bai tap di truyen.pdf

7/29/2019 bai tap di truyen.pdf

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cng bi tp sinh hc 12 luyn thi i hc

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I. DI TRUYN HC PHN T.1. Tnh s nucltit ca ADN hoc ca gena. i vi mi mch ca gen :- Trong ADN, 2 mch b sung nhau , nn s nu v chiu di ca 2 mch bng nhau

A1 + T1 + G1 + X1 = T2 + A2 + X2 + G2 = 2

N

A1 = T2 ; T1 = A2 ; G1 = X2 ; X1 = G2b. i vi c 2 mch :- S nu mi loi ca ADN l s nu loi c 2 mch :

A =T = A1 + A2 = T1 + T2 = A1 + T1 = A2 + T2G =X = G1 + G2 = X1 + X2 = G1 + X1 = G2 + X2Ch :khi tnh t l %

%A = % T =

2

2%1% AA

2

2%1% TT = ..

%G = % X =

2

2%1% GG

2

2%1% XX

=.c. Tng s nu ca ADN (N)Tng s nu ca ADN l tng s ca 4 loi nu A + T + G+ X . Nhng theo nguyn tc b sung(NTBS) A= T, G=X . V vy, tng s nu ca ADN c tnh l :

N = 2A + 2G = 2T + 2X hay N = 2( A+ G)

Do A + G =2

N hoc %A + %G = 50%

d. Tnh s chu k xon ( C )Mt chu k xon gm 10 cp nu = 20 nu . khi bit tng s nu ( N) ca ADN :

N = C . 20 => C =20

N

e. Tnh khi lng phn t ADN (M ) :Mt nu c khi lng trung bnh l 300 vc. khi bit tng s nu suy ra

M = N x 300 vcf. Tnh chiu di ca phn t ADN ( L ) :

L =2

N. 3,4A0

n v thng dng :1 micrmet = 10 4 angstron ( A0 )1 micrmet = 103 nanmet ( nm)1 mm = 103 micrmet = 106 nm = 107 A0

2. Tnh s lin kt Hir v lin kt Ha Tr Pa. S lin kt Hir ( H )

H = 2A + 3 G hoc H = 2T + 3Xb. S lin kt ho tr ( HT )

S lin kt ho tr ni cc nu trn 1 mch gen :2

N- 1

Trong mi mch n ca gen, 2 nu ni vi nhau bng 1 lk ho tr, 3 nu ni nhau bng 2 lin

kt ho tr 2

N nu ni nhau bng2

N - 1

S lin kt ho tr ni cc nu trn 2 mch gen : 2(2

N - 1 )


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Do s lin kt ho tr ni gia cc nu trn 2 mch ca ADN : 2(2

N - 1 )

S lin kt ho tr ng photpht trong gen ( HT-P)Ngoi cc lin kt ho tr ni gia cc nu trong gen th trong mi nu c 1 lk ho tr gn thnh

phn ca H3PO4 vo thnh phn ng . Do s lin kt ho tr P trong c ADN l :HT-P= 2(

2

N- 1 ) + N = 2 (N 1)

3. C ch nhn i ca AND.a. Tnh nucltit mi trng cung cpQua 1 ln t nhn i ( t sao , ti sinh , ti bn )+ Khi ADN t nhn i hon ton 2 mch u lin kt cc nu t do theo NTBS : AADN ni viTT do v ngc li ; GADN ni vi X T do v ngc li . V vy s nu t do mi loi cn dngbng s nu m loi n b sung

Atd=Ttd= A = T ; Gtd= Xtd= G = X

+ S nu t do cn dng bng s nu ca ADNNtd= NQua nhiu t t nhn i ( x t )+ Tnh s ADN con- 1 ADN m qua 1 t t nhn i to 2 = 21 ADN con- 1 ADN m qua 2 t t nhn i to 4 = 22 ADN con- 1 ADN m qua3 t t nhn i to 8 = 23 ADN con- 1 ADN m qua x t t nhn i to 2x ADN conVy : Tng s ADN con = 2x- D t t nhn i no, trong s ADN con to ra t 1 ADN ban u, vn c 2 ADN conm mi ADN con ny c cha 1 mch c ca ADN m . V vy s ADN con cn li l c c 2mch cu thnh hon ton t nu mi ca mi trng ni bo .

S ADN con c 2 mch u mi = 2x 2+ Tnh s nu t do cn dng :- S nu t do cn dng th ADN tri qua x t t nhn i bng tng s nu sau cng trongcc ADN con tr s nu ban u ca ADN mTng s nu sau cng trong trong cc ADN con : N.2x

S nu ban u ca ADN m :NV vy tng s nu t do cn dng cho 1 ADN qua x t t nhn i :

N td= N .2x N = N( 2X-1)

S nu t do mi loi cn dng l:

A td= T td = A( 2X-1)G td= X td = G( 2

X-1)+ Nu tnh s nu t do ca ADN con m c 2 mch hon tan mi :

Ntd hon ton mi = N( 2X- 2)

A td hon ton mi= T td = A( 2X-2)

G td hon ton mi = X td = G( 2X2)

4.Tnh s nucltit ca ARN:

- ARN thng gm 4 loi ribnu : A ,U , G , X v c tng hp t 1 mch ADN theoNTBS . V v s ribnu ca ARN bng s nu 1 mch ca ADN


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rN = rA + rU + rG + rX =2

N

- Trong ARN A v U cng nh G v X khng lin kt b sung nn khng nht thit phi bngnhau . S b sung ch c gia A, U , G, X ca ARN ln lt vi T, A, X, G ca mch gc

ADN . V vy s nucltit mi loi ca ARN bng s nu b sung mch gc ADN .rA = T gc ; rU = A gcrG = X gc ; rX = Ggc

* Ch : Ngc li , s lng v t l % tng loi nu ca ADN c tnh nh sau :+ S lng :

A = T = rA + rUG = X = rR + rX+ T l % :

% A = %T =2

%% rUrA

%G = % X = 2

%% rXrG

5. Tnh khi lng ARN (MARN)Mt nucltit c khi lng trung bnh l 300 vc, nn:

MARN = rN . 300vc =2

N. 300 vc

6. Tnh chiu di v s lin kt ho tr (lin kt phosphodieste) p ca ARNa. Tnh chiu di :- ARN gm c mch rN ribnu vi di 1 nu l 3,4 A0 . V vy chiu di ARN bng chiu

di ADN tng hp nn ARN - V vy LADN =LARN = rN . 3,4A0 =

2

N. 3,4 A0

b. Tnh s lin kt ho tr P:+ Trong chui mch ARN : 2 ribnu ni nhau bng 1 lin kt ho tr , 3 ribnu ni nhaubng 2 lin kt ho tr Do s lin kt ho tr ni cc ribnu trong mch ARN l rN 1+ Trong mi ribnu c 1 lin kt ho tr gn thnh phn axit H3PO4 vo thnh phn ng .Do s lin kt ha tr loi ny c trong rN ribnu l rN

Vy s lin kt ho tr P ca ARN:HTARN= rN 1 + rN = 2 .rN -1

7. Tnh s nucltit cn dnga . Qua 1 ln sao m :Khi tng hp ARN, ch mch gc ca ADN lm khun mu lin cc ribnu t do theo NTBS :

AADN ni U ARN ; TADN ni A ARNGADN ni X ARN ; XADN ni G ARN

V vy :+ S ribnu t do mi loi cn dng bng s nu loi m n b sung trn mch gc ca ADN

rAtd= Tgc ; rUtd= AgcrGtd= Xgc; rXtd= Ggc

+ S ribnu t do cc loi cn dng bng s nu ca 1 mch ADN

rNtd = 2

N

b. Qua nhiu ln sao m ( k ln )


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Mi ln sao m to nn 1 phn t ARN nn s phn t ARN sinh ra t 1 gen bng sln phin m ca gen .

S phn t ARN = S ln sao m = K+ S ribnu t do cn dng l s ribnu cu thnh cc phn t ARN . V vy qua K ln

sao m to thnh cc phn t ARN th tng s ribnu t do cn dng l:

rNtd = K . rN+ Suy lun tng t, s ribnu t do mi loi cn dng l :

rAtd = K. rA = K . Tgc; rUtd = K. rU = K . Agc

rGtd = K. rG = K . Xgc; rXtd = K. rX = K . Ggc

8. Tnh s b ba m ha- s axit amin+ C 3 nu k tip nhau trn mch gc ca gen hp thnh 1 b ba m gc , 3 ribnu k tipca mch ARN thng tin ( mARN) hp thnh 1 b ba m sao . V s ribnu ca mARN bngvi s nu ca mch gc , nn s b ba m gc trong gen bng s b ba m sao trong mARN .

