Ky Yeu Hoi Thao Dbbb Mon Sinh Hoc-2010

7/31/2019 Ky Yeu Hoi Thao Dbbb Mon Sinh Hoc-2010

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HI CC TRNG THPT CHUYN

KHU VC DUYN HI V NG BNG BC B

K YUHI THO KHOA HC, LN TH III

MN SINH HC(TI LIU LU HNH NI B)


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H NAM, THNG 11 NM 2010

MC LC

STT NI DUNG TRANG

1 Li ni u 5

2Chng 1. C s vt cht ca hin tng di truyn v bin d

Trn Hong Xun - Trng THPT chuyn Bc Ninh

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3

Chng 2. Tnh quy lut ca hin tng di truyn

A. C s l thuyt

Nguyn Th Hng - THPT chuyn Bin Ho, tnh H Nam

B. Bi tp qui lut di truyn

Nguyn Th Hi - Trng THPT chuyn Thi Bnh

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71

4Chng 3. Di truyn hc qun th

L Huy Chin - Trng THPT chuyn Nguyn Tri, tnh Hi Dng

85

5Chng 4. ng dng di truyn hc

Lng Th Lin -Trng THPT chuyn Trn Ph, Tp. Hi Phng

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6Chng 5. Di truyn hc ngi

L Th Thu Hin - Trng THPT chuyn Nguyn Tri, tnh Hi Dng

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7Chuyn : Tin ho

Chng 1. Bng chng tin ha

Nguyn Th Nm Trng THPT chuyn Hng Yn

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8Chng 2. Nguyn nhn v c ch tin ha

Lu Th Yn - Trng THPT chuyn L Hng Phong , Tnh Nam nh

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9Chng 3. S pht sinh v pht trin ca s sng trn tri t

Phm Th Vit Hoa - Trng THPT chuyn Lng Vn Tu, tnh Ninh Bnh

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10Chuyn : Sinh l ng vt

A. Sinh L ng vt : Sinh l tun hon

V Cng Ngha - Trng THPT chuyn H Long, tnh Qung Ninh

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B. Sinh l Ni tit

T Th Thu Hin - Trng THPT chuyn Vnh Phc

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LI NI U

Th k XXI l th k casinh hc. Cha bao gi Sinh hcli pht trin mnh m nhnhng nm cui th k XX vu th k XXI ny. Cc s liuthng k cho thy c vi nm,kin thc v sinh hc li tnggp i . Vy lm th no hcsinh c th nm bt c nhngkin thc rt mi v rt kh casinh hc ngy nay ? c bit lnhng ni dung kin thc dnh

cho hc sinh cc trng Chuyn. gp phn vo vic

gii quyt nhng kh khn trn,cc trng Chuyn cng tchc cc cuc hi tho v t


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DI TRUYN HC

CHUYN DI TRUYN HC

chuyn th ti liu cng gp phnquan trng. gip cho thy vtr cc trng Chuyn c cmt s ti liu tham kho hu chtrong vic bi dng, rn luynnng khiu cho hc sinh gii,hc sinh c ti liu t nng

cao kin thc, thc hnh luyntp, hi cc trng Chuyn khuvc ng bng duyn hi Bc B ra mt cun k yu hi thoca cc trng Chuyn khu vcng bng duyn hi Bc Bnm 2010 vi ba chuyn chnh l: Di truyn, Tin ha vSinh l ng vt. Cc chuyn ny do cc thy c gio ca cctrng Chuyn tham gia sontho; l: Trng THPTchuyn Bc Ninh, TrngTHPT chuyn Bin Ho (H

Nam), Trng THPT chuynNguyn Tri (Hi Dng),Trng chuyn Trn Ph (Hi

Phng), Trng THPT chuynHng Yn, Trng THPTchuyn L Hng Phong (Namnh), Trng THPT chuyn LQu n (Thi Bnh), TrngTHPT chuyn Lng Vn Ty(Ninh Bnh), Trng THPTchuyn Qung Ninh, v Trng

THPT chuyn Vnh Phc. Hyvng rng cun k yu ny sphn no gip cho nhng hcsinh yu thch mn sinh hc cctrng chuyn hc tp c hiuqu cao, khng ch khi lnh hikin thc mi m cn n tpcng c kin thc, lc t hc, to su kin thc, c th giip c mt s hin tngthng gp trong thc t cucsng v mt s cu hi trong cc

bi thi quc gia v quc t.Chng ti mong nhn c

NHM SINH HC

TRNG THPT CHUYN BIN HO, TNH H NAM


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CHUNG 1. C S VT CHT CA HIN TNG DI TRUYN V BIN D

A - C S VT CHT CA HIN TNG DI TRUYN V BIN D CP PHN T

I. CU TRC CA AXIT NUCLEIC

1. Cu trc ADN

a) Cu tao ha hc ca ADN

- ADN lun tn ti trong nhn t bo v c mt c ti th, lp th. ADN cha cc nguynt ha hc ch yu C, H, O, N v P.

- ADN l i phn t, c khi lng phn t ln, chiu di c th t ti hng trmmicromet khi lng phn t c t 4 n 8 triu, mt s c th t ti 16 triu vC.

- ADN cu to theo nguyn tc a phn, mi nucleotit c ba thnh phn, trong thnh

phn c bn l baznitric. C 4 loi nuleotit mang tn gi ca cc baznitric, trong A vG c kch thc ln, T v X c kch thc b.

- Trn mch n ca phn t ADN cc n phn lin kt vi nhau bng lin kt ho tr llin kt c hnh thnh gia ng C5H10O4 ca nucleotit ny vi phn t H3PO4 canucleotit k tip. Lin kt ho tr l lin kt rt bn m bo cho thng tin di truyn trn mimch n n nh k c khi ADN ti bn v phin m.

- T 4 loi nucleotit c th to nn tnh a dng v c th ca ADN cc loi sinh vt

bi s lng, thnh phn, trnh t phn b ca nucleotit.b) Cu trc khng gian ca ADN (M hnh Oatxn v Crick)

+ ADN l mt chui xon kp gm 2 mch n (mch polinucleotit) qun song songquanh mt trc tng tng trong khng gian theo chiu t tri sang phi (xon phi) nhmt thang dy xon: tay thang l phn t ng (C5H10O4) v axit photphoric sp xp xen knhau, mi bc thang l mt cp baznitric ng i din v lin kt vi nhau bng lin kthiro theo nguyn tc b sung (NTBS). l nguyn tc A ca mch n ny c kch thcln b sung vi T ca mch n kia c kch thc b v ni vi nhau bng 2 lin kt hiro.

G ca mch n ny c kch thc ln b sung vi X ca mch n kia c kch thc b vni vi nhau bng 3 lin kt hiro v ngc li.

+ Trong phn t ADN, do cc cp nucleotit lin kt vi nhau theo NTBS m bo chochiu rng ca chui xon kp bng 20 , khong cch gia cc bc thang trn cc chuixon bng 3,4 , phn t ADN xon theo chu k xon, mi chu k xon c 10 cp nucleotit,c chiu cao 34 .

- ADN ca mt s virut ch gm mt mch polinucleotit. ADN ca vi khun v ADNca lp th, ti th li c dng vng khp kn.

c) Tnh c trng ca phn t ADN

cu n s c c o n t n n.Xin trn trng cm n!


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+ ADN c trng bi s lng, thnh phn trnh t phn b cc nucleotit, v vy t 4 loinucleotit to nn nhiu loi phn t ADN c trng cho mi loi.

+ ADN c trng bi t l

+ ADN c trng bi s lng, thnh phn trnh t phn b cc gen trong tng nhm genlin kt.

2. Cu trc ARN

- ARN l mt a phn t c cu to t nhiu n phn.

- C 4 loi ribonucleotit to nn cc phn t ARN: Aenin, Uraxin, Xitozin, Guanin, mi nphn gm 3 thnh phn: baznitric, ng riboz (C5H10O5) v H3PO4.

- Trn phn t ARN cc ribonucleotit lin kt vi nhau bng lin kt ho tr gia ngC5H10O5 ca ribonucleotit ny vi phn t H3PO4 ca ribonucleotit k tip.

- C 3 loi ARN: rARN chim 70-80%, tARN chim 10-20%, mARN chim 5-10%.- Mi phn t mARN c khong 600 n 1500 n phn, tARN gm 80 n 100 n

phn, trong tARN ngoi 4 loi ribonucleotit k trn cn c 1 s bin dng ca cc baznitric(trn tARN c nhng on xon ging cu trc ADN, ti cc ribonucleotit lin kt vinhau theo NTBS (A-U, G-X). C nhng on khng lin kt c vi nhau theo NTBS vcha nhng bin dng ca cc baznitric, nhng on ny to thnh nhng thu trn. Nhcch cu to nh vy nn mi tARN c hai b phn quan trng: b ba i m v on mangaxit amin c tn cng l adenin.

- Phn t rARN c dng mch n, hoc qun li tng t tARN trong c ti 70% sribonucleotit lin kt vi nhau theo nguyn tc b sung. Trong t bo nhn s c 3 loirARN (23S, 5S v 16S); sinh vt nhn tht c ti 6 loi rARN (28S, 23S, 18S, 16S, 5,8S,5S) vi s ribonucleoti tt 120 n 5000/1 phn t.

- Ngoi ba loi ARN tn ti trong cc loi sinh vt m vt cht di truyn l ADN th nhng loi virut vt cht di truyn l ARN th ARN ca chng c dng mch n, mt viloi c ARN 2 mch.

II. CU TRC PROTEIN

1. Cu trc ho hc

- L hp cht hu c gm 4 nguyn t c bn C, H. O, N thng c thm S v i lc c P.

- Thuc loi i phn t, phn t ln nht di 0,1 micromet, phn t lng c th t ti1,5 triu vC.

- Thuc loi a phn t, n phn l cc axit amin.

- C 20 loi axit amin to nn cc protein, mi axit amin c 3 thnh phn: gc cacbon (R),nhm amin (-NH2), nhm cacboxyl (-COOH), chng khc nhau bi gc R. Mi axit amin c

kch thc trung bnh 3.


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- Trn phn t protein, cc axit amin lin kt vi nhau bng lin kt peptit l lin ktgia nhm amin ca axit amin ny vi nhm cacboxyl ca axit amin bn cnh cng nhau mti mt phn t nc. Nhiu lin kt peptit to thnh mt chui polipeptit. Mi phn t

protein c th gm mt hay mt s chui polipeptit cng loi hay khc loi.

- T 20 loi axit amin to nn khong 10 14 1015 loi protein c trng cho mi loi. Ccphn t protein phn bit vi nhau bi s lng thnh phn, trnh t phn b cc axit amin.

2. Cu trc khng gian

C 4 bc cu trc khng gian

- Cu trc bc I: do cc axit amin lin kt vi nhau bng lin kt peptit, ng u mchpolipeptit l nhm amin, cui mch l nhm cacboxyl.

- Cu trc bc II: c dng xon tri, kiu xon anpha, chiu cao mt vng xon 5,4 A0,vi 3,7 axit amin/1 vng xon cn chui bta mi vng xon li c 5,1 axit amin. Cnhng protein khng c cu trc xon hoc ch cun xon mt phn ca polipeptit.

- Cu trc bc III: l hnh dng ca phn t protein trong khng gian ba chiu, do xoncp II cun theo kiu c trng cho mi loi protein, to thnh nhng khi hnh cu.

- Cu trc bc IV: l nhng protein gm 2 hoc nhiu chui polipeptit kt hp vi nhau.V d, phn t hemoglobin gm 2 chui anpha v 2 chui bta, mi chui cha mt nhnhem vi mt nguyn t Fe.

3. Tnh c trng v tnh nhiu dng ca protein

- Protein c trng bi s lng thnh phn, trnh t phn b cc axit amin trong chuipolipeptit. V vy, t 20 loi axit amin to nn 1014 1015 loi protein rt c trng v adng cho mi loi sinh vt.

- Protein c trng bi s lng thnh phn trnh t phn b cc chui polipeptit trongmi phn t protein.

- Protein c trng bi cc kiu cu trc khng gian ca cc loi protein thc hin cc

chc nng sinh hc.III. C CH TNG HP ADN, ARN V PROTEIN

1. C ch tng hp ADN

- Di tc ng ca enzim ADN polimeraza, cc lin kt hiro trn phn t ADN b

ct, 2 mch n ca ADN tch nhau ra, trn mi mch n cc nucleotit ln lt lin kt vicc nucleotit t do ca mi trng theo NTBS. Kt qu t mt phn t ADN m hnh thnh2 phn t ADN con, trong mi ADN con c mt mch l nguyn liu c, mt mch l

nguyn liu mi c xy dng nn, theo nguyn tc bn bo ton.


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- S tng hp ADN l c s hnh thnh NST, m bo cho qu trnh phn bo nguyn phn,gim phn, th tinh xy ra bnh thng, thng tin di truyn ca loi c n nh. cp t

bo v cp phn t qua cc th h. Nh con sinh ra ging vi b m, ng b t tin.

2. C ch tng hp mARN

- Di tc dng ca enzim ARN polimeraza. Cc lin kt hiro trn mt on phn tADN ng vi mt hay mt s gen ln lt b ct t, qu trnh lp rp cc ribonucleotit tdo ca mt trng ni bo vi cc nucleotit trn mch m gc ca gen (mch 3 5) theo

NTBS A-U, G-X xy ra. Kt qu to ra cc mARN c chiu 5-3. Sau 2 mch gen lilin kt vi nhau theo NTBS. S tng hp tARN v rARN chng theo c ch trn.

- sinh vt trc nhn s phin m cng mt lc nhiu phn t mARN, cc mARN cs dng ny tr thnh bn phin m chnh thc. Cn sinh vt nhn chun s phin m tng

mARN ring bit, cc mARN ny sau phi c ch bin li bng cch loi b cc onv ngha, gi li cc on c ngha to ra mARN trng thnh.

- S tng hp ARN m bo cho qu trnh dch m chnh xc t bo cht to nn ccprotein cn thit cho t bo.

3. C ch tng hp protein

Gm 2 giai on:

Giai on 1: Tng hp ARN chuyn thng tin di truyn t gen sang sn phm protein

Giai on 2: Tng hp protein t bo cht gm 4 bc c bn.+ Bc 1: Hot ho cc axit amin: Cc axit amin c hot ho bng ngun nng lng

ATP ri mi axit amin c gn vo mt tARN i vo riboxom thnh dng lin tc.

