc n l hon ton tng ng. Nhit lng trong c th ngi c chia lm hai loi l nhit lng c bn (hay nhit lng s cp) v nhit lng tch cc (hay nhit lng th cp). Nhit lng c bn xut hin ngay sau khi c th hp th thc n v tiu th xy thc hin phn ng xy ho ng thi gii phng ra nhit lng. V d khi c th hp th 1 phn t gam (tc 1M) glucose, lp tc xy ra phn ng xy ho ng v gii phng ra 678 KCal (nhit lng c bn). C th s s dng nhit lng c bn vo cc hot ng sng, nu cn d s c tch lu vo ATP. Phn nhit lng tch lu vo cc hp cht cao nng gi l nhit lng tch cc. Trong c th sng, nhit lng c bn v nhit lng tch cc c lin quan vi nhau. Nu nhit lng c bn nhiu m c th s dng t th nhit lng tch cc s tng ln. Nu nhit lng c bn khng c th khng nhng nhit lng tch cc bng khng m c th phi phn gii ATP, gii phng ra nng lng cung cp cho cc hot ng sng. trng thi sinh l bnh thng, c th sng s duy tr mi tng quan nht nh gia nhit lng c bn v nhit lng tch cc. mc t bo, c khong 50% nng lng ca cht dinh dng c tch lu vo ATP.

VI. Phng php nhit lng k gin tip v nguyn tc hot ng ca c th sngPhng php o nhit lng ca Lavoadie v Laplace dng trong th nghim chng minh tnh ng n ca nh lut I nhit ng hc khi p dng vo h sinh vt, gi l phng php nhit lng k gin tip. C s ca phng php ny l da vo lng kh xy tiu th hoc lng kh

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CO2 do c th thi ra ng vt mu nng (ng vt c v v ngi), c lin quan cht ch vi nhit lng cha trong thc n. V d: Qu trnh xy ha glucose, phn ng din ra nh sau: C6H12O6 + 6O2 = 6CO2 + 6H2O + 678 KCal (180gam) (134,4l) (134,4l) T phn ng trn cho thy c xy ho hon ton 1 phn t gam glucose th cn phi tiu th 6 phn t gam xy ng thi thi ra 6 phn t gam kh CO2 v gii phng ra 678 KCal. iu kin tiu chun, mi phn t gam cht kh u cha 22,4 lt. Do vy 6 phn t gam xy hoc CO2 u cha: 6 x 22,4 lt = 134,4 lt. T suy ra, c th c tiu th 1 lt O2 xy ho hon ton mt phn t gam glucose ng thi thi ra 1 lt CO2 th km theo gii phng mt nhit lng l: 678 KCal: 134,4 lt = 5,047 KCal/lt v gi l ng lng nhit ca xy. Da vo phng php nhit lng k gin tip, c th xc nh c s thi nhit ca bt k ng vt mu nng no thng qua s lt xy tiu th (hoc s lt CO2 thi ra). T phn ng xy ha glucose trn v sau ny p dng chung cho Gluxit khi xy ho hon ton s gii phng ra nhit lng c tnh theo cng thc: (1.4) Q(KCal) = s lt O2 ( hoc s lt CO2) x 5,047 Khi xy ha Protein, nhit lng gii phng ra c tnh theo cng thc: Q(KCal) = s lt O2 x 4,46 (1.5) Khi xy ho Lipit, nhit lng gii phng ra c tnh theo cng thc: Q(KCal) = s lt O2 x 4,74 (1.6) Mi quan h gia thc n, s lt O2 tiu th v s lt CO2 thi ra cng ng lng nhit ca xy c th hin qua bng 1.1.Bng 1.1: ng lng nhit ca xy i vi cc loi thc n.

Thc n Gluxit Protein Lipit

S lt O2 cn xy ho 1 gam thc n 0,83 0,97 2,03

s lt CO2 thi ra sau khi xy ho 1g thc n 0,83 0,77 1,42

ng lng nhit ca xy 5,047 4,46 4,74

i vi thc n hn hp gm c Gluxit, Protein v Lipit khi b xy ho, nhit lng gii phng ra c tnh theo cng thc:

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Q(KCal) = s lt O2 x 4,825 (1.7) Phng php nhit lng k gin tip cn c th xc nh c nhit lng gii phng ra khi xy ho thc n thng qua: Thng s h hp l t l kh CO2 trn kh O2. Thng s h hp cng thay i tu thuc vo loi thc n c xy ho. - i vi phn ng xy ho glucose C6H12O6 + 6O2 = 6CO2 + 6H2O S lt kh CO2 Thng s h hp = S lt O2 = 6 x 22,4 lt 6 x 22,4 lt =1

Thng s h hp ca glucose c s dng cho c Gluxit. - i vi phn ng xy ha Lipit c thng s h hp bng 0,7, i vi Protein bng 0,8 cn vi thc n hn hp c gi tr nm trong khong t 0,85 n 0,9. Thng s h hp c lin quan vi ng lng nhit ca xy, th hin qua bng 1.2.Bng 1.2: Thng s h hp (TS h hp) v ng lng nhit ca xy (LN ca xy)

TS h hp

0,7

0,75

0,8

0,85

0,9

0,95

1 5,05

LN ca xy 4,686 4,739 4,801 4,862 4,924 4,985

Khi xy ho thc n, bng cch o lng kh O2 tiu th v lng kh CO2 thi ra (n v l lt), tnh c thng s h hp. Da vo bng 1.2, ly gi tr ng lng nhit ca xy tng ng vi thng s h hp nhn vi s lt O2 tiu th s bit c nhit lng gii phng (cn gi l lng nhit trao i hay tr s trao i nng lng). V d: Nu thng s h hp l 0,85 th c ng lng nhit ca xy l 4,862 v bit c th tiu th 20 lt O2 th tr s trao i nng lng s l: 4,862 x 20 lt O2 = 97,24 KCal

VII. Phn bit nguyn tc hot ng ca c th sng vi my nhitMy nhit l ng c dng nhin liu t nh du Diedel, xng sinh cng nh my n t, xe my, my bm nc... Trong c hc, hiu sut s dng nng lng ca mt ci my nhit, c tnh theo cng thc: T T = 2 1 (1.8) T2

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: Hiu sut ca my nhit (%) T1: Nhit Kelvin trng thi u (oK) T2: Nhit Kelvin trng thi cui (oK) Cng thc chuyn i gia nhit Kelvin v nhit bch phn (oC): T(oK) = t(oC) + 273 (1.9) Gi s hiu sut s dng nng lng ca mt my nhit t trung bnh l 33% = 33/100 1/3 Nu ta cng gi s c th sng hot ng ging nh mt my nhit, tc l cng c hiu sut s dng nng lng l 33% ? Nhit ban u ca c th ngi l 37oC, theo cng thc (1.9) tnh ra: T1 = 37 + 273 = 310oK 1 Thay = 33% v T1 = 310oK vo cng thc (1.8) s c: 31 T2 310 o K = 3.(T2 - 310oK)=T2 3 T2 2T2 = 930oK T2 = 465oK p dng cng thc (1.9) ta c: 465 = t2(oC) + 273 t2 = 465 - 273 = 192oC Kt qu trn cho thy c th sng hot ng khng ging nh mt my nhit. i vi c th sng, Protein b bin tnh ngay nhit t 40oC n 60oC cn 192oC th khng c mt sinh vt nhn chun no c th sng c. iu khng nh c th sng hot ng khng ging nh mt my nhit m hot ng theo nguyn l ca cc qu trnh sinh hc.

VIII. nh lut II nhit ng hcnh lut I nhit ng hc ch cho bit v s bin i gia cc dng nng lng khc nhau, cho php xc nh biu thc ch r s lin quan v lng gia cc dng nng lng khc nhau khi xut hin trong mt qu trnh cho trc. Song nh lut I nhit ng hc khng cho bit qu trnh khi no c th xy ra hoc khng xy ra v chiu hng din bin ca qu trnh nu xy ra th theo chiu hng no? nh lut II nhit ng hc xc nh c chiu hng t din bin ca mt qu trnh cng nh cho bit qu trnh t din bin n khi no th dng li v cho php nh gi kh nng sinh cng ca cc h nhit ng khc nhau. nh lut II nhit ng hc c ba cch pht biu.

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Cch pht biu th nht cn gi l tin Clausius a ra 1850: "Nhit khng th t ng truyn t vt lnh sang vt nng". T suy ra rng nhit ni ring cn nhng qu trnh nhit ng ni chung ch c th t din ra nu xy ra s truyn nng lng t mc cao n mc thp, tc l theo chiu gradien. Gradien ca mt thng s c trng cho mt tnh cht no v trng thi ca h (nh nng ) c xc nh bng hiu s gi tr ca thng s ti hai im chia cho khong cch gia hai im . V d: Gi s ti v tr 1 c nng l C1 cn v tr 2 c nng l C2 v C1 > C2 th hiu s nng s l C1 - C2 cn hiu s khong cch l x2 - x1. t C = C1 - C2 v x = x2 - x1 th gradien nng c xc nh: C1 > C2 C C1 C 2 = (1.10) x x 2 x1 x1 x2

Gradien l mt i lng vect, c gi tr v hng v gi tr ln. Khi so snh mt t bo sng vi mt vt v sinh nh mt ht ct ta thy r ngay rng trong t bo sng duy tr nhiu loi gradien khc nhau. Gradien mng duy tr in th tnh v in th hot ng, gradien nng duy tr nng , gradien p sut thm thu duy tr lng nc trong t bo... Nu t bo cht th cc loi gradien cng b trit tiu. Nu xt mc gradien th s sng ca t bo lun km theo s tn ti ca cc loi gradien. Cch pht biu th hai do Thomson pht trin tin ca Clausius "Khng th c mt qu trnh bin i chuyn ton b nhit lng thnh cng". Theo cch pht biu ca Thomson th hiu sut hu ch ca qu trnh bao gi cng nh hn 1 (tc < 1). iu ny c ngha trong t nhin khng c mt qu trnh no c th chuyn ton b nhit lng c cung cp thnh cng hu ch. i vi cc qu trnh din ra trong h thng sng c tun theo cch pht biu ca Thomson hay khng? Vn ny s cp n phn sau. Cch pht biu th ba trn c s kin ca Planck, cho rng Entropi l mt tiu chun y v cn thit xc nh tnh thun nghch v khng thun nghch ca bt c qu trnh vt l no din ra trong thin nhin. nh lut II nhit ng hc pht biu nh sau: "i vi h c lp, mi qu trnh trong t nhin u din bin theo chiu tng ca entropi". Vy entropi l

15

g? hiu r i lng ny ta xt v d v nguyn l hot ng ca my nhit. Theo hnh 1.1, nguyn l hot ng ca my n nh sau: My ch c kh nng sinh cng A (tc bnh quay) khi c cung cp nng Ngun cung cp nhit T1 lng l xng. Khi xng b t chy c nhit l T1 v gii phng nhit lng Q1 l Q1. Mt phn ca nhit lng Q1 dng sinh cng, phn cn li My sinh truyn cho ngun nc lm lnh my l cng Q2, dn n lm tng nhit ca nc l T2. y Q1>Q2 v T1>T2 Q2 Theo (1.8) th hiu sut hu ch ca qu trnh thun nghch c xc nh theo T2 Ngun thu nhit cng thc: A Q1 Q 2 T1 T2 = = = (1.11) Hnh 1.1: Nguyn l Q1 Q1 T1 hot ng ca my n Cn bng phng trnh ta c: T1(Q1-Q2)=Q1(T1-T2) T1Q1-T1Q2=Q1T1-Q1T2 Q Q (1.12) Gin c ta c: T1Q2=Q1T2 hay 1 = 2 T1 T2

T vt l hc cho bit s thay i entropi ca mt h c xc nh theo cng thc: Q S = (1.13) T S: s thay i entropi ca h. Q: Nhit lng cung cp cho h (calo) T: Nhit Kelvin (oK) ca h i vi qu trnh bin thin v cng nh, ta c: Q (1.14) dS = T n v ca entropi l Cal/M. Entropi l mt hm trng thi nn n ch ph thuc vo trng thi u v trng thi cui cng ca h. Cng thc (1.12) c th biu din qua hm entropi nh sau:

16

S1 =

Q1 Q ; S2 = 2 T1 T2

S1: Entropi trng thi u S2: Entropi trng thi cui i vi qu trnh thun nghch theo cng thc (1.12) ta c: S1=S2 S = Const (hng s) (1.15) Trong mt h nu ch xy ra cc qu trnh thun nghch th h lun duy tr trng thi cn bng nn entropi ca h l khng i. i vi qu trnh khng thun nghch th S >

ch lm thay i entropi ca h m cn lm thay i entropi ca mi trng xung quanh do s ma st v ta nhit. Thc nghim xc nh i vi mt qu trnh khng thun nghch th entropi ca h trng cui (tc S2) bao gi cng ln hn so vi entropi ca h trng thi u (tc S1). Do vy: S2-S1>0 (1.16) Trong mt h xy ra cc qu trnh khng thun nghch th entropi ca h bao gi cng tng ln. Do vy, nu l h c lp th cc qu trnh xy ra trong h s tin trin theo chiu tng ca entropi v entropi ca h s t gi tr cc i trng thi cn bng nhit ng. Tnh chung cho c qu trnh thun nghch v khng thun nghch th s thay i entropi ca h c th vit nh sau: S 0 (1.17) i vi qu trnh thay i entropi v cng nh (gi l qu trnh vi phn) th: dS 0 (1.18) (Du bng dng cho qu trnh thun nghch cn du ln hn dng cho qu trnh khng thun nghch). H c lp Vt th V d 1: H c lp gm vt th A v A vt th B. Vt th A c nhit l TA cn vt th B c nhit l TB. Gi s TA>TB th theo nh lut II Q nhit ng hc vt A s truyn cho vt B mt nhit lng l Q. S thay i entropi ca vt A do mt nhit l: B Q cn s thay i entropi dSA= TA Vt th

Q v nhit lng cung cp cho h khng T

17

ca vt B do nhn nhit l dSB=

Q TB (du tr ch entropi ca vt A b gim). S thay i entropi ca ton h c xc nh: Q Q Q(TA TB ) = dS=dSA+dSB= TB TA TA .TB

V TA>TB nn TA-TB>0 suy ra dS>0 V d 2: Ngm cc nc Cc nc Lp cch c nhit l 0oC (tc nhit hon (0oC) tnh ra nhit Kelvin l: ton T1=273+0oC=273oK) vo trong thng du c nhit l 100oC. 100oC Thng du H c lp (tc T2=273+100=373oK). Nhit s t ng truyn t du sang cc nc qui ra nhit lng l Q. Thc nghim xc nh c Q=80cal. Du cung cp nhit nn s thay i entropi ca du l: 80 = 0,214 cal/ (b gim i). dSD= 373 Cc nc nhn nhit nn s thay i entropi ca nc l: 80 =0,293cal/. dD= 273 S thay i entropi ca h (gm du v cc nc ) s l: dSh= dSD+dS = 0,293cal/ - 0,214cal/d = 0,079cal/ > 0 Nhn xt: i vi h c lp, qu trnh truyn nhit t din bin theo chiu tng ca entropi. iu ny minh chng cho tnh ng n ca nh lut II nhit ng hc.

IX. Tnh cht thng k ca nh lut II nhit ng hcnh lut II nhit ng hc c tnh cht thng k nn chnh xc ph thuc vo s phn t c trong h. Trng thi t trt t (tc c entropi ln) c xc sut cao hn trng thi c trt t cao (tc c entropi nh). Gia entropi v xc sut nhit ng c mi quan h nh th no?

18

Nm 1896 Boltzmann l ngi u tin tm c mi lin quan gia entropi v xc sut nhit ng qua phng trnh: S = klnW (1.19) S: Entropi ca h k: Hng s Boltzmann bng 1,38.10-16ec/ W: Xc sut nhit ng hiu r tnh cht thng k ca nh lut II nhit ng hc, ta xt v d: Trong 1 bnh c 6 phn t c nh s t s 1 n s 6. Ta ngn bnh thnh 2 phn: pha bn phi khng c phn t no c. - Thi im ban u: Bn tri Bn phi 1 23 5 4 6

Thnh ngn

Sau khi b thnh ngn do chuyn ng nhit ca cc phn t nn cc phn t s chuyn ng trong ton b th tch ca bnh to nn tt c c 26 = 64 - Kiu phn b 5 phn t bn tri bnh cn 1 phn t bn phi bnh (c 6 kiu phn b). 2 2 1 1 1 23 5 4 6 3 5 4 6 5 4 6 3

Kiu 1

Kiu 2

Kiu 3

1 3 5

2 4 6

1 3

2 4 6 5

1 3 5

2 4 6

Kiu 4

Kiu 5

Kiu 6

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Xc sut phn b ca phn t trong bnh c trnh by bng 1.3. Xc sut nhit ng l s kiu c th xy ra ca mi kiu phn b cho nn xc sut nhit ng lun lun ln hn 1 hoc nh nht l bng 1. Xc sut ton hc (k hiu l w) bao gi cng nh hn mt v bng 1 l ln nht. Xc sut ton hc l kh nng xy ra s kiu trn ton b s kiu phn b.Bng 1.3: S phn b ca cc phn t gia hai phn bnh v xc sut phn b ca chng.

S phn t Xc sut Xc sut nhit ng W ton hc w Bn tri bnh Bn phi bnh 6 0 1 1/64 5 1 6 6/64 4 2 15 15/64 3 3 20 20/64 2 4 15 15/64 1 5 6 6/64 0 6 1 1/64 Kt qu bng 1.3 cho thy s phn b cn bng mi bn c 3 phn t l c xc sut nhit ng ln nht, tc W=20 nn theo cng thc (1.19) th entropi cng c gi tr ln nht. Nh vy, trng thi cn bng l trng thi c xc sut nhit ng ln nht ng thi cng c entropi ln nht nn l trng thi d xy ra nht. Trong 64 kiu phn b ch c 20 kiu tun theo nh lut II nhit ng hc, tin trin theo chiu hng tng ca entropi v t ti entropi ln nht cn 42 kiu (gm 6+15+15+6) cng tin trin theo chiu tng ca entropi nhng cha t ti gi tr entropi ln nht. c bit chiu hng tin trin c 6 phn t cng mt pha ca bnh c xc sut nhit ng v entropi ging vi trng thi ban u (tc entropi khng tng ln). iu ny khng ng vi nh lut II nhit ng hc l qu trnh lun din ra theo chiu tng ca entropi. Khi s phn t trong bnh cng tng ln, v d l 50 phn t th tng s kiu phn b s l 250, l 1 con s cc k ln. Khi kh nng tin trin c c 50 phn t mt pha ca bnh cng ch chim 1 kh nng (tc W=1) so vi 250 kh nng v nh vy trng hp xy ra khng ng vi nh lut II nhit ng hc s cng t i

1 50 . chnh l tnh thng k ca nh lut II nhit ng hc, tc l 2 khi s phn t trong h cng ln th tnh ng n ca nh lut II nhit ng hc s cng cao.

X. nh lut II nhit ng hc p dng vo h sinh vt

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H thng sng l mt h thng m, lun xy ra qu trnh trao i vt cht v nng lng vi mi trng ngoi. Theo cch pht biu ca Thomson: "Khng th ch to c ng c vnh cu loi hai" l ng c c hiu sut hu ch l 100%, khi p dng vo h thng sng l hon ton ng n. Thc nghim xc nh mi qu trnh din ra trong h thng sng u c hiu sut hu ch nh hn 100% (xem bng 1.4).Bng 1.4: Hiu sut ca mt s qu trnh sinh vt

TT 1 2 3 4 5

Qu trnh sinh vt Qu trnh glicoliz (tiu glucoza) Qu trnh co c Qu trnh oxy ha photphorin ha Qu trnh quang hp Qu trnh pht quang ca vi khun

Hiu sut () 36% 30% 55% 75% 90%

Qu trnh quang hp ca thc vt c hiu sut 75% c ngha l cy xanh c hp th 100calo t nng lng nh sng mt tri th c 75 calo c s dng vo tng hp cht (phn nng lng c ch) cn 25 calo ta nhit si m c th hay pht tn nhit ra mi trng xung quanh (phn nng lng v ch). * Vai tr ca entropi i vi h c lp, nh lut II nhit ng hc khng nh mi qu trnh din bin u din ra theo chiu tng ca entropi v t gi tr cc i khi t n trng thi cn bng nhit ng th dng hn. C th sng l mt h m cho nn khng th p dng nh lut II nhit ng hc trc tip ln c th sng. nh lut II nhit ng hc ch c th p dng vo h sinh vt nu xem h bao gm c c th sng v mi trng sng. Ta hy xt s dS ) c lin quan ti s thay i thay i entropi ca h (k hiu l dt dSC entropi ca c th sng (k hiu l ) v s thay i entropi ca mi dt dSm trng (k hiu l ) nh th no? Entropi l mt hm trng thi v c dt tnh cht cng tnh (tc l entropi ca h bng tng entropi ca cc thnh phn nm trong h) nn s thay i entropi ca h s c xc nh theo cng thc:

21

dS dSC dSm = + (1.20) dt dt dt dSC = 0 khi cc qu trnh din ra trong c th sng u l qu Thnh phn dt trnh thun nghch. Song trn thc t phn ln cc qu trnh din ra c dSC th sng l nhng qu trnh bt thun nghch, do vy >0 dt dS m = 0 khi gia c th sng v mi trng khng xy ra qu Thnh phn dt trnh trao i vt cht v nng lng. iu ny khng xy ra v tri vi thc t. Qu trnh trao i vt cht v nng lng gia c th sng vi mi trng l qu trnh bt thun nghch (nht l qu trnh trao i nhit) do dSC dSm dSm dS >0. Nh vy, khi >0 v >0 th >0. iu ny l ph vy dt dt dt dt hp vi nh lut II nhit ng hc. Trong c th sng xy ra trng hp khi qu trnh ng ha din ra vi dSC cng mnh hn so vi qu trnh d ha th g th nng gc t do s c xc nh theo cng thc: w n= o (2.41) gf Gii phng trnh (2.40) s thu c nghim: w n = o e t 1 (2.42)

(

)

Nu tch s .t rt ln hn 1 th nghim c tnh theo cng thc:

40

n=

w o t e

(2.43)

* c im ca phn ng dy chuyn Phn ng dy chuyn c cc c im sau: - Phn ng dy chuyn c thi gian tim n. Trong thi gian ny ch yu to trung tm hot ng u tin. - Phn ng dy chuyn c hai gii hn nng . gii hn nng gc t do thp th gc t do d tng tc vi thnh bnh hay vi phn t cht c ch nn phn ng khng tin trin c. gii hn nng gc t do qu cao th cc gc t do d tng tc vi nhau gy ra hin tng t mch nn lm cho tc phn ng tin trin chm li. - Tc phn ng dy chuyn ny nhnh khng tun theo nh lut Arrhenius. Khi nhit tng th tc phn ng dy chuyn ny nhnh tng gp bi so vi nh lut Arrenius. Taruxop theo c ch ca phn ng dy chuyn ny nhnh gii thch s tc dng ca tia phng x ln c th sng. Khi chiu x, nu xt v mt nng lng th c gi tr rt thp nhng li gy ra hiu ng sinh hc cao (gi l hiu ng nghch l nng lng). iu ny c gii thch l do tia bc x gy ra hin tng ion ha, to ra cc trung tm hot ng l nhng gc t do v chnh cc gc t do gy ra phn ng dy chuyn ny nhnh, dn ti hiu ng tn thng sinh vt b chiu x. Taruxop cng cho rng cc cht c tc dng hn ch tn thng do tia x gy ra (gi l cc cht bo v phng x) l do n ngn chn c phn ng dy chuyn ny nhnh. Phn ng min dch ngi khi tim khng nguyn, sau thi gian bnh (thng t 3 n 21 ngy) khng th cha xut hin trong mu, sau l giai on nng khng th tng theo hm s m, khng t l thun vi lng khng nguyn a vo c th. Theo E. Manuel, phn ng to khng th din ra theo kiu phn ng dy chuyn ny nhnh. V d khi tim vaccine phng bnh bch cu, a vo c th ngi ch 0,36mg nhng sau thi gian bnh, lng khng th trong mu sut hin ln gp 1 triu ln so vi lng khng nguyn. E. Manuel cng cho rng c th khi b nhim cc c t nh nc c ca rn, b cp, iperit, rixin... th phn ng to khng th xy ra theo kiu phn ng dy chuyn ny nhnh.

XII. Nhit v tc phn ngTheo l thuyt ng hc ca cc qu trnh ho hc th tc phn ng bao gi cng ph thuc vo nng cht tham gia vo phn ng v ph thuc vo nhit . Khi nhit tng th tc phn ng cng tng. Ban

41

u c gii thch do tng s ln va chm ca cc nguyn t hay phn t tham gia vo phn ng ho hc. Trn c s tnh ton xc sut va chm, nu va chm no gia cc nguyn t hay phn t cng dn ti phn ng th tc phn ng s tng ln t 102 n 106 ln so vi tc thc t o c nu ta tng nhit trong khong t 0oC n 100oC. T cc nh nghin cu cho rng ch mt phn trong s cc va chm mi c kh nng to phn ng ho hc. Phn ng ch c th xy ra khi cc nguyn t hay phn t t c mt gi tr nng lng nht nh gi l nng lng hot ha. Nng lng hot ha l gi tr nng lng nh nht m nguyn t hay phn t cn phi t c mi c th tham gia vo phn ng. Cc nh nghin cu p dng hm phn b Maxwell - Boltzmann gii thch mi lin quan gia tc phn ng v nng lng hot ha. Maxwell - Boltzmann thit lp c cng thc : (2.44) N =N. e N : S phn t t gi tr nng lng bng hoc ln hn nng lng hot ha. N: Tng s phn t ban u trong mt n v th tch. E: Nng lng hot ha ca phn t v R l hng s kh T: Nhit tuyt i (nhit Kelvin) Cng thc (2.44) biu din trn th (hnh 2.4) gi l hm phn b mv 2 Boltzmann. th biu din s phn t N theo vn tc v (E= ). 2*N N

*

E RT

T1 T1) th ng cong phn

42

b Boltzmann dch chuyn v pha bn phi, tc l s lng phn t c EEh tng ln (phn gch ngang hnh 2.4b) v vy tc phn ng tng ln. Nhiu phn ng, nht l cc phn ng ha sinh, tuy c nng , nhit v nng lng hot ha ging nhau nhng tc phn ng vn khc nhau. Nh vy, khng phi mi va chm gia cc nguyn t hay phn t c nng lng bng hay ln hn nng lng hot ha u dn ti phn ng. iu ny c gii thch do tc phn ng cn lin quan ti cu trc hnh hc ca cc phn t va chm, gi l yu t lp th (k hiu l p).

