Download MCB 421 HOMEWORK #4 ANSWERS FALL 2006 Page 1 of 3

HOMEWORK #5 ANSWERS
FALL 2011
Due Sept 27th
Page 1 of 3
Questions 1-2
1. An amber mutation in phage T4 can grow on strains carrying sup-1 but not on strains containing sup-2,
even though both sup-1 and sup-2 are amber suppressors. Suggest an explanation for this result.
ANSWER: sup-1 and sup-2 are both amber suppressors -- due to a mutation in the gene encoding a
tRNA which allows recognition of the UAG codon. However these two mutations affect two different
tRNA genes such that, although both mutant tRNAs recognise amber codons, they insert different
amino acids (because they are charged with the amino acid that charges each of the two different
wild-type tRNAs). For example, sup-1 might be a mutated tRNA-leu gene whereas sup-2 might be a
mutated tRNA-tyr gene. Thus, the supression in sup-1 would insert either leucine and the
suppression in sup-2 would insert a tyrosine at the position in the protein corresponding to the amber
codon. If the inserted amino acid is not similar in size and/or charge to the amino acid at that
position in the wild type protein, the resulting amino acid substitution may interfere with the
structure and function of the resulting protein. If suppression is not efficient enough an insufficient
amount of active protein might be made.
2. Oosawa and Simon isolated numerous intragenic pseudorevertants of an Ala-19 to Lys mutation in the
tar gene of E. coli. Several different pseudorevertants obtained are shown below.
a.) Based upon second-site suppressors #1-3, what can you conclude about the role of amino acid
19 in the Tar protein?
ANSWER: Insertion of a variety of different amino acids (including Ala, Gln, Ile, and Thr)
at position 19 of the Tar protein does not disrupt its function. However, not all amino acids
are permissive at this site. The results indicate the site may play a role in protein structure
but probably do not directly affect the active site of the protein.
b.) When the second-site suppressors were backcrossed into the wild-type gene tar gene, most of
the resulting proteins (which now only have the suppressor mutation) were functional. Based upon
these results, what can you conclude about the allele specificity of the suppressor mutations #4-8
and what does this tell you about the effect of these mutations on the Tar protein?
Mutant # Original mutation Second-site suppressor
1
Ala-19 to Lys
Lys-19 to Gln
2
Ala-19 to Lys
Lys-19 to Ile
3
Ala-19 to Lys
Lys-19 to Thr
4
Ala-19 to Lys
Val-17 to Glu
5
Ala-19 to Lys
Trp-192 to Arg
6
Ala-19 to Lys
Gly-271 to Ala
7
Ala-19 to Lys
Asp-288 to Val
8
Ala-19 to Lys
Thr-303 to Ile
ANSWER: Since the protein remained functional when the suppressor mutation was present
but the wild-type amino acid was present at position 19, the results indicate that the
suppressor can function with both Lys-19 and Ala-19 (amino acids with very different
properties) and hence they are not allele specific suppressors. This suggests that the effect of
the suppressor mutation on the Tar protein is independent of the amino acid at position 19,
and they probably act by causing increased protein activity (possibly by increasing stability
of the protein).
HOMEWORK #5 ANSWERS
FALL 2011
Due Sept 27th
Page 2 of 3
End Questions Due Sept 27th. Rest are to ponder for the exam.
3. In E. coli, the malK operon encodes three genes: malK, lamB, and malM. It is activated by the MalT
protein which is transcribed from the malT gene. The malT gene is unlinked to the MalK operon. The
operon is transcribed from a regulatory region as shown below.
The malK gene is required for the cell to use the disaccharide maltose as a C-source. The lamB gene
encodes the LamB protein. It is a trimeric outer membrane protein that allows diffusion of maltose (a
glucose-glucose disaccharide) and maltodextrins (long polymers of glucose) into the periplasm. LamB is
required for diffusion of the lengthy maltodextrins across the outer membrane. However, maltose is small
enough to cross using another porin called OmpF. In addition to forming a pore, LamB is also required for
adsorption of phage lambda to the cell surface. The function of malM gene product is unknown (and not
relevant to this problem).
If one selects mutants that are resistant to infection by lambda, but still able to use maltose as a C-source,
all of the mutations map in the lamB gene. Thus LamB mutants cannot grow when maltodextrins as the
only C-source but they can grow with maltose as the sole C-source because it diffuses into the cell via
OmpF.
a. Starting with a wild-type E. coli strain, you isolate 5-BrU induced mutants that are resistant to infection
by phage lambda. (5-BrU causes GC to AT and AT to GC substitutions).
