Download (A genetic code dictionary is on the last page of the exam

5). (14 Points).
(A genetic code dictionary is on the last page of the exam. You will probably need to
use it).
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? (6 Points).
[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. (6 Points).
[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.]
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?
(2 Points).
[5BU causes bidirectional transitions so the UAG codon would be substituted by
CUG (Leu), UGG (Trp) and UAA (Ochre). No. Only UAA codons could be
generated by HA].