T LIU LM THI TUYN SINH I HC NHNG BI TP HAY. KH CP NHT: 24-7-2011 I. C ch di truyn v bin (10) 8-2-2

Cu 1. Mt gen m ho lin tc vi khun m ho loi prtin A, sau khi b t bin iu khin tng hp phn t prtin B. Phn t prtin B t hn prtin A mt axit amin v c 3 axit amin mi. Gi s khng c hin tng d tha m, th nhng bin i xy ra trong gen t bin l: A. B thay th 15 cp nucltit. B. Mt 3 cp nucltit v thay th 12 cp nucltit. C. Mt 3 cp nucltit thuc phm vi 4 cdon lin tip nhau trn gen. D. Mt 3 cp nucltit thuc phm vi 5 cdon lin tip nhau trn gen. Gii V d Khi cha gb t binATG XAG TAA GXX.(mch 1) .TAX GTX ATT XGG.(mch 2) Mt 3 cp nucltit thuc phm vi 4 cdon lin tip nhau trn gen. Sau khi mt 3 cp nu GX; AT; XG cu trc m s thy i ATX AGT AGX.(mch 1) TAG TXA TXG .....(mch 2) Cu 2. 1 gen c 3 alen: alen A1 m ha cho enzim E1 xc tc cho qu trnh tng hp sc t , alen A2 tng hp 1 loi enzim ko c hot tnh; alen A3 tng hp protein B l protein gy cht. Enzim E1 c th gi tc phn ng mc bnh thng ngay c khi nng n gim xung mt na. Dng t bin no sau y c gi l t bin ln? A. t bin A1 thnh A2 B. t bin A2 thnh A3 C. t bin A1 thnh A3 D. t bin A2 thnh A1 Gii A2 tng hp 1 loi enzim ko c hot tnh, v vy A2 xm nh mt alen ln. Nn t bin A1 thnh A2 c gi l t bin ln Cu 3 Mt loi c b NST 2n= 24, c bao nhiu loi giao t bnh thng c to ra t qu trnh gim phn ca cc t bin th ba (2n+1) loi ny? A. 2^11 B. 2^12 C. 2^ 24 D. 2^ 23 Cu 4: C th c kiu gen cho ra bao nhiu loi giao t, nu CD v (cd) lin kt hon ton cn AB (ab) c hon v v khng phn li trong k sau II? A. 8. B. 12. C. 16. D. 24 GIAI a. Xet cap gen (AB//ab) Dau // the hien 2 NST tronh cung cap dong dang chua 2 cap gen AB & ab O pha G1 tu nhan doi thanh (AB//AB) va (ab//ab) 1

Do trao doi cheo o ky truoc 1 nen co hoan vi gen. Ket qua lan phan bao giam phan I cho 4 loai te bao co bo NST n kep nhu sau (AB//aB) & (Ab//ab) hoac (AB//Ab) & (aB//ab) Vi khng phn li trong k sau II nen ket thuc ky sau 2 se cho cac kieu giao tu: (AB//aB) ; (Ab//ab) ; (AB//Ab) ; (aB//ab) => 4 kieu - Xet cap gen CD//cd: Vi lien ket hoan toan nen cho 2 kieu giao tu CD = cd. Vay so kieu giao tu = 4 x 2 = 8 {(AB//aB) : (Ab//ab): (AB//Ab): (aB//ab)} { CD : cd} = (AB//aB)CD : (Ab//ab)CD: (AB//Ab)CD: (aB//ab)CD :(AB//aB)cd : (Ab//ab)cd: (AB//Ab)cd: (aB//ab)cd Cu 6: gii ci mt loi ng vt (2n = 24), trong bn cp NST ng dng c cu trc ging nhau, gim phn c trao i cho n xy ra 2 cp NST, s loi giao t ti a l A. 1024. B. 4096. C. 16384. D. 16. Gii: C 4 cp ng dng c cu trc ging nhau, vy cn li c 8 cp ng dng c cu trc khc nhau. c s loi giao t ti a (theo yu cu ca ) th bt buc trong gim phn c trao i cho n xy ra 2 trong s 8 cp ng dng c cu trc khc nhau; khng xy ra trao i cho n 2 trong s 4 cp ng dng c cu trc ging nhau Vy s kiu giao t khc nhau v thnh phn NST ca b m l: 42 x 2(8-2) = 42 x 26 = 24 x 26 =1024 . Chn p n A Cu 7:Trong qu trnh dch m tng hp chui polypeptit, axit amin th (p+1) c lin kt vi axit amin th p ca chui pooly peptit hnh thnh lin kt pptt mi hp bng cch: A. Gc COOH ca axit amin th p+1 kt hp vi nhm NH2 ca axit amin th p. B. Gc COOH ca axit amin th p kt hp vi nhm NH2 ca axit amin th p+1. C. Gc NH2 ca axit amin th p+1 kt hp vi nhm COOH ca axit amin th p. D. Gc NH2 ca axit amin th p kt hp vi nhm COOH ca axit amin th p+1. Cu 8: Tt c cc loi ARNt u c mt u gn axit amin khi vn chuyn to thnh aminoacyltARN l u c 3 ribnucltit ln lt: A. .XXA-3OH B. .AXX-3OH C. .XXA-5P D. AXX-5P Cu 9: Chiu c m di truyn m gc, m sao v i m ln lt nh sau: A. 3OH->5P; 5P->3OH; 3OH->5P. B. 3P->5OH; 5OH->3P; 3P->5OH. B. 5P->3OH; 3OH->5P; 3OH->5P. C. 3OH->5P; 5P->3OH;5P->3OH. Cu 10- Cho b NST 2n = 4 k hiu AaBb (A, B l NST ca b; a, b l NST ca m). C 200 t bo sinh tinh i vo gim phn bnh thng hnh thnh giao t, trong : - 20% t bo sinh tinh c xy ra hin tng bt cho ti 1 im cp nhim sc th Aa, cn cp Bb th khng bt cho. - 30% t bo sinh tinh c xy ra hin tng bt cho ti 1 im cp nhim sc th Bb, cn cp a th khng bt cho. - Cc t bo cn li u c hin tng bt cho ti 1 im c 2 cp nhim sc th Aa v Bb S t bo tinh trng cha hon ton NST ca m khng mang gen trao i cho l: A. 50 B. 75 C. 100 D. 200 2

Cu 11. ngi c s chuyn on tng h xy ra gia NST s 13 v NST s 18 .T bo gim phn sinh giao t s c ti a bao nhiu loi giao t khc nhau v ngun gc b m ca 2 cp NST ny. A. 8 B. 16 C. 20 D. 24 1. Gii ca HS : xt 2 NST k hiu: 13+13- 18+18- (du + l NST ca b. du l NST ca m) Sau khi chyn on tng h s cho t bo c b NST 4 kh nng sau: (a) 13+*13- 18+*18- ; trao i gia 13+ v 18+ (b) 13+*13- 18+18-*; trao i gia 13+ v 18(c) 13+13-* 18+*18-; trao i gia 13- v 18+ (d) 13+13-* 18+18-* trao i gia 13- v 18Vi kh nng (a) khi gim phn s cho 4 kiu giao t 13+*18+*; 13+*18- ; 13- 18+*; 13- 18- ; Vi kh nng (b) khi gim phn s cho 4 kiu giao t 13+*18+; 13+*18-*; 13- 18+; 13- 18-*; Vi kh nng (c) khi gim phn s cho 4 kiu giao t 13+18+*; 13+18-; 13-* 18+*; 13-* 18-; Vi kh nng (d) khi gim phn s cho 4 kiu giao t 13+18+; 13+18-*; 13-* 18+; 13-* 18-*; Vy s cho ti a 4+4+4+4 =16 kiu giao t 2. P N (ca Dng) Quy c cp NST s l A v a ;cp NST s 13 l B v b. chuyn on gia cp 18 v 13 : 1. A chuyn on vi B to ra 2 NST mi l : AB v BA k gia ca gim phn NST c dng AABaa BBAbb To ra 4 giao t bnh thng : AB ;Ab ;aB ;ab 5 giao t t bin : AB BA ;AB B ;AB b ;ABA ; aBA ; 2.A chuyn on vi b to ra 2 NST mi l : Ab v bA To ra 5 giao t t bin mi 3 .a chuyn on vi b to ra 2 NST mi l : aB v bA (SAI) a chuyn on vi b to ra 2 NST mi l ab v ba (NG) To ra 5 giao t t bin mi 4. a chuyn on vi B to ra 2 NST mi l : aB v BA To ra 5 giao t t bin mi 3

Vy tng s giao t c tao ra l : 5 x 4 + 4 = 24 p n D :24 + 1 BUI NHU. 3. p n ca T Quy c cp NST s 18 l A v a ;cp NST s 13 l B v b. chuyn on gia cp 18 v 13 : 1. A chuyn on vi B to ra 2 NST mi l : AB v BA k trc I ca gim phn NST c dng: AABaa BBAbb k gia I ca gim phn NST c 2 kh nng Trn: {AABBBA {AABbb Di: {aa bb ; {aa BBA k cui I s c 4 kh nng: AABBBA, AABbb, aa bb; aa BBA k cui II s c: : AABBBA =>AB, ABA, ABB, ABBA, AABbb => Ab, ABb aa bb => ab A aa BB => aB, aBA Kt qu: To ra 4 giao t bnh thng : AB ;Ab ;aB ;ab V 5 giao t t bin : AB BA ;AB B ;AB b ;ABA ; aBA 2.A chuyn on vi b to ra 2 NST mi l : Ab v bA k trc I ca gim phn NST c dng: AAbaa BB bbA k gia I ca gim phn NST c 2 kh nng Trn: {AAbBB {AAbbbA Di: {aa bbA ; {aa BB k cui I s c 4 kh nng: AAbBB, AAbbbA, aa bbA ; aa BB k cui II s c: : AAbBB => AB, AbB AAbbbA => Ab, AbA, Abb,AbbA aa bbA => ab, abA aa BB => aB Kt qu: To ra 4 giao t bnh thng : AB ;Ab ;aB ;ab (trng vi 1) V 5 giao t t bin : AbB, AbA, Abb,AbbA , abA 3 . a chuyn on vi b to ra 2 NST mi l : ab v ba k trc I ca gim phn NST c dng: AAaab BB bba k gia I ca gim phn NST c 2 kh nng Trn: {AA BB {AA bba Di: {aab bba ; {aab BB k cui I s c 4 kh nng: AA BB, aab bba, AA bba, aab BB k cui II s c: : AA BB => AB aab bba => a b, aba , ab b, ab ba AA bba => Ab, Aba aab BB => aB, ab B Kt qu: To ra 4 giao t bnh thng : AB ;Ab ;aB ;ab (trng vi 1) V 5 giao t t bin : aba , ab b, ab ba , Aba, ab B 4. a chuyn on vi B to ra 2 NST mi l : aB v Ba To ra 5 giao t t bin mi Vy tng s giao t c tao ra l : 5 x 4 + 4 = 24 p n D :24 Cu 12. mt sinh vt nhn s, on u gen cu trc c trnh t cc nucltit trn mch b sung l 5...ATGTXXTAXTXTATTXTAGXGGTXAAT...3 Tc nhn t bin lm cp nucl ootit th 16 G - X b mt th phn t prtein tng ng c 4

tng hp t gen t bin c s axit amin l a 5. b 4. c 9. d 8 B SUNG 5'...ATG TXX TAX TXT ATT M GC 3'...TAX AGG ATG AGA TAA 123 456 789101112 131415 ARNm 5' AUG UXX UAX UXU AUU XTA GXG GTX AAT...3 GAT XGX XAG TTA...5' 161718 UAG XGG UXA AU

V t bin dch khung nn xut hin b UAG l b kt thc, do phn t prtein (hon chnh) tng ng c tng hp t gen t bin c s axit amin l 4 Cu 13. 33/ Ht phn ca loi A c 8 nhim sc th, t bo r ca loi B c 24 nhim sc th. Cho giao phn gia loi A v loi B c con lai F1. C th F1 xy ra a bi ha to c th lai hu th c b nhim sc th trong t bo giao t l a 20. b 16. c 32. d 40. 58/ Trn mARN axit amin Asparagin c m ha bi b ba 5'GAU3' (b sung ), tARN mang axit amin ny c b ba i m l a 3XUA 5. b 3 XTA 5. c 5 XUA 3. d 5 XTA3' II.Tnh quy lut ca hin tng di truyn (10) 8-2-2

Cu 1. 1 loi thc vt, khi cho cc cy hoa t th phn, i con thu c 56,25% cy hoa : 37,5% cy hoa tm: 6,25% cy hoa trng. Ngi ta ly ht phn ca cc cy hoa F1 th phn cho cc cy hoa trng F1, t l kiu hnh thu c F2 l: A. 9 hoa : 6 hoa tm: 1 hoa trng B. 4 hoa : 4 hoa tm: 4 hoa trng C.1 hoa : 2 hoa tm: 1 hoa trng D. 2 hoa : 1 hoa tm: 1 hoa trng Gii (P)2 -> F1 = 56,25% cy hoa : 37,5% cy hoa tm: 6,25% cy hoa trng. = 9 cy hoa : 6 cy hoa tm: 1 cy hoa trng. Suy ra tng tc b tr: A-bb, aaB- = hoa tm; A-B- = hoa ; aabb = hoa trng Kiu gen ca P : AaBb Kiu gen ca F1 = 1AABB: 2AABb: 1Aabb : 2AaBB: 4AaBb: Aabb: 1aaBB: 2aaBb: 1aabb Kiu gen nhng cy hoa F1: 4AaBb : 2AaBB: 2aaBb: 1AABB Khi choht phn cc cy hoa F1 th phn vi cc cy hoa trng F1 ta c s lai trong qun th: 4/9 (AaBb x aabb) + 2/9 (AaBB x aabb) +2/9 (ABb x aabb) + 1/9 (AABB x aabb) 4/9(AaBb:Aabb: aaBb: aabb) + 2/9(AaBb: aaBb) + 2/9(AaBb:Aabb) +1/9AaBb = 1/9 AaBb: 1/9Aabb: 1/9aaBb: 1/9aabb: 1/9AaBb: 1/9 aaBb + 1/9AaBb:1/9Aabb +1/9AaBb = 4/9 AaBb : 2/9 Aabb: 2/9aaBb: 1/9aabb = 4 : 4 tm: 1 trng p n sai trt lt. Cu 2 Cho bit gen A tri hon ton so vi gen a trng; sc sng tham gia th tinh hnh thnh hp t ca giao t mang gen A gp i giao t mang gen a; sc sng ca hp t v ca phi pht trin thnh c th con mang gen AA = 100%, mang gen Aa = 75%, mang gen aa = 50%. B v m mang gen d hp th t l kiu hnh ca i con F1 (mi sinh) s l: A. 7 A- : 1 aa B. 7 A- : 2 aa C: 14 A-: 1aa D. 15 A-: 1aa Cu 3 5

C 1 c th mang 2 cp gen cho 4 kiu giao t t l bng nhau. Nu c th t phi th i con F1 s c s nhm kiu gen l: A. 9 B. 10 C. 9 hoc 10. D. 16 Cu 4 V trt t khong cch gia 3 gen A, B v C ngi ta nhn thy nh sau: A------------20-----------------B---------11----------C. H s trng hp l 0,7. Abc abc Nu P : x th % kiu hnh khng bt cho ca F1: aBC abc A. 70,54% B. 69% C. 67,9% D. khng xc nh c Gii: Abc abc X aBC abc Abc aBC F1: = = Kiu hnh khng bt cho abc abc abC ABc = = Kiu hnh bt cho kp abc abc abc ABC = = Kiu hnh bt cho I (Ab/aB) abc abc AbC aBc = = Kiu hnh bt cho II (bC/Bc) abc abc P: T l bt cho kp l thuyt: 20% x 11% = 2,2% T l bt cho kp thc t: 2,2% x 0,7 = 1,54% Khong ch gia 2 gen A v B l 20%. Khong cch ng vi t l cc c th c th xy ra bt cho gia cc gen A v B, trong cc c th c th bt cho n v cho kp. Nh vy bt cho n A/B 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 B v C (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% Vy t l kiu gen, kiu hnh ca i con F1 l : Abc aBC F1: = = 70,54% /2 = 35,27% (Kiu hnh khng bt cho) abc abc abC ABc = = 1,54% /2 = 0,77% (Kiu hnh bt cho kp) abc abc abc ABC = = 18,46% /2 = 9,23% (Kiu hnh bt cho I-Ab/aB) abc abc AbC aBc = = 9,46% / 2 = 4,73% (Kiu hnh bt cho II-bC/Bc) abc abc 6

Cu 5 Ad BE Ad BE x aD be aD be Mi gen mi tnh tri hon ton, tn s hon v gen ca c th c v ci bng nhau: f1 (A/d) = 0,2, f2 (B/E) = 0,4; th i F1 c t l kiu hnh A-B-D-E- chim t l: A.30,09% B.42,75% C.56,25% D.75% Nu P : Cu 6 Cho bit chiu cao cy do 5 cp gen PLL tc ng cng gp. Nu P thun chng khc nhau n cp gen tng ng, i F2 c s c th c kiu hnh chiu cao trung bnh chim t l l: A. 252/1024 B. 1/8 C. 1/16 D. Cu 7 Cho bn gen nh sau: A..20..B..10..D Nu h s ph hp 0,8 th tn s hon v kp thc t l: A. B. C. D. Cu 8] i con F1 c 4 nhm kiu hnh phn ly theo t l vit di dng ti gin l a : b : c : d. Nu a + b + c + d = 2n (a, b, c, d, n l s nguyn ln hn 0) iu chng t A. Cc gen PLL B. Cc gen lin kt hon ton C. Cc gen hon v vi tn s c bit 1/2k (k l s nguyn) D. Cc gen PLL hoc gen lin kt hon ton hoc hon v vi tn s c bit 1/2k Cu 9. Trong php lai AaXBDY x AaXBDXbd, bit tn s hon v gen l 20%. Nu mi gen quy nh mt tnh trng v cc tnh trng di truyn theo quy lut tri hon ton th t l cc c th c (XY) c kiu hnh mang 3 tnh trng tri so vi tng s c th c c sinh ra i con l? A. 52,5% B. 15% C. 75% D. 30% Gii P: AaXBDY x AaXBDXbd ~ (Aa x Aa) (XBDY x XBDXbd ) F1 : (0,75 A-: 0,25aa) ( 0,2 XBDY: 0,2 XbdY : 0,05XBdY: 0,05 XbDY : 0,5XBD X??) Vy t l cc c th c (XY) c kiu hnh mang 3 tnh trng tri (A-XBDY) so vi tng s c th c c sinh ra i con l: (0,75 x 0,2)/ (0,75 x ) = 0,015/ 0,100 = 15% Cu 10: Kiu gen ca c chp khng vy l Aa, c chp c vy l aa. Kiu gen AA lm trng khng n. Tnh theo l thuyt, php lai gia cc c chp (P) khng vy s cho t l kiu hnh i F2 l A. l c chp khng vy : 2 c chp c vy. B. 3 c chp khng vy : l c chp c vy. C. 75% c chp khng vy : 25% c c vy. D. 4 c chp khng vy : l c chp c vy 7

Gii P: Aa (khng vy) x Aa (khng vy) F1: 1AA : 2Aa : 1aa V trng c kiu gen AA khng n c, nn F1: 2Aa (khng vy) : 1 aa (c vy) Tn s alen p (A) ca F1 = 2/6 = 1/3 Tn s alen q (a) ca F1 = 2/3 Vy F2 c: AA= (1/3)2 = 1/9 khng n Aa = 2. 1/3. 2/3 = 4/9 aa = (2/3)2 = 4/9 Vy i F2 c 1Aa : 1aa. p n sai trt lt. Cu 11: Cho cy hoa qu trn t th phn, ngi ta thu c i con c t l kiu hnh phn li: 510 cy hoa , qu trn: 240 cy hoa , qu di: 242 cy hoa trng, qu trn: 10 cy hoa trng, qu di. T kt qu ca php lai ny, kt lun no c rt ra l ng A. Alen qui nh mu hoa v alen qui nh qu trn cng thuc 1 NST B. Alen qui nh mu hoa v alen qui nh qu di cng thuc 1 NST C. Alen qui nh mu hoa v alen qui nh qu trn lin kt khng hon ton D. Alen qui nh mu hoa v alen qui nh qu trn lin kt hon ton Gii: P: (, trn)2 F1: 510 , trn: 240 , di: 242 trng, trn: 10 trng, di. = 510/1002 , trn: 240/1002 , di: 242/1002 trng, trn: 10/1002 trng, di ~ 51% , trn: 24% , di: 24% trng, trn: 1% trng, di Xt mu sc: F1 = (510 + 240) : (242 + 10) trng ~ 3 : 1 trng Xt hnh dng: F1= (510 + 242) trn: (240 + 10) di ~ 3 trn : 1 di Ta thy F1 mu sc x F1 hnh dng = (3 : 1 trng) x (3 trn : 1 di) 51% , trn: 24% , di: 24% trng, trn: 1% trng, di Alen qui nh mu hoa v alen qui nh qu trn lin kt khng hon ton (hon v gen) Tn s hon v gen 0,2; kiu gen ca P l d hp t cho: Ab/aB Chn p n C. C th gii nhanh bng cahs s dng quy tc s T ca Nguyn T (xem file nh km) Cu 12. rui gim, tnh trng thn xm tri hon ton so vi tnh trng thn en, cnh di tri hon ton so vi cnh ngn. Cc gen quy nh mu thn v chiu di cnh cng nm trn 1 nhim sc th v cch nhau 40 cM. Cho rui gim thun chng thn xm, cnh di lai vi rui thn en, cnh ct; F1 thu c 100% thn xm, cnh di. Cho rui ci F1 lai vi rui thn en, cnh di d hp. F2 thu c kiu hnh thn xm, cnh ct chim t l a 20%. b 10%. c 30%. d 15%. Quy c: A: Xm, a: en; B: di, b: ct P: AB/AB thn xm, cnh di x ab/abthn en, cnh ct F1: AB/ab thn xm, cnh di Ci F1: AB/ab thn xm, cnh di x Gt Gt 0,5 aB Gt 8 0,5 ab c aB/abthn en, cnh di d hp

0,3 AB

0,15 AB/aB

0,15 AB/ab

0,2 Ab 0,2 aB 0,3 ab

0,1 Ab/aB 0,1 aB/aB 0,15 aB/ab

0,1 Ab/ab thn xm, cnh ct 0,1 aB/ab 0,15 ab/ab

Bi tp: Trong mt php lai th ci thun chng c mu mt v lng dng hoang di vi th c c mu mt m v lng xm, ngi ta thu c F1 tt c u c mu mt v lng hoang di. Cho cc th ci F1 giao phi vi nhau th thu c F2 c t l phn li kiu hnh nh sau: Tt c th ci F2 u c mt v lng hoang di. Cc th c F2 c t l phn li kiu hnh nh sau: 45% mt v lng mu hoang di. 45% mt mu m v lng mu xm 5% mt mu hoang di v lng mu xm. 5% mt mu m v lng mu hoang di. Hy gii thch kt qu trn v vit s lai t P n F2. Bit rng mi gen quy nh mi tnh trng. GII Tm tt : Pt/c: ci mt v lng dng hoang di X c mt m v lng xm F1: 100% mt v lng hoang di. F1 X F1 -> F2: -Th ci: mt v lng hoang di. -Th c; 45% mt v lng mu hoang di. 45% mt mu m v lng mu xm 5% mt mu hoang di v lng mu xm. 5% mt mu m v lng mu hoang di. (% tnh theo gii) Hay: Pt/c: ci mt v lng dng hoang di X c mt m v lng xm F1: 100% mt v lng hoang di. F1 X F1 -> F2: -50% th ci mt v lng hoang di. 45%/2 = 22,5% th c mt v lng mu hoang di. 9

45%/2 = 22,5% th c mt mu m v lng mu xm 5%/2 =2,25% th c mt mu hoang di v lng mu xm. 5%/2 =2,25% 5%/2 =2,25% mt mu m v lng mu hoang di. (% tnh theo tng th c 2 gii) BC 1: Tch ring tng tnh xt km theo gii tnh: a. Xt mu mt km theo gii tnh: Pt/c: ci mt hoang di X c mt m F1: 100% mt hoang di. F1 X F1 -> F2: -Th ci: mt hoang di. -Th c: 45% + 5% mt mu hoang di: 45% + 5% mt mu m = 50% mt mu hoang di: 50% mt mu m (% theo gii) Tnh theo tng th ta c F2: 2 ci: mt mu hoang di : 1 c mt mu hoang di: 1 c mt mu m Ta thy trong cng mt th h con lai F2 m c, ci biu hin kiu hnh khc nhau, chng t mu mt di truyn LKGT. Nu tch ring mu mt ta thy F2 = 3 mt mu hoang di: 1 mt mu m Suy ra mu mt hoang di l tri (A) so vi mu mt m (a) V th con c l XY con ci l XX, nn ta c kiu gen ca P, F1, F2 v mu sc mt: Pt/c: XAXA F1: XAXa : XAYa? F2: 1 XAXA : 1 XAXa : 1 XAYa? : 1 XaYa? 2 ci mu mt hoang di : 1 ci mu hoang di : 1 ci mu mt m b. Xt mu lng km theo gii tnh: Pt/c: ci lng dng hoang di X c lng xm F1: 100% lng hoang di. F1 X F1 -> F2: -Th ci: lng hoang di. -Th c: (45% + 5%) lng mu hoang di : (45% + 5%) lng mu xm. Lp lun tng t s di truyn ca mu mt, ta rt ra c kt lun: Mu lng cng di truyn LKGT. Mu lng hoang di l tri (B) so vi mu lng xm (b) Kiu gen ca P, F1, F2 v mu lng: Pt/c: XBXB F1: XBXb : XBYb? F2: 1 XBXB : 1 XBXb : 1 XBYb? : 1 XbYb? 10 x XbYb? x XaYa?

2 ci mu lng hoang di : 1 ci mu lng hoang di : 1 ci mu lng xm BC 2: Kt lun kiu gen ca P v 2 tnh Ta thy, khi xt v 2 tnh mu mt v mu lng, kiu gen ca p l Pt/c: XABXAB F1: XABXab x XabYab? ( Ch Y c th mang hoc khng mang gen ab x XABYab?

Nu NST Y c mang gen ab, th n phi nm trn on tng ng ca cp XY, c bt cho xy ra tng ng v con dc F1 cng cho nhiu loi giao t: XAB XAb = Xab = XaB = YAB = = YAb = Yab YaB = (1-f)/2 = f/2

Vy con ci F2 s phn tnh nh con c F2 Chng t gen AB; ab ch nm trn NST X m khng nm trn NST Y Ta c kiu gen ca P: Pt/c: XABXAB F1: XABXab x XabY x XABY ca con

Gi f l tn s hon v gen gia 2 cp gen AB/ab nm trn cp NST XX ch cho 2 kiu giao t = XAB = Y =1/2 F2: 1/2 XABX?? : (1-f)/2 XABY : (1-f)/2 XabY : f/2 XAbY : f/2 XaBY Ta c f/2 = 5%/2 => f= 5% Kt lun: Mu mt hoang di (A) > mu mt m (a) Mu lng hoang di (B) > mu lng xm (b) Kiu gen ca Pt/c: XABXAB S lai kim chng: (Em t lm) x XabY Hon v gen tn s 5% xy ra th ci F1

Ci F1; cn con c F1 trn on XY khng tng ng th khng th xy ra bt cho, nn con c F1

III.Di truyn hc qun th (3) 3-0-0 Em c c hc 1 s cng thc v ng dng qui lut di truyn ca Menden vo Ton qun th, nhng em ko hiu my cng thc c t u. Thy gii thch gim em vi: 1. Tn s alen ca c th c v ci trong qun th p = {p(ci) + p(c)}/2 2. C t bin gen: x l tn s t bin thun y l tn s t bin nghch Sau 1 th h th: tn s tng i alen tri l p' = p + (yq - xp) q' = q + (xp - yq) 3. Di- nhp gen: Qun th A c 1 nhm c th c kch thc nhm nhp c (ngha l g) l m, di c sang qun th B, t qun th B chuyn thnh qun th C, th: pC = pB - m( pB - pA) 4. Xy ra chn lc t nhin 11

* Tn s alen sau 1 th h chc lc s l h s chc lc alen th: q' = {q(1 - sq)} / (1 - s.q^2) * Nu KG ng hp ln gy cht th sau 1 th h chn lc th: q' = q/(1+q) * Nu trong qun th cc th d hp u th hn c th ng hp th tn s alen ln ca qun th trng thi cn bng l: q' = s/ (s + s') s: h s chn lc ca c th ng hp tri s' h s chn lc ca c th ng hp ln Cu 1: Trong mt qun th bm gm 900 con, tn s alen quy nh cu t chuyn ng nhanh ca mt enzim p = 0,7 v tn s alen quy nh cu t chuyn ng chm q = 0,3. C 90 con bm t qun th ny nhp c n mt qun th c q = 0,8. Tn s alen ca qun th mi l A. p = 0,7 v q = 0,3. B. p = 0,75 v q = 0,25. C. p = 0,25 v q = 0,75. D. p = 0,3 v q = 0,7. Gii: Ta c: Gi qun th 1 l qun th gm 900 con, tn s alen p'A = 0,7, q'a = 0,3. Qun th 2 l qun th c 90 con bm t qun th 1n nhp c c q''a = 0,8 v p''A = 0,2 Gi N l s c th qun th 1, M l s c th qun th 2 ban u khi cha xut, nhp c. S c th qun th 1 xut c l 90 = 90/900 = 10%N = 0,1N Tn s kiu gen ca qun th 1 p'2AA : 2. p'.q'Aa : q'2aa = 0,49AA : 0,42Aa : 0,09aa S cc th qun th 1 xut c tng ng tng kiu gen: AA = 0,49 x 0,1N = 0,049N Aa = 0,42 x 0,1N = 0,042N aa = 0,09 x 0,1N = 0,009N Tn s kiu gen ca qun th 2 khi ch c nhp c: p''2AA : 2. p''.q'' Aa : q''2aa = 0,04AA : 0,32Aa : 0,64aa S cc th qun th 2 ban u tng ng tng kiu gen: AA = 0,04M; Aa = 0,32M, aa = 0,64M Tng s c th ca qun th 2 sau khi c nhp c: M + 0,1N S cc th qun th 2 tng ng tng kiu gen sau khi c nhp c: AA = 0,04M + 0,049N Aa = 0,32M + 0,042N aa = 0,64M + 0,009N Vy tn s kiu gen ca qun th 2 sau khi c nhp c l: pA = {2 (0,04M + 0,049N) + 0,32M + 0,042N}/2 (M + 0,1N) = (0,4M + 0,14N)/2 (M + 0,1N) = = (0,2M + 0,07N)/(M + 0,1N) = = (0,2M + 0,07 x 900)/(M + 0,1 x 900) = (0,2 M + 63)/(M + 90) pA = (0,2M + 63)/(M + 90) qa = (0,8M + 17)/(M + 90) => pA/qa = (0,2M + 63)/(0,8M + 17) V khng bit c M l bao nhiu nn khng th tnh pA v qa. Vy sai Cu 2: ngi, bnh m mu do mt alen ln nm trn NST gi tnh X quy nh. Xt mt qun th mt hn o c 100 c th trong c 50 ph n v 50 n ng, hai ngi n ng b bnh m mu. Nu qun th trng thi cn bng th tn s ngi ph n bnh thng mang gen gy bnh l A. 4%. B. 7,68%. C. 96%. D. 99,84%. 12

Gii: Kiu gen ca n b: X X : X X : X X Kiu gen n ng: XAY: XaY Gi p1 l tn s alen A , q1 l tn s alen a ca n ng Gi p2 l tn s alen A , q2 l tn s alen a ca n b Gi p l tn s alen A , q l tn s alen a ca qun th ngi Qun th trng thi cn bng di truyn nn p1 = p2, q1 = q2 p = p1/3 + 2p2/3 = 3p1/3 = 3p2/3 = p1 = p2 q = q1/3 + 2q2/3 = 3q1 /3 = 3q2/3 = q1 = q2 Ta c tn s kiu gen trong qun th n ng: p2XAXA : 2pqXAXa : q2XaXa n b: pXAY: qXaY Suy ra: q = 2/50 = 0,04 => p = 1- 0,04 = 0,96 Vy tn s ngi ph n bnh thng mang gen gy bnh l: 2pq XAXa = 2 x 0,96 x 0,04 =0,0768 = 7,68% Cu 3. 1 qun th ngu phi, xt 2 gen: gen th nht c 3 alen, nm trn on ko tng ng ca NST gii tnh X, gen th 2 c 5 alen, nm trn NST thng, trong trng hp ko xy ra t bin, s kiu giao phi ti a c th xy ra trong qun th l: A. 18225 B. 1500 C. 4050 D. 1350 Gii Xt gen th nht c 3 alen A, a1, a2 nm trn on ko tng ng ca NST gii tnh X. Kiu gen XX: 6 kiu gen = XAXA, Xa1Xa1, Xa2Xa2, XAXa1, XAXa2; Xa1Xa2 Kiu gen XY: 3 kiu gen = XAY; Xa1Y; Xa2Y Xt gen th hai c 5 alen B, B1, B2, B3, B4 nm trn NST thng. Mi gii u c t hp ca 5 phn t chp 2 c lp kiu gen = t hp ca 6 phn t chp 2 khng lp kiu gen = 15. Xt c 2 cp gen th c th XX c 6 x 15 = 90 kiu gen khc nhau, c th Xy c 3 x 15 = 45 kiu gen khc nhau Vy c kiu giao phi ti a c th xy ra trong qun th l= 90 x 45 = 4050 Cu 4 Cho bit gen A quy nh mt , gen a quy nh mt trng, gen nm trn NST X, gii c l XY. mt qun th cn trng ngu phi, gii c c 10% con mt trng, gii ci c 1% con mt trng, cn li l nhng con mt . T l kiu gen chung c hai gii: A. 0,45 XAY : 0,05 XaY : 0, 405 XAXA : 0,09 XAXa : 0,005 XaXa B. 0,9 XAY : 0,1XaY : 0,81 XAXA : 0,18 XAXa : 0,01 XaXa C. 0,45 XAY : 0,05 XaY : 0,81 XAXA : 0,18 XAXa : 0,01 XaXa D.0,9 XAY : 0,1XaY : 0, 405 XAXA : 0,09 XAXa : 0,005 XaXa Cu 5 Mt qun th c tn s alen A l 0,6. Gi s ban u qun th ang t trng thi cn bng di truyn. Sau mt s th h giao phi thy tn s kiu gen aa l 0,301696. Bit trong qun th xy ra ni phi vi h s l 0,2. Tnh s th h giao phi? Gii 13A A A a a a

