cell bio ch 17-20

17-1 Identify the cytoskeletal structures (black lines) depicted in the epithelial cells shown in Figure Q17-1. Figure Q17-1

(A) microtubules (B) intermediate filaments (C) actin

17-2 Which of the following statements about the cytoskeleton is false?

Covalent bonds between protein monomers hold together cytoskeletal filaments.

17-3 Indicate which of the three major classes of cytoskeletal elements each statement below refers to. A. monomer that binds ATP B. includes keratin and neurofilaments C. important for formation of the contractile ring during cytokinesis D. supports and strengthens the nuclear envelope E. their stability involves a GTP cap F. used in the eukaryotic flagellum G. a component of the mitotic spindle H. can be connected through desmosomes I. directly involved in muscle contraction J. abundant in filopodia

A. actin B. intermediate filaments C. actin D. intermediate filaments E. microtubules F. microtubules G. microtubules H. intermediate filaments I. actin J. actin

17-4 Which of the following statements about the cytoskeleton is true?

Actin filaments and microtubules have an inherent polarity, with a plus end that grows more quickly than the minus end.

17-5 Rank the following cytoskeletal filaments from smallest to largest in diameter (1 = smallest in diameter, 4 = largest) ______ intermediate filaments ______ microtubules ______ actin filament ______ myofibril Intermediate Filaments

__2___ intermediate filaments (10 nm diameter) __3___ microtubules (25 nm) __1___ actin filament (5-9 nm) __4___ myofibril (1-2 μm)

17-6 Which of the statements below about intermediate filaments is false?

Each filament is about 10 μm in diameter.

17-7 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. Intermediate filaments are elongated fibrous proteins with an N-terminal globular _________________ region and a C-terminal globular _________________ region; these regions flank the elongated rod domain. The α-helical region of the rod interacts with the α-helical region of another monomer in a _____________________ configuration to form a dimer. ______________ dimers will line up to form a staggered tetramer. ______________ strands of tetramers come together and twist together to form the _________________ nm filament. The ___________________ domains are exposed on the surface of the intermediate filament, allowing for interaction with cytoplasmic components. antiparallel four tail β barrel globular ten coiled-coil head trim

Intermediate filaments are elongated fibrous proteins with an N-terminal globular head region and a C-terminal globular tail region; these regions flank the elongated rod domain. The α-helical region of the rod interacts with the α-helical region of another monomer in a coiled-coil configuration to form a dimer. Two dimers will line up to form a staggered tetramer. Eight strands of tetramers come together and twist together to form the ten nm filament. The globular domains are exposed on the surface of the intermediate filament, allowing for interaction with cytoplasmic components.

17-8 All intermediate filaments are of similar diameter because ____________.

the central rod domains are similar in size and amino acid sequence.

17-9 Intermediate filaments help protect animal cells from mechanical stress because ____________.

filaments in each cell are indirectly connected to the filaments of a neighboring cell through the desmosome, creating a continuous mechanical link between cells.

17-10 Intermediate filaments are made from elongated fibrous proteins that are assembled into a ropelike structure. Figure Q17-10 shows the structure of an intermediate filament subunit. You are you an interested in how intermediate filaments are formed, and create intermediate filament subunit whose α-helical region is twice as long as that of a normal intermediate filament by duplicating the normal α-helical region while keeping a globular head at the N-terminus and a globular tail at the C-terminus; you call this subunit IFαd. If you were to assemble intermediate filaments using IFαd as the subunit, which of the following predictions describes the most likely outcome? Figure Q17-10

Filaments assembled using IFαd will form dimers that are twice as long as dimers assembled from normal intermediate filaments.

17-11 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. Intermediate filaments are found mainly in cells that are subject to mechanical stress. Gene mutations that disrupt intermediate filaments cause some rare human genetic diseases. For example, the skin of people with epidermolysis bullosa simplex is very susceptible to mechanical injury; people with this disorder have mutations in their __________________ genes, which code for the intermediate filament found in epithelial cells. These filaments are usually connected from cell to cell through junctions called __________________s. The main filaments found in muscle cells belong to the __________________ family; people with disruptions in these intermediate filaments can have muscular dystrophy. In the nervous system, ___________

Intermediate filaments are found mainly in cells that are subject to mechanical stress. Gene mutations that disrupt intermediate filaments cause some rare human genetic diseases. For example, the skin of people with epidermolysis bullosa simplex is very susceptible to mechanical injury; people with this disorder have mutations in their keratin genes, the intermediate filament found in epithelial cells. These filaments are usually connected from cell to cell through junctions called desmosomes. The main filaments found in muscle cells belong to the vimentin family; people with disruptions in these intermediate filaments can have muscular dystrophy. In the nervous system, neurofilaments help strengthen the extremely long extensions often present in nerve cell axons; disruptions in these intermediate filaments can lead to neurodegeneration. People who carry mutations in the gene for plectin, an important protein for cross-linking intermediate filaments, have a disease that combines symptoms of epidermolysis bullosa simplex, muscular dystrophy, and neurodegeneration. Humans with progeria, a disease that causes premature aging, carry mutations in a nuclear lamin.

