TN S TRAO I CHO KP Phng php gii loi bi tp c tn s trao i cho kp (F1 d hp 3 cp gen x phn tch) 1. Bi ton thun: Bit KQ Fb X trnh t gen VD A_o a_vang B_tron b_det D_ ngot d_chua

F1 di hp v ca 3 cp gen co kiu hinh la cy hoa o, qua tron va ngot. em cy F 1 lai phn tich i con FB thu c kt qua nh sau: 426 cy hoa o, qua det va ngot; 104 cy hoa vang, qua det va ngot; 46 cy hoa o, qua det va chua; 15 cy hoa o, qua tron va ngot; 424 cy hoa vang, qua tron va chua 108 cy hoa o, tron va chua 41 cy hoa vang, qua tron va ngot. 12 cy hoa vang, qua det va chua

Xac inh khoang cach gia cac gen trn nhim sc th? Bc 1: Da vo 2 lp KH ln nht (Lp KH ko do TC) VD: T 2 lp KH 426 cy hoa o, qua det va ngot A-bbD=> KG F1:AbD aBd

424 cy hoa vang, qua tron va chua aaB-dd

Bc 2: T 2 lp KH nh nht (do TC kp) => gen no nm gia VD: 15 cy hoa o, qua tron va ngot A-B-DQuan st cng vi F1 A v D lun i cng nhau A v d lun i cng nhau B khc so vi F1 nn suy ra B phi nm gia Bc 3: Tnh tn s trao i cho n ( th thut da vo 2 lp KH thp tip theo) VD: Ta co kiu gen cua 2 kiu hinh do trao di cheo n tai 1 im co ti l bang nhau: - cy hoa vang, qua det va ngot : - cy hoa o, tron va chua:abD giao t F1: abD abd

12 cy hoa vang, qua det va chua aabbdd

ABd giao t F1: ABd abd

Nhn thy: a tai t hp vi bD; A tai t hp vi Bd im trao i cheo lam hoan vi A va a 15 + 12 104 + 108 + Tn s trao i cheo: f = = 0,1918 2 426 + 424 + 104 + 108 + 46 + 41 + 15 + 12 Khoang cach gia A-b la 19,18 cM Ta co kiu gen cua 2 kiu hinh do trao di cheo n tai 1 im co ti l bang nhau:Abd giao t F1: Abd abd aBD - cy hoa vang, qua tron va ngot: giao t F1: aBD abd

- cy hoa o, qua det va chua:

Nhn thy: d tai t hp vi Ab; D tai t hp vi aB im trao i cheo lam hoan vi D va d 15 + 12 46 + 41 + Tn s trao i cheo: f = = 0,0855 2 426 + 424 + 104 + 108 + 46 + 41 + 15 + 12 1

Khoang cach gia b-D la 8,55 cM Vy khoang cach gia cac gen trn nhim sc th la: A 19,18 cM b Bc 4: Tnh h s trng lp + f trao i cho thc t=

8,55 cM

D

TLKHnho KH

+ f trao i cho LT= f n A/a x f n D/d + h s trng lp C =fthucte fLT

VD: + f trao i cho thc t=

15 +121176

+ f trao i cho LT= f n A/a x f n D/d= 0,1918 x 0,0855= 2. Bi ton ngc: Bit trnh t gen T l cc loi giao t hoc KH ca Fb VD: ABD/abd , khong cch A v B = 0,3cM , B v D= 0,2cM. Cho bit h s trng hp l 0,7. Tnh t l cc loi giao t to thnh? HD: T gt: tn s trao i cho kp l thuyt l 0,3.0,2 = 0,06 H s trng hp= Tn s trao i cho kp thc t Tn s trao i cho kp l thuyt

Tc l 0,7 =

Tn s trao i cho kp thc t 0,06

Suy ra tn s trao i cho kp thc t = 0,7 x 0,06 = 0,042. t l giao t trao i cho kp AbD = aBd = 0,042/2 = 0,021 Tn s trao i cho n gia A v B l: 0,3 0,042/2 = 0,279 t l giao t aBD = Abd = 0,279/2 = 0,135 Tn s trao i cho n gia B v C l 0,2 0,042/2 = 0,179 t l giao t ABd = abD = 0,179/2 = 0,0895 t l giao t lin kt hon ton ABD = abd = (1-0.042-0.279-0.179)/2=0.25 Trao i cho kp Bai 7 A_o a_vang B_tron b_det D_ ngot d_chua

