Bacterial Problem Set
1. λ lysogens are immune to superinfection by λ, but not with other lambdoid
phages such as 434. λimm434 (a lambdoid hybrid phage that has mostly λ genes
except for cI and the surrounding “immunity” region of 434 phage) also can grow
on E. coli(λ).
a. If you take a 1ml culture of E. coli(λ) (this is wild-type E. coli) and superinfect
with λimm434 at an MOI of 1, the tube will clear after about an hour. Describe
how you would determine whether the released phages from this infection are
λimm434 or λ (induced from the prophage). Do you expect there will be more
λimm434, more λ, or equal numbers of both phages in the lysate? Way more
λimm434.
b. You perform infection experiments using wild-type E. coli and E. coli(λ),
infecting them at an MOI = 1 with either λ, 434, λimm434, or λimm434 Ram (an
amber mutant in the R gene) and get the following result:
λ
E. coli
+
E. coli(λ)
+ = lysis, - = no lysis
Infecting Phage
434
λimm434
+
+
+
+
λimm434 Ram
+
Explain, in detail, why λimm434 Ram is able to lyse the λ lysogen but not wild-type
E. coli. It doesn’t lyse wild-type because it is deficient in R, one of the two
lysin genes. The reason why infection of the lysogen leads to lysis is because
the R gene on the prophage is transcriptionally activated and translated,
complementing the mutant. The reason why R is transcribed, as well as all of
the late genes, is because Q is supplied in trans by λimm434. The important
thing here is that while PR and PL are directly repressed by cI, PR’ is not and
RNAP is able to bind and initiate transcription there, even in the prophage
state. Do you expect there are viable phages in the lysate from the infection of E.
coli(λ) and, if so, what phage genotype will be present? λimm434
Bacterial Problem Set
2. As mentioned in class, a number of interesting alleles of the genes in the SOS pathway in E.
coli have been identified. Describe how you would screen/select for alleles of recA that are
normal at 32C but activated at 42C (recAtif, tif = temperature-induced filamentation), alleles of
lexA that are uninducible (lexAind) by UV, and lexA alleles that are defective (lexAdef – lexA is an
essential gene). You can assume any molecular genetic tool is available (except access to the
recA and lexA alleles in question, of course). For each, describe a secondary test you would
perform that would increase your confidence that you have the desired allele. Indicate whether
your approach is a screen or selection, and whether you think each resulting allele is dominant or
recessive.
Multiple answers are possible. Examples:
recAtif - Start with a plasmid with recA on it, PCR mutagenize, and transform into wildtype E. coli that contains a LexA-controlled lacZ reporter (e.g. a din-lacZ fusion). Grow
colonies at 37C, replica plate and grow at 42C, pick the colonies from the 37C that
didn’t grow at 42C (tif alleles can’t grow at the NPT). Grow these strains at 37C, shift to
42C, monitor -galactosidase activity, and mutants that give rise to high activity will likely
be true tif alleles. It is possible that the mutation that causes the tif phenotype is not linked
to the plasmid, so harvesting the plasmid and re-transforming into the same strain and
repeating the experiments would be a good idea. Screen. Dominant.
lexAind – Start with a plasmid with lexA, PCR mutagenize, transform into recAtif mutant
and plate at 42C. A good secondary experiment would be to take that plasmid and
transform into the din-lacZ strain and show that it prevents induction in response to DNA
damage. It would also be important to transform the plasmid into a  lysogen strain and
show that it does not inhibit phage induction in response to DNA damage (i.e. it doesn’t
have an effect on RecA activity). Selection. Dominant.
lexAdef – This is a little trickier because lexA is an essential gene in wild-type E. coli, so you
can’t just propose to knock the gene out. Starting with the din-lacZ strain, mutagenize,
plate on X-gal plates, and screen for blue colonies. Complementation test by transforming
the blue mutants with a plasmid that contains lexA would discriminate between mutations
in lexA and any other gene. Screen. Recessive.