Tn s alen a l 0,4. Do qun th t trng thi cn bng nn cu trc ca qun th l: 0,301696AA+ 0,48Aa + 0,16aa = 1. Sau mt s th h giao phi, tn s aa l: 0,301696 => Tn s kiu gen aa tng l: 0,301696 - 0,16 = 0,141696 => Tn s Aa gim l: 0,141696 x 2 = 0,283392. Tn s Aa sau n th h giao phi l: 2pq(1 - f)n = 0,48(1 - f)n = 0,48.0,8n Tn s Aa gim l: 0,48 0,48.0,8n = 0,283392 n = 4. Vy h s giao phi l 4. Cu 6 Cho 2 qun th 1 v 2 cng loi, kch thc ca qun th 1 gp i qun th 2. Qun th 1 c tn s alen A = 0,3, qun th 2 c c tn s alen A = 0,4. Nu c 10% c th ca qun th 1 di c qua qun th 2 th tn s alen A ca qun th 2 s l: A. 0,3833 B. 0,3933 C. 0,3733 D. 0,3633 Cu 7 Cho 2 qun th 1 v 2 cng loi, kch thc ca qun th 1 gp i qun th 2. Qun th 1 c tn s alen A = 0,3, qun th 2 c c tn s alen A = 0,4. Nu c 10% c th ca qun th 1 di c qua qun th 2 v 20% c th ca qun th 2 di c qua qun th 1 th tn s alen A ca 2 qun th 1 v qun th 2 ln lt l: A. 0,35 v 0,4 B. 0,3 v 0,4 C. 0,4 v 0,3 D. bng nhau = 0,35 Cu 8 Bnh m mu - lc ngi lin kt vi gii tnh. Mt qun th ngi trn o c 50 ph n v 50 n ng trong c hai ngi n ng b m mu - lc. Tnh t l s ph n mang gen bnh. A. 7,68% B. 7,48% C. 7,58% D. 7,78% Hng dn: Vy tn s alen ln gy bnh m mu - lc l q = 4% Tn s n ng mc bnh l: 2/50 = 0,04 Tn s alen tri p l 1 0,04 = 0,96 Vy tn s ph n mang gen bnh l: 2pq = 2 (0,04) (0,96) = 0,0768 hay 7,68% Cu 9 Chng ta cng i tm li gii cho bi ton "Hon v gen trong qun th" Gi s th h xut pht P c 100% s c th c kiu gen d hp t cng bn (AB/ab) v tn s hon v gen l 20%. Bit rng mi din bin ca NST t bo sinh tinh v sinh trng l nh nhau; mi gen mi tnh, tri hon ton; sc sng ca giao t, ca hp t l nh nhau; khng chu p lc ca t bin. Tm t l kiu gen, t l kiu hnh ca i con F1, F2,..., Fn. Xt trong 2 trng hp: a) T phi nghim ngt. b) Ngu phi. T tm cng thc tng qut khi tn s hon v gen bng f . 14

Cu 10 Cho 2 qun th 1 v 2 cng loi, kch thc ca qun th 1 gp i qun th 2. Qun th 1 c tn s alen A = 0,3, qun th 2 c c tn s alen A = 0,4. Nu c 10% c th ca qun th 1 di c qua qun th 2 v 20% c th ca qun th 2 di c qua qun th 1 th tn s alen A ca 2 qun th 1 v qun th 2 ln lt l: A. 0,35 v 0,4 B. 0,31 v 0,38 C. 0,4 v 0,3 D. bng nhau = 0,30 GII Gi N l s lng c th ca qun th 2, vy s lng cc th ca qun th 1 l 2N. -Tn s alen ban u ca qun th 1: p1A = 0,3, q1a = 0,7 -T l kiu gen ca qun th 1 khi cha di nhp: 0,09AA: 0,42Aa: 0,49 aa - S lng c th tng ng vi tng kiu gen ca qun th 1 l: AA = 0,09 x 2N = 0,18N Aa = 0,42 x 2N = 0,84N aa = 0,49 x 2N = 0,98N -S lng c th ca qun th 1 di c qua qun th 2 l: 2N x 10% = 0,2N. -Trong s lng tng ng vi tng kiu gen ca qun th 1 di c qua qun th 2 l: AA = 0,18N x 0,1 = 0,018N Aa = 0,84N x 0,1 = 0,084N aa = 0,98N x 0,1 = 0,098N -Tn s alen ban u ca qun th 2: p2A = 0,4, q2a = 0,6 -T l kiu gen ca qun th 2 khi cha di nhp: 0,16AA: 0,48 Aa: 0,36 aa -S lng c th tng ng vi tng kiu gen ca qun th 2 khi cha di nhp l: AA = 0,16 N Aa = 0,48 N aa = 0,36 N S lng c th ca qun th 2 di c qua qun th 1 l: N x 20% = 0,2N. Trong s lng tng ng vi tng kiu gen ca qun th 2 di c qua qun th 1 l: AA = 0,16N x 0,2 = 0,032N Aa = 0,48N x 0,2 = 0,096N aa = 0,36N x 0,2 = 0,072N Kt qu sau ln di nhp th nht: - Qun th 1: S lng c th tng ng vi tng kiu gen ca mi qun th 1 sau ln di, nhp u tin: AA = 0,18N - 0,018N + 0,032N = 0,194N Aa = 0,84N - 0,084N + 0,096N = 0,852N aa = 0,98N - 0,098N + 0,072N = 0,954N Vy tn s alen ca qun th 1 sau ln di, nhp u tin: p'1 = {(2 x 0,194N) + 0,852N}/2N x 2 = (0,388N + 0,852N)/4N = 1,24N/4N = 0,31 q'1 = 1 - 0,31 = 0,69 - Qun th 2: S lng c th tng ng vi tng kiu gen ca mi qun th 2 sau ln di, nhp u tin: AA = 0,16N - 0,032N + 0,018N = 0,146N Aa = 0,48N - 0,096N + 0,084N = 0,468N aa = 0,36N - 0,072N + 0,098N = 0,386N Vy tn s alen ca qun th 2 sau ln di, nhp u tin: p'2 = {(2 x 0,146N) + 0,468N}/2N = (0,292N + 0,468N)/2N = 0,76N/2N = 0,38 q'2 = 1-0,38 = 0,62 Cu 11. Trong mt qun th, 90% alen locut Rh l R. Alen cn li l r. Bn mi tr em ca qun th ny i n mt trng hc nht nh. Xc sut tt c cc em u l Rh dng tnh s l: 15

A. 400^81 B. 0,99^40 C. 40^0,75 D. 1 0,81^40 Cu 12: mui st xut huyt Aedes aegypti, b gy bnh thng c mu trng c. Tnh trng mu sc thn b gy do mt gen trn nhim sc th thng quy nh. Mt t bin ln gen ny lm cho thn b gy c mu en. Trong phng th nghim, ngi ta cho giao phi ngu nhin 100 cp mui b m, thu c 10000 trng v cho n thnh 10000 b gy, trong s c 100 b gy thn en. Do mun loi b t bin ny ra khi qun th, ngi ta loi b i tt c cc b gy thn en. Gi s rng khng c t bin mi xy ra. Tn s alen ca qun th mui khi loi b b gy thn en l A. p = 0,91 v q = 0,09. B. p = 0,90 v q = 0,10. C. p = 0,80 v q = 0,20. D. p = 0,81 v q = 0,19. Gii: Gi gen A: trng c; gen a: en P: 100 ci ? X 100 c? F1: 10000 b trong c 100 b thn en aa Gi p l tn s alen A, q l tn s alen a trong qun th b gy F1 khi cha loi b gy thn en aa: Ta c q2 = 100/10.000 = 1/100 => q = 0,1; p = 0,9 Tn s kiu gen ca qun th F1 khi cha loi b gy thn en aa: p2AA : 2.p.q Aa : q2 aa = 0,92 AA : 2 x 0,9 x 0,1 Aa : 0,12 aa = 0,81 AA : 0, 18 Aa : 0,01 aa Tn s kiu gen ca qun th mui F1 khi loi b b gy thn en aa l: 0,81/ (0,81 + 0,18) AA : 0,18 / (0,81 + 0,18) Aa = 0,81/0,99 AA : 0,18/0,99 Aa Vy tn s alen ca qun th mui khi loi b b gy thn en l: p'A = {(2 x 0,81) + 0,18}/ (2 x 0,99) = (1,62 + 0,18) / 1,98 = 1,8/1,98 = 0,90909090 = 0,91 q'a = 0,18/1,98 =0,09090909... = 0,09 Vy chn p n A Cu 14. Qun th c cu trc 0.4 AABb: 0.4AaBb : 0.2aabb qun th t phi sau 3 th h tnh t l c th mang gen ng hp tri ? Gii: P0: 0,4 AABb: 0,4AaBb : 0,2aabb V cc gen PLL nn ta c th xem P: (0,4 AA: 0,4Aa : 0,2aa) (0,8Bb : 0,2 bb) Vi A,a t th phn c F3 c: 0,4 AA + 0,2 aa + 0,4 [{1-(1/2)3}/2 AA + (1/2)3 Aa + {1-(1/2)3}/2 aa] = 0,4 AA + 0, 2 aa + 0,4 ( 7/16 AA + 1/8 Aa + 7/16 aa) = 0,4 AA + 0,2 aa + 2,8/16 AA + 0,4/8 Aa + 2,8/16 aa = (0,4 + 0,175) AA + (0,2 +0,175) aa + 0,05 Aa = 0,575 AA + 0,050Aa + 0,375 aa Vi B,b t th phn c F3 c: 0,8 [{1-(1/2)3}/2 BB + (1/2)3 Bb + {1-(1/2)3}/2 bb] + 0,2 bb = 0,8 ( 7/16 BB + 1/8 Bb + 7/16 bb) +0,2 bb = 0,35 BB + 0,1 Bb + 0,35 bb + 0,2 bb = (0,35 BB + 0,1 Bb + 0,2 bb) Vy F3 mang gen ng hph tri l: 0,575 AA x 0,35 BB = 0,575 x 0,35 AABB = 0.20125 AABB= = 20,125% AABB Cu 15: Qun th c kiu gen 0.2AABb : 0.2AaBb : 0.3aaBB : 0.3aabb. Nu qun th ngu phi th th t l ng hp ln sau 1 th h l ? Tn s alen A: p = [(0,2 x 2) + (0,2 x 1)]/2 = 0,3 Tn s alen a: q = 1-0,3 = 0,7 Tn s alen B: p = [(0,2 x 1) + (0,2 x 1) +(0,3 x 2)]/2 = 0,5 16

Tn s alen b: q = 1-0,5 = 0,5) F1 ta c: (0,09AA: 0,42 Aa : 0,49 aa) (0,25BB :0,5 Bb : 0,25 bb) = Vy t l ng hp ln sau 1 th h l: 0,49 x 0,25 = 0.1225 = 12,25% IV.ng dng di truyn hc (4) 3-1-1

Cu 1. ci to ging heo Thuc Nhiu nh Tng, ngi ta dng con c ging i Bch lai ci tin vi con ci tt nht ca ging a phng. Nu ly h gen ca c i Bch lm tiu chun th th h F4 t l gen ca i Bch trong qun th l: A. 50%. B. 75%. C. 87,5%. D. 93,25%. Gii F1: 1/2 B + 1/2TNT F2: 3/4 B + 1/4TNT F3: 7/82 B + 1/8TNT F4: 15/16 B + 1/16TNT Nu ly h gen ca c i Bch lm tiu chun th th h F4 t l gen ca i Bch trong qun th l: 15/16 = 0.9375 % Khng c p n no trng c. +++++ Cu 2: chuyn 1 gen ca ngi vo vi khun Ecoli nhm to ra nhiu sn phm ca gen ngi trong t bo vi khun ngi ta phi ly mARN ca gen ngi cn chuyn cho phin m ngc thnh ADN ri mi gn ADN ny vo plasmid v chuyn vo vi khun. Ngi ta cn phi lm nh vy l v: A. Gen ca ngi qu ln nn khng vo c t bo vi khun B. Nu khng lm nh vy th gen ca ngi khng th phin m c trong t bo vi khun C. Nu khng lm nh vy th gen ca ngi khng th dch m c trong t bo vi khun D. Nu khng lm nh vy th sn phm ca gen s khng bnh thng, khng c gi tr s dng

V.Di truyn hc ngi

(3)

2-1-1

Cu 1: ngi, bnh m mu do mt alen ln nm trn NST gi tnh X quy nh. Xt mt qun th mt hn o c 100 c th trong c 50 ph n v 50 n ng, hai ngi n ng b bnh m mu. Nu qun th trng thi cn bng th tn s ngi ph n bnh thng mang gen gy bnh l A. 4%. B. 7,68%. C. 96%. D. 99,84%. Gii: Kiu gen ca n b: XAXA : XAXa : XaXa Kiu gen n ng: XAY: XaY Gi p1 l tn s alen A , q1 l tn s alen a ca n ng Gi p2 l tn s alen A , q2 l tn s alen a ca n b Gi p l tn s alen A , q l tn s alen a ca qun th ngi Qun th trng thi cn bng di truyn nn p1 = p2, q1 = q2 p = p1/3 + 2p2/3 = 3p1/3 = 3p2/3 = p1 = p2 q = q1/3 + 2q2/3 = 3q1 /3 = 3q2/3 = q1 = q2 Ta c tn s kiu gen trong qun th n ng: p2XAXA : 2pqXAXa : q2XaXa n b: pXAY: qXaY Suy ra: q = 2/50 = 0,04 => p = 1- 0,04 = 0,96 Vy tn s ngi ph n bnh thng mang gen gy bnh l: 2pq XAXa = 2 x 0,96 x 0,04 =0,0768 = 7,68% 17

Cu: Cho P thun chng giao phn thu c F1: 100% hoa , cho F1 t th phn thu F2: 9 hoa : 7 hoa trng. Chn ngu nhin 2 cy hoa thuc th h lai F2 cho giao phn, tnh xc sut thu c th h F3 cy hoa ng hp v cc gen ln. Gii Ta c mu sc di truyen do 2 cp gen tng tc b tr: A-B-: ; A-bb, aaB-, aabb: Trng F1: AaBb F2: 1AABB: 2AaBB: 1aaBB: 2AABb: 4AaBb: 2aaBb: 1AAbb: 2Aabb: 1aabb i F2 hoa gm 4 loi kiu gen khc nhau: 1/9 AABB: 2/9 AaBB: 2/9 AABb: 4/9 AaBb Mun F3 c aabb th chn b m F2 c kiu gen AaBb Xc sut chn 2 cy F2 hoa c kiu gen AaBb l: (4/9)2 Xc sut F2 mang gen AaBb cho ra F3 c kiu gen aabb l 1/16 Vy xc sut "Chn ngu nhin 2 cy hoa thuc th h lai F2 cho giao phn, thu c th h F3 cy hoa ng hp v cc gen ln" l : (4/9)2 x 1/16 = (16/81) x (1/16) = 1/81

VI.Bng chng tin ho

(1)

1-0-0

VII.C ch tin ho

(7)

5-2-2

Cu 1: Trong mt qun th bm gm 900 con, tn s alen quy nh cu t chuyn ng nhanh ca mt enzim p = 0,7 v tn s alen quy nh cu t chuyn ng chm q = 0,3. C 90 con bm t qun th ny nhp c n mt qun th c q = 0,8. Tn s alen ca qun th mi l A. p = 0,7 v q = 0,3. B. p = 0,75 v q = 0,25. C. p = 0,25 v q = 0,75. D. p = 0,3 v q = 0,7. Gii: Ta c: Gi qun th 1 l qun th gm 900 con, tn s alen p'A = 0,7, q'a = 0,3. Qun th 2 l qun th c 90 con bm t qun th 1n nhp c c q''a = 0,8 v p''A = 0,2 Gi N l s c th qun th 1, M l s c th qun th 2 ban u khi cha xut, nhp c. S c th qun th 1 xut c l 90 = 90/900 = 10%N = 0,1N Tn s kiu gen ca qun th 1 p'2AA : 2. p'.q'Aa : q'2aa = 0,49AA : 0,42Aa : 0,09aa S cc th qun th 1 xut c tng ng tng kiu gen: AA = 0,49 x 0,1N = 0,049N Aa = 0,42 x 0,1N = 0,042N aa = 0,09 x 0,1N = 0,009N Tn s kiu gen ca qun th 2 khi ch c nhp c: p''2AA : 2. p''.q'' Aa : q''2aa = 0,04AA : 0,32Aa : 0,64aa S cc th qun th 2 ban u tng ng tng kiu gen: AA = 0,04M; Aa = 0,32M, aa = 0,64M Tng s c th ca qun th 2 sau khi c nhp c: M + 0,1N S cc th qun th 2 tng ng tng kiu gen sau khi c nhp c: AA = 0,04M + 0,049N Aa = 0,32M + 0,042N aa = 0,64M + 0,009N Vy tn s kiu gen ca qun th 2 sau khi c nhp c l: pA = {2 (0,04M + 0,049N) + 0,32M + 0,042N}/2 (M + 0,1N) = (0,4M + 0,14N)/2 (M + 0,1N) = 18

= (0,2M + 0,07N)/(M + 0,1N) = = (0,2M + 0,07 x 900)/(M + 0,1 x 900) = (0,2 M + 63)/(M + 90) pA = (0,2M + 63)/(M + 90) qa = (0,8M + 17)/(M + 90) => pA/qa = (0,2M + 63)/(0,8M + 17) V khng bit c M l bao nhiu nn khng th tnh pA v qa. Vy sai Cu 3: T mt qun th ca 1 loi cy c tch ra thnh 2 qun th ring bit. Hai qun th ny ch tr thnh hai loi khc nhau trong trng hp no: A. gia chng c s sai khc v thnh phn kiu gen B. gia chng c s khc bit ng k v cc c im hnh thi C. gia chng c s khc bit v tn s alen D. gia chng c s khc bit ng k v thi gian ra hoa Gii: V khc nhau v tn s alen s dn n khc nhau v tn s kiu gen, cn khc nhau v tn s kiu gen ch chc khc nhau v tn s alen. Tn s elen l vn gen ca qun th. Mt qun th trng thi cn bng v 1 gen gm 2 alen A v a, trong P(A) = 0,4. Nu qu trnh chn lc o thi nhng c th c kiu gen aa xy ra vi p lc S = 0,02. Cu trc di truyn ca qun th 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 VII.S pht sinh v pht trin s sng trn Tri t (2) 2-0-0 Cu 1: S sng u tin trn tri t ch c hnh thnh khi c s xut hin ca: A. mt cu trc c mng bao bc, c kh nng trao i cht, sinh trng v t nhn i B. mt cu trc c mng bao bc, bn trong c cha AND v protein C. mt tp hp cc i phn t gm AND, protein, lipit D. mt cu trc c mng bao bc, c kh nng trao i cht v sinh trng

IX.Sinh thi hc c th (1) 1-0-0 0 Bi 1. Su xm hi ng c im gy cht gii hn di l 9,60C, gii hn trn l 42 C. Thi gian trung bnh ca mt chu k sng l 42 ngy. 1/ Xc nh tng nhit hu hiu v s la trn 1 nm ca loi su trn. Bit nhit MT l 23,60C. 2/ Xc nh s la/nm nu nhit MT l 260C. Gii Theo gi thit "Su xm hi ng c im gy cht gii hn di l 9,60C" suy ra ngng nhit pht trin l 9,60C. 1/ Xc nh tng nhit hu hiu v s la trn 1 nm ca loi su trn. Bit nhit MT l 23,60C. Ta c cng thc: T = (x-k)n x = nhit mi trng; k = ngng nhit pht trin; n= s ngy ca 1 chu k sng sinh vt T = (23,6 - 9,6)42 = 14 x 42 = 5880C/ngy -Tng nhit hu hiu T= 5880C/ngy -S la trong nm: 365/42 = 8,69 la 19

2/ Xc nh s la/nm nu nhit MT l 260C. Ta c: 588 = (26 - 9,6) n Suy ra n = 35,85 ngy S la trong nm: 365: 35,85 = 10,18 la. Ch cho bit nhit im gy cht gii hn di l 9,60C, gii hn trn l 420C t ta i tm ngng nhit pht trin

X. Sinh thi hc qun th

(2.5)

2-1-0

XI. Qun x sinh vt (2,5) 2-0-1 Cu 1.Trong qun x,nhm loi no tuy c thng gp v phong ph thp nhng li c vai tr lm tng tnh a dng cho qun th? A. loi c trng B. loi u th C. loi th yu D. loi ngu nhin

XII. H sinh thi, sinh quyn v bo v mi trng

(4) 50

3-1-1

047: Cho 2 qun th 1 v 2 cng loi, kch thc ca qun th 1 gp i qun th 2. Qun th 1 c tn s alen A = 0,3, qun th 2 c c tn s alen A = 0,4. Nu c 10% c th ca qun th 1 di c qua qun th 2 v 20% c th ca qun th 2 di c qua qun th 1 th tn s alen A ca 2 qun th 1 v qun th 2 ln lt l: A. 0,35 v 0,4 B. 0,31 v 0,38 C. 0,4 v 0,3 D. bng nhau = 0,30 GII Gi N l s lng c th ca qun th 2, vy s lng cc th ca qun th 1 l 2N. -Tn s alen ban u ca qun th 1: p1A = 0,3, q1a = 0,7 -T l kiu gen ca qun th 1 khi cha di nhp: 0,09AA: 0,42Aa: 0,49 aa - S lng c th tng ng vi tng kiu gen ca qun th 1 l: AA = 0,09 x 2N = 0,18N Aa = 0,42 x 2N = 0,84N aa = 0,49 x 2N = 0,98N -S lng c th ca qun th 1 di c qua qun th 2 l: 2N x 10% = 0,2N. -Trong s lng tng ng vi tng kiu gen ca qun th 1 di c qua qun th 2 l: AA = 0,18N x 0,1 = 0,018N Aa = 0,84N x 0,1 = 0,084N aa = 0,98N x 0,1 = 0,098N 20

-Tn s alen ban u ca qun th 2: p2A = 0,4, q2a = 0,6 -T l kiu gen ca qun th 2 khi cha di nhp: 0,16AA: 0,48 Aa: 0,36 aa -S lng c th tng ng vi tng kiu gen ca qun th 2 khi cha di nhp l: AA = 0,16 N Aa = 0,48 N aa = 0,36 N S lng c th ca qun th 2 di c qua qun th 1 l: N x 20% = 0,2N. Trong s lng tng ng vi tng kiu gen ca qun th 2 di c qua qun th 1 l: AA = 0,16N x 0,2 = 0,032N Aa = 0,48N x 0,2 = 0,096N aa = 0,36N x 0,2 = 0,072N Kt qu sau ln di nhp th nht: - Qun th 1: S lng c th tng ng vi tng kiu gen ca mi qun th 1 sau ln di, nhp u tin: AA = 0,18N - 0,018N + 0,032N = 0,194N Aa = 0,84N - 0,084N + 0,096N = 0,852N aa = 0,98N - 0,098N + 0,072N = 0,954N Vy tn s alen ca qun th 1 sau ln di, nhp u tin: p'1 = {(2 x 0,194N) + 0,852N}/2N x 2 = (0,388N + 0,852N)/4N = 1,24N/4N = 0,31 q'1 = 1 - 0,31 = 0,69 - Qun th 2: S lng c th tng ng vi tng kiu gen ca mi qun th 2 sau ln di, nhp u tin: AA = 0,16N - 0,032N + 0,018N = 0,146N Aa = 0,48N - 0,096N + 0,084N = 0,468N aa = 0,36N - 0,072N + 0,098N = 0,386N Vy tn s alen ca qun th 2 sau ln di, nhp u tin: p'2 = {(2 x 0,146N) + 0,468N}/2N = (0,292N + 0,468N)/2N = 0,76N/2N = 0,38 q'2 = 1-0,38 = 0,62 021: 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: Gi f1, f2 ln lt l tn s trao i cho thc t ca 2 gen Xy/xY v yz/YZ. f1, f2 tnh theo%. Khong cch trn bn gen ca 2 gen Xy = f1 + f1.f2 Khong cch trn bn gen ca 2 gen yz = f2 + f1.f2 Ta c cng thc: - H s trng hp C = (tn s trao i cho kp thc t)/ (tn s trao i cho kp l thuyt) = O/E - Tn s trao i cho l thuyt l: E = f1.f2 - Khong cch thc t trn bn gen ca 2 gen = (tn s trao i cho n thc t ca 2 gen) + (tn s trao i cho kp thc t ca 2 gen) = f + O Vy ta c phng trnh: 0,7 = O/E . Suy ra O = 0,7E = 0,7 f1.f2 Vy ta c: 20 = f1 + O = f1 + 0,7 f1.f2 (1) => 0,7 f1.f2 = 20-f1 11 = f2 + O = f2 + 0,7 f1.f2 (2 )=> 0,7 f1.f2 = 11-f2 Kiu hnh khng bt cho ca F1 l kiu hnh do giao t bnh thng Xyz v xYZ kt hp vi giao t xyz thnh kiu hnh (Xyz/xyz) v (xYZ/xyz) T ta tnh c kt qu nh trn. 21

Ad BE Ad BE x aD be aD be Mi gen mi tnh tri hon ton, tn s hon v gen ca c th c v ci bng nhau: f(A/d) = 0,2, f(B/E) = 0,4; th i F1 c t l kiu hnh A-B-D-E- chim t l: A. 30,09% B. 42,75% C. 56,25% D. 75% Gii: Ad BE Ad BE Ad Ad BE BE P: x =P: ( x )( x ) aD be aD be aD aD be be Ad Ad - Vi P : x f(A/d) = 0,2 aD aD => F1: % aadd = % ad x % ad = 0,1 x 0,1 = 0,01 % A-dd = 0,25 - 0,01 = 0,24 % A-D- = 0,75 - 0,24 = 0,51 ( nh lut v php lai 2 cp gen d hp ca Nguyn T) BE BE - Vi P : x be be F1: % bbee = 0,3 x 0,3 = 0,09 % B-ee = 0,25 - 0,09 = 0,16 % B-E- = 0, 75 - 0,16 = 0,59 Ta c: % A-B-D-E- = (%A-D-) (%B-E-) = 0,51 x 0,59 = 0,3009 =30,09%

022: Nu P :

024: Cho bit mu sc qu di truyn tng tc kiu: A-bb, aaB-, aabb: mu trng; A-B-: mu . Chiu cao cy di truyn tng tc kiu: D-ee, ddE-, ddee: cy thp; D-E-: cy cao. Ad BE Ad BE P: x v tn s hon v gen 2 gii l nh nhau: f(A/d) = 0,2; f(B/E) = 0,4. aD be aD be i con F1 c kiu hnh qu , cy cao (A-B-D-E-) chim t l: A. 30,09% B. 20,91% C. 28,91% D. S khc Gii: Ad BE Ad BE Ad Ad BE BE P: x =( x )( x ) aD be aD be aD aD be be Tng t nh cu trn ta c % A-B-D-E- = (%A-D-) (%B-E-) = 0,51 x 0,59 = 0,3009 =30,09%

002: ngi c s chuyn on tng h xy ra gia NST s 13 v NST s 18. T bo gim phn sinh giao t s c ti a bao nhiu loi giao t khc nhau v ngun gc b m ca 2 cp NST ny. A. 8 B. 16 C. 20 D. 24 Gii Quy c cp NST s 18 l A v a ;cp NST s 13 l B v b. chuyn on gia cp 18 v 13 : 1. A chuyn on vi B to ra 2 NST mi l : AB v BA k trc I ca gim phn NST c dng: AABaa BBAbb k gia I ca gim phn NST c 2 kh nng Trn: {AABBBA {AABbb Di: {aa bb ; {aa BBA k cui I s c 4 kh nng: AABBBA, AABbb, aa bb; aa BBA k cui II s c: : AABBBA =>AB, ABA, ABB, ABBA, AABbb => Ab, ABb aa bb => ab 22

aa BBA => aB, aBA Kt qu: To ra 4 giao t bnh thng : AB ;Ab ;aB ;ab V 5 giao t t bin : AB BA ;AB B ;AB b ;ABA ; aBA 2.A chuyn on vi b to ra 2 NST mi l : Ab v bA k trc I ca gim phn NST c dng: AAbaa BB bbA k gia I ca gim phn NST c 2 kh nng Trn: {AAbBB {AAbbbA Di: {aa bbA ; {aa BB k cui I s c 4 kh nng: AAbBB, AAbbbA, aa bbA ; aa BB k cui II s c: : AAbBB => AB, AbB AAbbbA => Ab, AbA, Abb,AbbA aa bbA => ab, abA aa BB => aB Kt qu: To ra 4 giao t bnh thng : AB ;Ab ;aB ;ab (trng vi 1) V 5 giao t t bin : AbB, AbA, Abb,AbbA , abA 3 . a chuyn on vi b to ra 2 NST mi l : ab v ba k trc I ca gim phn NST c dng: AAaab BB bba k gia I ca gim phn NST c 2 kh nng Trn: {AA BB {AA bba Di: {aab bba ; {aab BB k cui I s c 4 kh nng: AA BB, aab bba, AA bba, aab BB k cui II s c: : AA BB => AB aab bba => a b, aba , ab b, ab ba AA bba => Ab, Aba aab BB => aB, ab B Kt qu: To ra 4 giao t bnh thng : AB ;Ab ;aB ;ab (trng vi 1) V 5 giao t t bin : aba , ab b, ab ba , Aba, ab B 4. a chuyn on vi B to ra 2 NST mi l : aB v Ba To ra 5 giao t t bin mi Vy tng s giao t c tao ra l : 5 x 4 + 4 = 24 p n D :24 023: Cho bit chiu cao cy do 5 cp gen PLL tc ng cng gp. Nu P thun chng khc nhau n cp gen tng ng, i F2 c s c th c kiu hnh chiu cao trung bnh chim t l l: A. B. 1/8 C. s khc D. Gii: Vi 5 cp gen tc ng cng gp th i F2 c t l kiu hnh tun theo h s ca cc s hng trong trin khai nh thc Newton (a+b)10. Vy ta c: C100: C101:C102:C103:C104 :C105 :C106: C107: C108: C109 :C1010 (c l t hp n phn t chp p khng lp; n =10; p = 0,1,2,...,10 . V khng c phn mn ton vit cng thc t hp) Ta c c th c kiu hnh chiu cao trung bnh chim t l (C105)/210 = (10!)/(5!)(10-5)! = (10.9.8.7.6) / 5.4.3.2 = 252/1024 = 0,24609375 ~ 1/4 V vy p n ng l C, cn gn ng l A. Sinh hc ly gn ng thi em ... 77777777777777 1,: Cho cy hoa , qu trn t th phn, ngi ta thu c i con c t l kiu hnh phn li: 510 cy hoa , qu trn: 240 cy hoa , qu di: 242 cy hoa trng, qu trn: 10 cy hoa trng, qu di. T kt qu ca php lai ny, kt lun no c rt ra l ng A. Alen qui nh mu hoa v alen qui nh qu trn cng thuc 1 NST 23

B. Alen qui nh mu hoa v alen qui nh qu di cng thuc 1 NST C. Alen qui nh mu hoa v alen qui nh qu trn lin kt khng hon ton D. Alen qui nh mu hoa v alen qui nh qu trn lin kt hon ton

