Cell Biology Exam 4

mitogen

An extracellular signal molecule that stimulates cell proliferation.

mitosis

Division of the nucleus of a eukaryotic cell.

survival factor

Extracellular signal molecule that must be present to suppress apoptosis.

Sister chromatid separation occurs because __________ are destroyed by the APC/C. A) securins B) cohesins C) kinetochores D) condensins

A) securins

M cyclin

Regulatory protein that binds to mitotic Cdk to form M-Cdk, the protein complex that triggers the M phase of the cell cycle.

anaphase-promoting complex (APC/C)

A protein complex that triggers the separation of sister chromatids and orchestrates the carefully timed destruction of proteins that control progress through the cell cycle; the complex catalyzes the ubiquitylation of its targets.

programmed cell death

A tightly controlled form of cell suicide that allows cells that are unneeded or unwanted to be eliminated from an adult or developing organism; the major form is called apoptosis.

apoptosis

A tightly controlled form of programmed cell death that allows excess cells to be eliminated from an adult or developing organism.

Which of the following statements about the cell cycle is FALSE? A) Once a cell decides to enter the cell cycle, the time from start to finish is the same in all eukaryotic cells. B) An unfavorable environment can cause cells to arrest in G1. C) A cell has more DNA during G2 than it did in G1. D) The cleavage divisions that occur in an early embryo have short G1 and G2 phases.

A) Once a cell decides to enter the cell cycle, the time from start to finish is the same in all eukaryotic cells.

Mitogens are A) extracellular signals that stimulate cell division. B) transcription factors important for cyclin production. C) kinases that cause cells to grow in size. D) produced by mitotic cells to keep nearby neighboring cells from dividing.

A) extracellular signals that stimulate cell division.

Which of the following does not occur during M phase in animal cells? A) growth of the cell B) condensation of chromosomes C) breakdown of nuclear envelope D) attachment of chromosomes to microtubules

A) growth of the cell

Which of the following statements are correct? Explain your answers. A. Cells do not pass from G1 into M phase of the cell cycle unless there are sufficient nutrients to complete an entire cell cycle. B. Apoptosis is mediated by special intracellular proteases, one of which cleaves nuclear lamins. C. Developing neurons compete for limited amounts of survival factors. D. Some vertebrate cell-cycle control proteins function when expressed in yeast cells. E. The enzymatic activity of a Cdk protein is determined both by the presence of a bound cyclin and by the phosphorylation state of the Cdk.

A. False. There is no G1 to M phase transition. The statement is correct, however, for the G1 to S phase transition, in which cells commit themselves to a division cycle. B. True. Apoptosis is an active process carried out by special proteases (caspases). C. True. This mechanism is thought to adjust the number of neurons to the number of specific target cells to which the neurons connect. D. True. An amazing evolutionary conservation! E. True. Association of a Cdk protein with a cyclin is required for its activity (hence its name cyclin-dependent kinase). Furthermore, dephosphorylation at specificsites on the Cdk protein is required for the cyclin-Cdk complex to be active.

If cells are grown in a culture medium containing radioactive thymidine, the thymidine will be covalently incorporated into the cell's DNA during S phase. The radioactive DNA can be detected in the nuclei of individual cells by autoradiography: radioactive cells will activatea photographic emulsion and be labeled by black dots when looked at under a microscope. Consider a simple experiment in which cells are radioactively labeled bythis method for only a short period (about 30 minutes). The radioactive thymidine medium is then replaced with one containing unlabeled thymidine, and the cells are grown for some additional time. At different time points after replacement of the medium, cells are examined in a microscope. The fraction of cells in mitosis (which can be easily recognized because the cells have rounded up and their chromosomes are condensed) that have radioactive DNA in their nuclei is then determined and plotted asa function of time after the labeling with radioactive thymidine (Figure Q18-14). A. Would all cells (including cells at all phases of the cell cycle) be expected to contain radioactive DNA after the labeling procedure? B. Initially, there are no mitotic cells that contain radioactive DNA (see Figure Q18-14). Why is this? C. Explain the rise and fall and then rise again of the curve. D. Estimate the length of the G2 phase from this graph.

A. Only the cells that were in the S phase of their cell cycle (i.e., those cells making DNA) during the 30-minute labeling period contain any radioactive DNA. B. Initially, mitotic cells contain no radioactive DNA because these cells were not engaged in DNA synthesis during the labeling period. Indeed, it takes about two hours before the first labeled mitotic cells appear. C. The initial rise of the curve corresponds to cells that were just finishing DNA replication when the radioactive thymidine was added. The curve rises as more labeled cells enter mitosis; the peak corresponds to those cells that had just started S phase when the radioactive thymidine was added. The labeled cells then exit from mitosis, and are replaced by unlabeled mitotic cells, which were not yet in S phase during the labeling period. After 20 hours, the curve starts rising again, because the labeled cells enter their second round of mitosis. D. The initial two-hour lag before any labeled mitoticcells appear corresponds to the G2 phase, which is the time between the end of S phase and the beginning of mitosis. The first labeled cells seen in mitosis were those that were just finishing S phase (DNA synthesis) when the radioactive thymidine was added.