Bacterial Problem Set
3. You perform a screen for new clear plaque mutants in , you mutagenize WT:GFP (GFP is
inserted into a non-essential region of the  genome) with EMS and plate on WT K12 E. coli,
and screen for clear plaques. Out of 10,000 plaques, you find 50 clear mutants. These mutants
fall into four complementation groups:
Comp. Group
1
2
3
4
#mutants
11
15
11
13
Since you know that cI , cII-, and cIII-mutants form clear plaques, you perform
complementation tests with one mutant from each complementation group with known  mutants
(no GFP), assaying for cloudy plaques. (+ = cloudy plaques, lysogen formation)
1
2
3
4
WT
+
+
+
+
cI+
+
+
cII+
+
+
cIII+
+
+
a. What gene is likely mutated in the mutants of complementation groups 1-3? (OK, this is
easy, 1=cIII, 2=cI, 3=cII
b. From this analysis, can you determine if any of the mutations are dominant or recessive?
They are all recessive.
You isolate multiple lysogens from each of the plaques in the complementation tests, colony
purify, and do two experiments. First, you test for GFP expression and get the following results.
(w = all white colonies, g = all green colonies, g/w = mix of green and white colonies)
WT cIcIIcIII1
2
3
4
g/w
w
g/w
g/w
g
NA
g
g
g/w
w
NA
g/w
NA
w
g/w
g/w
You then treat each lysogen with UV light to test for prophage induction. Here is the results (+
means phages produced):
1
2
3
4
WT
+
+
+
+
cI+
NA
+
-
cII+
+
NA
+
cIIINA
+
+
+
Bacterial Problem Set
c. Why are all the lysogens produced from co-infections with cI- green? The trick here is
that the cI mutants are not going to stably integrate into the genome. Any cI- genomes that
are integrated into the genome during the co-infection (cI protein supplied by the GFP
phage) will give rise to lytic growth upon outgrowth of the bacterium as the cI+ phage is no
longer around. The cI+ (GFP+) phages that integrate, however, will give rise to stable
lysogens.
d. Posit an explanation for why no phages were produced from the lysogens produced from the
co-infections of 4 + cI-? What gene is likely mutated in 4? Describe an experiment to test
this. #4 is an int mutant, that phage genome is stuck in the chromosome and can’t be
packaged.
Bacterial Problem Set
4. After getting shut out in your first quarter rotation picks, you end up in the laboratory of a
new UCSF professor, Dr. Walter White. For your rotation project, the cunning but mad PI hands
you four mysterious vials labeled Skyler, Jesse, Marie, and Hank. Dr. White claims each
contains one of four strains of E. coli but none of them are  lysogens. Just to be sure, you
attempt to confirm that no prophages exist by performing the following three experiments: 1.
You treat a culture of each strain with UV light and test for prophage induction, and 2. You
perform PCR on chromosomal DNA from each strain using oligos specific to  genes. 3. You
spot 105 wild-type  particles (from a stock grown on wild-type E. coli K-12) in the middle of a
lawn of each strain to test for immunity. The results are as follows:
Expt. 1 and 2
Strain
Prophage induction
Skyler
Jesse
Marie
Hank
+
PCR product
+
Expt. 3
Strain: Skyler
huge plaque
Jesse
few single plaques
Marie
Hank
huge plaque
no plaque
Finally, you make stocks of phage grown on each strain (for the Hank strain, you use the phage
induced after UV treatment) and test all of your phages for growth on new lawns of K-12 and the
three strains and get the following results:

phage stocks
(host strain)


(K-12)
Skyler
Jesse
(Marie)
(Hank)
K-12
+
+
+
Test Strain
Skyler
Jesse
+
+
+
+
+
+
-
(+ = huge plaque, - = none or few single plaques)
Marie
+
+
+
Hank
-
Bacterial Problem Set
Did Dr. White lie to you? Yes, Hank is a lysogen. Give a reasonable explanation for the
differences between the five E. coli strains (i.e. what didn’t Dr. White tell you about the
strains?). The differences in plaquing efficiency are due to different restriction modification
(RM) systems. K12, Marie, and Hank have the same RM system, Jesse has a different
system that restricts growth to phage that has only been grown on itself, and Skyler has no
RM system that will restrict growth of phage derived from any of the other hosts. It may
be tempting to speculate that Jesse has a CRISPR array that includes sequence in one of
the spacers, and that escape mutants (mutations in the targeted sequence) arose that were
immune. However, this would not explain why this phage does not grow on K12.
