Answer Key to Part One
Problem Set 1: Replication
You have discovered a new organism from the scorched earth around Chernobyl. It is a
previously uncharacterized strain of bacteria. You like names. You name it Bacillus
radii. You like DNA polymerase. You wish to purify a DNA polymerase activity.
Briefly describe how you would assay for DNA polymerase activity in crude extracts.
What is your read out? What kind of template would you use? How would this affect
your result?
Couple of possible answers. Depends which activity they want to purify. The easiest one
is a PolI type activity. Could select for this with a nicked template. Lots of people said
that using a primed circular ssDNA template would select for a PolIII type activity. This
isn’t really true: PolI could also act on such a template. It would allow purification of
PolIII but sheer abundance would still bias towards PolI.
Assuming that your activity is essential for DNA replication in Bacillus radii, briefly
describe a genetic strategy that, in conjunction with the assay you have designed, could
create mutants for your DNA polymerase. Nobody has offered to pay for a ski trip. No
one even offered to buy you a drink. Your in vitro assay is time consuming and fills your
radioactive waste bucket in no time. You want to design your strategy to enrich for
replication mutants to the greatest possible extent prior to embarking on in vitro work.
I would screen for ts lethal mutants. I would then screen through the ts mutants for
mutants which fail to incorporate radiolabeled nucleotide at the restrictive temperature.
You could do this by selecting against cells that replicate at the restrictive temp. i.e. by
incubating with BUdr and then UV irradiating. Basically, any assay that distinguishes
those BR that can’t incorporate nucleotides is OK. I would then assay these mutants in
vitro to see if they are Pol mutants.
Give one reason why it is sometimes difficult to purify an activity by fractionation of
extracts.
Activity may be sensitive to dilution. May be a large, weakly associated complex which
falls into different fractions. May require some post translational modification which is
lost on fractionation. etc.
You manage to clone two different polymerase enzymes. You don’t pass up the
opportunity to name them. You name them PolX and PolY. You wish to characterize
them further. Design an experiment to assay fidelity in vitro. You wish to be
quantitative and as unbiased as possible.
Two good answers: 1) Preincubate with only 3 of 4 dNTPs. Use a template that contains
an initial tract complementary to those 3 dNTPs and then finishes with a long tract of
dNTP comp to the missing NTP. Pol will stall at the first template position
complementary to the missing dNTP. Add radiolabeled dNTP (not the missing one) to
reaction to a known specific activity. Rate of incorporation of radiolabel = rate
incorporation incorrect nucleotide. Repeat for each type of missing dNTP
2) Incubate the polymerase with template and dNTP mix. The dNTP mix would be
labeled with tritium. I would dope in a 32P labeled nucleotide analogue. The ratio of
32P to tritium activity would give a rate of misincorporation. I would probably use
several different nucleotide analogues to get a range of values.
You are curious to see how accurately BR replicates its genome. You realize that it will
be difficult to be quantitative and unbiased so you just decide to determine the relative
mutation rates of BR and E. coli. How might you make this comparison?
I would transform both strains with a plasmid containing a reporter gene like LacZ
closely linked to a selectable marker. The gene would be identical in both bacteria. I
would then simply compare the rate of blue-white conversion.
A more facile assay would be to use an auxotrophic or drug resistance marker with an
internal nonsense mutation. This way you can screen far more colony forming units and
only a tiny fraction of them will grow on plates. This # is inversly proportional to
fidelity.
Does your assay underestimate or overestimate the mutation rate? Give one reason why.
This underestimates mutation frequency because not all mutations will lead to gene
inactivation (or activation in the 2nd example).
You find that BR is 100 fold more susceptible to spontaneous mutations than E. coli.
You also determine that it is resistant to high doses of UV radiation, gamma radiation and
DNA modifying agents. What kind of DNA damage does UV cause?
Thymidine dimers.
What kind of damage does gamma radiation cause?
dsDNA breaks.
Name two mechanisms by which BR could deal with lesions in DNA during replication
in order to overcome the damage caused by each of these types of damage.
Can use a low fidelity polymerase to read through damaged bases. Can use
recombinational repair to bypass dsDNA breaks.