S b ba mt m =3.2

N =3

rN

+ Trong mch gc ca gen cng nh trong s m sao ca mARN th c 1 b ba m kt thckhng m ho a amin . Cc b ba cn li co m ho a.amin

S b ba c m ho a amin (a.amin chui polipeptit)=3.2

N- 1 =

3

rN- 1

+ Ngoi m kt thc khng m ha a amin , m m u tuy c m ha a amin , nhng a aminny b ct b khng tham gia vo cu trc prtin

S a amin ca phn t prtin (a.amin pr hon chnh )=3.2

N- 2 =

3

rN- 2

9. Tnh s lin kt peptitS lin kt peptit hnh thnh = s phn t H2O to raHai axit amin ni nhau bng 1 lin kt pptit , 3 a amin c 2 lin kt peptit ..chuipolipeptit c m l a amin th s lin kt peptit l :

S lin kt peptit = m -1

10. Tnh s axit amin cn dngTrong qu tnh gii m, tng hp prtein, ch b ba no ca mARN c m ho a amin th mic ARN mang a amin n gii m .Gii m to thnh 1 phn t prtein:

Khi ribxm chuyn dch t u ny n u n ca mARN hnh thnh chui polipeptitth s a amin t do cn dng c ARN vn chuyn mang n l gii m m u v cc mk tip , m cui cng khng c gii . V vy s a amin t do cn dng cho mi ln tng hpchui polipeptit l :

S a amin t do cn dng : S aatd =3.2

N- 1 =

3

rN- 1

Khi ri khi ribxm, trong chui polipeptit khng cn a amin tng ng vi m m u .Do, s a amin t do cn dng cu thnh phn t prtin (tham gia vo cu trc prtin thc hin chc nng sinh hc) l :S a amin t do cn dng cu thnh prtin hon chnh :

S aap = 3.2N - 2 = 3rN - 2


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BI TP P DNG.1. Trn mt mch ca gen c 150 anin v 120 timin. Gen ni trn c 20% guanin. S lngtng loi nucltit ca gen l:A. A = T = 180; G = X =270 B. A = T = 270; G = X = 180C. A = T = 360; G = X = 540 D. A = T = 540; G = X = 3602. Mt gen c chiu di 1938 ngstron v c 1490 lin kt hir. S lng tng loi nucltitca gen l:A. A = T = 250; G = X = 340 B. A = T = 340; G = X = 250C. A = T = 350; G = X = 220 D. A = T = 220; G = X = 3503. Mt gen di 0,408 micrmet v c t l tng loi nucltit bng nhau. Phn t mARN dogen tng hp c cha 15% uraxin v 20% guanin. S lin kt hir ca gen ni trn l:A. 3900 lin kt B. 3600 lin kt C. 3000 lin kt D. 2400 lin kt4. Trn mt mch ca gen c 25% guanin v 35% xitzin. Chiu di ca gen bng 0,306micrmetS lng tng loi nucltit ca gen l:

A. A = T = 360; G = X = 540 B. A = T = 540; G = X = 360C. A = T = 270; G = X = 630 D. A = T = 630; G = X = 2705. Chui plipeptit c iu khin tng hp t gen c khi lng 594000 n v cacbon chabao nhiu axit amin?A. 328 axit amin B. 329 axit amin C. 330 axit amin D. 331 axit amin6. Mt gen c khi lng phn t l 72.104 vC. Trong gen c X = 850. Gen ni trn t nhni 3 ln th s lng tng loi Nu t do mi trng cung cp l :A.ATD = TTD = 4550, XTD = GTD = 3850 B.ATD = TTD = 3850, XTD = GTD = 4550C.ATD = TTD = 5950, XTD = GTD = 2450 D.ATD = TTD = 2450, XTD = GTD = 59507. Trong mt on phn t AND c khi lng phn t l 7,2.105 vC, mch 1 c A1 + T1 =60%, mch 2 c G2 X2 = 10%, A2 = 2G2. Nu on AND ni trn t nhn i 5 ln th s

lng tng loi Nu mi trng cung cp l :A.ATD = TTD = 22320, XTD = GTD = 14880 B.ATD = TTD = 14880, XTD = GTD = 22320C.ATD = TTD = 18600, XTD = GTD = 27900 D.ATD = TTD = 21700, XTD = GTD = 24800

8. Mt gen c s lin kt hir l 3450, c hiu s gia A vi mt loi Nu khng b sung l20%. Gen ni trn t nhn i lin tip 5 t th s lng tng loi Nu mi trng cungcp cho qu trnh t nhn i trn ca gen l :A.ATD = TTD = 13950, XTD = GTD = 32550 B.ATD = TTD = 35520, XTD = GTD = 13500

C.ATD = TTD = 32550, XTD = GTD = 13950 D.ATD = TTD = 13500, XTD = GTD = 355209. Phn t ADN gm 3000 nucltit c s T chim 20%, thA. ADN ny di 10200 vi A=T=600, G=X=900B. ADN ny di 5100 vi A=T=600, G=X=900

C. ADN ny di 10200 vi G=X=600, A=T=900D. ADN ny di 5100 vi G=X=600, A=T=90010. Nu nui cy ADN trong mi trng c nit phng x 15N, ri chuyn sang mi trng(ch c 14N), th 1 ADN 1o t sao 5 ln lin tip, s sinh ra s mch n cha 15N l

A. 4 B. 64 C. 2 D. 12811. Mt gen c hiu s gia guanin vi anin bng 15% s nuclotit ca gen. Trn mch thnht ca gen c 10% timin v 30% xitzin. Kt lun sau y ng v gen ni trn l:A. A2 = 10%, T2 = 25%, G2= 30%, X2 = 35%. B. A1 = 7,5%, T1 = 10%, G1= 2,5%, X1 =30%.C. A1 = 10%, T1 = 25%, G1= 30%, X1 = 35%. D. A2 = 10%, T2 = 7,5%, G2= 30%, X2 =2,5%.12. Mt gen c khi lng 540000 vC c 2320 lin kt hidr. S lng tng loi nucltit nitrn bng:A. A = T = 380, G = X = 520. B. A = T = 520, G = X = 380.


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C. A = T = 360, G = X = 540. D. A = T = 540, G = X = 360.

13. C mt phn t ADN c khi lng bng 75.107 n v cacbon v t l 32

A

G t nhn i

3 ln.

S lng tng loi nucltit mi trng cung cp cho qu trnh nhn i ni trn l:A. G = X = 3,5.106, A = T = 5,25.106. B. G = X = 3,25.106, A = T = 5,5.106.C. G = X = 3,25.106, A = T = 5,5.105. D. G = X = 3,5.105, A = T = 5,25.105.14. Trn mt mch ca phn t ADN c s nucltit cc loi: A=60; G=120; X=80; T=30. Mtln nhn i ca phn t ADN ny i hi mi trng ni bo cung cp cho tng loinucltit ca gen l:A. A = T = 90; G = X = 200. B. A = G = 180; T = X = 110.C. A = T = 180; G = X = 110. D. A = T = 150; G = X = 140.15. Mt phn t mARN gm hai loi nucltit A v U th s loi b ba phin m trong mARNc th l:A. 8 loi. B. 6 loi. C. 4 loi. D. 2 loi.

