CH 12
How is it that a single nuclear pore complex can efficiently transport proteins that possess different kinds of nuclear localization signal?
A single nuclear pore complex can transport proteins with quite different kinds of nuclear localization signal because transport is mediated by a variety of nuclear import receptors that are encoded by a family of related genes. Each family member—each gene product—is specialized for transport of a group of nuclear proteins that share structurally similar nuclear localization signals. At the same time, all family members share common features that allow them to interact with nuclear pore complexes. Thus, nuclear import receptors act as adaptors between proteins with diverse nuclear localization signals and the uniform population of nuclear pore complexes.
If you think of the protein as a traveler, what kind of vehicle would best describe the sorting receptor: a private car, a taxi, or a bus? Explain your choice.
A taxi is the closest analogy. Anyone who has the fare—the sorting signal—is taken on the journey. A private car implies a specific relationship between the traveler (the sorted protein) and the vehicle (the sorting receptor); namely, that there are specific sorting receptors for each different kind of protein. In reality, all the proteins destined for the ER, for example, use the same sorting receptor. A bus would imply that travelers—sorted proteins—are carried in groups. Sorting receptors handle proteins one at a time.
A. Why do you think the substrate accumulates at the nuclear periphery, as is seen in the absence of GTP or with importin alone in the presence of GTP?
A. Figure 12-7A shows that in the absence of GTP, but in the presence of importin, labeled substrate accumulates at the nuclear periphery. A reasonable inference is that importin facilitates the binding of the substrate to the nuclear pore complexes and that this binding is independent of GTP.
A. Why doesn't NES-BSA accumulate to a higher concentration in the nucleus than in the cytoplasm in these experiments?
A. Fluorescent NES-BSA does not accumulate in the nucleus in these experiments because no gradient of Ran-GTP can be established. Since RanQ69L does not hydrolyze its bound GTP, it will ultimately be present in the same concentration on both sides of the nuclear membrane. Once the concentration of NES-BSA in the nucleus becomes equal to that out- side the nucleus, RanQ69L-GTP/ Crm1 will transport NES-BSA at equal rates in both directions.
A. In a typical growing cell that is dividing once every 24 hours, the equivalent of one new plasma membrane must transit the ER every day. If the ER membrane is 20 times the area of a plasma membrane, what is the ratio of plasma membrane proteins to other membrane proteins in the ER?
A. If the equivalent of one plasma membrane transits the ER every 24 hours and individual membrane proteins remain in the ER for 30 minutes (0.5 hr), then at any one time, 0.021 (0.5 hr/24 hr) plasma membrane equivalents are present in the ER. Since the area of the ER membrane is 20 times greater than the area of the plasma membrane, the fraction of plasma membrane proteins in the ER is 0.021/20 = 0.001. Thus, the ratio of plasma membrane proteins to other membrane proteins in the ER is 1 to 1000. Out of every 1000 proteins in the ER membrane, only 1 is in transit to the plasma membrane.
A. How would examining the two nuclei in a heterokaryon answer the question? What results would you expect if the protein were a true nuclear protein? What would you expect if it were a shuttling protein?
A. In a heterokaryon with two nuclei, one expressing the GFP-tagged protein and the other not, it is possible to decide whether the protein is a true nuclear protein or a shuttling protein. If it is a nuclear protein, the GFP should remain associated with a single nucleus. If it is a shuttling protein, the GFP will redistribute to both nuclei.
Although the vast majority of transmembrane proteins insert into membranes with the help of dedicated protein-translocation machines, a few proteins can insert into membranes on their own. Such proteins may provide a window into how membrane insertion occurred in the days before complex translocators had evolved. You are studying a protein that inserts itself into the bacterial membrane independent of the normal translocation machinery. This protein has an N-terminal, 18-amino-acid hydrophilic segment that is located on the outside of the membrane, a 19-amino-acid hydrophobic transmembrane segment flanked by negatively and positively charged amino acids, and a C-terminal domain that resides inside the cell (Figure 12-1A). If the protein is properly inserted in the membrane, the N-terminal segment is exposed externally where it can be clipped off by a protease, allowing you to quantify insertion. To examine the roles of the hydrophobic segment and its flanking charges, you construct a set of modified genes that express mutant proteins with altered charges in the N-terminal segment, altered lengths of the hydrophobic segment, and combinations of the two (Figure 12-1B). For each gene you measure the fraction of the total protein that is cleaved by the protease, which is the fraction that was inserted correctly (Figure 12-1B). To assess the contribution of the normal membrane potential (positive outside, negative inside), you repeat the measurements in the presence of CCCP, an ionophore that eliminates the charge on the membrane (Figure 12-1B).
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A. Why did you use RanQ69L-GTP instead of Ran-GTP in these experi- ments? Could you have used Ran-GppNp instead of RanQ69L-GTP to achieve the same purpose?
A. Ran is a GTPase and will slowly convert GTP to GDP. Thus, if you had prepared Ran-GTP to start with, by the time you did the experiments you would have had an undefined mixture of Ran-GTP and Ran-GDP, which would have confused the results. By using a form of Ran that cannot hydrolyze GTP, you guaranteed that Ran was in its Ran-GTP conformation. Either RanQ69L-GTP, as was used here, or Ran-GppNp would have served equally well in the experiments described in this problem.
B. Which, if any, of the RNAs are normally exported from the nucleus?
B. Both mRNA and tRNA are fully exported from the nucleus, whereas U1 snRNA and U5 snRNA are only partially exported. U6 snRNA is retained within the nucleus.
B. If in the same cell the Golgi membrane is three times the area of the plasma membrane, what is the ratio of Golgi membrane proteins to other membrane proteins in the ER?
B. In a cell that is dividing once per day, the equivalent of one Golgi apparatus also must transit the ER every 24 hours. If the membrane of the Golgi apparatus is three times the area of the plasma membrane, three times as many Golgi apparatus membrane proteins will be present in the ER. Therefore, the ratio of Golgi apparatus membrane proteins to other membrane proteins in the ER is 3 to 1000.
B. In the presence of the membrane potential (minus CCCP), is the hydrophobic segment or the N-terminal charge more important for insertion of the protein into the membrane? Explain your reasoning.
