MCB 421 Exam #3
Fall 2007
There are 9 questions, a genetic code, and an amino
acid wheel
Be sure your name is on each page
1. (12 points) When developing a new genetic system, it is desirable to have both a
shuttle vector and also a suicide vector.
a. (4 Points). Describe a use of each vector.
Answer: 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 possible.
b. (4 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?
Answer: 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). What are two simple methods that could be used to introduce DNA into an
organism that has not been used previously for genetics?
Answer: Electroporation or conjugation.
2. (8 points) Compare replicative and non-replicative transposons with regard to:
Replicative
Non-replicative
Target site duplication
Yes
Yes
Transposition intermediate
Co-integrate
Complete transposon
Resolution
Ability to make insertion mutations
Recombination
between
duplicate
transposons
Yes
No
Yes
3. (10 points) You have identified a new transposon, carried on an R factor plasmid, that
carries resistance to the antibiotic neomycin. You have transposed it onto phage lambda
and have several independent isolates. Next you isolate DNA from a few of the
independent isolates and plan to perform experiments designed to determine of the drug
resistance element is flanked by indirect repeats of an IS element.
a. (3 points) How could you determine if the new transposon is flanked by inverted
repeats (without sequencing DNA)?
Answer: Do a snapback experiment like we discussed in class. Denature the
lambda::neo DNA to single strands by heat or OH- and allow DNA to reanneal (by
rapidly decreasing the temperature) by intramolecular interactions. If the neo gene
is flanked by inverted repeats you would see a stem structure of double stranded
DNA and a loop of single stranded DNA emerging from the loop in EM analysis.
b. (3 points) What would you expect to find if the neo gene is flanked by an IS element
in direct repeat orientation?
Asnwer: In the snapback experiment you would expect to observe single stranded
DNA (and possible duplex DNA where the entire phage DNA reannealed to reform
a duplex) and no stem-and-loop structures like those described above).
c. (4 points) What is a likely mechanism for formation of the transposon in Nature? How
was it made?
Answer: The IS element transposed independently to sites on each side of the neo
gene to form the transposon. The orientation of the IS elements is in an indirect
repeat orientation.
4. (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. The resulting strain contains a disrupted gene which is
inactive.
a. (5 points) Why does the integration event inactivate the gene? Use a drawing to
illustrate your answer.
b. (5 points) What would happen if you made a mistake and cloned the promoter region
and the adjacent DNA encoding the first 50 amino acids of the protein into the suicide
vector? Use a drawing to illustrate your answer.
5. (10 Points) Given a S. typhimurium pyr::MudJ (KanR) mutant (i.e., a pyrimidine
auxotroph) and phage P22 grown on a random pool of Tn10 insertions in the S.
typhimurium chromosome, how could you isolate a Tn10 (TetR) insertion near (i.e.
"linked to") the pyr mutation?
ANSWER: Transduce a pyr- recipient with the pool of Tn10 insertions, selecting for
TetR transductants. Replica print the TetR transductants onto minimal plates
without pyrimidines and with tetracycline. Colonies that grow are potential
transductants that brought in a copy of pyr+ with a closely located ("linked") Tn10
insertion. You can also select transductants that grow on minimal tet plates (pur+
will grow)
6. (10 points) A new mutant was isolated that is StrR and unable to use acetate as a
carbon source (ace). To determine where the mutation maps, it was mated with the four
different StrS ace+ 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 acetate 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 ace mutation map?
[Indicate the map position in minutes.]
ANSWER: About 95 min, in the region transferred early between Hfr#1 and
Hfr#3.
7. (X points) The put genes are required for growth on proline as a carbon or nitrogen
source. By transduction between P22 grown on a donor strain with a Tn10 insertion to the
right of the put operon, and a recipient strain with a Tn10 insertion to the left of the put
operon, Muro-Pastor and Maloy (1995. J. Biol. Chem. 270: 9819-9827) isolated a tandem
chromosomal duplication of the put operon. The donor and recipient strains are shown
below (abc and xyz simply represent genes on the left or right hand side of the put operon
shown for orientation):
a. (6 points) Draw a diagram showing how you could select for the chromosomal
duplication. Indicate what phenotype(s) you would select for.
b. (4 points) If the cells containing the duplication are grown in the absence of
tetracycline, tetS cells can arise by homologous recombination. Draw the structures of
the 2 types of segregates between the abc and xyz regions.
ANSWER:
8. (20 Points). Site-directed mutagenesis was done to determine if a "helix-turn-helix"
motif in the PutA protein is required for DNA-binding.
a. (6 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 oligo is shown below.:
3’ CGA TAC TAA CGG CGG AGA AGG CGA 5’
5’ GCT ATG ATT GTC GAC TCT TCC GCG 3’
ala met ile val asp ser ser ala
The bottom strand oligo must be complementary to the top strand oligo as shown below:
5’ GCT ATG ATT GCC GCC TCT TCC GCT 3’
ala met ile ala ala ser ser ala
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. (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).
D. (2 Points). Describe one potential problem with your answer to part (C).
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.
9. (10 points) You have isolated a novel operon, the vir 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
ßgalactosidase) for each of the two genes, virF and virG. 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
virF
virG
galK operon fusion
Galactose kinase activity
-superoxide
+superoxide
40
40
5
45
lacZ gene fusion
B-galactosidase activity
-superoxide
+superoxide
40
400
50
450
a. What do the results indicate about the regulation of the virF and virG genes? Briefly
explain your answer.
ANSWER: virF is not transcriptionally regulated by superoxide because we see no
increase in reporter activity in response to superoxide in the operon fusion. virF is
regulated translationally in response to superoxide because we see a marked
increase in reporter gene acitivity in the gene fusion in response to superoxide. Since
there was no increase in the operon fusion, we know that the entire difference must
be due to translational regulation. virG is regulated transcriptionally in response to
superoxide because we see an increase in reporter 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.