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Problem Set 5, 7.06, Spring 2003
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1. In order to please your demanding thesis advisor, you've
completed an extensive fractionation and biochemical purification of
proteins localized to the mitochondria, the chloroplasts, the
peroxisomes, and, because you've gotten so good at purifying
proteins, the nucleus, too. Everything was going fine, and you have
tubes of protein extract from each of the organelles fresh out of the
centrifuge and ready to be labelled (you're so good at the purification
that you never label tubes until you are ready to put them in the
freezer, but simply rely on memory to tell which is which while you are
performing the purification). But on the way back from the centrifuge,
the new (and somewhat clumsy) post-doc in the lab bumps into you
and tubes go flying. You groan and say to yourself, "Great now I have
to start the purification all over again, just because I didn't keep my
tubes labelled. How can I ever sort this mess out?" The new postdoc, eager to redeem himself, suggests that you run a little of each
extract out on an SDS-PAGE gel and sequence some of the proteins
from each tube. That way you'll be able to tell relatively easily which
tubes contain proteins from which organelles (and maybe finally learn
the lesson of why it's important to ALWAYS LABEL YOUR TUBES!).
a. Assuming that the new post-doc is correct and this approach is
possible, why would this type of experiment help you to sort out
what type of proteins are in each tube?
After sequencing some of the proteins from each tube, you
could use a computational biology/bioinformatics approach to
search for target signal sequences in each set of proteins
sequenced. Once you knew which target sequences
predominated in the proteins from each tube, you could make a
pretty solid judgement as to which tube contained proteins
from the mitochondria vs. the nucleus, etc.
b. You try the approach and obtain the following partial polypeptide
sequences from 4 specific proteins you sequenced (listed using 1 letter
amino acid code from N to C termini, sequence may be from either N
or C terminus). Using these sequences, tell to which organelle the
proteins would be targeted and give a short explanation of how you
determined this.
1) MSTSVSASSTVTASTVGPCIPVNQFWQNMKARHKCGHE-Chloroplasts, N-terminal region rich in S, T, and small
hydrophobic residues (A, V)
2) MSKRTKPKRLNGQPCSVMLFYGPKSFCGHESTPYWFQ-Mitochondria, N-terminal region with 3-5 nonconsectutive K, R,
and some S, T
Problem Set 5, 7.06, Spring 2003
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3) MSLQFWGQNDVILAQNTGPCKHRRKGYWPCFMLDESTLLLLL-Nucleus, internal region with 5 basic amino acids (K, H, R) is
NLS, patch of L is part of NES
SKL sequence at extreme C-terminus
c. Do all of the proteins you've sequenced enter their target
organelles in the folded state? If not, which ones enter in an unfolded
state and how is accomplished? 3 and 4 enter the nucleus and
peroxisomes respectively in their folded state. 1 and 2 enter
the chloroplasts and mitochondria respectively in their
unfolded states with the help of chaperone proteins like
cytosolic Hsc70 and MSF, which prohibit the proteins from
being folded until they are in their proper organelle.
d. What forms of energy do each of these proteins that you've
sequenced require to be imported into their respective organelles?
1 requires ATP hydrolysis (and possibly some energy or
conformational change from Toc34 protein binding GTP and
influencing Toc75 channel; additionally there may be some
influence by the pH gradient across the thylakoid membrane on
the uptake of some proteins into the thylakoid lumen)
2 requires ATP hydrolysis and a energy from the proton motive
force (electrochemical H+ gradient) across the inner
mitochondrial membrane
3 requires ATP hydrolysis and GTP hydrolysis by Ran GTPase
4 requires ATP hydrolysis
2. You discovered 2 new proteins (Y and Z) that are encoded by
nuclear genes, translated in the cytosol and then may be imported
into the mitochondria. You have the cDNAs that encode these
proteins. You want to know if these proteins are taken up by the
mitochondria and where in the mitochondria these proteins would be
a. How could you determine if the proteins were imported by the
mitochondria? Design an experiment and don't forget the controls!