B. In the presence of the membrane potential (minus CCCP), the hydro- phobic segment is more important than the N-terminal negative charge. Construct 2, which has neither negative charge, is inserted as efficiently as construct 1, which has both negative charges. In addition, when both negative charges are present, the amount of inserted protein decreases as the length of the hydrophobic segment is reduced (compare constructs 1, 3, and 4).
B. In a standard nuclear import assay with added cytoplasm and GTP, proteins with a nuclear localization signal accumulate to essentially 100% in the nucleus. How is it that the standard assay allows 100% accumulation in the nucleus?
B. In the standard nuclear import assay, proteins enter the nucleus and cannot readily escape. Because of its high concentration inside the nucleus, Ran-GTP binds to the nuclear import receptors and displaces the transport substrate. Ran-GTP then accompanies the import receptor back outside the nucleus, where a Ran-GAP (Ran-GTPase-activating protein) rapidly converts Ran-GTP to Ran-GDP. Ran-GDP is then specifically transported back into the nucleus, where it is rapidly converted to Ran-GTP by a Ran-GEF (Ran-guanine exchange factor). Maintenance of nuclear Ran in its Ran-GTP form ensures that no import receptors will be available to carry the transport substrate back out of the nucleus, thereby allowing essentially 100% of the transport substrate to accumulate in the nucleus.
C. If the membranes of all other compartments (lysosomes, endosomes, inner nuclear membrane, and secretory vesicles) that receive membrane proteins from the ER are equal in total area to the area of the plasma membrane, what fraction of the membrane proteins in the ER of this cell are permanent residents of the ER membrane?
C. If the areas of the membranes of all the rest of the compartments are equal to the area of the plasma membrane, then the ratio of membrane proteins bound for these compartments to the membrane proteins in the ER is 1 to 1000. Summing the contributions from all compartments, the ratio of membrane proteins, in transit, to proteins that are permanent residents of the ER membrane is 5 to 1000. Thus, 99.5% of the membrane proteins in the ER are permanent residents. As this problem illustrates, the sorting of proteins to various cellular membranes represents a substantial purification from the mix of proteins in the ER.
C. In the absence of the membrane potential (plus CCCP), is the hydrophobic segment or the N-terminal charge more important for insertion of the protein into the membrane? Explain your reasoning.
C. In the absence of the membrane potential (plus CCCP), the hydrophobic segment is still the most important determinant of insertion efficiency: construct 2 with no negative charges is inserted as efficiently as construct 1. As the hydrophobic helix is made shorter, insertion comes to depend much more on the presence of the negative charges. Construct 3 is only about half as efficient at insertion when CCCP is added, and construct 4, with a slightly shorter hydrophobic sequence, is absolutely reliant on the membrane potential for insertion. Thus, in the absence of a membrane potential—presumably the case in the earliest cells—a sufficiently long hydrophobic segment may have been adequate to acterm-21complish insertion of a protein into a membrane. In the presence of a membrane potential, a second feature—the distribution of charges around a transmembrane segment—would have been available for translocator-independent insertion of membrane proteins.
C. Is the export of any of the RNAs inhibited by leptomycin B? What does your answer imply about export of this collection of RNAs?
C. Leptomycin B decreases the export of U1 snRNA and blocks the export of U5 snRNA from the nucleus, but has no effect on mRNA or tRNA export. Thus, mRNA and tRNA must be exported by a mechanism that is not affected by leptomycin B. This result implies that there are multiple export pathways (multiple nuclear export receptors) and that leptomycin B affects only one or a subset of them.
C. What other protein or proteins would you predict the Ran-GDP import factor would bind in order to carry out its function?
C. The small protein—the presumptive import factor—that binds to Ran- GDP is known as NTF2. In addition to binding tightly to Ran-GDP, NTF2 binds to the FG-repeats present in the nucleoporins of the nuclear pore complex. It is the progressive movement of the NTF2-Ran-GDP complex through the FG-repeat gel in the nuclear pore that allows Ran-GDP to be delivered to the nucleus. In the nucleus, the Ran-GEF converts Ran-GDP to Ran-GTP, causing it to dissociate from NTF2. NTF2 then recycles to the cytoplasm to bring in another Ran-GDP.
When cells are treated with drugs that depolymerize microtubules, the Golgi apparatus is fragmented into small vesicles and dispersed through- out each cell. When such drugs are removed, cells typically recover and grow normally. If cells that have recovered from such treatment are examined by electron microscopy, they are found to contain a perfectly normal-looking Golgi apparatus. Does this mean that the Golgi apparatus has been synthesized anew from scratch? If not, how do you suppose it might have happened?
Cells do not synthesize the Golgi apparatus from scratch after the drugs that caused its fragmentation are removed. The Golgi apparatus reassembles itself from the dispersed vesicles. Once the microtubule skeleton of the cell has been reestablished, the vesicles apparently use it to track back to their normal location in the cell, where they fuse with one another to re-form the Golgi apparatus. This same process of fragmentation and reassembly occurs in normal cells at each cell division, when the cytoskeleton breaks down in preparation for mitosis.
matrix space
Central subcompartment of a mitochondrion, enclosed by the inner mitochondrial membrane.
cytoplasm
Contents of a cell that are contained within its plasma membrane but, in the case of eukaryotic cells, outside the nucleus.
cytosol
Contents of the main compartment of the cell, excluding the nucleus and membrane-bounded compartments such as endoplasmic reticulum and mitochondria.
D. How might you confirm that the factor you have identified is necessary for promoting the nuclear uptake of Ran?
D. The information in the problem says only that cytoplasm passed over a Ran-GDP column is depleted of some factor that is essential for nuclear uptake, and the experimental results shown in Figure 12-9 indicate that the small protein later identified as NTF2 binds to Ran-GDP. The inference is that NTF2 is the critical factor necessary for nuclear uptake of Ran- GDP. To prove that NTF2 is the import factor, you would need to show that purified or recombinant NTF2 can promote uptake of Ran-GDP into nuclei. The authors of this study went even further. Using information from the crystal structure of the NTF2-Ran-GDP complex, they mutated the glutamate at position 42 in NTF2 to lysine, thereby disrupting a key salt bridge between the two proteins. This NTF2E42K mutant no longer promoted nuclear uptake of Ran-GDP. These additional experiments demonstrate that NTF2 is necessary for nuclear uptake of Ran-GDP.
Why do eukaryotic cells require a nucleus as a separate compartment when prokaryotic cells manage perfectly well without?
Eukaryotic gene expression is more complicated than prokaryotic gene expression. In particular, prokaryotic cells do not have introns that interrupt the coding sequences of their genes, so that an mRNA can be translated immediately after it is transcribed, without further processing. In fact, in prokaryotic cells, ribosomes start translating most mRNAs before transcription is finished. This would have disastrous consequences in eukaryotic cells, because most RNA transcripts have to be spliced before they can be translated. The nuclear envelope separates the transcription and translation processes in space and time. A primary RNA transcript is held in the nucleus until it is properly processed to form an mRNA, and only then is it allowed to leave the nucleus so that ribosomes can translate it.
T/F: Import of proteins into mitochondria and chloroplasts is very similar; even the individual components of their transport machinery are homologous, as befits their common evolutionary origin.
False. Although import of proteins is similar, the components of the import machinery in chloroplasts and mitochondria are not related. The functional similarities appear to have arisen by convergent evolution, reflecting the common requirements for translocation across a double- membrane system.
T/F: To avoid the inevitable collisions that would occur if two-way traffic through a single pore were allowed, nuclear pore complexes are specialized so that some mediate import while others mediate export.
False. Individual nuclear pores mediate transport in both directions. It is unclear how pores coordinate this two-way traffic so as to avoid head-on collisions.
T/F: The biological membranes that partition the cell into functionally distinct compartments are impermeable.
False. Lipid bilayers by themselves are impermeable to hydrophilic molecules, but biological membranes, which contain proteins in addition to the bilayer, are not. Cellular membranes contain various transport proteins that make them selectively permeable, allowing certain small molecules and particular proteins to cross. It is this selective permeability that establishes the unique chemical identity of each compartment.
Nuclear pore complex (NPC)
Large multiprotein structure forming a channel through the nuclear envelope that allows selected molecules to move between nucleus and cytoplasm.
organelle
Membrane-enclosed compartment in a eukaryotic cell that has a distinct structure, macromolecular composition, and function.
mitochondria
Membrane-enclosed organelles, about the size of bacteria, that carry out oxidative phosphorylation and produce most of the ATP in eukaryotic cells
Ran
Monomeric GTPase present in both cytosol and nucleus that is required for the active transport of macromolecules into and out of the nucleus through nuclear pore complexes
gated transport
Movement of proteins through nuclear pore complexes between the cytosol and the nucleus.
TOM complex
Multisubunit protein assembly that transports proteins across the mitochondrial outer membrane.
Components of the TIM complexes, the multisubunit protein translocators in the mitochondrial inner membrane, are much less abundant than those of the TOM complex. They were initially identified using a genetic trick. The yeast Ura3 gene, whose product is an enzyme that is normally located in the cytosol where it is essential for synthesis of uracil, was modified so that the protein carried an import signal for the mitochondrial matrix. A population of cells carrying the modified Ura3 gene in place of the normal gene was then grown in the absence of uracil. Most cells died, but the rare cells that grew were shown to be defective for import into the mitochondrial matrix. Explain how this selection identifies cells with defects in components required for import into the mitochondrial matrix. Why don't normal cells with the modified Ura3 gene grow in the absence of uracil? Why do cells that are defective for mitochondrial import grow in the absence of uracil?
Normal cells that carry the modified Ura3 gene make Ura3 that gets imported into mitochondria. It is therefore unavailable to carry out an essential reaction in the metabolic pathway for uracil synthesis. These cells might as well not have the enzyme at all, and they will grow only when uracil is supplied in the medium. By contrast, in cells that are defective for mitochondrial import, Ura3 is prevented from entering mitochondria and remains in the cytosol where it can function normally in the pathway for uracil synthesis. Thus, cells with defects in import into the mitochondrial matrix can grow in the absence of added uracil because they can make their own.
T/F: ER-bound and free ribosomes, which are structurally and functionally identical, differ only in the proteins they happen to be making at a particular time.
True. Ribosomes all begin translating mRNAs in the cytosol. The mRNAs for certain proteins encode a signal sequence for the ER membrane. After this sequence has been synthesized, it directs the nascent protein, along with the ribosome and the mRNA, to the ER membrane. Ribosomes translating mRNAs that do not encode such a sequence remain free in the cytosol.
T/F: Each signal sequence specifies a particular destination in the cell.
True. Stretches of amino acids, typically 15-60 residues long, serve as sorting signals for most proteins in the cell. Signal sequences that specify particular cellular destinations—import into ER, import into nucleus, etc.—have characteristic features that allow their interaction with appropriate sorting receptors, which guide the proteins to their correct compartment.
T/F: The nuclear membrane is freely permeable to ions and other small molecules under 5000 daltons
True. The nuclear membrane allows free passage of ions and small molecules because it is perforated with numerous nuclear pore complexes—3000-4000 in a typical mammalian cell—each of which has one or more open aqueous channels through which small water-soluble molecules can passively diffuse.
Is it really true that all human cells contain the same basic set of membrane-enclosed organelles? Do you know of any examples of human cells that do not have a complete set of organelles?
While the vast majority of cells in the human body do have a complete set of membrane-enclosed organelles, certain specialized cells do not. A prime example is the red blood cell. At a late stage in its development, the precursor of the red blood cell—the reticulocyte—jettisons all of its internal membrane-enclosed organelles, leaving just the plasma-membrane- enclosed cytosol. The cells that make up the lens of the eye, which lack mitochondria, are similar. But in a way, these are exceptions that prove the rule; these cells are derived from cells that do carry the complete set of membrane-enclosed organelles.