F1 di hp v ca 3 cp gen co kiu hinh la cy hoa o, qua tron va ngot. em cy F 1 lai phn tich i con FB thu c kt qua nh sau: 426 cy hoa o, qua det va ngot; 104 cy hoa vang, qua det va ngot; 46 cy hoa o, qua det va chua; 15 cy hoa o, qua tron va ngot; 424 cy hoa vang, qua tron va chua 108 cy hoa o, tron va chua 41 cy hoa vang, qua tron va ngot. 12 cy hoa vang, qua det va chua

Xac inh khoang cach gia cac gen trn nhim sc th? Bai 7 * F1 di hp 3 cp gen 1 gen quy inh 1 tinh trang P thun chung, F1 ng tinh hoa o tri so vi hoa vang; qua tron tri so vi qua det; qua ngot tri so vi qua chua. 2

Quy c: A_o B_tron D_ ngot a_vang b_det d_chua * Xac inh kiu gen F1 Kt qua phep lai phn tich co 8 kiu hinh khac nhau cac gen lin kt khng hoan toan vi nhau, co xay ra trao i cheo kep (HS co th chng minh thm khng th la phn li c lp hoc khng th la trng hp 2 gen lin kt) - Ta co 2 kiu hinh co ti l ln nht la cac kiu hinh mang gen lin kt, o la:AbD giao t lin kt cua F1 la: AbD abd aBd cy hoa vang, qua tron va chua kiu gen: giao t F1 la: aBd abd AbD Vy kiu gen F1 la: aBd

cy hoa o, qua det va ngot. kiu gen la :

* Xac inh vi tri cua gen trn nhim sc th: - Ta co 2 kiu hinh co ti l nho nht la 2 kiu hinh co c do F1 trao i cheo kep:ABD giao t F1 la: ABD abd abd cy hoa vang, qua det va chua kiu gen: giao t F1 la: abd abd

cy hoa o, qua tron va ngot kiu gen:

Nhn thy: Trong 2 giao t trao i cheo kep co B tai t hp vi AD; b tai t hp vi ad gen b nm gia AD, gen B nm gia ad trong kiu gen F1. A b D . . a B d * Xac inh khoang cach gia cac gen trn nhim sc th: Ta co kiu gen cua 2 kiu hinh do trao di cheo n tai 1 im co ti l bang nhau: - cy hoa vang, qua det va ngot : - cy hoa o, tron va chua:abD giao t F1: abD abd

ABd giao t F1: ABd abd

Nhn thy: a tai t hp vi bD; A tai t hp vi Bd im trao i cheo lam hoan vi A va a 15 + 12 104 + 108 + Tn s trao i cheo: f = = 0,1918 2 426 + 424 + 104 + 108 + 46 + 41 + 15 + 12 Khoang cach gia A-b la 19,18 cM Ta co kiu gen cua 2 kiu hinh do trao di cheo n tai 1 im co ti l bang nhau:Abd giao t F1: Abd abd aBD - cy hoa vang, qua tron va ngot: giao t F1: aBD abd

- cy hoa o, qua det va chua:

Nhn thy: d tai t hp vi Ab; D tai t hp vi aB im trao i cheo lam hoan vi D va d 15 + 12 46 + 41 + Tn s trao i cheo: f = = 0,0855 2 426 + 424 + 104 + 108 + 46 + 41 + 15 + 12 Khoang cach gia b-D la 8,55 cM Vy khoang cach gia cac gen trn nhim sc th la: A 19,18 cM b 8,55 cM D Bai 7 3

Cy co kiu gen di hp v 3 cp gen giam phn cho giao t vi s lng nh sau: ABD = 414 Abd = 70 aBd = 28 abD = 1 abd = 386 aBD = 80 AbD = 20 ABd = 1 Xac inh trt t cac gen trn nhim sc th va tinh khoang cach gia 3 lcut (theo n vi ban )? Bai 7 Xac inh trt t cac gen va tinh khoang cach gia 3 locut: - Tng s giao t thu c: 414+386+70+80+28+20+1+1=1000 - Nhn thy kiu gen trn giam phn cho 8 loai giao t co ti l khac nhau cac gen lin kt khng hoan toan. - Co 2 giao t chim ti l cao nht va bng nhau o la 2 giao t khng trao i cheo Kiu gen cua c th nay la:ABD abd

