Chapter_17_Solutions - Chapter 17 The Cell Cycle OVERVIEW...

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Unformatted text preview: Chapter 17 The Cell Cycle OVERVIEW OF THE CELL CYCLE DEFINITIONS 17–1 17–2 17–3 17–4 17–5 Interphase Cell cycle Start or restriction point G1 Budding yeast 17 In This Chapter OVERVIEW OF THE CELL CYCLE THE CELL-CYCLE CONTROL SYSTEM S PHASE MITOSIS CYTOKINESIS CONTROL OF CELL DIVISION AND CELL GROWTH A397 A402 A405 A407 A420 A424 TRUE/FALSE 17–6 False. Although a number of cells equivalent to an adult human is replaced about every three years, not all cells are replaced at the same rate. Blood cells and cells that line the gut are replaced at a high rate, whereas cells in most organs are replaced more slowly, and neurons are rarely replaced. True. The G1 phase is the most critical time for cell growth. Its length can vary greatly depending on external conditions and extracellular signals from other cells. 17–7 THOUGHT PROBLEMS 17–8 The gaps between S phase and M phase are required partly to permit cells the time they need to grow and double their mass of proteins and organelles. The two gaps also provide time for the cell to monitor the internal and external environment to ensure that conditions are suitable and preparations are complete before beginning S phase and mitosis. Enzymes for most metabolic reactions function in isolation; that is, their enzymatic competence does not depend on critical interactions with other proteins. So long as the enzyme folds properly and its small molecule substrate is present, the reaction will proceed. By contrast, cell-cycle proteins must interact with many other proteins to form the complexes that are critical for coordinated progression through the cell cycle. The ability of many human cell-cycle proteins to interact with yeast components implies that the binding surfaces responsible for these interactions have been preserved through more than a billion years of evolution. That’s remarkable. The constancy of the length of S phase in haploid versus diploid and diploid versus tetraploid organisms may not be so surprising. If one assumes that particular chromosomes and regions within chromosomes have a defined order and timing of replication, then halving or doubling the number of chromosomes would not be expected to alter the schedule. Moreover, the ratio of genes encoding the replication machinery (DNA 17–9 17–10 A397 A398 Chapter 17: The Cell Cycle polymerases, helicases, initiation factors, etc.) to the amount of DNA would not change. By contrast, the DNA of different species might well be expected to have different ratios of critical genes to DNA content, which could account for the correlation seen in Table 17–1. Reference: Prescott DM (1975) Reproduction of Eucaryotic Cells, pp 85–86. New York: Academic Press. 17–11 A yeast mutant that is defective in a cell-cycle gene can be isolated if its phenotype is conditionally defective—that is, it can be isolated if the product of the mutant gene fails to function under one set of conditions, but functions adequately under another. Most conditional cell-cycle mutants are temperature sensitive: the mutant protein fails to function at high temperature, but performs well enough at low temperature to allow the cell to divide. To isolate the wild-type gene, you would first grow a population of the Cdc mutant cells at 25°C. These cells would then be transfected with the DNA library and grown at 37°C. Cells that receive a plasmid expressing a DNA that does not correspond to the mutant gene will not be able to form a colony at the restrictive temperature. By contrast, cells that receive the wild-type version of the mutant gene will be able to progress through the cell cycle normally at the restrictive temperature and thus will form a colony. Plasmid DNA isolated from such colonies can be sequenced to determine the identity of the gene. Among the DNAs isolated in experiments like this one, it is not uncommon to find unexpected genes in addition to the one corresponding to the mutant gene. Occasionally, expression of a second gene at higher than normal levels, as often occurs from plasmids, can compensate for the decreased function of the mutant gene. Characterization of such suppressor genes can provide insight into the biological functions of both genes (see Problem 17–36). The key feature of fertilized eggs from Xenopus is the stockpile of proteins they contain. Fertilization of Xenopus eggs triggers a rapid series of divisions that produce 4096 cells in about 7 hours. These divisions are coordinated almost entirely by proteins that were present in the original egg. Mammalian cells do not contain the same concentrations of these proteins, making their isolation much more difficult. In addition, because they are large, it is a simple matter to inject eggs with test substances to analyze their effects on cellcycle progression. 17–12 17–13 17–14 A. During the second thymidine block, all the cells will accumulate at the beginning of S phase, since they cannot synthesize DNA. Thus, upon release of the second block the synchronized population will begin S phase. B. The first thymidine block halts all cells that are in S phase. Cells that are not in S phase traverse the cell cycle normally until they reach the beginning of S phase, where they stop. Since G2 + M + G1 is 15 hours long, the presence of thymidine for 18 hours should be sufficient for all the cells not originally in S phase to reach the beginning of S phase. Thus at the end of the first thymidine block, the majority of the population will be at the beginning of S phase, but the rest of the population will be distributed throughout S phase. The release of the first thymidine block for 10 hours allows the entire population to move through S phase, but does not allow any of the population to reenter S phase. When the second thymidine block is applied, none of the cells should be in S phase. Application of the second thymidine block for 16 hours allows the entire population to move through the cell cycle until they reach the beginning of S phase, where they accumulate. (In reality, a thymidine block does not completely stop DNA synthesis: it slows it to a fraction of its normal rate. Thus a double thymidine block does not result in the entire population accumulating exactly at the G1–S boundary. The population is actually distributed within the first bit of S phase.) References: Bootsma D, Budke L & Vos O (1964) Studies on synchronous division of tissue culture cells initiated by excess thymidine. Exp. Cell Res. 33, 301–309. OVERVIEW OF THE CELL CYCLE Bostock CJ, Prescott DM & Kirkpatrick JB (1971) An evaluation of the double thymidine block for synchronizing mammalian cells at the G1-S border. Exp. Cell Res. 68, 163–168. Rao PN & Johnson RT (1970) Mammalian cell fusion: I. Studies on the regulation of DNA synthesis and mitosis. Nature 225, 159–164. Xeros N (1962) Deoxyriboside control and synchronization of mitosis. Nature 194, 682–683. A399 CALCULATIONS 17–15 Since you examined 25,000 cells and found 3 in mitosis, the mitotic index is 3/25,000, which equals 0.00012. If mitosis is 30 minutes (0.5 hours) long and the frequency of cells in mitosis is 0.00012, then 0.5 hours is 0.00012 of the length of the cell cycle. Thus, the cell cycle is 0.5/0.00012 = 4167 hours in length, on average, which is nearly half a year. The overall length of the cell cycle is equivalent to the time it takes for the entire population of cells to double in number. To find the length of the cell cycle, select any two points on the graph in Figure 17–1, between which the number of cells has doubled. The time separating those two points is the length of the cell cycle. For example, the first two data points in Figure 17–1 are at 3 ¥ 105 cells (10 hours) and 6 ¥ 105 cells (30 hours). Since the population of mouse L cells doubled in 20 hours (30 hours – 10 hours), the length of the cell cycle is 20 hours. 17–16 DATA HANDLING 17–17 A. The two different temperature-shift experiments demonstrate that Cdc101 is blocked at 37°C in the G1 phase of the cell cycle. The experiment in which the cells were first incubated at the restrictive temperature shows that the mutational block precedes or coincides with the hydroxyurea block. If this were not the case, the cells would have divided when they were shifted to 20°C in the presence of hydroxyurea. The experiment in which the cells were first blocked with hydroxyurea shows that the hydroxyurea block occurs after the mutational block, because the cells divided once when they were shifted to 37°C in the absence of hydroxyurea. Together, the two experiments indicate that the mutational block in Cdc101 is in the G1 phase of the cell cycle. In the first experiment, the Cdc101 mutants accumulate in G1 at 37°C; when shifted into hydroxyurea medium at 20°C, they move to S phase, but are stopped there by the hydroxyurea block, and thus do not divide. In the second experiment, in the presence of hydroxyurea at 20°C the cells are blocked in S phase; when hydroxyurea is removed and the cells are shifted to 37°C, they progress normally through G2 and M before they are blocked in G1. Therefore, they undergo one round of cell division. B. The results with Cdc102 indicate that it is blocked at 37°C in S phase. The first experiment shows that the mutational block precedes or coincides with the hydroxyurea block, or the cells would have divided when they were shifted to 20°C. The second experiment shows that the hydroxyurea block precedes or coincides with the mutational block, or the cells would have divided when they were shifted to 37°C. Together, the two experiments indicate that the mutational block and the hydroxyurea block coincide. Since hydroxyurea and the Cdc102 mutation both affect the same phase of the cell cycle, the order of treatment makes no difference; the cells remain trapped in S phase and therefore do not divide. Reference: Hartwell LH (1978) Cell division from a genetic perspective. J. Cell Biol. 77, 627–637. A400 Chapter 17: The Cell Cycle Figure 17–47 The execution point and the point of arrest for the gene product affected in your mutant (Answer 17–18). execution point increasing bud size at time of temperature shift point of arrest in cell cycle 17–18 A. The execution point for your temperature-sensitive mutant is marked on Figure 17–47. Cells that had not reached the execution point in the cell cycle when the temperature was raised grew to the characteristic landmark morphology (large buds) but did not divide. Cells that were beyond the execution point when the temperature was raised divided and then stopped at the landmark morphology during the next cell cycle. B. The characteristic landmark morphology defines the time at which the cell stops its progress through the cell cycle, as indicated for your mutant in Figure 17–47. The landmark morphology is clearly different from the morphology at the execution point. Therefore, the execution point and the point of cell-cycle arrest do not correspond in your mutant. At first glance, it may seem odd that the execution point and the point of arrest do not coincide. An analogy may make the situation clearer. The addition of engine mounts to the chassis is an early step in the assembly of an automobile. Without engine mounts the engine cannot be added and a complete car cannot be built. In the absence of engine mounts, assembly of other parts of the automobile can continue until a point is reached at which all further assembly depends on the engine mounts. In this case the normal execution point for engine-mount addition is early, but the arrest point for assembly is relatively late, with a characteristic landmark morphology, which resembles a complete automobile (until one looks under the hood). Reference: Hartwell LH (1978) Cell division from a genetic perspective. J. Cell Biol. 77, 627–637. 17–19 A. Only the cells that were in the S phase of their cell cycle—those that were making DNA—during the 30-minute labeling period would contain any radioactive DNA. B. The first mitotic cells to appear contained no radioactive DNA because they were not engaged in DNA synthesis during the labeling period; they were in G2. It takes about 3 hours before the first labeled mitotic cells appear because it takes that long for cells to progress from the tail end of S phase to mitosis. 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 to a peak that corresponds to those times when all of the mitotic cells were in S phase during the time of labeling. The curve then falls when the labeled cells exit mitosis, being replaced by unlabeled mitotic cells that 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 ascending curve passes through 50% labeled mitoses at 3 hours, which corresponds to the length of the G2 phase. The initial 3-hour lag before OVERVIEW OF THE CELL CYCLE labeled mitotic cells appear corresponds to the time between the end of the S phase and the beginning of mitosis (Figure 17–48). The first labeled cells seen in mitosis were those that were just finishing S phase when the radioactive thymidine was added. The length of S phase can be estimated from the width of the first peak at 50% labeled mitoses, which is about 10.5 hours in this experiment (Figure 17–48). The overall length of the cell cycle is the time between the 50% points on the two ascending curves, which is about 27 hours (Figure 17–48). The total cell cycle minus G2, S, and M is equal to G1. Thus, G1 is 13 hours long [27 – (3 + 0.5 + 10.5)]. Reference: Baserga R & Wiebel F (1969) The cell cycle of mammalian cells. Intern. Rev. Exp. Pathol. 