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
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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
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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
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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
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259. DNK molekulasida 28450 H bog’i bor. Undagi A umumiy
nukleotidlarni 25 % ini tashkil etadi. Ushbu molekulada nechta G
bor ?
Dostları ilə paylaş: