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MCB 421 Exam #2
Fall 2007
There are 6 questions
Be sure your name is on each page
1. 3. (10 points) You have a strain of Salmonella typhimurium that contains the plasmid
pBR322. pBR322 carries a gene than encodes resistance to ampicillin (AmpR). The size
of pBR322 is about 5 kb. You grow wild type P22 and P22 HT on the strain and prepare
lysates. The titer of each lysate is 109 pfu/ml. When you infect an ampicillin-sensitive S.
typhimurium with the lysate made with wild-type P22 you obtain no AmpR colonies.
However, when you infect ampicillin-sensitive S. typhimurium with the P22 HT lysate,
you obtain 500 AmpR colonies.
a. (10 points) Why does the P22 HT lysate transduce AmpR while the wild type P22
lysate does not?
Answer: Wild-type P22 requires a pac site to package DNA into phage heads.
Apparently pBR322 does not contain a pac site or any good pac-like sites. P22 HT
does not require a pac site to package DNA so it can package the pBR322 DNA and
transduce it into a recipient. [P22 induces rolling circle replication of the plasmid so
that concatomers are actually packaged into phage capsids].
2. (6 points) A series of five mutants
of
phage
T4
were
isolated.
Complementation tests were performed by spotting a mixture of the two phage mutants to
be tested (about 106 of each) onto a lawn of bacteria at either 25°C or 40°C. The results
are summarized below (+ = complete lysis; − = no lysis or only a few plaques in the spot).
1
2
3
4
5
1
+
25°C
2
+
-
3
+
-
4
+
+
+
-
5
+
+
+
-
1
2
3
4
5
1
-
40°C
2
3
-
4
+
+
+
+
5
+
+
+
+
-
(a) (3 points) How many complementation groups do these mutants affect and which
mutants are in the same complementation groups?
ANSWER: Two complementation groups: group I = mutants 1,2,3 and group
II = mutants 4,5
(b) (3 points) Describe the nature of the mutation in each mutant (for example,
nonconditional, temperature sensitive, cold sensitive)?
ANSWER:
Mutant 1 = TS (temperature sensitive)
Mutant 2,3,5 = nonconditional (or null would be acceptable)
Mutant 4 = CS (cold sensitive)
3. (15 points) Imagine you have isolated a mutant strain of Salmonella typhimurium that
has a temperature sensitive mutation in the dnaA gene (dnaATS), You have isolated a
Tn10 insertion (encodes resistance to tetracycline) that is 25 % linked to the dnaATS
mutation by P22 HT mediated transduction.
a). (5 points) How could you transfer the dnaATS mutation to a new S. typhimurium
strain? How would you show that the new strain contained the dnaATS allele without
DNA sequencing? Be sure to include the composition of the media and temperatures of
each step. What frequency would you predict for transfer of the dnaATS allele in your
experiment?
Answer: Grow the dnaATS Tn10 strain at 30o, infect with P22 HT and make a lysate.
Use the lysate to infect the second strain at 30o and plate out on (LB) plates
supplemented with tetracycline. Incubate at 30o until colonies form. Streak or
replica plate the colonies onto a new plate and incubate at 42o. The colonies that
contain the dnaATS allele will not grow. You would expect that 25 % of the original
TetR colonies would be TS.
b). (10 points) Your lab mate has also isolated a temperature sensitive mutant. He claims
that the mutation is also in the dnaA gene. How could you determine if his mutation was
identical or different from your original dnaATS mutation?
Answer: (If the mutation in your lab mate’s strain is at the same site as your
mutation, you will not be able to obtain dnaA+ colonies when one strain is the donor
and the other is the recipient in a transduction cross). Use your P22 HT lysate
grown on your dnaATS Tn10 strain and infect the recipient. Select for TetR at 42o.