2:: chuyn 1 gen ca ngi vo vi khun Ecoli nhm to ra nhiu sn phm ca gen ngi trong t bo vi khun ngi ta phi ly mARN ca gen ngi cn chuyn cho phin m ngc thnh AND ri mi gn AND ny vo plasmid v chuyn vo vi khun. Ngi ta cn phi lm nh vy l v: A. Gen ca ngi qu ln nn khng vo c t bo vi khun B. Nu khng lm nh vy th gen ca ngi khng th phin m c trong t bo vi khun C. Nu khng lm nh vy th gen ca ngi khng th dch m c trong t bo vi khun D. Nu khng lm nh vy th sn phm ca gen s khng bnh thng, khng c gi tr s dng 3: T mt qun th ca 1 loi cy c tch ra thnh 2 qun th ring bit. Hai qun th ny ch tr thnh hai loi khc nhau trong trng hp no: A. gia chng c s sai khc v thnh phn kiu gen B. gia chng c s khc bit ng k v cc c im hnh thi C. gia chng c s khc bit v tn s alen D. gia chng c s khc bit ng k v thi gian ra hoa 4: S sng u tin trn tri t ch c hnh thnh khi c s xut hin ca: A. mt cu trc c mng bao bc, c kh nng trao i cht, sinh trng v t nhn i B. mt cu trc c mng bao bc, bn trong c cha AND v protein C. mt tp hp cc i phn t gm AND, protein, lipit D. mt cu trc c mng bao bc, c kh nng trao i cht v sinh trng 5. ci to ging heo Thuc Nhiu nh Tng, ngi ta dng con c ging i Bch lai ci tin vi con ci tt nht ca ging a phng. Nu ly h gen ca c i Bch lm tiu chun th th h F4 t l gen ca i Bch trong qun th l: A. 50%. B. 75%. C. 87,5%. D. 93,25%. Cu 14: Kiu gen ca c chp khng vy l Aa, c chp c vy l aa. Kiu gen AA lm trng khng n. Tnh theo l thuyt, php lai gia cc c chp (P) khng vy s cho t l kiu hnh i F2 l A. l c chp khng vy : 2 c chp c vy. B. 3 c chp khng vy : l c chp c vy. C. 75% c chp khng vy : 25% c c vy. D. 4 c chp khng vy : l c chp c vy

Cu 20: Mt gen m ho lin tc vi khun m ho loi prtin A, sau khi b t bin iu khin tng hp phn t prtin B. Phn t prtin B t hn prtin A mt axit amin v c 3 axit amin mi. Gi s khng c hin tng d tha m, th nhng bin i xy ra trong gen t bin l: A. B thay th 15 cp nucltit. B. Mt 3 cp nucltit v thay th 12 cp nucltit. C. Mt 3 cp nucltit thuc phm vi 4 cdon lin tip nhau trn gen. D. Mt 3 cp nucltit thuc phm vi 5 cdon lin tip nhau trn gen. 45: Mt qun th trng thi cn bng v 1 gen gm 2 alen A v a, trong P(A) = 0,4. Nu qu trnh chn lc o thi nhng c th c kiu gen aa xy ra vi p lc S = 0,02. Cu trc di truyn ca qun th 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 24

C. 0,1613 AA: 0,4830 Aa: 0,3455 aa Gii:

D. 0,1610 AA: 0,4875 Aa: 0,3513 aa

- Qun th cn bng di truyn, nn ta c: pA + qa = 1 qa = 1 0,4 = 0,6 - Cu trc di truyn ca qun th cn bng l: (0,4)2AA + 2(0,4 x 0,6)Aa + (0,6)2aa = 1 0,16AA : 0,48Aa : 0,36aa -Sau khi chn lc th t l kiu gen aa cn li l: 0,36 (1 S) = 0,36(1 0,02) = 0,3528. Mt khc, tng t l cc kiu gen sau chn lc l: 0,16 + 0,48 + 0,36 (1 S) = 0,16 + 0,48 + 0,3528 = 0,9928 - Vy cu trc di truyn ca qun th khi xy ra chn lc l: (0,16/0,9928) AA : (0,48/0,9928) Aa : (0,3528/0,9928) aa 0.1612 AA : 0.4835 Aa : 0.3553 aa TR LI: Cu 15: mui st xut huyt Aedes aegypti, b gy bnh thng c mu trng c. Tnh trng mu sc thn b gy do mt gen trn nhim sc th thng quy nh. Mt t bin ln gen ny lm cho thn b gy c mu en. Trong phng th nghim, ngi ta cho giao phi ngu nhin 100 cp mui b m, thu c 10000 trng v cho n thnh 10000 b gy, trong s c 100 b gy thn en. Do mun loi b t bin ny ra khi qun th, ngi ta loi b i tt c cc b gy thn en. Gi s rng khng c t bin mi xy ra. Tn s alen ca qun th mui khi loi b b gy thn en l A. p = 0,91 v q = 0,09. B. p = 0,90 v q = 0,10. C. p = 0,80 v q = 0,20. D. p = 0,81 v q = 0,19. TR LI: Cu 2: Ngi ta lai cy c hoa vi cy c hoa trng v thu c F1ton cy c hoa . Cho cc cy F1 t th phn ngi ta thu c i F2 c t l phn li kiu hnh l 3 hoa : mt hoa trng. Sau ngi ta ly ngu nhin mt cy hoa cho t th phn. Xc sut cy ny cho ra ton hoa l bao nhiu? A.1/3 B.1/2 C. 1/9 D. 1/4 Cu 3. Mt ngi n ng bnh thng c b m bnh thng nhng c ngi em trai b bnh phninkt niu ly mt ngi v bnh thng sinh ra trong mt gia nh c b m bnh thng nhng c c em gi b bnh phninkt niu. Xc sut cp v chng ny sinh ra ngi con b bnh l bao nhiu A. 0,11 B. 0,25 C. 0,14 D. 0,32 Cu 4. Gi s trong mt qun th ngi, tn s ngi thun tay tri l 1%. Xc sut hai v chng 25

u thun tay phi sinh ra ngi con thun tay tri s l bao nhiu? Bit rng tnh trng thun tay phi l tri so vi tnh trng thun tay tri. A. 0,0160. B. 0,0081. C. 0,0018. D. 0,0025 Nu bn no qun tm ti bi tp xc sut cho k thi i hc 2010 hy gii i nh, mnh s trao i li gii v cc cu tip theo vi cc bn no quan tm. -----------------------------------------------Cu 1: Tnh trng bch tng ngi l tnh trng ln do gen a quy nh. Nu b m u d hp h sinh c 5 ngi con th kh nng h sinh 2 trai bnh thng, 2 gi bnh thng, 1 trai bch tng l bao nhiu Gii P: XXAa x XYAa (v gen nm trn NST thng) ~ (XX x XY) (Aa xAa) F1: (1/2XX : 1/2 XY) (3/4 A- : 1/4 aa) = 3/8 XXA-: 1/8XXaa: 3/8XYA-: 1/8XYaa H sinh 5 ngi con 5 ln sinh khc nhau, cng xem nh sinh 1 lc 5 ngi con khc trng (a sinh khc trng). - Xc sut ln sinh 1 con trai bnh thng l: p = 3/8 - Xc sut ln sinh 1 con gi bnh thng l: q = 3/8 - Xc sut ln sinh 1 con trai b bnh l: r = 1/8 - Xc sut ln sinh 1 con gi b bnh l: s = 1/8 (0 p, q, r, s 1 ; p + q = 0,5; r = s = 0,5) Cu 1: Tnh trng bch tng ngi l tnh trng ln do gen a quy nh. Nu b m u d hp h sinh c 5 ngi con th kh nng h sinh 2 trai bnh thng, 2 gi bnh thng, 1 trai bch tng l bao nhiu Gii P: XXAa x XYAa (v gen nm trn NST thng) ~ (XX x XY) (Aa xAa) F1: (1/2XX : 1/2 XY) (3/4 A- : 1/4 aa) = 3/8 XXA-: 1/8XXaa: 3/8XYA-: 1/8XYaa H sinh 5 ngi con 5 ln sinh khc nhau, cng xem nh sinh 1 lc 5 ngi con khc trng (a sinh khc trng). - Xc sut ln sinh 1 con trai bnh thng l: p = 3/8 - Xc sut ln sinh 1 con gi bnh thng l: q = 3/8 - Xc sut ln sinh 1 con trai b bnh l: r = 1/8 - Xc sut ln sinh 1 con gi b bnh l: s = 1/8 (0 p, q, r, s 1 ; p + q = 0,5; r = s = 0,5) - Xc sut ln sinh 1 con trai bnh thng l: p = 3/8 - Xc sut ln sinh 1 con gi bnh thng l: q = 3/8 - Xc sut ln sinh 1 con trai b bnh l: r = 1/8 - Xc sut ln sinh 1 con gi b bnh l: s = 1/8 Vy h sinh c 5 ngi con th kh nng h sinh 2 trai bnh thng, 2 gi bnh thng, 1 trai bch tng l: 2 2 81 2430 13 3 1 chnh l C52 C32 C1 = 30. 15 = = 0.07415771484375 = 0,7415% 32768 2 8 8 8 Em nn p dng cng thc tnh h s ca trin khai (a+b+c+d)n q r s p H s ca apbqcrds l Cn Cn p Cn p q Cn p q r p = 2, q = 2, r = 1, s = 0 Cu 2: S lng NST ca 1 loi vi khun l: A.2n 26

B.1 C. 2 D.khng xc nh V vt liu di truyn ca vi khun l 1 phn t ADN dng vng, khng c histon. Cha hnh thnh NST. Cha c nhn chnh thc, nn s lng NST ca 1 loi vi khun l khng xc nh

Cu 3: Nhn t hn ch s giao phi t do gia cc qun th trong cng mt loi l A.cch li a l B.cch li di truyn C.cch li c hc D.cch li tp tnh Cu 4: Mt cp b m sinh bn ngi con 1 c nhm mu A, 1 c nhm mu B, 1 c nhm mu O, 1 c nhm mu AB. Xc sut cp b m trn sinh bn ngi con u nhm mu AB. A.0,390625% B.0,15625% C.1% D.0% Kiu gen ca b m l: IAio x IBio F1: 1/4 IAIB: 1/4 IAio: 1/4 IBio : 1/4 ioio Vy x sut sinh 4 ngi con c nhm mu AB l (1/4)4 = 1/64 Cu 5: Con mui ht mu ngi v cc ng vt khc th kiu quan h ny l k sinh - vt ch hay sinh vt n ny - n sinh vt khc Tr li: - Con mui ht mu ngi cngh con a , con vt ht mu ngi. khng phi l quan h k sinh - vt ch v co mui u c trong c th con ngi 100%. Con giun a, con sn x mt th sng k sinh trong c th ngi nn l quan h k sinh - vt ch. Nh vy con mui ht mu ngi v cc ng vt khc th kiu quan h ny khng phi l k sinh - vt ch m l quan h sinh vt n ny - n sinh vt khc. - Ring k sinh trng st rt plasmodium do mui truyn, n k sinh trong t bo hng cu, th k sinh trng ny v con ngi l quan h: k sinh - vt ch Why_Why_Why_150793@yahoo.com Cu 1: Tnh trng bch tng ngi l tnh trng ln do gen a quy nh. Nu b m u d hp h sinh c 5 ngi con th kh nng h sinh 2 trai bnh thng, 2 gi bnh thng, 1 trai bch tng l bao nhiu Gii P: XXAa x XYAa (v gen nm trn NST thng) ~ (XX x XY) (Aa xAa) F1: (1/2XX : 1/2 XY) (3/4 A- : 1/4 aa) = 3/8 XXA-: 1/8XXaa: 3/8XYA-: 1/8XYaa H sinh 5 ngi con 5 ln sinh khc nhau, cng xem nh sinh 1 lc 5 ngi con khc trng (a sinh khc trng). - Xc sut ln sinh 1 con trai bnh thng l: a = 3/8 - Xc sut ln sinh 1 con gi bnh thng l: b = 3/8 - Xc sut ln sinh 1 con trai b bnh l: c = 1/8 - Xc sut ln sinh 1 con gi b bnh l: d = 1/8 Vi 5 ln sinh ta c: (a+b+c+d)5 = (a+b+c+d)3 (a+b+c+d)2 ng vi trin khai c s m a2b2c (a+b+c+d)5 = (a+b+c+d)3 (a+b+c+d)2 - Ta c: (a+b+c+d)3 = (a+b+c+d)2 (a+b+c+d) = 27

(a2+b2+c2+d2+2ab+2ac+2ad+2bc+2bd+2dc) (a+b+c+d) = a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a2b+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c - Vy: (a+b+c+d)5 = (a+b+c+d)3 (a+b+c+d)2 = (a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a2b+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (a2+b2+c2+d2+2ab+2ac+2ad+2bc+2bd+2dc) = (1) [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a2b+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (a2)] + (2) [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + 2 3 a b+b +bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (b2)] + (3) [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + 2 3 a b+b +bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (c2)] + (4) [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a2b+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (d2)] + (5) [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a2b+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2ab)] + (6) [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + 2 3 a b+b +bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2ac)] + (7) [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + 2 3 a b+b +bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2ad)] + (8) [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a2b+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2bc)] +

28

(9) [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a2b+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2bd)] + (10) [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a2b+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2cd)] = [a5+a3b2+a3c2+a3d2+2a4b+2a4c+2a4d+2a3bc+2a3bd+2a3dc + a b+a b +a2bc2+a2bd2+2a3b2+2a3bc+2a3bd+2a2b2c (1)+2a2b2d+2a2bdc + a4c+a2b2c(2)+a2c3+a2cd2+2a3bc+2a3c2+2a3cd+2a2bc2+2a2bcd+2a2dc2+ a4d+a2b2d+a2c2d+a2d3+2a3bd+2a3cd+2a3d2+2a2bcd+2a2bd2+2a2d2c ] +4 2 3

[a3b2+ab4+ab2c2+ab2d2+2a2b3+2a2b2c(3)+2a2b2d+2ab3c+2ab3d+2ab2dc + a b +b +b3c2+b3d2+2ab4+2ab3c+2ab3d+2b4c+2b4d+2b3dc + a2b2c(4)+b4c+b2c3+b2cd2+2ab3c+2b2ac2+2b2acd+2b3c2+2b3cd+2b2dc2 + a2b2d+b2d+b2c2d+b2d3+2ab3d+2b2acd+2b2ad2+2b3cd+2b3d2+2b2d2c] +2 3 5

[a3c2+ab2c2+ac4+ad2c2+2a2bc2+2a2c3+2a2c2d+2abc3+2abc2d+2adc3 + a2bc2+b3c2+bc4+bc2d2+2ab2c2+2abc3+2abdc2+2b2c3+2b2c2d+2bdc3 + a2c3+b2c3+c5+c3d2+2abc3+2ac4+2ac3d+2bc4+2bc3d+2dc4 + a2c2d+b2c2d+c4d+c2d3+2abc2d+2ac3d+2ac2d2+2bc3d+2bc2d2+2d2c3] + [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a2b+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (d2)] + [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc(5)+2abd+2adc + a b+b +bc2+bd2+2ab2+2abc(6)+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc(7)+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2ab)] +2 3

[(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a b+b +bc2+bd2+2ab2(8)+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2ac)] +2 3

[(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a2b+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2ad)] + [(a3+ab2+ac2+ad2+2a2b(9)+2a2c+2a2d+2abc+2abd+2adc + a2b(10)+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2bc)] + [(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a b+b +bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2bd)] +2 3

29

[(a3+ab2+ac2+ad2+2a2b+2a2c+2a2d+2abc+2abd+2adc + a2b+b3+bc2+bd2+2ab2+2abc+2abd+2b2c+2b2d+2bdc + a2c+b2c+c3+cd2+2abc+2ac2+2acd+2bc2+2bcd+2dc2 + a2d+b2d+c2d+d3+2abd+2acd+2ad2+2bcd+2bd2+2d2c) (2cd)] Ta c h s ca a2b2c = (1)+(2)+(3)+(4)+(5)+(6)+(7)+(8)+(9)+(10) = 2a2b2c (1)+a2b2c(2)+2a2b2c(3) +a2b2c(4)+[(2abc(5)+2abc(6)+2abc(7))(2ab)]+[2ab2(8) (2ac)]+[(2a2b(9)+ a2b(10)) (2bc)] = 2a2b2c (1) + a2b2c(2) + 2a2b2c(3) + a2b2c(4) + 4a2b2c(5) + 4a2b2c(6) + 4a2b2c(7) + 4a2b2c] + 4a2b2c(9) + 2a2b2c = 28 a2b2c Vy xc sut sinh 2 gi bnh thng, 2 trai bnh thng v 1 trai b bnh l: 28 x (3/8)2 x (3/8)2 x (1/8) ==2268/32768 =0.0692138671875 =6,92138671875% = Tha thy, thy ch em cch gii bi ny vi Bi 1: Cho cu trc di truyn ca qun th nh sau :0,4AABb : 0,4AaBb : 0,2 aabb. Ngi ta tin hnh cho qun th trn t th phn bt buc qua 3 th h .T l c th mang hai cp gen ng hp tri l bao nhiu? Gii: P: 0,4AABb : 0,4AaBb : 0,2 aabb V cc gen PLL nn ta c th xem P: (0,4AA : 0,4Aa : 0,2 aa) (0,4Bb : 0,4Bb : 0,2bb) = (0,4AA : 0,4Aa : 0,2 aa) (0,8Bb : 0,2bb) Vi cp gen

C 3 s lai t th phn P vi tn s nh sau: 0,4 (AABb x AABb) + 0,4 (AaBb x AaBb) + 0,2 ( aabb x aabb) F3 =

Bi 2 xt 3 gen lin kt theo trt t nh sau A 30 B 20 D . Mt c th d hp v 3 gen AbD/aBd c lai vi c th abd/abd, gi s rng tn s ca c th c trao i cho kp l tch cc tn s trao i cho n. T l kiu hnh A- B- D theo l thuyt l bao nhiu? Tha thy, em khng khng r v c ch trao i cho ti hai im ng thi v hai im khng ng thi l nh th no .Thy c chuyn bi dng v phn ny cho em hc hi vi ? Vo 07:09 Ngy 04 thng 5 nm 2011, tu nguyen vit: C g cn trao i em gi E-mail cho thy.

EM L MT HC SINH KHI A,NN THY C TRNG EM T U T MN SINH,NHNG EM MUN THI I HC KHI B,XIN THY C C TH CH EM CCH LM CC BI TP NY,EM RT CM N QU THY C! Cu 4: rui gim, mu sc ca thn, chiu di ca cnh v mu sc ca mt u do mt gen gm 2 alen quy nh. Bit rng gen quy nh mu sc thn v gen quy nh chiu di cnh cng nm trn mt nhim sc th thng, gen quy nh mu mt nm trn nhim sc th gii tnh. S kiu gen ti a c trong qun th khi ch xt n 3 cp gen ny l 30

A. 27.

B. 30.

C. 45.

D. 50.

Cu 6: Mt th lc bi c kiu gen AAAaaa, trong qu trnh gim phn nu khng c trao i cho xy ra th cho bao nhiu t hp giao t tham gia th tinh? A. 3. B. 6. C. 20. D. 60. Cu 7: Cy t bi c kiu gen AAaaBBbb, trong qu trnh gim phn hnh thnh giao t nu khng c trao i cho xy ra th t l loi giao t AaBb trong nhng giao t tham gia th tinh l A. 16/36. B. 1/36. C. 4/6. D. 4/36. Cu 8: ng, th tam nhim khi gim phn cho 2 loi giao t, giao t d bi (n + 1) v giao t bnh thng (n). Ht phn d bi khng sc cnh tranh vi ht phn bnh thng nn khng tham gia th tinh, cn non d bi vn tham gia th tinh bnh thng. Nu R quy nh mu v r quy nh mu trng th t l kiu hnh i con trong trng hp b c kiu gen RRr v m c kiu gen Rrr l A. 1 : 1trng. B. 11 : 1trng. C. 3 : 1trng. D. 5 : 1trng. Cu 9: Cho cu trc di truyn qun th nh sau: 0,4AABb : 0,4AaBb : 0,2aabb. Ngi ta tin hnh cho qun th trn l qun th t th phn bt buc qua 3 th h. T l c th mang hai cp gen ng hp tri l A. 112/640 . B.161/640 . C.49/256 . D. .7/640 Cu 10: Cho cu trc di truyn qun th nh sau: 0,2AABb : 0,2 AaBb : 0,3aaBB : 0,3aabb. Nu qun th trn giao phi t do th t l c th mang 2 cp gen ng hp ln sau 1 th h l A. 12,25%. B. 30%. C. 35%. D. 5,25%. Cu 11: ong mt, c 7 mu sc mt khc nhau, mi mu do mt gen ln chi phi, gm c 7 alen khc nhau, nhng mu ny l: gch: a, vng cam: av, ng: an, kem: ak, trng: at, en: ab, thm: ac. Nu ch xt ring 7 alen ln ny th s kiu gen ti a trong qun th l A. 7. B. 28. C. 35. D. 49. Cu 12: Xt 3 gen lin kt theo trt t sau: A 30 B 20 C Mt c th d hp v 3 gen AbC/aBc c lai vi abc/abc, gi s rng tn s ca cc th c trao i cho kp l tch cc tn s trao i cho n. T l kiu hnh A-B-C- theo l thuyt l A. 0,06. B. 0,03. C. 0,12. D. 0,07. Cu 13: ngi, bnh m mu do mt alen ln nm trn NST gi tnh X quy nh. Xt mt qun th mt hn o c 100 c th trong c 50 ph n v 50 n ng, hai ngi n ng b bnh m mu. Nu qun th trng thi cn bng th tn s ngi ph n bnh thng mang gen gy bnh l A. 4%. B. 7,68%. C. 96%. D. 99,84%. Cu 14: Trong mt qun th bm gm 900 con, tn s alen quy nh cu t chuyn ng nhanh ca mt enzim p = 0,7 v tn s alen quy nh cu t chuyn ng chm q = 0,3. C 90 con bm t qun th ny nhp c n mt qun th c q = 0,8. Tn s alen ca qun th mi l A. p = 0,7 v q = 0,3. B. p = 0,75 v q = 0,25. C. p = 0,25 v q = 0,75. D. p = 0,3 v q = 0,7. Cu 15: mui st xut huyt Aedes aegypti, b gy bnh thng c mu trng c. Tnh trng mu sc thn b gy do mt gen trn nhim sc th thng quy nh. Mt t bin ln gen ny lm cho thn b gy c mu en. Trong phng th nghim, ngi ta cho giao phi ngu nhin 100 cp mui b m, thu c 10000 trng v cho n thnh 10000 b gy, trong s c 100 b gy thn en. Do mun loi b t bin ny ra khi qun th, ngi ta loi b i tt c cc b gy thn en. Gi s rng khng c t bin mi xy ra. Tn s alen ca qun th mui khi loi b b gy thn en l A. p = 0,91 v q = 0,09. B. p = 0,90 v q = 0,10. C. p = 0,80 v q = 0,20. D. p = 0,81 v q = 0,19. Cu 16: rui gim, gen A quy nh thn xm l tri hon ton so vi alen a quy nh thn en, gen B quy nh cnh di l tri hon ton so vi alen b quy nh cnh ct. Hai cp gen ny cng nm trn mt cp nhim sc th thng. Gen D quy nh mt l tri hon ton so vi alen d quy nh mt trng. Gen quy nh mu mt nm trn nhim sc th gii tnh X, khng c alen tng ng trn Y. 31

(AB/ab)X^D X^d X (AB/ab)X^D Y Php lai: cho F1 c kiu hnh thn en, cnh ct, mt chim t l 15%.

Tnh theo l thuyt, tn s hon v gen l A. 20%. B. 18%. C. 15%. D. 30%. < ="-" ="text/; =utf-8">< name="ProgId" ="Word.">< name="Generator" ="Microsoft Word 11">< name="Originator" ="Microsoft Word 11"> Cu 17: mt loi thc vt, gen A quy nh thn cao, alen a quy nh thn thp; gen B quy nh qu mu , alen b quy nh qu mu vng; gen D quy nh qu trn, alen d quy nh qu di. Bit rng cc gen tri l tri hon ton. Cho giao phn cy thn cao, qu mu , trn vi cy thn thp, qu mu vng, di thu c F1 gm 10000 cy, trong c 500 cy thn cao, qu di, ; 2000 cy thn thp, qu di, . Trong trng hp khng xy ra t bin, s lai no di y cho kt qu ph hp vi php lai trn? A Aa(BD/bd) X aa(bd/bd) B (AB/ab)Dd X (ab/ab)dd C (AD/ad)Bb X (ad/ad)bb D (Ad/aD)Bb X (ad/ad)bb http://www.thpt-lequydon-quangtri.edu.vn/forum/forum_posts.asp?TID=571&get=last#1539 EM cng L MT HC SINH KHI A,NN THY C TRNG EM T U T MN SINH,NHNG EM MUN THI I HC KHI B,XIN THY C C TH CH EM CCH LM CC BI TP NY,EM RT CM N QU THY C! Cu 4: rui gim, mu sc ca thn, chiu di ca cnh v mu sc ca mt u do mt gen gm 2 alen quy nh. Bit rng gen quy nh mu sc thn v gen quy nh chiu di cnh cng nm trn mt nhim sc th thng, gen quy nh mu mt nm trn nhim sc th gii tnh. S kiu gen ti a c trong qun th khi ch xt n 3 cp gen ny l A. 27. B. 30. C. 45. D. 50. Cu 6: Mt th lc bi c kiu gen AAAaaa, trong qu trnh gim phn nu khng c trao i cho xy ra th cho bao nhiu t hp giao t tham gia th tinh? A. 3. B. 6. C. 20. D. 60. Cu 7: Cy t bi c kiu gen AAaaBBbb, trong qu trnh gim phn hnh thnh giao t nu khng c trao i cho xy ra th t l loi giao t AaBb trong nhng giao t tham gia th tinh l A. 16/36. B. 1/36. C. 4/6. D. 4/36. Cu 8: ng, th tam nhim khi gim phn cho 2 loi giao t, giao t d bi (n + 1) v giao t bnh thng (n). Ht phn d bi khng sc cnh tranh vi ht phn bnh thng nn khng tham gia th tinh, cn non d bi vn tham gia th tinh bnh thng. Nu R quy nh mu v r quy nh mu trng th t l kiu hnh i con trong trng hp b c kiu gen RRr v m c kiu gen Rrr l A. 1 : 1trng. B. 11 : 1trng. C. 3 : 1trng. D. 5 : 1trng. Cu 9: Cho cu trc di truyn qun th nh sau: 0,4AABb : 0,4AaBb : 0,2aabb. Ngi ta tin hnh cho qun th trn l qun th t th phn bt buc qua 3 th h. T l c th mang hai cp gen ng hp tri l A. 112/640 . B.161/640 . C.49/256 . D. .7/640 Cu 10: Cho cu trc di truyn qun th nh sau: 0,2AABb : 0,2 AaBb : 0,3aaBB : 0,3aabb. Nu qun th trn giao phi t do th t l c th mang 2 cp gen ng hp ln sau 1 th h l A. 12,25%. B. 30%. C. 35%. D. 5,25%. Cu 11: ong mt, c 7 mu sc mt khc nhau, mi mu do mt gen ln chi phi, gm c 7 alen khc nhau, nhng mu ny l: gch: a, vng cam: av, ng: an, kem: ak, trng: at, en: ab, thm: ac. Nu ch xt ring 7 alen ln ny th s kiu gen ti a trong qun th l A. 7. B. 28. C. 35. D. 49. 32

Cu 12: Xt 3 gen lin kt theo trt t sau: A 30 B 20

C

Mt c th d hp v 3 gen AbC/aBc c lai vi abc/abc, gi s rng tn s ca cc th c trao i cho kp l tch cc tn s trao i cho n. T l kiu hnh A-B-C- theo l thuyt l A. 0,06. B. 0,03. C. 0,12. D. 0,07. Cu 13: (cu ny em c gng lm nhng cha ra) ngi, bnh m mu do mt alen ln nm trn NST gi tnh X quy nh. Xt mt qun th mt hn o c 100 c th trong c 50 ph n v 50 n ng, hai ngi n ng b bnh m mu. Nu qun th trng thi cn bng th tn s ngi ph n bnh thng mang gen gy bnh l A. 4%. B. 7,68%. C. 96%. D. 99,84%. Cu 14: Trong mt qun th bm gm 900 con, tn s alen quy nh cu t chuyn ng nhanh ca mt enzim p = 0,7 v tn s alen quy nh cu t chuyn ng chm q = 0,3. C 90 con bm t qun th ny nhp c n mt qun th c q = 0,8. Tn s alen ca qun th mi l A. p = 0,7 v q = 0,3. B. p = 0,75 v q = 0,25. C. p = 0,25 v q = 0,75. D. p = 0,3 v q = 0,7. Cu 15: mui st xut huyt Aedes aegypti, b gy bnh thng c mu trng c. Tnh trng mu sc thn b gy do mt gen trn nhim sc th thng quy nh. Mt t bin ln gen ny lm cho thn b gy c mu en. Trong phng th nghim, ngi ta cho giao phi ngu nhin 100 cp mui b m, thu c 10000 trng v cho n thnh 10000 b gy, trong s c 100 b gy thn en. Do mun loi b t bin ny ra khi qun th, ngi ta loi b i tt c cc b gy thn en. Gi s rng khng c t bin mi xy ra. Tn s alen ca qun th mui khi loi b b gy thn en l A. p = 0,91 v q = 0,09. B. p = 0,90 v q = 0,10. C. p = 0,80 v q = 0,20. D. p = 0,81 v q = 0,19 a ch gmail ca em l: dailoivictory@gmail.com BT: Mt c th c c kiu gen AB/Ab *De/dE khi em lai phn tch th h lai, kiu gen AB/ab*De/de chim 19%. Nh vy 10 000 t bo sinh tinh trng gim phn sinh giao t th c t bo xy ra hon v gen l? Gii: Ta c: P :AB/Ab *De/dE x ab/ab *de/de = (AB/Ab x ab/ab) (De/dE x de/de) Fa: {1/2 AB/ab : 1/2Ab/ab} x {/2 DE/de : f/2 de/de : (1-f)/2 De/de : (1-f)/2 dE/de} Ta c tn s kiu gen AB/ab*De/de = {1/2} x {(1-f)/2} = {(1-f)/4 = 0,19 Suy ra: 1-f = 0,76 ; f = 0,24 Ta c cng thc tnh tn s hon v gen f: f = (s TB sinh tinh trng gim phn c HVG) / (2 x s TB sinh tinh trng gphn sinh giao t) Vy: = s TB sinh tinh trng gim phn c HVG = 0,19 x 2 x 10.000 = 3800 t bo BT: Mt ngi chn chn Vizon cho cc con chn ca mnh giao ph ngu nhin. Trung bnh 9% s chn c lng rp. ng khng cho cc chn lng rp giao phi vi nhau na. (Tnh trng lng rp l do NST thng c alen ln quy nh).T l chn lng rp th h sau l? Gii: Gi p l tn s alen ca lng khng rp (A) Gi q l tn s alen ca lng rp (a) Gi s qun th chn cn bng di truyn, ta c tn s kiu gen ban u nh sau: p2AA : 2pq Aa : q2 aa q2 = 0,09 => q = 0,3 ; p =1 - 0,3 = 0,7 ta c tn s kiu gen ban u nh sau: 0,49 AA : 0,42 Aa : 0,09 aa V ng khng cho cc chn lng rp giao phi vi nhau na nn ch c chn lng khng rp giao phi vi nhau Ta c: 0,49 AA : 0,42 Aa = (0,49/0,91) AA : (0,42/0,91) Aa = 7AA : 6 Aa 33

Tn s alen A : p' = {(7 x 2) + 6}/ 2 x (7+6) = 20/ 26 = 10/13 Tn s alen a : q' = 1- (10/13) = 3/13 T l chn lng rp th h sau l: (10/13)2 AA: 2 x (10/13) x (3/13) Aa : (3/13)2 aa = 100/169 AA : 60/169/Aa : 9/169 aa Ch : vi d kin nh trn m hi :"T l chn lng rp trong qun th th h sau?" th p s l khc, v phi tnh n nhng con chn kiu gen aa th h P c sng. : Vi 2 gen khng alen cng nm trn 1 nhim sc th, cc gen c th lin kt khng hon ton, mt gen c nhiu alen, th trong qun th c bao nhiu nhm kiu gen. Gii Gii s gen A c n alen, gen B c m alen + Xt trng hp gen B ch c 2 alen, gen A c n alen: - Nu n = 2, ta c s nhm kiu gen: 10 2 - Nu n = 3, s nhm kiu gen: 10 + ( C 3 1) (10-3) = 24 Nu n = 4, s nhm kiu gen: Nu n = 5, s nhm kiu gen: Nu n = n, s nhm kiu gen: 10 + ( C 4 1) (10-3) = 45 10 + ( C 5 1) (10-3) = 73 10 + ( C n 1) (10-3) = 10 + [7 ( C n 1)]2 2 2 2

+ Xt trng hp gen A ch c 2 alen, gen B c nhiu alen: tng t ta c 2 2 Nu n = 2, m = m, s nhm kiu gen: 10 + ( C m 1) (10-3) = 10 + [7( C m 1)] + Nu gen A c n alen, gen B c m alen, tng s nhm kiu gen l:

Thy i, bi ny gii theo cch no? C 4 gen I, II, III, IV. gen I c 2 alen, gen II c 3 alen. 2 gen ny trn NST X, khng c trn Y. Gen III c 3 alen, gen IV c 4 alen. 2 gen ny cng trn (mt-b sung ) NST thng. S kiu gen ti a l bao nhiu? Gen I (A): c 2 alen A1, A2; gen II (B): c 3 alen B1, B2, B3 nm trn NST X, khng c trn Y Chng minh: 1. Xt cc gen I, II nm trn NST X: a. C th XX c s kiu gen = 23 -Vi cc alen A1,A2&B1,B2 ta c cc nhm kiu gen: A1B1(1), A1B2(2), A2B1(3), A2B2(4), A1B1(5), A1B1(6), A2B2(7), A2B2(8), A1B1(9), A1B2(10), A1B1 A1B2 A2B1 A2B2 A1B2 A2B1 A2B1 A1B2 A2B2 A2B1 -Vi cc alen A1,A2&B1,B3ta c thm cc nhm kiu gen: A1B1(1), A1B3(11), A2B1(3), A2B3(12), A1B1(13), A1B1(6), A2B3(14), A2B3(15), A1B1(16), A1B3(17), A1B1 A1B3 A2B1 A2B3 A1B3 A2B1 A2B1 A1B3 A2B3 A2B1 - Vi cc alen A1,A2&B2,B3 ta c thm cc nhm kiu gen: A1B2(2), A1B3(11), A2B2(4), A2B3(12), A1B2(18), A1B2(19), A2B3(20), A2B3(21), A1B2(22), A1B3(23), A1B2 A1B3 A2B2 A2B3 A1B3 A2B2 A2B2 A1B3 A2B3 A2B2 b. C th XY c 6 nhm kiu gen -Vi cc alen A1,A2&B1,B2 ta c cc nhm kiu gen: A1B1(1), A1B2(2), A2B1(3), A2B2(4), Y Y Y Y 1 2 1 -Vi cc alen A ,A &B ,B3ta c thm cc nhm kiu gen: 34

