Taasggatss a 20 b 22 c 24 d 25 Ishlanishi
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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 _____________________________ 246. 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 purin asoslari sonini toping. A) 600 B) 320 C) 1200 D) 960 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 - Purin yoki Pirimidin DNK dagi barcha N larning yarmini tashkil etadi.Bu esa DNK ning 1 ta ipidagi yoki i-RNK dagi N larga teng bo’ladi. Javob: C
247. 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 adeninli nukleotidlar sonini aniqlang. A) 240 B) 208 C) 312 D) 160 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 Javob: B _____________________________ 248. 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 vodorod bog’lar 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 2) A+T=A(T) -bitta zanjirdagi A va T yig’indisi A yoki T ga teng: 80+116=196 A-T G+S=G(S) -bitta zanjirdagi G va S yig’indisi G yoki S ga teng: 200+32=232 G-S 3) H=2A+3G H=2*196+3*232 H=392+696=1088 Javob: C
249. DNK molekulasining ma'lum qismida 1380 ta vodorod bog’lari bo‘lib, undagi sitozinli nukleotidlar 90 juftni tashkil etadi. Shu DNK bo’lagi asosida translyatsiya jarayonida hosil bo’lgan polipeptid molekulasidagi monomerlar sonini aniqlang. A) 400 B) 200 C) 460 D) 230 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 3) AK=N/3=600/=200AK Javob: B _____________________________ 250. 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‘ orqali bog’lanadi) A) 2400 B) 12398 C) 1199 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=1200N - Purin yoki Pirimidin DNK dagi barcha N larning yarmini tashkil etadi.Bu esa DNK ning 1 ta ipidagi yoki i-RNK dagi N larga teng bo’ladi. 5) 2N=2*1200=2400N –DNK dagi barcha N lar 6) F=N-2=2400-2=2398F Javob: D
251.
AUGCGAAUCCGACGGCUU nukleotidlar ketma- ketligiga ega i-RNK molekulasiga komplementar DNK qo‘sh zanjiridagi A, G, C, T nukleotidlar sonini (a), vodorod bog‘lar sonini (b) aniqlang. A) a-A va T 8 ta, G va C 10 ta; b-46 B) a-A va T 4 ta, G va C 15 ta; b-26 C) a-A va T 20 ta, G va C 8 ta; b-44
1) i-RNK da: AUGSGAAUSSGASGGSUU DNK 1-ipda: TA
SGST TAGGSTGSS GAA
DNK 2-ipda: 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: D
252. 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 adenin 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 _____________________________ 253. 5000 nukleotiddan iborat DNK fragmentining markaziy qismiga 8000 juft nukleotiddan iborat bo’lgan regulyator gen birikdi. Shu DNK asosida sintezlangan i-RNK zanjiridagi fosfodiefir bog’ini aniqlang? A) 10500 B) 6499 C) 10499 D) 20999 Ishlanishi: 1) 2500+8000=10500N -DNK ning 1 ta ipidagi yoki i-RNK dagi N lar. 2) F=N-1=10500-1=10499F Javob: C _____________________________ 254. Ximoptirsinogen fermenti 245 ta aminokislotadan iborat bo’lib, uni sintezlagan DNK molekulasi tarkibida A 20 %. Ushbu fragmentdagi G-S orasidagi vodorod bog’larni A-T orasidagi vodorod bog’lariga nisbati nechiga teng. A) 3,25 B) 1,5 C) 2,5 D) 2,25 Ishlanishi: 1) N=AK*3 N=245*3=735N*2=1470N -2 ta ipdagi N lar 2) 1470---100% A---20% A=(1470*20)/100=294N A-294=294-T 3) 294----20% G---30% G=(294*30)/20=441 G-441=441-S 4) H=2A+3G H=2*294+3*441=588+1323=1911 5) 3G/2A=1323/588=2,25 Javob: D _____________________________ 255. DNK molekulasida 1380 ta H bog’ mavjud bo’lib, undagi S nukleotidi soni 180 ta. Shu DNK asosida sintezlangan i-RNK dagi fosfodiefir bog’i sonini aniqlang. A) 1199 B) 599 C) 509 D) 1019
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 3) F=N-1=600-1=599F Javob: B
256. DNK molekulasida 1380 ta H bog’ mavjud bo’lib, undagi S nukleotidi soni 180 ta. Shu DNK asosida sintezlangan oqsildagi aminokislota sonini aniqlang. A) 200 B) 300 C) 500 D) 100 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 3) AK=N/3 AK=600/3=200AK Javob: A _____________________________ 257. DNK fragmentining bir zanjirida nukleotidlar quyidagi ketma – ketlikda joylshgan AAGTSTASGTATAAGTSTASGTAT. Gen uzunligini aniqlang (nm). (nukleotidlar orasidagi masofa 0,34 nm ga teng) A) 4,08 B) 8,16 C) 16,32 D) 8,96 Ishlanishi: 1) AAGTSTASGTATAAGTSTASGTAT L=N*0,34nm L=24*0,34nm=8,16nm Javob: B
258. DNK molekulasida 880 ta guanin nukleotidi mavjud bo`lib, ular barcha nukleotidlarning 22 % ni tashkil etadi. Adenin nukleotidlari sonini aniqlang. A) 1120 B) 2240 C) 1760 D) 880 Ishlanishi: 1) G-22%=22%-S 44%=G+S 100-44=56 2) 56%=A+T 56/2=28 A-28%=28%-T 880---22% A---28% A=(880*28)/22=1120N Javob: A
259. DNK molekulasida 28450 H bog’i bor. Undagi A umumiy nukleotidlarni 25 % ini tashkil etadi. Ushbu molekulada nechta G bor ?
