Isolations Cloning And Translation Of Plasmid Dna Biology Essay


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The objective of this experiment was to clone a kanamycin gene into the MCS of a pUC18 plasmid, and then to transform cells with the plasmids. Purified pUC18 and pKan plasmid samples were obtained. A 0.7 % agarose gel was prepared, and the wells loaded with the plasmid samples. Restriction endonucleases were used to cut a kanamycin resistance gene from a pKan plasmid. DNA ligases were used to ligate the kanamycin resistance gene on to the multiple cloning site of the pUC18 plasmid. Escherichia coli (strain DH5α) were then transformed with plasmids. The presence of the kanamycin resistance gene in the pUC18 was determined using the indirect (pUC18 selection) and direct selection methods. The results from the gel image were inadequate. Zero colony counts were recorded on the kanamycin plates for the indirect selection method. Zero colony counts were recorded on the kanamycin/ carbenicillin plate for the direct selection method. In Conclusion it can be said that although the kanamycin gene should have been inserted into the pUC18 plasmid, the results from both selection methods indicate that it wasn't.


DNA cloning is a process in which a certain piece of DNA is replicated several times [1]. This process in essence involves isolating the gene or DNA fragment of interest, and transferring it to another molecular of DNA [1]. In order for the cloning process to begin, the DNA of interest has to be cut at precise locations [1]. Specific endonucleases are used for this process. After which a small molecule of DNA is chosen, that has restriction sites that are complementary to the DNA of interest and is capable of self-replication [1]. These small molecules of DNA are called cloning vectors (phages, plasmids, Yeast Artificial Chromosomes, or Bacterial Artificial Chromosomes can be used) [1]. The two pieces of DNAs (the vector and DNA of interest) can be joined together by using a DNA ligase [1]. The newly formed composite DNA molecule is called a recombinant DNA [1]. The recombinant DNA can then be introduced into a host cell by a process of transformation [1]. Once transformed multiple copies of the host cell can be produced, and in doing so multiple copies of the DNA are also produced [1].

Bacterial DNA can carry genes for antibiotic resistance [2]. The antibiotic resistance gene can either be on the chromosomes or on other external chromosomal pieces of DNA (e.g. plasmids) [2]. The pUC18 is a cloning vector plasmid that contains an ampicillin resistance gene [2]. On the other hand the pKan plasmid contains a kanamycin resistance gene [2]. The pUC18 plasmids are extremely useful for transformation with an Escherichia coli host cell [2]. The pUC18 plasmid consist of an origin of DNA replication, pBR322 derived ampicillin resistance gene, and a lacZ gene of E.coli [2]. The lacZ gene is part of something called the lac operon [1]. The lac operon in essence consists of the lacZ, lacY, and lacA genes [1]. The combination of the three genes allows the cell to utilize lactose [1]. When sufficient quantity of lactose is available, the cell is able to utilize the lactose by producing the enzyme beta-galactosidase [2]. pUC18's lacZ gene contains a collection of different restriction enzyme recognitions sites [2]. This site within the lacZ gene is called a Multiple Cloning Site (MCS). The MCS of the pUC18 plasmids can be recognized by a number of different enzymes; hence cuts can be made at various different places [2]. In gene cloning experiments, X-gal (5-bromo-4-chloro-3-indolyl,-D-galactoside) is used to indicate the presence of the lacZ gene, and hence indicates whether or not a cell is producing the enzyme beta-galactosidase [2][3]. This indication is given by a blue coloration of the colonies growing on a medium containing X-gal [2]. Beta-galactosidase cleaves X-gal into D-galactoside and 5-bromo-4-chloro-3-indole [3]. The actual presence of 5-bromo-4-chloro-3-indole is what causes the colonies to true blue [3].

The pKan plasmid contains the kanamycin resistance gene. In this experiment the kanamycin resistance of the pKan plasmid will be cloned into the MCS of the pUC18 plasmid [2]. This new recombinant DNA will then be transformed into an E.coli strain DH5α host cell [2]. A brief overview of the isolation, cloning and transformation processes are given above [2]. This process in the end will yield an E. coli strain that is resistance to both ampicillin and kanamycin [2]. As mentioned earlier, the multiple cloning sites (MCS) of the pUC18 plasmid is located with its lacZ gene [2]. This means that when the kanamycin resistance gene is inserted into the multiple cloning sites, the lacZ gene is disturbed [2]. This alters the production of beta-galactosidase [2]. Hence the E.coli cells are not able to utilize X-gal on a growth media, producing white colonies instead of blue [2]. The presence of white colonies can be used as an indication for insertion of the kanamycin gene in pUC18 plasmid [2]. A kanamycin/ampicillin selective media can also be used to make sure that the pUC18 plasmid has the kanamycin gene inserted into it [2].

