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MCB421 HOMEWORK #1
FALL 2011
Page 1 of 6
1. CARP is an essential cofactor for many cellular processes. Because it is not transported, exogenous
CARP cannot supplement mutants unable to synthesize intracellular CARP. Five independent mutations
were obtained that affect the synthesis of CARP. The properties of the mutations are described in the table
below (where + indicates growth on rich medium, - indicates that no growth on rich medium, and -/+
indicates weak growth on rich medium).
Row
#
1
2
3
4
5
6
7
8
9
10
11
12
Mutation
car
car-601
car-602
car-603
car-604
car-606
car-607
car-601 car-602
car-601 car-603
car-601 car-604
car-601 car-607
car-604 car-607
30°C
42°C
+
+
+
+
-/+
+
+
-
+
+
-/+
+
+
Growth temperature
30  42°C
42  30°C
+
+
+
+
+
+
+
+
+
+
a. Note the properties of nad-601, nad-602, nad-603, nad-604, nad-606, and nad-607 in the above Table.
Indicate both whether the mutant has a conditional phenotype (temperature sensitive, cold sensitive, or nonconditional) and whether the allele is likely to be due to a missense, nonsense, frameshift, deletion, or
insertion mutation? Briefly explain your answers.
ANSWER:
car -601 Ts, missense (Probably AA substitution that destabilized protein)
car -602 Ts, missense
car -603 Ts, missense
car -604 Cs, missense
car -606 Leaky, nonconditional (probably a missense mutation because gene product retains some activity)
car -607 Cs, missense
All of these mutations are probably missense because Ts and Cs mutations usually arise due to single amino acid
substitutions, and the leaky mutation retains some activity so it is clearly not due to a complete gene disruption.
b. Interpret the results for each pair of double mutants in rows # 8-12. If you are not able to determine the
order of the reactions catalyzed by some of the gene products from the data given, suggest a likely reason
for this result.
ANSWER:
8:Cannot interpret gene order because both mutations are Ts
9:Cannot interpret gene order because both mutations are Ts
10:Mutation nad-604 (Cs) must act before nad-601 (Ts)
MCB421 HOMEWORK #1
FALL 2011
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11:Mutation nad-601 (Ts) must act before nad-607 (Cs)
12:Cannot interpret gene order because both mutations are CS.
2. You have LB (rich medium) cultures of two E. coli strains. One is a temperature sensitive (TS) leucine
auxotroph and the second contains a TS mutation in the rpoA gene that codes for the alpha subunit of RNA
polymerase.
a.) Which mutation is most likely to affect an essential function? Why?
Answer: A functional rpoA is required for transcription and hence for growth under any
conditions. Functional leucine biosynthesis is expendable when leucine may be obtained from
the environment.
b.) Using a simple agar plate test, how could you distinguish the two strains? Be sure to include the
composition of the media you would use and the temperature(s) you would incubate the plates.
Answer: The two strains can only be distinguished at 42 on rich medium or minimal medium
with leucine. Under these conditions, the TS Leu auxotroph will grow and the TS rpoA mutant
will not. Under all other conditions, the two strains will display the same growth phenotype (see
table).
Media Composition and
growth temperature
Minimal, 30
Minimal, 42
Minimal + Leu or Rich, 30
Minimal + Leu or Rich, 42
TS Leucine
auxotroph
+
+
+
TS rpoA mutant
+
+
-
c.) Which class (i.e. point mutant, deletion, insertion, et.) are these mutants likely to be? Why?
Answer: The mutations are likely to be point mutations, or more specifically missense mutations.
TS mutants are altered function mutants, and so require a change in the amino acid sequence.
Generally speaking, a missense point mutation is the only way to change an amino acid sequence
without destroying protein function completely.
3. The trp operon contains five genes that are required to synthesize the amino acid tryptophan. Seven
independent mutants were isolated that affected the biosynthesis of tryptophan. The growth properties of
the mutants and strains with two mutations are shown in the Table below. The plates used were minimal
medium lacking tryptophan. A (+) means growth, a (-) means no growth and a (+/-) means weak growth.