+ Bc 2: M u chui polipeptit: C s thanh gia ca riboxom, b ba m u AUG,tARN mang axit amin m u tin vo riboxom i m ca n khp vi m m u trmmARN theo NTBS.

+ Bc 3: Ko di chui polipeptit: tARN vn chuyn axit amin th nht tin vo riboxom

i m ca n khp vi m th nht trn mARN theo NTBS, mt lin kt peptit c hnh thnhgia axit amin m u vi axit amin th nht. Riboxom chuyn dch sang b ba th 2 y tARNaxit amin m u ra ngoi. Lp tc tARN axit amin th 2 tin vo riboxom i m ca n lprp vi m b ba trn mARN theo NTBS. C tin hnh theo phng thc cho n tn b batip gip vi b ba kt thc chui polipeptit lc ny c cu trc

aaMD aa1 aa2 aan vn cn gn vi tARN axit amin th n.

+ Bc 4: Kt thc chui polipeptit: Riboxom chuyn dch sang b ba kt thc lc nyngng qu trnh dch m 2 tiu phn ca riboxom tch nhau ra tARN axit amin cui cng


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c tch khi chui polipeptit. Mt enzim khc loi b axit amin m u gii phng chuipolipeptit.

Trn mi mARN cng lc c th c nhiu riboxom trt qua vi khong cch l 51 102 , ngha l trn mi mARN c th tng hp nhiu protein cng loi.

S tng hp protein gp phn thc hin chc nng biu hin tnh trng, cung cp nguynliu cu to nn cc bo quan v m nhn nhiu chc nng khc nhau.

IV. CHC NNG CA ADN, ARN V PROTEIN

1. Chc nng ca ADN

+ Cha thng tin di truyn, thng tin c trng cho mi loi bi trnh t phn b ccnucleotit trn phn t ADN

+ C kh nng nhn i chnh xc truyn thng tin di truyn qua cc th h.+ Cha cc gen khc nhau, gi chc nng khc nhau.

+ C kh nng t bin to nn thng tin di truyn mi.

2. Chc nng ca cc loi ARN c tng hp t ADN

- Chc nng ca mARN: bn phin thng tin di truyn t gen cu trc, trc tip tham giatng hp protein da trn cu trc v trnh t cc b ba trn mARN.

- Chc nng ca tARN: vn chuyn lp rp chnh xc cc axit amin vo chui polipeptit

da trn nguyn tc i m di truyn gia b ba i m trn tARN vi b ba m phin trnmARN.

- Chc nng ca rARN: lin kt vi cc phn t protein to trn cc riboxom tip xc vimARN v chuyn dch tng bc trn mARN, mi bc l mt b ba nh m lp rpchnh xc cc axit amin vo chui polipeptit theo ng thng tin di truyn c quy nh tgen cu trc.

3. Chc nng ca protein

- L thnh phn cu to ch yu cht nguyn sinh hp phn quan trng xy dng nn ccbo quan, mng sinh cht

- To nn cc enzim xc tc cc phn ng sinh ho.

- To nn cc hoocmon c chc nng iu ho qu trnh trao i cht trong t bo, c th.

- Hnh thnh cc khng th, c chc nng bo vv c th chng li cc vi khun gy bnh.

- Tham gia vo chc nng vn ng ca t bo v c th.

- Phn gii protein to nng lng cung cp cho cc hot ng sng ca t bo v c th.


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-Tm li protein m nhn nhiu chc nng lin quan n ton b hot ng sng ca tbo, quy nh tnh trng ca c th sng.

V. S IU HO HOT NG CA GEN

C ch iu ho sinh tng hp protein ca gen rt phc tp, c s khc bit r rt gia sinhvt trc nhn v sinh vt nhn chun. Sau y l c ch iu ho sinh vt trc nhn:

- Trong t bo c rt nhiu gen cu trc, khng phi cc gen u phin m, tng hpprotein ng thi. S iu ho hot ng ca gen c thc hin qua c ch iu ho. Vonm 1961, F.Jacop v J.Mono pht hin s iu ho hot ng ca gen E.coli

- Mt m hnh iu ho bao gm cc h thng gen sau:

+ Mt gen iu ho (R), gen ny lm khun sn xut mt loi protein c ch c tc dngiu chnh hot ng ca mt nhm gen cu trc qua tng tc vi gen ch huy.

+ Mt gen ch huy (O) nm k trc nhm gen cu trc, l v tr tng tc vi cht c ch.

+ Mt gen khi ng (P) nm trc gen ch huy v c th trm ln mt phn hoc ton bgen ny, l v tr tng tc ca ARN polimeraza khi u phin m.

+ Mt nhm gen cu trc lin quan vi nhau v chc nng, nm k nhau cng phin mto ra mt si mARN chung i vi sinh vt trc nhn, cn sinh vt nhn chun phin mch to ra 1 mARN ring bit.

Mt operon ch gm c gen ch huy v cc gen cu trc do n kim sot.

- C ch iu ho din ra nh sau:

Gen iu ho ch huy tng hp mt loi protein c ch, protein ny gn vo gen ch huy(O) lm ngn cn hot ng ca enzim phin m. V vy c ch hot ng tng hp ARNca cc gen cu trc. Khi trong mi trng ni bo c cht cm ng, cht ny kt hp vi

protein c ch lm v hiu ho cht c ch, khng gn vo gen ch huy. Kt qu l gen chhuy lm cho nhm gen cu trc chuyn t trng thi c ch sang trng thi hot ng. Qutrnh phin m li xy ra.

C ch iu ho sinh vt nhn chun rt phc tp n nay cn nhiu vn cha r.VI. M DI TRUYN, C IM CA M DI TRUYN

1. Khi nim m b ba

C 3 nucleotit cng loi hay khc loi ng k tip nhau trn phn t ADN hoc trn mARNm ho cho 1 axit amin hoc lm nhim v kt thc chui polipeptit gi l m b ba.

2. M di truyn l m b ba

- Nu mi nucleotit m ho 1 axit amin th 4 loi nucleotit ch m ho c 4 loi axit amin.


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- Nu c 2 nucleotit cng loi hay khc loi m ho cho 1 axit amin th ch to c 4 2 =16 m b ba khng m ho cho 20 loi axit amin.

- Nu theo nguyn tc m b ba s to c 43 = 64 m b ba m ho cho 20 loiaxit amin.

- Nu theo nguyn tc m b bn s to c 4 4 = 256 b m ho li qu tha. Vy vmt suy lun l thuyt m b ba l m ph hp.

Trong nghin cu, khi thm bt 1, 2, 3 nucleotit trong gen, ngi ta nhn thy m b bal m ph hp v xc nh c c 64 b ba c s dng m ho axit amin. Trong c Metionin ng vi m m u TAX l tn hiu bt u s tng hp chui polipeptit. Ba

b ba cn li ATT, ATX, AXT l m kt thc.

Hai mi loi axit amin c m ho bi 61 b ba. Nh vy mi axit amin c m ho

bi 1 s b ba. V d, lizin ng vi 2 b ba AAA, AAG, mt s axit amin c m ho binhiu b ba nh alanin ng vi 4 b ba, lxin ng vi 6 b ba.

3. Nhng c im c bn ca m di truyn

- M di truyn c c theo mt chiu 5 3 trn phn t mARN.

- M di truyn c c lin tc theo tng cm 3 nucleotit, cc b ba khng c gi ln nhau.

- M di truyn l c hiu khng mt b ba no m ho ng thi 2 hoc mt s axitamin khc nhau.

- M di truyn c tnh thoi ho c ngha l mi axit amin c m ho bi mt s b bakhc loi tr metionin, Triptophan ch c m ho bi mt b ba). Nh m gen m boc thng tin di truyn v xc nhn trong b ba 2 nucleotit u l quan trng cn nucleotitth 3 c th linh hot . S linh hot ny c th khng gy hu qu g. Nhng cng c th gynn s lp rp nhm cc axit amin trong chui polipeptit.

- M di truyn c tnh ph bin: tt c cc loi sinh vt u c m ho theo mt nguyn tcchung (cc t m ging nhau), iu ny phn nh ngun gc chung ca cc loi.

- M di truyn c m m u, c m ko di chui polipeptit v m kt thcVII. T BIN GEN

1. Khi nim

t bin gen l nhng bin i trong cu trc phn t ca gen lin quan ti mt hay mts cp nucleotit xy ra ti mt im no ca phn t ADN biu hin cc dng: mt,thm, thay th 1 cp nucleotit.

2. Nguyn nhn v c ch

a) Nguyn nhn


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- t bin gen pht sinh do tc nhn gy t bin l ho trong ngoi cnh hoc ri lontrong cc qu trnh sinh l, ho sinh ca t bo gy nn nhng sai st trong qu trnh t saoca ADN hoc trc tip bin i cu trc ca n.

- t bin gen ph thuc vo loi tc nhn, liu lng, cng ca tc nhn, c imcu trc ca gen.

b) C ch

S bin i ca mt nucleotit no thot u xy ra trn mt mch ca ADN di dngtin t bin. Lc ny enzim sa cha c th sa sai lm cho tin t bin tr li dng banu. Nu sai st khng c sa cha th qua ln t sao tip theo nucleotit lp sai s lin ktvi nucleotit b sung vi n lm pht sinh t bin gen.

3.S biu hin ca t bin gen- t bin gen khi pht sinh s c ti bn cng vi s ti bn ca phn t ADN

mang t bin.

- Nu t bin pht sinh trong gim phn s to t bin giao t qua th tinh i vo hpt. t bin tri s biu hin ngay kiu hnh ca c th mang t bin. t bin ln s ivo hp t dng d hp qua giao phi lan truyn dn trong qun th, tri qua nhiu th hc nhn ln ngy mt nhiu, ti mt thi im no cc t bin ln trong cc giao t

gp g nhau trong giao phi, hnh thnh t hp ng t ln, lc ny kiu hnh t bin lnmi xut hin.

- Khi t bin xy ra trong nguyn phn, chng s pht sinh mt t bo sinh dng ric nhn ln trong mt m. Nu l t bin tri s biu hin mt phn ca c th, to nnth khm.

- t bin soma c th di truyn bng sinh sn sinh dng nhng khng th di truyn quasinh sn hu tnh.

- t bin cu trc ca gen i hi mt s iu kin mi biu hin trn kiu hnh ca cth. V vy cn phi phn bit t bin l nhng bin i trong vt cht di truyn vi th tbin l nhng c th mang t bin biu hin kiu hnh

4. Hu qu ca t bin gen

- S bin i trong dy nucleotit ca gen dn n bin i trong dy ribonucleotit camARN lm bin i dy axit amin ca protein tng ng. Cui cng biu hin thnh mt

bin i t ngt, gin on v mt hoc mt s tnh trng no trn mt hoc mt s t c

th trong qun th.


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- t bin gen gy ri lon trong qu trnh sinh tng hp protein, c bit l t bin cc gen quy nh cu trc ca cc enzim nn a s t bin thng c hi cho c th, cngc nhng t bin gen trung tnh, mt s t bin li c li.

B C S VT CHT CA HIN TNG DI TRUYN V BIN D CP T BOI. CU TRC V TNH C TRNG CA NST

1. Khi nim NST

NST l th vt cht di truyn tn ti trong nhn t bo b bt mu bng thuc nhum

kim tnh, c s lng, hnh dng, kch thc, cu trc c trng: NST c kh nng t nhn

i, phn li, t hp n nh qua cc th h.

2.Cu trc ca NST

- virut, th n khun, NST ch l mt phn t ADN trn. sinh vt c nhn, NST cutrc phc tp.

- Sau khi nhn i mi NST c 2 cromatit, mi cromatit c 1 si phn t ADN m c mtna nguyn liu c v mt na nguyn liu mi c ly t mi trng t bo. Cc cromatitny ng xon cc i k gia nn chng c hnh dng v kch thc c trng. Mi NSTc 2 cromatit nh nhau tm ng ti eo th nht. Mt s NST cn c eo th 2 l ni tnghp rARN. Cc rARN tch t li to nn nhn con. Lc bc vo phn bo, NST ngng hot

ng, nhn con li ti hin.- NST c nhiu hnh dng khc nhau: hnh ht, hnh que, hnh ch V, hnh mc. mt s

loi sinh vt trong vng i c tri qua giai on u trng c xut hin cc NST vi kchthc ln hng nghn ln gi l NST khng l (nh u trng rui gim v cc loi thuc b2 cnh). in hnh l NST c hnh ch V vi 2 cnh kch thc bng nhau hoc khc nhau.Chiu di ca NST t 0,2 n 50 m, chiu ngang 0,2 n 2 m.

- NST c cu to bi ADN v protein. Phn t ADN qun quanh khi cu protein to

nn nucleoxom. Mi nucleoxom gm 8 phn t protein histon to nn khi hnh cu dt phangoi c bao bc bi 1 vn xon ADN khong 146 cp nucleotit. Cc nucleoxom ni vinhau bng cc on ADN v mt protein histon H1. Mi on c khong 15 100 cpnucleotit. T hp ADN vi histon trong chui nucleoxom to thnh si c bn c chiungang 100 , si c bn cun xon th cp to nn si nhim sc c chiu ngang 250 300. Si nhim sc tip tc ng xon to nn ng rng vi b ngang 2000 , cui cng hnhthnh cromatit c ng knh ti 6000.

- Do c cu trc xon cun nh vy nn chiu di ca NST c rt ngn 15000 20000 ln so vi chiu di phn t ADN, NST di nht ca ngi cha phn t ADN di 82


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mm, sau khi xon cc i k gia ch di 10 m. S thu gn cu trc khng gian nhth thun li cho s phn li, t hp cc NST trong chu k phn bo.

3. Tnh c trng ca NST

Mi loi sinh vt u c b NST c trng:- c trng v s lng, hnh dng, kch thc v cu trc. nhng loi giao phi, t

bo sinh dng mang b NST lng bi (2n), NST tn ti thnh cp tng ng, trong mt NST c ngun gc t b, mt NST c ngun gc t m. T bo giao t cha b NSTn bi.

V d: ngi 2n = 46, n = 23; ng 2n = 20, n = 10; la 2n = 24, n = 12;

u H Lan 2n = 14, n = 7

- c trng bi s lng, thnh phn, trnh t phn b cc gen trn mi NST.

