Download Spr01Exam II Answer Key

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Answer Key_
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Bacterial Genetics, BIO
4443/6443
Spring Semester 2001
Exam II
1.) What is the primary difference between conjugative plasmids and mobilizable plasmids? What
genes are typically found on conjugative plasmids that are not found on mobilizable plasmids?
(6pts)
Conjugative plasmids carry all the genes necessary for conjugation to occur
Mobilizable plasmids carry the genes for plasmid transfer but generally lack the genes needed for pilus
formation.
Conjugative plasmids encode the genes for pilus formation/function. Mobilizable plasmids usually lack
this set of gene products.
2.) What general role do the products of the finO and finP genes have in the conjugation of the F
plasmid? What fortuitous role did these genes play in the discovery of conjugation? (5pts)
Repress pilus formation/ inhibit conjugation from occurring all the time.
The original F plasmid was a fin mutant. Thus, the original F plasmid expressed the conjugational genes
all the time so conjugation was much more frequent and easily identified in cells harboring the plasmid.
3.) Naturally transformable cells take up DNA from outside the cell. Exactly why it would be
advantageous for a cell to take up non-self DNA is something that is still debated extensively.
Describe three possible roles (or functions) that transformation has been proposed to play in the
bacterial life cycle AND for each possibility, describe one piece of evidence that supports or fails to
support that idea. (6pts)
* Several responses are possible. See text/lecture notes for
others.
Nutrition:
+cells can degrade DNA and
reuse the bases
-most cells that salvage DNA
degrade it before taking it up,
its safer
Repair:
+some recombination genes
are induced by DNA damage,
suggestive perhaps.
-the survival of competent
cells is not significantly
different from noncompetent
cells
To Promote Recombination:
+”sexual” reproduction carries
advantages through faster
adaptation and variant disp of
seleterious mutations.
-”sexual” reproduction is
costly from the viewpoint of
energy resources and risky as
far as losing your own genetic
information
4.) Most recombination models propose that enzymes are required to catalyze at least four
fundamental functions (or steps) during the recombination process. In general, what steps are
required during the recombination process? (6pts)
Initiation (strand breakage, processing DNA ends)
Strand Invasion (homologous pairing)
Branch Migration
Resolution
5.) Name and describe one gene product in E.coli that is associated with performing each step in
the recombination process. (6pts)
Initiation: RecBC, degrades DNA from a Double Strand Break until it reaches a chi site. Then it unwinds
the DNA to generate a 3’ DNA end for strand invasion
Strand Invasion: RecA, binds single stranded DNA and pairs it with homologous duplex DNA.
Branch Migration: RuvAB or RecG, promote branch migration at Holliday junctions
Resolution: RuvC, endonuclease that specifically cleaves/resolves Holliday junctions
*Multiple genes are possible for this answer.
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6.) Some Insertion Sequences transpose by using a Replicative mechanism of transposition. Other
Insertion Sequences utilize a Cut and Paste mechanism. Describe two observations that
differentiate between these two mechanisms of transposition? (5pts)
Cut and Paste transposition does not form a cointegrate molecule as an intermediate, Replicative
transposition does.
Cut and Paste transposition does not require a resolvase activity, Replicative transposition does.
Following Cut and Paste transposition, only the target DNA has a copy of the transopson. Following
Replicative transposition , both the target and donor sequences have a copy of the transposon
Cut and Paste transposition leaves a double strand break in the donor sequence which must then be repaired.
Replicative transposition does not leave any strands unsealed.
7.) We are trying to knockout the chromosomal copy of the hisA gene in a naturally competent
strain of Bacillus subtilis. We have constructed various substrates (shown below) that contain the
ampicillin resistance gene (beta lactamase) inserted into the middle of the hisA gene.
You mix each of the DNA substrates together with your competent cells. For each
substrate, describe why you would, or would not, expect to obtain ampicillin resistant, histidine
deficeint transformants? (8pts)
ampR
hisA
You add linear single stranded
DNA containing the ampR
gene inserted into the middle
of the hisA gene
You add linear double stranded
DNA containing the ampR
gene inserted into the middle
of the hisA gene
Naturally competent cells bind double stranded
DNA, and if there is an DNA end, take it inside
tin a linear fashion, degrading one strand as it
enters. Recombination can then occur between
the linear single stranded DNA and homologous
sequences on the chromosome. So...
