Taasggatss a 20 b 22 c 24 d 25 Ishlanishi
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Ishlanishi: 1) 1-DNK=x*0,75; 2-DNK=x; 3-DNK=x/0,8; 2) x*0,75+x+x/0,8=420 0,6x+0,8x+x=336 2,4x=336 x=140 -bu 2-DNK dagi t-RNK lar soni 3) D1=D2*0,75 D1=140*0,75 D1=105 4) 105-1=104P Javob: C _____________________________ 180. 3 xil nomalum nukleotidli DNK malekulasi mavjud bo`lsa.Ulardan oqsil sintezlanishi uchun jami 420 ta t-RNK sarflangan. 2-DNK malekulasidan oqsil sintezlanishi uchun sarflangan t-RNKlar soni 1-DNKdan oqsil sintezlanishi uchun sarflangan t-RNKlar sonidan 0.75 marta kam, 3-DNKdan oqsil sintezlanishi uchun sarf bo`lgan t-RNKlar sonidan 0.8 marta ko`p bo`lsa, 2-DNKdagi peptid bog`lar sonini aniqlang. A)174 B)139 C)104 D)149
1) 1-DNK=x*0,75; 2-DNK=x; 3-DNK=x/0,8; 2) x*0,75+x+x/0,8=420 0,6x+0,8x+x=336 2,4x=336 x=140 -bu 2-DNK dagi t-RNK lar soni 3) 140-1=139P Javob: B
181. 3 xil nomalum nukleotidli DNK malekulasi mavjud bo`lsa.Ulardan oqsil sintezlanishi uchun jami 420 ta t-RNK sarflangan. 2-DNK malekulasidan oqsil sintezlanishi uchun sarflangan t-RNKlar soni 1-DNKdan oqsil sintezlanishi uchun sarflangan t-RNKlar sonidan 0.75 marta kam, 3-DNKdan oqsil sintezlanishi uchun sarf bo`lgan t-RNKlar sonidan 0.8 marta ko`p bo`lsa, 3-DNKdagi aminokislatalar sonini aniqlang.
1) 1-DNK=x*0,75; 2-DNK=x; 3-DNK=x/0,8; 2) x*0,75+x+x/0,8=420 0,6x+0,8x+x=336 2,4x=336 x=140 -bu 2-DNK dagi t-RNK lar soni 3) D3=D2/0,8 D3=140*0,8 D3=175AK Javob: A
182. 3 xil nomalum nukleotidli DNK malekulasi mavjud bo`lsa.Ulardan oqsil sintezlanishi uchun jami 420 ta t-RNK sarflangan. 2-DNK malekulasidan oqsil sintezlanishi uchun sarflangan t-RNKlar soni 1-DNKdan oqsil sintezlanishi uchun sarflangan t-RNKlar sonidan 0.75 marta kam, 3-DNKdan oqsil sintezlanishi uchun sarf bo`lgan t-RNKlar sonidan 0.8 marta ko`p bo`lsa, 1-DNKdagi aminokislatalar sonini aniqlang. A)175 B)140 C)105 D)150
1) 1-DNK=x*0,75; 2-DNK=x; 3-DNK=x/0,8; 2) x*0,75+x+x/0,8=420 0,6x+0,8x+x=336 2,4x=336 x=140 -bu 2-DNK dagi t-RNK lar soni 3) D1=D2*0,75 D1=140*0,75 D1=105AK Javob: C
183. 3 xil nomalum nukleotidli DNK malekulasi mavjud bo`lsa.Ulardan oqsil sintezlanishi uchun jami 420 ta t-RNK sarflangan. 2-DNK malekulasidan oqsil sintezlanishi uchun sarflangan t-RNKlar soni 1-DNKdan oqsil sintezlanishi uchun sarflangan t-RNKlar sonidan 0.75 marta kam, 3-DNKdan oqsil sintezlanishi uchun sarf bo`lgan t-RNKlar sonidan 0.8 marta ko`p bo`lsa, 2-DNKdagi aminokislatalar sonini aniqlang. A)175 B)140 C)105 D)150
