A+G=N 240+960=
1200 N -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=