Simply: Transcribe and translate each cDNAs in the presence
of 35S-Met. Add to the translation mix respiring mitochondria
purified from yeast and wait a while to allow the peptides to be
taken up by the mitochondria. Then you could: Treat the
mixture with a protease to degrade any proteins that have not
been imported into the mitochondria, re-isolate the
mitochondria and run the proteins from the mitochondria on an
SDS-PAGE gel, followed by autoradiography. If the proteins
were taken into the mitochondria, they will be present in the
Problem Set 5, 7.06, Spring 2003
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mitochondrial prep. If the proteins were not taken up by the
mitochondria, they will have been degraded by the protease
and will not be present in the mitochondrial prep. A good
control would be to run the translation mixture alone on the gel
to make sure that the protein was made properly. An
alternative to the experiment described is similar and involves
performing the experiment as described, but omitting the
protease treatment and then separating out the supernatant
from the mitochondria over 5 minute increments. If the
proteins were taken up by the mitochondria, there would be
less and less of them found in the supernatant as time
progressed and more and more in the mitochondrial fraction.
If the proteins were not taken up, the amount of protein would
remain contstant in the supernatant over time and not be found
in the mitochondrial fraction.
b. It is formally possible that your proteins are normally targeted to
the mitochondria, but that in this assay, they may not be imported.
Why might that be and how could you test this?
Mitochondrial proteins are imported in an unfolded state.
Normally in the cell, keeping the proteins unfolded requires
chaperone proteins like Hsc70. You didn't add any purified
chaperone proteins to the translation mixture, so it's possible
that your new proteins folded spontaneously prior to import
and were thus unable to be imported. To test this, you would
perform the above experiment both with and without
chaperone proteins being added to the translation mix.
(Another less likely possibility is that your experiment became
prematurely contaminated with proteases and that the protein
was degraded before it was able to be imported. If this were
the case, the alternative experiment described in a., where you
perform a time course and separate the supernatant from the
mitochondria then look at the proteins in each fraction by SDSPAGE and autoradiography would be used. By this experiment,
you could see if the protein was disappearing from the
supernatant and not being taken up by the mitochondria, which
would be suggestive of your protein being degraded by a nasty
protease that you didn't add intentionally.
Yet another possibility is that the mitochondria are not
activated and require ATP and/or a proton motive force; see d.
c. You tried a very simple pilot experiment where you had protein X
alone and then protein X added to mitochondria, and you observed
the following result for protein X:
Problem Set 5, 7.06, Spring 2003
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Lane 1: protein X alone
Lane 2: protein X with mitochondria
(not purified from supernatant)
Why does the lower band appear in lane 2?
When protein X is imported into the mitochondria, its matrixtargeting signal is removed from its N-terminus, yielding a
smaller version of protein X.
d. Being a good scientist, you repeated the simple experiment, but
realized afterward that the yeast (from which the mitochondria for
the experiment were taken) were treated with Dinitrophenol (DNP),
an agent which uncouples electron transport from proton
movement. What result did you see in this experiment and why?
Fill in lanes 3 and 4 on the gel with the expected bands (using lanes
1 and 2 as a guide).
The import of proteins into the mitochondria depends
on ATP and the electrochemical proton gradient, so
disrupting the proton gradient would lead to no
import of protein X. (and If protein X can't get
imported, it certainly doesn't get cleaved, so no protein
X cleavage band.) If you treated the samples in 3
and 4 with protease, there would be no band in either.
e. You've heard from a reliable source (MCB, 4th ed.) that even though
a mitochondrial protein can't be imported when there is no proton
motive force, it can
1 still 2bind to
3 the 4receptors of the outer
Problem Set 5, 7.06, Spring 2003
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mitochondrial membrane. Design an experiment to show that this
is the case.
Translate protein X in vitro as before, add to mitochondria
which have been treated w/ Nitric Oxide (no proton motive
force). Allow the reaction to proceed for a while, then separate
the mitochondria from the supernatant. Keep the supernatant
to run on SDS-PAGE and keep a portion of the intact
mitochondria for the gel. But for the rest of the mitochondria,
gently mash them up and purify them into membranes (both
inner and outer will be in the same fraction) vs.
matrix/intermembrane space fractions, then run each of these
samples on the gel also. If protein X is bound to the
mitochondrial membrane receptors, it will be present in the
whole mitochondria fraction, and the mitochondrial membrane
fraction, but not in the matrix/intermembrane space fraction.
