Taasggatss a 20 b 22 c 24 d 25 Ishlanishi
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_____________________________ 209. Teskari transkripsataza fermenti yordamida i-RNK dan DNK sintezlandi i-RNK da G=25% bo`lib, DNK qo`sh zanjirining 15% ni guanin tashkil qilasa i-RNKda A va T necha foizni tashkil qiladi?
1) i-RNK da: U-25% DNK da: G-15%=15%-S G+S=30% 2) 100%-30%=70% A+T=70% Javob: A _____________________________ 210. DNK qo`sh zanjirida 720 ta vodorod bog`I bo`lib guanin umumiy nukleotidlardan 5 marta kam bo`lsa umumiy nukleotidlar sonini toping. A)600 B)650 C)400 D)450 Ishlanishi: 1) 2A+3G=720 G=2A+2G/5 2A+2G=5G 2A=5G-2G 2A=3G 2) 3G+3G=720 6G=720 G=120 3) 2A+3*120=720 2A=720-360 2A=360 A=180 4) N=(A+G)*2 N=(180+120)*2=600N Javob: A _____________________________ 211. A=(G+S) holatda umumiy vodorod bog`lar soni 420 ta bo`lsa DNK fragmenti uzunligini aniqlang.(qo`sh. Nuk. Orasidagi masofa 0.34nm) A)61,2 B)44,3 C)51 D)54,4 Ishlanishi: 1) 2A+3G=420 A=(G+S) S=G A=2G 2) 2*2G+3G=420 4G+3G=420 7G=420 G=60 3) 2A+3*60=420 2A=420-180 2A=240 A=120 4) L=(A+G)*0,34 L=(120+60)*0,34=61,2nm Javob: A
212. Teskari transkripsataza fermenti yordamida i-RNK dan DNK sintezlandi i-RNK da U=10% bo`lib, DNK qo`sh zanjirining 15% ni timin tashkil qilasa i-RNKda A-necha foizni tashkil qiladi? A)20 B)30 C)25 D)35 Ishlanishi: 1) i-RNK da: U-10% DNK da: T-15%=15%-A A+T=30% 2) A+U=A+T A+10=15+15 A=30-10 A=20% Javob: A
213. Teskari transkripsataza fermenti yordamida i-RNK dan DNK sintezlandi i-RNK da U=30% bo`lib, DNK qo`sh zanjirining 25% ni timin tashkil qilasa i-RNKda A-necha foizni tashkil qiladi?
1) i-RNK da: U-30% DNK da: T-25%=25%-A A+T=50% 2) A+U=A+T A+30=25+25 A=50-30 A=20% Javob: A
214. DNK qo`sh zanjirida 720ta vodorod bog`I bo`lib guanin umumiy nukleotidlardan 5 marta kam bo`lsa G-S orasidagi va A-T orasidagi vodorod bog`lar farqini aniqlang. A) 0 B)20 C)25 D)35 Ishlanishi: 1) 2A+3G=720 G=2A+2G/5 2A+2G=5G 2A=5G-2G 2A=3G 2) 3G+3G=720 6G=720 G=120 3) 2A+3*120=720 2A=720-360 2A=360 A=180 4) H=2A-3G H=360-360=0H Javob: A _____________________________ 215. DNK qo`sh zanjirida 126 ta vodorod bog` bo`lib sitozin purin asosidan 1,6 marta kam bo`lsa, umumiy nukleotidlarni aniqlang. A)104 B)98 C)96 D)108
1) 2A+3G=126 S=A+G/1,6 S=G A+G=1,6G A=1,6G-G A=0,6G 2) 2*0,6G+3G=126 1,2G+3G=126 4,2G=126 G=30 3) 2A+3*30=126 2A=126-90 2A=36 A=18 4) N=(A+G)*2 N=(18+30)*2=96N Javob: A
