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MCB 421
HOMEWORK 7
Due 10/20
FALL 2011
1. 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. 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. 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 42 o.
(Actually if you selected the tetR colonies at 30 o and streaked or replica plated them and grew at 42 o,
the frequency of colonies that could grow at 42 o 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)
2. A temperature sensitive mutation was isolated in the gene encoding NAD kinase from Salmonella
[Cheng and Roth. 1994. J. Bacteriol. 176: 4260-4268]. Two factor crosses were done to map the nadF
mutation against nearby markers. The lip gene encodes enzymes required for the biosynthesis of lipoate
and the cam gene encodes resistance to chloramphenicol. Given the results shown below, draw a genetic
linkage map showing the most likely gene order and the coinheritance frequencies with arrows pointing
toward the selected marker. [You do not need to refer to the reference to answer the following questions.]
Donor
Recipient
lip- nadF+
lip+ nadF-
Selected
phenotype
NadF+
cam+ nadF-
cam- nadF+
Cam+
lip- cam+
lip+ cam-
Cam+
The co-transduction frequencies are:
Cross #1- # nadF+ + #lip- =
Total (500)
Cross #2-
#cam+
5%
+ # nadF- = 10 %
Recombinants
LipLip+
NadFNadF+
LipLip+
# obtained
25
475
20
180
4
496
MCB 421
HOMEWORK 7
Due 10/20
FALL 2011
Total (200)
Cross #3-
#cam+ + # lipTotal (500)
= 0.8 %
3. 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
Lambda on K
Lambda on K (P1)
E. coli K
1
E. coli K (P1)
0.001
1
1
a. What is the likely explanation for this result? Why?
Answer: The P1 prophage strain 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 these phages 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 few plaques found when the K-grown phage infects the P1 lysogen are
rare lambda chromosomes that became P1 modified before the P1 restriction enzyme degrades the
DNA). 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.
b. 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 nonradioactive 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
phage 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
MCB 421
HOMEWORK 7
Due 10/20
FALL 2011
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. 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.
4. Phages Phi 21, and Phi 80 were tested for the ability to infect two strains of E. coli. As a control, phage
-
+
lambda was grown on E. coli K-12 r m was tested. The results are shown in the table below. “Clear”
indicates that the plaques were clear while “Turbid” indicates the plaques were turbid. (-) indicates that
plaques were not made.
E. coli B
Phage
-
+
r m
r
E. coli K-12
+
+
m
r
-
+
m
r+
m+
Phi 21
clear
clear
clear
-
lambda
+
-
+
+
Phi 80
-
-
-
-
a. What can you infer about the lifestyle of phage Phi 21 based on the type of plaques formed?
Answer: Phi 21 appears to be a lytic phage because it makes clear plaques).
b. Suggest a likely explanation for the growth of each phage (Phi 21, Phi 80 and lambda) on the E. coli B
and K-12 strains.
(Phi 21 comes from E. coli B ( because it is not restricted by B but is restricted by K-12. Lambda
comes from K-12 because it is restricted by B but not by K-12. Phi 80 does not grow on either B or
K12 even when they are r- so it is likely that it cannot absorb to B or K-12 or else it can’t undergo
morphogenesis in B and K-12).
c. Suggest two different reasons why Phi 80 can’t grow on E. coli B or E. coli K.
(The E. coli strains might lack a receptor for Phi 80 or lack a gene product needed for Phi 80
morphogenesis).
d. Draw the predicted results from a one-step
growth curve for each of the two possibilities
in the following figure. Also include the
results expected for infection of a sensitive cell
by Phi80. (The Y axis should be PFU / ml and
the X axis should be TIME (in minutes).
MCB 421
HOMEWORK 7
Due 10/20
FALL 2011
(See handout on lytic growth of phage).
Questions to Ponder
5. E. coli K and B have different restriction and modification specificities, indicated by the subscript K or
B respectively. A variety of mutations were isolated in the two restriction systems with the phenotypes: rm+ or r-m-. The ability of
E. coli strains to grow on each of the
indicated host strains and a merodiploid strain carrying the genes for both restriction systems is shown in
the table below. [+ = lysis; - = no lysis].
]
Phage
E. coli strain
rB+mB+ r +m +
rB-mB+
rK-mK+
rB+mB+/rK-mK+
K K
+ +
B mB )
+ +
K mK )
- +
B mB )
- +
K mK )
+ + - +
B mB /rK mK )
a.
+
-
+
+
+
-
+
+
+
-
+
-
+
+
+
-
+
+
+
-
+
+
+
+
+
Explain the results for the growth of each phage on each of the 5 strains.
ANSWER:
+
+
+
• rB+mB+ = only phage modified by mB+
B mB
B mB
(rB+mB+/rK-mK+)]
+
+
+
• rK+mK+ = only phage modified by mK+
K mK
K mK
(rB+mB+/rK-mK+)]
• rB-mB+ = no functional restriction for B or K so all the phage will grow
• rK-mK+ = no functional restriction for B or K so all the phage will grow
• rB+mB+ / rK-mK+ = will restrict growth of any phage not modified by mB+
+
+
- +
[
K mK
k mk )] ; will not restrict growth of phage previously modified by
+
+
+
+ +
+
mB+
B mB
B mB ),
B mB /rK mK )]
b.
Why were no r+m- mutants isolated?
ANSWER: Such mutants would probably be lethal because the host chromosome would be
degraded.
6. Phage lambda is unable to grow lytically on wild-type Salmonella typhimurium.
a.
Suggest two possible reasons for these results.
ANSWER: Either (i) lambda may not be able to adsorb to S. typhimurium or (ii) S.
typhimurium may be missing some host function required for maturation of lambda (i.e.
lambda would adsorb to the cells but would not lyse them).
MCB 421
b.
HOMEWORK 7
Due 10/20
FALL 2011
How would a one-step growth curve distinguish between these two possibilities? [Show a drawing
of the expected results for both of the possiblities listed above.]
ANSWER: