back to Molecular Biology Homepage
back to main lab report keys page
1. (2 pts) Summarize your colony data in a neatly organized, clearly labeled table. Include a description of each transformation that includes what DNA was transformed and how much vector and/or insert was present in each ligation.
The primary criterion for full credit here was that the information necessary to interpret your numbers was present in your table. This means, for example, that there was an indication of whether the vector DNA in each transformation was or was not subjected to restriction enzyme digestion (and ideally, with what enzymes). Whether or not ligase was present should also have been indicated, as well as what "100µL" and "1 ml" referred to.
2. (5 pts) Interpret your control data. Go through each of your controls one at a time (your ligation controls as well as your transformation controls.) What was the purpose of each one, and what do your data tell you? Include a calculation of the "transformation efficiency" of your cells (in colonies per microgram of circular vector transformed--calculate this from your pCITE vector transformation control.)
Transformations 1 and 2 are digestion/ligation controls. #1 shows you whether any of your vector molecules remain uncut--if so, you should get colonies on this plate (as all of you did). #2 shows you whether any of your vector molecules were cut by only one enzyme, and thus had compatible "sticky" ends that could anneal together and then be rejoined by ligase. The colonies on the #2 plates represent uncut vector molecules (as in #1) plus singly-cut vector molecules that were successfully religated.
Transformations 6 and 7 are transformation controls. #6 is a positive control. Colonies on these plates demonstrate that addition of circular vector encoding ampicillin resistance will result in bacterial colonies on amp plates. #6 also allows you to measure your "transformation efficiency," as discussed below. #7 is a negative control. A lack of colonies on these plates demonstrates that your starting culture was made up of amp-sensitive cells, and that your solutions were not contaminated with plasmid DNA encoding amp resistance. If your #7 plates are colony-free, then you can be reasonably confident that the colonies you see on your other plates resulted from DNA that you added and not from some other source.
Transformation efficiency is a calculation of the # of colonies that your competent cells give you per microgram of circular vector added. This can only be calculated from transformations in which you know the amount of circular vector present. In this case, this is transformation #6 (only). Count the number of colonies (total) that resulted from the 10 ng of DNA you added (this means adding all of your colonies for this transformation together)--this is the efficiency per 10 ng. Multiply by 100 to get the efficiency per microgram. (Alternatively, for those with high numbers, multiply the number of colonies on your 100 µl plate by approximately 1100).
Why calculate this number? It gives you an idea of how "good" your competent cells are. Particularly since you prepared more competent cells than you needed for this experiment and stored the extras, you'll have an idea whether to use these cells in future experiments or not. For example, the efficiency of the cells prepared in 312 this semester ranges from 300 to 52,000 colonies per microgram of pCITE vector. In experiments in which the amount of circular DNA is a limiting factor (which it often is), you'd want to use "highly competent" cells.
NOTE: be careful of the word "anneal," which has a specific meaning in molecular biology. It means the base-pairing of complementary strands of nucleic acid. In a ligation reaction, the sticky ends of fragments anneal to each other. Ligase has no effect on this process! What ligase does is catalyze the formation of a covalent phosphodiester bond between fragments that have annealed.
3. (1 pt) Interpret your experimental results. Do you have colonies on your "vector plus insert" plates? If so, given your answers in #2, do you think they are likely to be what you want?
In short, no. These colonies are very likely to represent bacterial cells that have taken up religated vector fragments, without insert. We're hoping that some small proportion of them represent cells that have taken up recombinant plasmids, but the control data are not promising.
4. (2 pts) Propose your next step. Provide a clear explanation why you think this step is appropriate given your results. Your options include:
For every group, the control data suggest that restriction enzyme digestion was incomplete. The most logical approach would be to repeat the restriction digestions of both vector and insert (with the addition of some control reactions and/or additional monitoring to assess whether digestion was progressing), then to repeat the ligations, and then to repeat the transformations. For some groups, repeating the digestion reactions would mean repeating the PCR first in order to obtain enough fragment.
As we discussed together, this would take a large amount of time. Instead, we are proceeding with the last suggestion above, choosing colonies from ligation #5 in the hopes that some of them will contain recombinant plasmid. Time will tell whether this hope was justified. The reason for choosing #5 is to increase the odds of obtaining constructs with insert as opposed to religated vector.
back to main lab report keys page
back to Molecular Biology Homepage
Created by:
bkbaxter@lclark.edu
Updated: 23 Oct 00