216. DNK tarkibida 960 ta guanin nukleotidi bo’lib, u umumiy nukleotidlarni 40% tashkil qiladi. Shu DNK dan hosil bo’lgan oqsildagi tarkibidagi aminokislota sonini aniqlang. A) 3360 B) 2640 C) 800 D) 400
1) G-960------40% A-x----10% A-x=(960*10)/40=240 2) A+G=960+240=1200N 3) 1200N/3=400AK Javob: D
217. DNK tarkibida 960 ta guanin nukleotidi bo’lib, u umumiy nukleotidlarni 40% tashkil qiladi. Shu DNK dagi vodarod bog’lar sonini aniqlang A) 3360 B) 2640 C) 2380 D) 400 Ishlanishi: 1) G-960------40% A-x----10% A-x=(960*10)/40=240 2) A+G=960+240=1200N 3) H=2A+3G H=(2*240)+(3*960)=3360H Javob: A
218. DNK molekulasining 153 nm uzunlikdagi qismida deleytsiyadan so’ng 12 juft nukleotid yo’qoldi. Mutatsiyadan keyin hosil bo’lgan DNK dan sintezlangan oqsildagi aminokislota sonini aniqlang. A) 292 B) 146 C) 150 D) 148 Ishlanishi: 1) N=L/0,34nm L=153/0,34nm=450N -1 ta ipdagi N lar 2) 450-12=438N Deletsiyadan so’ng 3) 438N/3=146AK Javob: B _____________________________ 219. DNK molekulasining 153 nm uzunlikdagi qismida deleytsiyadan so’ng 12 juft nukleotid yo’qoldi. Mutatsiyadan keyin hosil bo’lgan DNK dan sintezlangan oqsildagi peptid bog’lar sonini aniqlang. A) 292 B) 145 C) 147 D) 149 Ishlanishi: 1) N=L/0,34nm L=153/0,34nm=450N -1 ta ipdagi N lar 2) 450-12=438N Deletsiyadan so’ng 3) 438N/3=146AK 4) 146-1=145P Javob: B
220. DNK molekulasining og’irligi 90000 ga teng bo’lsa, undagi nukleotidlar sonini aniqlang. (bitta nukleotid og’irligi 300 deb olinsin) A) 600 B) 300 C) 900 D) 100
1) N=M/300 N=90000/300=300N -2 ta ipdagi N lar Javob: B _____________________________ 221. DNK molekulasining og’irligi 90000 ga teng bo’lsa, shu DNK dan sintezlangan oqsildagi peptid bog’lar sonini aniqlang. (bitta nukleotid og’irligi 300 deb olinsin) A) 99 B) 199 C) 149 D) 49
1) N=M/300 N=90000/300=300N -1 ta ipdagi N lar 2) 300N/2=150N -1 ta ipdagi N lar 3) 150/3=50AK 4) 50-1=49P Javob: D
222. DNK molekulasining og’irligi 90000 ga teng bo’lsa, shu DNK dan sintezlangan oqsildagi aminokislota sonini aniqlang. (bitta nukleotid og’irligi 300 deb olinsin) A) 100 B) 200 C) 150 D) 50
1) N=M/300 N=90000/300=300N -1 ta ipdagi N lar 2) 300N/2=150N -1 ta ipdagi N lar 3) 150/3=50AK Javob: D _____________________________ 223. Ma’lum bir DNK bo’lagidan 34 ta aminokislotaga ega oqsil sintezlandi. Mutatsiya natijasida 3 ta aminokislota hosil bo’lmadi. Ushbu oqsil sinteziga javobgar bo’lgan t-RNK sonini aniqlang. A) 34 B) 31 C) 1 D) 2
1) 34-3=31AK 2) 31 ta AK=31 ta t-RNK Javob: B