17-12 Match the type of intermediate filament with its appropriate location. lamins _________ A. nerve cells neurofilaments _________ B. epithelia vimentins _________ C. nucleus keratins _________ D. connective tissue

lamins ____C____ neurofilaments ____A____ vimentins ____D____ keratins ____B____

17-13 Keratins, neurofilaments, and vimentins are all categories of intermediate filaments. Which of the following properties is not true of these types of intermediate filaments?

Phosphorylation causes disassembly during every mitotic cycle.

17-14 Phosphorylation of nuclear lamins regulates their assembly and disassembly during mitosis. You add a drug to cells undergoing mitosis that inhibits the activity of an enzyme that dephosphorylates nuclear lamins. What do you predict will happen to these cells? Why?

Cells should become arrested in mitosis. Normally, the lamins are phosphorylated during mitosis, causing disassembly of the nuclear envelope. At the end of mitosis, the nuclear lamins are dephosphorylated, causing the lamins to reassemble. Inhibition of this last step should therefore prevent the nuclear lamins from reassembling after mitosis.

17-15 You are studying nuclear lamins in yeast. Using recombinant DNA technology, you alter the coding sequence of a nuclear lamin gene such that the gene now codes for a nuclear lamin protein that can no longer be phosphorylated when the nuclear envelope is broken down during mitosis. What do you predict would happen if the yeast cell only had the altered nuclear lamin gene (and not the unaltered version)?

Nuclear lamins will no longer disassemble properly during mitosis.

17-16 You are interested in understanding the regulation of nuclear lamina assembly. To create an in vitro system for studying this process you start with partly purified nuclear lamina subunits to which you will add back purified cellular components to drive nuclear lamina assembly. Before you start doing experiments, your instructor suggests that you consider what type of conditions would be most amenable to the assembly of the nuclear lamina from its individual subunits in vitro. Which of the following conditions do you predict would be most likely to enhance the assembly of the nuclear lamina?

addition of protein kinase inhibitors Microtubules

17-17 Place the following in order of size, from the smallest to the largest. A. protofilament B. microtubule C. α-tubulin D. tubulin dimer E. mitotic spindle

C, D, A, B, E

17-18 In the three cell outlines in Figure Q17-18, indicate the arrangement of the microtubules, showing clearly their free and attached ends. On each figure, indicate the plus end and the minus end for one of the microtubules. Figure Q17-18

See Figure A17-18. Figure A17-18

17-19 Which of the following statements about the structure of microtubules is false?

α-Tubulin and β-tubulin are covalently bound to make the tubulin dimer that then assembles into protofilaments.

17-20 Which of the following statements about the function of the centrosome is false?

Microtubules emanating from the centrosome have alternating polarity such that some have their plus end attached to the centrosome while others have their minus end attached to the centrosome.

17-21 Which of the following statements about microtubules is true?

Motor proteins move in a directional fashion along microtubules by using the inherent structural polarity of a protofilament.

17-22 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. Microtubules are formed from the tubulin heterodimer, which is composed of the nucleotide-binding __________________ protein and the __________________ protein. Tubulin dimers are stacked together into protofilaments; __________________ parallel protofilaments form the tubelike structure of a microtubule. __________________ rings are important for microtubule nucleation and are found in the __________________ , which is usually found near the cell's nucleus in cells that are not undergoing mitosis. A microtubule that is quickly growing will have a __________________ cap that helps prevent the loss of subunits from its growing end. Stable microtubules are used in cilia and flagella; these microtubules are nucleated from a ____

Microtubules are formed from the tubulin heterodimer, which is composed of the nucleotide-binding β-tubulin protein and the α-tubulin protein. Tubulin dimers are stacked together into protofilaments; thirteen parallel protofilaments form the tubelike structure nucleation and of a microtubule. γ-Tubulinrings are important for microtubule are found in the centrosome, which is usually found near the cell's nucleus in cells that are not undergoing mitosis. A microtubule that is quickly growing will have a GTP cap that helps prevent the loss of subunits from its growing end. Stable microtubules are used in cilia and flagella; these microtubules are nucleated from a basal body and involve a "nine plus two" array of microtubules. The motor protein dynein generates the bending motion in cilia; the lack of this protein can cause Kartagener's syndrome in humans.

17-23 The hydrolysis of GTP to GDP carried out by tubulin molecules ________________.

allows the behavior of microtubules called dynamic instability.

17-24 The microtubules in a cell form a structural framework that can have all the following functions except which one?

strengthening the plasma membrane

17-25 Your friend discovers a protein that she names EBP. EBP binds to microtubule plus ends, and she hypothesizes a role for EBP in increasing dynamic instability. To determine the function of EBP, she examines its effect on microtubules. She polymerizes microtubules from purified centrosomes in a Petri plate and determines the number of shrinking microtubules over a three-minute time interval for different concentrations of EBP. The data she obtained are shown in Figure Q17-25. Figure Q17-25 Is this result consistent with her hypothesis? Explain.

