Name:
MCB 421
EXAM #3
December 1, 2011
There are 6 Questions.
Be sure your name is on all pages.
Name:
1. (7 Points) Transposable elements are useful tools for performing genetic experiments.
a. (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.
b. (2 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
Name:
2. (11 points)
a. (3 Points) Plasmids often have oriT and oriV sites. What is the function of each?
Answer: 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.
b. (3 Points) Do the sites function in cis or in trans (or both)?
Answer: 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.
c. (5 points) What is the amino acid in Lambda Int that forms a bond with the DNA during
recombination? What group is formed on the DNA?
Answer: Lambda Int uses a tyrosine to form a 5’ phosphotyrosyl bond with the DNA. A
3’OH group results on the DNA.
Name:
3. (24 Points). A protein involved in the excision of CTnDOT is Xis2c. It is a small basic DNA binding
protein. In order to study it further, we will first attempt to purify it.
Sequence of Xis2c (Top Strand)
M K V I T M E S S V F R S L T E Q I A E
1
ATGAAAGTGATAACAATGGAAAGTTCCGTATTTCGGTCATTGACGGAACAGATAGCGGAG
60
61
I A A H V R A V S G D K R A A S P D R L
ATTGCGGCACACGTCCGTGCCGTCTCCGGCGACAAGAGAGCGGCATCGCCCGACAGGTTG
120
121
L T T R E A A H L L N V S T R T L Q R M
CTTACCACACGCGAGGCGGCGCACCTGCTTAACGTGAGTACCCGCACCCTCCAGCGTATG
180
181
R S E Q R I G Y V V L R G K C R Y R Q S
CGCAGCGAGCAACGCATCGGCTATGTCGTGCTTCGTGGCAAATGCCGCTACCGACAGTCT
240
241
E I D R L L D E C T V N E D A A T P Q E
GAAATCGACCGCCTGCTTGATGAATGTACCGTCAACGAGGACGCCGCGACACCGCAGGAG
300
301
L K R N H T L R T G G T P K G R R T *
CTGAAACGCAACCACACGCTACGAACGGGAGGCACACCCAAAGGAAGGAGGACATAG
357
a. (6 points) Using the attached pET vector map on the last page of the exam, design a set of primers to
clone PCR products into the NdeI (CATATG) and HindIII (AAGCTT) site of the pET-30a vector. These
primers will be used in PCR. Be sure to indicate the 5’ and 3’ end of the primers. For brevity limit the
oligos to 15 basepairs. (Note the sequence of the pET vector shows only 1 strand of the DNA sequence).
Answer: Each primer should include the RE site, an additional 5-10 bp to allow restriction, and
part of Xis2c sequence.
Example
5’ GGATCCATATGAAAGTGA 3’ forward
5’ GGATCAAGCTTCTATGTCCT 3’ reverse
b. (2 Points) Describe a method that could help you purify a large amount of the mutant protein for
biochemical studies.
Answer: 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).
c. (2 Points) Describe one potential problem with your answer to part (b).
Answer: 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.
d. (6 points) In addition, site-directed mutagenesis was done to determine if a putative "helix-turn-helix"
motif in the Xis2c protein is required for DNA-binding. You need to change the highlighted glycine to a
glutaminek. Specify the sequence of the primers to be used in construction of this new plasmid. (For
brevity, limit the primer to 10bp). Use the codon table at the end of the exam.
Answer:
5’ GTGCTTCGTCAAAAATGCC 3’
5’ CACGAAGCAGTTTTTACGG 3’
could also be CAG
Name:
e. (8 points) Briefly describe how you would use Quick Change procedure to isolate mutant plasmids.
Please include all steps of the procedure.
ANSWER: 2 Points for each concept below
1. The Quick-Change procedure uses complementary oligos that have the desired mutation(s)
incorporated into their sequences.
2. 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. The DNA strands are allowed to anneal.
When the annealed DNA is introduced into a recipient cell, the nicks 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.
3. 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.