One important biological effect of a large dose of ionizing radiation is to halt cell division. A. How does this occur? B. What happens if a cell has a mutation that prevents it from halting cell division after being irradiated? C. What might be the effects of such a mutation if the cell is not irradiated? D. An adult human who has reached maturity will die within a few days of receiving a radiation dose large enough to stop cell division. What does that tell you (other than that one should avoid large doses of radiation)?

A. Radiation leads to DNA damage, which activates a regulatory mechanism (mediated by p53 and p21; see Figure 18−15) that arrests the cell cycle until the DNA has been repaired. B. The cell will replicate damaged DNA and thereby introduce mutations in the daughter cells when the cell divides. C. The cell will be able to divide normally, but it will be prone to mutations, because some DNA damage always occurs as the result of natural irradiation caused, for example, by cosmic rays. The mechanism mediated by p53 is mainly required as a safeguard against the devastating effects of accumulating DNA damage; this mechanism is not required for the natural progression of the cell cycle in undamaged cells. D. Cell division in humans is an ongoing process that does not cease upon reaching maturity, and it is requiredfor survival. Blood cells and epithelial cells in the skinor lining the gut, for example, are being constantly produced by cell division to meet the body's needs; each day, your body produces about 1011 new red blood cells alone.

Are the statements below TRUE or FALSE? Explain your answer. A. Statement 1: Generally, in a given organism, the S, G2, and M phases of the cell cycle take a defined and stereotyped amount of time in most cells. B. Statement 2: Therefore, the cell-cycle control system operates primarily by a timing mechanism, in which the entry into one phase starts a timer set for sufficient time to complete the required tasks. After a given amount of time has elapsed, a molecular "alarm" triggers movement to the next phase.

A. True B. False, process do not occur but time but completeness.

Which of the following statements are correct? Explain your answers. A. Centrosomes are replicated before M phase begins. B. Two sister chromatids arise by replication of the DNA of the same chromosome and remain paired as they line up on the metaphase plate. C. Interpolar microtubules attach end-to-end and are therefore continuous from one spindle pole to the other. D. Microtubule polymerization and depolymerization and microtubule motor proteins are all required for DNA replication. E. Microtubules nucleate at the centromeres and then connect to the kinetochores, which are structures at the centrosome regions of chromosomes.

A. True. Centrosomes replicate during interphase, before M phase begins. B. True. Sister chromatids separate completely only at the start of anaphase. C. False. The ends of interpolar microtubules overlap and attach to one another via proteins (including motor proteins) that bridge between the microtubules. D. False. Microtubules and their motor proteins play no role in DNA replication. E. False. To be a correct statement, the terms "centromere" and "centrosome" must be switched.

What do you suppose happens in mutant cells that cannot degrade M cyclin? always express high levels of p21? cannot phosphorylate Rb?

All three types of mutant cells would be unable to divide. The cells: A. would enter mitosis but would not be able to exit mitosis. B. would arrest permanently in G1 because the cyclin-Cdk complexes that act in G1 would be inactivated. C. would not be able to activate the transcription of genes required for cell division because the required transcription regulators would be constantly inhibited by unphosphorylated Rb.

Compare the changes in the mitotic spindle that underlie chromosome segregation during anaphase A and anaphase B, and delineate the driving forces responsible for each process.

Anaphase begins abruptly with the breakage of the cohesin linkages that hold together the sister chromatids in a duplicated chromosome. In anaphase A, the kinetochore microtubules shorten and the attached chromosomes move poleward. In anaphase B, the spindle poles themselves move apart, further segregating the two sets of chromosomes. The driving force for the movements of anaphase A is thought to be provided mainly by the loss of tubulin subunits from both ends of the kinetochore microtubules. The driving forces in anaphase B are thought to be provided by two sets of motor proteins—members of the kinesin and dynein families—operating on different types of spindle microtubules.

An antibody that binds to myosin prevents the movement of myosin molecules along actin filaments (the interaction between actin and myosin is described in Chapter 17). How do you suppose the antibody exerts this effect? What might be the result of injecting this antibody into cells (A) on the movement of chromosomes at anaphase or (B) on cytokinesis? Explain your answers.

Antibodies bind tightly to the antigen(in this case myosin) to which they were raised. When bound, an antibody can interfere with the function of the antigen by preventing it from interacting properly with other cell components. (A) The movement of chromosomes at anaphase depends on microtubules and their motor proteins and does not depend on actin or myosin. Injection of an anti-myosin antibody into a cell will therefore have no effect on chromosome movement during anaphase. (B) Cytokinesis, on the other hand, depends on the assembly and contraction of a ring of actin and myosin filaments, which forms the cleavage furrow that splits the cell in two. Injection of an anti-myosin antibody will therefore block cytokinesis.

mitotic spindle

Array of microtubules and associated molecules that forms between the opposite poles of a eukaryotic cell during mitosis and pulls duplicated chromosome sets apart.

Why do you think apoptosisoccurs 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?

As apoptosis occurs on a large scale in both developing and adult tissues, it must not triggeralarm reactions that are normally associated with cell injury. Tissue injury, for example, leads to the release of signal molecules that stimulate the proliferation of surrounding cells so that the wound heals. It also causes the release of signals that can cause a destructive inflammatory reaction. Moreover, the release of intracellular contents could elicit an immune response against molecules that are normally not encountered by the immune system. Such reactions would be self-defeating if they occurred in response to the massive cell death that occurs in normal development.

A mutant yeast strain stops proliferating when shifted from 25°C to 37°C. When these cells are analyzed at the two different temperatures, using a machine that sorts cells according to the amount of DNA they contain, the graphs in Figure 1 are obtained. Which of the following would NOT explain the results with the mutant? A) inability to initiate DNA replication B) inability to begin M phase C) inability to activate proteins needed to enter S phase D) inappropriate production of a signal that causes the cells to remain in G1

B) inability to begin M phase

Progression through the cell cycle requires a cyclin to bind to!a Cdk because A) the cyclins are the molecules with the enzymatic activity in the complex. B) the binding of. a cyclin to Cdk is required for Cdk enzymatic activity. C) cyclin binding inhibits Cdk activity until the appropriate time in the cell cycle. D) without cyclin binding, a cell-cycle checkpoint will be activated.

B) the binding of. a cyclin to Cdk is required for Cdk enzymatic activity.

1. Consider the following statement: "All present-day cells have arisen by an uninterrupted series of cell divisions extending back in time to the first cell division." Is this strictly true?

Because all cells arise by division of another cell, this statement is correct, assuming that "first cell division" refers to the division of the successful founder cell from which all life as we know it has derived. There were probably many other unsuccessful attempts to start the chain of life.

Roughly, how long would it take a single fertilized human egg to make a cluster of cells weighing 70 kg by repeated divisions, if each cell weighs 1 nanogram just after cell division and each cell cycle takes 24 hours? Why does it take very much longer than this to make a 70 kg adult human?

Because the cell population is increasing exponentially, doubling its weight at every cell division, the weight of the cell cluster after N cell divisions is2N × 10-9 g. Therefore, 70 kg (70 × 103 g) = 2N × 10-9 g, or 2N = 7 × 1013. Taking the logarithm of both sides allows you to solve the equation for N. Therefore,N = ln (7 × 1013) / ln 2 = 46; that is it would take only46 days if cells proliferated exponentially. Cell division in animals is tightly controlled, however, and most cells inthe human body stop dividing when they become highly specialized. The example demonstrates that exponential cell proliferation occurs only for very brief periods, even during embryonic development.

A small amount of cytoplasm isolated from a mitotic cell is injected into an unfertilized frog oocyte, causing the oocyte toenter M phase (see Figure 18−7A). A sample of the injected oocyte's cytoplasm is then taken and injected into a second oocyte, causing this cell also to enter M phase. The process is repeated many times until, essentially, none of the original protein sample remains, and yet, cytoplasm taken from the last in the series of injected oocytes is still able to trigger entry into M phase with undiminished efficiency. Explain this remarkable observation.

Before injection, the frog oocytes must contain inactive M-Cdk. Upon injection of the M-phase cytoplasm, the small amount of the active M-Cdk inthe injected cytoplasm activates the inactive M-Cdk by switching on the activating phosphatase (Cdc25), which removes the inhibitory phosphate groups from the inactive M-Cdk (see Figure 18−17). An extract of the second oocyte, now in M phase itself, will therefore contain as much active M-Cdk as the original cytoplasmic extract, and so on.

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?

Both sister chromatids could end up in the same daughter cell for any of a number of reasons. (1) If the microtubules or their connections with a kinetochore were to break during anaphase, both sister chromatids could be drawn to the same pole, and hence into the same daughter cell. (2) If microtubules from the same spindle pole attached to both kinetochores, the chromosome would be pulled to the same pole. (3) If the cohesins that link sister chromatids were not degraded, the pair of chromatids might be pulled to the same pole. (4) If a duplicated chromosome never engaged microtubules and was left out of the spindle, it would also end up in one daughter cell. Some of these errors in the mitotic process would be expected to activate a checkpoint mechanism that delays the onset of anaphase until all chromosomes are attached properly to both poles of the spindle. This "spindle assembly checkpoint" mechanism should allow most chromosome-attachment errors to be corrected, which is one reason why such errors are rare. The consequences of both sister chromatids ending up in one daughter cell are usually dire. One daughter cell would contain only one copy of all the genes carried on that chromosome and the other daughter cell would contain three copies. The altered gene dosage, leading to correspondingly changed amounts of the mRNAs and proteins produced, is often detrimental to the cell. In addition, there is the possibility that the single copy of the chromosome may contain a defective gene with a critical function, which would normally be taken care of by the good copy of the gene on the other chromosome that is now missing.

You engineer yeast cells that express the M cyclin during S phase by replacing the gene regulatory sequences of the M cyclin gene with! those of the S cyclin gene. Keeping in mind that yeast cells have one common Cdk that binds to all cyclins, which of the following outcomes is LEAST likely during this experiment? A) There will be both M cyclin-Cdk and S cyclin-Cdk complexes in the cell during S phase. B) Some substrates that are normally phosphorylated in M phase will now be phosphorylated in S phase. C) G1 cyclin-Cdks will be activated earlier in G1. D) S-Cdk targets will be phosphorylated during S phase.

C) G1 cyclin-Cdks will be activated earlier in G1.

