7.28 Spring 2001
Exam One
Name ______________________________
Question 1 _____/24 points
Question 2 _____/34 points
Question 3 _____/22 points
Question 4 _____/20 points
______________________________
Total _____/100 points
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7.28 Spring 2001
Exam One
Name ______________________________
Question 1 (24 Points). You are studying the initiation of chromosomal DNA
replication in a novel bacterium that grows at 75°C. To this end you have isolated
temperature sensitive replication mutants in twelve different proteins/genes
derived from this strain. You call these mutations hot1-hot12. Your hope is to use
these mutations to identify proteins that act during initiation of DNA replication.
1A (5 Points). How could you reduce the number of mutants to investigate?
Describe the experiment that you would perform to focus on initiation factors and
which type of mutants you would choose for further investigation.
( 3 p ts ) M e as ur e 3 H dTT P inc or p or at i on i n a n as s ay t o m o n it or s l o w s t o p a nd
f as t s t op mu t an ts . G r ow ts m ut a nts at p er m is s iv e te m p er at ur e ( 7 5C) a n d
m o n i to r i nc o r p or at i o n o f 3 H dT T P af t er s h if t in g to t h e n on - p er m is s iv e
t e m pe ra t ur e .
( 2 p ts ) S l ow s to p m u tat i o ns ar e t h os e w h o c on t i n ue t o inc or p ora t e 3 H dTT P
f or a s h or t w h i l e af t er s h if t in g t o t he no n - p erm i s s iv e t e m pe ra t ure . T h es e
a r e c e l ls w it h m u ta t io ns i n g e n es r e qu i re d f or i n i ti a t io n o f DN A r e pl ic at i o n.
F as t s to p m u ta n ts ar e r e qu ir e d f or e l on g at i on .
Using your approach described above you narrow the field to 3 candidate
mutations (hot4, hot6, hot9). You decide to take two approaches to study the
proteins encoded by these mutations.
(1) You will clone the genes that are mutated and over-express the proteins
they encode.
(2) Your technician will use biochemical complementation to purify the activity
that complements the ability of cell extracts derived from the mutant cells
to replicate a plasmid containing the chromosomal origin.
You quickly clone the genes that complement the mutant strains and purify
the proteins they encode. You decide to first test whether one of more of the
proteins associate with the origin using a Gel Shift Assay. You get the results shown
on the next page:
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7.28 Spring 2001
Exam One
Name ______________________________
1B (5 Points). Based on the results above, what can you conclude about the
association of the three factors with the origin?
(1 pt) hot6 binds the origin.
(2 pts) hot4 binds if hot6 is present.
(2 pts) hot9 never binds.
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7.28 Spring 2001
Exam One
Name ______________________________
After several months of biochemistry, your technician assures you that he has
purified factors that complement each of the three mutant extracts. He tests each of
the fractions in the same assay and he gets the following results.
1C (5 Points). How can you explain the difference between your technician’s
results and the results that you obtained with the proteins you purified.
(2 pts) hot9 now binds if hot6 is present, and will form a complex with hot6
and hot4 to bind the origin.
(3 pts) One explanation for hot9 now binding would be that the technician
purified a complex of proteins that complemented hot9 mutants. This
complex of proteins included another protein that is responsible for hot9
binding with the origin (or hot6). The other protein in this complex might
be encoded by a second gene, and therefore was not purified using the
cloning approach.
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7.28 Spring 2001
Exam One
Name ______________________________
Based on your ideas concerning the different results obtained by you and your
technician, you ask your technician to test whether the hot4 complementing factor
also complements the defects observed in extracts derived from any other extract.
He finds that the hot4 complementing factor also complements the defects in
extracts derived from hot 8 and hot 12, two mutations that you eliminated in your
initial analysis of the mutants (Question 1A).
1D (5 Points). How can you explain the ability of the hot4 complementing factor to
complement replication in extracts derived from mutations that you have determined
to effect either initiation (hot4) or elongation (hot 8 and hot 12).
(2 pts) The technician purified a multisubunit complex.
(3 pts) One component of this complex could be responsible for initiation
while a different component is responsible for elongation.
(Note: Other interpretations were accepted, including a single hot4 protein
having activities for elongation and initiation as long as an explanation of
how hot4 could be mutated for both of these activities. e.g. hot4, hot8, and
hot12 were actually a single gene mutated in different regions responsible
for elongation and initiation.)
1E (4 Points). Assuming that your technician has purified the hot4 complementing
factor to homogeneity, briefly describe how you would test your hypothesis
experimentally.
(2 pts) Assuming a multisubunit complex, you can run the proteins purified
by cloning and biochemically on a denaturing gel.
(2 pts) A multisubunit complex would yield >1 band on this gel, while a
single protein would give a single band.
