Download 7.06 Problem Set Four, 2006

7.06 Problem Set Four, 2006
1. Explain the molecular mechanism behind each of the following events that occur during the
cell cycle, making sure to discuss specific proteins that are involved in making each event
happen.
(a) During G1 phase, cells pass through the Restriction Point if growth factors are present in the
cell medium.
Growth factors are sensed by RTKs, which then become activated and thereby activate, in
linear order, GRB2, Sos, Ras, Raf, MEK, and finally MAP kinase. MAP kinase therefore
dimerizes and enters the nucleus, where it regulates transcription factors that control genes
that are necessary for the progression of cells through the Restriction Point (i.e. genes
necessary to promote the occurrence of events in the cell cycle).
(b) During S phase, origins of replication fire.
When we say that origins “fire,” what we mean is the DNA replication initiates from that
origin. This occurs because the origin replication complex ORC recognizes and binds to
origins of replication. ORC acts as a landing pad for Cdc6 and Cdt1. These two proteins
lead to the recruitment of MCM proteins, which allow for the melting of the DNA at
origins. Once the DNA is melted into single-stranded DNA, a replication bubble is formed,
to which DNA polymerase is recruited.
(c) During S phase, each origin of replication never fires more than once.
The “firing” of origins that is described above triggers disassembly of the pre-replicative
complex (the proteins bound at the origin named above). A functional pre-RC cannot be
formed at the origin again until cells have progressed through the cell cycle to the next G1
phase. This is because Cdt1, an essential part of the pre-RC, is degraded after origins fire,
and is not resynthesized until cells have proceeded through M phase.
(d) During prophase of mitosis, chromosomes become condensed.
Once cells enter S phase, the cells start to accumulate mitotic cyclins (B-type cyclins).
These cyclins associate with CDKs, and upon activation of the mitotic cyclin/CDK
compexes at the G2/M boundary, these complexes phosphorylate condensin complexes.
Condensins are proteins that organize and compact chromatin into tightly wound
structures. Phosphorylated condensins associate with chromosomes and condense
chromosomes.
(e) During prophase of mitosis (in many cell types, but not in yeast), the nuclear membrane
breaks down.
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Once cells enter S phase, the cells start to accumulate mitotic cyclins (B-type cyclins).
These cyclins associate with CDKs, and upon activation of the mitotic cyclin/CDK
compexes at the G2/M boundary, these complexes phosphorylate nuclear membrane
proteins, including lamins and some nuclear integral membrane proteins. Once
phosphorylated, the nuclear lamin tetramer disassembles. Phosphorylation of some of the
nuclear integral membrane proteins reduces their affinity for chromatin. Together, these
two factors contribute to nuclear membrane breakdown.
(f) During prophase and metaphase of mitosis, the mitotic spindle forms.
Once cells enter S phase, the cells start to accumulate mitotic cyclins (B-type cyclins).
These cyclins associate with CDKs, and upon activation of the mitotic cyclin/CDK
compexes at the G2/M boundary, these complexes phosphorylate proteins that are
associated with microtubules. This promotes the reorganization and polymerization of the
mitotic spindle and asters. The spindle and asters are organized around two centers called
MTOCs (microtubule organizing centers).
(g) During prophase and metaphase of mitosis, chromosomes attach to microtubules.
Every chromosome has a region of DNA called the centromere. This region contains a
series of repeats of the same DNA sequence over and over again. This centromeric DNA is
bound by proteins, and a complex is thereby formed called a kinetochore. Some
kinetochore proteins have affinity for microtubules. Thus chromosomes become attached
to microtubules by protein-protein interactions between the microtubules and kinetochore
proteins.
(h) During metaphase of mitosis, chromosomes become aligned.
Chromosome alignment at the metaphase plate occurs because of the opposing forces
generated by (1) the bipolar mitotic spindle that attaches each sister kinetochore to
microtubules associated with opposite poles of the spindle and (2) the “cohesive” force that
holds sister chromatids together.
