Công thức giải nhanh sinh học cực hay.pdf

7/26/2019 Cng thc gii nhanh sinh hc cc hay.pdf

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www.viettriedu.net Gia S Vit Tr301/28 Dng B Trc,Q8 Hotline:01656 156 268 0968 922 955LTH 2014 ti Trung Tm, Bo m Tng im >=15

LTH 2014 Khi A,A1,B,D Bomim thi H thp nht15 im 1

MT S CNG THC SINH HC C BN

1) Tng s nucltit :N =300

m

m = N x 300v.C ( m : khi lng ca gen)

2) Chiu di ca phn t ADN(gen) : L =2

N

x 3,4 A0 N =4,3

2 L

(1A0 =10-4 m =10-7mm)

3) S lin kt hyr ca phn t ADN(gen) : H = 2A + 3G4) S lin kt ha tr : *Gia cc nucltit : N 2

*Trong c phn t ADN : 2(N 1)

5) S vng xon (Chu k xon) : C =20

N

N = C x 20

6)Gi A1, T1, G1,X1 l cc nucltit trnmch 1Gi A2, T2, G2,X2 l cc nucltit trn mch 2: Theo NTBS gia 2 mch ta c :

A1= T2

T1= A2G1= X2

X1= G2*V mt s lng : A = T = A1+ A2= T1+ T2 G = X = G1+ G2= X1+ X2

*V mt t l % : A% = T% =2

1( A1% + A2%) =

2

1( T1% + T2%)

G% = X% =2

1( G1% + G2%) =

2

1( X1% + X2%)

A% + T% + G% + X% = 100%

A1+ T1 + G1 + X1= 100% ; A2+ T2 + G2 + X2= 100%

7) S phn t ADN(gen) con to ra sau n ln nhn i : 2n8) S nucltit mi loi mi trng cung cp cho gen nhn i n ln l :

A = T = (2n1)Agen G = X = (2n1)Ggen

9) Quan h gia gen v mARN :rN=2

1N (rN: Tng s nu trn mARN)

rN= Am+ Um+ Gm+ Xm

Agc= Um

Tgc= AmGgc= XmX gc= Gm

*V mt s lng : Agen= Tgen= Am+ Um Ggen= Xgen= Gm+ Xm

*V mt t l % : A% = T% =2

1( Am% + Um%) G% = X% =

2

1( Gm% + Xm%)

* Chiu di ARN: LARN =L =2

N

x 3,4 A0= rN x 3,4 A0

* Khi lng mARN: rN x 300v.C10) S lin kt hyr b ph v khi gen nhn i n ln l : (2 n1)H

11) S lin kt hyr b ph v khi gen nhn i n ln l : 2 n.H12) S b ba mt m :

32 x

N

=3

rN
http://www.viettriedu.net/http://www.viettriedu.net/http://www.viettriedu.net/


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13) S axitamin mi trng cung cp cho mt phn t prtin :32 x

N- 1=

3

rN

-1

14) S axitamin ca mt phn t prtin hon chnh :32 x

N- 2=

3

rN

-2

15) S lin kt peptit : S axitamin 116) S phn t nc b loi ra khi hnh thnh chui polypeptit : S axitamin217)Khi lng phn t prtin: S axitamin x 110v.C18) T l cc loi giao t ca cc dng t bin s lng NST

Kiu gen T l cc loi giao t

Th bnh thng

AAAA AA

AAAa 1AA : 1Aa

AAaa 1AA : 4Aa : 1aa

Aaaa 1Aa : 1aa

aaaa aa

Th t binAAA 1AA :1A

AAa 1AA : 2Aa : 2A : 1a

Aaa 2Aa : 1aa : 1A : 2a

aaa 1aa : 1a

19) Nguyn phn v gim phnCc yu t

Phn boS NST S tm ng S crmatit

NGUYN PHNKT 2n(kp) 2n 2n x 2 = 4nKG 2n(kp) 2n 2n x 2 = 4nKS 4n(n) 4n 0KC 2n(n) 2n 0

GIM PHNKT1 2n(kp) 2n 2n x 2 = 4nKG1 2n(kp) 2n 2n x 2 = 4nKS1 2n(kp) 2n 2n x 2 = 4nKC1 n(kp) n 2n

KT2 n(kp) n 2nKG2 n(kp) n 2nKS2 2n(n) 2n 0KC2 n(n) n 0

1 t bo sinh tinh (2n) 4 tinh trng (n)

1 t bo sinh trng (2n) 1 trng chn (n) : c kh nng th tinh3 th nh hng (n) : tiu bin

Gim phn

Gim phn


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Gii thiu phng php xc nh bi ton thuc quy lut di truynPhn li c lp

Trong qu trnh lm cc bi ton lai, vn quan trngl vit c s lai, vit c s lai, phi xcnh c bi ton thuc quy lut di truyn no.

nhn din bi ton thuc quy lut di truyn Phn lic lp, chng ta c th cn c vo cc c s sau :

Trng hp 1:

Da vo iu kin nghim ng ca quy lut Menen m u bi cho:

- Mi tnh trng do mt gen quy nh.

- Mi gen nm trn mt nhim sc th hay cc cp gen nm trn cc cpnhim sc th tng ng khc nhau.

Khi bi cho cc iu kin trn, chng ta c th bit ngay quy lut ditruyn chi phi l quy lut Menen.

Trng hp 2 :

Da vo t l phn ly kiu hnh i con.

- Nu lai mt cp tnh trng, mi tnh trng do mt gen quy nh cho kiuhnh l mt trong cc t l sau : 100% ; 1 : 1; 3 : 1; 1 : 2 : 1 (tnh trng trung

gian); 2 : 1 (t l gy cht). Nhn tch cc cp tnh trng cho kt qu gingu bi.

- Khi lai hai hay nhiu cp tnh trng cho kiu hnh l mt trong cc t lsau : (1 : 1)n; (3 : 1)n; (1 : 2 : 1)n....

Trng hp 3:

Nu bi ch cho bit t l ca mt kiu hnh no con lai.

- Khi lai mt cp tnh trng, t l mt kiu hnh c bit bng hoc l bis ca 25% (hay 1/4).


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- Khi lai hai cp tnh trng m t l mt kiu hnh c bit bng, hoc lbi s ca 6,25% hoc 1/16; hay khi lai n cp tnh trng m t t l cakiu hnh bit cho php xc nh c s loi giao t ca b hoc m

c t l bng nhau hoc l c s ca 25%.

Quy tc s T (total) - Nhn dng nhanh php lai chi phi

1. Quy tc:Gi T l tng s h s cc s hng trong chui tng quan ti gin,nguyn ca t t kiu gen hay t t kiu hnh i con F1trong php lai 2 tnh.

a. Nu T = 2k( k l s nguyn t nhin): Th c hin tng :

a1. Hoc phn ly c lp;a2. Hoc lin kt gen hon ton;a3. Hoc hon v gen vi tn s c bit: 1/2, 1/4, 1/8, 1/16,...1/2 n (n l s nguyntnhin).

b. Nu T 2k: cc gen hon v vi tn s f (0 < f


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1. Thuc quy lut lin kt gen hon ton

- Nu bi cho hoc c th xc nh c y cc yu t sau y :

- Lai hai hay nhiu cp tnh trng, c quan h tri - ln.

