Cac Dang Bai Tap Di Truyen Hoc Quan the on Thi Dh Hsg12

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<p>S GIO DC V O TO THANH HA</p> <p>TRNG THPT L VN LINH</p> <p>SNG KIN KINH NGHIM TI:"PHNG PHP GII MT S DNG BI TP PHN DI TRUYN HC QUN TH "</p> <p>H v tn tc gi: Trnh Khc Hi Chc v: Ph t chuyn mn T: L - Ha - Sinh - Cng ngh Mn: Sinh hc SKKN thuc lnh vc phng php</p> <p>NM HC: 2010 20111</p> <p>MC LC Ni Dung A. t vn B. Gii quyt vn I. C s ca phng php gii mt s dng bi tp phn di truyn hc qun th II. Cc dng thng gp Dng 1 Dng 2 Dng 3 III. T chc thc hin C. Kt lun Trang 1 2 2 2 2 15 17 21 22</p> <p>A. T VN 2</p> <p>Trong vic hc tp phn di truyn hc qun th SH 12, vic gii bi tp c mt ngha rt quan trng. Ngoi vic rn luyn k nng vn dng, o su v m rng kin thc hc mt cch sinh ng; bi tp cng c dng n tp, rn luyn mt s k nng v hc l thuyt, nm vng kin thc. Thng qua gii bi tp, gip hc sinh rn luyn tnh tch cc, tr thng minh, sng to, bi dng hng th trong hc tp. Vic la chn phng php thch hp gii bi tp li cng c ngha quan trng hn. Mi bi tp c th c nhiu phng php gii khc nhau cng nh c nhng dng bi tp c nhng phng php gii c trng. Nu bit la chn phng php hp l v nm vng cc dng bi tp c bn thng gp, s gip hc sinh nm vng hn bn cht ca cc hin tng, c ch sinh hc. Qua qu trnh ging dy i tr, qua dy bi dng hc sinh n thi i hc, bi dng hc sinh gii nhiu nm v vic tham kho nhiu ti liu, ti tch lu c mt dng bi tp v mt s phng php gii bi tp sinh hc. Vic vn dng cc cng thc c chng minh gii cc dng bi tp sinh hc phn di truyn hc qun th t ra c nhiu tc dng, c bit l khi cc k thi ngy nay chuyn i sang phng php TNKQ. Trong trng hp ny, vic nm c cc dng bi tp v cc cng thc tng qut th hc sinh s c phng php gii hp l, tit kim c rt nhiu thi gian. Mt s tc gi khc cng cp n cch lm ny trong mt s ti liu tham kho. Tuy nhin, cng mi ch dng li vic gii mt s bi tp n l m cha c tnh khi qut, cha c tnh a dng v cc dng bi tp. Chnh v vy, ti vit ti ny nhm khi qut vic vn dng cc cng thc gii mt s dng bi tp sinh hc phn di truyn hc qun th. Thng qua ti mun gii thiu vi cc thy c gio v hc sinh mt s phng php gii bi tp sinh hc rt c hiu qu. Vn dng c phng php v cc dng bi tp ny s gip cho qu trnh ging dy v hc tp phn di truyn hc qun th c thun li hn rt nhiu. ti c vit da trn c s a ra cng thc v mt s v d in hnh khc nhau. Trong vic vn dng cng thc c vai tr quan trng. T chc ging dy mt lp, nh gi vic vn dng, p dng phng php v cc cng thc ny sau khi c hc tp. So snh kt qu lm bi vi mt lp khc khng c gii thiu vn dng cc cng thc v cc dng bi tp in hnh trong hc tp. Trn c s kt qu thu 3</p> <p>c, nh gi c u im v khi qut thnh phng php chung cho mt s dng bi tp sinh hc phn kin thc ny.