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MCB 421 HOMEWORK #4 ANSWERS
FALL 2011
Page 1 of 4
Due Sept 27th
You need to only answer questions 1-3.
1. You have obtained five phage T4 plaque morphology mutants that form tiny plaques on E. coli whereas
the wild-type phage forms large plaques. The mutants are designated r1, r2, r3, r4, and r5. In order to try to
determine the type of mutation present in the r gene of each phage, you perform a reversion analysis of the
mutants and measure the frequency that you get wild-type plaques (Don’t worry about how this was done).
The results are shown in the Table below.
Mutagen X is an unknown mutagenic compound.
(+) means that revertants are frequent
(-) means revertants are very rare
(+/-) means that the frequency is intermediate
Mutagen
Mutant
r1
r2
r3
r4
r5
5BU
+
+
+
-
EMS
+/+/+
-
NG
+/+/+
-
ICR191
+
-
X
+
+
-
HA
+
-
a. What is the likely type of mutation (ie. missense, nonsense, frameshift, insertion, or deletion) in each
mutant phage? Why?
ANSWER: 1, 3, 4 are missense or nonsense because they revert with mutagens that make base
substitutions. 2 is a fs because it reverts with ICR191, a mutagen that favors frameshifts. 5 is likely a
deletion or insertion. (Possibly a fs that can’t be reverted by a second site mutation.)
b. Can any specific predictions be made about the base changes made by any of the mutagens?
ANSWER: 4 is reverted by a GC to AT transition because HA is specific for GC to AC. Mutagen X
must revert AT to GC because it doesn’t revert HA induced mutations and HA doesn’t revert any X
induced mutations. A mutagen that is bidirectional (5BU, revert both HA and X as expected. 4 is
reverted by HA and EMS and NG that prefer GC to AT changes,1 and 3 could be reverted by AT to
GC because EMS and NG prefer GC to AT.
c. Which mutagens would most likely produce revertants that are TS or CS? How could such revertants
occur?
ANSWER: All but ICR191 make base substitutions so revertants that are TS or CS must not be true
revertants. Thus secondary site substitutions, either in the original mutant codon or elsewhere in the
gene would be able to make TS or CS mutants. ICR most likely will not make TS or CS because they
usually make inactive proteins.
2. Mutations in genes that repair DNA damage often result in a “mutator” phenotype. Such mutator strains
accumulate mutations at a much higher frequency than that observed in wild-type cells.
a.
Some investigators do not like to use mutator strains to isolate mutations in genes within the
bacterial chromosome. Instead they use mutator strains to isolate mutations in phage chromosomes or in
plasmids. Suggest a reason for the preference. If you did use a mutator strain, what would be a high priority
MCB 421 HOMEWORK #4 ANSWERS
FALL 2011
Page 2 of 4
Due Sept 27th
for the next step in your project? (You may not know how to do what should be done, but what would you
propose to do?)
ANSWER: Use of a mutator strain results in many mutations on the bacterial chromosome, but
ideally you want to study a single mutation. Plasmids and phage chromosomes are used with mutator
strains since they are easy to remove from the mutator strain and transfer to a non-mutator strain. If
you used the mutator strain bacterial chromosome you should perform a backcross. (ie. Move the
desired mutation to the parent strain by generalized transduction)
b.
You grow phage lambda lytically on a wild-type strain and on a mutD strain. You then check the
phage particles from each infection for phage mutants that affect the morphology of the phage plaques on
normal bacteria. Which host strain would give you the highest yield of mutants? Why?
ANSWER: The mutD strain, which contains a mutated DNA pol III proofreading subunit, increases
the mutation rate 1000x.
c.
Often, investigators grow multiple independent cultures inoculated from independent colonies of
mutator strains for mutagenesis. Why?
ANSWER: This reduces the chances of isolating sibling mutations.
3. You need to isolate a temperature sensitive (TS) tryptophan auxotroph. Such a mutant would be able to
synthesize the amino acid tryptophan at 30 C but unable to synthesize tryptophan at 42C.
a. What mutation type would most likely give rise to this TS mutant?
ANSWER: Missense mutation
b. Design a protocol (starting with a mutagen) that would allow you to isolate such a mutant. Be sure to use
an ENRICHMENT, to give the COMPOSITION OF THE MEDIA at each step and give the
TEMPERATURE you would use at each step. (Hint: This is very similar to the example discussed in class
except that the temperature of the steps needs to be specified).
ANSWER: Begin with a mutagenized culture. The mutagen may be any that is capable of causing a
missense mutation. The culture should then be grown in minimal media at 30C. Both trp+ cells and
the desired mutant will be able to grow under these conditions. Growing on minimal medium at this
step also eliminates any non-conditional auxotrophs. After outgrowth, the population may be
enriched for TS Trp auxotrophs by incubating at 42C in minimal media with penicillin. under these
conditions, the desired mutant will be unable to grow, but most other cells will continue growing,
leading to death by penicillin. This step may be repeated after outgrowth in minimal media at 30, if
desired. Following enrichment, the cells should be plated on minimal media at 30C. The resulting
plates may then be screened for the desired mutant by replicating to one minimal media + Trp plate
and one minimal without Trp plate. Both plates should be grown at 42C. Any colony that grows on
the Trp plate but not on the no Trp plate is a TS tryptophan auxotroph. This is the ideal protocol,
but others are possible. However, certain steps are essential. There must be a penicillin enrichment to
increase the proportion of the population with the desired mutation. At some point, either in the
outgrowth phases or upon plating the enrichment, the population must be made to grow in minimal
media at 30. This is to prevent the isolation of mutants that are Trp- rather than TrpTS. Trp- cells
will be unable to grow in these conditions, while TrpTS ones will. Also, the screening step must
include a minimal + Trp plate at 42. If you only compare minimal at 30 to minimal at 42, then you
can just as easily isolate TS mutants in essential genes as TS mutants in Trp synthesis. Finally, using
rich media in this protocol is probably a poor idea, since this would allow for the growth of many
auxotrophs, not just Trp ones.