What are 6 different types of 5-BU induced mutants would you expect to find? Would any be suppressed
by an amber suppressor? If so, which ones. Would any be highly likely to be dominant? If so which ones?
malT region- missense and amber in malT gene. Some ambers suppressed missense not suppresed.
Promoter down mutation in promoter of malT not ever suppressible and always dominant.
MalK operonMutations in the activator or promoter sites of the malK operon causing lower expression of MalK
and thus lower expression of the structural genes. These would also be dominant. None suppressible.
Polar mutation in malK, some suppressible. Missense and nonsense in lamB. None of the missense
suppressed, some nonsense suppressed.]
b. Assume your lamB mutant contains a nonsense mutation. How would you isolate lamB revertants?
Demand growth on maltodextrins as sole C-source.
c. What are three different types of revertants you would be likely to obtain? Which ones would be resistant
or sensitive to lambda infection?
True revertants -lambda sensitive, second site mutations in the nonsense codon that make a missense
substitution that inserts an acceptable amino acid residue into LamB –probably also lambda
sensitive, unlinked amber suppressor – lambda sensitive, Bypass suppressor – lambda resistant].
4. You want to make a collection of amber mutants of phage PhiX184. You have at your disposal an E. coli
strain that lacks any nonsense suppressors (supo) and a variety of strains that carry known amber
suppressors supE (Gln), supD(Ser) and supF(Tyr).
a. Using a simple genetic assay how could you isolate PhiX174 mutants that contain amber mutations
suppressed by supE?
HOMEWORK #5 ANSWERS
FALL 2011
Due Sept 27th
Page 3 of 3
Mutagenize the phage with a mutagen that causes transitions and plate with the supE strain as the
indicator. After plaques form, test them for growth on both the supo and supE strain. Amber mutants
suppressed by supE will form plaques on E. coli supE but not E. coli supo.
Assume you successfully isolated amber mutants in the experiment above. You find that the amber mutants
are suppressed by supE at both 30o and 42o C. You test the ability of other suppressors to suppress the
amber mutants. The suppression tests are done at 30o and 42o C and the results shown in the table below are
shown.
30°
Mutant
1
2
3
supD
+
+
42°
supF
+
supD
+
+
supF
-
b. For each of the 3 amber mutants, explain the results with the suppressor-containing strains.
[Mutant 1- It is not suppressed by supD or supF at any temperature. Therefore, Ser and Tyr are not
acceptable amino acid residues at the position of the nonsense codon in the protein. (Gln is acceptable
because it forms plaques on the supE strain).
Mutant 2- It is suppressed by supD but not supF at both temperatures. Therefore the protein is
active at either temperature with either Gln or Ser at the position of the nonsense codon in the
protein. Tyr is not an acceptable amino acid for function.]
Mutant 3- It is suppressed by supD at both temperatures so Gln and Ser are acceptable amino acids
at both temperatures. The pattern with supF is different. Insertion of Tyr makes the protein
temperature sensitive.]
c. As we discussed in class, 5BU causes transitions and HA is specific for the CG to TA transition. You
decide to isolate revertants of the original amber mutant #1. If you used 5BU as the mutagen, would you
expect to isolate mutagen-induced revertants (assume only 1 mutation per phage chromosome)? If so what
amino acid residues would be found at the position of the amber codon in revertant proteins? Would HA
give you revertants?
5BU causes bidirectional transitions so the UAG codon would be substituted by UUG (Leu), UGG
(Trp) and UAA (Ochre). No. Only UAA codons could be generated by HA.
5. a. Would you expect a bypass suppressor to be dominant or recessive to the wild-type gene? Explain
your answer with regard to the molecular mechanism involved.
Dominant because it is expressed independently of the wild type gene.
b. What does allele-specific mean? What does it tell you if a suppressor is allele-specific?
An allele-specific suppressor is a second-site mutation that repairs the mutant phenotype but only in
strains with certain, specific mutations at the first-site. (Interaction suppressors are usually allele
specific).
c. Would you expect an amber nonsense suppressor mutation to be dominant or recessive to the wild-type
tRNA gene? Explain your answer with regard to the molecular mechanism involved.
The suppressor would be dominant (expressed) because it is expressed independently of the wild type
tRNA. Thus both sense and amber codon can be decoded