Suppose I want to scale up the procedure to isolate DNA from a much larger E. co

Question

Suppose I want to scale up the procedure to isolate DNA from a much larger E. coli culture than 1.5 ml.   A) Please indicate how much Solution II and Solution III you would need if you added 15 ml of Solution I and B) What is the volume relationship between solution I, solution II and solution III such that regardless of the amount of culture, if you calculate the volume of solution 1, the rest can be determined quickly?

Break open (lyse) the cells containing the DNA of interest. This is done using a solution (cleverly called Solution 1) that contains glucose, Tris, and EDTA. Tris is a buffering agent used to maintain a constant pH (-8.0). EDTA is a divalent ion chelator that helps protect the DNA from degradative enzymes (called DNAases). DNAases require Mg2 for activity, so the EDTA binds the Mg2 ions preventing this cofactor from being available in the solution for the DNAases. Once the cells are resuspended in Solution 1, you then add a second solution called Solution 2 which contains NaOH and SDS (a detergent). Since NaOH is a strong base it causes the solution to become highly alkaline. The alkaline mixture ruptures the cells, and the detergent breaks apart the lipid membrane and solubilizes cellular proteins. NaOH also denatures the DNA into single strands. After step 2 the cells will be lysed Once the cells are lysed, you now need to get rid of the lipids and proteins present in the sample Precipitation of the protein is aided by the addition of a salt such as ammonium or sodium acetate. At this step you will add a solution called guess what Solution 3. Solution 3 contains a mixture of acetic acid and potassium acetate. The acetic acid neutralizes the pH, allowing the DNA strands to renature The potassium acetate salt precipitates the SDS from solution, along with the cellular debris (e.g proteins). The E. coli chromosomal DNA, a partially renatured tangle at this stage will get trapped in the precipitate. Only the plasmid DNA will remain in solution The next step is to separate the plasmid DNA from the cellular debris and E. coli chromosomal DNA. This is accomplished by placing the tubes containing your sample into a microcentrifuge and centrifuging the tubes for 10 minutes. The insoluble cell debris will pellet to the bottom of the tube, and the plasmid DNA will remain in solution The next step is to carefully transfer the supernatant to a fresh microcentrifuge tube using a pipetman with a clean tip. Be very careful to avoid taking any of the white precipitate during the transfer. It is better to leave a little supernatant behind to avoid taking the precipitate The next step is to concentrate the DNA into a smaller volume. This is accomplished by precipitation of the DNA with Isopropanol. Isopropanol effectively precipitates nucleic acids, but is much less effective with protein precipitation. Aquick precipitation can not only concentrate the DNA, but also provide another level of purification from protein contaminants. If allowed to site for longer time, proteins will also precipitate. Thus, this step is relatively quick to prevent protein precipitation After incubation of the DNA in isopropanol for 10 minutes at room temperature, the tubes containing the DNA are again placed in the microcentrifuge and centrifuged for 10min. After centrifugation, a milky white pellet should be at the bottom of the tube. Carefully pour off the supernatant without dumping out the pellet. Drain the tube on a paper towel and allow the tube to dry for 15 minutes Your TA will tell you when it is sufficiently dry When the dry pellet is ready, add 20 microliters of a solution called TE to the tube. TE stands for Tris EDTA. Again the Tris is the buffer and the EDTA is to protect the DNA from degradation 1. 2. 3. 4. 5. 6. 7. 8.

Explanation / Answer

For plasmid isolation using alkaline lysis method,
Volume of solution I to be added for 1.5 mL of culture = 150 uL
Volume of solution II to be added for 1.5 mL of culture = 300 uL
Volume of solution III to be added for 1.5 mL of culture = 225 uL

For 15 mL bacterial culture,
Sol I = 1500 uL = 1.5 mL
Sol II = 3000 uL = 3 mL
Sol III = 2250 uL = 2.25 mL

The ratio between Sol I: Sol II: Sol III = 1 : 2 : 1.5

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