1). (10 Points). E. coli can use maltose as a carbon source. The malE and malF genes are required
for transport of maltose into the cell. The malE gene encodes a periplasmic protein and the malF
gene encodes a cytoplasmic membrane protein.
TnphoA gene fusions were isolated in the malE and malF genes in a host deleted for the
chromosomal phoA gene. Some TnphoA insertions in malE and malF express alkaline phosphatase
activity and some do not express alkaline phosphatase activity.
A). (5 points). List 2 reasons why some malE::TnphoA insertions might not express alkaline
phosphatase activity.
ANSWER: Either inserted in the wrong orientation or wrong reading frame. (Note that this
protein is exported into the periplasm, so TnphoA insertions will be exported into an
environment where S-S bonds can form.)
B). (5 points). One of the malF::TnphoA insertions did not express alkaline phosphatase enzyme
activity but reacted with antibody against alkaline phosphatase. What is the most likely explanation
for this phenotype?
ANSWER: The join-point between the integral membrane protein and the phoA fusion places the
phoA protein on the inside of the cytoplasmic membrane -- thus, the PhoA protein is made but it
is unable to properly form S-S bonds and dimerize, so it remains inactive.
2). (10 Points).
You are studying an interesting operon, the sitAB operon, that contains two genes required for
virulence in Salmonella. You have constructed operon fusions with the galK gene (encoding
galactose kinase) and gene fusions with the lacZ gene (encoding ß-galactosidase) for each of the
two genes, sitA and sitB. The expression of each fusion was assayed with or without exposure to
superoxide (superoxide is an oxygen radical that pathogens can encounter in hosts). The results are
shown in the following table.
What do the results indicate about the regulation of the sitA and sitB genes? Briefly explain your
answer.
(5 Points). (sitA is regulated ten-fold at the level of transcription because we see an increase in
reporter (GalK) gene activity in the operon fusion. It is not regulated translationally
because, although we see a difference in the gene fusion, the ratio is the same as in the
operon fusion, so we know that the effect can be entirely accounted for by transcriptional
regulation.
(5 Points). sitA is not regulated transcriptionally because there is no difference in reporter gene
activity in response to superoxide in the operon fusion. sitA is regulated translationally in response
to superoxide because we see a ten-fold increase in reporter gene activity in response to
superoxide. Since there is no increase in the operon fusion, we know that the entire difference is
due to translational regulation).
R
3). 10 Points). A new mutant was isolated that is Str and unable to use proline as a
carbon source (put). To determine where the mutation maps, it was mated
S
+
with the four different Str put Hfr donor strains shown below.
[Arrowheads indicate the location and direction of transfer from each different Hfr.]
A). (2 Points). What is the selection for exconjugants in this experiment?
ANSWER: Growth on proline as a sole C-source.
B.) (2 Points). What is the counterselection against the donor cells in this experiment?
ANSWER: Streptomycin resistance.
C). (6 Points). Given the results in the following table, where does the put mutation
map? [Indicate the approximate map position in minutes.]
ANSWER: About 40 min or near that region, in the region transferred early between Hfr-2 and
Hfr-1 but closer to the origin of Hfr 2.
4). (20 Points). 8. Site-directed mutagenesis was done to determine if a "helix-turn-helix" motif in the
PutA protein is required for DNA-binding
A). (5 Points). Given the following sequence, what oligonucleotides could you use to obtain a
double mutant that changes both val → ala and asp → ala? Specify the 5’ end of your
oligonucleotide. You are using the Quick-change technique to enrich for your mutant.
Although longer oligos would actually be used in a real lab experiment, design the oligos you
would use from the DNA sequence shown below. [A genetic code table attached to the back
of the exam.]
ANSWER: One example of a top strand oligi is shown below.:
3’ CGA TAC TAA CGG CGG AGA AGG CGA 5’
B). (10 Points). Briefly describe how the Quick Change procedure enhances isolation of the
desired recombinants.
ANSWER: 2 Points for each concept below
The Quick-Change procedure uses complementary oligos that have the desired mutation(s)
incorporated into their sequences. In this case there needs to be information included to
make 2 consecutive Ala codons so at least 2 bases have to be changed.
