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MCB 421
EXAM #3
December 8, 2009
There are 7 Questions.
Be sure your name is on all pages.
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1.(15 points)
a. (3 points) Describe the principles behind microarray analysis and RT-PCR for measuring
mRNA levels in cells. Are either or both quantitative? If not what must be done to make one
quantitative?
Answer: Microarrays use hybridization to measure levels of mRNA in cells. These
measurements are, in theory, quantitative. RT-PCR uses labeled DNA copies of mRNA to
measure levels of mRNA and is not quantitative unless you do RT-qPCR. RT-qPCR is
needed for quantitative measurements.
b. (3 points) What is an advantage of microarray analysis over gene (operon or protein) fusions?
Answers: Several answers possible:
Microarrays can measure expression of several genes on a single chip whereas fusions
measure expression of one gene at a time.
Microarrays can find unsuspected genes of interest whereas fusions are usually made to
genes known to be of interest.
c. (3 points) What is an advantage of having gene fusions for genetic analyses over microarrays
or RT-PCR?
Answer: Gene fusions (operon or protein) can be used to screen or select for regulatory
mutants on plates while microarrays or TR-PCR are usually not suitable for efficient
screens.
d. (3 points) Name two limitations of using genomic or metagenomic data to draw conclusions
about the genes in an organism or community of organisms.
Answer: Annotation can be incorrect. Size of orf can be wrong. Very small orfs will be
missed. (There are probably more).
e. (3 points) Why is it necessary to treat samples with DNAase for RT-PCR experiments?
Answer: Cellular DNA must be destroyed to be sure it is not amplified during the RT-PCR
reaction. Otherwise it would contribute to the apparent signal.
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2. (24 Points). A protein involved in the excision of CTnDOT is Orf2d. It is a small basic
DNA binding protein. In order to study it further, we will first attempt to purify it.
a. (6 points) Using the attached guide 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 given pet
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 Orf2d sequence.
Example 5’ GGATCCATATGAAAGTGA 3’ forward
5’ GGATCAAGCTTTTATGTCCTC 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 Cterminus 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 "helix-turn-helix"
motif in the Orf2d protein is required for DNA-binding. You need to change the highlighted G
into an A. Specify the sequence of the primers to be used in constriction of this new plasmid.
(For brevity, limit the primer to 10bp).
Answer:
5’ ACACACGAGG 3’
5’ CCTCACGTGT 3’
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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. 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 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.
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3. (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. (2 points) 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 recombinases
nicks the oriT and starts transfer. Other proteins involved are helicases to pump the DNA,
misc enzymes to form relaxasome.
b. ( 3 points) 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 mate.
c. (10 points) Dr. Farrand discussed the only two accepted by him experimental ways to
definitively determine host range of a plasmid during his lecture. Choose one of the ways and
diagram how you would set up the experiment of mating pCar33 into E. coli B-12, a strain of
Salmonella 215, and a strain of Bacteroides theta. Include cell types, media, and a diagram of all
constructs you might use.
1. Transform into the E. coli B strain a mobilizable plasmid with a broad host range such
as RP4. Mate the cells to each of the recipients and plate on selective media (can be
determined by the student). If pCar33 is capable of mating and transferring DNA into the
recipient, RP4 will co-mobilize. After the mating is complete RP4 could be detected by
plating on selective media or isolation.
2. Insert a transposon on pCar33. If the plasmid can conjugated and transfer DNA into
the recipient, the transposon can also move into the recipient. Once it is insde the cell, it
can transpose into the chromosome for example and be detected by selectable media.
d. (5 points) Finally, during one of your matings of pCar33 into a new strain of E. coli B-17, very
few colonies are recovered <10-8. After careful thought, you realize you may have a case of
zygotic induction. What does this mean?
Answer: Your strain of E. coli B containing pCar33 must have been a lambda lysogen.
The new strain of E. coli B-17 must not have lambda. When pCar33 transfers into B-17, no
C1 repressor is there to force lambda into a lysogenic lifecycle. Most of the time, the virus
goes lytic and kills the cells. However, it is possible to also choose lysogenic at a low
frequency and form lysogens again.
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4. (12 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. (4 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:
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c. (4 points) Why is it important to select with tet and put? What would happen if you only
selected with kanamycin? Draw out the resulting product.
The bottom product from part (a) then you can get the product like the bottom one below. Other
possible products but must have putA::kan.
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5. (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. Why can this be a problem?
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6. (6 points)
a. (3 points) Why is conjugation often better than electroporation or transformation to introduce
DNA into the cells?
Answer: It avoids the restriction modification system by entering the cell as single
stranded DNA that is then replicated and modified.
b. (3 points) Why is the power of 10 rule important when starting a new genetic system?
Answer: When you introduce DNA into a cell for homologous recombination, homologous
recombination occurs usually at 10-4 frequency. If you cannot efficiently get DNA into the
cell at a high enough frequency to absorb the low frequency of homologous recombination,
you will never be able to do genetics in the strain since you cannot obtain enough colonies
to be significant.
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7. (13 points) Many different mutations have been isolated in the putA gene of Salmonella
enterica. A few of these mutations are shown on the following deletion map of the putA
structural gene. The promoter is at the left, pointing rightward into the putA gene. Based on the
relative location of the mutations and the results in the table below, explain the phenotypes
observed for each of the single and double mutants shown. [The putA1992::MudJ and
putA1993::MudK insertions map in the same deletion interval.]
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Grade on This Exam
Final Grade
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