Exam 3 Study Guide 11

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You are studying the synthesis and destruction of cyclins in dividing clam embryos. You suspect that colchicine, a drug that binds to tubulin and arrests cells in mitosis, might work by inhibiting the normal destruction of cyclin at the metaphase-anaphase transition. To test this idea, you add 35S-methionine to a suspension of fertilized clam eggs, divide the suspension in two and add colchicine to one. You take duplicate samples at 5-minute intervals, one for analysis of cyclin by gel electrophoresis and the other for analysis of mitotic chromosomes by fixing and staining the cells. As shown in Figure 3, untreated cells alternated between interphase and mitosis every 30 minutes, whereas colchicine-treated cells entered mitosis normally but remained there for hours. Moreover, colchicine treatment abolished the normal disappearance of cyclin that precedes the metaphase-anaphase transition. To get a clearer picture of how colchicine inhibits cyclin destruction, you repeat the experiment, but add the protein synthesis inhibitor emetine just before cells enter mitosis. As shown in Figure 4, emetine-treated cells in the absence of colchicine entered and exited mitosis normally, and divided into two daughter cells, which then remained indefinitely in interphase. In the presence of colchicine, the emetine-treated cells stayed in mitosis for about 2 hours and then decondensed their chromosomes and re-formed nuclei without dividing. In the presence and absence of colchicine, the exit from mitosis coincided with the disappearance of cyclin. A. What effect does colchicine have on cyclin synthesis and destruction? Can these effects explain how colchicine causes metaphase arrest? B. How does inhibition of protein synthesis eventually reverse the metaphase arrest produced by colchicine?

A Colchicine has no apparent effect on the synthesis of cyclin, since the amount of cyclin continues to increase in colchicine-treated cells. Colchicine does, however, have a dramatic effect on cyclin destruction. Colchicine strongly delays the destruction of cyclin but does not completely block it. Two observations in these experiments point to the continued destruction of cyclin, albeit with significantly altered kinetics. As shown in Figure 3, the amount of cyclin levels off in the presence of colchicine, suggesting that the rate of synthesis is balanced by the rate of destruction. As shown in Figure 4, when cyclin synthesis is blocked by emetine, cyclin is eventually destroyed in the presence of colchicine. The stabilization of cyclin certainly could account for the metaphase arrest produced by colchicine. As long as cyclin levels stay high, MPF will also stay high and keep the cell in M phase. B Inhibition of protein synthesis blocks the synthesis of new cyclin. Because colchicine delays cyclin destruction but does not prevent it, cyclin is eventually destroyed. When cyclin levels fall, MPF activity declines as well. In the absence of MPF activity, cells exit mitosis. In the presence of colchicine the exit from mitosis is rather peculiar: the absence of microtubules prevents normal chromosome segregation and cell division.

EGF stimulates the proliferation of many types of epithelial cells by binding to EGF receptors on their surface. The role of the EGF receptor in propagating the proliferation signal is being clarified by study of the receptor itself. The mouse fibroblast A-431 cell line, which fortuitously carries enormously increased numbers of EGF receptors, makes their characterization much easier. Consider the following set of experiments. 1. Plasma membrane preparations from A-431 cells contain many proteins as shown on the SDS gel in Figure 9A. However, when 125 I-EGF is added to such a preparation in the presence of a protein cross-linking agent, two proteins become labeled (Figure 9B, lane 1). When excess unlabeled EGF is included in the incubation mixture, the labeled band at 170 kd disappears (Figure 9B, lane 2). 2. If the membrane preparation is incubated with gamma32P-ATP, several proteins, including the 170 kd protein, become phosphorylated. This reaction is significantly stimulated by including EGF in the incubation mixture. 3. When antibodies specific for the 170 kd protein are used to precipitate the protein and the incubation with gamma32P-ATP is repeated with the precipitate, the 170 kd protein is phosphorylated in an EGF-stimulated reaction (Figure 9C, lanes 3 and 4). 4. If the antibody-precipitated protein is first run on the SDS gel and then renatured in the gel, subsequent incubation with gamma-32P-ATP in the presence and absence of EGF yields the same pattern shown in Figure 9C. A. Which of these experiments demonstrates most clearly that the 170-kd protein is the EGF receptor? B. Is the EGF receptor a substrate for an EGF-stimulated protein kinase? How do you know? C. Which experiments show that the EGF receptor is a protein kinase? D. Is it clear whether the EGF receptor is a substrate for its own protein kinase activity?