A1B1(1), A1B3(5), A2B1(3), A2B3(6) Y Y Y Y - Vi cc alen A1,A2&B2,B3 ta c thm cc nhm kiu gen: A1B2(2), A1B3(5), A2B2(4), A2B3(6), Y Y Y Y 2. Gen III (D): c 3 alen D1, D2, D3, Gen IV(E): c 4 alen E1, E2, E3, E4 cng nm trn 1 NST thng a. Xt D1, D2 v E1, E2 ta c 10 nhm kiu gen b. Xt D1, D2 v E1, E3 c. Xt D1, D2 v E1, E4 d. Xt D1, D2 v E2, E3 e. Xt D1, D2 v E2, E4 f. Xt D1, D2 v E3, E4 a. Xt D1, D3 v E1, E2 b. Xt D1, D3 v E1, E3 c. Xt D1, D3 v E1, E4 d. Xt D1, D3 v E2, E3 e. Xt D1, D3 v E2, E4 f. Xt D1, D3 v E3, E4 a. Xt D2, D3 v E1, E2 b. Xt D2, D3 v E1, E3 c. Xt D2, D3 v E1, E4 d. Xt D2, D3 v E2, E3 e. Xt D2, D3 v E2, E4 f. Xt D2, D3 v E3, E4

10 + [7 ( C n 1)] + [7( C m 1)] = 10 + [7 ( C n + 7 (C n +2

2

2

2

C

2 m

2)] = 10 + [ -14 + 7 ( C n +

2

C

2 m

)=

C

2 m

)-4

21/5/2011 =============15/5/2011 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. xyz Xyz Nu P : x th t l % kiu hnh khng bt cho ca F1 l: xyz xYZ A. 70,54% B. 69% C. 67,9% D. khng xc nh c 034: Bnh m mu - lc ngi lin kt vi gii tnh. Mt qun th ngi trn o c 50 ph n v 50 n ng trong c hai ngi n ng b m mu - lc. Tnh t l s ph n mang gen bnh. A. 7,68% B. 7,48% C. 7,58% D. 7,78% 21/5/2011 TR LI CHO EM 35

Qun th ngu phi c cu trc di truyn 0,25BB : 0,5Bb : 0,25bb. Bit kiu gen BB khng c kh nng sinh sn. Sau 1 th h tn s kiu gen BB =? Gii P: 0,25BB : 0,5Bb : 0,25bb V BB khng c kh nng sinh sn nn ta c tn s cc php ngu phi P nh sau: P: [0,5/(0,5 + 0,25) Bb : 0,25/(0,5 + 0,25) bb]2 = [0,5/(0,75) Bb : 0,25/(0,75) bb]2 = [2/3 Bb : 1/3 bb]2 = 4/9 (Bb x Bb) + 4/9 (Bb x bb) + 1/9 (bb x bb) F1 = 4/9 (1/4BB:1/2Bb:1/4bb)+ 4/9(1/2Bb:1/2 bb)+1/9bb = (4/36 BB : 8/36Bb : 4/36bb) +(8/36Bb :8/36bb)+ 4/36bb = 4/36 BB : 16/36Bb : 16/36bb = 1/9 BB : 4/9Bb : 4/9bb + phi ni th ny : Sau 1 th h tn s kiu gen BB ca F1 l bao nhiu? NU HI VY TH P S L: 1/9 BB +Nu ni Sau 1 th h tn s kiu gen BB ca qun th l bao nhiu? Th phi xt 2 trng hp: 1) Sau 1 th h m P khng c ai na th p s nh trn = 1/9 BB 2) Sau 1 th h m P CN SNG CHUNG VI F1 th p s li khc: 0,25 BB : 0,75(1/9 BB : 4/9Bb : 4/9bb ) = 1/4BB : 3/36BB : 12/36Bb: 12/36bb = 12/36 BB: 12/36Bb: 12/36bb Vy tn s kiu gen BB trong qun th hin thi s l: 1/3BB Nguyn T, 0914252216 21/5/2011 Thy gii gip em bi tp ny. Thnh tht cm n Thy!

I

1

2

3

4

5

II

1

2

3

4

III IV

1 1

2

3 2

4

s ph h trn, I1 v II2 b bnh nhc c, do gen ln trn NST X quy nh. III2 b bch tng do gen ln trn NST thng quy nh (vi t l ngi mc bnh l 1/10000). Kh nng b bnh ngi IV1 v IV2 l kng r. Kh nng ngi IV1 mang gen bnh (khng b bnh ) l: A. 1/4 B. 1/2 C. 2/3 D. 99/100 KIU GEN CC NHN VT: 36

Quy c: A: Khng b nhc c; a: b nhc c; gen nm trn NST X B: Khng b bch tn; b: b bch tng; gen nm trn NST thng; tn s alen b l cn bc 2 ca 1/10000 = 0,01; tn s alen B l cn bc 2 ca 1- 0,01 =0,99 Tn s kiu gen ca bch tng trong qun th l: (0,99)2 BB : 2 x 0,99 x 0,01 Bb: (0,01)2bb = 0,9801 BB: 0,0198Bb: 0,0001bb Kiu gen gen ca cc nhn vt: III2 b bch tng do gen ln trn NST thng quy nh , nhng c b bnh nhc c khng? V trn ph h thy k hiu III2 c b bnh nhc c nh II2 v I1. ni III2 b bch tng th phi k hiu gch cho khc i phn bit vi bnh nhc c. V vy ti xt 2 trng hp: a) III2 b bch tng v khng b nhc co? b) III2 b bch tng v b nhc co? Ni tm li CHA R, CHA CHUN Trng hp a: III2 b bch tng v khng b nhc co: Kiu gen l: XAYbb Kiu gen ca nam III2 l: XAYbb; Kiu gen ca n III1 l: XAX-BTh kiu gen ca n IV1: XAX-Bb = XAXaBb = Ta c s lai F3: n III1: XAX-B- x nam III2: XaYB- = (XAX- x : XaY) (B- x B-) F4: ( XAXa: XaX- : XAY: X-Y) ( 0,9801 BB: 0,0198Bb: 0,0001bb)

I1: XaY; I2: XAX- I3: XAY; I4: XAY; I5: XAXII2: XaY suy ra: I2: XAX- l XAXa, Suy ra II1: XAX-, Suy ra III2: bbXaY 23/5/2011 TR LI CHO EM Cu 1: Mt qun th 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, qun th trng thi cn bng di truyn. Tn s ph n bnh thng nhng mang gen gy bnh l A. 0.0384. B. 0.0768. C. 0.2408. D. 0.1204. Gii: Kiu gen ca n b: XAXA : XAXa : XaXa Kiu gen n ng: XAY: XaY Gi p1 l tn s alen A , q1 l tn s alen a ca n ng Gi p2 l tn s alen A , q2 l tn s alen a ca n b Gi p l tn s alen A , q l tn s alen a ca qun th ngi chung cho c 2 gii Qun th trng thi cn ng di truyn nn p1 = p2, q1 = q2 Ta c: p = p1/3 + 2p2/3 = 3p1/3 = 3p2/3 = p1 = p2 37

q = q1/3 + 2q2/3 = 3q1 /3 = 3q2/3 = q1 = q2 Tn s kiu gen trong qun th: n ng: p2XAXA : 2pqXAXa : q2XaXa n b: pXAY: qXaY Suy ra: q = 4/100 = 0,04 => p = 1- 0,04 = 0,96 Vy tn s ngi ph n bnh thng mang gen gy bnh l: 2pq XAXa = 2 x 0,96 x 0,04 =0,0768 = 7,68% Cu 2: Kt qu ca lai phn tch mt th d hp v 3 cp gen (Mm, Nn v Pp) nm trn 1 cp nhim sc th tng ng c cho nh sau: MNP = 3 (c th) mNP + Mnp = 10 (c th) MNp + mnP = 16 (c th) MnP + mNp = 71 (c th) Tt c cc nhn nh sau y da trn s liu thc t l ng, ngoi tr: A. khong cch 2 gen N v M l 19% B. trt t ba gen l: M-P-N C. khuyt mt kiu hnh D. tn s trao i cho kp l 3% Gii: Tng s c th: 3 + 10 + 16 + 71 =100 S c th nmp = 0 l chuyn bnh thng, c th xy ra, v tn s hon v kp qu b. Nn p n C l ng. Nhng ta khng nh rng, d; vo s liu thc t th s cc th Fa (i con ca php lai phn tch) sinh ra do hon v kp l 3. MNP + mnp = 3 (c th) Vy giao t sinh ra do hon v kp l MNP v mnp Tn s trao i cho kp l 3%, nn p n D l ng. V kiu hnh MnP + mNp = 71 (c th) chim t l ln, l kiu hnh sinh ra do giao t bnh thng, nn giao t bnh thng phi cha ng thi 3 gen khng alen: M, n, P hoc m, N, p. Suy ra kiu gen ca c th em lai phn tch s l 1 trong 3 kh nng: (1) mNp ; (2) Npm ; (3) Nmp ; MnP nPM nMP Ch c kiu gen nh trng hp (1) mNp mi cho giao t hon v kp l MNP v mnp Vy nhn nh B. trt t ba gen l: M-P-N l sai. Ta chn p n B Ni thm: Trn bn gen ta c: Khong cch gia 2 gen = tn s hon v n ca 2 gen + tn s hon v kp, nn p n A l ng Cu 3: loi mo nh, cp alen D,d quy nh mu lng nm trn NST gii tnh X (DD: lng en, Dd: tam th, dd: lng vng).Trong mt qun th mo thnh ph Lun n ngi ta ghi c s liu v cc kiu hnh sau: -Mo c: 311 lng en, 42 lng vng -Mo ci: 277 lng en, 7 lng vng, 54 tam th Tn s cc alen D v d trong qun th trong iu kin cn bng ln lt l: A. 0,654 v 0,34 B. 0,726 v 0,274 C. 0,893 v 0,107 D. 0,85 v 0,15 Gii: mo con c XY, con ci XX -Mo c: 311 + 42 = 353 con -Mo ci: 277 + 7 + 54 = 338 con Tng: 353 + 338 = 691 con Gi p1 l tn s alen A , q1 l tn s alen a ca mo c 38

Gi p2 l tn s alen A , q2 l tn s alen a ca mo ci Gi p l tn s alen A , q l tn s alen a ca qun th mo chung cho c 2 gii. Ta c cng thc: p = p1/3 + 2p2/3 ; q = 1- p = q1/3 + 2q2/3 p1 = 311/353 = 0,8810 q1 = 42/353 = 0,1190 p2 = [ (277 x 2) + 54] / (338 x 2) = 608/676 = 0,8994 q2 = [ (7 x 2) + 54] / (338 x 2) = 68/676 = 0,1006 Tn s cc alen D v d trong qun th trong iu kin cn bng ln lt l: p = p1/3 + 2p2/3 = 0,8810/3 + (2 x 0,8994)/3 = 0,2936 + 0,5996 = 0,8932 q = q1/3 + 2q2/3 = 0,1190/3 + (2 x 0,1006)/3 = 0,0397 + 0,0670 = 0,1067 Vy chn C. 0,893 v 0,107 Cu 4: C c dc c 2n = 24 NST. C mt th t bin, trong cp NST s 1 c 1 chic b mt on, mt chic ca NST s 5 b o 1 on, NST s 3 lp 1 on. Khi gim phn nu cc cp NST phn li bnh thng , khng trao i cho (thm ca NT) th trong s cc loi giao t c to ra, giao t t bin c t l A. 25%. B. 12,5%. C. 75%. D. 87,5%. hi c th nh sau: giao t mang 3 NST t bin c t l Nu mang c 3 NST t bin th ()3 = 1/8 = 12,25% Cu 5: Cho php lai AaBbDdEe x AaBbDdEe. Tt c cc em u cng quy nh 1 tnh trng di truyn theo kiu tng tc cng gp. T l phn li kiu hnh xut hin trong php lai trn s l: A. 1: 4: 6: 4: 1 B. 1:8:28:56:70:56:28:8:1 C. 9:3:3:1 D. (3:1)4 27-5-2011 Ni dung cu hi: ngi bnh bch tng do gen ln nm trn NST thng quy nh. Mt qun th ngi cn bng di truyn, t l ngi mang gen d hp trong s ngi bnh thng l 1%. Tnh xc sut 1 cp v chng khng c quan h huyt thng sinh ra a con trai mc bnh? Gii: 2 2 p AA + 2pq Aa + q aa Aa/ A- = 1/100 = 0,01 2pq/ p2 + 2pq = 1/100 2pq/ p(p + 2q) = 1/100 2q/p+2q = 0,01 (1) p+q=1 (2) p = 0,995; q = 0,005 Vy: Xc sut 1 cp v chng khng c quan h huyt thng sinh ra a con trai mc bnh: l q2 = (0,005) = 0, 000025 Ch : 1 cp v chng khng c quan h huyt thng: C ngha l chn bt k 2 c th khc xa nhau v ngun gc t tin, b m ch khng phi khc nhau v kiu gen. C th c nhiu ngi hiu nhm rng: khng c quan h huyt thng l khc kiu gen, v vy c 3 kh nng v s lai: - hoc AA x Aa; - hoc AA x aa; - hoc Aa x aa Nu hiu theo cch ny th p s s khc. NGY CP NHT: 29-5-2011 CNG THC TNH TN S ALEN TRONG TRNG HP DI NHP GEN I. TRNG HP QUN TH A CHUYN SANG QUN TH B. TNH TN S ALE QUN TH B. 39

1. Bi ton tng qut: Cho bit gen A c 2 alen A, a. C 2 qun th I v II t trng thi cn bng di truyn v gen . Kch thc ca qun th I, II ln lt l N1, N2 Gi p1, q1 l tn s alen A, a ca qun th I; p2, q2 l tn s alen ca qun th II. Nu s cc th t qun th I di nhp qua qun th II chim t l x (x tnh theo % , 0 x 1) th tn s alen A, a ca qun th mi II s l bao nhiu? Tr li: p(A) = (x.p1.N1 + p2. N2)/ x.N1 + N2 q(a) = (x.q1.N1 + q2. N2)/ x.N1 + N2 p(A) + q(a) = 1 Nu N1 = N2 = N ta c: p(A) = (x..p1.N + p2. N)/ x.N + N = N (x.p1 + p2) / N (x +1) = p(A) = (x.p1 + p2) / (x +1)

2. Cho bit gen A c 2 alen A, a. C 2 qun th I v II t trng thi cn bng di truyn v gen . Gi p1, q1 l tn s alen A, a ca qun th I; p2, q2 l tn s alen ca qun th II. Nu s c th t qun th I di nhp qua qun th II chim t l x (x tnh theo % , 0 x 1) v s c th t qun th II di nhp qua qun th I chim t l y (y tnh theo % , 0 y 1) th tn s alen A, a ca qun th mi I v II s l bao nhiu Gii: 3. NGY CP NHT: 31-5-2011 Mt qun th c cu trc di truyn nh sau: 4AABb:4AaBb:2aabb. Qua 3 th h t th phn th t l c th mang 2 cp gen ng hp tri l: A. 112/640. B. 161/640. C. 49/256. D. 7/640. p n l B. Em nh thy gii gip. Em cm n thy. Gii Qua 3 th h t th phn ngha l P, F1 v F2 u t th phn v cho i F3. y bt ta tnh F3 P: 4 AABb:4 AaBb:2 aabb = 2/5 AABb: 2/5AaBb:1/5 aabb V A, a v B, b phn ly c lp nn ta xt ring tng cp gen ri tch i s. - Xt cp gen A, a ta c: P = 2/5 AA : 2/5 Aa : 1/5 aa => F3 = 2/5 AA : 2/5 (7/16 AA: 1/8 Aa : 7/16aa) : 1/5 aa = (2/5 + 14/80) AA : 2/40 Aa : 14/80 aa : 1/5 aa = 46/80 AA : 4/80Aa : 30/80 aa Xt cp gen B, b ta c: P: 4/5 Bb : 1/5 bb F3 = 4/5 (7/16 BB: 1/8 Bb : 7/16 bb) + 1/5 bb = (28/80 BB : 4/40 Bb : 28/80 bb) + 1/5 bb = = 28/80 BB : 8/80 Bb : 44/80 bb V A,a c lp vi B, b nn Qua 3 th h t th phn th t l c th mang 2 cp gen ng hp tri AABBl 46/80 x 28/80 = 1288/6400 = 0.20125 Nh vy khng c p n ng.