The rough ER is the site of synthesis of many classes of membrane proteins. Some of these proteins remain in the ER, whereas others are sorted to compartments such as the Golgi apparatus, lysosomes, and the plasma membrane. One measure of the difficulty of the sorting problem is the degree of "purification" that must be achieved during transport from the ER to the other compartments. For example, if membrane proteins bound for the plasma membrane represented 90% of all proteins in the ER, then only a small degree of purification would be needed (and the sorting problem would appear relatively easy). On the other hand, if plasma membrane proteins represented only 0.01% of the proteins in the ER, a very large degree of purification would be required (and the sorting problem would appear correspondingly more difficult). What is the magnitude of the sorting problem? What fraction of the membrane proteins in the ER are destined for other compartments? A few simple considerations allow one to answer these questions. Assume that all proteins on their way to other compartments remain in the ER for 30 minutes on average before exiting, and that the ratio of proteins to lipids in the membranes of all compartments is the same.
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To test the hypothesis that the directionality of transport across the nuclear membrane is determined primarily by the gradient of the Ran- GDP outside the nucleus and Ran-GTP inside the nucleus, you decide to reverse the gradient to see if you can force the import of a protein that is normally exported from the nucleus. You add a well-defined nuclear export substrate, fluorescent BSA coupled with a nuclear export signal (NES-BSA), to the standard permeabilized cell assay. Sure enough, it is excluded from the nuclei (Figure 12-3A). Now you add Crm1, the nuclear export receptor that recognizes the export signal, and RanQ69L-GTP, a mutant form of Ran that cannot hydrolyze GTP. With these additions, the tagged BSA now enters the nuclei (Figure 12-3B). Unlike conventional nuclear import, which concentrates imported proteins in the nucleus, the concentration of NES-BSA in the nucleus in the import assay is no higher than in the surrounding cytoplasm.
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Your advisor is explaining his latest results in your weekly lab meeting. By fusing his protein of interest to green fluorescent protein (GFP), he has shown that it is located entirely in the nucleus. But he wonders if it is a true nuclear protein or a shuttling protein that just spends most of its time in the nucleus. He is unsure how to resolve this issue. Having just read an article about how a similar problem was answered, you suggest that he make a heterokaryon by fusing cells that are expressing his tagged protein with an excess of cells that are not expressing it. You tell him that in the presence of a protein synthesis inhibitor to block new synthesis of the tagged protein, he can resolve the issue by examining fused cells with two nuclei. He gives you a puzzled look and asks, "How does that help?" You tell him what he has so often told you: "Think about it."
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A. Which of the two N-terminal negative charges is the more important for insertion of the protein in the presence of the normal membrane potential (minus CCCP)? Explain your reasoning.
A. The negative charge nearest to the transmembrane segment is the more important. In the presence of a normal-length hydrophobic segment, neither N-terminal negative charge is essential for membrane insertion, as shown by the results with constructs 1 and 2. Proteins with shortened hydrophobic segments, however, depend on the negative charges for proper insertion, as shown by comparison of constructs 4 through 7. Construct 5 with the single negative charge adjacent to the hydrophobic segment is inserted nearly as efficiently as construct 4, which has both negative charges. By contrast, construct 6 with the single negative charge near the N-terminus is not inserted at all.
A. What portion of the nucleoplasmin molecule is responsible for localization in the nucleus?
A. The portion of nucleoplasmin responsible for localization to the nucleus must reside in the tail. The nucleoplasmin head does not localize to the nucleus when injected into the cytoplasm, and it is the only injected component that is missing a tail.
Imagine that you have engineered a set of genes, each encoding a protein with a pair of conflicting signal sequences that specify different compartments. If the genes were expressed in a cell, predict which signal would win out for the following combinations. Explain your reasoning. A. Signals for import into nucleus and import into ER. B. Signals for import into peroxisomes and import into ER. C. Signals for import into mitochondria and retention in ER. D. Signals for import into nucleus and export from nucleus.
A. The protein would enter the ER. The signal for import into the ER is located at the N-terminus of the protein and functions before the internal signal for nuclear import is synthesized. Once the protein entered the ER, the signal sequence for nuclear import could not function because it would be prevented from interacting with cytosolic nuclear import receptors. B. The protein would enter the ER. Once again, the N-terminal signal for ER import would function before the internal signal for peroxisome import is synthesized. The peroxisome import signal could not function once the protein was sequestered in the ER. C. The protein would enter the mitochondria. In order to be retained in the ER, the protein must first be imported into the ER. Without a signal for ER import, the ER retention signal could not function. D. A protein with signals for both nuclear import and nuclear export would shuttle between the cytosol and the nucleus. Unlike the other pairs of signals, these signals are not necessarily in conflict. A number of cellular proteins, whose function requires shuttling in and out of the nucleus, are designed in just this way.
Why do yeasts with the pNL+ plasmid proliferate in the presence of glucose but die in the presence of galactose, whereas yeasts with the pNL- plasmid proliferate in both media?
A. The rationale for the experiment is that the restriction nuclease EcoRI will cleave the cell's DNA into pieces if it gains access to the nucleus, thereby killing the cell. In glucose-containing medium, the hybrid gene is not transcribed and the fusion protein is not made; thus, both types of yeast proliferate perfectly well. In the presence of galactose, the hybrid gene is expressed. In yeasts carrying the pNL- plasmid, the hybrid protein, although expressed, cannot enter the nucleus because it lacks an NLS and thus does no harm to the cell. By contrast, in yeasts carrying the pNL+ plasmid, the hybrid protein, which has a functional NLS, enters the nucleus and cuts up the cell's DNA; thus, the cells die.
A. Is Tim23 an integral component of the inner or outer mitochondrial membrane? Explain your reasoning.
A. Tim23 appears to be an integral component of both mitochondrial membranes. In intact mitochondria, a small portion of Tim23 is digested by the protease, indicating that a segment of Tim23 is exposed outside mitochondria. This result implies that a portion of Tim23 extends through the outer mitochondrial membrane. The digested segment of Tim23 must be at the N-terminus because the remaining portion is still recognized by antibodies specific for the C-terminus (see Figure 12-13B, lane 2). In mitoplasts, a larger N-terminal segment of Tim23 is digested by the protease but the C-terminal portion is still protected, indicating that it is in the inner membrane or inside the mitoplasts (see Figure 12-13B, lane 3). In combination, these results indicate that Tim23 must extend through both mitochondrial membranes.
A. How good was your injection technique? Did you actually inject into the nucleus? Did you rip apart the nuclear envelope when you injected the RNAs? How do you know?