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 gamma-32P-ATP. Transfer of the phosphate from the gamma 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, the activity of the protein kinase is stimulated by EGF. 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 is 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.

Frog oocytes mature into eggs when incubated with progesterone. This maturation is characterized by disappearance of the nucleus (termed germinal vesicle breakdown) and formation of a meiotic spindle. The requirement for progesterone can be bypassed by microinjecting 50 nl of egg cytoplasm directly into a fresh oocyte (1000 nl), which then matures normally (Figure 2). The control experiment of microinjecting cytoplasm from untreated oocytes into other oocytes causes no maturation, as expected. MPF activity in the egg cytoplasm is responsible for maturation. Progesterone-induced maturation requires protein synthesis, as indicated by its sensitivity to cycloheximide; however, MPF-induced maturation does not. By placing progesterone-stimulated oocytes into cycloheximide at different times after stimulation, it can be shown that maturation becomes cycloheximide independent (no longer inhibited by cycloheximide) a few hours before the oocytes become eggs. In addition, the time at which the oocytes become cycloheximide independent corresponds to the appearance of MPF activity. Is synthesis of MPF itself the cycloheximide-sensitive event? To test this possibility, you transfer MPF serially from egg to oocyte to test whether its activity diminishes with dilution. You first microinject 50 nl of cytoplasm from an activated egg into an immature oocyte as shown in Figure 2; when the oocyte matures into an egg, you transfer 50 nl of its cytoplasm into another immature oocyte; and so on. Surprisingly, you find that you can continue this process for at least 10 transfers, even if the recipient oocytes are bathed in cycloheximide! Moreover, the apparent MPF activity in the last egg is equal to that in the first egg. A. What dilution factor is achieved by 10 serial transfers of 50 nl into 1000 nl? Do you consider it likely that a molecule might have an undiminished biological effect over this concentration range? B. How can MPF activity, which is due to a protein, be absent from immature oocytes yet 11.6 Figure 3 Figure 4 appear in activated eggs, even when protein synthesis has been blocked by cycloheximide? C. Propose a means by which MPF might maintain its activity through repeated serial transfers. D. Propose a role for the cycloheximide-sensitive factor that is required for the appearance of MPF activity in a progesterone-stimulated oocyte.

A Since each transfer accomplishes a twentyfold 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 posttranslational modifications, such as protease cleavage or phosphorylation. MPF is activated by phosphorylation. 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, such as trypsin-mediated cleavage of trypsinogen to produce more trypsin. In the case of MPF, however, active MPF functions as a protein kinase that activates its inactive precursor by phosphorylation. Note that it is not necessary that MPF activates itself; for example, it could activate another protein kinase, which in turn activates the precursor to MPF. Nevertheless, the principle is the same. D Since there is no detectable MPF activity in an immature oocyte, MPF cannot be the source of the original activation event. Presumably, a protein synthesized in response to progesterone stimulation (therefore cycloheximide sensitive) is responsible, directly or indirectly, for the initial activation of MPF

A common first step in characterizing cell-division-cycle (cdc) mutants is to define the phase of the cell cycle at which the mutational block stops the cell's progress. Temperature-sensitive cdc mutants are particularly useful because they grow normally at one temperature (the permissive temperature) but express a mutant phenotype when grown at a higher temperature (the restrictive temperature). One method for characterizing temperature-sensitive cdc mutants uses the drug hydroxyurea, which blocks DNA synthesis by inhibiting ribonucleotide reductase (which provides deoxyribonucleotide precursors). Hydroxyurea blockade of DNA synthesis can be reversed simply by changing the incubation medium. Consider the following results with the hypothetical mutants cdc101 and cdc102. You incubate a culture of a yeast cdc101 mutant at its restrictive temperature (37 C) for 2 hours (the approximate length of the cell cycle in yeasts) so that its mutant phenotype is expressed. Then you transfer it to medium containing hydroxyurea at the permissive temperature (20 C). None of the cells divide. You now reverse the order of treatment. You incubate cdc101 at 20 C for 2 hours in medium containing hydroxyurea and then transfer it to medium without hydroxyurea at 37 C. The cells undergo one round of division. You repeat these two experiments with the cdc102 mutant. The cells do not divide in either case. A. In what phase of the cell cycle is cdc101 blocked at the restrictive temperature? Explain the results of the reciprocal temperature-shift experiments. B. In what phase of the cell cycle is cdc102 blocked at the restrictive temperature? Explain the results of the reciprocal temperature-shift experiments.

A The reciprocal temperature-shift experiments demonstrate that cdc101 is blocked at 37 C in the G1 phase of the cell cycle. The first experiment shows that the mutational block precedes 11.24 Figure 18 or coincides with the hydroxyurea block, or the cells would have divided when they were shifted to 20 C in the presence of hydroxyurea. The second experiment 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. Thus in the first experiment, at 37 C the cdc101 mutants accumulate in G1; 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.

One critically important aspect of cell cycle control isensuring that transitions such as entry into mitosis occur rapidly and completely. The regulation of MPF by Wee1 tyrosine kinase and Cdc25 tyrosine phosphatase gives a good appreciation of how this control can be achieved. The balance of the activities of Weel and Cdc25 determines the state of phosphorylation of tyrosine 15 in the Cdc2 component of MPF. When tyrosine 15 is phosphorylated, MPF is inactive; when tyrosine 15 is not phosphorylated, MPF is active (Figure 5). Just as the activity of MPF itself is controlled by phosphorylation, so too are the activities of Weel kinase and Cdc25 phosphatase. The regulation of these various activities can be studied in extracts of frog oocytes. In such extracts Weel kinase is active and Cdc25 phosphatase is inactive. As a result MPF is inactive because its Cdc2 component is phosphorylated on tyrosine 15. MPF in these extracts can be rapidly activated by addition of okadaic acid, which is a potent inhibitor of serine/threonine protein phosphatases. Using antibodies specific for each component, it is possible to examine their phosphorylation state by changes in mobility upon gel electrophoresis (Figure 6). (Phosphorylated proteins generally run slower than their nonphosphorylated counterparts.) A. Based on the results with okadaic acid decide whether the active forms of Wee1 kinase and Cdc25 phosphatase are phosphorylated or nonphosphorylated. In Figure 6 indicate the phosphorylated forms of Weel and Cdc25 and label the arrows to show which are protein kinases and which are protein phosphatases. B. Are the protein kinases and phosphatases that control Weel and Cdc25 specific for serine/threonine residues or tyrosine residues? How do you know? C. How does addition of okadaic acid cause an increase in phosphorylation of Weel and Cdc25, but a decrease in phosphorylation of Cdc2? D. If you assume that Cdc25 and Weel are targets for phosphorylation by Cdc2 kinase in active MPF, can you explain how the appearance of a small amount of active MPF would lead to its rapid and complete activation?

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 as shown in Figure 18. Since this change activates MPF, 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. Knowing which forms are phosphorylated allows you to label the arrows that correspond to the kinases and phosphatases that control Weel and Cdc25 phosphorylation. B. The protein kinases and phosphatases that control phosphorylation of Weel and Cdc25 must be specific for serine/threonine residues because they are affected by okadaic acid, which is specific for serine/threonine phosphatases. C Okadaic acid has no direct effect on Cdc2 phosphorylation because it is phosphorylated on a tyrosine residue. Tyrosine phosphatases are unaffected by okadaic acid. The decrease in Cdc2 phosphorylation is a consequence of the change in activation of Weel kinase and Cdc25 phosphatase. D As soon as some active MPF appears it would begin to phosphorylate Weel and Cdc25, inactivating the kinase and activating the phosphatase. The resultant decrease in Weel kinase activity and increase in Cdc25 phosphatase activity would lead to dephosphorylation (and 11.25 Figure 19 activation) of more MPF. This in turn would further decrease the activity of Weel kinase and further increase the activity of Cdc25 phosphatase, leading to still more MPF activity. Thus the initial appearance of a little MPF activity would rapidly lead to its complete activation. This sort of activation is referred to as a positive feedback loop. 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

Which of the following statements about the human M-phase promoting Cdk (Cdk1) and associated cyclin (cyclin B) are true? A. Cyclin levels oscillate throughout the cell cycle, peaking in mitosis, but Cdk1 levels remain constant. B. Cdk1 levels oscillate throughout the cell cycle, peaking in mitosis, but cyclin B levels remain constant. C. Both Cdk1 and cyclin B levels oscillate throughout the cell cycle, peaking in mitosis. D. Both Cdk1 and cyclin B levels remain constant through the cell cycle. E. Both Cdk1 and cyclin B levels remain constant through the cell cycle, but their activity oscillates.

A. Cyclin levels oscillate throughout the cell cycle, peaking in mitosis, but Cdk1 levels remain constant.

All of the following statements correctly describe M-Cdk, EXCEPT: A. M-Cdk causes the cell to enter S phase and begin DNA replication. B. M-Cdk has two subunits, a protein kinase and a cyclin-type protein. C. M-Cdk only becomes active during M-phase. D. M-Cdk triggers many events by phosphorylating other proteins.

A. M-Cdk causes the cell to enter S

A. M-Cdk causes the cell to enter S phase and begin DNA replication.

Programmed cell death occurs ________________. A. by means of an intracellular suicide program B. rarely and selectively only during animal development C. only in unhealthy or abnormal cells D. only during embryonic development

A. by means of an intracellular suicide program

Cyclins are proteins involved in regulation of A. cell-cycle protein kinases B. circadian rhythms C. synthesis of cAMP D. membrane circulation via exocytosis and endocytosis E. the cycling of tubulin subunits through microtubules

A. cell-cycle protein kinases

The isolation of conditional cell cycle mutations in yeast helped identify many of the key proteins involved in cell cycle control. Many of these mutations were given the name cdc (cell division cycle) and are temperature sensitive. A cdc temperature sensitive mutant will: A. grow normally at the permissive temperature and arrest at a specific phase in the cell cycle at non-permissive temperature. B. grow normally at non-permissive temperature and arrest at a specific phase in the cell cycle at permissive temperatures. C. grow slower at permissive temperature and speed up through the cell cycle at non-permissive temperatures. D. grow normally at permissive temperature and have an extended period of time in G1 of the cell cycle at non-permissive temperatures. E. grow normally at permissive temperature and go backwards through the cell cycle at non-permissive temperatures.

A. grow normally at the permissive temperature and arrest at a specific phase in the cell cycle at non-permissive temperature.

A. if the statement is associated with phorbol ester treatment of HL-60 cells only; B. if the statement is associated with expression of E7 oncoprotein in HL-60 cells only; C. if the statement is associated with both A and B; D if the statement is associated with neither A nor B. Increases the amount of E2F transcription factor-bound RB protein.

A. if the statement is associated with phorbol ester treatment of HL-60 cells only;

One of the Nobel laureates in Physiology and Medicine of 2008, Harald Zur Hausen, discovered that certain types of the human papilloma virus (HPV)1 are the causative agents of cervical cancer. HPV type 16 (HPV-16) is among the most common and best studied human oncogenic viruses. Two oncoproteins (E6 and E7) encoded by the viral genome are responsible for the transformation of normal epithelial cells into tumor cells. Experiments he designed were able to analyze the effects of E7 protein at the molecular level, namely, its relationship to the retinoblastoma protein (RB). Figure 16 shows the results of a preliminary experiment. Cells of a human leukemia cell line (HL-60) were cultured without (Samples 1 and 3) or in the presence of phorbol ester (Samples 2 and 4), and labeled with 35S-methionine (Samples 1 and 2) or 32P-phosphate (Samples 3 and 4). Cell extracts were immunoprecipitated with an antibody specific for the retinoblastoma protein (anti-RB), SDS-polyacrylamide gel electrophoresis was performed followed by autoradiography. A. if the statement is supported by the data or information; B. if the statement is contradicted by the data or information; C. if the statement is neither supported nor contradicted by the date or information All RB protein molecules are phosphorylated in untreated HL-60 cells.

A. if the statement is supported by the data or information;

One of the Nobel laureates in Physiology and Medicine of 2008, Harald Zur Hausen, discovered that certain types of the human papilloma virus (HPV)1 are the causative agents of cervical cancer. HPV type 16 (HPV-16) is among the most common and best studied human oncogenic viruses. Two oncoproteins (E6 and E7) encoded by the viral genome are responsible for the transformation of normal epithelial cells into tumor cells. Experiments he designed were able to analyze the effects of E7 protein at the molecular level, namely, its relationship to the retinoblastoma protein (RB). Figure 16 shows the results of a preliminary experiment. Cells of a human leukemia cell line (HL-60) were cultured without (Samples 1 and 3) or in the presence of phorbol ester (Samples 2 and 4), and labeled with 35S-methionine (Samples 1 and 2) or 32P-phosphate (Samples 3 and 4). Cell extracts were immunoprecipitated with an antibody specific for the retinoblastoma protein (anti-RB), SDS-polyacrylamide gel electrophoresis was performed followed by autoradiography. A. if the statement is supported by the data or information; B. if the statement is contradicted by the data or information; C. if the statement is neither supported nor contradicted by the date or information. The phorbol ester triggers dephosphorylation of the RB protein in HL-60 cells.

A. if the statement is supported by the data or information;

Which of the following is a known mechanism by which p53 activation or inactivation can lead to a block 11.14 Figure 12 (arrest) of the cell cycle? A. p53 acts as a transcription factor inducing the synthesis of p21, a Cdk inhibitor. Inhibition of Cdk's causes the cell cycle to stall. B. Activation of p53 increases the translational control of DNA repair enzymes. The activity of the DNA repair enzymes causes the cell cycle to stall. C. When the amount of p53 mRNA increases in the cell, the p53 mRNA is antisense to cyclin mRNA causing silencing of cyclin. The subsequent decrease in cyclin content in the cell causes the cell cycle to stall. D. The activity of the proteasome is increased by p53. The decrease in ubiquinated proteins causes the cell cycle to stall. E. The loss of p53 reduces the kinase activity of Wee1. The loss of the activator phosphorylation event on Cdk causes the cell cycle to stall.

A. p53 acts as a transcription factor inducing the synthesis of p21, a Cdk inhibitor. Inhibition of Cdk's causes the cell cycle to stall.

8 The total length of the cell cycle of cells in a cell culture is 24 hours. M phase is found to be 2 hours, S phase is found to be 9 hours and G2 is found to be 4.5 hours. How long is G1? A. Insufficient data to make a determination B. 8.5 hours C. 10 hours D. 4.5 hours E. 1 hour

B. 8.5 hours

A human cell's incorporation of radioactive uracil indicates the rate at which it is progressing through S phase. A. True B. False

B. False

Tagging a protein such as a cyclin or p21 with ubiquinone leads to its degradation by a proteasome. A. True B. False

B. False

B. G1

When activated by the signal, the platelet-derived growth factor (PDGF) receptor phosphorylates itself on multiple tyrosines (as indicated below by the circled Ps; the numbers next to these Ps indicate the amino acid number of the tyrosine). These phosphorylated tyrosines serve as docking sites for proteins that interact with the activated PDGF-receptor. These proteins are indicated in the figure below, and include the proteins A, B, C, and D. The PDGF-receptor is activated when it binds to the signal, PDGF. One of the cellular responses to PDGF is an increase in DNA synthesis, which can be measured by incorporation of radioactive thymidine into the DNA. To determine whether protein A, B, C, and/or D are responsible for activation of DNA synthesis, you construct mutant versions of the PDGF-receptor that retain one or more tyrosine phosphorylation sites. You express these mutant versions in cells that do not make a PDGF-receptor. In these cells, the various versions of the PDGF-receptor are expressed normally and in response to PDGF binding, become phosphorylated on whichever tyrosines remain. You measure the level of DNA synthesis in cells that express the various mutant receptors and obtain the data shown in Figure 10. Which, if any, of these proteins inhibit DNA synthesis? A. Protein A B. Protein B C. Protein C D. Protein D E. None of the proteins

B. Protein B

Transcription of gene X is controlled by transcription factor A. Gene X is only transcribed when factor A is phosphorylated. Data on the tissue distribution of factor A and the activities of a protein kinase and a protein phosphatase specific for factor A are presented in the table below. Of these three tissues, gene X will most likely have the highest levels of transcription in A. muscle tissue B. heart tissue D. brain tissue E. brain and heart tissues

B. heart tissue

A. if both assertion and reason are true statements and the reason is a correct explanation of the assertion; B. if both assertion and reason are true statements but the reason is not a correct explanation of the assertion; C. if the assertion is true but the reason is a false statement; D. if the assertion is false but the reason is a true statement; E. if both assertion and reason are false statements. Phorbol ester treatment must have inhibited the proliferation of HL-60 cells, because it stimulated the binding of RB protein to the E7 oncoprotein.

