Download MCB421 FALL2005 EXAM#1 ANSWERS MCB421 EXAM1 Page 1

MCB421
FALL2005
Average
Std. Dev.
Min.
Max.
(10pts)
EXAM#1 ANSWERS
62
14.6
28
84
1. Briefly explain the expected results for the Luria-Delbruck fluctuation test if
mutants arise:
(a) Spontaneously in a population of cells before exposure to a selective agent:
ANSWER:
Wide fluctuation in mutants from cells grown in different tubes (i.e.,
independent samples) relative to cells grown in the same tube. That is, there is
great variance relative to the mean number of mutants.
(b) Due to adaptation after exposure to the selective agent:
ANSWER:
Relatively little fluctuation in mutants from cells grown in different tubes (i.e.,
independent samples) relative to cells grown in the same tube. That is, the
number of mutants on each plate approximately equals the mean number of
mutants.
2. In the following table, briefly diagram or indicate the common properties of each type
of mutation. (1 point for each box; 15 points total)
Mutation
Missense
Nonsense
Frameshift
Deletion
Insertion
Effect on
DNA
Base
substitution
Base
substitution
resulting in a
stop codon
Insertion or
deletion of 1 or 2
base pairs
Loss of
multiple
base pairs
Addition of
multiple base
pairs
Effect on
Protein
Substituted
amino acid
Truncated
polypeptide;
inactive
protein
Altered amino
acid sequence
downstream of
mutation;
usually
truncated
polypeptide;
inactive protein
Usually
absent;
usually
inactive
protein
May insert extra
amino acid (if in
frame) or cause
premature
truncation;
usually inactive
protein
Effect on
Phenotype
May result in
loss of
function or
conditional
phenotype
Usually null
Null
Null
Usually null; if
insertion is in
frame and doesn’t
cause premature
termination and is
in a permissive
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site, sometimes
protein remains
functional
MCB421 EXAM1
MCB421
FALL2005
phenotype
Phenotype
EXAM#1 ANSWERS
termination and is
in a permissive
site, sometimes
protein remains
functional
3. Resistance to the toxic tryptophan analog 5-methyl tryptophan (5-MT) can occur in
two ways: (i) specific missense mutations in the trpE gene (the first step in tryptophan
biosynthesis) which makes the enzyme insensitive to feedback inhibition; and (ii)
mutations that inactivate the trpR gene (the repressor which represses the trp operon
when the intracellular concentration of tryptophan is high). The trpR mutants synthesize
a large amount of tryptophan so that the 5-methyl tryptophan is not toxic. That is, the
concentration of tryptophan is much higher than the concentration of 5-MT.
(3pts.)
a). Which class of mutants would you expect to be more common and why?
ANSWER:
The trpR mutations would be most common because a null mutation anywhere
within the gene could inactivate the protein. Only a limited number of sites in
trpE would be likely to produce a functional protein, which is insensitive to
feedback inhibition.
(3pts.)
b). Indicate whether each of the two types of mutation is dominant or recessive
to the wildtype allele of that gene. [Explain your logic.]
ANSWER:
A tprR null mutation is a simple loss of function mutation, which would be
recessive to the wild-type allele. That is, trpR- /trpR+ would be able to repress
the trp operon, which would prevent the cell from accumulating high
concentrations of tryptophan, thus resulting in sensitivity to the tryptophan
analog.
A trpE mutant insensitive to feedback inhibition ("trpE*") would be dominant
to wild-type. That is, trpE* /trpE+ would have one copy of the gene which is
insensitive to feedback inhibition, allowing accumulation of a high intracellular
tryptophan concentration and thus resulting in resistance to the tryptophan
analog.
MCB421 EXAM1
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MCB421
FALL2005
EXAM#1 ANSWERS
3. (Cont.) You introduce a copy of several amber suppressors into cells that have one of
the trpR mutations and score for sensitivity or resistance to 5-MT and obtain the
following results. The suppressors and the tRNA gene encoding them are listed in the
Table. (S = sensitive. R = resistant)
supE (amber) glutamine
supF (amber) tyrosine
supL (ochre) lysine
trpT- Su9 (opal) tryptophan
trpR1
R
R
R
R
trpR2
S
S
R
R
trpR3
R
S
R
R
trpR4
R
R
S
R
trpR5
R
R
R
S
For each trpR mutant specify:
(3pts.) c). The class of mutation (substitution, frameshift, deletion, or insertion) that
could be present.
ANSWER:
trpR1 – any of the 4 mutations listed above
trpR2 – amber
trpR3 – amber
trpR4 – ochre
trpR5 – opal
(3pts.)
d). Whether the trpR mutation is likely to be a nonsense mutation and why.
ANSWER: trpR2, trpR3, trpR4, and trpR5 are nonsense because they
are suppressed by amber, ochre, and opal suppressors.
(3pts.)
e). In cases where suppression occurs, state what structure/function you can
predict for the TrpR repressor?