The oligos are annealed to denatured template plasmid DNA containing your gene of
interest and the DNA is copied by a DNA polymerase to give completed complementary full
length complementary single strands containing the mutation.
When the annealed DNA is introduced into a recipient cell, the nick are sealed by host
DNA ligase.to make a double stranded plasmid.
The actual formation of the desired plasmid is fairly rare so an enrichment is included. The
template is grown in a dam+ host that modifies A residues in the GATC sequence. The newly
synthesized DNA will not contain modified A residues so the double stranded DNA is unmodified.
The DNA is digested by DpnI which cleaves ONLY modified GATC sites. Thus, the parental
template is degraded by the enzyme and in not infective after electroporation. On the
other hand, the newly synthesized DNA is not modified so it is not cleaved by the DpnI
enzyme. Thus. It is replicated when it gets into cells.
C). (3 Points). Describe a method that could help you purify a large amount of the mutant protein for
biochemical studies.
One could clone the gene downstream of the T7 promoter to express large amount of the protein or
one could construct a variant that contains a His-tag on the N or C-terminus to aid in purification on a
nickel column (or both).
D). (2 Points). Describe one potential problem with your answer to part (C).
Expression of the protein from the T7 promoter might result in inclusion bodies which will be
insoluble and inactive. The His-tag could alter or abolish activity of the protein.
5). (10 Points). A map of the S. typhimurium chromosome is shown below. You have a collection of
S. typhimurium strains that have Tn10 insertions in each of the genes shown on the map. They also
carry a mutation that make them resistant to the antibiotic rifampicin. It is possible to construct an
Hfr at any desired Tn10 insertion in S. typhimruium using a F'(Ts) lac+ Tn10 plasmid. (The plasmid is
F with a mutation that makes it temperature sensitive for replication. It also has a copy of the lac
operon and a copy of Tn10). You have a strain of S. typhimurium that carries the plasmid and is
sensitive to rifampicin.
Using a F'(Ts) lac+ Tn10 donor and any mutant recipient you choose, draw a diagram
showing how you could select for an Hfr located at 23 min on the S. typhimurium
chromosome? [Recall that the lac genes are absent from S. typhimurium i.] Include the
composition of the media used and the temperature you would use at each step.
(5 Points for media and temperature and 5 Points for selection design).
ANSWER: To construct the strain carrying the F', mate with F- pyrC::Tn10 RifRrecipient, selecting
for Lac+ RifR at 30°C. (Lac+ selects for transfer ofthe F' and RifR counterselects against the donors.)
Then select for theHfr as shown below.
6). (10 Points).
A). (2 Points). How many cyles of PCR does it take before one obtains discrete endpoints defined by
the primers?
After 2 cycles
B). (2 Points). If you were planning to make mutants with PCR, what type of DNA polymerase or
reaction conditions would you use?
Use an error-prone DNA polymerase like Taq or ( alter the ratio of dNTPs in the reaction).
C). (2 Points). If you were planning to clone a gene, what type of DNA polymerase or conditions
would you use?
Use a polymerase that is relatively error-free like Pfu.
D). (2 points for each answer). What are 2 variables in PCR reactions that can affect the amount and
specificity of PCR fragments?
The concentration of Mg and the temperature that the primers anneal are critical for yield and
specificity of product.
7). (10 Points).
When developing a new genetic system, it is desirable to have both a shuttle vector and also a suicide
vector.
A). (4 Points Total, 2 Points Each). Describe a use of each vector.
A shuttle vector can be used to introduce a gene into an organism (by electroporation of conjugation).
A suicide vector can be used to integrate into the chromosome by HR between homologous
sequences on the plasmid and in the chromosome.
Other answers posible.
B). (2 Points). Suppose you made a transposon insertion using your new genetic system and the
bacterium has an observable phenotype. How would you show that the phenotype is due to your
insertion?
Isolate DNA and reintroduce it into a wild type version of your organism (and select for the AbR
associated with the transposon). It should have the same phenotype as the original strain.