7, 1–30. 17–20 In outline, the length of G2 can be derived from the data in Figure 17–4A, the length of S from the data in Figure 17–4B, and the length of G1 from the overall length of the cell cycle minus (M + S + G2). 3H-thymidine is incorporated only during S phase. Thus no label will be present in cells undergoing mitosis until the cells that were near the end of S phase when the label was added have traversed the G2 portion of the cell cycle. The appearance of the first labeled mitotic cells at 3 hours after 3Hthymidine addition (see Figure 17–4A) suggests that G2 is 3 hours, long. The majority of mitotic cells, however, did not become labeled until 4 hours after addition of label (and a few did not become labeled until more than 5 hours after addition of label). This variation suggests that there is some variability in the length of G2: if the length of G2 were precisely defined, 100% of mitoses would be labeled when labeling was first observed. Thus, the length of G2 in these cells is 3–4 hours. The length of S phase can be deduced from the number of silver grains over labeled mitotic cells (see Figure 17–4B). Cells that were at the very end of S phase when label was added will have incorporated very little 3H-thymidine, and thus will have very few silver grains. Since labeled mitotic cells begin to appear at 3 hours that marks the end of S phase. Cells that were at the beginning of S phase will have incorporated much more label, and thus will have many more silver grains. The important realization is that cells at the beginning of S phase and cells that were in G1 will have the same number of silver grains, since they incorporated label for the same length of time (that is, throughout S phase). Thus the beginning of S phase is the point in Figure 17–4B at which the number of silver grains per labeled cell reaches a plateau, which is about 10 hours before mitosis. Since the first labeled mitotic cells appeared at about 3 hours, S phase must be about 7 hours long. Given that M phase is 30 minutes, S phase is 7 hours, G2 is 3–4 hours, and the overall length of the cell cycle is 20 hours, G1 must be 8.5–9.5 hours long. Reference: Stanners CP & Till JE (1960) DNA synthesis in individual L-strain mouse cells. Biochim. Biophys. Acta 37, 406–419. 17–21 A. The relationships between cell fluorescence and position in the cell cycle are indicated in Figure 17–49A. Because Hoechst 33342 binds to DNA, cellular fluorescence is proportional to DNA content. The peak with the lowest fluorescence corresponds to cells in G1, which are diploid. The peak with the highest fluorescence corresponds to cells in G2 and M, which have finished replication and are tetraploid (and thus have twice the fluorescence of G1 cells). Cells in S phase, which are replicating their DNA, are between diploid and tetraploid and thus have intermediate levels of fluorescence. B. The distributions of fluorescence for cells treated with agents that block the cell cycle in G1, S, and M phases are shown in Figure 17–49B, C, and D. Cells Figure 17–49 Relationship between fluorescence and the cell cycle (Answer 17–21). (A) Distribution of fluorescent cells among phases of the cell cycle for a normal population of dividing cells. (B) A cell population blocked in G1. (C) A cell population blocked in S. (D) A cell population blocked in M. A401 100 mitotic cells (% labeled) 75 G2 50 length of cell cycle 25 S G2 + M + G1 0 0 5 10 15 20 25 time (hours) 30 Figure 17–48 Lengths of phases of the cell cycle deduced from pulse labeling (Answer 17–19). (A) G1 number of cells G2 + M S 0 (B) 1 2 number of cells 0 (C) 1 2 number of cells 0 (D) 1 2 number of cells 0 1 2 relative fluorescence per cell A402 Chapter 17: The Cell Cycle blocked in either G1 or in M form sharp distributions because all the cells have the same amount of DNA (diploid for G1 and tetraploid for M). Cells treated with an inhibitor that blocks in S phase give a biphasic distribution. Cells that were in S phase at the time the inhibitor was added give a broad distribution because the cells are distributed through all stages of replication. Cells that were in other phases of the cell cycle, however, pile up at the beginning of S phase, giving a sharp peak with a DNA content very close to that of G1 cells. THE CELL-CYCLE CONTROL SYSTEM DEFINITIONS 17–22 17–23 17–24 17–25 17–26 17–27 17–28 17–29 Cyclin-dependent kinase (Cdk) Checkpoint Anaphase-promoting complex or cyclosome (APC/C) M-Cdk Cyclin Metaphase-to-anaphase transition Cell-cycle control system Cyclin-Cdk complex TRUE/FALSE 17–30 False. Although cyclin–Cdk complexes are indeed regulated by phosphorylation and dephosphorylation, they can also be regulated by the binding of Cdk inhibitor (CKI) proteins. Moreover, the rates of synthesis and proteolysis of the cyclin subunits are extremely important for regulating Cdk activity. True. If the length of the cell cycle were shorter than it takes for the cell to double in size, the cell would get progressively smaller with each division; if it were longer, the cells would get bigger and bigger. 17–31 THOUGHT PROBLEMS 17–32 At different transitions in the cell cycle in budding yeast, the single Cdk (Cdk1) binds to different cyclins. These cyclins not only activate Cdk1, but also influence its target specificity. As a result, each cyclin–Cdk1 complex phosphorylates a different set of substrate proteins. Even though vertebrates use multiple Cdks, they use a variety of cyclins to target the cyclin–Cdk complexes to different substrates at different stages of the cell cycle. It would be much easier, and much more reliable, to design a system in which each sensor sends out a negative signal that prevents the curtain going up. In that way the curtain will stay down until the last person takes a seat, and the last inhibitory signal is eliminated. To design a positive signaling system would be much more difficult. To detect the last person to be seated, the system would need to be able to detect a tiny increment over an already large signal (the difference between 199 and 200 for a theater with 200 seats). The consequence of sensor failure also favors a negative signaling system. In a positive signaling system, if a single sensor failed, the show could not go on. By contrast, in a negative signaling system, if a single sensor failed, the show could still go on. 17–33 THE CELL-CYCLE CONTROL SYSTEM Cells operate their checkpoint mechanisms in much the same way. They tend to use negative intracellular signals to arrest the cell cycle, rather than to withold positive signals that normally stimulate cell-cycle progression. A403 DATA HANDLING 17–34 A. Since each transfer accomplishes a 20-fold dilution (50 nL/1000 nL), 10 transfers yield a dilution factor of 2010, which is equal to 1013. It is unreasonable for a molecule to have an undiminished biological effect over this range of dilution. B. The appearance of MPF activity in the absence of protein synthesis suggests that an inactive precursor of MPF is being activated. In principle, activation could involve one of several kinds of post-translational modifications such as protease cleavage or changes in protein phosphorylation. It is now known that MPF is held in an inactive state by phosphorylation of two key residues on its Cdc2 subunit. MPF is activated when those inhibitory phosphates are removed by the protein phosphatase, Cdc25. C. In order for MPF to propagate its activated state through serial transfers, it must be able to activate itself. If it were a protease, for example, active MPF might activate its inactive precursor by cleavage, in the same way that trypsin-mediated cleavage of trypsinogen produces more trypsin. Since MPF is a protein kinase, it is reasonable to suggest that it operates via phosphorylation to trigger the activation of the inactive pre-MPF. In principle, MPF could activate pre-MPF directly by adding one or more phosphates. Or it could act indirectly through other kinases or phosphatases to alter the phosphorylation status of pre-MPF and thereby activate it. As it turns out, MPF triggers the addition of activating phosphates to Cdc25, which then removes the inhibitory phosphates from pre-MPF, activating it. This sets up an auto-amplification cycle that rapidly activates all the MPF, thereby stimulating egg maturation in each serial transfer. Progesterone induces egg maturation by setting up the same auto-amplification cycle, but initiates it in a different way. Progesterone stimulates translation of the mRNA for Mos, which is the frog’s MAP kinase kinase kinase. Expression of Mos activates the MAP kinase cascade, which inhibits the Wee1-like protein kinase that is responsible for adding inhibitory phosphates to pre-MPF. Inhibition of this kinase allows a small amount of preMPF to become active, thereby triggering auto-amplification of MPF activity and subsequent egg maturation. References: Wasserman WJ & Masui Y (1975) Effects of cycloheximide on a cytoplasmic factor initiating meiotic maturation in Xenopus oocytes. Exp. Cell Res. 91, 381–388. Karaiskou A, Jessus C, Brassac T & Ozon R (1999) Phosphatase 2A and Polo kinase, two antagonistic regulators of Cdc25 activation and MPF autoamplification. J. Cell Sci. 112, 3747–3756. 17–35 A. Colchicine has no apparent effect on the synthesis of cyclin B, which continues to accumulate in colchicine-treated cells. Colchicine does have a dramatic effect on cyclin B destruction. Colchicine delays the destruction of cyclin B, but does not completely block it. Two observations in these experiments point to the continued destruction of cyclin B, albeit with significantly slower kinetics. The amount of cyclin B reaches a plateau in the presence of colchicine, suggesting that its rate of synthesis must be balanced by its rate of destruction (see Figure 17–7). In addition, when its synthesis is blocked by emetine, cyclin B is eventually destroyed in the presence of colchicine (see Figure 17–8). The stabilization of cyclin B certainly could account for the metaphase arrest produced by colchicine. As long as cyclin B levels stay high, M-Cdk activity will remain high, keeping the cell in M phase. A404 Chapter 17: The Cell Cycle B. Inhibition of protein synthesis blocks the synthesis of new cyclin B. Because colchicine delays its destruction but does not prevent it, cyclin B is eventually destroyed. When cyclin B decreases, M-Cdk activity declines as well. In the absence of M-Cdk activity, cells exit mitosis. In the presence of colchicine, the exit from mitosis is aberrant because chromosomes cannot segregate normally without microtubules. Reference: Hunt T, Luca FC & Ruderman JV (1992) The requirement for protein synthesis and degradation, and the control of destruction of cyclins A and B in the meiotic and mitotic cell cycles of the clam embryo. J. Cell Biol. 116, 707–724. 17–36 At the restrictive temperature, the mutant Cdk1 binds Cln1 and Cln2 with low affinity. High copy-number plasmids express much more protein than is expressed from a single chromosomal gene, or from a gene on a single-copy plasmid. By overexpressing Cln1 or Cln2, it is possible to overcome the low affinity of Cdk1, permitting a sufficient amount of Cln1–Cdk1 or Cln2–Cdk1 complex to form to allow progression through the cell cycle. Reference: Hadwiger JA, Wittenberg C, Richardson HE, de Barros Lopes M & Reed SI (1989) A family of cyclin homologs that control the G1 phase in yeast. Proc. Natl Acad. Sci. U.S.A. 86, 6255–6259. 17–37 A. When the promoter was turned on and Cln3 was expressed at high levels, many fewer cells were in G1 and many more cells were in S and G2. Overexpression of Cln3 might reasonably be expected to increase the amount of active G1-Cdk, thereby pushing cells through Start and committing them to DNA replication. If cells were constantly stimulated in this way, fewer of them would be expected to be in G1, and more would be expected to be in S and G2. B. If the cells were being forced through G1 more quickly than normal, they would have less time to grow before cell division. As a result, the cells would be smaller than normal. Reference: Richardson HE, Wittenberg C, Cross F & Reed SI (1989) An essential G1 function for cyclin-like proteins in yeast. Cell 59, 1127–1133. 