Download Problem Set #6 4/19/2002 7.06 Spring `02 Page: 1 of 8

Problem Set #6
4/19/2002
7.06 Spring ‘02
Question 1
a) You have just joined a laboratory that is engaged in defining the nuclear transport
machinery in yeast. Your advisor, who is known for her extraordinarily clever ideas, has
given you a project with enormous potential. In principle, it would allow a genetic
selection for conditional-lethal mutants in the nuclear transport apparatus.
pNL+
pNL-
nuclear
protein
GAL
EcoRI
GAL
signal
present
nuclear
protein
EcoRI
signal
absent
Fig 1-1. Two plasmids for investigating nuclear localization in yeast.
She gave you the two plasmids shown in Fig 1-1. Each plasmid consists of a hybrid gene
under the control of a regulatable promoter. The hybrid gene is a fusion between a gene
whose product is normally imported into nucleus and the gene for the restriction enzyme
EcoRI. The plasmid pNL+ contains a functional nuclear localization signal; the plasmid
pNL- contains a nonfunctional signal. The promoter, which is from the yeast GAL1 gene,
allows transcription of the hybrid gene only when the sugar galactose is present in the
growth medium.
Table 1-1. Results of Growth Experiments with Yeast Carrying the Plasmids pNL+ or pNL-.
Plasmid
pNL+
pNL-
Glucose Medium
Growth
Growth
Galactose Medium
Death
Growth
Following her instructions, you introduce the plasmids into yeast (in the absence of
galactose) and then assay the transformed yeast in medium containing glucose and in
medium containing galactose. Your results are shown in Table 1-1. You don’t remember
what your advisor told you to expect, but you know you will be expected to explain
these results at the weekly lab meeting. Why do yeasts with the pNL+ plasmid grow in
the presence of glucose but die in the presence of galactose?
EcoRI is a restriction enzyme that cleaves DNA based on a particular sequence.
Introducing the restriction enzyme into a cell’s nucleus, will result in many
double strand breaks of the host’s chromosomes and will ultimately kill the
cell.
b) Now that you understand why plasmid pNL+ (Fig 1-1) kills cells in galactose-containing
medium, you begin to understand how your advisor intends to exploit its properties to
select mutants in the nuclear transport machinery. You also understand why she
emphasized that the desired mutants would have to be conditionally lethal. Since
nuclear import is essential to the cell, a fully defective mutant could never be grown and
thus would not be available for study. By contrast, conditionally lethal mutants can be
grown perfectly well under one set of conditions (permissive conditions); under a
different set of conditions (restrictive conditions), however, the cells exhibit the defect,
which can then be studied.
Page: 1 of 8
Problem Set #6
4/19/2002
7.06 Spring ‘02
With this overall strategy in mind, you design a selection scheme for temperaturesensitive (ts) mutants in the nuclear translocation machinery. You want to find mutants
that grow well at low temperature (the permissive condition) but are defective at high
temperature (the restrictive condition). You plan to mutagenize cells containing the pNL+
plasmid at low temperature and then shift them to high temperature in the presence of
glactose. You reason that at the restrictive temperature the nuclear transport mutants
will not take up the killer protein encoded by pNL+ and, therefore, will not be killed.
Normal cells, however, will transport the killer protein into the nucleus and die. After one
or two hours at the high temperature to allow selection against normal cells, you intend
to lower the temperature and put the surviving cells on nutrient agar plates containing
glucose. You anticipate that the nuclear translocation mutants will recover at the lower
temperature and form colonies.
When you show your advisor your scheme, she is very pleased at your initiative.
However, she sees a critical flaw in your experimental design that would prevent
isolation of nuclear translocation mutants, but she won’t tell you what it is—but she
hints that is has to do with the killer protein made during the high-temperature phase of
the experiment. What is the critical flaw in your original experimental protocol?
The critical flaw in your original protocol is that the killer protein, which was
made at the high temperature but denied access to the nucleus in the
translocation mutants, will still be present when the cells are shifted to low
temperature. When nuclear translocation resumes at the low temperature, the
killer protein ensured that no new killer protein would be made (by shifting to
low temperature in the presence of glucose) but did not take into account the
killer protein that had been made in the presence of galactose before the
temperature shift.
c) How might you modify your protocol to correct this flaw?
You need to modify your experimental protocol so that the previously made
killer protein is rendered inactive before nuclear import is allowed to resume.
There are many possible ways to accomplish this. For example, you might try
leaving the cells at high temperature in the presence of glucose (so there is no
new synthesis) for increasing periods of time to allow the previously made
killer protein to be inactivated by normal degradation processes. You might
also try to increase its rate of degradation by engineering its N-terminus so
that it carries a destabilizing amino acid. Such a modification could turn the
killer protein into a very short-acting molecule, which would disappear very
rapidly in the absence of new synthesis. Alternatively, at the high temperature
you might try to make a temperature-sensitive mutant in EcoRI that is active at
the high temperature and inactive at the low temperature. Such a coldsensitive protein would be active when nuclear transport was blocked in the
import mutant and inactive when nuclear transport resumed.
Page: 2 of 8
Problem Set #6
4/19/2002
7.06 Spring ‘02
d) Assuming for the moment that your original protocol would work, can you think of any
other types of mutants (not involved in nuclear transport) that would survive your
selection scheme?
There are several types of mutants that would be expected to survive the
selection protocol without affecting the nuclear-transport machinery. For
example, mutants that cannot transport galactose into the cell would be unable
to turn on the hybrid gene and thus would survive the shift to high temperature
in the presence of galactose. Other mutants in which EcoRI was not expressed
from the hybrid gene would also survive the selection scheme. These mutants
might not include mutations in the promoter (which would prevent expression
of the hybrid gene), mutations in the EcoRI portion of the hybrid gene (which
would render EcoRI inactive), and mutations in the nuclear localization signal
in the hybrid gene (which would prevent import of EcoRI into the nucleus).
Genetic selection schemes are rarely specific for the precise mutant you are
seeking. In some cases, other classes can be predicted and screened out by an
additional step. However, the real beauty of genetic selection schemes, the
aspect that makes them like treasure hunts, is that they often give classes of
mutants that you did not foresee. Sometimes these classes turn out to be trivial
and uninteresting; however, they can also be extremely informative leading
you to insights that you did not anticipate.
Question 2
You have isolated several cell lines that are defective in their ability to add carbohydrate to
exported proteins. Using an easily purified protein that carries only N-linked complex
oligosaccharides, you have analyzed the sugars in the N-linked oligosaccharides that are
added in the different mutant cells. Each mutant is unique in the kinds and numbers of
different sugars contained in its oligosaccharides (Table 2-1). (NOTE: the original table
contained several mistakes and need to be corrected to match the class text.)
Table 2-1. Analysis of the Sugars Present in the N-linked Oligosaccharides from wild-type
Cells and from Mutant Cell Lines Defective in Oligosaccharide Processing.
Numbers indicate the no. of sugar monomers in the oligosaccharide.
Cell Line
Man
GlcNAc
Gal
NANA
Glc
Wild-type
3
5
3
3
0
Mutant A
3
5
0
0
0
Mutant B
5
3
0
0
0
Mutant C
9
2
0
0
3
Mutant D
9
2
0
0
0
Mutant E
5
2
0
0
0
Mutant F
3
3
0
0
0
Mutant G
8
2
0
0
0
Mutant H
9
2
0
0
2
Mutant I
3
5
3
0
0
Abbreviations: Man = mannose; GlcNAc = N-acetylglucosamine; Gal = galactose; NANA =
N-acetylneuraminic acid, or sialic acid; Glc = glucose.
Page: 3 of 8
Problem Set #6
4/19/2002
7.06 Spring ‘02
a) Arrange the mutants in the order that corresponds to the step in the pathway for
processing N-linked oligosaccharides:
Mutants: C -> H -> D -> G -> E -> B -> F -> A -> I -> Wild-type
b) Which of these mutants are defective in processing events that occur in the ER? Which
mutants are defective in processing steps that occur in the Golgi?
ER Mutants:
C, H, D, G
Golgi Mutants:
E, B, F, A, I
c) Which of the mutants are likely to be defective in a processing enzyme that is directly
responsible for modifying N-linked oligosaccharides? Which mutants might not be
defective in a processing enzyme but rather, in another enzyme that affects
oligosaccharide processing indirectly?
