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What factors can affect the determination of the rate of mutagenesis? How do they affect it?

I really have no idea. Help is appreciated.

2007-07-17 14:18:52 · 3 answers · asked by Anonymous in Science & Mathematics Biology

3 answers

Fall 2006, W2501
Transposition Mutagenesis
I. Introduction
This experiment allows us to look at “jumping genes,” a phenomenon that first came to
light in Barbara McClintock’s work with corn. Later it was found that this phenomenon occurs
across species — one such species is the bacterium E. coli. Fundamentally there are two parts
to this experiment: conjugation, which, in this experiment, will allow a specific plasmid to
move from one bacterium to another; and transposition mutagenesis, a process that, here,
induces a gene to move from the plasmid into the bacterial chromosome itself, and in so doing
has the potential to “jump” into a gene we can assay, i.e., the “jumping” gene can interrupt the
expression of a known gene on the bacterial chromosome.
In the first part of the experiment, we will mate a donor strain of E. coli with a recipient
strain of E. coli. The donor strain contains two different plasmids, a “helper” plasmid and an
artificially constructed plasmid named pVJT128 (for Valerie J. Thompson, a graduate student at
Columbia Physicians & Surgeons). The helper plasmid contains the tra (for transfer) genes
that, when expressed, allow a pilus to develop through which a strand of DNA can travel (just
as in the bacterial conjugation experiment you just performed). However, this helper plasmid
has a mutated oriT (origin of transfer) which makes it impossible for the helper plasmid itself to
be transferred to the recipient strain. The other plasmid, pVJT128, on the other hand, has an
intact oriT and can readily be transferred. In addition, pVJT128 has a chloramphenicol (an
antibiotic) resistant gene (conferring chloramphenicol resistance on the E. coli cell that contains
this plasmid) and has built into it a kanamycin resistance transposon (kanamycin is another
antibiotic). The gene conferring kanamycin resistance, however, is “silent” until it is transposed
from the plasmid to the larger bacterial chromosome in such a way that it can be expressed.
The recipient strain contains no plasmids, but is nalR (resistant to naladixic acid, which
is another antibiotic). This is a different situation than that with the donor strain because in this
case the antibiotic resistance lies on the bacterial chromosome itself, not on a plasmid, and is
thus more stable. By mating the donor and the recipient, we are allowing the pVJT128 plasmid
to be transferred from the donor to the recipient by means of the helper plasmid. If our mating
is successful, we should be able to select for those recipient colonies that received the plasmid
pVJT128 and are therefore both chloramphenicol and naladixic acid resistant.
In the second part of the experiment we will do a transposition mutagenesis procedure
in which we induce the previously silent kanamycin gene to “jump” from the plasmid pVJT128
into the bacterial chromosome of the recipient. If induction is successful, the kan gene will be
expressed, and the E. coli cell will now be kanamycin resistant. We should be able to select for
these cells by plating them on a medium that contains kanamycin. At the same time we will try
to discover where the kanamycin gene “jumped.” We will be looking at the expression of the
gene coding for β-galactosidase, the enzyme that splits lactose into its component parts, glucose
and galactose. When lactose is present in a medium, the enzyme β-galactosidase is produced,
which breaks the lactose molecule into glucose and galactose. When there is no lactose present
in the medium, the gene is not expressed, the enzyme is not made, and lactose is not metabolized.
It turns out that we can substitute another compound for lactose so that the lactose is not “used
up.” The substitute compound is X-gal, a compound consisting of pigment X and galactose. The
enzyme β-galactosidase recognizes the bond between pigment X and galactose as if it were lactose,
and splits X-gal into galactose and pigment X, which is blue in color and turns the colony
blue. Thus if the gene is being expressed, β-galactosidase is being made, and we should see blue
colonies because pigment x is being liberated from its bond with galactose. If, however, the
kanamycin gene has been transposed into the middle of the β-gal gene, then β-galactosidase will
not be made, the X-gal compound will not be split, the blue pigment will not be liberated, and the
colony will look white. These white colonies will provide the evidence that the kanamycin gene
has “jumped” into the middle of, and deactivated, the β-gal gene and will thus be a sign of mutagenesis.
We will examine the mechanics of this process in class.
