Ishlanishi: 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
Ishlanishi: 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