Gene Expression

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12 Δεκ 2012 (πριν από 4 χρόνια και 6 μήνες)

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Exercise 9: Gene Expression


Learning Objectives:



Enable students to observe the experimental process called bacterial transformation,



Demonstrate the relationship between the genetic constitution of an organism and its physical
attributes,



Enable
students to observe the change in phenotype caused by the uptake and expression of a known
plasmid sequence, and



Reinforce the need for sterile technique when working with bacteria.


Introduction:

The blueprint of life itself is found in DNA. But, life wit
hout its supporting
molecules is not possible. DNA
is transcribed into RNA. As RNA the blueprint that was in the DNA can now be read by the translational
machinery to convert this blueprint message into proteins, the language of the cell.
It is the protein
s
expressed in a cell that determines the physical and biochemical

structural

properties
and function
of a cell. It
is the molecular machinery that transfers the information from DNA to RNA to a protein capable of
performing a function within the cell that

is
essential

to
all
life. This involve
s

several complex pathways each
of which is subject to regulation. The first step of this cellular pathway is the synthesis of an RNA molecule
from the DNA template. This process is termed transcription. Upon exiting
the nucleus, a mature RNA can be
translated into a polypeptide sequence. This process is known as translation. These pathways happen, with
some modification, in both eukaryotes and prokaryotes.
As much as this process is regulated by the cell it can
also
be manipulated by researchers in biology, particularly in prokaryotes.


When we talk about DNA what is typically referred to is nuclear DNA, but there are other forms in which DNA
is found, like plasmids.
Plasmids are small, circular DNA molecules that exi
st apart from the chromosome
(
s
)

in most bacterial species
, in the nucleoid region
.
P
lasmids are not essential for survival of the host bacteria
, in
most cases
. However,
when bacteria are placed into certain environments, plasmids could give them that
extra

advantage that allows bacteria to survive and reproduce in these environments. Plasmids can carry
genes

that, when expressed,

help bacteria survive.

For
instance
,

some plasmids can have

genes that
grant
bacteria

resistance to

certain

antibiotics. A bacte
rial cell containing such a plasmid can live and multiply in
the presence of the
antibiotic drug
.
A
ntibiotic
-
resistant

bacteria like

Escherichia coli (E. coli)
isolated in many
parts of the world contain plasmids that carry the genetic information for prot
ein products that interfere with
the action of many different antibiotics. In this laboratory, you will introduce a plasmid that contains an
ampicillin
-
resistance gene into
E. coli.


Biotechnology
is the technological application of biological systems or
their derivatives to make/modify
products or processes for a specific use. Biotechnology plays a vital role in health care to manufacture
hormones (
i.e.

insulin), antibiotics and vaccines, in agriculture to produce disease resistant crops, and in
environme
ntal preservation to make biodegradable products. Today, the most commonly used form of
biotechnology is
genetic engineering
. This field involves the direct manipulation of genes
, through

the
movement
of DNA

from one organism to another
or

from one species

to another
,

to impart a particular
characteristic
, such as,
pesticide/herbicide resistance, a longer shelf life, and increased nutritional value of
agricultural crops to an organism of interest. Genetic engineering includes techniques such as
transformati
on
and
cloning
. While the latter process creates multiple copies of a desired gene, transformation
, discovered by
Frederick Griffith,
alters the genetic code of a cell through the uptake, incorporation and expression of a
foreign gene provided by a “donor”

cell.



In order for transformation to be successful, three

conditions are required: 1) a host into which the foreign

DNA can be inserted, 2) a means of delivering the DNA

into the host cells, and 3) a way to identify and select
for

the transformed cells.

You will use the bacterium
Escherichia coli
as the host

organism for the current
experiment.
E. coli
are gram
-
negative

bacteria that form the normal flora of the gut. In the

digestive

system,
E.
coli
benefit their host by producing vitamin K as

well as
preventing other pathogenic bacteria from
establishing

residence. However, certain strains of
E. coli
are pathogenic

and result in food poisoning,
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gastrointestinal and urinary tract

infections, neonatal meningitis and pneumonia.
E. coli
is used

extensively

in
biotechnology because it has only one chromosome

composed of 5 million base pairs (less than 0.2% of the

human genome), a short reproduction time (cell division

every 20 minutes) and a fairly rapid growth rate.


Delivery of foreign DNA into the host ce
ll is mediated

by a
vector
. Commonly used vectors in genetic
engineering

include viruses and plasmids. In today’s experiment

you will use a
plasmid
to transport the gene
of interest into

the host
E. coli
cells. Because the chances of a successful transform
ation are

small, an
experimental setup that will allow researchers to

identify transformed cells is crucial. One way to separate

the transformed from non
-
transformed cells is by “tagging”

the plasmid with a selectable marker. This is done
by adding

a gene
to the plasmid that confers some type of selective

advantage
like antibiotic resistance

to the
host cells
. For example,

a plasmid containing a gene for

ampicillin resistance (
pAMP
)
can be used
to transform
the
E. coli

bacterium into an ampicillin resistant

strain. Through the

acquisition of this gene,
E. coli

will
become resistant to

ampicillin (an antibiotic similar to penicillin capable of

killing the bacteria) enabling the
bacterial cells to grow in

its presence.


One plasmid that you will use is called
pUC18
. Plasmid
pUC18

contains only 2,686 nucleotide pairs (molecular
weight = 2 x 10
6
). The small size of this plasmid makes it less susceptible to physical damage during handling.
In addition, smaller plasmids generally replicate more efficiently in bacteria and produce larger numbers of
plasmids per cell. Plasmid
pUC18

contains an ampici
llin
-
resistance gene that enables
E. coli
to grow in the
presence of the antibiotic. Bacteria lacking this plasmid, or bacteria that lose the plasmid, will not grow in the
presence of ampicillin. The ampicillin
-
resistance gene codes for the enzyme beta
-
lac
tamase (penicillinase),
which inactivates ampicillin and other
antibiotics in the beta
-
lactam family of antibiotics
.
The
lux

operon is
found in the luminescent bacterium
Vibrio fischeri
and contains two genes that code for luciferase (the
enzyme that catal
yzes the light
-
emitting reaction) and several genes that code for enzymes that produce the
luciferins (the substrates for the light
-
emitting reaction).

The second plasmid’s, the
lux

plasmid, MW is
approximately 4.5 x 10
6
.


In the laboratory, plasmids can b
e introduced into living bacterial cells by a process known as
transformation. When bacteria are placed in a solution of calcium chloride (CaCl
2
), they acquire the ability to
take in plasmid DNA molecules.
It increases the cell competency, the cell’s ability to pick up a plasmid.
This
procedure provides a means for preparing large amounts of specific plasmid DNA, since one transformed cell
gives rise to clones that contain exact replicas of the parent plasmi
d DNA molecule. Following growth of the
bacteria in the presence of the antibiotic, the plasmid DNA can be readily isolated from the bacterial culture.


General
Procedure

Summary

In this exercise, plasmid
lux
and a control plasmid (
pUC18
) will be introduce
d into
E. coli
by

transformation. There are four basic steps to the procedure:


Treat bacterial cells with CaCl
2

solution in order to enhance the uptake of plasmid DNA. Such

CaCl
2
-
treated cells are said to be “competent.” (This step should be performed by
the instructor

before or during the laboratory session.)

Incubate the competent cells with plasmid DNA. Select those cells
that have taken up the plasmid DNA by growth on an ampicillin
-
containing

medium. Examine the cultures in
the dark.



Procedure


A.
Preparation of competent cells (These steps were performed by the instructor)

1. Place a vial of CaCl
2

solution and the tube of
E. coli
in the ice bath.

2. Using a sterile pipet, transfer
590

µL

CaCl
2

solution to the tube containing
50

µL of
the bacteria.

3
. Tap the
tube

with the tip of your index finger to mix the solution.

4
. Incubate the cells for
at least
10 minutes on ice. The cells are then called competent because they

can take
up DNA from the medium. If desired, the cells can be stored in the CaCl
2

solution for

12

24 hours.

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B. Uptake of DNA by competent cells


Note:
There are 2 plasmids involved in this experiment each group will only use 1 of the 2 plasmids.
The
instructor will assign which plasmid your group will use.


