The extracted plasmid DNA is pUC19. Look up the plasmid map and include in your pre-lab as well as your report.
a. What is the location of the BamH1 restriction site? Xmn1 restriction site? (sequence location by number). How many of these sites exist each?
b. If single digestions were performed (one restriction enzyme) with BamH1 and Xmn1, respectively, how many fragments would form, and what would be the sizes of each of these fragments?
c. If a double digestion with both BamH1 and Xmn1 is performed, how many fragments would form, and what would be the sizes of each of these fragments?
d. What gene(s) do(es) the plasmid have that allows for selection post-transformation?
e. Will the selectivity in part d change at all when foreign DNA is inserted into either of the two restriction sites, BamH1 or Xmn1? Why?
Homework Answers
- BamHI site starts from nucleotide 417. There is one site in pUC19.
XmnI site starts from nucleotide 2298. There is one site in pUC19.
- Restriction digestion with BamHI will produce a single fragment 2686bp long. It will linearize the plasmid.
Restriction digestion with XmnI will produce a single fragment 2686bp long. It will linearize the plasmid.
- Restriction digestion with BamHI and XmnI will produce 2 fragments. 1885bp and 801 bp long.
- It has ampicillin resistance gene (AmpR) that will allow only those cells containing pUC19 plasmid to grow in ampicillin containing medium.
The lacZ gene produces a lacZ fragment which along with the host counterpart can be induced in the presence of IPTG to hydrolyse X-gal, giving rise to blue colonies. Cells with no pUC19 produces white colonies. Thus, only blue colonies are selected post transformation.
- When the foreign DNA is inserted in the XmnI site, ampicillin resistance will be affected because the XmnI site is present within the AmpR gene. The sequence of AmpR gene will be mutated.
When the foreign DNA is inserted in the BamHI site, LacZ will be affected because the BamHI site is present within the LacZ gene. The sequence of LacZgene will be mutated and all cells with foreign DNA insert will form white colonies.
Add Answer to:
The extracted plasmid DNA is pUC19. Look up the plasmid map and
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restriction enzymes. Using agarose gel electrophoresis, the DNA
fragments can be separated on a gel, based on their lengths. In
order to see the fragments, a stain is typically added to the gel.
The size of each fragment can be determined by comparing each one
to a DNA molecular weight marker of known size.
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number of the...
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