- t nht mt c th em lai d hp cc cp gen

- T l con lai ging vi t l ca lai mt cp tnh trng ca quy lut Menen(100%; 1 : 2 : 1; 3 : 1; 2 : 1; 1 : 1).

- C th em lai d hp3 cp gen tr lnnm trn hai cp nhim sc thtng ng:

- Nu kiu gen t th phn cho con lai 16 t hp.

- Nu kiu gen lai phn tch cho t l con lai l 1: 1 : 1 : 1; ... (t hn so vitrng hp phn ly c lp)

Trong trng hp , c th suy ra rng c th d hp 3 cp gen ch to 4 loigiao t ngang nhau tc phi c hin tng lin kt gen hon ton.

2. Thuc quy lut hon v gen :

Nu php lai t hai cp tnh trng tr ln, c quan h tri - ln cho t l kiuhnh i con khng phi l t l ca quy lut phn ly c lp v quy lutlin kt gen hon ton.

V d : C th d hp 2 cp gen quy nh hai cp tnh trng t giao phi cmt kiu hnh no i con chim t l 20,5%.

PHP LAI TNG NG

I. nh ngha: Hai php lai gi l tng ng (hay cn gi l lai tng ng) khicc dng b, m c nhng b NST v b gen tng hp. Khi b v m hon i


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cho nhau cc cp gen tng ng m kt qui con kiu gen khng thay i (ch cp n kiu gen).

Lai tng ng tun theo nh lut phn ly c lp ca Menden v tnh cht cacon lai tng ng ging nhau trong php lai thun nghch

V d:

P1: AABB x aabb v P2: AAbb x aaBB

Php lai P1 v P2 l 2 php lai tng ng.

P3: AaBb x aabb v P4: Aabb x aaBb l 2 php lai tng ng.

II. iu kin c php lai tng ng:

1. Cc gen phi phn ly c lp. Nu cc gen lin kt, hon v th khng bao gic php lai tng ng

V d: P: AB/AB x ab/ab khc vi P: Ab/Ab x aB/aB mc d b m hon icho nhau cc cp gen tng ng.

2. B v m khc nhau t nht 2 cp gen:

V d: Php lai P1: AABB x AAbb khng c php lai tng ng vi n

3. Gen m b v m hon i chonhau phi nm trn NST thng, nu nmtrn NST gii tnh th khng th thnh lp php lai tng ng.

V d:

- P: AAXBXB x aaXbYb tng ng vi P: aaXBXBx AAXbYb

- P: AAXBXB x aaXbY tng ng vi P: aaXBXBx AAXbY

- P: AAXBXB x aaXbYb khng tng ng vi P: AAXbXb x aaXBYB

- P: AAXBXB x aaXbY khng tng ng vi P: AAXbXb x aaXBY

III. Cng thc tnh s php lai tng ng:

1. Nu b v m khc nhau n cp gen alen th ta c c 2 n-1php lai tng

ng vi nhau.


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V d: B v m khc nhau 3 cp gen alen th ta c 23 = 4 php lai tng ngvi nhau

M BCp gen alen th nht AA aaCp gen alen th hai BB bbCp gen alen th ba DD dd

Ta c 4 php lai tng ng vi nhau:

P1: AABBDD x aabbdd

P2: AAbbDD x aaBBdd

P3: aaBBDD x AAbbdd

P4: aabbDD x AABBdd

2. Nu ta c 1 php lai gc m b v m khc nhau n cp gen alen th ta stm thm c (2n-1-1) php lai tng ng vi php lai gc cho.

V d:Nu ta c 1 php lai gc m b v m khc nhau 3 cp gen alen th ta s tm

thm c 23-1

-1 = 3 php lai tng ng vi php lai gc.

Nu ta c php lai gc: P1: AABBDD x aabbdd

Th ta s tm thm c 3 php lai tng ng vi n:

P2: AAbbDD x aaBBdd

P3: aaBBDD x AAbbdd

P4: aabbDD x AABBdd

IV. Cch thit lp php lai tng ng:

a. Dng php nhn i s:

(AA : aa) (BB : bb) (DD : dd) = (AABB: AAbb: aaBB: aabb) (DD : dd) =

AABBDD: AAbbDD: aaBBDD: aabbDD:AABBdd: AAbbdd: aaBBdd: aabbdd

(1) (2) (3) (4) (4) (3) (2) (1)

Ta c 4 php lai tng ng khi rp:


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P1: (1) x (1) = AABBDD x aabbdd

P1: (2) x (2) = AAbbDD x aaBBdd

P1: (3 x (3) = aaBBDD x AAbbdd

P1: (4) x (4) = aabbDD x AABBdd

b. Dng s nhnh (tng t nh vit kiu giao t)

V. ng dng v ngha ca vic nghin cu php lai tng ng:

1. tm ht nghim kiu gen ca b m khi gii bi ton nghch.

V d 1: u H Lan gen A qui nh ht vng, gen a qui nh ht xanh; gen Bqui nh v ht trn, gen b qui nh v ht nhn, cc gen phn ly c lp.

Tm kiu gen ca P nu F1 ng tnh ht vng, v trn.

Tr li:

Ccc kh nng v kiu gen ca P:

P1: AABB (vng, trn) x AABB (vng, trn)

P2: AABB (vng, trn) x AABb (vng, trn)

P3: AABB (vng, trn) x AAbb (vng, nhn)

P4: AABB (vng, trn) x AaBB (vng, trn)

P5: AABB (vng, trn) x AaBb (vng, trn)

P6: AABB (vng, trn) x Aabb (vng, nhn)

P7: AABB (vng, trn) x aaBB (xanh, trn)

P8: AABB (vng, trn) x aaBb (xanh, trn)

P9: AABB (vng, trn) x aabb (xanh, nhn)

Ngoi ra ta c thm 4 nghim tng ng rt ra t P5, P6, P8, P9.

P10: AABb (vng, trn) x AaBB (vng, trn) # P5: AABB (vng, trn) xAaBb(vng, trn)


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P11: AAbb (vng, nhn) x AaBB (vng, trn) # P6: AABB (vng, trn) x Aabb(vng, nhn)

P12: AABb (vng, trn) x aaBB(xanh, trn) # P8: AABB (vng, trn) x aaBb(xanh, trn)

P13: AAbb (vng, nhn) x aaBB (xanh, trn) # P9: AABB (vng, trn) x aabb(xanh, nhn)

Vy c 13 trng hp khc nhau v kiu gen ca P nu F1 ng tnh ht vng,v trn.

V d 2: Cho bit mu sc ca lng chut di truyn bi 2 cp gen phn ly c lp,

tng tc t ch bi cp alen ln (aa > B).

A-bb: Mu en;

aaB-, aabb: mu trng;

A-B-: mu xm

Nu F1ng tnh xm th kiu gen, kiu hnhca P s nh th no.