XIII. S ph thuc ca tc phn ng ha sinh vo nhit S ph thuc ca tc phn ng vo nhit c m t qua phng trnh Arrenius. K= p.z. e RT (2.45) K: Tc phn ng p: Yu t lp th Eh: Nng lng hot ha z: S ln va chm R: Hng s kh bng 1,987 KCal/M. T: Nhit tuyt i Phng trnh Arrenius c th vit di dng logarit: E lnK = ln(p.z) - h (2.46) RT E 1 t y = lnK ; x = ; a = h ; b = ln(p.z) T R Phng trnh (2.46) c th vit di dng: y = a.x + b (2.47) y l phng trnh bc nht, th ca n c dng ng thng (hnh 2.5). Da vo th xc nh c: Eh tg = Eh = R.tg R Bit tg s xc nh c nng lng hot ha. Eh

73

Chng 4

IN TR CA T BO V MI. in tr ca t bo v m.Khi nghin cu tnh cht in ca h thng sng, cc qu trnh in sinh hc ca c th, ta cn phi nm vng cc kin thc vt l ng dung vo cc h thng sng. Ngy nay, vic ng dng cc hin tng in sinh vt kho st, thm d v iu tr trong Y hc l vic lm tng i ph bin. Xc nh cc thng s in ca t bo, m, si c... lm c s cho vic chn on v iu tr bnh. Vic tr liu bng phng php ny nhm gii thch s hot ng ca c th hoc h thng sng, trng thi trao i cht bnh thng cng nh phn tch trong cc tnh trng bnh l khc nhau. Nh vy bng phng php l sinh da trn hin tng in hc, khoa hc a ra c nhiu i lng c th, cc thng s trng thi c trng cho c th. Hn na, ngy nay nh vo cc thit b vi k thut hin i cung cp cho ta nhiu hng nghin cu mi trn t chc sng. Thc nghim cho thy rng, in tr hoc dn in ca cc i tng sinh vt c gi tr hng nh trong cng mt iu kin xc nh. l cc thng s c trng cn thit trong xt nghim cn lm sng, cc ch s ht sc c bn trong nghin cu y-sinh hc. V vy nghin cu tnh cht in nh o in dn, in th... ca cc t bo v m c mt ngha tht s quan trng. Vic nghin cu thng s in nhm mc ch tm hiu mt s c tnh, cu trc cng nh s nh hng ca cc tc nhn vt l ln cc c quan, t chc sinh vt. Vic xc nh gi tr in tr xut, in dn xut, tnh dn in hay ni chung l cc thng s in ca t bo v m cho php ta nh gi c trng thi sinh l v chc nng hot ng ca cc i tng kho st. Ngoi ra, t cc thng s ghi o c, ngi nghin cu cng c th suy on ra c cu trc ca cc t chc sng mt cch d dng. Bng nhiu phng php khc nhau ca l sinh hc, ta c th nhn bit v bn cht ca t chc sinh vt, m khng cn phi ph hu cu trc ca i tng kho st. Ngay t th u k th XIX, cc nh khoa hc bit cch chn phng php nghin cu ng theo hnh thc o in. Bng cch d dng

74nht l o in tr cng nh xc nh dn in ca mu, t bo, m sng, c, si thn kinh. ng thi cng c th a ra c nhiu thng s lin quan m nhiu cng trnh nghin cu sau ny cp n. l cc ti nghin cu v in tr, in dn, tnh dn in trn cc ng vt, hoc ngay c trn cc t bo thc vt ... Nhiu kt qu thc nghim trc y chng minh cho ta thy rng: Vi cc t bo ng vt hoc thc vt cng nh i vi cc m sng, di tc ng ca dng mt chiu th in tr sut c gi tr trung bnh vo khong 106 107 /cm. Nn di tc dng ca dng in mt chiu cho thy, dn in ca cc i tng sinh vt gn ging nh tnh dn ca mt cht bn dn in. Cn di tc dng ca dng in xoay chiu th gi tr in dn o c c gi tr nh hn rt nhiu. xc nh c thng s v in tr thun, in tr khng ca cc h thng sng l mt vic lm khng n gin. Thng thng, ta gp phi nhng kh khn v phc tp trong khi o v: - i tng sng l mt h a pha v t chc khng ng nht v cu trc. - Th tch t bo khng c nh m c th bin i tu theo trng thi sinh l ca i tng khi kho st. - B mt t bo c mt lp v protin bao bc, lp mng bo v t bo c in dn rt ln. - Ngoi ra, dng in i vo m ch yu chy qua lp gian bo c dn in tt v bn thn n cha nhiu loi ion vi nng rt cao. - Cc vi in cc lm tn thng mng. Nhng s liu ng tin cy l o in tr ca cc t bo c kch thc nh vi hnh dng d tnh ton v th huyn ph. ph hp vi kt qu nghin cu t thc nghim, khi s dng cng thc Maxwell o thng s in tr. Phng trnh ny c dng nh sau:r1 1 r = r1 +2 r r1 1 r2 r1 +2 r2

(4.1)

Trong , r : in tr sut ca dung dch. r1 : in tr sut ca mi trng.

75r2 : in tr sut ca t bo. : l t s gia th tch ca t bo vi ton b th tch ca dung dch. Vi phng php xc nh nh trn, thc nghim cho ta thy rng in tr sut ca hng cu c gi tr khong 1012 /cm. Nh vy hng cu l mt vt th c c trng ca mt cht in mi. Ngy nay t cc kt qu nghin cu c, ngi ta a ra nhiu phng php khc nhau xc nh cc thng s in ca i tng kho st tu theo yu cu. Vic ng dng trong lm sng cng c mt ngha ht sc to ln. Ngoi ra, c s l thuyt cng c xy dng mt cch hon chnh, cng nh gii thch c cho nhiu c ch hnh thnh mt cch hp l vic ng dng hin tng in trong nghin cu y-sinh hc.

II. in tr ca t bo v m di nh hng dng in mt chiu. xc nh gi tr v in tr thun trn c th ngi v ng vt, ta c th tin hnh theo mt s phng php n gin. o in tr bng cch dng in cc t ti hai v tr khc nhau trn i tng nghin cu, c m phng nh (hnh 4.1) di y:

G

A

-

+

Hnh 4.1: o in tr ti hai v tr xc nh ca c th.

76Kt qu thc nghim cho ta thy rng: Vi cc t bo, m sng khc nhau th gi tr ghi nhn c cng khc nhau. Tht vy, khi cho dng in mt chiu i qua cc i tng nghin cu trn th cng dng in o c khng phi l mt hng s. Gi tr o c gim mt cch lin tc theo thi gian n mt gi tr xc nh no , mc d hiu in th ca ngun cung cp l khng i. Theo nh lut Ohm, cng dng in (I) s bin i mt cch tuyn tnh theo hiu s in th p t vo mch o. Vi ngun in mt chiu th cng dng (I) l khng i, nhng y kt qu ghi nhn c th hon ton c s khc bit. Tht vy, kt qu thc nghim cho ta thy rng dng in ghi o c trn i tng sng khng phi l hng s m gi tr ca n b thay i nhiu theo thi gian. Cng dng in l mt hm s bin i theo thi gian nh (hnh 4.2): I I = Const.

I = f (t) tHnh 4.2: S ph thuc cng dng in (I) theo thi gian (t).

Nguyn nhn c s gim dng in nh nu trn l do hin tng phn cc xy ra trong i tng nghin cu. Khi cho dng in mt chiu i qua, th ngay trn bn thn h thng sng xut hin mt sut in ng (E). Hin tng phn cc P(t) tng n mt gi tr xc nh no v pht trin theo chiu hng ngc li so vi s phn cc ban u. S phn cc ton phn ny lm hiu in th ton mch gim, do dng in ngoi cng gim theo. p dng nh lut OHM cho mt mch kn, trong trng hp ny ta c:

I=

U P(t ) R

(4.2)

77Trong : U: l hiu in th p t vo i tng kho st I : l cng dng in qua on mch R : l in tr ca t bo, m sng m ta nghin cu. Tng t nh vy, trong h thng sng v c s phn cc khi cho dng in chy qua, nn ta cng c th xc nh c gi tr in dung C tng ng theo hng s in mi ca cc i tng nghin cu bng biu thc: C= Vi: : hng s in mi S : din tch b mt bn cc d : khong cch gia 2 bn cc. Trong t bo gm in dung tnh v in dung phn cc (Cp) ca cc loi t bo, m sng, c ...di tc dng ca dng in:

S 4d

(4.3)

Cp =Trong :

Idt0

t

(4.4)

R(I 0 I t )

I0, It : Cng dng in ti im t = 0 v ti thi im t bt k I : Cng dng in ca on mch. h thng sng, trong trng thi bnh thng th in dung phn cc c gi tr ln, cn khi i tng ny b tn thng hay hoi t th kt qu o c v in dung li c gi tr nh hn nhiu. in dung phn cc ca cc i tng sinh vt ni chung c gi tr trung bnh vo khong t 0,1 F/cm2 n 10 F/cm2 , cn i vi cc si c th in dung ca n c th ln n khong 40 F/cm2.

78III. S bin i in tr theo tn s dng xoay chiu. Di tc dng ca dng in xoay chiu th tr khng ca t bo, m sng ph thuc nhiu vo tn s () ca ngun pht. Do , thnh phn in khng ca i tng sinh vt nh cm khng, dung khng v tng tr mch s ph thuc nhiu vo tn s ca ngun. Khi mt trng thi sinh l no , ng vi mt tn s nht nh ca ngun xoay chiu th in tr ca t bo v m sng c gi tr khng i. T nm 1920, nhiu cng trnh nghin cu kho st s bin i ca in tr di tc dng ca ngun xoay chiu cho thy: in tr ca hu ht cc loi m, c, t bo sng s b thay i di cc tn s khc nhau. c tuyn bin i ca in tr theo tn s () c dng nh (hnh 4.3) di y:

R

1

2

Hnh 4.3: S bin i in tr ca t bo m theo tn s.

T th trn ta thy: Vi < 1 v > 2 : in tr tng i n nh. 1 < < 2 : in tr gim khi tn s tng. Quy lut ny hon ton xy ra v bin i l nh nhau vi mi loi t bo, m v nhiu i tng khc. c bit, kt qu xy ra cng tng t i vi cc si c mc d gi tr tr khng tuyt i ca c l khc nhau. Cc gi tr 1 2 c gi tr thay i trong khong t 102 108 Hz.

79a s cc m sng, in tr cc tiu hay tr s dn in cao ng vi tn s khong 106 Hz, cn i vi si thn kinh th gi tr ny tng ng vi tn s vo khong 109 Hz. T thc nghim cho thy, dn in ca t bo v m thay i ch yu tn s thp. tn s cao tnh cht ny t thay i hn v tn s thp hin tng phn cc din ra chm chp v r rt hn nhiu. S ph thuc dn in vo tn s l i lng c trng biu din cho i tng l t bo hoc m sng. i vi cc t bo, m b tn thng tu theo mc , tnh cht ny cng gim dn theo, n khi t bo cht th tnh cht cng khng cn na. Ngha l in tr ca t bo khi b tn thng khng cn ph thuc nhiu vo tn s. hiu r iu , ta kho st in tr ca mt s m thc vt trng thi sinh l bnh thng v b tn thng vi cc mc khc nhau di tc dng nhit, nh (h.4.4):R (1)

(2) (3(1)

(4)

Hnh 4.4: Bin i in tr m thc vt theo tn s trng thi sinh l bnh thng v khi b tn thng.

(1): Trng thi bnh thng. (2): un 500C trong 2 pht (3): un 500C trong 4 pht (4): un 1000C trong 20 pht.

T c trng bin i ca tng tr theo tn s ca dng in xoay chiu trn, ta thy ti nhit cao nu thi gian tc dng cng di th mc tn thng cng ln, ng thi s ph thuc ca in tr theo tn s t hn. Da trn c s nghin cu v cc thng s in c th d dng xc nh, nh gi cng nh m t i tng theo cc ch s sinh hc ca

80mt t chc sng trng thi sinh l bnh thng hay b bnh l. Ngy nay ngi ta thng biu din ch s sinh hc qua c trng bin i ca in tr theo tn s R() cho cc loi t bo, m sng cng nh xc nh cho nhiu i tng nghin cu sinh vt khc.

