4/26/2013

Bi Tp Sinh hc * Ti liu luyn thi i hc nm 2014

Ngi bin tp : Trng Tn Ti ST: 0902651694

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Vi lng say m b mn sinh hc ti su tm v chn lc ra nhng dng bi tp hay v thng xuyn xut hin trong cc thi th v thi chnh thc ca b gn y , y l b ti liu c tham kho t rt nhiu ti liu chuyn sinh v nhng phng php m ti bit , ti cng a vo trong cun ti liu ny. Mong rng y l ngun ti liu hu ch cho cc em c th n thi tt b mn sinh hc .

Tn Ti Mi th t thc mc xin gi v hotmail : taitt@hotmail.com.vn hoc yahoo : zhangjincai94 facebook : https://www.facebook.com/tai.truongtan

I,PHNG PHAP GII BAI TP DI TRUYN PHAN T :

Cu 1: Mt gen cha on mch c trnh t nucltit l . . . A- G - X - T - T - A - G - X - A . . . .

Xc nh trnh t nucltit tng ng trn mch b sung. Hng dn : Theo NTBS cc nucltit trn gen lin kt vi nhau theo nguyn tc A lin kt

vi T, G lin kt vi X Vy: Mch c trnh t: . . . A - G - X - T - T - A - G - X - A . . . . Mch b sung l: . . . T - A - G - A - A - T - X - G - A . . . Cu 2: Mt gen cha on mch b sung vi on mch mang m gc c trnh

t nucltit l: . . . A - G - X - T - T - A - G - X - A . . . .

Xc nh trnh t cc rib nucltit c tng hp t on gen ny.

Hng dn Khi bit mch b sung => Xc nh mch gc => xc nh ARN (theo

nguyn tc b sung)

Gii - Theo NTBS: Cc nucltit trn gen lin kt vi nhau theo nguyn tc A lin kt

vi T, G lin kt vi X Trong qu trnh phin m cc nucltit trn gen lin kt vi cc

nucltit mt trng theo nguyn tc: A mch gc lin kt vi U mi trng T mch gc lin kt vi A mi trng G mch gc lin kt vi X mi trng X mch gc lin kt vi G mi trng Theo bi ra: mch b sung ca gen: . . . A - G - X - T - T - A - G - X - A . . . . => Mch gc ca gen: . . . T - X - G - A - A - T - X - G - T . . . .

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=> ARN . . . A - G - X - U - U - A - G - X - A . . . .

Cng thc c bn cn nh: 1. Tnh chiu di gen: lgen = 3.4.N/2

2. N = 2l/3,4= A+T+G+X = 2A + 2G

3. A=T; G=X. => A+G = T+X

4. %A=%T; %G=%X. => %A+%G = %T+%X=50%.

5. S chu k xon: (C) = N/20 6. S b ba m ha =N/6

6.Tnh s axit amin: 6.1. S axitamin trong chui plipeptit c tng hp (gen phin m 1 ln, 1

ribxm trtt qua khng lp li:) : N/6 - 1 6.2. S axitamin mi trng cung cp trong dch m khi gen phin m 1 ln, 1

ribxm trt qua khng lp li: N/6-2

6.3. Gen phin m k ln. Trn mi phn t mARN tham gia dch m c n Ribxmcng trt qua m ln. S axit amin mi trng cung cp l: k. n. (m+1)(N- 1)

6.4. Gen phin m k ln. Trn mi phn t mARN tham gia dch m c n Ribxmcng trt qua, lp li m ln. S axit amin mi trng cung cp l:

7. S Lin kt hir ca gen: H = 2A + 3G ( lk) 8. Khi lng phn t ADN (gen): MAD N = N . 300 ( vC). 9. S lin kt phtphoieste 9.1. S lin kt phtphoieste trn mt mch = s lin kt phtphoieste trn

ARN = N -1.

9.2. S lin kt phtphoieste trn c phn t ADN = 2N - 2. 10. S gen con c to ra sau k ln ti bn: 2k. 11. S gen con c 2 mch hon ton mi c to ra sau k ln ti bn: 2k - 2. 12. S nucltit trong cc gen con khi gen ti bn k ln: N. 2k 13. S nucltit mi trng cung cp khi gen ti bn k ln: N. (2k-1) 14. S nucltit trn cc phn t mARN khi gen phin m k ln: k.N/2 15. S lin kt peptit trn chui plipeptit = s axitamin trong phn t prtin -1 16. S nu tng loi tng mch v c gen: A1 = T2 %A1 = % T2

T1 = A2 % T1 = % A2

G1 = X2 % G1 = % X2

X1 = G2 % X1 = % G2

=> A = T = A1 + A2 = T1 + T2 = A1 + T1 = A2 + T2

G=X = G1 + G2 = X1 + X2 = G1 + X1 = G2 + X2

17. Phin m: (n phn ca ARN l rNu) - Gi s nu tng loi ca ARN l rA, rU, rX, rG th - Theo NTBS:

rA = Tmch gc. % rA = % Tmch gc

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rU = Amch gc. % rU = % Amch gc.

rX = Gmch gc % rX = % Gmch gc

rG = Xmch gc % rG = % Xmch gc

V Amch gc + Tmch gc = Agen = Tgen

=> rA + rU = Agen = Tgen

rG + rX = Ggen = Tgen

18. Khi lng ARN: Ngen/2. 300VC 19. S Lk hir b ph hy: Hph hy = Hgen. (2k 1). 20. S LK hir hnh thnh: Hht = H. 2k 21. S rib nucltit (rNu) mi trng cung cp cho gen phin m K ln: rAmt = rA. K = Tgc . K

rUmt = rU. K = Agc . K

rXmt = rX. K = Ggc . K

rGmt = rG. K = Xgc . K

22. S lin kt peptit c hnh thnh khi cc axit amin lin kt nhau = s phn t H2O = s aa -1.

23. s on mi ARN=s on okazaki+s n v ti bn.2 Cu 1: Trong t bo nhn s, xt mt gen di 4080 A0, c 560 Anin. Mch n

th nht ca gen c 260 Anin v 380 Guanin, gen ny phin m cn mi trng ni bo cung cp 600 Uraxin.

1. Tnh s lng tng loi nucltit trn gen. 2. Tnh s lng tng loi nucltit trn mch mang m gc ca gen. 3. Tnh s lng nuclotit tng loi trn mARN do gen phin m.

Hng dn: 1. Tnh s lng tng loi nucltit trn gen. - N = = 4080x2/3,4 = 2400

- A = T = 560 => G = X = (2400 -2x560):2 = 640.

2. Tnh s lng tng loi nucltit trn mch mang m gc ca gen. Theo NTBS, A1 = T2 = 260

G1 = X2 = 380.

X1 = G2 = Ggen - G1= 640 - 380 = 260.

T1 = A2 = A - A1 = 560 - 260 = 300.

Do Umtcc = Agc= 600 => mch 2 l mch gc. 3. Tnh s lng nuclotit tng loi trn mARN do gen phin m. Do mch 2 l mch gc nn trn mARN c A = Tgc = 260; U = Agc = 300; G = Xgc = 380; X = Ggc = 260. Cu 2: Mt gen c 450 Anin v 1050 Guanin. Mch mang m gc ca gen c

300 Timin v 600 Xitzin.

1. Tnh s lng tng loi: rA, rU, rG, rX trn phn t ARN c tng hp t gen ny.

2. Tnh chiu di gen. 3. Tnh s chu k xon ca gen. 4. Tnh s axitamin mi trng cung cp to ra 1 chui plipeptit.

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Hng dn: 1. S lng tng loi: rA, rU, rG, rX trn phn t ARN c tng hp t gen

ny.

Ag = Tbs = 300

Tg = Abs = A - Ag = 450 -300 = 150.

Xg = Gbs = 600

Gg = Xbs = G - Gbs = 1050 - 600 = 450

Vy rA = Tg = 300; rU = Ag = 150; rG = Xg = 600; rX = Gg = 450 2. Chiu di gen. N = A + T + G + X = 2A + 2G = 3000.

Lgen = N/2x3,4 = 5100A0.

3. S chu k xon ca gen. C = 150 4. S axitamin mi trng cung cp to ra 1 chui plipeptit = 3000/6 - 1 =

499

Cu 3: Phn t mARN trng thnh c to ra cha 20%U, 10%A, 40%X v 450G. Cc on intron b ct b c tng chiu di l 30,6m, trong c t l G = 2U = 3X = 4A.

1. Tnh s nucltit trn gen tng hp mARN trn. 2. Tnh s lng tng loi nucltit trong phn t mARN s khai tng ng. 3. Tnh t l mi loi nucltit trn mch m gc ca gen.

Hng dn: 1. Tnh s nucltit trn gen tng hp mARN trn. %U + %A + %X + %G = 100% => %G = 30%.

=> S nucltit trn mARN = 450x100/30 = 1500. S nucltit trn on b ct b l 306000/3,4 = 90.000. => S nucltit trn gen l 1500x2 + 90.000x2 = 93.000 (nu) 2. Tnh s lng tng loi nucltit trong phn t mARN s khai tng ng. S nucltit tng loi trn cc on intron l: G = 43200; U = 21600; X = 14400; A = 10800

S nucltit tng loi trn mARN trng thnh: A = 150; U = 300; X = 600; G = 450.

S nucltit tng loi trn mARN s khai A = 10800 + 150 = 10950;

U = 21600 + 300 = 21900;

X = 600 + 14400 = 15000;

G = 43200 + 450 = 43650;

3. S lng nucltit trn mch m gc = N/2 = 91500. A = rU = 21900 => %A = 21900/91500*100 = 23,9

T = rA = 10950 => %T = 10900/91500*100 = 11,9

G = rX = 15000 = %G = 15000/91500*100 = 16,4

X = rG = 100 -%A-%T-%G-%X = 47,8

Cu 5: Phn tch thnh phn ha hc ca mt axit nuclic cho thy t l cc loi nucltit A = 20%; G = 35%; T = 20% v s lng X = 150.

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1. Axit nuclic ny l ADN hay ARN, cu trc mch n hay kp? 2. Tnh s lin kt photphodieste trn axit nucltit trn. 3. Tnh chiu di axit nucltit trn. Hng dn: 1. Do trn axit nucltit c A, T, G, X => ADN. V %A %T => mch n. Vy, Axit nuclic ny l ADN c cu trc mch n. 2. S lin kt photphodieste trn axit nuclic trn: X = 150, chim 30% => N = 150/30*100 = 500. S lin kt photphodieste = 500-1=499.

II, DI TRUYN QUN TH:

A. QUN TH NI PHI (T th phn, t phi) Xt 1 gen gm 2 alen A v a. Gi s QT ban u c 100%Aa vi n: s th h

t phi.

T l KG d hp qua n ln t phi = 1

2

n

T l KG ng hp mi loi (AA = aa) qua n ln t phi =

11

2

2

n

*Ch : Nu qun th ban u khng phi l 100% Aa m c dng: xAA + yAa + zaa = 1 qua n th h t phi th ta phi tnh phc tp hn. Lc ny, t l KG Aa, AA, aa ln lt l:

Aa =

1

2

n

. y AA = x +

11

2

2

n

. y aa = z +

11

2

2

n

. y

B. QUN TH NGU PHI: ( nh lut Haci-Vanbec ) Ta c: xAA + yAa + zaa = 1 ; Nu gi p l tn s alen A, q l tn s alen a

th:

pA = x + 2

y; qa = z +

2

y

1. Ni dung nh lut: Khi xy ra ngu phi, qun th t trng thi cn bng theo nh lut Haci-

Vanbec. Khi tho mn ng thc: p2AA + 2pqAa + q2aa = 1, QT cn

bng p + q = 1 2. Kim tra s cn bng ca qun th :

Nu p2 x q2 = 2

2

2

pq

qun th cn bng.

Nu : p2 x q2 # 2

2

2

pq

Qun th khng cn bng

3. Xc nh s loi kiu gen ca qun th:

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- S kiu gen ={ r ( r + 1 ) /2 }n ( r : s alen thuc 1 gen (lcut), n : s gen khc nhau, trong cc gen phn li c lp).

- Nu c r ca cc locut khc nhau th tnh tng locut theo cng thc nhn kt qu tnh tng locut.

- Nu gen nm trn cng mt NST th tng kiu gen l: rn(rn +1)/2. - Nu gen nm trn NST gii tnh th tng kiu gen l: r(r+2)/2 + r....

4. Trng hp gen a alen: V d: Qun th Ngi: ( 1 gen c 3 alen Ngi c 4 nhm mu: A, B, AB,

O )

Gi : p(IA); q(IB), r(i) ln lt l tn s tng i cc alen IA, IB, IO . Ta c : p + q + r = 1

Nhm

mu

A

B

AB

O

Kiu gen

IA IA

+

IA

IO

IB IB

+

IB

IO

IA

IB

IO

IO

Tn s kiu

gen

p2 +

2

pr

q2

+ 2

pr

2pq

r2

C. GEN TRN NST GII TNH

i vi 1 locus trn NST gii tnh X c 2 alen s c 5 kiu gen: A AX X , A aX X , a aX X , AX Y , aX Y

Cc c th ci c 2 alen trn NST X v vy khi xt trong phm vi gii ci th tn

s cc kiu gen A AX X , A aX X , a aX X c tnh ging trng hp cc alen trn NST thng, c ngha l tn s cc kiu gen trng tha cn bng Hacdi Vanbec l:

p2 A AX X + 2pq A aX X + q2 a aX X = 1.