1) 1-DNK=x*0,75; 2-DNK=x; 3-DNK=x/0,8; 2) x*0,75+x+x/0,8=420 0,6x+0,8x+x=336 2,4x=336 x=140 -bu 2-DNK => t-RNK=AK Javob: B _____________________________ 184. 3 xil nomalum nukleotidli DNK malekulasi mavjud bo`lsa.Ulardan oqsil sintezlanishi uchun jami 420 ta t-RNK sarflangan. 2-DNK malekulasidan oqsil sintezlanishi uchun sarflangan t-RNKlar soni 1-DNKdan oqsil sintezlanishi uchun sarflangan t-RNKlar sonidan 0.75 marta kam, 3-DNKdan oqsil sintezlanishi uchun sarf bo`lgan t-RNKlar sonidan 0.8 marta ko`p bo`lsa, 1-DNK malekulasini uzunligini aniqlang.(10 ta nukleotid oralig`I 3.4 nm hisoblanadi)
1) 1-DNK=x*0,75; 2-DNK=x; 3-DNK=x/0,8; 2) x*0,75+x+x/0,8=420 0,6x+0,8x+x=336 2,4x=336 x=140 -bu 2-DNK dagi t-RNK lar soni 3) D1=D2*0,75 D1=140*0,75 D1=105 105*3=315 4) 315*0,34nm=107,1nm Javob: A _____________________________ 185. 3 xil nomalum nukleotidli DNK malekulasi mavjud bo`lsa.Ulardan oqsil sintezlanishi uchun jami 420 ta t-RNK sarflangan. 2-DNK malekulasidan oqsil sintezlanishi uchun sarflangan t-RNKlar soni 1-DNKdan oqsil sintezlanishi uchun sarflangan t-RNKlar sonidan 0.75 marta kam, 3-DNKdan oqsil sintezlanishi uchun sarf bo`lgan t-RNKlar sonidan 0.8 marta ko`p bo`lsa, 2-DNK malekulasini uzunligini aniqlang.(10 ta nukleotid oralig`I 3.4 nm hisoblanadi) A)107.1 B)142.8 C)178.5 D)163.2 Ishlanishi: 1) 1-DNK=x*0,75; 2-DNK=x; 3-DNK=x/0,8; 2) x*0,75+x+x/0,8=420 0,6x+0,8x+x=336 2,4x=336 x=140 -bu 2-DNK dagi t-RNK lar soni 3) 140*3=420 4) 420*0,34nm=142,8nm Javob: B _____________________________ 186. 3 xil nomalum nukleotidli DNK malekulasi mavjud bo`lsa.Ulardan oqsil sintezlanishi uchun jami 420 ta t-RNK sarflangan. 2-DNK malekulasidan oqsil sintezlanishi uchun sarflangan t-RNKlar soni 1-DNKdan oqsil sintezlanishi uchun sarflangan t-RNKlar sonidan 0.75 marta kam, 3-DNKdan oqsil sintezlanishi uchun sarf bo`lgan t-RNKlar sonidan 0.8 marta ko`p bo`lsa, 3-DNK malekulasini uzunligini aniqlang.(10 ta nukleotid oralig`I 3.4 nm hisoblanadi) A)107.1 B)142.8 C)178.5 D)163.2 Ishlanishi: 1) 1-DNK=x*0,75; 2-DNK=x; 3-DNK=x/0,8; 2) x*0,75+x+x/0,8=420 0,6x+0,8x+x=336 2,4x=336 x=140 -bu 2-DNK dagi t-RNK lar soni 3) D3=D2/0,8 D3=140*0,8 D3=175 175*3=525 4) 525*0,34nm=178,5nm Javob: C _____________________________ 187. 3 xil nomalum nukleotidli DNK malekulasi mavjud. DNK malekulalarida jami 798 ta adenin nukleotidi bor. DNK malekulalaridagi adenin nukleotidlari foizlari quyidagicha 1-DNK da jami nukleotidlarni 20% ni, 2- DNK da jami nukleotidlarni 30%ni, 3- DNK da jami nukleotidlarni 40% ni tashkil qiladi. 