Week 6 Study Problems

Question 1

Describe three mechanisms that regulate the activity of cyclin-dependent kinases?

Answer 1

An active cyclin-dependent kinase (Cdk) has to be bound to a cyclin and phosphorylated at an activating site and have no phosphate at an inhibitory site.

One way a Cdk is regulated is by binding to a cyclin to form a heterodimer. A Cdk is inactive if it is NOT bound to a cyclin. Although Cdks are present throughout the cell cycle, specific cyclins are expressed only during a particular phase: G1, S, G2 or M.

A second way a Cdk can be regulated is by a Protein Kinase transferring a phosphate onto the activating site of Cdk.

A third way a Cdk can be regulated is by a Protein Kinase transferring a phosphate onto the inhibitory site of Cdk.

A fourth way a Cdk can be regulated is by a Cdk-inhibitor protein binding to the Cdk- cyclin heterodimer. Binding of the inhibitor inactivates the Cdk.

A fifth way a Cdk can be regulated is by the addition of ubiquitin to the cyclin. Ubiquitinated-cyclin is then degraded by the proteasome.

Question 2

What are the four cyclin-Cdk complexes that ensure events of the cell cycle happen in the proper sequence?

Answer 2

For the exam you don’t have to know the specific cyclin and Cdk names, but recognize that G1-Cdk, and the other 3, represents a cyclin-Cdk heterodimer.

G1-Cdk (cyclinD:Cdk4)

G1/S-Cdk (cylcinE:Cdk2) S-

Cdk (cyclinA:Cdk2)

M-Cdk (cyclinB:Cdk1)

Question 3

What is the cell doing during the different stages of the cell cycle?

Answer 3

During G1 the cell is increasing synthesis of macromolecules for growth, and checking for growth factor signals (such as mitogens) and checking the degree to which the cell is stressed. During S the cell copies its two genomes (for a diploid, 2N) so the cell is busy doing DNA replication. The result of replication are sister chromosomes (sister chromatids), so 4 genomes (4N). During G2 the cell checks to be sure there are no breaks in the DNA and DNA replication is complete, and prepares for Mitosis. During M the cell segregates the sister chromosomes, pulling and pushing the sisters to opposite poles, so that each daughter cell gets complete copies of both genomes (each daughter returns to 2N).

Question 4

What are cell cycle checkpoints and how are they important?

Answer 4

Cell cycle checkpoints are signaling pathways that regulate the activity of the cyclin-Cdk heterodimers. Active cyclin-Cdk complexes turn ‘on’ events needed to transition from one phase to the next. Inhibiting cyclin-Cdk complexes prevents the cell, or delays the cell, from transitioning to a next phase. Checkpoints are important because they ensure a cell will only transition when events of a phase are completed.

For example, if a cell does not have enough resources to copy its genomes, then the cell needs to wait before transitioning to S phase, otherwise the cell will have replication problems that can increase mutation. Similarly, if DNA replication is not completed before entering mitosis, then chromosomes can break and segregate randomly resulting in aneuploidy. Another example is if the DNA is damaged, replication needs to be delayed, otherwise copying of damaged DNA results in increased mutation.

Question 5

5. Describe the organization of three cell cycle checkpoints.

Answer 5

1) The G1 checkpoint prevents a cell from progressing to S phase until G1 events are completed. The organization of the checkpoint includes receptors for growth factors (mitogens) and signaling pathways that remove inhibitory phosphate from, and transfer activating phosphate onto the G1- and G1/S-Cdks. The checkpoint also includes retinoblastoma (Rb) proteins, which bind to transcription regulators to prevent the expression of many genes encoding proteins needed for S-phase. Overcoming the G1 checkpoint involves active G1- and G1/S-Cdks transferring phosphate onto Rb. Phosphorylated Rb changes conformation so that transcription regulators are released and can then turn ‘on’ expression of many genes.

The G1 checkpoint also prevents cells from entering S-phase if there is too much DNA damage, such as double strand breaks. There are proteins that bind to broken DNA and initiate signaling. One major target of the signaling is p53, which in the absence of signaling is continuously sent to the proteasome for degradation. Signaling results in p53 being phosphorylated, the p53 escapes degradation and, being a transcription regulator, binds to DNA and turns ‘on’ many genes. One gene turned ‘on’ encodes p21, which can bind to, and inhibit the G1/S-Cdk and the S-Cdk. While the cell cycle is delayed by p21, DNA repair proteins fix the broken DNA.

