B)17600 C)3200 D)34600
Ishlanishi: 1) DNK dagi jami G-S ni topamiz: har doim A miqdori T ga, G
miqdori S ga teng. T-A=
1600 T+A=1600+1600=3200 –bu jami A va T
2) 12800-3200=9600 –bu jami G va S 9600/2=4800 G-S=
4800 A=25% bo’lgani uchun T-G-S lar ham 25% dan bo’ladi, ya’ni
A=T=G=S=A-1600
2) H=2A+3G H=(2*1600)+(3*4800)=3200+14400=
17600 Javob: B ____________________________
60. DNK molekulasida 1950 ta vodorod bog’i bo’lib, bu
molekula tuzilishida 300 ta adenin ishtirok etgan. Nuklein kislotani
tarkibidagi guaninlar sonini toping?
A) 14400 B) 900
C) 450 D) 600
Ishlanishi: 1) DNK dagi jami G-S ni topamiz: har doim A miqdori T ga, G
miqdori S ga teng. T-A=
300 2A+3G=1950
(2*300)+3G=1950
600+3G=1950
3G=1350
G=
450 Javob: C ____________________________
61. DNK fragmentida 222 ta A nukleotidi mavjud bo’lib, u
umumiy nukleotidlarning 10 % ni tashkil qilishi ma’lum. Shu
fragmentga restriktaza fermenti yordamida ishlov berilgandan so'ng
A-T juftligi 9,91 % ga; G-S juftligi esa 25 % ga kamaydi. Dastlabki
va ishlov berilgandan so’ng DNK fragmentining uzunligini toping
(nm).
A) 377,4; 181,22 B) 377,4; 806
C) 377,4; 294,44 D) 754,8; 588,88
Ishlanishi: 1) A-222-10%=10%-222-T 10%+10%=20%
100%-20%=80% G-40%=40%-S
2) 222---10%
x---40% x=(222*40)/10=888
3) 222+888=1110 –bu 1ta zanjirdagi nukleotidlar
4) L=N*0,34nm L=1110*0,34=
377,4nm 5) 222---100% 100-9,91=90,09%
x---90,09% x=(222*90,09)/100=200
6) 888---100% 100-25=75%
x---75% x=(888*75)/100=666
7) 200+666=866 L=N*0,34=866*0,34=
294,44nm J:C ____________________________
62. Endigina sintezlangan DNK molekulasida 2300 ta nukleotid
bor. Shu DNK tarkibidagi eski fosfat kislota qoldig’ini toping
A) 2300