Describe how mismatched nucleotides are post-replicatively repaired in prokariotes
including, how mismatch is detected and how the daughter strand nucleotide is
selectively excised.
You undertake a kinetic analysis of polymerization by PolX and PolY and determine the
rate constants illustrated below when incorporating the correct nucleotide or an incorrect
nucleotide.
2
E.DNAN + dNTPa
1
E.DNAN+1 + dNTPb
E.DNAN+2
3
E.DNAN + dNMPa
Rate Constant
Rate when dNTPa is Correct
PolX
PolY
Rate when dNTPa is wrong
PolX
PolY
1
300
200
0.2
0.02
2
300
200
0.9
0.008
3
0.06
0.05
5
3
Which polymerase has a lower fidelity? Describe two ways this difference in fidelity
could come about.
PolX has lower fidelity. The important pararmeter is the ratio of Rates 2 and 3.
Individually, they are not really informative.
PolX might have a looser active site and so accept incorrect base pairs more readily.
PolX might be more tolerant of perturbations in the DNA helix downstream of the active
site.
PolX has a low processivity. Design an assay to determine polymerase processivity.
I would use a long, primed circular ssDNA template and add the polymerase in the
absence of nucleotides. I would have a huge excess of template and determine how long
the products of a synthesis reaction incubated just suffient time to completely replicate
the template if it were completely processive. I would then check the size of my
radiolabeled product on a gel. Dilution is the 2nd method of choice.
In a microarray analysis, PolX is found to be induced upon UV irradiation and exposure
to DNA modifying agents. What is a possible function for PolX? Why do its
biochemical qualities suit it for this task?
A low fidelity polymerase which can synthesise through DNA lesions.
These next questions are a bit trickier. You may wish to help each other out.
You want to see if BR synthesizes DNA in a canonical fashion and decide to look for
Okazaki fragments in vivo. How do you know your products are Okazaki fragments and
not just fragmented DNA?
The key point is that Okazaki fragments are RNA/DNA hybrids:
To confirm are Okazaki fragments: could use 32P TTP and 3H UTP and check for
fragments with both. Could chemically protect 5’ of strands, RNAse digest and then label
with a 5’ kinase. Could use a 5’ to 3’ DNAse and check if your fragment is protected.
Could check for a size shift following RNAse treatment. Could isolate chains, hydrolyse
with micrococcal nuclease and check for presence of NTPs in the chain by TLC.
It’s also OK to repeat the Okazaki experiment taking a pulse chase approach.
You determine that PolY does make Okazaki fragments. You wonder: does the lagging
strand PolY operate processively from fragment to fragment? That is to say, on
completion of an Okazaki fragment, does the lagging strand polymerase remain
associated with the fork, transfer to the start of the next OF and then continue or does the
lagging Pol dissociate on OF completion and a new pol associate at the beginning of the
next OF? Design an assay to test this in vitro. Discuss any assumptions or problems with
your assay.
Some people designed assays that looked for evidence of Pol dissociation and by
inference would suggest that: If Pol can be shown to dissociate, Pol is not processive; If
Pol can be shown not to dissociate, Pol is processive. This is not a good design because
it is looking for a negative result. It is very likely that, even if Pol were processive, there
would still be some low level of dissociation i.e. these things are rarely absolute.
The two best ways to carry out this experiment are:
1) Competition assays: preincubate to create replication fork and then challenge with an
excess of competing template. These assays are fine but it should be stated what the
caveats are. When using heparin as a competition, is it likely that Pol will have the same
affinity for heparin as for a replication fork?
2) Dilution assay: This is good as you only need make one assumption: that the complex
can withstand dilution. Can use agents like PEG to help. Get the fork going, dilute many
many times. Does lagging strand synthesis continue?
A note about the template: You can’t use ssDNA, this will not make a fork. You can use
dsDNA but you will need all initiation factors. You can use a nicked, circular dsDNA
template with an overhanging section of ssDNA at the nick.
OF length (bp)
The Okazaki fragments are short by prokariotic standards (200bp). You are curious as to
why the Okazaki fragments are so short: what determines their length? You decide to
test if it is the concentration of primase. You place the primase gene on two different
promoters and achieve one very high expressing and one very low expressing strain. You
achieve the results below:
Primase Concentration
Okazaki Fragment Length
Endogenous (350nM)
200 bp
Low (100nM)
260 bp
Very Low (10nM)
500 bp
You make the initial conclusion that primase concentration determines Okazaki length.
You now wish to quantitate your model. You put primase on a titratable promoter and
check OF length at a range of primase concentrations. You achieve this result:
600
500
400
300
200
100
0
0
200
400
600
[Primase] (nM)
Are these data consistent with primase determining OF length? Why or why not?
No. withing 100nM either side of physiological concentrations (350nM), the OF length
is not dependent on [Primase]. The [Primase] can be made limiting when the system is
stretched to extreems but this doesn’t necessarily inform as to the mechanisms of control.
Answer Key to Problem 2. DNA Replication & Protein Machines… (50 points)
Becky’s Problems: E-mail any questions to me at [email protected]
Note: I realize the answers I wrote don’t actually fit into the space I left for you guys. In
an attempt to write extremely thorough answers, my explanations have become longer
than anything I expected you to write.
During a screen of conditional mutants, you come across a strain of E. coli that is
defective for replication. Though DNA replication is normal at both RT and 37°C,
you find that your mutant of interest is unable to replicate its DNA above 42°C.
You perform in vitro assays to determine the extent of DNA synthesis when the
presynthesis and synthesis reactions are performed at a variety of temperatures.
With your data, you are able to plot the following graph:
2.1 ( 5 points) Based on the graph shown above, do you think that your mutated gene is
normally involved in the presynthesis or synthesis phases of DNA replication?
Why?
The basics:
The gene is involved in both the presynthesis and synthesis parts of the reaction. On the
42/37 plot, the presence of a lag before DNA synthesis occurs indicates that no
presynthesis occurs at 42 (and instead, presynthesis complexes are actually forming
during the lag at 37C). The fact that synthesis never occurs at 42 (on the 37/42 or 42/42
plots) shows that the protein is also involved in synthesis.
Some extra explanation:
If the protein had been involved in presynthesis only, you would have expected the 37/42
plot to show an initial burst of synthesis (as synthesis would proceed from the
presynthesis complexes formed during the 37 incubation), but DNA synthesis would then
plateau due to problems re-initiating DNA replication in subsequent rounds of synthesis.
This is generally expected b/c proteins in complexes help stabilize one another, so
temperature inactivation of a protein already in a complex often takes a much longer
period of time (so you think of the TS as preventing formation of new complexes).
However, it is formally possible that the nature of the mutation causes complexes to fall
apart - so if you explained this possibility and answered that the mutation was involved
only in initiation, you still got the points J Good thinking.
If the protein had been involved in synthesis only, typically the 42/37 plot would have
shown burst synthesis, and not a lag period. Presynthesis complexes would have formed
normally during the 42 incubation, and then synthesis at 37 would have proceeded
normally. Again, there are alternative explanations which are formally possible. If the
nature of the mutation caused your protein to be immediately degraded (or permanently
misfold) at the restrictive temp, the lag phase seen on the 42/37 plot could be attributed
to a period of time needed to synthesis new, functional protein.
Luckily, a friendly postdoc in your lab already has purified the various components
of the replication fork from a wildtype strain of E.coli.
3.2a (4 points) What does “95% pure” mean in terms of your purified components? How
do you assay for protein purity?
To say that your helicase is 95% pure means that 95% of the total protein content of your
sample is the desired protein (helicase) and 5% is contamination. This is a physical
measure of relative protein abundance, NOT activity. There is no standard for 100%
activity for any enzyme – your 5% of protein ‘contamination’ could be responsible for all
of the activity you see in a fraction. Instead, you look for the presence of a protein to
consistently and tightly co-fractionate with the activity throughout purification (so the
protein should become a higher percentage of the fraction as the activity increases during
purification). The standard way to measure purity is to run your fraction out on an SDSPAGE gel and stain with Coomassie or silver stain and then using densitometry (or
eyeballing) the relative amount of your protein in the sample. If it is a monomer, or has
identical subunits, your protein of interest should be a single, major band which
constitutes 95% of the protein in the lane. In actuality, you’d like any ‘contaminating’
protein to constitute less than 1% of the protein visible in the gel, b/c you have to assume
that there are proteins in the gel that stained below the limit of detection. Also, keep in
mind that SDS-PAGE gels will disassociate any multi-subunit proteins, and interpretation
of ‘purity’ may be complicated if the proteins is made up of subunits of different sizes.
Some people also mentioned using mass spec, which also works J
3.2b. (3 points) How would you use these purified components to determine in vitro
which replication fork machinery is defective in your mutant?
In vitro complementation assay: You have a crude protein extract from your ts mutant,
which is unable to produce DNA in your in vitro assay. Adding one purified, wt
component at a time (and later in combination), you see if your crude+protein X mix can
now synthesize DNA at the restrictive temperature (42C). If you get replication, protein
X is the wt version of the protein mutated in your ts mutant. [don’t need details on
replication assay here]
3.2c (10 points) Design an assay to detect replication in vitro. What are the expected
results of this assay if replication does/does not occur?
To assay DNA polymerase activity, monitor the incorporation of radioactive dNTPs into
a polymer. For our purposes, you’d most likely be assaying replication in a crude
protein extract from E. coli, but the assay would also work on purified proteins as long as
you include all the necessary cofactors. The template is circular, unnicked dsDNA –
ssDNA will not test helicase function unless the rxn continues into rolling-circle
replication (you’d need to detect your product on a gel to determine if it’s longer than the
size of the circular DNA you added). You can assay for incorporation of the
radiolabelled dNTPs using acid pptn of the DNA and filter binding. Detection (by
scintillation counter) of hot dNTPs on the filter indicates the nucleotides were
incorporated into DNA polymers (and synthesis occurred). If synthesis doesn’t occur, all
radioactivity will be found in the waste. Alternatively, you can run your rxn out on a gel
and use autoradiography to detect bands of DNA – which would run at the size of your
original template if you used a dsDNA circle.
3.3a (3 points) You’re excited when you find that a wild-type helicase allows you to get
replication at 42°C (woohoo!). However, you decide that you need to be able to run
DNA synthesis assays from purified components of your mutant (no more mooching off
your friendly postdoc). What technique would you use to purify your mutant helicase?
Please give some examples of purification methods.
Fractionation purification …. Some possible traits that could be used to purify your
protein are size, salt, DNA binding, other affinities …. (many possible answers)
3.3b. (10 points) Throughout your purification, you rely on a helicase assay to determine
which fraction contains the desired activity (and presumably your desired protein).
Describe your helicase assay… what is your template? Do any other cofactors need to
be added to the reaction, and if so, what? How do you separate and detect the products of
the reaction? What would the data look like?
A helicase assay monitors the change in gel mobility of a radiolabeled oligonucleotide.
The template is a radiolabeled oligonucleotide that is hybed to a ssDNA circle. If the
helicase is functional, the oligo will be separated from the ssDNA circle, and will run
faster on a gel, and the bands can be detected by autoradiography. In addition to the
template and the putative helicase, a proper buffer, ATP, and Mg2+ are needed to ensure
proper helicase function. Helicases have the natural ability to self-load onto ssDNA
(though it isn’t very efficient), so it actually isn’t necessary to use an ORI and provide
DnaA and DnaC. Though I didn’t require any controls, some good controls to run would
be +/- wt helicase, and boiling the template to separate the strands.
While celebrating your isolation of the ts helicase, you accidentally leave the replication
assay out on your bench, instead of incubating it at 37°C. You find that replication
doesn’t occur at 25°C. You then remember that the priming step of DNA replication is
cold sensitive, and doesn’t function at room temperature. Just to be thorough, you decide
to run a reciprocal shift assay for DNA synthesis (with your newly purified helicase!),
using all combinations of 25°C, 37°C, and 42°C for the presynthesis incubation and the
synthesis steps of elongation. Your template for the DNA synthesis assay is a dsDNA
circle, which contains and ORI. Assume all proteins required for DNA synthesis
(including your ts mutant helicase) are used in every reaction condition
3.4a. Fill out the chart below. For each row, the reaction is incubated at the temperature
indicated in column 1 for a sufficient amount of time, then dNTPs are added and the
temperature is shifted to the temperature indicated in column 2 for the duration of the
experiment. For each situation, please indicate to what degree DNA replication occurs -completely, not at all, or partially (please state the intermediate that results, or the DNA
product of the partial reaction). Also note the reaction intermediate that could be seen
using EM.