IV. Tng tr ca t bo v m. c trng nh lng cho mi lin h k trn, ng thi tin vic kho st, Tarutxop a ra h s K cho vic nh gi i tng nghin cu s nhanh v c ngha hn, thay v dng th biu din s bin i ca in tr nh trnh by trn. H s ny c xc nh nh sau:k= R 1 R 2

(2.5)

Trong 1 , 2 c chn tu theo i tng nghin cu. V du: Khi kho st trn mt dy thn kinh, thc nghim cho thy rng in tr c gi tr cc tiu tn s 2 = 109 Hz v 1 = 104 Hz nn:K= R10 4 R109

(2.6)

Gi tr K cng ln s ph thuc ca in tr vo tn s ny cng mnh v ngc li gi tr ny nh khi t bo, m b cht. Trong cc t bo v m bnh thng, h s K ph thuc vo v tr ca i tng theo bc thang tin ho ca s sng. V d: H s K ca gan ng vt c v khong 9 10 H s K ca gan ng vt mu lnh khong 2 3 Cn trong c th ngi hoc ng vt, h s K ph thuc vo cng trao i cht ca tng loi m. - cc m sng c cng trao i cht mnh m nh gan, lch th h s K c gi tr ln. - Cn mt s c quan c cng trao i cht thp hn (c) th h s K li c gi tr nh. Cc gi tr R1, R2 c tin hnh trong cng nhit , iu kin th nghim nh nhau, v thi gian rt gn nhau. Nn t s o c tng i n nh khi m v t bo trng thi sinh l bnh thng ( tn s cao in tr ca c t thay i trong mt thi gian di, trong khi in tr vng tn s ngun in thp b thay i nhiu hn)

81S thay i in tr lm cho tnh dn in ca t bo v m b thay i theo, ngun gc chnh l do kh nng xy ra hin tng phn cc ca chng. Do h s K cn c gi l h s phn cc

V. p dng phng php o in trong chn on v iu tr bnh.1. Trong chn on v iu tr bnh. Ngy nay, phng php o in tr bng dng in th ng c s dng rng ri trong vic nghin cu qu trnh din bin xy ra t bo, m sng v cc t chc sinh hc di tc nhn ca cc yu t vt l, ho hc cng nh cc qu trnh bin i trong trng thi bnh l. V s dng vi ngun in p c gi tr nh, do tc nhn ny khng gy nn nhng bin i nhiu v tnh cht l ho trong t chc sng cng nh khng lm tn thng n i tng nghin cu l cc t chc sinh vt. Thc nghim cho ta thy rng, tnh trng bnh l bin i c lin quan nhiu n s thay i v gi tr in tr, in dn v cc thng s in in ca t bo. Khi mt t chc sinh hc b vim nhim th ngay trong giai on u, th tch t bo b trng ln, khong khng gian gia cc t bo b thu hp li, in tr ca m tng ln rt nhanh. Du hiu d dng thy nht l di tc dng ca dng in c tn s thp, in tr ca m lc ny c xc nh ch yu bi in tr ca cc thnh phn cht dch khong gian bo.

giai on u trong qu trnh vim, cc thnh phn v nht l cu trc t bo cha c s thay i ng k. Th tch t bo bin i khng ng k, nn gi tr in dung ca n cha c s thay i nhiu. Hin nhin rng: Khi tng gi tr in tr m in dung vn gi nguyn khng thay i th l du hiu ca s trng t bo, ngc li nu gim in tr m vn gi nguyn gi tr in dung th l du hiu ca s gim th tch ca t bo. giai on sau trong qa trnh vim nhim, ta thy c s thay i v cu trc t bo mt cch r rt, hin tng vim lm tng thm qua mng t bo. V vy lm thay i tnh dn in ca t bo v m sng, dn n s gim gi tr in dung v in tr. Ngy nay hin tng in sinh hc c ng dng nhiu cho vic chn on, iu tr ng dng trong y hc da vo cc tham s trn. Chng hn, p dng phng php o in tr xc nh nhn p, o chnh

82lch in tr gia na mt bn phi v bn tri, o in c tai, kho st phn ng c bng cch o sut Galvanic. Xc nh mc tn thng t bo di cc tc nhn thng qua cc gi tr o in dn, o gi tr in tr thun v tr khng ca da...

2. Phn cc trong h thng sng.Nhiu nh khoa hc cho rng, s phn cc cc ion ch yu xy ra trn b mt ca mng t bo. Quan nim phn cc mng xc nh l t bo sng ging nh mt vt th, m bn thn n c cha cc cht in phn trong cu trc thnh phn ca h. Cc cht in ly thng nm trng thi t do. Tht vy, khi trng thi sinh l bnh thng, t bo khng cho mt s cht i qua chng hn nh i vi ion Na+ m ch thm mt chiu i vi ion K+. Khi c in trng bn ngoi, cc ion trong t bo s di chuyn v cc cc in tng ng, tu thuc vo du ca in tch. Cui cng cc ion tri du s tp trung hai pha u cc ca mng t bo. Tng ng vi s tch t in tch ion pha bn trong mng, th mi trng bn ngoi mng t bo pha i din cng c s tch t cc in tch ion mang du ngc li. Nh vy, hai pha cc ca mng s xut hin hai lp in tch b mt. Hin tng to thnh trn t bo c m t ging nh hai bn cc ca mt t in v cng gi vai tr tch in trong cht in mi. y l gi thuyt a ra gii thch v s xut hin gi tr in dung ca mt t bo. in dung ph thuc vo hng s in mi ca t bo, cn gi tr in tr xut hin c xc nh thng qua in tr b mt ca n. Ngy nay bng phng php ng v phng x kim chng c iu . Mng t bo khng ch thm i vi ion K+ m cn thm ng thi i vi c nhiu loi ion khc nhau. Trong trng thi sinh l bnh thng hay qu trnh trao i cht bnh thng, ta thy lun lun xy ra hin tng vn chuyn cc loi ion Na+ - K+ qua li hai phi ca mng. Bng nhiu kt qu thc nghim khc nhau, ta d dng thy rng s phn cc trong t bo khng ch xut hin trn b mt ca mng m cn xy ra trong ton b th tch ca t bo. Tht vy, do hin tng in hng cc vt dn di tc dng ca in trng ngoi, lm cho cc in tch t do dch chuyn. Cc phn t tch in tng ng c s phn b in tch tr li trn pha i din trong cc t bo, nn cc lp bn trong t bo cng c s tch in vi du in tch ph hp.

83

E

P (t)Hnh 4.5: S hnh thnh sc in ng phn cc.

T bo sinh vt c xem nh l mt h a pha, mi pha c tnh cht ho l khc nhau, do in tr v in dn ca n cng khc nhau. ng thi cc cht khi thm nhp vo t bo sng cng tn ti cc pha ho tan hon ton khc nhau. S phn cc xy ra trong ton b th tch ca t bo, ngay trn b mt v k c cc thnh phn bn trong t chc sng. c bit s phn cc trong ni bo ng mt vai tr ht sc quan trng. a s cc dch sinh vt trong h thng sng l cc dung dch in ly, cha cc ion v cc i phn t tch in. Do , c ch phn cc trong h thng sng c lin quan vi tnh cht ca mt dch a in phn trong dung dch in ly ca i a s cc phn t sinh vt. Trong c th, a s cc t chc u c cc mng ngn cch gia cc c phn vi nhau hoc cc t chc ring bit. Theo thc t c th c cu trc phc tp, ging nh bn thn chng c chia ra thnh nhiu ngn c lp. S dch chuyn ca cc ion qua cc vch ngn v th cng khng cn t do di tc dng ca in trng ngoi. Cc ion tch t trn t bo, m sng, trong cc t chc sinh vt to ra mt qu trnh phn cc in di tc dng ca in trng ngoi nh (hnh 4.6) di y:

84

Hnh 4.6: S cc tnh ho t chc sng. Ngy nay cc s liu thc nghim cho thy rng, qu trnh phn cc h thng sng c xc nh khng nhng bi s phn cc ca cc ion m cn c xc nh bi s phn cc (c tnh lng cc in) cc cht hu c. Cc phn t protein, nucleotid u c gi tr moment lng cc kh ln, hng s in mi ca dung dch in ly cng b thay i nhiu theo tn s dng in bn ngoi. i a s cc phn t nm trong thnh phn nguyn sinh cht u cha cc nhm phn cc, di tc dng ca cc yu t bn ngoi gy tn thng trong nguyn sinh cht. Do , trng thi thay i thng hnh thnh hng lot cu trc mi c s nh hng khc nhau. S phn cc in mi trong h thng sng cng c th do s dch chuyn in tch trong thnh phn cu trc, m bao gm s phn ln l cc phn t hu c. Qu trnh ny thng c gi l s phn cc in trong cu trc v m.

VI. Mt s phng php in diin di l phng php phn tch da trn s dch chuyn cc in tch, phn t nhim in di tc dng ca trng lc in khng i. Cc thnh phn khc nhau trong dch sinh vt khi nhim in s mang mt in lng c ln v du in tch khc nhau. Tu theo loi phn t m cc ht nhim in s dch chuyn vi tc nhanh hay chm di tc dng ca in trng ngoi. Cc phn t sinh vt d dng nhim in v tch in, nn da vo tnh cht ny xy dng mt s phng php in di. Ngy nay, phng php in di c p dng rng ri v ng dng nhiu trong cc nghin cu y - sinh hc. Thng thng, ta s dng phng php in di tch chit cc thnh phn albumin, globulin trong huyt thanh; in di protein (phc cht lipide vi protein); in di glucoprotein (xc nh cc thnh phn ca protein v glucide trong huyt thanh).

85Nm 1937, Tiselius l ngi u tin a ra phng php in di trong mi trng t do kho st s hin din ca protein trong huyt thanh. Sau Tiselius, Cremer ta thy c nhiu nh khoa hc khc nh Durrum xy dng l thuyt hon chnh v hin tng ny mt cch tng minh v pht hin thm nhiu ch tiu o t khc trong cc thit b mi ca mnh. Nm 1950, cc nh sn xut a ra mt loi thit b mi hn, l s phn tch da trn phng php in di hin th v kho st thng tin bnh l bng cch ghi li cc c tuyn bin i ca n trn bng giy ch th. Ngy nay c nhiu cng trnh nghin cu y-sinh hc p dng phng php in di, da trn nguyn tc dch chuyn ion tch chit v phn tch nhiu thnh phn khc nhau ca dch sinh vt. Cc hng nghin cu bng phng php in di thng tp trung trong ba loi ch yu l: - Pht hin s dch chuyn ca cc phn t bng cch kho st di knh hin vi (in di t bo). - in di trong dung dch t do (xc nh theo linh ng ca cc ht nhim in trong dung dch). - in di trn cc cht gi (phng php in di kho st trn bng giy).CU HI N TP CHNG 4 1.in tr ca t bo v m ? 2. in tr ca t bo v m i vi dng in mt chiu? 3. in tr ca t bo v m i vi dng in xoay chiu? 4. Tng tr ca t bo v m ? 5. S dng phng php o n trong chn an v iu tr bnh? 7. Gii thch hin tng phn cc trong h thng sng? 8. Phng php in di: nguyn l v ng dng?

86

Chng 5

IN SINH HCI. M u:T lu chu u, ngi ta tin hnh nhng th nghim l th khm ph v cc kh nng lm xut hin dng in trn c th ng vt. T khi nim v in ng vt mi xut hin v c chng minh s tn ti ca n. Mt s loi c sinh sng sng v bin c b phn bo v c bit pht in nh c tr in, c ui in, chnh in...Ngc dng lch s v s pht hin ra cc dng in t sinh vt trn cho thy, t rt lu ngi Ai Cp gp phi v lm quen vi nhng hin tng in ny. Mt tnh cht c trng ca t bo ng vt l gia chng v mi trng bn ngoi lun lun tn ti mt s chnh lch in th. o hiu in th trn cc loi t bo khc nhau th s chnh lch ny vo khong 0,1 V. c bit c mt s loi c in c th sinh ra cc xung in rt cao n khong 600V, vi dng in c hng trm mA. Tnh cht in sinh hc c Dr. Louis De Galvanie khm ph. Sau , ti ny thu ht nhiu nh khoa hc khc quan tm v u t vo vic nghin cu mt cch l th. Tuy nhin sau hn 100 nm, k t nhng pht hin u tin di s ghi nhn ca cc nh khoa hc, con ngi vn cha gii thch c c ch hnh thnh hin tng in sinh vt mt cch r rng. Cc kt qu thc nghim vn cn ng khung trong vic m t hin tng. Trong vi thp k gn y, nh cc phng tin ghi o c nhy cao, chnh xc, cng nh cc thit b in t hin i...ngi ta mi khm ph c nhiu qui lut hnh thnh dng in ca t bo. T kt qu thc nghim o c bng cc phng php khc nhau nh ng v phng x, ng hc phn t, hin vi in t, ho t bo..., cc nh khoa hc cho thy bn cht ca dng in sinh hc. Vic xy dng c s l thuyt v gii thch c ch ca vic hnh thnh dng in sinh hc cn c nhiu hn ch. S d nh vy l v khi nghin cu hin tng in sinh vt thng gp phi mt s gii hn sau: - Tc bin i tn hiu trn i tng nghin cu thay i qu nhanh, trong khi cc gi tr o c thng rt nh, nn yu cu v thit b nghin cu phi l cc dng c ghi o tht nhy v c chnh xc tht cao.

