HNG DN CHM CHNH THC THPT

Bi 01: Khi lng phn t ca 1 prtin hon chnh (cu trc bc 1) l 21780 vC a. Xc nh s lng nucltt cc loi ca mARN quy nh tng hp prtin trn. Bit rng trong mARN ny c U = 120 v s nucltt cc loi trong mch khun ca gen tng hp ra phn t mARN trn theo th t(T:A:G:X) tng ng vi t l (1:2:3:4)

b. Xc nh chiu di, khi lng phn t, chu k xon v t l phn trm nucltt cc loi ca gen tng hp prtin trn ?

CCH GII KT QU IM

* Cch gii thng thng: a. S lng mARN tham gia tng hp prtin: - s aa trong phn t pr: 21780/110=198 - S tARN tham gia l: 198+1=199 * S nu trn mARN - Goi mch 1 l mch m gc ca gen ta c: T1=Am, G1=Xm, A1= Um, X1=Gm

Theo gi thuyt T1:A1:G1:X1= Am: Um: Xm: Gm= 1:2:3:4

M Um =120 ->Am/Um=1/2

Um/Xm =2/3

Um/Gm=2/4

- > Am=120/2=60 - Xm=3x120/2=180 Gm=2x120=240

b. Chiu di: - s lng nu trn mARN = Am+Um+Xm+Gm = 60+120+180+240 = 600

- Chiu di ca gen: L= 600x3.4AO = 2040 AO

S nu ca gen l 600x2=1200 Vy khi lng ca gen l: 1200x300vC = 360000vC Theo NTBS

Am = 60 =T1=A2

Um=120=A1=T2

Xm=180=G1=X2

Gm=240=X1=G2

Am=60

Xm=180

Gm=240

Um = 120

L= 2040 AO

M= 360000vC

%A =%T = 15%

%G = %X = 35%

Chu k xon = 60

-> A=A1+A2= 120+60=180 =T

G=G1+G2= 180+240= 420 = X

%A =%T=(180x100%)/1200 = 15%

%G = %X = 50% - 15% = 35%

Chu k xon = 1200/20= 60

Bi 2: Mt on phn t ADN cha 2 gen - Gen th nht di 0,51 micrmet c t l tng loi nucltt trn mch n th nht nh sau: A:T:G:X = 1:2:3:4 - Gen th hai bng na chiu di gen th nht v c s lng tng loi

nucltt trn mch n th hai l: A432

XGT

a. Tnh s lng v t l % tng loi nucltt trn mi mch n ca tng gen?

b. Tnh s lng v t l % tng loi nucltt trn ADN?

CCH GII KT QU IM

a. s lng v t l % tng loi nu trn mi mch n ca tng gen: - Trn gen th nht: s lng nu = 0,51x10000A0/3.4= 1500 nu T2=A1= 1500/(1+2+3+4)= 150 = 10%

A2=T1= (1500x2)/(1+2+3+4)= 300= 20%

X2=G1=(1500x3)/(1+2+3+4)= 450= 30%

G2=X1=(1500x4)/(1+2+3+4)= 600= 40%

Trn gen 2:

Chiu di ca gen 2= gen 1. Nn c s Nu l: 1500/2 = 750.

A2=T2/2=G2/3=X2/4 =A2+2A2+3A2+4A2= 750

->10A2=750 ->A2=750/10 = 75

T1=A2=75=10%

A1=T2=75x2= 150=20%

X1=G2=75x3=225=30%

G1=X2=75x4=300=40%

b. S lng v % tng loi nu ca ADN: A=T=150+300+75+150=675

G=X=450+600+225+300=1575

%A=%T= %15%1002)1575675(

675x

x

%G=%X= 50% -15% = 35%

- Trn gen th nht: s lng nu T2=A1=150 = 10%

A2=T1= 300= 20%

X2=G1= 450= 30%

G2=X1= 600= 40%

Trn gen 2:

A2=T1=750=10%

A1=T2=150=20%

X1=G2=225=30%

G1=X2=300=40%

b. S lng v % tng loi nu ca ADN: A=T=675

G=X=1575

%A=%T=15%

%G=%X=35%

Bi 3: Ba hp t ca mt loi, lc cha nhn i s lng NST n trong mi t bo bng 20. Hp t 1 c s t nguyn phn bng 1/4 s ln nguyn phn ca hp t 2. Hp t 2 c s t nguyn phn lin tip bng 50% s t nguyn phn ca hp t 3. S lng NST n lc cha nhn i trong tt c cc t bo con sinh ra t 3 hp t bng 5480. a. Tnh s t nguyn phn lin tip ca mi hp t ? b. S lng NST n c to ra t nguyn liu mi trng t bo cung cp cho mi hp t nguyn phn l bao nhiu

CCH GII KT QU IM

. S ln nguyn phn ca mi hp t: - Gi s ln nguyn phn ca hp t 1 l n th s ln nguyn phn ca hp t 2 l 4n, hp t 3 l 8n

- S tb con c to ra do 3 hp t s l:

20

5480 =274

ta c 2n +2

4n +2

8n =274

gii ra ta c n = 1 vy s ln nguyn phn ca ht 1 l 1, hp t 2 l 4, ht 3 l 8 b. S NST n do mt cung cp - hp t 1= (21-1)x20=20 - hp t 2= (24-1)x20=300 - hp t 1= (28-1)x20=5100

Bi 4: 10 t bo sinh dc ca mt c th nguyn phn lin tip mt s t i hi mi trng ni bo cung cp nguyn liu to ra 2480 NST n mi tng ng. Cc t bo con u chuyn qua vng sinh trng bc vo vng chn, gim phn to nn cc giao t, mi trng t bo cung cp thm nguyn liu to nn 2560 NST n. Hiu sut th tinh ca giao t l 10% nn to ra 128 hp t lng bi a. Tm b NST lng bi ca loi ? b. Xc nh gii tnh ca cc c th to nn cc giao t trn ? c. Hp t c chia thnh 2 nhm A v B c s lng bng nhau. Mi hp

t trong nhm A c s t nguyn phn gp 2 ln s t nguyn phn ca nhm B. Cc hp t trong mi nhm c s t nguyn phn bng

nhau. Tng s NST n c trong ton b cc t bo con sinh ra t 2 nhm bng 10240 NST n lc cha nhn i. Tm s t nguyn phn ca mi hp t trong mi nhm t bo ?

CCH GII KT QU IM

a. B NST lng bi vng chn mi tb sinh dc c 1 ln nhn i k trung gian ca ln phn bo 1 nn s lng NST mi trng cung cp= s NST trong cc tb ban u -> s lng NST n trong cc tb trc khi gim phn = 2560 - S lng NST n trong 10 tb=2560-2480=80 -> B NST lng bi ca loi l 2n = 80/10 = 8 b. Hiu sut th tinh l 10% to ra 128 hp t -> s giao t = 128x100/10=1280 s lng tb sinh dc con bc vo vng chn = 2560/8=320. Nu tb ny to trng s to ra 320 trng nhm tb trn l gii tnh c c. Gi s ln nguyn phn ca mi hp t trong nhm B l k -> s ln nguyn phn ca hp t nhm A = 2k

theo gi thuyt (2kx64 + 22k x64)x8 =10240 ->2

k + 2

2k = 10x240/8x64 = 20

-> k = 2

-> s ln nguyn phn ca hp t nhm A l 4, nhm B l 2

Bi 05: Mt gen di 0,4080 micromet c 3120 lin kt hyro. Gen ny b t bin di hnh thc thay th 1 cp baz nitric ny bng mt cp baz nitric khc.

a. Nu t bin khng lm cho s lin kt hyro ca gen thay i th s lng tng loi nucltt ca gen b t bin bng bao nhiu ?

b. Nu s t bin lm cho s lin kt hyro ca gen b thay i v khi gen mi t sao 5 t lin tip th mi trng ni bo phi cung cp tng loi nucltt l bao nhiu ?

CCH GII KT QU IM

a. S Nu mi loi ca gen: - S Nu ca gen ban u: (0.4080x104 : 3.4)x2 =2400.

- Gi x l s Nu loi G ca gen ban u. Th s Nu loi A =T = (2400/2) x Ta c:

3x + 2((2400/2) - x) =3120

-> x + 2400 = 3120 -> x = 720

-> G =X = 720

A = T =1200- 720 = 480

- B khng lm thay i s lin kt Hidro -> y l dng thay cp A-T = T-A, hoc G-X = X-G hoc t bin o v tr ca 1 cp Nu. -> S Nu ca gen B G =X = 720

A = T = 480

b. Khi t bin lm thay i s lien kt Hiro, gen B nhn i 5 ln cn mi trng cung cp: * TH1: Thay cp A-T = G-X -> S lin kt Hidr tng ln 1 lin kt. -> A=T = 480-1 = 479

G=X = 720+1 =721

- S Nu mi loi mi trng cung cp: A=T = 479x(2

5 - 1)= 14849

G = X = 721x(25 - 1)= 22351

* TH 2: Thay cp G-X = A-T -> G=X = 720-1 = 719

A= T = 480 +1= 481.

Sau 5 ln nhn i, mi trng cung cp s Nu l:

A=T = 481x(25 - 1)= 14911

G = X = 719x(25 - 1)= 22289

G =X = 720

A = T = 480

- B khng lm thay i s lin kt Hidro -> y l dng thay cp A-T = T-A, hoc G-X = X-G hoc t bin o v tr ca 1 cp Nu. -> S Nu ca gen B G =X = 720

A = T = 480

* TH1: Thay cp A-T = G-X -> S lin kt Hidr tng ln 1 lin kt. A=T = 479x(2

5 - 1)=

14849

G = X = 721x(25 - 1)=

22351

* TH 2: Thay cp G-X = A-T

Sau 5 ln nhn i, mi trng cung cp s Nu l: A=T = 481x(2

5 - 1)=

14911

G = X = 719x(25 - 1)=

22289

Bi 06: Mt gen ch huy tng hp 5 chui plypeptt huy ng t mi trng ni bo 995 axit amin cc loi. Phn t mARN c tng hp t gen trn c: Am= 100

Um = 125. Gen b t bin dn n hu qu tng s nucltt trong gen

khng thay i nhng t l X

T59,57%

a. Tm s lng s nucltt mi loi ca gen sau t bin thay i nh th no so vi trc khi b t bin ? b. Gen sau t bin t nhn i 3 ln. Xc nh tng s mch n c to ra t nucltt t do ca mi trng ni bo ?

CCH GII KT QU IM

a. S nu mi loi ca gen trc v sau b: - Xt gen ban u: s nu ca gen s l: ((995/5)+1)x2x3 = 1200 s nu mi loi: A=T= Am + Um = 100+125 = 225

G=X= 1200/2 -225=375

T l: T/X = 225/375= 0.6 - Khi gen t bin T/X= 59.57% B khng lm thay i s lng nu nhng lm thay i t l T/X -> y l dng b thay th T l T/X gim, chng t b thay cp A-T = cp G-X

Gi x l s cp nu b thay th ta c:

%57.59375

225

x

x

xX

xT

- > 1.5957x 1.62 -> x =1 suy ra b thay th cp A-T bng cp G-X Vy A=T= 225-1 = 224 G=X=375+1=376

b. S mch n mi c to ra t nucltt ca mt ni bo: (2

3 x 2)-2 = 14

Bi 07: Trao i cho - hon v gen c th xy ra trong qu trnh gim phn hnh thnh c giao t c v ci (hon v hai bn) hoc ch qu trnh hnh thnh mt trong hai loi giao t (hon v mt bn). Xt php lai hai c th d hp t u v hai cp gen (A v B) quy nh hai cp tnh trng t-ng phn nm trn mt cp nhim sc th. Bit tn s hon v gen l 8%. Hy xc nh t l kiu hnh ca th h F1?

CCH GII KT QU IM