- Nhn thy 2 giao t co ti l nho nht la abD va ABd la 2 giao t do trao i cheo kep ng thi tao ra, y ta thy D tai t hp vi ab va d tai t hp vi AB alen D nm gia A va B A D B - Giao t Abd va aBD co c la do trao i cheo n vi co ti l khng phai la ln nht va khng phai la nho nht. Trong 2 giao t nay A tai t hp vi bd con a tai t hp vi BD Trao i cheo xay ra im gia A va D Tn s trao i cheo n : fA/D =70 + 80 +1 151 = = 0,151 414 + 386 + 70 + 80 + 28 + 20 +1 +1 1000

Khoang cach gia A va D la 15,1 cM - Giao t aBd va AbD co c la do trao i cheo n vi co ti l khng phai la ln nht va khng phai la nho nht. Trong 2 giao t nay B tai t hp vi ad, b tai t hp vi AD Trao i cheo xay ra im gia B va D Tn s trao i cheo n : fB/D =28 + 20 +1 49 = = 0,049 414 + 386 + 70 + 80 + 28 + 20 +1 +1 1000

Khoang cach gia A va D la 4,9 cM Vy trt t cac gen va khoang cach gia 3 locut nh sau: A 15,1 cM D 4,9 cM B Cu 7: rui gim gen A : thn xm, a : thn en; B : cnh thng, b : cnh ct; D : mt , d : mt nu. Cho F1 mang 3 cp gen d hp lai phn tch c FB : 451 xm, thng, : 449 en, ct, nu : 152 xm, ct, nu : 148 en, thng, : 113xm, ct, : 112 en, thng, nu : 18en, ct, : 19 xm, thng, nu. Cho bit cc cp gen nm trn nhim sc thng. Xc nh kiu gen F1 v tnh khong cch gia cc gen. Cch gii F1 lai phn tch cho 8 kiu hnh phn thnh 4 nhm nn xy ra hin tng lin kt khng hon ca 3 gen. 2 kiu hnh chim t l ln : xm, thng, v en, ct, nu to ra do 2 giao t lin kt hon ton ABD v abd. Nn 3 gen A,B,D cng nm trn 1 NST v 3 gen a,b,d cng nm trn 1 NST. Kiu hnh en, ct, v xm, thng, nu chim t l nh nht do trao i cho 4 im

kp to ra t 2 loi giao t abD v ABd. Suy ra trt t 3 gen trn l A-D-B, nn kiuADB adb 152 +148 1 18 +19 Khong cch AD = + ( )= 21,7852% 1462 2 1462

gen ca F1 l :

3,00

Khong cch DB =

113 +112 1 18 +19 + ( )= 16,6553% 1462 2 1462

Khong cch AB = 21,7852% + 16,6553% = 38,4405% A 21,7852% D 16,6553% B 38,4405%

2,00

Trao i cho giai on 4 cromatid, tc trao i cho xy ra sau khi nhim sc th t nhn i (cn gi l giai on 4 si). V nguyn tc c th cho rng trao i cho c th xy ra c khi nhim sc th cha t nhn i (giai on 2 si)

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Phn tch b bn c th gii quyt vn ny. Phn tch b bn l php phn tch di truyn hc nghin cu bn sn phm trc tip ca gim phn khi t bo lng bi d hp v mt gen hay nhiu gen lin kt phn chia gim phn. Nm 1925, C. Bridge v I. Anderson chng minh trao i cho cc cromatid rui gim. Tc gi s dng dng rui c nhim sc th X mang thm on nhim sc th Y (X,XY) d hp v cc gen ca nhim sc th X: f (forked) - lng phn nhnh, g (garnet) - mt rc. Khi cho lai con ci ny vi con c bnh thng th chng truyn trc tip 2 nhim sc th X cho th h sau v ch mt na th h con ca chng sng st. Khi y mt phn c th con i sau t php lai ny l ng hp theo cc gen ca nhim sc th X. 6

