Download 1) - life.illinois.edu

1). (10 Points). Transposable elements are useful tools for performing genetic
experiments.
(5 Points). List 5 different uses of transposon insertions.
Transposons can be found at a large number of sites in a chromosome (providing linked,
selectable markers).
Interrupted genes (almost) always are inactive.
The phenotype of the insertion is completely linked to drug resistance in genetic crosses.
Insertion mutations can revert by precise excision and loss of drug resistance.
Insertions are usually polar.
Transposons provide a portable region of homology.
Mud phages and other elements can be used to make operon and gene fusions.
Others possible.
(5 Points). Describe 2 different ways that have been used to introduce transposons into
cells.
Phages that contain the transposon but cannot replicate in the host
Suicide plasmids (introduced by conjugation or electroporation)
EZ Tn (introduced by electroporation)
Others possible
2). (15 Points).
i). (5 Points). Plasmids often have oriT and oriV sites. What is the function of each?
oriT sites are sites used for nicking of DNA in preperation for DNA transfer to a
recipient. oriV sites are used for initiation of bidirectional DNA replication.
ii). (5 Points). Do the sites function in cis or in trans (or both)?
oriT sites can function in both cis and trans. The site is the important for nickase protein
recognition. OriV sites only work in cis because they are the sites for initiation of DNA
synthesis.
iii). (5 Points). E. coli mutants that have a temperature sensitive mutation in the dnaA
gene (dnaATS) can initiate chromosomal DNA synthesis at 30o C but not at 42o C. Thus
cells are inviable at 42o C. If one grows cells containing a dnaATS mutation and the F
plasmid at 42o C, some rare colonies grow. What event likely accounts for this
observation? What is an obvious genetic prediction from your answer?
Most likely F has integrated into the host chromosome to form an Hfr. The host
chromosome is replicated by initiation of DNA synthesis at the oriV site. The Hfr should
transfer chromosomal genes to an F- strain in a conjugation experiment. This is a real
technique for isolation of Hfs called integrative suppression.
3). (20 Points). CTnDOT is a conjugative transposon originally found in Bacteroides. Work on the
element produced the DNA sequence (attDOT) of the region involved in site-specific recombination of the
element. The DNA sequence of the attDOT site is on the last page of the exam. (Just rip it off to use).
Analysis of the sequence attDOT and bacterial (attB) sequences showed that the recombination occurs
between attDOT and attB by staggered cleavages seven base apart on each att site. The sites of cleavage in
attDOT are shown between the D and D’ sites in the sequence.
In vitro experiments indicated that the IntDOT integrase, which catalyzes the reaction, binds to two classes
of sites in attDOT. One class, called core type sequences, are represented by the D and D’ sequences on
the sequence. These sites are thought to be sites where IntDOT binds to cleave the DNA during strand
exchange. A second class of sites, called arm sites, are labeled as R1, R2, R2’, L1 and L2 in boxes in the
sequence. These sites are thought to be bound by IntDOT to form nucleoprotein complexes required for
either integrative or excisive recombination. If that hypothesis is correct, mutations of the sites should
affect either the integration or excision reaction because the protein won’t recognize the mutant site. For
example, mutation of L1 might abolish integration but have no effect on excision etc.
In order to test this hypothesis, you decide to mutate the L1 site and test the mutant attDOT site in
integration and excision assays. Assume that you have attDOT cloned on a pUC-based vector and that you
have assays for both integration and excision of any mutant you make.
i). (10 Points). You decide to use the Quickchange method to change all six basepairs in the box in the L1
site using oligonucleotide mutagenesis. Design the sequences of the oligonucleotides showing the changes
in the L1 site and the six bases on the 3’ and 5’ sides of the box. Make sure to clearly label the 5’ and 3’
ends of the oligos. In a real experiment how long would the oligos be?
Oligo #1
5’ TAATAGXXXXXXATTTAG 3’
Oligo #2
5’ CTAAATZZZZZZCTATTA 3’
X and Z should be bases that differ from ones in the L1 site. The oligos should be around 50 bases long.
ii). (5 Points). Why is the template grown in a dam+ cell?
To modify the A in GATC sequences in the plasmid template.
iii). (5 Points). Why is the DNA treated with DpnI after the DNA synthesis step?
DpnI cleaves dam-modified DNA but not newly synthesized, unmodified DNA. Thus the parental DNA
template is degraded by the enzyme but newly synthesized DNA is resistant. This enriches for mutants.