3) The G2 checkpoint prevents cells from entering M-phase (mitosis) if there is too much broken DNA or gaps (incomplete DNA replication). Proteins that bind DNA with double strand breaks or single strands (unreplicated DNA; gaps) initiate signals that activate the Protein Kinase, Wee1, but inhibit the Protein Phosphatase, Cdc25. When active, Wee1 transfers a phosphate onto the inhibitory site of the M-Cdk. When active, Cdc25 catalyzes removal of the inhibitory phosphate from M-Cdk. When M-Cdk is active, then the cell transitions into M-phase.
4) Another checkpoint is the spindle assembly checkpoint. The organization of this checkpoint involves an undefined signal that is generated when sister chromosomes are NOT under tension. After the nuclear envelope breaks down, spindle microtubules bind to kinetochores, a protein complex formed on chromosome centromeres. Each sister in a pair has a kinetochore. When microtubules are bound to each sister’s kinetochore, then the sisters are under tension. When microtubules are bound to only one of the sister’s kinetochores, then the sisters are not under tension, and the checkpoint signal is generated. The signal inhibits the anaphase-promoting complex (APC). When all sister chromosomes are under tension, then the APC is active and transfers ubiquitin onto a protein called securin. The ubiquitin-tagged securin is degraded by the proteasome. Degradation of securin releases its partner, an enzyme called separase, which catalyzes cleavage of the cohesins that hold the sisters together. Once cleaved, the sisters can segregate to opposite poles.

Question 6

What mechanism triggers separation of sister-chromatids?

Answer 6

Cohesins are protein complexes that hold sister chromosomes together. For sisters to separate, and segregate to opposite poles, the cohesins need to be cleaved (cut). To cut the cohesins, the anaphase-promoting complex tags the protein securin for degradation. Degradation of securin releases its partner, an enzyme called separase, which catalyzes cleavage of the cohesins. Once cleaved, the sisters can segregate to opposite poles.

Question 7

For the following proteins, state how the protein contributes to the cell cycle?

a. Cdc6

Answer 7

contributes by binding to the origin recognition complex and keeping it in a pre- replicative state. After Cdc6 is phosphorylated by S-Cdk, and subsequently degraded, the machinery for DNA replication can assemble and start DNA replication.

Question 8

For the following proteins, state how the protein contributes to the cell cycle?

c. p53

Answer 8

contributes as a transcription regulator that integrates signals coming from DNA damage, or cellular stress, and if needed, turns ‘on’ expression of many genes coding proteins that prevent the cell from entering S-phase. One of the genes it turns ‘on’ is p21. If there is much DNA damage, p53 can turn on expression of genes coding proteins that initiate cell death.

Question 9

For the following proteins, state how the protein contributes to the cell cycle?

d. p21

Answer 9

contributes as a protein that binds to cyclin-Cdk complexes and inhibits their activity. The inhibition prevents the cell from transitioning into S-phase. p21 is turned ‘on’ in response to too much DNA damage, so p21 delays DNA replication until the damage is repaired.

Question 10

For the following proteins, state how the protein contributes to the cell cycle?

b. Cohesins

Answer 10

contribute by holding sister chromatids together until the anaphase- promoting complex initiates anaphase.

Question 11

For the following proteins, state how the protein contributes to the cell cycle?

e. Cdc25 phosphatase

Answer 11

contributes by removing inhibitory phosphate from the M-Cdk, which when active transitions a cell into mitosis.

Question 12

For the following proteins, state how the protein contributes to the cell cycle?

f. Securin

Answer 12

contributes by binding separase and keeping it inactive until all sister chromatids are under tension.

Question 13

For the following proteins, state how the protein contributes to the cell cycle?

g. Separase

Answer 13

contributes by cleaving/cutting cohesins so that sister chromatids can be segregated to opposite poles.

Question 14

For the following proteins, state how the protein contributes to the cell cycle?

h. Anaphase-promoting complex

Answer 14

contributes by transferring ubiquitin onto securin, which is then degraded, releasing separase and thus initiating anaphase.

Question 15

For the following proteins, state how the protein contributes to the cell cycle?
i. Retinoblastoma (Rb)

Answer 15

contributes by binding transcription regulators so that many genes, required for transition to S-phase, are kept ‘off’ until the G1- and G1/S-Cdks are active.

Question 16

Q18-2 see graph

Answer 16

see graph

Question 17

Why do you think apoptosis occurs by a different mechanism from the cell death that occurs in cell necrosis? What might be the consequences if apoptosis were not achieved in so neat and orderly a fashion, whereby the cell destroys itself from within and avoids leakage of its contents into the extracellular space?

Answer 17

Cell necrosis results in the cell rupturing and its cell contents spilling out into the extracellular matrix. This can cause damage to adjacent cells and stimulate the immune system to cause inflammation, all of which damages the tissue. Apoptosis occurs by a mechanism that cuts-up many of the cells proteins, which results in a dismantling of the cell into a condensed, shrunken blob. In addition, the surface of the cell is altered so that phagocytic cells, such as macrophages and neutrophils, recognize the dead cell and remove it by phagocytosis.
If apoptosis were not achieved, then as described above, there would be more damage to tissues caused by rupturing cells and the subsequent inflammatory response by the immune system.

Question 18

Ionizing radiation causes double strand breaks in DNA. One important biological effect of a large dose of ionizing radiation is to halt cell division.

A. How does this occur?

Answer 18

Proteins recognize, by selective binding, the double strand breaks and then send signals that phosphorylate p53. Once phosphorylated, p53 turns ‘on’ the expression of many genes including one that codes for p21. The p21 protein binds to, and inhibits the G1- and G1/S-Cdks so that the cell cycle is delayed and there is time to repair the damaged DNA before DNA replication is initiated.

Question 19

Ionizing radiation causes double strand breaks in DNA. One important biological effect of a large dose of ionizing radiation is to halt cell division.

What happens if a cell has a mutation that prevents it from halting cell division after being irradiated?

Answer 19

If the cell cannot delay the cell cycle when there are double strand breaks, then DNA replication will initiate and replication forks will encounter the broken DNA. Such encounters increase the mutation rate, in particular rearrangements.

Question 20

C. What might be the effects of such a mutation if the cell is not irradiated?

Answer 20

If the cell is not irradiated, then there will be less DNA damage, so there will be less problems with DNA replication relative to after being irradiated. Even without irradiation there are background levels of DNA damage, so with out proper control of cell division the mutation rate will increase slightly (less so than after irradiation, but more than in a normal cell with checkpoint functioning), and the mutant cell is likely to take less time to go through the cell cycle.

Question 21

Q-18-16:

Answer 21

See picture

Question 22

As mitosis comes to an end, the cyclin concentration decreases very rapidly as does Cdk activity. How does this happen?

Answer 22

The cyclin is degraded. Without the cyclin, the Cdk is no longer active. Cyclin is degraded by being covalently modified with ubiquitin. Proteins modified with ubiquitin are recognized by the proteasome and degraded.

Question 23

Rarely, both sister chromatids of a replicated chromosome end up in one daughter cell. How might this happen? What could be the consequences of such a mitotic error?

Answer 23

If spindle microtubules attach to only one of the sister chroosome’s kinetochores and the spindle assembly checkpoint is NOT activated, then both sisters will move to one pole, and that daughter with have an extra chromosome copy while the other daughter will be haploid for that chromosome. Having an extra chromosome or missing a chromosome is called aneuploidy, and is a consequence of mitotic errors.
Mitotic errors can also occur if spindle microtubules from a single pole attach to both kinetochores of the sisters, or if sister chromosomes, or pieces of chromosomes, never attach to the spindle microtubules.

Question 24

In his highly classified research laboratory, Dr. Lawrence M. is charged with the task of developing a strain of dog-sized rats to be deployed behind enemy lines. In your opinion, which of the following strategies should Dr. M. pursue to increase the size of rats?

A. Block all apoptosis

Answer 24

Blocking all apoptosis would likely not work. Apoptosis is required for the development of organs and tissues, and for the removal of damaged cells. Without apoptosis, damaged cells would die by necrosis and thus cause excessive tissue damage directly and by stimulating inflammation. Improper development would also cause problems.

Question 25

In his highly classified research laboratory, Dr. Lawrence M. is charged with the task of developing a strain of dog-sized rats to be deployed behind enemy lines. In your opinion, which of the following strategies should Dr. M. pursue to increase the size of rats?

B. Block p53 function

Answer 25

Blocking p53 function would likely not work. Without p53, cells enter S-phase with too much DNA damage and so the rate of mutation increases. The animal would likely have many tumors, if it could complete development at all.