87i tng nghin cu thng c kch thc ht sc nh (vo c kch thc t bo). iu kin nghin cu, phi c tin hnh vi phng php nh th no khng lm nh hng n trng thi sinh l ca i tng kho st.

Trc khi tm hiu v cc loi in th sinh vt, ta lu rng cc dch th hai pha trong v ngoi mng t bo l cc dung dch in phn (electrolytic solutions). Nng trung bnh ca cc anion c gi tr khong 155 mEq/l, ng thi c xut hin mt nng tng ng ca cc loi cation pht trin theo pha ngc li. Theo c ch vn chuyn vt cht qua mng sinh hc ta thy c s phn b tr li ca cc anion v cation hai pha mng. ng thi vi qu trnh vn chuyn tch cc, th c c s khuych tn ca cc ion vi cc thm khc nhau. Kt qu cui cng l trong ton b qu trnh h c s chnh lch nng ion hai pha mng, do lm xut hin mt hiu s in th mng (membranne potential). Hai yu t c bn c lin quan n s hnh thnh hiu th mng sinh hc c ngha quyt nh l: - S khuych tn nhng ion qua mng do s chnh lch nng ca cc loi ion hai pha mng. - S vn chuyn tch cc ca nhng ion qua mng khi chuyn dch t pha (phase) ny sang pha khc, to thnh mt cn bng mi l s cn bng c bit ca cc ion. Vi mt s c im nu trn th mc ch v yu cu khi nghin cu hin tng in sinh vt l: Hiu c bn cht ca cc loi in th sinh vt c bn nh loi in th ngh, in th tn thng, in th hot ng...Ngoi ra cn nm vng v cch ghi o, iu kin th nghim, cc giai on xut hin. Xy dng l thuyt ph hp gii thch s hnh thnh cc loi in th trn. Gii thch v cc kt qu ghi nhn c, k c cc mi quan h gia chng. Tm hiu mt s ng dng in sinh hc ca cc cng trnh nghin cu trong Y-Sinh hc. a ra mt s ng dng hin tng in trong cng tc chn on, thm d chc nng, cng nh cc ng dng iu tr bnh trong Y hc. Vic nghin cu cc hin tng in sinh vt v k thut ghi o cc thng s lin quan c mt ngha ht sc quan trng. c bit, ngy nay vi cc thit b khoa hc hin i, vic ng dng hin tng in

88trong Y hc, xt nghim trn cn lm sng c s dng kh ph bin. Do ta cn phi nm k phng php ghi o, hiu r bn cht ca cc loi in th sinh vt c bn.

II. in th mng v in th pha. 1. in th pha.* Gradient in th (the electric gradient) trng thi bnh thng ta thy c s phn b khng u ca cc ion hai pha ca mng sinh vt. Do c s chnh lch nng , cc ion ny s khuych tn qua mng t ni c nng cao n ni c nng thp hn. Di nh hng ca gradien nng , cc ion c khuynh hng tin ti trng thi cn bng mi, ng thi hnh thnh gia n mt lp in tch kp ngay trong dch sinh vt. nh hng ca in trng ngoi s lm xut hin gradien in th do s phn cc ca mng. Cc ion dng c xu hng ri mi trng c n pha m, ngc li ion m dch chuyn v mi trng c th dng hn. S chuyn di ca cc ht mang in s dng li khi lc tc dng ln cc ion di nh hng ca gradien in th cn bng vi gradien nng . Trong gradien nng pht trin theo hng ngc li so vi gradien in th nh (hnh 5.1) di y:

Hnh 5.1: nh hng ca gradien in th v gradien nng

: Dng chuyn dch di tc dng gradient in th : Dng chuyn dch di tc dng gradient nng

89* in th in cc (electrode) in th in cc l hiu th c hnh thnh gia in cc v dung dch in phn. Hay ni cch khc l in th xut hin lp in kp khi in cc nhng vo dung dch in phn. V d: C th ly v d bng cch kho st trng hp hiu th ny khi s dng in cc bc (Ag) trong dung dch nitrat bc (AgNO3). Nu gi: ic l th ho hc ion kim loi trong in cc idd l th ho hc ca ion trong dung dch. Ta thy: - Nu ic < idd , di nh hng ca gradient th ho hc lm cc ion bc (Ag+) chuyn dch vo kim loi lm thanh kim loi tch in dng. Chuyn ng ca cc ion ny dng li khi h thng trng thi cn bng in ho. Thanh kim loi tch in dng v xung quanh s c mt lp ion NO3- bao bc to thnh lp in kp. trng thi cn bng in ho, s chnh lch in th ho hc gia in cc (c) v dung dch (dd) s c gi tr bng hiu s in th in ho ca lp in tch kp ny. Ta c: ZF = idd - ic (5.1) Trong : : in th ca in cc i vi dung dch Z : Ho tr ca cc ion t do F : Hng s Faraday - Nu ic > idd in cc s tch in m v xung quanh c lp ion Ag+ v cc ion Ag+ s ri khi thanh kim loi i vo dung dch. in cc b ho tan dn dn trong dung dch, hin tng chuyn dch ion ch dng li khi t ti trng thi cn bng in ho. - Nu ic = idd Trong trng hp ny s chuyn dch ca ion theo hai hng l cn bng nhau, nn th in cc bng khng. Da vo biu thc tnh cng thm thu phi thc hin khi tng nng dung dch ln 1gam/mol, ngha l lm thay i p sut thm thu ion t P1 n P2 tnh hiu th in cc. Ta c: P A = RT ln 2 (5.2) P1

90Trong A: l cng thm thu R: Hng s kh l tng. T: Nhit tuyt i ca mi trng. Ngoi ra, cng ca lc in trong qu trnh chuyn ho lm thay i nng ion trong dung dch. Cng ny c xc nh trong trng th nng t nh in l: A = UF (5.3) Cng ca qu trnh in ho v cng thm thu trong mi qu trnh cn bng nhau, nn t (5.2) v (5.3) ta c: P2 (5.4) = UF P 1 RT P2 (5.5) U = ln F P 1 Theo phng trnh Vant Hoff th p sut thm thu t l thun vi nng nn: RT lnU= RT C 2 ln F C1

(5.6)

Trong th nghim trn th C1, C2 chnh l nng ion trong dung dch (Cdd) v trong in cc (Cc). Vi Z l ho tr ca ion kim loi, nn tng qut ta c th vit li hiu in th ion nh sau:Ui

=

C RT ln c ZF C dd

(5.7)

Vi Ui l hiu in th xut hin do s chnh lch in th ca cc ion to thnh. y chnh l phng trnh Nernst xc nh gi tr hiu s in th in ho (electrochemical potential different) ca cc dung dch in ly.* Hiu in th pha. Nu nhng hai in cc cng kim loi vo hai bnh ng dung dch cha cng mt cht in ly vi ho tan C1 v C2 khc nhau. mi in cc, sau khi t n trng thi cn bng s xut hin mt in th m

91ln ca n s t l vi t s nng cc ion kim loi trong in cc v trong dung dch . V nng ion kim loi trong hai dung dch l khc nhau nn gi tr in th xut hin trn mi in cc cng khc nhau, gia chng tn ti mt hiu in th Uc gi l hiu in th nng .

U c = U c1 U c2RT C c RT C c ln ln Uc = ZF C1 ZF C 2

(5.8) (5.9) (5.10)

Uc =

RT C2 ln ZF C1

Vi C1, C2 l cc nng ion kim loi trong dung dch. Do , khi nghin cu trn thn nhit ngi bnh thng, nhanh chng cho vic tnh ton ngi ta thay vo cc gi tr hng nh, cng thc c xc nh li l:

UC =

C 61 log 2 Z C1

(5.11)

Nh trn ta thy di nh hng ca gradient nng v gradient in th m cc ion khuych tn qua mng vi tc khc nhau. Ni cch khc mi loi ion c d? linh ng khc nhau. Do s khuych tn ca mi loi ion khc nhau, nn gia hai lp dung dch c nng khc nhau th lp in tch kp xut hin cng c gi tr khc nhau. Hiu in th hnh thnh nn do s khych tn ca tng loi ion, to thnh cc pha ho tan trong mi dung dch. Hiu in th ny c gi l hiu in th pha khuych tn (UKT) c xc nh :

U KT =

RT + C 2 ln ZF + + C1

(5.12)

vi + , - l linh ng ca cc ion dng v ion m Hiu in th pha (khuych tn) tn ti trong mt thi gian rt ngn cho n khi c s phn b ng u tr li cc ion m v ion dng

92trong ton b th tch dung dch. i vi dch sinh vt c dng l mt h in ly rt phc tp, nn vic ng dng cng thc ny vo h thng sng cng gp phi rt nhiu kh khn v tn ti nhiu i phn t ion ho khc nhau.

2. in th mng* Cn bng Gibbs - Donnan Trong chng trc trnh by v trng thi cn bng GibbsDonnan, nn phn ny ta ch nhc n mt s tnh cht c lin quan v hin tng in m thi, l: -Trong c th ngi v ng vt c cc protein (R+) dng mui, n l cc i phn t khng lt c qua mng. Mc du cc phn t ny khng qua c mng nhng n ng mt vai tr ht sc quan trng, l lm nh hng nhiu n tc dng ca p sut thm thu ln mng. -Do s phn phi tr li cc ion khi trng thi cn bng ng c hnh thnh, nn hai pha mng c s chnh lch nng cc ion (c kh nng khuych tn c) qua mng. - Mt s ion khc cn li m khng c kh nng chuyn dch t pha ny n pha kia c, th s to thnh mt s cn bng c bit. chnh l cn bng Donnan. Thc nghim cho thy cn bng Gibbs-Donnan khng nhng ph thuc vo bn cht dung dch, tnh thm chn lc ion, kch thc ca mng m cn ph thuc nhiu vo loi in tch ca cc ion trong h sinh vt.

hiu r bn cht s phn b cc loi ion trong cn bng trn, ta kho st th nghim di y: Dung dch Protein cho vo bnh th nht RCl, c ion R+ l cc proteine mang in tch dng c kch thc ln khng qua c mng ngn cch gia hai bnh. Bnh th 2 cha dung dch mui NaCl, cc ion Na+ v Cl- c th d dng qua mng. Bnh th nht c nng C1, bnh th hai c nng C2. trng thi ban u cc ion c phn b trong mi bnh nh sau: [R+]1 = [Cl-]1 = C1 [Na+]2 = [Cl-]2 = C2 (5.13) C th biu din s phn b cc loi ion hai bnh (1) v (2) cha dung dch RCl v NaCl c nng ban u l (C1) v (C2) . Hai bnh (1) v (2) ngn cch nhau bng mt mng gia nh m hnh di y:

93

(RCl) (NaCl)R+

(C1) (C2) (C1) (C2)

Na+ ClCl-

(2)

(1)

Hnh 5.2 : S phn b ion hai pha mng trong trng thi ban u.

Gi s sau mt thi gian ngn, s c mt lng ion Na+ v Cl- i qua mng vi nng X. Nh vy c s di chuyn ca Na+ v Cl- t bnh (2) sang bnh (1). By gi ta thy nng ca mi loi ion hai pha mng c phn b tr li trong qu trnh cn bng mi trng thi tip theo, c m t nh s di y:

(RCl) (NaCl)R+ + Na (C2-X) Cl(C2-X) (C1) (X) Na+ (C1+X) Cl-

(2)

(1)

Hnh 5.3 : S phn b ion hai pha mng trng thi cn bng mi trnh tnh trng tch t in tch dng v mt pha, nn s di chuyn thng xy ra mt cch ng thi vi c hai loi ion Na+ v Cl- . S dch chuyn ca cc ion ch tm dng li khi tin n trng thi cn bng : [Na+]1 [Cl-]1 = [Na+]2 [Cl-]2 (5.14)

94Cn bng mi ny c gi l trng thi cn bng Donnan (Donnan equilibrium). c im ca trng thi ny l khi cn bng din ra, trong mi ngn u c s trung ho v in:[R+]1 + [Na+]1 = [Cl-]1 [Na+]2 = [Cl-]2 (5.15)

trng thi sau l trng thi m h tin ti s cn bng ng. Khi trng thi cn bng xy ra, nu thay cc gi tr (5.15) vo phng trnh (5.14) ta c: (C2 - X )2 = (C1 + X ) X2 C2 X = C1 + 2C 2

(5.16)

Kho st trong cc trng hp c bit, t phng trnh trn ta *Nu nng ion R+ hay Cl- lc u rt b (C1 >C2 ) so vi mi trng bn ngoi, th c cht khuych tn qua mng xem nh khng ng k: X 0 (5.18) *Nu nng dung dch ho tan gia hai mi trng cn bng nhau (C1 = C2 ) th : X = C2 / 3 (5.19) T cc trng hp trn ta c nhn xt l: Khi t bo tip xc vi dung dch cht in ly c cng loi v vi gc proteine l i phn t ion chnh ca t bo, th trong tt c cc trng hp u c mt lng cht nht nh i vo t bo. Di nh hng ca qu trnh vn chuyn lm cho p sut thm thu pha bn trong ca t bo lun lun c gi tr ln hn so vi mi trng xung quanh.* Hiu in th trng thi cn bng Gibbs - Donnan th gia hai pha mng lun lun tn ti mt s chnh lch in th. Hiu in th xut hin l do c s

thy:

95phn b khng ng u cc ion trng thi cn bng Donnan. Hiu in th c gi l hiu in th mng (Um). Theo th d trn ta thy, khi cn bng din ra th phng trnh cn bng Gibbs -Donnan cho ta: [Na+]1 [Cl-]1 = [Na+]2 [Cl-]2 C th vit li theo t s nng ion nh sau: [Cl ]1 [ Na + ] 2 = [Cl ] 2 [ Na + ]1(5.20)

Lc , phng trnh hiu s in th in ho ca Nernst c vit li cho hiu in th mng l:

RT [ Na + ] 2 RT [Cl ]1 Um = = ln ln ZF [ Na + ]1 ZF [Cl ] 2

(5.21)

III. in th tnh.Trong c th ng vt, trn cc t bo, m sng thng xut hin v tn ti nhiu loi in th khc nhau. Cc loi in th ny c cng ngun gc nh nhau nhng tu theo nguyn nhn xut hin, phng php o c v iu kin th nghim m ta c th phn chia ra thnh nhiu loi c tn gi khc nhau. l cc loi in th c bn nh in th ngh, in th tn thng, in th hot ng, in th ti ch. in th tnh hay cn gi l in th ngh. l in th c trng cho trng thi sinh l bnh thng ca i tng sinh vt. Ni cch khc, in th ny cng c trng cho tnh cht in ca h thng sng trng thi trao i cht bnh thng. in th tnh chnh l hiu in th bnh thng tn ti hai pha mng, c xc nh bng cch ghi o s chnh lch hiu th gia t bo cht v dch ngoi bo. C th tin hnh th nghim nh di y.1. Th nghim. kho st s bin i dng in v o hiu in th mng ca mt t bo (m sng hay mt si thn kinh...) no , thng thng ta hay s dng phng php ghi o vi in cc ni bo. Th nghim c tin hnh nh hnh 5.4 (a,b,c) di y:

96

(b (c ) ) Hnh 5.4: Ghi o in th ngh. a) t hai in cc pha ngoi mng sinh hc. b) t mt in cc bn ngoi v mt vi in cc xuyn qua mng. c) Cm hai vi in cc xuyn qua mng. Ghi o bng cch t hai in cc trn b mt si thn kinh, ta thy kim in k ng h o dng in khng lch khi im khng. iu chng t khng c s chnh lch in th gia chng (hnh 5.4a). Nu t mt in cc pha bn ngoi mng v mt vi in cc cm xuyn qua mng, ta thy gia hai in cc ny c xut hin mt hiu in th (hnh 5.4b). Cn khi chc c hai vi in cc xuyn qua mng th ta cng thy kim in k vn ch gi tr khng. iu chng t gia hai in cc khng c mt s chnh lch in th no. (Hnh 5.4c). Kt qu th nghim trn cho thy: Gia mt ngoi t bo khng b tn thng v mi trng bn ngoi khng c s chnh lch in th. Ngc li gia phn bn trong t bo v mi trng bn ngoi lun lun tn ti mt hiu in th no . S chnh lch in th ny c gi l in th ngh hay in th tnh ca mng (Resting membrane potential). 2. c im. in th ngh c hai c im nh sau: - Mt trong t bo sng lun lun c gi tr in th m so vi mt bn ngoi. Ni cch khc chiu in th ngh l khng i. - Bnh thng in th ngh c gi tr in th bin i rt chm theo thi gian. Bng cc phng php v k thut ghi o tt, ta c th duy tr dng in ny trong mt thi gian di. ln in th gim chm theo thi gian. Gi tr ny ch gim i khi chc nng ca t bo, hay ca si c bt u xut hin.

(a )

973.Cc yu t nh hng n in th ngh. in th ngh c trng cho trng thi sinh l bnh thng ca h thng sng. Nu thay i trng thi sinh l s lin quan n trng thi chc nng ca h. Do bt k yu t no lm nh hng n qu trnh trao i cht bnh thng ca n cng u nh hng n in th ngh ca h, chng hn nh: - Di tc dng ca dng in bn ngoi. - Gi tr in th b thay i khi lm thay i thnh phn ion ca mi trng. - S tc ng ca mt s c t ln h thng sng cng lm bin i nhanh in th mng. - Khi thay i lng oxy trong mi trng cng s lin quan n qu trnh h hp ca m, c..., do s lm nh hng n in th ngh. - ...

cc loi t bo khc nhau th in th ngh cng c gi tr khc nhau. Gi tr ny thay i trong khong t -10mV n -100mV. S chnh lch in th tn ti gia cc phn khc nhau trong mt h sinh vt cng l mt trong nhng yu t c trng cho c th sng.

IV. in th tn thng.in th tn thng l hiu in th xut hin do s chnh lch in th gia vng b tn thng v vng khng b tn thng. S tn thng xy ra c th do nhiu nguyn nhn khc nhau (nh di tc ng c hc, nhit, in, hoc ho hc ...) u lm xut hin s chnh lch in th. Loi in th ny c cng dng nh nhau trn cc i tng sinh vt. c trng c bn ca in th tn thng l: - Gi tr ca hiu in th gim dn v bin i chm theo thi gian. - in th tn thng ph thuc nhiu vo iu kin kho st v phng php ghi o. - ln in th b nh hng nhiu tu thuc vo iu kin sinh l ca cc i tng nghin cu.

1. i tng ng vt.Thc nghim cho thy rng, trng thi sinh l bnh thng th cc thnh phn ion mt trong mng t bo (m, c...) v pha bn ngoi c s phn b n nh. Cn gia cc v tr khc nhau mi trng bn ngoi khng b tn thng so vi mi trng xung quanh s khng c s

98chnh lch no v in th. Ni cch khc, trng thi sinh l bnh thng ta thy c s phn b in tch ban u hai pha mng sinh hc. Nu khi cc t bo (m) b tn thng, s lm nh hng n qu trnh vn chuyn cht, m c th l s trao i cc cht qua mng t bo. Ni tm li, s tn thng i tng sng m c th nh t bo (m, c,..) lm thay i trng thi chc nng ca t bo hay s lm thay i trng thi sinh l bnh thng ca cc i tng nghin cu.

2. i tng thc vt.Kho st tnh cht in trn i tng thc vt cng cho thy c nhiu im tng t nh ng vt, l: - C s chnh lch in th gia vng b tn thng v vng khng b tn thng. - in th tn thng c gi tr m. - in th ny tn ti trong mt khong thi gian ngn. - Gi tr in th gim nhanh theo thi gian v tu thuc vo iu kin th nghim, ph thuc vo khong cch gia cc vng kho st. - Kh nng xut hin in th ny ch khu tr ti vng b thng tn.

3. Cc yu t nh hng.Thc nghim chng t rng, cc yu t no lm nh hng n qu trnh trao i cht bnh thng ca t bo v m u lm thay i gi tr in th tn thng nh: - nh hng ca nhit mi trng. - Thay i thnh phn mi trng, nht l i vi Oxy lin quan nhiu trong qu trnh trao i cht. - S tc ng ca cc trng lc in bn ngoi (in trng, t trng..) lin quan n s chuyn dch ca cc ion qua mng. - S tc ng ca cc c t vo mi trng c lin quan n s thay i iu kin sinh l bnh thng.

V. in th hot ng.in th hot ng l s dao ng nhanh ca in th mng. Dao ng in mng xut hin trong cc t bo thn kinh, c, v mt s t bo khc khi c sng hng phn truyn qua. Do dng in lm xut hin in th ny cn c gi l dng in hng phn. Tt c t bo sng u c c tnh d b kch thch, tc l c kh nng chuyn t iu kin sinh l bnh thng trng thi tnh sang trng thi hot ho. Di nh hng

99ca tc nhn kch thch no , t bo s d dng b thay i tnh cht ho l ca mng. Khi c sng hng phn truyn n, du hiu in tch hai pha mng t bo b o ngc hn li so vi gi tr in th ngh lc ban u. Hiu in th ny xut hin l do c s chnh lch v gi tr in th gia hai pha ca mng. Lc ny gi tr ca in th mt bn ngoi s m hn so vi in th mt bn trong ca n. xc nh in th hot ng, thng thng ta s dng cc k thut ghi o vi in cc ni bo. C th tin hnh kho st s xut hin in th hot ng bng hai phng php nh di y:

1. Phng php 2 pha.C th tin hnh kho st trn si thn kinh c kch thch ti v tr (1), hai in cc t ti hai v tr (2) v (3) trn mt si thn kinh. Theo di s bin i gi tr in th ca chng qua mt in k G nhy ni gia hai in cc nh hnh (5.5):U = 0 (a) (b) U = 60V

(1) (2) (c)

(3) U = 0 (d)

(1) (2)

(3) U = 60V

(1) (2)

(3)

(1) (2)

(3)

Hnh 5.5: Ghi o in th hot ng hai pha a) Kch thch ti v tr (1). b) Sng hng phn truyn n v tr (2). c) Sng hng phn nm gia v tr (2) v (3).

d) Sng hng phn truyn n v tr (3).

100Nu dng mt tc nhn no kch thch si thn kinh ti v tr (1); th theo quan nim c in s c mt sng hng phn mang in tch m truyn dc theo si thn kinh. - Khi sng kch thch lan truyn n v tr (2) th gia hai in cc t ti v tr (2) v (3) s xut hin mt gi tr hiu in th U no , khong 60 mV (hnh 5.5b) - Sng kch thch lan dn v v tr (3) th hiu in th ny gim dn v tin gn n gi tr in p bng khng (U = O mV) khi sng hng phn trong vng gia v tr (2) v (3). Khi sng kch thch tin ti v tr (3) th hiu in th gia hai cc bin i v pha in th m (hnh 5.5c). - Khi sng kch thch truyn n v tr (3) th in th m ny t gi tr in p ti hn (Uth) (Uth = -60 mV) nh hnh 5.5d. - Khi sng ri khi v tr (3) th hiu in th gia hai in cc tr v li gi tr U bng khng nh ban u. Theo di c tuyn bin i theo thi gian ta c dng in th hot ng nh (hnh 5.6):U Ub

0 Ud

t

Hnh 5.6: c tuyn bin i ca in th hot ng hai pha theo thi gian

2. Phng php mt pha.* Phng php ghi o. Phng php mt pha l phng php ghi o in th hot ng bng cch dng mt in cc t ti v tr (2) v mt vi in cc khc cm xuyn qua mng t v tr (3). Sau kch thch ti v tr (1) v kho st sng hng phn kch thch truyn dc theo i tng nghin cu (t bo, si c, ...) nh (hnh 5.7):

101U = 60V (a) (b) U = 0

(1) (2) (c)

(3)

U = 60V

(1) (2)

(3)

(1) (2)

(3)

Hnh 5.7: S ghi o in th hot ng mt pha trn si thn kinh. a) Kch thch ti v tr (1). b) Sng kch thch truyn n v tr (2). c) Sng kch thch truyn n v tr (3).

