Biology, asked by KIRTIRAJ8154, 1 year ago

Method for seperation of plasmid dna from large bacterial chromosomal dna

Answers

Answered by mannasusanabr04
0

please following this protocol step.

Materials Required:

Lysis buffer (40mM tris-acetate PH 7.8, 20mM sodium-acetate, 1mM EDTA, 1% SDS), chloroform, 5 MNacl, 70% EtOH, TE buffer.

Protocol:

1- Harvest bacteria colony into 1.5 ml Eppendorf tube.

2- resuspend cell pellet in 200 µl of lysis buffer (40 mM Tris-acetate PH 7.8, 20µl sodium – acetate, 1mM EDTA, 1% SDS) by vigorous pipetting.

3- Add 66 µl of 5M NaCl solution (to remove most protein and cell debris).

4- Centrifuge at 12000 rpm for 10 min at 4℃.

5- Transfer the clear supernatant into new vial, add equal volume of chloroform.

6- Shake tube gently at least 50 times until a milky solution completely formed.

7- Centrifuge 12000 rpm for 3min.

8- Transfer the supernatant to another vial

9- Add equal volume of cold absolute ethanol to precipitate DNA.

10-wash DNA with 70% Ethanol

11-dry DNA and dissolve in 50µl TE buffer or double distil water.

The yield of DNA around 50-150ug per 109 cells

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