A. Your technique is excellent, as judged by the lack of labeled RNA in the cytoplasm immediately after injection (see Figure 12-11). Thus, you must have hit the nucleus and not done irreparable harm to it.
Mitochondria normally provide cells with most of the ATP they require to meet their energy needs. Mitochondria that cannot import proteins are defective for ATP synthesis. How is it that cells with import-defective mitochondria can survive at all? How do they get the ATP they need to function?
Although it is true that mitochondria normally generate the majority of the cell's ATP, they are not the sole source. Two steps in the glycolytic pathway for glucose catabolism, which occurs in the cytosol, also generate ATP. If glucose is provided, yeast (and many other cells) can survive on the ATP they generate by glucose metabolism. It is this process that allows cells to survive in the absence of oxygen, which is required for ATP production by mitochondria. In the absence of oxygen, or when mitochondria are defective, the end product of glucose metabolism in yeast is ethanol.
Nuclear localization signals are not cleaved off after transport into the nucleus, whereas the signal sequences for import into other organelles are often removed after import. Why do you suppose it is critical that nuclear localization signals remain attached to their proteins?
At each mitosis, the contents of the nucleus and the cytosol mix when the nuclear envelope disassembles. When the nucleus reassembles, the nuclear proteins must be selectively re-imported. If the nuclear localization signals were removed upon import, the proteins would be trapped in the cytosol after the next mitosis. By contrast, the contents of other organelles never mix with the cytosol. At mitosis, organelles such as the Golgi apparatus and the ER break up into vesicles, which retain the luminal contents of their larger parents. Because of this, their resident proteins have to be imported only once, and their signal sequences are therefore dispensable.
B. Which of the many proteins eluted from the two different affinity col- umns is a likely candidate for the factor that promotes nuclear import of Ran-GDP?
B. Since the Ran-GDP column removed the nuclear import factor, whereas the RanQ69L-GTP column did not, you are looking for a protein that is present in lane 1 but not in lane 2 (see Figure 12-9). One such protein is present, between the 7 kd and 14 kd markers. Note that the RanQ69L- GTP column binds a set of proteins between the 97 kd and 116 kd markers that the Ran-GDP column does not. They are members of the importin family of nuclear import receptors, which are evidently not required for the nuclear uptake of Ran-GDP
B. To the extent these data allow, define the order of events that leads to uptake of substrate into the nucleus.
B. These data, especially the two-step incubation in Figure 12-7B, suggest that importin first binds the substrate to the nuclear periphery (presumably the nuclear pore complexes). Bound substrate is then acted on by Ran-GTP to promote uptake into the nucleus. These data do not define how Ran-GTP promotes uptake. Additional experiments have led to the current view that Ran-GTP in the nucleus binds to importin, displacing the substrate and thereby accomplishing the final step of nuclear uptake: release into the nucleus. Ran-GTP then normally accompanies the receptor out of the nucleus, where it is converted by a Ran-GAP to Ran-GDP.
B. How do these experiments distinguish between active transport, in which a nuclear localization signal triggers transport by the nuclear pore complex, and passive diffusion, in which a binding site for a nuclear component allows accumulation in the nucleus?
B. These experiments suggest that the nucleoplasmin tail carries a nuclear localization signal and that accumulation in the nucleus is not the result of passive diffusion. The observations involving complete nucleoplasmin or fragments that retain the tail do not distinguish between passive diffusion and active import; they say only that the tail carries the important part of nucleoplasmin—be it a localization signal or a binding site. The key observations that argue against passive diffusion are the results with the nucleoplasmin heads. They do not diffuse into the cytoplasm when they are injected into the nucleus, nor do they diffuse into the nucleus when injected into the cytoplasm, suggesting that the heads are too large to pass through the nuclear pores. Since the more massive forms of nucleoplasmin with tails do pass through the nuclear pores, passive diffusion of nucleoplasmin is ruled out.
T/F: The two signal sequences required for transport of nucleus-encoded proteins into the mitochondrial inner membrane via the TIM23 complex are cleaved off the protein in different mitochondrial compartments.
False. Only one of the two signal sequences is cleaved. The N-terminal signal is cleaved off the imported protein when it reaches the mitochondrial matrix. The second signal—a very hydrophobic sequence at the new N-terminus—directs the protein to the inner membrane through either the TIM23 complex or the OXA complex. The second signal is not cleaved; it anchors the protein in the inner membrane.
All cytosolic proteins have nuclear export signals that allow them to be removed from the nucleus when it reassembles after mitosis
False. Resident proteins of the cytosol do not have nuclear export signals. They are efficiently excluded from reassembling nuclei by the mechanism of reassembly. The nuclear envelope is initially closely applied to the surface of the chromosomes, excluding all proteins except those bound to the mitotic chromosomes. Once the envelope is complete, other residents of the nucleus are imported via their nuclear localization signals.
T/F: Like the lumen of the endoplasmic reticulum (ER), the interior of the nucleus is topologically equivalent to the outside of the cell.
False. The interior of the nucleus and the cytosol communicate through the nuclear pore complexes, which allow free passage of ions and small molecules. The cytoplasm and the nucleus are said to be topologically equivalent because the outer and inner nuclear membranes are continuous with one another, so that the flow of material between the nucleus and cytosol occurs without crossing a lipid bilayer. By contrast, the lumen of the ER and the outside of the cell are each separated from the cytosol by a layer of membrane and thus are topologically distinct from the cytosol, but they are topologically equivalent to each other.
nuclear lamina
Fibrous meshwork of proteins on the inner surface of the inner nuclear membrane.
thylakoid
Flattened sac of membrane in a chloroplast that contains the protein subunits of the photosynthetic system and of the ATP synthase.
B. Why did you suggest that a protein synthesis inhibitor would be needed in this experiment?
It is critical in this experiment to block synthesis of new GFP-tagged protein. If new protein were made, it would enter both nuclei, regardless of whether it was a nuclear protein or a shuttling protein.