How does S-Cdk help guarantee that replication occurs only once during each cell cycle? A) It blocks the rise of Cdc6 concentrations early in G1. B) It phosphorylates and inactivates DNA helicase. C) It phosphorylates and inactivates Cdc6. D) It promotes the assembly of a prereplicative complex.

C) It phosphorylates and inactivates Cdc6.

At the end of DNA replication, the sister chromatids are held together by the A) kinetochores. B) securins. C) cohesins. D)histones.

C) cohesins.

A cell that is terminally differentiated will A) replicate its DNA. B) reenter the cell cycle only once a year. C) dismantle the cell-cycle control system. D) arrest after S phase.

C) dismantle the cell-cycle control system.

The concentration of mitotic cyclin (M cyclin) A) rises markedly during M phase. B) is activated by phosphorylation. C) falls toward the end of M phase as a result of ubiquitylation and degradation. D) is highest in G1 phase.

C) falls toward the end of M phase as a result of ubiquitylation and degradation.

What is the main molecular difference between cells in a G0 state and cells that have simply paused in G1?

Cells at G0 are perfectly healthy and can continue to the next phase when needed, for example liver cells usually stay at G0 and if a part of the liver was cut off they would leave G) and start creating more liver cells, Cells at G1 are paused for other factors and they need to be fixed before continuing the cycle.

Compare the rules of cell behavior in an animal with the rules that govern human behavior in society. What would happen to an animal if its cells behaved as people normally behave in our society? Could the rules that govern cell behavior be applied to human societies?

Cells in an animal must behave for the good of the organism as a whole—to a much greaterextent than people generally act for the good of society as a whole. In the context of an organism, unsocial behavior would lead to a loss of organization and possibly tocancer. Many of the rules that cells have to obey would be unacceptable in a human society. Most people, for example, would be reluctant to kill themselves for the good of society, yet our cells do it all the time.

A population of proliferating cellsis stained with a dye that becomes fluorescent when it binds to DNA, so that the amount of fluorescence is directly proportional to the amount of DNA in each cell. To measurethe amount of DNA in each cell,the cells are then passed througha flow cytometer, an instrumentthat measures the amount of fluorescence in individual cells. The number of cells with a given DNA content is plotted on the graph below. Indicate on the graph where you would expect to find cells that are in G1, S, G2, and mitosis. Which is the longest phase of the cell cycle in this population of cells?

Cells in peak B contain twice as much DNA as those in peak A, indicating that they contain replicated DNA, whereas the cells in peak A contain unreplicated DNA. Peak A therefore contains cells that are in G1, and peak B contains cells that are in G2 and mitosis. Cells in S phase have begun but not finished DNA synthesis; they therefore have various intermediate amounts of DNA and are found in the region between the two peaks. Most cells are in G1, indicating that it is the longest phase of the cell cycle (see Figure 18−2).

spindle pole

Centrosome from which microtubules radiate to form the mitotic spindle.

sister chromatid

Copy of a chromosome, produced by DNA replication, that remains bound to the other copy.

Recall when cytokinesis takes place with respect to mitosis.

Cytokinesis, the process by which the cytoplasm is cleaved in two, com- pletes M phase. It usually begins in anaphase but is not completed until after the two daughter nuclei have re-formed in telophase. Whereas mitosis depends on a transient microtubule-based structure, the mitotic spindle, cytokinesis in animal cells depends on a transient structure based on actin and myosin filaments, the contractile ring. Both the plane of cleavage and the timing of cytokinesis, however, are determined by the mitotic spindle.

What would be the most obvious outcome of repeated cell cycles consisting of S phase and M phase only? A) The cells would not be able to replicate their DNA. B) The mitotic spindle could not assemble. C) The cells would get larger and larger. D) The cells produced would get smaller and smaller.

D) The cells produced would get smaller and smaller.

The G1 DNA damage checkpoint A) causes cells to proceed through S phase more quickly. B) involves the degradation of p53. C) is activated by errors caused during DNA replication. D) involves the inhibition of cyclin-Cdk complexes by p21.

D) involves the inhibition of cyclin-Cdk complexes by p21.

Cdk (cyclin-dependent protein kinase)

Enzyme that, when complexed with a regulatory cyclin protein, can trigger various events in the cell-division cycle by phosphorylating specific target proteins.

growth factor

Extracellular signal molecule that stimulates a cell to increase in size and mass. Examples include epidermal growth factor (EGF) and platelet-derived growth factor (PDGF).

telophase

Final stage of mitosis in which the two sets of separated chromosomes decondense and become enclosed by a nuclear envelope.

prophase

First stage of mitosis, during which the duplicated chromosomes condense and the mitotic spindle forms

Why do you suppose cells have evolved a special G0 phase to exit from the cell cycle, rather than just stopping in G1 and not moving on to S phase?

For multicellular organisms, the control of cell division is extremely important. Individual cells must not proliferate unless it is to the benefit of the whole organism. The G0 state offers protection from aberrant activation of cell division because the cell-cycle control system is largely dismantled. If, on the other hand, a cell just paused in G1, it would still contain all of the cell-cycle control system and could readily be induced to divide. The cell would also have to remake the "decision" not to divide almost continuously. To re-enter the cell cycle from G0, a cell has to resynthesize all of the components that have disappeared.

G1 phase

Gap 1 phase of the eukaryotic cell cycle; falls between the end of cytokinesis and the start of DNA synthesis.

