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 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
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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
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) 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
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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-
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