16. Mt mch n ca phn t ADN c trnh t cc nucltit nh sau:.A T G X A T G G X X G X .Trong qu trnh nhn i ADN mi c hnh thnh t on mch ny s c trnh tA..T A X G T A X X G G X G. B.A T G X A T G G X X G XC.U A X G U A X X G G X G. D.A T G X G T A X X G G X T.17. Mt gen thc hin 2 ln phin m i hi mi trng cung cp nucltit cc loi: A =400; U = 360; G = 240; X = 280. S lng tng loi nucltit ca gen:A. A = T = 380; G = X = 260. B. A = T = 60; G = X = 520.C. A = T = 360; G = X = 240. D. A = T = 180; G = X = 240.18. Mt gen c chiu di l 4080 A0 c nucltit loi A l 560. Trn mt mch c nucltit A =260; G = 380, gen trn thc hin mt s ln phin m cn mi trng ni bo cung cp

nucltit U l 600. S lng cc loi nucltit trn mch gc ca gen l:A. A = 260; T = 300; G = 380; X= 260. B. A = 380; T = 180; G = 260; X = 380.C. A = 300; T = 260; G = 260; X = 380. D. A= 260; T = 300; G = 260; X = 380.19. Mt gen c 20% anin v trn mch gc c 35% xitzin. Gen tin hnh phin m 4 ln v s dng mi trng tng s 4800 ribnucltit t do. Mi phn t mARN c to ra ccha 320 uraxin. S lng tng loi ribnucltit mi trng cung cp cho phin m l:A. rA = 640, rU = 1280, rG = 1680, rX = 1200. B. rA = 480, rU = 960, rG = 1260, rX = 900.C. rA = 480, rU = 1260, rG = 960, rX = 900. D. rA = 640, rU = 1680, rG = 1280, rX =1200.. 20. Mt phn t mARN trng thnh c chiu di 5100A0, phn t prtin tng hp tmARN c:A. 498 axit amin. B. 600 axit amin. C. 950 axit amin. D. 499 axit amin.21. Cho bit cc b ba i m tng ng vi cc loi axit amin nh sau:AXX: triptphan GAA: lxin UGG: thrnin. XGG: alanin UUX: lizin.Trt t cc axit amin ca on m u ca mt chui plipeptit sau khi c tng hp nhsau:...Lizin-alanin-thrnin-lxin-triptphan...on m u ca phn t mARN dch m chui plipeptit ni trn c trt t cc b baribnuclotit l:A....UUX-XGG-UGG-GAA-AXX.... B...AAG-GXX-AXX-XUU-UGG...C...UAA-AAG-GXX-AXX-XUU-UGG... D...AUG-AAG-GXX-AXX-XUU-UGG...

II. PHN TON DI TRUYN THEO QUI LUT PHN LI V PHN LI C LP.A. TNH S LOI V THNH PHN GEN GIAO T1. S loi giao t :


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Khng tu thuc vo kiu gen trong KG m tu thuc vo s cp gen d hp trong :+ Trong KG c 1 cp gen d hp 21 loi giao t

+ Trong KG c 2 cp gen d hp 22loi giao t+ Trong KG c 3 cp gen d hp 23 loi giao t+ Trong KG c n cp gen d hp 2n loi giao t2 . Thnh phn gen (KG) ca giao t :Trong t bo (2n) ca c th gen tn ti thnh tngcp tng ng , cn trong giao t (n) ch cn mang 1

gen trong cp+ i vi cp gen ng hp AA ( hoc aa) : cho 1loi giao t A ( hoc 1 loi giao t a )+ i vi cp gen d hp Aa:cho 2 loi giao t vi t lbng nhau giao t A v giao t a+ Suy lun tng t i vi nhiu cp gen d hp

name trn cc cp NST khc nhau , thnh phn kiugen ca cc loi giao t c ghi theo s phnnhnh ( s Auerbac ) hoc bng cch nhn i s

V d : Kiu gen :AaBbDd cho giao t (s trn)

B. TNH S KIU T HP , KIU GEN , KIU HNH V CC T L PHN LI ICON1. S kiu t hp :

S kiu t hp = s loi giao t c x s loi giao t ci

Ch : + Bit kiu t hp => bit s loi giao t c , giao t ci => bit c cp gend hp trong kiu gen ca cha m

+ Kiu t hp khc nhau nhng c th a n kiu gen ging nhau => s KG< s kiu t hp .

S loi giao t v t l phn li v kiu gen(KG) , kiu hnh (KH):S di truyn ca cc gen l c lp vi nhau => s t hp t do gia cc cp gen cng nhgia cc cp tnh trng .V vy , kt qa v kiu gen cng nh v kiu hnh i con ctnh nh sau :

+ T l KG chung ca nhiu cp gen = cc t l KG ring r ca mi cp gennhn vi nhau

=> S KG tnh chung = s KG ring ca mi cp gen nhn vi nhau+ T l KH chung ca nhiu cp tnh trng = cc t l KH ring r ca mi cptnh trng nhn vi nhau

2. Kiu gen chung ca nhiu loi tnh trngCho lai 2 cy cha r KG v KH vi nhau thu c F1 gm : 3 cy trn ;3 bu dc ;1vng trn ; 1 vng bu dc . Cc cp gen nm trn cc cp NST khc nhau .Tm KG 2 cythuc th h P

Xt ring tng cp tnh trng :+ F1gm (3+3) : ( 1 + 1) vng = 3 : 1 vng ( theo L ng tnh ) =>P : Aa x Aa+ F1gm (3 +1 ) trn : (3 + 1 ) bu dc = 1 trn : 1 bu dc ( lai phn tch d hp ) => P : Bb xbb

Xt chung : Kt hp kt qa v KG ring ca mi loi tnh trng trn => KG ca Pl: AaBb x AaBb .

A

a

B

b

B

b

D

d

d

d

D

D

D

d

ABD

ABd

AbD

Abd

aBD

aBd

abD

abd


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3. Cc php lai nn tng gii quyt bi tp hiu qu.Cctrng

hp

Cc php lai mt cp gen T l kiu gen T l kiu hnh

1 AA x AA 100% AA 100%2 AA x Aa 1AA:1Aa 100%3 AA x aa 100% Aa 100%4 aa x aa 100% aa 100%5 Aa x Aa 1AA :2Aa: 1aa 3:16 Aa x aa 1Aa: 1aa 1:1

Cctrnghp

Kiu hnh Gii thch v kiu hnh

1 A Mang mt tnh trng tri2 aa Mang mt tnh trng ln3 AB Mang hai tnh trng tri4 Abb Mang tnh trng th nht tri, tnh trng th hai ln5 aaB Mang tnh trng th nht ln, tnh trng th hai tri6 aabb Mang hai tnh trng ln7 ABD Mang ba tnh trng tri.8 ABdd Mang hai tnh trng u tri, tnh trng th hai ln

4. Bng cng thc tng qut.

S cpgen dhp F1

S lngcc loi giaot F1

T l phn liKG F2 S lngcc loikiu gen F2

T l phn likiu hnh F2

S lngcc loikiu hnh F2

123n

2122

23.2n

(1+2+1)1(1+2+1)2(1+2+1)3...(1+2+1)n

3132333n

(3 + 1)1(3 + 1)2(3 + 1)3.(3 + 1)n

212223.2n

C. CCH NHN NH QUY LUT DI TRUYN .1. Cc phng php xc nh nhanh kiu gen ca b m.

Cctrnghp

Kiu hnh ca b m P Kiu gen ca b m P

1B m thun chng khc nhau mt cp tnhtrng tng phn hoc b m thun chngkhc nhau mt cp gen tng phn

P: AA x aa

2B m thun chng khc nhau hai cp tnhtrng tng phn hoc b m thun chng

khc nhau hai cp gen tng phn

P: AABB x aabbP: AAbb x aaBB

3B m thun chng khc nhau 3 cp tnhtrng tng phn hoc b m thun chngkhc nhau 3 cp gen tng phn

P: AABBDD x aabbddP: AABBdd x aabbDDP: AAbbDD x aaBBdd


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P: AAbbdd x aaBBDD

4 B m thun chng v c F1 xc nh kiugen Aabb P: AAbb x aabb

5 B m thun chng v c F1 xc nh kiugen aaBb P: aaBB x aabb

2. Phng php tch ring tng loi tnh trng.- Tm t l phn tnh v KH th h con i vi mi loi tnh trng- Nhn t l KH ring r ca loi tnh trng ny vi vi t l KH ring ca loi tnh

trng kia . Nu thy kt qu tnh c ph hp vi kt qa php lai => 2 cp gen quy nh 2loi tnh trng nm trn 2 cp NST khc nhau, di truyn theo quy lut phn li c lp caMenden ( tr t l 1:1 nhn vi nhau )

V d : Cho lai 2 th c chua : qu thn cao vi qu thn thp thu c 37,5%qu thn cao : 37,5% qu thn thp :12,5% qu vng thn cao , 12,5% qu vng thnthp . Bit rng mi tnh trng do 1 gen quy nh

Gii+ Xt ring tng tnh trng th h con( 37,5% + 37,5% ) : ( 12,5% + 12,5% ) vng = 3 : 1 vng( 37,5% + 12,5% ) cao : ( 37,5 % + 12,5% ) thp = 1 cao : 1 thp+ Nhn 2 t l ny ( 3 : 1 vng ) ( 1 cao : 1 thp ) = 3 cao : 3 thp : 1 vng cao : 1 vngthp . ph hp vi php lai trong bi . Vy 2 cp gen quy nh 2 cp nm trn 2 cp NSTkhc nhau .

BI TP P DNG.1. T l ca kiu gen aaBbdd to ra t php lai aaBbDd x AabbDd l bao nhiu?A. 3,125% B. 6,25% C. 56,25% D. 18,75%

2. mt loi cn trng, gen qui nh tnh trng nm trn nhim sc th thng v di truyntheo hin tng tri hon ton.Gen A: thn xm; gen a: thn enGen B: mt ; gen b: mt vngGen D: lng ngn; gen d: lng diCc gen ni trn phn li c lp v t hp t do trong gim phn.Php lai no sau y khng to ra kiu hnh thn en, mt vng, lng di con lai?A. AaBbDd x aaBbdd B. Aabbdd x aaBbDdC. AaBBdd x aabbdd D. aabbDd x aabbDd3. mt loi cn trng, gen qui nh tnh trng nm trn nhim sc th thng v di truyntheo hin tng tri hon ton.

Gen A: thn xm; gen a: thn enGen B: mt ; gen b: mt vngGen D: lng ngn; gen d: lng diCc gen ni trn phn li c lp v t hp t do trong gim phnT l ca loi hp t A- B- D- to ra t php lai AaBbDd x AaBbDD l:A. 6,35% B. 18,75% C. 37,5% D. 56,25%4. mt loi cn trng, gen qui nh tnh trng nm trn nhim sc th thng v di truyntheo hin tng tri hon ton.Gen A: thn xm; gen a: thn enGen B: mt ; gen b: mt vngGen D: lng ngn; gen d: lng diCc gen ni trn phn li c lp v t hp t do trong gim phnB m c kiu gen, kiu hnh no sau y sinh ra con lai c 50% thn xm, mt v 50%thn xm, mt vng?


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A. AAbb (thn xm, mt vng) x aaBb (thn en, mt )B. AaBB (thn xm, mt ) x aabb (thn en, mt vng)C. Aabb (thn xm, mt vng) x AaBB (thn xm, mt )D. aaBB (thn en, mt ) x aaBb (thn en, mt )5. mt loi cn trng, gen qui nh tnh trng nm trn nhim sc th thng v di truyntheo hin tng tri hon ton.Gen A: thn xm; gen a: thn enGen B: mt ; gen b: mt vngGen D: lng ngn; gen d: lng diCc gen ni trn phn li c lp v t hp t do trong gim phn6. Php lai no sau y c kh nng to ra nhiu bin d t hp nht?A. AaBbDD x AaBbDd B. AABBDD x aabbddC. AabbDd x AabbDd D. AaBbDd x AaBbDd7. mt loi cn trng, gen qui nh tnh trng nm trn nhim sc th thng v di truyntheo hin tng tri hon ton.

Gen A: thn xm; gen a: thn enGen B: mt ; gen b: mt vngGen D: lng ngn; gen d: lng diCc gen ni trn phn li c lp v t hp t do trong gim phnT l kiu hnh c to ra t php lai AaDd x aaDd l:A. 3 thn xm, lng ngn : 1 thn xm, lng di : 3 thn en, lng ngn : 1 thn en, lng diB. 1 thn xm, lng di : 1 thn en, lng ngnC. 1 thn xm, lng ngn : 1 thn xm, lng di : 1 thn en, lng ngn : 1 thn en, lng diD. 3 thn en, lng di : 1 thn en, lng ngn8. mt loi cn trng, gen qui nh tnh trng nm trn nhim sc th thng v di truyntheo hin tng tri hon ton.

Gen A: thn xm; gen a: thn enGen B: mt ; gen b: mt vngGen D: lng ngn; gen d: lng diCc gen ni trn phn li c lp v t hp t do trong gim phnT hp ba tnh trng ni trn, s kiu gen c th c loi cn trng c nu l:A. 36 kiu B. 27 kiu C. 21 kiu D. 16 kiu9. mt loi thc vt bit rng:A-: thn cao, aa: thn thpBB: hoa , Bb: hoa hng, bb: hoa trngHai tnh trng, chiu cao ca thn vo mu hoa di truyn c lp vi nhau.Php lai Aabb x aaBb cho con c t l kiu hnh no sau y?

A. 50% thn cao, hoa hng : 50% thn thp, hoa hngB. 50% thn cao, hoa trng : 50% thn thp, hoa trngC. 25% thn cao, hoa hng : 25% thn cao, hoa trng : 25% thn thp, hoa hng : 25% thnthp, hoa trngD. 75% thn cao, hoa hng : 25% thn thp, hoa trng10. mt loi thc vt bit rng:A-: thn cao, aa: thn thpBB: hoa , Bb: hoa hng, bb: hoa trngHai tnh trng, chiu cao ca thn vo mu hoa di truyn c lp vi nhau.T l ca loi kiu hnh thn thp, hoa hng to ra t php lai AaBb x aaBb l:A. 18,75% B. 25% C. 37,5% D. 56,25%11. mt loi thc vt bit rng:A-: thn cao, aa: thn thpBB: hoa , Bb: hoa hng, bb: hoa trng


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Hai tnh trng, chiu cao ca thn vo mu hoa di truyn c lp vi nhau.T l ca loi hp t AAbb c to ra t php lai AaBb x AaBb l:A. 6,25% B. 12,5% C. 18,75% D. 25%12. mt loi thc vt bit rng:A-: thn cao, aa: thn thpBB: hoa , Bb: hoa hng, bb: hoa trngHai tnh trng, chiu cao ca thn vo mu hoa di truyn c lp vi nhauCon lai c t l kiu hnh 75% thn cao, hoa hng : 25% thn thp, hoa hng c to ra tphp lai no sau y?A. AaBb x AaBb B. AABb x aaBb C. AaBB x Aabb D. AABB x aabb13. T l kiu gen ca php lai AaBbDd x AaBbDd c trin khai t biu thc no sau y?A. (1 : 2 : 1) (1 : 2 : 1) (1 : 2 : 1) B. (1 : 2 : 1) (3 : 1)C. (3 : 1) (3 : 1) (3 : 1) D. (1 : 2 : 1) (3 : 1) (1 : 1)14. iu khng ng khi ni v kiu gen AaBBDd l:A. Th d hp

B. To 4 loi giao t vi t l ngang nhauC. Lai phn tch cho 4 kiu hnh vi t l ngang nhau (nu mi gen qui nh mt tnh trng)D. To giao t aBD c t l 12,5%15. F1 d hp hai cp gen li vi nhau, mi gen qui nh mt tnh trng v khng c hintng di truyn trung gian th F2 c:A. 6 kiu hnh khc nhau B. T l kiu hnh l 3 : 3 : 1 : 1C. 14 t hp D. C 9 kiu gen16. mt loi, gen D qui nh hoa , tri hon ton so vi gen d qui nh hoa trng.Php lai to ra hin tng ng tnh con lai l:A. P: DD x dd v P: Dd x dd B. P: dd x dd v P: DD x DdC. P: Dd x dd v P: DD x dd D. P: Dd x dd v P: DD x DD

III. QUI LUT DI TRUYN THEO MORGAN.A. TNH S LOI V THNH PHN GIAO T1. Cc gen lin kt hon ton :

Trn 1 cp NST (1 nhm gen)Cc gen ng hp t 1 loi giao tV d :

Ab

Ab 1 loi giao t Aa ;ABd

ABd AbdNu c 1 cp gen d hp tr ln 2 loi giao t t l tng ngV d :

Ab

AB AB = Ab ;ab

AB AB = ab ;abd

ABD ABD = abdTrn nhiu cp NST ( nhiu nhm gen ) nu mi nhm Gen c t nht 1 cp gen d hpS loi giao t = 2n vi n = s nhm gen (s cp NST)

* Tm thnh phn gen mi loi giao t: dng s phn nhnh hoc nhn i s l mi loigiao t ca mi nhm gen ny phi hp kiu vi cc loi giao t ca nhm gen kia

V d : C th c KGab

AB.de

DE 4 loi giao t : AB.DE : AB.de : ab .DE : ab.de

V s nhm gen l 2 s loi giao t 22 = 4 loi giao tCc php lai trong lin kt gen hon ton.