</p> <p>B. GII QUYT VN I. C S CA "PHNG PHP GII MT S DNG BI TP PHN DI TRUYN HC QUN TH" C s ca phng php l cc cch xc nh tn s cc alen cc loi qun th, p dng nh lut Haci - Vanbec i cc cc gen trn nhim sc th thng v nhim sc th gii tnh cng nh xt s thay i tn s cc alen ca qun th di p lc ca cc nhn t tin ho. II. CC DNG THNG GP</p> <p>Dng 1: Cch tnh tn s cc len, tn s kiu gen v xc nh cu trc di truyn ca cc loi qun th I. Xt 1 gen gm 2 alen trn NST thng 1. Cch xc nh tn s alen, tn s kiu gen v cu trc di truyn ca qun th Xt 1 gen gm 2 alen, alen tri (A) v alen ln (a)Khi , trong QT c 3 KG khc nhau l AA, Aa, aa. Gi N l tng s c th ca QT D l s c th mang KG AA H l s c th mang KG Aa R l s c th mang KG aa Khi N = D + H + R Gi d l tn s ca KG AA d = D/N h l tn s ca KG Aa h = H/N r l tn s ca KG aa r = R/N (d + h + r = 1) Cu trc di truyn ca QT l Gi p l tn s ca alen A q l tn s ca alen a 4 d AA : h Aa : r aa</p> <p>Ta c:</p> <p>p=</p> <p>2D + H h =d+ ; 2N 2</p> <p>q=</p> <p>2R + H h =r+ 2 2N</p> <p>VD1: Xt QT gm 1000 c th, trong c 500 c th c KG AA, 200 c th c KG Aa, s cn li c kiu gen aa . a. Tnh tn s cc alen A v a ca QT. b. Tnh tn s cc KG ca QT, t suy ra cu trc di truyn ca QT. Gii: a. Ta c S c th c kiu gen aa = 1000 (500 + 200) = 300 Tng s alen trong qun th = 2x1000 = 2000 Tn s alen A = Tn s alen a =2 x500 + 200 = 0,6 2 x1000 2 x300 + 200 = 0,4 2 x1000</p> <p>b. Tn s cc kiu gen - Tn s kiu gen AA = - Tn s kiu gen Aa = - Tn s kiu gen aa =500 = 0,5 1000 200 = 0,2 1000 300 = 0,3 1000</p> <p>=&gt; Cu trc di truyn ca qun th l Tnh tn s cc alen A, a ca qun th Gii</p> <p>0,5 AA : 0,2 Aa : 0,3 aa</p> <p>VD2: Mt qun th c cu trc di truyn l 0,7 AA : 0,2 Aa : 0,1 aa</p> <p>Ta c: Tn s alen A = 0,7 + 0,2/2 = 0,8 Tn s alen a = 0,1 + 0,2/2 = 0,2 VD3: Mt qun th sc gm 1050 sc lng nu ng hp t, 150 sc lng nu d hp t v 300 sc lng trng. Bit tnh trng mu lng do mt gen gm hai alen quy nh. Tnh tn s cc kiu gen v tn s cc alen trong qun th. Gii: Ta c tng s sc trong qun th = 1050 + 150 + 300 = 1500 5</p> <p>Quy c: A: lng nu A: lng trng Tn s cc kiu gen c xc nh nh sau 1050/1500 AA + 150/1500Aa + 300/1500 aa = 1 Hay 0,7 AA + 0,1 Aa + 0,2 aa = 1 T suy ra: Tn s cc kiu gen AA, Aa v aa ln lt l 0,7, 0,1 v 0,2 Tn s alen A = 0,7 + 0,1/2 = 0,75 Tn s alen a = 0,2 + 0,1/2 = 0,25 2. Cu trc di truyn ca cc loi qun th 2.1. Cu trc di truyn ca qun th t phi (ni phi) QT t phi l cc QT thc vt t th phn, QT ng vt t th tinh, QT ng vt giao giao phi gn. a. Nu qun th khi u ch c 1 KG l Aa (P0: 100% Aa) S th h t phi 0 1 2 3 n Suy ra: T l th d hp Aa cn