- c trng bi cc tp tnh hot ng ca NST ti sinh, phn li, t hp, trao i on,t bin v s lng, cu trc NST.

II. C CH HNH THNH CC LOI B NST T T BO 2n

1. C ch hnh thnh b NST n

- Mt nhm t bo sinh dng cc c th trng thnh c tch ra lm nhim v sinhsn, gi l t bo sinh dc s khai. Cc t bo ny ln lt tri qua 3 giai on:

+ Giai on sinh sn: nguyn phn lin tip nhiu t to ra cc t bo sinh dc con.

+ Giai on sinh trng: cc t bo tip nhn nguyn liu t mi trng ngoi to nncc t bo c kch thc ln (k c nhn v t bo cht).

+ Giai on chn: cc t bo sinh tinh trng, sinh trng bc vo gim phn gm 2 lnphn bo lin tip to ra cc giao t n bi.

+ Giai on sau chn: thc vt khi kt thc gim phn mi t bo n bi hnh thnh tt bo sinh dc c tip tc nguyn phn 2 t to ra 3 t bo n bi hnh thnh ht phnchn. Mi t bo n bi mi t bo sinh dc ci li nguyn phn 3 t to ra 8 t bo n

bi hnh thnh non.

- Gim phn I:

+ k trung gian ADN nhn i, mi cp NST tng ng nhn i thnh cp NST

tng ng kp.

+ k trc I: NST tip tc xon li, k ny ti mt s cp NST tng ng c xy ratrao i on giu 2 cromatit khc ngun gc. Cui k trc I, mng nhn bin mt, thoi tv sc bt u hnh thnh.


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+ k gia I: thoi t v sc hnh thnh xong. Cc NST tng ng kp tp trung thnhcp trn mt phng xch o v ni vi thoi t v sc ti tm ng theo nhiu kiu sp xp.

+ k sau I: mi NST dng kp trong cp tng ng kp phn li v 2 cc t bo, hnhthnh cc t bo c b NST n trng thi kp.

+ k cui I: to 2 t bo con cha b NST n trng thi kp, khc nhau v ngungc, cht lng NST.

- Gim phn II: ln ny, k trung gian tri qua rt ngn k gia II, cc NST n trng thi kp trong mi t bo tp trung trn mt phng xch o ni vi thoi t v sc. Ksau II, mi cromatit trong mi NST n trng thi kp phn li v 2 cc. K cui II to racc t bo n bi. T mt t bo sinh tinh trng to ra 4 tinh trng, t 1 t bo sinh trngto ra 1 trng v 3 th nh hng.

2. C ch hnh thnh b NST 2n

- Qua nguyn phn:

+ k trung gian: mi NST n tho xon cc i dng si mnh, ADN nhn i to ra cc NST kp.

+ K trc: NST xon li, cui k trc mng nhn mt, thoi v sc bt u hnh thnh.

+ K gia: thoi v sc hnh thnh xong, NST kp tp trung trn mt phng xch o nivi dy t v sc ti tm ng.

+ K sau: mi cromatit trong tng NST kp tch nhau qua tm ng phn chia v 2 cc tbo.

+ K cui: cc NST n gin xon cc i, mng nhn hnh thnh, mi t bo cha bNST lng bi (2n)

- Qua gim phn khng bnh thng:

Cc t bo sinh tinh trng hoc sinh trng nu b tc ng ca cc nhn t phng x, hohc lm ct t thoi t v sc hoc c ch hnh thnh thoi t v sc trn ton b b NST s

to nn cc giao t lng bi.- Qua c ch th tinh:

S kt hp gia tinh rng n bi v trng n bi qua th tinh s to nn hp t lngbi (2n).

3. C ch hnh thnh b NST 3n, 4n

- T bo 2n gim phn do ri lon phn bo (thoi t v sc b ct hoc c hnh thnh)xy ra trn tt c cc cp NST s to nn giao t 2n. Giao t ny kt hp vi giao t bnhthng n s to nn hp t 3n.


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- Cc giao t khng bnh thng 2n kt hp vi nhau s to nn hp t 4n.

- Ngoi ra dng 3n cn c hnh thnh trong c ch th tinh kp ( thc vt) do nhnthu cp 2n kt hp vi mt tinh t n trong ht phn chn to nn ni nh 3n.

- Dng t bo 4n, cn c hnh thnh do nguyn phn ri lon xy ra trn tt c cc cpNST sau khi nhn i.

III. NGHA SINH HC V MI LIN QUAN GIA NGUYN PHN, GIM PHN,TH TINH.

1. ngha sinh hc ca nguyn phn, gim phn v th tinh

a) ngha ca nguyn phn v gim phn

- Nguyn phn: n nh b NST qua cc th h t bo ca cng mt c th, tng nhanhsinh khi t bo m bo phn ho m, c quan to ra c th.

- Gim phn: m bo s kt tc vt cht di truyn n nh tng i qua cc th h.

b) ngha ca th tinh

Phc hi li b NST lng bi do s kt hp gia giao t c (n) vi giao t ci (n). Mtkhc trong th tinh do s phi hp ngu nhin ca cc loi giao t khc gii tnh m cngto nn nhiu kiu hp t khc nhau v ngun gc v cht lng b NST lm tng tn s cc

loi bin d t hp.2. Mi lin quan gia nguyn phn, gim phn v th tinh trong qu trnh truyn t

thng tin di truyn

- Nh nguyn phn m cc th h t bo khc nhau vn cha ng cc thng tin di truynging nhau, c trng cho loi.

- Nh gim phn m to nn cc giao t n bi khi th tinh s khi phc li trng thilng bi.

- Nh th tinh kt hp b NST n bi trong tinh trng vi b NST n bi trongtrng hnh thnh b NST 2n, m bo vic truyn thng tin di truyn t b m cho conci n nh tng i.

- Nh s kt hp 3 qu trnh trn m to iu kin cho cc t bin c th lan rng chmchp trong loi c dp biu hin thnh kiu hnh t bin.

IV. T BIN CU TRC NST

1. Khi nim


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t bin cu trc NST l nhng bin i t ngt trong cu trc ca NST do tc nhn

gy t bin lm thay i cu trc NST to ra nhng tnh trng mi.

2. Nguyn nhn

Do tc nhn gy t bin l ho trong mi trng hoc nhng bin i sinh l ni bo lmph v cu trc NST nh hng ti qu trnh ti bn, tip hp, trao i cho ca NST.

3. C ch v hu qu

t bin cu trc NST gm cc dng : mt on, lp on, o on, chuyn on.

- Mt on: Mt on NST b t ra lm gim s lng gen trn NST. on b mt cth pha ngoi hoc pha trong ca cnh. t mt on thng gim sc sng hoc gycht. V d, mt on cp 21 ngi gy ung th mu.

- Lp on: Mt on NST no c lp li mt ln hay nhiu ln lm tng s lnggen cng loi. t bin lp on c th do on NST b t c ni xen vo NST tngng hoc do NSt tip hp khng bnh thng, do trao i cho khng u gia cccromatit. t bin lp on c th lm tng cng hay gim st sc biu hin tnh trng. Vd, lp on 16A rui gim lm cho mt hnh cu thnh mt dt, cng lp nhiu on mtcng dt.

- o on: Mt on NST b t ri quay ngc li 180 v gn vo ch b t lm thay

i trt t phn b gen trn NST. on b o ngc c th mang tm ng hoc khng. tbin o on NST t nh hng ti sc sng ca c th v vt cht di truyn khng b mti. S o on NST to nn s a dng gia cc ni trong phm vi mt loi.

- Chuyn on: Mt on NST ny b dt ra v gn vo mt NST khc hoc c 2 NSTkhc cp cng t mt on no ri li trao i on b t vi nhau, cc on trao i cth tng ng hoc khng tng ng. Nh vy c th thy c 2 kiu chuyn on lchuyn on tng h v chuyn on khng tng h. s chuyn on lm phn b li cc

gen trong phm vi mt cp NST hay gia cc NST khc nhau to ra nhm gen lin kt mi.Chuyn on ln thng gy cht hoc lm mt kh nng sinh sn. Ngi ta gp s chuynon nh u la, chui, u trong thin nhin. Trong thc nghim ngi ta chuyn genc nh nit ca vi khun vo h gen hng hng to ra ging hng hng c nit caotrong du.

V. T BIN S LNG NST

1. Khi nim


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t bin s lng NST l hin tng b NST ca loi tng ln mt s nguyn ln b n bi

(to th a bi) hoc tng ln hay gim i mt hay mt s cp NST s to nn th d bi.

2. Th d bi

Th d bi gm c: th ba nhim, th a nhim, th mt nhim, th khuyt nhim. Cc tbin d bi a phn gy nn hu qu c hi ng vt. V d, ngi c 3 NST 21, xut hinhi chng ao, tui sinh ngi m cng cao t l mc hi chng ao cng nhiu

Ngoi ra, cn gp hi chng XXX, XO, XXY, OY u gy nn hu qu c hi.

3. Th a bi

C 2 dng a bi : a bi chn v a bi l

- a bi chn c hnh thnh bng c ch nguyn phn ri lon trn ton b b NST 2n s

to nn dng 4n, hoc do kt hp gia 2 loi giao t lng bi khng bnh thng vi nhau.- a bi l c hnh thnh l do s kt hp gia giao t 2n khng bnh thng vi giao

t n hnh thnh th a bi l 3n.

- C th a bi c hm lng ADN tng gp bi dn ti trao i cht tng cng, c tha bi t bo kch thc ln, c quan sinh dng, sinh sn to, chng chu tt vi iu kin

bt li ca mi trng.

- C th a bi l khng c kh nng sinh sn hu tnh v qu trnh gim phn b tr ngi.Mun duy tr phi nhn bng con ng sinh sn sinh dng.

- Th a bi kh ph bin thc vt, ng vt giao phi thng rt t gp.

H THNG CU HI N TP

Cu 1. Ti sao ni ADN l c s vt cht v c ch di truyn cp phn t? Tnh ctrung v a dng ca ADN c th hin ch no?

Cu 2: S ging nhau v khc nhau c bn v cu trc gia ADN v mARN nhng im

no?Cu 3: ADN v protein ging v khc nhau v cu trc nhng im no? Nhng chcnng c bn ca prtin ?

Cu 4: im ging nhau va khc nhau gia c ch tng hp ADN sinh vt nhn chun viE.coli. ngha ca c ch tng hp ADN

Cu 5: Mi quan h gia ADN v prtin trong cu trc v c ch di truyn.Tnh c trngcu prtin do yu t no quy nh?


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Cu 6: Ti sao ni t bin gen thng c hi, t c li, tn s thp, nhng li l ngunnguyn liu ch yu ca chon lc t nhin?.

Cu 7: im ging v khc nhau c bn gia nguyn phn v gim phn. ngha ca chngtrong di truyn v tin ho..

Cu 8: Trnh by c ch hnh thnh cc dng t bo n, 2n, 3n, 4n t dng t bo 2n. nghaca vic hnh thnh cc loi t bo ni trn

Cu 9: Cc c ch sinh hc xy ra nh th no i vi 1 cp NST tng ng cp tbo?.

Cu10. Cu trc v chc nng ca NST thng v NST gii tnh ging nhau v khc nhau im no?

Cu11. Trnh by cc loi bin d lm thay i v s lng, cu trc NST?

Cu12. im ging v khc nhau c bn gia bin d t hp v bin d t bin. Vai tr cacc loi bin d trong tin ho v chn ging.

CC BI TP RN K NNG

Bi 1. Mt gen c chiu di 5100 ngstron. Hiu s phn trm gia adenin vi mt loinucleotit khc bng 10% s nucleotit ca gen. Trn phn t mARN c tng hp t gen c 10% uraxin. mt mch n ca gen c 16% xitozin, s timin bng 150 nucleotit.

a) T l phn trm v s lng tng loi nucleotit ca gen.

b) T l phn trm v s lng tng loi ribonucleotit ca mt phn t mARN bng baonhiu?

c) Nu gen sao m 6 ln v trn mi phn t mARN c 10 riboxom trt qua khnglp li th s lng axit amin m mi trng ni bo cung cp cho qu trnh tng hp proteinl bao nhiu?

d) Nu thi gian gii m mt axit amin la 0,1 giy, thi gian tip xc ca mt phn tmARN vi cc riboxom l 58,1 giy, khong cch gia cc riboxom k tip khong baonhiu ngstron?

Bi 2: Mt gen di 5100 ngtron. Khi gen t sao lin tip hai t, mi trng ni bo cung cp 2700 anin. Phn t mARN c tng hp t gen c 600 adenin v 240guanin. vn tc gii m l 10 axit amin/ giy. Tnh t lc ribxm th nht trt qua phn tmARN cho n khi ht phn t mARN l 55,6 giy.

a) Xc nh s lng tng loi nucltit trong ton b cc gen c hnh thnh sau hai

t t sao lin tip


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b) Xc nh s lng tng loi nucltit mi mch n ca gen

c) Tnh khong cch theo ngtron gia ribxm th nht vi ribxm cui cng khichng ang tham gia gii m trn mt phn t mARN

Bi 3: mch th nht ca gen c t l cc nucltit l 10% adenine, 20% timin v 25%xitozin. Phn t mARN c sao t gen c 20% uraxin.

a) Tnh t l tng loi nucltit ca gen v tng loi ribnucltit ca phn t mARN

b) Nu gen di 0,306 micromet th n cha boa nhiu lin kt hidro?

c) Mt phn t mARN sinh ra t gen c chiu di ni trn v c mt s riboxom cnghot ng trong qu trnh gii m, thi gian riboxom th nht trt qua ht phn t l 30giy v thi gian tnh t lc bt u c s gii m n khi riboxom cui cng trt qua ht

phn t mARN l 35,4 giy. Hi c bao nhiu riboxom tham gia vo qu trnh gii m?Bit rng cc riboxom cch u nhau mt khong bng 61,2

Bi 4: Mch th nht ca gen c 240 timin, hiu s gia guanin vi adenine bng 10% snucleotit ca mch. mch th hai, hiu s gia adenin v xitozin bng 20% s nucleotit camch. Khi gen tng hp phn te mARN th mi trng ni bo cung cp 360 uraxin.

a) T l phn trm v s lng tng loi nucleotit ca gen v ca tng mch l bao nhiu?

b) Hy tnh chiu di ca phn t mARN, t l phn trm v s lng mi loiribonucleotit ca n.

c) Trn mi phn t mARN c 8 riboxom cng gii m, tnh t lc riboxom bt uftrt trn phn t mARN th thi gian riboxom th nht trt qua ht sphn t l 20giy, cn riboxom cui cng th phi cn n 26,3 giy mi hon tt vic gii m. khongcch u gia cc riboxom l bao nhiu ngtron? Bit rang cc riboxom trt vi vn tc

bng nhau

Bi 5: Trn mt phn t mARN c mt s riboxom trt qua vi khong cch u bngnhau. Riboxom th nht trt qua ht phn t mARN ht 50 giy. Tnh t lc riboxom

th nht trt qua v tip xc vi phn t mARN th riboxom cui cng ppji mt 57,2giy mi hon thnh vic i qua phn t mARN . Bit rng phn t prtin th hai ctng hp chm hn phn t prtin th nht 0,9 giy.

