MCB 421 HOMEWORK #6
FALL 2009
1. (10 points) Complementation analysis was done on six mutants that lack methoinie synthetase
activity. The order of the mutational sites is not known. The results are shown below:
Mrg-1
Mrg-2
Mrg-3
Mrg-4
Mrg-5
Mrg-6
Mrg-1
-
Mrg-2
-
Mrg-3
+
-
Mrg-4
+
-
Mrg-5
+
+
-
Mrg-6
+
+
-
a. (2 points) How many complementation groups are represented?
ANSWER: At least two complementation groups: Group 1 = 2, 4;
Group 2 = 3, 5,6 ; Ungrouped = 1
b. (6 points) Suggest two explanations for the results of Mrg-1. What two categories do these
mutations fall under?
ANSWER: Mrg-1 fails to complement all of the other mutants. This could be either due to
a trans-dominant negative phenotype of this mutant (e.g. due to a missense mutation that
poisons threonine synthetase) or a cis-dominant negative phenotype caused by the mutation
(e.g. due to an amber mutation that prevents expression of downstream genes). In either
case, it is impossible to determine whether the thr-6 mutation is within one of the two
complementation groups described by the other mutations or whether it is in a different
complementation group.
c. (2 points) What is a simple experiment that would provide more information on the
mechanism of Mrg-1?
Answer: Complement to WT. If it does, its cis. If not trans.
MCB 421 HOMEWORK #6
FALL 2009
2. (8 points) 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
E. coli K-12
Phage
r- m+
r+
m+
r- m+
r+
m+
Phi 21
clear
clear
clear
-
lambda
+
-
+
+
Phi 80
-
-
-
-
a. ( 2 points) 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. ( 6 points) 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 (1pt) because it is not restricted by B but is restricted by K-12
(1pt). Lambda comes from K-12 (1pt) because it is restricted by B but not by K-12 (1pt).
Phi 80 does not grow on either B or K12 even when they are r- (1pt) 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).
MCB 421 HOMEWORK #6
FALL 2009
3. (20 points) Phage lambda is unable to grow lytically on wild-type Salmonella typhimurium.
a. (4 points) Suggest two possible reasons for these results.
Answer: It cannot bind or is defective in phage production.
To test each of these two pssibilites Salmonella typhimurium cells were infected with phage
lambda. Every 15 minutes cells were removed from the culture and pelleted. The supernatant
was removed and tested for phage. In addition, the cell pellet was lysed and tested for phage.
b. (8 points) Show a drawing of the expected results from the plaque forming assay of the
supernatant fraction for each of the two possibilities. (WT has been provided for you.
c. (8 points) Show a drawing of the expected results from the plaque forming assay of the
infected cell fractions for each of the two possibilities. (WT has been provided for you.
MCB 421 HOMEWORK #6
FALL 2009
4. (10 pints) The lacZ gene encodes an enzyme called B-galactosidase. B-galactosidase must
form a tetramer to be active. A. Ullman and J. Monod isolated two mutants of the lacZ shown
below. One of the mutants (am-90) was a nonsense mutant that contained an amber codon at the
site in the gene that encodes amino acid residues #90. The second contains a deletion (M15) that
deletes the indicted region of the lacZ gene.
A suppressor free strain of E. coli containing either mutation is Lac-. However, a suppressor free
strain containing a copy of each gene is Lac+. What is a specific molecular explanation for this
observation.
Answer: Alpha complementation. Since the protein needs to be a tetramer to be active,
each mutant can produce the part of the protein the other lacks. When they come together,
they will form a functional structure.
MCB 421 HOMEWORK #6
FALL 2009
5. (22 points) Phage HK51 integrates into the E. coil chromosome between the proB and
phoE genes. HK51 specialized transducing phage were isolated which carry the phoE
gene.
a. (5 points) Draw a diagram of the prophage map in a HK51 lysogen.
b. (5 points) If a HK5 lysogen is infected by λC1- or λ vir, it was found that a λC1- cannot
form plaques while λ vir can form plaques. What does this indicate?
Answer: The cell has immunity to C1- but not vir.
c. ( 2 points) Would your transducing particle contain cos? Why?
Answer: Yes to transducer you will need a cos site to package the virus.
d. (10 points) Draw a diagram showing how the LFT lysate could be formed following infection
of E. coli with HK51.
MCB 421 HOMEWORK #6
FALL 2009
6. (10 points)A deletion map of the maltose transport operon from Salmonella typhimurium
is shown below [from Schneider, E., L. Bishop, E. Schneider, V. Alfandary, and G. F.
Ames. 1989. J. Bacteriol. 171: 5860-5865]. The region removed by each deletion is
shown by the box below the genes and the corresponding allele number is indicated
above each line.
ANSWER: See shaded areas in the figure below. [Note that this question is similar, but not
quite the same as the one posted on the web, so simply copying the answer from the web
without analyzing the results will yield an incorrect answer.
The map position of several mal mutations was determined by mapping against these deletions.
The results are shown in the following table. Given these results, indicate the map position for
each of the mutations on the deletion map. Indicate precisely which deletion interval the
mutations map in. If it is not possible to determine whether a mutation maps in a single interval,
be sure to comment on this in your answer and suggest a reason that might cause such a result.
[Hint: use a ruler to draw the endpoints of the deletion intervals on the map, then you can
identify each deletion interval with a number or letter.]
MCB 421 HOMEWORK #6
FALL 2009
7. (20 points) P22 is a lambdoid phage that has a repressor gene (CII) that encodes a repressor
(CII) that functions in a manner similar to the lambda CI repressor. P22 also has a cro gene that
codes for a Cro protein that functions similarly to the lambda Cro protein. However, the
immunity properties of the two phages are different. Thus lambda CI repressor and Cro proteins
are specific for lambda OL and OR operation sites and P22 CII repressor is specific for P22 O L
and OR sites. The positions in the genome of P22 genes are designated on the phage maps
shown below.
P22 also has a region labeled lys that encodes lysis functions, a region call cap that codes for
phage capsids and a region called tail that codes for P22 tail fiber. The region called X is
relevant to the data presented below.
a. (5 points) When P22 infects a cell lysogenic for phage lambda, it can grow lytically and
produce P22 progeny. During lytic growth, it was also observed that lambda phage particles
were also formed and released from the cells into the medium. In a control experiment, where
cells were not infected by P22, no lambda phages particles are released. What is a simple
explanation of this result? What molecular evens would be required for release?
Answer: P22 induced lambda out of its lysogenic phase into lytic growth. This could be
because it either induces/supresses Cro or CI.
b. ( 5 points)
Answer: The X region codes for a protein that is involved in the formation of active lambda
particles. This protein induces virus formation.
c. Overexpression of lambda CI inhibits the signal to enter the lyric cycle. This protein must
compete with X. With more C1 than X, the phage stays lytic.
d. The mechanism of induction or suppression of this virus is shared between different types of
viruses.