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205. DNK qo`sh zanjirida 126 ta vodorod bog` bo`lib sitozin
purin asosidan 1,6 marta kam bo`lsa, DNK uzunligini aniqlang.
A)16,32 B)32,64 C)18,36 D)36,72
Ishlanishi:
1) 2A+3G=126
S=A+G/1,6 S=G
A+G=1,6G
A=1,6G-G
A=0,6G
2) 2*0,6G+3G=126
1,2G+3G=126
4,2G=126
G=30
3) 2A+3*30=126
2A=126-90
2A=36
A=18
4) L=(A+G)*0,34 L=(18+30)*0,34=16,32nm Javob: A
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206. DNK qo`sh zanjirida 510 ta vodorod bog` bo`lib sitozin
purin asosidan 2,333 marta kam bo`lsa, DNK uzunligini aniqlang.
A)71,4nm B)44 C)88 D)22
Ishlanishi:
1) 2A+3G=510
S=A+G/2,333 S=G
A+G=2,333G
A=2,333G-G
A=1,333G
2) 2*1,333G+3G=510
2,666G+3G=510
5,666G=510
G=90
3) 2A+3*90=510
2A=510-270
2A=240
A=120
4) L=(A+G)*0,34 L=(120+90)*0,34=71,4nm Javob: A
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207. DNK qo`sh zanjirida 510 ta vodorod bog` bo`lib sitozin
purin asosidan 2,333 marta kam bo`lsa, DNK fosfodiefir bog`lar
sonini aniqlang.
A)418ta B)510ta C)390ta D)260ta
Ishlanishi:
1) 2A+3G=510
S=A+G/2,333 S=G
A+G=2,333G
A=2,333G-G
A=1,333G
2) 2*1,333G+3G=510
2,666G+3G=510
5,666G=510
G=90
3) 2A+3*90=510
2A=510-270
2A=240
A=120
4) N=(A+G)*2 N=(120+90)*2=420
5) F=N-2 F=420-2=418F Javob: A
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208. DNK qo`sh zanjirida 510 ta vodorod bog` bo`lib sitozin
purin asosidan 2,333 marta kam bo`lsa, DNK umumiy nukleotidlar
sonini aniqlang.
A)420 B)480 C)360 D)540
Ishlanishi:
1) 2A+3G=510
S=A+G/2,333 S=G
A+G=2,333G
A=2,333G-G
A=1,333G
2) 2*1,333G+3G=510
2,666G+3G=510
5,666G=510
G=90
3) 2A+3*90=510
2A=510-270
2A=240
A=120
4) N=(A+G)*2 N=(120+90)*2=420N Javob: A
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209. Teskari transkripsataza fermenti yordamida i-RNK dan
DNK sintezlandi i-RNK da G=25% bo`lib, DNK qo`sh zanjirining
15% ni guanin tashkil qilasa i-RNKda A va T necha foizni tashkil
qiladi?
A)70 B)50 C)60 D)80
Ishlanishi:
1) i-RNK da: U-25%
DNK da: G-15%=15%-S G+S=30%
2) 100%-30%=70% A+T=70% Javob: A
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210. DNK qo`sh zanjirida 720 ta vodorod bog`I bo`lib guanin
umumiy nukleotidlardan 5 marta kam bo`lsa umumiy nukleotidlar
sonini toping.
A)600 B)650 C)400 D)450
Ishlanishi:
1) 2A+3G=720
G=2A+2G/5
2A+2G=5G
2A=5G-2G
2A=3G
2) 3G+3G=720
6G=720
G=120
3) 2A+3*120=720
2A=720-360
2A=360
A=180
4) N=(A+G)*2 N=(180+120)*2=600N Javob: A
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211. A=(G+S) holatda umumiy vodorod bog`lar soni 420 ta
bo`lsa DNK fragmenti uzunligini aniqlang.(qo`sh. Nuk. Orasidagi
masofa 0.34nm)
A)61,2 B)44,3 C)51 D)54,4
Ishlanishi:
1) 2A+3G=420
A=(G+S) S=G
A=2G
2) 2*2G+3G=420
4G+3G=420
7G=420
G=60
3) 2A+3*60=420
2A=420-180
2A=240
A=120
4) L=(A+G)*0,34 L=(120+60)*0,34=61,2nm Javob: A
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212. Teskari transkripsataza fermenti yordamida i-RNK dan
DNK sintezlandi i-RNK da U=10% bo`lib, DNK qo`sh zanjirining
15% ni timin tashkil qilasa i-RNKda A-necha foizni tashkil qiladi?
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