B. if both assertion and reason are true statements but the reason is not a correct explanation of the assertion;

B. if the statement is associated with expression of E7 oncoprotein in HL-60 cells only;

One of the Nobel laureates in Physiology and Medicine of 2008, Harald Zur Hausen, discovered that certain types of the human papilloma virus (HPV)1 are the causative agents of cervical cancer. HPV type 16 (HPV-16) is among the most common and best studied human oncogenic viruses. Two oncoproteins (E6 and E7) encoded by the viral genome are responsible for the transformation of normal epithelial cells into tumor cells. Experiments he designed were able to analyze the effects of E7 protein at the molecular level, namely, its relationship to the retinoblastoma protein (RB). Figure 16 shows the results of a preliminary experiment. Cells of a human leukemia cell line (HL-60) were cultured without (Samples 1 and 3) or in the presence of phorbol ester (Samples 2 and 4), and labeled with 35S-methionine (Samples 1 and 2) or 32P-phosphate (Samples 3 and 4). Cell extracts were immunoprecipitated with an antibody specific for the retinoblastoma protein (anti-RB), SDS-polyacrylamide gel electrophoresis was performed followed by autoradiography. A. if the statement is supported by the data or information; B. if the statement is contradicted by the data or information; C. if the statement is neither supported nor contradicted by the date or information. The RB protein is induced by phorbol ester in HL-60 cells.

B. if the statement is contradicted by the data or information;

A mutant yeast strain stops proliferating when shifted from 25°C to 37°C. When these cells are analyzed at the two different temperatures, using a machine that sorts cells according to the amount of DNA they contain, the graphs in Figure 13 are obtained. Which of the following would NOT explain the results with the mutant? A. inability to initiate DNA replication B. inability to begin M phase C. inability to activate proteins needed to enter S phase D. inappropriate production of a signal that causes the cells to remain in G1 E. absence of S-cyclin

B. inability to begin M phase

You have isolated a strain of mutant yeast cells that divides normally at 30°C but cannot enter M phase at 37°C. You have isolated its mitotic cyclin and mitotic Cdk and find that both proteins are produced and can form a normal M-Cdk complex at both temperatures. Which of the following temperature-sensitive mutations could not be responsible for the behavior of this strain of yeast? A. inactivation of a protein kinase that acts on the mitotic Cdk kinase B. inactivation of an enzyme that ubiquitylates M cyclin C. inactivation of a phosphatase that acts on the mitotic Cdk kinase D. a decrease in the levels of a transcriptional regulator required for producing sufficient amounts of M

B. inactivation of an enzyme that ubiquitylates M cyclin

Which of the following would INHIBIT the onset of mitosis? A. binding of M Cyclin to Cdk B. phosphorylation of Cdk by Wee1 C. phosphorylation of Wee1 by Cdk D. dephosphorylation of Cdk by Cdc25

B. phosphorylation of Cdk by Wee1

In a recent study of the control of cell division (Groth et al. 2008. Science 318:1928-1931), the researchers used synchronously dividing cultured human cells. Cells were synchronized using a technique called a double-thymidine-block, which is performed by incubating dividing cells twice in the presence of a high concentration of thymidine (phosphate-lacking nucleotide), with an intervening incubation without thymidine. During the initial thymidine treatment, cells in mid-S-phase immediately arrest and other cells will stop dividing when they reach the G1/S boundary. Incubation in medium lacking thymidine is then allowed long enough for the cells to pass through S-phase before a second thymidine treatment arrests all cells at the G1/S boundary. Placing the cells in medium lacking thymidine releases the 'block' and cells begin to divide synchronously. At different time points after releasing the block, FACS (fluorescence activated cell sorting) can be used to count cells and measure their DNA content. This was done for synchronously and asynchronously dividing cells. Results for 3 samples are shown in Figure 14, in which number of cells is plotted on the y-axis and cell DNA content is on the x-axis. Which sample or samples (A, B, C) best presents the expected results for asynchronously dividing cells? A. A B. B C. C D. A and C E. A and B