4). (10 Points).
i). (5 Points). You are part of a team that is doing a metagenomic analysis of the bacterial population of
the human colon. You notice that among the sequences are partial sequences of the tetracycline resistance
gene tetQ. You also notice fragments of sequence data that are identical to the rteA gene found
immediately downstream of tetQ on CTnDOT. You have the original sample of DNA from which the
sequencing was done. How could you use PCR to show that the tetQ gene is carried on CTnDOT? What
control would you do?
Since the sequence of CTnDOT is known, you could design PCR primers to amplify possible CTnDOT
DNA in the sample. The best way to do this is to design primers that bind in tetQ DNA and in adjacent
DNA of CTnDOT. A control would be a sample containing known CTnDOT. You should get identical
size fragments from each sample. (Oligos that amplify just tetQ or CTnDOT are not as useful because the
DNA fragments may not be derived from the same templates in the sample).
ii). (5 Points). Assume that the original sample was collected in a way that preserves RNA. How could
you determine whether tetQ is being expressed?
Quantitative RT PCR using oligos that amplify tetQ would show that the mRNA was present in the
sample.
5). (10 Points). The man operon is required for E. coli to grow on the carbohydrate mannitol as a sole
carbon source. In the absence of mannitol a man::MudJ(lac, KanR) operon fusion expresses very low
levels of -galactosidase. However, when mannitol is added to the medium, the fusion expresses very high
levels of -galactosidase.
i). (4 Points). How could you use this man::MudJ (lac, KanR) fusion to isolate mutations that affect
regulation of the man operon?
Mutagenize the strain and look for colonies that showed increased or decreased levels of -gal by screening
on MacLac plates or plates supplemented with X-gal.
ii). (2 Points). If you isolated a mutant with a mutation that resulted in decreased expression of Gal,
what could you conclude about the mechanism of regulation of the wild type man operon?
(Credit for either answer). Because MudJ makes operon fusions the results indicate that the mutation acts
at the level of transcription. This could be a mutation that affects the promoter resulting in a decrease in
transcription of the operon. Alternatively it could affect a protein that activates the operon “in trans” most
likely by interacting with RNA polymerase at the promoter site.
iii). (4 Points). How could you tell whether the regulatory mutation is within or near the
man operon or elsewhere in the chromosome. Would the results help in your hypothesis
on the mechanism of regulation of the man operon?
Backcross the mutant with the wild-type parent strain containing the native man operon.
This is done by making a P1 lysate of the mutant strain and using it to transduce the
man+ strain to KanR. Then screen for LacZ expression.
a). If the regulatory mutation is unlinked, all of KanR transductants should retain the
wild-type srl::MudJ phenotype.
b). If the regulatory mutation is linked to the srl operon you should see 2 classes of
KanR transductants. One will have the wild type srl::MudJ phenotype and the second
will have the phenotype of the regulatory mutation in the srl::MudJ (original mutant
phenotype).
Thus, a promoter mutation will be linked to the srl::KanR. An activator mutation may or
may not be linked to the srl::KanR depending upon where it maps in the chromosome
relative to the srl::MudJ.
6). (10 Points).
If one clones a DNA fragment containing an internal sequence of a gene into a suicide
vector, the construct can be integrated into the bacterial chromosome by homologous
recombination.
i). (5 Points). If you cloned an internal fragment of the trpA gene into a suicide vector
and integrated it into the chromosome what would the resulting integrant look like? Use
a diagram to illustrate your answer. Would the integrant be Trp+ or Trp-? Why?
The integrant will contain two disrupted and truncated copies of trpA. One copy would contain the 5’ end
of trpA with the 3’ end missing and the other copy would contain the 3’ end of the gene with the 5’ end
missing. It would be Trp-.
ii). (5 Points). If you made a mistake and the cloned DNA contained the trpA promoter
and 5’ coding region of the protein, what would the resulting integrant look like? Would
it be Trp+ or Trp-? Why?
The integrant would contain a truncated copy if the DNA containing the promoter and 5’
coding region of trpA and a second copy containing the promoter and an intact copy of
trpA. Thus the resulting integrant would be Trp+.
7). (15 Points). Consider the two bacterial chromosomes shown below. The lacZ33
allele is a missense mutation that makes the cell unable to use lactose as a carbon source.