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7.28 Spring 2001
Exam One
Name ______________________________
Question 2 (34 points total). You are studying what limits the rate of E. coli
replication fork movement. You decide to test the hypothesis that the rate of sliding
clamp (-protein) loading by the -complex is rate limiting. To test this hypothesis
you want to prevent the -complex from diffusing away from the replication fork.
You decide to test your hypothesis by fusing the gene encoding the largest
subunit of the -complex (dnaX) to the gene encoding the DNA polymerase III
catalytic subunit (dnaE) so that the DNA polymerase and the -complex will be
covalently linked.
Because you are worried that fusing the dnaE and dnaX proteins could inhibit
the enzymatic activity of the polymerase, you want to assay for both activities and
compare them to the normal DNA Pol III Holoenzyme without fused proteins.
2A (4 points). Describe an assay you would use to determine whether the DNA
polymerase activity of the modified DNA Pol III Holoenzyme is altered. Assume you
have access to any radiolabeled DNA or dNTPs that you need.
Since the question asks for a polymerase activity assay, either a dNTP
incorporation assay or a primer extension assay was good. We did not ask
for a test of processivity, so template challenge was unnecessary, but OK. A
complete answer included:
-a primer/template junction
-radiolabeled primer or radiolabeled dNTPs
-Mg++
-filter binding or gel electrophoresis to detect radio labeled product
NOTE: Polymerases do NOT require ATP. Points were not taken off, as it
wouldn't hurt anything to have ATP around, but it is unnecessary.
You are also worried that the fusion alters the ability of the -complex within
the DNA Pol III Holoenzyme to function. To test this possibility you want to develop
a sliding clamp loading assay based on the Template Association assay described in
class.
2B (3 points). What molecule would you radiolabel to perform this assay?
Beta sliding clamp. (You want to test whether the beta clamp is
being loaded by the gamma clamp loader, not whether the gamma clamp
loader itself is associating. Partial credit for mentioning that you want to
label the protein of interest since it is small and will elute in different
fractions depending on whether it associates with the large DNA or not. )
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7.28 Spring 2001
Exam One
Name ______________________________
2C (3 Points). Given that the clamp loader requires a primer‧template junction to
load the sliding clamp on DNA, draw the structure of the DNA template that you
would use to perform your assay.
Large circular ssDNA with an annealed primer for the clamp to recognize. Partial credit
was given for simply drawing a primer-template junction, but the entire point of this
assay is that the large circular unlabeled DNA elutes from the gel filtration column at a
different place from the small labeled protein on its own. Many people also tried drawing
the fork as it looks in vivo. This is not a template that you could make and use in an
assay.
2D (6 Points). Draw the result of the assay you would expect if the clamp loader
was active or inactive. Make sure that you label the axes of your plot.
Active
Inactive
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7.28 Spring 2001
Exam One
Name ______________________________
You are happy to find that both the DNA polymerase and the clamp loading
activities of the modified DNA Pol III Holoenzyme were the same as the unmodified
Holoenzyme when assayed separately from one another. To determine whether
covalently linking these enzymes allows the activities to work well together, you
want to assay the rate of replication fork movement (where both activities must
act simultaneously as opposed to questions 2A-D in which you have assayed the
activities separately).
You first address this question in vitro, using an OriC-containing plasmid as a
template for the DNA replication reaction.
2E (6 Points). Your advisor sees that you are using the OriC plasmid and (without
thinking carefully) argues that you should use the simpler ssDNA template and an
annealed primer. Why do you need to perform this assay with a DNA template
containing OriC rather than using a ssDNA template with a single primer? Enlighten
your advisor, below.
To set up a replication fork with both leading and lagging strands, you
need a double-stranded origin template, not simply a primer-template
junction. The full holoenzyme consists of leading and lagging polymerases
connected by tau and made more processive by the clamp and clamp loader
proteins, all of which can only function together in a fork on doublestranded
DNA. Partial credit was given for a discussion of dnaA/B/ and C
loading at the origin, but the key point is that assaying the holoenzyme
activities requires a fork to be set up.
You perform a time course/replication assay using radiolabeled dTTP, other
dNTPs, Mg+2, a 5000 bp plasmid containing a single OriC sequence, dnaA, dnaB,
dnaC, primase, topoisomerase and either the modified or unmodified DNA Pol III
Holoenzyme. You separate the products on a denaturing gel and obtain the following
autoradiograms:
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7.28 Spring 2001
Exam One
Name ______________________________
2F (5 Points). You do not observe any products greater than ~2500 bases in your
assay. Explain why not. What additional proteins would you need to add to your
assay to see such molecules?