You would not obtain recombinants because
natuarally competent cells do not bind single
stranded DNA .
You would obtain recombinants. See above.
You add a circular double
stranded plasmid containing
the ampR gene inserted into
the middle of the hisA gene
You would not obtain recombinants because
although the DNA can bind to the cell, there is
no DNA end so that it can be taken up into the
cell. Plasmids do not enter naturally competent
cells.
You add linear double stranded
DNA. This time, however,
the hisA gene remains intact
and complete and the complete
ampR gene is inserted after
the hisA gene.
You would not obtain recombinants because
the his gene is completely intact and functional.
So, if recombination occurs, one functional
copy would simply replace another functional
copy.
***This last one was hard.
Most mutations that make E. coli resistant to the antibiotic rifampicin are found in the rpoB gene of
the RNA polymerase. You have just isolated a new rifR mutant and to your suprize, your
conjugational mapping experiment suggests that the mutation is on the opposite side of the
chromosome from the rpoB gene. Your mutation does map nearby both pro and lac, however.
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Since this represents a possibly novel mechanism of bacterial resistance, you decide to map your
rifR mutation in a little more detail and set up a transduction experiment.
To the best of your knowledge, your mutant is wild type for all other genes besides the rifR
mutation. You infect the mutant rifR pro+ lac+ E.coli with P1 bacteriophage and make a P1 lysate.
You then use the lysate to infect a rifS pro- lac- recipient and select for rifR transductants.
Of the rifR transductants, you find that 10% are pro+. and 60% are lac+.
8.)What are the two possible orders that these three genes could be in? (4pts)
rif cotransduces with lac much more frequently than with pro so lac must be much closer to rif than pro.
So it could be either...
rif----lac-------------------pro
or
pro-------------------rif----lac
9.) You repeat the infection, this time selecting for lac+ transductants.
Of the lac+ transductants, 60% are also rifR but none are pro+.
Where does the rifR gene map in relation to pro and lac? (4pts)
lac and pro cotransduce less frequently than do rif and pro. So lac must be further away from pro than rif
and therefore
pro-------------------rif----lac
must be the correct order.
One year later...
Following Stanford’s dominating performance at the NCAA championships, 1,345 fans
were hospitalized and seven eventually died from a hemolytic fever which set in within hours after
the game ended. Health officials quickly zeroed in on the arena’s hot dogs as the culprit.
Scientists were easily able to isolate a bacteria from the hot dogs that appears almost identical to the
common nonpathogenic strain of E.coli that is normally found in our intestines. When culturing
the bacteria, scientists found that the pathogenic strain is resistant to penicillin, a problem that
delayed effective treatment for some patients. Additionally, rather than forming normal round,
white-ish, colonies on plates, this pathogenic strain grows into red, pussy, mucoid colonies.
Hot dog sales around the country have taken a serious plunge and the high profile case has
both scientists and the public worried about where or how this E.coli strain became pathogenic.
Although the colonies look very different from E.coli, the initial genotyping hasn’t shown any
genetic difference between this strain and normal E.coli.
Based on the penicillin resistance and pathogenicity, you speculate that this variant of E. coli
may have acquired a plasmid that confers these phenotypes.
The pathogenic strain of E. coli seems to grow well on minimal plates that contain only
glucose. You decide to make use of a common trpB- hisA- leuC- nalR (nalidixic acid resistant)
strain of E. coli that you have in your lab and test whether the pathogenic strain contains a selftransmissible plasmid.
10.) Which cell will serve as the donor in your test? What is the genotype of the donor? (Include
trpB, hisA, leuC, nal, pen, and red (for red,mucoid) (3pts)
The pathogenic strain is the donor. Its genotype would be trpB+
hisA+ leuC+ nalS penR red+
11.) Which cell will serve as the recipient in your test? What is the genotype of the recipient?
(Include trpB, hisA, leuC, nal, pen, and red (for red,mucoid) (3pts)
The common lab strain is the recipient. Its genotype would be trpB-
hisA- leuC- nalR penS red12.) Describe how you would test to see if the pathogenic strain contains a self-transmissible
plasmid. Be sure to include which supplements and/or antibiotics would be used in your selection
plates (9pts).