Additional questions to ponder (don’t turn these in):
MCB 421 HOMEWORK #4 ANSWERS
FALL 2011
Page 3 of 4
Due Sept 27th
4. Strain FC40 is unable to grow on lactose as a sole carbon source (Lac -). A Luria-Delbruck fluctuation
test was done by plating 108 cells from 10 independent cultures (#1-10) onto minimal lactose plates, and at
the same time plating 108 cells from a single culture (#11) onto 10 minimal lactose plates. After a long
incubation, the following results were obtained.
Independent Cultures
(Culture #)
1
2
3
4
5
6
7
8
9
10
Number of
Colonies
22
18
19
24
20
23
21
22
21
17
Culture 11
(Plate #)
1
2
3
4
5
6
7
8
9
10
Number of
Colonies
17
24
23
26
20
22
21
24
20
19
Based upon these results, would you conclude that the mutation to Lac + is random or adaptive? (Briefly
explain your answer.)
[The variance between the number of Lac+ colonies from individual cultures equals that seen with
multiple samples of a single culture. These results suggest that the phenotype is not likely to be due to
a random, spontaneous mutation. (5 points for this) (A random mutation which would show much
greater variance between individual cultures.) The simplest interpretation of these results is that the
mutation to Lac+ is an example of “adaptive or induced mutagenesis” (5 points for this). Note – your
answer must describe the comparison of the variance between cultures #1-10 vs the multiple platings
of culture #11. When given a problem like this, you need to evaluate the data given, do not simply say
that it must be random mutagenesis because I said in lecture that mutagenesis is random!]
5. Strains with null mutations in the leuA, leuB or leuC genes are leucine auxotrophs. You decide to isolate
o
revertants of a leuA mutant at 30 C.
a. How would you isolate the revertants? What medium would you use? Is this a selection or a screen?.
ANSWER: Plate cells on media without leucine at 30°C and look for growth. This is a selection
b. You isolate a leuB revertant and find that the reversion is a mutation at a second site in the leuB gene. In
addition, the revertant is temperature sensitive for leucine synthesis. How would you demonstrate that the
revertant is temperature sensitive for leucine synthesis?
ANSWER:
-
leuB LB
-
leuB Minimal + Leu
-
leuB Minimal
+
leuB (revertant) LB
+
leuB (revertant) Minimal + Leu
+
leuB (revertant) Minimal
30°C
42°C
+
+
+
+
-
-
+
+
+
+
+
-
c. What is the most likely explanation for the temperature sensitive phenotype?
MCB 421 HOMEWORK #4 ANSWERS
FALL 2011
Page 4 of 4
Due Sept 27th
ANSWER: The second site mutation allows proper protein folding at the permissive temperature
(30°C), thus creating a functional LeuB protein, and allowing leucine biosynthesis to take place. At
the non-permissive temperature (42°C) the leuB revertant produces an unstable protein that does not
allow leucine biosynthesis to occur.
d. It is possible to isolate leucine auxotrophs that do not have mutations in the leuA, B, or C genes. Some
have mutations in the leuS gene, which codes for the leucyl-tRNA synthetase protein. The enzyme is not
required for leucine biosynthesis but is responsible for amino acylation of leu-tRNA for protein synthesis.
What type of mutation would the leuS mutant likely contain? Would you expect the mutation to be a null
mutation?
ANSWER: Its possible to get mutants that make leaky tRNA synthetases that result in auxotrophy.
They are usually mutants that can't charge efficiently unless the intracellular concentration of the
correct amino acid is high (by supplementing in the plates). Dr. Gardner actually got trpS mutants
by looking for trp auxotrophs.
The leuS gene codes for the enzyme that charges leu-tRNA. So the mutation makes a leaky LeuS
protein that can't charge the tRNAs efficiently enough to allow leucine insertion into proteins. In
other words, the cell can't make enough Leu to help the mutant LeuS enzyme charge tRNA. (It’s like
a mutation that changes the Km of the LeuS protein for leucine). When leucine is supplied for
growth, the cellular concentration increases so the defective enzyme charges more of the tRNAs to
allow for growth.
It would be a missense mutant that could affect the catalytic site of the enzyme so it charges less
efficiently.
e. You isolate a revertant of the leuS mutant and find that the mutation maps at a site adjacent to the leuS
gene. What is the likely mechanism for the revertant phenotype?
The mutation is likely in the promoter for the LeuS gene that increases expression of the gene. The
increased expression of the defective gene gives the cell enough charged tRNA to overcome the effects
of the mutant.