MCB 421
Homework 10
Due Thur Nov 20
Fall 2011
1. 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.
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.
a. 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.
b. Why is the template grown in a dam+ cell?
To modify the A in GATC sequences in the plasmid template.
c. 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.
MCB 421
Homework 10
Due Thur Nov 20
Fall 2011
2. 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.
MCB 421
Gene
virF
virG
Homework 10
Due Thur Nov 20
galK operon fusion
Galactose kinase activity
-superoxide
+superoxide
40
5
40
45
Fall 2011
lacZ gene fusion
B-galactosidase activity
-superoxide
+superoxide
40
50
400
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
activity 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.
3. 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 ßgalactosidase) 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.
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.
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
MCB 421
Homework 10
Due Thur Nov 20
Fall 2011
response to superoxide. Since there is no increase in the operon fusion, we know that the
entire difference is due to translational regulation).
4. You have isolated a new gene, yssA, from Salmonella. After obtaining the DNA
sequence of the gene and performing a GenBank search you find that the gene has
homology to known Type III topoisomerases from a variety of bacteria. It has a tyrosine
residue that is part of the active site of all known Type III topoisomerases. In described
topoisomerases, this residue is required for activity. Assume that you have a method to
isolate YssA protein in small quantities and that you have detected topoisomerase activity
in YssA isolates. Assume that you have a simple method to assign a positive or negative
phenotype for isolated YssA protein (i.e. you can say whether a given version of the
YssA protein is functional or not).
a. Explain how you would use the Quickchange PCR method of site-directed mutagenesis
to test whether the conserved tyrosine residue is required for YssA function. Draw a
diagram showing where each primer would anneal on a plasmid containing yssA. Show
the location of the mutagenic nucleotide on each primer in your diagram.
ANSWER: Your diagram should show a plasmid containing yssA and two
anticomplementary primers annealing on either side of the plasmid within the yssA
gene. The mutagenic nucleotide should be in the middle of each primer.
b. What would you want to do with the Quickchange PCR reaction after PCR and before
transformation? Why?
ANSWER: digest with DpnI to eliminate residual template DNA.
c.
If the relevant Tyr codon is UAU, suggest a one-nucleotide change you could
make to change the amino acid using the above protocol. What new amino acid would
this result in? Do you think this change in amino acid would definitively test the
importance of the Tyr residue? Why or why not?
ANSWER: The codon can be changed to Phe, Ser, Cys, His, Asn, or Asp. Regardless
of the amino acid, it cannot be a definitive test. Each of these amino acids has
something in common with or similar to Tyrosine. Phe has an aromatic group, Ser
has a hydroxyl, Cys has a thiol (sometimes similar to hydroxyl), His, Asn, and Asp
can sometimes act as proton donors or acceptors, as can Tyr.
d. Describe a genetic experiment that could help you purify a large amount of the protein
for biochemical studies.
Put the gene on an overexpression vector, perhaps one with a T7 promoter system.
A His tag could also be attached if you so desire.
e. What is one possible problem with the experiment described in part d?
MCB 421
Homework 10
Due Thur Nov 20
Fall 2011
Answers: Several possible answers – overexpression could be toxic, His-tag could
affect activity, etc.