Download MCB 421 Exam #1 (A)

MCB 421 Exam #1 (B)
Fall 2006
There are 9 questions and 1 supplement on last page.
Answer all 9 questions.
Be sure your name is on each page
1). (8 points).
A Luria-Delbruck fluctuation test was done to determine the rate of mutation to
Dehydroproline resistance (a toxic proline analog) in E. coli. Twenty tubes of rich
medium were each inoculated with a few wild-type cells and the cultures grown to 5 x
109 cells / ml. A 0.1 ml sample of each culture was then plated on minimal medium to
detect DHPR mutants. The results are shown in the following table.
Culture #
1
2
3
4
5
6
7
8
9
10
# DHPR mutants
12
10
18
14
17
16
11
22
11
9
Culture #
11
12
13
14
15
16
17
18
19
20
# DHPR mutants
9
3
10
20
14
21
12
13
12
11
A). (2 points). From the data shown in the table is the resistance to dehydroproline due
to spontaneous or induced mutation?
B). (3 points). Why?
[Induced mutations. A superficial (non-statistical) analysis of the data shows that
there is a very low variance indicating that the appearance of mutations was
approximately the same on all plates indicating that each bacterium in the culture
has a equal and small probability to undergo induces mutagenesis due to the
exposure to dihydroproline. ]
C). (3 points). If you picked a colony from the plate supplemented with dehydroproline
and grew it in LB broth for several generations, would the cells be resistant or sensitive to
dehydroproline?
[The cells would all (or almost all) be resistant to dehydroproline because the
mutat=ion should be stable even in the absence of dehydroproline.]
2). (12 points).
Consider the following experiment. Your ultimate goal is to isolate tryptophan
auxotrophs but you did not review my lecture notes carefully and did the following
experiment.
You mutagenize 50 cultures of wild type E. coli with a EMS, chemical mutagen. The
cells are then grown for several generations in LB broth at 37o C to recover from the
treatment with mutagen. Next the cells are centrifuged and resuspended in minimal
medium supplemented with penicillin and grown several for several generations at 37o C.
Next, the cells from each of the 20 cultures are resuspended in LB broth, grown for a few
generations and the minimal medium-penicillin step is repeated.
At the end of the second penicillin step, 100 cells from each of the 20 cultures are plated
on LB plates and incubated overnight at 37o C until colonies are formed.
The colonies are then replica plated to minimal plates and grown at 37o C.
Approximately 95% of the colonies that grew on the master LB plates also grow on the
Minimal plates while about 5% do not grow on minimal plates.
A). (2 points). Why do most colonies grow on the minimal plate?
[They come from cells that are still prototrophs that escaped killing by the
penicillin.]
B). (2 points). Why don’t some of the colonies grow on the minimal plate?
[They are auxotrophs that require supplements included in LB broth. This would
also include tryptophan auxotrophs.]
C). (1 point). Would any of the colonies be tryptophan auxotrophs?
[Probably because there are 50 x 5 = 250 auxotrophs represented in the experiment
and several independent cultures.]
D). (2 points). How could you identify tryptophan auxotrophs?
[Replica plate or streak the auxotrophs from the LB plate onto min + trp plates.
Trp auxotrophs will grow while other auxotrophs will not grow.]
E). (2 points). How could you have modified the protocol to get only Trp auxotrophs?
[Grow cells in Min + Trp at the steps where LB was used.]
F). (1 point). Is the experiment a selection, enrichment or a screen?
[Enrichment]
G). Why were 50 tubes rather than 1 tube used for mutagenesis?
[Prevent isolation of siblings.]
3). (9 points).
A). (3 points). 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). (3 points). 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). (3 points). 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.]
4). (10 points).
Complementation analysis was done on six mutants that lack theonine synthetase
activity. The order of the mutational sites is not known. The results are shown
below:
thr-1 thr-2 thr-3 thr-4 thr-5 thr-6
thr-1
+
+
thr-2
+
+
thr-3
+
+
thr-4
thr-5
thr-6
A). (2 points).
How many complementation groups are represented?
[At least two complementation groups: Group 1 = thr-1, thr-4, thr-5; Group 2
= thr-2, thr-3; Ungrouped = thr-6.]
B). (6 points).
Suggest two explanations for the results for thr-6.
[ thr-6 fails to complement all of the other mutants. This could be either due
to a trans-dominant negative phenotype of this mutant (e.g. due to a missense
mutation that poisons threonine synthetase) or a cis-dominant negative
phenotype caused by the mutation (e.g. due to an amber mutation that
prevents expression of downstream genes). In either case, it is impossible to
determine whether the thr-6 mutation is within one of the two
complementation groups described by the other mutations or whether it is in
a different complementation group.]
C). (2 points).
How could you distinguish between your explanations?