Cc php lai thng dng T l kiu hnh

AB/AB x ab/ab 100% (AB)Ab/Ab x aB/aB 100% (AB)AB/ab x AB/ab 3 (AB): 1 (aabb)


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Ab/aB x Ab/aB 1 (Abb): 2 (AB): 1 (aaB)Ab/aB x AB/ab 1 (Abb): 2 (AB): 1 (aaB)AB/ab x ab/ab 1 (AB): 1 (aabb)Ab/aB x ab/ab 1 (Abb): 1 (aaB)

Ab/Ab x ab/ab 1 (Abb): 1 (aabb)aB/aB x ab/ab 1 (aaB): 1 (aabb)AB/Ab x AB/Ab 3 (AB): 1 (Abb)Ab/ab x aB/ab 1 (AB): 1(Abb): 1(aaB): 1 (aabb)

2. Cc gen lin kt khng hon ton .Mi nhm gen phi cha 2 cp gen d hp tr ln mi pht sinh giao t mang t hp

gen cho (giao t HVG) trong qu trnh gim phnTrng hp 2 cp gen d hp :* S loi giao t : 22 = 4 loi t l khng bng nhau

Thnh phn gen :

+ 2 loi giao t bnh thng mang gen lin kt t l mi loi giao t ny > 25% .+ 2 loi giao t HVG mang t hp gen cho nhau do 2 gen tng ng i ch ,t l mi loi giao t ny < 25% .

V d : C th c KGab

AB lin kt khng hon ton to giao t :

+ 2 loi giao t bnh thng t l cao l : AB = ab > 25% .+ 2 loi giao t HVG t l thp l: Ab = aB


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2) Khong cch tng i gia cc gen trn cng 1 NST+ Tn s HVG th hin khong cch tng i gia 2 gen: 2 gen cng nm xa

nhau th tn s HVG cng ln v ngc li cc gen cng nm xa nhau th tn s HVG cngnh .

+ Da vo tn s HVG => khong cch gia cc gen => v tr tng i (locut)trong nhm gen lin kt .

Qui c : 1 cM (xentiMorgan) = 1% HVG

C. TNH TN S HON V GEN1. Trong php lai phn tch :

Tn s HVG p = (S c th hnh thnh do TC : Tng s c th nghin cu )x100%

V d : Lai phn tch rui ci thn xm cnh di thuc KG iaB

Ab c th h lai gm 376

con xm ngn : 375 con en di : 124 con xm di : 125 con en ngn

GiiXm di v en ngn l 2 KH do TC to ra :

=> Tn s HVG =125124375376

125124

. 100 = 25%

2. Trong cc php lai khc . gii bng i s

+ t P : Tn s HVG => t l giao t HVG l2

p

t l giao t BT l2

1 p

+ Davo loi KH m bi cho bit lp t l :T l cc KG lm nn KH theo n s p = S c th thu c KH bit c : Tng s c th thuc

V d : Cho cy thn cao ht di c KGaB

Ab t th phn F1 thu c 4000 cy , trong c

260 cy thp trnGii

+ t p = Tn s HVG => t l giao t HVG l2

p

+ F1 thu c cy thp trn (ab

ab) c t l

4000

160 =100

4

T l KG lm nn cy thp trn l (2

p )2 .ab

ab

=> phng trnh (2

p )2 =100

4 c p = 40%

BI TP P DNG.

1. Trong trng hp lin kt hon ton, php laiaB

Ab xaB

Ab c s kiu t hp l

A. 4. B. 6. C. 8. D. 16.


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2. Trong trng hp lin kt hon ton v mi gen quy nh mt tnh trng, php laiaB

Ab x

aB

Ab c t l phn li kiu hnh l:

A. 3: 1. B. 1:2: 1. C. 3:3:1: 1. D. 9:3:3: 1.3. Xt php lai sau:

aB

AB xAb

AB (Cho bit mi cp gen quy nh mt cp tnh trng

v tri hon ton, c trao i cho gia cc crmatt tng ng). S loi kiu hnh i conl:A. 1. B. 2. C. 3. D. 4.

4. Mt t bo sinh tinh c kiu genab

AB tin hnh gim phn (khng c hon v gen), s loi

giao t c to ra l:A. 1 loi. B. 2 loi. C. 4 loi. D. 6 loi.

5. Mt t bo sinh tinh c kiu gen abAB

tin hnh gim phn (c hon v gen), s loi giao tc to ra l:

A. 1 loi. B. 2 loi. C. 4 loi. D. 6 loi.

6. Php laiab

AB xab

AB . Nu cc cp tnh trng di truyn tri hon ton v b m u c

hon v gen vi tn s 20% th kiu hnh ln chim t l:A. 6,25%. B. 40%. C. 16%. D. 10%.

7. Php laiab

AB xab

AB . Nu cc cp tnh trng di truyn tri hon ton v c th ci c hon

v gen vi tn s 40% th tng s c th c kiu hnh khc b m chim t l:A. 50%. B. 65%. C. 35%. D. 10%.

8. Cho cy d hp v hai cp gen t th phn, i con F1 c 4 loi KH vi t l: 51% cy cao, hoa : 24% cy cao, hoa trng : 24% cy thp, hoa : 1% cy thp, hoa trng. (cho bit mi cptnh trng do mt cp gen quy nh)B. Tn s hon v gen l:

A. 1%. B. 10%. C. 20%. D. 40%.

9. Mt c th c kiu genab

AB . Nu xy ra hon v gen vi tn s 20% th loi giao t Ab

chim t l:A. 0,2. B. 0,3. C. 0,4. D. 0, 1.

10. Trong trng hp cc tnh trng di truyn tri hon ton v c b v m u c hon vgen vi tn s 40% th php lai

aB

Ab xab

AB , kiu hnh mang hai tnh trng tri c t l:

A. 48%. B. 30%. C. 56,25%. D. 36%.

11. Hai gen A v B cng nm trn mt cp NST v tr cch nhau 20cMB. C thab

AB khi

gim phn s cho giao t AB vi t l:A. 50%. B. 40%. C. 20%. D. 10%.

12. C thab

AB t th phn, nu xy ra s hon v gen c b v m th s kiu t hp giao t

i con l:A. 4 kiu. B. 8 kiu. C. 16 kiu. D. 100 kiu.


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13. C th P d hp cc cp gen gim phn c hon v gen Aa to ra loi A B d chim t l

17,5%

A. Kiu gen ABdabd

, tn s hon v 35%. B. Kiu gen ABdabd

, tn hon v 30%.

C. Kiu gen AbDaBd

, tn s hon v gen 35%. D. Kiu gen AbDaBd

, tn s hon v gen 30%.

14. Bit A-: qu trn, aa: qu di; B-: qu ngt, bb: qu chua.Php lai sau y cho t l con lai c 3 qu trn, chua: 3 qu di, ngt: 1 qu trn, ngt: 1 qudi, chua l:A. AB/ab (tn s hon v 25%) x ab/ab. B. Ab/aB (tn s hon v 25%) x ab/ab.C. AB/ab (tn s hon v 50%) x ab/ab. D. Ab/aB (tn s hon v 50%) x ab/ab.

15. Mt c th c kiu genab

AB . Nu xy ra hon v gen vi tn s 20% th loi giao t AB

chim t l:

A. 0,2. B. 0,3. C. 0,4. D. 0,1.

IV. QUI LUT TNG TC GEN1. Thng k cc php lai thng dng.Mi kiu tng tc c 1 t l KH tiu biu da theo bin dng ca (3:1)2 nh sau :Kiu h tr c 3 t l KH : 9: 3:3:1 ; 9:6: 1 ; 9: 7 .H tr gen tri hnh thnh 4 KH : 9:3:3:1A-B- A-bb aaB- aabb thuc t l : 9:3:3:1H tr gen tri hnh thnh 3 KH : 9:6: 1A-B- ( A-bb = aaB- ) aabb thuc t l : 9:6:1H tr gen tri hnh thnh 2 KH : 9:7

A-B- ( A-bb = aaB- = aabb ) thuc t l : 9:7Kiu t ch c 3 t l KG : 12:3:1 ; 13:3 ; 9:4:3t ch gen tri hnh thnh 3 KH : 12:3:1(A-B- = A-bb) aaB- aabb thuc t l : 12:3:1.t ch gen tri hnh thnh 2 KH : 13: 3(A-B- = A-bb = aabb ) aaB- thuc t l : 13:3t ch gen ln hnh thnh 3 KH : 9:4:3A-B- ( A-bb = aabb ) aaB- thuc t l : 9:4:3Tc ng cng gp ( tch lu) hnh thnh 2 KH : 15 :1A-B- ( A-bb = aabb ) aaB-