Tr li:

C cc kh nng v kiu gen ca P:

P1: AABB (xm) x AABB (xm)

P2: AABB (xm) x AABb (xm)

P3: AABB (xm) x AAbb (en)

P4: AABB (xm) x AaBB (xm)

P5: AABB (xm) x AaBb (xm)

P6: AABB (xm) x Aabb (en)

P7: AABB (xm) x aaBB (trng)

P8: AABB (xm) x aaBb (trng)

P9: AABB (xm) x aabb (trng)


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LTH 2014 Khi A,A1,B,D Bomim thi H thp nht15 im10

Ngoi ra ta c thm 4 nghim tng ng rt ra t P5, P6, P8, P9.

P10: AABb (xm) x AaBB (xm) # P5: AABB (xm) x AaBb (xm)

P11: AAbb (en) x AaBB (xm) # P6: AABB (xm) x Aabb (en)

P12: AABb (xm) x aaBB (trng) # P8: AABB (xm) x aaBb (trng)

P13: AAbb (en) x aaBB (trng) # P9: AABB (xm) x aabb (trng)

2. ngha ca vic nghin cu:

Ta thy i con F1, F2, Fn c kiu gen, kiu hnh hon ton ging nhau nhng chc

g cng mt ngun gc b, m. T thy rng c 2 ngi no khng cquan h huyt thng vi nhau nhng h c th ging nhau mt hoc vi cp tnhtrng no l chuyn thng tnh.

V vy ng vi thy ging nhau m nhn b con.

Cng c trng hp khng h ging nhau m li cng chung huyt thng.

V d: P: AABBCCDDEE x aabbccdddee

F1: AaBbCcDdEe

F2, Fn: Qun th c 3nkiu gen khc nhau.

T thy rng th gii sng v cng a dng, phong ph. Trn i ny c rtnhiu ngi khng cng ni ging nhng li ging ta; cng c nhiu ngi rtkhc ta nhng cng chung ni ging vi ta.

TM HIUCC BC GII NHANH MT BI TON V LAI 1, 2 HAY

NHIU CP TNH TRNG TNG PHN, S TUNG TC GIA CC

GEN KHNG ALEN


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Vic nhn dng cc quy lut di truyn l vn quyt nh cho vic gii

nhanh v cc bi ton lai. nhn dng cc quy lut di truyn phi da vo cc

iu kin c th ca bi ton.+ i vi cc bi ton lai v 1, 2 hoc nhiu cp tnh trng phn ly c lp th

ta da vo:

- Cc iu kin v tnh trng gen quy nh

- Kt qu ca php lai xc nh

+ i vibi ton v tng tc gia cc gen khng allen th ta da vo:- Da vo cc iu kin v php lai

- Kt qu phn tch i con qua cc php lai

I.Cch nhn dng quy lut di truyn:

1. Trng hp bi ton xc nh t l phn ly kiu hnh i con:

1.1. Da vo kt qu phn ly kiu hnh ca i con:

1.1.1. Khi lai 1 tnh trng:

Tm t l phn tch v KH th h con i vi loi tnh trng t xc

nh quy lut di truyn chi phi.

+ 3:1 l quy lut di truyn phn tch tri ln hon ton.

+ 1:2:1 l quy lut di truyn phn tch tri khng hon ton (xut hin tnh

trng trung gian do gen nm trn NST thng hoc gii tnh.

+ 1:1 hoc 2:1 t l ca gen gy cht.

+ 9:3:3:1 hoc 9:6:1 hoc 9:7 l tnh trng di truyn theo tng tc b tr.

+ 12:3:1 hoc 13:3 l tnh trng di truyn theo quy lut tng tc t ch tri.

+ 9:3:4 l tng tc t ch do gen ln.

+ 15:1 l tng tc cng gp kiu khng tch ly cc gen tri.

1.1.2. Khi lai 2 hay nhiu cp tnh trng:

+ Tm t l phn tch v kiu hnh th h con i vi mi loi tnh trng.


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+ Nhn t l KH ring r ca loi tnh trng ny vi t l KH ring ca loi

tnh trng kia.

Nu thy kt qu tnh c ph hp vi kt qu php lai th c th kt lun 2

cp gen quy nh 2 loi tnh trng nm trn 2 cp NST khc nhau, di truyn

theo nh lut phn li c lp ca Menden (tr t l 1:1 nhn vi nhau).

V d: Cho lai hai 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-thn thp. 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 vng-thp, ph hp vi php lai trong bi. Vy 2 cp gen quy

nh 2 tnh trng nm trn 2 cp NST khc nhau.

1.2. Da vo kt qu phn ly kiu hnh trongphp lai phn tch:

Da vo kt qu ca php lai xc nh t l v loi giao t sinh ra ca cc

c th cn tm.

+ Nu t l KH 1:1 th l s di truyn 1 tnh trng do 1 gen chi phi

+ Nu t l KH 3:1 th c di truyn theo quy lut tng tc gen, trong tnh

trng c 2 kiu hnh.

- Tng tc b tr 9:7

-Tng tc t ch 13:3

- Tng tc cng gp 15:1

+ Nu c t l KH 1:2:1 th tnh trng c di truyn theo quy lut tng tc

gen trong trng hp tnh trng c 3 kiu hnh.


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-Tng tc b tr 9:6:1

-Tng tc t ch ln 9:3:4

-Tng tc t ch tri 12:3:1

+ T l KH 1:1:1:1 l s di truyn tng tc b tr 1 tnh trng c 4 kiu

hnh 9:3:3:1 hoc l lai 2 cp tnh trng tun theo nh lut phn ly c lp c t l

kiu hnh l 9:3:3:1.

2.Nu bi khng xc nh t l phn li kiu hnh ca i con m ch cho

bit 1 kiu hnh no con lai.+ Khi lai 1 cp tnh trng, t l 1 kiu hnh c bit bng hoc l bi s ca

25% (hay4

1).

+ Khi lai 2 cp tnh trng m t l 1 kiu hnh c bit bng hoc l bi s

ca 6.25% (hay16

1), hay khi lai n cp tnh trng m t t l ca KH bit cho

php xc nh c s loi giao t ca b (hoc m) c t l bng nhau v bng

25% hoc l c s ca 25%.

lcc bi ton thuc nh lut Menden.

V d:Cho lai 2 cy u thun chng khc nhau 2 cp tnh trng tng phn,

F1thu c ton b cy thn cao - hoa . Cho F1tp giao F2thu c 16000 cy

trong c 9000 cy thn cao - hoa . Hai cp tnh trng trn b chi phi bi quy

lut di truyn.

A. Tng tc t ch C.Tng tc b tr

B. Phn li c lp D. Tng tc cng gp

Gii:


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T l cy cao- thu c th h F2l16000

9000=

16

9= 56.25% l bi s ca

6.25% l bi ton thuc nh lut Menden

=> Chn p n B

3.Tnh trng do 1 hay 2 gen quy nh? Xc nh kiu gen tng ng ca c

th lai:

Ty vo s t hp i con ca tng php lai v tnh tri ln hon ton hay

khng hon ton th h lai.

+ Php lai hai c th d hp (thng l cho F1 giao phi vi nhau) cho s t

hp khng qu 4 th thng do 1 gen quy nh; s t hp hn 4 nhng khng qu

16 thng do 2 gen quy nh.