Cc c th c ch c 1 alen trn X nn tn s cc kiu gen gii c p AX Y + qaX Y =1. (Khi xt ch trong phm vi gii c).

V t l c : ci l 1: 1 nn t l cc kiu gen trn mi gii tnh phi gim i mt na khi xt trong phm vi ton b qun th, v vy trng thi cn bng qun th Hacdi Vanbec, cng thc tnh kiu gen lin quan n locus gen trn NST trn NST X ( vng khng tng ng) gm 2 alen l:

0.5p2 A AX X + pq A aX X + 0.5q2 a aX X + 0.5p AX Y + 0.5q aX Y = 1.

Phn 2: PHNG PHP GII CC BI TP a. BI TP QUN TH T PHI 1. Dng 1:

Cho thnh phn kiu gen ca th h P (th h xut pht) 100% d hp Aa qua n th h t phi tm thnh phn kiu gen ca th h Fn

*Cch gii:

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Qun th P Sau n th h t phi thnh phn kiu gen thay i nh sau T l th ng hp tri AA trong qun th Fn l

AA = 2

2

11

n

T l th d hp Aa trong qun th Fn l

Aa = n

2

1

T l th ng hp ln aa trong qun th Fn l

aa = 2

2

11

n

*V d: Qun th ban u 100% c th c kiu gen d hp. Sau 3 th h t th phn thnh phn kiu gen ca qun th nh th no?

Gii nhanh: Sau n th h t phi thnh phn kiu gen thay i nh sau (Vi n=3) T l th ng hp tri AA trong qun th Fn l

AA = 2

2

11

n

=

3

2

2

11

= 0,4375

T l th d hp Aa trong qun th Fn l

Aa = n

2

1=

3

2

1

= 0,125

T l th ng hp ln aa trong qun th Fn l

aa = 2

2

11

n

=

3

2

2

11

= 0,4375

2. Dng 2:

Cho thnh phn kiu gen ca th h P qua n th h t phi tm thnh phn kiu gen ca th h Fn

*Cch gii: Qun th t phi c thnh phn kiu gen ca th h P ban u nh sau:xAA +

yAa + zaa

Qun th P Sau n th h t phi thnh phn kiu gen thay i nh sau T l th ng hp tri AA trong qun th Fn l

AA = x + 2

y.2

1y

n

T l th d hp Aa trong qun th Fn l

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Aa = y.2

1n

T l th ng hp ln aa trong qun th Fn l

aa = z + 2

y.2

1y

n

* V d 1: Qun th P c 35AA, 14Aa, 91aa =1 Cc c th trong qun th t phi bt buc qua 3 th h tm cu trc ca qun th

qua 3 th h

Gii: Cu trc ca qun th P 0,25AA + 0,1Aa + 0,65aa Cu trc ca qun th qua 3 th h

AA = x + 2

y.2

1y

n

= 0,25 + 2

1,0.2

11,0

3

= 0,29375

T l th d hp Aa trong qun th Fn l

Aa = y.2

1n

= 1,0.

2

13

= 0,0125

T l th ng hp ln aa trong qun th Fn l

aa = z + 2

y.2

1y

n

= 0,65 + 2

1,0.2

11,0

3

= 0,69375

Vy cu trc ca qun th qua 3 th h 0,29375 AA + 0,125 Aa + 0,69375 aa = 1

*V d 2 : Qun th t th phn c thnh phn kiu gen th h P l 0,8Bb + 0,2bb = 1. Sau 3 th h t th phn cu trc ca qun th nh th no?

Gii: T l th ng hp tri AA trong qun th F3 l

BB = x + 2

y.2

1y

n

= 2

8,0.2

18,0

0

3

= 0,35

T l th d hp Aa trong qun th F3 l

Bb = y.2

1n

= 8,0.

2

13

= 0,1

T l th ng hp ln aa trong qun th F3 l

bb = z + 2

y.2

1y

n

= 2

8,0.2

18,0

2,0

3

= 0,55

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Vy cu trc ca qun th qua 3 th h t th phn l: 0,35 BB + 0,1 Bb + 0,55 bb = 1

*V d 3 : Qun th t th c thnh phn kiu gen th h P l 0,4BB + 0,2 Bb + 0,4bb = 1. Cn bao nhiu th h t th phn c c t l ng hp tri chim 0,475 ?

Gii:

T l th ng hp tri BB trong qun th Fn l

BB = x + 2

y.2

1y

n

= 2

2,0.2

12,0

4,0

n

= 0,475

n=2

Vy sau 2 th h BB = 0,475. b. BI TP QUN TH NGU PHI 1. Dng 1: T cu trc di truyn qun th chng minh qun th t trng thi cn bng

hay khng, qua bao nhiu th h qun th t trng thi cn bng. * Cch gii 1: - Gi p l tn s tng i ca alen A - Gi q l tn s tng i ca alen a p+q = 1

Cu trc di truyn ca qun th khi t trng thi cn bng: p2 AA + 2pqAa + q2 aa

Nh vy trng thi cn bng ca qun th phn nh mi tng quan sau: p2 q2 = (2pq/2)2

Xc nh h s p2, q2, 2pq Th vo p2 q2 = (2pq/2)2 qun th cn bng. Th vo p2 q2 # (2pq/2)2 qun th khng cn bng. * Cch gii 2: - T cu trc di truyn qun th tm tn s tng i ca cc alen. C tn s

tng i ca cc alen th vo cng thc nh lut. - Nu qun th ban u cho nghim ng cng thc nh lut (tc trng cng

thc nh lut) suy ra qun th cn bng - Nu qun th ban u cho khng nghim ng cng thc nh lut (tc

khng trng cng thc nh lut) suy ra qun th khng cn bng * V d 1: Cc qun th sau qun th no t trng thi cn bng QT1: 0,36AA; 0,48Aa; 0,16aa

QT2: 0,7AA; 0,2Aa; 0,1aa

Cch gii 1: QT1: 0.36AA; 0.48Aa; 0.16aa

- Gi p l tn s tng i ca alen A - Gi q l tn s tng i ca alen a Qun th t trng thi cn bng khi tho mn p2AA + 2pqAa + q2 aa = 1

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v khi c c p2 q2 = (2pq/2)2 . qun th 1 c p2 = 0.36 , q2 = 0.16, 2pq = 0.48 0.36 x 0.16 = (0.48/2)2 vy qun th ban u cho l cn bng. Cch gii 2: QT2: 0,7AA; 0,2Aa; 0,1aa

- Gi p l tn s tng i ca alen A - Gi q l tn s tng i ca alen a p = 0,7 + 0,1 q = 0.1 +0.1

Qun th t trng thi cn bng khi tho mn p2AA + 2pqAa + q2 aa Tc 0,82 AA + 2.0,8.0,2Aa + 0,22 aa = 0,7AA + 0,2Aa + 0,1aa vy qun th

khng cn bng. *V d 2: Qun th no trong cc qun th di y t trng thi cn bng

Qun th

Tn s kiu gen AA

Tn s kiu gen Aa

Tn s kiu gen aa

1 1 0 0

2 0 1 0

3 0 0 1

4 0,2 0,5 0,3

Gii nhanh Qun th 1: Nu cn bng th p2 q2 = (2pq/2)2 =>1 x 0 = (0/2)2 => qun th

cn bng. Qun th 2: Nu cn bng th p2 q2 = (2pq/2)2 =>0 x 0 (1/2)2 => qun th

khng cn bng. Qun th 3: Nu cn bng th p2 q2 = (2pq/2)2 =>0 x 1 = (0/2)2 => qun th

cn bng. Qun th 4: Nu cn bng th p2 q2 = (2pq/2)2 =>0,2 x 0,3 = (0,5/2)2 =>

qun th khng cn bng. 2. Dng 2: T s lng kiu hnh cho cho xc nh cu trc di truyn ca qun th

(cho s lng tt c kiu hnh c trong qun th). Cch gii: Cu trc di truyn ca qun th - T l kiu gen ng tri = s lng c th do kiu gen ng tri qui

nh/Tng s c th ca qun th -T l kiu gen d hp = s c th do kiu gen d hp quy nh/ Tng s c

th ca qun th - T l kiu gen ng ln = S c th do kiu gen ln quy nh/ Tng s c th

ca qun th. * V d 1: g, cho bit cc kiu gen: AA qui nh lng en, Aa qui nh lng

m, aa qui nh lng trng. Mt qun th g c 410 con lng en, 580 con lng m, 10 con lng trng.

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a. Cu trc di truyn ca qun th ni trn c trng thi cn bng khng? b. Qun th t trng thi cn bng vi iu kin no? c. Xc nh cu trc di truyn ca qun th khi t trng thi cn bng? Gii: a. Cu trc di truyn ca qun th c xc nh da vo t l ca cc kiu gen: Tng s c th ca qun th: 580 + 410 + 10 =1000 T l th ng hp tri AA l 410/1000 = 0,41 T l th d hp Aa l 580/1000 = 0,58 T l th ng hp ln aa l 10/1000 = 0.01 Cu trc di truyn ca qun th nh sau: 0.41 AA + 0.58aa + 0.01aa

Cu trc ny cho thy qun th khng trng thi cn bng v 0,41 x 0,01 = (0,58/2)2

=> 0,0041 = 0.0841.

b. iu kin qun th t v tr cn bng di truyn khi qu trnh ngu phi din ra th ngay th h tip theo qun th at s cn bng di truyn

c. Tn s alen A l 0,41 + 0,58/2 = 0.7 Tn s ca alen a l 1 - 0.7 = 0,3 Sau khi qu trnh ngu phi xy ra th cu trc di truyn ca qun th th h

sau l

(0,7A:0,3a) x (0,7A:0,3a) => 0,49AA + 0,42Aa + 0,09aa

Vi cu trc trn qun th t trng thi cn bng v tho mn (0,9)2 AA + 2(0,7 x 0,3) Aa + (0,3)2 aa

* V d 2: Mt qun th sc c s lng nh sau 1050 con lng nu ng hp, 150 con lng nu d hp, 300 con lng trng, mu lng do mt gen gm 2 alen qui nh. Tm tn s tng i ca cc alen?

Gii: Tnh trng lng nu l tri do A quy nh Tnh trng lng trng l ln do a quy nh T l th ng hp tri AA l 1050/1500 = 0,7 T l th d hp Aa l 150/1500 = 0,1 T l th ng hp ln aa l 300/1500 = 0,2 Vy cu trc di truyn ca qun th l: 0,7AA; 0,1Aa; 0,2aa 3. Dng 3: T s lng kiu hnh cho cho xc nh cu trc di truyn ca qun th

(ch cho tng s c th v s c th mang kiu hnh ln hoc tri). Cch gii:

- Nu t l kiu hnh tri=> kiu hnh ln = 100% - Tri. - T l kiu gen ng ln = S c th do kiu gen ln quy nh/ Tng s c th

ca qun th. + T t l kiu gen ng ln => Tn s tng i ca alen ln tc tn s ca

q => Tn s tng i ca alen tri tc tn s p.

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+ p dng cng thc nh lut p2 AA + 2pq Aa + q2 aa = 1 => cu trc di truyn qun th.

* V d 1: Qun th ngu phi c thnh phn kiu gen t trng thi cn bng vi 2 loi kiu hnh l hoa (do B tri hon ton quy nh) v hoa trng(do b quy nh). T l hoa 84%. Xc nh cu trc di trun ca qun th?

Gii: - Gi p tn s tng i ca alen B - q tn s tng i alen b - %hoa trng bb = 100%- 84%= 16%=q2 => q = 0,4 => p = 0,6 p dng cng thc nh lut p2 BB + 2pq Bb + q2 bb = 1 => cu trc di truyn qun th :0.62 BB + 2.0,6.0,4 Bb + 0,42 bb = 0,36BB +

0,48Bb + 0,16bb = 1

* V d 2: b A qui nh lng en, a: lng vng. Trong mt qun th b lng vng chim 9% tng s c th ca n. Bit qun th t trng thi cn bng. Tm tn s ca gen A?

Ch gii nhanh: Qun th t trng thi cn bng aa = 9% = q2 => q = a = o,3 => p = A= 0,7 * V d 3: Qun th ngi c tn s ngi b bch tng 1/10000. Gi s qun th

ny cn bng( bit bch tng do gen ln nm trn nhim sc th thng quy nh)

a. Tnh tn s cc alen? b. Tnh xc sut 2 ngi bnh thng trong qun th ly nhau sinh ra

ngi con u lng b bch tng?

Gii nhanh: a. Tnh tn s cc alen ? A: bnh thng (khng bch tng), a: bch tng Qun th cn bng aa = q2 = 1/10000 = > a = q = 0,01 => A = p = 0,99 b. Tnh xc sut 2 ngi bnh thng trong qun th ly nhau sinh ra ngi

con u lng b bch tng?

- B d hp (Aa) xc sut pqp

pq

2

22

- M d hp (Aa) xc sut pqp

pq

2

22

- Xc sut con b bnh 4

1

Vy xc sut 2 ngi bnh thng trong qun th ly nhau sinh ra ngi con

u lng b bch tng l: pqp

pq

2

22

x pqp

pq

2

22

x 4

1

th p=0,01 , q= 0,99 => pqp

pq

2

22

x pqp

pq

2

22

x 4

1 = 0,00495

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* V d 4: Trong mt qun th cn bng c 90% alen lcus Rh l R. Alen cn li l r. C 40 tr em ca qun th ny n mt trng hc nht nh . Xc sut tt c cc em u l Rh dng tnh l bao nhiu?