In summary the main objectives of this experiment is to clone a kanamycin gene into the MCS of a pUC18 plasmid, and then to transforms a cell with the plasmids. The hypothesis is that a kanamycin resistance gene will be inserted onto the MCS of the pUC18 plasmid, and as a result the cells will be resistant to both antibiotics.

Materials and Methods:

The following materials and methods are taken from: Hausner, M., & Jong, M. (2010). Experiments in Biotechnology (BLG888 ed.). Toronto: Ryerson University. Pg 7-19


Bacterial plasmids, restriction enzymes, solutions and media used: Overnight cultures of DH5α/ pUC18 and MM294/pKan (5x10mL) were used. DNA solution kit that was used consist of solution 1 (glucose/Tris/EDTA to which lysozymes were added), solution 2 (SDS/NaOH), and solution 3 (KOAc). Enzymes RNAase (5mg/ml) and DNA ligase were used. Isopropanol and ethanol were used. TE buffer used contained 10Mm TRIS and 0.1 mM EDTA. Tris borate buffer that was used contained (TBE)(1X)10.8g Tris, 5.5g Boric acid, 10 mM EDTA, and up to 1000 ml distilled water. DNA loading dye and Ethidium bromide solution were used. The plasmids pUC18 and pKan were used. The restriction enzymes that were used were BamHI (Bacillus amyloliquefaciens H.) and HinDIII (isolated from Haemophilus influenza). 5M ammonium acetate was used. Phenol:chloroform:isoamyl was used. 50mM EDTA was used. 5 x ligation and restriction buffers were used. TE buffer that was used contained 10Mm Tris, 0.1 Mm EDTA. Cell culture of E. coli strain DH5α was used. 50 ml of LB broth and 3 sterile saline tubes. 2 LB plates, 8 LB + carbenicillin (carb), and 3 LB + carbenicillin (carb) + kanamycin (kan) plates were used. X-gal solution was used. 1 plate of LB+ kanamycin (kan).


Preparation of the plasmid DNA: pUC18 and pKan plasmid were prepared over a period of three days (three weeks). Two centrifuge tubes with the culture sample were centrifuged for 10 minutes and supernatant discarded. 100µl of solution 1 was added followed by 10µl of RNase. After 20 minutes solution 2 was added. Five minutes later ice cold solution 3 was added, which was centrifuged 10 minutes later for 10 minutes. 400µl of the supernatant was extracted to a clean tube, to which 400µl of isopropanol was then added and was left for 30 minutes at -20oC. The DNA sample was then centrifuged and the pellet speed vac. The dry pellet was re-suspended in 20µl of TE buffer. A gel was prepared with accordance to steps in the lab manual. The DNA samples were then loaded on to the wells and the electrophoresis apparatus ran. The gel images were taken to see presence of the pUC18 and pKan plasmids.

Endonuclease restriction digestion of the plasmids and ligation of the kanamycin fragment to pUC18: Two centrifuge tubes were prepared from 10µl of pUC18 and 10µl of pKan plasmids. To each tube restriction buffers, restriction enzymes and sterile water were added (refer to the lab manual for details). The prepared tubes were centrifuged and left in a water bath. 5µl of EDTA was added to each tube. 100µl of TE buffer and Phenol:chloroform:isoamyl were added. The tubes were then pulse centrifuged and top layer remove and transferred to new tubes (A1 and B1). 100µl of Phenol:chloroform:isoamyl was added, top layer removed and transferred to new tubes again (A2 and B2). Ammonium acetate and ethanol were added to tubes A2 and B2. The tubes were centrifuged, supernatant discarded, pellet speed vacuumed, and finally re-suspended in TE buffer. Tube C and D were prepared with accordance to the lab manual. The new tubes were then centrifuged and incubated.

Transformation of an ampicillin sensitive E.coli Strain: The first five steps to prepare the cell culture of DH5α for transformation were done by the lab staff. Details on the steps can be found in the lab manual. Four centrifuge tubes were prepared. Tube 1 contained uncut DNA plasmids, tube 2 contained DNA sample from tube C, tube 3 contained DNA sample from tube D, and tube 4 contained sterile water. The pre-prepared cells were then added to the tubes and heat shocked. LB broth was added to each tube and incubated for 20 minutes.

X gal was spread evenly on the 8 LB+ carb plates. 100µl from tubes 1, 2, and 4 were spread on 3 of the LB+carb+X-gal plates. 100µl from tube three was then plated on the remaining five LB+carb+X-gal plates. Tube 3 was also plated on to 3 LB+carb+kan plate. A dilution series (using 0.1µl from the previous) was prepared from tube 3 using 3 sterile saline tubes. 10 µl from dilution 2 and 100 µl from dilution 3 were spread plated onto 2 LB plates. Colonies from each plate were counted. Blue and white colonies from tube 3 plates were then streaked on to a LB+Kan plate. Results from the LB+Kan plates were then recorded.

Additional details can be found in the lab manual: Hausner, M., & Jong, M. (2010). Experiments in Biotechnology (BLG888 ed.). Toronto: Ryerson University


Figure 1: 0.7 % agarose gel digest showing the presence of the pUC18 and pKan plasmids. Lane 3 and 2 were used by Abbas and Jamie.

The figure above shows the 0.7% agarose gel image showing the presence of pUC18 and pKan plasmids. If banes appeared in the respective lanes, the plasmid samples would be used in the next part of the experiment. The image above shows bands appearing for lane 3 (pKan), but none for lane 2 (pUC18). This indicates the presence of the pKan plasmid but absences of the pUC18 plasmid. Hence due to inadequate results, additional plasmid sample were prepared by the lab staff.

In total results from all 14 plates were recorded.

Indirect Method:

Table 1: Results for colony counts for the indirect (pUC18) selection method on LB+ carb+ X-gal plates

Plate 1

Plate 2

Plate 3

Plate 4

Plate 5

Tube 1



Tube 2


Tube 4


Tube 3

40 Blue/ 5 White

55 Blue/ 15 White

79 Blue/ 22 White

65 Blue/ 3 White

54 Blue/ 12 White

The results for tube 1, 2, 3, and 4 plated on the 8 LB+ carb+ X-gal plates are shown above. Tube 1 contained an uncut plasmid which explained the high number of colonies for plate 1. Tube 2 contained a cut pUC18 plasmid, which can be explained by only 5 colonies. Tube 4 contained only sterile water; hence zero colonies appeared on the plates. Tube 3 was plated on 5 plates, showing an average of 59 blue colonies and 11 white colonies.

Direct Method:

No colonies were obtained from the three plates of LB + carb + kan plates.

Competent Cell and Percentage Transformation Calculation:

The dilution series was prepared from tube 3, as indicated in the materials and methods section. Dilution 2 had a 100 colonies and dilution 3 had 30 colonies. The CFU (colony forming unit) calculations and values are shown below.

CFU = (# of colonies) x (dilution factor) / (volume plated)

CFU for dilution 2 = 100 x 104/ 0.1 = 10000000 cells/ml

CFU for dilution 3 can't be calculated because it doesn't fall between the 30-300 colony limit.

Table 2: Percentage transformation of colonies using competent cells (CFU)

Plate 1

Plate 2

Plate 3

Plate 4

Plate 5


Percentage Transformation

For total colonies (%)













In order to calculate percentage transformation, calculations from CFU are need. Percentage transformation can be calculated using the total colonies (i.e.. plate 1: 40 blue+5 white =45 total).

Percentage Transformation = (Transformed cells per ml /competent cells (CFU) )x 100%

So for example for the average of 70 total colonies; =0.007%


As mentioned in the result section the agarose gel image was inadequate. Lane 2 and 3 in figure 1 represent the pUC18 and pKan plasmids respectively. Clear bands were seen for the pKan plasmid however this is not the case for the pUC18 plasmid. In order for the plasmids to show up, they had to be extracted from their respective E. coli strains(pUC18 (DH5α) and pKan (MM294). The presence of bands on the pKan lane proves that there is actual extraction from the cells. The presence of multiple bands could indicate the presence of multiple size plasmids of pKan. The fact that no bands were seen for pUC18 could be as a consequence of inadequate extraction from the E. coli cells (DH5α). Experimental procedural error could have resulted in this. Both strains of microbes would have been genetically engineered to only contain the plasmid of interest; hence the risk of contamination is reduced. The selection methods for the experiments were divided into indirect (pUC18 selection) and direct selection methods.