2. Tng qut n cp gen tc ng cng gp => t l KH theo h s mi s hng trong trin khaica nh thc Newton ( A+a)n .+ A: S alen tri, a s alen ln3. Bng tng hp cc php tng hpCc kiu tngtc AaBb x AaBb

AaBb x aabbAabb x aaBb

AaBb x AabbAaBb x aaBb

AaBb x AABbAaBb x AaBB

Tngtc b

tr

9:3:3:19A-B- 3A-bb 3aaB- 1aabb

1A-B-1A-bb1aaB-1aabb

+3A-B-3A-bb1aaB- 1aabb+3A-B-3aaB-1A-bb 1aabb

+6A-B-3A-bb+6A-B-3aaB-

9:6:1 9A-B- 3A-bb= 3aaB- 1A-B- 1A-bb= 1aaB-

+3A-B-3A-bb=1aaB-1aabb



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1aabb 1aabb +3A-B-3aaB-=1A-bb1aabb

9:7 9A-B-3A-bb=3aaB-=1aabb1A-B-1A-bb=1aaB- = 1aabb

+3A-B-3A-bb=1aaB-=1aabb+3A-B-3aaB-=1A-bb=1aabb


Tngtckiut ch

12:3:19A-B- = 3A-bb 3aaB- 1aabb

1A-B- = 1A-bb 1aaB- 1aabb

+3A-B-=3A-bb1aaB-1aabb+3A-B-=1A-bb3aaB1aabb

+6A-B-=3A-bb+6A-B-3aaB-

13:39A-B- = 3A-bb= 1aabb 3aaB-

1A-B-=1A-bb=1aabb 1aaB-

+3A-B-=3A-bb =1aabb1aaB-+3A-B-=1A-bb=1aabb3aaB-

+6A-B-=3A-bb+6A-B-3aaB-

9:3:4 9A-B- 3A-bb 3aaB-=1aabb

1A-B-1A-bb1aaB-=1aabb

+3A-B-3A-

bb=1aabb1aaB-+3A-B- 1A-bb=1aabb3aaB-


BI TP P DNG.1. Cho mt php lai gia hai ging g thun chng mu long trng khc nhau v ngun gc,F1 ng lot long mu. F2 thu c 180 lng mu , 140 lng trngTnh trng mu long g c di truyn theo quy lut

A. Phn li c lp B. Tng tc gen theo kiu b trC. Tng tc gen theo kiu t ch D Tng tc gen theo kiu cng gp

2. Khi lai gia ch nu vi ch trng thun chng ngi ta thu c ton ch trng, F2 thuc 37 trng, 9 en, 3 nu. Tnh trng mu long cho c di truyn theo quy lutA. Phn li c lp B. Tng tc b tr C. Tng tc t ch D. Tng tccng gp3. Khi lai gia ch trng thun chng vi ch nu ngi ta thu c ton ch trng, F2 thuc 74 trng, 18 en, 6 nu. Xc nh kiu gen ca 2 ging b m thun chng?A. AAbb x aabb B. aaBB x aabb C. AABB x aabb D. AAbb x aaBB4. ng c 3 gen (mi gen gm 2 alen) phn li c lp, tc ng qua li vi nhau hnhthnh chiu cao cy. cho rng c mi gen tri lm cy ln i 20 cm. ngi ta tin hnh lai cythp nht vi cy cao nht c chiu cao 210 cm. Chiu cao cy F1 l?A. 60 cm B. 90 cm C. 120 cm D. 150 cm5. B long trng ca g c xc nh bi hai cp gen khng alen di truyn c lp. mtcp, gen tri xc nh b long mu, gen ln xc nh b long trng. cp kia gen tri t chmu, gen ln khng t ch mu.Lai gia cc g trng vi nhau, i sau c 1275 g controng c 315 c mu, cn li l trng. Xc nh kiu gen ca cha m v cc con g c mu.Bit A: long mu, a: Lng trng ;gen B:t ch mu, b khng t ch muKiu gen ca b m lA. AABB x aabb B. AAbb x aaBBC. AaBb x AaBb D. AaBb x aabb6. Li b qu vng vi b qu trng i con cho qu trng . Khi lai cc cy con qu trng vinhau c 204 qu trng, 53 qu vng, 17 qu xanh. G i Aa, Bb l hai cp gen chi phi s ditruyn tnh trng trn. Kiu gen ca b m lA. AaBb x AaBb B. AABB x aabbC. AAbb x aaBB D. AaBb x aabb


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7. Chiu cao ca ngi c xc inh bi mt s cp gen khng alen di truyn c lp. Cccp gen tc ng theo kiu cng gp. Nu b qua nh hng ca mi trng v gii hn chc 3 cp gen xc inh tnh trng ny th ngi ng hp ln c chiu cao 150Cm, ngi caonht 180cm. Xc nh chiu cao ca ngi d hp c 3 cp genA. 160 B. 165 C. 170 D. 1758. Khi cho cy P t th phn, ngi ta thu c F1 c 225 cy c qu dt, 150 cy c qu trnv 25 cy c qu di.Nu cho cy cp P ni trn lai vi cy c mang kiu gen Aabb th t l kiu hnh thu c con lai bng:A. 2 qu dt: 1 qu trn: 1 qu di.. B. 6 qu dt: 1 qu trn: 1 qu di.C. 3 qu dt: 4 qu trn: 1qu di. D. 15 qu dt: 1 qu di.T l kiu hnh c to ra t php lai AaBb x aaBb l:A. 3 qu trn: 3 qu di: 2 qu bu dc. B. 3 qu trn: 4 qu di: 1 bu dc.C. 3 qu di: 3 qu bu dc: 2 qu trn. D. 3 qu trn: 4 qu bu dc: 1 qu di.

V DI TRUYN LIN KT VI GII TNHA. XC NH KIU NST GII TNH1. Trong thin nhin , gp 1 s kiu NST gii tnh nh sau : XX, XY , XO ( XX l nggiao t , XY hoc XO l d giao t )- c XY , ci XX : ngi , ng vt c v , rui gim - c XX , ci XY : cc loi chim , bm tm , ch nhi , b st .- c XO ; ci XX : b xt , chu chu , rp- c XX ; ci XO : b nhy .2. Nu c th c cp trong bi khng nu loi no => kiu NST gii tnh c th xcnh theo 2 cch :

- Da vo c th mang tnh trng ln th h c t l phn tnh 3:1 . V tnh trng ny

d xut hin c th XY => gii tnh ca c th thuc NST gii tnh XY- Dng cch loi suy , ln lt th tng kiu NST gii tnh => kiu no cho kt quph hp vi bi th nhnV d : cho 1 c th mang 1 cp gen d hp quy nh cnh thng giao phi vi c th khcc F1 gm 256 con cnh thng : 85 con cnh cong ( ch ton con c )=> - Cp gen d hp quy nh cnh thng => cnh thng l tnh trng tri, cnh cong ltnh trng ln .

- F1 c t l 3 cnh thng : 1 cnh cong . Nhng tnh trng ln cnh cong ch biu hin con c => NST gii tnh ca con c l XY , con ci XX .

B. DI TRUYN LIN KT VI GII TNH

1. Khi nim : l hin tng di tuyn cc tnh trng m c gen xc nh chng nm trn ccNST gii tnh

2. Gen trn NST X ( Qui lut di truyn cho )a. Th nghim : Moocgan lai rui gim mt vi mt trng-Lai thun :

+ P : (mt ) x ( mt trng )+ F1 : 100% mt + F2 : 3 mt : 1 mt trng ( ton con c )

-Lai nghch :+ P : (mt trng ) x ( mt )+ F1 : 1 mt : 1 mt trng+ F2 : 1 mt : 1 mt trng : 1 mt : 1 mt trng

Gii thch :


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- F1 ng lot mt , theo nh lut ng tnh th mt tri hon ton so vi mttrng .

