MCB 421 Exam #3
December 7, 2010
There are 7 questions. Answer all 7 questions. There is a DNA
sequence at the end that will be used for question 3.
Be sure your name is on all pages.
1). (20 Points).
There are two mechanisms for transposition used by bacterial transposable elements:
replicative (Tn3) and non-replicative (Tn5 and Tn10). Compare and contrast the two
mechanisms with respect to:
A). Host DNA sequences adjacent to the ends of the element:
ANSWER: Both transposon types form small duplications of adjacent host DNA
sequences.
B). Formation of co-integrants:
ANSWER: Replicative transposons form co-integrants. Non-replicative
transposons do not form co-integrants.
C). Resolvase enzyme:
ANSWER: Used by replicative transposons but not by non-replicative transposons.
D). Requirement for DNA synthesis:
ANSWER: Both use DNA polymerase I to fill in gaps caused by transposition.
E). Structure of DNA in intermediates of the reaction:
ANSWER: Non-replicative transposons form excised loops bound by transposase.
Replicative transposons form co-integrant structures.
2). (15 Points).
Microarrays, proteomics and RTq-PCR are techniques for measuring various aspects
of gene expression. For each of the 3 techniques, describe the application with
respect to measuring:
A). Transcription regulation
B). Translation regulation
C). Post- translational modification of a protein.
Answers:
A). Microarrays and RTq-PCR can be used to measure transcription because
they measure RNA levels. Proteomics only measure the amount of protein
made so effects at the transcriptional and translation levels can affect the end
result.
B). None of the methods directly measures translational regulation.
C). None of the methods directly detects post-translational modification of a
protein.
3). (15 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 homework (just remove it from the back). 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 R1 might abolish integration but have no effect on excision
etc.
In order to test this hypothesis, you decide to mutate the R1 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.
A). You decide to use the Quickchange method to change all six basepairs in the box
in the R1 site using oligonucleotide mutagenesis (see sequence handout). The site
should be changed to an EcoRI site. (The sequence of the EcoRI site is: 5’-GAATTC3’). Design the sequences of the oligonucleotides showing the changes in the R1 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?
ANSWER:
Oligo #1
5’ ---GAACAAGAATTCTTGTGG--- 3’
Oligo #2
5’ ---CCACAAGAATTCTTGTTC ---3’
The oligos should be around 50 bases long.
B). Why is the template grown in a dam+ cell?
ANSWER:
To modify the A in GATC sequences in the plasmid template.
C). Why is the DNA treated with DpnI after the DNA synthesis step?
ANSWER:
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). There is a group of genes, A through F, mutations in which affect
the phenomenon of interest to you. To see possible pathways within the group, you
run epistatic analysis. The results are as follows the values are “deficiency factors”,
derived by dividing the value obtained for WT cells by the values obtained for the
corresponding mutant):
WT
A
B
C
D
E
F
G
WT
1
10
10
10
1000
10
10
10
A
B
C
D
10
1000
10
1000
1000
10
1000
10
1000
1000
10
1000
10
10
1000
1000
10
1000
1000
1000 10
1000 1000 10
1000 10
1000 10
Derive as many conclusions as you can.
Answer:
1) D in the central gene
2) A, C and F work together
3) B, E and G work together
E
F
G
5). (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
5
45
40
40
lacZ gene fusion
B-galactosidase activity
-superoxide
+superoxide
50
450
40
400
What do the results indicate about the regulation of the virF and virG genes? Briefly
explain your answer.
ANSWER:
virF 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.
virG is not transcriptionally regulated by superoxide because we see no increase in
reporter activity in response to superoxide in the operon fusion. virG 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
6). (20 Points).
You recently discovered a new plasmid from an environmental isolate of E. coli B,
which you named pCar33. It carries resistance to ampicillin.
A). What DNA sequence would be required for the plasmid to transfer by
conjugation? What class of enzymes mediates this process?
Answer: It would require an oriT. To start the transfer process, a tyrosine
recombinase nicks the oriT and starts transfer. Other proteins involved are
helicases to pump the DNA, misc enzymes to form relaxasome.