C. C

In a recent study of the control of cell division (Groth et al. 2008. Science 318:1928-1931), the researchers used synchronously dividing cultured human cells. Cells were synchronized using a technique called a double-thymidine-block, which is performed by incubating dividing cells twice in the presence of a high concentration of thymidine (phosphate-lacking nucleotide), with an intervening incubation without thymidine. During the initial thymidine treatment, cells in mid-S-phase immediately arrest and other cells will stop dividing when they reach the G1/S boundary. Incubation in medium lacking thymidine is then allowed long enough for the cells to pass through S-phase before a second thymidine treatment arrests all cells at the G1/S boundary. Placing the cells in medium lacking thymidine releases the 'block' and cells begin to divide synchronously. At different time points after releasing the block, FACS (fluorescence activated cell sorting) can be used to count cells and measure their DNA content. This was done for synchronously and asynchronously dividing cells. Results for 3 samples are shown in Figure 14, in which number of cells is plotted on the y-axis and cell DNA content is on the x-axis. Using this system Groth et al. investigated the role of the protein Asf1 in the cell cycle. In one study they repressed Asf1 expression via RNA interference by treating the cells with siAsf1, a small RNA that targets Asf1 mRNA. As a control, cells were treated with siGFP, which would repress expression of the GFP gene. The results for asynchronous cells are shown in Figure 15. siAsf1 treatment most likely causes accumulation of cells in which stage of the cell cycle? A. Entering S B. Entering G0 C. Leaving S D. Entering M E. Leaving Metaphase

C. Leaving S

Which of the following descriptions is consistent with the behavior of a cell that lacks a protein required for a checkpoint mechanism that operates in G2? 11.18 A. The cell would be unable to enter M phase. B. The cell would be unable to enter G2. C. The cell would enter M phase under conditions when normal cells would not. D. The cell would pass through M phase more slowly than normal cells.

C. The cell would enter M phase under conditions when normal cells would not.

Figure 12 shows that injection of M-phase cytoplasm into an oocyte triggers the cell to enter mitosis. What would be the expected effect on cell division if cytoplasm from a G2-phase cell were injected into an M-phase cell? A. DNA replication would begin again. B. All cyclin-CDK complexes will be activated. C. There will be little or no effect. D. Mitosis will immediately stop

C. There will be little or no effect.

The results of the experiment carried out to analyze the function of the E7 protein are shown in Figure 17. The human leukemia cells were cultured in the absence (Samples 1 to 3) or presence of phorbol ester (Samples 4 to 6). Cell extracts were prepared and aliquots were incubated with agarose beads to which E7 protein molecules had been covalently attached. After incubation the beads were pelleted by centrifugation and Western blot analysis after SDS-polyacrylamide gel electrophoresis was performed using the anti-RB antibody with total cell extracts (Samples 1 and 4), the supernatants (Samples 2 and 5), and the sediments (Samples 3 and 6). What was the aim of using this E7-agarose bead protocol? A. To study the expression of RB protein 11.17 B. To determine the rate of phosphorylation of RB protein C. To analyze E7-RB interactions. D. To purify the RB protein E. To dephosphorylate the E7 protein.

C. To analyze E7-RB interactions.

An example of a constitutively ON mutation that affects the cell division cycle would be: A. a mutation in p53 that constitutively enhances its DNA binding activity. B. a mutation in a CKI that enhances its binding to a CDK. C. a mutation that constitutively increases Cdc25 phosphatase activity. D. a mutation in a CDK that blocks ATP binding. E. a mutation in a CDK that blocks cyclin binding.

C. a mutation that constitutively increases Cdc25 phosphatase activity.

Cyclin dependent kinases which control progression through cell cycle checkpoints are activated by which of the following: A. release from binding by cyclins. B. phosphorylation by Cdk activating protein kinase. C. binding to cyclin and phosphorylation by a Cdk activating protein kinase. D. phosphorylation by a tyrosine kinase. E. increased transcription of CDK genes.

C. binding to cyclin and phosphorylation by a Cdk activating protein kinase.

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

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

Which of the following statements is FALSE? A. Mitotic Cdk must be phosphorylated by an activating kinase (CAK) before it is active. B. Phosphorylation of mitotic Cdk by the inhibitory kinase (Wee1) makes the Cdk inactive, even if it is phosphorylated by the activating kinase. C. Active M-Cdk phosphorylates the activating phosphatase (Cdc25) in a positive feedback loop. D. The activating phosphatase (Cdc25) removes all phosphates from mitotic Cdk so that M-Cdk will be active.

D. The activating phosphatase (Cdc25) removes all phosphates from mitotic Cdk so that M-Cdk will be active.

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

D. The cells produced would get smaller and smaller.

Cultured (grown in vitro) mouse fibroblasts may enter a resting state (G0 ) that can be reversed by a number of agents, including serum. The upper panel (A) of Figure 11 shows the changes in cGMP concentration and cAMP concentration after the addition of serum to a resting culture of mouse fibroblasts. Figure 11B shows that the resting state has been reversed by addition of serum as measured by an assay for DNA synthesis, the number of nuclei in which radioactively labeled thymidine is found, and, by change in the number of cells. The data support all of the following statements, EXCEPT: A. Decreased concentrations of cAMP is correlated with the onset of mitosis. B. cGMP concentration increased transiently during G1. C. cAMP concentration correlates inversely with cGMP concentration until S phase D. The ratio of cAMP concentration to cGMP concentration is constant throughout the cell cycle. E. A decrease in cGMP concentration precedes S phase.