A) 5690 B) 11380 C) 8850 D) 2276 Ishlanishi: 1) A=T => 25%=A-T=G-S=25% 2A+3G=H 2A+3A=28450 5A=28450 A=5690 Javob: A _____________________________ 260. DNK fragmentining bir zanjirida nukleotidlar quyidagi ketma – ketlikda joylshgan AAGTSTASGTATAAGTSTASGTAT. Ushbu fragmentdagi adenin va timin azot asoslarining foizini aniqlang. A) 50 B) 16,67 C) 66,67 D) 33,33 Ishlanishi: 1) AAGTSTASGTATAAGTSTASGTAT A-T=16 G-S=8 16+8=24 2) 24---100% 16---x x=(16*100)/24=66,67% Javob: C
261. DNK molekulasida 30000 ta adenin nukleotidi bor. Replikatsiya jarayoni sodir bo`ldi. Nechta adenin va timin nukliotidi bor.
A) А-15000 Т-15000 B) А-60000 Т-60000 C) А-30000 Т-30000 D) А-60000 Т-30000 Ishlanishi: 1) 30000A*2=60000A A=T Javob: B _____________________________ 262. DNK molekulasining birinchi ipida umumiy nukleotidlarni 4,5 foizini tashkil etuvchi G joylashgan bo'lib, qolgan G ikkinchi ipda joylashgan. Ikkinchi ipda yana 162 ta S bor. Shu DNK molekulasida nechta nukleotid bor A) 900 B) 2800 C) 3600 D) 1800 Ishlanishi: 1) har doim A miqdori T ga, G miqdori S ga teng. A-4,5%-162=162-4,5%-T 4,5+4,5=9% 100%-9%=91% -bu foiz G va S ga tegishli, shuning uchun: G-45,5%=45,5%-S 162 (A-T) ---- 4,5% x (G-S) ---- 45,5% x=(162*45,5)/4,5=1638 Demak, G-1638=1638-S 2) 1 ta zanjirdagi jami nukleotidlarni topish uchun A ga G ni qo’shamiz: 162+1638=1800 3) 2N=2*1800=3600 Javob: C
263. DNK molekulasining birinchi ipida umumiy nukleotidlarni 4,5 foizini tashkil etuvchi G joylashgan bo’lib, qolgan G ikkinchi ipda joylashgan. Ikkinchi ipda yana 162 ta sitozin bor. Shu DNK molekulasining bir ipidan sintezlangan i-RNK ning uzunligi necha nm. A) 112 B) 900 C) 612 D) 1213 Ishlanishi: 1) har doim A miqdori T ga, G miqdori S ga teng. A-4,5%-162=162-4,5%-T 4,5+4,5=9% 100%-9%=91% -bu foiz G va S ga tegishli, shuning uchun: G-45,5%=45,5%-S 162 (A-T) ---- 4,5% x (G-S) ---- 45,5% x=(162*45,5)/4,5=1638 Demak, G-1638=1638-S 2) 1 ta zanjirdagi jami nukleotidlarni topish uchun A ga G ni qo’shamiz: 162+1638=1800 3) Nukleotidlar orasidagi masofa-0,34 nm yoki 3,4 A 0
H=1800*0,34=612nm Javob: C
264. DNK molekulasida 4836 ta A va 8423 ta G bor. Ushbu molekulani uzunligi va H bog’lar sonini aniqlang. A) 45080,6 A; 34941 B) 45980 A; 8672 C) 34941 A; 2843 D) 13259 A; 45080 Ishlanishi: 1) har doim A miqdori T ga, G miqdori S ga teng. A-4836=4836-T G-8423=8423-S 2) 1 t a zanjirdagi jami nukleotidlarni topish uchun A ga G ni qo’shamiz: 4836+8423=13259 3) Nukleotidlar orasidagi masofa-0,34 nm yoki 3,4 A 0
H=13259*3,4 A 0 =45080,6 A 0