G2 phase

Gap 2 phase of the eukaryotic cell cycle; falls between the end of DNA synthesis and the beginning of mitosis.

The lifetime of a microtubule in a mammalian cell, between its formation by polymerization and its spontaneous disappearance by depolymerization, varies with the stage of the cell cycle. For an actively proliferating cell, the average lifetime is 5 minutes in interphase and 15 seconds in mitosis. If the average length of a microtubule in interphase is 20 μm, how long will it be during mitosis, assuming that the rates of microtubule elongation due to the addition of tubulin subunits in the two phases are the same?

If the growth rate of microtubules is the same in mitotic and in interphase cells, their length is proportional to their lifetime. Thus, the average length of microtubules in mitosis is 1 μm (= 20 μm × 15 s/300 s).

phragmoplast

In a dividing plant cell, structure made of microtubules and membrane vesicles that guides the formation of a new cell wall

The Golgi apparatus is thought to be partitioned into the daughter cells at cell division by a random distribution of fragments that are created at mitosis. Explain why random partitioning of chromosomes would not work.

In a eukaryotic organism, the genetic information that the organism needs to survive and reproduce is distributed between multiple chromosomes. It is therefore crucial that each daughter cell receives a copy of each chromosome when a cell divides; if a daughter cell receives too few or too many chromosomes, the effects are usually deleterious or even lethal. Only two copies of each chromosome are produced by chromosome replication in mitosis. If the cell were to randomly distribute the chromosomes when it divided, it would be very unlikelythat each daughter cell would receive precisely one copyof each chromosome. In contrast, the Golgi apparatus fragments into tiny vesicles that are all alike, and by random distribution it is very likely that each daughter cell will receive an approximately equal number of them.

Liver cells proliferate excessively both in patients with chronic alcoholism and in patients with liver cancer. What are the differences in the mechanisms by which cell proliferation is induced in these diseases?

In alcoholism, liver cells proliferate because the organ is overburdened and becomes damaged by the large amounts of alcohol that have to be metabolized. This need for more liver cells activates the control mechanisms that normally regulate cell proliferation. Unless badly damaged and full of scar tissue, the liver will usually shrink back to a normal size after the patient stops drinking excessively. In liver cancer, in contrast, mutations abolish normal cell proliferation control and, as a result, cells divide and keep on dividing in an uncontrolled manner, which is usually fatal.

Explain how Rb blocks cell proliferation and how mitogens reverse this inhibition.

In the absence of mitogens, dephosphorylated Rb protein holds specific transcription regulators in an inactive state. Mitogens binding to cell-surface receptors activate intracellular signaling pathways that lead to the formation and activation of G1-Cdk and G1/S-Cdk complexes. These complexes phosphorylate, and thereby inactivate, the Rb protein, releasing the transcription regulators needed to activate the transcription of genes required for entry into S phase. Rb is abundant in the nuclei of all vertebrate cells, where it binds to par- ticular transcription regulators and prevents them from turning on the genes required for cell proliferation. Mitogens release the Rb brake by triggering the activation of G1-Cdks and G1/S-Cdks. These complexes phosphorylate the Rb protein, altering its conformation so that it releases its bound transcription regulators, which are then free to activate the genes required for entry S phase.

The polar movement of chromosomes during anaphaseA is associated with microtubule shortening. In particular, microtubules depolymerize at the ends at which they are attached to the kinetochores. Sketch a model that explains how a microtubule can shorten and generate force yet remain firmly attached to the chromosome.

It is still not certain what drives the poleward movement of chromosomes during anaphase.In principle, two possible models could explain it(Figure A18-21). In the model shown in (A), microtubule motor proteins associated with the kinetochore dash toward the minus end of the depolymerizing microtubule, dragging the chromosome toward the pole. Although this model is appealingly simple, there is little evidence that motor proteins are required for chromosome movement during anaphase. Instead, current experimental evidence greatly supports the model outlined in (B). In this model, chromosome movement is driven by kinetochore proteins that cling to the sides of the depolymerizing microtubule (see Figure 18-23). These proteins frequently detach from— and reattach to—the kinetochore microtubule. As tubulin subunits continue to dissociate, the kinetochore must slide poleward to maintain its grip on the retreating end of the shrinking microtubule.

M-Cdk

Protein complex that triggers the M phase of the cell cycle; consists of an M cyclin plus a mitotic cyclin-dependent protein kinase (Cdk).

interphase

Long period of the cell cycle between one mitosis and the next. Includes G1 phase, S phase, and G2 phase.

One of the functions of M-Cdk is to cause a precipitous drop in M cyclin concentration halfway through M phase. Describe the consequences of this sudden decrease and suggest possible mechanisms by which it might occur.

Loss of M cyclin leads to inactivation of M-Cdk. As a result, the M-Cdk target proteins become dephosphorylated by phosphatases, and the cells exit from mitosis: they disassemble the mitotic spindle, reassemble the nuclear envelope, decondense their chromosomes, and so on. The M cyclin is degraded by ubiquitin-dependent destruction in proteasomes, and the activation of M-Cdk leads to the activation of APC/C, which ubiquitylates the cyclin, but with a substantial delay. As discussed in Chapter 7, ubiquitylation tags proteins for degradation in proteasomes.