li 1 (1/2)1 (1/2)2 (1/2)3 (1/2)n T l th ng hp (AA+aa) to ra 0 1 - (1/2)1 1 - (1/2)2 1 - (1/2)3 1 - (1/2)n T l mi th ng hp AA hoc aa0</p> <p>[1 - (1/2)1] : 2 [1 - (1/2)2] : 2 [1 - (1/2)3] : 2 [1 - (1/2)n] : 2</p> <p>- Sau mi th h t phi, t l th d hp Aa gim mt na so vi th h trc - Khi n th t l th d hp Aa = lim [(1/2)n] = 0 T l mi th ng hp AA = aa = lim [1 - (1/2)n] : 2] = 1/2 Cu trc di truyn ca QT th h xut pht P0 l : 0 AA : 1 Aa : 0 aa Cu trc di truyn ca QT th h n l Pn:1/2 AA : 0 Aa : 1/2 aa hay 0,5 AA : 0Aa : 0,5aa b. Nu qun th t phi khi u c cu trc di truyn l P0: d AA : h Aa : r aa S th h Aa (d + h + r = 1) T l mi KG trong QT AA aa 6</p> <p>t phi 0 1 2 3 n Ch :</p> <p>h (1/2)1. h (1/2)2. h (1/2)3. h (1/2)n. h</p> <p>d d + [h - (1/2)1 . h] : 2 d + [h - (1/2)2 . h] : 2 d + [h - (1/2)3 . h] : 2 d + [h - (1/2)n . h] : 2</p> <p>r r + [h - (1/2)1 . h] : 2 r + [h - (1/2)2 . h] : 2 r + [h - (1/2)3 . h] : 2 r + [h - (1/2)n . h] : 2</p> <p>- Qu trnh t phi lm cho QT dn dn phn thnh cc dng thun c kiu gen khc nhau. - Cu trc di truyn ca QT t phi bin i qua cc th h theo hng gim dn t l d hp, tng dn t l ng hp nhng khng lm thay i tn s cc alen. VD: Cho 2 QT: QT1: 100% Aa QT2: 0,7AA + 0,2 Aa + 0,1 aa = 1 a. Tnh tn s cc alen A v a mi QT. b. Xc nh t l th d hp cn li v t l mi th ng hp to ra mi QT sau 5 th h t phi. Gii: a. - QT1: Tn s alen A = a = 1/2 = 0,5 - QT2: Tn s alen A = 0,7 + 0,2/2 = 0,8 Tn s alen a = 0,1 + 0,2/2 = 0,2 b. - QT1: T l th d hp cn li sau 5 th h t phi l 1/25 = 0,03125 T l mi th ng hp to ra l AA = aa = [1 - (1/2)5] : 2 = 0,484375 - QT2: T l th d hp cn li sau 5 th h t phi l 0,2x1/25 = 0,00625 T l th ng hp AA to ra l = 0,7 + [0,2 - (1/2)5 . 0,2] : 2 = 0,796875 T l th ng hp aa to ra l = 0,1 + [0,2 - (1/2)5 . 0,2] : 2 = 0,196875 * Ch : Nu qu trnh ni phi din ra yu th vic xc nh thnh phn KG ca QT c xc nh nh sau Gi H1 l tn s th d hp Aa b gim i do ni phi qua mt th h. 7</p> <p>F l h s ni phi Ta c F = (2pq H1)/2pq T suy ra Tn s KG AA = p2 + pqF = p2 (1 - F) + pF Tn s KG Aa = H1 = 2pq (1 - F) Tn s KG aa = q2 + pqF = q2 (1 - F) + qF 2.2 Cu trc di truyn ca qun th ngu phi a. Qun th ngu phi - L QT m cc c th trong QT la chn bn tnh giao phi mt cch ngu nhin. - QT ngu phi c th duy tr tn s cc alen v tn s cc KG qua cc th h duy tr s a dng di truyn. - QT giao phi to ra v s bin d t hp, v vy lm cho QT a hnh v KG, dn n a hnh v KH. b. Trng thi cn bng di truyn ca qun th Qun th t trng thi cn bng di truyn nu c tn s cc kiu gen tho mn cng thc p2AA + 2pq Aa + q2 aa = 1 Trong p l tn s alen A q l tn s alen a (p + q = 1) Hoc Qun th c cu