Are proteins imported into mitochondria as completely unfolded polypeptide chains, or can the translocation apparatus accommodate fully or partially folded structures? That is, is the protein sucked up like a noodle, or is it swallowed whole, as a python devours its prey? It is possible to engineer cysteine amino acids into barnase and then cross-link them to make disulfide bonds either between C5 and C78 or between C43 and C80 (Figure 12-12). Import of N95-barnase (see Problem 12-70) was tested in the presence and absence of disulfide cross-links at these two positions. Its import was unaffected by either cross-link. By contrast, import of N65-barnase was blocked by the C5-C78 cross-link but unaffected by the C43-C80 cross-link. Do these results allow you to distinguish between import of extended polypeptide chains or of folded structures? Why or why not?
If barnase could be imported in its native folded configuration, the cross-links should have no effect on import; however, import of N65- barnase was blocked by the C5-C78 cross-link. On the other hand, if import required a completely unfolded, extended polypeptide chain, the presence of either cross-link should have blocked import of both N65- barnase and N95-barnase. Thus, it seems that the mitochondrial import machinery cannot import completely folded structures, but it doesn't require that the protein be in a fully extended configuration either. The ability of the import machinery to accommodate cross-links indicates that it can pass at least two, side-by-side polypeptide chains. The authors of the original study point out that a key event in mitochondrial import is destabilization of the N-terminus of the imported protein. Thus, import of N65-barnase, which occurs at a low rate, is completely blocked when the N-terminus is stabilized by a cross-link. They also show that while the N95 extension attached to dihydrofolate reductase (DHFR) allows efficient import, import of N95-DHFR is blocked by methotrexate (see Problem 12-67).
To aid your studies of protein import into mitochondria, you treat yeast cells with cycloheximide, which blocks ribosome movement along mRNA. When you examine these cells in the electron microscope, you are surprised to find cytosolic ribosomes attached to the outside of the mitochondria. You have never seen attached ribosomes in the absence of cycloheximide. To investigate this phenomenon further, you prepare mitochondria from cells that have been treated with cycloheximide and then extract the mRNA that is bound to the ribosomes associated with the mitochondria. You translate this mRNA in vitro and compare the protein products with similarly translated mRNA from the cytosol. The results are clear-cut: the mitochondria-associated ribosomes are translating mRNAs that encode mitochondrial proteins. You are astounded! Here, clearly visible in the electron micrographs, seems to be proof that protein import into mitochondria occurs during translation. How might you reconcile this result with the prevailing view that mitochondrial proteins are imported only after they have been synthesized and released from ribosomes?
Import of mitochondrial proteins occurs post-translationally. Normally, translation is much faster than mitochondrial import, so that proteins completely clear the ribosome before interacting with the mitochondrial membrane. By blocking protein synthesis with cycloheximide, you have made the rate of translation artificially slower than the rate of import. Since the signal peptide for protein import into mitochondria resides at the N-terminus, some of the partially synthesized mitochondrial proteins, which are still attached to ribosomes, will be able to interact with the mitochondrial membrane. The attempted import of even one such protein will attach the ribosome and the mRNA (and all other ribosomes translating the same mRNA molecule) to the mitochondrial membrane.
Discuss the following statement: "The plasma membrane is only a minor component of most eukaryotic cells."
In terms of its functional importance to a cell, the plasma membrane is anything but minor. It is the boundary that separates the cell from the outside world, it controls selective entry and exit of molecules, and it is the principal site at which intercellular communications are received. Only in terms of its surface area and mass is it a minor component, accounting for 2-5% of all the membranes in a eukaryotic cell.
What is the fate of a protein with no sorting signal?
In the absence of a sorting signal, a protein will remain in the cytosol.
Describe in a general way how you might use radiolabeled proteins and proteases to study import processes in isolated, intact mitochondria. What sorts of experimental controls might you include to ensure that the results you obtain mean what you think they do?
Incubate the radiolabeled proteins with isolated mitochondria under conditions you wish to test, allow a sufficient time for import, and then treat the mixture with a protease. Proteins that are not imported will be digested by the protease. Proteins that have been imported will be resist- ant to the protease. Protease-resistant proteins could be assayed by re-isolating the mitochondria and measuring the counts associated with them. Alternatively, they could be assayed by solubilizing the entire mixture and separating the proteins by gel electrophoresis. Protease-resist- ant proteins would run at the same position as untreated proteins. These analyses assume that proteins are protease-resistant because they are sequestered inside mitochondria, meaning they have been imported. You would need to include several controls before you could make this conclusion. You would need to know that the protease is work- ing, which could be measured by leaving the mitochondria out of the incubation mixture. You would need to know that the protein is stable in the absence of the protease, which you could assay by leaving the protease out of the incubation mixture. You would need to know that protease-resistant proteins are in the mitochondria, which could be assayed by solubilizing the mitochondria with a detergent to show that protease-resistant proteins now become protease-sensitive. Appropriate controls are essential for informative research into any biological problem.
Protein synthesis in a liver cell occurs nearly exclusively on free ribosomes in the cytosol and on ribosomes that are bound to the ER membrane. (A small fraction of total protein synthesis is directed by the mitochondrial genome and occurs on ribosomes in the mitochondrial matrix.) Which type of protein synthesis—in the cytosol or on the ER—do you think is responsible for the majority of protein synthesis in a liver cell? Assume that the average density and lifetimes of proteins are about the same in all compartments. Explain the basis for your answer. Would your answer change if you took into account that some proteins are secreted from liver cells?
One way to approach this problem is to compare the relative volumes of the compartments that are served by cytosolic and ER protein synthesis. Assuming that the average density and lifetimes of proteins are about the same in all compartments—a reasonable first approximation—their relative volumes would provide a rough estimate of the amount of protein synthesis. The compartments served by cytosolic protein synthesis, which include the cytosol, nucleus, mitochondria, and peroxisomes, account for more than 80% of the cell volume. The compartments that depend on ER protein synthesis—the ER, Golgi apparatus, endosomes, and lysosomes—account for less than 20% of the cell volume. On this basis, then, one would conclude that cytosolic protein synthesis is responsible for the majority of cellular protein synthesis. In cells that do not secrete large amounts of protein, the majority of protein synthesis is likely to occur in the cytosol. One of the main functions of liver cells, however, is to export proteins such as albumin, which makes up about half of the total serum protein. The fraction of liver ribosomes engaged in synthesizing albumin is probably less than 10%. At this level, a liver cell would still be carrying out the majority of its protein synthesis on cytosolic ribosomes, but in other, specialized secretory cells (like those of the pancreas), ER protein synthesis may exceed cytosolic protein synthesis.
mitochondrial hsp70
Part of a multisubunit protein assembly that is bound to the matrix side of the TIM23 complex and acts as a motor to pull the precursor protein into the matrix space.