Also, a small portion will be in the supernatant, as not all
protein X will bind to the mitochondria.
3. Predict the arrangement of each of the following proteins with
respect to the ER membrane and lumen.
Problem Set 5, 7.06, Spring 2003
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Problem Set 5, 7.06, Spring 2003
7 of 13
4. You are studying a human secreted protein that is 200 amino acids in length that
has the following properties:
• An arginine residue at amino acid 47 and a lysine residue at position 74
• No other arginine or lysine residues
• Only 3 cysteine residues at positions 36, 58, and 143
• Only 3 asparagine residues at positions 30, 52, and 110
• There are no O-linked sugars in this protein
• A signal sequence at the N-terminus for insertion into the ER followed by a
cleavage site at amino acid residue 23
You have the cDNA that encodes for the protein, so you use it to in-vitro translate
the protein in a cell-free system without microsomes present. In addition, you use
recombinant DNA technology to express the protein in cultured CHO cells, which do
not normally express the protein. You then immunoprecipitate the secreted protein
from the extracellular medium and incubate the protein as indicated for each lane
below. After incubation, you run the protein on a non-reducing SDS-PAGE gel (i.e. no
reducing agent added) and silver stain to visualize the bands. (Notes: Trypsin is a
protease that cleaves proteins after arginine and lysine residues; BME= bMercaptoethanol which is a reducing agent; PNGase F is an endoglycosidase that
cleaves all Asn-linked sugars; the average molecular weight of an amino acid is 110
Daltons (Da)). Shown below is the gel:
Lanes 1-2: in-vitro translated proteins
Lane 1: no trypsin + BME
Lane 2: with trypsin + BME
Lanes 3-7: protein immunoprecipitated from extra-cellular medium of
wild type cells
Lane 3: no trypsin
Lane 4: with trypsin
Lane 5: with trypsin + BME
Lane 6: with trypsin + PNGase F
Lane 7: with trypsin + BME + PNGase F
Problem Set 5, 7.06, Spring 2003
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Lanes 8-12: protein immunoprecipitated from the extracellular
medium of cells lacking protein disulfide isomerase (PDI-)
Lane 8: no trypsin
Lane 9: with trypsin
Lane 10: with trypsin + BME
Lane 11: with trypsin + PNGase F
Lane 12: with trypsin + BME + PNGase F
Using lanes 1-7, answer the following questions:
Trypsin cleaves the protein made in the absence of microsomes into a
5.2kDa fragment, a 3kDa fragment, and a 13.9kDa fragment (lane 2)
which derive, respectively, from the N- terminus, Middle, and Cterminus of the protein, as determined by the positions of the lysine
and arginine residue.
Trypsin cleaves the protein secreted in the medium into three
fragments when any disulfide bonds are reduced (lane 5). The size of
the 13.9kDa fragment derived from the c terminus is unaffected. The
smallest fragment of 2.6kDa can only come from the N- terminus, and
thus the ~3 kDa fragment derives from the middle of the protein.
a) Is there any disulfide linkage in the secreted protein? If so, which cysteines
are involved? Explain your answer.
The cysteine residues at position 58 and position 143 form a disulfide
bridge, linking the 3kDa and the 13.9kDa fragments together. This
can be seen by comparing lane 4 with lane 5.
b) Is the protein a glycoprotein? If so, which residue(s) are glycosylated?
Explain your answer.
Yes. The sugar is attached to the asparagine residue at position 52, in
the middle section of the protein. A comparison of lanes 5 and 7 (both
run after reduction of the disulfide bond) shows that the apparent
molecular weight of the ~3 kDa fragment is reduced by removal of
the oligosaccharide.
You now express your protein in cultured CHO cells that are mutant for
protein disulfide isomerase (PDI) (i.e. PDI does not function in these cells).
You discover that your protein is not secreted from these mutant CHO cells.
c) In which cellular compartment of the PDI- CHO cells would you find your
protein? Why?