2-DNK malekulasidagi adenin nukleotidlar soni 1-DNK malekulasidagi adenin nukleotidlari sonidan 0.5 marta kam, 3- DNK malekulasidagi adeninlar sonidan 0.6 marta ko`p bo`lsa,1-DNK malekulasidagi timin va sitozinlar sonini yig`indisini aniqlang A)315 ta B)420 C)525 D)675 Ishlanishi: 1) 1-DNK=0,5x; 2-DNK=x; 3-DNK=x/0,6; 2) 0,5x+x+x/0,6=798 0,3x+0,6x+x=478,8 1,9x=478,8 x=252 -bu 2-DNK dagi A lar soni 3) D1=D2*0,5 D1=252*0,5 D1=126 A-20%=20%-A 4) 126---20% x---30% x=(126*30)/20=189 S-30%=30%-G 5) S+T=126+189=315 Javob: A _____________________________ 188.3 xil nomalum nukleotidli DNK malekulasi mavjud. DNK malekulalarida jami 798 ta adenin nukleotidi bor. DNK malekulalaridagi adenin nukleotidlari foizlari quyidagicha 1-DNK da jami nukleotidlarni 20% ni, 2- DNK da jami nukleotidlarni 30%ni, 3- DNK da jami nukleotidlarni 40% ni tashkil qiladi. 2-DNK malekulasidagi adenin nukleotidlar soni 1-DNK malekulasidagi adenin nukleotidlari sonidan 0.5 marta kam, 3- DNK malekulasidagi adeninlar sonidan 0.6 marta ko`p bo`lsa,2-DNK malekulasidagi timin va sitozinlar sonini yig`indisini aniqlang A)315 ta B)420 C)525 D)675
1) 1-DNK=0,5x; 2-DNK=x; 3-DNK=x/0,6; 2) 0,5x+x+x/0,6=798 0,3x+0,6x+x=478,8 1,9x=478,8 x=252 -bu 2-DNK dagi A lar soni A-30%=30%-A 3) 252---30% x---20% x=(252*20)/30=168 S-20%=20%-G 4) S+T=252+168=420 Javob: B _____________________________ 189. 3 xil nomalum nukleotidli DNK malekulasi mavjud. DNK malekulalarida jami 798 ta adenin nukleotidi bor. DNK malekulalaridagi adenin nukleotidlari foizlari quyidagicha 1-DNK da jami nukleotidlarni 20% ni, 2- DNK da jami nukleotidlarni 30%ni, 3- DNK da jami nukleotidlarni 40% ni tashkil qiladi. 2-DNK malekulasidagi adenin nukleotidlar soni 1-DNK malekulasidagi adenin nukleotidlari sonidan 0.5 marta kam, 3- DNK malekulasidagi adeninlar sonidan 0.6 marta ko`p bo`lsa,3-DNK malekulasidagi timin va sitozinlar sonini yig`indisini aniqlang A)315 ta B)420 C)525 D)675
1) 1-DNK=0,5x; 2-DNK=x; 3-DNK=x/0,6; 2) 0,5x+x+x/0,6=798 0,3x+0,6x+x=478,8 1,9x=478,8 x=252 -bu 2-DNK dagi A lar soni 3) D3=D2/0,6 D3=252/0,6 D3=420 A-40%=40%-A 4) 420---40% x---10% x=(452*30)/20=105 S-10%=10%-G 5) S+T=420+105=425 Javob: C _____________________________ 190. 