As mentioned in the materials and method section, cells from tube 1 were streaked on to a plate. The cells were transformed with undigested pUC18 plasmids. The colonies were too many to count and were all blue. The high number of colonies could simply occur because of the stable natural of the undigested pUC18 plasmid. The undigested pUC18 plasmids contain an uninterrupted lacZ gene, capable of producing beta-galactosidase. Beta-galactosidase is hence able to utilize X-gal on the plates and produce the large number of blue colonies. Since the cells were carbenicillin resistance (due to the pUC18 plasmids), they were able to grow on the plates. Cells were transformed with digested pUC18 plasmids from tube 2. Cells from tube 2 formed too few colonies (only 5) when compared to tube 1 (TMTC). This is due to the unstable nature of the digested pUC18 plasmids. These plasmids were digested with HinDIII and BamHI, and it possible that not all of them had an opportunity to re-ligate properly. The restriction enzymes could have cut up the lacZ gene or the carbenicillin (ampicillin) gene making it difficult for the plasmid to come back to its original conformation and survive on the X-gal+carb plate. A large majority of the pUC18 could have been cut in to smaller fragments rendering then inactive. Tube 3 initially contained the digested pUC18 and pKan plasmids. Cells were then transformed with the content of this tube. Since the transformation process is not perfect, there is no way to know what plasmid the cell took up. Hence it can be assumed that cells were transformed with either only the pUC18 plasmids, the pUC18 plasmids with the kanamycin gene, pKan and Puc18 or in some case only the pKan plasmid. Five plates were spread plated with these cells and presence of blue and white colonies were noted. As the results indicate a mixture of both blue and white colonies were obtained with an average of about 59 blue colonies and 11 white colonies. Blue colonies would hypothetically contain cells (plasmids) with an intact lacZ (producing beta-galactosidase) gene justifying the blue color. The white colonies would have there lacZ gene disturbed (not producing beta-galactosidase), because another piece of DNA would have been inserted into the MCS. However the production of white colonies doesn't dictate the insertion of the kanamycin gene into the pUC18 plasmids. It is highly possible that another gene or DNA fragments from the pKan plasmid got inserted in the pUC18 plasmids. Confirmation of this was performed by streaking white colonies onto a kanamycin plate. The fact that no colonies grew, indicated that the kanamycin gene was in fact not inserted. This proves that the results are false positive because white colonies appeared on the X gal plates, but didn't on the kanamycin plates. This means that the white colonies weren't transformed with what we wanted. Finally it is noted that when tube 4 was streaked on to a plate, no growth occurred. This seems logical as the cells in this tube were only transformed with sterile water, which means no plasmids were present. The cells would not have contained plasmids with the carbenicillin resistance gene, and hence did not survive on the carbenicillin plates.

The direct method results were recorded from the LB + carb + kan plates. No growth was observed in any of the plate, which proved to be highly contradictory to our hypothesis. Presence of white colonies on the indirect method plates but none on the direct method plates was suprising. White colonies were assumed to have pUC18 plasmids with both kanamycin and carbenicillin resistance genes. Hence its inability to grow on the carb + kan plates was surprising because white colonies grew on the X-gal plates. However as mentioned earlier it could be possible that another fragment of DNA was inserted into MCS besides the kanamycin gene. The fact that white colonies also didn't appear when they were streaked on to a kanamycin plate, ties in with these results.

Both direct and indirect methods have their own advantages and disadvantages. Indirect method involves multiple steps and hence in many cases can be time consuming. More plates are involved in the indirect methods, making it difficult to keep track sometimes, also adding to cost. However the indirect method helps to indentify the false positive/false negative results. The indirect selection method helps to make a comparison between the cut and uncut pUC18 plasmids. Comparison of the colonies shows the effect of restriction enzymes of the activity of the pUC18 plasmids. Moreover the indirect method is much more selective. This is because it first shows which colonies have an insertion in the multiple cloning site through the blue/white screening method. Then the plating of these white colonies on to a kanamycin plate helps to confirm that it was a kanamycin resistance gene that was actually inserted (on the MCS). The direct method is very concise involving only one plate, which save both time and money. This selection method has no chance of giving false negative/false positive results. The direct selection method selects for cells that have been transformed with pUC18 plasmids, and have a kanamycin resistance gene in their MCS. Since the pUC18 plasmid already has an ampicillin resistance gene (carbenicillin in this case), the insertion of kanamycin resistance gene allows it to survive on a LB+carb+kan plate. A problem comes when the plasmids don't have the necessary gene inserted in their MCS. So in this case for example it could be possible that the plasmid doesn't contain the kanamycin gene so the kanamycin antibody kills it, even though the carbenicillin resistance gene is there. Another technicality comes when a cell transformed contains both pUC18 and pKan at the same time. Because this selection method only selects for cells that have both carbenicillin and kanamycin resistance, it is difficult to tell whether the cell selected has both plasmids (pUC18 and pKan) or only a pUC18 (with the kanamycin gene). Therefore although more time consuming the indirect method is more useful.

Some of the experimental errors that occurred could have been due to improper spreading techniques. The process of cell transformation that was used was through heat shock. It could be possible to use other cell transformation technique such as electroporation.

In Conclusion it can be said that although the kanamycin gene should have been inserted into the pUC18 plasmid, the results indicate that it wasn't.

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