To Isolate Plasmid Dna from a Bacteria Culture and Set up a Restriction Enzyme Digest for the Plasmid Dna That Had Been Isolated
Autor: Maryam • January 2, 2018 • 1,510 Words (7 Pages) • 585 Views
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* 1 Absorbance unit (260 nm) = 50 µg/mL
Take into consideration the dilution of your sample for the UV reading.
UV absorbance calculation
Plasmid 1
As 1 Absorbance unit (260 nm) = 50 µg/mL,
Then, 0.3701 x 50 µg/mL x 200 (dilution factor)
= 3702 µg/mL
Plasmid 2
As 1 Absorbance unit (260 nm) = 50 µg/mL,
Then, 0.5373 x 50 µg/mL x 200 (dilution factor)
= 5374 µg/mL
260/280 ratio calculation
Plasmid 1
0.3701/0.2400
= 1.542
Plasmid 2
0.5373/0.3426
= 1.568
Table 4. Electrophoretic mobilities of molecular weight marker.
Base Pairs (bp)
Log10 bp
Electrophoretic mobility (mm)
4000
3.602
59
2000
3.301
68
1250
3.100
76
800
2.903
82
500
2.699
89
300
2.477
96
200
2.301
103
100
2.000
109
Table 5. Estimated molecular-weights of plasmid fragments present in plasmids 1 and 2.
Electrophoretic mobility (mm)
Log10 Bp
Estimated number of base pairs
Probable identity of the plasmid fragment
100 (plasmid 2)
2.347
222
unknown
85 (plasmid 1)
2.809
645
VEGF
- Distance travelled for plasmid 1 is obtained from group 15, third lane.
In order to obtain the Log10 Bp for plasmid 1, refer to standard curve (figure 2). From the curve, equation obtained is y = -0.0308x + 5.4274. Substitute electrophoretic mobility measured, in the third lane, which is 85mm.
Y= -0.0308(85) + 5.4274
= 2.809
- Distance travelled for plasmid 2 is obtained from group 15, second lane.
In order to obtain the Log10 Bp for plasmid 2, refer to standard curve (figure 2). From the curve, equation obtained is y = -0.0308x + 5.4274. Substitute electrophoretic mobility measured, in the third lane, which is 62mm.
Y= -0.0308(62) + 5.4274
= 3.218
Table 6.
UV method: (µg/mL)
SYBR Green method (µg/mL)
Plasmid #1
3702
(2µL)= 92.05 µg/mL
(20µL)= 154.83 µg/mL
Plasmid #2
5374
(2µL)= 294.03 µg/mL
(20µL)= 100.61 µg/mL
Calculation for SYBR Green plasmid concentration
Plasmid 1
From graph, y= 310.38x
Plasmid DNA 1 (2µL) has absorbance value of 11427
Substitute absorbance value into equation,
Plasmid DNA 1 (2µL)=
11427= 310.38x
x= 36.82 ng/mL
36.82 ng/mL → 0.03682 µg/mL
0.03682 µg/mL x 50 (dilution factor) x 50 (dilution factor)
= 92.05 µg/mL
Plasmid DNA 1 (20µL) has absorbance value of 192226
Substitute absorbance value into equation,
Plasmid DNA 1 (20µL)=
192226= 310.38x
x= 619.32 ng/mL
619.32 ng/mL→0.6193 µg/mL
0.6193 µg/mL x 5 (dilution factor) x 50 (dilution factor)
= 154.83 µg/mL
Plasmid DNA 2 (2µL) has absorbance value of 36505
Substitute absorbance value into equation,
Plasmid DNA 1 (2µL)=
36505= 310.38x
x= 117.61 ng/mL
117.61
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