4. The DNA is digested by DpnI which cleaves ONLY modified GATC sites. Thus, the parental
template is degraded by the enzyme and is 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.
Name:
4. (10 points) You are studying an interesting operon, the virST 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, virS
and virT. 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.
Gene
virS
virT
galK operon fusion Galactose kinase
activity
- superoxide
+ superoxide
40
40
5
50
lacZ gene fusion B-galactosidase
activity
- superoxide
+ superoxide
40
400
50
500
a. (5 points) What do the results indicate about the regulation of the virS and virT genes? Briefly explain
your answer.
Answer: virT 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. virS is not
regulated transcriptionally because there is no difference in reporter gene activity in response to
superoxide in the operon fusion. virS 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).
b. (5 points) How could you isolate a regulatory mutant that affects expression of virT?
Answer: Mutagenize the virT::galK or virT:lacZ fusion and screen for colonies that have increased
GalK or LacZ in the absence of superoxide. Other possible answers.
Name:
5. (20 points) You have two strains of S. typhimurium shown below. Both strains carry an insertion of
Tn10 in the same relative orientations as designated by the arrow.
Tn10
trpA-
cys-
#1
Strain 1
#2
trpA+
cys-
Tn10
Strain 2
Strain #1 contains a mutation in the trpA gene (trpA-) that makes the cell a tryptophan auxotroph. The
strain has a mutation in the cys region (cys- #1) that causes a requirement for the amino acid cysteine.
Strain #2 contains a wild-type trpA gene (trpA+) and an independently isolated Cys auxotroph called cys#2).
You need to perform a complementation experiment to determine if the cys- #1 and cys- #2 mutations are
in the same or different genes. To do this, you decide to use P22 HT –mediated transduction to construct
a duplication of the cys region.
A. (14 points) Draw a diagram to show how you could use generalized transduction to construct the
duplication. Be sure to CLEARLY show the points of the crossovers and the medium used for the
experiment.
P22 HT on #2
#2
trpA+
cys-
Tn10
Strain 2
#1
trpA-
Tn10
cys-
Strain 1
trpA+ Tc+ cys
trpA+
cys- 2
Tn10
cys- 1
(Medium: Min (glucose) minus tryptophan plus tetracycline and cysteine.)
Name:
B. (3 points) How would you determine if the two cys mutants complement?
(Streak on minimal plates with Tet minus Cys. If the cells grow, the mutations complement. If not
they are in the same gene. (Note: Need Tet to prevent colonies that segregate the duplication).
C. (3 points) Do the copies of Tn10 need to be in the same orientation relative to the bacterial
chromosome in both strains for this technique to work? Why?
Answer: They need to be in the same orientation. If they are not, transduction will produce a
double stranded break in the chromosome and the cells would be inviable.)
Name:
6. (20 points) You have constructed the plasmid pJG1 shown below. This contains a copy of the
thrA gene containing an internal deletion of the thrA genes as shown. The thrA gene codes for
an enzyme for threonine biosynthesis.
tet
A
B
C
D
sacB
thrA
oriR6G
pJG1
A
D
BC
The plasmid has an oriR6G origin of replication, a tetracycline cassette, and the sacB gene. The
Tet protein makes the cell resistant to tetracycline and the SacB protein makes E. coli sensitive to
sucrose.
a. (20 points) Draw out the genetic events that would allow you to move the deletion into the E.
coli chromosome. Specify the media.
tet
sacB
oriR6G
A
A
BC
B
pJG1
D
C
D
thrA
Tet
A
BC
plasmid
D
sac
B
thrA+
C
A
B
C
D
oriR6G
Deletion
WT
A
tet
A
BC
D
thrA-
D
Name:
The crossover to integrate the plasmid in shown to be between A or B. If the crossover event
occurs between A the colonies will be thrA+. If the crossover event occurs between C and D, the
chromosome will contain the deletion. (Opposite if the plasmid integrated between C and D).
Plate the cells on min + thr +sucrose. Check the colonies for growth on threonine. The deletion
will require threonine.
Name: