Brindan Thiyagarajah

A61

Adam Cooke

Jared Gatto

Transformation, PCR, & Electrophoresis Lab

Introduction

These two sessions of lab demonstrated several key techniques and aspects of biology,

and DNA in general, such as PCR and Electrophoresis. The experiments involved the

transformation of E.Coli cells in different environments, essentially manipulation of the genes,

and then using such techniques such as PCR and electrophoresis to observe these changes. The

first half of the lab required the actual transformation of the E.Coli cells; this was done through

the application of certain independent variables: Ampicillin, and plasmids. Essentially, the point

of the ampicillin was to kill off the E.Coli cells, and leave behind ampicillin resistant mutants,

which then would be grown, the plasmids on the other hand were meant to “house” the E.Coli

from the ampicillin, and allowed cells without the resistant gene to survive in the ampicillin

environment. Ultimately, with these variables set in place, the growth of these E.Coli colonies

were observed, in four different scenarios: #1- no ampicillin and no plasmids (control group),

#2- yes ampicillin and no plasmids, #3- no ampicillin and yes plasmids, #4- yes ampicillin and

yes plasmids. The E.Coli was mixed in with a nutrient broth to promote growth, afterwards

placed in these different environments, and then were left to incubate until the next lab. The

second lab (Lab #10), is where the grown colonies of transformed cells would be observed in

order that the introduction of the resistant gene is confirmed. How is it observed exactly? Well,

since the gene sequence is known, two techniques known as PCR and Gel Electrophoresis can be

used. To start off, Polymerase Chain Reaction (PCR) is a method of analysis, where sequences of

DNA are copied through DNA synthesis and replication. In Lab 10, the E.Coli cells are going to

go through the same process of synthesis and replication; they will be broken in ssDNA and then

synthesized with primer and Taq polymerase, but since Helicase and Topoisomerase aren’t

available, the E.Coli DNA will need to be heated to the point where the DNA denatures into

single strands, and then is synthesized by the primer and Taq polymerase from there on. In the

experiment, a PCR premix, was distributed, which contained the Taq polymerase, essential salts

and buffers, and dNTPS (dATP, dCTP, dGTP, and dTTP). The Taq polymerase and dNTPs,

work along with the primer (phage m13 oligonucleotides) which is then later added, to assist Taq

polymerase. As it’s known, DNA polymerase works as an “extender”, it cannot start off a

sequence, so we have the primer in the solution to set a base template (5’

GTAAAACGACGGCCAGT 3’ and 5’ CAGGAAACAGCTATGAC 3’) where the polymerase

can work off of. The second half of the lab was the use of electrophoresis to identify the number

of DNA that has migrated from a well, located on the gel. By using UV rays, several DNA

“bands” can be seen, which show how far the DNA has migrated. Essentially the further strand

the more DNA there is. But how is one supposed to tell the number just by looking? Molecular

weight standards that contains several DNA of known value and high conductivity are used in

comparison with the DNA that is being measured; the molecular weight standard will be ahead

of the E.Coli DNA due to its conductively, and once the procedure is done the E.Coli DNA will

be compared to that of the Molecular weight standard so that an estimate number can be

produced. Knowing the number, will allow us to know whether or not the DNA contains the

resistant gene. By using these techniques scientists are able to discover new things about

biological life every day. Just humans alone have 24,000 genes, and electrophoresis and PCR

enables humans to find out what a particular gene is capable of, whether it’s detrimental,

provides resistance, etc. It is because of electrophoresis, and PCR, a scientist is able to identify

and isolate specific genes and find out its’ purpose, even potentially creating a vaccine, for

example is a possible outcome.

Hypothesis

Part A:

It is hypothesized that plate one containing neither ampicillin nor plasmids will lead to an

immense overgrowth of cells to the point there aren't any colonies but a densely collected and

large abundance of cells; reason being because there is no ampicillin being introduced to the

environment, allowing the E.Coli cells to replicate freely without dying to the ampicillin. As for

plate two, there will potentially be no cells at all because the ampicillin that is present will kill

them all, except for the very few cells that have an ampicillin resistant gene. Plate three, will be

very similar to plate one simply because there is no ampicillin present so the cells can replicate

freely in the nutrient broth, but they will not utilize the plasmids because that requires an

immense amount of energy, whereas E.Coli cells likes to be energy efficient; it is quite possible

that plate three will contain a dense collection of E. Coli cells along with plasmids. As for plate

four, it is predicted that it will have some growth of E.Coli, depending on the colonies found on

the plate; for example, if the plate contains ampicillin resistant E.Coli they will replicate without

the use of a plasmid simply because it consumes too much energy, and as a result the plate will

have colonies of ampicillin resistant E.Coli cells, but you will also have colonies where E.Coli

cells without the resistance gene, and in turn will have to use the plasmids in order to survive in

the ampicillin. The colonies for the E.Coli cells which have to use the plasmids will probably be

less in number simply because they will consume more energy to replicate compared to a cell

with the resistance gene. And one also has to take into account that not all the E.Coli cells will

transform along with the plasmids.

Part B:

It is hypothesized that in the PCR process and Gel electrophoresis that +amp and –amp

samples of E.Coli will naturally have a ratio of lower concentration of cell, to higher

concentration of cells. Reason being for this hypothesis is due to the fact that, in the +amp tube

there will be E.Coli cells which were once exposed to ampicillin, and the cells that are in tube are

either are or descendants of the surviving E.Coli cells that didn’t die off to the ampicillin; hence

it will consist of E.Coli that utilized plasmids, and/or E.Coli that has the ampicillin resistant

gene. It is also predicted that a majority of the cells in the +amp tube utilized the plasmids to

survive in the ampicillin, and as a result they will consume more energy which in turn will slow

down the replication process. However, the –amp quite likely will have a higher cell count

simply because the cells will not be killed off and will be able to replicate at a normal pace, and

as for the plasmids they simply won’t be used by the cells because they are simply inept for

energy consumption. With these predictions at hand, it can also be assumed that in

electrophoresis the +amp will form a band that exhibits a color under an ultra violet light, simply

because the cell expresses the gfp gene, and then this will be visible under ultra violet. The fact

that in the +amp solution the cell is transformed further proves that a color will be emitted,

because the ampicillin will force many cells to transform in order that they may survive, and as a

result these transformed cells will express GFP which is apparent under UV lighting. Knowing

this, it can also be assumed that in the –amp solution, there will be no GFP being expressed

because there is no need for the cells to transform in the solution, therefore GFP will not be

produced nor expressed.

Methods & Materials

Lab 9

Materials: 50 mM CaCl2 Microfuge tubes, Assorted Micro-pipettes, E.Coli culture nutrient agar

plates, inoculation loops, ice, water bath (42oC), 250 μL Nutrient broth, four nutrient agar plates

In Lab 9, start of by taking two microfuge tubes containing 50mM CaCl2, and label one –

pGreen and the other +pGreen. Next step is to pipet 10 μL of pGreen plasmid into the +pGreen

tube, once that’s completed, an inoculation loop is required to transfer E.Coli cultures from agar

plates into the microfuge tubes. Mix the content of the tube, and then ice it for 15 minutes, while

placing the inoculation loop back into its sterile sleeve. Once the 15 minutes have elapsed, you

need to heat shock the cells in order that the cells’ pores open up and allow the plasmids in, and

to do so place the tubes into 42oC water bath for 50 seconds precisely, and afterwards the tubes

need to be dropped into an ice-water bath for 2 minutes; this temperature change will cause a

rapid shrinking that acts like a vacuum for the plasmids to be absorbed into the cell. Once

completed, take 250 μL of nutrient broth, and add it into both microfuge tubes by using a p200

micropipette (125 μL at a time), and then let the solution sit for 10 minutes in room temperature.

While the solution is idly sitting at room temperature, four nutrient agar plates should be

obtained, each should be labeled respectively: #1-amp/-p, #2+amp/-p, #3–amp/+p, and

#4+amp/+p. Using the inoculation loop and micropipette, pipette 100 μL of –pGreen tube into

plate #1, then spread it with the loop: repeat the step for plate #2. As for plates #3 and #4, deposit

100 μL of the +pGreen tube content into each plate respectively, and use a new inoculation loop

end to spread it throughout each plate. Take the plates, bundle them together, and put them in the

fridge in order that the cultures on the plates may incubate.