[ Do a complementation analysis with the WT and thr-6. If the thr-6 mutation
is cis-dominant, the phenotype would be wild type. If thr-6 is transdomionant, the phenotype would be Thr-.]
5). (12 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
30°C
Growth temperature
42°C
30  42°C
42  30°C
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
A). (7 points).
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 nonconditional) 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). (5 points).
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
6). (16 points).
Phage T4 forms wild-type plaques on both E. coli K-12 (+) and E. coli B. Phage
geneticists isolated mutants they designated “rII” for rapid lysis. rII mutants form large
plaques (larger than wild-type plaques) on E. coli B but cannot grow on E. coli K-12 (+)
so no plaques are formed. These facts are shown in the Table below.
Phage genotype
Wild type
rII
E. coli K12(+)
normal plaques
no plaques
E. coli B
normal plaques
r-type plaques
You decide to isolate an rII mutant using proflavin as a mutagen. Proflavin induces
frameshift mutations when phage are grown in cells treated with proflavin.
A). (2 points). Which E. coli strain would you use for the mutagenesis? Why?
[E. coli B because rII mutants can grow in E. coli B but cannot grow in E. coli K-12
(+) and would be lost.]
B). (3 points). How would you identify rII mutants? Is this a selection or screen? Why?
[By plating mutagenized phage on a B strain and identifying plaques with rII
morphology. This is a screen because all types may grow and the desired mutant
can be identified by screening many plaques.]
Assume you were successful in isolating an rII mutant. Next, you decide to isolate a
revertant that restores the activity of the protein encoded by the rII gene.
C). (3 points). How would you isolate the revertant? Which E. coli strain would you use
and how would you identify the revertants? Is this a selection or a screen? Why?
[Revertants could be isolated by plating a pool of rII phage on K12 (+); only
revertants will be able to form plaques so 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)]
D). (4 points). What are the two most likely types of mutations that could restore
activity?
[A true revertant or a frameshift suppressor (pseudorevertant). The original rII
mutant must be a frameshift mutant since we used proflavin as the original
mutagen. Thus, they can only be reverted by true reversion or a frameshift
mutation at a second site.]
NOTE: Version B had part E deleted so everyone got credit for E and F.
E). (2 points). How could you distinguish the two types of revertants using a back cross?
(ie. crossing your revertant with wild type T4). Draw diagrams showing the predicted
results and phenotype for each type of cross.
[A true revertant, when crossed with a wild type phage, will produce only wild type
progeny. A revertant with an intragenic frameshift suppressor will produce some
rII type progeny when crossed with wild type T4. Recombination between the sites
of the 2 frameshift mutations will produce phage that contain a single frameshift
each. One will have the original rII mutation and the other will carry the
suppressor.]
F). (2 [points). Which strain would you use for the cross and analyzing the progeny of
the cross? Why?
[E. coli B, because both phage types can grow on B and the genotypes distinguished
on B]
7). (12 points).
Van Way et al. [2000. J. Mol. Biol. 297: 7-24] took advantage of amber suppressors
to determine the role of specific amino acid residues in the flagellar motor protein,
MotB. Amber mutations were isolated at many sites throughout motB. The
mutants were then tested for suppression in isogenic genetic backgrounds with
either supo, supE, or supFamber suppressors. The results are shown in the Table
below.
Allele
wild-type
Q4Am
Q57Am
Q94Am
Q100Am
Q112Am
Q124Am
Q145Am
Q271Am
Q281Am
Mobility relative to wild-type:
supE
supF
supo
100%
80%
60%
0%
80%
100%
0%
100%
100%
0%
50%
80%
0%
10%
10%
0%
0%
0%
0%
20%
30%
0%
100%
100%
80%
100%
100%
100%
100%
100%
This table was prepared from the data in Figure 6 of Van Way et al. [2000]. In
the allele column, the mutations are shown as follows: Q, the one letter symbol
for gln, indicates the wild-type codon that was mutated, the number indicates
the position of that particular amino acid in the protein, and Am indicates the
change to a stop codon.
A). (6 points).
What is the advantage and disadvantage of using suppressor mutations to make multiple
amino acid substitutions at a particular position?
ANSWER:
2 points Advantage = a single mutation will allow multiple amino acid
substitutions
any 2 2 points each Disadvantages = different suppressors have different
suppression efficiencies, so the amount of substituted product may differ
depending upon the sup tRNA; the suppression efficiency depends upon
codon context so you cannot predict the level of suppression a priori;
suppression is never 100% efficient so some truncated product will
always be produced
B). (2 points). Amber mutations after position 271 in the protein do not eliminate
motility. Suggest a possible explanation for this result.
ANSWER: The simplest explanation of this result is that the C-terminus
of the protein is not required for motility.
