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Learning Module - Experiment 5A

Transformation of E. coli XL-1 Blue and DM700 with pTrc99A/topAcysB DNA

In this lab, you will use the competent cells made in Experiment 4A and the recombinant DNA made in Experiment 4B to produce bacteria containing the pTrc99A/topAcysB plasmid.

First you will combine DNA and competent cells and incubate on ice:

DNA binds to the surface of the bacterium as a calcium-phosphate-DNA complex

The DNA is usually in excess (there are several DNA molecules per cell) in the transformation mixture. There is an equilibrium between DNA bound to the cell surface and DNA that is free in solution, and it takes about 20 minutes to reach this point. The DNA that is bound to the surface is not bound tightly, and any disturbance to the tube during this incubation can cause the DNA to "fall off" the cell surface, leading to low transformation efficiency.

Then you expose the bacteria to a "heat shock":

The heat shock technique is illustrated.

The uptake of DNA at this stage is one of molecular biology's "magical mysteries". No one really knows how the DNA gets inside the cell, we just know that it works. It doesn't work with untreated bacteria, that is, bacteria that have not been made competent, so it must have something to do with the treatment in calcium chloride that makes the cell more permeable to DNA. Some cells may take in more than one plasmid, some may take in none, and some will take in only a single plasmid, even though several may be bound to the surface at the time of the heat shock. Supercoiled plasmids are most easily taken up, yielding transformation efficiencies in the range 106 - 1010 transformants/μg of DNA. Relaxed circular DNA (nicked or covalently closed) gives efficiencies about 10- to 100-fold less than supercoiled, and linear DNA is down by another factor of 10- to 100-fold from relaxed circular DNA. In which topological form will your recombinant pTrc99A/topAcysB plasmid be? How about the positive control DNA?

The cells are then incubated in a nutrient broth at 37 C to allow them to recover before plating on selective media:

DNA replication inside a bacterial cell.

Cells that have not taken up a plasmid will not grow on the selective media when plated. Cells that have taken up a relaxed plasmid will immediately repair any nicks in the plasmid by using E. coli DNA ligase, and then introduce negative supercoils by using DNA gyrase. The plasmid is now ready to participate in cellular proccesses such as replication and transcription.

When you get your plates back in Period 6, you should enter the colony count data in the Data and Results section of Experiment 5A. You should construct a table like this for entering the data:

Data table layout

You need to calculate the amount of DNA added to the "Ligation" tubes and the "Tube 2" control. Do this by reviewing the protocol used in Period 4B and calculate the amounts of pTrc99A DNA in each tube (you should have already done this for the Experiment 5B analysis). Transformation efficiencies are usually based on the amount of vector DNA present in the transformation reaction. The amounts of DNA used in the positive controls are known. The number of colonies (if any) on the "negative control" needs to be subtracted from all colony counts, and the CFU/μg from "Tube 2" needs to be subtracted from the CFU/μg calculated for the "ligation" plates. If none of your plates have colony counts that are statistically significant, you should divide the total number of colonies on the three "ligation" plates by the total amount of DNA used in the three transformations with the ligation reaction. Prepare a similar table for the DM700 cells. In the end, you should report a transformation efficiency for XL-1 Blue with pTrc99A/topAcysB (positive control), XL-1 Blue with ligation reaction, DM700 positive control, and DM700 ligation.