DNK-testlar_270 ta masalani ishlanishi
1. Quyidagi ketma-ketlikka ega bo‘lgan
DNK zanjiridagi
vodorod bog‘lar sonini toping.
ATTGSSTAGG
TAASGGATSS A) 20 B) 22 C) 24
D) 25
Ishlanishi:
Bitta zanzirdagi A-T va G-T lar sonini topiladi:
1) A-T lar 5 ta; G-S lar ham 5 ta
A va T o’rtasida 2 ta hamda G va S o’rtasida 3 ta vodorod bog’
bor
2) (5*2)+(5*3)=
25
Javob: D
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2. Agar DNKdagi qo‘shni nukleotidlar orasidagi masofa 0,34 nm
ekanligini
hisobga olsak, 90 ta aminokislotadan
iborat oqsilni
kodlovchi DNK fragmentidagi
nukleotidlar sonini va DNK
uzunligini toping.
A) 270 ta, 91,8 nm
B) 540 ta, 91,8 nm
C) 270 ta, 94 nm D) 90 ta, 91,8 nm
Ishlanishi:
1 AK=3 N -ya’ni 1 ta aminokislotani 3ta nukleotid kodlaydi
1) 90 AK=270 N bu nukleotidlar
i-RNKdagi nukleotidlar soni
va bu DNK ning bitta zanjiriga to’g’ri
keladi
2) 1 ta zanjirda – 270 N
2 ta zanjirda – 270 N jami nukleotidlar –
540
N
3) L=N*0,34nm ya’ni, 270*0,34=
91,8 nm Javob: B
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3. Agar DNK uzunligi 183,6 nm ga teng bo‘lib, undagi qo‘shni
nukleotidlar orasidagi masofa 0,34 nm ga teng bo‘lsa,
ushbu DNK
fragmentida nechta nukleotidlar bo‘ladi
va ular nechta
aminokislotalarni kodlaydilar?
A) 540 ta, 180 ta
B) 1080 ta, 180 ta
C) 536 ta, 1800 ta D) 5480 ta, 180 ta
Ishlanishi:
1) L=N*0,34 => N=L/0,34 => 183,6nm/0,34nm=540N
2) 1 ta zanjirda - 540N;
2 ta zanjirda –
1080N
3) i-RNK DNK ni bitta zanjiridan axborot oladi: i-RNK da ham
540 ta nukleotid bo’ladi. Shunga ko’ra,
bu nukleotidlar nechta
aminokislotani kodlashini topamiz :
540/3=