- Khi cha kch thch, gia in cc (2) v vi in cc (3), c xut hin mt s chnh lch in th, l in th ngh ca si thn kinh. in th ny c gi tr khong - 60mV n - 100mV. - Khi kch thch ti v tr (1), sng hng phn lan truyn n v tr (2) th hiu in th ny tng dn ln t gi tr in th m n gi tr khng. Hiu th ny tng nhanh v t ti gi tr cao nht ti in th khng (U = 0) khi sng hng phn n v tr (2) (hnh 5.7b). - Khi sng hng phn truyn t v tr (2) n (3) th hiu in th hot ng mt pha gim tr li v in th ngh nh lc u (-80mV) Vy in th hot ng mt pha chnh l s bin i nhanh chng ca in th ngh di tc dng ca mt tc nhn kch thch no . Dng in th hot ng mt pha bin i theo thi gian trong th nghim trn mt si thn kinh, c biu din nh (hnh 5.8):

102

U(mV) 0 -40

-80 a c b Hnh 5.8: c tuyn bin i ca in th hot ng mt pha theo thi gian. * Cc giai on hnh thnh. Khong vi thp nin tr li y, nh cc thit b ghi o hin i, in th hot ng mt pha c biu din mt cch t m, chnh xc hn. S hnh thnh in th hot ng c chia ra lm nhiu giai on nh (hnh 5.9). o trn si thn trc khng l ca thn kinh c mc, ta thy in th ngh c gi tr khong -60mV phn nh ca in th hot ng c gi tr khong 50mV. U (mV) 60 40 20 Cm vi O -20 -40 -60 -80 Hnh 5.9: Cc giai on bin i ca in th hot ng A C Din cc Kch thch

t

B

A

B

103in th hot ng c cc giai on bin i l: + Giai on kh cc (Depolarization), on AA. Lc ny hiu in th hai pha mng bin i t gi tr in th ngh (U ngh) n im c in th bng khng (U = 0 mV) + Giai on qu kh cc, on ABB. Trong giai on ny hiu in th hai pha mng vt qu gi tr in th khng, tip tc bin i v pha c in th dng. + Giai on phn cc li (Repolarization), on BC. l giai on m hiu in th hai pha mng gim tr li v gi tr in th ngh. + Giai on qu phn cc, on CD. Giai on ny ng vi lc hiu in th hai pha mng c gi tr m hn in th ngh. Nu kch thch c cng ln ta nhn thy rng: - Trong thi gian xut hin pha ln (nhnh ln) in th mng vt qu gi tr in th khng, ta thy c s o cc ca in th mng. - Trong pha xung (nhnh xung), mng c s phn cc li. in th hot ng pha ny ph thuc vo khong cch gia hai in cc v ph thuc nhiu vo tc dn truyn hng phn. Cc nghin cu ca Erlange v Gatse chng minh rng: -in th hot ng ghi c t mt thn kinh l tng cc in th lan truyn trn cc si t c cu to nn si trc thn kinh (hnh 5.10)U (mV) 100

50

0

0,4

1,20,4

2,4

t (ms)

Hnh 5.10: in th hot ng ca t c v si thn kinh.

T c tuyn trn, tc gi gii thch cho ta thy rng:-C s tng ng gia gi tr in th hot ng ghi o c trn si ca thn kinh ca

104mo (nt t) v in th hot ng xut hin ghi o c trn tng si t thn kinh tng hp nn si trc thn kinh (nt lin).

VI. Bn cht ca in th tnh v in th tn thng. gii thch v c ch, bn cht v ngun gc ca cc loi in th sinh vt, ta da vo mt s gi thuyt, cc l thuyt ion cng nh mt s cch l gii khc ca cc nh khoa hc: C nhiu quan im khc nhau gii thch v s hnh thnh in th sinh hc. Tuy nhin l thuyt m ang c nhiu nh khoa hc chp nhn v c c s vng chc hn c, l L thuyt ion mng. Theo thuyt ny, trong qu trnh hnh thnh in th sinh vt th cc ion (c bit l cc ion Na+, K+, Cl-) trong dch ni bo v bn ngoi t bo ng vai tr quyt nh. Cho n nay l thuyt ny vn chim nhiu u th trong vic gii thch hin tng in sinh vt. Da vo l thuyt trn, ta c th gii thch v s hnh thnh cc loi in th sinh vt c bn. Trc khi gii thch c ch hnh thnh in th ngh v in th tn thng, ta kho st s phn b cc loi ion chnh nh hng n hiu th mng. trng thi bnh thng, c th xc nh c gi tr in th tnh tng ng vi s phn b nng ion hai pha mng. Chng hn nh s phn b ion trong t bo c Mamalian nh (bng 5.1):Bng 5.1: Nng mt s loi ion trong t bo c Mamalian

Nng ion dch ngoi bo [ion]0 (M/cm3) Cations: Na+ 145 K+ 4 + -5 H 3,8.10 .. .. .. ion khc 5

Nng ion dch ni bo [ion]i (M/cm3) Cations Na+ 12 K+ 155 H+ 13.10-5 .. .. ..

[ion]0 / [ion]i

m (mV)

12,1 1/39 1/3,4

66 -97 -32

105Anion ClHCO3.. .. .. Ion khc Anion ClHCO3.. .. .. A-

120 27 7

4 8 155

30 3,7

-90 -32

in th

0

-90

1/30

-90

1. Nhn xt: Kho st cc thnh phn tng t nh trn nhiu i tng nghin cu khc nhau, nh thn kinh ch, tim chut cng, c xng ch .. .., ta thy c s phn b khng ng u ca cc loi ion hai pha mng. c bit, i vi cc loi ion Na+, K+, Cl- cho thy t l gia cc ion ny thng l: - Ion K+ trong t bo ln hn bn ngoi khong vi chc ln. - Ion Na+ bn ngoi ln hn bn trong rt nhiu. - Ion Cl- bn ngoi ln hn bn trong khong 30 ln. 2. L thuyt ion mng Bernstein l ngi u tin cho rng, in th tnh l kt qu ca s phn b khng u cc ion hai pha mng t bo. trng thi tnh, mng khng thm ion Na+ v Cl- m ch cho cc ion K+ lt qua. Hin tng vn chuyn cc cht xy ra, do c s phn b khng u c ba loi ion ny hai pha mng t bo. Ngoi ra mng c tnh bn thm v tnh thm ca mng i vi tng loi ion l khc nhau, l yu t c bn to nn in th tnh. in th tnh c cc gi tr khc nhau tu thuc vo i tng nghin cu.

Boyler v Conwey pht trin thm quan im ca Bernstein, bng cch chng minh cho ta thy rng mng thm ng thi i vi c ion K+ v Cl-. trng thi tnh, cc ion Na+, K+, Cl- c phn b tr li ti hai pha mng. Qu trnh vn chuyn v c ch hot ng ging nh s phn b ca cc ion trong trng thi cn bng Donnan. Do , in th tnh (US) cng c xc nh bng t s nng ca cc loi ion khuych tn qua mng. Hn na do tnh bn thm ca mng i vi tng loi ion m c s phn b li cc in tch ca chng hai pha mng. Thc nghim cho thy pha trong mng tch in m cn pha ngoi mng tch in dng.

106Khi cn bng Donnan in th tnh ca mng t bo ng vt c th xc nh bi cng thc:

US =

Trong : [K+]0 v [Cl-]0 l cc nng ion mi trng bn ngoi t bo. [K+]i v [Cl-]i l cc nng ion pha bn trong t bo. Goldmann a ra gi thuyt l: - Mng t bo c tnh ng nht v cu trc v in trng tc dng ln mng ti mi v tr l khng i. - Dung dch in ly ca cc dch sinh vt c coi nh l dung dch l tng. - Mng c tnh cht bn thm nhng khng phi hon ton tuyt i, ngha l c th cho ion ny qua cn ion khc th khng th qua c. c trng cho kh nng dch chuyn ca cc ion qua c mng nhiu hay t, ta dng i lng h s thm (P) cho tng loi ion. - Cc ion Natri cng c tham gia vo qu trnh hnh thnh nn in th tnh ny. Do cng thc v in th tnh c Goldmann xc nh li nh sau: RT PK [ K + ] 0 + PNa [ Na + ] 0 + PCl [Cl ] i US = ln (5.23) F PK [ K + ] i + PNa [ Na + ] i + PCl [Cl ] 0 Trong : PK, PNa, PCl, l h s thm i vi cc ion Kali, Natri v Clo. Nhng nghin cu gn y xc nh v s hin din ca ion Natri tham gia vo qu trnh hnh thnh in th mng. Bng phng php ng v phng x nh du cho ta kt qu nh sau: -Mng t bo thm tt i vi ion K+ , Cl- v t thm i vi ion Na+ (tc dng ion Natri vo khong 14. 10-12 mol/ cm2 /sec). T cc kt qu thc nghim, Hodgkin v Keynes khng nh mt cch chc chn rng: -Mng t bo thm i vi ion Natri, mc du t nhng khng th b qua c. Tm li thuyt ion mng chim nhiu u th v vic gii thch bn cht s hnh thnh in th tnh. Kt qu o t thc nghim hon ton ph hp vi l thuyt. Thc vy, gi tr tnh ton t l thuyt gn ng vi kt qu o c t thc t. Tuy nhin l thuyt cng rc ri v cng khng

RT [ K + ]0 RT [Cl ]i = ln ln ZF [ K + ]i ZF [Cl ]0

(5.22)

107th d dng gii thch trn nhiu i tng khi m iu kin th nghim kh phc tp. Tht vy l thuyt ion mng cho rng, cc ion u khuych tn qua mng di nh hng ca Gradient. Tuy nhin khi nghin cu trn mt s ng vt th cc ion Kali khng phi hon ton trng thi t do m c th cn lin kt vi cc cht bn trong t bo. Kt qu l thuyt v thc nghim khi kho st trn c ch theo ba phng php c xc nh nh (hnh 5.13).UM (mV) 0 -20 -40 -60 -80 100 120 mM/ l 0, 1 5 5 10 50 100 Kt qua thc nghim Tnh theo Goldmann Tnh theo Nernst

[K ]O

+

Hnh 5.13: Gi tr in th tnh ca c ch theo cc php o

Nu lm thay i nng ion Kali mi trng bn ngoi c ch th in th tnh o c c s sai khc nhau t nhiu so vi cc kt qu tnh ton t l thuyt.

VII. Bn cht ca in th hot ng.Tt c t bo sng u c c tnh d b kch thch, ngha l c nhiu kh nng chuyn t trng thi tnh sang trng thi hot ho di nh hng ca cc tc nhn. S bin i cc thng s c trng cho trng thi, thc ra l do b thay i tnh thm ca mng. Tuy nhin thut ng t bo d hng phn thng thng hay c s dng i vi cc loi t bo thn kinh, c.. ..ngha l cc i tng ny c kh nng p ng ngay di tc dng ca ngun kch thch. p ng thay i do kch thch thng c biu hin bng s xut hin

108mt in th hot ng. V bn cht v c ch hnh thnh in th kh phc tp, da vo l thuyt ion mng ta mi c th gii thch mt cch hp l nht.1. S kh cc v ti phn cc. - Ta bit rng trng thi ngh, c s phn b cc loi ion hai pha mng lm cho bn trong mng tch in m v pha bn ngoi mng tch in dng. in th chnh l gi tr cu in th ngh ca t bo trng thi bnh thng (hnh 5.14a). - Khi mng t bo c kch thch th t bo trng thi hng phn. Theo Bernstein v mt s tc gi khc cho rng mng t bo thm vi mt s loi ion no . Khi tnh thm ca mng i vi nhng ion Na+ t nhin tng, th nhiu ion Na+ thm t ngoi vo pha trong mng, mang lng in tch dng vo pha trong. Trng thi ngh bnh thng bin mt, pha trong mng c gi tr in th dng hn so vi gi tr in th m lc bnh thng. S phn cc tr li trong lc ny c gi l in th bin i (reversal potention) v giai on ny c gi l giai on kh cc (hnh 5.14b).

a) Trng thi ngh

b)Trng thi kh cc

c)Trng thi phn cc li

Hnh 5.14: S phn b in tch cc giai on ca in th hot ng

- Ngay lp tc sau khi c s kh cc khong mt phn trm giy (milisecond), mng hu nh thm hon ton i vi cc ion Na+. Do mt cn bng ion th bm Na+ v K+ xut hin a ion Na+ quay tr li. V vy to thnh to s cn bng mi ca cc ion gia hai pha mng. S phn cc lc ca mng ging nh s phn b ion lc ban u, nn giai on ny c gi l giai on phn cc li (hnh 5.14c).

109

2. S thay i tnh thm ca mng. Theo Hodgkin hay Huxley th gia in th mng v nhng ion i qua mng c mi lin quan vi nhau. Cc ion i qua mng tu thuc vo tnh thm ca mng, nn da vo tnh cht ny i vi cc loi ion hoc ni mt cch khc l c th da vo s thay i v dn in bi cc ion. in dn thay i lm in th mng (Um) cng thay i theo khi i tng sinh vt trong trng thi hot ng.

* Kho st th biu din dn in ca ion Na+ v ion K+ , tng ng nh khi kho st c trng tnh thm ca mng, ta c kt qu nh (hnh 5.15). Ta thy mi s kh cc mng u lm tng tnh thm ca mng i vi ion Na+ . Khi s kh cc t ti mt gi tr no (ngng kh cc) th tnh thm ca mng i vi ion Na+ t nhin tng vt ln. Tng ng khi dn in ca mng i vi cc ion Na+ cng tng ln hng ngn ln. S gia tng ny ch tm thi trong sut thi gian rt ngn, khong mt phn nh ca mili giy (ms). Cn i vi ion K+ ta thy lc mng trng thi ngh, dn in ca ion K+ ln gp khong 100 ln i vi ion Na+. Nhng trong giai on u hnh thnh in th hot ng, dn ion K+ ch tng ln khong 30 ln n 40 ln trong khi dn i vi ion Na+ li tng ln hng ngn ln. V tnh thm ion K+ xy ra tr hn v ko di trong mt thi gian lu hn so vi s gia tng ca ion Na+. Cc giai on bin i trong in th hot ng thng khng ng b nhau, nn ta phi kho st in th hot ng da trn s thay i v t s thm gia chng, ngha l da vo gi tr: PNa / PK Hoc tng ng vi s thay i v tnh thm, ta cng c th kho st s bin i ph thuc theo t s dn in: Na / K

11060 40 in th 10 0 10 in th hoat ng 20 0 20 40 60 80 100

1

0, 1 0,0 1 0,00 1 10 0 10

Ty s dn

PNa PK

1

0, 1 0,0 1 0,00 1

1, 5 Hnh 5.15: Bin i dn Na+, K+ mng t bo tng ng vi s

0

0, 5

1

ms

hnh thnh in th hot ng (theo Hodgkim v Huxley).