II. Protocol, Step 1: Conjugation
Mating Procedure
Use sterile technique to perform the following procedure:
1. Briefly vortex the donor and recipient tubes to resuspend the cells.
2. Put 500 μl each of the donor and the recipient into a new tube (this is the mating tube). Mix
gently by inverting the tube several times (don’t vortex here, just mix gently but thoroughly).
3. Incubate at 37 degrees C for 60 minutes. Make sure the tube is pressed securely in place in
the floater. The hinge of the tube should fit in the tongue of each hole in the floater. Since you
will have only one tube, balance it with another, empty, tube, or with the tube from another
group.
4. During incubation, make serial dilutions, as we did during the conjugation experiment, of the
recipient cells to 10-7, and plate the 10-5, 10-6, 10-7 dilutions of each onto nutrient agar (4 red
stripes), using 100 μl of each tube to spread onto each plate.
2
5. Plate 100 μl of the 100 donor tube onto a Nal plate (1 black and 1 blue stripe).
6. Plate 100 μl of the 100 recipient tube onto a Cm plate (1 black and 1 red stripe).
7. At the end of the 60-minute incubation, vortex the mating tube for 60 seconds (to break the
mating pili).
8. After vortexing each mixture, make serial dilutions of the mating mixture to 10-3, and plate
100 μl straight from the mating tube (100) and 100 μl of each dilution (10-1, 10-2, 10-3) onto
Cm + Nal plates (1 blue and 2 red stripes). You will have a total of 4 plates for this step.
9. Place your plates, agar side up, in the large 37 degree C incubator on the side of the room.
10. Next week we will see what happened.
III. Protocol, Step 2: Transposition Mutagenesis
In this part of the experiment, we will use colonies from the Cm + Nal plate, 100 dilution,
which we assume to be cells that have received the plasmid pVJT128 from the donor, and
“induce” the kan gene to “jump” from the plasmid DNA into the bacterial chromosome (this is
the transposition part). As we do this, we will try to see where the kan gene is landing by
selecting for colonies whose β-gal gene has been interrupted (this is the mutagenesis part).
Transposon Mutagenesis Procedure
Remember, of course, to do all of this procedure sterilely.
1. To create the inducing broth, add 1 μl of IPTG and 1 μl of Cm to 1000 μl of nutrient broth.
Then put a loopful of cells from the 100 Cm + Nal plate. Take only enough cells to create a
solution with a skim-milk appearance; a very cloudy solution will contain too many cells later
on. The IPTG in the inducing broth is a lactose substitute and is used to induce the genes that
will transpose the silent kan gene from the plasmid to the bacterial chromosome. The chloramphenicol
is added to provide the “pressure” that helps the cells keep pVJT128.
2. Vortex briefly to resuspend your cells. The resulting mixture should be uniformly cloudy.
3. Incubate the suspension in a 37 degree C water bath for 60 minutes.
3
4. At the end of the incubation period, take your tube out of the water bath and vortex it
briefly. Put 100 μl of the mixture onto an X-gal + kan plate. This will be the 100 plate.
5. Make serial dilutions of the mutagenesis mixture to 10-7.
6. Spread 100 μl from each of the 10-1, 10-2, 10-3 dilution tubes onto X-gal + kan plates (one
black stripe on the side).
7. Spread 100 μl from each of the 10-4, 10-5, 10-6, and 10-7 plates onto nutrient agar plates
(four red stripes on the side).
8. Label your plates on the agar side with your name and section, and incubate them, agar side
up, in the large incubator on the side wall on the shelf marked with your section. Next week
we will look at the results.
9. To clean up, put your inducing tube, dilution tubes, and saline tube in the hazardous waste
basket by the side of the room. Put the micropipettors, floaters, tips, tubes, and test tube racks
on the front desk. Leave the vortexers on your bench. Wipe the bench down with EtOH and
remember to wash your hands.
4
W2501, Fall 06
Transposition Post-Lab Questions and Lab Report Guidelines
Post-Lab Questions
1a) What is a transposon?
b) What is the transposon in this experiment?
2a) In the first part of the experiment, why did we put the donor on a plate containing naladixic
acid?
b) Why did we put the recipient on a plate containing chloramphenicol?
3) What is the function of the tnp gene in pVJT128?
4) What is the purpose of the lac regulatory region that was engineered into pVJT128?
5) Give three reasons why kan is not expressed by pVJT128?
6a) What is the purpose of the helper plasmid in the donor?
b) Can the helper plasmid be transferred to the recipient? Why or why not?
c) Can pVJT128 be transferred from our recipient strain back to our donor strain? Why or
why not?
7) Why did we choose an E. coli strain with a lactose operon in its bacterial chromosome as
the host for our transferred pVJT128?
5
Transposition Mutagenesis Lab Report Guidelines