ANSWER:
trpR2 – Gln and Tyr can make an active protein
trpR3 – Tyr can make an active protein
trpR4 – Lys can make an active protein
trpR5 – Trp can make an active protein
(2pts.)
f). Which mutant would most likely be reverted by ICR191? Why?
ANSWER: trpR1 because it could be a frameshift mutation.
MCB421 EXAM1
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MCB421
FALL2005
EXAM#1 ANSWERS
4. Salmonella Enteritidis expresses a specific fimbriae called Sef which facilitate
attachment of the bacteria to macrophages during infections. A large number of
mutations were isolated that prevent expression of sef genes, and complementation
analysis was done to determine the number of genes disrupted by the mutations. For each
mutation, one copy of the genes was present on the chromosome and a second copy of
the genes was present on a plasmid integrated into the chromosome at a different site.
The results are shown in the table below.
(− indicates no Sef fimbriae produced and + indicates that Sef fimbriae are produced.)
Copy
#1
+
sef
sef-51
sef-52
sef-53
sef-54
sef-55
sef-56
sef-57
sef-58
sef-59
sef-60
(3pts.)
Copy #2
sef-51
sef-52
sef-53
sef-54
sef-55
sef-56
sef-57
sef-58
sef-59
sef-60
+
+
+
+
+
−
+
+
+
+
−
−
−
−
+
−
−
+
+
−
−
−
+
+
−
−
−
−
−
−
−
−
−
+
+
−
−
−
−
+
+
+
+
−
+
−
−
+
−
+
+
−
+
+
−
−
+
+
+
+
−
+
−
+
−
a). Based upon these results, how many complementation groups were detected?
ANSWER: 4 complementation groups.
(3pts.)
b). Indicate which mutations fall into each complementation group:
ANSWER: Group #1 = sef-52, sef-55, sef-57; group #2 = sef-53 and sef-59;
group #3 = sef-54; group #4 = sef-58, sef-60; Note that sef-51 and sef-56
cannot be placed into a complementation group.
c). Suggest a likely explanation for the complementation behavior of the sef-51
mutation. Would you expect this mutant to produce CRM? [Briefly explain your answer.]
(3pts.)
ANSWER: sef-51 is recessive to the wild-type in trans, indicating that this
phenotype is NOT likely to be due to a trans-dominant negative (negativedominant) mutant. Thus, this phenotype is likely to be due to a cis-acting
mutation that prevents expression of all of the sef genes. Not likely to
produce CRM because mutations in a gene resulting in cis-dominance are
usually not made or truncated and thus rapidly degraded.
MCB421 EXAM1
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MCB421
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EXAM#1 ANSWERS
d). Suggest a likely explanation for the complementation behavior of the sef-56
mutation. Would you expect this mutant to produce CRM? [Briefly explain your answer.]
(3pts.)
ANSWER: sef-561 is dominant to the wild-type in trans, indicating that
this phenotype is likely to be due to a trans-dominant negative (negativedominant) mutant gene product. Likley to produce CRM because such
mutations are typically due to missense mutations.
5. You have isolated 4 independent tryptophan auxotrophs of E. coli. To characterize the
types of mutations that cause the auxotrophies, you perform reversion analyses. You
select revertants without any mutagen (spontaneous) and in the presence of mutagens.
You use common mutagens we discussed in class and newly discovered chemicals called
“X” and “Y”. The data are shown in the table below.
Number of Revertants / 108 cells
Mutant
trp1
trp2
trp3
trp4
5-BrU
300
643
9
3
Spontaneous
20
23
16
6
ICR191
7
3
603
5
X
500
26
6
5
Y
30
603
7
3
a. How would you have designed the experiment to ensure that all 4 auxotrophs are
independent? What type of medium would you use to select the revertants?
(3pts.)
ANSWER: Isolate mutants from several independent cultures and
save only one from each. Use minimal media + glucose.
(3pts.)
b. Based on the reversion data, what type of mutation is present in each auxotroph?
How do you know?
ANSWER:
trp1 – base substitution (transition) because 5-BrU causes GC ⇔ AT
trp2 - base substitution (transition) because 5-BrU causes GC ⇔ AT
trp3 – frameshift because ICR191 is specific for frameshift mutations
trp4 – insertion because you get revertants, but the reversion
frequency is not increased in the presence of mutagens
(3pts.)
c. Are compounds X and Y mutagens? How do you know? Can you determine if
they have any specificities for mutations they cause?
ANSWER: Yes, because you get an increased reversion rate, over the
rate for spontaneous revertants. One causes GC → AT and the other
causes AT → GC. However, you cannot tell which one causes which
direction (if you compare to 5-BrU).
MCB421 EXAM1
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MCB421
(3pts.)
FALL2005
EXAM#1 ANSWERS
d. Which auxotrophs would most likely be capable of suppression by intragenic
suppressors? Intergenic suppressors?
ANSWER: All four could be suppressed by intragenic suppressors. Only trp3
would not be able to be suppressed by intergenic suppressors.