Gen iu khin vic tng hp cc phn t prtin ni trn c mch 1 cha 10% adeini v20% guanine, mch 2 cha 15% adenine. Qu trnh sao m ca gen i hi mi trngni bo cung cp 150 uraxin v 155 adenin gp phn tng hp mt phn t mARN

a) Tnh chiu di ca gen

b) Tnh s lng tng loi ribnucltit ca mt phn t mARN


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c) S riboxom tham gia vo qua trnh gii m trn mt phn t mARN l baonhiu? Bit rng mi riboxom ch trt qua mt ln.

d) Khong cch gia 2 riboxom k tip v khong cch gia riboxom th nht viriboxom cui cng tnh theo ngtron l bao nhiu?

e) Ton b qu trnh gii m ni trn cn bao nhiu axit amin ca mi trng ni bov trong tt c cc prtin hon chnh cha bao nhiu axit amin?

Bi 6: mt loi mt t bo sinh dc 2n thc hin s nguyn phn lin tip mt s ln, ihi mi trng ni bo cung cp nguyn kiu hnh thnh nn 4826 nhim sc th nmi. cc t bo con sinh ra t ln nguyn phn cui cng u gim nhim bnh thng cho256 tinh trng cha nhim sc th gii tnh Y

a) B nhim sc th lng bi ca loi ? s ln nguyn phn lin tip ca t bo sinh

dc 2n u tin ? to ra cc t bo con 2n c bao nhiu thoi dy t v sc c hnhthnh trong cc ln nguyn phn y ?

b) Nu c 3 tinh trng c th tinh vi 3 trng khc nhau to ra cc hp t th c bao nhiucromait trong cc t bo sinh dc ci sinh ra cc trng , vo lc m cc t bo bt u thchin s phn bo gim nhim? Cc hp t va c hnh thnh c bao nhiu NST n?

c) Gi thit rng s gim phn xy ra trong cc t bo sinh dc ca c th ci, mi cptng ng u gm hai nhim sc th cu trc khc nhau, mi cp nhim sc th tngng u gm hai nhim sc th cu trc khc nhau, s trao i on ch xy ra mt cpnhim sc th thng,s t bin d bi ch xy ra cp nhim sc th gii tnh. Kh nngc th ci c th cho bao nhiu loi trng?

Bi 7: gi thit trong cc cp nhim sc th tng ng ca mt b nhim ssc th lng biu cha c cp gen d hp t v hin tng trao i on ti mt im ch xy ra vi mtcp nhim sc th tng ng. Cho bit khng c hin tng t bin v s loi giao t csinh ra t cc iu kin trn l 32.

Gi thit trung bnh mi k trong phn bo nguyn phn ht 5 pht, hai ln phn bo (k

trung gian) ht 10 pht, qu trnh nguyn phn din ra lin tc, cc t bo con sinh ra utip tc nguyn phn.

a) B nhim sc th lng bi ni trn trng thi cha nhn i l bao nhiu ?

b) hp t thc hin c qu trnh nguyn phn th mi trng ni bo cung cpnguyn kiu tng ng vi bao nhiu NST n vo cc thi im:

- Kt thc 20 pht

- Kt thc 32 pht

- Kt thc 100 pht


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Bit rng khi hp t bc vo k trc c tnh l thi gian bt u

Bi 8: g khi quan st mt t bo sinh dc c ang k gia ca nguyn phn, ngi tam c 78 nhim sc th kp

a) T bo nguyn phn 5 t lin tip i hi mi trng cung cp nguyn liu to ra bao nhiu nhim sc th n mi ?

b) Loi t bo ny gim phn bnh thng, kh nng nhiu nht c th cho bao nhiuloi tinh trng trong trng hp khng c hin tng trao i on gia cc nhim sc thkp trong cp tng ng? iu kin cho s loi tinh trng nhiu nht l g?

c) Gi thit rng c 1000 t bo sinh tinh trng gim phn bnh thng, hiusut th tinh ca tinh trng l 1/1000, cn ca trng l 20%, mi tinh trng ch th tinh vimt trng. Xc nh s t bo sinh trng.

d) Cc hp t c to thnh nguyn phn lin tip nhiu t vi s ln bng nhau,mi trung ni bo cung cp nguyn kiu to ra 2184 nhim sc th n mi. Xc nhs t bo con sinh ra v s t nguyn phn ca mi hp t.

Bi 9: Mt t bo sinh duc c 2n v mt t bo sinh dc ci 2n u nguyn phn mt s tbng nhau (cc t bo con sinh ra u tip tc nguyn phn). Gi thit rng cc t bo consinh ra t t nguyn phn cui cng u gim nhim cho tng s 80 giao t binh thng.Cho bit s lng nhim sc th n trong cc giao t c nhiu hn s lng nhim sc thn trong cc giao t ci l 192.

1. Loi tn l g?

2. M t hnh dng v s lng b nhim sc th lng b trong loi

Bi 10: Mt g mi c mt s trng, nhng khi p ch c 12 trng n thnh g con.Cc hp t n thnh g con c 936 nhim sc th n trng thi cha t nhn i. S trngcn li khng n thnh g con. S tinh trng c sinh ra phc v cho g giao phi c624000 nhim sc th n. Gi thit s tinh trng c trc tip th tinh vi cc trng nitrn chim t l 1/1000 so vi tng s tinh trng c hnh thnh. Mi tinh trng th tinh

vi 1trng

a) S trng c th tinh

b) Trng g khng n thnh g con c b nhim sc th nh th no?

c) S g trng v g mi trong n g con ni trn c tun theo t l 1: 1 khng?

GI P N CHO CC CU HI V BI TP TNG PHN

A- PHN : CU HI


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Cu1. Ti sao ni ADN l c s vt cht v c ch di truyn cp phn t ? Tnh ctrung v a dng ca ADN c th hin ch no?

Tr li

1. ADN l c s vt cht v c ch di truyn cp phn t :

- ADN l thnh phn chnh ca nhim sc th (NST), m nhim sc ht l c s vtcht ca tnh di truyn cp t bo, v vy ADN l cp di truyn cp phn t.

- ADN cha thng tin di truyn c trng cho mi loi s lng, thnh phn v trtt sp xp cc nucltit.

- ADN c kh nng t nhn i, m bo cho NST hnh thnh qu trnh nguyn phn,gim phn din ra bnh thng, thong tin di truyn ca loi c n nh cp t bo vcp phn t.

- ADN cha cc gen, mi gen thc hin mt chc nng di truyn khc nhau thng quac ch dch m v phin m.

- ADN c kh nng t bin v cu trc: mt, thm, thay th, o v tr cc nucltit tonn cc alen mi.

- Nhiu bng chng chng minh vai tr mang thng tin di truyn ca axit nucltit:

+ Kh nng hp th tia ngoi t cc i bc sng 260 nm.

+ Th nghim bin np ca F.Griffith (1928), ca O.T.Avery, C.M.Macleod(1994) vFraenket-Conrat, Singer (1957) chng minh c axit nucltit l vt cht mang thng tin

di truyn.2. Tnh c trng v tnh a dng ca ADN

- c trng bi s lng, thnh phn v trnh t sp xp cc nucltit. V vy t 4 loinucltit to nn tnh c trng v tnh a dng ca ADN.

- c trng bi t l A+T/G+X cho mi loi.

c trng bi s lng, thnh phn v trnh t phn b cc gen trn phn t ADN.

Cu 2: S ging nhau v khc nhau c bn v cu trc gia ADN v mARN ?

Tr li

1. im ging nhau v cu trc ADN v mARN- u cu to theo nguyn tc a phn.

- Mi n phn u gm 3 thnh phn c bn trong thnh phn quan trng nht lbaznitric.

- Trn mch n ADN v trn phn t mARN cc n phn c lin kt vi nhau bilin kt ho tr bn vng.

- u c cu to xon.

- c trng bi s lng, thnh phn v trt t phn b cc n phn.

2. im khc nhau gia cu trc ADN v mARN


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ADN mARN- i phn t c kch thc v khi lngrt ln- C cu trc mch kp

- Xy dng t 4 loi nucltit- C baznitric Timin- Trong mi nucltit c ngxiribza (C5H10O4)- Lin kt ho tr trn mch n ca ADNl lin kt gia ng C5H10O4 canucltit vi phn t H3PO4 ca nucltit

bn cnh. Nhiu lin kt ho tr to nn

chui plinucltit.

- a phn t c khi lng v kch thcrt b.- C cu trc mch n

- Xy dng t 4 loi ribnucltit- C baznitric U l dn xut ca T- Trong mi ribnucltit c ng ribza(C5 H10 O5)- Lin kt ho tr trn mch mARN l linkt c hnh thnh gia ng C5H10 O5ca ribnucltit ny vi phn t H3 PO4ca ribnucltit bn cnh. nhiu lin kt

ho tr to nn chui pliribnucltit.

Cu 3: ADN v protein ging v khc nhau v cu trc nhng im no? Nhng chcnng c bn ca prtin ?

Tr li

1. im ging nhau v cu trc gia ADN v prtin

u c cu to theo nguyn tc a phn

u c cu to t cc nguyn t, phn t C, H, O, N.

Cc n phn c lin kt vi nhau bng cc lin kt ho hc.

u c cu trc xon.

u c c trng bi s lng, thnh phn v trnh t phn b cc n phn.

u l 2 thnh phn c bn to nn cu trc nhim sc th.

2. im khc nhau gia cu trc ca ADN v prtin

ADN Prtin

- ADN c cu trc xon kp, gm 2mch n.

- ADN l i phn t chiu di ti hngtrm micrmet, khi lng phn t t 4triu n 8 triu, thm ch c th ti 16triu .vC

- ADN c cu to t 4 loi nucltit,

- Prtin c cu to xon, mc xontu thuc vo cc bc cu trc

- Prtin cng l i phn t c kchthc b hn ADN, phn t prtin lnnht cng ch ti 0,1 micrmt, khilng phn t 1,5 triu vC

- Prtin c cu to t 20 loi


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thnh phn c bn ca mi nucltit kbaznitric.

- Lin kt trn mi mch n ADN l

lin kt phtphoieste (gia ng C5 H10O4 ca nucltit ny vi phn t H3 PO4ca nucltit bn cnh). nhiu lin kt

photphoieste t thnh mchpolinucltit.

- Trn mch kp phn t ADN cc cpnucltit gia 2 mch n lin kt vi

nhau theo nguyn tc b sung (NTBS) Alin kt vi T bng 2 lin kt hir, G linkt vi X bng 3 lin kt hir, v ngcli to nn cu trc ADN chiu rngkhong 20, khong cch mi bc thang

bng 3,4. Mi chu k xon gm 10 cpnucltit c chiu cao 34 (xon phi)

- Mi phn t ADN gm nhiu gen

- Cu trc ho hc ca phn t ADN quy

nh cu trc ho hc ca cc prtintng ng.

axitamin, thnh phn c bn ca miaxitamin l gc cacbon (R)

- Trong phn t prtin cc axitamin lin

kt vi nhau bng lin kt peptit (gianhm amin ca axit amin ny vi nhmccbxin ca axit amin bn cnh cngnhau gii phng 1 phn t nc). Nhiukin kt peptit to thnh chui plipeptit.Mi phn t prtin c th gm 1 hoc 1s chiui plipeptit.

- Trn phn t prtin t nn 4 bc cu

trc khng gian: bc 1 cc axit amin kinkt vi nhau bng lin kt pepit; bc 2xon theo hnh l xo (xon anpha)( xon tri), chiu cao mi vng xon5,4, vi 3,7aa/vng. Trong chui xon

bta mi vng xon c 5,1aa; cu trcbc 3 l hnh dng phan t prtin trongkhng gian 3 chiu to thnh nhng khihnh cu cu trc bc 4 l nhng prtingm 2 hay nhiu plipeptit lin kt vinhau. V d phn t hmglbin c 2chui anpha v 2 chui bta, mi chuicha mt nhn km vi mt nguyn tFe

- Mi phn t prtin gm nhiu chuiplipeptit

- cu trc ho hc ca prtin ph thucvo cu trc ho hc ca cc gen trnphn t ADN

3. Nhng chc nng c bn ca prtin

- Kin to nn cc bo quan, cc t chc c quan ca c th, quy nh cc c im hnhthi, gii phu ca c th.

- To nn cc enzim, m bn cht l prtin xc tc cc phn ng sinh ho tng hp vt

cht di truyn, cc bn phin m, tham gia tng hp prtin v cc hp cht hu c quantrng khc. hin nay bit khong 3500 loi enzim.


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- To nn cc khng th c chc nng bo v c th chng li cc vi khun gy bnh.

- Cc hoocmn phn ln l prtin c chc nng iu ho cc qu trnh trong t bo v cth.

- Nhiu loi prtin tham gia vo chc nng vn ng t bo c th- Khi thiu ht gluxit, lipit, prtin t phn hu gii phng nng lng di dng ATP

cung cp cho qu trnh hot ho cc nguyn liu.

Prtin m nhn nhiu chc nng lin quan n ton b hot ng sng ca t bo, quynh cc tnh trng, tnh cht ca c th sng.

Cu 4: im ging nhau v khc nhau gia c ch tng hp ADN sinh vt nhn chun viE.coli. ngha ca c ch tng hp ADN

Tr li1. im ging nhau gia c ch tng hp ADN ca sinh vt nhn chun v E,coli :

- u da trn khun mu ca ADN m

- Cn nguyn kiu l ribnucltit, cc nucltit

- Cn c ezim xc tc m xon, tch 2 mch n lp rp cc nucltit

- Cn tng hp on mi to ra nhm 3OH

- C mt mch tng hp gin on (mi on l mt oan okazaki)

- u da vo nguyn tc b sung khi lp rp cc nucltit trn khun mu ca tngmch n ADN m.