trc di truyn dng d AA : h Aa : r aa s t cn bng di truyn nu tho mn biu thc dr = (h/2)2 VD1: QT no sau y t cn bng DT QT1: 0,36AA + 0,60 Aa + 0,04 aa = 1 QT2: 0,64AA + 0,32 Aa + 0,04 aa = 1 QT3: 0,7AA + 0,2 Aa + 0,1 aa = 1 QT4: 0,36AA + 0,48 Aa + 0,16 aa = 1 Gii: p dng 1 trong 2 cng thc trn ta thy QT c cu trc di truyn t cn bng l QT2 v QT4 VD2: Mt QT ngu phi cn bng di truyn c tn s cc alen A/a = 0,3/0,7. 8</p> <p>Xc nh cu trc di truyn ca QT. Gii: Cu trc di truyn ca qun th l 0,09AA + 0,42 Aa + 0,49 aa = 1 VD3: Chng bch tng ngi do t bin gen ln trn NST thng gy nn. Tn s ngi bch tng trong QT ngi l 1/10000. Bit qun th t cn bng di truyn. Xc nh tn s cc alen v cu trc di truyn ca QT. Gii: T gi thuyt suy ra: Tn s ngi bch tng trong qun th l q2 = 1/10000 = 0,0001 --&gt; q = 0,01 --&gt; Tn s alen ln (b) gy bch tng = 0,01 --&gt; Tn s alen tri (B) l p = 1 - 0,01 = 0,99 --&gt; Cu trc di truyn ca qun th l 0,992 BB + 2x0,99x0,01 Bb + 0,012 bb = 1 Hay 0,9801 BB + 0,0198 Bb + 0,0001 bb = 1 c. nh lut Haci Vanbec - Ni dung: i vi qun th ngu phi, trong nhng iu kin nht nh th thnh phn KG v tn s cc alen c duy tr n nh qua cc th h. - Chng minh nh lut: Xt mt gen vi 2 alen, trong qun th c 3 kiu gen AA, Aa, aa vi cc tn s tng ng l d, h, r. Trong qun th, s ngu phi din ra gia cc c th c cng hay khc kiu gen vi nhau. Nh vy, trong qun th c nhiu cp lai khc nhau. Tn s ca mi kiu lai bng tch cc tn s ca hai kiu gen trong cp lai. V d: AA x AA = d.d = d2. Kt qu ngu phi trong qun th c phn nh bng di y Kiu lai AA x AA AA x Aa Aa x AA 9 Tn s kiu lai d2 2dh AA d2 dh dh Th h con Aa</p> <p>aa</p> <p>AA x aa aa x AA Aa x Aa Aa x aa aa x Aa aa x aa Tng</p> <p>2dr</p> <p>2dr</p> <p>h2 2hr</p> <p>1/4h2</p> <p>1/2h2 hr</p> <p>1/4h2 hr</p> <p>r2 (d+h+r)2 = 1</p> <p>(d+1/2h)2 = p2</p> <p>r2 2(d+1/2h)(r+1/2h)=2pq (r+1/2h)2 = q2</p> <p>T bng trn ta thy, phn th h con c sn sinh ra t mt trong 9 kiu lai tng ng vi tn s ca mi kiu lai, v d: Aa x Aa = h2 th th h lai c c 3 kiu gen AA, Aa, aa vi cc tn s tng ng l 1/4h2, 1/2h2, 1/4h2. Qua bng trn cn cho thy th h con, t l ca AA l p2, ca Aa l 2pq, ca aa l q2. Nh vy, qua ngu phi tn s cc kiu gen qun th khi u l d, h, r thnh p2, 2pq, q2 tng ng th h tip theo. T tn s ca cc kiu gen c th xc nh c tn s alen th h sau: Gi thit p1 l tn s ca alen A th h con th: p1 = p2 + 1/2(2pq) = p2 + pq = p (p+q) = p Vi tn s ca alen a cng xc nh tng t nh trn. Qun th p2 : 2pq : q2 khi ngu phi tip theo th (pA+qa)x(pA+qa) = p2AA : 2pq Aa : q2 aa T cho thy tn s tng i ca mi alen v tn s cc kiu gen c khuynh hng khng i qua cc th h khi c s ngu phi din ra. - ng