(14 pts.)
6. The malK operon of E. coli contains three structural genes- malK, lamB, and malM.
The genes are transcribed from a promoter site (P), located to the left of the malK gene as
shown below. The malT gene, which is unlinked to the malK operon, encodes a protein
that activates expression of the malK operon.
The MalK protein is required for utilization of maltose as a carbon source. The LamB
protein is the receptor for phage lambda. The function of MalM is not known.
malT
P
malK
lamB
malM
a. How would you isolate a mutant resistant to infection by phage lambda? Is it a
selection, enrichment, or a screen?
ANSWER: Plate a lawn of bacteria on rich media and then overlay with an
excess of phage lambda (high MOI to ensure there are enough phage to infect all
the bacteria plated). Look for colonies that grow and do not lyse. These bacteria
will be resistant to infection by phage lambda.
b. Describe 4 types of base-pair substitution mutations would you expect to find in the
malK operon that cause resistance to infection by phage lambda.
ANSWER:
(i) P down
(ii) malK nonsense that are polar on lamB
(iii, iv) lamB missense and non-sense that produce inactive proteins
(deletions, insertions and frameshifts are not base-pair susbstitutions)
(Mutations in malM have no effect)
c. Would any of the mutants be cis- or trans-dominant? Why?
ANSWER: malK nonsense mutants would be polar and cis-dominant. Promoter
defective mutants would be cis-dominant.
MCB421 EXAM1
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MCB421
FALL2005
EXAM#1 ANSWERS
d. Would any of the mutants be suppressed by informational suppressors? Why?
ANSWER: malK nonsense and lamB nonsense could be suppressed by informational
suppressors.
e. Describe 2 types of base-pair substitution mutations that you would expect to find in
or near the malT gene that would cause resistance to infection by phage lambda.
Describe the mechanism behind each mutant.
ANSWER: P down so MalT is not made at a high enough concentration to activate
the malK operon. So the concentration of LamB is low. MalT- Missense or nonsense
mutation that makes protein that can’t activate.
f. Would any of these mutants be suppressed by informational suppressors? Why?
ANSWER: MalT nonsense could be suppressed.
g. Would any of these mutations be dominant or recessive in an experiment where the
wild-type malT gene is introduced in trans into the cell?
ANSWER: malT+ would be dominant to malT down promoter mutant because the
wild-type MalT protein expressed by the new malT+ copy could activate the operon.
A malT mutant could be recessive if it is a loss of function mutant. Some could be
trans-dominant by making a protein that forms inactive multimers with the wild
type protein so that the complex can’t activate transcription.
(10 pts.)
7. The trpEDCBA operon contains 5 structural genes transcribed in the direction trpE to
trpA (see diagram below). You have isolated a tryptophan auxotroph that affects the trpE
gene. You find that the mutant shows polarity on the expression of the downstream
trpDCBA genes.
P
trpE
trpD
trpC
trpB
trpA
Your next step is to isolate mutants that are able to grow in the presence of indole. The
last reaction of the tryptophan biosynthesis is the conversion of indole to tryptophan,
catalyzed by the TrpB and TrpA proteins.
Describe 5 types of intragenic and intergenic mutations could potentially be isolated as
revertants:
MCB421 EXAM1
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MCB421
FALL2005
EXAM#1 ANSWERS
ANSWER:
Intragenic
-True revertants
-Same base to different base
-Different base in same codon
(Other intragenic events will not compensate for the nonsense mutation in trpE)
Intergenic
-Mutation that creates a mutant tRNA that can read thru the nonsense codon
-Mutation in Rho (that could be defective in transcription termination thus relieving
polarity)
-Mutation in a subunit of RNA polymerase (that doesn’t interact well with Rho thus
relieving polarity)
(10 pts.)
8. Fred Sanger and his collaborators [Nature. (1977) 265: 687-95] determined the DNA
sequence of the bacteriophage Phi X 184. At the time it was the first genome that was
sequenced. One of the interesting observations was that two of the genes (A and B)
overlapped as shown below.
B
H
a).
A
E
C
D
F
G
The protein products of the A and B genes have completely different
amino acid sequences. How could this be explained?
ANSWER: The DNA of the coding sequences of A and B overlap in the C terminus
of A and the N terminus of B. The proteins are translated in different reading
frames so that the amino acid sequences of both proteins have no common amino
acid sequences.
b).
They isolated a mutation that made the A gene product temperature
sensitive. Interestingly, the B protein was absent in a suppressor-free strain
but active in a strain containing supE (coding for a mutant form of a gene
encoding a tyrosine tRNA). How could this be explained?
ANSWER: The mutation that makes the A protein TS apparently made a mutation
in the overlapping codon of the B protein that created a nonsense mutation in the
coding sequence of the B protein. When supE is introduced, it suppresses the
nonsense mutation so it becomes phenotypically B+.
MCB421 EXAM1
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MCB421
MCB421 EXAM1
FALL2005
EXAM#1 ANSWERS
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