Question 26

In his highly classified research laboratory, Dr. Lawrence M. is charged with the task of developing a strain of dog-sized rats to be deployed behind enemy lines. In your opinion, which of the following strategies should Dr. M. pursue to increase the size of rats?

C. Overproduce growth factors, mitogens or survival factors

Answer 26

This approach would have the best chance of succeeding, depending on what specific growth factors, mitogens or survival factors were overproduced. Overproduction of certain factors could increase the number of cells, although the increase would have to be for all tissues and organs.

Question 27

In his highly classified research laboratory, Dr. Lawrence M. is charged with the task of developing a strain of dog-sized rats to be deployed behind enemy lines. In your opinion, which of the following strategies should Dr. M. pursue to increase the size of rats?

D. Obtain a taxi driver’s license and switch careers.

Answer 27

Maybe a little drastic. Although the ultimate goal is improbable, the research might uncover interesting mechanisms that govern body size and development.

Question 28

What do suppose happens in mutant cells that

A. cannot degrade M-cyclin?

Answer 28

If M-cyclin is not degraded, then the nuclear envelope would likely not reassemble. To reassemble, phosphates have to be removed from pore proteins and lamins. Active M-Cdk would keep phosphorylating any lamins or pore proteins that are dephosphorylated by phosphatases. This would prevent a cell from exiting M-phase and transitioning back to G1.

Question 29

What do suppose happens in mutant cells that

always express high levels of p21?

Answer 29

The cell would be stuck in G1, and unable to transition into S-phase. p21 binds to, and inhibits the G1- and G1/S-Cdks.

Question 30

What do suppose happens in mutant cells that

cannot phosphorylate RB?

Answer 30

The cell would be stuck in G1, and unable to transition into S-phase. Rb (retinoblastoma protein) binds to transcription regulators and thus prevents the expression of many genes needed for S-phase. Rb releases the transcription regulators when it is phosphorylated by G1- and G1/S-Cdks. If it cannot be phosphorylated, then Rb does NOT release the transcription regulators.

Question 31

Describe in general how each compartment gets a distinct set of proteins?
(Include the following in your description: signal sequences, pores or translocators, vesicle transport.)

Answer 31

1) The nucleus gets a set of proteins that each has a nuclear localization signal (NLS). The nuclear proteins are synthesized to completion on ribosomes in the cytosol. The nuclear transport receptor binds to proteins (cargo) with a NLS and binds to a nuclear pore. The receptor with its fully folded cargo pulls itself through the pore. Once inside, a GTP-bound Ran protein binds the receptor causing it to change conformation and release the cargo, so delivering the protein into the nuclues.
2) The cytosol gets its proteins by their complete synthesis on ribosomes in the cytosol, and then the proteins stay in the cytosol because they lack signal sequences.
3) The matrix of the mitochondria gets a set of proteins that each has a signal sequence that is recognized by a receptor protein embedded in the mitochondrial outer membrane. After complete synthesis in the cytosol, the signal sequence binds to the receptor and is then transferred to a protein translocator, which is in two components: one spans the outer and the other the inner membrane. The protein is then unfolded and pulled through the protein translocator. The chloroplast and peroxisimes get their proteins by a similar mechanism.
4) The endoplasmic reticulum (ER) gets a set of proteins that have a signal sequence that is recognized by a protein called the signal-recognition particle (SRP). The SRP binds to a protein receptor embedded in the ER membrane. Binding brings the partially synthesized protein and the ribosome to the ER. Synthesis of the protein is then completed at the ER: synthesis pushes the protein through a translocator and into the lumen of the ER or additional signal sequences cause transfer to stop and results in a transmembrane domain. Multiple start-stop transfer signals results in multiple transmembrane domains.
5) Proteins synthesized into the ER lumen, or into the membrane, remain at the ER if they have ER retention signals, or are packaged into vesicles and delivered to the Golgi apparatus, which is how the golgi gets its proteins. However, proteins moving through the golgi do not necessarily stay. Many proteins are packaged into vesicles at the trans- golgi and delivered to the endosome or to the plasma membrane. The former is how the endosome, and once mature, lysosome, get their proteins, and the latter is how the extracellular matrix gets its proteins.

Question 32

How is import of a protein into the nucleus directional? Does directional import require an input of energy?