One note---b/c I didn’t specify the nature of the ts mutation in the helicase, these answers
will vary depending on whether or not you assume the mutant helicase can be loaded
onto the ORI at 42. I didn’t care if you assumed binding or not, but you had to be
consistent with your answers.
Also –the cold sensitive step for presynthesis is DnaB loading – which isn’t reversible if
you drop the temp.
If you assume that the ts helicase can’t bind at 42 and is non functional at 42
Col 1 Col 2
Extent of DNA replication EM intermediate
25
25
None
DnaA bound
25
37
Complete, with lag
all intermediates – DnaA alone,
DnaA+DnaB, open bubble, forks, theta
structure, circular DNA
25
42
None
DnaA bound*
37
25
Partial; one round
Mostly circular DNA, but possible to see
all intermediates depending on when you
look
37
37
Complete
All intermediates
37
42
None
Primed bubble
42
25
None;
DnaA*
42
37
Complete (with lag)
All intermediates
42
42
None
DnaA bound*
**If you assume that the ts helicase can bind at 42, but is still non-functional at 42, these
answers would be “DnaA + DnaB”.
3.4b. (1 point) Besides the proteins that comprise the replication machinery, what
cofactors do you need to include in the beginning reaction to allow the presynthesis
process to occur?
ATP (to power helicase loading and motor function), Mg2+, rNTPs for priming(though
technically a substrate). If you had two of these, I gave you credit.
3.5. (5 points) Because you have invested so much time in with your mutant strain, you
decide to look for suppressor mutations. You further mutate your ts mutant, and you look
for isolates that are now able to undergo DNA replication at 42°C. Given your original
mutation was in a helicase, give 3 possible means of suppressing this mutation.
Many possible answers (especially since I never specified what the original mutation
was)…
Generally, I was looking for:
-reversion of the original mutation
-Second-site suppression: point mutations within the helicase that allow for the
stabilization of the helicase hexamer (conceivable if the original ts mutation was caused
by destabilization btwn helicase subunits).
-Allele specific complementation mutants: if the ts defect was due to interactions btwn
helicase and another component of the rep. machinery, mutations in the other
components could suppress the original mutation (ex. t dimer, DnaA, DnaC, primase).
General things to keep in mind: If the mutation was due to helicase’s interaction with
primase, it could be conceivable that the helicase actually functions fine at 42C, but
doesn’t stimulate primase so the replication defect would actually be due to problems
with priming.
Also, many people suggested a mutation in the promoter of DnaB could allow for overexpression of DnaB, which may overcome the defect at 42 just by saturation of the
system. This could conceptually work if the problem is a binding equilibrium –then a
huge excess may overcome the poor binding. However, this type of mutation won’t
suppress a mutation where the helicase misfolds, etc… having lots of a dead helicase
doesn’t help.
In any of these cases, you should think about what your suppressing mutation is going to
do to the function of your helicase at the permissive temps… could increased binding
affect helicase turnover? Could constitutive over-expression of helicase cause problems
for the cells? Etc…
Answer Key to #3 – Rachel’s problem ([email protected])
Note – Most of these problems are meant to test your understanding of the class material.
The last part of each problem (**) is trickier; these problems are meant to stretch your
brain a little.
3.1) 30 points Chromatin is a many-splendored thing
You are interested in characterizing the establishment of silencing at the silent mating
type locus HMR in S. cerevisiae. You know that these loci require both cis and transacting factors in order to be silenced. You decide to investigate the neighboring flanks of
HMR for possible sequences that could confer silencing.
3.1a) 4 pts
You decide to look for origins of replication in these regions. Describe
how you would map these origins biochemically/physically and genetically (give one
example technique of each), and name some drawbacks to each technique.
