Recombinant DNA Lab
Questions 1 & 2
LB-PlasmidPrediction: No growth
Reason: No heat, no DNA Observed Result: There appears to be significant growth. |
LB+PlasmidPrediction: Growth
Reason: No heat, DNA Observed Result: There appears to be significant growth. |
LB/Amp-PlasmidPrediction: Growth
Reason: Heat, no DNA Observed Result: There appears to be no growth. |
LB/Amp+PlasmidPrediction: Growth
Reason: Heat and DNA Observed Result: There appears to be little growth. |
Questions 3-10
3. The LB-/Amp bacteria did not appear to grow because the bacteria may have been destroyed by ampicillin.
4.
- The LB+Plasmid (Positive Control) was "lawn."
- The LB-Plasmid (Positive Control) was "lawn."
- The LB+/Amp Plasmid (Experimental) had 2 colonies.
- The LB-Plasmid had no colonies.
5.
a. LB+Plasmid and LB-Plasmid:
The bacteria transformed in both plates to the point where they were lawns.
b. LB/Amp-Plasmid and LB-Plasmid:
The LB/Amp-plasmid showed no growth while the LB-Plasmid had lawn transformation.
c. LB/Amp+Plasmid and LB/Amp-Plasmid:
The LB/Amp+Plasmid had two bacterial colonies and the LB/Amp-Plasmid showed no
growth.
d. LB/Amp+ Plasmid and LB+Plasmid:
The ampicillin-treated positive plasmid had two visible colonies and the other transformed.
6. The transformed colonies allow for the identification of bacteria that has taken up the plasmid, so any transformation whether lawn or subtle will show which bacteria successfully had taken up the plasmid.
7. There was no apparent phenotype due to the lack of color present in the transformed colonies. The pGreen was not present.
8. The first plate to inspect to conclude that the transformation occurred successfully would be the LB/Amp+Plasmid because the marker that shows the selection will only grow in a plasmid treated with ampicillin. Also, the LB/Amp+Plasmid is the easiest to observe due to the fact that it only has two colonies.
9.
a. Total mass: 10 x 0.005 = 0.05 ug
b. Cell suspension volume: 250 calcium cholride + 250 LB + 10 green capped liquid =
0.510 uL
c. Fraction spread: 10 (Suspension Spread)/510 (Total Suspension) = 0.0196
d. Mass of Plasmid Spread: 0.05 (Total Mass) x 10/510 (Fraction
Spread) = 0.00098
e. Transformation efficiency: 2 (colonies)/ 0.00098 (Mass Plasmid Spread) = 2040.81633 or
204081633000 x 10^-8
10. The transformation efficiency could have been influenced by the colony size that was added to the liquid solution. A greater amount of cell receptors comes with a greater colony size, so there would be better efficiency. A lesser colony would have lesser cell receptors, so there would not be as great of efficiency. Another factor that may have influenced the transformation efficiency is the amount of time the plasmids were in the hot or cold water. If the plasmids were in the hot water longer, then the growth of the colonies would be more efficient. If the plasmids were in the cold water longer, the growth could be stunned and the colonies would not be as efficient in transformation. In the experiment, the plasmids that were in the hot water were heated longer because the bath was not at the desired temperature, thus affecting the heat shock of the plasmids. The time the plasmids were outside of the water was also erroneous because the timing was not accurate.
4.
- The LB+Plasmid (Positive Control) was "lawn."
- The LB-Plasmid (Positive Control) was "lawn."
- The LB+/Amp Plasmid (Experimental) had 2 colonies.
- The LB-Plasmid had no colonies.
5.
a. LB+Plasmid and LB-Plasmid:
The bacteria transformed in both plates to the point where they were lawns.
b. LB/Amp-Plasmid and LB-Plasmid:
The LB/Amp-plasmid showed no growth while the LB-Plasmid had lawn transformation.
c. LB/Amp+Plasmid and LB/Amp-Plasmid:
The LB/Amp+Plasmid had two bacterial colonies and the LB/Amp-Plasmid showed no
growth.
d. LB/Amp+ Plasmid and LB+Plasmid:
The ampicillin-treated positive plasmid had two visible colonies and the other transformed.
6. The transformed colonies allow for the identification of bacteria that has taken up the plasmid, so any transformation whether lawn or subtle will show which bacteria successfully had taken up the plasmid.
7. There was no apparent phenotype due to the lack of color present in the transformed colonies. The pGreen was not present.
8. The first plate to inspect to conclude that the transformation occurred successfully would be the LB/Amp+Plasmid because the marker that shows the selection will only grow in a plasmid treated with ampicillin. Also, the LB/Amp+Plasmid is the easiest to observe due to the fact that it only has two colonies.
9.
a. Total mass: 10 x 0.005 = 0.05 ug
b. Cell suspension volume: 250 calcium cholride + 250 LB + 10 green capped liquid =
0.510 uL
c. Fraction spread: 10 (Suspension Spread)/510 (Total Suspension) = 0.0196
d. Mass of Plasmid Spread: 0.05 (Total Mass) x 10/510 (Fraction
Spread) = 0.00098
e. Transformation efficiency: 2 (colonies)/ 0.00098 (Mass Plasmid Spread) = 2040.81633 or
204081633000 x 10^-8
10. The transformation efficiency could have been influenced by the colony size that was added to the liquid solution. A greater amount of cell receptors comes with a greater colony size, so there would be better efficiency. A lesser colony would have lesser cell receptors, so there would not be as great of efficiency. Another factor that may have influenced the transformation efficiency is the amount of time the plasmids were in the hot or cold water. If the plasmids were in the hot water longer, then the growth of the colonies would be more efficient. If the plasmids were in the cold water longer, the growth could be stunned and the colonies would not be as efficient in transformation. In the experiment, the plasmids that were in the hot water were heated longer because the bath was not at the desired temperature, thus affecting the heat shock of the plasmids. The time the plasmids were outside of the water was also erroneous because the timing was not accurate.