3. Suppose Dr. White gives you the following additional information about the genotypes of
Hank and Skyler:
Hank = leu+, lac+, gal+, bio-, trp+, his-, recA-, cys+, met+
Skyler = leu-, lac-, gal-, bio+, trp-, his+, recA+, cys-, met(lac and gal are required for growth on plates containing either lactose (Lac) or galactose (Gal)
as the sole carbon source, respectively, and the others are auxotrophic markers). White says the
only way he will write a positive rotation evaluation is if you are able to do the impossible, make
a his+ trp+ strain. Luckily, you packed your handy E. coli chromosome map with the position
of these markers indicated (the entire chromosome is 100 min.):
leu lac
met
91’
gal
2’ 8’
bio
17’
17.5’
28’
62’
61’
cys
recA
trp
45’
his
After mixing cells of strains Hank and Skyler together, you find met+ his+ cells at high frequency.
You surmise that one of your strains is an Hfr strain and you perform a disrupted mating
experiment and plate the cells on different types of media. You get the following interesting
results, where + indicates the growth of at least one colony on the indicated media:
media:
mating
time
(min)
His- Met-
His- Leu-
His- Lac+
His- Gal+
His- Trp-
Bacterial Problem Set
0
1
2
3
4
5
10
15
20
25
30
35
40
45
50
55
60
+
+
+
+
+
+
+
+
-
+
+
+
+
+
-
+
+
+
-
+
+
-
-
a.
Which strain is the donor strain and which is the recipient? Hank = donor, Skyler =
recipient. From these results, can you determine if the donor molecule is an F’ plasmid or an
Hfr chromosome? Well, it could be an F’ that contains the region from met-gal. The only
problem with this is that it would be the biggest F’ ever - ~1.2 Mbp. So it has to be an Hfr.
If I forget to mention size restrictions in plasmids, then I wouldn’t take points off if they
say F’, but they should get the next question correct. If not, what experiment would you do to
determine this? There are at least two easy experiments one could do. First, one could
conjugate the donor with a recA- recipient (recombination is required for Hfr-mediated
genetic exchange, not for F’). The other way is to measure the frequency of exchange – Hfr
is lower than F’.
b.
Assuming the donor strain is Hfr, draw a map showing the position of the integrated
plasmid and indicate the direction of transfer. 4 min counter-clockwise to the met gene,
transfer proceeding clockwise.
c.
Propose a reasonable explanation for why no recombinants were ever found after 35 min
or longer of mating and why no Trp+ His+ recombinants were ever obtained. The answer is that
there is an integrated prophage at the λ attB site (between gal and bio!) in Hank but not
Skyler. When this region of the lysogen chromosome is transferred to the recipient, λ is
induced because there is no cI (newly replicated DNA w/o any proteins is delivered into
recipient cells and there is no repressor). This phenomenon is called zygotic induction. λ
replication ensues and kills the cell, hence no recombinants. How would you test this idea?
Checking for phage induction in Hank after DNA damage is not a correct answer – the
strain is recA- so it won’t induce the prophage. There are several ways to check this
(superinfection resistance, pcr). Perhaps the cleverest way would be to lysogenize Skyler
w/ λ (one could use the λ released from the mating) and perform conjugal transfer with
Hank and show that the bio, trp, his markers now can be transferred into Skyler.
d.
Design a scheme that would allow you to create the his+ trp+ strain. At least two ways
are possible. The first could be done with available reagents - lysogenize Skyler w/ λ and
Bacterial Problem Set
then mate, as described above. The second would be to use a transducing phage, such as
P1, grow on Hank and transduce Skyler, selecting for his and trp.