* V dKhi lai F1d hp c F2phn ly t l 11: 2: 2: 1 (tng c 16 t hp)

th chc chn khng phi l 1 gen quy nh.

+ Php lai phn tch F1: nu cho s t hp khng qu 4 nhng khng phi

1:1, lc ny li do 2 gen quy nh...

* V dKhi lai phn tch c 3 : 1 xanh (4 t hp) th cng chc chn

khng phi l 1 gen.

+ Lai F1vi 1 c th bt k: s t hp ti a khi lai hai c th d hp vinhau, t c th loi tr cc kh nng khng ng.

*V dKhi lai hai c th bt k v tnh trng A m cho con ti 8 t hp th

chc chn tnh trng do 2 gen quy nh, trong 1 c th d hp c 2 gen, 1 c th

d hp 1 gen (thng l d hp v ng hp ln gen cn li)...

4. Gen ny c gy cht khng?

Du hiu ca kiu ny l s t hp i con khng chn, c th l 3, 7,..

thay v 4, 8... y l 1 du hiu t gp nhng vnphi ngh n.


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Nu i con phn ly t l c bit VD 2:1 th gn nh c th chc chn l

gen gy cht, v thng l gy cht trng thi ng hp tri.

5. Cc trng hp ring:

+ Da vo kt qu phn li kiu hnh ca F1 lai vi c th khc. cn ch

nhng t l c bit sau y: 7:1; 4:3:1; 6:1:1; 5:3 y l t l ca tnh trng ny

sinh do tng tc gen, ty tng trng hp c th m xc nh chnh xc tnh trng

c xt, di truyn theo quy lut di truyn no.

+ Trng hp ng tri da vo iu kin nh: 1 tnh trng c qui nh

bi 1 cp gen c 3 alen, IA= IB> IO. S kiu gen ti a l 6, s kiu hnh ti a l

4.

V d: Mu lng ca mt loi c mo chu s kim sot ca dy a allen

xp theo th t tnh tri gim dn l: R1 (lng ) > R2 (lng en) > R3 (lng

xm). Hy xc nh Kiu gen ca c lng , lng en v lng xm.

Gii:

Dy a allen xp theo th t tnh tri gim dn l: R1(lng ) > R2(lng

en) > R3(lng xm)

KG ca c lng c th l: R1R1; R1R2; R1R3

KG ca c lng en c th l: R2R2; R2R3

KG ca c lng xm c th l: R3R3

II. Phng php gii bi tp:

Ty tng yu cu ca bi ton m ta c cc phng php gii khc nhau.

1. Trong php lai 1, 2 hay nhiu cp tnh trng tng phn:thng gp

2 dng chnh

- Dng ton thun: cho bit tnh trng (hay gen) tri, ln t tm t l phn

tch i sau

- Dng ton nghch: cho bit kt qu i con t tm kiu gen ca b m


16/37



1.1.Tnh s loi v thnh phn gen giao t:

1.1.1. S loi giao t:Ty thuc vo s cp gen d hp trong kiu gen

+ Trong KG c 1 cp gen d hp21

loi giao t+ Trong KG c 2 cp gen d hp22loi giao t

+ Trong KG c 3 cp gen d hp23loi giao t

Vy trong KG c n cp gen d hp2nloi giao t

V d:Kiu gen AaBbCcDd c kh nng to ra bao nhiu loi giao t?

Ta xt kiu gen trn c 4 cp gen d hp, vy s loi giao t l 2n=24=16

1.1.2.Thnh phn gen (KG) ca giao tTrong t bo (2n) ca c th gen tn ti thnh tng cp tng ng, cn

trong giao t (n) ch cn mang 1 gen trong cp.

+ i vi cp gen ng hp AA (hoc aa): cho 1 loi giao t A (hoc 1 loi

giao t a)

+ i vi cp gen d hp Aa: cho 2 loi giao t vi t l bng nhau gm giao

t A v giao t a+ Suy lun tng t i vi nhiu cp cp gen d hp nm trn cc cp NST

khc nhau, thnh phn kiu gen ca cc loi giao t c ghi theo s phn

nhnh (s Auerbac) hoc bng cch nhn i s.

V d: Cho bit thnh phn gen mi loi giao t ca kiu gen

sau:AaBBDdeeTa c s sau:

A a

B B

D d D d

E e e e

KG ca giao t l :ABDE Abde aBDe aBde


17/37



V d: Trong iu kin gim phn bnh thng, c th AaBbCcDD sinh ra cc loi

giao t no?A. ABCD v abcD

B.ABCD, ABcD, AbCD, AbcD

C. ABCD, AbcD, aBCD, AbcD, abCD, AbCd, abcD, AbcD

D. ABCD, AbcD, AbCD, AbcD, aBCD, abCD, abcD, AbcD.

Gii:KG ang xt d hp 3 cp allen => s giao t c th to ra l 23=8

V khng cha gen ln d.

Chn p n D

1.2.Tnh s kiu t hp, kiu gen, kiu hnh v cc t l phn li i con (dng

ton thun)

1.2.1. S kiu t hp:

Mi loi giao t c t hp t do vi cc loi giao t ci to thnh nhiu

kiu t hp trong cc hp t. V vy s kiu t hp gia cc loi giao t c v ci

l:

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

Kiu t hp khc nhau nhng c th a n kiu gen ging nhau

=> s KG s kiu t hp

V d:Nu cy m c 3 cp gen d hp, 3 cp gen ng hp, cy b c 2 cp

gen d hp, 4 cp gen ng hp ln.

S kiut hp giao t i F1l:

A. 16 B.32 C.64 D.128

Gii:

+ Cy m c 3 cp gen d hp => c 23loi giao t

+ Cy b c 2 cp gen d hp => c 22loi giao t


18/37



=> S kiu t hp giao t i F1l 23x 22= 32

Chn p n B

1.2.2 S kiu gen, kiu hnh i con:

S di truyn ca cc cp gen l c lp vi nhau, v vy s t hp t do gia

cc cp gen cng nh gia cc cp tnh trng. V vy, kt qu v kiu gen cng

nh v kiu hnh i con c xc nh:

+ T l kiu gen chung ca nhiu cp gen = Tch cc t l kiu gen ring l ca

mi cp gen.S kiu gen tnh chung = Tch s cc kiu gen ring ca mi cp gen

+ T l kiu hnh chung ca nhiu cp tnh trng = Tch cc t l kiu hnh

ring l ca mi cp tnh trng.

S kiu hnh tnh chung = Tch s kiu hnh ring ca mi cp tnh

trng

V d1: Cho gi thuyt sau:

A: ht vng a: ht xanh

B: ht trn b: ht nhn

D: thn cao d: thn thp

Cc cp gen ny di truyn c lp nhau. Ngi ta tin hnh php lai gia 2 c th

c kiu gen: AabbDd lai vi AaBbdd.

Xc nh s kiu gen v s kiu hnh chung ca con lai.