Gii nhanh: Tn s tng i ca alen R =p= 0,9 => tn s alen r=q = 0,1 Rh dng c kiu gen RR, Rr tn s ca 2 nhm kiu gen trn l RR= p2= 0,92 = 0,81, Rr = 2pq = 2.0,9.0,1 = 0,18.

Tn s 1 hc sinh c Rh dng l: 0,81+0,18 = 0,99 Xc sut 40 hc sinh c Rh dng l (0,99)40

c. BI TP GEN A ALEN * V d: Gi thit trong mt qun th ngi, tn s ca cc nhm mu l: Nhm A = 0,45 Nhm B = 0,21

Nhm AB = 0,3 Nhm O = 0,004

Xc nh tn s tng i ca cc alen qui nh nhm mu v cu trc di truyn ca qun th?

Gii: -Gi p l tn s tng i ca alen IA. - Gi q l tn s tng i ca alen IB

- Gi r l tn s tng i ca alen IO

Nhm

mu

A B AB O

Kiu gen

Kiu hnh

IAIA

+IAIO

p2 + 2pr

0,45

IBIB +

IBIO q2 +

2qr

0,21

IAIB

2pq

0,3

IOIO

r2

0,04

T bng trn ta c: p2 + 2pr + r2 = 0,45 + 0,04

=> (p + r)2 = 0,49 => p + r = 0,7

r2 = 0,04 => r = 0,2

Vy p = 0,7 - 0,2 = 0,5 => q = 0,3 Cu trc di truyn ca qun th c xc nh l: (0,5 IA + 0,3 IB + 0,2IO) (0,5 IA + 0,3 IB + 0,2IO) = 0,25IAIA + 0,09IBIB + 0,04 IOIO

+ 0,3IAIB + 0,2IAIO + 0,12IBIO

d,Bi tp xc nh s kiu gen ti a trong qun th Xc nh tng s KG, s KGH, KGDH trong trng hp nhiu cp gen PLL,

mi gen c 2 hoc nhiu alen a. Tng qut a1)Trng hp gen nm trn NST thng xc nh tng s KG, s KGH, KGDH trong trng hp nhiu cp gen

PLL, mi gen c 2 hoc nhiu alen, GV cn phi cho HS thy r: * Vi mi gen:

0902651694

Phn tch v chng minh s KGDH, s KGH, s KG ca mi gen, ch ra mi quan h gia 3 yu t vi nhau v vi s alen ca mi gen:

- S alen ca mi gen c th ln hn hoc bng 2 nhng trong KG lun c mt ch 2 trong s cc alen .

- Nu gi s alen ca gen l r th s kiu gen d hp = Cr2 = r( r 1)/2 - S kiu gen ng hp lun bng s alen = r - S KG ti a trong qun th i vi mt gen c r alen = s KGH + s KGDH

= r +r( r 1)/2 = r( r + 1)/2 * Vi nhiu gen: Do cc gen PLL nn kt qu chung = tch cc kt qu ring V vy GV nn gi cho HS lp bng sau:

GE

N

S ALEN/G

EN

S KI

U GE

N

S KG NG

HP

S K

G

D H

P

I 2 3 2 1

II 3 6 3 3

III 4 10 4 6

n r r( r +

1)/

2

r r( r 1)

/2

( Lu : thay v tnh r( r + 1)/2, c th tnh nhanh 1 + 2 + 3 + +r ) 2) Trng hp gen nm trn NST gii tnh X(khng c alen tng ng trn

Y)

*Trn gii XX = r( r + 1)/2 (V cp NST tng ng nn ging nh trn NST thng)

* Trn gii XY = r ( v alen ch c trn X,khng c trn Y) Vy tng s KG ti a trong QT = r( r + 1)/2 + r

-Nu ch c gen nm trn nhim sc th Y khng c alen tng ng nm trn X

-S kiu gen ti a trong qun th i vi 1 gen =r Nu trng hp trn X v Y u c alen tng ng(nm trn on tng ng)

th cng nh NST thng -Nu 2 gen cng nm trn mt cp nhim sc th: s alen ca hai gen = tch cc

alen ca tng gen

-Trng hp gen nm trn nhim sc th thng khc nhau

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Cu1: Gen I,II v III c s alen ln lt l 2,3 v 4.Tnh s kiu gen ti a c th c trong qun th cc trng hp:

1/ 3 gen trn nm trn 3 cp NST thng. A. 124 B. 156 C. 180

D. 192

2/ Gen I v II cng nm trn mt cp NST thng,gen III nm trn cp NST thng khc

A. 156 B. 184 C. 210

D. 242

Cu2: ngi, gen qui nh dng tc do 2 alen A v a trn nhim sc th thng qui nh ; bnh mu kh ng do 2 alen M v m nm trn nhim sc th X on khng tng ng vi Y.Gen qui nh nhm mu do 3alen : IA ; IB (ng tri )v IO (ln).S kiu gen v kiu hnh ti a trong qun th i vi 3 tnh trng trn :

A. 90 kiu gen v 16 kiu hnh B. 54 kiu gen v 16 kiu hnh C. 90 kiu gen v 12 kiu hnh D. 54 kiu gen v 12 kiu hnh -Trng hp gen nm trn nhim sc th thng v nhim sc th gii

tnh X

Cu1: ngi gen a: qui nh m mu; A: bnh thng trn NST X khng c alen trn NST Y. Gen quy nh nhm mu c 3 alen IA, IB,IO. S kiu gen ti a c th c ngi v cc gen ny l:

A. 27 B. 30 C. 9 D. 18

Cu2: S alen ca gen I, II v III ln lt l 3, 4 v 5.Bit cc gen u nm trn NST thng v khng cng nhm lin kt. Xc nh trong QT:

S KG H v tt c cc gen v d hp tt c cc gen ln lt l: A. 60 v 90 B. 120 v 180 C. 60 v 180 D. 30 v 60

Cu2b S KG H v 2 cp gen v d hp v 2 cp gen ln lt l: A. 240 v 270 B. 180 v 270 C. 290 v 370 D. 270 v 390

Cu2c S KG d hp A. 840 B. 690 ` C. 750 D. 660

1) S Kg H tt c cc gen = 3.4.5 =60 ; S Kg d hp tt c cc gen = 3.6.10 =180

2) S Kg H 2 cp, d hp 1 cp = (3.4.10+4.5.3+3.5.6) =270 S Kg d hp 2 cp, H 1 cp = (3.6.5+6.10.3+3.10.4) =390 3) S KG d hp = (6.10.15) (3.4.5) = 840 Cu3: Gen I c 3 alen, gen II c 4 alen , gen III c 5 alen. Bit gen I v II nm

trn X khng c alen trn Y v gen III nm trn Y khng c alen trn X. S KG ti a trong QT

A. 154 B. 184 C. 138 D. 214

s Kg trn XX= 3.4(3.4+1) = 78 s Kg trn XY = 3.4.5 = 60 Tng s Kg = 78+60= 138

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Cu4: S alen tng ng ca gen I, II, III v IV ln lt l 2, 3, 4 v 5. Gen I v II cng nm trn NST X on khng tng ng vi Y, gen IV v V cng nm trn mt cp NST thng.S kiugen ti a trong QT:

A. 181 B. 187 C. 231 D. 237

Bi tp v gen hai hay nhiu gen cng nm trn mt nhim sc th Bi 1: ngi gen A Quy nh mt nhn mu bnh thng, alen a quy nh bnh

m mu v lc; gen B quy nh mu ng bnh thng, alen b quy nh bnh mu kh ng. Cc gen ny nm trn NST gii tnh X khng c alen tng ng trn Y. Gen D quy nh thun tay phi, alen d quy nh thun tay tri nm trn NST thng. S KG ti a v 3 locut trn trong QT ngi l:

A.42 B.36 C.39

D.27

Gii : Cc gen ( AaBb ) nm trn NST gii tnh X khng c alen tng ng trn Y: c 14 KG

S KG nm trn Y l 4 : XABY, XabY, XAbY, XaBY S KG nm trn X l 10: XABXAB, XaB XaB , XAB XaB, XABXAb, XaB Xab , XAb

Xab,

XAbXAb, Xab Xab , XAB Xab, XAb XaB

Gen nm trn NST thng ( D v d ) c: (2(2+1) : 2 )1 = 3 KG Vy: QT Ngi c s loi KG ti a v 3 locut trn l: 14 x 3 = 42 Chn

A

Cch2: Cc gen ( AaBb ) nm trn NST gii tnh X khng c alen tng ng trn Y : S alen ca hai gen l : 2.2=4

-S kiu gen trn NST gi tnh X l : (4+1).4 /2= 10 -S kiu gen trn NST Y=4.Vy s kiu gen ti a trn cp XY =10+4=14 -S kiu gen ti a trong qun th l : 14.3=42 Cu 2: Trong QT ca mt loi th, xt hai lcut: lcut mt c 3 alen l A1, A2,

A3; lcut hai c 2 alen l B v b. C hai lcut u nm trn on khng tng ng ca nhim sc th gii tnh X v cc alen ca hai lcut ny lin kt khng hon ton. Bit rng khng xy ra t bin, tnh theo l thuyt, s KG ti a v hai lcut trn trong QT ny l:

A.18 B. 36 C.30 D. 27

Gii: + Ta coi cp NST XX l cp NST tng ng nn khi vit KG vi cc gen lin

kt vi cp NST XX s ging vi cp NST thng nn ta c 21 loi KG ti a khi xt hai lcut: lcut mt c 3 alen l A1, A2, A3; lcut hai c 2 alen l B v b.ng vi trng hp cp XX l:

1

1

A B

A B ,

1

1

A b

A b ,

1

1

A B

A b

1

2

A B

A B ,

1

2

A b

A b ,

1

2

A B

A b

1

2

A b

A B ,

1

3

A b

A B ,

2

3

A b

A B

2

2

A B

A B ,

2

2

A b

A b ,

2

2

A B

A b

1

3

A B

A B ,

1

3

A b

A b ,

1

3

A B

A b

0902651694

3

3

A B

A B ,

3

3

A b

A b ,

3

3

A B

A b

2

3

A B

A B ,

2

3

A b

A b ,

2

3

A B

A b

(C th vit cc cp gen lin kt vi cp XX: 1 1A A

B BX X .....)

+ Vi cp XY l cp khng tng ng nn c ti a 6 loi KG khi xt hai lcut: lcut mt c 3 alen l A1, A2, A3; lcut hai c 2 alen l B v b l:

1A

BX Y , 2A

BX Y , 3A

BX Y1A

bX Y , 2A

bX Y , 3A

bX Y

Nu khng xy ra t bin, tnh theo l thuyt, s KG ti a v hai lcut trn trong QT ny l:21 + 6 = 27 loi KG

p n l: D. 27 HD :C 2 alen A va B cng nm trn 1 NST X nn chng ta xem t hp 2 alen

ny l mt gen (gi l gen M) Khi d gen M c s alen bng tch s 2 alen ca A v B=3x2=6 alen..

gii XX s KG s l 6(6+1)/2=21 KG ( ADCT nhu NST thuong r(r+1)/2 trong do r l s alen

- gii XY S KG= r=S alen=6. Vy s KG ti a v hai lcut trn trong QT ny l: 21+6 = 27 p n D Cu3: ngi, xt 3 gen: gen th nht c 3 alen nm trn NST thng, cc gen

2 v 3 mi gen u c 2 alen nm trn NST X (khng c alen trn Y). Cc gen trn X lin kt hon ton vi nhau. Theo l thuyt s kiu gen ti a v cc lcut trn trong qun th ngi l

A. 30 B. 15 C. 84 D. 42

Bi tp v qun th ni phi Bi4: Mt QT thc vt th h XP u c KG Aa. Tnh theo l thuyt TL KG

AA trong QT sau 5 th h t th phn bt buc l: A.46,8750 % B.48,4375 % C.43,7500 %

D.37,5000 %

Gii TL KG AA = (( 1 ( 1/2 )5 ) : 2 ) = 31/ 64 = 48,4375 % Chn B

Bi5: 1 QT c 0,36AA; 0,48Aa; 0,16aa. Xc nh CTDT ca QT trn qua 3 th h t phi.

A.0,57AA : 0,06Aa : 0,37aa B.0,7AA : 0,2Aa ;

0,1aa

C.0,36AA : 0,24Aa : 0,40aa D.0,36AA : 0,48Aa :

0,16aa

Gii TL KG Aa qua 3 th h t phi = ( 1 / 2 )3 x 0,48 = 0,06.TL KG AA = 0,36 + (0,48 0,06)/2 = 0,36 + 0,21 = 0,57.

TL KG aa = 0, 16 + 0,21 = 0,37.

