MCB421 HOMEWORK #1FALL 2011
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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
/ Mutation / Growth temperature# / 30°C / 42°C / 30 42°C / 42 30°C
1 / car / + / + / + / +
2 / car-601 / + / -
3 / car-602 / + / -
4 / car-603 / + / -
5 / car-604 / - / +
6 / car-606 / -/+ / -/+
7 / car-607 / - / +
8 / car-601 car-602 / + / - / + / +
9 / car-601 car-603 / + / - / + / +
10 / car-601 car-604 / - / - / - / +
11 / car-601 car-607 / - / - / + / -
12 / car-604 car-607 / - / + / + / +
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 non-conditional) and whether the allele is likely to be due to a missense, nonsense, frameshift, deletion, or insertion mutation? Briefly explain your answers.
ANSWER:
car -601Ts, missense (Probably AA substitution that destabilized protein)
car -602Ts, missense
car -603Ts, missense
car -604Cs, missense
car -606Leaky, nonconditional (probably a missense mutation because gene product retains some activity)
car -607Cs, 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)
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 andgrowth temperature / TS Leucine
auxotroph / TS rpoA mutant
Minimal, 30 / + / +
Minimal, 42 / - / -
Minimal + Leu or Rich, 30 / + / +
Minimal + Leu or Rich, 42 / + / -
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.
Row #
/ Mutation / Growth temperature30°C / 42°C / 30 42°C / 42 30°C
1 / trp+ / + / + / + / +
2 / trp101 / - / -
3 / trp102 / -/+ / -/+
4 / trp103 / - / +
5 / trp104 / - / +
6 / trp105 / + / -
7 / trp106 / + / -
8 / trp107 / - / +
9 / trp103 trp104 / - / +
10 / trp103 trp105 / + / -
11 / trp103 trp106 / + / -
12 / trp105 trp106 / + / -
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:
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 30C and 42C and look for colonies that can grow. These will be revertants of trp101.
- 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.]
.
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 K-12 strains.
Phage / Strain infectedE. 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 AAUGCX GCX
#1 =
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