(i) At the metaphase-to-anaphase transition of mitosis, sister chromatids become separated.
Sister chromatids associate with each other along the length of the chromosomes through a
protein complex called cohesin, and sister chromatid separation initiates anaphase. Prior
to anaphase, the protein securin binds and inhibits a protease named separase. Once
chromosomes align on the metaphase plate, the cdc20/APC complex ubiquitinates securin,
resulting in degradation of securin by the 26S proteasome. Separase is activated (because
it is free from its inhibitor securin), and so separase cleaves cohesin complexes. This results
in sister chromatid separation.
(j) During anaphase of mitosis, sister chromatids are pulled to opposite poles.
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Chromosomes are segregated to opposite poles during anaphase due to a combination of
two things: 1) motor proteins that are part of the kinetochores allow the chromosomes to
travel along the microtubules (anaphase A), and 2) the two MTOCs (microtubule
organizing centers) move further and further away from each other (anaphase B).
2. Below are listed four different mechanisms of regulating of cyclin/CDK activity. For each
mechanism:
(i) Explain how that mechanism regulates cyclin/CDK activity.
(ii) Describe an experiment you could do to show that cyclin/CDKs are indeed regulated
in that way. Be sure to mention the result you would get that would demonstrate that such a type
of regulation is indeed occurring.
(a) Phosphorylation by Wee1
(i) Phosphorylation of cyclin/CDKs by Wee1 is inhibitory. These phosphorylation events
are located near the ATP binding sites of the CDKs, and block ATP binding due to
repulsive interactions.
(ii) Do an in vitro kinase assay to examine the kinase activity of cyclin/CDKs. Put in a test
tube: active cyclin/CDKs, one of their substrates (such as histone H1, a protein involved in
chromatin structure), and radiolabeled ATP. Incubate, and then run on an SDS-PAGE
gel. Expose the gel to film. If the cyclin/CDKs were not pre-incubated with Wee1, then you
should see a band on the film that is about the size of the substrate protein. If the
cyclin/CDKs were pre-incubated with Wee1, you should not see that band on the film.
(b) Dephosphorylation by Cdc25
(i) De-phosphorylation of cyclin/CDKs by Cdc25 is activating. Cdc25 removes the
repulsive phosphates at the ATP binding site, allowing ATP binding to the CDK.
(ii) Repeat the kinase assay from above, but before adding cyclin/CDKs to the in vitro
kinase assay, incubate them with active Cdc25. The cyclin/CDKs that were inactive in the
assay previously (after phosphorylation by Wee1) should now be active.
(c) Binding by the inhibitor p21
(i) p21 binds to cyclin/CDKs and prevents them from accessing their substrates.
(ii) Repeat the kinase assay from above, using cyclin/CDK complexes that are active, but
then do the kinase assay with p21 present. Cyclin-CDKs should be able to phosphorylate
their targets only in the absence of p21.
(d) Degradation of cyclins by the 26S proteasome
(i) Cyclins are ubiquitinated and targeted for proteasomal degradation at times during the
cell cycle when they are not supposed to be active.
(ii) Grow a culture of cells, and synchronize the cells such that they are at the point in the
cell cycle right before the cyclin you are testing is about to be degraded. Then release the
culture so that it is growing again (synchronously), but split the released culture in half.
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Leave half the cells untreated, and treat the other half with proteasome inhibitors. Take
samples at different time points, and do Western blotting on the different samples, using an
antibody against your cyclin. You should see that the band corresponding to the cyclin
remains around longer in the half of the culture to which you added proteasome inhibitors.
3. The surprise success of this year’s Oscar-winning documentary, March of the Penguins, has
spurred a sudden resurgence in penguin research in molecular biology labs across the globe. Of
particular concern is the threat posed to penguins from increased exposure to UV radiation,
brought about by the expanding hole in the protective ozone layer over Antarctica. As a
concerned scientist and avid penguin lover, you decide to study the mechanisms that penguin
cells employ to protect themselves from UV-induced DNA damage.
