Hay cho bi~t thai gian th8 M va t6c dQ sinh tnrong rieng cua loai sinh v~t tren?. Hay xac dinh bQ nhi€m s~c th~ luong bQi cua loai, bb[r]
(1)so GlAo Dl)C vA DAo T~O I LAo CAl CUQC TID cAP TiNH Gill ToAN TREN MAy TiNH CASIO, VINACAL NAM HOC 2012-2013 Mon thi: Sinh hoc (Chuong trinh THPT) Ngay thi: 17/0112013 Huang ddn g6m co 10 cdu, C"au 1: (5 ditern A ) Ch ba ng sau: Thai gian So Ian phan chi a , 0 30 60 90 trang 2n 2u 21 22 2.1 Sotebaocuaquanth~ a Hay cho bi~t thai gian th8 M va t6c dQ sinh tnrong rieng cua loai sinh v~t tren? b Gia su cay mQt hrong 5.104 te bao VSV tren vao moi tnrong dinh duong A, nhan thAy pha can bang dat diroc sau gic 30 phut voi t6ng s6 te bao hi 2048.1 05 t~ bao Lieu vi khuan tren co ; p haat h ay kh"ong.? qua p h a tiem Cach giai K~t qua Di~m a Tic bang s6 Ii~u tren ta thfiy thai gian the h~ cua loai sinh v~t tren la g= 30 phut T6c dQ sinh tnrong rieng u= Ilg = 110,5 = b Sau gio 30 phut = 390 phut chung VSV tren da thuc hien s6 Iftn phan chia la: Ap dung cong thirc: N, = Nox2n, suy 2n = NtINo = 4096, suy n=12 T6ng thai gian thuc t~ cftn cho 121ftn phan chia la: 12x30= 360 phut, V~y chu VSV co trai qua pha tiem phat voi thai gian la: 390 - 360 = 30 phut g= 30 phut, J.l=2 I Co trai qua pha ti~m phat voi thai gian la 30 phUt Cau (5 di~m): mQt loai, qua trinh phat sinh giao nr n8u co trao d6i cheo W mQt di8m tren c~p nhi€m s~c th~ wong d6ng thi s6 loai giao til t6i da co th8 dat diroc la 32 a Hay xac dinh bQ nhi€m s~c th~ luong bQi cua loai, b Trong vung sinh san cua 6ng dan sinh due a ca th~ due co s6 t€ bao sinh due So' khai nguyen phan Iftn lien tiep Co 87, 5% te bao chuyen sang vung chin tro te bao sinh tinh Trong s6 cac t~ bao tinh trung tao chi co 25% s6 tinh trung chua X va 12,5 % s6 tinh trung chua Y tham gia thu tinh; cac tinh trung thu tinh tao 168 hQ'P tu, Tfnh s6 te bao sinh due So' khai d\lc da sinh cac loai tinh trung Cach giai Ket qua Diem a Xac dinh bQ NST luang bQi 2n cua loai: Loai co n c~p NST rna m6i c~p NST co cAu true khac nhau, co k c~p NST rna rn6i c~p eo trao d6i doan tai mQt di~m (k<n), 2n+k so; IO~lgIaO ill = n Ta co: + = 32, suy n=4 2n= 2,0 V~y b9 NST IUO'ng b9i cua loai 2n = b Goi s6 t~ bao sinh due due So' khai la k Ta c6 s6 t€ bao sau lfin nguyen phan la: k x 25=32k (te bao) S6 te bao chuyen sang vung chin tro te bao sinh tinh la: 0,875 x 32k = 28k (t€ bao) M6i te bao sinh tinh giilm phan cho giao tl'rX va giao tir Y S6 giao til X t~o = 2x 28k = 56k ( giao tir) S6 giao tu' Y t~o = 2x 28k = 56k ( giao tu) - 1- (2) x M~t khac, so giao tir tham gia thu tinh tao hop tir la: 0,25 x 56k =14 k (giao tir) s6 giao tir Y tham gia thu tinh