Download LS1a Fall 06 Problem Set #9 (100 points total) all questions

LS1a Fall 06
Problem Set #9
(100 points total)
all questions including the (*extra*) one should be turned in
1. (23 points total) If dividing cells are grown in a culture medium containing
radioactive thymidine, the thymidine will be covalently incorporated into the cell’s
DNA during replication. The radioactive DNA can be detected in the nuclei of
individual cells by autoradiography (i.e., by placing a photographic film over the
cells, radioactive cells will activate the film and show up as black dots when looked
at under a microscope).
a) During what phase of the cell cycle will the radioactive thymidine be
incorporated? Why?
(2 point) Radioactive thymidine will be incorporated during S phase,
since this is the phase of the cell cycle where DNA is replicated.
Consider an experiment in which cells are radioactively labeled by this method for
only a short period of time (about 30 minutes). The radioactive thymidine medium
is then replaced with medium containing unlabeled thymidine, and the cells are
grown for some additional time. At different time points after replacement of the
medium, cells are examined in a microscope. The fraction of cells in mitosis (which
can be easily recognized because their chromosomes are condensed) that have
radioactive DNA in their nuclei is then determined and plotted as a function of time
after the labeling with radioactive thymidine.
b) Would cells at all phases of the cell cycle contain radioactive DNA immediately
after the labeling procedure? Why?
(4 points) No, only those cells in S phase during the cell cycle would
contain radioactive DNA immediately after the labeling procedure.
c) Why are there initially no mitotic cells that contain radioactive DNA?
(4 points) At t=0, the only cells that have radioactive DNA are in S
phase. The mitotic cells at t=0 contain no radioactive DNA. It takes
about two and a half hours before the first labeled mitotic cells appear.
d) Explain the rise and fall and then rise again of the curve.
(5 points) The only cells that are labeled are those in S phase during
the 30 minute treatment. The initial rise of the curve corresponds to
cells that were just finishing DNA replication when the radioactive
thymidine was added. The curve rises as more of those cells enter
mitosis to a peak that corresponds to times when all of the mitotic cells
were in the S phase during the time of labeling. The labeled cells then
exit mitosis, being replaced by unlabeled mitotic cells that were not yet
in S phase during the labeling period. After 20 hours the curve starts
rising again, because the cells labeled in the first 30 min pulse are
entering a second round of mitosis.
e) Estimate the length of G2 for these cells
(4 points) The initial lag before the cells enter mitosis corresponds to
the G 2 phase (approximately 2.5 hours), which is the time between the
end of S phase and the beginning of mitosis.
f) Estimate the length of the total cell cycle for these cells
(4 points) Around 26-28 hours (measure time from 50% of labeled
cells entering mitosis for the first time to the time when 50% of
labeled cells enter mitosis for the second time).
2. (28 points total) In lecture you learned about the experiments of Tim Hunt in sea
urchin eggs that lead to the discovery of cyclin. The synchronous division of these
cells during the early embryonic development facilitates the analysis of the
population of cells since the cells are essentially all in the same biochemical “state.”
You decide to identify if similar factors are involved in regulating cell division in frog
embryos. So you perform experiments like those of Tim Hunt. You chemically trick
the frog eggs into thinking they had been fertilized, incubate the eggs in
radioactively labeled methionine, collect the eggs every 10 minutes, lyse the cells
and separate the proteins according to their molecular weight and then use
photographic film to detect the amount of proteins that are radioactive. Radioactive
proteins emit particles that will expose film, so the more the film is exposed the
more radioactive protein present.
a) Only proteins that have been synthesized after fertilization will be detected by
this procedure. Briefly explain why?
(3 points) Incubation of the eggs in radioactively labeled methionine
occurs after fertilization. Therefore only proteins synthesized after
fertilization will incorporate this labeled Met and be detected.
The results of your experiment are shown on the gel below. You can clearly see the
pattern for 4 different proteins.
Time after fertilization
(min)
10
20
30
40
50
60
70
80
90
100
110
120
Protein A
Protein B
Protein C
Protein D
b) The time it takes the eggs to complete a round of division during you experiment
is 60 minutes. Which protein(s) of do you hypothesize might be cyclin(s)?
(2 points) Protein B
c) During which time point(s) does mitosis end?
(3 points) The concentration of cyclin falls rapidly at the end of mitosis
so 60 (or between 50 and 60 minutes) and 110 (or between 100 and
110 minutes)
d) Would your answer to b) change if a round of cell division took 120 minutes?
Why or why not?
(3 points) Yes, then Protein D would be a cyclin candidate since it
increases during interphase and is destroyed at the end of mitosis (the
cell cycle).
e) For this experiment why is it necessary to synchronize the fertilization of the
cells?
(4 points) If the cells were not synchronized they would be at all
different stages of the cell cycle at every time-point. You could not
monitor the rise and fall of individual proteins in a cell as this cell goes
through its own cell cycle since there is no easy way to distinguish this
cell’s proteins from those of other cells at different stages.
f) Assume that it takes the eggs 60 minutes to divide (complete one cell cycle). On
the following diagram please draw in expression patterns for the following
proteins.
(5 points-1 point for wt, 2 points for each mutant)
Time after fertilization
(min)
10
20
30
40
50
60
70
80
90
100
110
Wild type Cyclin
Mutant cyclin that
cannot be ubiquitinated
Mutant cyclin that
cannot bind to Cdk-1
g) What would be the consequence(s) of a cyclin that could not be ubiquinated on
the progression of the cell cycle?
(4 points) The destruction of cyclin occurs through ubiquitin-mediated
proteolysis so if cyclin could not be ubiquinated it would not be
destroyed. The rapid elimination of cyclin is required for the cell to exit
mitosis, so these cells would be stuck in mitosis.
h) What would be the consequence(s) of a cyclin that could not bind to Cdk-1 on
the progression of the cell cycle?
(4 points) Since cyclin cannot bind to Cdk-1 neither an inactive or
active cyclin-Cdk1 complex can form. An active cyclin-Cdk-1 complex
is necessary to trigger entry into mitosis, so these cells would be stuck
in G2.
3. (16 points) Cell survival depends on the accurate transmission of the cell's genetic
material to its daughters. Briefly explain the consequences of the following
scenarios on cell division.
a) Chromosomes fail to condense during mitosis.
(4 points) Chromosomes are intertwined with one another and cannot
be properly segregated during anaphase. Cell will fail to divide
properly.
b) Mutation of Ser-22 and Ser-392 to alanines in lamin protein prevents the nuclear
lamina from disassembling during mitosis.
(4 points) Nucleus cannot be disassembled, chromosomes are
inaccessible to microtubules, and cell cannot divide.
c) The centrosome fails to duplicate and a monopolar spindle forms during mitosis.
120
(4 points) Single spindle pole is unable to line up the chromosomes at
the midline of the cell, chromosomes are unable to be segregated to
opposite sides of the cell, and cell division will be halted by spindle
checkpoint.
d) The rate of microtubule catastrophe does not increase during mitosis.
(4 points) Microtubule search and capture will occur much more slowly
and the rate of mitosis will be decreased.
4. (33 points) You have been studying the filament systems of the cytoskeleton in
fibroblast cells. You are interested in migration of your fibroblast cells. You have
read in Alberts (p. 592-598) that actin filaments are essential for cell crawling.
a) Actin filaments like microtubules have polarity. What does this mean?
(3 points) The filaments are made of asymmetric subunits so the
polymer has an inherent direction embedded in its structure and two
different ends.
b) The two ends of an actin filament are called the plus end and the minus end.
What are the two ends of a microtubule called?
(1 point) plus end and the minus end
c) Which end of the microtubule is anchored at centrosome?
(1 point) minus end
d) From which end does a microtubule preferentially grow?
(1 point) plus end
Actin filaments can grow by the addition of actin monomers at either end. However, the
rate of growth is faster at the plus end than at the minus end.
e) When cells are crawling the minus end of the actin filament is anchored in the
cell’s interior by binding to other proteins, and the plus end is located close to the
plasma membrane. How might addition of monomers aid in cell crawling?
(3 points) Growth of actin filaments at the plasma membrane pushes
the membrane forward.
f) You treat your cell with a toxin called cytochalasin that prevents actin
polymerization. What effect would this treatment have on cell migration?
(3 points) Cell migration would be inhibited.
Similar to the polymerization of tubulin, each free actin monomer is tightly bound to a
nucleotide triphosphate, in this case ATP (tubulin uses GTP), which is hydrolyzed to ADP
soon after the incorporation of the actin monomer into the filament. Nucleotide
hydrolysis as with microtubules promotes depolymerization of actin filaments.
g) You take some of your cells and rupture their membranes to isolate their contents
(make a cell extract). You split your cell extract into 2 tubes and add ATP to
Tube 1 and non-hydrolyzable ATP to Tube 2. Assuming both ATP and nonhydrolyzable ATP bind actin monomers with similar affinities, would the average
length of actin filaments differ between Tube 1 and Tube 2? Briefly explain why
or why not?
(4 points) The average length of an actin filament in Tube 2 would be
longer than the average length of an actin filament in Tube 1. If ActinATP cannot hydrolyze to actin-ADP, depolymerization of actin filaments
would be inhibited.
One day you notice that your cells have become infected with bacteria. Interestingly,
the bacteria have long “tails” and in each case where a bacterial “tail” is observed you
find that the bacteria are able to move. You isolate the bacteria and begin doing
experiments with cell extracts from non-infected fibroblast cells. Figure 1 shows the
phase contrast microscope images that you collect after 30 minutes of incubation of the
bacteria with the extracts of the cells.
A
B
C
D
Figure 1: Phase contrast analysis of bacteria incubated in fibroblast whole cell
extracts.
A) Unmodified extract; B) Extract depleted of ß-tubulin; C) Extract depleted of
keratin(component of intermediate filaments); D) Extract depleted of actin.
h) What do the results in Figure 1 suggest about the composition of the bacterial
“tail”?
(4 points) Actin is either a component of the “tail” or necessary for its
formation. ß-tubulin and keratin are not needed for formation of the
bacterial “tail”.
i)
Hypothesize how the component identified in (h) might facilitate motility?
(4 points) One hypothesis (others are acceptable): If the tail is made
of actin, addition of actin monomers between the bacteria and the tail
could “drive” the bacteria forward. Therefore, actin polymerization is
moving the bacteria.
To test your hypothesis you decide to do another experiment. Figure 2 shows the phase
contrast microscope images that you collect after 30 minutes of incubation of the
bacteria with the designated cell extracts. You again observe that only bacteria with
tails were mobile.
A
B
Figure 2: Phase contrast analysis of bacteria incubated in modified fibroblast whole
cell extracts.
A) Unmodified extract; B) Extract containing a 100X excess of ADP
j) Does the result shown in Figure 2B agree with your hypothesis? Briefly explain
why or why not.
(4 points) Yes. Since actin polymerization is driving the movement of
the bacteria, treatment with ADP would inhibit actin polymerization
and therefore inhibit bacterial movement.
k) Do you think the migration of the fibroblast cells would be affected by bacterial
infection? Briefly explain why or why not.
(5 points) Yes. Movement of both requires actin monomers and actin
polymerization, assuming there is a finite pool of actin in the cell actin
monomers that are recruited for polymerization of the bacterial tail
cannot be used for addition to actin filaments involved in cell
migration.