The proC61 allele is a missense mutation that makes the cell a proline auxotroph. Tn10
is a transposable element that encodes resistance to the antibiotic tetracycline.
a). (5 Points). Given the two strains shown below, how could you construct a strain that
has a tandem duplication with one copy of the lacZ33 allele and one copy of the lacZ+
allele. Draw a diagram showing the recombination events and indicate how you would
select or screen for the tandem duplication.
b). (5 Points). How could you select for cells that maintain the duplication?
c). (5 Points). In the absence of selection, the duplication is unstable and segregates at a
frequency of about 1% per generation. Two classes of proC61 segregants are obtained.
Draw a diagram showing how the segregation occurs and indicate the phenotypes of the
two types of proC61 segregants.
Transduce strain (ii) with a lysate grown on strain (i) selecting for TetR Pro+ Lac+. The
resulting duplication will be made.
Prevention of segregation could be done by selecting for tetR or growth on lactose or
growth in absence of proline.
The segregation by homologous recombination on either side of the lacZ33 allele will
yield segregants with the proC61 allele. Depending upon the side of lacZ33 the
recombination event occurs the segregants will be lac+ or
lacZ33.
8). (10 Points).
You have discovered a new transposon called Tn88. DNA sequence analysis indicated
that it contains inverted repeats of an IS element called IS88. In the diagram below, the
copies are called IS88L and IS88R. The element also contains a gene that confers
resistance to the antibiotic tetracycline (tetR). You introduced a copy of the lacZ gene
(lacZ+) into one copy of the element and transposed it onto phage lambda containing
amber mutations in the lambda cI, int, and O genes (so the phage can’t grow lytically or
integrate into the chromosome by site specific recombination in a suppressor free strain).
The element is called Tn88-lacZ+ shown on the top of the diagram below.
You also isolate a mutant that is identical to the lambda-Tn88-lacZ+ except that it
contains an amber mutation in the lacZ gene. This phage is called lambda-Tn88-lacZam
and is shown below. Thus the two phages are completely isogenic except for the
presence or absence of the lacZam mutation.
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Next you want to determine if the mechanism of transposition is “replicative” or “cut and
paste”. To do this, you repeat the experiment done by Bender and Kleckner for the
analysis of transposition of Tn10.
You perform the following experiment.
Phage DNA from each phage is purified by phenol extraction of phage particles. Equal
molar amounts of the 2 phage DNAs are mixed and denatured by treatment with base
followed by neutralization and reannealing of the DNA. The DNA is then packaged into
lambda phage particles with an in vitro packaging kit to produce infective phage.
Cells containing a chromosomal deletion of the lac operon and lacking any nonsense
suppressors are infected with the phage and colonies that are tetracycline resistant on LBtetracycline-Xgal plates are selected. When you examine several hundred colonies, you
find that all the colonies are completely white or completely blue. None of the colonies
contain blue or white sectors.
i). (5 Points). Which mechanism of transposition does this experiment support? Why?
(Use a diagram or precise language to explain your answer).
The result supports replicative transposition because a transposition event will transfer an
element containing a single strand of Tn88-lacZ which is replicated by DNA synthesis to
produce either lacZ+ or lacZ- cells. The other product is lost when the phage is lost from
the cell.
ii). (5 Points). The experiment does not prove that heteroduplex phages were actually
made during the experiment. For example, if phage DNA was not denatured by base, no
heteroduplexes would be formed and that would produce the results above.
How could you show that heteroduplex phages were actually formed by a simple genetic
test? (Hint: It should have been a control for the experiment described above).
An easy way to test this is to infect a strain containing a lac deletion and a lambda
lysogen and selecting for TetR colonies. They will arise by homologous recombination.
(This will be much more frequent than transposition.) The heteroduplexes will make
sectored colonies because the two cells formed after the first cell division will contain lac
and lac+ chromosomes, respectively.
Alternatively
Infect cells that contain a suppressor that suppresses the int amber mutation and the cI
amber mutation (but not the lacZam mutation) and select for tetR lysogens. The
lysogens will be much more frequent than transposition events and would be made by
site-specific recombination. Colonies from phages that contained heteroduplexes will be
sectored because replication of the lysogen will produce a lacZ+ cell and a lacZ- cell after
the first cell division. The resulting colony will theoretically be half blue and half white.
If heteroduplexes were not made lysogens will be entirely white or blue because the DNA
was not denatured and all the DNA molecules were homoduplex.