Cc th ng hp ny ch c th xut hin do trao i cho giai on 4 si trong on gen - tm ng. Trao i cho nhiu ln Trao i cho gia 2 chromatid c th xy ra nhiu ln: 2, 3, 4 ln ... Nu 2 trao i cho xy ra trn cng 2 chromatid on gia 2 gen nh du th sn phm cui cng u c kiu cha m, nn khng pht hin c. Kiu trao i cho ny ch c th pht hin c khi s dng thm mt gen nh du th ba nm gia 2 gen ny. Nu xc sut trao i cho gia A v C v gia B v C tng ng vi x v y, th xc sut xy ra trao i cho i l: 0,2 x 0,1 = 0,02 (2%) Nhiu (Interference) v trng hp (Coincidence) Thng th s trao i cho mt ch lm gim xc sut trao i cho th hai gn k n. l hin tng nhiu. nh gi kt qu ngi ta dng h s trng hp H s trng hp = (% trao i cho i quan st c)/(% trao i cho i theo l thuyt) S trng hp + nhiu = 100% = 1 V d: Bit khong cch A-B = 10 n v (10%) v B-C = 20 n v (20%), nu khng c nhiu th tn s trao i cho i theo l thuyt l 0,1 x 0,2 = 0,02 hay 2%. Gi s quan st c tn s trao i cho i l 1,6%. S trng hp = 1,6/2,0 = 0,8. iu ny cho thy trao i cho i ch xyra c 80% v s nhiu 1,0 - 0,8 = 0,2 (hay 20%) Bi 6 : Xt 3 cp gen d hp nm trn cng mt cp nhim sc th thng. ABD a. Nu mt c th c trnh t sp xp cc gen trn cp nhim sc th ny l , khong abd cch tng i trn nhim sc th gia gen A vi gen B l 20 cM ; gia gen B vi gen D l 15 cM v trong gim phn xy ra c trao i cho n ln trao i cho kp th theo l thuyt c th ny to ra giao t AbD c t l l bao nhiu? b. Nu qu trnh gim phn mt c th to ra 8 loi giao t vi thnh phn alen v t l nh sau : ABD = abd = 2,1% ; AbD = aBd = 12,95% ; ABd = abD = 28,5% v Abd = aBD = 6,45% th trnh t sp xp cc gen trn cp nhim sc th ny v khong cch tng i gia chng l bao nhiu cM? Gii : Cch gii Kt qu a. AbD l giao t sinh ra do trao i cho kp nn t l = 20% . 15% : 2 = a. 1,5%. (0,2 im) 1,5% (0,4 im) b. AdB//aDb (0,2 im) b. - Kiu gen : AdB//aDb (0,4 im) - Khong cch A, a vi D, - Khong cch tng i gia cp gen A, a vi D, d : d = 28 cM. = 12,95% .2 + 2,1% = 28 % = 28 cM. (0,2 im) (0,2 im) 7

- Khong cch tng i gia cp gen D, d vi B, b : = 6,45% .2 + 2,1% = 15% = 15 cM (0,2 im)

- Khong cch D, d vi B, b = 15 cM (0,2 im)

* VD 2 : ABD/abd co A-B=0,3. B-D=0,2. Cho bit h s trng hp l 0,7. Tnh t l cc loi giao t to thnh?

- TST kp = (0,3.0,2).0,7 = 0,042 - Tkp to nn 1 lp vi 2 loi giao t (b HV ti im nm gia): AbD = aBd = 0,042/2=0,021 - T n th I to nn 1 lp vi 2 loi giao t : aBD = Abd = (0,3 - 0,042)/2=0,129 - T n th II to nn 1 lp vi 2 loi giao t : ABd= abD = (0,2 - 0,042)/2=0,079 2 loi gt bnh thng (khng b TC) : ABD = abd = [1- (0,042+0,3-0,042+0,2-0,042)]/2 = 0,027 Cu 4: ABD/abd , khong cch A v B = 0,3cM , B v D= 0,2cM. Cho bit h s trng hp l 0,7. Tnh t l cc loi giao t to thnh? HD: T gt: tn s trao i cho kp l thuyt l 0,3.0,2 = 0,06 H s trng hp= Tn s trao i cho kp thc t Tn s trao i cho kp l thuyt

Tc l 0,7 =

Tn s trao i cho kp thc t 0,06

Suy ra tn s trao i cho kp thc t = 0,7 x 0,06 = 0,042. t l giao t trao i cho kp AbD = aBd = 0,042/2 = 0,021 Tn s trao i cho n gia A v B l: 0,3 0,042 = 0,258 t l giao t aBD = Abd = 0,258/2 = 0,129 Tn s trao i cho n gia B v C l 0,2 0,042 = 0,158 t l giao t ABd = abD = 0,158/2 = 0,079 t l giao t lin kt hon ton ABD = abd = (1-0.042-0.258-0.158)/2=0.271 Cu 5 : l giao t c to t trao i cho n gia a/b V trt t khong cch gia 3 gen X, Y v Z ngi ta nhn thy nh sau: X------------------20-----------------Y---------11----------Z. H s trng hp l 0,7. Nu P :xyz Xyz x xyz xYZ

th t l % kiu hnh khng bt cho ca F1 l:

