Lab Report E. Colo Transformation
Title: E. Coli Transformation with a Plasmid DNA Containing the GFP Gene
Introduction:
Bacterial transformation is the process of bacteria taking in and expressing exogenous DNA. This has led to many other discoveries. In order for bacterial transformation to occur the bacteria must be in a certain physical state to be able to take in DNA. This is called competency and it allows the cell membrane to be permeable so DNA can pass through. Currently researchers are studying the transformation of E. Coli, which was used in this experiment. It is not normally in a state of competency so the researchers treat it with chloride salts and heat shocking to induce competency. (EDVOTEK, 5-6)
The GFP gene is the green fluorescent …show more content…
Transformation Efficiency Calculations:
•Equation:
(Number of transformants/ µg of DNA) X (final volume at recovery(ml)/volume plated(ml)
= number of transformants per µg
•Example (+DNA)/(+AMP) (Figure 1)
(1000 transformants/ 0.05 µg) X (0.50 ml/0.25 ml)
20,000 X 2= 40,000 (4 X 10^4) transformants per µg
•(+DNA)/(+AMP) (Figure 3)
(1 transformant/ 0.05 µg) X (0.50 ml/0.25 ml)
20 X 2 = 40 (4 X 10¹) transformants per µg
Discussion:
In this experiment bacterial transformation was tested and only one of the four plates was expected to express signs of bacterial transformation. Plates (-DNA)/(-AMP) and (-DNA)/(+AMP) were the controls of this experiment. This is because there was no DNA added to them so they would show the difference between bacterial transformation and no bacterial transformation. Plate (-DNA)/(-AMP) had regular bacterial growth, while plate (-DNA)/(+AMP) had no growth. Plates (+DNA)/(+AMP) and (+DNA)/(-AMP) were the plates being tested because they were the ones that were given the DNA. This means that they could potentially transform the bacteria depending on the whether the bacteria is competent or not.(EDVOTEK, 5) Plate (+DNA)/(+AMP) showed no signs of bacterial transformation because the GFP gene was not being expressed and few bacterial colonies grew. Plate (+DNA)/(-AMP) showed normal bacterial growth with no signs of bacterial …show more content…
The example (Figure 1) had a transformation efficiency of 40,000 transformants per µg. This is a relatively high transformation efficiency and is what is typically expected as a result for this particular experiment. However, the (+DNA)/(+AMP) plate from this experiment (Figure 3) as much lower than what was considered accurate. The transformation efficiency of this plate was 40 transformants per µg. Due to the huge gap between the expected transformation efficiency and the one that was shown from this experiment it can be assumed that many errors occurred throughout the experiment inhibiting the bacterial transformation
Introduction:
Bacterial transformation is the process of bacteria taking in and expressing exogenous DNA. This has led to many other discoveries. In order for bacterial transformation to occur the bacteria must be in a certain physical state to be able to take in DNA. This is called competency and it allows the cell membrane to be permeable so DNA can pass through. Currently researchers are studying the transformation of E. Coli, which was used in this experiment. It is not normally in a state of competency so the researchers treat it with chloride salts and heat shocking to induce competency. (EDVOTEK, 5-6)
The GFP gene is the green fluorescent …show more content…
Transformation Efficiency Calculations:
•Equation:
(Number of transformants/ µg of DNA) X (final volume at recovery(ml)/volume plated(ml)
= number of transformants per µg
•Example (+DNA)/(+AMP) (Figure 1)
(1000 transformants/ 0.05 µg) X (0.50 ml/0.25 ml)
20,000 X 2= 40,000 (4 X 10^4) transformants per µg
•(+DNA)/(+AMP) (Figure 3)
(1 transformant/ 0.05 µg) X (0.50 ml/0.25 ml)
20 X 2 = 40 (4 X 10¹) transformants per µg
Discussion:
In this experiment bacterial transformation was tested and only one of the four plates was expected to express signs of bacterial transformation. Plates (-DNA)/(-AMP) and (-DNA)/(+AMP) were the controls of this experiment. This is because there was no DNA added to them so they would show the difference between bacterial transformation and no bacterial transformation. Plate (-DNA)/(-AMP) had regular bacterial growth, while plate (-DNA)/(+AMP) had no growth. Plates (+DNA)/(+AMP) and (+DNA)/(-AMP) were the plates being tested because they were the ones that were given the DNA. This means that they could potentially transform the bacteria depending on the whether the bacteria is competent or not.(EDVOTEK, 5) Plate (+DNA)/(+AMP) showed no signs of bacterial transformation because the GFP gene was not being expressed and few bacterial colonies grew. Plate (+DNA)/(-AMP) showed normal bacterial growth with no signs of bacterial …show more content…
The example (Figure 1) had a transformation efficiency of 40,000 transformants per µg. This is a relatively high transformation efficiency and is what is typically expected as a result for this particular experiment. However, the (+DNA)/(+AMP) plate from this experiment (Figure 3) as much lower than what was considered accurate. The transformation efficiency of this plate was 40 transformants per µg. Due to the huge gap between the expected transformation efficiency and the one that was shown from this experiment it can be assumed that many errors occurred throughout the experiment inhibiting the bacterial transformation