B). You attempt to mate pCar33 into a closely related species of E. coli B-12, a strain
of Salmonella 215, and a strain of Bacteroides theta. After mating, you only find
colonies when pCar33 was mated into the E. coli B -12 recipient. What can you
conclude about its host range? Why was this experiment not conclusive?
Answer: It has the ability to mate and replicate in E. coli B-12. It may not be
able to mate or replicate in the other two strains. You cannot assume it lacks
the ability to transfer DNA to 215 or Bt..
C). What genetic event would be required for pCar33 to form an Hfr? How could
the Hfr be used to order genes in a genome?
Answer: If the plasmid integrated into the host chromosome, it could form an
Hfr. The Hfr can transfer chromosomal genes beginning from the oriT to a
recipient. Chromosomal genes will also be transferred as a function of how far
they are from the oriT. Thus, donor genes enter the recipient in a timedependent manner so that genes can be mapped by time of entry into the
recipient.
D). During one of your matings of pCar33 into a new strain of E. coli B-17, very few
colonies of E. coli B-17 are recovered <10-8. After careful thought, you realize you
may have a case of zygotic induction. What does this mean? How would you verify
your hypothesis?
Answer: Your strain of E. coli B containing pCar33 may have been a lambda
lysogen (or a lysogen for some other prophage. The prophage could be on the
plasmid or the chromosome). The new strain of E. coli B-17 must be a nonlysogen. When pCar33 transfers DNA into B-17 (either plasmid DNA or Hfr
transfer of the chromosome), no cI repressor is present in the E. coli B-17 cells
to repress the lambda OL and OR sites to promote the lysogenic lifecycle.
Most of the time, after synthesis of the donor DNA to dsDNA in the recipient
the virus goes lytic and kills the cells. (However, it is possible to also choose
lysogeny at a low frequency and form lysogens again).
If zygotic induction occurred, the E. coli B-17 cells would be lysed and phage
would be released into the medium. You could do a standard plaque-forming
assay using E. coli B-17 as the indicator. You should observe plaques.
7) (10 Points). Bacteroides fragilis is a major bacterium in the human
colon. It can use sialic acid as a carbon and energy source. The gene
products of the nanLET operon are required for utilization of sialic acid.
(See diagram below). In order to study the function of the nanL gene
Brigham et al [(2009) J. Bacteriology 191:3629 – 3638] made a mutant
containing a deletion of the nanL gene by the method described below:
A B. fragilis strain carrying a deletion of the thyA gene (thyA) but
otherwise wild type was used. Since the thyA gene is non-functional, this
strain requires thymine in order to grow.
A suicide plasmid called pMBD5 containing a copy of the thyA+ gene and
DNA that flanks the 5’ and 3’ ends of nanL gene (labeled a and b on the 5’
side and y and z on the 3’side, (see diagrams above and below) was
introduced into the B. fragilis thyA strain. The plasmid cannot replicate in
B. fragilis so selection for cells that no longer require thymine results in cells
with the plasmid integrated into the chromosome.
A). Diagram a genetic event that results in integration of the plasmid into
the B. fragilis chromosome. Be sure to clearly label your diagram.
(Answer in diagram below)
In many bacteria, including B. fragilis, incubation of thyA mutant cells
with the antibiotic trimethroprim and a high concentration of thymine (don’t
worry about the mechanism) results in the selection for cells that contain the
thyA allele and lose the thyA+ allele. Thus, all survivors of this selection
contain the thyA allele. When they performed the selection on the strain
made in (A), they found that some of the survivors of this selection
contained a deletion of the nanL gene.
B). Diagram the genetic event that lead to this phenotype. Again, be sure
your diagram is clear.
(Answer in diagram below)
C). Since the genome sequence of B. fragilis is known, how would you use
that information to show that the nanL gene is deleted?
ANSWER:
You could design PCR primers that amplify the ab / yz region and a set
of primers that amplify the nanL gene and show that the ab / yz set act
primers for PCR while the ones for nanL do not. (A good control would
be to amplify the nanL gene in the parent chromosome to show that the
primers are functional).
The event shown below results from crossover between a b sequences to
integrate the plasmid. Segregation by recombination between y z
sequences will delete the plasmid and thyA gene. However, if the
plasmid integrated by recombination between y z sequences,
recombination between a b sequences would give a nanL deletion.