3 xil nomalum nukleotidli DNK malekulasi mavjud. DNK malekulalarida jami 798 ta adenin nukleotidi bor. DNK malekulalaridagi adenin nukleotidlari foizlari quyidagicha 1-DNK da jami nukleotidlarni 20% ni, 2- DNK da jami nukleotidlarni 30%ni, 3- DNK da jami nukleotidlarni 40% ni tashkil qiladi. 2-DNK malekulasidagi adenin nukleotidlar soni 1-DNK malekulasidagi adenin nukleotidlari sonidan 0.5 marta kam, 3- DNK malekulasidagi adeninlar sonidan 0.6 marta ko`p bo`lsa,1-DNK malekulasidagi kimyoviy bog`lar sonini aniqlang A)1447 B)819 C)1008 D)1846
1) 1-DNK=0,5x; 2-DNK=x; 3-DNK=x/0,6; 2) 0,5x+x+x/0,6=798 0,3x+0,6x+x=478,8 1,9x=478,8 x=252 -bu 2-DNK dagi A lar soni 3) D1=D2*0,5 D1=252*0,5 D1=126 A-20%=20%-A 4) 126---20% x---30% x=(126*30)/20=189 S-30%=30%-G 5) S+T=126+189=315 315*2=630 -DNK dagi archa N lar 6) H=2A+3G H=(2*126)+(3*189)=819 -H bog’ Javob: A
191. 3 xil nomalum nukleotidli DNK malekulasi mavjud. DNK malekulalarida jami 798 ta adenin nukleotidi bor. DNK malekulalaridagi adenin nukleotidlari foizlari quyidagicha 1-DNK da jami nukleotidlarni 20% ni, 2- DNK da jami nukleotidlarni 30%ni, 3- DNK da jami nukleotidlarni 40% ni tashkil qiladi. 2-DNK malekulasidagi adenin nukleotidlar soni 1-DNK malekulasidagi adenin nukleotidlari sonidan 0.5 marta kam, 3- DNK malekulasidagi adeninlar sonidan 0.6 marta ko`p bo`lsa,2-DNK malekulasidagi kimyoviy bog`lar sonini aniqlang A)1447 B)819 C)1008 D)1846
1) 1-DNK=0,5x; 2-DNK=x; 3-DNK=x/0,6; 2) 0,5x+x+x/0,6=798 0,3x+0,6x+x=478,8 1,9x=478,8 x=252 -bu 2-DNK dagi A lar soni A-30%=30%-A 3) 252---30% x---20% x=(252*20)/30=168 S-20%=20%-G 4) S+T=252+168=420 420*2=840 –DNK dagi barcha N lar 5) H=2A+3G H=(2*252)+(3*168)=1008 -H bog’ Javob: C _____________________________ 192. 3 xil nomalum nukleotidli DNK malekulasi mavjud. DNK malekulalarida jami 798 ta adenin nukleotidi bor. DNK malekulalaridagi adenin nukleotidlari foizlari quyidagicha 1-DNK da jami nukleotidlarni 20% ni, 2- DNK da jami nukleotidlarni 30%ni, 3- DNK da jami nukleotidlarni 40% ni tashkil qiladi. 2-DNK malekulasidagi adenin nukleotidlar soni 1-DNK malekulasidagi adenin nukleotidlari sonidan 0.5 marta kam, 3- DNK malekulasidagi adeninlar sonidan 0.6 marta ko`p bo`lsa, 3-DNK malekulasidagi kimyoviy bog`lar sonini aniqlang A)2203 B)1155 C)1008 D)1846
1) 1-DNK=0,5x; 2-DNK=x; 3-DNK=x/0,6; 2) 0,5x+x+x/0,6=798 0,3x+0,6x+x=478,8 1,9x=478,8 x=252 -bu 2-DNK dagi A lar soni 3) D3=D2/0,6 D3=252/0,6 D3=420 A-40%=40%-A 4) 420---40% x---10% x=(452*30)/20=105 S-10%=10%-G 5) S+T=420+105=525 525*2=1050 –DNK dagi barcha N lar 5) H=2A+3G H=(2*420)+(3*105)=1155 -H bog’ Javob: B