224. DNK zanjiridagi nukleotidlar orasidagi masofa 0,34 nm ga teng va ular o'zaro fosfodiefir bog'lari orqali bog'lansa, 153 nm uzunlikdagi DNK qo'sh zanjirida 12 juft nukleotid deletsiyaga uchragan bo'lsa, avvalgi va mutatsiyaga uchragan DNK fragmenti asosida transkripsiya bo'lgan i-RNKdagi nukleotidlar orasidagi fosfodiefir bog'lar sonini aniqlang. A) 450; 438 B) 450; 444 C) 449; 437 D) 448; 436 Ishlanishi: 1) N=L/0,34nm L=153/0,34nm=450N -1 ta ipdagi N lar 2) 450-12=438N Deletsiyadan so’ng 3) 450N-1=449P 4) 438N-1=437P Javob: C
225. DNKning bir zanjirida 180 ta triplet kod bo’lib, shundan 30 tasi adeninli nukleotid. Shu DNK asosida transkripsiya bo’lgan i- RNKda guaninli nukleotidlar soni DNKning o‘sha zanjirdagi adeninli nukleotidlaridan 4 marta ko‘p, sitozinlar soni guanindan 2 marta kam bo’lsa, DNK qo‘sh zanjiridagi purin va pirimidin asoslar nisbatini toping. A) 1 purin (360A+180G) : 1 pirimidin (180C+360T) B) 2 purin (360A+180G) : 1 pirimidin (90C+180T) C) 5 purin (330A+120G) : 1 pirimidin (60C+30U) D) 5 purin (330A+120G) : 1 pirimidin (60C+30T) Ishlanishi: 1) N=AK*3 N=180*3=540N -1 ta ipdagi N lar 2) i-RNK da: G=4A G=4*30=120; S=G/2; S=120/2=60 U=30; A=540-(G+S+U) A=540-210=330 3) DNK da: U+A=A(T) 30+330=360A(360T) G+S=G(S) 120+60=180G(180S) 4) Purin=A+G Purin=360A+180G Pirimidin=S+T Pirimidin=180S+360T Javob: A _____________________________ 226. DNK ning qo'sh zanjirida G nukleotidlari 40%ni tashkil etadi. DNK dagi gen asosida sintezlangan oqsil 400 ta aminokislotadan iborat bo'lsa, shu DNK qo'sh zanjiridagi pirimidin asoslari sonini toping. A) 600 B) 960 C) 1200 D) 320 Ishlanishi: 1) N=AK*3 N=400*3=1200N*2=2400N -2 ta ipdagi N lar 2) 2400---100% G-x---40% G-x=(2400*40)/100=960N G-960=960-S 3) 960----40% A-x---10% A-x=(960*10)/40=240 A-240=240-T 4) A+G=960+240=1200N Purin 5) S+T=960+240=1200N Pirimidin Javob: C _____________________________ 227. iRNK molekulasida A-20%, G-10%, U-40% bo‘lsa, teskari transkripsiya jarayonida sintezlangan DNK qo‘sh zanjiridagi guaninning foizini ko'rsating. A) 30 B) 10 C) 20 D) 40
1) i-RNK da: A-20%; G-10%; U-40%; S-? S=100-(A+G+U)=100-(20+10+40)=100-70=30%-S 2) DNK da A(T)=(U+A)/2 A(T)=(20+40)/2=30%A(T) G(S)=(G+S)/2 G(S)=(10+30)/2=20%G(S) Javob: C
228. DNK ning qo‘sh zanjirida G nukleotidlari 40%ni tashkil etadi. DNK dagi gen asosida sintezlangan oqsil 400 ta aminokislotadan iborat bodsa, shu DNK qo‘sh zanjiridagi timin nukleotidlari sonini toping.