- Kt qu u to ra nhng ADN con ging ADN m theo nguyn tc bn bo ton.

2. im khc nhau gia tng hp ADN sinh vt nhn chun vi E.coli

Tng hp ADN E.coli Tng hp ADN sinh vt nhn chun


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- Ton b ADN ch c mt n v ti bn

- s tng hp xy ra trn hai phiu tibn

- ADN- plimraza gm 3 loi enzim cchc nng khc nhau u tham gia tng hpADN, trong ADN- plimraza III ngvai tr ch yu.

- c nhiu n v ti bn

- S tng hp xy ra trn nhiu n v ti

bn, nhng n v ti bn no c nhiuG==X c tng hp trc, nhiu ATc tng hp sau.

- ADN- plimraza c 5 loi enzim:ADN- plimraza tham gia ti bnADN ca ti th. Enzim phi , enzim l2 enzim chnh tham gia tng hp ADN.Pli v kiu n v b c hot tnh

c sa

3. ngha ca c ch tng hp

Nh kh nng tng hp, ADN truyn t c thng tin di truyn ca n qua cc thh t bo v qua cc th h khc nhau ca loi.

Cu 5: Mi quan h gia ADN v prtin trong cu trc v c ch di truyn.Tnh c trngcu prtin do yu t no quy nh?

Tr li

1. Mi quan h gia ADN v prtin trong cu trc di truyn

- Prtin v ADN l hai thnh phn c bn cu trc nn nhim sc th, trong nhim sc thADN v prtin c t l tng ng. Prtin lin kt vi cc vng xon ca ADN gi chocu trc ADN n nh, thng tin di truyn trn ADN c iu ho.

- Prtin ca ADN t hp vi nhau to nn cht nhim sc hnh thnh nhim sc th. Phnt ADN qun quanh mt khi hinh cu dt (gm 8 phan t prtin hton) to nn nuclxom.

Cc nuclxm ni vi nhau bng cc on ADN di 15-100 cp nucltit v mt phn tprtin histon. t hp ADN vi histon trong chui nuclexom to thnh cc si c bn.

- t bo c nhn ribxm gm c mt ht ln v mt ht b. Ht ln gm 45 phn tprtin v 3 phn t rARN. Lc tng hp prtin hai ht ny lin kt vi nhau, tip xc vimARN v chuyn dch tng bc trn mARN thc hin qu trnh dch m.

- Cu trc ho hc ADN quy nh cu trc ho hc ca prtin (trnh t phn b ccnucltit tren ADN quy nh trnh t cc axit amin trn phn t prtin)

2. Mi quan h gia ADN v prtin trong cc c ch di truyn


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- ADN l khun mu tng hp mARN t quy nh cu trc prtin

- ADN cha nhiu gen cu trc, mi gen cu trc mang thng tin v mt loi prtin.

- Prtin c ch c tng hp t khun mu ca gen iu ho gn vo gen vn hnh cn tr

hot ng ca enzim phin m, do vy gen cu trc c duy tr trng thi khng hot ng.- Prtin tham gia to nn cc enzim tham gia vo tng hp ADN, ARN, prtin

- Prtin cn tham gia to nn cc yu t m u, ko di, kt thc qu trnh sinh tng hpprtin t bn phin m mARN.

- Prtin to nn thoi t v sc, cc dy t ni vi cc nhim sc th tm ng, m bocho sn phn li nhanh v chnh xc n inh vt cht di truyn cp t bo.

3. Tnh c trung ca prtin do cc yu t sau quy nh:

- V cu trc ho hc: do gen quy nh.- v cu to khng gian: do chc nng sinh hc ca cc lai prtin trong t bo quy nh.

Cu 6: Ti sao ni t bin gen thng c hi, t c li, tn s thp, nhng li l ngunnguyn liu ch yu ca chon lc t nhin?

Tr li

1. Khi nim t bin gen

t bin gen l nhng bin i t ngt trong cu trc phn t ca gen lin quan ti mthoc mt vi cp nucltit xy ra ti mt im no ca phn t ADN. Dn ti lm xuthin mt vi c im mi trn c th sinh vt.

2. Ni t bin gen thng c hi, t c li, tn s thp, nhng li l ngun nguyn liu chyu ca chon lc t nhin v:

- Phn ln t bin gen trong t nhin l c hi cho c th sinh vt v chng ph v miquan h hi ho trong kiu gen, trong ni b c th, gia c th vi mi trng c hnhthnh qua chn lc t nhin. Trong mi trng quen thuc, a s th t bin thng t ra

c sc sng km hoc km thch nghi so vi dng gc. t vo iu kin mi, n c th t rathch nghi hn, c sc sng cao hn. Th d trong mi trng khng c DDT th dng ruimang t bin khng DDT sinh trng chm hn dng rui bnh thng, nhng khi phunthuc DDT th li c li cho rui.

- Tuy t bin thng c hi nhng phn ln alen t bin l alen ln, xut hin mt giaot no , alen ln s i vo hp t v tn ti bn cnh alen tri tng ng th d hp, do khng biu hin thnh kiu hnh. Tri qua nhiu th h giao phi cc alen ln c nhnln, c iu kkin gp g nhau trong giao phi to nn th ng hp v biu hin thnh kiu

hnh. Gi tr thch ng ca mt t bin c th thay i tu t hp gen. mt t bin nm


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trong t hp ny c hi nhng tn ti trong t hp gen khc li tr nn c li. V vy c thni tnh li, hi ca t bin gen ch l tng .

- Nh qu trnh giao phi cc t bin c pht tn trong qun th to ra v s bin d thp. C th ni t bin l ngun nguyn kiu s cp cn bin d t hp l ngun nguynliu th cp cho chn lc t nhin do u to nn vn gen ca qun th. S tin hokhng ch s dng cc t bin mi xut hin m cn huy ng kho d tr cc gen t bin pht sinh t lu nhng tim n trong trng thi d hp.

Cu 7: im ging v khc nhau c bn gia nguyn phn v gim phn. ngha ca chngtrong di truyn v tin ho.

Tr li

1. im ging nhau gia nguyn phn v gim phn

- C s nhn i ca nhim sc th m thc cht l s nhn i ca ADN k trung gian.

- Tri qua cc k phn bo tng t nhau

- u c s bin i hnh thi nhim sc th theo chu k ng v tho xon m bo chonhim sc th nhn i v thu gon cu trc tp trung trn mt phng xch o k gia.

- ln phn bo II ca gim phn ging phn bo nguyn phn.

- u l c ch sinh hc m bo n nh vt cht di truyn qua cc th h.

2. im khc nhau gia nguyn phn v gim phnNguyn phn Gim phn

- Xy ra mt ln phn bo gm 5 k

- Mi nhim sc th tng ng c

nhn i thnh 2 nhim sc th kp, minhim sc th kp gm 2 crmatit

- k trc khng xy ra trao i chogia 2 crmatit cng ngun gc

- Ti k gia cc nhim sc th tp trung

- Xy ra 2 ln phn bo liln tip. ln phnbo I l phn bo gim phn, ln phn bo IIl phn bonguyn phn

- Mi nhim sc th tng ng c nhni thnh mt cp nhim sc th tng ng

kp gm 4 crmatit to thnh mt th thngnht.

- k trc I mt s cp nhim sc th cxy ra hin tng tip hp v xy ra trao ion gia 2 crmatit khc ngun gc, tonhm gen lin kt mi.

- ti k gia I cc nhim sc th tp trung


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thnh tng nhim sc th kp

- k sau ca nguyn phn c s phn licc crmatit trong tng nhim sc th kp

v 2 cc ca t bo

- Kt qu mi ln phn bo to ra 2 t bocon c b nhim sc th lng bi n nh

- Xy ra trong t bo sinh dng v m tdbo sinh duc s khai

thnh tng nhim sc th tng ng kp.

- k sau I ca gim phn c s phn li ccnhim sc th n trng thi kp trong

tng cp nhim sc th tng ng kp to ra cc t bo con c b nhim sc thn trng thi kp khc nhau v ngungc nhim sc th

- Kt qu qua 2 ln phn bo to ra cc tbo giao t c b nhim sc th gim i mtna khc bit nhau v ngun gc v chtlng nhim sc th

- Xy ra t bo sinh dc sau khi cc t bo kt thc giai on sinh trng

2. ngha ca nguyn phn v gim phn

- ngha ca nguyn phn

+ n nh b nhim sc th ca loi qua cc th h t bo ca cng mt c th

+ Tng nhanh sinh khi t bo, m bo s phn ho m, c quan to lp nn mt c thhon chnh

+ To iu kin cho cc t bin t bo sinh dn c th nhn ln qua cc th h t bo,to nn th khm.

- ngha ca gim phn

+ Gim b nhim sc th trong giao t, nh vy kkhi th tinh khi phc c trng thilng bi ca loi trong qu trnh to giao t

+ Trong gim phn c xy ra hin tng phn li c lp, t hp t do ca cc nhim scth, s trao i on ti k trc I ca gim phn to nn nhiu loi giao t khc nhau v

ngun gc v cht lng.. y l c s to nn cc bin d t hp cung cp nguyn liu choqu trnh chn lc t nhin, to nn tnh a dng ca sinh gii.

+ Nh gim phn cc t bin c nhn ln dn trong qun th, trong loi biu hinthnh kiu hnh t bin.

Cu 8: Trnh by c ch hnh thnh cc dng t bo n, 2n,3n,4n t dng t bo 2n. nghaca vic hnh thnh cc loi t bo ni trn

Tr li

1. C ch hnh thnh cc dng t bo


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a. C ch hnh thnh t bo n:

- cc c th trng thnh c mt nhm t bo sinh dng c tch ra lm nhim vsinh sn, gi l t bo sinh dc s khai. Cc t bo ln lt tri qua 3 giai on:

+ Giai on sinh sn: nguyn phn lin tip nhiu t to ra cc t bo sinh dc con+ Giai on sinh trng: cc t bo tip nhn nguyn liu mi trng ngoi to nn cc t

bo con c kch thc ln (k c nhn v t bo cht)

+ Giai on chn: cc t bo sinh tinh trng, sinh trng bc vo gim phn gm 2 lnphn bo lin tip: ln 1: gim phn, ln 2: nguyn phn to ra cc giao t n bi

- Gim phn I:

+ k trung gian ADN nhn i, mi cp nhim sc th tng ng nhn i thnh cp

tng ng kp.+ k trc I: nhim sc th tip tc xon li, k ny ti mt s cp nhim sc th c xy

ra trao i on gia 2 crmatit khc ngun gc trong cp tng ng. Cui k trc I, mngnhn mt, bt u hnh thnh dy t v sc.

+ k gia I: thoi t v sc hnh thnh xong. Cc nhim tng ng kp tp trung thnhcp trn mt phng xch o ni vi dy t v sc ti tm ng

+ k cui I: to 2 t bo con cha b nhim sc th trang thi kp, khc nhau vngun gc, cht lng nhim sc th

- Gim phn II: ln phn bo ny ging phn bo nguyn phan, k trung gian tri qua rtngn k gia II, cc nhim sc th n trng thi kp trong mi t bo tp trung trn mt

phng xch o ni vi dy t v sc. k sau II, mi crmatit trong mi nhim sc th n trngj thi lp phn li v hai cc. K cui II to ra cc t bo n bi. T mt t bo sinhtinh trng to ra 4 tinh trng, t 1 t bo sinh trng to ra 1 trng v 3 th nh hng.

b. C ch hnh thnh t bo 2n:

- Qua nguyn phn

+ k trung gian mi nhim sc th n tho xon cc i dng si mnh, ADN nhni to ra cc nhim sc th kp

+ K trc: nhim sc th tip tc xon li, cui k trc mng nhn mt, thoi v sc btu hnh thnh.

+ K gia: thoi v sc hnh thnh xong, nhim sc th kp tp trung trn mt phng xcho ni vi dy t v sc ti tm ng

+ K sau: mi crmatit trong tng nhim sc th kp tp trung trn mt phng xch o

ni vi dy t v sc ti tm ng


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+ K sau: mi crmatit trong tng nhim sc th kp tch nhau qua tm ng phn chia v2 cc t bo

+ K cui: cc nhim sc th n gin xon cc i, mng nhn hnh thnh, mi t bocha b nhim sc th lng bi (2n)

Nhn i phn chia2n 4n 2n

- Qua gim phn khng bnh thng

Cc t bo sinh tinh trng hoc sinh trng nu bi tc ng ca cc tc nhn phng x, hohc ... lm ct t dy t v sc trn ton b b nhim sc th s to nn cc giao t lng

bi.

- Qua c ch th tinh:

S kt hp gia tinh trng n bi v trng n bi qua th tinh s to nn hp t lngbi (2n)

c. C ch hnh thnh dng t bo 3n, 4n:

- T bo 2n gim phn khng bnh thng xy ra trn tt c cc cp nhim sc th s tonn giao t 2n. Giao t ny kt hp vi giao t binh thng n s to nn hp t 3n.

- Giao t khng bnh thng 2n kt hp vi nhau s to nn hp t 4n

- Ngoi ra dng 3n cn c hnh thnh trong c ch th tinh kp thc vt do nhn thcp 2n kt hp vi mt tinh t n to nn ni nh 3n.

- dng t bo 4n cn uc hnh thnh do nguyn phn ri lon xy ra trn tt c cc cpnhim sc th sau khi nhn i.

2. ngha ca vic hnh thnh cc dng t bo n, 2n ,3n, 4n

a. ngha vic hnh thnh dng t bo n

b. ngha vic hnh thnh dng t bo 2n, 3n, 4n:

- K tc vt cht di truyn n nh trong sinh sn sinh dng v sinh sn hu tnh- Nu t bo 2n c hnh thnh do gim phn ri lon l c s to nn cc dng 3n, 4n

qua th tinh hnh thnh nhng c trng mi v cc tnh trng hnh thi, sinh l, ho sinh choc th.

- thc vt dng 3n cn c hnh thnh do nhn th cp kt hp vi 1 tinh t to nnni nh 3n

- Vic to nn cc t bo c b nhim sc th khng bnh thng v s lng, y cng l

mt hng tin ho c bn thc vt trong t nhin.