tao hop tir la: 0,125 x 56k =7 k ( giao nr) Ta co 14k + 7k = 168; 21k = 168, suy k = 168/21=8 (te bao) S6 t~ bao = 3,0 Ciill 3: ( diem) KMo sat luu hrong tim va the tich tam thu, tam tnrong (] mQt s6lmli dQng v~t, nguoi ta thu duqe b'ang so~ rieu sau day: Lofti Nguoi Ch6 Chi s6 The tich tam thu ( ml) 60 580 15 The tich tam tnrong (ml) 120 1160 25 4,5 Luu hrong tim ( llt/phut) 34,8 1,2 , a Tinh nhip tim ella mal loai? b Tfnh thai gian mQt ehu ky tim cua ml)i loai? e N~u mQt ehu ky tim ti l~ thoi gian cua pha co tam nhi: pha co tam thk pha dan chung = 1: 3: thi mQt ehu ki tim thai gian nghi cua tam thit, thai gian nghi cua mm nhi (] m6i loai bing bao nhieu? A A Bo K€tqui Cach giii a Nhip tim cua moi loai: - Nhip tim cua loai Nguoi= 4500: (120-60) = 75 (nhip) - Nhip tim cua loai Bo= 34800: (1160-580) = 60 (nhjp) - Nhjp tim cua loai Ch6= 1200: (25-15) = 120 (nhjp) , b Thai gian ehu kl tim cua m6i loai: - Loai Nguoi: 60175 = O,S giay Di~m - Nhip tim cua loai Nguoi= 75 (nhip) - Nhip tim ella loai Bo= 60 (nhjp) - Nhip tim ella loai Ch6= 120 (nhjp) Thai glan ehu ki tim: -Ngum: 0,8 giay - Loai Bo: giay - Loai Ch6:0,5 giay - Loai Bo: 60/60 = giay, - Loai Ch6: 601120 = 0,5 giay c Thai gian nghi ngoi cua tam nhi va tam thAt chu ki tim cua ml)i 10l1ilit: - LOlli ngiroi: pha co tam nhi:O,8xl/8= 0,1 giay pha co tam thit: 0,8x3/S = 0,3 giay, pba dan chung: 0,8x4/8 = 0,4 giay Suy thai gian nghi ngoi cua tam nhl = 0,8 - 0,1 = 0,7 giay thai gian nghi ngoi cua tam ~hit = 0,8 - 0,3 = 0,5 giay - Loai Bo: pha co tam nhi: lx1l8= 0,125 giay pha eo tam thAt: 1x3/8 = 0,375 giay pha dan ehung: lx4/8 = 0,5 giay Suy thai gian nghi ngoi cua tam nhi =1 - 0,125 = 0,875 giay thoi gian nghi ngoi cua tam thAt= - 0,375 = 0,625 giay - Loai Ch6: pha eo tam nhi: 0.5x1/8= 0,0625 giay pha eo tam thAt: Ix3/8 = 0,1875 giay pha dan chung: 1x4/8 = 0,25 giay Suy thai gian nghi ngoi cua tam nhi =0,5- 0,062=0,4375 giay, thai gian nghi ngoi cua tam thftt=0,5-O, 1875=0,3125 giay Thai gian nghi ngoi cua mm nhi va tam th~t ehu kl tim cua m6i loai 1<\: -Loai Ngiroi: Tam nhi=0,7 s Tam thit=0,5 s 0.5 0.5 0.5 0.5 0.5 0.5 -Loai Bo: Tam nhi=0,875 s Tam thftt=0,625 s -Loai Ch6: Tam nhi=0,4375 s Tam thfrt=0,3125 s Ca.u 4: ( diem) MQt gen c§.u true & sinh v~t nhan sa tlJ lien tiep dot doi hoi moi tnrong nQi bao cung cftp 18600 guanin Ti I? nucleotit guanin va mQt Ioai nucleotit khong bd sung la 2/3 M6i gen deu phien rna IAn -2- (3) a Tlnh 56 luong nucleotit tirng loai cua gen? b S6 hrong nucleotlt m6i loai moi tnrong cung cAp eho gen t6ng hQP mARN, bi~t ring phan tli mARN co ti l~ A:U:G:X;= 8:4:2:1 c N~u m6i phan tir mAR~ co 10 riboxom tnrot qua