IntroductionThis tube. 3. Insert a plasmid DNA into the

IntroductionThis practical is to show a bacterial transformation by plasmid DNA. By having this plasmid DNA,  it will actually change the bacterial cells genetic traits. The plasmid DNA (pGLO) contains the gene of interest which is the Green fluorescent Protein (GFP). It is also resistant to the antibiotic ampicillin. The gene of interest will be activated only when arabinose is present in the environment. Bacteria with the gene of interest will cause them to glow a neon green colour under ultraviolet light. We will be learning to insert the plasmid DNA into bacterial cells and observe how the transformed cells will look like.We will be viewing them under ultraviolet light. The transformed solution that we will be using will be calcium chloride solution. Ca2+ cations will neutralize the repulsive negative charges of DNA phosphate backbone and the phospholipids of the cell membrane. This will then allow the entry of DNA into the bacteria cells.Procedures Collect all equipments required for the experiment and label 2 micro test tubes, +pGLO and -pGLO with our group name (JR). (Yellow tube – LB broth, Orange tube – CaCL2 solution)2. A single colony of bacteria are placed in the abovementioned microtubes using a sterile loop. Make sure to tilt the micro test tube when mixing it with the bacteria to ensure that all bacteria has been transferred over into the tube. 3. Insert a plasmid DNA into the +pGLO tube using a loop. Place the 2 micro tubes into the beakers filled with ice. Ensure that the tubes have been fully immersed in the beaker. 4. Using a permanent marker, label 4 agar plates with: Label one LB/amp plate: +pGLOLabel the LB/amp/ara plate: +pGLOLabel the other LB/amp plate: -pGLOLabel the LB plate: -pGLO5. After the 10 minutes have passed, immediately place the tubes for heat shock for 50 seconds. Ensure that the tubes are ¾ immersed in the hot water.  6. As soon as the 50 seconds are over, place the tubes back into the beaker with ice for 2 minutes to achieve best results for the experiment. 7. Use a pipette to insert 250 ?l of LB nutrient broth into both tubes and leave it at room temperature for 10 minutes. 8. Pipette 100? of the +pGLO micro test tube into the LB/amp plate (+pGLO) and LB/amp/ara plate (+pGLO) agar plate respectively and 100? of the -pGLO micro test tube into LB/amp plate (-pGLO) andLabel the LB plate (-pGLO) agar plates respectively. 9. Use a loop to spread the suspensions onto the surface of the agar plate lightly but evenly. Repeat this for all 4 agar plates (LB/amp plate (+pGLO), LB/amp/ara plate (+pGLO), LB/amp plate (-pGLO), LB plate (-pGLO))10. Once done, tape the plates together and label our group name (JR). Leave it for incubation at 37°C for a day. ResultsTable 1.1 : Genetic Transformation Plate results (after 1 day of incubation)Observation (under UV light)Observation LB/amp plate (+pGLO): White colonies of bacteriaLB/amp/ara plate (+pGLO): Fluorescent green colonies under UV lightLB/amp plate (-pGLO): No growth of cellsLabel the LB plate (-pGLO): Bacterial lawnObservations after 4 days  DiscussionBased on our observation, the bacteria plates were able to produce the expected results. The plate that had LB/amp/ara gave the bacteria colonies a neon green light under ultraviolet light due to the simple sugar arabinose in the plasmids.The colonies on the plate will form due to ampicillin resistance gene of the plasmids too. +pGLO plasmids DNA which are resistance to ampicillin able to survive and grow on the transformation plates such as LB/amp plate. This also mean that -pGLO plasmids which are sensitive to ampicillin will fail to grow on the LB/amp plate. There were mass of bacteria cells found on the LB plate (-PGLO). This was due to the LB broth which is a rich nutrient that is used for the growth of bacteria. Hence, due to this rich nutrient, mass of cell will be grown. There are possible sources of error in this experiment. The process of pre-chilling to water bath and back to chilling of the tubes might not be fast enough to get the best transformation results. The transformed cell suspension might not be evenly spread around the surface of the agar plate. Questions1. Describe how you could use two LB nutrient agar plates, some E. coli, and some ampicillin to determine how E. coli cells are affected by ampicillin. On which of the plates would you expect to find bacteria most like the original untransformed E. coli colonies you initially observed? Explain your prediction.Answer (1st part) : One of the LB nutrient plate can have some E. coli and ampicillin. Another plate can have some E. coli and no ampicillin. After incubation, there will be differences shown on both of the plates. The growth of E. coli should be compared. There should be lesser bacteria colonies on that plate if ampicillin negatively affects the growth of E. coli. There should be equal amount of colonies on both plates if E.coli did not get affected by ampicillin. Answer (2nd part): It can be found on the LB(-pGLO) plate. Everything is allowed to grow on this plate and there is no plasmid added. Furthermore, the E.coli will be allowed to grow as there is no ampicillin too. Hence, the bacteria on LB(-pGLO) plate will be same as the untransformed E.coli colonies.2. If there are any genetically transformed bacterial cells, on which plate(s) would they most likely be located? Explain your prediction.Genetically transformed bacterial cells would most likely be located on the +pGLO LB/amp and the +pGLO LB/amp/ara plates. This is because both of these plates contain the plasmid which allows the transformation.3. Which plates should be compared to determine if any genetic transformation has occurred?  Why?We should compare the +pGLO LB/amp and the -pGLO LB/amp plates. The plasmid which provides resistance to the antibiotic in the plate was given to the +pGLO plate but not the -pGLO plate so the bacteria in the +pGLO plate will survive. This proves that a genetic transformation has occurred. 4. What is meant by control plate? What purpose does a control serve?The control plate is a guide to show the contrasting results obtained in the experiment. In this experiment, the plates used were:LB/amp plate (+pGLO)LB/amp/ara plate (+pGLO)LB/amp plate (-pGLO)Label the LB plate (-pGLO)Plates -pGLO were the control plates as green fluorescence and ampicillin resistance was not present in it. 5. Describe the evidence that indicates whether your attempt at performing a genetic transformation was successful or not successful.Our attempt at performing a genetic transformation was successful. This can be seen from the colonies on the plate (LB/amp/ara) that were formed due to ampicillin resistance gene of the plasmids under the the presence of UV light. As expected, the results were: LB/amp plate (+pGLO) which had white colonies of bacteria, LB/amp/ara plate (+pGLO) which has a presence of fluorescent green colonies under UV light, LB/amp plate (-pGLO) that had no growth of cells and the LB plate (-pGLO) that consists of bacterial lawn. Thus, we had achieved the aim of our experiment.  ConclusionOur experiment was a success as we were able to cultivate our bacteria cells to glow under the fluorescent light as expected. We were able to gain new insight about bacterial transformation by plasmid DNA. By comparing the plates with the control, we were able to learn and prove that the pGLO plasmid bacteria caused the bacteria to become resistant to the ampicillin. We can also prove that the arabinose must be present in order for the bacteria to fluoresce green.  References