17–38 In the normal situation, Wee1 adds phosphates to Cdk1 to inactivate it, and Cdc25 removes those phosphates, activating Cdk1. In principle, different types of mutation in any one of these genes might produce the giant and tiny temperature-sensitive strains of yeast. For example, a mutation in the Cdk1 gene that led to an inactive protein at high temperature would generate a yeast strain like giant that was unable to proceed through the cell cycle. A mutation in the same gene that prevented Cdk1 from being inactivated by Wee1 at the restrictive temperature would lead to a constitutively active Cdk1 that would continually push the cell through the cell cycle, as in the tiny strain. You could imagine analogous mutations in the Wee1 and Cdc25 genes. An inactive form of Wee1 (or a hyperactive form of Cdc25) would leave Cdk1 permanently active, generating cells with a short cell cycle, like the tiny strain. A hyperactive version of Wee1 (or an inactive form of Cdc25) would generated an inactive Cdk1, would produce cells that were unable to progress through the cell cycle, as in the giant strain. In various screens, inactivating mutations in Wee1 and Cdc25 have been found, and they have the expected phenotypes, but mutations that lead to hyperactive forms of these proteins have not been isolated. The mutations described in this problem correspond to two alleles of Cdc2 (the original name for the Cdk1 gene) discovered by Paul Nurse in 1974. Nurse had been very excited by his discovery of the Wee1 gene, which, when mutated, gave rise to small yeast cells that entered mitosis at a smaller-than-usual size compared to wild-type cells. To see if any other genes could be identified with similar properties, he set up a genetic screen based on looking at the S PHASE mutants in the microscope, hunting for small cells. After identifying 49 new temperature-sensitive, small-cell strains, all of which proved to be allelic to the original Wee1 gene, he was on the point of giving up when the 50th mutant strain turned out not to map in the Wee1 locus. Instead, it corresponded to a different, but already known gene, Cdc2. Existing mutations in Cdc2 interfered with cell-cycle progression, giving rise to large cells under restrictive conditions. Because this gene could be mutated to inactivity or hyperactivity, Nurse concluded that it must encode a really important cellcycle regulator. Many genes can be mutated so that they block cell-cycle progression, but this was the only one that could apparently be made more active. Time proved this conclusion to be absolutely correct. References: Nurse P (1975) Genetic control of cell size at cell division in yeast. Nature 256, 547–551. Nurse P & Thuriaux P (1980) Regulatory genes controlling mitosis in the fission yeast Schizosaccharomyces pombe. Genetics 96, 627–637. A405 S PHASE DEFINITIONS 17–39 17–40 17–41 17–42 Cohesin Origin recognition complex (ORC) Mcm protein Geminin TRUE/FALSE 17–43 17–44 True. Origins of replication are licensed for replication by the binding of large complex of initiator proteins, called the prereplicative complex. True. The origin recognition complex serves as a scaffold at origins of replication in eucaryotic cells around which other proteins are assembled and activated to initiate DNA replication. THOUGHT PROBLEMS 17–45 S-Cdk could initiate the firing of replication origins—directly or indirectly— by activating origin-binding proteins via phosphorylation events. Such binding proteins might bind to prereplicative complexes at different times in S phase as a result of their accessibility or surrounding DNA sequences, thereby accounting for the characteristic firing times for different origins. Alternatively, S-Cdk could modify all origins at the same time, preparing them for the subsequent binding of a key initiation factor, the timing of which determines when origins fire. Preventing re-replication is not in conflict with the role of S-Cdk in promoting origin firing. When DNA replication is initiated, S-Cdk phosphorylates ORC and Cdc6, inhibiting them and preventing reassembly of the preRC until after the next mitosis. Reference: Bell SP & Dutta A (2002) DNA replication in eukaryotic cells. Annu. Rev. Biochem. 71, 333–374. 17–46 Cohesins must be present during S phase because it is only while DNA is being replicated that sister chromatids can be reliably identified by the cellular A406 Chapter 17: The Cell Cycle machinery that links them together. Once sister chromatids have separated, it is impossible for a nonspecific DNA-binding protein like cohesin to tell which chromosomes are sisters. And it would be virtually impossible for any protein to distinguish sister chromatids from homologous chromosomes. If sister chromatids are not kept together after their formation, they cannot be accurately segregated to the two daughter cells during mitosis. Reference: Uhlmann F & Nasmyth K (1998) Cohesion between sister chromatids must be established during DNA replication. Curr. Biol. 8, 1095–1101. 17–47 In cells that fully condense their chromosomes at mitosis, as vertebrate cells do, much of the cohesin is released from the chromosome arms at the start of mitosis, when chromosome condensation begins. With most of the cohesin out of the way, the cell can coil individual sister chromatids into separate domains. The small amount of cohesin that remains is sufficient to hold sister chromatids together until anaphase, when the residual cohesins are degraded. DATA HANDLING 17–48 When a G1 cell is fused with an S-phase cell, DNA replication occurs in the G1 nucleus, presumably triggered by S-Cdk activity in the S-phase cell. Fusion of a G2 cell with an S-phase cell, however, does not cause DNA replication in the G2 nucleus, indicating that G2 nuclei are refractory to the effects of S-Cdk. Thus cells that have completed S phase cannot replicate their DNA again, consistent with a block to re-replication. Fusion of a G2 cell with a G1 cell does not drive the G1 nucleus into S phase, indicating that S-Cdk in G2 cells is no longer capable of triggering entry of G1 nuclei into S phase. Reference: Johnson RT & Rao PN (1971) Nucleo-cytoplasmic interactions in the achievement of nuclear synchrony in DNA synthesis and mitosis in multinucleate cells Biol. Rev. Camb. Philos. Soc. 46, 97–155. 17–49 A. Sister chromatids behave as expected for wild-type cells: they are stuck together in small budded cells and separated into the mother and daughter cells at mitosis. The single spot of fluorescence in the small-budded cell in Figure 17–12 likely results from two sites of binding that are close together on the paired sister chromatids. The spot is brighter than the individual spots in cells that have two separated spots, and the size of the bud indicates that the site should have been replicated (by comparison with the size of the small-budded Scc1ts cell with two spots). B. In one of the small-budded cells from Scc1ts, it is clear from the presence of two spots of fluorescence that sister chromatids have already separated. Although the sister chromatids are separated in the large-budded cells, they are abnormal because they have both remained in the mother cell. C. Prematurely separated sister chromatids prevent the formation of a normal spindle apparatus, and thus prevent normal segregation of chromosomes into the mother and daughter cells. It is likely that they trigger the spindleattachment checkpoint; that is, that unpaired sister chromatids behave as if they are unattached. Reference: Michaelis C, Ciosk R & Nasmyth K (1997) Cohesins: chromosomal proteins that prevent premature separation of sister chromatids. Cell 91, 35–45. 17–50 A. Both Scc1 and Eco1 must be present during replication. Scc1ts and Eco1ts cells that are grown at 37°C maintain viability until they begin to enter S phase (at about 1.5 hours), at which time viability drops dramatically. B. Scc1 is required after S phase is completed, but Eco1 is not. When Scc1ts cells were allowed to complete S phase at 25°C and then were shifted to 37°C, they lost viability rapidly. Thus, Scc1 is needed in a functional form beyond MITOSIS S phase. By contrast, when Eco1ts cells were allowed to complete S phase at 25°C before the shift to 37°C, they maintained their viability and began to proliferate. Thus, Eco1 is required in a functional form only during S phase; its function is entirely dispensable thereafter. C. Scc1 must be functional in order for the cohesin complex to be laid down during S phase and for sister-chromatid cohesion to be maintained until mitosis. Thus, functional Scc1 is required during S, G2, and M, until mitosis. This behavior is understandable since Scc1 is a component of the cohesin complex; in its absence sister-chromatid cohesion cannot be established or maintained. D. Eco1 behaves differently from Scc1. It is required during S phase when the cohesin complex is laid down, but not beyond S phase when the complex must be maintained. This behavior suggests that Eco1 in some way helps to establish the cohesin complex. Other studies suggest that Eco1 interacts with PCNA at the replication fork and may help to load the cohesin complex onto the daughter duplexes as they are generated by replication. References: Skibbens RV, Corson LB, Koshland D & Hieter P (1999) Ctf7 is essential for sister chromatid cohesion and links mitotic chromosome structure to the DNA replication machinery. Genes Dev. 13, 307–319. Toth A, Ciosk R, Uhlmann F, Galova M, Schleiffer A & Nasmyth K (1999) Yeast cohesin complex requires a conserved protein, Eco1p(Ctf7), to establish cohesion between sister chromatids during DNA replication. Genes Dev. 13, 320–333. A407 MITOSIS DEFINITIONS 17–51 17–52 17–53 17–54 17–55 17–56 17–57 17–58 17–59 17–60 17–61 Microtubule flux Anaphase B Spindle assembly checkpoint Centrosome Anaphase A Metaphase plate Interpolar microtubules Telophase Condensin Separase Astral microtubule TRUE/FALSE 17–62 True. Microtubules nucleated by the centrosomes grow outward toward the chromosomes in a highly dynamic process, alternately growing and shrinking. When they eventually attach to the kinetochore of a chromosome, they become stabilized and are referred to as kinetochore microtubules. False. Equal and opposite forces that tug the chromosomes toward the two spindle poles would tend to position them at random locations between the poles. The poleward force on each chromosome is opposed by a polar 17–63 A408 Chapter 17: The Cell Cycle ejection force that pushes the chromosome away from the pole. As a chromosome approaches a spindle pole, the ejection force is thought to increase, thereby raising the tension on the kinetochore, temporarily shutting off the poleward force, allowing the chromosome to be pushed away from the pole. This balance of forces at each pole tends to position the chromosomes at the midpoint between the poles—the metaphase plate. 17–64 False. Kinetochore microtubules polymerize at their plus ends up to anaphase, at which point they begin to depolymerize. Prior to anaphase, kinetochore microtubules maintain a fairly constant length by treadmilling, with addition to the plus ends being balanced by removal at the minus ends. At anaphase, coincident with sister-chromatid separation, kinetochore microtubules begin to depolymerize at their plus ends as well, and therefore become shorter, moving the chromosomes toward the spindle poles. False. The five stages of mitosis occur in strict sequential order, but cytokinesis begins during anaphase and continues through the end of M phase. 17–65 THOUGHT PROBLEMS 17–66 Prophase (see Figure 17–14E), prometaphase (see Figure 17–14D), metaphase (see Figure 17–14C), anaphase (see Figure 17–14A), telophase (see Figure 17–14F), and cytokinesis (see Figure 17–14B). The sharp activation of M-Cdk at the end of G2 is due to the events shown in Figure 17–50. Cyclin B accumulates gradually by steady synthesis. As it accumulates, it will bind to Cdk1 molecules to form M-Cdk. M-Cdk is phosphorylated at two sites: an activating site by CAK and an inhibitory site by Wee1 kinase (Figure 17–50). The inhibitory phosphate keeps the M-Cdk largely inactive, but it is slowly removed by the phosphatase Cdc25. After a certain threshold level has been reached, active M-Cdk initiates two positive feedback loops that trigger its explosive activation. In one loop active M-Cdk stimulates Cdc25 phosphatase, causing removal of the inhibitory phosphate from inactive molecules of M-Cdk. In the second loop active M-Cdk inhibits phosphorylation by Wee1, thereby preventing further addition of the inhibitory phosphate. Wee1 kinase is already maximally active to ensure that the inhibitory phosphate is added to M-Cdk to prevent its premature activation by the Cdk-activating kinase (CAK). By contrast, Cdc25 phosphatase has a low—but finite— activity in order to activate some M-Cdk to initiate the rapid activation of MCdk that characterizes entry into mitosis. By blocking the low activity of Cdc25 phosphatase, the DNA replication checkpoint mechanism ensures that all M-Cdk will remain in its inactive form. 17–67 17–68 Cdc25 POSITIVE FEEDBACK M-cyclin Cdk-activating kinase CAK P P Wee1 Cdk inactive M-Cdk Cdk-inhibitory kinase inactive M-Cdk P Cdc25 P active M-Cdk POSITIVE FEEDBACK Figure 17–50 The activation of M-Cdk (Answer 17–67). Inhibitory phosphate is shown in black; activating phosphates are shown in white. MITOSIS 17–69 Astral, kinetochore, and interpolar microtubules all radiate from the spindle poles, with their plus ends directed outwards. Astral microtubules, which radiate in all directions, act as ‘handles’ for orienting and positioning the spindle in the cell. In addition, those in contact with the cell cortex aid in separation of the spindle poles during anaphase. Kinetochore microtubules link chromosomes to the spindle and are responsible for movement of the sister chromatids to the poles. Interpolar microtubules interdigitate at the equator of the spindle and are responsible for the symmetrical, bipolar shape of the spindle. Their movement relative to one another pushes the poles apart at anaphase. Overexpression of catastrophe factors would destabilize microtubules, reducing their average length and leading to a shorter than normal mitotic spindle. Overexpression of MAPs would have the opposite effect: the microtubules would be more stable, hence longer, leading to a longer than normal mitotic spindle. The two types of motors operate on the interpolar microtubules. Plus-enddirected motors tend to push the spindle poles apart by decreasing the overlap, whereas minus-end-directed motors tend to pull the poles together by increasing the overlap (Figure 17–51). When the two motor domains of a kinesin-5 molecule attach to different microtubules, they will each move toward the plus end. When the kinesin-5 motors reach the plus ends, they can move no farther. At that point, they will link the plus ends tightly together. Multiple copies of kinesin-5 would tie multiple microtubules together by their plus ends, leaving the minus ends free. Thus, the astral array generated by the plus-end-directed kinesin-5 motor proteins will have the plus ends of the microtubules at the center and the minus ends at the periphery. Reference: Hyman AA & Karsenti E (1996) Morphogenetic properties of microtubules and mitotic spindle assembly. Cell 84, 401–410. 