Gii:

Ta xt cc php lai c lp :

Kiu gen kiu hnh

Aa x Aa =AA: 2Aa: aa 3 vng: 1 xanh

Bb x bb = Bb: bb 1 trn: 1 nhn


19/37



Dd x dd = Dd: dd 1 cao: 1 thp

Vy:S t hp 1 cp gen d hp Aa cho ra 3 KG (Aa x Aa =1AA: 2Aa: 1aa )

S t hp 2 cp gen 1 bn d hp bn kia ng hp cho ra 2 KG

(Bb x bb = 1Bb : 1bb; Dd x dd = 1Dd : 1dd)

Tl KG chung l: (1AA : 2Aa : 1aa)(1Bb : 1bb)(1Dd : 1dd)

= AABbDd ; AABbdd ; AAbbDd ; Aabbdd....

S kiu gen tnh chung: 3.2.2 = 12

Lp lun tng t:

S t hp 1 cp gen d hp Aa cho ra 2KH (3 vng: 1 xanh)

S t hp 2 cp gen 1 bn d hp bnkia ng hp cho ra 2 KH

T l KH tnh chung: (3 vng : 1 xanh)(1 trn : 1 nhn)(1 cao : 1 thp)

S kiu hnh tnh chung: 2.2.2 = 8

1.2.3. Tnh t l phn ly i con:

T l phn li kiu gen i con= Tch cc t l kiu gen ring l ca mi

cp gen.

V d1: Du ty: genR (tri khng hon ton)quy nh tnh trng qu

Gen r (ln khng hon ton) quy nh tnh trng qu trng

Gen Rr quy nh qu hng

Gen H quy nh tnh trng cy cao (tri)

Gen h quy nh tnh trng cy thp (ln)

2 cp gen ny nm trn 2 cp NST khc nhau. Khi cho lai 2 cy du ty d hp

v hai cp gen trn F1c t l kiu di truyn l:

A. 9 : 3 : 3 : 1 B. 3 : 6 : 3 : 1 : 2 : 1

C. 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1 D. C 3 trn u sai


20/37



Gii:

P: RrHh x RrHh

Ta xt 2 php lai c lp nhau (do cc gen phn li c lp)

Rr x Rr = 1RR : 2Rr : 1rr.

Hh x Hh = 1HH : 2Hh : 1hh.

T l phn li kiu gen i con l: (1 : 2 : 1) (1 : 2 : 1)

= 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1

=> Chn p n C

V d 2:php lai AaBbccDdee x AabbccDdEe s sinh ra kiu gen aabbccddee

chim t l bao nhiu ?(Vi 5 cp gen nm trn 5 cp NST khc nhau, cc tnh

trng u tri hon ton.)

Gii:

trng hp ny ta xt 5 php lai c lp nhau:

Aa x Aa4

3A- +

4

1aa

Bb x bb2

1B- +

2

1bb

cc x cc 1cc

Dd x Dd4

3D- +

4

1dd

Ee x ee2

1E- +

2

1ee

Vy kiu gen aabbccddee sinh ra i con chim t l l:

4

1

x2

1

x 1 x4

1

x2

1

=64

1


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V d 3: Cho lai 2 c th AaBbCc, vi 3 cp gen nm trn 3 cp NST khc nhau,

cc tnh trng u tri hon ton.

a.

T l kiu di truyn c th d hp 2 cp gen, cp gen cn li ng hp:

A.64

1 B.

64

8 C.

64

24 D.

64

32

b. T l kiu di truyn c th d hp 1 cp gen, 2 cp cn li ng hp:

A.64

1 B.

64

8 C.

64

24 D.

64

32

Gii:

Ta xt 3 php lai c lp nhau:

Aa x Aa4

1AA +

4

2Aa +

4

1aa

Bb x Bb 4

1

BB + 4

2

Bb + 4

1

bb

Cc x Cc4

1CC +

4

2Cc +

4

1cc

a. C th d hp 2 cp gen, cp gen cn li ng hp l : AaBbCC; AaBbcc;

AaBBCc; AabbCc; AABbCc; aaBbCc

M t l ca tng kiu gen l : 42

x 4

2

x 4

1

= 64

4

Tng t cho cc kiu hnh cn li

Vy t l kiu di truyn c th d hp 2cp gen, cp gen cn li ng hp l:

(4

2x

4

2x

4

1) x 6 =

64

4x 6 =

64

24

Chn p n C


22/37



b. C th d hp 1 cp gen, 2 cp cn li ng hp l: AaBBCC; AabbCC;

Aabbcc; AaBBcc; AABbCC; AABbcc; aaBbCC; aaBbcc; AABBCc;

AAbbCc; aaBBCc; aabbCc

M t l ca tng kiu gen l:4

2 x

4

1 x

4

1=

64

2

Tng t cho cc kiu hnh cn li

Vy t l kiu di truyn c th d hp 1 cp gen, 2 cp cn li ng hp l:

(4

2 x

4

1 x

4

1) x 12 =

64

2x 12 =

64

24

Chn p n C

T l phn li kiu hnh= Tch cc t l kiu hnhring l ca mi cp gen.

V d1: C th d hp kiu gen AaBb tp giao s cho F1phn tnh kiu hnh theo t

l no, nu cc gen ny phn ly c lp v gen A tri khng hon ton?

A. 9 : 3 : 3 : 1 C. 6 : 3 : 3: 2: 1: 1B. 27: 9 : 9: 9: 3: 3: 3:1 D. 9 : 3 : 4

Gii:

Ta xt 2php lai c lp nhau(do cc gen phn li c lp)

Aa x Aa = 1AA : 2Aa : 1aa. V gen A tri khng hon ton, lc kiu gen

AA, Aa, aa quy nh 3 KH khc nhau =>Cho ra 3 kiu hnh

Bb x Bb = 1BB : 2Bb : 1bb. V gen B tri hon ton, lc kiu gen BB v

Bb c cng 1 KH =>Cho ra 2 kiu hnh (3B-, 1bb)

T l phn li kiu hnh i con l: (1 : 2 : 1) (3 : 1) = 6 : 3 : 3: 2: 1: 1

Chn p n C

V d 2:Cho lai phn tch c th ci d hp 4cp gen nm trn 4 cp NST khc

nhau, t l kiu kiu hnh i F1l:

A. 1 : 1 : 1 : 1


23/37



B. 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1

C. 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1: 1 : 1 : 1 : 1

D. 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1

Gii:

C1: Tng t lp lun v d 1

C2: Trong php lai phn tch th 1 c th ng hp ln lai vi c th khc ( c th

c kiu hnh tri kim tra kiu gen).

Vy c th ng hp cho ra 1 loi giao tC th em lai phn tch c 4 cp gen d hp => s loi giaot c to ra

l: 24= 16

S t hp giao t to ra l 1 x 16 = 16

Xt cc p n trn, ch c p n D l c 16 t hp

Chn p n D

CNG THC TNG QUTKhi so snh lai mt cp tnh trng v lai hai cp tnh trng ta thy rng trong

lai mt cp tnh trng F2phn li thnh 2 loi kiu hnh theo t l 3 : 1, trong khi

lai 2 cp tnh trng chng phn li thnh 4 loi kiu hnh theo t l 9 : 3 : 3 : 1. T l

ny ng vi bnh phng ca biu thc (3 + 1)

(3 + 1)2= 9 + 3 + 3 + 1

Mt cch tng t trong lai 3 cp tnh trng s phn li kiu hnh F2cho

8 loi kiu hnh ng vi:

(3 + 1)3= 27 + 9 + 9 + 9 + 3 + 3 + 3 + 1

T c th nu nhn xt khi qut: Trong lai n cp tnh trng th t l

phn li kiu hnh F2ng vi cng thc (3 + 1)n.


