This experiment is discussed:
You want to clone a cDNA into an expression vector so you can make large amounts of the protein in E. coli. The cDNA is flanked by BamHI sites and you plan to insert it at the BamHI site in the vector.
The manual recommends that you cleave the vector DNA and then treat it with alkaline phosphatase to remove the 5' phosphates. The next step is to mix the treated vector with the BamHI-cut cDNA fragment and incubate with DNA ligase.
After ligation the DNA is mixed with bacterial cells that have been treated to make them competent to take up DNA. Finally, the mixture is spread unto culture dishes with a solid growth medium that contains an antibiotic that kills all cells that have not taken up the vector. The vector allows cells to survive because it carries a gene for resistance to the antibiotic.
The cloning manual also suggests four controls:
Control 1: plate bacterial cells that have not been exposed to any vector onto the culture dishes.
Control 2: Plate cells that have been transformed with the vector that has not been cut.
Control 3: Plate cells that have been transformed with vector that has been cut (but not treated with alkaline phosphatase) and then incubated with DNA ligase (in the absence of the cDNA fragment).
Control 4: Plate cells that have been transformed with vector that has been cut and treated with alkaline phophatase and then incubated with DNA ligase in the absence of the cDNA fragment).
Your results are listed in the table below:
Sample Resulting colonies
Control 2 >1000
Control 3 435
Control 4 25
Experimental sample 34
You pick 12 colonies from the experimental sample, prepare plasmid DNA from them, and digest the DNA with BamHI.
Nine colonies yield a single band the same size as the linearized vector, but three colonies have in addition, a fragment the size of the cDNA you wanted to clone.
a. What is the point of control 1?
b. What is the point of control 2?
c. What is the point of control 3?
d. What is the point of control 4?
e. Explain the results of your experiment.
f. Why did the manual recommend treating the vector with alkaline phosphatase?
Please remember that if you disagree with the answers provided, it is best to provide your own. Good Luck!
For control 1: These bacterial cells were not exposed to the vector, and no colonies were seen growing in the culture medium. The culture medium contains an antibiotic that only the cell carrying a vector will be protected from. Therefore, the point of the first control is make sure that cells lacking vector will not grow in your culture medium. If cells lacking vector did grow, then your antibiotic check would be pointless.
For control 2: Here you have bacterial cells that have been exposed to an uncut vector, and less than 1000 colonies grow. The point of cutting the vector in the first place was to allow the insertion of your cDNA. The vector is a means to transfect your bacterial cells with your gene of interest. Here it seems that your vector is not cut and not exposed to the cDNA. When bacterial cells are added to the culture medium (containing antibiotic) the cells that took up your vector will survive and multiple to colonies. This is because it is not your cDNA that is protective in the medium, but rather your vector that contains your antibiotic resistance gene. As long as cells contain that vector, they will survive in the culture medium. The point of control 2 was to make sure your vector ...
This is a sample experiment designed to understand the purpose of control groups in a cDNA vector study. Designing appropriate control groups in experiments is important for understanding the true meaning of experimental group results. In this example, E.coli are used for the expression of a cDNA segment via the use of a viral vector. Four control groups are described, and the purpose of those control groups is discussed. The results of the experiment are given and an explanation for why those results were found is provided.