Download Biol 207 Dr. Locke`s section WS9 Page 1 Workshop 9 Biol207

Workshop 9 Biol207
Please try the sample lab exam (Lab Book) in preparation for the final lab exam. We can
discuss any questions during the workshop.
1. Your project is to clone the lacZ gene from a wild type E. coli strain.
The first step in the project is to produce a genomic library from E. coli. To produce this
library, genomic DNA from a wild type E. coli will be digested using restriction enzymes
into fragements which will then be ligated into a plasmid vector to produce recombinant
DNA molecules.
The second step involves transforming the recombinant DNA into a mutant (lacZ -) E. coli
host strain that can no longer metabolize lactose (lac mutation).
The third step is to identify the rare clones that can complement mutations in the lac
operon and restore the ability to metabolize lactose.
In the first step you must plan how to construct the genomic library from wild type E. coli.
The following vector pAT1 (3kb) is provided.
a) Which feature required for cloning is missing from the following diagram of the
plasmid?
AmpR
pAT1
3 kbp
BamH I
EcoRI
TetR
The pAT1 vector DNA was treated with a variety of restriction enzymes and
ligations and then used to transform ampicillin sensitive E. coli that were then plated out on
Ampicillin containing plates to determine the number of Amp R transformants. The results of
these procedures are in the following table (Table 1, Expts#1-6).
In another experiment the pAT1 vector DNA was digested with BamH I and EcoR I,
separated on an agarose gel, and the 2.2 kb long BamH I – AmpR-EcoR I fragment was
isolated (purified) from the agarose gel (Expts#7-8). The 0.8 kb B-Tet-E DNA fragment
was discarded.
Table 1: pAT1 DNA digested, ligated, and transformed into Ampicillin sensitive E. coli
Experiment DNA
Treatment
Total colonies on
#
ampicillin plates
1
pAT1
undigested
3000
2
pAT1
BamH I
0
3
pAT1
EcoR I
0
4
pAT1
BamH I+EcoR I
0
5
pAT1
BamH I and then
250
ligated
6
pAT1
EcoR I+BamH I and
100
then ligation reaction
Biol 207 Dr. Locke’s section WS9
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8
then ligation reaction
isolated 2.2 kbp band not treated
isolated 2.2 kbp band ligation reaction
0
0
b) Based on these transformation results, are the BamH I and EcoR I restriction
enzymes actively digesting the pAT1 DNA?
c) Is the DNA ligase reaction working? How do you know?
d) Why were no transformants recovered in Expt#8, which used ligase?
After, doing these control experiments, the lab used up all the available EcoR I
restriction enzyme, but still retained the purified digested vector DNA.
e) Could the genomic E. coli DNA be digested with the one of the following
enzymes (see Table 2 below) instead of EcoR I and still ligated to the 2.2 kbp fragment.
The EcoR I enzyme cuts at G/AATTC (/slash represents cut site).
Table 2. Other enzymes
Enzyme Recognition
Sequence
Bcl I
T/GATCA
Hpa I
GTT/AAC
Mfe I
C/AATTG
f) If BamH I cuts at G/GATCC and the second enzyme (Mfe I) also cuts at a 6 base
pair recognition sequence, what is the average E. coli genomic DNA fragment size
expected based solely on chance (assume equal frequencies of A, C, G, and T)?
g) Using your answer from part “F”, and if the E.coli genome is 4.5 Mbp, how many
clones would you need to test to make sure you had 99% probability of finding the gene
you seek?
To test how many transformants they are getting per plasmid molecule they
transformed 0.02 µg of pAT1 into 100 µL of E. coli. Then they added 900 µL of LB and
incubated the mixture for 45 min at 37°C. They did a 1/10 dilution three times before a 100
µL aliquot was added to an ampicillin plate. After incubating the plate overnight at 37°C
they had 78 colonies.
(Helpful notes - the molecular weight per base pair is 660 g/mol and Avogadro’s number is
6x1023.)
h) What is the ratio of plasmid molecules to transformants?
i) How many grams of plamid DNA would they need to get enough transformants to
test each clone at least once with 99% certainty in the genomic library?
You will now proceed to use BamH I and the enzyme in part “e” to build the
genomic library of the wild type E coli in the vector pAT1.
A lacZ- E. coli host strain is transformed with recombinant DNA from the E. coli
genomic DNA ligated into the vector. These transformants are plated on an ampicillin IPTG
X-gal plates. X-gal is converted into a blue coloured compound by β-galactosidase. IPTG
is an analogue of lactose and induces the lac operon.
Biol 207 Dr. Locke’s section WS9
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Most of the resulting colonies are white, but rare blue colonies are recovered.
In the controls (Expt#1-8), containing ony the original pAT1 DNA, was also
transformed into the lacZ- strain. After spreading the transformants on the same type of
plates, only white colonies are recovered.
j) What are the two desired phenotypes seen in a transformant that has the lac+
operon?
k) The transfomants were plated out on a Petri dish. If a petri dish can hold 500
transformants, how many plates must be screened to have a 99% probability of finding a
lacZ+ clone?
The same recombinant DNA library is transformed into an E. coli strain that is
mutant in lac I (lac I ) and constitutively expresses β-galactosidase. The transformants
are plated on ampicillin X-gal plates without IPTG. Almost all of the colonies are blue, but
some rare white colonies appear.
l) Can you explain this result?
m) If IPTG had been added to the media, would white colonies have been
recovered in this screen?
n) If the E. coli host strain had been an Oc constitutive mutation, would there have
been any white colonies?
Biol 207 Dr. Locke’s section WS9
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