24/37



Cng thc phn tnh chung trong nh lut phn ly c lp ( trng hp c

tnh tri hon ton) i vi c th c n cp gen d hp phn li c lp, khi

AaBb...Nn t th.

F1 F2

Kiu genS kiu

giao t

S kiu

t hp

giao t

Sloi

kiu gen

T l kiu

gen

S loi

kiu hnh

T l kiu

hnh

Lai 1 tnh

Lai 2 tnh

Lai 3 tnh

...............

Aa

AaBb

AaBbCc

...............

21

22

23

..............

21 x 21

22x 22

23x 23

..............

31

32

33

..............

(1:2:1)1

(1:2:1)2

(1:2:1)3

...............

21

22

23

..............

(3:1)1

(3:1)2

(3:1)3

...............

Lai n tnh AaBbCc... 2n 2nx 2n 3n (1:2:1)n 2n (3:1)n

Tng qut hn, nu mt cy d hp v n cp allen giao phn vi cy d hp

v m cp allen th ta c:

+ Cy d hp v n cp allen c 2nloi giao t

+ Cy d hp v m cp allen c 2mloi giao t

Do => Tng s hp t = 2nx 2m= 2n+m

- T l cy c kiu hnh tri =mk

4

3

- T l th ng hp ton tri =mnmn

2

1

2

1*

2

1

- T l th ng hp ton ln =mnmn

2

1

2

1*

2

1

Tm s kiu gen ca mt c th:


25/37



Mt c th c n cp gen nm trn n cp NST tng ng, trong c k cp

gen d hp v m=n-k cp gen ng hp. S kiu gen c th c ca c th tnh

theo cng thc:mm

n

knkn

n CCA 22

Trong : A l s kiu gen c th c ca c th

n l s cp gen

k l s cp gen d hp

m l s cp gen ng hp

V d:Trong c th c 4 cp gen nm trn 4 cp NST tng ng, c th b c 3

cp gen d hp, 1 cp gen ng hp. cn m th ngc li. C bao nhiu kiugiao

phi c th xy ra?

A. 64 B.16 C.256 D.32

Gii:

C1: Gii theo cch lit k cc kiu gen c th c ca c th b m sau nhn livi nhau:

+ Xt c th b: c 3 cp gen d hp, 1 ng hp => cc kiu gen c th c:

AaBbCcDD AaBbCcdd

AaBbCCDd AaBbccDd

AaBBCcDd AabbCcDd

AABbCcDd aaBbCcDd

Vy c tt c l 8 trng hp c th xy ra

+ Xt c th m: c 1 cp d hp, 3 cp ng hp=> cc kiu gen c th c:

AaBBCCDD AabbCCDD

AaBBCCdd AabbCCdd

AaBBccDD AabbccDD

AaBBccdd Aabbccdd


26/37



Nu ta gi nh Aa l cp gen d hp cn 3 cp gen cn li ng hp th ta lit k

c 8 kiu gen, sau ta thay i vai tr d hp cho 3 cp gen cn li. Lc , s

kiu gen c th c ca c th m l:8 . 4 = 32

Suy ra, s kiu giao phi l: 8. 32 = 256

chn p n C

C2:p dng cng thc tnh:

S kiu gen c th c ca c th b l:

8242

!1!.14

!42

111

4

CA

S kiu gen c th c ca c th m l:

32842

!3!.34

!42

333

4

CB

Suy ra, s kiu giao phi l: 8 . 32 = 256 chn p n C

1.3. Tm kiu gen ca b m (dng ton nghch):

1.3.1. Kiu gen tnh ring ca tng loi tnh trng:

Xt ring kt qu i con lai F1ca tng tnh trng

1.3.1.1. F1ng tnh:+ Nu b m (P) c kiu hnh khc nhau th F1nghim ng nh lutng

tnh ca Menden => tnh trng biu hin F1 l tnh trng tri v th h P u

thun chng: AA x aa

+ Nu P cng kiu hnh v F1 mang tnh trng tri th 1 trong 2P c KG

ng hp tri AA, P cn li c th l AA hoc Aa.


27/37



+ Nu P khng r kiu hnh v F1 mang tnh trng tri, th 1 trong 2P l

ng hp tri AA, P cn li mang KG ty : AA, Aa, aa.

1.3.1.2. F1phn tnh:

+ F1phn tnh theo t l 3:1

- F1 nghim ng nh lut phn tnh ca Menden => tnh trng4

3 l tnh

trng tri,4

1l tnh trng ln v P u d hp Aa x Aa.

- Trong trng hp tri khng hon ton th t l F1l 1:2:1.- Trong trng hp gen gy cht trng thi ng hp th t l F1l 2:1

+ F1phn tnh theo t l 1:1

F1 l kt qu c trng ca php lai phn tch th d hp => th 1Pc KG d

hp Aa, P cn li ng hp aa.

+ F1phn tnh khng r t l:

Da vo c th mang tnh trng ln F1l aa => P u cha gen ln a, phi

hp vi kiu hnh ca P suy ra KG ca P.

1.3.2. Kiu gen tnh chung ca nhiu loi tnh trng:

1.3.2.1. Trong php lai khng phi l php lai phn tch:

Kt hp kt qu v KG ring ca tng loi tnh trng vi nhau.V d:c chua

A: qu a: qu vng

B: qu trn b: qu bu dc

Cho lai 2 cy c chua cha r KG v KH vi nhau th thu c F1gm: 3 -

trn, 3 -bu dc, 1 vng-trn, 1 vng-bu dc. Tm KG 2 cy thuc th h P.

Gii:


28/37



+ Xt chung tng cp tnh trng:

F1gm)11(

)33(

= 3 : 1 vng (theo L ng tnh) => P: Aa x Aa

F1gm)13(

)13(

= 1 trn : 1 bu dc (lai phn tch d hp)

P: Bb x bb

+ Xt chung trong KG: Kt hp kt qu v KG ring ca mi tnh trng

trn, suy ra kiu gen ca P l AaBb x Aabb.

T t hp giao t i con, bin lun suy ra s giao t c to thnh trong

pht sinh giao t ca c th b m, t suy ra KG ca c th b m cn tm.

V d1: C th c d hp hai cp gen AaBb nm trn 2 cp NST khc nhau. Cho

bit 2 loi tnh trng trn tri hon ton.Cho lai c th trn vi c th ci. F1thu

c t l kiu hnh 3:3:1:1. Kiu gen ca c th ci l:

A. AaBb C.aaBb

B.Aabb D.B v C ng

Gii:

F1thu c c t l kiu hnh l: 3:3:1:1

- Do s t hp ca F1l: 3 + 3 + 1 + 1= 8 t hp giao t- M c th c d hp hai cp gen (AaBb)=> cho 4 loi giao t

Suy ra c th ci s cho 2 loi giao t.