You have made a peptide that contains a functional mitochondrial import signal. Would you expect the addition of an excess of this peptide to affect the import of mitochondrial proteins? Why or why not?
Peptides with mitochondrial import signals would be expected to compete with mitochondrial proteins for binding to the translocation machinery. Thus, an excess of such peptides should reduce or abolish import of mitochondrial proteins.
mitochondrial precursor protein
Protein encoded by a nuclear gene, synthesized in the cytosol, and sub- sequently transported into mitochondria
signal sequence
Protein sorting signal that consists of a short continuous sequence of amino acids.
signal patch
Protein sorting signal that consists of a specific three-dimensional arrangement of atoms on the folded protein's surface.
nuclear import receptor
Protein that binds nuclear localization signals and facilitates the transport of proteins with these signals from the cytosol into the nucleus through nuclear pore complexes.
How do you suppose that proteins with a nuclear export signal get into the nucleus?
Proteins with a nuclear export signal also have a nuclear localization signal. Thus, such proteins typically shuttle between the nucleus and cytoplasm.
Barnase is a 110-amino-acid bacterial ribonuclease that is often used as a model for studies of protein folding and unfolding. It forms a com- pact folded structure that has a high energy of activation for unfolding (about 85 kJ/mole). Can such a protein be imported into mitochondria? To the N-terminus of barnase, you add 35, 65, or 95 amino acids from the N-terminus of pre-cytochrome b2, all of which include the cytochrome's mitochondrial import signal. N35-barnase is not imported, N65-barnase is imported at a low rate, and N95-barnase is imported very efficiently into isolated mitochondria. None of these N-terminal extensions have any measurable effect on the stability of the barnase domain. If these proteins are denatured before testing for import, they are all imported at the same high rate. How do you suppose that longer N-terminal extensions facilitate the import of barnase?
Since each modified barnase includes an import signal and the length of the N-terminal extension does not affect the stability of the barnase domain, the dependence of import on the length of the extension presumably reflects some process inside mitochondria. The most likely possibility is that only the longer extensions can span both mitochondrial membranes and project into the matrix. There they can be bound by the mitochondrial hsp70, which can use the hydrolysis of ATP to help drive import. Presumably, the 95-amino-acid extension is long enough to be efficiently engaged by hsp70, whereas the 65-amino-acid extension must be less efficiently bound. Hsp70 and the energy of ATP hydrolysis are required for import of barnase because of its extremely stable folded structure. If the protein is first denatured, all three N-terminal extensions can facilitate its import at the same high rate because the unfolded protein does not hinder entry into the matrix.
peroxisome
Small membrane-bounded organelle that uses molecular oxygen to oxidize organic molecules.
Nuclear export signal
Sorting signal contained in the structure of macromolecules and complexes that are transported from the nucleus to the cytosol through nuclear pore complexes.
nuclear localization signal
Sorting signal found in proteins destined for the nucleus and which enable their selective transport into the nucleus from the cytosol through the nuclear pore complexes.
A typical animal cell is said to contain some 10 billion protein molecules that need to be sorted into their proper compartments. That's a lot of proteins. Can 10 billion protein molecules even fit into a cell? An average protein encoded by the human genome is 450 amino acids in length, and the average mass of an amino acid in a protein is 110 daltons. Given that the average density of a protein is 1.4 g/cm3, what fraction of the volume of a cell would 10 billion average protein molecules occupy? Consider a liver cell, which has a volume of about 5000 μm3, and a pancreatic exocrine cell, which has a volume of about 1000 μm3.
Ten billion average proteins would occupy a volume of 589 μm3, where aa is amino acid. This value represents about 12% of the volume of a liver cell (589/5000) and about 59% of the volume of a pancreatic exocrine cell (589/1000). Thus, 10 billion protein molecules would fit in a typical animal cell. This calculation obviously gives only the crudest of estimates. The number of protein molecules expressed in a cell is not constant. On average, smaller cells would be expected to make fewer proteins than larger cells; thus, the volume occupied by proteins would be roughly constant. Some specialized cells might be very different. For example, a large droplet of triglycerides takes up most of the volume of a fat cell, which therefore has a lower fraction of its total cell volume occupied by protein (although if one excludes the volume of the triglyceride droplet from the calculation, a fat cell is probably not different from other cells). The values calculated here bracket the accepted value of about 20%.
If the enzyme dihydrofolate reductase (DHFR), which is normally located in the cytosol, is engineered to carry a mitochondrial targeting sequence at its N-terminus, it is efficiently imported into mitochondria. If the modified DHFR is first incubated with methotrexate, which binds tightly to the active site, the enzyme remains in the cytosol. How do you suppose that the binding of methotrexate interferes with mitochondrial import?
The binding of methotrexate to the active site prevents the enzyme from unfolding, which is necessary for import into mitochondria. Evidently, methotrexate binds so tightly that it locks the enzyme into its folded con- formation and prevents chaperone proteins from unfolding it.
The lipid bilayer, which is 5 nm thick, occupies about 60% of the volume of a typical cell membrane. (Lipids and proteins contribute equally on a mass basis, but lipids are less dense and therefore account for more of the volume.) For liver cells and pancreatic exocrine cells, the total area of all cell membranes is estimated at about 110,000 μm^2 and 13,000 μm^2, respectively. What fraction of the total volumes of these cells is accounted for by lipid bilayers? The volumes of liver cells and pancreatic exocrine cells are about 5000 μm^3 and 1000 μm^3, respectively.
The lipid bilayers in all the membranes in liver and pancreatic exocrine cells would have a volume of 330 μm3 and 39 μm3, respectively, as calculated below for a liver cell: Thus, lipid bilayers would account for 6.6% (330/5000) of the volume of a liver cell, and 3.9% (39/1000) of the volume of a pancreatic exocrine cell. For typical cells, then, about 5% of their volume is occupied by lipid bilayers.
stroma
The matrix space of a chloroplast.
inner membrane
The membrane of a mitochondrion that encloses the matrix and is folded into cristae.