Follow the Lab Report Guidelines for the Title, Abstract, and Introduction sections. For
this report only, combine the Results and the Discussion sections into one section, called the
Results and Discussion section. Please use the following questions and comments as a guide to
the form that this section should take.
1) This experiment was divided into two parts. Diagram or outline the procedure used in each
part. Briefly explain what each step accomplished or what information it provided. You have
probably already completed this step in the flowchart you turned in. You can use that flowchart
for this part.
2) Create a table showing your results for Part I and Part II of this experiment.
3) What was the frequency (or rate) of conjugation?
To answer this question, divide the number of cells in the mating tube that received the
plasmid pVJT128 (to calculate this number, use the number of colonies you found on the cm +
nal plates, and work backwards to get the number of cells in the mating tube) by the number of
recipient cells that you estimate were in the mating tube (use the nutrient agar plates).
Remember to take into account any spontaneous reversions that you may have seen in the
donor (Nal plate) or the recipient (Cm plate). Be sure to show calculations or samples of calculations
where appropriate.
4) What was the efficiency of transposition?
To calculate this, divide the number of successfully transposed cells by the total number
of cells in the incubation tube. The number of transposed cells can be found from the number
of blue plus white colonies on the X-gal + kan plates. The total number of possible cells that
could have been transformed can be found from number of colonies found on the nutrient agar
plates (remember that the cells in the incubation tube were taken from a nal + cm plate and
were therefore the recipients that had received the plasmid pVJT128 and so are potentially
6
Fall 06
transposable). Of course you will have to calculate backwards to the original tube, as you did
in question 3, to get the total number in the incubation tube.
5) What was the frequency (or rate) of mutagenesis?
This is calculated by dividing the number of cells (in the incubation tube) in which the
X-gal gene was interrupted (represented by white colonies) by the total number of cells that
were transformed (i.e., represented by blue + white colonies). If your rate is zero, then report
that.
6) List three factors that could affect our determination of the rate of mutagenesis and explain
why how each would affect it.
7) Based on your experiment, how would you characterize the size of the β-gal gene (a large
fraction of the bacterial genome, a small fraction)? On what do you base your decision?
Note: Please include your own raw data, and don't forget to include a reference section, if
appropriate.
7
The Streptococcus faecalis transposon Tn917 was introduced into Bacillus subtilis by transformation of competent cells with the plasmid pAM 1::Tn917 and was tested for transposition activity by selection for insertions into the temperate phage SPß . Insertions were obtained at a frequency indicating relatively efficient movement of the element, and Southern hybridization analysis of a particular insertion confirmed it to be the result of a genuine transposition event. A restriction fragment from pAM 1::Tn917 containing the transposon sequences was ligated into a temperature-sensitive plasmid (pBD95), and transpositions into the B. subtilis chromosome were selected by requiring the transposon drug resistance to be maintained at temperatures nonpermissive for plasmid replication. Insertions have been recovered at many chromosomal sites, including ones that produced auxotrophy of different kinds and ones that produced various different sporulation-defective phenotypes, indicating good prospects for the use of Tn917 as a tool for insertional mutagenesis in B. subtilis.

2007-07-21 14:08:20 · answer #1 · answered by krys 1 · 0 1

It's really great that you want to do well. And, obviously, you usually reach your goals. That's just great! But, learning to deal with failure (although I'd hardly call this failure) is an important part of maturing. When you get to college, you'll find that there will be courses that will be even harder to ace. If you get an occasional B, don't worry, noone but you will even notice. Remember, that B in Advanced Biology is worth more than an A in the regular Biology class. So cut yourself some slack and have a little fun over the summer.

2016-03-15 06:03:31 · answer #2 · answered by Anonymous · 0 1

UV light increases the rate of mutation. http://www.genetics.org/cgi/content/full/149/3/1173

More info on other factors here:
http://www.ias.ac.in/jbiosci/december1999/article17.htm

2007-07-17 15:22:03 · answer #3 · answered by ecolink 7 · 0 0

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