If the mutations are at different sites, you should obtain some colonies that grow at
42o because they have a dnaA+ allele made by recombination (a crossover between
the mutant sites). If the mutations are at the same site, you would expect no colonies
at 42o. (Actually if you selected the tetR colonies at 30o and streaked or replica
plated them and grew at 42o, the frequency of colonies that could grow at 42o would
depend upon the distance between the mutant sites. If close together, they would be
rare, but detectible if enough colonies were screened. If at the same site, no
temperature resistant colonies would be found)
4. (30 points). The diagram below shows the immunity region of phage λ. The genes
are:









cI – codes for cI repressor. The cI857 allele is temperature sensitive.
cro – codes for Cro protein.
N – codes for the transcription anti-terminator.
cII – codes for cI activator.
cIII – codes for the cIII protease inhibitor.
OL 1,2, and 3 and OR 1,2, and 3 are operators.
PL and PR are the rightward and leftward promoters.
PRE is the promoter for repressor establishment.
PRM is the promoter for repressor maintenance.
Calef et al. isolated a mutated lysogen that deleted most lambda DNA on the right and
left sides of the immunity region as shown below. The mutant also has the cI857 TS
allele (regions not shown below are deleted):
The lysogen had always been maintained at 30° C previously to the experiment
described below. They grew the lysogen at 30° C until cells reached mid-log growth.
They split the culture in half and grew one at 30° C and the other at 42° C for several
hours. They then lysed the cells from each culture as indicator bacteria for the ability of
wt λ and λ imm434 to form plaques to 30° C. They obtained the results shown in the
Table:
Indicator Cells
Grown at 30° C
Grown at 42° C
Cells Infected with wt
no plaques
clear plaques
Infected with limm434
normal,turbid plaques
normal , clear plaques
The indicator cells grown at 42° C were returned to 30°C and grown for several
generations and checked again for plaquing by wt λ and λ imm434. They obtained the
same results as with the cells grown at 42° and not shifted back to 30° (i.e. they still
obtained clear plaques with wt λ and turbid plaques with λ imm434).
a. (5 points) Why doesn't wt λ form plaques on the indicator lysogen grown at 30° C?
ANSWER: At 30°C, the cI857 repressor in the lysogen is active and will bind to the
right and left operators of the infecting phage. This will prevent transcription of any
phage proteins and thus prevent lysis.
b. (5 points) Why can λ imm434 (as opposed to wt λ, above) form plaques on the
indicator lysogen grown at 30°? Why are the plaques turbid?
ANSWER: λ imm434 has the immunity region from phage 434, so the λ repressor is
unable to bind its operators and repress transcription. So, the infecting phage may
lyse the indicator. The plaques are turbid because the λ imm434 will eventually start
making its own cI-like repressor (probably once the MOI becomes high) and form
lysogens.
c. (10 points) Why does wt λ form clear plaques on the indicator strain if it is grown at
42°C?
ANSWER: λ is able to form plaques on the lysogen at 42° because the cI857
repressor is inactive at 42° and so does not prevent lysis by the infecting phage. The
lack of cI in the recipient lysogen leads to accumulation of cro protein. Cro protein
prevents transcription of cI, both in the lysogen and in the incoming phage. This
ensures that incoming phage will only be able to grow lytically, and so the plaques
are clear.
d. (10 points) Why doesn't growth of the indicator lysogen at 30°C for several hours
restore the original phenotype (no plaques with wt λ) of indicator cells once they have
been grown at 42°?
Answer: Once cI is deactivated at 42°, transcription of cro protein starts. Once cro
protein is present, it prevents transcription of cI, even when the temperature is
shifted back down to 30°C. Thus, the lysogen loses immunity even when shifted back
to 30°.
5. (25 points) Arber and Dussoix (1962, J. Mol. Biol. 5:37 – 49) performed the
following series of experiments to study restriction and modification in E. coli. In one
experiment they grew phage lambda on an E. coli K strain and an E. coli K strain
lysogenic for phage P1- E. coli K (P1). They used the lysate obtained to perform plaque
forming assays on either E. coli K or E. coli K (P1) as the indicator strains. They
obtained the data in the table below.
Relative efficiency of PFU / ml
Lysate
E. coli K
E. coli K (P1)
Lambda on K
1
0.001
Lambda on K (P)
1
1
a, (5 points) What is the likely explanation for this result? Why?
Answer: The P1 prophage has a restriction-modification system. The system is
different from the K system. When lambda is grown on the K host, the DNA is K
modified but not P1 modified. Thus when this phage infects the K strain, they make
plaques efficiently because the DNA is not restricted. However, the P1 restriction
system degrades the DNA and the frequency of plaques decreases by 1,000-fold.
The lambda phage grown on the E. coli K (P1) host is modified for both K and P1
sites so they are resistant to both K and P1 restriction systems. OR the few plaques
found when the K-grown phage infects the P1 lysogen are rare chromosomes that
became P1 modified before the P1 restriction enzyme degrades the DNA.
b. (10 points) In a second experiment, they labeled lambda DNA of phage growing in E.
coli K (P1) with 32P so that the newly synthesized DNA in the phage was radioactive.
They then infected E. coli K cells with the radioactive phage in medium containing
nonradiactive phosphorus and found that the progeny phage that form plaques on E. coli
K are resistant to inactivation by radioactive decay. However, the progeny that were able
to form plaques on E. coli K (P1) were destroyed as time passed. What is the explanation
for this result?
Answer: The vast majority of DNA synthesized during growth in the E. coli K strain
is not radioactive because the phages were made in non-radioactive medium.
However, a (rare) phage that is able to form a plaque on E. coli K (P1) must contain
one strand that has the original P1 modification so it can prevent the P1 restriction
enzyme from destroying its DNA. This strand will also contain 32P from growth in
the presence of the radioisotope and decay of the isotope over time emits radiation
that destroys the infectivity of the phage DNA (i. e. radioactive decay destroys the
chromosome.
They also performed another experiment using density labeling of phage DNA. They
grew lambda on E. coli K (P1) in medium containing deuterated water (D2O). D is a
heavy isotope of hydrogen. Incorporation of D instead of H into DNA makes it denser
than DNA made in the presence of H2O. They used the resulting lysate to infect E. coli
K in medium containing normal H2O and used density gradient centrifugation to separate
the resulting light phage form the more dense phage (Don’t worry about how this was
done). They found that the phages that formed plaques on the E. coli K (P1) strain were
heavy. Light phages (containing only hydrogen) could only form plaques on E. coli K.
c. (10 points) What is the explanation for this result?
Answer: The observation that the higher density phages (meaning that the DNA
they contain is labeled with D) form plaques on E. coli K (P1) suggests that one
strand of their DNA is modified. This indicates that these phage contain a strand of
DNA from the original “heavy” phage and a complementary “light” strand made
from DNA replication containing hydrogen in the E. coli K host. Thus, these phages
can form plaques on E. coli K (P1) because the hemimethylated P1 site in the
original parental strand prevents cleavage by the P1 restriction enzyme. The vast
majority of the phage growing in E. coli K in the presence of H20 are light because
all of their DNA is light. Since they are growing in E. coli K, their DNA is not P1
modified. Thus the light phage can only make plaques on E. coli K.
6. (14 points) Compare and Contract Generalized and Specialized Transduction for the
indicated properties.
Generalized Transduction
Site-Specific Transduction
Mechanism of Formation of
Transducing Particles
Packaging Chromosomal
DNA
Aberrant Excision
Method of DNA Packaging
Headful/pac
Headful/cos
Markers for Transducing
Any (or many)
Bacterial + phage
DNA in capsids of
Transducing Particle
Bacterial Only
Bacterial + phage
Mechanism of Transduction
HR
HR or site specific
Enzymes/protein required
RecA mediated
Int/RecA
Susceptibility to host
modification
Susceptible
Susceptible