40

Em l hc sinh lp 12, em c c mt s v chuyn tn s hon v gen ca Thy. Em va gp mt bi tp phn phn t nhng khng gii c, nh Thy gip em Thy nh. Em cm n Thy nhiu! Bi tp nh sau: Mt phn t ARN c 4 loi nu theo t l sau: A : U : G : X = 1 : 2 : 3 : 4. T l cc loi b ba c cha 2 nu loi A l bao nhiu ? Ta c tn s cc nu: A =1/10; U = 2/10; G = 3/10; X = 4/10 T chc m b ba ng vi trin khai a thc: (A + U + G + X)3 = A3 + U3 + G3 + X3 + 3A2U + 3A2G + 3A2X + 3U2A + 3U2G + 3U2X +3G2A + 3G2U + 3G2X + 3X2A + 3X2G + 3X2U + 6XAG + 6XAU + 6AGU + 6GUX B c 2 A bao gm: 2 A+1U, 2A+1G ; 2A+1 X Tn s b ba 2A + 1U = 3 x (1/10)2 x 2/10 = 3 x 2/1000 = 6/1000 Tn s b ba 2A + 1G = 3 x (1/10)2 x 3/10 = 3 x 3/1000 = 9/1000 Tn s b ba 2A + 1X = 3 x (1/10)2 x 4/10 = 3 x 4/1000 = 12/1000 Vy: T l cc loi b ba c cha 2 nu loi A l : 6/1000 + 9/1000 + 12/1000 = 27/1000 10/6/2011 C c dc c b nhim sc th 2n = 12. Dng th ba ca loi ny c kh nng to ra bao nhiu loi giao t khng bnh thng v s lng nhim sc th? Cho rng s kt hp v phn li ca cc nhim sc th din ra hon ton ngu nhin 11-6-2011 P thun chng, F1 ng tnh. Cho F1 lai vi F1 cho ra F2. F2 c:

con ci: 3 en, x, 1 en, x. c: 30% en, x; 17.5% xm, x; 25% trng, x; 5.4% en, quon; 23.7% xm, quon; 8.4% trng, quon; 2.1% en, thng; 8.8% xm, thng; 1.6% trng, thng. Gii

Bin lun kt qu tm P Xt Mu sc, F2 c:

con ci: 4 en. con c: 30% en + 5.4% en + 2.1% en = 37,5 % en 17.5% xm + 23.7% xm + 8.8% xm = 50% xm 25% trng + 8.4% trng + 1.6% trng = 35 % trng Ta thy 37,5 % + 50% + 35 % =122,5% > 100%

Xt hnh dng, F2 c:

con ci: 4 x. c: 30% + 17.5% + 25% = 72,5% x 5.4% + 23.7% + 8.4% quon = 37,5% quon 2.1% + 8.8% = 1.6% thng = 12,5% thng 72,5% + 37,5% + 12,5% = 122,5% > 100% Vy sai ri

em tn la Huynh Bach Hu, vung tau.tel 01678356484. thay chi gium e may bai nay nha: 41

cau 1: hp t binh thng cua mt loai nguyn phn binh thng 5 t, moi trng a cug cp nguyn liu tng ng vi 496 cromatit. s lng NST trong t bao sinh dng cua th 3 kep loai trn luc cha nhn i? cu 2: kt qua xet nghim t bao hoc cua 1 ngi cho thy co 2 th barr, chung to ngi o la: a nam klinefelter b nu binh thuong c siu n d nam xxy cau 3:f1 di 2 cap gen lai voi co the khac=>F2 fan li theo kiu hinh 6 den: 1 xam: 1 nau, quy lut di truyn o day la? a cong gop 15:1 b at che 12:3;1 c at che 13:3 d bo sung 9:6:1 cau 4:o loai meo D,d quy dinh mau lng nm tren x(DD: den, Dd: tam the,dd: vang)trong 1 quan the meo co meo duc:311 den, 42 vang.Meo cai :277 den, 7 vang, 54 tam the. tan so D va d trong dk can bang la? cau 5:o ca c dc (2n=24) co cac dang the 3 ca 12 cap NST. cac the 3 nay co s lng NST trong t bao x ma va co kiu hinh nh the nao?(giong hay khac nhau) cau6: tren mARN co l=5100 Ao, co 5 riboxom tham gia dich ma 1 lan vi vn tc nh nhau =102Ao/s va cach u nhau 1 khoang =81.6Ao. thoi gian hoan tat qua trinh tong hop tren ARN ? cau 7: 1 gen co 2 alen A,a gen khac co 3 alen 2 gen nay cung nam tren NST thuong. gen khac co 4 alen nam tren NST gioi tinh.y k co alen.so to hop toi da cua ca 3 gen tren la? cau8:o nguoi kha nang cuon luoi do 1 gen troi nam tren NST thuong quy dinh.trong qun th dat cn bang co 64% nguoi cuon luoi.1 cap v chng binh thuong sinh nguoi con 1 k the cun li , nguoi 2 cun li. dua 2 ket hon vi ngi cun li kha nng sinh ra ngi con cun li la? cau 9: 1 loai co b NST lng bi 2n=12.1 hop t cua loai nay nguyn phn 3 ln lin tip cac t bao con co tng s NST =104.hop tu tren phat trin thanh a the 1 b the 3 nhim c th k d the 4 cau10: 1 loai luong boi co ADN gm 6x10^9 cap NU.O KI DAU CUA NGUYEN PHAN T BAO nay co ham lng ADN ? cau 11: o loai su co nhiet do ngng cua s phat trien la 5 d C, thoi gian 1 vong i 30 do C la 20 ngay.khi nhiet do 25 thi thoi gian 1 vong i cua loai nay? cau 12:NST co cac gen voi chieu dai nhu nhau. nst bi dut 1 doan ng vi 30mARN , chim 20% tng s gen cua NST. khi do, ADN dt bin nhn doi 1 lan da s dung 240.000 NU tu do.chieu dai cua gen la? cau 13: mi tinh trang do 1 gen quy dinh, tri la troi hoan toan.khi (AB/ab)dD X (AB/ab)dd, f= 20% xay ra voi ca 2 ben.ti le kieu 42

hinh aaB-D-? cau 14: 1 t bao sinh duong nguyen phan 1 so t va tng s t bao con lan luot tao ra la 30. so lan nguyen phan va so te bao tao ra sau nguyen phan? cau 15: s nhom gen lien ket cua 1 loai luong boi la 8. trong loai co th xut hin ti da bao nhieu dang the 3 kep? cau 16: 1 loai sinh vat co 2n=10 NST co the hinh thanh duoc ti da: a 1 the 4 b 2 loai the 4 c 5 loai the 4 d 10 la loai the 4 cau 17: c th cai co 1 cap NST trao doi doan tai 1 im, con c giam phan binh thng.qua thu tinh tao ra 512 kieu to hop. loai co bo NST co cu truc khac nhau.b NST cua loai 2n=? t bi c KG AAaaBBbb trong qu trnh gim phn hnh thnh giao t nu khng c trao i cho xy ra th t l loi giao t AaBb trong nhng giao t tham gia th tinh l? --cau 18: 4 te bao sinh giao t c cua ruoi giam giam phan thuc t cho nhiu nht bao nhiu loai tinh trung?. cu truc cac cap NST cua t bao sinh giao t c khac nhau THAY CHI E GIAI CAC BAI NAY NHA.Thy vui long giai cu th giup e va nu co cach giai nhanh thy chi gium e.e cam on thy nhiu.

11-6-2011 P thun chng, F1 ng tnh. Cho F1 lai vi F1 cho ra F2. F2 c:

con ci: 3 en, x, 1 en, x. c: 30% en, x; 17.5% xm, x; 25% trng, x; 5.4% en, quon; 23.7% xm, quon; 8.4% trng, quon; 2.1% en, thng; 8.8% xm, thng; 1.6% trng, thng. Gii

Bin lun kt qu tm P Xt Mu sc, F2 c:

con ci: 4 en. con c: 30% en + 5.4% en + 2.1% en = 37,5 % en 17.5% xm + 23.7% xm + 8.8% xm = 50% xm 25% trng + 8.4% trng + 1.6% trng = 35 % trng Ta thy 37,5 % + 50% + 35 % =122,5% > 100%

Xt hnh dng, F2 c:

con ci: 4 x. c: 30% + 17.5% + 25% = 72,5% x 5.4% + 23.7% + 8.4% quon = 37,5% quon 2.1% + 8.8% = 1.6% thng = 12,5% thng 72,5% + 37,5% + 12,5% = 122,5% > 100% 43

Vy sai ri

em tn la Huynh Bach Hu, vung tau.tel 01678356484. thay chi gium e may bai nay nha: cau 1: hp t binh thng cua mt loai nguyn phn binh thng 5 t, moi trng a cug cp nguyn liu tng ng vi 496 cromatit. s lng NST trong t bao sinh dng cua th 3 kep loai trn luc cha nhn i? cu 2: kt qua xet nghim t bao hoc cua 1 ngi cho thy co 2 th barr, chung to ngi o la: a nam klinefelter b nu binh thuong c siu n d nam xxy cau 3:f1 di 2 cap gen lai voi co the khac=>F2 fan li theo kiu hinh 6 den: 1 xam: 1 nau, quy lut di truyn o day la? a cong gop 15:1 b at che 12:3;1 c at che 13:3 d bo sung 9:6:1 cau 4:o loai meo D,d quy dinh mau lng nm tren x(DD: den, Dd: tam the,dd: vang)trong 1 quan the meo co meo duc:311 den, 42 vang.Meo cai :277 den, 7 vang, 54 tam the. tan so D va d trong dk can bang la? cau 5:o ca c dc (2n=24) co cac dang the 3 ca 12 cap NST. cac the 3 nay co s lng NST trong t bao x ma va co kiu hinh nh the nao?(giong hay khac nhau) cau6: tren mARN co l=5100 Ao, co 5 riboxom tham gia dich ma 1 lan vi vn tc nh nhau =102Ao/s va cach u nhau 1 khoang =81.6Ao. thoi gian hoan tat qua trinh tong hop tren ARN ? cau 7: 1 gen co 2 alen A,a gen khac co 3 alen 2 gen nay cung nam tren NST thuong. gen khac co 4 alen nam tren NST gioi tinh.y k co alen.so to hop toi da cua ca 3 gen tren la? cau8:o nguoi kha nang cuon luoi do 1 gen troi nam tren NST thuong quy dinh.trong qun th dat cn bang co 64% nguoi cuon luoi.1 cap v chng binh thuong sinh nguoi con 1 k the cun li , nguoi 2 cun li. dua 2 ket hon vi ngi cun li kha nng sinh ra ngi con cun li la? cau 9: 1 loai co b NST lng bi 2n=12.1 hop t cua loai nay nguyn phn 3 ln lin tip cac t bao con co tng s NST =104.hop tu tren phat trin thanh a the 1 b the 3 nhim c th k d the 4 cau10: 1 loai luong boi co ADN gm 6x10^9 cap NU.O KI DAU CUA NGUYEN PHAN T BAO nay co ham lng ADN ? cau 11: o loai su co nhiet do ngng cua s phat trien la 5 d C, thoi gian 1 vong i 30 do C la 20 ngay.khi nhiet do 25 thi thoi gian 1 vong i cua loai nay? 44

cau 12:NST co cac gen voi chieu dai nhu nhau. nst bi dut 1 doan ng vi 30mARN , chim 20% tng s gen cua NST. khi do, ADN dt bin nhn doi 1 lan da s dung 240.000 NU tu do.chieu dai cua gen la? cau 13: mi tinh trang do 1 gen quy dinh, tri la troi hoan toan.khi (AB/ab)dD X (AB/ab)dd, f= 20% xay ra voi ca 2 ben.ti le kieu hinh aaB-D-? cau 14: 1 t bao sinh duong nguyen phan 1 so t va tng s t bao con lan luot tao ra la 30. so lan nguyen phan va so te bao tao ra sau nguyen phan? cau 15: s nhom gen lien ket cua 1 loai luong boi la 8. trong loai co th xut hin ti da bao nhieu dang the 3 kep? cau 16: 1 loai sinh vat co 2n=10 NST co the hinh thanh duoc ti da: a 1 the 4 b 2 loai the 4 c 5 loai the 4 d 10 la loai the 4 cau 17: c th cai co 1 cap NST trao doi doan tai 1 im, con c giam phan binh thng.qua thu tinh tao ra 512 kieu to hop. loai co bo NST co cu truc khac nhau.b NST cua loai 2n=? cau 18: 4 te bao sinh giao t c cua ruoi giam giam phan thuc t cho nhiu nht bao nhiu loai tinh trung?. cu truc cac cap NST cua t bao sinh giao t c khac nhau THAY CHI E GIAI CAC BAI NAY NHA.Thy vui long giai cu th giup e va nu co cach giai nhanh thy chi gium e.e cam on thy nhiu.

12-6-2011 11-6-2011 P thun chng, F1 ng tnh. Cho F1 lai vi F1 cho ra F2. F2 c:

con ci: 3 en, x, 1 en, x. c: 30% en, x; 17.5% xm, x; 25% trng, x; 5.4% en, quon; 23.7% xm, quon; 8.4% trng, quon; 2.1% en, thng; 8.8% xm, thng; 1.6% trng, thng. Gii

Bin lun kt qu tm P Xt Mu sc, F2 c:

con ci: 4 en. con c: 30% en + 5.4% en + 2.1% en = 37,5 % en 17.5% xm + 23.7% xm + 8.8% xm = 50% xm 25% trng + 8.4% trng + 1.6% trng = 35 % trng Ta thy 37,5 % + 50% + 35 % =122,5% > 100% 45

Xt hnh dng, F2 c:

con ci: 4 x. c: 30% + 17.5% + 25% = 72,5% x 5.4% + 23.7% + 8.4% quon = 37,5% quon 2.1% + 8.8% = 1.6% thng = 12,5% thng 72,5% + 37,5% + 12,5% = 122,5% > 100% Vy sai ri

em tn la Huynh Bach Hu, vung tau.tel 01678356484. thay chi gium e may bai nay nha: cau 1: hp t binh thng cua mt loai nguyn phn binh thng 5 t, moi trng a cug cp nguyn liu tng ng vi 496 cromatit. s lng NST trong t bao sinh dng cua th 3 kep loai trn luc cha nhn i? cu 2: kt qua xet nghim t bao hoc cua 1 ngi cho thy co 2 th barr, chung to ngi o la: a nam klinefelter b nu binh thuong c siu n d nam xxy cau 3:f1 di 2 cap gen lai voi co the khac=>F2 fan li theo kiu hinh 6 den: 1 xam: 1 nau, quy lut di truyn o day la? a cong gop 15:1 b at che 12:3;1 c at che 13:3 d bo sung 9:6:1 cau 4:o loai meo D,d quy dinh mau lng nm tren x(DD: den, Dd: tam the,dd: vang)trong 1 quan the meo co meo duc:311 den, 42 vang.Meo cai :277 den, 7 vang, 54 tam the. tan so D va d trong dk can bang la? cau 5:o ca c dc (2n=24) co cac dang the 3 ca 12 cap NST. cac the 3 nay co s lng NST trong t bao x ma va co kiu hinh nh the nao?(giong hay khac nhau) cau6: tren mARN co l=5100 Ao, co 5 riboxom tham gia dich ma 1 lan vi vn tc nh nhau =102Ao/s va cach u nhau 1 khoang =81.6Ao. thoi gian hoan tat qua trinh tong hop tren ARN ? cau 7: 1 gen co 2 alen A,a gen khac co 3 alen 2 gen nay cung nam tren NST thuong. gen khac co 4 alen nam tren NST gioi tinh.y k co alen.so to hop toi da cua ca 3 gen tren la? cau8:o nguoi kha nang cuon luoi do 1 gen troi nam tren NST thuong quy dinh.trong qun th dat cn bang co 64% nguoi cuon luoi.1 cap v chng binh thuong sinh nguoi con 1 k the cun li , nguoi 2 cun li. dua 2 ket hon vi ngi cun li kha nng sinh ra ngi con cun li la? cau 9: 1 loai co b NST lng bi 2n=12.1 hop t cua loai nay nguyn phn 3 ln lin tip cac t bao con co tng s NST =104.hop tu tren phat trin thanh a the 1 b the 3 nhim c th k d the 4 46

cau10: 1 loai luong boi co ADN gm 6x10^9 cap NU.O KI DAU CUA NGUYEN PHAN T BAO nay co ham lng ADN ? cau 11: o loai su co nhiet do ngng cua s phat trien la 5 d C, thoi gian 1 vong i 30 do C la 20 ngay.khi nhiet do 25 thi thoi gian 1 vong i cua loai nay? cau 12:NST c