Answer 32

Protein import into the nucleus is directional because the nuclear transport receptor binds to and moves through a nuclear pore only after binding cargo. Once inside the nucleus, the receptor releases the cargo only after binding to Ran- GTP (a monomeric G-protein, but NOT involved in signaling).
Binding to RAN-GTP also puts the receptor in a conformation that allows it to move through the pore in the opposite direction: out. The receptor returns to its original conformation only after Ran hydrolyzes the GTP to GDP, which causes Ran to release the receptor. The receptor can then bind another protein. Thus, GTP hydrolysis drives the work of conformation changes in the nuclear transport receptor that result in directional import of the receptor’s cargo proteins.

Question 33

How does a ribosome get located to the endoplasmic reticulum?

Answer 33

A ribosome gets located to the ER if the protein it is translating has an ER signal sequence. As soon as the signal sequence is synthesized, the signal-recognition particle binds to it, and also binds to a receptor embedded in the ER membrane, thus recruiting the attached ribosome to the ER.

Question 34

Describe the mechanism of how a protein gets into the ER lumen compared to how a protein gets embedded in the ER lipid bilayer.

Answer 34

A protein that gets synthesized into the ER lumen has a single, N-terminal signal sequence. The signal sequence binds to the protein translocator and continued synthesis pushes the polypeptide chain through the translocator into the lumen. An enzyme, called a peptidase, then catalyzes hydrolysis that cuts off the signal sequence, releasing the new protein into the lumen.

In contrast, a protein that gets embedded into the lipid bilayer has either NO N-terminal signal sequence, but one or more internal signal sequences, or has a N-terminal signal sequence, in addition to one or more internal signal sequences. Each internal signal sequence in turn binds to the translocator, displacing the previous signal sequence, which, if internal, floats into the lipid bilayer as a transmembrane domain.

Question 35

If a protein with a single transmembrane domain is embedded in the ER lipid bilayer and transported to the plasma membrane, will the N-terminus (+H3N-) or the C-terminus (-COO-) be located in the cytosol? Explain your answer

Answer 35

It depends. If the protein has a N-terminal signal sequence plus an internal signal sequence, then the C-terminus will be located in the cytosol. If the protein only has one internal signal sequence (NO N-terminal signal sequence) then the N-terminus will be in the cytosol.

Question 36

How do vesicles form?

Answer 36

Coat proteins assemble, by binding to adaptins, at a membrane and bend the membrane into a sphere. The vesicle remains continuous with the membrane by a constricted region. The GTP-binding protein, dynamin, then ‘pinches’ the constricted region to release the vesicle. ‘Pinching’ happens by dynamin changing conformation as it binds GTP and then catalyzes hydrolysis of GTP.

Question 37

How do vesicles take the correct cargo?

Answer 37

Proteins, or other cargo, in the lumen of the ER or golgi sacs, or just outside the cell, are recognized by transmembrane proteins referred to as cargo receptors. Cytosolic proteins called adaptins recognize the cargo receptors, as well as other proteins embedded in the membrane, and help gather the cargo to a patch of membrane. The adaptins are then recognized by the coat proteins, which bend the membrane into a vesicle that then contains the gathered cargo.

Question 38

How do vesicles go to the correct membrane?

Answer 38

There is a gene family coding for Rab proteins and another family coding for tethering proteins. A family means multiple genes encoding proteins that share much identity in amino acid sequence, but not complete identity. Some of the differences in amino acids are at binding sites so that each member of the family has different binding selectivity.

Each membrane, ER, cis-golgi, trans-golgi, endosome and the plasma membrane, has a specific set of Rab proteins associated with the membrane’s cytosolic surface. When a vesicle forms, the Rab proteins are gathered onto the surface of the vesicle. Each type of Rab protein binds to a specific tethering protein. Each membrane has its own distinct tethering proteins, so that a given vesicle will only fuse with a membrane that has the matching tethering protein. After fusing with the target membrane, the Rab proteins are recycled back to their original membrane.

Question 39

9. How do vesicles fuse with the target membrane?

Answer 39

A vesicle fuses with a target membrane when its Rab proteins bind to tethering proteins on the target membrane. Vesicles also have a protein called v-SNARE that binds to a protein called t-SNARE that is on target membranes. The binding of SNAREs brings the vesicle in contact with the target membrane so that the lipid bilayers flow together completing fusion.

Question 40

How is it that oligosaccharides covalently bonded to proteins (glycoproteins) are exposed on the outer surface of the plasma membrane and not the cytosolic surface?