1. Label one small Eppendorf
tube “C” (for control plasmid DNA) or one tube “lux” (for plasmid
lux
DNA)

2. Place both tubes in an ice bath.

3. Using a sterile micropipette, add 5 µL control plasmid to the tube labeled “C”
or

5 µL plasmid
lux

to the
tube labeled “lux”. Make sure to kee
p all tubes in the ice until instructed otherwise

(concentration =
0.005μg/μL)
.

4.
Gently tap the tube of competent cells with the tip of your index finger to ensure that the cells are in
suspension.

5. Using a sterile transfer pipet, add
70

µL of the competent cells to each of the two tubes.

6. Tap each of these tubes with the tip of your index finger to mix these solutions, and store both tubes on ice
for 15 minutes.
During this time, one member of the group should obtain
one

additional tu
be.

Add
35

µL
of competent cells to each tube and label the tubes “NP” (no plasmid).

Every group will have a no plasmid
tube assigned to them.

7.
HEAT SHOCK:
Transfer
all
the tubes to a water bath preheated to 37ºC, and allow them to sit in the bath
for 5
minutes.

Make sure you are able to identify which tubes below to your group before you place them
in the water bath.

8. Use a sterile pipet to add
275

µ
L nutrient broth to
the control and lux

tube
s

and

150 µL of nutrient broth
into the no plasmid tube.

9.
I
ncubate the tubes at 37
°
C

for 45 minutes.

Use this time to make your prediction (Table
9.
1) and answer
questions.


C. Selection of cells that have taken up the plasmid by growth on an ampicillin containing medium


Note: There are a total of 6 plates per
trial. Each group will be working with 3 plates.

The instructor will
assign the plates your group will work with.


1.
Each group will obtain 3 agar plates from the instructor. Label the
plates as indicated in Figure 1. Keep in
mind the instructor assigned
your group the
treatments

you will be working with.

2. Using a sterile pipet, remove
1
3
0 µL

mixed
bacterial suspension from the “C
” tube,

remove the lid from the “
Control
” plate, and dispense the bacteria onto the agar. Use a

cell spreader

to
spread the ba
cteria evenly onto the agar surface.


-

Use of the cell spreader:



a. Dip the cell spreader in ethanol. Pass the spreader across the flame of the ethanol lamp. Make sure
you only pass it through the flame and not keep it in the flame. Once the ethanol has

burned off keep the
spreader still for about 30 seconds. This allows the spreader to cool down before you start spreading the
cells. Once the spreader has cooled use the spreader to evenly distribute the cell suspension over the
entire surface of the plat
e. Return the cell spreader to the ethanol contained
without flaming

and repeat
the same procedure until you have plated al the bacteria.

3. Transfer
1
3
0 µL
bacterial suspension from the “
lux
” tube to the “
lux
” plate and spread

these cells onto the
agar
surface as described in the previous step.

4. Cells from the
tubes

that did not contain plasmids (NP) should be plated onto two plates

(
NP) as described
in steps 2

3.

5. Replace the lids on the plates, and leave the plates at room temperature until the liq
uid has been

absorbed
(about 10

minutes).

6. Invert the plates and incubate them
at 37
°
C
.


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Figure
9.
1. Petri Dish Label

Same figure as in the lab manual (Figure 10.3 pp.110)


D. Examine cultures in the dark


1.

Retrieve your group’s plates from the
refrigerator.

2.

Open each plate, one by one, to determine if
E. coli

growth occurred. If growth occurred, note the growth
type (lawn or colonial). Record your results

in Table 2
.

3.

Allow at least 3 minutes for the eyes to adjust to the dark in a light
-
free ro
om. View your plates and the
plates of your classmates in the dark and then in the light. Record your results in the following table.
Were the results as expected? Explain possible reasons for variations from expected results

(Table 2)
.


4.

Calculate Transfor
mation Efficiency and answer questions.



Data Analysis:


Table 1.
PREDICTIONS:

Treatment

Expected Result

(Growth or No Growth)

Bioluminescence (yes
or no)

Reason For Expectations


LB
c






LB/Amp
c






LB
NP






LB/Amp
NP






LB/Amp
lux






LB
lux











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Figure 9.2. Experimental Predictions





Figure
9.3
. (a) Lawn Growth, (b) Colonial Growth. Note: The colonial growth exhibited in this figure is
seen in two large patches. Each patch holds hundreds of individual colonies.