- Xt tt c p n trn c 2 p n B v C, c th ci aaBb v Aabb khi

gim phn to ra 2 loi giao t.

chn p n D

Vng

Trn

Bu dc


29/37



V d 2:Mt loi thc vt gen A quy nh cy cao, gen a- cy thp; gen B qu ,

gen b- qu trng. Cc gen di truyn c lp. i lai c mt loi kiu hnh cy thp

qu trng chim 1/16. Kiu gen ca cc cy b m l

A. AaBb x Aabb. B. AaBB x aaBb.

C. Aabb x AaBB. D. AaBb x AaBb.

Gii:

C1: Da vo s lai suy ra p n ng

Kiuhnhcy thp qu trng l kiu hnh do gen ln quy nh (theo quy c) do

kiu gen ca n l: aabb

Xt ln lt cc php lai c th th b m trn:

Php lai p n C v B u khngto ra c th c KG l aabb.

Php lai p n A v D to ra c th c KG l aabb

Xt p n A. php lai AaBb x Aabb s to ra aabb chim t l l8

1

Php lai AaBb x AaBb p n D s to ra aabb chim t l l 161

=> chn p n D

C2: Da vo t hp giao t ca php lai

C th con thu c chim t l16

1 , t , suy ra s t hp ca php lai trn l 16

t hp.

Xt ln lt cc php lai:

Php lai AaBb s cho 22loi giao t

Aabb s cho 21loi giao t

=> S t hp giao t c to ra l 22* 21= 8

Tng t:

+ AaBB x aaBb = 21* 21= 4 t hp giao t

+ Aabb x AaBB = 21* 21= 4 t hp giao t

+ AaBb x AaBb = 22* 22= 16 t hp gaio t

=> chn p n D


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1.3.2.2. Trong php lai phn tch:

Khng xt ring tng loi tnh trng m phi da vo kt qu ca php lai

xc nh t l v thnh phn gen ca mi loi giao t sinh ra suy ra KG ca c

th .

V d :C th c d hp hai cp gen AaBb nm trn 2 cp NST khc nhau.

Cho bit 2 loi tnh trng trn tri hon ton.Cho lai c th trn vi c th ci. F1

thu c t l kiu hnh 1:1:1:1. Kiu gen ca c th ci sinh ra l:

A. AaBb C.aaBb

B. Aabb D.aabb

Gii:

F1thu c c t l kiu hnh l: 1:1:1:1

- Do st hp ca F1l: 1 + 1 + 1 + 1= 4 t hp giao t

- M c th c d hp hai cp gen (AaBb) => cho 4 loi giao tSuy ra c th ci s cho 1 loi giao t.

- Xt tt c p n trn :

+ p n A, c th ci AaBb khi gim phn to ra 4 loi giao t

+ C 2 p n B v C, c th ci aaBb v Aabb khi gim phn to ra 2

loi giao t.

+ Ch c p n D, c th aabb khi gim phn to ra mt loi giao t.=> Chn p n D

2. Tng tc gia cc gen khng alen:

Mi kiu tng tc c 1 t l KH tiu biu da theo bin dng ca (3:1) 2 nh

sau:

2.1. Cc kiu tng tc gen:2.1.1. Tng tc b tr c 3 t l KH: 9:3:3:1; 9:6:1; 9:7.


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2.1.1.1. Tng tc b tr gen tri hnh thnh 4 KH: 9:3:3:1

A-B- A-bb aaB- aabb thuc t l: 9:3:3:12.1.1.2. Tng tc b trgen tri hnh thnh 3 KH: 9:6:1

A-B- (A-bb = aaB-) aabb thuc t l 9:6:1

2.1.1.3. Tng tc b tr gen tri hnh thnh 2 KH: 9:7

A-B- (A-bb = aaB- = aabb) thuc t l 9:7

2.1.2. Tng tc t ch c 3 t l KH: 9:3:4; 12:3:1; 13:3

2.1.2.1. Tng tc t ch gen tri hnh thnh 3 KH: 12:3:1(A-B- = A-bb) aaB- aabb thuc t l 12:3:1

2.1.2.2. Tng tc t ch gen tri hnh thnh 2 KH: 13:3

(A-B- = A-bb = aabb) aaB- thuc t l 13:3

2.1.2.3. Tng tc t ch gen ln hnh thnh 3 KH: 9:3:4

A-B- aaB- (A-bb = aabb) thuc t l 9:3:4

2.1.3. Tc ng cng gp (tch ly) hnh thnh 2 KH: 15:1

(A-B- = A-bb = aaB-) aabb

Tng qut n cp gen tc ng cng gp => t l KH theo h s mi s hng

trong khai trin ca nh thc Newton (A+a)n.

=> Tng tc b tr km theo xut hin tnh trng mi

Tng tc t ch ngn cn tc dng ca cc gen khng alen.

Tng tc cng gp mi gen gp phn nh nhau vo s pht trin.

2.2. Dng ton thun:

+ Cho bit kiu tng tc tm t l phn li i con

V d : nga s c mt ca 2 gen tri Av B cng kiu gen qui nh lng

xm, gen A c kh nng nh ch hot ng ca gen B nn gen B cho lng mu en

khi khng ng cng vi gen A trong kiu gen.Nga mang 2 cp gen ng hp

ln cho kiu hnh lng hung. Cc gen phn li c lp trong qu trnh di truyn.

Tnh trng mu lng nga l kt qu ca hin tng no?


32/37



A.tc ng cng gp C. Tc ng c ch

B.Tri khng hon ton D. Tc ng b tr

Gii:

Theo gen A c kh nng nh ch hot ng ca gen B, gen B ch biu

hn kiu hnh khi khng ng cng vi gen A trong cng 1 kiu gen.

Hay ni cch khc l gen A t ch hot ng ca gen tri B

Suy ra, Tnh trng mu lng nga l kt qu ca hin tngtng tc t ch

=> chn p n: C

+ Cho bit kiu gen (kiu hnh) ca b m tm t l phn li v kiu gen v kiu hnh

i con.

V d1:Lai hai dng b thun chng qu trn c F1ton qu dt; F2gm 271

qu dt : 179 qu trn : 28 qu di. S di truyn hnh dng qu tun theo quy lut

di truyn no?

A. Tng tc t ch B. Tng tc cng gpC. Tri khng hon ton D. Tng tc b tr

Gii:

Xt t l KH i con l: 271 qu dt : 179 qu trn : 28 qu di

9 qu dt: 6 qu trn : 1 qu di

Quy lut di truyn chi phi l: Tng tc b tr

=> Chn p n D

Ch :i vi cc bi ton dng ny, ta coi s nh nht nh 1 n v,

ri chia cc s ln hn vi n

V d2:Cho lai hai dng vt thun chng lng vng vi lng xanh, c F1ton

mu hoa thin l (xanh-vng).F2gm 9/16 mu thin l : 3/16 lng vng : 3/16

lng xanh : 1/16 lng trng. Tnh trng ny di truyn theo quy lut:

A.Phn li c lp C.Tri khng hon ton

B.Tng tc gen D. Lin kt gen


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Gii:

Tl phn tnh v KH th h F2l: 9:3:3:1

M y l kt qu ca php lai ca hai c th v mt cp tnh trng tngphn .

