Abituriyentlarga qulayliklari

J: 150

107. DNKda timindan 850taga sitozindan 900taga ko`p vodorod bog` bor bo`lsa DNKni uzunligini aniqlang.
(nukleotidlar orasidagi masofa 0,34nm) J:153 nm

X—850 = T

X - 900 = S
(x - 850) ×2+ (x- 900) × 3= x
2x - 1700 + 3x – 2700 = x
5x – 4400 = x
4x = 4400
X= 1100
1100 - 850 = 250; 1100 – 900 = 200
250 × 2 = 500 A – T
200 × 2 = 400 G – S
450 × 0.34 = 153 nm
Javob 153 nm

107. 2ta DNK da 1560 ta nukleotid bor T yig`indisi 262ta 1-chi DNKda 15%T 2-chisida 30% G bor bo`lsa 1-chi DNK uzunligini hisoblang. J:170nm

0.15x + (1560-x) × 0.2 = 262
0.15x + 312 – 0.2x = 262
0.05x = 50
X= 1000 / 2 = 500
500 × 0.34 = 170 nm

108. DNK fragmentida 188 ta kimyoviy bog’ mavjud bo’lib, undagi H bog’ va fodfodiefir bog’ ayirmasi 32 ga teng bo’lsa T necha foizni tashkil etadi.

A) 37.5 % B) 24 % C) 36 % D) 12.5 %
Yechimi:
x--- vodorod bog’ y-- fosfodiefir bog’(F)
{x + y = 188| y=78 F bog’
{x - y = 32| x= 110 H bog’
78+2 = 80 jami nukleotid
2x + 2y = 80| y= 30 G yoki S
2x + 2y = 110| x= 10 A yoki T
80 ta --- 100 %
10 ta ---- x = 12.5 %
Javob: D

• 
  EcoRI 10 bog' kesadi. 2 tasi G-A orasidagi fosfodiefir.
  

8 tasi A va T orasidagi H bog'lar.

• 
  HaeIII faqat G-S orasidagi fosfodiefir bog'ni kesadi 2 ta.
  

• 
  BamH1 12 ta bog' kesadi shundan 2 tasi fosfodiefir G-G orasidagi. Qolgan 10 tasi H bog', 6 tasi S-G orasida
  

4 tasi A-T orasida.