Conveniently, Morgan Freeman, while narrating the documentary, cultured some penguin cells
and brought them back to your lab where you can grow and study them. However, as much as
you appreciate his generous contribution to your scientific endeavors, you decide that it would be
much easier to carry out a genetic screen to hunt for mutants that cannot protect themselves from
DNA damage in a simpler system, Schizosaccharomyces pombe. You begin by performing a
genetic screen in fission yeast to identify UV-sensitive mutants.
(a) Describe how you would perform such a screen.
First, mutagenize the yeast (EMS, UV, etc.), and plate on rich media. Then replica plate
the cells, briefly irradiate the replica plates with UV, and allow the cells to grow and
recover. Colonies that appear on the untreated plates but fail to grow on the UV-treated
plates contain your mutant cells of interest. You should perform this screen at various
temperatures so that you can isolate any temperature-sensitive mutants if they exist.
Lastly, use cloning by complementation to identify the mutated genes.
From your initial UV-sensitivity screen, you manage to pull out a cold-sensitive mutant version
of the Chk1 kinase. At room temperature (25°C) the mutant exhibits a wild-type phenotype, but
shifting to 16°C produces the mutant phenotype of sensitivity to UV. You make the analogous
mutation in the Chk1 homologue in the penguin cells (i.e. you mutate the same amino acid in the
same way in the penguin homolog of this protein). You then grow an asynchronous culture of
these mutant penguin cells at room temperature, and then shift to 16°C in the presence of UV
irradiation. You take samples of cells from each of these steps and perform FACS analysis on
each of the samples.
(b) Draw the FACS profiles you would expect to see for the Chk1 mutant with and without UV
treatment:
-- at room temperature, and
-- at 16°C.
Also draw the FACS profiles you would expect to see for wild-type cells under the same
conditions. Make sure to label the axes of your FACs profiles completely.
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Without UV treatment – You should see a regular FACS profile for wild-type and the
Chk1 mutant at both the permissive and non-permissive temperature.
With UV treatment, at room temperature: both wild-type and the Chk1 mutant will stall at
the G2 DNA damage checkpoint, producing the following FACS profile. These cells are
stalled at the entry into M phase (as they should be), and are trying to repair the DNA
damage caused by the UV before entering M phase.
At 16 degrees + UV: wild-type cells will stall at the G2 DNA damage checkpoint as usual.
The Chk1 mutants will continue to progress through the cell cycle because they cannot
trigger the G2 DNA damage checkpoint. Thus the mutants produce a “regular” FACS
profile.
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(c) Morgan Freeman has isolated a few mutants that he would like you to introduce into your
newly developed penguin cell line. Predict how the following mutations will affect cell cycle
progression in:
-- an otherwise wild-type cell line treated with UV, and
-- your Chk1 mutant cell line, shifted to 16°C and treated with UV.
Assume that you are replacing the endogenous gene with these mutant versions.
(i) a Y15D amino acid substitution in Cdk2 (i.e. amino acid 15 is changed from Y to D)
Replacing Y15 (which is normally phosphorylated by Wee1, thereby inactivating Cdk2)
with D (aspartate) yields an inhibitory “phosphorylated” residue that will keep Cdk2
inactive all the time. This will prevent mitotic entry because the mitotic cyclin/Cdk2
complexes cannot be active. You will see the same effect in wild-type cells and Chk1
mutant cells. This is because the Chk1 gene acts upstream in the DNA damage checkpoint
pathway from Cdk2. Thus a double mutant cell with mutations in Chk1 and Cdk2 will
show the phenotype resulting from the single mutation in Cdk2.
The wild-type pathway is:
Damage  Chk1 --] Cdc25  cyclin/Cdk2  mitotic entry
(ii) a Y15A amino acid substitution in Cdk2
The alanine residue prevents phosphorylation of Cdk2 by Wee1, which yields a
consitutively active MPF (because phosphorylation of Cdk2 by Wee1 is normally
inhibitory). This mutant will cause premature entry into mitosis. You will see the same
effect in wild-type cells and Chk1 mutant cells. This is because the Chk1 gene acts
upstream in the DNA damage checkpoint pathway from Cdk2. Thus a double mutant cell
with mutations in Chk1 and Cdk2 will show the phenotype resulting from the single
mutation in Cdk2.
The wild-type pathway is:
Damage  Chk1 --] Cdc25  cyclin/Cdk2  mitotic entry
(iii) overexpression of Wee1
Wee1 phosphorylation of Cdk2 is inhibitory, and prevents entry into mitosis.
Overexpression of Wee1 in wild-type cells will prevent mitotic entry. Just as above, you
will see the same effect in wild-type cells and Chk1 mutant cells. This is because the Chk1
gene acts upstream in the DNA damage checkpoint pathway from Cdk2, which is the
protein that is inactivated by overexpression of Wee1.
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4. You are interested in studying the cell mechanisms that control the separation of sister
chromatids at the metaphase-to-anaphase transition of the cell cycle.
(a) What mutant phenotype would you expect to see if there was no cohesion between sister
chromatids but the kinetochores of the sister chromatids were able to attach to microtubules?
Assume that you are able to mark chromosomes such that you can visualize their duplication and
segregation.
Chromosome alignment at the metaphase plate in the wild-type situation occurs because of
the opposing forces generated by (1) the bipolar mitotic spindle that attaches each sister
kinetochore to microtubules associated with opposite poles of the spindle and (2) the
“cohesive” force that holds sister chromatids together. In the absence of cohesion, the
separation of sister chromatids to opposite poles would not be properly regulated. That is,
sister chromatids would segregate randomly, rather than one to each pole. You would see
a greater number of cells where chromosomes were lost (both sister chromatids went to the
other daughter cell) or chromosomes were duplicated (the cell inherited both sister
chromatids).
(b) In budding yeast, loss of sister chromatid cohesion depends on the separase Esp1 and is
accompanied by the dissociation of the cohesin complex subunit Scc1 from the chromosomes.
To study the mechanism by which Esp1 causes Scc1 to dissociate from chromosomes, you
perform an in vitro Scc1 dissociation assay.
You arrest wild-type yeast in metaphase by nocadozole treatment, and harvest the cell extract.
You then isolate chromatin from the extract using a protocol by which chromatin can be pelleted
by centrifugation and separated from the soluble (non-chromatin associated) proteins. This
results in a separation of the chromatin (C) and supernatant preparations (S). Note that
“addition” refers to the addition of an excess of purified recombinant protein to the cell extract
before centrifugation. “Input” refers to the total material, before any treatment or centifugation.
The reactions are analyzed by Western blot using an antibody that recognizes Scc1.
Explain the results for cell extracts treated in three different ways before centrifugation: (A), (B),
and (C). Make sure to explain why the band is the size it is, and why the band is in the
chromatin pellet or in the supernatant, for each cell extract.
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(A)
(B)
(C)
wild-type wild-type wild-type
–
Input
C
S
Esp1
C
S
C
Esp1 +
Pds1
S
Cell extract
Addition
C
Note that the cell extracts have their own separase/Esp1 protein in them, but this separase
protein is being held inactive by securin/Pds1 because the cells are frozen at metaphase,
when cohesin/Scc1 is still protected from cleavage.
(A) In the absence of active separase, Scc1 is chromatin-associated and not cleaved
because there is no active separase to cleave the Scc1 cohesin protein.
(B) In the presence of an excess of separase (which overwhelms the cell’s
endogenous securin), Scc1 is not chromatin-associated because it has been cleaved by
separase. Thus the antibody detects as a faster-migrating Scc1 protein in the soluble
fraction.