Qui c : W : mt ; w : mt trng .Nu gen nm trn NST thng th F2 ( trong php lai thun ) mt trng phn b c giic v ci . Thc t F2 mu mt trng ch c con c , V vy gen qui nh mu mt rui phi nm trn NST gii tnh X , khng c alen trn Y .S lai :+ Php lai thun :P : XWXW ( mt ) x XwY( mt trng )G : XW , XW ; Xw , YF1 : XWXw , XWY ( 100% mt )F1x F1 : XWXw x XWYG: XW , Xw ; XW , YF2 : XWXW : XWXw :XWY : XwY

3 mt : 1 mt trng

+ Php lai nghch:P : XwXw ( mt trng ) x XWY( mt )G : Xw ; Xw , YF1 : XWXw (50% mt ) , XwY ( mt trng)F1x F1 : XWXw x XwYG : : XW , Xw ; Xw , YF2 : XWXw : XwXw : XWY : XwY

1 mt : 1 mt trng : 1 mt : mt trngVy, php lai thun gen ln trn X do b truyn cho con gi v biu hin chu traib. Ni dung nh lut :- Di truyn cho : tnh trng ca b truyn cho con ci (gi), tnh trng ca m truyn cho con

c- Hai php lai thun nghch cho: kt qu khc nhau .

3. GEN TRN NST Y :( quy lut di truyn thng )-NST Y a s loi hu nh khng mang gen , nn hu nh gen trn NST X

him c gen tng ng trn Y . Tuy nhin , 1 s loi ng vt , NST Y cng mang gen .- NST Y ngi c on mang gen tng ng vi gen trn X , nhng cng c

on gen trn Y m khng c gen tng ng trn XV d : ngi tt dnh ngn tay s 2 v s 3 do gen ln (a) trn NST Y gy ra v ch

biu hin nam giiP : XX x XYa

G : X ; X , YaF1 : XX ( 50% gi BT ) : XYa ( 50% trai dnh ngn )Ni dung di truyn thng : tnh trng qui nh bi gen nm trn NST Y di truyn 100% cc cp NST gii tnh XY ( 100% con trai )4. ngha : hin tng di truyn lin kt vi gii tnh c ng dng phn bit gii choc gii ci giai on sm pht trin ca c th lc hnh thi gii tnh cha th hin ra kiuhnh, nh c th chn lc tng nng sut .

V d : ga ngi ta s dng gen tri A trn NST gii tnh X xc nh lng vn , phnbit trng, mi khi mi n. G trng con XA XA c lng vn u r hn so vi con mi XA Y.

BI TP P DNG.


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ngi bnh mu kh ng do gen ln (h) lin kt vi gii tnh gy ra. Trong mt gia nhb b mu kh ng m bnh thng, c 2 con : Con trai mu kh ng, con gi bnh thng.Kiu gen ca m phi nh th no?A. XHXH B. XHXh C. XhXh D. XHXH hoc XHXh

ngi bnh mu kh ng do gen ln (h) lin kt vi gii tnh gy ra. Kiu gen v kiu hnhca b m phi nh th no cc con sinh ra vi t l 3 bnh thng ; 1 mu kh ng l contraiA. XHXH x XhY B. XHXh x XhYC. XHXh x XHY D. XHXH x XHY g A: Lng sc vn; a: long trng . Cc gen nm trn NSt gii tnh X Lai g mi trng vig trng sc vn F1 c g mi trng. Kiu gen ca b m lA. XaY x XAXA B. XaY x XAXaC. XAY x XaXa D. XAY x XAXaBnh mu kh ng ngi do gen ln trn NSt gii tnh XB mu kh ong ly m mu ng bnh thng sinh c con trai con gi bnh thng.

Nhng ngi con gi ny ly chng bnh thng. Kt lun no sau y l ngA. Chu ca h khng th mc bnh B. Chu ca h chc chn mc bnhC. H c th c chu trai mc bnh D. H c th c chu gi mc bnhBnh mu kh ng ngi do gen ln trn NSt gii tnh XB mu kh ng, m bnh thng ng ngoi mc bnh. Xc nh t l sinh ra nhng achu kho mch trong gia nh?A. 25% B. 50% C. 0 % D. 100%

VI. DI TRUYN HC NGIBI TP P DNG ngi, bnh bch tng do gen ln nm trn nhim sc th thng qui nh. Nu b v m

u l th d hp th xc sut sinh con b bch tng l bao nhiu?A. 12,5% B. 25% C. 37,5% D. 50%Ngi ph n mang kiu gen d hp Aa ly chng thun tay phi. Xc xut ngi ph nny sinh ra a con thun tay tri l:A. 12,5% B. 25% C. 50% D.75%T l kiu gen sinh ra t IAIO x IBIO l:

A.P: IAIB : IAIO : IBIO : IOIO B. P: IAIO : 2IAIB : IBIOC. P: IAIB : IOIO D. P: IAIO : 2IOIO : IBIOTrong 1 gia nh b v m u c tm vc thp, tc xon sinh c cc a con mang cckiu hnh khc nhau, trong c a c tm vc cao, tc thng. Bit mi tnh trng do 1 gentrn nhim sc th thng qui nh, cc gen phn li c lp v khng xy ra t bin.

T l xut hin nhng a con c tm vc thp, tc thng t cp b m trn l bao nhiu?A. 0,5625 B. 0,1875 C. 0,125 D. 0,0625

30. ngi, bnh m mu do gen ln nm trn nhim sc th gii tnh X qui nh, gen triqui nh kiu hnh nhn mu bnh thng. Cho 1 s ph h sau y:

I K hiu


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1 2 nam bnh thng

II nam b m mu1 2 3

: n bnh thngPht biu sau y ng v I2 l:A. Mang i gii tnh XY v cha gen triB. Mang i gii tnh XX v l th d hpC. Mang i gii tnh XX v l th ng hp triD. Mang i gii tnh XX v l th ng hp lnC 2 anh em sinh i cng trng, ngi anh ci v mu A sinh a con mu B, ngi emci v mu B sinh a con mu A.Kiu gen, kiu hnh ca 2 anh em sinh i ni trn l:A. IAIB (mu AB) B. IAIA hoc IAIO (mu A)C. IBIB hoc IBIO (mu B) D. IOIO (mu O)

VII. DI TRUYN HC QUN THBI TP P DNGCho bit: AA: lng en, Aa: lng m, aa: lng trng. Mt qun th g c 410 con lng en,580 con lng m v 10 con lng trng.Tn s tng i ca mi alen A v a l:A. 0,7A; 0,3a B. 0,3A; 0,7a C. 0,42A; 0,48a D. 0,48A; 0,42a

Cng theo d liu ca cu 6, t l kiu gen ca qun th g ni tn khi t trng thi cn bngl:A. 0,09 AA : 0,42 Aa : 0,49 aa B. 0,49 AA : 0,42 Aa : 0,09 aa

C. 0,1764 AA : 0,5932 Aa : 0,2304 aa D. 0,2304 AA : 0,5932 Aa : 0,1764 aaCho bit P: 100% Aa. Sau cc th h t phi (ni phi), t l kiu gen F3 l:A. 0,125 AA : 0,4375 Aa: 0,4375 aa B. 0,4375 AA : 0,4375 Aa : 0,125 aaC. 0,4375 AA : 0,125 Aa : 0,4375 aa D. 0,4 AA : 0,1 Aa : 0,5 aa mt n g ni th chung, im ngu nhin 100 con thy c 9 con lng trng,11 con mtrng en cn li l lng en. Nu gi gen D quy nh mu en l tri khng hon ton, thkiu gen DD -> en, Dd -> m, dd -> trng. Trong trng hp ny , tn s (f) ca D v d lA. f(D)= 0,7; f(d)= 0,3 B. f(D)= 0,91; f(d)= 0,09C. f(D)= 0,855; f(d)= 0,145 D. f(D)= 0,8; f(d)= 0,2 mt qun th giao phi: gen A c 2 alen, cn gen B c 3 alen. Nu 2 gen ny phn li c lp,s th tinh v gim phn bnh thng th i sau co s kiu gen l

A. 8 B. 15 C. 18 D. 36mt qun th giao phi 120 c th kiu gen BB, 400 c th Bb v 480 c th bb. Nu gi p ltn s alen b, l.A. p= 0,32; q= 0,68 B. p= 0,68; q= 0,32 C. p= 0,12; q= 0,48 D. p= 0,36; q=0,64


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VIII. T BIN NHIM SC THt bin NST

.