PDI catalyzes proper disulfide-bond formation. In the absence of PDI,
the protein forms inappropriate disulfide linkages and the unfoldedprotein response is triggered through IRE1. Only properly folded
proteins are moved to the Golgi complex. Improperly folded proteins
are bound to ER chaperones in the rough ER, and subsequently
degraded if they are irreversibly misfolded. So you would expect to
find your protein in the ER.
Problem Set 5, 7.06, Spring 2003
9 of 13
You repeat your immunoprecipitation from total cell lysate made from the
PDI- mutant CHO cells and incubate your immunoprecipitated protein as
indicated. The results of your experiments are shown in lanes 8-12.
Using lanes 8-12, answer the following questions:
d) Is there any disulfide linkage in the protein? If so, which cysteines are
involved? Explain.
In the absence of PDI, the Cys36 and Cys 58 inappropriately form a
disulfide linkage, and therefore the 2.6kDa fragment and the 3kDa
fragment become linked. This is apparent in the tryptic digest alone
(lane 9) and the trypsin + PNGase F incubation (lane 11). The trypsin
+ BME reaction products are unchanged.
e) Is the protein glycosylated? If so, which residue(s) are involved?
Yes, Asn 52 in the protein is glycosylated since the initial N-linked
glycosylation occurs in the rough ER.
5. You are evaluating the early steps in translocation and processing of the
secretory protein prolactin. By a new experimental approach, you can use
truncated prolactin mRNAs to control the length of the nascent prolactin
polypeptides that are synthesized. When prolactin mRNA that lacks a chaintermination (stop) codon is translated in vitro, the newly synthesized
polypeptide ending with the last codon included on the mRNA will remain
attached to the ribosome, thus allowing a polypeptide of defined length to
extend from the ribosome.
You have generated a set of mRNAs that encode segments of the N-terminus
of prolactin of increasing length, and each mRNA can be translated in vitro by
a cytosolic translation extract containing ribosomes, tRNAs, aminoacyl-tRNA
synthetases, GTP, and translation initiation and elongation factors. When
radio-labeled amino acids are included in the translation mixture, only the
polypeptide encoded by the added mRNA will be labeled. After completion of
translation, each reaction mixture was resolved by SDS-PAGE, and the
labeled polypeptides were identified by autoradiography.
The autoradiogram depicted below shows the results of an experiment in
which each translation reaction was carried out either in the presence (+) or
absence (-) of microsomal membranes.
Problem Set 5, 7.06, Spring 2003
10 of
Size of labeled
Size of mRNA (in codons)
a) Based on the gel mobility of peptides synthesized in the presence or
absence of microsomes, deduce how long the prolactin nascent chain must be
in order for the prolactin signal peptide to enter the ER lumen and to be
cleaved by signal peptidase. (Note that microsomes carry significant
quantities of SRP weakly bound to the membranes.)
The prolactin nascent chain must be 130 codons. When the truncated
mRNA is 130 codons in length, some cleavage product begins to
appear in the presence of microsomes. Proteins encoded by shorter
mRNA’s do not get cleaved even in the presence of microsomes since
the nascent polypeptide does not extend into the microsome far
enough for signal peptidase to cleave the signal peptide.
b) Given this length, what can you conclude about the conformational state of
the nascent prolactin polypeptide when it is cleaved by signal peptidase? The
following lengths will be useful for your calculation: the prolactin signal
sequence is cleaved after amino acid 31; the channel within the ribosome
occupied by a nascent polypeptide is about 150 angstroms (A) long; a
membrane bilayer is about 50 A thick; in polypeptides with an a-helical
conformation, one residue extends 1.5 A, whereas in fully extended
polypeptides, one residue extends about 3.5 A.
In order for the signal peptide to be cleaved, the cleavage site
(located at amino acid 31) must be in the lumen of the microsome. If
the nascent prolactin polypeptide were in an a-helical conformation,
then at a length of 130 codons, the signal peptide would just be
entering the lumen of the microsome which is (130-31=99; 99 X 1.5=
148.5A). This would explain why some, but not all of the 130 codon
mRNA’s can be cleaved by signal peptidase. If the nascent prolactin
polypeptide were in a fully extended conformation, even at 110
codons, the mRNA would be long enough to pass through the
membrane and be cleaved by signal peptidase, but the data suggests
that this does not occur. Therefore, the nascent prolactin polypeptide
probably is in an a-helical conformation.