17–73 Centrosome duplication is semi-conservative. The pair of centrioles in the centrosome separates, and each serves to nucleate synthesis of a new centriole. As a consequence, each new centrosome consists of one old and one new centriole. Thus, centrosome duplication is analogous to DNA replication: a new duplex consists of one old strand and one newly replicated strand. Reference: Stearns T (2001) Centrosome duplication: A centriolar pas de deux. Cell 105, 417–420. 17–74 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 A409 17–70 17–71 17–72 (A) PLUS-END-DIRECTED (+) (-) )-( )+( (B) MINUS-END-DIRECTED (+) (-) )-( )+( Figure 17–51 Operation of motors on the interpolar microtubules in the spindle (Answer 17–71). (A) Plus-end-directed motors. (B) Minus-end-directed motors. Arrows indicate the direction of movement of the microtubules. A410 Chapter 17: The Cell Cycle other spindle pole attaches to the kinetochore of its partner sister chromatid, which places the duplicated chromosome under mechanical tension by pulling forces from both poles. The tension ensures that both microtubules remain attached to the chromosome. The position and orientation of the chromosome in the cell at the time the nuclear envelope breaks down may be the key factors that determine the spindle pole to which the chromatid will be pulled. A kinetochore is most likely to become attached to the spindle pole toward which it is facing. 17–75 There are 46 human chromosomes, each with two kinetochores—one for each sister chromatid—thus, there are 92 kinetochores in a human cell at mitosis. Both sister chromatids could end up in the same daughter cell for any of a number of reasons. If the microtubules or their connections with a kinetochore were to break during anaphase, both sister chromatids could be drawn to the same pole, hence the same daughter cell. If microtubules from the same spindle pole attached to both kinetochores, the chromosome would be pulled to the same pole. If the cohesins that link sister chromatids were not degraded, the pair of chromatids might be pulled to the same pole. If a 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 engage a checkpoint mechanism—for example, the spindle-attachment checkpoint—and allow most such errors to be corrected, which is one reason why such errors are so rare. As a consequence of this error, one daughter cell would contain only a single 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 cell with a single copy of the chromosome may be defective for a critical gene, a defect that was hidden by the presence of a second, good copy of the gene on the other chromosome. The kinetochore microtubules and the overlap microtubules treadmill by constantly adding tubulin subunits to their plus ends and removing an equivalent number from their minus ends. This balanced addition and removal results in a movement of subunits toward the poles (the minus ends of the microtubules), but leaves the overall appearance of the spindle unchanged. Microtubules that are not directly connected to the centrosomes are held in place by minus-end-directed motors, which link them to the other microtubules and tend to ‘walk’ them toward the spindle pole. Nocodazole arrests cells in M phase of the cell cycle. By preventing microtubule polymerization—hence spindle formation—nocodazole triggers the spindle-attachment checkpoint, which inhibits the APC/C ubiquitin ligase so that the metaphase-to-anaphase transition cannot occur. In the absence of benomyl, the majority of spindles form normally and the spindle-attachment checkpoint plays no role. As a consequence, Mad2 is irrelevant. In the presence of benomyl, however, cells that are defective for Mad2 cannot stop cell-cycle progression, with the result that chromosomes are segregated incorrectly, causing the cells to die. Reference: Li R & Murray AW (1991) Feedback control of mitosis in budding yeast. Cell 66, 519–531. 17–76 17–77 17–78 17–79 17–80 17–81 This experiment shows that for microtubules to remain attached to kinetochores, tension has to be exerted. Tension is normally achieved by the opposing pulling forces from the two 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 MITOSIS connections will break, and microtubules from the opposing spindle pole will have another chance to attach properly. 17–82 In anaphase A the sister chromatids separate and the chromosomes move toward the poles by the shortening of the kinetochore microtubules. Anaphase A depends on the action of motors at the kinetochore. In anaphase B, which overlaps with anaphase A, the spindle poles move apart, separating the chromosomes still farther. Two sets of motors cooperate to accomplish spindle separation. Plus-end-directed motors acting on overlap microtubules push the spindle poles apart. This movement is accompanied by microtubule growth at their plus ends. Minus-end-directed motors that link the cell cortex to the astral microtubules act to pull the spindle poles apart. At a gross level this analogy appears valid. Chromosomes move to the spindle poles tethered on a microtubule line, much as fish move to a fishing pole on a fishing line. But in detail the analogy fails. A fishing line is shortened at the end opposite the fish, whereas a microtubule is shortened by disassembly at the end attached to the chromosome. The new nuclear envelope reassembles on the surface of the chromosomes, effectively preventing cytosolic proteins from being trapped between the chromosomes and the envelope. Cytosolic proteins are also excluded as clusters of chromosomes coalesce to form the complete nucleus. During this process nuclear pores are incorporated into the envelope. They permit selective import of nuclear proteins, causing the nucleus to expand while maintaining its characteristic protein composition. The events occur in the following order: duplication of the centrosome (F), separation of centrosomes (J), condensation of chromosomes (D), breakdown of nuclear envelope (C), attachment of microtubules to chromosomes (B), alignment of chromosomes at the spindle equator (A), separation of sister chromatids (K), elongation of the spindle (G), re-formation of nuclear envelope (I), decondensation of chromosomes (E), and pinching of cell in two (H). A411 17–83 17–84 17–85 CALCULATIONS 17–86 The dose of caffeine required to interfere with the DNA replication checkpoint mechanism is much higher than the amount imbibed by even the most excessive drinkers of coffee and colas. The concentration of caffeine in a cup of coffee is about 3.4 mM. g [caffeine] = 100 mg ¥ ¥ mole ¥ 1000 mL 150 mL 1000 mg 196 g L = 3.4 ¥ 10–3 M = 3.4 mM Since the concentration in a cup is less than the 10 mM required to interfere with the DNA replication checkpoint mechanism, you cannot get a higher concentration by drinking it and diluting it in the water volume of the body. If you assume for the purposes of calculation that the caffeine is not metabolized or excreted (but that all the liquid is), then you can ask how many cups of coffee would you need to drink (at 100 mg of caffeine per cup) to reach a concentration of 10 mM in 40 L of body water. The answer is: you would need to drink 784 cups of coffee! 17–87 The average length of chromosomes in base pairs and millimeters, and the average number of base pairs carried per microtubule, are given in Table 17–8. The length of DNA carried by each microtubule varies about 50-fold from S. cerevisiae to Haemanthus. Although this represents a fair amount of variability, it is much more constant than the average length of chromosomal DNA, which spans three to four orders of magnitude over the same range of organisms. A412 Chapter 17: The Cell Cycle Table 17–8 Average lengths of chromosomes and average number of base pairs per microtubule in a variety of organisms (Answer 17–87). TYPE OF ORGANISM Yeast Yeast Protozoan Fly Human Plant SPECIES S. cerevisiae S. pombe Chlamydomonas Drosophila Homo sapiens Haemanthus AVERAGE LENGTH OF CHROMOSOME (bp) 0.9 ¥ 106 4.7 ¥ 106 5.8 ¥ 106 4.3 ¥ 107 1.4 ¥ 108 6.1 ¥ 109 AVERAGE LENGTH OF CHROMOSOME (mm) 0.3 1.6 2.0 15 47 2100 AVERAGE DNA (bp)/ MICROTUBULE 0.9 ¥ 106 1.6 ¥ 106 5.8 ¥ 106 4.3 ¥ 106 5.6 ¥ 106 5.1 ¥ 107 Reference: Bloom K (1993) The centromere frontier: kinetochore components, microtubule-based motility, and the CEN–value paradox. Cell 73, 621–624. DATA HANDLING 17–88 A. The state of phosphorylation of Wee1 and Cdc25 is the result of the balance between the protein kinase and protein phosphatase activities that regulate them. By inhibiting the protein phosphatases, okadaic acid causes Wee1 and Cdc25 to accumulate in their phosphorylated forms (Figure 17–52). Since this change activates M-Cdk, Wee1 and Cdc25 must have originally been present in the extract in their nonphosphorylated forms. Thus active Wee1 kinase is nonphosphorylated, as is inactive Cdc25 phosphatase (Figure 17–52). Knowing which forms are phosphorylated allows you to label the arrows that correspond to the kinases and phosphatases that control Wee1 and Cdc25 phosphorylation (Figure 17–52). B. The protein kinases and phosphatases that control phosphorylation of Wee1 and Cdc25 must be specific for serine/threonine side chains because they are affected by okadaic acid, which inhibits only serine/threonine phosphatases. C. Okadaic acid has no direct effect on Cdk1 phosphorylation because it is phosphorylated on a tyrosine side chain. Tyrosine phosphatases are unaffected by okadaic acid. The decrease in Cdk1 phosphorylation is a consequence of the change in activities of Wee1 kinase, which becomes less active, and of Cdc25 phosphatase, which becomes more active. D. As soon as some active M-Cdk appears, it would begin to phosphorylate Wee1 and Cdc25, inactivating the kinase and activating the phosphatase. The resultant decrease in Wee1 kinase activity and increase in Cdc25 phosphatase activity would lead to dephosphorylation (and activation) of more M-Cdk. This in turn would further decrease the activity of Wee1 kinase and further increase the activity of Cdc25 phosphatase, leading to still more MCdk activity. Thus the initial appearance of a little M-Cdk activity would rapidly lead to its complete activation. POSITIVE FEEDBACK M-Cdk active Wee1 P kinase inactive Wee1 kinase active Cdc25 P phosphatase active Cdc25 phosphatase inactive P M-Cdk inactive protein phosphatase protein phosphatase Figure 17–52 Control of M-Cdk activity by Wee1 kinase and Cdc25 phosphatase (Answer 17–88). Inhibitory phosphates are shown in black; activating phosphate is shown in white. MITOSIS (A) COHESIN A413 TOPO II catenane (B) CONDENSIN Figure 17–53 Cohesin- and condensinmediated topological changes (Answer 17–89). (A) Cohesin-mediated catenane formation. Arrowheads indicate sites at which topoisomerase II could catalyze duplex crossing to generate the indicated molecule. (B) Condesin-mediated knot formation. The relationship between the initial product, formed by duplex crossing at the arrowheads, and the trefoil knot is illustrated in discrete steps to try to make the topology clearer. TOPO II knot This sort of activation is referred to as a positive feedback loop (Figure 17–52). It is a common means of regulation when it is advantageous for a system to flip rapidly from one state to another without lingering in the intermediate states. Reference: Kumagai A & Dunphy WG (1992) Regulation of the Cdc25 protein during the cell cycle in Xenopus extracts. Cell 70, 139–151. 17–89 A. One fundamental difference in the proposed roles for cohesin and condensin is the number of duplexes that are involved. Cohesin is proposed to link sister chromatids together; that is, to link two duplexes. By contrast, condensin is proposed to operate on a single duplex, causing it to be coiled, and thus condensed. These experiments are in perfect accord with this fundamental difference. Incubation with cohesin generates catenanes, which must be produced by topoisomerase II action on juxtaposed, separate duplexes. Incubation with condensin generates knots, which arise by topoisomerase II action on juxtaposed parts of a single duplex. B. Figure 17–53A shows a plausible mechanism by which cohesins might bring two duplexes close enough together to allow topoisomerase II to link them. C. Figure 17–53B shows one way that condensins could organize a single circle of DNA so that topoisomerase could tie it into a knot with a single duplexcrossing event. Remarkably, the particular type of trefoil knot that is generated (a mirror-image could have been produced in principle) places severe constraints on the type of coiling that condensin introduces. If you got this part of the problem correct, pat yourself on the back! References: Kimura K, Rybenkov VV, Crisona NJ, Hirano T, & Cozzarelli NR (1999) 13S condensin actively reconfigures DNA by introducing global positive writhe: Implications for chromosome condensation. Cell 98, 239–248. Losada A & Hirano T (2001) Intermolecular DNA interactions stimulated by the cohesin complex in vitro: Implications for sister chromatid cohesion. Curr. Biol. 11, 268–272. 