Mutations that only effect processing enzymes are those which remove a sugar
group from the oligosaccharide. While mutations that are deficient in the
addition of a new sugar to an existing oligosaccharide could be either in the
processing enzyme or an upstream step. For instance, you can’t tell the
difference between a mutation in the sugar transporter or a mutation in the
enzyme that adds the sugar based solely on the oligosaccharide composition
analysis.
Those likely to be defective in a processing enzyme only:
C, H, D, G, B, F (Note: F is unlikely to be a transporter
mutant because a deficiency in the transporter for UDPGlcNAc would result in the same phenotype as mutant
E.)
Those that could be defective in other processes:
E, A, I.
Page: 4 of 8
Problem Set #6
4/19/2002
7.06 Spring ‘02
Question 3
Transport from ER to the Golgi apparatus involves the formation of COPII-coated vesicles.
The process is initiated by the recruitment of a small GTP-binding protein called Sar1 to the
ER membrane. The vesicles are then transported to and fused with the cis-Golgi membrane.
This fusion process is regulated by another small GTP-binding protein called Rab1.
You add the following drug to cells.
a) A drug that inhibits the guanine exchange factor (GEF) required for the exchange of
bound GDP with GTP on Sar1
b) A drug that inhibits the GEF required for the exchange of bound GDP with GTP on
Rab1
c) A drug that greatly slows down the hydrolysis of GTP on Sar1
d) A drug that inhibits the GTPase accelerating protein (GAP) for Rab1
e) Nocodazole
What effects do you expect on COPII-coated vesicle formation, trafficking, and fusion with
target membrane?
(NOTE: Please do not worry about the details about this question)
a) Vesicles will not be able to form on ER membrane. Recruitment of the coat
protein COPII to budding vesicles requires Sar-GTP.
b) The Rab protein family is thought to accelerate vesicle fusion. Inhibition of
Rab1 will likely lead to COPII vesicle accumulation outside the Golgi membrane.
c) Formation of vesicles is not affected but dissociation of the coat protein COPII
is inhibited. Since the dissociation of the coat is required to expose the SNARE
proteins etc.., the vesicles cannot be directed to the right place.
d) Hydrolysis of GTP on Rab1 is thought to accelerate COPII-vesicle fusion to the
target membrane. In the presence of the drug, vesicles will accumulate on the
surface of cis-Golgi because the fusion process is much slowed down.
e) COPII vesicles are transported along microtubules between ER and Golgi.
Nocodazole depolymerizes microtubules. Vesicles cannot be transported efficiently
across long-distance.
Question 4
The following is a hypothetical pedigree of a family with Leigh syndrome. The phenotypes of
3 offspring marked with (?) are not available. It is believed that one subclass of the disease
–mitochondrial inherited Leigh syndrome (MILS)—is due to mutation of a mitochondria
encoded gene.
Page: 5 of 8
Problem Set #6
4/19/2002
7.06 Spring ‘02
a) Based on the pattern of inheritance, is it consistent that the patients have MILS?
Yes. Because mitochondria are maternally inherited, MILS is inherited from
the mother.
b) Is it consistent that the patients suffer from another subclass of Leigh syndrome that
arises from a nuclear mutation?
Yes. There are several possibilities. The mutation can be a dominant –all
affected individuals are heterozygous for the mutation. It can also be
recessive, if both fathers (a, b) are +/- as the question did not specify that
this is a rare disease and that they are unrelated. It is also possible that we
are seeing a dominant maternal effect that requires the mother to carry
wild-type copies of the gene.
c) If 1 of the 3 offspring marked with the (?) is now diagnosed to be affected while the
other 2 are normal, which mode of inheritance will the new data support?
It will support the nuclear inheritance model. The mutation can be either
recessive or dominant depending on the prevalence of the disease. Note
that if it is MILS or if there is a nuclear inherited dominant maternal effect,
all the 3 offspring are likely to be affected.
Question 5:
You are interested in exocytosis and endocytosis in a line of cultured liver cells that secrete
albumin and take up transferrin. To distinguish between these events, you add transferrin
tagged with colloidal gold to the medium, and then after a few minutes you fix the cells,
prepare thin sections, and react then with ferritin-labeled antibodies against albumin.