1) N=AK*3 N=400*3=1200N*2=2400N -2 ta ipdagi N lar 2) 2400---100% G-x---40% G-x=(2400*40)/100=960N G-960=960-S 3) 960----40% A-x---10% A-x=(960*10)/40=240 A-240=240-T Javob: A _____________________________ 229. Ma’lum bir oqsil sintezida ishtirok etuvchi DNK qo'sh zanjirida nukleotidlar soni 846 taga teng. Hujayra kimyoviy modda ta’sirida mutatsiyaga uchrashi natijasida 12 juft nukleotid yo'qolgan bo'lsa, DNKning bitta zanjirida nechta fosfodiefir bog’lari qoladi? (Nuklein kislota molekulasida nukleotidlar o'zaro fosfodiefir bog'lari yordamida bog'lanadi.) A) 833 B) 410 C) 416 D) 820 Ishlanishi: 1) 846N-24=822N -2 ta ipdagi N lar 2) 822/2=411 -1 ta ipdagi N lar 3) F=N-1 F=411-1=410F Javob: B _____________________________ 230. DNK ning qo'sh zanjirida G nukleotidlari 40% ni tashkil etadi. DNK dagi gen asosida sintezlangan oqsil 400 ta aminokislotadan iborat bo’lsa, shu DNK qo‘sh zanjiridagi fosfodiefir bog’lari sonini toping, (nukleotidlar o'zaro fosfodiefir bog’i orqali bog’lanadi) A) 2400 B) 2399 C) 1198 D) 2398
1) N=AK*3 N=400*3=1200N*2=2400N -2 ta ipdagi N lar 2) 2400---100% G-x---40% G-x=(2400*40)/100=960N G-960=960-S 3) 960----40% A-x---10% A-x=(960*10)/40=240 A-240=240-T 4) A+G=N 240+960=1200 -1 ta ipdagi N lar 5) F=(N-1)2N-2 F=(2*1200)-2=2398F Javob: D _____________________________ 231. DNK ning qo'sh zanjirida G nukleotidlari 40%ni tashkil etadi. DNK dagi gen asosida sintezlangan oqsil 400 ta aminokislotadan iborat bo'lsa, shu DNK qo'sh zanjiri asosida sintezlangan i-RNKdagi fosfodiefir bog'lari sonini toping, (nukleotidlar o'zaro fosfodiefir bog' orqali bog'lanadi) A) 2398 B) 1199 C) 2399 D) 2400
1) N=AK*3 N=400*3=1200N*2=2400N -2 ta ipdagi N lar 2) 2400---100% G-x---40% G-x=(2400*40)/100=960N G-960=960-S 3) 960----40% A-x---10% A-x=(960*10)/40=240 A-240=240-T 4) A+G=N 240+960=1200 -1 ta ipdagi N lar 5) F=(N-1)2N-2 F=1200-1=1199F Javob: B _____________________________ 232. DNK ning zanjirida 960 ta guanin nukleotidi bo‘lib umumiy nukleotidlarning 40% ini tashkil etadi. Undagi gen asosida sintezlangan oqsildagi peptid bog’lari sonini toping. A) 390 B) 400 C) 398 D) 399
1) G-960------40% A-x----10% A-x=(960*10)/40=240 2) A+G=960+240=1200N 3) 1200N/3=400AK 4) Peptid=AK-1 Peptid=400-1=399P Javob: D _____________________________ 233. Oqsil molekulasida 41 ta peptid bog’i bor. Ushbu DNK bo‘lagi asosida sintezlangan oqsildagi aminokislotalar sonini aniqlang.
A) 40 ta B) 41 ta C) 43 ta D) 42 ta Ishlanishi: 1) AK=Peptid+1 AK=41+1=42 AK Javob: D _____________________________ 234. iRNK molekulasida A-20%, C-30%, U-40% bo’lsa, teskari transkripsiya jarayonida sintezlangan DNK qo‘sh zanjiridagi sitozinning foizini ko‘rsating. A) 10 B) 30 C) 20 D) 40
1) i-RNK da: A-20%; G-10%; U-40%; S-? S=100-(A+G+U)=100-(20+10+40)=100-70=30%-S 2) DNK da A(T)=(U+A)/2 A(T)=(20+40)/2=30%A(T) G(S)=(G+S)/2 G(S)=(10+30)/2=20%G(S) Javob: C
235.
AUGCGAAUCCGACGGCUU nukleotidlar ketma- ketligiga ega i-RNK molekulasiga komplementar DNK qo‘sh zanjiridagi nukleotidlar orasidagi fosfodiefir bog’lar sonini (a), A, G, C, T nukleotidlar sonini (b) va ular orasidagi vodorod bog’lar sonini (c) aniqlang. Polinukleotid zanjiridagi nukleotidlar o‘zaro fosfodiefir bog’lari orqali bog’langan. A) a-36; b-A va T 20 ta, G va C 8 ta; c-44 B) a-17; b-A va T 8 ta, G va C 10 ta; c-46 C) a-34; b-A va T 16 ta, G va C 20 ta; c-46 D) a-36; b-A va T 8 ta, G va C 10 ta; c-46 Ishlanishi: 1) i-RNK da: AUGSGAAUSSGASGGSUU DNK 1-ipda: TA
SGST TAGGSTGSS GAA
i-RNK da: ATGSGAAT
S SGASGGSTT 2) 1 ta ipda: A+T=16N; G+S=20N
3) F=2N-2 F=36-2=34F 4) H=2A+3G H=2*8+3*10=16+30=46H Javob: C _____________________________ 236. DNK ning qo‘sh zanjirida G nukleotidlari 40% ni tashkil etadi. DNK dagi gen asosida sintezlangan oqsil 400 ta aminokislotadan iborat bo’lsa, shu DNK qo‘sh zanjiri asosida sintezlangan i-RNK dagi nukleotidlar sonini toping. A) 600 B) 960 C) 2400 D) 1200 Ishlanishi: 1) N=AK*3 N=400*3=1200N*2=2400N -2 ta ipdagi N lar 2) 2400---100% G-x---40% G-x=(2400*40)/100=960N G-960=960-S 3) 960----40% A-x---10% A-x=(960*10)/40=240 A-240=240-T 4) A+G=N 240+960=1200N -DNK ning 1 ta ipidagi yoki i- RNK dagi N lar Javob: B _____________________________ 237. i-RNK zanjirida 80 ta uratsil bor, shu RNK zanjiridan teskari transkripsiyalangan DNK ning 1 ta zanjiridagi sitozinlar soni RNK dagi uratsillar sonidan 2,5 marta ko'p, guaninlar soni esa 2,5 marta kam. DNKning shu zanjiridagi timin miqdori undagi G va sitozin nukleotidlar miqdorining o'rtachasiga teng bo'lsa, i-RNK molekulasining uzunligini aniqlang. (Nukleotidlar orasidagi masofa 0,34 nm ga teng.) A) 145,52 B) 291,04 C) 357,68 D) 369,92 Ishlanishi: 1) DNK da: A=80; S=2,5U=2,5*80=200; G=U/2,5=80/2,5=32; T=(S+G)/2=(200+32)/2=116 i-RNK da: U=80; G=200; S=32; A=116 2) DNK da: A+S+G+T =>i-RNK da: U+G+S+A 80+200+32+116=428 3) L=N*0,34nm=428*0,34nm=145,52nm Javob: A
238. DNK ning qo'sh zanjirida G nukleotidlari 40%ni tashkil etadi. DNK dagi gen asosida sintezlangan oqsil 400 ta aminokislotadan iborat bo'lsa, shu DNK qo'sh zanjiridagi sitozin nukleotidlari sonini toping. A) 240 B) 600 C)960 D) 120 Ishlanishi: 1) N=AK*3 N=400*3=1200N*2=2400N -2 ta ipdagi N lar 2) 2400---100% G-x---40% G-x=(2400*40)/100=960N G-960=960-S J: C _____________________________ 239. DNK molekulasining ma'lum qismida 1380 ta vodorod bog’lari bo’lib, undagi sitozinli nukleotidlar 90 juftni tashkil etadi. Shu DNK bo‘lagi asosida transkripsiya jarayonida hosil bo’lgan i- RNK molekulasidagi nukleotidlar sonini aniqlang. A) 700 B) 645 C) 1200 D) 600 Ishlanishi: 1) 2A+3G=H => S=G => 2A+3*180=1380 2A=1380-540 2A=840 A=420 2) A+G=N => 180+420=600N DNK ni 1 ta ipi yoki i-RNK dagi N lar Javob: D