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Cu 9: Cc c ch sinh hc xy ra nh th no i vi 1 cp NST tng ng cp TB?

Tr li

C nhiu c ch sinh hc xy ra i vi 1 cp NST tng ng. Mi c ch u c

ngha khc nhau gp phn n nh b NST ca loi.1. C ch nhn i ca nhim sc th:

Thc cht mi NST n trong cp NST tng ng l s nhn i ADN trn NST vo ktrung gian. Nh mi NST n to ra 1 NST kp.

2. C ch trao i on:

k trc I c xy ra hin tng tip hp v trao i cho gia 2 crmatit khc ngungc trong cp NST tung ng. S trao i cho gp phn to ra nhiu kiu giao t, to s

a dng, phong ph ca loi.3. C ch phn li:

- Trong nguyn phn, cc NST trong cp NST tng ng phn li ng u v micc ca TB gp phn to b NST 2n trong cc TB con.

- Trong gim phn I bnh thng, mi NST trong cp NST tng ng phn li v 2cc ca TB to ra b NST n bi th kp. mi TB con c 1 kiu sp xp NST khcnhau trn mt phng xch o k gia I. Ti k sau II ln phn bo II, mi crmatit trongtng NST kp tch nhau qua tm ng, kt qu mi giao t ch cha 1 NST n trong cptng ng.

4. C ch t hp t do ca NST:

Ti k gia I do mi TB c 1 kiu sp xp nn khi phn li s to ra cc loi giao t khcnhau.

- Nh th tinh phi hp ngu nhin cc giao t c vi giao t ci to li cp NSTtng ng, n nh th h sau.

5. C ch t bin d bi th:

Do nguyn nhn phng x, ho cht, c hc, sc li tm, sc nhit hoc qu trnh trao icht ni bo b ri lon. Cc tc nhn lm t day t v sc, hoc c ch hnh thnh day tv sc xy ra trong nguyn phn s to nn TB cha c 2 NST, TB khng cha NST ca cptng ng. Nu xy ra trong gim phn s to nn giao t di bi n-1, n+1. Cc giao t nykt hp vi giao t bnh thng to hp t cha 3 NST trong cp tng ng . Th dngi mc bnh ao c 3 NST th 21. Cc t bin d bi thng gy hu qu c hi.

6. C ch t bin cu trc NST:


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Do nguyn nhn bn trong TB nh ri lon trao i cht ni bo, bin i sinh l, hosinh hay do tc nhn phng x, ho hc tc ng vo NST lc chng ang nhn i s tonn cc t bin mt on, lp on, chuyn on, xy ra trong phm vi 1 cp NST. Cc t

bin cu trc NST thng gy ra hu qu c hi. t bin mt oan thng lm gim sc

sng hoc gy cht. ngi, mt on NST 21 gy ung th mu. t bin lp onthng gy hu qu khc nhau, hoc tng cng hoc gim bt mc biu hin ca tnh trng. rui gim lp on 2 ln NST X lm cho mt li thnh mt dt. i mch ,lp onlm tng hot tnh enzim amilaza rt c ngha trong cng nghip SX bia. o on thngt nh hng ti sc sng ca c th v vt cht di truyn khng b mt mt. Chuyn on cs phn b li cc gen gia cc NST khc nhau, 1 s gen trong nhm lin kt ny chuynsang nhm lin kt khc. S chuyn on thng gy cht hoc mt kh nng sinh sn. Tmli t bin cu trc NST th s lm ri lon s lin kt cc cp NST tng ng trong gim

phn, lm thay i t hp gen trong giao t dn ti bin i kiu gen v kiu hnh.Cu 10: Cu trc v chc nng ca NST thng v NST gii tnh ging nhau v khc nhau im no?

Tr li

1. im ging nhau ca NST thng v NST gii tnh v cu trc

- Trong TB sinh dng u tn ti thnh cp, mi cp NST gm 2 NST n thuc 2ngun gc. Trong TB giao t tn ti tng chic.

- Cc thnh phn c bn cu trc nn NST thng v NST gii tnh u l ADN vprtein (ch yu l histon)

- n v c bn to nn NST u l nuclxm, mi nuclxm gm 8 phn t prtein

histon pha ngoi bc bi431 vng xon gm khong 146 cp nucltit. Cc nuclxm ni

vi nhau bng cc on ni ADN (c kch thc 15-100 cp nucltit v 1 phn t H1

(Histn H1) to nn si c bn c chiu rng 100 0A . Si c bn tip tc xon thnh si

nhim sc. Si nhim sc ng xon tip thnh crmatit.- C kch thc, hnh dng c trng cho mi loi.

- u cha cc nhm gen lin kt hon ton hay lin kt khng hon ton.

- C kh nng t bin lm thay i s lng v cu trc NST

2. im khc nhau c bn

- NST thng gm nhiu cp NST, lun tn ti thnh tng cp NST tng ng. Cn

NST gii tnh ch tn ti c 1 cp khi th ng dng gii c, khi li ng dng gii citu thuc vo loi.


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-Gen trn NST thng phn b thnh tng cp gen tng ng. Gen trn NST gii tnhXY phn b thnh nhiu vng. V d ngi c 3 vng gen:

+ Vng gen c c trn X v Y

+ Vng gen ch c trn X+ Vng gen ch c trn Y

- i b phn cc gen phn b trn NST thng ch c 1 s gen c phn b trn NSTgii tnh.

3. im ging nhau v chc nng gia NST thng vi NST gii tnh.

- u gp phn quy nh tnh c trng b NST loi.

- u c kh nng nhn i, phn li, t hp t do, trao i on trong nguyn phn, gimphn v th tinh m bo di truyn n nh tng i qua cc th h.

- Cc gen trn NST c kh nng iu ho, tng hp ARN, ch huy tng hp prtein, hnh thnh tnh trng c trng cho loi.

- Cc t bin s lng, cu trc NST u c th hnh thnh cc tnh trng khng bnhthng, nh hng ti s tn ti v pht trin ca c th.

4. im khc nhau v chc nng:

- NST thng mang gen xc nh vic hnh thnh cc tnh trng thng, NST gii tnhquy dnh hnh thnh tnh trng c, ci sinh vt, cc tnh trng sinh dc ph v gen xcnh hnh thnh 1 s tnh trng lin quan n gii tnh, tun theo quy lut di truyn cho (gennm trn X), tun theo quy lut di truyn thng (gen nm trn Y).

- Cc t bin gen ln xut hin trn NST thng biu hin tnh trng chm hn. Cn gent bin ln trn NST X, Y c th biu hin kiu hnh ngay trong i c th v di truyn choth h sau.

Cu 11. Trnh by cc loi bin d lm thay i v s lng, cu trc NST?

Tr li

1. t bin s lng NST:

- S thay i s lng NST 1 hay 1 s cp NST to nn th d bi, ton b cp NSTto nn th a bi.

- C 2 nguyn nhn gy ra t bin s lng NST l tc nhn l ho trong ngoi cnhhoc MT ni bo lam cn tr s phn li ca cc cp NST k sau qu trnh phn bo.

- C 2 loi t bin s lng NST:


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+ Th d bi (lch bi): TB sinh dng ng nh cha 2 NST mi cp tng ng thli c th cha 3 NST ( th 3 nhim), nhiu NST (th a nhim), 1 NST (th 1 nhim) hocthiu hn 1 NST (th khuyt nhim). Cc t bin ny thng gy ra nhng hu qu c hi,v d 3 NST 21 ngi gy bnh ao. Dng XO biu hin hi chng tcn, XXY hi

chng claiphent.+ Th a bi, ngi ta chia thnh 2 loi: a bi chn (4n, 6n, ..) v a bi l (3n, 5n, .).

- Cc th a bi, d bi l ngun nguyn liu cho chn lc t nhin c ngha trong chnging cy trng.

2. Bin d lm thay i cu trc NST:

- Trao i on xy ra k trc I gim phn gia 2 cromatit khc ngun gc trong cptng ng, chng t ra cc on tng ng hoc khng tng ng v chuyn i cho

nhau. Trao i on NST dn ti sp xp li gen trong phm vi 1 cp NST to ra nhm genlin kt mi, lm thay i cu trc NST.

- t bin cu trc NST bao gm mt on, o on, lp on NST, chuyn on NST.Cc t bin ny lm thay i cu trc NST cc mc khc nhau.

Cu 12. im ging v khc nhau c bn gia bin d t hp v bin d t bin. Vai tr cacc lo bin d trong tin ho v chn ging.

Tr li

1. im ging- C 2 u l bin d di truyn, qua giao t vo hp t di truyn qua cc th h.

- C 2 u thuc bin d v hng c th c li, c hi hay trung tnh.

C th xut hin nhng bin d mi cha c b m hoc t tin.

- u l nhng bin d c lin quan n bin i vt cht di truyn.

2. im khc nhau:

- V nguyn nhn: Bin d t hp xut hin nh qu trnh giao phi, bin d t bin xuthin do tc ng ca MT trong v ngoi c th.

- V c ch: bin d t hp pht sinh do c ch phn li c lp, t hp t do, hon v gentrong qu trnh to giao t v s kt hp ngu nhin gia giao t c v ci trong qu trnhth tinh. Cn bin d t bin pht sinh do ri lon qu trnh phn bo hoc do ri lon qutrnh ti sinh NST, lm thay i s lng, cu trc vt cht di truyn.

- V tnh cht biu hin:

+ Bin d t bin biu hin 1 cch t ngt, ngu nhin, c bit, khng nh hng. Phnln t bin l c hi, tn s t bin c li l rt thp. Cn bin d t hp da trn c s t


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hp li cc gen vn c b m v t tin, v th c th lm xut hin cc tnh trng choc cha c th h trc. S phn li c lp, s hon v gen, tng tc qua li gia ccgen sp xp li tnh trng vn c hoc xut hin t hp kiu hnh mi th h sau.

+ Bin d t hp c th d on c quy m xut hin, tn s xut hin nu bit trcc kiu di truyn ca b m. Cn bin d t bin khng th ch ng xc nh trc khnng xut hin i con vi loi t bin v tn s t bin l bao nhiu.

+ Bin d t hp thng c gi tr khng c sc nh t bin, chng thng xut hin mc nh, a dng to ra ngun bin d thng xuyn, v tn cho chn lc t nhin.

3. Vai tr

- Bin d t hp l ngun ngyn liu bin d di truyn th cp cung cp cho tin ho.Nh cc bin d ny m tri qua lch s di t 1 vi loi ban u c th to ra nhiu loi mi.

Trong chon ging da trn c ch xut hin bin d t hp xut cc phng php laiging nhm nhanh chng to ging c gi tr.

- Bin d t bin l ngun nguyn liu bin d di truyn s cp cung cp cho tin ho.c bit t bin gen l ngun nguyn liu c bn. Trong chn ging da trn c ch xuthin, gi tr ca cc loi t bin, ngi ta xy dng cc phng php gy t bin nhmnhanh chng to ra cc t bin c gi tr, gp phn to ra ging cy trng v sinh vt cnng sut, phm cht cao, thch nghi tt.

B -PHN : BI TP

Bi 1. Mt gen c chiu di 5100 ngstron. Hiu s phn trm gia adenin vi mt loinucleotit khc bng 10% s nucleotit ca gen. Trn phn t mARN c tng hp t gen c 10% uraxin. mt mch n ca gen c 16% xitozin, s timin bng 150 nucleotit.

e) T l phn trm v s lng tng loi nucleotit ca gen.

f) T l phn trm v s lng tng loi ribonucleotit ca mt phn t mARN bng baonhiu?

g) Nu gen sao m 6 ln v trn mi phn t mARN c 10 riboxom trt qua khnglp li th s lng axit amin m mi trng ni bo cung cp cho qu trnh tng hp proteinl bao nhiu?

h) Nu thi gian gii m mt axit amin la 0,1 giy, thi gian tip xc ca mt phn tmARN vi cc riboxom l 58,1 giy, khong cch gia cc riboxom k tip khong baonhiu ngstron?

GII

a) T l % v s lng tng loi nucleotit


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- S nucleotit ca genl 5100

N= 2 . = 2 . = 3000Nu3,4 3,4

- T l % tng loi nucltit ca gen:N

A + G = = 50% (1)2

A G = 10% (2)=> A = T = 30% ; G = X = 20%- S lng tng loi nucltit ca gen

A = T = 3000 . 30% = 900Nu

G = X = 3000 . 20% = 600Nub) T l % v s lng tng loi ribnucltit3000

- S uraxin ca mARN : rU = . 10% = 150Nu2

- S nucltit tng loi mi mch:T1 = A2 = 150 ; A1 = T2 = A A2 = 900 150 = 750Nu

300X1 = G2 = . 16% = 240Nu

2G1 = X2 = G G2 = 600 240 = 360Nu

- U ca mARN c tng hp t A ca mch gc

rU = A gc => rU = A2 = 150. Vy mch 2 l mch gc

- T l % v s lng tng loi ribnucltit

750rA = T2 = 750 => . 100% = 50%; rU = 150 >10%

1500

360rX = G2 = X1 = 240 > 16%; rG = X2 = 360 > . 100% = 24%1500

c) S axit amin t doS phn t prtin : kn = 6.10 = 60- S axit amin t do cn dng:

rN 1500aatd = S P ( 1) = 60 ( 1) = 29940

3 3

d) Khong cch gia cc ribxm


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l- Vn tc trt ca ribxm : v = .3.3,4 = 102 / s

0,1- Thi gian k t RB1 tip xc cho n RB cui cng ht tip xc mARN :

l lT = t + t = + ( n- 1) v v

5100 l=> Phng trnh + (10-1) = 58,1

102 102=> l = 91,8

Bi 2: Mt gen di 5100 ngtron. Khi gen t sao lin tip hai t, mi trng ni bo cung cp 2700 anin. Phn t mARN c tng hp t gen c 600 adenin v 240guanin. vn tc gii m l 10 axit amin/ giy. Tnh t lc ribxm th nht trt qua phn tmARN cho n khi ht phn t mARN l 55,6 giy.

d) Xc nh s lng tng loi nucltit trong ton b cc gen c hnh thnh sau hait t sao lin tip

e) Xc nh s lng tng loi nucltit mi mch n ca gen

f) Tnh khong cch theo ngtron gia ribxm th nht vi ribxm cui cng khichng dang tham gia gii m trn mt phn t mARN