The key point from the gel was that the Okazaki fragments (about
500-1000 bases) are not being ligated together or to the leading strands
(2500 bases). This was because DNA Ligase, Polymerase I exonuclease and
RNAse H were all missing. However, many of you noticed that we forgot to
include SSB as well. And, although it is not a protein, and therefore
received less credit, we also forgot ATP. Full credit involved mentioning 2
proteins missing from the reaction and a reasonable explanation.
You are disappointed that there is no strong difference between the two
polymerases. You notice, however, that both reactions go from incomplete to
complete by the 15 second time-point and suspect that both enzymes are so fast
that they are completing the entire template in less than 15 seconds.
2G (3 Points). Because of the difficulty of performing the experiment, you cannot
take a time point sooner than 15 seconds. How could you change the template to
increase your chances of seeing a difference between the two polymerases? Briefly
explain your logic.
The simplest answer is to make the template longer so that it will
take more than 15 seconds for the wild-type polymerase to replicate it.
Although no-one followed the logic through, we learned that the E. coli
polymerase moves at a rate of about 1000 bp/second. Therefore, the
template should be longer than 15,000 bases.
Using your new template, the in vitro studies indicate that your hypothesis
is correct and the replication fork moves twice as fast using the modified DNA Pol
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III Holoenzyme in the test tube. As a final test of your hypothesis, you replace the
normal copies of the dnaE and dnaX genes in E. coli with the fused version of these
genes and determine what the effect is on the rate of cell growth.
2H (4 Points). You are surprised to find that the cells grow much slower than wild
type cells, however, if the gene encoding the -protein is over-expressed the cells
grow faster than wild type. How can you explain these findings. (Hint: the complex is required both to put the sliding clamp on and to take it off the DNA).
A complete answer involved explaining both why the fused proteins
make the cells grow slower AND why over-expressing beta makes them
grow faster. One answer is that fusing gamma to the fork keeps it from
going back and removing beta clamps from behind the fork (at the end of
Okazaki fragments, for example). Therefore, the cell runs out of beta
clamps and grows slower. If beta is over-expressed, that solves this
problem and allows the polymerase to move faster than wild-type as it does
7.28 Spring 2001
in vitro.
Exam One
Name ______________________________
Question 3 (22 Points). Using a replicator cloning approach, you have isolated a
DNA fragment that contains a replicator derived from the yeast, B. rewski The
fragment that you identified is 5 kb in length and has the following restriction map.
Your initial studies indicate that there is only one origin in this region. You
want to identify the site of the origin of replication in the fragment. Based on DNA
sequence, you think that the replication origin is located in fragment C in the map.
3A (3 Points). What restriction enzyme would you cut the genomic DNA with prior
to performing a 2D gel analysis of replication intermediates to identify the origin?
Briefly explain your choice.
EcoRI- so that probes will be able to bind.
Or
HindIII- so that suspected origin in C will be off-center in the fragment (but
this answer will get you into trouble when you try and use probe 1 or 2 in
part b)
Note: Answers based on the size of the restriction fragments do not make
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sense for this assay. You are detecting only one fragment with a
radiolabeled probe, and each fragment will have a range of sizes depending
3B (3 Points). Which of the radiolabeled probes (shown by the dark bars above the
restriction map) would you select? Briefly explain your choice.
If digest with RI- probe 2- will bind to region that includes C where you
suspect there’s an origin.
If digest with HindIII- neither probe will work, as they will both recognize
multiple fragments, giving multiple patterns on the same gel.
7.28 Spring 2001
Exam One
Name ______________________________
You perform 2D gel replication intermediate mapping experiments using
different restriction enzyme digests and different probes in each case. You get the
following results.
3C (5 Points). Based on these data, what conclusions concerning the location of
the origin in the fragment can you make?
The origin is located asymmetrically at one end of either fragment A or
fragment B because probe 1 gives a bubble-Y transition.
The origin is NOT in fragment C. The pattern given by probe 2 could be
EITHER a bubble OR a Y arc. As stated in class, there is no way to tell the
difference just by looking, which is why we look for a bubble-Y transition as
seen with probe I. More credit was given if you realized this, but the
question did state that there was only one origin. 11
3D (6 Points). Describe one additional 2D gel experiment that you could do to
further clarify or confirm your conclusion. Explain your reasoning.
There are many possibilities. An example: Cut with EcoRI and HindIII,
probe with a probe specific to the B region, and look for a bubble-Y
transition, since we know from part C that if the origin is in B it’s at the end
of B, not in the center.
Note that using the probes given would not work, as stated above. More
credit was given if you at least realized the problem. The question may not
have been altogether clear in stating that you could design a new probe.