Figure 18−5 shows the rise of M cyclin concentrationand the rise of M-Cdk activity in cells as they progress through the cell cycle. It is remarkable that the M cyclin concentration rises slowly and steadily, whereas M-Cdk activity increases suddenly. How do you think this difference arises?

M cyclin accumulates gradually as it is steadily synthesized. As it accumulates, it will tend to form complexes with the mitotic Cdk molecules that are present. The Cdk in these complexes is inhibited by phosphorylation (see Figure 18-10). After a certain threshold level has been reached, M-Cdk is activated by the phosphatase Cdc25. Once activated, M-Cdk acts to enhance the activity of the activating phosphatase; this positive feedback leads to the complete activation of M-Cdk (see Figure 18−17). Thus,M cyclin accumulation acts like a slow-burning fuse, which eventually helps trigger the explosive self-activation of M-Cdk. The precipitous destruction of M cyclin terminates M-Cdk activity, and a new round of M cyclin accumulation begins.

centrosome

Microtubule-organizing center that sits near the nucleus in an animal cell; during the cell cycle, this structure duplicates to form the two poles of the mitotic spindle.

cell-cycle control system

Network of regulatory proteins that govern the orderly progression of a eukaryotic cell through the stages of cell division.

caspase

One of a family of proteases that, when activated, mediates the destruction of the cell by apoptosis.

S phase

Period during a eukaryotic cell cycle in which DNA is synthesized.

M phase

Period of the eukaryotic cell cycle during which the nucleus and cytoplasm divide.

chromosome condensation

Process by which a duplicated chromosome becomes packed into a more compact structure prior to cell division.

centrosome cycle

Process by which the centrosome duplicates (during interphase) and the two new centrosomes separate (at the beginning of mitosis) to form the poles of the mitotic spindle.

cytokinesis

Process by which the cytoplasm of a plant or animal cell divides in two to form individual daughter cells.

kinetochore

Protein complex that assembles on the centromere of a condensed mitotic chromosome; the site to which spindle microtubules attach.

condensin

Protein complex that helps configure duplicated chromosomes for segregation by making them more compact.

cohesin

Protein complex that holds sister chromatids together after DNA has been replicated in the cell cycle.

G1-Cdk

Protein complex whose activity drives the cell through the first gap phase of the cell cycle; consists of a G1 cyclin plus a cyclin-dependent protein kinase (Cdk).

S-Cdk

Protein complex whose activity initiates DNA replication; consists of an S cyclin plus a cyclin-dependent protein kinase (Cdk).

G1/S-Cdk

Protein complex whose activity triggers entry into S phase of the cell cycle; consists of a G1/S cyclin plus a cyclin- dependent protein kinase (Cdk).

Cdk inhibitor proteins (CKIs)

Protein that binds to and inhibits cyclin-Cdk complexes, primarily involved in the control of G1 and S phases.

Consider the events that lead to the formation of the new nucleus at telophase. How do nuclear and cytosolic proteins become properly re-sorted so that the new nucleus contains nuclear proteins but not cytosolic proteins?

Recall from Figure 18−30 that the new nuclear envelope reassembles on the surface of the chromosomes. The close apposition of the envelope to the chromosomes prevents cytosolic proteins from being trapped between the chromosomes and the envelope. Nuclear proteins are then selectively imported through the nuclear pores, causing the nucleus to expand while maintaining its characteristic protein composition.

G1 cyclin

Regulatory protein that helps drive a cell through the first gap phase of the cell cycle and toward S phase.

G1/S cyclin

Regulatory protein that helps to launch the S phase of the cell cycle.

S cyclin

Regulatory protein that helps to launch the S phase of the cell cycle.

cyclin

Regulatory protein whose concentration rises and falls at specific times during the eukaryotic cell cycle; cyclins help control progression from one stage of the cell cycle to the next by binding to cyclin-dependent protein kinases (Cdks).

Bcl2 family

Related group of intracellular proteins that regulates apoptosis; some family members promote cell death, others inhibit it.

anaphase

Stage of mitosis during which the two sets of chromosomes separate and are pulled toward opposite ends of the dividing cell.

metaphase

Stage of mitosis in which chromosomes are properly attached to the mitotic spindle at its equator but have not yet segregated toward opposite poles.

prometaphase

Stage of mitosis in which the nuclear envelope breaks down and duplicated chromosomes are captured by the spindle microtubules; precedes metaphase.

aster

Star-shaped array of microtubules emanating from a centrosome or from a pole of a mitotic spindle.

contractile ring

Structure made of actin and myosin filaments that forms a belt around a dividing cell, pinching it in two.

What might be the consequences if a cell replicated damaged DNA before repairing it?

The cell would replicate its damaged DNA and therefore would introduce mutations to the two daughter cells when the cell divides. Such mutations could increase the chances that the progeny of the affected daughter cells would eventually become cancer cells.

The shortest eukaryotic cell cycles of all—shorter eventhan those of many bacteria—occur in many early animal embryos. These so-called cleavage divisions take place without any significant increase in the weight of the embryo. How can this be? Which phase of the cell cycle would you expect to be most reduced?