(C) In the presence of separase and equal amounts of its inhibitor securn (Pds1),
separase is bound by securin, and thus Scc1 is chromatin-associated and not cleaved.
(c) You generate various mutations in genes involved in the metaphase-to-anaphase transition in
budding yeast. You perform each of your studies such that you have replaced the endogenous
wild-type form of the gene with the mutant form of the gene.
In each case, predict what would go wrong during the cell cycle, and explain your reasoning.
Your choices are: wild-type segregation of sister chromatids, random segregation of sister
chromatids, OR metaphase arrest.
(i) A mutant form of securin (Pds1) that cannot be ubiquitinated.
If securin cannot be ubiquitinated, it will never be degraded and thus will never relieve
inhibition of separase. Therefore, separase will be unable to cleave any cohesin, and sister
chromatids will arrest at the metaphase plate.
(ii) A mutant form of separase (Esp1) that cannot bind securin.
Without being sequestered by securin, separase will be free to cleave cohesin such that
there will not be any sister chromatid cohesion. This would occur even before metaphase
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(which is when cohesin is normally supposed to be cleaved). Thus you would see random
segregation of sister chromatids because of the reasoning in part (a).
(iii) A mutant form of Cdc20 that cannot bind to the APC.
If there is no interaction between the APC and the specificity factor Cdc20, securin (pds1)
will not be ubiquitinated and targeted for proteasome-mediated degradation. Thus the
sister chromatids will arrest at the metaphase plate, because separase is always inactive,
and thus cohesion is never cleaved.
(d) The spindle-assembly checkpoint prevents cells from completing mitosis (entering
anaphase) before all of the chromosomes are properly attached to the mitotic spindle. Genes
involved in the spindle-checkpoint were identified in a genetic screen in which mutant yeast
were isolated that were hyper-sensitive to benomyl, a microtubule depolymerizing drug.
Outline in steps how this screen was actually done.
First, mutagenize the yeast (EMS, UV, etc.), and plate on rich media. Then replica plate
the cells on plates containing benomyl. Make sure to put in a concentration of benomyl
that does not kill wild-type cells. Colonies that appear on the untreated plates but fail to
grow on the benomyl plates contain your mutant cells of interest. Lastly, use cloning by
complementation to identify the mutated genes.
(e) Explain the theory behind why the screen described in part (d) led to the identification of
mutants that could not properly undergo the spindle checkpoint.
Benomyl slows down microtubule assembly. If microtubule assembly was affected in wildtype yeast, the spindle-assembly checkpoint would result in cell cycle arrest until all of the
chromosomes were properly attached to the mitotic spindle. In this genetic screen, you
would look for yeast that proceeded to anaphase despite the presence of kinetochores that
were not properly associated with spindle microtubules. These mutants did not trigger the
spindle assembly checkpoint. This premature entry into anaphase would lead to random
segregation of chromosomes and cell death.
(f) Mad2 (Mitotic assembly defective #2) was identified using the strategy described above in
part (d). Mad2 localizes to kinetochores that are unattached to microtubules. How do you think
this localization pattern was discovered?
You can visualize Mad2 localization either by immunofluorescence with an anti-Mad2
antibody, or by visualizing the localization of a Mad2-GFP translational fusion protein in
live cells. You would have to examine Mad2’s localization in cells in which some
kinetochores are not attached to the spindle (i.e. conditions under which the spindle
checkpoint has been triggered). You could try to watch a lot of wild-type cells, waiting to
see a lagging chromosome, onto whose kinetochore Mad2 will be attached. However you
could also enrich for cells with unattached kinetochores, either by using kinetochore
mutants, or by treating cells with a drug like benomyl.
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5. (Please note that this question is on Meiosis from Lecture #11, which is
after exam #2. This material will therefore not be covered on
exam #2. You will be tested on Lecture #11 later in the semester.)