A. 70,54% B. 69% C. 67,9% D. khng xc nh c Cch tnh hon ton tng t bi tp trn : (lu y l php lai phn tch v vy t l % kiu hnh khng bt cho ca F1 = % t l giao t ko bt cho lin kt ) Cu 6 : ng gen A mm xanh, a mm vng; B mm m, b mm bng; D l bnh thng, d l b ca. Khi lai phn tch cy ng d hp v c 3 cp gen th thu c kt qu bng 3. 8

Bng 3. Kt qu ca php lai ng Giao t ca P Khng trao i ABD cho (TC) abd TC n on I Abd aBD TC n on II ABd abD TC kp on I v II AbD aBd Khong cch gia a-b v b-d ln lt l A. 17,55 & 12,85 B. 16,05 & 11,35 KG ca Fa ABD abd abd abd Abd abd aBD abd ABd abd abD abd AbD abd aBd abd Tng cng S c th 235 505 270 62 122 60 40 88 48 7 11 4 726 100 1,5 12,1 % s c th 69,6

16,8

C. 15,6 & 10,06

D. 18,3 & 13,6

Hd : T bng kt qu trn ta nhn thy y l php lai phn tch Ch : tn s ca trao i cho n.= Tn s trao i cho - tn s trao i cho kp Giao t Abd,aBD l giao t c to t trao i cho n gia a/b - Tn s trao i cho a/b = TC n on I + TC kp on I v II = 16,8 +1,5 = 18,3% Khong cch gia a-b l 18.3 Giao t Abd, abD l giao t c to t trao i cho n gia a/b Tn s trao i cho b/d = TC n on II + TC kp on I v II = 12,1 +1,5 = 13,6% Khong cch gia a-b l 13.6 p n D Mt s vn l lin quan n trao i cho kp : Nhiu (Interference) v trng hp (Coincidence) Thng th s trao i cho mt ch lm gim xc sut trao i cho th hai gn k n. l hin tng nhiu. nh gi kt qu ngi ta dng h s trng hp H s trng hp = (% trao i cho kp quan st c)/(% trao i cho kp theo l thuyt) S trng hp + nhiu = 100% = 1 Cu 021: V trt t khong cch gia 3 gen X, Y v Z ngi ta nhn thy nh sau: X------------------20-----------------Y---------11----------Z. H s trng hp l 0,7. Nu P : (Xyz/xYZ) x (xyz/xyz) th t l % kiu hnh khng bt cho ca F1 l: A. 70,54% B. 69% C. 67,9% D. khng xc nh c Gii: T l bt cho kp l thuyt = Tch khong cch trn bn gen X/Y v Y/Z = 20% x 11% = 2,2%

9

H s trng hp C= T l bt cho kp thc t (O)/T l bt cho kp l thuyt (E) Suy ra: T l bt cho kp thc t: 2,2% x 0,7 = 1,54%

Khong ch gia 2 gen X v Y l 20%. Khong cch ng vi t l cc c th c th xy ra bt cho gia cc gen X v Y, trong cc c th c th bt cho n v cho kp. Nh vy bt cho n X/Y l 20% = bt cho I + bt cho kp Suy ra bt cho I = 20% -1,54% =18,46% Tng t t l bt cho c th ci th xy ra bt cho gia Y v Z (bt cho II) Bt cho B/C l 11%-1,54% = 9,46% Vy tng s c th c th xy ra bt cho l : 18,46% + 9,46% + 1,54% = 29,46% Suy ra tng s cc th khng xy ra bt cho l : 100% - 29,46% = 70,54% Cu 7. (2,0 im) Xt 4 gen lin kt trn mt nhim sc th, mi gen qui nh 1 tnh trng. Cho mt c th d hp t 4 cp gen (AaBbCcDd) lai phn tch vi c th ng hp t ln, FB thu c 1000 cc th gm 8 phn lp kiu hnh nh sau: Kiu hnh S lng Kiu hnh S lng aaBbCcDd Aabbccdd AaBbCcdd 42 43 140 aaBbccDd AabbCcdd AaBbccdd 6 9 305 310