Same figure as in the lab manual (Figure 10.6 pp.112)



Based on your predictions, state your
scientific,
null and alternative hypotheses regarding what you
expect to see if the
bacteria plated on the ampicillin rich medium were successfully transformed.

Scientific:



H
o
:



H
a
:


LB
c

LB/Amp
c

LB
NP

LB
lux

LB/Amp
lux

LB/
Amp
NP

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Questions:


1.

What role did the following compounds or steps play in bacterial transformation?

a.

Calcium Chloride



b.

Heat Shock



c.

Agar



d.

LB Broth



e.

Ampicillin



f.

Pl
a
smid



g.

Aseptic Techniques



2.

What are you selecting for in this experiment? (i.e., what allows you to identify which bacteria have
taken up the plasmid?)




3.

What does the phenotype of the transformed colonies tell you?





4.

What one plate
would you first inspect to conclude that the transformation occurred successfully?
Why?





5.

In nature, DNA uptake by different organisms can impart advantages as well as disadvantages to the
host organism.

a.

When would genetic transformation be advantageous
to a host organism?




b.

When would genetic transformation be maladaptive to a host organism? What consequences
would there be for the host cells?




c.

Can you think of a case where the uptake of foreign DNA would be advantageous for the
organism, but
disruptive for the surrounding ecosystem?

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Table
9.2
.
Results:



Treatment

Observed Growth Type

Bioluminescence (yes
or no)

Reasoning for observed
results


LB
c







LB/Amp
c







LB
NP







LB/Amp
NP







LB/Amp
lux







LB
lux

































LB
c

LB/Amp
c

LB
NP

LB
lux

LB/Amp
lux

LB/Amp
NP

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Transformation Efficiency
: (Use the left side of the division to calculate the first transformation efficiency for
the
LB/Amp
c
and use the right for the second set of calculation for the
LB/Amp
lux
)
.

Transformation efficiency calculations result in very large numbers. It is normally written in scientific
notation. For example, if the calculated transformation efficie
ncy is 1000 ba
cteria/
μ
g of DNA it should be
reported as 1 x 10
3

transformants/
μ
g. Suppose that an efficiency is calculated as 5000 bacteria/
μ
g of DNA.
This would be reported as 5 x 10
3

transformants/
μ
g.

6.

The total amount (µg) of plasmid DNA used can be calculated with
the following formula.


µg DNA = concentration (µg/ µL) of DNA x volume of DNA (µL)











7.

Calculate the total volume of cell suspension prepared in the control DNA tube.


Total volume (µL) = amount (µL) of plasmid + amount (µL) of LB









8.

Since
only a portion of the control DNA solution was added to the LB/Amp
C

plate, you will need to
calculate the fraction of DNA spread onto this plate using the formula below:













(

)















(

)



















9.

Using the values obtained for questions 3
-
5, determine the actual amount of DNA (µg) present on
your plate


Total amount (µg) of DNA = µg of DNA x Fraction of DNA spread




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10.

Calculate transformation efficiency






















































11.

Repeat questions 3
-
7 to calculate the transformation efficiency for the













12.

Compare and contrast the number of colonies on each of the following pairs of plates. What does
each pair of results tell you about the experiment?

a.

LB
C

and LB
NP




b.

LB/Amp
NP

and LB
C




c.

LB/Amp
C

and LB/Amp
NP




d.

LB/Amp
C

and LB
C




e.

LB/Amp
lux

and LB
NP




f.

LB/Amp
lux

and LB/Amp
NP




g.

LB/Amp
C

and LB/Amp
lux




h.

LB/Amp
lux

and LB
C



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13.

Do your results agree with your predictions in Table
9.
1?

Explain.








14.

What factors could have
affected transformation efficiency?








15.

What transformation efficiency would you expect for all the other treatments?








16.


Why do the cells transformed with
pUC18

and plasmid
lux
grow in the presence of ampicillin?








17.

Name one enzyme that is produced by cells transformed with plasmid
lux
that is not produced by the
cells transformed by
pUC18
.








18.


Remembering that plasmid size will affect the efficiency of transformation, which plate would be
expected to show the few
est colonies?








19.

How would you improve or extend this experiment?