** Kin thc b sung :t bin lch bi ngi :Ngi c 3 NST s 21 Hi chng Down ( ao) : c ngn , gy rng , khe mt xch , si n ,v sinhCp NST s 13,15 : 3 chic NST sc mi , tha ngn , cht yuCp NST s 23 : + XXX : ( siu n ) : bung trng , d con khng pht trin , kh

c con+ XO : ( tcn ) : n ,ln , c ngn , tr tu chm pht trin , v

sinh+ XXY :( Claiphent ) : nam , chn tay di , m mu , si n , v

sinh+ OY : c l b cht

B. Cch vit giao t cc th a bi :Qui c gen : A : qu a: Qu vng

Th t bi ( 4n )Qu : AAAA Giao t : AAQu : AAAa Giao t : AA : Aa

Cu trc

Mt on : ngi mt on nhim sc th 21 gy ung th

Lp on : rui gim, lp on 16A hai ln trn nhim sc th X

lm cho mt hnh cu tr thnh mt dt

o on : rui gim ngi ta pht hin c 12 o ontrn NST s 3, lin quan ti kh nng thch nghi vi nhng iukin nhit khc nhau trong mi trng

Chuyn on : mt on nhim sc th ny b t ra v gn vomt NST khc, hoc 2 nhim sc th khc cp cng t mt onno ri trao i on b t vi nhau, cc on trao i c th ltng ng hoc khng

t bin NST

S lung

Lch bi

2n -2 : th khng

2n 1 : th m t

2n +1 : th ba

2n + 2 : th t

a bi chn : 4n,6n,8n

a bi

D a bi : 2n +2n (chab NST ca 2 loi

a bi l: 3n, 5n,7n


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Qu : AAaa Giao t : 1/6 AA : 4/6Aa : 1/6 aaQu : Aaaa Giao t : Aa : aaQu vng : aaaa Giao t : aaTh tam bi : ( 3n )Qu : AAA Giao t : AA : AQu : AAa Giao t : 1/6AA : 2/6 Aa : 2/6A: 1/6 aQu : Aaa Giao t : 1/6 Aa : 2/6 aa : 2/6 a : 1/6 AQu vng : aaa Giao t : aa : a

BI TP P DNG1/ Dng t bin cu trc nhim sc th t nh hng n sc sng ca sinh vt l:Chuyn on. B.Mt on.Lp on. D.o on.2/ Th a bi t gp ng vt v: t bo ng vt s lng NST thng ln.

Th a bi ng vt thng gy cht, hoc c ch xc nh gii tnh b ri lon v nhhng n qu trnh sinh sn. t bo ng vt khng bao gi xy ra hin tng khng phn li ca cc cp NST tngKhng c gii thch no ng3/ Hu qu ca t bin lp on l:Tng hoc gim bt s biu hin cng ca tnh trng.Lm tng cng s hot ng ca tnh trng.Tng s lng gen trn NST.A v C4/ Nhng dng t bin cu trc nhim sc th l:

Mt mt hoc mt s cp nhim sc th.. B.Mt on, o on, lp on, chuyn on.Thm mt hoc mt s cp nhim sc th. D. Mt, thm, thay th, o v tr ca mt cpnucltit5/ Th mt nhim c b nhim sc th thuc dng:2n-1 B.n + 1 C.2n+1 D.n 16/ Hin tng lp on NST s dn n:Lm tng hay gim cng biu hin ca tnh trng.Lm tng kch thc ca t bo, lm c th c kch thc ln hnKhng nh hng n kiu hnh do khng mt vt cht di truynGy cht cho sinh vt7/ ngi 2n = 46 NST, s lng NST t bo ngi b bnh ao l:A. 45 B. 47 C. 48 D. 928/ Hi chng ao ngi c nhng biu hin c bn l:Khe mt xch, li dy v di. B. C th pht trin chm, si n v thngv sinhC ngn, gy rng v dt, ngn tay ngn D. Tt c cc biu hin trn9/ Cc dng t bin ch lm thay i v tr ca gen trong phm vi 1 nhim sc th l:Chuyn on v lp on trn 1 NST B. Mt on v lp onnhim sc tho on v lp on trn 1 nhim sc th d. o on v chuyn ontrn 1 nhim sc th

10/ C ch pht sinh t bin lp on l:Mt on no ca NST thm vo.Mt on no ca NST t ra v gn vo v tr khc.


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Mt on no ca NST lp li mt hay nhiu lnTrao i cho khng cn gia 2 NST tng ng lm tng s lng gen trn NST11/ Ngi con trai c NST gii tnh k hiu l XXY, mc hi chng no sau y:A. Siu n. B. Tooc n (Turner).C. ao (Down D. Claiphent (Klinefelter).12/ c chua 2n = 24 nhim sc th, s nhim sc th th tam bi l:

A. 25 B. 48 C. 27 D. 3613/ Trng hp trong t bo ca c th sinh vt c 2 cp NST, mi cp tng ln mt chic gil:A. Th tam nhim B. Th tam nhim kpC. Th t nhim D. Th mt nhim kp14/Trong trng hp ri lon phn bo 2, cc loi giao t c to ra t c th mang kiugen XAXa lA. XAXA, XaXa v 0. B. XA v Xa. C. XAXA v 0. D. XaXa v 0.15/ .Trong trng hp ri lon phn bo 2, cc loi giao t c to ra t c th mang kiu

gen XY khi nhim sc th kp XX khng phn ly lA. XX, XYv 0. B. XX , Yv 0. C. XY v 0. D. X, Y v 0.16/ .Trong trng hp ri lon phn bo 2, cc loi giao t c to ra t c th mang kiugen XY khi nhim sc th kp XX khng phn ly lA. XX, XYv 0. B. XX , Yv 0. C. XY v 0. D. X, YY v 0.17/.Tng s nhim sc th ca b lng bi bnh thng mt loi c s lng 22, trong tbo c th A cp th 5 v cp th 6 u c 4 chic, c th l tht bi. B. th bn kp . C. a bi chn. D. th tam nhim kp.18/ T bo 2n mang kiu gen Aa khng hnh thnh thoi v sc trong nguyn phn dn n tora kiu gen no sau y t bo con?

A. AAAA B. aaaa C. AAaa D. Aaa19/ .Mt ph n c c 47 nhim sc th trong c 3 nhim sc th X. Ngi thuc thba. B.tam bi. C. a bi l. D. n bi lch.20/ .Mt ph n c 47 nhim sc th trong c 3 nhim sc th X. Ngi b hi chngTc n. B. ao. C. siu n. D. Claiphent.161.S tng mt s nguyn ln s NST n bi ca mt loi l hin tngt a bi. B.tam bi. C.t bi. D.d a bi.162.Cc a bi lch c ngha trongA. tin ho, nghin cu di truyn. B.chn ging, nghin cu di truyn.chn ging, tin ho, nghin cu di truyn. D.chn ging, tin ho.164.S th tinh gia 2 giao t (n+1) s to nn

A. th 4 nhim hoc th ba nhim kp. B. th ba nhim.C. th 1 nhim. D. th khuyt nhim.

176.Mt loi c b nhim sc th 2n = 14. Mt c th ca loi trong t bo c 21 nhim sc thc th thuc thA. d bi. B. tam nhim. C. tam bi. D. a bilch.177.Mt loi c b nhim sc th 2n = 24. Mt cc th ca loi trong t bo c 48 nhim scth c th thuc thA. t bi. B. bn nhim. C. d bi. D. a bilch.178.Th a bi c hnh thnh do trong phn boA. mt s cp nhim sc th khng phn ly. B. tt c cc cp nhim sc th khng phnly.


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C. mt cp nhim sc th khng phn ly. D. mt na s cp nhim sc th khng phnly.179.S kt hp gia giao t n vi giao t 2n ca loi to thtam nhim. B.tam bi. C.ba nhim kp. D. tam nhim kp.180.S kt hp gia 2 giao t 2n ca loi to thbn nhim. B.t bi. C.bn nhim kp. D.d bi lch.181.D a bi l hin tng trong t bo cha b nhim sc thlng bi ca loi. B. lng bi ca 2 loi.C.ln hn 2n. D.n bi ca 2 loi.*182.Nu k hiu b nhim sc th ca loi th nht l AA, loi th 2 l BB th song nh bi lAABB. B.AAAA. C.BBBB. D. AB.183.Nu k hiu b nhim sc th lng bi ca loi th nht l AA, loi th 2 l BB, t a bigmAABB v AAAA. B. AAAA v BBBB.C.BBBB v AABB. D. AB v AABB.

184.Nu k hiu b nhim sc th lng bi ca loi th nht l AA, loi th 2 l BB, th d tbi lA. AABB. B. AAAA. C. BBBB. D. AB.185.S kt hp gia giao t 2n ca loi A vi giao t 2n ca loi B to tht bi. B. song nh bi th. C.bn nhim. D. bn nhim kp.

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