C). (2 points). Some amber mutations, such as Q112Am, cannot be suppressed by supE
or supF, and other amber mutations, such as Q94Am, are suppressed to different extents
by supE or supF. Suggest a possible explanation for these result.
ANSWER: Some positions in the protein may not tolerate either of the
amino acid substitutions, and some positions may tolerate one amino acid
substitution but not another.
D). (2 points). Some amber mutations, such as Q57Am, are suppressed equally well by
supE or supF. Suggest a possible explanation for this result.
ANSWER: Some positions in the protein are tolerant of many different
amino acid substitutions. Such positions probably do not play a direct
role in the function of the protein.
8). (12 points).
In E. coli dam mutants display a mutator phenotype. That is, they have a higher
spontaneous mutation frequency relative to dam+strains. The dam gene encodes an
enzyme that methylates the adenine residue in the DNA sequence 5—G-A-T-C-3’. dam
mutants fail to methylate the adenine residue in that sequence. In addition, dam mutants
are killed by 2-aminopurine (2-AP). This is because this adenine analog incorporates into
DNA, but is then recognized as a mismatch and is excised, somehow causing formation
of lethal double strand breaks in the DNA of the dam mutants (but not in WT cells).
A). (4 points). How could you show that the dam mutants have a mutator phenotype
using a simple genetic test?
[Check for spontaneous StrR, phage resistance, or other easily selectable resistance
marker relative to the spontaneous resistance by an isogenic dam+ strain.]
Glickman and Radman [(1980) P. N. A. S. 77:1063-1067] carried out a suppressor
analysis of the sensitivity to 2-AP exhibited by dam mutants. When they isolated
suppressors, they found that all of the mutations mapped in several genes not linked to
dam (don’t worry about how this was done). They also found that when they moved the
suppressor mutations to a new strain (dam+) the new strains also had a mutator
phenotype.
B). (4 points). How did they likely isolate the suppressors of the original dam mutation?
[Plate cells on plates supplemented with 2-AP so that resistant mutants are able to
grow. (A mutagen is not necessary since this is a selection).]
C). (4 points). What is the likely cellular function of the suppressors with respect to their
mutator phenotypes?
[The mutants likely encode enzymes that are involved in removing mismatches in
DNA caused by 2-AP incorporation. Thus, the DNA is not repaired and the double
strand breaks do not accumulate. (In fact most of the suppressors mapped in the
mutH, mutL and mutS genes that are involved in mismatch correction. This helped
define the proteins that are required for mismatch correction).]
9). (9 points).
This question will require consultation with the Genetic Code Dictionary which is located
on the last page of the Exam.
The starting strain has a deletion of the proB gene and is suppressor-free (supo). A
plasmid that carries the wild type proB+ gene makes the strain proline independent. You
treat your plasmid with a mutagen like 2 AP and isolated a proB mutant. DNA sequence
analysis showed that codon 64, which is normally a tyrosine, is replaced by an amber
codon.
Next, you introduce a copy of the supD amber suppressor (inserts Serine) and find that
the strain is still a proline auxotroph.
A). (1 point). What is the likely explanation for this result?
[Replacement of tyrosine with serine does not restore activity of the protein (or the
suppresor is not efficient enough to produce enough copies of the protein to restore
activity.]
You next select a Pro+ revertant using hydroxylamine (HA) as the mutagen. HA is
specific for G:C to A:T transitions.
After isolating such a Pro+ revertant, you replace the copy of supD with the wild-type
supo allele. You now find that this strain becomes a proline auxotroph.
B). (2 points). Could the reversion event affect codon 64? What are two acceptable
molecular explanations for the reversion event that produced the HA induced revertant?
(A G:C to A:T mutation is not a sufficient answer).
[No, because the UAG codon cannot be mutated to a sense amino acid using HA
because the UAG to UAA change makes an ochre codon. A second site suppressor
in proA that restores activity of the ProA protein with the serine at position 64
(inserted by supD). must have occurred. (The results rule out intergenic
suppressors because only the proA gene on the plasmid was mutagenized). Thus the
mutation has to be elsewhere in proA. This could be allele specific or non allele
specific from the data shown.]
C). (4 points). What would you predict would happen of you changed codon 64 back to
a tyrosine in the revertant (Hint: Two answers)?
[If the interaction between the suppressor amino acid residue and the Ser at codon
64 is allele specific, the cell would be a proline auxotroph when Tyr is prsent at
codon 64 because the Tyr and second site suppressor could not interact.
Alternatively, if the second site suppressor improved that activity of the protein the
Tyr would be acceptable and the cells would be Pro+.]
D). (2 points). How could you test your hypothesis by purely genetic methods?
[Introduce other known amber suppressors that insert other amino acid residues at
codon 64 to see if they are allele-specific or not. This could be tested by looking for
growth in the absence of proline.]