1). (15 Points). Mutations in the proA, proB, or proC gene are proline auxotrophs.
Revertants of proB and proC mutants were isolated.
a). (5 Points). How would you select for revertants?
ANSWER: Demand growth on minimal medium without proline (i.e. select for
prototrophy)
b). (5 Points). Some intragenic suppressors of the proC mutants were found to be
temperature sensitive.
Suggest a likely explanation for this result.
ANSWER: The temperature sensitive properties of some intragenic
suppressors is a consequence of the second amino acid substitution in the
protein (i.e. a second missense mutation) that affects protein folding and
stability.
c. (5 Points). Intergenic suppressors were found that restore prototrophy of a deletion
mutant that
removes the proB gene. Suggest a likely mechanism for this suppression.
ANSWER: The results indicate that in the absence of ProB another pathway
can substitute for this gene product to bypass the ProB reaction (i.e. the
intergenic suppression is due to a bypass suppressor).
2). (22 Points). NADP is an essential cofactor for many cellular processes.
Because it is not transported, exogenous NADP cannot supplement mutants
unable to synthesize intracellular NADP. Five independent mutations were
obtained that affect the synthesis of NADP. 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
nad
nad-601
nad-602
nad-603
nad-604
nad-606
nad-607
nad-601 nad-602
nad-601 nad-603
nad-601 nad-604
nad-601 nad-607
nad-604 nad-607
30°C
+
+
+
+
-/+
+
+
-
Growth temperature
42°C
30  42°C
+
+
+
-/+
+
+
+
+
+
+
42  30°C
+
+
+
+
+
a). (12 Points). Note the properties of nad-601, nad-602, nad-603, nad-604, nad606, 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:
nad-601
nad-602
nad-603
nad-604
nad-606
nad-607
Ts, missense (Probably AA substitution that
destabilized protein)
Ts, missense
Ts, missense
Cs, missense
Leaky, nonconditional (probably a missense
mutation because gene product retains some
activity)
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). (10 Points). Interpret the results for each pair of double mutants in rows # 812. 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.
3). (18 Points). You are working on the mechanism of biosynthesis of the amino acid
leucine in E. coli. Leucine is an essential amino acid in proteins so bacteria either must
synthesize leucine or transport it in from the medium. You have isolated several
auxotrophs that require leucine for growth. You find that there are three proteins
required for synthesis of leucine called LeuA, LeuB, and LeuC (don’t worry about how
this was done). They are encoded by three genes called leuA, leuB and leuC. You
perform some experiments and find:
1). Mapping studies (don’t worry about how this was done) indicate that the gene
order is: leu Promoter—leuA—leuB—leuC, You have three mutants, one in each of the
genes.
2). The LeuA and leuB mutants revert spontaneously and with 5 BrU and 2 AP.
They are not revertible by acridines. The leuC mutant has never been observed to revert
either spontaneously or with mutagens.
3). You assay each of the mutants for LeuA, LeuB and LeuC activity (don’t worry
about how this was done) and find the following:
Mutant
#1
LeuA activity
0
LeuB activity
0
LeuC activity
0
#2
+
0
0
#3
+
+
0
a). (6 Points). What is the likely type of mutation leading to the phenotypes the three
mutants? Why do some mutants lack more than one activity? Why does one mutant lack
all 3 activities?
We know that the #1 and #2 mutants are base substitutions from the
reversion data. Mutant #3 is likely a deletion from the reversion data.
Mutant #1 is probably a nonsense mutant in leuA because this would
explain the null phenotype for LeuB activity due to polarity. We can’t
tell whether the mutant #2 is a missense or nonsense mutant from the
data so far. (Thus Mutant #1 is a nonsense mutant, mutant #2 is a base
substitution that could be either a missense or nonsense mutant, and #3
is a deletion).
You perform some further experiments in your mutant strains. You introduce some
amber suppressors into your strains and assay for LeuA, LeuB and LeuC activities. The
supE and supD suppressors insert Tyr or Ser, respectively. The results are shown in the
Table below:
Strain
Mutant #1
Mutant #1
Mutant #2
Mutant #2
Mutant #3
Mutant #3
Suppressor
supE
supD
supE
supD
supE
supD
LeuA activity
+
0
+
+
+
+
LeuB activity
+
0
0
0
0
0
LeuC activity
0
0
0
0
0
0
b). (12 Points). Explain the phenotypes of each mutant / amber suppressor
combination. Be sure to explain the phenotype of each protein for each
combination, and to say what type of mutation you think is present in each of your
original three mutant. How do you know? Your answer should be consistent with
the one from the first question.
Mutant #1, supE- This confirms our answer above that mutant #1 is an amber
mutant. We know this because it is suppressed by inserting a Tyr in the
protein at the amber codon. It also suppresses the polarity on LeuB.
Mutant #1, supD- LeuA activity is not restored by inserting serine (or the
efficiency of suppression is too low). We know that translational readthrough
is occurring because polarity on LeuB is relieved.
Mutant #2, supE- We don’t learn anything new here. Mutant #2 could still be
a missense or non-sense mutant.
Mutant #2, supD- Same.
Mutant #3, supE- No new information.
Mutant #3, supD- Same
4). (10 Points). Lee et al (J. Mol. Biol. (2005) 345:475 – 485) isolated a series of
mutants that made amino acid substitutions in the phage lambda integrase. The integrase
is an enzyme that catalyzes site-specific recombination.
One mutant changed residue Arg 30 to Ala (Arg30Ala). The mutant was defective for
recombination.
A second mutant changed residue Asp 71 to Ala (Asp71Ala). This mutant was also
defective for recombination.
They used in vitro mutagenesis to construct the double mutant Ala30 and Ala71. The
double mutant was inactive also.
They proposed that the Arg and Asp residues at positions 30 and 71 in the wild-type
protein form an interacting ion pair in the structure of the protein because of the opposite
charges of the amino acid side chains involved.
a). (5 Points). If their prediction is correct, why is the double mutant Ala30 – Ala71
inactive?
The Ala side chains fail to form the ion pair so the protein cannot catalyze
recombination.
b). (5 Points). What specific mutants would you make to test the model? What
prediction could you make from your mutants? Assume you can use in vitro mutagenesis
technology to make any mutants you desire.
If the Arg and Asp form an ion pair (opposite charges attract), then it should be
possible to reverse the positions of the residues in the protein and retain activity.
Thus, you could make a double mutant that has an Asp at position 30 and an Arg at
position 71. You would predict that this interaction is allele specific. Thus the
Asp30 – Asp71 and Arg30 – Arg71 proteins would be predicted to be inactive.
5). (16 Points). The DNA sequence of a portion of the phage T4 rII region is shown
below. The sequence is the coding or “m-RNA – like” strand. The rII genes of four
proflavin-induced mutants (#1 to #4) were sequenced and the base pair insertions
are shown in the sequence below. (#1 is a deletion of the indicated G, #2 is an
insertion of a G between the C and T residues, #3 is an addition of a C between the
two Ts and #4 contains a deletion of the indicated C). The following double mutants
were constructed and tested for their ability to form plaques on E. coli K (the strain
that does not allow growth of rII mutants) and the results are shown in the table.
For the experiments you perform for this question you have the wild-type T4 and
the mutant phages available as well as E. coli B and E. coli K strains.
A genetic Code dictionary is included at the end of the exam. Just rip it off to use.
Double mutant combination
Phenotype in
E. coli K
E. coli B
T4 +
+
+
#1 + #2
+
+
#1 + #3
-
r
#3 + #4
+
+
#2 + #3
-
r
+ = Wild type plaques
- = No plaques formed
r = r-type plaques made
a). (8 Points). For each mutant combination, provide a detailed explanation for the
phenotype.
First- compensating frameshifts restore the correct reading frame.
Second- The –G apparently places an UGA codon in the new reading frame.
