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1. (3 pts) Last week's lab handout included maps of the two cloning vectors we're using, pCITE-2a and pCITE-4a. Consult these maps to answer the following question.
a) Draw a diagram of your construct, with your PCR product cloned into your pCITE vector. Model this diagram after the one included on the pCITE handout.
Obviously the correct answer for this question varied from group to group. Criteria for full credit included clear labels including the name of your plasmid construct and of all of the coding regions, arrows to show the orientation of each coding region, and numbers to show the location and size of each relevant segment.
b) Below your diagram, give the nucleotide sequence of each of the fusion points between the pCITE vector and your insert. Include 20 basepairs on either side of each restriction site that you used in cloning.
Everyone identified the fusion points correctly. The only stumbling block here was identifying the 5' to 3' strand at the end of the coding region. The top strand (i.e. the coding strand, or the strand that runs 5' to 3' in the direction of transcription) is constant from the beginning to the end of the gene and is the complement of the reverse PCR primer.
c) Below the nucleotide sequence of your fusion points, provide the appropriate amino acid translation. Position the reading frames so that the ATG of the NcoI site encodes methionine, and the CTC of the AvaI/XhoI site encodes leucine. Do not translate past the stop codon or before the start codon, as the cell will not.
This was trickiest if you had an "S tag" construct. The S tag is an N-terminal protein fusion, which means that the relevant start codon (i.e. the first methionine codon after the beginning of transcription) is at the beginning of the S tag, not at the beginning of your gene.
The start of transcription is labeled "T7 start" on your pCITE vector map.
The start of translation in the pCITE vector is indicated by the amino acid sequence provided at the bottom of the map, which begins with Met. Notice that this site is *after* the NcoI site in vector pCITE-2a, and thus is removed in the cloning that we did (so that the first Met codon of the PCR product becomes the start codon). In vector pCITE-4a, however, the first Met codon comes well *before* the NcoI site. Translation will thus begin in the pCITE-4a vector sequence (which encodes the S tag) and read through into the insert.
Remember, in a eukaryotic system (such as the one we'll be using for in vitro translation) the start codon is the first Met codon after the beginning of the mRNA.
The S tag constructs were engineered so that translation would stop immediately before the XhoI/AvaI site. In the His tag constructs translation should proceed through this site into the downstream vector sequence, which encodes the His tag.
2. (5 pts) Next week you'll be transforming your ligation mixes into an ampicillin-sensitive E. coli strain. (The "Ap" gene on the pCITE vectors encodes ampicillin resistance.) You'll plate your transformations on plates containing ampicillin, incubate your plates overnight, and return to count your colonies the next day. Interpret the following possible results:
[full-credit student answers]
a) equal, large numbers of colonies per plate for all five ligation mixtures.
If equal, large numbers of colonies per plate grow for all five ligation reactions then the plasmid vectors may never have been cut in the first place by the restriction enzymes. If this is the case then it is probably best to assume that our favorite gene is not in these vectors.
NOTE: if there were no amp in the plates or if the E. coli were already amp resistant before transformation, you would observe lawns of E. coli on your plates, not individual colonies.
b) no colonies for ligation #1. Equal, large numbers of colonies per plate for ligations 2-5.
If no colonies appear for ligation #1 but large numbers of colonies appear for the other four ligation reactions this may indicate that one of the restriction enzymes was not functionaing properly. Ligase is needed to attach sticky ends which would make sense for why #1 shows nothing. However, with number two we were hoping that few colonies would appear because that would mean both restriction enzymes were functioning. But since there are large numbers of colonies for #2 one of the enzymes was not functioning because the vector was able to be ligated back together by the ligase because of complementary ends.
c) No colonies for ligation #1. A few colonies for ligation #2. Several hundred colonies per plate for ligations 3-5 (approximately equal numbers for each of these three ligations).
This tells us that both enzymes were working as expected. Number one shows no colonies because there was no ligase to complete the vectors which means they were cut in the first place. #2 shows a few colonies because restriction enzymes are very efficient in the middle of DNA strands but less efficient at ends. This means that occasionally a plasmid was only cut once by an enzyme because the other was unable to cope with the close end at that moment. Since there are few we know that for the msot part both restriction enzymes did their job adequately. Large numbers of colonies for #3 - #5 indiates that our gene was indeed ligated into the plasmids.
d) No colonies for any of the ligations. (provide at least two possible explanations for this one, please.)
Perhaps transformation did not occur for any of the bacteria. [Perhaps the transformation buffers were not correctly made, or the cells were not kept sufficiently cold during preparation, or the heat shock temperature was not right, or the cells were harvested at the wrong stage of growth, etc.]
Another possibility is incubation time. If you do not allow the bacteria to incubate for an hour or so [at least 30 minutes] before plating them then they will most likely die as they have not yet had a chance to express resistance to ampicillin.
Yet another possibility is that ligase was not working at all and therefore no plasmids were intact upon insertion into bacteria.
[Another possibility is that the vector is not what we thought it was, and does not in fact encode ampicillin resistance.]
3. (2 pts) The next step after E. coli transformation is to choose colonies to grow up into cultures. You'll grow small (5 ml) cultures from each of 5 - 10 colonies and then isolate plasmid DNA from these cultures.
If you got result 2c above, which ligation mix would you choose as the source of colonies for further analysis? Why?
[full-credit student answer] If I had the result described in 2c, I would use the ligation #3 as the source of colonies for further analysis because the [molar] concentrations of the insert and the vector are equal. In the 2x and 5x ligations, it is possible that the gene could have been inserted more than once into some of the vectors.
4. (1 pt extra credit) Why did I suggest that the groups cloning LTV1 use twice as much insert as the other six groups?
[full-credit student answer] LTV1 is roughly twice the size of the other inserts. Therefore, if the concentrations were calculated using the same amount of mass, then the actual molar concentration of the LTV1 gene will be half what it needs to be to start with a one to one ratio of insert to vector.
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Created by:
bkbaxter@lclark.edu
Updated: 17 Oct 00