l~n de t6ng hQP protein Tfnh s6 hrong axit amin moi tnrong cung cap eho qua trlnh tbng hQP protein tren cac mARN? Cach giii a Tinh so hrong nucleotit tung loai ella gen: S6 nuleotit loai G la: G = Gmt/(25 -1) = 18600/31 = 600 nu s6 nucleotit A=T = 3/2G = 600x3/2 = 900 nu b S6 nucleotit tren phan nr mARN = A+G = 1500 nu s6 nucleotit nrong loai cua mARN: Am:: 1500 xS/15 = SOOnu Urn = 1500x4/15 = 400 nu Om= 1500x2/15 = 200nu Xm = 1500xll15 = 100 nu se phan nr mARN diroc tao = 25 x2 = 64 phan nr 86 nucleotit tirng loai moi tnrong cung cAp: Amt = Am X 64 = 800x64 = 51200 nu Umt = Urn X 64 = 400 x64=25600nu Gmt= Gm X 64 = 200x64 = 12800nu Xmt = Xm X 64 = 100x64=6400 nu c axit amin moi tnrong eung cAp eho t6ng hop phan nr protein = 1500/3-1=499 a.a S6 phan tli protein dugc tao thanh: = 25x2xl0=640 phan tir Tang s6 axit amin rnoi tnrong cung cAp cho t6ng hop protein: = 640 X 499 = 319360 a.a K~t qua Diem G=X = 600 nu A=T=900nu 1 s6 nu tirng loai moi truong eung A cap: Amt = 51200 nu Umt = 25600nu Gmt= 12800nu Xmt = 6400 nu se 56 a.a moi tnrong cung cAp = 319360 a.a 1 Can (5 di~m) Xet mQt e~p gen ella mQt loai tll ph6i, a) Th~ h~ ban d~u ella mQt quAn th~ co ph~n ki8u gen la 300 AA + 600 Aa + 100 aa Qua nhieu th~ h~ tv ph6i, quAn th~ da: phan hoa cac dong thuAn ve ki€u gen AA va aa Tfnh ti I~ cac dong thuan v8 gen kieu gen AA va aa hinh quin b) MQt quan th~ khac ella loai co ph~n ki~u gen th~ h~ ban dAu la 0,36AA + 0,64Aa Do khong thich nghi voi di~u ki~n s6ng, tAt colcac ca th€ mang kieu gen aa cac the h~ d~u ch~t Tfnh ti I~ kieu gen cua quAn the sau th~ h~ the a a Cach giiii K~tqui a) Khi quan the phan hoa cac dong thuan thi ti I~ Aa = 0, ti 1~ a Aa ban d.1u chia d8u cho cac kieu gen d6ng hQP, nen ti 1~ cac kiSu TI dong thu~n v~ gen hinh quAn the nay: Kieu gen AA = 0,6 Ti I~ dong thuan ve _ Ki€u en AA = 300 + 600 : 06 Ki~u gen aa = 0,4 g 300+600+100' f)i~m .I~ = _ Kieu 1,0 = en aa = 100 + 600 : 04 g 300 + 600 + 100 ' b) Ti I~ kieu gen cua qu~n the sau th~ h~ : _ Ti I~ kieu gen AA = - Ti l~ ki~u gen Aa = 1,0 b Ti l~ kieu gen AA = 0,6190 = 13/21 ~0,6190 0,36 + 0,64: (0,36 + 0,64: 4) +0,64: 0,64: (0,36 + 0,64: 4) +0,64: = 8/21 ~0,3810 1,5 Ti 1~ ki8u gen Aa 0,3810 = 1,5 Cau 6: (5 diem) d ru6i giftm, gen A qui dinh mit do, gen a qui dinh mit triing ; gen B qui dinh canh xe va gen b qui djnh canh thirong Phep lai gitta ru6i gifim cai m~t do, canh xe vai ru6i giftm dllc m~t do, canh xe -3- (4) da thu ?uQ'~ F I ru6i cai 100% mit do" canh xe ; ru6i due g6m eo10~ dire mit go, canh due mat n:ang, can~ xe,: 10% due ma! do, canh xe : 10% d\fc mat trang, canh thuong b o my va tan so hoa gen cua oan v! gen neu co Cach giii Ketqui HS Bien lu~n tim quy lu~t di truyen: => c~p tfnh trang mau mit va dang canh cua ru6i gifun di truyen lien k~t khong holm toan tren NST gioi tinh X (khong co P: x=x= x XABy alen tren NST gioi tinh Y) - FIco 40% due mit do, canh thuong (XADy) : 40% d\lc mit trang, canh xe (XaBy) sinh til giao tir lien k~t cua ru6i gifun - Tfin s6 hOan vi cai ~ ki~u gen cai (1 P Ia XAbXaB gen: 20% - F I co 10% dire mit do, canh xe (XABy): 10% dgc mit trang, canh thirong (Xaby) sinh tir giao tir hoan vi gen cua ru6i gifun cai ¢ tAn s6 hoan vi gen = 10% + 10% = 20% - Ki~u gen cua ru6i giAm d\lc mit do, canh xe (1 P hi XABy A 'A A thuong : 40% Xac dinh ki6u A' Diem 1,0 0,5 1,5 1,5 0,5 Call 7: (5 diem) : Xet qua trinh phan baa cua te bao A, B, C, D cua l loai: - T6 bao A nguyen phan mQt s6 dot tao cac t~ bao bAng lfin s6 NST d<JJ1trong bQ dan boi cua loai T~ bao B nguyen phan mQt s6 dot tao cac bao co t6ng s6 NST dan gAp IAns6 NST don cung nguon g6c bQ 2n cua loai, T8 bao C nguyen phan da lAy tir moi tnrong nQi bao 210 NST d<JJ1 - T€ bao D nguyen phan dot Tang s6 NST dan cua bQ don bQi tAt cit cac t~ bao sinh tir dot nguyen phan cu6i cung cua t~ bao tren la 1134 a Xac dinh bQ NST 2n cua loai b s6 IAnnguyen phan cua t~ bao B, C c T8 bao A nguyen phan, s6 t8 bao duoc tao tnroc ti~n hanh dot nguyen phan cu6i cung chi co mQt s6 tS bao tham gia vao dot nguyen phan cu6i cung, s6 l~i khong phan bao Xac dinh s6 hrong t8 bao cua t8 bao tham gia vao dot nguyen phan cu6i cung Cach giii Dap an Di~m a GQi 2n Iii bQ NST hrong bQi cua loai - T6ng s6 NST don co cac t8 bao sinh tir t~ bao Ala: 0~5 2n x 2n=4n2 - T6ng s6 NST dan co cac t8 bao sinh tu t~ bao B lit: 0,5 8n - T6ng s6 NST d<JJ1co cac t€ bao sinh til t6 bao Cia: 0~5 210 + 2n - T6ng s6 NST dan co cae t8 bao sinh tir t8 bao D la: 0,5 a 2n= 14 256n Theo d~ bai ta co: 4n2 + 8n + 210 + 2n + 256n = 1134 x 1,0 q 2n = 14 b - s6 t€ bao sinh til dQt nguyen phan eu6i eimg etia t~ bao b s6 dgt nguyen 0~5 B: 8n : 2n = t~ bao pMn eua t~ bao B: dgt => s6 dQ'1nguyen phan etia t~ bao B: dgt - s6 TB sinh til dQt nguyen phan eu6i eung etia TB C; s6 dQt nguyen 0,5 (210+14) : 14 = 16 TB phan ctia tb C: => s6 dgt nguyen philn cua tb C: c GQi x: s6 tb sinh tU tb A tham gia vao dgt nguyen phan " cung , cum G9i y: s6 tb sinh tu tb A khOng tham gia vao dQ'1nguyen phan c s6 tb cua tb A tham gia dqt 0,5 cu6i cimg Ta co: x + y = 2k ( k: s6 dQt nguyen phan eua tb A hinh tMnh x, y tb) nguyen phan cu6i cung la 2x + y = 14 => x = 0.5 te (5) Cau (5 di8m) : Co 10 phan tft glucozo qua giai doan dirong phan, 50% san p}Jfun tiep tuc di vao chu trinh Crep X~c flinh nang hrong (Keal) duQ'c tao k~t thuc qua trinh ho hap t8 bao sinh vlftt , pong he n h~an t h IJC; b'~ let ng, IATP giai 73k , ca nang 1trong, NADPH "'"3ATP, IF ADHz "'" 2ATP) Cach giiii Dap an Di~m - S6 ATP tao tir qua trlnh dtrong phan va chu6i chuyen electron ho hAp la: 2ATP x 10 + 2(NADPH) x xl O = 80 ATP - S6 ATP tao chu trinh Crep va chu6i chuyen electron hO hAp la: 2ATP x + 8(NADPH) x x + 2(FADHz) x x = 150 ATP - S6 ATP "'"80 + 150"'" 230 ATP * Nang hrong tao "'"(230 x 7,3) = 1679 kcal 1,5 tu * Nang luong tao 1679 kcal 1,5 1,0 1,0 Cau 9: (5 di~m) MQt song co hai quin the 6c sen: quin the 16'0 (quan the chfnh) phia tren va quAn th~ nho nam cu6i dong tren mQt hon dao (quin the dao) Do mroc chay xuoi nen 6c chi di chuyen diroc tlr quin th6 chinh den quin th6 dao rna khong di chuyen ngiroc lai Xet mQt gen gbm hai alen: A va a d quin the chinh co PA=1, quan the dao co PA= 0,6 Do di cu, qu!n th~ dao tro qudn the moi, co 12% s6 ca the la cua quan the chinh a Tinh tAn s6 urong d6i cua cac alen quin the moi sau di cu b Quan th~ moi sinh san Vi mQt li nao xay qua trinh dQt bien: A -7 a, voi t6c dQ la 0,3% Khong co dQt bien ngiroc ~ so; t irong d,l;·, , th,l; - Ti10h tan 01 cua cac a en 0' e h ~ f.l: e mm lep th eo cuaquanx th,l Cach ghli K~tqua Diem a - Ta co: Quan the chinh co pA= 1, quan the dao: pA= 0,6 QuAn the chinh di cu den quan the dao va chiem 12% quan the moi V~y quan the dao chiem 88% quan th~ moi, a pmai= 0,648 - QuAn the moi & dao (sau di cu) co tAn s6 nrong d6i cua cac qmoi=0,352 alen la: Pm&i= 12% x + 88% x 0,6 = 0,648 1,5 qmm= 1- pmoi = 1- 0,648 = 0,352 1,5 a A b - TAn s6 dQt bien: A a la: 0,3% b pA = 0,646 - TAn s6 cac alen sau dQt bien la 1,0 pA= 0,648 - (0,3% x 0,648) = 0,646 qa= 0,354 1,0 qa = - 0,646 = 0,354 CaD 10 : (5 diem) Chieu cao cay c~p gen phan ly dQc l~p, tac dQng cQng gQP quy dinh Sl,I'co m~t m6i alen trQi t6 hop gen lam tang chieu cao cay len 5cm Cay thAp nhAt co chi~u cao = 150cm Cho cay co e~p gen di hop tlJ thu phein Xac dinh: - Tin s6 xuat hi~n t6 hop gen co alen trQi, alen trci, - KM nang co duoc mot cay co chieu cao 165cm Cach giai K~t qua Diem Tan s6 xuat hi~n : Tan so xuat hi~n: - to - t6 hgp gen co alen trQi = C2na / 4n = C61 /43 = 6/64 hQP gen co alen trQi 1,5 - t6 hgp gen co alen trQi = Czna / 4n = C64 / 43 = 15/64 =6/64 1,5 - Cay co chi~u cao 165cm hon cay thAp nhit la: - t6 hqp gen co alen 165cm -150cm = 15cm trQi = 15/64 => co alen trQi ( x 5cm = 15cm ) Kha nang co dugc * V~y khA nang co dugc mQt cay co chi~u cao 165cm la: cay co chi~u cao 3 C6 /4 = 20/64 165cm 1a 20/64 2,0 * * * H€t Chu y: HQc sinh lam thea cach khac dUng vftn cho dj~m t6i da -5 - (6)