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FALL 2011
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Row #
Mutation
30°C
1
2
3
4
5
6
7
8
9
10
11
12
trp+
trp101
trp102
trp103
trp104
trp105
trp106
trp107
trp103 trp104
trp103 trp105
trp103 trp106
trp105 trp106
+
-/+
+
+
-
Growth temperature
42°C
30  42°C
+
-/+
+
+
+
42  30°C
+
+
+
+
+
+
-
a). Why were plates lacking tryptophan used for these experiments? What result would you expect if
plates supplemented with tryptophan were used?
ANSWER:
To determine if these mutations affected the synthesis of tryptophan; the mutants must synthesize
tryptophan or else they will not survive. If plates supplemented with tryptophan were used, all the
mutants should be able to grow.
b). Note the properties of trp101, trp102, trp103. trp104. trp105. trp106, and trp107. Indicate whether the
mutant has a conditional phenotype (temperature sensitive, cold sensitive, null) and whether the allele is
likely to be due to a missense, nonsense, frameshift, deletion, or insertion mutation? Briefly explain your
answers.
ANSWER:
trp101………null mutation; could be insertion, deletion, frameshift,
nonsense or missense
trp102………leaky, missense; missense mutation because gene
product retains some activity
trp103………cold sensitive, missense
trp104………cold sensitive, missense
trp105………temperature sensitive, missense
trp106………temperature sensitive, missense
trp107………cold sensitive, missense
c). Interpret the results for each pair of double mutants in rows 9 to 12. If you are unable to determine the
order of some of the gene products in the tryptophan biosynthesis pathway from the data given, indicate
which gene products fall into this category.
ANSWER:
MCB421 HOMEWORK #1
FALL 2011
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trp103trp104………can’t tell
trp103trp105………trp105  trp103
trp103trp106………trp106  trp103
trp105trp106………can’t tell
d). How would you select for a revertant of trp101?
ANSWER:
Plate on media without tryptophan at 30C and 42C and look for colonies that can grow. These will
be revertants of trp101.
4.
A diagram of the E. coli lac operon is shown below. The three gene products can be assayed
independently. Many mutations in the upstream region of the lacZ gene (labeled 1) are strongly
polar on expression of the lacY and lacA genes, but mutations near downstream end of the lacZ
gene (labeled 2) are much less polar. Propose a molecular explanation for this observation. [Hint:
think about the coupling of transcription and translation and the role of rho factor.
[See notes and handouts for mechanism of polarity. Nonsense mutations near the 3’ end of a
gene are often less polar than nonsense mutations near the 5’ end. This is likely because
termination of translation in mRNA near the 3’ end of the gene occurs close to a ribosome
binding site and AUG codon for the next downstream gene. Ribosomes can reinitiate translation
and thereby protect the mRNA from binding Rho thus inhibiting transcription termination.
Nonsense codons further upstream also lead to dissociation of the ribosome but there is no
ribosome binding site and AUG nearby to reinitiate translation and prevent Rho binding. Thus
as the RNA polymerase continues transcription, naked RNA is exposed that binds Rho and
promotes transcription termination.]
Starting with a mutant with a mutation in region #1 of the lacZ gene, Beckwith and his collaborators
isolated a mutant that was still LacZ- but expression of LacY and LacA was normal. When the secondary
mutants were mapped (don’t worry how) they were found to be unlinked to the Lac operon. They called
these mutants “polarity suppressors”. What is a likely mechanism for the action of the polarity
suppressor?
[Polarity requires Rho factor so the suppressors could affect the action of Rho in premature
transcription termination that causes polarity. The mutation doesn’t correct the lacZ mutation so
the strain is still LacZ-. {The suppressor mutation could be in a tRNA gene which allows translation
of the lacZ mRNA containing the non-sense codon. We will encounter these informational
suppressors later in the course}.