Answer 40

Enzymes that catalyze addition of sugars to proteins are located in the lumen of the endoplasmic reticulum and the lumen of the golgi sacs, but not in the cytosol. As transmembrane proteins are transported from ER lumen to the plasma membrane, the lumen facing domains of the proteins are modified by the ER and golgi enzymes. When the vesicle transporting the proteins fuses with the plasma membrane, the lumen face always becomes the outer surface of the plasma membrane.

Question 41

How does the structure of a Golgi apparatus relate to the order of sugar modifications found on glycoproteins?

Answer 41

The Golgi apparatus is organized into a stack of sacs (cisterna) that do not share a continuous lumen. Thus, each sac, from the cis face to the middle sacs to the trans face, can hold different modifying enzymes in each of their lumens. As a glycoprotein first enters the cis face, it can be modified, and then sequentially modified as it passes, by vesicle transport, into and out of each subsequent lumen.

Question 42

Consider a transcription regulator that has a nuclear localization signal.

A. In what cell compartment is the protein synthesized?

Answer 42

Cytosol on free ribosomes.

Question 43

How does the protein get into the nucleus?

Answer 43

The nuclear transport receptor selectively binds to proteins with a nuclear localization signal, and then binds to the nuclear pore and facilitates transport of its cargo, in this case the transcription regulator, into the nucleus.

Question 44

C. Compare the following two mutants: (1) a mutation in the nuclear localization signal of the transcription regulator that prevents it from binding to the nuclear transport receptor, and (2) a mutation in the nuclear transport receptor that prevents it from binding any protein’s nuclear localization signal.

State the consequence for each mutant and which mutant is more likely to disrupt cell function.

Answer 44

In the case of (1), the transcription regulator will remain in the cytosol. The subsequent consequence to the cell depends on what genes the transcription regulator normally turns ‘on’, which in the mutant remain ‘off’ because the regulator cannot enter the nucleus.

In the case of (2), the transcription regulator and all other proteins intended to be in the nucleus remain in the cytosol. No proteins enter the nucleus, at least not efficiently (some may slowly by chance bump their way through a pore).

Mutant (2) is more likely to disrupt function, because the consequence includes that of mutant (1) plus the fact that no other proteins can get in the nucleus. Such a mutation would likely be lethal.

Question 45

The budding of clathrin-coated vesicles from eukaryotic plasma membrane fragments can be observed when adaptins, clathrin, and dynamin-GTP are added to the membrane preparation.

A. What would you observe if you omitted (A) adaptins, (B) clathrin, or (C) dynamin?

Answer 45

Adaptins bind cargo receptors and coat proteins. In the absence of adaptins, the initial forming of a vesicle cannot happen.

In the absence of clathrin (a coat protein), the initial forming of a vesicle does not happen. Coat proteins are needed to deform the membrane into a sphere (vesicle).

Dynamin is needed to ‘pinch’ off a vesicle from the membrane. In the absence of dynamin, vesicles would form but remain attached to the membrane by a narrow, constricted region.

Question 46

Q15-5: The budding of clathrin-coated vesicles from eukaryotic plasma membrane fragments can be observed when adaptins, clathrin, and dynamin-GTP are added to the membrane preparation.

B. What would you observe if the plasma membrane fragments were from a prokaryotic cell?

Answer 46

No vesicles would form because the adaptins selectively bind cargo receptors found in eukaryotic membranes. Prokaryotes lack endocytosis and the ability to form vesicles and do not have any proteins with similarity to cargo receptors, so nothing for adaptins to bind to.

Question 47

Q15-15: Consider a protein that contains an ER signal sequence at its N-terminus and a nuclear localization sequence in its middle. What do you think the fate of this protein would be? Explain your answer

Answer 47

Protein synthesis begins with the N-terminus. As the first part of the protein is synthesized, a signal-recognition particle will bind to the ER signal sequence, and then to the SRP receptor at the ER. The protein will then be synthesized into the ER lumen. Once inside the ER lumen, the protein will not have access to the nucleus, so its nuclear localization sequence serves no function.

Question 48

What would happen to proteins bound for the nucleus if there were insufficient energy to transport them?

Answer 48

The nuclear transport receptor would move some proteins into the nuclues. However, the nuclear transport receptors require binding to Ran-GTP to release their cargo and move back to the cytosol. With insufficient energy, the transport of proteins into the nucleus would stop once all the nuclear transport receptors moved into the nucleus.