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Mix the pathogenic, donor strain with the common, recipient strain. Incubate and allow transfer to occur (if it can).
Then plate the cells on bacterial plates containing naladixic acid and penicillin. The naladixic acid is to
counterselect against growth of the donor cells. It kills the donor cells. The penicillin selects against growth of any
recipients that did not receive the penR gene following transfer from the donor cells.
If any recombinant grow on the selective plates, then it would indicate that the donor probably contains a
conjugative plasmid that can transfer the penicillin resistance gene.
Based upon your results, you conclude that there probably is some form of a selftransmissible plasmid in the pathogenic strain.
Interestingly, you never are able to get the entire population to transfer the plasmid
successfully so you decide to examine your transconjugates more closely and you are suprized to
find that several of the transconjugates are now able to grow without the addition of histidine
leucine or tryptophan.
13.)
What does this most likely imply about the plasmid in the pathogenic strain of E.coli?(5pts)
Since the plasmid is also transferring chromosomal genes, it probably has integrated into the chromosome (i.e Its an
Hfr cell). It could also be an F’-like cell, however, the fact that transfer is not entirely efficeint would argue against
this.
You decide to map the position of this virulence gene(s) in the pathogenic strain. You repeat
your transfer experiment, allowing the mating to go on for several hours. (for simplicity, we will
assume that nal transfers extremely early). You obtain the following numbers of recombinants
individually.
10 cells out of every100 donor cells are trp+ nalR
53 cells out of every 100 donor cells are his+. nalR
5 cells out of every100 donor cells are leu+. nalR
44 cells out of every 100 donor cells are resistant to penicillin
<0.1 cells out of every 100 donor cells are pussy, red, or mucoid.
14.) What type of experiment would this be called? Is it a time of entry experiment or a gradient of
transfer experiment? (2pts)
gradient of transfer experiment
15.) Draw the map order of trp, his, leu, penR (penicillin resistance), and red (red/mucoid). (10pts)
his----pen----trp----leu----red
16.) If you wanted to know the distance between genes rather than their relative order, what type of
experiment could you do? (3pts)
a time of entry experiment
17.) What can you conclude about the penR gene and the red gene(s)? What does this imply about
the acquisition of the new traits in the pathogenic strain? (5pts)
The penR gene and the red gene(s) are not linked. They are located very far apart on the chromosome and
therefore the new traits in this pathogenic strain were probably acquired through multiple, independent events.
Clearly you need more information about the pathogenicity. Several scientists have shown
that the red, mucoid phenotype correlates perfectly with causing pathogenicity in mice. Using the
red/mucoid phenotype as your marker for pathogenicity, you mutagenize a red, mucoid culture and
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isolate 5 mutants that are no longer red or mucoid. After a lot of work, you also obtain F’ factors
for each mutant.
You transfer each F’ into each mutant and score for complementation of the red, mucoid phenotype.
The color and shape of each transconjugate carrying the F’ is shown.
Recipient mutant #1
F’ that
carries mutant
#1
#2
#3
#4
#5
White
Normal
Red
Mucoid
Red
Mucoid
White
Normal
Red
Mucoid
#2
#3
#4
#5
White
Normal
White
Normal
Red
Mucoid
Red
Mucoid
White
Normal
Red
Mucoid
Red
Mucoid
White
Normal
Red
Mucoid
White
Normal
18.) Which mutations belong to the same complement group(s). At least how many genes or
complementation groups do you suspect are involved in the red (red, mucoid) pathway? (10pts)
If the two mutations can restore the red mucoid phenotype, then these mutations are likely to be in separate
genes.
Mutation#1 complements (restores Red Mucoid) when present with mutation#2, #3, and #5 but not #4. So
#1 and #4 are probably mutations in the same gene.
Mutation#2 complements (restores Red Mucoid) when present with mutation#1, #4, and #5 but not #3. So
#2 and #3 are probably mutations in the same gene.
Mutation#5 complements every mutation except itself. So it is probably the only mutation isolated in that
gene.
There are 3 complementation groups: A(#1,#4), B(#2,#3), and C(#5) suggesting that at least 3 genes are needed to
produce the Red, Mucoid phenotype.
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