240. Oqsil zanjirining molekulyar massasi 12000. Bitta aminokislotaning o‘rtacha massasi 120 ga teng. Qo‘shni nukleotidlar orasidagi masofa 0,34 nm bo’lsa, quyidagilarni aniqlang. a) genning uzunligi (nm); b) DNK qo‘sh zanjiridagi nukleotidlar soni; c) i-RNKdagi nukleotidlar soni; d) purin va pirimidin asoslarining o‘zaro nisbati; 1) 102; 2) 34; 3) 4000; 4) 600; 5) 100; 6) 300; 7) 120; 8) 1:1; 9) 1:3; 10) 1:2 A) a-1; b-4; c-7; d-10 B) a-1; b-4; c-6; d-8 C) a-3; b-7; c-4; d-10 D) a-2; b-5; c-6; d-9
1) 12000/120=100AK; 100*3=300N a) L=300*0,34=102nm; b) 300*2=600N; c) i-RNK da-300N; d) 1:1 Javob: B _____________________________ 241. DNK molekulasida adenin va timin nukleotidlari orasida ikkita, guanin va sitozin nukleotidlari orasida uchta vodorod bog’i borligi ma’lum. DNK fragmentida 1040 ta vodorod bog’i bo’lib, ushbu fragmentda adenin nukleotidlari umumiy nukleotidlarning 20% ini tashkil etsa, ushbu molekuladagi jami nukleotidlar sonini aniqlang. A) 810 B) 800 C) 1000 D) 832 Ishlanishi: 1) A-20%=20%-T 40%-A+T 100-40=60% 60:2=30%-G-S 2) 30/20=1,5 G=1,5A 3) 2A+3G=H => 2A+3*1,5A=1040 2A+4,5A=1040 6,5A=1040 A=160 4) 2*160+3G=1040 3G=1040-320 3G=720 G=240 5) (A+G)*2=(160+240)*2=400*2=800N Javob: B
242. DNK ning qo‘sh zanjirida G nukleotidlari 40% ni tashkil etadi. DNK dagi gen asosida sintezlangan oqsil 400 ta aminokislotadan iborat bo‘lsa, shu DNK qo'sh zanjiridagi guanin va sitozin orasidagi vodorod bog'lari sonini toping.
1) N=AK*3 N=400*3=1200N*2=2400N -2 ta ipdagi N lar 2) 2400---100% G-x---40% G-x=(2400*40)/100=960N G-960=960-S 3) H=3G H=3*960 H=2880 Javob: A
243. i-RNK zanjirida 80 ta uratsil bor, shu RNK zanjiridan teskari transkripsiyalangan DNK ning 1 ta zanjiridagi sitozinlar soni RNK dagi uratsillar sonidan 2,5 marta ko'p, guaninlar soni esa 2,5 marta kam. DNKning shu zanjiridagi timin miqdori undagi guanin va sitozin nukleotidlar miqdorining o'rtachasiga teng bo'lsa DNK molekulasidagi nukleotidlar sonini aniqlang. A) 1052 B) 428 C) 1088 D) 856 Ishlanishi: 1) DNK da: A=80; S=2,5U=2,5*80=200; G=U/2,5=80/2,5=32; T=(S+G)/2=(200+32)/2=116 i-RNK da: U=80; G=200; S=32; A=116 2) DNK da: A+S+G+T =>i-RNK da: U+G+S+A 80+200+32+116=428 3) 2N=2*428=856 Javob: D
244. DNK molekulasida adenin va timin orasida ikkita, guanin va sitozin orasida uchta vodorod bog’i bor, qo'shni nukleotidlar orasi 0,34 nmga teng. Nuklein kislotalar molekulasida nukleotidlar o‘zaro fosfodiefir bog’lari orqali bog’lanadi. DNKning bitta zanjirida 209 ta fosfodiefir bog’lari bo’lsa va adenin nukleotidlari DNK qo‘sh zanjiridagi jami nukleotidlarning 20%ini tashkil etsa, DNK reduplikatsiyasi uchun nechta guanin nukleotidlari kerak bo’ladi? A) 126 ta B) 84 ta C) 63 ta D) 42 ta Ishlanishi: 1) 209+1=210N -1 ta ipdagi N lar 2) 210*2=420N -2 ta ipdagi N lar 3) A-20%=20%-T bo’lsa, G-30%=30%-S bo’ladi. 4) 420---100% G---30% G=(420*30)/100=126 Javob: A _____________________________ 245. DNK ning qo‘sh zanjirida G nukleotidlari 40%ni tashkil etadi. DNK dagi gen asosida sintezlangan oqsil 400 ta aminokislotadan iborat bo‘lsa, shu DNK qo‘sh zanjiri asosida sintezlangan i-RNK dagi nukleotidlar sonini toping. A) 600 B) 2400 C) 1200 D) 960 Download 0.58 Mb. Do'stlaringiz bilan baham: 
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