7.28 Spring 2001
Exam One
Name ______________________________
3E (5 Points). Based on the experiments described above, you find that the origin
is located in a single 500 bp fragment. You are nevertheless curious to determine if
the replicator is completely located within the same fragment. Briefly describe how
you would map the location of the replicator in the fragment
You want to know if the origin is “completely located within” the 500 base
pair fragment. The only way to test this is the Plasmid Replicator Assay:
clone the fragment into a vector containing a selectable marker, a
centromere, and no origin, transform into yeast (not bacteria, this is a yeast
origin), and select for the marker. If the cells can grow, the fragment
contained the replicator. Mutational mapping can then be used to map it
further, but the only way to test that the region is sufficient for origin
function is to clone it away from the flanking sequences into a plasmid.
Question 4 (20 Points Total). To identify proteins involved in DNA repair in a
newly identified bacterial strain, you decide to isolate mutant versions of this strain
that have a mutator phenotype. In the context of this analysis, you define mutant
strain as a mutator strain if it has at least a 10-fold higher frequency of mutations
than the starting strain, as determined from an in vivo reversion assay, similar to
that used in the Ames test. The starting test strain you use for the reversion assay
carries a missence mutation in a gene required for arginine biosynthesis; as a result,
this strain is an arginine auxotroph.
4A (6 Points). Name four genes (or the protein encoded by these genes) (use E.
coli nomenclature) that are likely to give the most elevated frequency of mutations
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(e.g. >100-fold increase) by this assay. Briefly justify your answers.
(4 pts) mutS, mutL, mutH, and proofreading exonuclease of PolIII.
(2 pts) An explanation of why the above mutations yield the MOST elevated
frequency of mutations is necessary. mutS, mutL, and mutH are involved in
mismatch repair. A mutation in any of these would cause a 1000x increase
in frequency of mutations. A mutation in the proofreading exonuclease of
PolIII would also cause a 1000x increase in frequency of mutations.
Genes necessary for BER and NER were not accepted because the
frequency of mutation in these mutants is significantly lower than in the
mutants above in the absence of a mutagen.
(Note: dam- was also accepted, though a mutation in this gene would cause
a 500x increase in mutation frequency, which is not as high as the
mutations mentioned above.)
7.28 Spring 2001
Exam One
Name ______________________________
As a result of this genetic screen, you isolate cells that constitutively express
proteins required for trans-lesion DNA synthesis (TLS). These cells express the
proteins in the absence of induction by a DNA damaging agent. The TLS proteins
themselves are, however, perfectly normal in this strain.
4B (4 Points). Why do these cells have a mutator phenotype? Explain your
answer.
(2pts) TLS is error prone and will cause the incorporation of incorrect
nucleotides during replication.
(2 pts) TLS genes are normally repressed. However, in these cells, TLS is
constitutively active. Due to this, TLS proteins are free to act and therefore
cause increased mutation.
In the process of characterizing the strain that constitutively expresses the
TLS proteins (used in part b), you notice that one clone derived from this strain has
an even HIGHER mutation rate that the parent strain. You observe this high
mutation frequency in many different types of in vivo assays, regardless of the
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nature of the “tester” allele used in experiment. Further analysis of this hypermutable
strain reveals that it carries a mutation in the umuC gene.
4C (5 Points). What characteristic of the UmuC (polymerase V) protein could be
altered in the mutant version to explain the phenotype? Briefly explain your answer.
(2 pts) UmuC is a low fidelity DNA polymerase and also has low
processivity. However, a mutation that increases the processivity of UmuC
will allow it to polymerize much longer than normal.
(3 pts) The longer UmuC polymerizes DNA, the more incorrect nucleotides
it incorporates, increasing the mutation frequency.
Note: Other explanations were accepted, such as a mutation in the
proofreading subunit of UmuC or a mutation making UmuC even more error
prone. However, UmuC does not possess proofreading activity.
Additionally, a mutation making it more error prone may be unlikely as it
incorporates
7.28
Springnucleotides
2001 almost randomly without consideration for proper
Exam One
Name ______________________________
base pairing.
4D (5 Points). Design an experiment to test your hypothesis for the nature of the
defect in the UmuC protein. Explain specifically what results you would expect in
your experiment with the mutant protein if your hypothesis was correct. Compare
these data with those you find in control experiments using the wild-type UmuC
protein.
(1 pt) A processivity/template challenge assay can test our hypothesis that
UmuC obtained a mutation increasing its processivity.
(1 pt) A proper substrate would include a long single stranded template
with a short complementary primer radioactively labeled at its 5’ end.
Incubate UmuC and labeled substrate in an equimolar ratio.
(1 pt) After incubation, add dNTPs and 1000x cold substrate (ssDNA
template and unlabeled primer). Allow reaction to proceed and run the
products out on a denaturing gel. Dry onto paper and expose to X ray film.
(2 pts) A mutant increasing processivity in UmuC will yield a long DNA
product, indicating its ability to polymerize a long stretch of DNA in a single
binding event. Wild type UmuC will yield a short DNA product, indicating
low processivity in this enzyme.
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