Third- Same as the first.
Fourth- Two single frameshifts still make an incorrect protein.
Some people simply re-stated what the phenotype was, and you were asked to
explain it. Feel lucky that I gave you any points if this is what you did.
b). (4 Points). Is it possible to deduce the translational reading frame for the wildtype protein? Explain your conclusion.
The -1 frameshift in mutant #1 apparently puts the TGA in the new reading
frame. Thus the reading frame for the wild type must be displaced by 1 basepair.
…AAT TGG CTG ATT GGC AAG… = Wild-type
…AAT TGC TGA (STOP)
Partial credit was given to those who identified the possible stop codon. Most
still didn’t sufficiently explain how this helped determine the reading frame
c). (4 Points). How would you show that the double mutant #1 + #2 actually
contained two r mutations (as opposed to containing a wild type r gene)? Explain
with the phages and bacteria you would use at each step.
Perform a back cross between the putative double mutant and WT T4 in E. coli
B so that the parent and possible r- recombinant phages can grow. Plate the
progeny of the cross on E. coli B. Most of the plaques will look wild type
because of the parental phages look wild type. If recombination occurs
between the sites of the two mutations in the double mutant and the wild type
phage the recombinants will contain either mutation #1 or #2. They will form
r plaques on E. coli B.
6). (10 Points). Guarente and Beckwith isolated mutants they called polarity
suppressors (psu). They were isolated as mutants that relieved polar effects of a
nonsense mutation in an operon. The mutations were mapped and were shown to be
unlinked to the original non-sense mutation. By relying on your knowledge of the
mechanism of polarity, propose two likely genes or gene products that could potentially
give rise to polarity suppressors.
(5 Points). (Informational Suppresor): Mutate anticodon of a t-RNA to read
nonsense codon (so that transcription and translation continue to be coupled).
(5 Points). (Bypass suppressor): A Mutation which downregulates or knocks out
Rho (so that premature transcription termination does not occur).
Partial credit was rewarded where appropriate. Some people mentioned an
alternate pathway as a bypass suppressor. This would give wild type phenotype in
some cases, but it is incorrect in that an alternate pathway will not relieve polarity
in an operon.
7). (9 Points). In E. coli, the wild-type glyT gene encodes a glycyl-tRNA that
decodes GGA codons. Thus, when the charged t-RNA reads a GGA codon, a glycine is
inserted into the growing protein chain. A mutant (glyT*) was isolated that changed
the anitcodon sequence of the glyT t-RNA so that it recognizes AGA codons. The
AGA codon normally codes for arginine.
One of the tryptophan auxotrophs isolated by Yanofsky contained a trpA missense
mutant that made a gly to arg substitution at position 20 of the TrpA protein. Thus,
the strain required tryptophan for growth. The strain was wild type for all other
genes. When he crossed in the glyT* allele into the mutant strain (don’t worry about
how that was done), it no longer required tryptophan for growth.
a). (3 Points). What does this experiment tell us about residue 20 of the trpA
protein?
Substitution of residue 20 with Arg makes the TrpA protein inactive in the
otherwise wild type background.
Many said this demonstrates gly is critical for folding and/or activity. You
can’t necessarily say that. All we know is that arg is unacceptable at residue
20.
b). (3 Points). Why does the new strain grow in the absence of tryptophan?
The glyT* tRNA occasionally inserts a gly at position 20 in response to the AGA
codon. This inserts a gly at position 20 so that the resulting TrpA protein
sequence is wild type. (The suppression is efficient enough to allow the
synthesis of enough TrpA protein to allow growth). Thus the strain does not
require tryptophan.
c). (3 Points). The strain with the glyT* allele grows much slower than a strain with
the wild type glyT allele. Why?
The t-RNA encoded by the glyT* strain decodes AGA codons in mRNAs
encoding other proteins. This means that other proteins will contain gly at
positions that would normally contain arg. This would result in inactive
proteins that would cause the cell to grow slowly.