dng nh lut Haci - Vanbec + Xt 1 QT c cu trc di truyn trng thi cn bng l P0: 0,36 AA : 0,48 Aa : 0,16 aa Suy ra: pA = t l % s loi giao t mang A ca QT = 0,6 qa = t l % s loi giao t mang A ca QT = 0,4 10</p> <p> th h ngu phi tip theo, cu trc di truyn ca QT c xc nh nh sau 0,6A 0,4a 0,6A 0,36 AA 0,24 Aa 0,4a 0,24 Aa 0,16 aa</p> <p> cu trc di truyn ca QT vn l: 0,36 AA : 0,48 Aa : 0,16 aa thnh phn KG v tn s alen khng thay i so vi th h trc. + Xt 1 QT c cu trc di truyn khng t cn bng l P0: 0,68 AA : 0,24 Aa : 0,08 aa Suy ra: pA = t l % s loi giao t mang A ca QT = 0,8 qa = t l % s loi giao t mang A ca QT = 0,2 th h ngu phi tip theo, cu trc di truyn ca QT c xc nh nh sau 0,8A 0,2a 0,8A 0,64 AA 0,16 Aa 0,2a 0,16 Aa 0,04 aa</p> <p> Cu trc di truyn ca QT th h tip theo t cn bng di truyn l: 0,64 AA : 0,32 Aa : 0,04 aa Nu th h xut pht QT khng t trng thi cn bng di truyn th ch qua 1 th h ngu phi QT s t cn bng (L giao phi n nh). VD: Cho QT c cu trc DT l 0,7AA + 0,2 Aa + 0,1 aa = 1 Xc nh cu trc DT ca QT sau 6 th h ngu phi. Gii: Ta c: Tn s alen A = 0,7+0,2/2 = 0,8 Tn s alen a = 0,1+0,2/2 = 0,2 Cu trc di truyn th h th nht (P1) l 0,64 AA + 0,32 Aa + 0,04 aa = 1 P1 t cn bng di truyn nn P6 cng c cu trc di truyn nh P1 * iu kin nghim ng nh lut Haci - Vanbec + QT phi c kch thc ln. + Cc c th phi ngu phi. + Sc sng v kh nng sinh sn ca c KG khc nhau phi nh nhau. + Khng c B (hoc B thun = B nghch), CLTN, di nhp gen 11</p> <p>II. Xt gen a alen nm trn NST thng V d: Gen quy nh tnh trng nhm mu ngi gm 3 alen l IA, IB, Io. Trong IA = IB&gt; Io. Gi p, q, r ln lt l tn s ca cc alen IA, IB, Io. (p + q + r = 1) S ngu phi to ra trng thi cn bng di truyn v tnh trng nhm mu nh sau (pIA : qIB : rIo)2 = p2IAIA : 2pq IAIB : q2IBIB : 2qr IBIo : r2IoIo : 2pr IAIo Kiu gen IAIA IAIo IBIB IBIo IAIB IoIo Tn s alen Io = r 2 = o Tn s alen IA: Ta c: p2+2pr+r2 = a + o (p+r)2 = a+o p=a+o - r = a+o o</p> <p>Tn s kiu gen p2 2pr q2 2qr 2pq r2</p> <p>Kiu hnh Mu A Mu A Mu B Mu B Mu AB Mu O</p> <p>Gi a, b, o ln lt l tn s kiu hnh ca cc nhm mu A, B, O</p> <p>Tn s alen IB = 1 p r hoc c th tnh tng t nh tnh tn s IA q2+2qr+r2 = b + o (q+r)2 = b+o q=b+o - r = b+o o</p> <p>Do p + q + r = 1 a + o - o + b + o - o + o = 1 T , suy ra cng thc p = 1 - b+o q = 1 - a+o r= o VD1: Mt qun th ngi t cn bng di truyn. Xt gen quy nh tnh trng nhm mu gm 3 alen l IA, IB v Io. Bit tn s cc alen IA , IB, Io ln lt bng 0,3; 0,5; 0,2. Xc nh cu trc di truyn ca qun th. Gii: Tn s cc kiu gen ca qun th c xc nh qua bng sau pIA = 0,3 qIB = 0,5 rIo = 0,2 12</p> <p>pIA = 0,3 qIB = 0,5 rIo = 0,2</p> <p>0,09IAIA 