The transformation of bacterial cells is a useful experiment to help develop an understanding of transformation by plasmid DNA. This experiment involved four different scenarios of bacterial cells on agar plates. The scenarios were as follows, one plate with a plasmid, one without and one plate with ampicillin and plasmid, and one with ampicillin and without plasmid. The transformation effectiveness was then determined by analyzing the amount of resulted colonies created.
Transformation is the introduction of foreign DNA, in this experiment by a plasmid, into a bacterial cell. Transformation is vital in molecular biology and the observable results of this experiment are evidence of the effectiveness of transformation. Since the bacteria, E. coli are not naturally competent cells the E. coli bacteria cells and pVIB plasmid must be mixed with calcium chloride. This solution then must undergo a heat shock in the water at 42 degrees to create a draft and sweep the plasmids into the pores of the bacterial membrane.
After preparing two bacterial cells, one with plasmid and one without, another agar plate is prepared where ampicillin is introduced. Ampicillin is an antibiotic that is known to treat bacterial infections, in this experiment ampicillin’s role is to kill all bacteria that did not undergo a transformation. The purpose of this lab was to develop an understanding and appreciate the results of transformation as well as focusing on the effects of ampicillin on the pVIB plasmid.
- 2 sterile 15-mL tubes
- 4 sterile transfer pipets
- 3-4 glass beads
- 4 sterile transfer loops
- 2 LB agar plates
- 2 LB agar plates with ampicillin
- pVIB plasmid
- 500 μL of CaCl2
- LB Broth
- beaker of 42 degrees water
- beaker full of ice
- waste beaker
Observations and Results
Table 1.0: Results
Reason: bacteria will line the Luria Broth
Observed Result: lawn
Reason: same as plate 1
Observed Result: Lawn
Prediction: no growth
Reason: ampicilin kills bacteria
Observed Result: no growth
Prediction: isolated colonies
Reason: selecting for the ones that took up plasmids
Observed Results: 2 colonies
LB+plasmid (positive control) – lawn
LB-plasmid (positive control) – lawn
LB/Amp+plasmid (experimental) – 2 colonies
LB/Amp-plasmid (negative control) – no growth
Analysis and Conclusions
My observed results (listed above) were very accurate with my prediction. As predicted the LB/Amp-plasmid showed no growth due to the fact that the ampicillin killed the bacterial cells and there was no plasmid present for resistance. As expected my group and I observed two isolated colonies within the LB/Amp+plasmid and finally, the LB-plasmid and LB+plasmid displayed lawn colony growths.
Both the LB+plasmid and the LB-plasmid plates exhibited lawn colonies. These results tell us that the experiment was performed efficiently because the bacteria was able to line the Luria broth. Also, because there was no ampicillin present there was a major growth of bacteria because the bacteria was replicating without being destroyed by other factors.
The LB/Amp-plasmid and the LB-plasmid had results that contrasted with one another. The LB/Amp-plasmid showed no growth due to the fact that the ampicillin killed the bacteria. However, the LB-plasmid displayed a lawn growth because there was no presence of ampicillin. These results clearly show what a major impact the presence of ampicillin in bacterial cells have.
The LB/Amp+plasmid and the LB/Amp-plasmid showed contrasting results as well. While the LB/Amp-plasmid showed no signs of growth, the LB/Amp+plasmid displayed isolated colonies. This is because the bacteria that took up the plasmid were able to withstand ampicillin. This data from the experiment shows us that the process of transformation by plasmids is very effective. Both the LB/Amp+plasmid and the LB+plasmid results showed signs of growth however they were too different.
The LB/Amp+plasmid showed isolated colonies because only the bacteria that took up the plasmid glowed. However, the LB+plasmid displayed a lawn growth because the bacteria were able to line the Luria broth efficiently. These outcomes demonstrate that ampicillin will only have an impact if the experiment is completed very accurately.
In this experiment, we are selecting the fluorescence that occurs when a bacteria has taken up a plasmid. To identify if the plasmid worked effectively my group and I were taken to a dark closet where we observed the plate to see if it would fluoresce. The fluorescence was an indication that the bacteria survived and have taken the plasmid and therefore because we had two isolated colonies, we performed the experiment efficiently. The phenotype of the colonies is evidence that they were transformed efficiently.
The fluorescence appearance shows that the bacteria took up the plasmid. The plate that I would first inspect to conclude that the transformation was done successfully is the LB/Amp+plasmid. This is because for a transformation to be done effectively the foreign DNA must be inserted into the bacterial cell by a plasmid. Obviously one would not look to the plates with –plasmid so the only two left are the LB+plasmid and the LB/Amp+plasmid.
The LB/Amp+plasmid would be the better choice because the ampicillin will kill all the bacteria not taken up by a plasmid. Therefore the plate will only fluoresce if the transformation was a success.
Total mass = volume x concentration
0.05 μg plasmid = 10 μL plasmid x 0.0005 μg/μL
250 μL LB + 250 μL CaCl2 + 10 μL plasmid DNA
= 510 μL cell suspension
Volume suspension spread/total volume suspension = fraction spread
100 μL cell suspension / 510 μL total suspension
= 0.196 (fraction spread)
Total mass plasmid x fraction spread = mass plasmid DNA spread
0.05 μg plasmid x 0.196
= 0.0098 μg
Colonies observed/plasmid mass spread = transformation efficiency
2 colonies / 0.0098 μg plasmid
= 2.041x 102 colonies/μg plasmid
I think that there are various factors that influence transformation efficiency. What I believe is the most important factor is the completion of the heat shock. If the heat shock were to be done incorrectly then the plasmids would be unable to enter the bacterial cells and therefore transformation would be unable to occur. For the heat shock to be performed properly the temperature must not be too high or too low as this would affect the rate at which the plasmids enter the pores of the bacteria.
Another factor that is important in this experiment is the amount of E. coli and pVIB plasmids that are transferred to the various plates. If the amount transferred is too much or too little the transformation may not even occur. The final factor that plays a major role in the efficiency of this experiment is the temperature at which the plates are incubated at. The plates must incubate at 30 degrees and if the temperature is even a few degrees off the bacteria will be unable to fluoresce. If this were to occur our data would be wrong because we would be unable to count the total number of colonies.
Thank you so much for this!!! this helped me with my bio lab so much. Love you guys.
This is great, and nicely detailed!
Dude. Thanks for this. I totally understand this now.