Lab 10

Materials: Ultra violet light, two 1.5 mL microfuge tubes, distilled water, inoculation loop, heat

block (95oC), ice, two .5 mL microfuge tubes, PCR Premix, Primer solution, PCR thermal

cycler, Agarose, gel casting tray, gel box, leads, power supply, TBE buffer, +amp and –amp

tubes, tracking dye, molecular weight standard, gel documentation system

The E.Coli cultures acquired in Lab 9, will now be observed, and so they must be taken

out of the refrigerator, and then placed underneath ultra violet light; whichever plate has colonies

emitting a fluorescent green color should be recorded.

PCR: For the PCR procedure, two 1.5 mL microfuge tubes should be obtained and then filled

with 50 μL of distilled water, then respectively labeled +amp and –amp. Taking a sterile

inoculation loop, open the plate with the fluorescent green colonies (+amp/+p #4 plate) and

acquire a colony on the inoculation loop, then transfer it to the +amp microfuge tube. Next take

plate #3, and repeat the steps prior, but this time place it in the –amp tube. Take these microfuge

tubes, and place them in a 95oC heat block for 5 minutes so the DNA can denature and disperse,

then cool the tubes in an ice-water bath. Next step is to obtain two more microfuge tubes, but that

are .5 mL each and containing 40 μL of PCR Premix, and label these respectively +amp and –

amp. Furthermore, take the PCR Premix tubes, and add 5 μL of the solutions that were

previously heated, into their appropriate tubes. In addition, to this solution, add 5 μL of primer

solution to both tubes. Finally, place the microfuge tubes into the PCR thermal cycler for 3

hours.

Gel Electrophoresis: For the gel electrophoresis, agarose gel will first need to be acquired from

the 60oC water bath. Then at the lab bench, a gel casting tray will be available, and the flood

gates of the casting tray must be secured in order to proceed to the next step. Place a comb at the

far end of the tray, and then place the tray on a leveled platform where it won’t be interfered

with. Pour the agarose onto the tray, and let it solidify which will take around 20 minutes. Next,

loosen the gates and take the tray and place it in the gel box so that the wells formed by the comb

are facing towards the cathode. Take 300 mL of TBE Buffer and pour it into the gel box, while

removing the comb from the gel. Take the +amp and –amp tubes, and add 10 μL of tracking dye

to each. Then take 25 μL of Molecular weight standard, pipette it into one of the wells, do the

same for the +amp and –amp (each sample should have their own respective well). Put the cover

on the gel box, connect the leads to the power supply, and set the voltage to 100V - 120V. Run

the experiment until the tracking dye has gone ¾ or more of the gel, which is approximately 45-

60 minutes. Once done, turn off the power supply, take out the leads, and take the gel out, and

put it on a weight tray. Take the gel, and place it under ultraviolet light and then DNA bands

should become apparent, and using the gel documentation system, take a print out of the gel.

Results

Figure 1: Table 1: Lab 9 Plate Variables

Plate Number Plate Content Fluorescent Number of Colonies

1 -amp/-p No TNTC

2 +amp/-p No 0

3 -amp/+p No TNTC

4 +amp/+p Yes 24

In figure 1, the table shows the relationship of ampicillin presence and plasmid presence,

to the number of colonies formed, and fluorescence. Plate 1 had no ampicillin and no plasmids,

so the cells were able to grow freely without persecution, and so they ended up amassing to a

large amount, so large that it wasn’t countable, but it wasn’t fluorescent because the cell did not

transform with plasmids which provide the fluorescent color. As for Plate 2, there was presence

of ampicillin, but there were no plasmids introduced to the solution, and so the E.Coli cells were

killed off by the ampicillin, hence no fluorescence and no colonies either. Plate 3, like Plate 1,

had no ampicillin, but Plate 3 did have plasmids which generally had no purpose for the cells,

mainly because they weren’t exposed to ampicillin so the cells weren’t at risk therefore they

didn’t need the plasmids to survive; ultimately, the cultures on Plate 3 replicated so much that it

was too numerous to count like Plate 1, and since the plasmids weren’t utilized by the cells, GFP

couldn’t be expressed, therefore no fluorescence. Finally, Plate 4, was exposed to ampicillin, and

plasmids, meaning that cells that accepted the plasmids were transformed and had the ability to

survive in the ampicillin, and at the same time these cells will also express GFP, meaning they

had fluorescent. The reason why Plate 4, had a countable amount of colonies is because not all

the cells transformed.

Figure 2: Gel Electrophoresis reading

In figure 2, each lane can be observed to contain molecular weight standard, +amp, and –

amp respectively. Using the molecular weight standard, the number of base pairs and molecular

weight of the samples can be determined. The bands on the gel also provide evidence that a gene

is present, for example +amp has bands visible under the UV lights meaning the gene is present,

whereas –amp isn’t apparent, or is barely distinguishable meaning the gene is not present. By

comparing the +amp with the MWS, it is estimated that the gfp gene has approximately 850 to

900 base pairs.

Figure 3.

2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

f(x) = − 0.371440517130496 x + 1.7060463872964R² = 0.972714344045802

Migration Coefficient vs. LogBP

LogBP

Mig

ratio

n Co

efficie

nt

In figure 3, the Log of the Base Pairs is being compared to the migration coefficient. The

migration coefficient was found by dividing distance of the dye front from the well (8.5cm) by

the MWS bands: each one of those values gave the migration coefficient, and as for the Base

pairs, the log was just simply taken of each base pair band on the MWS. Compiling the data

together, results in figure 3, and with the trend-line equation given in figure 3, “x” can be found:

“x” representing the number of base pairs found in the +amp DNA sample. By finding the

distance of the +amp band from the well (5.2 cm), and then dividing that by the dye front

distance from well (8.5 cm) you get the migration coefficient for the band on +amp. This

migration coefficient can then be plugged into the trend line formula as the y value, and then x

can be solved for. In this case, x = 2.9462 which is also the LogBP, and so to get the number of

base pairs simply do, 10X which in this case is 102.9462= 883.49 base pairs. The fact that the

data on figure 3 had also had an R2= .97, further strengthens how accurate the results are, simply

because the closer the R2 is to 1, the more accurate your results.

Discussion

The data values that were recorded and experimentally found, do happen to correlate as

well as agree with the hypotheses given prior: since the gfp gene is apparent in the +amp sample

electrophoresis bands, the hypothesis that the +amp would express the GFP because the cells had

transformed along with the plasmids holds veracity. Clearly, if this weren’t the case, there

wouldn’t be any DNA bands showing up under the UV light. The growth on the plates, had also

further proved the hypotheses correct; it was stated earlier that the plates each had differing

results because of the combination of variables, so for example plate one having no ampicillin

and no plastids reciprocated by having an abundance of E.Coli cultures growing on the plate,

whereas plate two had ampicillin but no plastids, and the result was that there would be no

cultures because the ampicillin would have killed them off, which happened to be the case. As

for the PCR product, the base pairs were relatively close to 750, but slightly off: the reasoning

behind the discrepancy can quite possible be the cause of human error, for example, if too many

plasmids were added to the +amp solution, then it is possible that more than intended cells

transformed, meaning more GFP, and then a higher base pair count.

Conclusion

In conclusion, it was seen that gfp gene was found to be in the +amp sample, compared to

the –amp which didn’t have the gfp gene, and the fact that these bands can be seen under UV

lights further goes to show that gfp gene is actually apparent in the +amp sample. Since it is

known that the +amp sample contains the gfp gene, it is also correct to assume that the

hypotheses given earlier about +amp and –amp were correct: in layman’s terms, it was

hypothesized that the +amp would have the gene whereas –amp wouldn’t, simply because on

culture decided to use the plasmids where the other didn’t.

Work Cited

- Lawless, Jeanne “BIOL 118 Lab Manual”