17–90 A. In the absence of functional Zyg1 protein kinase, the two centrioles of a centrosome can separate, but the individual centrioles cannot initiate synthesis of new centrioles. Thus, the centrosome cycle is interrupted between G1/S and S in Figure 17–16. At mitosis the individual centrioles can form a normal looking bipolar spindle, except that each pole contains a single centriole. Upon completion of mitosis, each daughter cell gets a single centriole. A414 Chapter 17: The Cell Cycle Because the individual centriole cannot initiate synthesis of a new centriole in the absence of Zyg1, at the next mitosis the single centriole will form a monopolar spindle. B. The mutant, nonphosphorylatable form of nucleophosmin prevents separation of the centriole pair that normally occurs just before S phase. Thus, the centrosome cycle is interrupted between G1 and G1/S in Figure 17–16. It is thought that G1/S-Cdk normally phosphorylates nucleophosmin, allowing it to dissociate from the centrosome as a prelude to separation of the centrioles. If nucleophosmin is not released, the centrosome cannot be duplicated and a monopolar spindle will form with a single centrosome at the pole. References: O’Connell KF, Caron C, Kopish KR, Hurd DD, Kemphues KJ, Li Y & White JG (2001) The C. elegans zyg-1 gene encodes a regulator of centrosome duplication with distinct maternal and paternal roles in the embryo. Cell 105, 547–558. Tokuyama Y, Horn HF, Kawamura K, Tarapore P & Fukasawa K (2001) Specific phosphorylation of nucleophosmin on Thr199 by cyclin-dependent kinase 2-cyclin E and its role in centrosome duplication. J. Biol. Chem. 276, 21529–21537. 17–91 The patterns of centriole duplication and splitting that account for the mercaptoethanol-induced abnormalities in cell division are diagrammed in Figure 17–54. In essence, treatment with mercaptoethanol allows the centrosome to split a second time without an intervening duplication of the centrioles (Figure 17–54A). (Centrosomes split the first time as the egg entered mitosis.) As a consequence, each of the spindle poles of the tetrapolar spindle has only one centriole, rather than the normal pair of centrioles. Evidently, a centrosome with a single centriole is a perfectly adequate spindle pole. The daughter cells from the four-way division of the egg receive a centrosome with a single centriole. During the next cell cycle, the centriole is duplicated to form a normal centrosome with a centriole pair (Figure 17–54A). The usual form of the centrosome upon entry into mitosis has two centriole pairs instead of one. Thus the centrosome in the daughter cells looks like a single spindle pole and indeed forms a monopolar spindle (Figure 17–54B). Most commonly, cell division is aborted and the cell traverses the cell cycle again, allowing the centrosome to be duplicated a second time so that it possesses two centriole pairs (Figure 17–54B, top). This second (A) add MSH remove MSH split (B) duplicate major pathway split normal mitosis minor pathway monopolar mitosis duplicate Figure 17–54 Duplication and splitting of centrosomes in mercaptoethanol-treated sea urchin eggs (Answer 17–91). (A) Mercaptoethanol-induced splitting of centrosomes. When MSH is removed, a tetrapolar spindle forms with each centriole serving as a spindle pole. (B) Abnormal cell divisions by cells that receive a single centriole. In most such cells (upper pathway) the centriole is duplicated twice—with an intervening aborted mitosis—to generate a normal pair of centrosomes and a normal bipolar spindle. In a few cells (lower pathway) the centrosome splits into single centrioles that form a bipolar spindle and allow cell division, but once again generate daughter cells with single centrioles. MITOSIS duplication puts the centrosome cycle back in step with the cell cycle and subsequent cell divisions occur normally (Figure 17–54B, top). In the rarer cases in which the daughter cells form a bipolar spindle, the centrosomes split to form two centrosomes each with a single centriole (Figure 17–54B, bottom). Although the splitting allows cell division to occur, it presents the daughter cells with the same problem as the parent: a centrosome with a single centriole (Figure 17–54B, bottom). In this case the centrosome cycle remains out of step with the cell-division cycle. Note that although the cell divisions can be restored to normal, eggs that undergo a four-way division develop abnormally because none of the cells receives a full complement of chromosomes. References: Maizia D, Harris PJ & Bibring T (1960) The multiplicity of mitotic centers and the time-course of their duplication and separation. J. Biophys. Biochem. Cytol. 7, 1–20. Sluder G & Rieder CL (1985) Centriole number and the reproductive capacity of spindle poles. J. Cell Biol. 100, 887–896. 17–92 A. A kinetochore microtubule is relatively stable because both of its ends are protected from disassembly: one by attachment to the centrosome, and the other by attachment to the kinetochore. This stabilization suggests that kinetochores cap the plus ends of the microtubules, thereby altering the equilibrium for subunit dissociation. B. Astral microtubules disassemble when tubulin is below the critical concentration for microtubule assembly. Under these conditions, the rate of addition does not balance the rate of dissociation, and microtubules get progressively shorter. The presence of kinetochore microtubules strengthens this interpretation relative to the other two possibilities mentioned. Neither detachment from the centrosome nor random breakage would explain the stability of kinetochore microtubules. C. The possible mechanisms of disappearance would be readily distinguished by a time course. At intermediate time points, the number and length of microtubules will be sensitive indicators of which mechanism operates. If microtubules detached, the number of microtubules per centrosome would decrease, but the length would remain the same. If microtubules depolymerized from the end, the number would remain constant and their lengths would decrease relatively uniformly. If the microtubules broke at random, the number would remain relatively constant, but the distribution of lengths would be very broad. Reference: Mitchison TJ & Kirschner MW (1985) Properties of the kinetochore in vitro. II. Microtubule capture and ATP-dependent translocation. J. Cell. Biol. 101, 766–777. 17–93 A. Dicentric plasmids are stable in bacteria because bacteria use a completely different mechanism to segregate their chromosomes. A bacterial chromosome is polarized so that its single origin of replication (OriC) is located at one pole of the bacterium. As soon as OriC sequences are replicated, one copy is translocated to the opposite pole, ensuring that the daughter chromosomes lie on either side of the plane of fission that separates the bacterium into daughter cells. This mechanism of cell division makes bacteria indifferent to the presence of centromeres on the plasmid DNA. B. Dicentric plasmids are unstable in yeasts for the same reason that dicentric chromosomes are unstable in higher eucaryotes. If the two centromeres attach to opposite poles, the spindle apparatus can exert enough force on the DNA molecule to break its phosphodiester backbone. Roughly half the time a plasmid would be expected to orient itself on the spindle so that its two centromeres are attached to opposite poles. Thus there is a very high probability that a plasmid will be broken at each cell division, hence the instability. A415 A416 Chapter 17: The Cell Cycle Ars1 Trp1 Cen4 Cen3 AmpR first break Cen4 Ars1 Ori Figure 17–55 A mechanism for generating monocentric and acentric plasmids from a dicentric plasmid in yeast (Answer 17–93). Dashed arrows indicate the direction of pull toward the spindle poles. Viability refers to the ability of the plasmid to grow in yeast under selective conditions (which requires Ars1 and Trp1). Ori AmpR Trp1 second break Cen3 Cen4 Ars1 Cen3 Trp1 AmpR Ori CIRCULARIZE Cen4 EXONUCLEASE, CIRCULARIZE CIRCULARIZE Cen3 Ori Ars1 Ars1 AmpR Trp1 monocentric, viable LARGE COLONIES Trp1 acentric, viable SMALL COLONIES monocentric, nonviable NO COLONIES C. Since monocentric plasmids are very stable, it seems most likely that the mechanism for deletion of centromeric sequences from dicentric plasmids relates to the breakage they suffer during mitosis. As illustrated in Figure 17–55, a circular plasmid must suffer two breaks to permit the centromeres to separate during mitosis. This breakage naturally separates the centromeres from one another onto linear fragments of the original plasmid. If the ends of a fragment join to make a circle, the resulting plasmid will contain a single centromeric sequence (Figure 17–55). Only those fragments that contain the yeast origin of replication (Ars1) and the selected marker (Trp1) can continue to grow in future generations in yeast. This mechanism does not readily account for the loss of both centromeres; rather, it predicts that one centromere will be retained. Once the dicentric plasmid is reduced to a monocentric plasmid, it should be stable. The loss of both centromeres probably involves a process other than simple breakage. One likely possibility is that the broken ends are digested by exonucleases, which occasionally remove the remaining centromeric sequence before the fragment circularizes (Figure 17–55). Reference: Mann C & Davis RW (1983) Instability of dicentric plasmids in yeast. Proc. Natl Acad. Sci. U.S.A. 80, 228–232. 17–94 A. The selection is for the gene product and not the gene. After the plasmid (gene) is lost from a cell, the cell can still grow and divide in the absence of histidine until the gene product is diluted so much that it can no longer meet the nutritional requirements of the cell. Only at that point does the cell stop dividing. This is presumably the reason why at any given instant MITOSIS only 5% to 25% of the cells in the culture have the plasmid; the majority of cells are ones that have lost the plasmid recently but have not yet stopped dividing. B. The pedigree analysis shows clearly that the plasmid does not segregate equally to mother and daughter cells. At each division where the plasmid is lost, it is lost from the daughter cell. Thus there is some asymmetry in the segregation such that the mother cell tends to hold onto the plasmid. C. Even if plasmids replicated only once per cell cycle, a large number of plasmids could accumulate in the mother cells due to asymmetric segregation. If a mother cell with one plasmid managed to hold onto all the replicated plasmids for five divisions, it would contain 32 (25) plasmid molecules without violating the rule of one replication per cell cycle. D. Plasmids with centromeres segregate like chromosomes, with one copy retained by the mother cell, and one copy going to the daughter cell. Since all cells have a plasmid, they (and their progeny) continue to grow under selective conditions. Thus the colony grows at a maximal rate. Plasmids without centromeres are segregated abnormally as shown by the pedigree analysis. Since only a fraction of the cells gives rise to progeny cells that continue to divide under selective conditions, the overall colony grows at a less than maximal rate because many individual cells stop dividing. As a consequence, colonies of cells containing plasmids with centromeres grow large, and colonies of cells containing plasmids without centromeres remain small. Reference: Murray A & Szostak JW (1982) Pedigree analysis of plasmid segregation in yeast. Cell 34, 961–970. 17–95 A. If the second centromere were active, it would indeed destabilize the plasmid as described in Problem 17–93. In your experiments, however, growth on galactose keeps the introduced Cen6 inactive by promoting transcription across it. B. Transcription requires that RNA polymerase be able to unwind DNA and separate the two strands over a short distance as it moves along the DNA. The stable assembly of a kinetochore physically prevents unwinding of the DNA by RNA polymerase. This has ample precedent in the activity of some gene regulatory proteins, which bind tightly to DNA sequences in the path of RNA polymerase and thereby block its progression along the DNA. It is not so clear how transcription inactivates a kinetochore. In experiments not described here, it has been shown that transcription apparently does not disturb the special arrangement of nucleosomes around centromeric DNA. In addition, the majority of transcripts terminate at the border of the Cen sequence. It may be that the approach of the transcriptional apparatus close to the edge of the centromere destabilizes microtubule attachment just enough to kill the kinetochore. C. Low fidelity of chromosome transmission can arise in several ways other than by disruption of kinetochore function. Mutations in tubulin genes and in microtubule-based motor proteins, which can interfere with attachment to kinetochores or with spindle function, also show elevated rates of chromosome loss. Mutations that affect DNA metabolism can also lead to chromosome loss by interfering with proper chromosome replication. Reference: Doheny KF, Sorger PK, Hyman AA, Tugendreich S, Spencer F & Hieter P (1993) Identification of essential components of the S. cerevisiae kinetochore. Cell 73, 761–774. 