Problem Set 5, 7.06, Spring 2003
11 of
The experiment described in part (a) is carried out in an identical manner
except that microsomal membranes are not present during the translation but
are added after translation is complete. In this case, none of the samples
shows a difference in mobility in the presence or absence of microsomes.
c) What can you conclude about whether prolactin can be translocated into
isolated microsomes post-translationally?
You can conclude that prolactin cannot be translocated into isolated
microsomes post-translationally since the polypeptide is never
cleaved even in the presence of microsomes.
Size of labeled
In another experiment, each translation reaction was carried out in the
presence of microsomes, and then the microsomal membranes and bound
ribosomes were separated from free ribosomes and soluble proteins by
centrifugation. For each translation reaction, both the total reaction (T) and
the membrane fraction (M) were resolved in neighboring gel lanes. The
autoradiogram is shown below.
Size of mRNA (in codons)
d) Based on the amounts of labeled polypeptide in the membrane fractions in
the autoradiogram, deduce how long the prolactin nascent chain must be in
order for ribosomes engaged in translation to engage the SRP and thereby
become bound to microsomal membranes.
The prolactin nascent chain must be at least 90 codons in length in
order for the ribosomes engaged in translation to engage the SRP and
become bound to microsomal membranes. When the mRNA is 90
codons long, some of the nascent polypeptide is present in the
membrane fraction. In a 110 codon mRNA, all of the nascent
polypeptide is membrane-associated because an equal amount of
nascent polypeptide is presence in the total reaction and membrane
fractions. In the 130 codon mRNA reactions, two bands are present
because a fraction of the nascent polypeptides have already entered
the microsomal lumens and have been cleaved by signal peptidase,
whereas a fraction have not yet entered the microsomal lumen. In
Problem Set 5, 7.06, Spring 2003
12 of
reactions involved 150 codon mRNA’s, all of the nascent chains have
been cleaved.
6. You identify a novel human protein and you have no idea what its function
is. However, you have made an antibody against it, and upon performing an
immunofluorescence experiment in cultured HeLa cells, you observe that the
protein is present in the nucleus during G1, but is present in the cytoplasm
throughout the rest of the cell cycle. You would like to know whether the
shuttling of the protein in and out of the nucleus is important for its function.
You reason that the protein must have an NES (nuclear export signal) and
you want to mutate the signal and examine the consequences.
In order to identify the NES sequence in your protein, you perform a
heterokaryon experiment. You subdivide your protein into four segments, A,
B, C, and D, and fuse each segment individually onto the C-terminus of the
human nucleoplasmin protein using recombinant DNA technology. The wild
type nucleoplasmin protein localizes to the nucleus at all times during the cell
cycle. You express each fusion protein in HeLa cells, and then fuse the
transgenic HeLa cells with Xenopus cells, creating a heterokaryon where both
the HeLa cell and Xenopus nuclei are in a common cytoplasm. Using an
antibody to the human nucleoplasmin protein (which does not recognize its
Xenopus homolog) you examine the localization of each fusion protein. You
obtain the following data:
Fusion protein
h-nucleoplasmin only
Detected in HeLa cell
Detected in Xenopus
cell nucleus
a) Which fragment(s), A, B, C, or D, contains the NES and how do you know?
Fragments A and C both contain an NES since the fusion of either A or
C to h-nucleoplasmin allows the fusion protein to be exported from
the HeLa cell nucleus and subsequently imported into the Xenopus
cell nucleus.
b) If you now wanted to identify the NLS in your protein and you did not have
the protein’s sequence, how could you do so?
You could fuse fragments A, B, C, and D to a human protein that you
know is localized exclusively to the cytoplasm. Then you would inject
mRNA encoding this fusion protein into Xenopus oocytes and observe
whether the fusion protein were able to be imported into the nucleus
or not using an antibody to the cytoplasmic protein. Only a fragment
of the protein that contained the NLS would support localization of
the fusion protein to the Xenopus nucleus.
Problem Set 5, 7.06, Spring 2003
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