D. The ratio of cAMP concentration to cGMP concentration is constant throughout the cell cycle.

Which of the following is NOT good direct evidence that the cell-cycle control system is conserved through a billion years of divergent evolution? A. A yeast cell lacking Cdk function can use the human Cdk to substitute for its missing Cdk during the cell cycle. B. The amino acid sequences of cyclins in plants are similar to the amino acid sequences of cylins in humans. C. The Cdk proteins in humans share conserved phosphorylation sites with the Cdk proteins in yeast. D. Yeast cells have only one Cdk, whereas humans have many Cdks.

D. Yeast cells have only one Cdk, whereas humans have many Cdks.

Two opposing processes that keep the number of somatic cells at the proper level are: A. mitosis and cytokinesis. B. apoptosis and interphase. C. cell division and cytokinesis. D. cell division and apoptosis

D. cell division and apoptosis

Levels of Cdk activity change during the cell cycle, in part, because ________________. A. the Cdks phosphorylate each other B. the Cdks activate the cyclins C. Cdk degradation precedes entry into the next phase of the cell cycle D. cyclin levels change during the cycle

D. cyclin levels change during the cycle

A. if both assertion and reason are true statements and the reason is a correct explanation of the assertion; B. if both assertion and reason are true statements but the reason is not a correct explanation of the assertion; C. if the assertion is true but the reason is a false statement; D. if the assertion is false but the reason is a true statement; E. if both assertion and reason are false statements. HL-60 cells were in the Go phase in the absence of phorbol esters, because their RB proteins were fully phosphorylated.

D. if the assertion is false but the reason is a true statement;

The Anaphase Promoting Complex (APC) A. controls G1-S transition B. phosphorylates M-phase cyclins, leading to exit from mitosis C. activates cohesins that "glue" sister chromatids together D. mediates ubiquitin-dependent protein degradation E. is activated in response to DNA damage

D. mediates ubiquitin-dependent protein degradation

The role of 'cyclin' in the regulation of the cell cycle would be best compared to: A. a digital watch that produces a precisely timed signal every few microseconds. B. a row of dominoes, that all fall sequentially after the first one is flipped. C. a light switch that alternates between on and off states. D. the accumulation of sand in an hourglass.

D. the accumulation of sand in an hourglass.

What determines the length of S phase? One possibility is that its length depends on how much DNA the nucleus contains. As a test, you measure the length of S phase in dividing cells of a lizard, a frog, and a newt, each one of which has a different amount of DNA. As shown in Figure 1, the length of S phase does increase with increasing DNA content. Even though these organisms are similar in that in that they are all cold blooded, they are different species. You recall that it is possible to obtain haploid embryos of frogs and repeat your measurements with haploid and diploid frog cells. Haploid frog cells have the same length S phase as diploid frog cells. Further research in the literature show that in plants, tetraploid strains of beans and oats have the same length S phase as their diploid cousins. Propose an explanation to reconcile these apparently contradictory results. Why is it that the length of S phase increases with increasing DNA content in different species but remains constant with increasing DNA content in the same species?

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 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 polymerases, helicases, initiation factors, etc.) to the amount of DNA would not change. By contrast, the DNA of different organisms might well be expected to have different ratios of critical genes to DNA content, which could account for the correlation seen in Figure 1.

Fission yeast respond to damaged DNA and unreplicated DNA by delaying entry into mitosis. You want to know how the signals from such DNA interact with the mitotic entry checkpoint. You screen a large number of mutant yeast strains and find six that do not delay mitosis in response to DNA damage, unreplicated DNA, or both, as shown in Figure 7. Which of the signaling pathways shown in Figure 8 is supported by your data? On the pathway you choose, indicate where each of the mutant genes acts.

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 both kinds of DNA. If they used independent pathways, each mutant would respond to only one or the other DNAs. As indicated in Figure 19, rad24 affects a part of the pathway that is specific to 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 part of the pathway that is common to signals from damaged DNA and unreplicated DNA.

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