17–96 A. For the sectoring assay to be meaningful, it is essential that all cells start with the same genetic endowment. The use of a conditional centromere on minichromosome A makes this possible. Growing the cells on galactose keeps the second centromere inactive so that minichromosome A is faithfully transmitted at each cell division. When the cells are spread on agar A417 A418 Chapter 17: The Cell Cycle plates containing glucose, the second centromere is activated because transcription from the Gal10 promoter ceases. B. Sectoring results when the minichromosome is broken. A dicentric minichromosome is more apt to break when it is pulled toward opposite spindle poles by attachment to two strong centromeres (minichromosome A) than when pulled by attachments to one strong and one weak centromere (minichromosome B). The weak centromere, which is known to reduce centromere function by an order of magnitude, either lets go in the tug-of-war, or perhaps does not attach in the first place. In either case, the result is less sectoring than observed for minichromosome A, which has two strong centromeres. C. The decrease in sectoring observed when minichromosome A is introduced into Ctf13 yeast, as compared with wild-type yeast, is fully consistent with the idea that the Ctf13 gene encodes a kinetochore protein. Presumably, the defective kinetochore protein encoded by Ctf13 weakens the kinetochore sufficiently, so that when the dicentric minichromosome is stretched between the spindle poles, one of the protein–DNA connections will break before the DNA breaks. In the wild-type strain, where the protein–DNA connection is stronger, the DNA breaks more often, resulting in a higher frequency of sectoring. D. The Ctf13 mutation enhances the loss of normal chromosomes and stabilizes dicentric minichromosomes for the same reason: it makes a slightly defective kinetochore. The defect in the kinetochore allows normal chromosomes to detach from the spindle microtubules occasionally and be lost. The defect also allows one centromere to detach when a dicentric minichromosome is pulled toward both spindle poles, thereby preventing breakage of the minichromosome and allowing it to be transmitted faithfully at a higher frequency than in a wild-type strain. Reference: Doheny KF, Sorger PK, Hyman AA, Tugendreich S, Spencer F & Hieter P (1993). Identification of essential components of the S. cerevisiae kinetochore. Cell 73, 761–774. 17–97 A. Xkid helps align chromosomes on the metaphase plate by moving the chromosome arms toward the plus ends of the interpolar microtubules. Outside the central region of the spindle, where the interpolar microtubules overlap, Xkid will propel a chromosome unidirectionally toward the center of the spindle. Inside the region of overlap Xkid motors will move a chromosome back and forth between the plus ends of the overlapping interpolar microtubules, confining the chromosome to the spindle equator. B. Plus-end-directed, microtubule-dependent motors that bind chromosome arms, as Xkid does, are plausible candidates for mediators of the astral ejection force, which pushes chromosomes away from the poles toward the spindle equator. Reference: Antonio C, Ferby I, Wilhelm H, Jones MJ, Karsenti E, Nebreda AR & Vernos I (2000) Xkid, a chromokinesin required for chromosome alignment on the metaphase plate. Cell 102, 425–435. 17–98 The separation of chromosomes is blocked in the presence of stable Xkid, even though the sister chromatids have been split apart. The continued function of Xkid, which constantly pushes the chromosomes away from the poles toward the equator, evidently is sufficient to oppose the normal force exerted by the shortening of the kinetochore microtubules. Note that not all aspects of anaphase appear to be blocked by stable Xkid. The spindle, for example, has increased in size, suggesting that the poles have moved apart. Reference: Funabiki H & Murray AW (2000) The Xenopus chromokinesin Xkid is essential for metaphase chromosome alignment and must be degraded to allow anaphase chromosome movement. Cell 102, 411–424. 17–99 Spindles with astral microtubules will tend to repel one another, keeping themselves out of each other’s way. When spindles with asters come close to MITOSIS one another, their astral microtubules will overlap, allowing them to be pushed apart by plus-end-directed motors such as kinesin-5. As a consequence, the nuclei that are generated are well distributed in the common cytoplasm. By contrast, spindles without astral microtubules will not be able to repel one another and their nuclei will be clustered. Reference: de Saint Phalle B & Sullivan W (1998) Spindle assembly and mitosis without centrosomes in parthenogenetic Sciara embryos. J. Cell Biol. 141, 1383–1391. 17–100 A. M-Cdk remains active in the presence of cyclin BD90 because cyclin BD90 is missing the destruction box and cannot be destroyed by APC/C. This socalled indestructible cyclin remains bound to Cdk1 and keeps it active. B. M-Cdk remains active in the presence of cyclin B13-110 for a different reason. Excess cyclin B13-110 overwhelms APC/C, which would otherwise degrade the normal cyclin B that is complexed with Cdk1 in M-Cdk. By keeping APC/C occupied, cyclin B13-110 prevents degradation of normal cyclin B and keeps M-Cdk activity high. C. The mutant forms of cyclin B, unlike normal cyclin B, both maintain a high level of M-Cdk activity in the presence of Ca2+. Cyclin BD90, however, allows normal separation of sister chromatids. Therefore sister-chromatid separation cannot depend directly on loss of M-Cdk activity, which suggests that dephosphorylation of a linker protein is not the critical event for sister-chromatid separation. Addition of cyclin B13-110 prevents sister-chromatid separation. Since cyclin B13-110 blocks APC/C, it seems likely that the linker protein that holds sister chromatids together may itself be a target of APC/C. Reference: Holloway SL, Glotzer M, King RW & Murray AW (1993) Anaphase is initiated by proteolysis rather than by the inactivation of maturation-promoting factor. Cell 73, 1393–1402. 17–101 A. Cells expressing uncleavable Scc1 are expected to behave the same as normal cells up to the point when the sister chromatids need to be separated during mitosis. In wild-type cells, this occurs 3 hours after the release from the G1 block, as shown by the reappearance of cells in the G1 phase of the cell cycle. Cells expressing uncleavable Scc1 are indistinguishable from wildtype cells up to the point of cell division. B. After 3 hours, the wild-type cells simply enter another cell cycle with the usual peaks corresponding to cells in G1 and G2 (or M). By contrast, cells expressing uncleavable Scc1 generate a set of peaks that do not correspond to cells with one or two times the normal DNA content. These peaks instead have DNA contents that range from less than the normal DNA content of a G1 cell to less than the normal DNA content of a G2 cell. These abnormal DNA contents arise because the cells are unable to separate their sister chromatids. When the normal—and equal—segregation of chromosomes is prevented, some cells get less than the correct number, while other get more, giving rise to the distribution shown in Figure 17–33. Reference: Uhlmann F, Lottspeich F & Nasmyth K (1999) Sister-chromatid separation at anaphase onset is promoted by cleavage of the cohesin subunit of Scc1. Nature 400, 37–42. 17–102 A. The 10 arrangements of chromosomes on tripolar spindles are illustrated schematically in Figure 17–56A. Upon separation of the sister chromatids and cell division, some of the arrangements yield cells that do not contain at least one chromosome, as indicated by shaded regions in Figure 17–56A. Of the 10 possible arrangements for tripolar eggs, 7 would give rise to three cells, each of which has at least one chromosome. Thus 70% of tripolar eggs will produce three cells that each carry at least one chromosome. A419 A420 Chapter 17: The Cell Cycle (B) TETRAPOLAR EGGS 0 0 2 2 0 0 0 0 0 1 3 3 1 0 0 0 1 1 1 0 1 2 0 2 2 0 2 1 1 1 1 0 1 2 0 2 1 0 0 1 0 0 2 1 1 1 0 0 1 0 0 2 1 1 1 1 1 0 1 2 0 1 0 1 1 0 1 0 2 0 0 1 0 0 1 0 0 2 0 1 0 2 0 0 3 0 2 1 0 0 2 0 0 1 2 3 0 0 0 3 0 0 0 3 0 0 0 2 1 (A) TRIPOLAR EGGS 3 (For tetrapolar eggs, there are twenty possible arrangements, eight of which give rise to four cells that each have at least one chromosome, as illustrated in Figure 17–56B. Thus 40% of tetrapolar eggs will give rise to four cells with at least one chromosome.) B. If the total number of chromosomes were the critical factor, 70% of tripolar eggs would be expected to develop into normal plutei. If the distribution of chromosomes were the critical factor, then (0.7)9, or 4%, of tripolar eggs would be expected to develop into normal plutei. Boveri found that 58 out of 695, or 8%, of tripolar eggs developed into normal plutei. These results agree remarkably well with expectations based on the idea that individual chromosomes carry only a portion of the total genetic information. (For tetrapolar eggs the expectations are 40% for total number of chromosomes and (0.4)9 or 0.03% for distribution of chromosomes. His observation that zero of 1170, or less than 0.09%, of tetrapolar eggs developed into normal plutei again matches the expectations based on the idea that individual chromosomes carry a fraction of the total genetic information.) References: Baltzer F (1967) Theodor Boveri: Life and Work of a Great Biologist. Berkeley: University of California Press. Boveri T (1902) Über mehrpolige Mitosen als mittel zur Analyse des Zellkerns. Verh. d. phys.-med. Ges. Würzburg, N.F. 35, 67–90. (Available in English translation, Foundations of Experimental Embryology, B Willier and J Oppenheimer eds. Englewood Cliffs, NJ: Prentice-Hall, 1964.) Boveri T (1907) Zellenstudien VI: Die Entwicklung dispermer Seeigeleier. Ein Beitrag zur Befruchtungslehre und zur Theorie des Kernes. Jenaische Zeitschr. Naturwissen. 43, 1–292. Figure 17–56 Arrangements of three chromosomes on tripolar and tetrapolar spindles (Answer 17–102). (A) Tripolar eggs. (B) Tetrapolar eggs. Lines show the positions of the metaphase plates of the spindles. Numbers indicate how many chromosomes are on a particular metaphase plate. Shaded portions indicate the positions of daughter cells that will not have at least one chromosome. CYTOKINESIS DEFINITIONS 17–103 Syncytium 17–104 Midbody 17–105 Cytokinesis 17–106 Phragmoplast 17–107 Contractile ring CYTOKINESIS A421 TRUE/FALSE 17–108 False. For most cells—typical cells—this statement is true. There are exceptions, however. Osteoclasts, for example, undergo mitosis without cytokinesis and become multinucleate. 17–109 False. The position of the mitotic spindle anticipates the position of the cleavage furrow: it is positioned centrally for symmetric cleavage and asymmetrically for asymmetric cleavage. THOUGHT PROBLEMS 17–110 The two cytoskeletal machines are the mitotic spindle and the contractile ring. The segregation of chromosomes and their distribution to daughter cells is accomplished by the bipolar mitotic spindle, which is composed of microtubules and a variety of microtubule-dependent motors and other proteins. The division of an animal cell into daughter cells by cytokinesis is accomplished by the contractile ring, which is composed of actin and myosin filaments and is located just under the plasma membrane. As the ring constricts, it pulls the membrane inward, ultimately dividing the cell in two. 17–111 The movement of chromosomes at anaphase depends on microtubules, not on actin or myosin. Injection of an antibody against myosin would therefore have no effect on chromosome movement during mitosis. Cytokinesis, on the other hand, depends on the assembly and contraction of a ring of actin and myosin II filaments, which forms the cleavage furrow that splits the cell into two. If the injected antibody interferes with myosin II, the contractile ring will not be able to initiate cleavage. 17–112 Nocodazole treatment disassembles microtubules, preventing formation of a spindle. Because nuclei break down and chromosomes condense, these events must be independent of aster formation by centrosomes, for example, and of any other microtubule-dependent process. Mitosis finally arrests because the unattached chromosomes trigger a signal that engages the spindle assembly checkpoint, which halts the cycle. Treatment with cytochalasin D does not affect mitosis because actin filaments are not involved in the process. Moreover, the disassembly of actin filaments does not trigger a cell-cycle arrest; the cell completes mitosis and forms a binucleate cell. Thus, there does not seem to be a cytokinesis checkpoint. 