The egg cells of many animals are bigand contain stores of enough cell components to last for many cell divisions. The daughter cells that form duringthe first cell divisions after fertilization are progressively smaller in size and thus can be formed without a need for new protein or RNA synthesis. Whereas normally dividing cells would grow continuously in G1, G2, and S phases, until their size doubled, there is no cell growth in these early cleavage divisions, and both G1 and G2 are virtually absent. As G1 is usually longer than G2 and S phase, G1 is the most drastically reduced phase in these divisions.

If fine glass needles are used to manipulate a chromosome inside a living cell during early M phase, it is possible to trick the kinetochores on the two sister chromatids into attaching to the same spindle pole. This arrangement is normally unstable, but the attachments can be stabilized if the needle is used to gently pull the chromosome so that the microtubules attached to both kinetochores (via the same spindle pole) are under tension. What does this suggest to you about the mechanism by which kinetochores normally become attached and stay attached to microtubules from opposite spindle poles? Is the finding consistent with the possibility that a kinetochore is programmed to attach to microtubules from a particular spindle pole? Explain your answers.

The experiment shows that kinetochores are not preassigned to one or other spindle pole; microtubules attach to the kinetochores that they are able to reach. For the chromosome to remain attached to a microtubule, however, tension has to be exerted. Tension is normally achieved by the opposing pulling forces from opposite spindle poles. The requirement for such tension ensures that if two sister kinetochores ever become attached to the same spindle pole, so that tension is not generated, one or both of the connections would be lost, and microtubules from the opposing spindle pole would have another chance to attach properly.

Describe how the synthesis and destruction of cyclins regulate progression from one phase of the cell cycle to the next.

The gradual increase in cyclin concentration stems from continued transcription of cyclin genes and synthesis of cyclin proteins, whereas the rapid fall in cyclin concentration is precipitated by a full- scale targeted destruction of the protein. At the G1-to-S transition, it uses Cdk inhibitors to keep cells from entering S phase and replicating their DNA. At the G2-to-M transition, it suppresses the activation of M-Cdk by inhibiting the phosphatase required to activate the Cdk. And it can delay the exit from mitosis by inhibiting the activation of APC/C, thus preventing the degradation of M cyclin.

Draw a detailed view of the formation of the new cell wallthat separates the two daughter cells when a plant cell divides (see Figure 18−35). In particular, show where the membrane proteins of the Golgi-derived vesicles endup, indicating what happens to the part of a protein in the Golgi vesicle membrane that is exposed to the interior of the Golgi vesicle. (Refer to Chapter 11 if you need a reminder of membrane structure.)

The membranes of the Golgi vesicles fuse to form part of the plasma membranes of the two daughter cells. The interiors of the vesicles, which are filled with cell wall material, become the new cell wall matrix separating the two daughter cells. Proteins in the membranes of the Golgi vesicles thus become plasma membrane proteins. Those parts of the proteins that were exposed to the lumen of the Golgi vesicle will end up exposed to the new cell wal

Sketch the principal stages of mitosis, using Panel 18−1 (pp. 628-629) as a guide. Color one sister chromatid and follow it through mitosis and cytokinesis. What event commits this chromatid to a particular daughter cell? Once initially committed, can its fate be reversed? What may influence this commitment?

The sister chromatid becomes committed when a microtubule from one of the spindle poles attaches to the kinetochore of the chromatid. Microtubule attachment is still reversible until a second microtubule from the other spindle pole attaches to the kinetochore of its partner sister chromatid, so that the duplicated chromosome is now put under mechanical tension by pulling forces from both poles. The tension ensures that both microtubules remain attached to the chromosome. The position of a chromatid in the cell at the time that the nuclear envelope breaks down will influence which spindle pole it will be pulled to, as its kinetochore is most likely to become attached to the spindle pole toward which it is facing.

Outline how apoptosis is regulated and initiated by the Bcl2 family of proteins.

The molecular machinery responsible for apoptosis, which seems to be similar in most animal cells, involves a family of proteases called caspases. The main proteins that regulate the activation of caspases are members of the Bcl2 family of intracellular proteins. Some members of this pro- tein family promote caspase activation and cell death, whereas others inhibit these processes. Two of the most important death-inducing family members are proteins called Bax and Bak. These proteins—which are acti- vated in response to DNA damage or other insults—promote cell death by inducing the release of the electron-transport protein cytochrome c from mitochondria into the cytosol. Other members of the Bcl2 family (including Bcl2 itself) inhibit apoptosis by preventing Bax and Bak from releasing cytochrome c. The balance between the activities of pro-apop- totic and anti-apoptotic members of the Bcl2 family largely determines whether a cell lives or dies by apoptosis. The cytochrome c molecules released from mitochondria activate initia- tor procaspases—and induce cell death—by promoting the assembly of a large, seven-armed, pinwheel-like protein complex called an apopto- some. The apoptosome then recruits and activates a particular initiator procaspase, which then triggers a caspase cascade that leads to apopto- sis

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? Block all apoptosis. Block p53 function. Overproduce growth factors, mitogens, or survival factors. Explain the likely consequences of each option.