Biochemical/Physical Mapping: The two general approaches discussed in lecture
were detection of nascent daughter strands by PCR and detection of replication bubbles
by 2D gel. A few important points: to detect nascent strands by PCR, you must size
fractionate the DNA first, then do multiple PCR (or quantitative PCR) reactions for each
fraction. You cannot just do PCR without size fractionating first, as you will get product
for every reaction. Similarly, quantitative PCR is not sensitive enough to detect the small
(less than 2-fold) changes resulting from replication. Drawbacks to both of these
techniques are that you need to have a general idea as to where the origins may be in
order to design oligos or probes for your 2D gel.
Genetic mapping: Make a library of non-replicating plasmids that contain different DNA
fragments, and introduce this library into bacteria. Look for maintenance of this plasmid
by maintenance of a marker on the plasmid, since only plasmids that contain a fragment
with an origin of replication will be maintained. Drawbacks to this method are that more
than replication may be necessary to maintain the plasmid (i.e., segregation, so it might
be a good idea to use CEN plasmids), and that the origin may behave differently on the
plasmid than it would in its chromosomal context.
3.1b) 3 pts You find sequences that look like origins of replication near the silencers
of the mating type loci (see above illustration). You decide to further characterize the
timing and efficiency of these loci. You decide to look at these origins by 2D gel after
synchronizing your cells in alpha factor and then releasing them. You obtain the
following data using specific probes for the two origins near HMR as well as for an
origin in a different location:
Characterize these origins as early/late, efficient/inefficient.
A- early and efficient
B- early and inefficient
C- late and inefficient
3.1 c) 4 pts
You find that one of the origins near HMR replicates late in S phase and
inefficiently. How would you test whether their chromosomal context affects their
replication timing and efficiency (see above diagram of HMR)?
Move the origin to another part of the chromosome (or to a plasmid) and assay its
replication timing and efficiency by 2D gel.
3.1 d) 6 pts You wonder whether progression through the cell cycle is important for
the establishment of silencing at HMR, and if so, which stage of the cell cycle may be
important. Diagram an experimental outline to test this hypothesis, using a sir3ts strain
(your experimental read-out is mRNA levels expressed from a gene within HMR,
HMRA1). You don’t have to provide as many details as a lab protocol would, but
provide enough details so that a proficient post-doc would be able to perform the
experiment based on your description.
1.Using a sir3ts strain, raise the temperature to 37C to inactivate Sir3 and relieve
silencing.
2. Arrest the cells at the restrictive temperature in alpha factor (this is important – if the
cells are non-synchronous at 37C and you lower the temperature and then try to arrest
them at different points of the cell cycle, some of the cells will go through an entire cell
cycle before arresting and your assay becomes useless).
3. Lower the temperature to the permissive temp and block aliquots of the cells at
different points of the cell cycle (i.e., either leave them in alpha factor and lower the
temp, or wash out alpha factor and then block in HU or Nocodazole or allow them to
progress through the cell cycle once and then re-arrest in alpha factor).
4. Assay silencing by looking at mRNA levels from HMRA1.
3.1 e) 6 pts
Show the expected results from part d if a cell cycle event (you
pick the phase) is necessary for establishment of silencing (i.e., what do the levels of
HMRA1 mRNA transcript look like at different points in the cell cycle? The simplest
way to represent this would be a Northern):
(if passage through S phase is required for establishment of silencing – and assuming that
the half-life of HMRA1 mRNA is very short):
RT
PT
alpha
HU
Noc
HMRA1 mRNA
If no cell cycle event is necessary for establishment of silencing:
RT
PT
alpha
HU
Noc
HMRA1 mRNA
3.1 f) 7 pts You find that passage through S phase is required for establishment of
silencing. You decide to test whether DNA replication may be the event needed for
establishment of silencing.
i)
How would you test whether firing of the origins flanking HMR is
necessary to establish silencing?
A reasonable answer is to delete the flanking origins and then repeat the above
experiment. When this experiment was done it was found that silencing was not
established in the absence of the flanking origins, but when they added back a
tethered ORC-GAL4 construct to the HMRE origin it was found that this was
sufficient to establish silencing.