Vy: qua 3 th h t phi QT trn c CTDT l: 0,57AA : 0,06Aa : 0,37aa Chn A

0902651694

Bi 6: Nu P TS cc KG ca QT l :20%AA :50%Aa :30%aa ,th sau 3 th h t th, TS KG AA :Aa :aa s l :

A.51,875 % AA : 6, 25 % Aa : 41,875 % aa B.57, 250 % AA : 6,25 %

Aa : 36,50 %aa

C.41,875 % AA : 6,25 % Aa : 51,875 % aa D.0,36 AA : 0,48 Aa :

0,16 aa

Gii : TS KG Aa = ( 1 / 2 )3 x 0,5 = 0,0625 = 6,25 % TS KG AA = 0,2 + (( 0,5 -

0,0625 ) /2 ) = 0,41875 = 41,875 %

TS KG aa = 0,3 + (( 0,5 - 0,0625 ) /2 ) = 0,51875 = 51,875 % Chn C Bi 7: QT t th phn c thnh phn KG l 0,3 BB + 0,4 Bb + 0,3 bb = 1. Cn bao nhiu th h t th phn th TL th H chim 0,95 ? A. n = 1 ; B. n = 2 C. n = 3 D. n = 4

Gii: Th H gm BB v bb chim 0,95 => TL th H BB = bb = 0,95 / 2 = 0,475 TL KG Bb = 0,4 ( 1 / 2 )n

TL KG BB = 0,3 + (( 0,4 - 0,4( 1 / 2 )n ) /2 ) = 0,475

0,6 + 0,4 ( 0,4( 1 / 2 )n ) = 0,475 x 2

0,4( 1 / 2 )n = 1 0,95 = 0,05 ( 1 / 2 )n = 0,05 / 0,4 = 0,125

( 1 / 2 )n = ( 1 / 2 )3 => n = 3 Chn C Bi 8: Xt QT t th phn c thnh phn KG th h P l: 0,3 BB + 0,3 Bb + 0,4 bb = 1.Cc c th bb khng c kh nng sinh sn, th

thnh phn KG F1 nh th no? A.0,25AA + 0,15Aa + 0,60aa = 1 B.0,7AA + 0,2Aa +

0,1aa = 1

C.0,625AA + 0,25Aa + 0,125 aa = 1 D.0,36AA + 0,48Aa +

0,16aa = 1

Gii: P : 0,3 BB + 0,3 Bb + 0,4 bb = 1.Cc c th bb khng c kh nng sinh sn

cc c th BB, bb khi t th phn : 0,3 BB : 0,3 Bb ch t 60 % , th : TL KG BB = ( 30 x 100 ) / 60 = 50 % = 0,5

TL KG bb = ( 30 x 100 ) / 60 = 50 % = 0,5

P: 0,5 BB + 0,5 bb = 1

Lc ny F1; TL KG Bb = ( 1 / 2 )1 x 0,5 = 0,25

TL KG BB = 0,3 + (( 0,5 0,25 )/2 ) = 0,625 TL KG bb = 0 + ((0,5 0,25 ) / 2) = 0,125 Vy: thnh phn KG F1 l 0,625BB + 0,25 Bb + 0,125 bb = 1 Chn C Bi 9: Mt QT XP c TL ca th d hp Bb bng 60%. Sau mt s th h t phi

lin tip, TL ca th d hp cn li bng 3,75%. S th h t phi xy ra QT tnh n thi im ni trn l bao nhiu?

A. n = 1 ; B. n = 2 C. n = 3 D. n = 4

Gii:

0902651694

TL KG Bb = ( 1 / 2 )n x 60 % = 3,75 %

( 1 / 2 )n x 3/5 = 3 / 80 (60 % = 60 /100 = 3/5 ; 3,75 % =375/10000

= 3/80 )

( 1 / 2 )n = 3/80 : 3/5 = 3/80 x 5/3 = 5/80 = 1/16 = ( 1 / 2 )4

( 1 / 2 )n = ( 1 / 2 )4 => n = 4 Chn D Bi 10: Mt QT Thc vt t th phn c TL KG th h XP: 0,45AA : 0,30Aa :

0,25aa. Cho bit c th c KG aa khng c kh nng sinh sn. Tnh theo l thuyt TL KG thu c F1 l:

A.0,525AA : 0,150Aa : 0,325aa B.0,7AA : 0,2Aa ;

0,1aa

C.0,36AA : 0,24Aa : 0,40aa D.0,36AA : 0,48Aa :

0,16aa

Gii: P : 0,45 AA : 0,30 Aa : 0,25 aa .Cc c th c KG aa khng c kh nng sinh sn

Cc c th AA, Aa khi t th phn : 0,45 AA : 0,30 Aa ch t 75 %, th :

TL KG AA = ( 45 x 100 ) / 75 = 60 % = 0,6

TL KG Aa = ( 30 x 100 ) / 75 = 40 % = 0,4

P: 0,6 AA + 0,4 Aa = 1

Lc ny F1; TL KG Aa = ( 1 / 2 )1 x 0,4 = 0,2

TL KG AA = 0,6 + (( 0,4 0,2 )/2 ) = 0,7 TL KG aa = 0 + ((0,4 0,2 ) / 2) = 0,1 Vy: TL KG F1 l : 0,7AA : 0,2Aa ; 0,1aa Chn B Bi 11: Xt mt QT thc vt c TP KG l 25% AA : 50% Aa : 25% aa. Nu tin

hnh t th phn bt buc th TL KG H th h F2 l A. 12,5%. B. 25%. C. 75%. D. 87,5%.

Gii: TL KG Aa = ( 1 / 2 )2 x 50 % = 12,5 %.

Nu tin hnh t th phn bt buc th TL KG H th h F2 l: 100 % - 12,5% = 87,5 % . Hay : TL KG AA = 25 % + (( 50 % 12,5 % ) /2 ) = 43,75 %

TL KG aa = 25 % + (( 50 % 12,5 % ) /2 ) = 43,75 % Vy : TL KG H th h F2 l: 43,75 % + 43,75 % = 87,5 % Chn D Bi 12: mt QT sau khi tri qua 3 th h t phi, TL ca th d hp trong QT

bng 8%. Bit rng th h XP, QT c 20% s c th H tri v cnh di l tnh tri hon ton so vi cnh ngn. Hy cho bit trc khi xy ra qu trnh t phi, TL KH no sau y l ca QT trn?

A. 36% cnh di : 64% cnh ngn. B. 64% cnh di : 36% cnh ngn.

C. 84% cnh di : 16% cnh ngn. D. 16% cnh di : 84% cnh ngn.

Gii : TL th d hp Aa th h XP: ( 1/2 )3 x Aa = 0,08 => Aa = 0, 64 = 64 %

Vy: TL KH cnh di : 64 % + 20 % = 84 %

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TL KH cnh ngn : 100 % - 84 % = 16 % Chn C Bi tp v nh lut hc i- van bc( Qun th ngu phi) -Vi mt gen c hai alen( A, a) th thnh phn kiu ca qun th trng thi cn

bng di truyn l: P2AA + 2pqAa + q2aa =1

*Trng hp c bit: - Qun th ng nht mt kiu gen 100% AA hay 100%aa th lun t trng thi

cn bng di truyn - qun th ch c AA v Aa hay aa v Aa th cha t trng thi cn bng di

truyn -Trng hp mt gen c hai alen nm trn Nhim sc th gii tnh X th cu trc

di truyn qun th l: P2XAXA + 2pq XAXa + p XAY + qXaY + q2 XaXa

=1

-T l giao t XA = p2 + 2pq + p -T l giao t Xa = q2 + q + 2pq

Bi tp vn dng: A.Bi ton v mt gen c hai alen: Bi 1: QT no sau y trng thi CBDT? A. QTI : 0,32 AA : 0,64 Aa : 0,04 aa. B.QT II: 0,04 AA : 0,64

Aa : 0,32 aa.

C. QT III: 0,64 AA : 0,04 Aa : 0,32 aa. D. QT IV: 0,64 AA : 0,32 Aa :

0,04 aa.

Gii: Dng cng thc p2AA x q2aa = ( 2pqAa / 2 )2 Xt QTI: 0,32 x 0,04 = ( 0,64 /2 ) 2 0,0128 khng bng 0,1024 Xt QTII: 0,04 x 0,32 = ( 0,64 /2 ) 2 0,0128 khng bng 0,1024 Xt QTIII: 0,64 x 0,32 = ( 0,04 /2 ) 2 0,2048 khng bng 0,0004 Xt QTIV: 0,64 x 0,04 = ( 0,32 /2 ) 2 0,0256 = 0,0256 => Chn D Bi 2.Mt QT bao gm 120 c th c KG AA, 400 c th c KG Aa, 680 c th

c KG aa. TS alen A v a trong QT trn ln lt l : A.0,265 v 0,735 B.0,27 v 0,73 C.0,25 v 0,75

D.0,3 v 0,7

Gii: Tng s c th trong QT : 120 + 400 + 680 = 1200 TS KG AA = 120 / 1200 = 0,1 : TS KG Aa = 400 / 1200 = 0,33

TS KG aa = 680 / 1200 = 0,57

Vy : pA = 0,1 + 0,33 / 2 = 0,265 ; qa = 0,57 + 0,33 / 2 = 0,735 chn A Bi 3: Gen BB Q hoa , Bb Q hoa hng, bb Q hoa trng. Mt QT c 300 c

th , 400 c th hoa hng v 300 c th hoa trng tin hnh giao phn ngu nhin. Nu khng c s tc ng ca cc nhn t tin ha th TP KG ca QT F1 l

A) 0,25 BB+0,50Bb+0,25bb=1. B) 0,36 BB+0,48Bb+0,16bb=1

C) 0,81 BB+0,18Bb+0,01bb=1. D) 0,49 BB+0,42Bb+0,09bb=1

Gii: -Tng s c th trong QT P: 300 + 400 + 300 = 1000

0902651694

TS KG BB = 300 / 1000 = 0,3; TS KG Bb = 400 / 1000 = 0,4

TS KG bb = 300 / 1000 = 0,3 => pA = 0,3 + 0,4 / 2 = 0, 5 ; qa = 0,3 + 0,4 / 2

= 0, 5

- Vy TP KG ca QT F1 l: 0,25 BB + 0,50Bb + 0,25bb = 1. chn A Bi 4: Bnh bch tng do gen ln nm trn NST thng Q. huyn A c 106

ngi, c 100 ngi b bnh bch tng. Xc sut bt gp ngi bnh thng c KG d hp l:

A)1,98. B)0,198. C)0,0198.

D)0,00198

Gii: Gi a l gen ln gy bnh bch tng KG aa: ngi b bnh bch tng Ta c : q2aa = 100 / 1000.000 => qa = 1/100 = 0,01

M : pA + qa = 1 => pA = 1- qa = 1 0,01 = 0,99 2pqAa = 2 x 0,01 x 0,99 = 0,0198 chn C Bi 5: Bit alen A quy nh lng xm l tri hon ton so vi alen a quy nh

lng trng, cc alen nm trn NST thng. Mt QT chut th h XP c 1020 chut lng xm H, 510 chut c KG d hp. Khi QT t TTCB c 3600 c th.

S dng d kin trn tr li cc cu hi a) v b) sau y: TS tng i ca mi alen l: A. A: a = 1/6 : 5/6 B. A: a = 5/6 : 1/6 C. A: a = 4/6 : 2/6 D

A: a = 0,7 : 0,3

b) S lng chut tng KG khi t TTCB: A. AA = 1000; Aa = 2500; aa = 100 B. AA = 1000; Aa =

100; aa = 2500

C. AA = 2500; Aa = 100; aa = 1000 D. AA = 2500; Aa =

1000; aa = 100

Gii: a)TS tng i ca mi alen l: Tng s c th chut trong QT th h XP: 1020 + 510 = 1530 => TS KG AA = 1020 / 1530 = 2 / 3 ; TS KG Aa = 510 / 1530 = 1 / 3

Vy : TP KG th h XP l 2/3 AA + 1/3 Aa = 1. TS tng i ca mi alen l: pA = 2/3 + ( 1/3 : 2 ) = 5 / 6 ; qa = 0 + ( 1/3 : 2 ) = 1 / 6 chn B b) Kt qu ngu phi gia cc c th th h P: P: ( 5/6A : 1/6 a ) x ( 5/6A : 1/6 a ) = 25AA : 10Aa : 1aa ( hay k pennett ) Vy: S lng chut tng KG khi t TTCB: KG AA = ( 25 : 36 ) 3600 = 2500 ; KG Aa = ( 10 : 36 ) 3600 = 1000

KG aa = ( 1 : 36 ) 3600 = 100 chn D Bi 6: n b c TP KG t CB, vi TS tng i ca alen quy nh lng en l

0,6, TS tng i ca alen quy nh lng vng l 0,4. TL KH ca n b ny nh th no ?

A)84% b lng en, 16% b lng vng. B)16% b lng en, 84% b lng vng. C)75% b lng en, 25% b lng vng. D)99% b lng en, 1% b lng vng. Gii: TS KG AA = 0,36 TS KG Aa = 2( 0,6 x 0,4 ) = 0,48; TS KG aa = 0,16

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TL KH b lng en l : 0,36 + 0,48 = 0,84 = 84 % TL KH n b lng vng: 0,16 = 16 % chn A Bi 7: QT giao phn c TP KG t TTCB, c hoa chim 84%. TP KG ca QT

nh th no (B Q hoa tri hon ton so b Q hoa trng)? A)0,16 BB + 0,48 Bb + 0,36 bb = 1. B)0,36 BB + 0,48 Bb + 0,16 bb =

1.

