E. Coli

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Big Idea 3 – Investigation (Lab) 8
Big Idea 3 – Investigation (Lab) 8
SUBCLONING
Recall how a gene of interest is
obtained (PCR), inserted into a
plasmid using restriction enzymes /
DNA ligase, and transformed into a
bacterium for copying or expression
to get protein…review chapter 20
powerpoint if you have forgotten.
Big Idea 3 – Investigation (Lab) 8
Big Idea 3 – Investigation (Lab) 8
Procedure
1. In this lab you will simply take an E. coli colony, put it in a tube containing the CaCl2 solution, and
either add pAMP plasmid (+) or not (-).
Tube 1 (-)
Tube 2 (+)
E. coli (1 colony)
240ul CaCl2 solution
10 ul water
E. Coli (1 colony)
240ul CaCl2 solution
10ul of Ampr plasmid (0.005 ug/ul)
pAMP plasmid
- Ampr gene coding for β-lactamse (enzyme that
cuts up ampicillin).
- ORI – origin of replication
- MCS – multiple cloning sites (restriction sites)
Big Idea 3 – Investigation (Lab) 8
Procedure
2. You now of course heat shock to transform the cells. The (-) tube is obviously a negative control.
The CaCl2 helps with the efficiency of the heat shock. Right after the heat shock, 250ul of liquid LB
(nutrient solution) is added to tubes so that cells can begin to recover and grow.
Tube 1 (-)
Tube 2 (+)
E. coli (1 colony)
240ul CaCl2 solution
10 ul water
250 ul liquid LB
E. Coli (1 colony)
240ul CaCl2 solution
10ul of Ampr plasmid (0.005 ug/ul)
250 ul liquid LB
Big Idea 3 – Investigation (Lab) 8
Procedure
3. Plate 100ul of the 500ul cell solution on agar plates both without and amp and with ampicillin to
kill untransformed cells. This is what you should see…If you grow either tube on straight LB agar
then tons of colonies are observed like plates 1 and 3. If you grow the cells from each tube on LB
agar +AMP then you should not see any cells for the (-) tube and only a dozen or two for the
transformed (+) cells….
Big Idea 3 – Investigation (Lab) 8
1.
Simple…just count the number of colonies you observed…this is how many cells got
transformed because each colony comes from a single cell undergoing multiple rounds of binary
fission.
Let’s say you counted 32 colonies…
2. Now we need to know how much DNA in ug it took to get those colonies you just counted.
Remember that you added 100ul of this solution:
Tube 2 (+)
E. Coli (1 colony)
240ul CaCl2 solution
10ul of Ampr plasmid (0.005 ug/ul)
250 ul liquid LB
Figure out how much DNA you added….
Big Idea 3 – Investigation (Lab) 8
Remember that you added 100ul of this solution:
Tube 2 (+)
E. Coli (1 colony)
240ul CaCl2 solution
10ul of Ampr plasmid (0.005 ug/ul)
250 ul liquid LB
1. You added 10ul at 0.005ug/ul. Therefore you have .05ug of plasmid in the tube.
2. The total volume is 500ul. Therefore you have 0.05ug/500ul or 0.0001ug/ul.
3. You took 100ul of this solution: 100ul x 0.0001ug/ul = .01ug of plasmid
It took .01 ug of plasmid to get those 32 colonies!
= 32/0.01 or 32 colonies per 0.01 ug of plasmid
= 3200 colonies per ug plasmid
http://www.sciencegateway.org/tools/transform.htm
Big Idea 3 – Investigation (Lab) 8

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