** ii) How would you test whether a replication fork actually needs to pass
through HMR in order to establish silencing?
Here you need to find a way to uncouple DNA replication from a concurrent cell
cycle event. Therefore it is not sufficient to follow the replication forks during the cell
cycle and HMRA1 mRNA levels at the same time (just because replication and silencing
happen at the same time does not mean that replication causes silencing). One way to do
this would be to pop out the HMR cassette from the chromosome and circularize it into a
non-replicating ring (or recombine it onto a non-replicating plasmid). Given the results
from the ORC tethering experiment, it is important to include a similarly tethered ORC
construct to the non-replicating HMR. Then repeat the experiment from 3.1 e) assaying
mRNA levels from this non-replicating HMR. (By the way, when this experiment was
performed it was found that replication is not necessary for establishment of silencing,
but passage through S and M phase is.)
3.2 20 points Phun with phosphorylation
3.2 a) 6 pts You are studying the role of an essential budding yeast protein, Bioreg1
(Brg1), in DNA synthesis. You isolate a temperature-sensitive mutation of this gene,
where cells at the restrictive temperature arrest with large buds and a 1C DNA content.
Thus you suspect that Brg1 plays a role in DNA replication, but you are not sure whether
it is important in initiation or elongation. How would you distinguish between the two
possibilities?
Perform a reciprocal shift experiment with hydroxyurea (HU). HU arrests cells
in S phase (i.e., it inhibits elongation). So using a Brg1ts strain you would first arrest
your cells with HU to allow initiation but not elongation to proceed, then raise the cells
to the restrictive temperature and release from HU arrest, then look for cell division (if
DNA replication does not complete, cells won’t divide). If you see cell division, that
means Brg1 is probably involved in initiation and not elongation. If there is no cell
division, that means that Brg1 is involved in elongation.
3.2 b) i. 6 pts
You notice that Brg1 contains consensus sites for phosphorylation
by Cdk1. You find that Cdk1 phosphorylates Brg1 in vitro. Showing that Brg1 is a Cdk1
substrate in vivo is very difficult, but what data could you get that would strongly suggest
that Cdk1 phosphorylates Brg1 in vivo?
To imply that Cdk1 phosphorylates Brg1 in vivo, you can either make a ts strain
of Cdk1 (you can’t delete Cdk1 as it is essential) and look for decrease in
phosphorylation of Brg1 at the restrictive temperature by Western using an antiphospho-Brg1 antibody. Similarly, you can create a Shokat-style Cdk1 mutant that can
be chemically inhibited and then assay Brg1 phosphorylation. Unfortunately, the other
Shokat method of assaying direct phosphorylation using transfer of a bulky ATP analog
cannot be done in vivo. Another good hint that Brg1 is phosphorylated by Cdk1 in vivo
once you know that it is phosphorylated in vitro is to look for concurrent expression of
Brg1 and activity or expression of Cdk1.
Note – while the presence of Cdk1 consensus sites on Brg1 is a good hint that
Cdk1 may phosphorylate Brg1, simply mutating these consensus sites and looking for
loss of phosphorylation tells you nothing about the kinase that is acting on Brg1, it just
tells you that these sites are actually phosphorylated. There may be another kinase that
acts on the same consensus sites.
** ii. 8 pts You found that Brg1 plays a role in replication initiation, and you want to
test whether phosphorylation of Brg1 by Cdk1 is important for this role. You make a
mutant strain of Brg1 where its Cdk1 phosphorylation sites are mutated, but this strain is
dead. You’ve found out that the Cdk1 phosphorylation sites are important for viability,
but how would you show that they are important for initiation?
Here you simply want to repeat the experiment from 3.2 a) but include the Brg1
allele that is mutated for its Cdk1 phosphorylation sites. If you have the Brg1ts allele
present, the strain will be alive at the permissive temperature. Then perform the
reciprocal shift experiment with HU; if the cells fail to divide after block with HU and
then release into the restrictive temperature, you have shown that the phospho-site Brg1
allele is insufficient to complement the Brg1ts allele in initiation of replication and
therefore the phospho-sites are likely to be important for initiation.