17–113 The membranes of the Golgi vesicles fuse to form the new 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 protein that were exposed to the lumen of the Golgi vesicle will end up exposed to the new cell wall (Figure 17–57). Figure 17–57 Fusion of Golgi-derived vesicles to form new cell wall (Answer 17–113). plant cell in cytokinesis A422 Chapter 17: The Cell Cycle CALCULATIONS 17–114 A. If the volume of the parental cell is 1, then the volume of each progeny cell is 0.5. (Note that it is not necessary to specify the units for the volume. Since we are calculating a fractional increase in surface area, the units drop out at the end.) The radius of the parent cell can be calculated as follows V = 4pr 3 3 Substituting for V = 1 and rearranging r3 = 3 ( 4π ( 3 r =( ( 4π 13 The surface area of the parental cell is A = 4pr 2 Substituting for r = 0.62, A = 4.83 The same calculation for a progeny cell, using 0.5 as the volume, gives r = 0.49. The surface area of a single progeny cell is 3.02, and the surface area of both progeny cells is 6.04. The fractional increase in surface area due to division is Increase = (6.04 − 4.83)/4.83 = 25% Thus, at cell division there is a 25% increase in the amount of plasma membrane. This increase is unlikely to cause the cell any problem for two reasons. First, most cells round up at cell division. Since a sphere has the lowest surface-to-volume ratio of any three-dimensional shape, it follows that cells have more plasma membrane before rounding up. Second, the internal pools of membrane in most cells are much greater than the plasma membrane and serve as a ready source of new plasma membrane. B. If there is a 25% increase in membrane at each cell division, then the total membrane area after a division is 1.25 times the membrane area before division. Thus, for 12 successive divisions with no increase in total cell volume, the total membrane area increases (1.25)12 = 15-fold. This increase is much more substantial than in a single cell division and suggests that eggs must have an extra large internal supply of membrane or that membrane synthesis keeps pace with cell division. DATA HANDLING 17–115 A. The chromosome-signaling hypothesis predicts that furrows will form only where chromosomes have been aligned. This prediction matches the result of the first division, but not the result of the second division, where three furrows are formed instead of the two that were expected. B. The polar-relaxation hypothesis predicts that there should be two furrows at the first division instead of just one. In the toroidal egg the spindles should relax a ring of the cell surface on either side of the equatorial plane. As a consequence, there should be two regions where the cell surface furrows: at the equatorial plane and opposite it on the other side of the torus. The events of CYTOKINESIS the second division match the expectations of the relaxation hypothesis. C. The aster-stimulation hypothesis predicts that there should be a single furrow at the first cell division and three furrows at the second cell division, which matches the experimental observations. A single furrow is expected at the first division because the spindle fibers from the two poles interact only at the equatorial plane. At the second division, however, the spindle fibers interact not only at the two equatorial positions, but also at the position of the third furrow. Reference: Rappaport R (1986) Establishment of the mechanism of cytokinesis in animal cells. Int. Rev. Cytol. 105, 245–281. 17–116 Plus-end-directed motors that carry vesicles as cargo could deliver them to the midbody, using the spindle microtubules as tracks. Fusion of the vesicles with each other and with the plasma membrane could complete cell separation (Figure 17–58). This sort of a mechanism would make cytokinesis in animal cells and plant cells much more similar than previously suspected. Reference: Skop AR, Bergmann D, Mohler WA & White JG (2001) Completion of cytokinesis in C. elegans requires a brefeldin A-sensitive membrane accumulation at the cleavage furrow apex. Curr. Biol. 11, 735–746. 17–117 In the absence of the psychosine receptor, the population of cells is distributed between 2N (unreplicated) cells and 4N (replicated) cells. In the presence of the receptor, but in the absence of psychosine, there is the same distribution. When treated with psychosine, however, prominent peaks corresponding to 8N and 16N cells appear. FACS analysis cannot determine whether the increase in DNA is due to multiple nuclei, or to a single nucleus with several copies of the genome. When the cells in these experiments were examined microscopically, they were shown to contain multiple nuclei. References: Im D-S, Heise CE, Nguyen T, O’Dowd BF & Lynch KR (2001) Identification of a molecular target of psychosine and its role in globoid cell formation. J. Cell Biol. 153, 429–434. Mitchison T (2001) Psychosine, cytokinesis, and orphan receptors: Unexpected connections. J. Cell Biol. 153, F1–F3. 17–118 Megakaryocyte precursors enter their first mitosis in exactly the same way as normal cells and proceed normally up to anaphase. Megakaryocytes can complete anaphase A, allowing sister chromatids to separate and move toward the spindle poles. They do not, however, begin the movements characteristic of anaphase B; that is, separation of the spindle poles. Because the spindle poles remain close together and cytokinesis is avoided, the chromosomes become enclosed within one nuclear envelope. Most importantly, the cell now contains two centrosomes. At the next division, the two centrosomes duplicate and divide, ultimately forming the four poles of the A423 cleavage furrow Figure 17–58 Potential mechanism for vesicle fusion in sealing off the intercellular canal and completing cytokinesis (Answer 17–116). A424 Chapter 17: The Cell Cycle tetrapolar spindle that organizes the condensed, replicated chromosomes. Once again, the cells proceed to anaphase A, and then exit mitosis and enclose all the chromosomes in a single nuclear membrane. Megakaryocytes must lack the motor activities that are responsible for the movements of anaphase B; namely, plus-end-directed motors that push the poles apart and the minus-end-directed motors that interact with the cell cortex and pull the spindle poles apart. It is unclear whether the motors are missing or simply inactive. The megakaryocytes must also be missing the pathways that trigger cytokinesis. References: Nagata Y, Muro Y & Todokoro K (1997) Thrombopoietin-induced polyploidization of bone marrow megakaryocytes is due to a unique regulatory mechanism in late mitosis. J. Cell Biol. 139, 449–457. Zimmet J & Ravid K (2000) Polyploidy: occurrence in nature, mechanisms, and significance for the megakaryocyte-platelet system. Exp. Hematol. 28, 3–16. CONTROL OF CELL DIVISION AND CELL GROWTH DEFINITIONS 17–119 G0 17–120 Growth factor 17–121 Telomere 17–122 Replicative cell senescence 17–123 Mitogen 17–124 E2F protein 17–125 ATM TRUE/FALSE 17–126 False. Serum-deprived cells continue through the current cell cycle until they reach the G1 checkpoint, which diverts them into G0. Cells can only enter G0 from G1. 17–127 False. Budding yeast resume their cell cycle, even if they are unable to repair the damage; evidently life with mutation is better than no life at all. By contrast, mammalian cells do not resume the cycle, but instead commit suicide by undergoing apoptosis. In multicellular organisms, the health of the organism takes precedence over the life of an individual cell. Cells with severe damage threaten the life of the organism, since genetic damage can lead to cancer and other lethal defects. 17–128 False. Organism senescence (aging) is distinct from replicative cell senescence, which occurs in the absence of telomerase. Aging is thought to depend largely on progressive oxidative damage to macromolecules. Strategies that reduce metabolism—for example, restricted caloric intake— decrease the production of reactive oxygen species, and can extend the lifespan of experimental animals. THOUGHT PROBLEMS 17–129 Mitogens stimulate cell division, primarily by relieving intracellular negative controls that otherwise block progress through the cell cycle. Growth factors CONTROL OF CELL DIVISION AND CELL GROWTH stimulate cell growth (an increase in cell mass) by promoting the synthesis of proteins and other macromolecules and by inhibiting their degradation. Survival factors promote cell survival by suppressing apoptosis. 17–130 A. Most of the cells in an adult human are in this class, having withdrawn from the cell cycle into G0. Liver cells, for example, remain quiescent for long periods, although they can grow and divide when the need arises. B. Nerve cells grow as they extend axons over long distances, but do not divide. Fat cells can accumulate large quantities of triglyceride, which causes them to increase in size (although this is not properly growth, per se). Oocytes grow to become very large cells prior to fertilization. C. This is the most rare category of cell, but the production of red blood cells is a good example. During production of red blood cells, precursor reticulocytes undergo five cell divisions with little increase in overall volume, ultimately generating very small red blood cells from a much larger precursor cell. D. Most cells in the human body grow and divide actively at some point during development, until we become adults. Even in adults some cells continue to grow and divide; most notably, intestinal cells and hematopoietic cells, which must constantly renew the lining of the gut and the cells in the blood, respectively. Most other cells grow and divide often enough to balance cell death. Reference: Dolznig H, Bartunek P, Nasmyth K, Mullner EW & Beug H (1995) Terminal differentiation of normal chicken erythroid progenitors: shortening of G1 correlates with loss of D-cyclin/cdk4 expression and altered cell size control. Cell Growth Differ. 6, 1341–1352. 17–131 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 partly or completely dismantled. If a cell just paused in G1, it would still contain all the cell-cycle machinery and might still be induced to divide. It would also have to remake the ‘decision’ not to divide almost continuously. To reenter the cell cycle from G0, a cell has to resynthesize the components that have disappeared, which is unlikely to occur by accident. 17–132 The on-demand, limited release of PDGF at a wound site triggers cell division of neighboring cells for a limited time, until 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 inappropriately express their own PDGF receptor, so that they can stimulate their own proliferation, thereby promoting the development of cancer. 17–133 A. Radiation leads to DNA damage, which activates ATM and ATR kinases, which phosphorylate and activate Chk1 and Chk2 kinases, which phosphorylate and stabilize p53, which induces expression of p21, which binds to and inactivates G1/S-Cdk and S-Cdk, which stops progression through the cell cycle. B. In the absence of a functional DNA damage checkpoint, the cell will replicate the damaged DNA, introducing mutations into the genomes inherited by the daughter cells. C. A checkpoint-deficient 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 processes (for example, by cosmic rays). The checkpoint mediated by p53 is mainly required as a safeguard against the devastating effects of DNA damage, but not for the natural progression of the cell cycle in undamaged cells. D. Cell division is an ongoing process that does not cease when we reach maturity. Blood cells, epithelial cells in the skin or lining the gut, and the cells of A425 A426 Chapter 17: The Cell Cycle the immune system, for example, are being constantly produced by cell division to meet the body’s needs. Our bodies produce about 1011 new blood cells each day. 17–134 A. Cells that cannot degrade M-phase cyclins would be unable to divide. The cells would enter mitosis, but would not be able to exit. B. Cells that always expressed high levels of p21 would be unable to divide. The cells would arrest permanently in G1 because their G1/S-Cdk and their SCdk would be inactivated. C. Cells that cannot phosphorylate Rb would be unable to divide. The cells would not be able to activate the transcription of genes required for entry into S phase because the required regulatory protein, E2F, would be sequestered by unphosphorylated Rb. 17–135 The gene for telomerase is turned off early in development and remains off in most cells in humans. Thereafter, each time a cell replicates its chromosomes, it fails to copy a short segment of the telomeric DNA at the very end of the chromosome (see Problem 5–66). As a result, the telomere becomes progressively shorter with each cell division; thus the length of the telomere is a rough gage of the number of times a cell has divided. When the telomere gets too short to function properly, it triggers a p53-dependent cell-cycle arrest. 17–136 Although telomerase activity is absent from most human somatic cells, the gene for the RNA component continues to be expressed. Thus, telomerase is inactive because expression of the protein component has been turned off. As a result, expression of the protein component alone is sufficient to reactivate telomerase. Reference: Weinrich SL, Pruzan R, Ma L, Ouellette M, Tesmer VM, Holt SE, Bodnar AG, Lichtsteiner S, Kim NW, Trager JB, Taylor RD, Carlos R, Andrews WH, Wright WE, Shay JW, Harley CB & Morin GB (1997) Reconstitution of human telomerase with the template RNA component hTR and the catalytic protein subunit hTRT. Nat. Genet. 17, 498–502. 