aabbccDd 145 aabbCcDd Xc nh trt t v khong cch gia cc gen. S lc cch gii

Kt qu

Trt t phn b v khong cch gia cc gen: 0,5 * Trt t phn b gia cc gen: - Nhn thy cp gen ln a lun i lin vi gen tri D trn cng 1 NST; cn gen tri A lun i lin vi gen ln d trn cng 1 NST suy ra 2 gen ny lin kt hon ton vi nhau. - Kt qu php lai thu c 8 phn lp kiu hnh vi t l khng bng nhau, chng t d xy ra trao i cho n ti 2 im khng ng thi v trao i cho kp trong qu trnh to giao t c th AaBbCcDd. - 2 phn lp kiu hnh chim t l thp nht l kt qu ca TC kp. Suy ra trt t phn b ca cc gen ca 2 phn lp ny l BbaaDdcc v bbAaddCc. - Hai phn lp kiu hnh c s lng c th ln nht mang gen lin kt Gi s kiu gen ca c th mang lai phn tch l * Khong cch gia cc gen :B 42 + 43 + 9 + 6 B A ) + f (kp) = - Tn s HVG vng A = f (n = 10% 1000BAdc baDC

0,5

10

d 140 + 145 + 9 + 6 d c ) + f (kp) = - Tn s HVG vng c = f (n = 30% 1000

0,5 0,5

- Hai phn lp kiu hnh mang gen lin kt chim t l: 305 + 310 615 60%. Vy BAd + Adc = 10% + 30% = 40%. Suy ra 2 = 1000 1000 gen Ad nm gia. Cu 8. (2,0 im) Trong mt c th gi nh, con ci thn b, lng trng, thng c lai vi con c thn mnh, lng en, qun to ra F1 thn mnh, lng trng, thng. Cho con ci F1 giao phi vi con c thn b, lng en, qun thu c i sau: Thn mnh, lng trng, thng 169 Thn mnh, lng en, thng 19 Thn mnh, lng en, qun 301 Thn b, lng trng, qun 21 Thn mnh, lng trng, qun 8 Thn b, lng en, qun 172 Thn b, lng en, thng 6 Thn b, lng trng, thng 304 Hy lp bn di truyn xc nh trt t cc gen v khong cch gia chng. S lc cch gii Kt qu phn li F2 di truyn lin kt, c hon v gen. Theo u bi, ta c: A/a: thn mnh/b; B/b: thn trng/en; C/c: lng thng/qun F2: aaB-C-; A-bbcc: khng xy ra ti t hp A-B-C-; aabbcc: trao i cho n (A vi B) A-bbC-; aaB-cc: trao i cho n (B vi C) A-B-cc; aabbC-: trao i chp kp (A, B, C) T kt qu trn trnh t sp xp cc gen: A B C, kiu genabc abc 169 + 172 + 6 + 8 100% = 35,5% f (A-B)= 1000 21 + 19 + 6 + 8 100% = 5,4% f (B-C)= 1000

Kt qu 0,5 0,5

F1:

aBB Abb

0,5

0,5 a (35,5) B (5,4) C Cu 8 : Cho F1 d hp t 3 cp gen lai phn tch, FB thu c nh sau : 165 cy c KG : A-B-D88 cy c KG: A-B-dd 163 cy c KG: aabbdd 20 cy c kiu gen: A-bbD86 cy c KG: aabbD18 cy c kiu gen aaB-dd (?) Bin lun v xc nh kiu gen ca cy d hp ni trn v lp bn v 3 cp gen ? - Kt qu lai phn tch cho ra 6 loi KH -> c th d hp to ra 6 loi giao t, 3 cp gen lin kt khng hon ton, trao i cho xy ra ti 2 im khng cng lc. - Xc nh 2 loi giao t cn thiu do TC kp l: A-bbdd v aaB-D-> trt t gen trn NST lBAD KG ca cy d hp l: BAD/bad - Khong cch gia cc gen: 11

+ Hai loi KG c t l ln: [ (165+ 163)/540] x 100% = 61% khong cch gia B v D l : 100% - 61% = 39% = 39cM khong cch AD l: [(88 + 86)/540]x100% = 32% = 32cM khong cch BA l : [(20 +18)/540]x100% = 7% = 7cM -> v bn gen.

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