T/F: The TOM complex is required for the import of all nucleus-encoded mitochondrial proteins.
True. Regardless of their final destination in the mitochondrion, all proteins that are synthesized in the cytosol (that is, all nucleus-encoded mitochondrial proteins) must first enter the TOM complex. After the TOM complex, the pathways of import diverge as proteins are sorted to their appropriate mitochondrial compartment.
List the organelles in an animal cell that obtain their proteins via gated transport, via transmembrane transport, or via vesicular transport.
The nucleus is the only compartment that receives its proteins by gated transport, which is carried out by nuclear pore complexes. The ER, mitochondria, and peroxisomes all receive their proteins by transmembrane transport, which is mediated by specific protein translocators that reside in the membrane of the target organelle. The Golgi apparatus, secretory vesicles, early and late endosomes, and lysosomes all obtain their proteins via vesicular transport, in which small vesicles ferry proteins from one compartment to another.
How might you use this system for a selection assay to isolate cells defective in nuclear transport?
The pNL+ plasmid possesses the basic features needed for a selection assay to isolate cells defective in nuclear transport. In the presence of galactose, yeast cells will express the hybrid protein. A normal cell will transport the protein into the nucleus where it will kill the cell by cleaving its DNA. By contrast, in a mutant cell defective for nuclear transport, the killer protein will be confined to the cytosol, where it can do no harm. Thus, in a culture of yeast cells carrying the plasmid, individual cells defective for nuclear transport would be expected to survive transfer to a galactose-containing medium. Actually building a working selection assay requires dealing with several additional considerations. For example, a cell that is permanently defective for nuclear transport would not be viable. Therefore, the mutants would need to be conditionally lethal; that is, with functional nuclear transport at low temperature, for instance, so you can grow the cells, and defective transport at high temperature, where you apply selection. After sufficient exposure to the hybrid protein to kill normal cells, the culture could then be shifted to a glucose medium at low temperature to permit the mutant cells to grow. An additional complication 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 will be imported and the mutant cells will die. Thus, you would need to modify the assay 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 (see Problem 6-87). 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, you might try to make a mutant of EcoRI that is active at the high temperature and inactive at the low temperature. Such a cold-sensitive protein would be active when nuclear transport was blocked in the import mutant and inactive when nuclear transport resumed.
B. To the extent the information in this problem allows, diagram the arrangement of Tim23 in mitochondrial membranes.
The pattern of protease sensitivity of Tim23 in mitochondria and mito- plasts suggests that Tim23 is arranged as shown in Figure 12-24. You may have noticed that the hydropathy plot does not predict a membrane-spanning segment at the N-terminus. The authors of the original study noticed the same thing. There is, however, a predicted propensity for β-sheet formation in the N-terminus (not shown). Thus, the authors suggest that Tim23 may span the outer membrane in a β-strand conformation, which is typical of certain outer membrane proteins such as porins.
Why do mitochondria need a special translocator to import proteins across the outer membrane, when the membrane has already has large pores formed by porins?
The pores formed by porins are large enough for all ions and metabolic intermediates, but not large enough for most proteins. The size cutoff for free passage through the pores of mitochondrial porins is roughly 10 kilo- daltons.
outer nuclear membrane
The portion of the nuclear envelope that is continuous with the endoplasmic reticulum and is studded with ribosomes on its cytosolic surface.
The broad-spectrum antibiotic leptomycin B inhibits nuclear export, but how does it work? In the yeast S. pombe, resistance to leptomycin B can arise by mutations in the Crm1 gene, which encodes a nuclear export receptor for proteins with leucine-rich nuclear export signals. To look at nuclear export directly, you modify the green fluorescent protein (GFP) by adding a nuclear export signal (NES). In both wild-type and mutant cells that are resistant to leptomycin B, NES-GFP is found exclusively in the cytoplasm in the absence of leptomycin B (Figure 12-10). In the presence of leptomycin B, however, NES-GFP is present in the nuclei of wild-type cells, but in the cytoplasm of mutant cells (Figure 12-10). Is this result the one you would expect if leptomycin B blocked nuclear export? Why or why not?
These results are more or less what you would expect if leptomycin B blocked nuclear export. In the absence of leptomycin B, NES-GFP is excluded from the nuclei, as shown by the dark areas that correspond to the positions of the nuclei in the DNA panels (see Figure 12-10). This result indicates that NES-GFP is efficiently exported from nuclei. (Of course, it could also mean that NES-GFP never entered the nuclei in the first place.) The same result is observed in leptomycin B-resistant cells in the presence of leptomycin B, as expected if it is without effect in the mutant cells. The presence of NES-GFP in the nuclei of wild-type cells treated with leptomycin B confirms that NES-GFP can enter the nucleus and that leptomycin B prevents its export. The presence of NES-GFP in the cytoplasm, as well, indicates either that NES-GFP doesn't enter the nucleus very well or that leptomycin B doesn't completely block nuclear export.
T/F: Some proteins are kept out of the nucleus, until needed, by inactivating their nuclear localization signals by phosphorylation.
True. Gene regulatory proteins in particular are subject to this kind of regulation, as a way of preventing gene activation (or repression) until the proper time.
As shown in Figure 12-2, the inner and outer nuclear membranes form a continuous sheet, connecting through the nuclear pores. Continuity implies that membrane proteins can move freely between the two nuclear membranes by diffusing through the bilayer at the nuclear pores. Yet the inner and outer nuclear membranes have different protein com- positions, as befits their different functions. How do you suppose this apparent paradox is reconciled?
Two aspects of protein function may contribute to the difference in the protein compositions of the inner and outer nuclear membranes. First, proteins that function in the inner membrane are usually anchored by their interactions with components of the nucleus such as chromosomes and the nuclear lamina, which is a protein meshwork underlying the inner nuclear membrane. Freely diffusing proteins that are anchored once they reach the inner membrane would accumulate there. Second, proteins that form the nuclear pore itself may restrict the free diffusion of other membrane proteins by virtue of their insertion into the lipid bilayer at the boundary between the inner and outer membranes. Any membrane protein that cannot pass through the ring of nuclear pore proteins would be restricted to the outer membrane.