C). (4 Points Total, 2 Points Each). What are two simple methods that could be used to introduce
DNA into an organism that has not been used previously for genetics?
Electroporation or conjugation.
8). (20 Points). You have received a clinical isolate of E. coli (strain A) that contains a plasmid or
conjugative transposon (CTn) that encodes resistance to tetracycline (TetR). The strain also contains
a plasmid that encodes resistance to ampicillin (AmpR). The strain is sensitive to all other
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8
antibiotics. You grow strain A and mix 10 cells with 10 cells of another E. coli strain (Strain B) that is
TetS, AmpS and contains a chromosomally-encoded mutation that makes the cells resistant to
streptomycin (StrR). After incubation for an hour, you plate dilutions of the mixture on LB + Str + Tet
plates, LB + Str + Amp plates and LB + Str +Tet + Amp plates and find obtain the results shown in the
Table below. You also did controls where strain A and strain B were incubated separately. The
results are shown in the table below:
A). (3 Points). Why are colonies obtained from the mixture of cells A and B but not from A or B
only cells?
(There must be some type of genetic exchange occurring where TetR and AmpR are being
transferred to strain B. [Str is not being transferred to strain A because if that were true, one would
expect all the StrR to be TetR and AmpR]. A cells are killed by Str and B cells are killed by Tet
and/or Amp).
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Next you take 5 colonies from the LB + Str + Tet plate, grow them up separately and mix 10 of each
8
with 10 of an E. coli strain (strain C) that contains a chromosomally encoded mutation that makes
cells resistant to nalidixic acid (nalR). The strain is sensitive to Str, Tet, and Amp. You find that all 5
mixtures give colonies about 1000 colonies that grow on LB + Nal + Tet, and no colonies that grow
on LB + Nal + Amp or on LB + Nal + Tet+ Amp plates.
When the same experiment is done with 5 colonies from the LB + Str + Amp plate, no colonies are
found on the LB + Nal + Tet, LB + Nal + Amp or LB + Nal + Tet + Amp plates.
When the experiment is repeated with five colonies from the LB + Str + Tet + Amp plate, there were
about 1000 colonies on the LB + Nal + Tet plate, about 100 colonies on the LB + Nal + Amp plate and
about 10 colonies on the LB + Nal + Tet + Amp plate.
B). (4 Points). What is the likely explanation for these results?
(The tetR plasmid/conjugative transposon in strain A is conjugative and can transfer a copy of itself
to strain B (or C). It can also mobilize the ampR plasmid into strain B (or
C). At low frequency a recipient can receive both the TetR and AmpR plasmids. However,
exconjugants that receive only the TetR plasmid can only transfer TetR to strain B (or C). Cells that
receive only the ampR plasmid cannot transfer to strain B (or C) because the TetR plasmid is not
present to mobilize it. Exconjugants that get both plasmids act like the original strain A in that they
can transfer the TetR plasmid by itself, the AmpR plasmid by itself or both.)
C). (3 Points for each answer). Explain the results in terms of:
ori sites on the plasmids/CTn.
(Both must have an oriT because they can be transferred. The plasmid carrying AmpR must have an
oriV to replicate in strain A. A conjugative plasmid would need an oriV to replicate. A CTn would
not have an oriV because it is integrated in the chromosome and is replicated as part of the host
chromosome).
tra genes on the plasmids/CTn.
(The TetR plasmid/CTn must have tra genes because it is conjugative. The AmpR plasmid probably
does not carry tra genes because it requires the TetR plasmid to mobilize it).
Nickase or Mob protein.
(Each plasmid likely has its own nickase / mob protein that is specific for the oriT).
D). (4 Points). How could you determine if the TetR is on a plasmid or CTn?
(Pulse field electrophoreses of some exconjugants should give you the answer. If TetR is on a
plasmid, all the restriction fragments of the donor and recipient will be the same because the
plasmid is extrachromosomal. If the tetR element integrates into the chromosome like a
conjugative transposon, one of the chromosomal bands will be lost and a new band (which has the
CTn in it) will be seen).