You are studying the process of meiosis in the yeast S. cerevisiae. You wish to see whether a
mutant that you have found is defective in the process of recombination or not.
You use a segment of chromosome #V (ChrV) to test, via Southern Blots, whether
recombination has occurred or not at this specific region. You choose this segment because it
contains a “recombination hot spot,” which is a site on the chromosome at which recombination
events are known to occur more frequently than at other sites in the area. You designate the
original MATa yeast strain as “Mom” and the original MATalpha strain as “Dad.” You choose
these two strains because they differ by the placement of NotI restriction sites within this region
of Chromosome V. The configurations of “Mom” and “Dad” at this region of ChrV are
illustrated in the diagram below. You generate three radiolabeled DNA probes (P1, P2, P3),
which are complementary to both Mom’s DNA and Dad’s DNA at the sites indicated on the
diagram.
2kb
6kb
MATa “Mom” ChrV
P1
P2
P3
MATalpha “Dad” ChrV
Not1 restriction site
3kb
4kb
Recombination hot spot
(a) You make two versions of your mom and dad strains – one version is wild-type at your gene
of interest, and the other version has a deletion of your gene of interest. Describe how you
would perform an experiment to check whether your mutant is defective in recombination or not.
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One must first generate a diploid to study meiosis. Cross the wild-type MATa and
MATalpha cells to generate a wild-type diploid. Also, cross the mutant MATa and
MATalpha cells to generate a mutant diploid. Induce each of your two diploid strains to
undergo synchronous meiosis. Collect samples of cells at various time points. Isolate DNA,
digest it with Not1 enzyme, perform Southern blotting and compare the results. If there is a
difference in the appearance of the Southern blot between the two strains, then your
mutant must be defective in recombination.
(b) Which probe (i.e. P1, P2, or P3) would you use to identify recombinants in your experiment
that you designed in part a), and why?
You would use probe P2, because it is the only one that allows you to distinguish between
crossover and non-crossover events, when using the Not1 restriction enzyme.
(c) You allow a wild-type diploid to undergo meiosis and collect samples at various time points
(0 hours through 10 hours after induction of meiosis). You then isolate DNA from the cell
population that exists after different amounts of time have passed. You digest the DNA from
each sample with Not1 and perform a Southern blot with probe P2. What is the identity of each
of the bands (i - vi) on the resulting blot?
0h 1 2 3 4 5 6 7 8 9 10h
10 kb
8 kb
6 kb
(i)
(ii)
(iii)
(iv)
4 kb
(v)
2 kb
(vi)
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(i)
(ii)
(iii)
(iv)
(v)
(vi)
One recombination product (9kb = 3 kb “Dad” + 6 kb “Mom”)
“Mom” DNA, no recombination ( 8 kb)
“Dad” DNA , no recombination (7 kb)
other recombination product (6 kb = 2 kb “Mom” + 4 kb “Dad”)
Double Strand breaks (3 kb, from “Dad”)
Double Strand Breaks (2 kb, from “Mom”)
“Double strand break” products (v and vi) are formed after cleavage has occurred at the
recombination hot spot, but before the recombination events have been resolved.
(d) Suppose your mutant is defective forming double strand breaks (DSBs). What would your
Southern blot look like?
In the absence of DSBs, recombination is impossible. One would only see “Mom” and
“Dad” DNA without recombination, and thus see bands only at 8kb and 7 kb respectively.
(e) You do the experiment you suggested in part a) using a diploid strain that lacks function in
your gene of interest, and see the following result. What do you conclude and why?
0h 1 2 3 4 5 6 7 8 9 10h
10h
10 kb
8 kb
6 kb
4 kb
2 kb
The mutant is able to forms DSBs but is unable to resolve and/or repair them. Thus, with
time, only DSBs are formed (at the loss of full length parental forms). The DSB bands start
to disappear because these broken DNA fragments are likely degraded by DNases in the
cell.
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