Unformatted text preview: his is actually
fairly simple. Cells are grown on agar plates containing two extra components. The first of these is a
chemical called IPTG which is an “inducer” for the enzyme (an inducer is part of the gene’s regulation
mechanism) and the chromogenic substrate X-GAL. When X-GAL is acted upon by a functional galactosidase (X-GAL is very similar in structure to galactose) the resulting product is blue. Thus
colonies of transformed cells are blue and easily recognized by color. But how can you know if DNA
has been cloned into the MCS of the pBluescript plasmid? What must be kept in mind here is that the
MCS is located within the coding sequence of the -galactosidase subunit gene. If DNA has been
added in this location, the gene would be “fatally” interrupted and could not produce a functional
enzyme subunit. Such plasmids, when introduced into E. coli, would not produce blue colonies when
grown on media with IPTG and X-GAL. Instead, these colonies would be white. Biology 05LA – Fall Quarter 2012 Lab 7 – Page 5 Procedure
Follow the instructions for the lab for transforming and plating these plasmids. The only
alteration is in the composition of the ampicillin plates. The two groups performing these
modifications will use ampicillin plates that contain IPTG and X-GAL. These have already been
added to the plates by the lab staff.
Questions you should ask:
1. From the numbers of blue and white colonies on the ampicillin plate, what do you think the ratio
pBluescript plasmid: pBluescript plasmid + insert were in the transformation mix?
2. Do you think the plasmids in the white colonies will be smaller or larger than the plasmids in the
blue colonies? Why?
The “blue-white” colony technology demonstrated here has proven very useful for isolating and
copying genes from a wide variety of organisms with the use of the prokaryote E. coli. The utility of
this technique for “cloning” genes is related to the very low frequency of being able to insert (ligate) a
sequence of DNA (the gene you wish to copy) into the MCS present in the -galactosidase sequence.
In practice, the frequency of insertion is 1 in 1000 or less. In should be evident that looking for
colonies of cells containing the DNA of interest among the many not containing this DNA is literally
like “looking for a needle in a haystack.” This selection is greatly simplified when all that needs to be
done is look for the small number of white colonies that are present among the numerous blue
colonies. As described above, the white colonies are comprised of cells unable to metabolize galactose
and thus contain the inserted gene. Once this selection has been done, the cells of the white colonies
are sub-cultured. The cloned gene can be extracted after the cells have multiplied to the number
required. Biology 05LA – Fall Quarter 2012 Lab 7 – Page 6 10 drops competent
E. coli cells FLOW CHART
__ +DNA Remember - each time
cells are dispensed or
something is added,
you must re-suspend
the cells!! ps
E. coli cells no DNA - + ____ drops
__ ml __ ml __ ml Ice for ___ min 42oC for ___min A ___
drops - ___ml saline D ___ml saline __ ml saline __ drops __ drops ___ml saline + C B __ drops Add ___ drops L broth
___oC for ___ min ___
AMP + DNA
LB - DNA
AMP - DNA
LB B C no
antibiotic Incubate all plates
For 16h @ ___oC D
antibiotic Biology 05LA – Fall Quarter 2012 Lab 7 – Page 7 WORKSHEET FOR CALCULATION OF TRANSFORMATION EFFICIENCY:
1. Calculate the dilution factor for the transformants:
Ask: How many transformed cells were generated from 3 drops of competent E. coli cells?
3 drops competent E. coli cells
1 drop plasmid
+5 drops sterile media
9 drops of diluted cells (total)
3 drops of the diluted cells were plated. Therefore 1/3 of the diluted cells were evaluated on each
antibiotic plate. Dilution Factor = # of drops plated on each plate
total # of drops of diluted cells 3 drops
9 drops 1
3 2. Calculate the number of transformed cells:
Counting the number of colonies on each of the antibiotic plates gives you the number of transformed
cells (Tdil) in 3 drops of your diluted cell mixture.
Therefore the total number of transformants (Ttotal) = Tdil x dilution factor.
Ttotal = Tdil x 3.
Make this calculation for the +AMP plate.
3. Calculate the dilution factor used for determining the number of viable E. coli cells used in each
Ask: How many total cells were the in 3 drops (= 0.1 ml) of competent cell suspension?
There are two dilution factors to consider here. First, cells are diluted in the A, B, and C tubes with
saline (tube dilution factor). Second, only a portion of the cells in the A, B, and C tubes are plated on
the petri plate for counting (plate dilution factor).
Total dilution = Tube dilution x Plate dilution.
Plate Tube Dilution Plate Dilution Total Dilution A Not plated 0.1 ml/10 ml = 1/100 1/100 = 10-2 B 1 ml/10 ml = 1/10 0.1 ml/10 ml = 1/100 1/1,000 = 10-3 C 1 ml/10 ml x 1/10 = 1/100 0.1 ml/10 ml = 1/100 1/10,000 = 10-4 D 1 ml...
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This note was uploaded on 08/27/2013 for the course BIO BIOL05LA taught by Professor Abbottl during the Fall '12 term at UC Riverside.
- Fall '12