GII

a) S lng tng loi nucltit trong tt c cc gen con:

L 5100- S nucltit ca gen N = 2. = 2. = 3000Nu

3,4 3,4- S nucltit tng loi ca mi gen:

2700

Atd = A (22

1) = 2700 => A = T = = 900Nu22 1N 3000

G = X = - A = - 900 = 600 Nu2 2

- S nucltit tng poi trong ton b gen con:

A = T = 900 . 22 = 3600Nu

G = X = 600 . 22 = 2400Nu


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b) s lng tng loi nucltit mi mch gen:

Quy c mch gc l mch th nht

T1 = A2 = rA = 660; A1 = T2 = A A2 = 900 600 = 240Nu

X1 = G2 = rG = 240; G1 = X2 = G G2 = 600 240 = 360Nuc) khong cch gia ribxm th nht vi ribxoom cui cng

- Vn tc trt ca RB: v = 10 . 3 . 3,4 = 102/s

- Thi gian k t RB1 tip xc khi n trt qua mARNl 5100

t = = = 50 giyv 102

- Thi gian k t RB1 trt qua mARN n khi RB cui cng trt ht mARN =

khong cch v thi gian gia RB1 vi RB cui cngt = T t = 56,5 50 = 5,6 giy

- Khong cch gia RB1 vi RB cui cng

102 . 5,6 = 571,2

Bi 3: mch th nht ca gen c t l cc nucltit l 10% adenine, 20% timin v 25%xitozin. Phn t mARN c sao t gen c 20% urin.

d) tnh t l tng loi nucltit ca gen v tng loi ribnucltit ca phn t mARN

e) nu gen di 0,306 micromet th n cha boa nhiu lin kt hidro?

f) mt phn t mARN sinh ra t gen c chiu di ni trn v c mt s riboxom cnghot ng trong qu trnh gii m, thi gian riboxom th nht trt qua ht phn t l 30giy v thi gian tnh t lc bt u c s gii m n khi riboxom cui cng trt qua ht

phn t mARN l 35,4 giy. Hi c bao nhiu riboxom tham gia vo qu trnh gii m?Bit rang cc riboxom cch u nhau mt khong bng 61,2

GII

a) t l tng loi nucltit v tng loi ribnucltit

- T l tng loi nucltit cau rmi mch gen

A1 = T2 = 10% ; T1 = A2 = 20%

X1 = G2 = 25% ; G1 = X2 = 100% - ( 10% + 20% + 25%) = 45%

- T l tng loi ribnucltit ca c gen

%A1 + %A2 10% + 20%

A = T = = = 15%


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2 2%G1 + %G2 45% + 25%

G = X = = = 35%2 2

- U ca mARN c sao t A gc ca gen

rU = A gc => rU = A2 = 20%

Vy mch 2 l mch gc v t l 20% tng loi ribnucltit cca mARN :

rA = T2 = 10%; rU = A2 = 20%

rG = X2 = 45%; rX = G2 = 25%

b) s lin kt hidro ca gen

L 3060- S nucltit ca gen: N = 2 . = 2 . = 1800Nu

3,4 3,4

- S nucltit tng loi ca gen:

A = T = 1800 . 15% = 270; G = X = 1800 . 35% = 630Nu

- S lin kt hidro ca gen:

H = 2A + 3G = 2 . 270 + 3 . 630 = 2430lin kt

c) S riboxom

l 3060- Vn tc trt ca RB: v = = = 102/s

t 30

- Thi gian tnh t lc RB1 trt qua mARN n khi RB cui cng trt qua mARN =khong cch v thi gian gia RB1 vi RB cui cng

t= T t = 35,4 30 = 5,4 giy

- S riboxom tham gia:(n )

l 61,2(n 1 ) = t => (n 1 ) = 5,4 => n = 10

v 102Bi 4: Mch th nht ca gen c 240 timin, hiu s gia guanin vi adenine bng 10% s

nucleotit ca mch. mch th hai, hiu s gia adenin v xitozin bng 20% s nucleotitca mch. Khi gen tng hp phn te mARN th mi trng ni bo cung cp 360uraxin.


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d) T l phn trm v s lng tng loi nucleotit ca gen v ca tng mch l baonhiu?

e) Hy tnh chiu di ca phn t mARN, t l phn trm v s lng mi loiribonucleotit ca n.

f) Trn mi phn t mARN c 8 riboxom cng giag m, tnh t lc riboxom bt uftrt trn phn t mARN th thi gian riboxom th nht trt qua ht sphn t l 20giy, cn riboxom cui cng th phi cn n 26,3 giy mi hon tt vic gii m. khongcch u gia cc riboxom l bao nhiu ngtron? Bit rang cc riboxom trt vi vn tc

bng nhau

GII

a) t l v s lng tng loi nucltit ca gen v ca tng mch n

- t l % tng loi nucltit mi mchG1 = A1 = 10% => X2 T2 = 10% => T2 = X2 10% (1)A2 X2 = 10% => A2 = 10% + X2 (2)X2 G2 = 20% => G2 = X2 20% (3)t (1) (2) (3) => (X2 10%) + ( 10% + X2) + (X2 20%) + X2 = 100% X2 = 30%Suy ra G1 = X2 = 30%; X1 = G2 = 10%

A1 = T2 = 20%; T1 = A2 = 40%

- S lng tng loi nucltit mi mch240

T1 = A2 = 240; A1 = T2 = . 20% = 120Nu40%

240 240G1 = X2 = . 30% = 180; G2 = X1 = . 10% = 60Nu

40% 40%- T l v s lng tng loi nucltit tng ng ca c gen:

%A1 + %A2 20% + 40%A = T = = = 30%

2 2A =T = A1 + A2 = 120 + 240 = 360

%G1 + %G2 30% + 10%G = X = = = 20%

2 2G =X = G1 + G2 = 180 + 60 = 240

b) Chiu di, t l % v s lng tng loi ribnucltit ca mARN- Chiu di ca mARN

Nl = L = . 3,4 = (210 + 240 + 180 + 60) . 3,4 = 2040 2


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- k s phn t mARN (nguyn, dng)- U ca mARN tng hp t A gc ca gen => rU = k . A gc

360- Nu mch 2 l mch gc : k = = 1,5 ( loi)

260

360Vy mch 1 l mch gc vi s ln sao m k = = 3120

- T l % v s lng ribnucltit tng loi ca mARN240 120

rA = T1 = 240 => = 40%; rU = A1 = 120 => = 20%600 60060 180

rG = X1 = 60% => = 10%; rG = G1 = 180 => = 30%600 600

b) Khong cch gia cc riboxom

l 2040- Vn tc trt ca RB: v = = = 102/ st 20

- Gi l : khong cch gia cc RBThi gian lc RB1 bt u trt cho n khi RB cui cng trt ht phn t mARN

l lT = t + t = t + (n 1 ) => 20 + (8 1 ) = 26,3

v 102l = 91,8

Bi 5: Trn mt phn t mARN c mt s riboxom trt qua vi khong cch u bngnhau. Riboxom th nht trt qua ht phn t mARN ht 50 giy. Tnh t lc riboxomth nht trt qua v tip xc vi phn t mARN th riboxom cui cng ppji mt 57,2giy mi hon thnh vic i qua phn t mARN . Bit rang phn t prtin th hai ctng hp chm hn phn t prtin th nht 0,9 giy.

Gen iu khin vic tng hp cc phn t prtin ni trn c mch 1 cha 10% adeini v20% guanine, mch 2 cha 15% adenine. Qu trnh sao m ca gen i hi mi trngni bo cung cp 150 uraxin v 155 adenin gp phn tng hp mt phn t mARN

f) tnh chiu di ca geng) Tnh s lng tng loi ribnucltit ca mt phn t mARN

h) S riboxom tham gia vo qua trnh gii m trn mt phn t mARN l baonhiu? Bit rng mi riboxom ch trt qua mt ln

i) khong cch gia 2 riboxom k tip v khong cch gia riboxom th nht viriboxom cui cng tnh theo ngtron l bao nhiu?

j) Ton b qu trnh gii m ni trn cn bao nhiu axit amin ca mi trng ni bo

v trong tt c cc prtin hon chnh cha bao nhiu axit amin?


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GII

a) Chiu di ca gen

- S nucltit loi A,T ca gen

A = T = rA + rU = 225 + 150 = 375- T l % tng loi nucltit ca gen

%A1 + %A2 10% + 15%A = T = = = 12,5%; G = X = 37,5%

2 2375. 100

- S nucltit ca gen: N = = 300012,5

b) S kng tng loi ribnucltit

- U ca mARN c tng hp t A gc3000rU = A gc => rU = A1 = . 10% = 150

2Vy mch mt l mch gc- T k % tng loi nucltit mch gcA1 = 10%; T1 = A2 = 15%; G1 = 30%

X1 = 2 . %G - %G1 = 2 . 37,5% - 30% = 45%

S ribnucltit tng loi ca mARN : rA = 225rU = 1503000 3000

rG = X1 = . 45% = 675; rX = G1 = . 30% = 4502 2

c) S riboxom: gi n : s riboxom

Thi gian k t RB1 tip xc mARN n khi RB cui cng trt qua ht mARN :

T = t + t = t + (n 1 ) t => 50 + (n 1 )0,9 = 57,2 => n = 9

d) Khong cch gia 2RB, gia RB1 vi RB cui cng

l 5100

- Vn tc trt ca RB: v = = = 102/ s

t 50

- Khang cch gia 2RB: l = t . v = 0,9 . 102 = 91,8

- Khong cch gia RB1 vi RB cui cng:

(n 1 ) l = (9 1) 91,8 = 734,4

e) S axit amin

- Tng s axit amin t do cn dng:


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rN 1500 aatd = S P ( - 1) = 9 . ( - 1 ) = 4491

3 3

- Tng s axit amin ca cc phn t prtin hon chnh:

rN 1500 aap = S P ( - 2 ) = 9. ( - 2) = 4482

3 3

Bi 6: mt loi mt t bo sinh dc 2n thc hin s nguyn phn lin tip mt s ln,

i hi mi trng ni bo cung cp nguyn kiu hnh thnh nn 4826 nhim sc th n

mi. cc t bo con snh ra t ln nguyn phn cui cng u gim nhim bnh thng cho

256 tinh trng cha nhim sc th gii tnh Y

d) B nhim sc th lng bi ca loi? s ln nguyn phn lin tip ca t bo sinh dc

2n u tin? to ra cc t bo con 2n c bao nhiu thoi dy t v sc c hnh thnh

trong cc ln nguyn phn y?

e) Nu c 3 tinh trng c th tinh vi 3 trng khc nhau to ra cc hp t th c bao nhiu

cromait trong cc t bo sinh dc ci sinh ra cc trng , vo lc m cc t bo bt u thc

hin s phn bo gim nhim? Cc hp t va c hnh thnh c bao nhiu NST n?

f) Gi thit rng s gim phn xy ra trong cc t bo sinh dc ca c th ci, mi cptng ng u gm hai nhim sc th cu trc khc nhau, mi cp nhim sc th tng

ng u gm hai nhim sc th cu trc khc nhau, s trao i on ch xy ra mt cp

nhim sc th thng,s t bin d bi ch xy ra cp nhim sc th gii tnh. Kh nng

c th ci c th cho bao nhiu loi trng?

GII

a) B nhim sc th lng bi - s ln nguyn phn - s thoi v sc- S lng tinh trng X = s lng tinh trng Y > Tng s tinh trng hnh thnh: 256

+ 256 = 512

S t bo con sinh ra sau cc ln nguyn phn : 512 : 4 = 128 = 2 7

s l n nguyn phn = 7

4826- B nhim sc th 2n = = 38

27

1


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- Mi t bo sau khi nguyn phn th hnh thnh 1 thoi v sc

+ Ln nguyn phn th 1: 1 t bo nguyn phn > 1 thoi v sc

+ Ln nguyn phn th 2: 2 t bo nguyn phn > 2 thoi v sc

+ Ln nguyn phn th 3: 4 t bo nguyn phn > 4 thoi v scVy tng s thoi v sc l tng mt dy cp s nhn vi s shng u a 1 = 1 thoi v sc

t bo ban u

S hng ca dy n = s ln nguyn phn x

Cng bi q = 2

qx 1Sn = a1. = 1(2x 1 ) = 27 1 = 127

q 1b) S cromatit s NST n

- S t bo sinh duc ci sinh ra trng: 3

- Bt u gim phn, mi nhim sc th n ca tng t bo sinh dc ci u tnhn i thnh 1 NST kp gm 2 cromatit > S cromait ca t bo sinh dc ci lc bygi: 2 .83 . 3 = 228

- S hp t to thnh: 3

- S nhim sc th n: 38 . 3 =114c) S loi trng

- C th c cho 2 lo tinh trng X v Y => C th c mang cp NST gii tnh XY. Vyc th ci mang cp NST gii tnh XX.

- 2n = 38 => 19 cp gm 18 cp NST thng v 1 cp NST gii tnh

+ 18 cp NST thpng trong c 1 cp T => 2 18+1

+ Cp NST gii tnh XX c t bin d bi => 2 loi giao t bt thng (XX v O)

Vy s loi trng c th c:

218+1 . 2 = 220 = 1048576

Bi 7: gi thit trong cc cp nhim sc th tng ng ca mt b nhim sc th lng biu cha c cp gen d hp t v hin tng trao i on ti mt im ch xy ra vi mtcp nhim sc th tng ng. Cho bit khng c hin tng t bin v s loi giao t csinh ra t cc iu kin trn l 32.