17–137 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 proliferation. Unless badly damaged, the liver will usually shrink back to a normal size after the patient stops drinking. In a liver tumor, in contrast, mutations abolish normal cell proliferation control, and as a result, cells divide and keep on dividing in an uncontrolled manner. DATA HANDLING 17–138 The results in Table 17–4 indicate that 3T3 cells require these factors in an ordered sequence. If the factors were required simultaneously, none of the pretreatments should have advanced entry into S phase. If the factors were required independently (that is, regardless of order), all of the pretreatments should have advanced the entry into S phase equally. Since the order of addition clearly makes a difference, the cells must respond sequentially to the factors. Since the pretreatment in experiment 3 (PDGF, EGF, and then IGF1) advances entry into S phase most markedly, the cells must respond to the factors in this order. From more extensive experiments, it appears that PDGF induces a state of competence in quiescent 3T3 cells that permits them to respond to EGF and IGF1. Exposure of competent 3T3 cells to EGF causes them to progress about 6 hours toward S phase; exposure of EGF-treated competent cells to IGF1 causes them to progress another 5 hours toward S phase. Reference: O’Keefe EJ & Pledger WJ (1983) A model of cell-cycle control: sequential events regulated by growth factors. Mol. Cell. Endocrinol. 31, 167–186. CONTROL OF CELL DIVISION AND CELL GROWTH 17–139 A. Experiment 1 shows that the 170-kd protein has a high-affinity binding site for EGF and therefore is most likely the EGF receptor. The control experiment of incubating the membrane preparation in the presence of excess unlabeled EGF demonstrates that EGF binding to the 170-kd protein is specific, not random. B. Experiments 2, 3, and 4 show that the 170-kd protein (the EGF receptor) becomes phosphorylated with radioactive phosphate in the presence of g32P-ATP Transfer of the phosphate from the g position in ATP demonstrates . that the EGF receptor is a substrate for a protein kinase. Since the labeling intensity of the 170-kd protein is increased in the presence of EGF, EGF stimulates the activity of the protein kinase. C. Experiments 3 and 4 show that the EGF receptor can transfer phosphate from ATP to protein; thus, it is a protein kinase. Experiment 3 is less convincing than experiment 4. In experiment 3, although the antibody is specific for the EGF receptor, it is not unreasonable to question whether other proteins, perhaps including a protein kinase, might have been trapped within the antibody precipitate. In experiment 4 the EGF receptor was first separated by molecular weight from other proteins that might contaminate the antibody precipitate. Only in the unlikely event that the contaminating protein kinase was also a 170-kd protein would experiment 4 lead you astray. D. With the caveat mentioned in part C, experiment 4 indicates convincingly that the EGF receptor is a substrate for its own protein kinase activity. Reference: Cohen S, Ushiro H, Stocheck C & Chinkers M (1982) A native 170,000 epidermal growth factor receptor-kinase complex from shed plasma membrane vesicles. J. Biol. Chem. 257, 1523–1531. 17–140 Cyclin D antibodies bind to cyclin D and make it unavailable for binding to its Cdk partners, thereby preventing formation of G1-Cdk, which is required for cells to progress through Start, the G1 checkpoint. Once Start has been passed—about 14 hours after addition of mitogenic growth factors—the cell is committed to enter S phase independent of any further requirement for G1-Cdk. Thus, after 14 hours antibodies to cyclin D do not affect the eventual entry of cells into S phase. Reference: Baldin V, Lukas J, Marcote MJ, Pagano M & Draetta G (1993) Cyclin D1 is a nuclear protein required for cell cycle progression in G1. Genes Dev. 7, 812–821. 17–141 A. Careful examination of the time-lapse pictures in Figure 17–42 shows that all the cells without buds arrested at the dumbbell stage, whereas all of the cells with buds formed viable colonies. The appearance of a bud corresponds with the beginning of S phase. Haploid cells that have partially or fully replicated their genomes are more resistant to x-ray-induced breaks because a break in one chromosome can be repaired by recombination with the intact sister chromatid. Haploid cells in G1 are especially sensitive to breaks because they contain no second intact copy of the chromosome with which to recombine. After replication, such a cell will contain two copies of the chromosome, but both will be broken at the same position. Thus, even in G2, a haploid cell that suffers a break in G1 will not have an intact chromosome with which to repair itself by homologous recombination. B. The observation that dumbbell-stage cells have a single nucleus and no spindle indicates that the cells are arrested in G2 prior to mitosis. The presence of a bud indicates that the cells have already passed the G1 checkpoint and such cells will complete S phase. C. Half the wild-type cells temporarily arrest in the dumbbell stage while they wait for damage to be repaired. A cell-cycle checkpoint senses damaged DNA and halts the cell cycle until the damage is repaired. When repair is complete, the cells enter mitosis and divide to produce viable colonies. The A427 A428 Chapter 17: The Cell Cycle nonviable cells died, either because they suffered damage too late to stop and divided with damaged chromosomes, or because they suffered so much damage that it could not be repaired. D. Rad52 cells remain arrested at the dumbbell stage because they are incapable of repairing their damaged chromosomes. The continued signal from the damaged DNA prevents the cells from passing the mitotic entry checkpoint. E. Very few Rad9 mutant cells arrest at the dumbbell stage because they are defective in their ability to sense DNA damage. In these cells, the mitotic entry checkpoint does not function. Division in the absence of repair leads to haploid cells that have broken chromosomes and cells that are missing pieces of chromosomes. Both situations lead to nonviable cells. Only a small fraction of cells (30%) manages to repair their chromosomes in the absence of a checkpoint delay. F. If Rad9 cells were artificially delayed in mitosis, the number of viable cells would increase. The artificial delay would allow the cells time to repair their damaged chromosomes, so that they then could complete mitosis with an intact genome. The important point is that Rad9 cells contain all the necessary enzymes required for DNA repair; they are defective only in sensing DNA damage. References: Hartwell LH & Weinert TA (1989) Checkpoints: controls that ensure the order of cell cycle events. Science 246, 629–634. Weinert TA & Hartwell LH (1988) The rad9 gene controls the cell cycle response to DNA damage in Saccharomyces cerevisiae. Science 241, 317–322. damaged DNA Rad24 Rad1 Hus1 Hus2 mitotic entry checkpoint unreplicated DNA Cdc2-3w Cdc2-F15 Figure 17–59 Branched signaling pathway from damaged and unreplicated DNA to the mitotic entry checkpoint (Answer 17–142). 17–142 Your results with mutant strains indicate that signals from damaged DNA and unreplicated DNA interact with the mitotic entry checkpoint via a branched pathway. If the two signals shared the same pathway, all the mutants would show mitotic delay in response to either kind of DNA. If they used independent pathways, each mutant would respond to only one or the other kind of DNA. As indicated in Figure 17–59, Rad24 affects a part of the pathway that is specific for damaged DNA, Cdc2-3w and Cdc2-F15 affect a part of the pathway that is specific for unreplicated DNA, and Rad1, Hus1, and Hus2 affect a common part of the pathway. Reference: Murray A & Hunt T (1993) The Cell Cycle: An Introduction, pp 143–144. New York: WH Freeman. 17–143 Because the extra mass of wing tissue is composed of normal-looking cells, Dpp must stimulate both cell division to increase the number of cells, and cell growth so that the extra cells are the right size. You would expect that Dpp stimulation would activate proteins such as PI 3-kinase, S6-kinase, Myc, and Ras, all of which stimulate cell growth. You would also expect Dpp to stimulate cell-cycle components such as cyclin E and Cdc25, which promote entry into S phase and mitosis, respectively, in order to drive progression through the cell cycle. Finally, you would expect Dpp to activate a set of transcription factors that control expression of wing-specific genes. Reference: Edgar BA & Lehner CF (1996) Developmental control of cell cycle regulators: A fly’s perspective. Science 274, 1646–1652. Prober DA & Edgar BA (2001) Growth regulation by oncogenes—new insights from model organisms. Curr. Opin. Genet. Dev. 11, 19–26. Zecca M, Basler K & Struhl G (1995) Sequential organizing activities of engrailed, hedgehog and decapentaplegic in the Drosophila wing. Development 121, 2265–2278. 17–144 These results suggest that density-dependent inhibition of cell growth can be completely ascribed to changes in cell shape. This correlation becomes apparent when the 3H-thymidine incorporation is expressed in a way that takes into account the difference in the number of cells per plate; for example, as cpm/1000 cells (Table 17–9). Note that these data do not distinguish CONTROL OF CELL DIVISION AND CELL GROWTH A429 Table 17–9 Incorporation of 3H-thymidine by cells grown on normal dishes and on poly(HEMA)-treated dishes (Answer 17–144). TYPE OF DISH Normal Normal Normal Poly(HEMA) Poly(HEMA) Poly(HEMA) CELL DENSITY (cells/dish) 60,000 200,000 500,000 30,000 30,000 30,000 CONFLUENCY subconfluent confluent confluent sparse sparse sparse CELL HEIGHT (mm) 6 15 22 6 15 22 3H INCORPORATION (cpm/1000 cells) 253 55 7 250 50 7 whether cell shape ‘controls’ cell proliferation, or whether both parameters are dependent on a third factor related in some way to the substrate. It is now thought that the area of contact with the substrate regulates the number of focal adhesions, hence the strength of signaling through focal adhesion kinase, for example, which promotes the survival, growth, and division of cells. Reference: Folkman J & Moscona A (1978) Role of cell shape in growth control. Nature 273, 345–349. 17–145 If the cell cycle did not increase in length, ts Wee1 cells would become smaller and smaller with each cell division until they could no longer carry out DNA synthesis and mitosis. It is likely that G1 becomes longer because of a size requirement to pass Start. In fission yeast, normal cells are born at a size that is adequate to pass Start and begin DNA synthesis (S phase) almost as soon as they finish mitosis. Although the mechanism is poorly understood, cells of most species measure their own size to make sure they have reached the proper threshold for entering S. If this size-determination mechanism were intact in the ts Wee1 cells, then the small cells generated by division at 37°C would be held in G1 until they grew to the proper size. Thus, the lengthening of G1 would be a consequence of the size requirement for entering S. Reference: Nurse P (1975) Genetic control of cell size at cell division in yeast. Nature 256, 547–551. 17–146 In order to generate more cells of the same size in the same amount of time, both the duration of the cell cycle and the rate of cell growth must have been altered by overexpression of cyclin D and Cdk4. To account for the observed results, the cells must grow faster and divide more quickly (shorter cell cycle). To generate the same size cells, the increase in growth rate must be exactly balanced by the decrease in cell-cycle time. These results suggest that cyclin D and Cdk4 in some way coordinate growth rate and cycle time. (It is interesting to note that flies defective for both copies of Cdk4 have a slower growth rate and a longer cell-cycle time, but maintain normal sized cells.) References: Meyer CA, Jacobs HW, Datar SA, Du W, Edgar BA & Lehner CF (2000) Drosophila Cdk4 is required for normal growth and is dispensable for cell cycle progression. EMBO J. 19, 4533–4542. Datar SA, Jacobs HW, Flor A, de la Cruz A, Lehner CF & Edgar BA (2000) The Drosophila cyclin D-Cdk4 complex promotes cellular growth. EMBO J. 19, 4543–4554. 17–147 Enlargement of cells that are no longer cycling indicates that the primary effect of cyclin D and Cdk4 is on cell growth. If the primary effect were on the cell cycle, the differentiated cells should have been the same size as normal cells. The surprising conclusion of this work is that the cyclin D–Cdk4 complex is a growth regulator in flies and may not be involved in cell-cycle progression. A430 Chapter 17: The Cell Cycle (As an aside, when the authors of this work overexpressed cyclin D and Cdk4 in the whole eye, they noted “…we found that all ommatidiae were enlarged, as was the entire eye, which bulged out of the head in an ominous fashion.”) References: Meyer CA, Jacobs HW, Datar SA, Du W, Edgar BA & Lehner CF (2000) Drosophila Cdk4 is required for normal growth and is dispensable for cell cycle progression. EMBO J. 19, 4533–4542. Datar SA, Jacobs HW, Flor A, de la Cruz A, Lehner CF & Edgar BA (2000) The Drosophila cyclin D-Cdk4 complex promotes cellular growth. EMBO J. 19, 4543–4554. ...
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This note was uploaded on 01/07/2011 for the course BIOLOGY 7.012 taught by Professor Ericlander during the Spring '04 term at MIT.

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