4) H=2A-T+3G-S => H=(2*4836)+(3*8423)=9672+25269=34941 Javob: A
265. DNK molekulasining birinchi ipida umumiy nukleotidlarni 4,5 foizini tashkil etuvchi A joylashgan bo’lib, qolgan A ikkinchi ipda joylashgan. Ikkinchi ipda yana 162 ta T bor. DNK ni uzunligini nanometrda aniqlang. A) 612 nm B) 816 nm C) 204 nm D) 306 nm Ishlanishi: 1) har doim A miqdori T ga, G miqdori S ga teng. A-4,5%-162=162-4,5%-T 4,5+4,5=9% 100%-9%=91% -bu foiz G va S ga tegishli, shuning uchun: G-45,5%=45,5%-S 162 (A-T) ---- 4,5% x (G-S) ---- 45,5% x=(162*45,5)/4,5=1638 Demak, G-1638=1638-S 2) 1 ta zanjirdagi jami nukleotidlarni topish uchun A ga G ni qo’shamiz: 162+1638=1800 3) Nukleotidlar orasidagi masofa-0,34 nm yoki 3,4 A 0
H=1800*0,34=612nm Javob: C
266. DNK molekulasi 200 ta timin va 100 ta sitozinga ega. Shu DNKdan hosil bo`lgan oqsil molekulasidagi peptid bog`lar sonini toping. A) 199 B) 99 C) 54 D) 299 Ishlanishi: 1) A=T; G=S; T+S=N => 200+100=300N DNK ni 1 ta ipi yoki i-RNK dagi N lar
2) AK=N/3 AK=300/3=100AK 3) P=AK-1=100-1=99 P Javob: D _____________________________ 267. DNK zanjiri ААТSАSGАТSSТ nukleotidlari ketma- ketligiga ega. Oqsil hosil bo`lishi uchun ishtirok etadigan t-RNK xillarini belgilang A) ААU, SАS, GАU, SSU B) ААU, GУG, SUА, АGG C) ААU, SАS, GАU, SSА D) UUА, GUG, SUА, GGА Ishlanishi: 1) DNK 1-ipda: T TAGTG S TAGGA DNK 2-ipda: ААТ SАS GАТ SS Т i-RNK da: UUAGUGSUAGGA Javob: D _____________________________ 268. UUAGSSGAU ushbu RNK dan teskari transkriptaza vositasida sintezlangan DNK da nechta H bog’lar bor
1) i-RNK da: UUAGSSGAU DNK 1-ipda: AAT SGGSTA DNK 2-ipda: T TAGS SGAT 2) A-T=5; G-S=4 H=2A+3G H=(2*5)+(3*4)=22 Javob: A
269. DNK molekulasida 880 ta guanin nukleotidi mavjud bo`lib, ular barcha nukleotidlarning 22 % ni tashkil etadi. Sitozin nukleotidlari soni DNK ning uzunligini aniqlang. A) 1120; 680 B) 2240; 880 C) 1760; 880 D) 880; 680 Ishlanishi: 1) G-22%=22%-S G-880=880-S 44%=G+S 100-44=56 2) 56%=A+T 56/2=28 A-28%=28%-T 880---22% A---28% A=(880*28)/22=1120N 3) A+G=880+1120=2000N -DNK ning 1 ta ipidagi N lar 4) L=N*0,34nm L=2000*0,34nm=680nm Javob: D
270. DNK molekulasida 28450 H bog’i bor. Undagi A umumiy nukleotidlarni 25 % ini tashkil etadi. Ushbu DNK ning uzunligini toping ? A) 1934,6 B) 3869,2 C) 9673 D) 5803,8
1) A=T => 25%=A-T=G-S=25% 2A+3G=H 2A+3A=28450 5A=28450 A=5690=T=G=S 2) 1 ta zanjirdagi jami nukleotidlarni topish uchun A ga G ni qo’shamiz: A+G=5690+5690=11380N 3) L=N*0,34nm L=11380*0,34nm=3869,2nm Javob: B
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