Nn suy ra tnh trng ny di truyn theo quy lut tng tc gen

Chn p n B

2.3.Dng ton nghch:Thng da vo kt qu phn tnh th h lai suy ra

s kiu t hp giao t v s loi b m => s cp gen tng tc.Sau khi xc nh c s cp gen tng tc, ng thi xc nh c kiu

gen ca b m v suy ra s lai c th c ca php lai thy t l KG thuc

dng no, i chiu vi kiu hnh ca bi d on kiu tng tc.

Thngthtng t l chn th h con bao gi cng l mt s chn bi n

l tch ca mt s chn vi mt s nguyn dng khc khi thc hin php nhn

xc sut trong qun th. T , suy ra s loi giao t ca b m.

+Khi lai F1x F1 to ra F2c 16 kiu t hp nh: 9:3:3:1; 9:6:1; 9:7; 12:3:1;

13:3, 9:3:4; 15:1. (16 = 4*4 => P gim phn cho 4 loi giao t)

+ Khi lai F1vi c th khc to ra F2c 8 kiu t hp nh: 3:3:1:1; 4:3:1;

3:3:2; 5:3; 6:1:1; 7:1. (8 = 4*2 => mt bn P cho 4 loi giao t, mt bn P cho 2

loi giao t)

+ Khi lai phn tch F1to ra F2c 4 kiu t hp nh: 3:1; 1:2:1; 1:1:1:1.

(4 = 4*1 => mt bn P cho 4 loi giao t, mt bn P cho 1 loi giao t)

V d 1:Khi lai cy hoa thun chng vi cy hoa trng thun chng, F1 thu

c 100% hoa . Cho lai F1vi cy hoa trng thun chng trn, F2thu c

3 hoa trng : 1 hoa . S di truyn tnh trng trn tun theo quy lut no?

Gii:

Pt/c, F1 thu c 100% hoa => tnh trng hoa tri hon ton so vihoa trng ( theo L ng tnh ca Menden).


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M tnh trng hoa trng l tnh trng do gen ln quy nh nn hoa trng ch

cho 1 loi giao t. Trong khi F2= 3+1= 4 kiu t hp, vy con lai F1phi cho 4

loi giao t => F1d hp 2 cp gen (AaBb), lc KG ca hoa trng thun chngl aabb, kiu gen ca cy hoa thun chng l AABB.

S lai:

Pt/c: AABB x aabb

hoa hoa trng

F1: AaBb

hoa

F1 x Pt/c(hoa trng): AaBb x aabb

hoa hoa trng

F2: 1AaBb 1Aabb 1aaBb 1aabb

M kt qu kiu hnh ca bi l 3 hoa trng: 1hoa . Ta xc nh c

trn KG aabb quy nh tnh trng hoa trng, AaBb quy nh tnh trng hoa .

T ta c th kt lun 2 KG cn li l Aabb v aaBb quy nh tnh trng hoa

trng.

Kt lun s di truyn tnh trng trn tun theo quy lut tng tc gen, kiu

tng tc b tr gen tri.

V d 2: u thm, s c mt ca 2 gen tri A, B trong cng kiu gen qui nh

mu hoa , cc t hp gen khc ch c 1 trong 2 loi gen tri trn, cng nh kiugen ng hp ln s cho kiu hnh hoa mu trng. Cho bit cc gen phn li c


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lp trong qu trnh di truyn. lai 2 ging u hoa trng thun chng, F1thu c

ton hoa mu . Cho F1giao phn vi hoa trng thu c F2phn tnh theo t l

37.5% : 62,5% trng. Kiu gen hoa trng em lai vi F1l:

A. Aabb hoc aaBb B. Aabb hoc AaBB

C. aaBb hoc AABb D. AaBB hoc AABb

Gii:

F2phn tnh c t l: 37.5% : 62,5% trng

= 3 : 5 trng = 8 thp = 4 giao t x 2 giao t.Theo gi thuyt th nhng cy hoa trng c th c l mt trong cc kiu gen

sau: AAbb Aabb

aaBB aaBb

aabb

Trong , Kiu gen AAbb, aaBB, aabb s gim phn cho 1 loi giao t

Kiu gen Aabb, aaBb gim phn cho 2loi giao t

Vy ch c KG Aabb, aaBb l tha mn, khi lai vi cy F1cho ra 8 t hp.

Do cy em lai s cho 2 loi giao t. nn cy em lai vi F1s c kiu gen

l: Aabb hoc aaBb.

=> Chn p n A

V d 3: Lai 2 dng b thun chng qu trn, thu c F1ton qu dt; cho F1t

th phn F2thu c 271 qu dt : 179 qu trn : 28 qu di. Kiu gen ca b m

l:

A. Aabb x aaBB C. AaBb x AaBb

B. AaBB x Aabb D. AABB x aabb

Gii:

XtF2c271 qu dt : 179 qu trn : 28 qu di

= 9 qu dt : 6 qu trn : 1 qu di

=>F2c 9+6+1= 16 t hp= 4 giao t * 4 giao t


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Suy ra F1d hp 2 cp gen : AaBb, c th b m thun chng v hai cp gen.

Quy c: A-B- : qu dt

A-bb v aaB-: qu trnAabb : qu di

Vy kiu gen b m thun chng l: Aabb x aaBB

=> chn p n A

V d 4: Ng, tnh trng kch thc ca thn do 3 cp alen (A1a1, A2a2, A3a3)

quy nh. Mi gen ln lm cho cy cao thm 10cm, chiu cao cy thp nht 80cm.

Nu F1ng lot xut hin kiu hnh Ng cao 110cm. Kiu gen ca P l:

A.A1A1A2A2A3A3 x a1a1a2a2a3a3 B.A1A1A2A2a3a3 x a1a1a2a2A3A3

C.A1A1a2a2A3A3 x a1a1A2A2a3a3 D.1 trong 4 trng hp ni trn.

Gii:

Theo bi suy ra, cy c chiu cao thp nht c kiu gen l ng hp tri

A1A1A2A2A3A3.

Mi gen ln lm cy cao thm 10cm

110 = 80+10+10+10

Suy ra F1xut hin 3 gen ln hay d hp t v 3 cp gen A1a1A2a2A3a3

By gi, da vo d kin bi cho:

+ Php lai: A1A1A2A2A3A3 x a1a1a2a2a3a3=> A1a1A2a2A3a3

+ Php lai: A1A1A2A2a3a3 x a1a1a2a2A3A3=> A1a1A2a2A3a3

+ Php lai: A1A1a2a2A3A3 x a1a1A2A2a3a3=> A1a1A2a2A3a3

=> chn p n ng l p n D

2.4.Tm li:Khi xt s di truyn ca 1 tnh trng, iu gip chng ta nhn bit

tnh trng c di truyn theo quy lut tng tc ca 2 gen khng alen l:


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+ Tnh trng c phn li KH th h sau theo t l 9:3:3:1 hay bin i

ca t l ny.

+ Tnh trng c phn li KH th h sau theo t l 3:3:1:1 hay bin ica t l ny.

+ Kt qu lai phn tch xut hin t l KH 1:1:1:1 hay bin i ca t l ny.

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