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Gi thit trung bnh mi k trong phn bo nguyn phn ht 5 pht, hia ln phn bo (ktrung gian) ht 10 pht, qu trnh nguyn phn din ra lin tc, cc t bo con sinh ra utip tc nguyn phn.

c) B nhim sc th lng bi ni trn trng thi cha nhn i l bao nhiu?

d) hp t thc hin c qu trnh nguyn phn th mi trng ni bo cung cpnguyn kiu tng ng vi bao nhiu NST n vo cc thi im:

- Kt thc 20 pht

- Kt thc 32 pht

- Kt thc 100 pht

Bit rng khi hp t bc vo k trc c tnh l thi gian bt u

GIIa) B NST lng bi

n = s cp NST trong b NST lng bi

m = s cp NST c trao i on ti mt im

- S loi giao t 2n+mARN

2n+1 = 32 = 25 => n = 4; 2n = 8

b) NST c cung cp- thi gian ca mt chu k nguyn phn : 4 .5 +10 = 30 pht

- S NST tng ng vi nguyn liu c cung cp:

+ thi im 20 pht

Cha qua 1 chu k nguyn phn nhng NST u t nhn i 1 ln: 8(21 1) = 8

+ thi im 32 pht

Qua 1 chu k nguyn phn v ang k trc ca chu k th 2: NST t nhn i 2 ln:8(22 1) = 24

+ thi im 100 pht

Qua 3 chu k nguyn phn v ang k gia ca chu k th 4: NST t nhn i 4 ln: 8(24 1) = 120

Bi 8: g khi quan st mt t bo sinh dc c dang k gia ca nguyn phan, ngi tam c 78 nhim sc th kp

e) T bo nguyn phn 5 t lin tip i hi mi trng cung cp nguyn liu to ra bao nhiu nhim sc th n mi ?


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f) Loi t bo ny gim phn binhd thng, kh nng nhiu nht c th cho bao nhiuloi tinh trng trong trng hp khng c hin tng trao i on gia cc nhim sc thkp trong cp tng ng? iu kin cho s loi tinh trng nhiu nht l g?

g) Gi thit rng c 1000 t bo sinh tinh trng gim phn bnh thng, hiu

1sut th tinh ca tinh trng l , cn ca trng l 20%, mi tinh trng ch

1000

th tinh vi mt trng. Xc nh s t bo sinh trng.

h) Cc hp t c to thnh nguyn phn lin tip nhiu t vi s ln bng nhau,mi trung ni bo cung cp nguyn kiu to ra 2184 nhim sc th n mi. Xc nhs t bo con sinh ra v s t nguyn phan ca mi hp t.

GIIa) S NST cung cp:

- k gia ca nguyn phan, mi NST trong b NST 2n ca t bo u t nhn ithnh mt NST kp. m c lc c 78 b NST kp th => 2n = 78

- S NST tng ng vi nguyn liu c cung cp:

2n (2x 1 ) = 78(25 1 ) = 2418

b) S loi tinh trng

- S loi tinh trng trong iu kin khng c T 2n = 239

- Cho s loi tinh trng nhiu nht trong trng hp khng c trao i on ni trn,khi mi cp NST phi cho 2 loi tinh trng => 39 cp NST cho 2 39 loi tinh trng

Mun vy 2 NST thuc cng mt cp phi c cu trc khc nhau

c) S t bo sinh trng

- S tinh trng hnh thnh: 4 . 1000 = 4000

- S t bo trng th tinh = s tinh trng th tinh: 4000.1/4000 = 4- S t bo sinh trng = s trng hnh thnh: 4.100/20 = 20

d) S t bo con - s t nguyn phn

- S hp t to thnh = s trng th tinh = s tinh trng th tinh = 4

Gi x l s t nguyn phn ca mi hp t

s nhim sc th tng ng mi trng ni bo cung cp

4 .2n(2x

1) = 2148 => phng trnh 4.78(2x

1 ) = 2148 ; gii ra ta c x = 3


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- S t bo con sinh ra: 4.23 = 32

Bi 9: Mt t bo sinh duc c 2n v mt t bo sinh duc ci 2n u nguyn phn mt s tbng nhau (cc t bo con sinh ra u tip tc nguyn phn). Gi thit rng cc t bo consinh ra t t nguyn phn cui cng u gim nhim cho tng s 80 giao t binh thng.Cho bit s lng nhim sc th n trong cc giao t c nhiu hn s lng nhim sc thn trong cc giao t ci l 192.

3. Loi tn l g?

4. M t hnh dng v s lng b nhim sc th lng b trong loi

GII

1. Tn loi

Gi 2n: B NST lng bi ca loi; x: s ln nguyn phnS giao t c: 4.2x; s giao t ci: 2x

- Tng s giao t: 4.2x + 2x = 80 (1)

- S NST trong cc giao t c nhiu hn s NST trong cc giao t ci:

n . 4 . 2x n . 2x = 192 (2) Gii ra ta c x = 4 => 2n = 8 ; y l b NST ca rui gim

c) M t b NST

Khi NST xon ti a:

- 3 cp NST thng: 2 cp hnh ch V, 1 cp hnh ht

- 1 cp NST gii tnh: 2 chic hnh que XX con ci, chic hnh que X, chc hnh mcY con c

Bi 10: Mt g mi c mt s trng, nhng khi p ch c 12 trng n thnh g con.Cc hp t n thnh g con c 936 nhim sc th n trng thi cha t nhn i. S trngcn li khng n thnh g con.

s tinh trng c sinh ra phuc v cho g giao phi c 624000 nhim sc th n. Gi

thit s tinh trng c trc tip th tinh vi cc trng ni tren chim t l 1/1000 so vitng s tinh trng c hnh thnh. Mi tinh trng th tinh vi 1trng

a) S trng c th tinh

b) Trng g khng n thnh g con c b nhim sc th nh th no?

c) S g trng v g mi trong n g con ni trn c tun theo t l 1: 1 khng?

GII

a) S trng c th tinh

- B NST lng bi ca g = s NST trong mi hp t : 2n = 936/12 = 78- S tinh trng c hnh thnh 62400 : (78 : 2) = 16000


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- S trng th tinh = s tinh trng th tinh 16000. 1/1000 = 16

b) B NST ca trng khng n

g trng th tinh hoc khng u c ra

- Trng khng c th tinh th lun lun khng n => b NST n = 39

- Trng c th tinh nhng khng gp iu kin thun li ca mi trng p th khngn => b NST 2n = 78

c) T l trng mi

- n g con c th gm: 6 g trng v 6 g mi tun theo t l 1:1

- n g con c th c s g trng khng bng s g mi khng tun theo t l 1:1. Bi vt l ny ch nghim ng trn s lng ln c th

NI DUNG TRNG TM TNG PHN - KINH NGHIM GING DY

A- NI DUNG TRNG TM TNG PHN-Kin thc

- Nu c nhng bng chng trc tip v gin tip ADN l vt cht di truyn.

- Trnh by c nhng din bin c th ca c ch sao chp ADN t bo nhn s vnhn thc, trong ch ti:

+ Vai tr ca cc ezim, cc protein.

+ Chiu tng hp ca mch mi.

+ Cc nguyn tc: bn bo tn, khun mu v nguyn tc b sung.

- Trnh by c mt s sai khc ca ADN ca t bo nhn thc so vi t bo nhn s.

- Trnh by c c im c ch ti bn ca axit amin mt s virut: X174, TMV,HIV v

- Trnh by c nguyn tc, phng php v ngha ca lai phn t.

- Pht biu nh ngha gen. Gii thch c cu trc ca gen sinh vt nhn s vnhn thc, dc bit l chc nng hay vai tr ca cc vng khi u, m ho v kt thc. Phn

bit c gen khng phn mnh v gen phn mnh. Nu c khi nim gen nhy hay ccyu t di truyn di ng cng nh vai tr v ngha ca chng.

- Gii thch c cc c im ca m di truyn. lp lun c v sao m di truyn trn lthuyt l m b ba. Trnh by c phng php thc nghim xc nh cc b ba m ho.

- Trnh by c nhng din bin c th ca c ch phin m. Nu c mt s cim ca t bo nhn thc khc vi t bo nhn s.

- Trnh by c nhng din bin chi tit ca c ch dch m. Phn tch c mi quanh gia ADN- protein- tnh trng.

- Trnh by c c ch iu ho hot ng ca gen sinh vt nhn s (theo m hnhca Mn v Jacp). Nu c khi nim pron. Nu c s khc bit gia c ch iuho dng tnh vi iu ho dng tnh ca gen sinh vt nhn s. Nu c mt s dc


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im ca c ch iu ho hot ng ca gen sinh vt nhn thc ( cc mc trc phinm, phin m v sau phin m)

- Trnh by c nguyn nhn, c ch, c im, biu hin, vai tr ca cc dng tbin gen. Trnh by c c ch sa sai nhng bin i ADN trong sao chp.

- Phn tch c cu trc siu hin vi ca NST. Gii thch c s bin i hnh thiNST qua cc k phn bo v cu trc NST c duy tr lien tc qua chu k t bo. Nu ckiu nhn v nhim sc .

- Trnh by c nguyn nhn c ch pht sinh, c im va vai tr ca cc dng tbin cu trc NST (mt on, lp on, o on v chuyn on) v s lng NST (lchbi v a bi)

- K nng:

- Lm c th nghim n gin v tch chit ADN.

- Bit lm tiu bn tm thi NST, xem tiu bn c nh nhn dng c mt vi tbin s lng NST di knh hin vi quang hc.

- Gii c cc bi tp di truyn phn t v t bo.

B -KINH NGHIM GING DY

- Phn nhm cho hc sinh lm chuyn tng phn v bo co.

- Thng xuyn kim tra v nh gi.

CHNG 2. TNH QUI LUT CA HIN TNG DI TRUYN

A. C S L THUYT

I. mt s khi nim v thut ng1. Phng php phn tch ging lai:

- Lai cc cp b m thun chng khc nhau v 1 hoc vi cp tnh trng, theo di ringcon chu ca tng cp b m.

- Dng ton thng k v l thuyt xc sut phn tch QLDT cc tnh trng ca b mcho cc th h sau.

2.Tnh trng l nhng c im c th v hnh thi, cu to sinh l ring cho 1 c th no ,

gip ta d dng phn bit n vi nhng c th khc. Vd:u thn cao, ht vng, ...


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a. Tnh trng tng ng l nhng biu hin khc nhau ca cng 1 loi tnh trng.VD:ht vng,ht xanh,ht en, ht trng.

b. Tnh trng tng phn l 2 tnh trng tng ng c biu hin tri ngc nhau.VD: ht muen v ht mu trng.

3. Cp gen tng ng:l cp gen nm v tr tng ng trn cpNST tng ng v quy nhcp tnh trng tng ng .VD: cp gen tng ng A,a nm v tr i din trn cp NST s 5 cy c chua quy nh dng qu khng mi hoc c mi.

4. Alen l nhng trng thi khc nhau ca cng 1 gen. Alen ny khc alen kia 1 hoc mt scp nu no , sn phm ca qu trnh t bin gen.

5. Kiu gen l t hp ton b cc gen trong t bo c th. Thng thng khi ni ti KG ca 1c th ngi ta ch xt 1 vi xp gen no lin quan vi cc tnh trng ang c quan

tm .VD:AABB, AaBB..6. Kiu hnh l ton b cc tnh trng ca c th. Thc t khi ni ti kiu hnh ca 1 c th, ngita ch xt 1 vi tnh trng m ngi ta quan tm .VD: u thn cao, hoa trng, ht vng .

7. Th ng hp: Th ng hp v mt gen no l trng hp 2 alen trong cp tng ngging nhau.VD:aa, BB,...

8. Th d hp: Th d hp v 1 gen no l trng hp 2 alen trong cp tng ng l khcnhau.VD:Aa, Bb,...

9. Ging thun chng l ging c c tnh di truyn ng nht v n nh, cc th h con chukhng phn ly, c KH ging b m. Trn thc t, ni ti ging thun chng l ni ti s thunchng v 1 hoc vi tnh trng no ang c nghin cu, cp gen quy nh tnh trng nyl ng hp.

10. Mt s k hiu thng dng:

P: cp b m xut pht,

x : K hiu cho php lai.

G: giao t.

F: th h con.

II. Quy lut phn ly

1. Phng php nghin cu khoa hc ca Menen

1.1. i tng nghin cu : u H Lan

1.2. To cc dng thun : Bng phng php t th phn bt buc qua nhiu th h

1.3. Theo di s di truyn ca tng tnh trng

1.4. S dng ton thng k v l thuyt xc sut

1.5. Tun th mt quy trnh nghin cu khoa hc c o


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- Quan st thu thp s liu: Lai cc dng thun chng khc nhau v kiu hnh tng phnri theo di t l phn li kiu hnh cc i con

- a ra gi thuyt khoa hc gii thch nhng g quan st c:

- Lm th nghim kim tra tnh ng n ca gi thuyt m mnh a ra: Lm th nghimlai phn tch kim tra tnh ng n ca gi thuyt

2. Ni dung.

2.1.Th nghim:SGK

2.2. Gii thch: Mi tnh trng c th do 1 cp nhn t DT quy nh m sau ny gi l gen.S phn ly v t hp ca cc nhn t DT chi phi s DT v biu hin ca cc cp tnh trngtng phn qua cc th h.

2.3. Quy lut phn ly: Mi tnh trng c quy nh bi 1 cp alen. Do s phn ly ng uca cp alen trong gim phn nn mi giao t ch cha 1 alen ca cp.

3.C s t bo hc

Trong t bo sinh dng, NST tn ti thnh tng cp, do gen cng tn ti thnh tngcp alen tng ng trn cp NST tng ng.

Khi gim phn mi NST trong cp phn ly v mi giao t. V vy mi giao t ch mang Ahoc a. S t hp ca cp NST tng ng trong th tinh hnh thnh F1 c KG Aa.

F1 gim phn hnh thnh nn 2 loi giao t A v ac to thnh vi xc sut ngang nhau.S th tinh ca 2 loi giao t c v ci to F2 c t l KG:1Aa:2Aa: 1aa. V A tri t honton alen ln a nn th ng hp tri v th d hp c KH nh nhau. Do F2 c t l 3 : 1trng, tnh trng ln c biu hin th ng hp ln.

4. C s phn t ca tnh tri - ln

4.1. C s phn t ca tnh trng ln

Mt alen t bin no c coi l ln khi c th d hp t v alen t bin c kiu hnhging nh kiu hnh ca c th c kiu gen ng hp tri. Ni cch khc, trng hp c thch cn 1 alen cng to ra sn phm duy tr 1 chc nng sinh hc nht nh th gen khi b t bin thng to ra cc alen ln. Xt trng hp sn phm ca gen l 1 E xc tc cho1 phn ng nht nh trong t bo, khi c 2 alen bnh thng th lng E c to ra trong t

bo s cao gp i so vi trng hp c th c kiu gen d hp. Tuy nhin trong nhiutrng hp, E khng hot ng vi tc cc i khi nng c cht trong t

Ky Yeu Hoi Thao Dbbb Mon Sinh Hoc-2010

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