5. Compare and contrast the number of colonies on each of the following pairs of plates. What does each pair of results tell you about the experiment?

LB+Plasmid and LB-Plasmid: Without the presence of ampicillin hindering the growth of the bacteria, the bacteria is able to reproduce to the point of growing a bacterial lawn with or without the pBlu plasmid.

LB/amp-Plasmid and LB-Plasmid: The petri dish containing LB/amp+plasmid grew a little bit because ampicillin hindered the growth of the bacteria. Tthe plasmid was incorporated into some of the bacteria, however, so smaller, less dense colonies were able to form. The petri dish containing LB-plasmid showed a lawn of bacterial growth because there was no ampicillin to prevent bacterial growth.

LB/amp+Plasmid and LB/amp-Plasmid: In the absence of the pBlu plasmid, the bacteria from the luria broth is unable to reproduce, and no colonies or bacterial lawn show. On the other hand, the presence of the pBlu plasmid allows smaller, less dense colonies to grow, but the plasmid is not incorporated into most of the bacteria’s DNA, so only a small amount of accumulation is present.

LB/amp+Plasmid and LB+Plasmid: The LB+plasmid petri dish demonstrated a lawn of bacterial growth because the bacteria was allowed to grow unhindered by ampicillin. The plasmid did not increase or decrease the amount of bacterial growth shown because the bacteria reproduced in a favorable environment. The LB/amp+plasmid demonstrated how the bacterial growth was hindered thanks to the ampicillin, but it was still able to reproduce.

LB/amp/X-gal-Plasmid and LB-Plasmid: The bacteria in the former petri dish was not able to reproduce as well as the bacteria in the latter petri dish because of the presence of ampicillin. Without ampicillin, it does not matter whether or not the bacteria contains the pBlu plasmid. However, in the presence of ampicillin the bacteria is unable to reproduce, even if it’s in the presence of X-gal. This is because x-gal is used to simply detect an enzyme that is produced when the pBlu gene is present in a particular bacterium’s genome. It does not encourage the growth of bacteria that do not have the plasmid.

LB/amp/X-gal+Plasmid and LB/amp/X-gal-Plasmid: The bacteria was unable to reproduce in the presence of amp and x-gal without the pBlu plasmid because, like the LB/amp-plasmid, the ampicillin blocked the reproduction of bacteria. Additionally, the colonies in the petri dish containing LB/amp/x-gal+plasmid were blue. This is because x-gal is used as an indicator of whether or not cells are producing an enzyme that breaks down certain products, and enzyme produced only when a certain gene is inserted into the DNA. In this case, it can be seen by the blue center of the colonies that the pBlu gene was primarily present on the interior of colonies. This is because bacteria with the pBlu gene are the likeliest to survive contact with ampicillin because of the enzyme they produce. Other bacteria are able to grip onto this blue bacteria for production and reproduce, but they will not be as auccessful as the bacteria with the pBlu gene.

LB/amp+Plasmid and LB/amp/X-gal+Plasmid: Both of the petri dishes containing these compounds had two colonies of bacteria because ampicillin prevented a bacterial lawn from growing, but some of the bacteria contained the plasmid, and those bacteria were able to reproduce. The colonies in the LB/amp/x-gal+plasmid petri dish seemed a little bit thicker, but this can be attributed to visual human error. The darkness of the blue color produced by x-gal may cause one to think that the bacteria is actually thicker than it is compared to the white, less opaque bacteria which do not (visibly, as indicated by x-gal) produce enzymes to break down ampicillin.

 

6. What are you selecting for in this experiment?

In this experiment, the bacteria with the pBlu plasmid are being selected for. They are identified by their blue color in the presence of x-gal. The bacteria that is white has not incorporated the plasmid into its DNA, and is therefore not the bacterial cell that this experiment is selecting for.

 

7. What does the phenotype of the transformed colonies tell you?

The phenotype of the transformed colonies are colonies that are blue in color. This means that these colonies contain the pBlu plasmid. The pBlu plasmid causes the bacteria to secrete enzymes to counteract the ampicillin. The creation of the blue color is a spontaneous reaction between the byproducts of the enzyme and the ampicillin interacting. Therefore, it is indicated that those particular bacteria, the bacteria that produces the ampicillin-fighting enzyme, have incorporated the pBlu gene into their own DNA. These bacteria have developed resistance to ampicillin, an antibiotic.

 

9. Transformation efficiency

a. Determine the total mass of plasmid used

10 uL x 0.005 ug/uL = 0.05 ug of plasmid

b. Calculate the total volume of cell suspension prepared

250 uL CaCl2 + 250 uL luria broth + 10 uL plasmid = 510 uL total volume

c. Calculate the fraction of the total cell suspension that was spread on the plate

 10 uL spread onto plate / 510 uL total suspension = 1.960784% of the whole suspension

d. Determine the total mass of plasmid spread

0.05 ug of plasmid spread x 0.01960784 (fraction of total cell suspension spread onto plate) = 0.00098039 ug spread onto plate

e. Determine the number of colonies per ug spread

2 colonies observed / 0.00098039 ug = 2.04 x 108 colonies spread per ug spread

 

10. What factors might influence transformation efficiency? Explain the effect of each factor you mention.

There are many factors that could affect transformation efficiency. One of these factors is the purity of the DNA. If, for some reason, the DNA is contaminated in preparation of the lab, the ability of the DNA to be incorporated into the cell could be affected. Also, the heat shock could affect the transformation efficiency. The heat shock must be done at a very specific time and temperature in order to ensure transformation efficiency. The water in this experiment had to be 42 degrees Celsius and the cells had to be held in there for exactly 90 seconds. If the water is too hot or the cells are held in the water for too long, proteins in the cell may denature and render the cell unable to divide. On the other hand, if the cells aren’t kept in the heat shock for too long or the temperature of the water is too cold, the cells won’t accept the plasmid DNA as easily, thus the transformation efficiency goes down. The length of time after transformation also affects the transformation efficiency because the longer the cells have to divide, the more cells there will be. If there are more cells dividing, then the likelihood of another bacterial colony growing is increased. Finally, the agar itself could affect the transformation efficiency of colonies. If the agar is a good quality, then more bacteria and more colonies will grow. If the agar is not a good quality, then less bacteria will grow.