Colloidal gold and ferritin are both electron dense and therefore readily visible when viewed
by electron microscopy; moreover, they can be easily distinguished from one another on the
basis of size and density.
a) Will this experiment allow you to identify vesicles in the exocytic and endocytic
pathways? How?
Vesicles on the endocytic pathway will be labeled with colloidal gold; vesicles on
the exocytic pathway will be labeled with ferritin.
Page: 6 of 8
Problem Set #6
4/19/2002
7.06 Spring ‘02
b) Not all the gold-labeled vesicles are clathrin coated. Why?
Clathrin-coated vesicles are uncoated within a few seconds after they pinch off
from the plasma membrane, so some will be caught with their coats off while
others will still have their coats on.
Question 6
You are studying a protein 175 amino acids in length that has the following properties:
• 2 lysine residues at positions 70 and 95
• no arginine residue
• 3 cysteine residues at positions 56, 80, and 161
• 3 asparagine residues at positions 40, 88, and 139
• there is no O-linked sugar on this protein
• At N-terminus a signal sequence for insertion into ER followed by a cleavage site at
residue position 30
You have the cDNA for the wild-type protein and mutant proteins as described below. In a
cell type that doesn’t normally express this protein, you inject a mixture of RNA that will
allow translation of the I) wild-type protein, II) protein with a deletion of the first 30 amino
acids at the N-terminus. You immunopreciate the protein using polyclonal antibodies that
specifically recognize it from the cytosolic extract and cis-Golgi lumen proteins. You run an
SDS-PAGE and stain proteins with silver to analyze the immunoprecipitated products.
Shown below is the gel:
(A CORRECTION has been made: lane 2, the 4.4 band should run at around 7.7 kD
as the signal sequence is not cleaved)
1
2
3
4
5
6
7
8
9
10
11
12
kDa
21
19
13
8.8
4.4
3.2
2.8
Lane1: wild-type in vitro translated protein
Lane2: wild-type in vitro translated protein + trypsin
Lanes 3-7: protein immunoprecipiated from cis-Golgi lumen
Lane3: no trypsin
Lane4: with trypsin
Lane5: with trypsin + endoglycosidase H (endoH)
Lane6: with trypsin + β−mercaptoethanol
Lane7: with trypsin + β−mercaptoethanol + endoH
Lanes 8-12: protein immunoprecipitated from cytosol
Lane8: no trypsin
Lane9: with trypsin
Lane10: with trypsin + endoglycosidase H (endoH)
Lane11: with trypsin + β−mercaptoethanol
Lane12: with trypsin + β−mercaptoethanol + endoH
Page: 7 of 8
Problem Set #6
4/19/2002
7.06 Spring ‘02
a) Where do you expect the following to localize when matured in an intact cell I) wildtype protein, II) protein with a deletion of the first 30 amino acids at the Nterminus?
I) The protein that will get secreted. Inside a cell, the matured form can be
found in secretary vesicles.
II) Without a signal sequence to direct it to the ER, the protein will stay in the
cytosol.
b) Is the wild-type protein a glycoprotein? If so, at what residue(s) does the protein
have sugar attached. Explain your answer.
Yes. The sugar is attached to the asparagines at position 88. A comparison of
lanes 6 –7 suggests that the 2.8 kD tryptic fragment is glycosylated. (Note:
trypsin cleaves after lysine and arginine. In this protein, it will generate a 40,
25, and 80 amino-acid long fragments that correspond to ~4.4, 2.8, and 8.8kD.
Note that the N-terminus signal sequence is cleaved off).
c) Is there any disulfide linkage in the wild-type protein? If so, which cysteine residues
are involved?
Yes, C56 and C161 are involved. Comparing lanes 4 to 6 (or lanes 5-7) suggest
a disulfide bond links the 4.4 and 8.8 kD fragments.
d) Draw out what you expect to see in lanes 8-12. Justify your answer briefly.
See the gel for drawing. Without the N-terminal ER targeting sequence, the
protein will remain in the cytosol. It will not get glycosylated. Disulfide linkage
also won’t form since the cytosol is reducing.
Page: 8 of 8