0,15IAIB 0,06IAIo</p> <p>0,15IAIB 0,25IBIB 0,10 IBIo</p> <p>0,06IAIo 0,10 IBIo 0,04 IoIo</p> <p> Qun th c cu trc di truyn trng thi cn bng l 0,09IAIA : 0,3 IAIB : 0,25IBIB : 0,2 IBIo : 0,04IoIo : 0,12 IAIo VD2: Tn s tng i ca cc nhm mu trong QT ngi l: Mu A: 0,45; B: 0,21; AB: 0,3; O: 0,04. Bit qun th t cn bng di truyn. a. Tnh tn s cc alen IA, IB v Io. b. Xc nh cu trc di truyn ca qun th. Gii: a. Gi tn s cc alen IA, IB v Io ln lt l p, q, r Ta c p = 1 - 0,21 + 0,04 = 0,5; q = 1 - 0,45 + 0,04 = 0,3; r = 0,04 = 0,2</p> <p>b. Cu trc di truyn ca qun th trng thi cn bng l 0,25IAIA : 0,3 IAIB : 0,09IBIB : 0,12 IBIo : 0,04IoIo : 0,2 IAIo III. Xt gen trn NST gii tnh 1. Xt gen trn NST gii tnh X (Khng c alen tng ng trn Y) Xt 1 gen trn NST gii tnh X gm 2 alen A v a Qu trnh ngu phi to ra 5 kiu gen nh sau: Gii ci: XAXA, XAXa, XaXa. Gii c: XAY, XaY. Gi N1 l tng s c th ci N2 l tng s c th c D l s lng c th mang kiu gen XAXA R l s lng c th mang kiu gen XAXa H l s lng c th mang kiu gen XaXa K l s lng c th mang kiu gen XAY L l s lng c th mang kiu gen XaY Gi p l tn s alen A, q l tn s alen a (p + q = 1) Ta c: p=2 xD + R + K 2 xN1 + N 2 2 xH + R + L 2 xN1 + N 2</p> <p>q=</p> <p>13</p> <p>- Cu trc di truyn ca qun th trng thi cn bng l 1/2(p2 XAXA : 2pq XAXa : q2 XaXa) : 1/2(p XAY : q XaY) 2. Xt gen trn NST gii tnh Y (Khng c alen tng ng trn X) - Xt 1 gen trn NST gii tnh Y gm 2 alen A v a Qu trnh ngu phi to ra 2 kiu gen gii c nh sau: XYA v XYa Gi N l tng s c th c K l s lng c th c mang kiu gen XYA L l s lng c th c mang kiu gen XYa Gi p l tn s alen A, q l tn s alen a (p + q = 1) Ta c: p=K N L N</p> <p>q=</p> <p>- Cu trc di truyn ca qun th trng thi cn bng l 1/2XX : 1/2 (p XYA : q XYa) 3. Xt gen nm trn vng tng ng ca NST X v Y Xt 1 gen gm 2 alen A v a nm trn vng tng ng ca X v Y. Gi p, q ln lt l tn s cc alen A v a. Khi cu trc di truyn ca qun th c xc nh nh trong trng hp gen nm trn NST thng. Ta c cu trc di truyn ca qun th trng thi cn bng di truyn l p2 (XAXA + XAYA) : 2pq (XAXa+ XAYa+ XaYA) : q2 (XaXa+ XaYa) VD1: loi mo nh, cp alen D v d quy nh tnh trng mu lng nm trn NST gii tnh X. DD: lng en; Dd: lng 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, 20 lng vng, 54 lng tam th. Bit qun th t cn bng di truyn. a. Hy tnh tn s cc alen D v d. b. Vit cu trc di truyn ca qun th. 14</p> <p>Gii a. p dng cng thc trn, ta c Tn s alen D = Tn s alen d =2 x 277 + 54 + 311 = 0,871 2 x351 + 353 2 x 20 + 54 + 42 = 0,129 2 x351 + 353</p> <p>b. Cu trc di truyn ca qun th 1/2(0,8712XDXD +2x0,871x0,129 XDXd +0,1292 XdXd)+1/2(0,871 XDY+0,129...</p>