111+ S pht trin ca giai on kh cc: Trong giai on ny, th dn ca ion Na+ tng ln hng ngn ln, ng thi khi dn ca ion K+ thay i khng ng k. Kt qu o c cho thy c s phn cc ngc tr li so vi ban u. Lc ny dn ca ion Na+ ln hn dn ca ion K+ khong 30 ln. Ni cch khc tnh thm ca mng i vi ion Na+ by gi ln hn nhiu so vi ion K+. V vy in th mng trong giai on ny c xc nh gn nh hon ton bi s khuych tn ca ion Na+ hn l do bi cc ion K+. Da vo cng thc tnh in th ion, ta c.

RT [ Na + ] 0 (3.24) ln U Na = ZF [ Na + ]i Vy khi t bo trng thi hng phn, mng t bo b kh i dn n lm in th ngh gim. S gim in th ngh lm cho cc ion Na+ chuyn ng theo hng gradient nng vo t bo mt cch mnh m hn trc. Dng in do cc ion ny to ra cng b kh cc mnh, chnh l giai on qu kh cc ca mng (hnh 5.14b).+ Giai on phn cc li: dn in ca ion Na+ ln hn dn in ca ion K+ ch trong khong thi gian vi mili giy (milisecond), nn giai on tip theo sau ta thy mng ging nh tr nn khng hot ng na. Tnh thm ca mng i vi ion Na+ li b c ch, cn tnh thm ca mng i vi ion K+ li tng ln. in th mng lc ny chu s nh hng nhiu bi ion K+. Tnh thm K+ gia tng tr nhng ko di lu hn, lng ion K+ khuych tn t trong ra ngoi t bo qua mng theo hng gradient nng mt cch mnh m lm cho mt trong t bo c gi tr m hn mt bn ngoi. Qu trnh pht trin theo khuynh hng tin ti cn bng, v in th lc ny c xc nh ch yu bi s tham gia ca ion K+, mng b tng nhanh qu trnh phn cc tr li hai pha mng. Hu qu ca giai on trn kt hp cng vi s hot ng ca bm + ion Na - K+ trong giai on phn cc li a mng tr v in th ngh ban u. Mng cng c gi tr in th m hn nhiu. ng thi vi s pht trin ca ion K+ lc ny khuych tn qua mng mt cch hon ton, lm cho mng c s phn cc nhiu hn. Do in th pha trong mng lc ny c gi tr m hn in th ngh bnh thng. Giai on hnh thnh ca in th hot ng ny chnh l giai on qu phn cc ca mng t bo.

112Da vo cng thc Nernst xc nh gi tr in th hnh thnh trong giai on phn cc li (ch yu do K+ to nn), ta c:

UK =

RT [ K + ] 0 ln ZF [ K + ] i

(3.25)

VIII. p dng phng php o in th trong chn on v iu tr bnh.Hin tng in ng, ngy nay c s dng mt cch ph bin trong Y-Sinh hc. Da vo cc quy lut t nhin, cc phng php vt l, l thuyt l sinh nghin cu h thng sng mt cch khoa hc. Tu theo ngun tc nhn c bn cht l ho khc nhau m lm thay i qu trnh phn b in tch hai pha mng t bo, thay i tnh thm, thay i p sut thm thu ...ca i tng sinh vt. Ni chung, cc yu t tc ng trn u lm thay i trng thi nhit ng, do lm nh hng qu trnh trao i cht, dn n s thay i trng thi chc nng ca cc c quan trong h thng sng. Tuy nhin di nh hng ca tc nhn l cc loi dng in ngoi c th xy ra nhiu tc dng khc nhau v nhiu mt ln c th ngi v ng vt. Chng hn nh: - Di tc dng ca dng in mt chiu, ta thy cc in tch-ion chu tc ng ca lc in trng. S phn b tr li cc in tch trn i tng kho st lm nh hng n qu trnh trao i cht cng nh trao i ion qua mng. Ni mt cch khi qut l c s thay i tnh cht ho l, cc ch tiu sinh hc ca nhng i tng sinh vt. Dng in mt chiu c ng dng nhiu trong Y hc lm ngun tc nhn kch thch c quan th cm thn kinh, o sut Galvanotonus, hoc s dng iu tr trong ion liu php (th nghim LeDue) - Di tc dng ca dao ng in t : Dng in cao tn c s dng rng ri trong y hc iu tr bnh. Da trn cc hiu ng nhit, hiu ng in - c khng mang bn cht nhit hoc hiu ng kch thch. Ngy nay trong vt l tr liu ng dng loi dng in ny iu tr cc bnh v xng, khp, kch thch c, thn kinh v phc hi chc nng... - Di tc dng ca ngun kch thch dng xung dng in hay xung in p vi cng nh v bin i trong khong thi gian ngn, tn s thch hp to cc kch thch in. Trong Vt l tr liu, ngi ta thng dng xung kch thch iu tr cc

113bnh thn kinh, mch mu, ri lon chuyn ho...Trong lm sng, ngi ta cn s dng ngun tc nhn kch thch to cc kch thch in, sc in (electric shock). thm d chc nng hot ng ca c, thn kinh, trong nghin cu y hc ngi ta thng a ra cc thng s nh thi tr (Chronaxy), thi gian p ng hoc xc nh v ngng kch thch (Rhobase) ca cc i tng nghin cu. - Da trn hin tng in ng lm thay i tnh cht ho l ca cc t bo, thay i qu trnh trao i ion cng nh trao i cht trong h thng sng. Tm li, ngy nay vic ng dng v hin tng in ng trong Y-Sinh c s dng kh ph bin v ng vai tr thit yu khng th thiu trong vic chn on, thm d v iu tr bnh.CU HI N TP CHNG 5 1. Gii thch c ch hnh thnh in th mng v hiu in th pha? 2. Gii thch c ch hnh thnh in th in cc? 3. Gii thch s xut hin trng thi cn bng Gibbs Donnan? 4. Gii thch s hnh thnh in th tnh v cc yu t nh hng ti in th tnh? 5. Gii thch s hnh thnh in th tn thng v cc yu t nh hng ti in th tn thng? 6. Gii thch s hnh thnh in th hat ng dng 2 pha? 7. Gii thch s hnh thnh in th hat ng dng 1 pha? 8. Bn cht ca in th tnh v in th tn thng? 9. L thuyt ion mng? 10. Bn cht s hnh thnh in th hat ng? 11. ng dng phng php o in th trong chn an v iu tr bnh?

114

Chng 6

IN NG HCI. Cc hin tng in ng hc.Cc t bo, cc t chc sng, cc c quan ca h thng sng l mt h keo d th phc tp bao gm nhiu pha khc nhau. Do tc ng ca in trng ngoi khng i lm xut hin s chuyn ng tng i gia cc pha trong h, ngc li nu cc pha c thnh phn cht ho tan khc nhau th di chuyn ng c hc ca cc ion cng s to nn trong h mt hiu in th no . Cc hin tng in xut hin trong qu trnh ny c gi l cc hin tng in ng v chng c phn thnh cc loi sau y nh: in di, in thm, in th chy, in th lng... Nm 1809 Reis l ngi u tin pht hin thy cc hin tng in ng hc khi nghin cu chuyn ng ca cc ht t st di tc dng ca dng in mt chiu. Qua th nghim ca mnh, Reis thy rng cc ht keo mang in tch cng c kh nng vn chuyn c trong in trng, ng thi cng vi qu trnh bin i th mi trng phn tn cng s chuyn ng theo. Nh vy trong th nghim trn, Reis pht hin ra hai hin tng c bit quan trng, l khi cc ht tch in s dch chuyn di tc dng ca in trng ngoi, l hin tng in di hay in thm.

1. in thm.L s chuyn ng ca cc mi trng phn tn ti pha in cc cng du vi din tch b mt ca pha phn tn. Da vo s dch chuyn ca cc ion trong in trng ta d dng phn tch mt hn hp polime sinh vt bng hin tng in thm hay in chuyn trn bng ghi. Dch sinh vt l cc dung dch in ly c nhiu thnh phn, vi ho tan khc nhau. Do di tc dng ca in trng ngoi, cc dung dch long v cc i phn t ion ho c tc dch chuyn khc nhau. Hin tng in thm d dng phn bit c l s chuyn ng ca dng cht lng, trong khi dng chuyn ng ca cc ht (phn t v cc i phn t ion ho) l do hin tng in di to nn. Qa trnh in thm c th xy ra trong nhiu trng hp v qua cc t chc sinh hc khc nhau, chng hn nh da ch, mng t bo, thnh ng mch, mao qun, vch mao mch...

115Da vo gi tr hiu s in th in ho ca lp in tch kp ( ), Smolukhovski a ra cng thc xc nh tc chuyn ng tng i ca cc ht gia hai lp l:

v=Trong :

E 4

(6.1)

l hng s in mi ca mi trng l h s ni ma st ca dung dch, E l cng in trng tnh ( v E c tnh theo n v mV )

2. in th chy.in th chy xut hin khi cht lng chuyn ng di tc dng ca p sut thu tnh qua cc mao qun, hoc cc l nh ca mng, m thnh l c mang in tch. Hin tng dch chuyn ca cc cht lm xut hin in th chy theo chiu hng ngc li so vi hin tng in thm. y s chuyn ng ca mi trng phn tn s to nn mt hiu in th trong bn thn h. thy r iu ny, ta tin hnh th nghim nh sau: -Dng mt bnh thu tinh hai ngn cha dung dch sinh l, ngn cch nhau bng mt mng da ch. y dung dch sinh vt l cc mi trng phn tn, cn mng da ch ng vai tr to ra cc pha phn tn. Hin tng xy ra c biu din nh (hnh 6.1) di y:

Hnh 6.1: in th chy xut hin qua cc l mng. Nu tng p sut na bnh pha bn tri th cht lng s chuyn ng v bn phi bnh, ng thi gia hai pha ca bnh s xut hin mt hiu in th. in th mi hnh thnh chnh l gi tr ca in th chy. S chnh lch in th hai bnh l hiu in

116th o c hai pha ca mng ngn cch hai dung dch. Nguyn nhn xut hin hiu in th trn l do trng thi cn bng tnh in b ph v. in th chy cng d dng kho st c khi tin hnh th nghim vi cc dch sinh vt v cho chuyn ng qua cc mng xp hay cc mng bn thm. Do , da vo hin tng trnh by trn, thng thng ngi ta hay s dng cc mng lc o xp ca cc i tng nghin cu l cc t chc sinh vt.

3. in th lng.in th lng l hiu s in th xut hin gia lp trn v lp di ca dung dch a pha trong qu trnh lng cc ht mang in ca cc pha phn tn di tc dng ca trng lc. Bn cht ca hin tng lm xut hin loi in th ny khc hn so vi hin tng lm xut hin hiu in th trong qu trnh in di. C th kho st hin tng xy ra trong qu trnh lng ca mu chng hn (hnh 6.2). Mu l mt dung dch keo, cc thnh phn hu hnh ca mu (hng cu, bch cu) c trng lng ring ln hn huyt thanh nn s lng xung y bnh. Trong qu trnh lng mu lm xut hin mt s chnh lch in th gia lp trn v lp di ca dch sinh vt. in th xut hin trong trng hp ny chnh l in th lng.

Hnh 6.2: in th lng trong qu trnh lng mu

Qu trnh chuyn ng lm xut hin in th lng ca mu c th gii thch nh sau: Hin tng lng mu lm cc ion dng tch ra khi s chuyn ng ca cc thnh phn hu hnh. Kt qu thc nghim cho ta thy r iu , cc lp di ca dung dch c in th m v mang in lng m, cn cc lp trn mang in th dng v c nhiu in tch dng.

117Tt c cc hin tng in ng u lin quan n s xut hin hiu in th gia pha phn tn v mi trng phn tn. in th ny cn c gi l in th in ng hoc Zeta in th ( - in th). Th in ng ch xut hin do qu trnh chuyn ng ca cc pha trong h d th. Hiu in th ny c hnh thnh trn ranh gii gia mng dung mi cc mng (gi l lp hp ph) trn b mt ca ht v ton b phn cht lng cn li ca dung dch.4. in di. in di l phng php phn tch da trn s dch chuyn cc in tch, phn t nhim in di tc dng ca trng lc in khng i. Thng thng, ta s dng phng php in di tch chit cc thnh phn albumin, globulin trong huyt thanh; in di protin (phc cht lipid vi protin); in di glucoprotin (xc nh cc thnh phn ca protin v glucid trong huyt thanh). Nm 1937, Tiselius l ngi u tin a ra phng php in di trong mi trng t do kho st s hin din ca p