Other possibilities include mutations in a subunit of RNA polymerase that don’t allow normal
interactions with Rho thus reducing termination and relieving polarity.
See the discussion of polarity under "Chromosomes, Genes, and Proteins" on the course web
supplement.]
MCB421 HOMEWORK #1
FALL 2011
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.
Phage T4 is a bacterial virus that lyses E. coli, producing a clearing in a lawn of sensitive bacteria. This
clearing is called a “plaque”. Benzer isolated a class of conditional mutations in phage T4 that depended
upon the strain of E. coli that the phage infected. T4 rII mutants grow on E. coli B strains but not E. coli K12 strains.
Phage Strain infected
E. coli K-12
E. coli B
T4
small plaques
small plaques
T4 rII
no plaques
large plaques
a. How could you select for revertants of T4 rII mutants?
[Revertants could be isolated by plating a pool of rII phage on K12 (λ+); only revertants will be able
to form plaques. This is a selection. Looking for wild type plaques on B would require a lot more
work since most of the plaques would be r type and wild type plaques would be exceedingly rare)]
b. How could you distinguish deletion mutations from point mutations?
Point Mutations often result in a protein that is folded much like the wild-type protein and thus often
is relatively stable. In contrast, deletion mutations are typically degraded rapidly by cellular
proteases since the protein is no longer functional. Can use antibodies to tell the difference.
c. If a mutation were stimulated to revert by proflavin (an intercalating agent), what would you
conclude about the mutation?
Proflavin induces frameshift mutations when phage are grown in cells treated with proflavin.
d. If a mutation were stimulated to revert by hydroxylamine, what would you conclude about the
mutation? [Hint: see MCF pg 192 or web supplement on mutagens.]
HA is specific for G:C to A:T transitions.
Streisinger et al. isolated a set of frameshift mutations and intragenic suppressors in the lysozyme gene of
phage T4. The sequence of part of the wild-type protein and the corresponding mutant region is shown
below for two different mutants. [From CSHL Symp. Quant. Biol. (1966) 31: 77-84]
Wild-type: Thr Lys Ser Pro Ser Leu Asn Ala Ala
Mutant 1: Thr Lys Val His His Leu Met Ala Ala
Wild-type: Thr Lys Ser Pro Ser Leu Asn Ala Ala
Mutant 2: Thr Lys Ser Val His His Leu Met Ala
Using the genetic code table, determine the likely nucleotide sequence of the wild-type and mutant DNA
sequences, noting the position of the two frameshift mutations in the mutant genes. [Note: a genetic code
table is shown on the inside cover of MCF.]
WT =
ACX AAA/G AGU CCA UCA CUU AAU GCX GCX
#1 =
MCB421 HOMEWORK #1
FALL 2011
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ACX AAA GUC CAU CAC UUA AUG GCX GCX
This mutant has a deletion of the A/G and an insertion of +1G between the U and G.
#2 is more difficult because it’s a 2 bp frameshift.
WT =
ACX AAA/G AGU CCA UCA CUU AAU GCX GCX
#2 =
ACX AAA/G AGU GUC CAU CAC UUA AUG – 2 BP GCX
Insertion of GU and deletion of CX
Farabaugh et al. measured the frequency of spontaneous mutations at different sites in the lacI gene of E.
coli [J. Mol. Biol. (1978) 126: 847-863]. Of 140 spontaneous mutations obtained, 37 were deletions. Of
these deletions, 18 removed a specific 4 bp sequence. The DNA sequence of the region surrounding this
deletion hot spot is shown below. Which base pairs are most likely to be deleted and why? [Hint: see
“Deletion mutations” in the Mutants and Mutations section of the course web supplement.]
TCGGCGCGTCTGCGTCTGGCTGGCTGGCATAAA
AGCCGCGCAGACGCAGACCGACCGACCGTATTT
GCGT or GCTG