C)0,25 BB + 0,50 Bb + 0,25 bb = 1. D)0,64 BB + 0,32 Bb + 0,04 bb =

1.

Gii : TL KH hoa : 84 % => TL KH hoa trng : 16 % = 0,16 TS KG bb = 0,16 => qb = 0,4

Theo nh lut Haci-Vanbec: pB + qb = 1 => pB = 1- qb= 1 - 0,4 = 0, 6 TS KG BB= 0,36 ; TS KG Bb = 2( 0,6 x 0,4 ) = 0,48

TP KG ca QT l : 0,36 BB + 0,48 Bb + 0,16 bb = 1. chn D Cu8: Mt gen c 2 alen, th h XP,TS alen A = 0,2 ; a = 0,8. Sau 5 th h

chn lc loi b hon ton KH ln ra khi QT th TS alen a trong QT l: A. 0,186 B. 0,146 C.

0,160 D. 0,284

p dng cng thc qn = q0/1+ nq0 = 0,8/1+5x0,8 = 0,16 Cu9: mo gen D nm trn phn khng tng ng ca nhim sc th X quy

nh mu lng en, gen ln a quy nh mu lng vng hung, khi trong KG c c D v d s biu hin mu lng tam th. Trong mt QT mo c 10% mo c lng en v 40% mo c lng vng hung, s cn li l mo ci. TL mo c mu tam th theo nh lut Hcdi-Van bc l bao nhiu?

A. 16% B. 2% C. 32% D. 8%

t gt Xa = 0,8 , XA = 0,2 CTDT: 0,04XAXA + 0,32XAXa + 0,64XaXa + 0,2XAY +0,8XaY

= 0,02XAXA + 0,16XAXa + 0,32XaXa + 0,1XAY

+0,4XaY

Bi10:Mt QT sc sng trong vn thc vt c 160 con c TS alen B = 0,9. Mt QT sc khc sng trong rng bn cnh c TS alen ny l 0,5. Do ma ng khc nghit t ngt, 40 con sc trng thnh t QT rng chuyn sang QT sc vn tm n v ha nhp vo QT vn, TS alen B sau s di c ny l bao nhiu ?

A. 0,70. B. 0,90. C. 0,75. D. 0,82.

Gii:Xt QT ban u: S allele B l: 0.9.160.2 = 288 ; s allele b l: (1-0,9).160.2 = 32

Xt nhm c th nhp c: S allele B = s alen a = 0,5.40.2 = 40 QT vn sau nhp c: S alen B = 288+40 = 328 ; s allele b = 40+32=72 TS alen B trong QT sau nhp c l: 328/(328+72) = 0,82 Cu11:Cho 2 QT 1 v 2 cng loi,kch thc QT 1 gp i QT 2. QT 1 c TS

alen A=0,3, QT 2 c TS alen A=0,4.

Nu c 10% c th ca QT 1 di c qua QT 2 v 20% c th ca QT 2 di c qua QT 1 th TS alen A ca 2 QT 1 v 2 ln lt l:

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A. 0,35 v 0,4 B. 0,31 v 0,38 C. 0,4 v 0,3 D. bng nhau v=0,35 gi N1 , p1 , v N2 , p2 ln lt l s lng c th (kch thc ) ca QT 1 v 2

v theo gt th

N1 =2 N1

TS alen p sau khi xut v nhp c 2 QT: * QT1: p(1) = [(p1x 9N1/10) +(p2x 2N2/10) ] / [9N1/10 +2N2/10] = 0,31

* QT2: p(2)= [(p1x N1/10) +(p2x 8N2/10) ] / [N1/10 +8N2/10] = 0,38 (p n B) Bi 12:Xt 4 gen mt qun th ngu phi lng bi: gen 1 quy nh mu hoa

c 3 alen A1; A2; a vi tn s tng ng l 0,5; 0,3; 0,2; gen 2 quy nh chiu cao cy c 2 alen (B v b), trong tn s alen B gii c l 0,6, gii ci l 0,8 v tn s alen b gii c l 0,4, gii ci l 0,2; gen 3 v gen 4 u c 4 alen. Gi thit cc gen nm trn NST thng. Hy xc nh:

a.S loi kiu gen ti a trong qun th. b.Thnh phn kiu gen v gen quy nh mu hoa khi qun th trng thi cn

bng di truyn. c.Thnh phn kiu gen v gen quy nh chiu cao cy F1 khi qun th ngu

phi v trng thi cn bng di truyn. d.Ly ngu nhin hai cy thn cao trong qun th trng thi cn bng cho lai

vi nhau Tnh xc sut sut xut hin cy thn thp i con ( gen B thn cao b thn thp)

a S KG trong QT: 6.3.10.10 = 1800 kiu gen

b

Thnh phn KG quy inh mu hoa khi QT t TTCB di truyn: 0,25A1A1 + 0,3 A1A2 + 0,2 A1a + 0,09 A2A2 + 0,12 A2a + 0,04 aa = 1

c

d

Thnh phn KG quy nh chiu cao cy F1 khi ngu phi: (0,6.0,8) BB + ( 0,6.0,2 + 0,8.0,4) Bb + ( 0,4.0,2)bb = 1

0,48 BB + 0,44 Bb + 0,08 bb = 1

Thnh phn KG quy nh chiu cao cy khi QT t TTCB di truyn: p B = 0,48 + 0,44/2 = 0,7 ; qb = 1- 0,7 = 0,3 0,49 BB + 0,42 Bb + 0,09 bb =

1

- i con xut hin cy thn thp th b, m thn cao u c kiu gen Bb -Xc sut b m c kiu gen Bb trong qun th trng thi cn bng

l=042/0.91=0.462

=>Xc sut i con xut hin cy thn thp =0.462x0.462x1/4=0.0533

Bi 13: Mt qun th ngu phi ban u phn ci tn s alen A l 0,8. Phn c tn s alen a l 0,4.

a. Xc nh cu trc di truyn ca qun th khi t cn bng di truyn? b. Gi s 1/2 s c th d hp khng c kh nng sinh sn, vy cu trc di truyn

ca qun th tip theo nh th no? GII a. Tn s alen ca qun th khi t cn bng l PA = (0,8 + 0.6 ) : 2 = 0,7 qa =

0,3

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Cu trc di truyn ca qun th khi t cn bng l: 0,49AA + 0.42Aa + 0.09aa = 1

b. Khi 1/2 s c th d hp khng c kh nng sinh sn th cu trc qun th tr thnh:

0,49/0,79AA + 0,21/0,79Aa + 0,09/0,70aa = 1 PA 0,73, qa 0,27 Vy cu trc ca qun th tip theo l: 0,5329AA + 0,3942Aa + 0,0729aa = 1 Bi 14: Xt mt gen c 2 alen A v alen a. Mt qun th sc gm 180 c th

trng thnh sng mt vn thc vt c tn s alen A l 0,9. Mt qun th sc khc sng khu rng bn cnh c tn s alen ny l 0,5. Do thi tit ma ng khc nghit t ngt 60 con sc trng thnh t qun th rng di c sang qun th vn thc vt tm thc n v ha nhp vo qun th sc trong vn thc vt.

a)Tnh tn s alen A v alen a ca qun th sc sau s di c c mong i l bao nhiu?

b) qun th sc vn thc vt sau s di c, gi s tn s t bin thun (Aa) gp 5 ln tn s t bin nghch (aA). Bit tn s t bin nghch l 10-5. Tnh tn s ca mi alen sau mt th h tip theo ca qun th sc ny.

c)Gi s tn s alen (a) ca qun th sc sng qun th rng l 0,2575 v 0,5625 qun th hn hp(sau nhp c), cho bit tc nhp c l 0,1. Tnh tn s ca alen (a) qun th sc vn thc vt ban u?

Ch : Cc kt qu tnh chnh xc ti 4 ch s thp phn sau du phy theo quy tc lm trn s ca n v tnh qui nh trong bi ton.

Ni dung gii S im

a) qun th vn thc vt s c th sc mang alen A l: 180 x 0,9=162 c th

qun th rng s c th sc mang alen A di c sang qun th vn thc vt l: 0,5x 60 = 30 c th.

Vy tng c th mang alen A ca qun th sc trong vn thc vt sau s di c l : 162 + 30 = 192 c th.

Tng s c th sc trong ng thc vt: 180 + 60 = 240 c th

Tn s alen A =

1920,8

240

, tn s alen a = 1- 0,8 = 0,2. b)pA = vq up = (10-5 x 0,2) (5.10-5 x 0,8) = -3,8.10-5 qa = up vq = (5.10-5 x 0,8) (10-5 x 0,2) = 3,8.10-5 Vy tn s ca alen A v alen a sau 1 th h l: pA=0,8 - 3,8.10-5 qa = 0,2 + 3,8.10-5

c) m = 0,1; qm = 0,2575; q = 0,5625.

Ta c phng trnh:

'( )

( )m

q qm

q q

'( ) (0,5625 0,1 0,2575)

0,5964(1 ) 1 0,1

mq mq xqm

Vy tn s alen (a) l: 0,5964

0,5

im

1 im

1 im

1 im

0,5

im

1 im

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Bi tp v mt gen c 3 alen Bi 15.Mu sc v c sn do mt gen c 3 alen kim sot: C1: nu, C2: hng, C3:

vng. Alen qui nh mu nu tri hon ton so vi 2 alen kia, alen qui nh mu hng tri hon ton so vi alen qui nh mu vng. iu tra mt qun th c sn ngi ta thu c cc s liu sau:Mu nu c 360 con; mu hng c 550 con; mu vng c 90 con. Bit qun th ny trng thi cn bng di truyn.

a. Hy xc nh kiu gen qui nh mi mu. b. Hy tnh tn s tng i ca cc alen trong qun th trn.

a. Cc kiu gen qui nh mi mu: C1C1, C1C2, C1C3: mu nu. C2C2, C2C3: mu hng.

C3C3: mu vng.

0,5

b. Gi p l tn s tng i ca alen C1, q l tn s tng i ca alen C2, r l tn s tng i ca alen C3.

Qun th cn bng c dng: (p+q+r)2 =

p2C1C1+q2C2C2+r2C3C3+2pqC1C2+2qrC2C3+2prC1C3

0,5

Tn s tng i mi loi kiu hnh: Nu = 360/1000= 0,36; Hng=550/1000=0,55; vng=90/1000=0,09.

0,5

Tn s tng i ca mi alen, ta c: Vng = 0,09 = r2 r=0,3.

Hng = 0,55=q2+2qr q=0,5

Nu = 0,35 = p2 + 2pq + 2pr p=0,2.

1

Bi 17:QT ngi c TL mu A chim 0,2125; mu B chim 0,4725; mu AB chim 0,2250; mu O chim 0,090. TS tng i ca mi alen l bao nhiu?

A)p(IA) = 0,25; q(IB) = 0,45; r(i) = 0,30 B)p(IA) = 0,35; q(IB) = 0,35; r(i) = 0,30

C)p(IA) = 0,15; q(IB) = 0,55; r(i) = 0,30 D)p(IA) = 0,45; q(IB) = 0,25; r(i) = 0,30

Gii : Gi : p(IA); q(IB), r(i) ln lt l TS tng i cc alen IA, IB, IO Ta c : p + q + r = 1 ( * ) Mu O chim 0,090 => r(i) = 0,30 TL mu A: IA IA + IA IO = 0,2125 => p2 + 2 pr = 0,2125

* p2 + 2 pr + r2 = ( p + r ) 2 = 0,2125 + 0,090 = 0, 3025 = ( 0,55 ) 2

( p + r ) 2 = ( 0,55 ) 2 => p + r = 0,55 => p = 0,55 0,30 = 0,25 T: ( * ) => q = 1 ( p + r ) = 1 - ( 0,25 + 0,30 ) = 0,45 Vy: TS tng i ca mi alen l : p(IA) = 0,25; q(IB) = 0,45; r(i) = 0,30 chn A

Bi 18: ngi, tnh trng nhm mu do 3 alen IA, IB v IO quy nh. Trong QT CBDT c 36% s ngi mang nhm mu O, 45% s ngi mang nhm A. V c nhm mu A ly chng c nhm mu B khng c quan h h hng vi nhau1/ Xc sut h sinh con mu O:

A. 11,11% B. 16,24% C.

18,46% D. 21,54%

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2: Nu h sinh a con u l trai mu O th kh nng sinh a con th 2 l gi c nhm mu khc b v m mnh l A. 44,44% B. 35,77% C. 42% D. 25%

Gi p, q, r ln lt l TS alen IA , IB , IO . V QT CB nn cu trc DT l: p2IAIA + q2IBIB +r2IOIO + 2pqIAIB + 2qrIBIO + 2prIAIO

T gt IA = 0,3 ; IB = 0,1 ; IO = 0,6

(A) p2IAIA + 2prIAIO x ( B) q2IBIB + 2qrIBIO (0,9) (0,36) (0,01) (0,12) TS IA = 3/5 ; IO = 2/5 IB =

7/13 ; IO = 6/13 1/ XS sinh con mu O = (2/5)(.6/13) = 12/65 (p n C) 2/ a con u mu O KG ca b,m: IAIO x IBIO do XS sinh con trai khc nhm mu b, m mnh = 1/2.1/2 = 25% (p n D)Bi5 ngi, gen quy nh nhm mu gm 3 alen: IA, IB, IO, trong IA v IB tri hon ton so vi IO, cn IA v IB ng tri. Qua nghin cu mt qun th ang trng thi cn bng di truyn xc nh c: t l ngi c nhm mu A chim 35%, nhm mu B chim 24%, nhm mu AB chim 40%, cn li l nhm mu O. a. Xc nh tn s tng i ca mi loi alen b. Mt ngi c nhm mu A kt hn vi mt ngi c nhm mu B. Tnh xc sut sinh con nhm mu O ca cp v chng ny. c. Nu ngha l lun v ngha thc tin ca nh lut Haci-Vanbe

a. Gi p l tn s tng i ca alen IA, q l tn s tng i ca alen IB, r l tn s tng i ca alen IO. Qun th cn bng c dng:

(p IA +q IB +r IO)2 = p2IA IA + q2 IBIB + r2 IOIO + 2pqIA IB + 2qrIBIO +

2prIAIO

Ngi nhm mu O chim 1%, r2 IOIO =1% r = 0,1. Ngi c nhm mu A chim 35% p2 + 2pr = 0,35, gii ra ta

c p=0,5, vy q=0,4.

0,25

0,25

0,5

b. Mt ngi c nhm mu A kt hn vi mt ngi c nhm mu B. h sinh con nhm mu O th kiu gen ca hai v chng ny phi l: IAIO x IBIO.

Xc sut ngi c nhm mu A c kiu gen IAIO l: prp

pr

2

22 =

0,285

Xc sut ngi c nhm mu B c kiu gen IBIO l: qrq

qr

2

22 =

0,333

M: IAIO x IBIO sinh ra con nhm mu O vi xc sut bng 0,25. Vy xc sut cn tm l: 0,285 x 0,333 x 0,25 = 0,0238

0,25

0,25

0,25

0,25

c. ngha l lun: - Phn nh trng thi cn bng di truyn ca qun th, gii thch v

sao trong thin nhin c nhng qun th c duy tr n nh qua thi gian di.

- y l nh lut c bn nghin cu di truyn hc qun th.

0,5

0,25

0,25

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ngha thc tin: Xc nh tn s tng i ca cc kiu gen v cc alen t t l cc kiu hnh.

Bi ton v hai hay nhiu cp gen Bi1: Mt QT ca 1 loi thc vt c TL cc KG trong QT nh sau: P: 0,35 AABb + 0,25 Aabb + 0,15 AaBB + 0,25 aaBb =1

Xc nh CTDT ca QT sau 5 th h giao phn ngu nhin Gii- Tch ring tng cp tnh trng, ta c: P : 0,35AA + 0,40Aa + 0,25aa = 1 v 0,15BB + 0,60Bb + 0,25bb = 1

- TST: A = 0,55 ; a = 0,45 B = 0,45 ; b = 0,55

TSKG F1 ,F2 ,F5 khng i v bng: 0,3025AA + 0,4950Aa + 0,2025aa = 1 0,2025BB + 0,4950Bb +

0,3025bb = 1

- Vy TSKG chung: (0,3025AA + 0,4950Aa + 0,2025aa)(0,2025BB + 0,4950Bb + 0,3025bb) = 1

= . (bn tnh gip TS 9 KG ny nh) Bi2: Cu trc DT ca mt QT nh sau: 0,2AABb : 0,2AaBb : 0,3aaBB :

0,3aabb. Nu QT trn giao phi t do th TL c th mang 2 cp gen H ln sau 1 th h l: A. 30% B. 5,25% C. 35% D. 12,25%

Gii:-Xt ring gen A: 0,2AA + 0,2Aa + 0,6aa -> A = 0,3 a=0,7 -> F1: 0,09AA+0,42Aa+0,49aa

- Xt ring gen B: 0,3BB +0,4Bb+0,3bb

-> B=0,5 b=0,5 -> F1: 0,25BB+0,50Bb+0,25bb

-Xt chung 2 gen: TL c th mang 2 cp gen H ln l: aabb=0,49 x 0,25=0.1225 = 12,25% Vy p n ng l D.

Bi ton xc sut trong di truyn hc qun th: Bi 1:Kh nng cun li ngi do gen tri trn NST thng qui nh, alen

ln Q ngi bnh thng. Mt ngi n ng c kh nng cun li ly ngi ph n khng c kh nng ny, bit xc sut gp ngi cun li trong QT ngi l 64%. Xc sut sinh a con trai b cun li l bao nhiu? GII: Ctrc DT tng qut ca QT: p2AA + 2pqAa + q2aa

Theo gt: q2 = 1- 64% = 36% q = 0,6 ; p = 0,4

Vy Ctrc DT ca QT l: 0,16AA + 0,48Aa + 0,36aa - Ngi v khng cun li c Kg (aa) TS a = 1 - Ngi chng b cun li c 1 trong 2 Kg: AA (0,16/0,64); Aa (0,48/0,64)

TS : A = (0,16 + 0,24)/0,64 = 0,4/0,64 = 0,625; a = 0,24/0,64 = 0,375

kh nng sinh con b cun li = 0,625 x 1 = 0,625 Vy XS sinh con trai b cun li = 0,625 x 1/2 = 0,3125 Bi2: ngi A-phn bit c mi v> a- ko phn bit c mi v. Nu trong

1 cng ng TS alen a=0,4 th xc sut ca mt cp v chng u phn bit c mi v c th sinh ra 3 con trong d 2 con trai phn bit c mi v v 1 con gi ko phn bit c mi v l?

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A.1,97% B.9,44% C.1,72% D.52%

cu trc DT ca Qt: p2AA + 2pqAa +q2aa v v chng phn bit (Bnh thng)) sinh con c phn bit v khng phn bit

mi v nn KG Aa x Aa vi XS = (2pq /p2+ 2pq)2 Xs sinh trai phn bit = 3/4.1/2 = 3/8 Xs sinh gi khng phn bit = 1/4.1/2 = 1/8 XS b m u bnh thng sinh 2 trai phn bit v 1 gi khng phn bit =3/8.3/8.1/8.C13.(2pq /p2+ 2pq)2 = 1,72%

Cu 3:Mt gen c 2 alen nm trn NST gii tnh X on khng tng ng vi Y, alen ln Q tnh trng bnh, alen tri Q tnh trng bnh thng. TL ngi b bnh trong QT ngi l 0,0208. Hai ngi bnh thng khng c quan h h hng kt hn vi nhau, cho rng QT c s CBDT v tnh trng trn. Xc sut sinh con b bnh ca cp v chng l

A. 1,92% B. 1,84% C. 0,96% D.

0,92%

Cu trc DT ca QT : p2/XAXA + 2pq/XAXa + q2/XaXa + p/XAY + q/XaY (p+q=1)

TS ngi b bnh = (q2 + q)/2 = 0,0208 q = 0,04 ; p = 0,96 XS 2 ngi bnh thng ly nhau sinh con bnh(m d hp) = 2pq/(p2 + 2pq) x 1 XS sinh con bnh = 2pq/(p2 + 2pq) x 1/4 = pq/2(p2 + 2pq) = 1,92% Cu 4: Chn 1 cu tr li ng nht trong cc phng n A,B,C,D Trong mt hn o bit lp c 5800 ngi sng, trong c 2800 nam gii.

Trong s ny c 196 nam b m mu xanh . Kiu m mu ny do 1 alen ln m nm trn NST gii tnh X. Kiu m mu ny khng nh hng n s thch nghi ca c th. Kh nng c t nht 1 ph n ca hn o ny b m mu xanh l bao nhiu?

A. 1 0,99513000 B. 0,073000 C. (0,07 x 5800)3000 D. 3000 x 0,0056 x 0,99442999

Bi gii: V y l o bit lp nn CTDT ca QT ny ang TTCB. XM l gen quy KH

bnh thng, Xm l gen quy nh bnh m mu lc, CTDT QT ny c dng:

Gii ci: p2 XMXM+2pq XMXm +q2 XmXm = 1 Gii c: p XMY+q XmY

+ Nam m mu c KG XmY chim TL q = 0,07 q2 XaXa =

0,0049 Xc sut 1 ngi n b bnh l 0,0049 Xc sut 1 ngi n khng b bnh l 1 0,0049 = 0,9951.

S lng n trn o l 5800-2800=3000 Xc sut c 3000 ngi n khng b bnh l (0,9951)3000. V bin c c t nht 1 ngi n b bnh l bin c i ca bin c c 3000 ngi

n u khng b bnh Xc sut c t nht 1 ngi n b bnh l:

0902651694

1 0,99513000 p n ng: A Cu5: Mt QT ngi c TS ngi b bnh bch tng l 1/10000. Gi s QT ny

CBDT.

-Hy tnh TS cc alen v TP cc KG ca QT. Bit rng, bnh bch tng l do mt gen ln nm trn NST thng quy nh.

Tnh xc sut 2 ngi bnh thng trong QT ny ly nhau sinh ra mt ngi con u lng b bnh bch tng.

Gii: Gi alen A quy nh tnh trng bnh thng, alen a quy nh bnh bch tng.

- pA l TS ca alen A, qa l TS ca alen a trong QT. - QT trng thi CBDT nn tha mn cng thc v TP KG sau: p2 AA + 2pq Aa

+ q2 aa = 1

- q2 aa = 1/10000 qa = 1/100. pA + qa = 1 pA = 1 1/100 = 99/100. TS KG AA = p2 =

(99/100)2

TS KG Aa = 2pq = 198/10000 TS KG aa = q2 = (1/100)

Ngi bnh thng c KG AA hoc Aa Hai ngi bnh thng ly nhau sinh ra ngi con b bnh bch tng th phi c

KG Aa.

TS ngi c KG d hp t (Aa) trong s nhng ngi bnh thng l: 2pq/ p2 + 2pq = 0,0198 / (0,9801 + 0,0198) = 0,0198/0,9999.

S lai P: Bnh thng x Bnh thng (0,9801/0,9999 AA + 0,0198/0,9999 Aa) (0,9801/0,9999 AA + 0,0198/0,9999

Aa)

TS cc alen : 0,0198/(0,9999x2) a 0,0198 /(0,9999x2) a

F1 : (0,0198/0,9999)2/4 (0,0198)2/4 aa Nh vy, xc sut sinh ngi con b bnh tng l (0,0198)2/4 Cu6: Mt QT ngi trn mt hn o c 100 ph n v 100 ngi n ng

trong c 4 ngi n ng b bnh mu kh ng. Bit rng bnh mu kh ng do gen ln nm trn NST gii tnh X khng c alen trn Y, QT trng thi CBDT. TS ph n bnh thng nhng mang gen gy bnh l

A. 0.0384. B. 0.0768. C. 0.2408. D. 0.1204.

Gii: Quy c: A bnh thng; a b bnh mu kh ng. Vi gen nm trn vng khng tng ng ca NST X th fXA, fXa nam v n

bng nhau. - gii nam: Ta c: 0,96 XAY : 0,04 XaY => fXa = 4/100 = 0,04; fXA = 1 -

0,04 = 0,96

- gii n: 0,962 XAXA : 2.0,96.0,04 XAXa : 0,042 XaXa => TS ph n bnh thng nhng mang gen gy bnh (XAXa) gii n l:

2.0,96.0,04/2 = 0,0768

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Vy, TS ph n bnh thng nhng mang gen gy bnh (XAXa) trong QT ngi l: 2.0,96.0,04/2 = 0,0384

Bi6:: Mt QT TTCB v 1 gen gm 2 alen A v a, trong P(A) = 0,4. Nu qutrnh chn lc o thi nhng c th c KG aa xy ra vi p lc S = 0,02. CTDT ca QT sau khi xy ra p lc chn lc:

A. 0,1612 AA: 0,4835 Aa: 0,3551 aa B. 0,16 AA: 0,48 Aa: 0,36 aa

C. 0,1613 AA: 0,4830 Aa: 0,3455 aa D. 0,1610 AA: 0,4875 Aa: 0,3513 aa

Gii Cu trc qun th AA=0,4^2=0,16 Aa=2*0,4*0,6=0,48

aa=0,6^2=0,36

qu trnh chn lc o thi nhng c th c KG aa xy ra vi p lc S = 0,02 aa=0,36-0,02*0,36=0,3528

cu trc qun th l aa=0,3528/(0,16+0,48+0,3528)=0,3551 Bi7:Cho CTDT ca QT nh sau: 0,4 AABb: 0,4 AaBb: 0,2 aabb. Ngi ta tin

hnh cho qu trnh trn t th phn bt buc qua 3 th h. TL c th mang hai cp gen H tri l.