193. Teskari transkripsataza fermenti yordamida i-RNK dan DNK sintezlandi i-RNK da U=20% bo`lib, DNK qo`sh zanjirining 15% ni timin tashkil qilasa i-RNKda A-necha foizni tashkil qiladi? A)10 B)15 C)20 D)30 Ishlanishi: 1) (i-RNK)U+A=DNK(A+T) 2) 20U+xA=15A+15T A=T 20+x=15+15 x=30-20 x=10 i-RNK-A=10% Javob: A _____________________________ 194. Teskari transkriptaza fermenti yordamida i-RNK dan DNK sintezlandi. i-RNK da U=55% bolib DNK qo`sh zanjirining 35% ni timin tashkil qilsa, i-RNK da necha foiz A bo`ladi? A)15 B)55 C)20 D)35 Ishlanishi: 1) (i-RNK)U+A=DNK(A+T) 2) 55U+xA=35A+35T A=T 55+x=35+35 x=70-55 x=15 i-RNK-A=15% Javob: A _____________________________ 195. DNK qo`sh zanjirida 180ta vodorod bog` bo`lib ulardan 1/3 G-S orasida bo`lsa DNK fragmenti uzunligini aniqlang(qo`shni nukleotidlar orasidagi masofa 0,34nm) A)27,2 B)17,8 C)30,6 D)71,4 Ishlanishi: 1) 3---180H 1---x x=180/3=60H -bu G va S orasodagi H bog’lar 2) 180-60=120H -bu A va T orasidagi H bog’lar 3) G va S orasida 3 ta H bog’ bo’lgani uchun: 60/3=20 G-S 4) A va T orasida 2 ta H bog’ bo’lgani uchun: 120/2=60 A-T 5) (A+G)=N 60+20=80 6) L=N*0,34nm L=80*0,34=27,2nm Javob: A _____________________________ 196. DNK qo`sh zanjirida 180ta vodorod bog` bo`lib ulardan 1/3 G-S orasida bo`lsa DNK fragmentidagi fosfodiefir bog`lar sonini aniqlang(qo`shni nukleotidlar orasidagi masofa 0,34nm) A)158 B)178 C)79 D)89 Ishlanishi: 1) 3---180H 1---x x=180/3=60H -bu G va S orasodagi H bog’lar 2) 180-60=120H -bu A va T orasidagi H bog’lar 3) G va S orasida 3 ta H bog’ bo’lgani uchun: 60/3=20 G-S 4) A va T orasida 2 ta H bog’ bo’lgani uchun: 120/2=60 A-T 5) (A+G)=N 60+20=80N –bu 1 ta ipdagi N lar soni 6) 2N=2*80=160 7) F=2N-2 F=160-2=158F Javob: A
197. A=(G+S) holatda umumiy vodorod bog`lar soni 126 ta bo`lsa DNK fragmenti uzunligini aniqlang.(qo`sh. Nuk. Orasidagi masofa 0.34nm)
1) 2A+3G=126 A=2G 2) 2*2G+3G=126 7G=126 G=18 3) 2A+3*18=126 2A=126-54 2A=72 A=36 4) (A+G)*0,34nm=(18+36)*0,34=54*0,34=18,36nm Javob: A
198. A=(G+S) holatda umumiy vodorod bog`lar soni 126 ta bo`lsa, DNK fragmentidagi fosfodiefir bog`lar sonini hisoblang.
1) 2A+3G=126 A=2G 2) 2*2G+3G=126 7G=126 G=18 3) 2A+3*18=126 2A=126-54 2A=72 A=36 4) (A+G)=(18+36)*2=54N 2N=54*2=108N 5) F=N-2 F=108-2=106F Javob: A _____________________________ 199. A=(G+S) holatda umumiy vodorod bog`lar soni 126 ta bo`lsa DNK dagi A-? T-? G-? S-?
C) A-36 T-36 G-36 S-36 D) A-18 T-18 G-18 S-18 Ishlanishi: 1) 2A+3G=126 A=2G 2) 2*2G+3G=126 7G=126 G=18 3) 2A+3*18=126 2A=126-54 2A=72 A=36 4) A-T=36; G-S=18 Javob: A
200. A=(G+S) holatda umumiy vodorod bog`lar soni 126 ta bo`lsa, DNK qo`sh zanjiridagi A-T orasidagi va G-S orasidagi vodorod bog`lar farqini aniqlang. A)18 B)36 C)20 D)24 Ishlanishi: 1) 2A+3G=126 A=2G 2) 2*2G+3G=126 7G=126 G=18 3) 2A+3*18=126 2A=126-54 2A=72 A=36 4) H=2A-3G H=(2*36)-(3*18)=72-54=18 Javob: A
201. DNK qo`sh zanjirida 198 ta vodorod bog`I bo`lib guanin umumiy nukleotidlardan 3,5 marta kam bo`lsa umumiy nukleotidlarni toping. A)154 B)88 C)132 D)110 Ishlanishi: 1) 2A+3G=198 G=2A+2G/3,5 2A+2G=3,5G 2A=3,5G-2G 2A=1,5G 2) 1,5G+3G=198 4,5G=198 G=44 3) 2A+3*44=198 2A=198-132 2A=66 A=33 4) N=(A+G)*2 N=(33+44)*2=77*2=154N Javob: A _____________________________ 202. DNK qo`sh zanjirida 198 ta vodorod bog`I bo`lib guanin umumiy nukleotidlardan 3,5 marta kam bo`lsa, DNK uzunligini aniqlang.(qo`shni nukleotidlar orasidagi masofa 0,34nm) A)26,18 B)52,36 C)67,32 D)33,66 Ishlanishi: 1) 2A+3G=198 G=2A+2G/3,5 2A+2G=3,5G 2A=3,5G-2G 2A=1,5G 2) 1,5G+3G=198 4,5G=198 G=44 3) 2A+3*44=198 2A=198-132 2A=66 A=33 4) L=(A+G)*0,34 N=(33+44)*0,34=26,18nm Javob: A _____________________________ 203. DNK qo`sh zanjirida 198 ta vodorod bog`I bo`lib guanin umumiy nukleotidlardan 3,5 marta kam bo`lsa G-S orasidagi va A-T orasidagi vodorod bog`lar farqini aniqlang. A)66 B)44 C)88 D)22 Ishlanishi: 1) 2A+3G=198 G=2A+2G/3,5 2A+2G=3,5G 2A=3,5G-2G 2A=1,5G 2) 1,5G+3G=198 4,5G=198 G=44 3) 2A+3*44=198 2A=198-132 2A=66 A=33 4) H=2A-3G H=(2*33)-(3*44)=66-132=66N Javob: A _____________________________ 204. DNK qo`sh zanjirida 126 ta vodorod bog` bo`lib sitozin purin asosidan 1,6 marta kam bo`lsa, umumiy nukleotidlarni aniqlang. A)96 B)98 C)104 D)108 Ishlanishi: 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=48*2=96N Javob: A
_____________________________ 205. DNK qo`sh zanjirida 126 ta vodorod bog` bo`lib sitozin purin asosidan 1,6 marta kam bo`lsa, DNK uzunligini aniqlang.
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) L=(A+G)*0,34 L=(18+30)*0,34=16,32nm Javob: A
206. DNK qo`sh zanjirida 510 ta vodorod bog` bo`lib sitozin purin asosidan 2,333 marta kam bo`lsa, DNK uzunligini aniqlang.
1) 2A+3G=510 S=A+G/2,333 S=G A+G=2,333G A=2,333G-G A=1,333G 2) 2*1,333G+3G=510 2,666G+3G=510 5,666G=510 G=90 3) 2A+3*90=510 2A=510-270 2A=240 A=120 4) L=(A+G)*0,34 L=(120+90)*0,34=71,4nm Javob: A
207. DNK qo`sh zanjirida 510 ta vodorod bog` bo`lib sitozin purin asosidan 2,333 marta kam bo`lsa, DNK fosfodiefir bog`lar sonini aniqlang. A)418ta B)510ta C)390ta D)260ta Ishlanishi: 1) 2A+3G=510 S=A+G/2,333 S=G A+G=2,333G A=2,333G-G A=1,333G 2) 2*1,333G+3G=510 2,666G+3G=510 5,666G=510 G=90 3) 2A+3*90=510 2A=510-270 2A=240 A=120 4) N=(A+G)*2 N=(120+90)*2=420 5) F=N-2 F=420-2=418F Javob: A _____________________________ 208. DNK qo`sh zanjirida 510 ta vodorod bog` bo`lib sitozin purin asosidan 2,333 marta kam bo`lsa, DNK umumiy nukleotidlar sonini aniqlang. A)420 B)480 C)360 D)540 Ishlanishi: 1) 2A+3G=510 S=A+G/2,333 S=G A+G=2,333G A=2,333G-G A=1,333G 2) 2*1,333G+3G=510 2,666G+3G=510 5,666G=510 G=90 3) 2A+3*90=510 2A=510-270 2A=240 A=120 4) N=(A+G)*2 N=(120+90)*2=420N Javob: A
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