The most likely approach to success (if that is what the goal should be called) is plan C, which should result in an increase in cell numbers. The problem is, of course, that cell numbers of each tissue must be increased similarly to maintain balanced proportions in the organism, yet different cells respond to different growth factors. As shown in Figure A18-28, however, the approach has indeed met with limited success. A mouse producing very large quantities of growth hormone (left)—which acts to stimulate the production of a secreted protein that acts as a survival factor, growth factor, or mitogen, depending on the cell type—grows to almost twice the size of a normal mouse (right). To achieve this twofold change in size, however, growth hormone was massively overproduced (about fiftyfold). And note that the mouse did not even attain the size of a rat, let alone a dog. The other two approaches have conceptual problems: A. Blocking all apoptosis would lead to defects in development, as rat development requires the selective death of many cells. It is unlikely that a viable animal would be obtained. B. Blocking p53 function would eliminate an important mechanism in the cell cycle that detects DNA damage and stops the cycle so that the cell can repair the damage; removing p53 would increase mutation rates and lead to cancer. Indeed, mice without p53 usually develop normally but die of cancer at a young age.

PDGF is encoded by a gene that can cause cancer when expressed inappropriately. Why do cancers not arise at wounds in which PDGF is released from platelets?

The on-demand, limited release of PDGF at a wound site triggers cell division of neighboring cells for a limited amount of time, until the PDGF is degraded. This is different from the continuous release of PDGF from mutant cells, where PDGF is made in an uncontrolled way at high levels. Moreover, the mutant cells that make PDGF often express their own PDGF receptor inappropriately, so that they can stimulate their own proliferation, thereby promoting the development of cancer.

What is the order in which the following events occur during cell division? A. anaphase B. metaphase C. prometaphase D. telophase E. mitosis F. prophase Where does cytokinesis fit in?

The order is F, C, B, A, D. Together, these five steps are referred to as mitosis (E). Cytokinesis is the last step in M phase, which overlaps with anaphase and telophase. Mitosis and cytokinesis are both part of M phase.

cell cycle

The orderly sequence of events by which a cell duplicates its contents and divides into two.

The balance between plus-end directed and minus-end directed motor proteins that bind to interpolar microtubules in the overlap region of the mitotic spindle is thought to help determine the length of the spindle. How might each type of motor protein contribute to the determination of spindle length?

The overlapping interpolar microtubules from opposite polesof the spindle have their plus ends pointing in opposite directions. Plus-end directed motor proteins cross-link adjacent, antiparallel microtubules together and tend to move the microtubules in the direction that will push the two poles of the spindle apart, as shown in the figure. Minus-end directed motor proteins also cross-link adjacent, antiparallel microtubules together but move in the opposite direction, tending to pull the spindle poles together (not shown).

Look carefully at the electron micrographs in Figure 18−38. Describe the differences between the cell that died by necrosis and those that died by apoptosis. How do the pictures confirm the differences between the two processes? Explain your answer.

The plasma membrane of the cell that died by necrosis in Figure 18−38A is ruptured; a clear break is visible, for example, at a position corresponding to the12 o'clock mark on a watch. The cell's contents, mostly membranous and cytoskeletal debris, are seen spilling into the surroundings through these breaks. The cytosol stains lightly, because most soluble cell components were lost before the cell was fixed. In contrast, the cell that underwent apoptosis in Figure 18−38B is surrounded byan intact membrane, and its cytosol is densely stained, indicating a normal concentration of cell components. The cell's interior is remarkably different from a normal cell, however. Particularly characteristic are the large "blobs" that extrude from the nucleus, probably as the result of the breakdown of the nuclear lamina. The cytosol also contains many large, round, membrane-enclosed vesicles of unknown origin, which are not normally seen in healthy cells. The pictures visually confirm the notion that necrosis involves cell lysis, whereas cells undergoing apoptosis remain relatively intact until they are phagocytosed and digested by another cell.

bi-orientation

The symmetrical attachment of a sister-chromatid pair on the mitotic spindle, such that one chromatid in the duplicated chromosome is attached to one spindle pole and the other is attached to the opposite pole.

Summarize the function of the cell-cycle control system and describe the transition points at which progression through the cycle is regulated.

To ensure that they replicate all their DNA and organelles, and divide in an orderly manner, eukaryotic cells possess a complex network of regulatory proteins known as the cell-cycle control system. This system guarantees that the events of the cell cycle—DNA replication, mitosis, and so on—occur in a set sequence and that each process has been com- pleted before the next one begins. The cell-cycle control system achieves all of this by employing a set of molecular brakes, sometimes called checkpoints, to pause the cycle at certain transition points. In this way, the control sys- tem does not trigger the next step in the cycle unless the cell is properly prepared. The cell-cycle control system regulates progression through the cell cycle at three main transition points: At the transition from G1 to S phase, the control system confirms that the environment is favorable for proliferation before committing to DNA replication. At the transition from G2 to M phase, the control system confirms that the DNA is undamaged and fully replicated, ensuring that the cell does not enter mitosis unless its DNA is intact. Finally, during mitosis, the cell-cycle control machinery ensures that the duplicated chromosomes are properly attached to a cytoskeletal machine, called the mitotic spindle, before the spindle pulls the chromosomes apart and segregates them into the two daughter cells.

p53

Transcription regulator that controls the cell's response to DNA damage, preventing the cell from entering S phase until the damage has been repaired or inducing the cell to commit suicide if the damage is too extensive; mutations in the gene encoding this protein are found in many human cancers.