HOMEWORK #5 ANSWERS
Questions 1-2
FALL 2011
Due Sept 27th
Page 1 of 3
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
2
3
Ala-19 to Lys
Ala-19 to Lys
Ala-19 to Lys
Lys-19 to Gln
Lys-19 to Ile
Lys-19 to Thr
6
7
4
5
8
Ala-19 to Lys
Ala-19 to Lys
Ala-19 to Lys
Ala-19 to Lys
Ala-19 to Lys
Val-17 to Glu
Trp-192 to Arg
Gly-271 to Ala
Asp-288 to Val
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 27 th . 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 operon-
Mutations 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 ( sup o ) 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 sup o and supE strain. Amber mutants suppressed by supE will form plaques on E. coli supE but not E. coli sup o .
Assume you successfully isolated amber mutants in the experiment above. You find that the amber mutants are suppressed by supE at both 30 o and 42 o C. You test the ability of other suppressors to suppress the amber mutants. The suppression tests are done at 30 o and 42 o C and the results shown in the table below are shown.
Mutant
1
30°
supD supF
- -
42°
supD supF
- -
2
3
+ -
+ +
+ -
+ - 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