A. 112

640 B.

161

640 C.

49

256 D.

7

640

- AABb x AABb ----> AABB = 0,4 x 1(AA) x [1/2(1-1/23)] BB = 7/40

- AaBb x AaBb -----> AABB = 0,4 x [1/2(1-1/23)] (AA) x [1/2(1-1/23)] BB

=49/640

----> Tng TL KG 2 cp H tri khi cho t th phn 3 th h : 7/40+49/640 = 161/640

Bi8: Mt QT c TS KG ban u: 0,4AA : 0,1aa : 0,5Aa. Bit rng cc c th d hp t ch c kh nng sinh sn bng 1/2 so vi kh nng sinh sn ca cc c th H t. Cc c th c KG AA v aa c kh nng sinh sn nh nhau. Sau mt th h t th phn th TS c th c KG d hp t s l:

A. 16,67% B. 12,25% C. 25,33%

D.15.20%

P: 0,4AA + 0,5Aa +0,1aa

Gi N l s c th sinh ra mi th h t kg d hp 2N l s c th sinh ra mi th h t kg HSau 1 th h t th ta c: Aa = N. 0,5.1/2 = 0,25N AA + aa = 2N. (0,4+0,1) +(0,5N- 0,25N)= 1,25N

TS kg Aa = 0,25/1,25 = 16,67% (A) Bi9: ngi, gen ln gy bnh bch tng nm trn nhim sc th thng, alen t

ri tng ng quynh da bnh thng. Gi s trong QT ngi, c trong 100 ngi da bnh thng th c mt ngi mang gen bch tng. Mt cp v chng c da bnh thng, xc sut sinh con bnh thng ca h l

A. 0,005%. B. 0,9925%. C. 0,0075%. D. 0,

9975%

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Gii: Cch gii bi ny gn nht nn tnh XS v chng bnh thng sinh con b bnh, sau tr ra ta c XS sinh con bnh thng:

Trong cc trng hp v v chng bthng ch c trng hp c cng KG Aa mi sinh con bnh.

- XS mt ngi trong QT nhng ngi bnh thng c KG : Aa = 1/100 XS c v v chng u c KG: Aa x Aa (=1/100 . 1/100 = 1/10.000) SDL: Aa x Aa 3/4 bthng / 1/4 bnh

XS sinh ngi con bnh = 1/4 .1/10000 = 0,0025% Vy XS sinh con bthng = 1 0,0025% = 0,9975% (p n ng l D) Bi 10: ngi, tnh trng nhm mu ABO do mt gen c 3 alen IA, IB, IO qui

nh. Trong mt qun th cn bng di truyn c 25% s ngi mang nhm mu O; 39% s ngi mang nhm mu B. Mt cp v chng u c nhm mu A sinh mt ngi con, xc sut ngi con ny mang nhm mu ging b m l bao nhiu?

A. 3/4. B. 119/144. C. 25/144. D. 19/24.

p n B. Tn s Io=0,5 ; IB = 0,3 ; IA = 0,2 T l IAIA trong qun th l : 0,04 ; IAIO = 2.0,5.0,2=0,2 T l IAIA trong s ngi nhm mu A = 0,04/(0,04+0,20) = 1/6 T l IAIo trong s ngi nhm mu A = 5/6

(OAAA II

6

5:II

6

1

) x (OAAA II

6

5:II

6

1

). Con mu O ch to ra t b m IAIo .

Con mu O c t l =

2

6

5

x 41

= Con ging b m = 1 - 14425

= 144119

Bi 11: Trong mt qun th ng vt c v, tnh trng mu lng do mt gen quy nh, ang trng thi cn bng di truyn. Trong , tnh trng lng mu nu do alen ln a quy nh, lng vng do alen A quy nh. Ngi ta tm thy 40% con c v 16% con ci c lng mu nu.Hy xc nh tn s tng i cc alen trong qun th ni trn?

Do qun th t trng thi cn bng di truyn m s biu hin tnh trng khng ng u 2 gii nn tnh trng mu lng do gene gm 2 allele nm trn NST gii tnh quy nh.

Xt 2 trng hp: *Nu XY l c, XX l ci: - Xt gii XX: C faa=0,16. Do qun th t trng thi cn bng nn fa=0,4,

fA=0,6

- Xt gii XY: C fa=0,4 -> fA=0,6 Vy qun th c cu trc (0,6A:0,4a)(0,6A:0,4a)=0,36AA:0,48Aa:0,16aa

-> fA=0,36+0,48/2=0,6; fa=0,4

*Nu XY l ci, XX l c:

- Xt gii XX: C faa=0,4. Do qun th t trng thi cn bng nn fa=

, fA=1-

Xt gii XY: C fa=0,16 -> fA=0,84

0902651694

Vy qun th c cu trc: ( 1- A: a)(0,16A: 0,84a) -> cha cn bng.(Loi)

Kt lun: fA=0,6; fa=0,4 III,Phng php gii bi tp lai : A. MT S KHI NIM, QUY C K HIU V THUT NG 1. Tnh trng: L c im v hnh thi, cu to, sinh l ring ca mt c th

no m c th lm du hiu phn bit vi c th khc. V d: Hoa mu tm, hoa mu trng, qu mu , ht mu vng

2. Alen: L trng thi khc nhau ca cng mt gen. Mi alen chim mt v tr xc nh trn mt cp nhim sc th tng ng gi l lcut.

V d: c chua lng bi c qu mu l tri so vi qu mu vng. C 2 alen: A qu mu ; a qu mu vng. C th dng k hiu khc l

B,C,D ch alen tri; b, c, d ch alen ln tng ng. 3. Cp alen: L 2 alen ging nhau hay khc nhau thuc cng mt gen nm trn

1 cp NST tng ng v tr tng ng trong b NST. V d: AA, Aa, aa. Cp gen ng hp: 2 alen c cu trc ging nhau. V d : AA. aa. BB, bb. Cp gen d hp : 2 alen c cu trc khc nhau. V d : Aa, Bb. 4. Kiu gen: T hp ton b cc gen trong t bo ca c th thuc 1 loi sinh

vt. 5. Kiu hnh: Tp hp ton b cc tnh trng ca c th. Kiu hnh thay i

theo giai on pht trin v iu kin mi trng. Trong thc t khi cp ti ging thun chng thng ch cp ti mt hay mt vi tnh trng.

6. Th ng hp: L c th mang 2 alen ging nhau thuc cng 1 gen, c gi l dng thun. V d: AA, aa, BB, bb, DD, dd.

7. Th d hp: L c th mang 2 alen khc nhau thuc cng 1 gen. V d: Aa, Bb, Dd. 8. Tnh trng tri: Tnh trng biu hin khi c kiu gen dng ng hp t

hoc d hp t. Thc t c tri hon ton v tri khng hon ton. 9. Tnh trng ln: Tnh trng ch xut hin khi kiu gen trng thi ng hp

ln.

B. LAI MT CP TNH TRNG (Mi gen quy nh mt tnh trng , quy

lut phn li)

Trng hp1. BIT TNH TRNG CA B M

XC NH ALEN TRI, LN, KIU GEN, KIU HNH Cc kin thc c bn .

- Quy c gen. - K hiu: P : Th h xut pht. F1, F2 : Th h th nht, th h th

hai - Th ng hp, th d hp, thun chng.

0902651694

- C ch gim phn hnh thnh giao t. - S lai .

Bi 1: c chua tnh trng qu mu l tri hon ton so vi qu mu vng l ln.

Vit k hiu alen quy nh qu mu v alen quy nh qu mu vng? Xc nh kiu gen cc cy c chua c th c?

Gii 1. Gi A l gen quy nh qu mu . Ta c: A: . a l alen quy ng qu mu vng. Ta c: a : vng. 2. Kiu gen ca cc cy c chua c th c: AA : qu mu . (ng hp tri) Aa : qu mu . (d hp) aa : qu mu vng. (ng hp ln) Qua bi ny cn ch th ng hp v d hp. Bit kiu hnh tnh trng ln

th c th suy ra kiu gen ca n. Bi 2: Lai c chua thun chng qu mu v c chua qu vng. Cho bit mu

tri hon ton so vi mu vng. Xc nh kiu gen ca cy b m em lai? Cho bit s loi giao t ca cc cy c chua trn? Xc nh kiu gen v kiu hnh F1? Gii 1. B m u thun chng nn c kiu gen l: C chua qu : AA. C chua qu vng : aa. 2. Kiu gen AA ch cho 1 loi giao t l A. Kiu gen aa ch cho 1 loi giao t l a. 3. T l kiu gen v kiu hnh F1 l: S lai: P : AA x aa GP : A a

F1 : Aa

KG: 100% Aa. KH: 100% qu mu . Bi 3: Cho c chua c qu mu l tri hon ton so vi qu mu vng. Xc

nh t l kiu gen, kiu hnh ca php lai sau F1? AA x AA.

AA x Aa.

Aa x Aa.

Aa x aa.

Gii a. P: AA x AA

GP: A A

F1 AA

KG: 100% AA. KH: 100% qu mu .

0902651694

b. P: AA x Aa

GP: A (A : a)

F1 : AA : Aa

KG: 50% AA : 50% Aa. KH: 100% qu mu . c. P: Aa x Aa

GP: (A : a); (A : a)

F1: Lp bng ta c

Giao

t A a

A AA Aa

a Aa aa

KG: 25% AA : 50% Aa : 25% aa. KH: 75% qu mu : 25% qu mu vng

d. P: Aa x aa

GP: (A : a) a

F1: Aa : aa

KG: 50% Aa : 50% aa. KH: 50% qu mu : 50% qu mu vng.

BNG GHI NH

STT PHP

LAI

K IU G EN F 1 K IU H NH F 1

1 AA x

AA

100% AA 100% q u

2 AA x

Aa

50% AA: 50 % Aa 100% q u

3 AA x

aa

100% Aa 100% qu

4 Aa x

Aa

25% AA:5 0% Aa:2 5%a a

Hay 1 : 2 : 1

75%q u :2 5%q u vn g

Hay 3 : 1

5

Aa x

aa

La i

phn

t ch

50% Aa:5 0%a a

Hay 1 : 1

50%q u :5 0%q u vn g

Hay 1 : 1

6 aa x aa 100% aa 100% qu vn g

0902651694

Trng hp 2. BIT KIU HNH, KIU GEN CA F1

XC NH KIU GEN, KIU HNH CA P

Cc kin thc c bn .

- C th mang kiu hnh ln c th suy ra kiu gen. - Phi nm chc bng ghi nh.

Bi 1: u H Lan, gen A quy nh qu mu vng tri hon ton so vi alen a quy nh qu mu xanh. Cho lai u H Lan c qu mu vng vi nhau, thu c t l kiu hnh l 3 vng : 1 xanh.

Xc nh kiu gen ca b m em lai. Nu kt qu cho t l 1 qu vng : 1 qu xanh th kiu gen v kiu hnh ca b m

phi nh th no?

Gii 1. Xc nh kiu gen ca b m: P: A x A

F1 3 vng ( A) : 1 xanh ( aa) F1 thu c cy c qu mu xanh => kiu gen l aa. 1 giao t a l ca b v 1 giao t a l ca m => kiu gen ca b m phi l Aa

v Aa.

2. Kt qu c t l 1 vng : 1 xanh => lai phn tch => Kiu gen, kiu hnh ca P l Aa , qu mu vng v aa qu mu xanh. ( xem bng ghi nh).

Bi 2: ngi, gen A quy nh mt en tri hon ton so vi gen a quy nh mt xanh. con sinh ra c ngi mt en, c ngi mt xanh th b m phi c kiu gen nh th no?

Gii Ngi mt en c kiu gen l A-. Ngi mt xanh c kiu gen l aa. C 2 trng hp: 1. P: A- x A-

F1 A- v aa

Kiu hnh mt xanh => kiu gen aa. Giao t a 1 l ca b v 1 l ca m. T kt qu F1 ta xc nh kiu gen P l Aa x Aa. 2. P: A- x aa.

F1: mt en v mt xanh

0902651694

Kiu hnh mt xanh => kiu gen aa. Giao t a 1 l ca b v 1 l ca m.=> Kiu gen A- P phi l Aa.

Kiu gen ca b m l Aa x aa.

C. LAI HAI V NHIU CP TNH TRNG (Mi gen quy nh mt tnh trng nm trn NSTthng, phn li c lp)

Trng hp 1. BIT KIU GEN CA B M XC NH T L GIAO T

Cc kin thc c bn .

- Giao t ch mang 1 alen i vi mi cp - gi n l s cp d hp, s loi giao t tun theo cng thc 2n loi.

Bi 1: Cho 2 cp gen nm trn hai cp NST tng ng khc nhau Hy xc nh t l giao t v vit cc lai giao t ca cc c th c kiu gen su

y: a. AABB

b. aaBB.

c. AAbb.

d. Aabb.

e. AaBb.

2. Xc nh kiu gen c th c ca cc c th trong cc trng hp : a. c th ch cho 1 loi giao t b. c th ch cho 2 loi giao t. c. c th ch cho 4 loi giao t. Gii: a. C th c kiu gen AABB ch to 1 loi giao t mang gen AB. ( 20 = 1). b. C th c kiu gen aaBB to ra 1 loi giao t aB. ( 20 = 1) c. C th c kiu gen AAbb to ra 1 loi giao t Ab. ( 20 = 1).. d. C th c kiu gen Aabb to ra 2 loi giao t Ab v ab. ( 21 = 2). e. C th c kiu gen AaBb to ra 4 loi giao t. ( 22 = 4). B AB c t l 1/4 ( 25%) A

b Ab c t l 1/4 ( 25%)

B aB c t l 1/4 ( 25%) a

b ab c t l 1/4 ( 25%). Gi n s cp gen d hp a. C th ch cho 1 loi giao t nn 2n = 1 n = 0 , c th ny khng mang

kiu gen d hp ch mang 2 kiu gen ng hp, do kiu gen c th l AABB hoc AAbb hoc aaBB hoc aabb

0902651694

b. C th cho 2 loi giao t nn 2n = 2 n = 1 do c th ny c 1 cp gen d hp, 1 cp gen ng hp. Kiu gen c th l AaBB hoc AABb, Aabb, aaBb.

c. C th cho 4 loi giao t, nn 2n = 4 n =2 do c th ny c 2 cp gen d hp, kiu gen l AaBb.

Bi 2: Xc nh s loi giao t v t l cc giao t trng hp cc gen phn li c lp, t hp t do, gim phn bnh thng cc c th c kiu gen sau:

AaBBDd.

AABbDd.

aaBbDdEe.

AaBbDd.

Gii 1. C th c kiu gen AaBBDd c 2 cp gen d hp nn ta c: 22 = 4 loi giao

t. Mi loi gt chim 1/4 2. C th c kiu gen AABbDd c 2 cp gen d hp nn ta c: 22 = 4 loi giao

t. Mi loi gt chim 1/4 3. C th c kiu gen aaBbDdEe c 3 cp gen d hp nn ta c: 23 = 8 loi giao

t. Mi loi gt chim 1/8 4. C th c kiu gen AaBbDd c 3 cp gen d hp nn ta c: 23 = 8 loi giao

t. Mi loi gt chim 1/8

Trng hp 2. BIT KIU GEN CA B M XC NH S T HP CA PHP LAI F1 V NGC LI

Cc kin thc c bn .

- Giao t ch mang 1 alen i vi mi cp - gi n1 l s cp d hp b, n2 l s cp d hp ca m, s t hp ca php

lai tun theo cng thc : S t hp lai = 2n1 x 2n2

Bi 1: Xc nh s t hp ca cc php lai sau: AaBB x AaBb.

AaBb x AABb.

AaBbDd x aaBbDd.

AaBbDd x AaBbDd

Gii n1 = 1. n2 = 2 . Ta c 21 x 22 = 2x4 = 8. Php lai AaBB x AaBb c 8 t hp. n1 = 2, n2 = 1. Ta c 22 x 21 = 4x2 = 8. Php lai AaBb x AABb c 8 t hp. n1 = 3, n2 = 2. Ta c 23 x 22 = 8x4 = 32 . Php lai AaBbDd x aaBbDd c 32 t

hp. n1 = 3, n2 = 3. Ta c 23 x 23 = 8x8 = 64. Php lai AaBbDd x AaBbDd c 64 t

hp. Bi 2: Gen A l tri so vi alen a l ln. B l tri so vi b l ln. Cc gen trn

nm trn cc nhim sc th khc nhau, trong qu trnh gim phn phn li bnh thng. Xc nh kiu gen ca b, m c th c trong cc trng hp sau:

C 8 t hp c hnh thnh.

0902651694

C 16 t hp c hnh thnh. Gii 1. C 8 t hp c hnh thnh => b cho 2 loi giao t, m cho 4 loi giao

t hoc ngc li. a. Trng hp 1: b cho 2 loi giao t, m cho 4 loi giao t. b cho 2 loi giao t => b ch c 1 cp d hp =>AaBB, AABb, aaBb,Aabb. m cho 4 loi giao t => m phi c 2 cp d hp => AaBb. b. Trng hp 2: b cho 4 loi giao t, m cho 2 loi giao t. b cho 4 loi giao t => b phi c 2 cp d hp => AaBb. m cho 2 loi giao t => m ch c 1 cp d hp =>AaBB, AABb, aaBb,Aabb.

Trng hp 3. BIT GEN TRI, LN, KIU GEN CA P XC NH KT QU LAI

Cch gii: + Quy c gen +Xc nh t l giao t ca P +Lp s lait l kiu gen, t l kiu hnh (c th dng php nhn xc

sut hoc s phn nhnh)

Bi 1: 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.

Cho cy thn cao, qu thun chng lai vi cy thn thp, qu trng. Xc nh t l kiu gen, kiu hnh F1?

Cho cy F1 t th phn, xc nh kiu gen, kiu hnh F2?

Gii Cy thn cao, qu thun chng nn c kiu gen l AABB. Cy thn thp, qu trng c kiu gen aabb ( mang tnh trng ln c 2 gen). Lp s lai Pt/c : AABB x aabb

GtP : AB ab

F1 : AaBb

Kt qu: Kiu gen : 100% AaBb. Kiu hnh : 100% thn cao, qu . 2. F1 x F1 : AaBb x AaBb

GtF1 : ( AB:Ab:Ab:ab) ( AB:Ab:Ab:ab)

Lp bng : Xc nh c t l kiu gen v t suy ra t l kiu hnh F2

Giao

t

AB Ab aB ab

A

B

1/16

AA

BB

1/16

AA

Bb

1/16

Aa

BB

1/16

Aa

Bb

0902651694

A

b

1/16

AA

Bb

1/16

Aab

b

1/16

Aa

Bb

1/16

Aa

bb

aB 1/16

AaB

B

1/16

AaB

b

1/16

Aa

BB

1/16

aaB

b

ab 1/16

AaB

b

1/16

Aab

b

1/16

aaB

b

1/16

aab

b

T l kiu hnh 9/16 thn cao, qu : 3/16 thn cao, qu trng : 3/16 thn thp, qu : 1/16 thn thp, qu trng. ( 9:3:3:1).

Bi 2: Trong trng hp mt gen quy nh mt tnh trng, gen tri l tri hon ton, cc gen phn li c lp, t hp t do. Xc nh t l phn li kiu gen, kiu hnh trong cc php lai sau:

AaBb x aaBb.

AaBb x aabb.

Gii 1. P1 : AaBb x aaBb

GP1 : (AB:Ab:aB:ab) (Ab:ab)

F1 :

Giao

t

AB Ab aB ab

A

b

1/8

AA

Bb

1/8

AA

bb

1/8

Aa

Bb

1/8

Aa

bb

ab 1/8

AaB

b

1/8

Aab

b

1/8

aaB

b

1/8

aab

b

T l kiu hnh 3 (A-B-) : 3 (A-bb) : 1 (aaB-) : 1 (aabb) = 3 : 3 : 1 : 1 2. P2 : AaBb x aabb

GP2 : (AB:Ab:aB:ab) (ab)

F1 :

Giao

t

AB Ab aB ab

ab 1/4

AaB

b

1/4

Aab

b

1/4

aaB

b

1/4

aab

b

0902651694

T l kiu hnh 1 (A-B-) : 1 (A-bb) : 1 (aaB-) : 1 (aabb) = ( 1 : 1 : 1 : 1 ) y l php lai phn tch.

Trng hp 4. T T HP CA PHP LAI XC NH KIU GEN CA P

Cch gii: + Quy c gen + T t hp lai Xc nh t l giao t ca P Kiu gen, kiu hnh ca P.

Bi 1: 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.

Xc nh kiu gen ca cc cy b m ? Nu thu c 25% cy thp, qu trng th kiu gen ca cy b, m phi nh th

no?

Gii Kiu hnh thn thp, qu trng kiu gen l aabb chim 1/16 php lai cho 16

t hp b v m phi c 2 cp gen d hp. Kiu gen ca cc cy b m l AaBb x AaBb.

Kiu hnh thn thp, qu trng chim 25% php lai cho 4 t hp cy b c 2 cp d hp ( cho 4 loi giao t) v cy m phi ng hp ln c 2 cp gen ( ch cho 1 loi giao t) hoc ngc li. y l php lai phn tch.

Kiu gen ca cc cy b m l AaBb x aabb.

Trng hp 5. T KIU GEN, KIU HNH CA P

XC NH T L KIU GEN, KIU HNH F1

Cch gii: + Quy c gen + S dng quy lut xc sut . p dng bng ghi nh

Bi 1: Trong php lai gia hai c th c kiu gen sau y:

AaBbCcDd x aaBbccDd ( cho bit cc cp gen quy nh cc tnh trng khc nhau nm trn cc cp NST

tng ng khc nhau). T l i con c kiu gen ging b? T l i con c kiu hnh ging m?

Gii 1. T l i con c kiu gen ging b:

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Xt tng cp gen, T l kiu gen Aa l 1/2, kiu gen Bb l 2/4, kiu gen Cc l 1/2, kiu gen Dd l 2/4

Ta c: 1/2x2/4x1/2x2/4 = 1/16 = 6,25%

2. T l i con c kiu hnh ging m: Xt tng cp gen, ta c: 1/2x3/4x1/2x3/4 = 9/64. Bi 2: Bit mt gen quy nh mt tnh trng, gen tri l tri hon ton, cc gen

phn li c lp v t hp t do. Theo l thuyt, xc nh kiu hnh tri v c ba cp tnh trng F1 trong php lai AaBbDd x AaBbdd

Gii T l kiu hnh tri v gen A l 3/4, v gen B l 3/4 , v gen D l 1/2. Do vy t l F1 c kiu hnh tri c 3 tnh trng l 3/4 x 3/4 x 1/2 = 9/32 Bi 3: Bnh bch tng v phninkt niu l do gen ln nm trn NST thng

qui nh, phn li c lp v t hp t do. Nu mt cp v chng u d hp v c 2 tnh trng ny th con ca h c kh nng mc bnh l bao nhiu phn trm?

Gii Gi gen A khng b bnh bch tng. Alen a b bnh bch tng. Gi gen B khng b bnh phninkt niu. Alen b b bnh phninkt niu. V chng u d hp v c 2 tnh trng ny nn u c kiu gen AaBb

P : AaBb x AaBb T l kiu hnh ln ( aa) l 1/4, t l kiu hnh ln bb l 1/4. Do vy t l F1 c kiu hnh ln c 2 tnh trng l1/4 x 1/4 = 1/16 = 6,25%. IV, phng php lm bi tp xc sut:

a. S lc v di truyn hc ngi . * Nhng bnh tt : + Lin quan n gen tri trn nhim sc th thng: ln + Lin quan n gen ln trn nhim sc th thng : bch tng, phnikt niu. +Lin quan n gen ng tri v tri ln: Nhm mu + Lin quan n gen ln trn nhim sc th X: m mu, teo c, mu kh ng * Nhng bi ton v tnh xc sut thng l nhng bi ton do mt gen qui nh,

tri ln hon ton, do vy n tun theo cc qui lut di truyn thng thng. * Cc em nm vng s phn li kiu gen, kiu hnh trong hai php lai c bn sau: + Mt gen qui nh mt tnh trng tri ln hon ton theo qui lut ca Menen: ** P t/c : AA x aa F: 100% kiu hnh tri v kiu gen Aa ** P t/c : AA x AA F: 100% kiu hnh tri v kiu gen AA ** P t/c : aa x aa F: 100% kiu hnh ln v kiu gen aa

** P : Aa x Aa F: 3

4 kiu hnh tri :

1

4 kiu hnh ln

v kiu gen l 1

4 AA :

1

2 Aa :

1

4 aa

** P t/c : Aa x aa F: 50% kiu hnh tri : 50% kiu hnh ln v kiu gen 50%Aa : 50% aa.

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* Gen ln qui nh tnh trng thng nm trn nhim sc th X khng c vng tng ng trn Y.

** P : XAXa x XAY F: KH : 3

4 Tri :

1

4 Ln ( Ch Nam mi c)

KG: 1

4 XAXA :

1

4 XAXa :

1

4 XAY :

1

4 XaY

** P : XAXa x XaY F: KH : 1

2 Tri :

1

2 Ln ( Ch Nam mi c)

KG: 1

4 XAXa :

1

4 XaXa :

1

4 XAY :

1

4 XaY

Khi lm bi ton v xc sut t nht cc em phi xc nh c kiu gen ca b m v xem n thuc trng hp no xc nh kiu hnh i con.

b. Nhng trng hp c th *Phng php gii Cc thy c v cc em c th tin hnh theo ba bc sau: Bc 1: Xc nh s xut hin kiu gen, kiu hnh i con nh th no. Bc 2: Tnh xc sut l con trai hay con gi theo yu cu ca u bi. Bc 3: Nhn cc xc sut tho mn yu cu bi ton c p s. * Mt vi v d v mt gen qui nh mt tnh trng v gen nm trn NST

thng. V d 1: ngi, bnh phnin kt niu do t bin gen gen ln nm trn NST

thng. B v m bnh thng sinh a con gi u lng b bnh phnin kt niu. Xc sut h sinh a con tip theo l trai khng b bnh trn l:

Bc1: *B m bnh thng sinh con u lng b bnh phnin kt niu c ngha l b m mng gen bnh trng thi d hp.

* Qui c A: bnh thng; a: bnh phnin kt niu. Kiu gen ca b m l Aa x Aa. Ta c: P: Aa x Aa

G A, a A, a

F : KG: 1

4 AA :

1

2 Aa :

1

4 aa.

KH: 3

4 bnh thng :

1

4 b bnh.

Bc 2: Xc sut sinh con trai l : 1

2

V sinh con trai hay con gi xc sut l :50% con trai : 50% con gi.

Bc 3: l con trai b bnh l: 3

4 .

1

2 =

3

8 .

Chn p n 3

8.

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V d 2 : Mt cp v chng c nhm mu A v u c kiu gen d hp v nhm mu. Nu h sinh hai a con th xc sut mt a c nhm mu A v mt a c nhm mu O l:

Bc1: Kiu gen ca b m l: IAIo x IAIa. * P : IAIo x IAIa

G : IA ; Io IA ; Io

F : KG 1

4 IAIA :

1

2 IAIo :

1

4 IoIo.

KH: 3

4 n