No, the data graphed in Figure Q17-25 are not consistent with her hypothesis. If EBP were to increase the dynamic instability of microtubules, you would expect an increase in Discover more both theducational number resources and start of shrinking earning by and microtubules selling your the own documents number at Schootex.com of growing microtubules. Since the graph shows a reduction in the number of shrinking microtubules, dynamic instability has evidently decreased. Further experiments would be needed to determine the net effect of these changes on microtubules and whether they become longer or shorter on average following EBP treatment.

17-26 You discover a protein, MtA, and find that it binds to the plus ends of microtubules in cells. The hypothesis that best explains this localization is ________________.

MtA binds to GTP-bound tubulin on microtubules.

17-27 You are curious about the dynamic instability of microtubules and decide to join a lab that works on microtubule polymerization. The people in the lab help you grow some microtubules in culture using conditions that allow you to watch individual microtubules under a microscope. You can see the microtubules growing and shrinking, as you expect. The professor who runs the lab gets in a new piece of equipment, a very fine laser beam that can be used to sever microtubules. She is very excited and wants to sever growing microtubules at their middle, using the laser beam. A. Do you predict that the newly exposed microtubule plus ends will grow or shrink? Explain your answer. B. What do you expect would happen to the newly exposed plus ends if you were to grow the microtubules in the presence of an analog of GTP that cannot be with a laser hydrolyzed, and you then severed the microtubules in the middle beam?

A. The newly exposed microtubule plus ends will most probably shrink if you sever the microtubules in the middle. This is because a microtubule grows by adding GTP-carrying subunits to the plus end. The GTP is hydrolyzed over time, leaving only a cap of GTP-carrying subunits at the plus end, with the remainder of the tubulin protofilament containing GDP-carrying subunits. Therefore, if you sever a growing microtubule in its middle, you will most probably create a plus end that contains GDP-carrying subunits. The GDP-carrying subunits are less tightly bound than the GTP-carrying subunits and will peel away from each other, causing depolymerization of the microtubule and shrinkage. B. If you were to polymerize the microtubules in the presence of a nonhydrolyzable analog of GTP and you then severed the microtubules with a laser, the newly exposed plus end would contain a GTP cap and so would probably continue to grow.

17-28 Which of the following statements regarding dynamic instability is false?

The newly freed tubulin dimers from a shrinking microtubule can be immediately captured by growing microtubules and added to their plus end.

17-29 Which of the situations below will enhance microtubule shrinkage?

addition of a drug that inhibits GTP exchange on free tubulin dimers

17-30 The graph in Figure Q17-30 shows the time course of the polymerization of pure tubulin in vitro. You can assume that the starting concentration of free tubulin is much higher than it is in cells. Figure Q17-30 A. Explain the reason for the initial lag in the rate of microtubule formation. B. Why does the curve level out after point C?

A. Before they can polymerize to form microtubules, tubulin molecules have to form small aggregates that act as nucleation centers. This aggregation step is slow because the molecules have to come together in the right configuration. This is why there is a lag phase before microtubules start to be formed. B. After point C, an equilibrium point has been reached, where the rates of polymerization and depolymerization are exactly balanced.

17-31 The graph in Figure Q17-31 shows the time course of the polymerization of pure tubulin in vitro. Assume that the starting concentration of free tubulin is higher than it is in cells. Figure Q17-31 Three parts of the curve are labeled above it as A, B, and C. You conduct a similar in vitro tubulin-polymerization experiment, only you include purified centrosomes in your preparation. When you plot your data, which part of your graph should be most dissimilar to the curve shown in Figure Q17-31?

A

17-32 Which of the following statements about organellar movement in the cell is false?

Only the microtubule cytoskeleton is involved in organellar movement.

17-33 Microtubules are important for transporting cargo in nerve cell axons, as diagrammed in Figure Q17-33. Notice that the two types of cargo are traveling in opposite directions. Which of the following statements is likely to be false? Figure Q17-33

The black cargo and the gray cargo are moving along microtubules of opposite polarity.

17-34 Do you agree or disagree with this statement? Explain your answer. Minus-end directed microtubule motors (like dyneins) deliver their cargo to the periphery of the cell, whereas plus-end directed microtubule motors (like kinesins) deliver their cargo to the interior of the cell.

Disagree. The plus ends of microtubules usually point toward the cell periphery, whereas the minus ends point toward the cell center. This is because the γ-tubulin in the centrosome serves to nucleate microtubule growth. Because the centrosomes are near the center of the cell, the minus ends of microtubules are located there. Therefore, a minus-end directed microtubule motor would direct its cargo toward the center of the cell, and a plus-end directed microtubule motor would direct its cargo toward the cell periphery.

17-35 Which of the following statements is correct? Kinesins and dyneins ____________________.

often move in opposite directions to each other.

17-36 Match the following labels to the numbered lines on Figure Q17-36. Figure earning by selling your own documents at Schootex.com A. minus end of microtubule B. tail of motor protein C. cargo of motor protein D. head of motor protein Which of the two motors in Figure Q17-36 is most probably a kinesin? Explain your answer.

A—4; B—2; C—1; D—3 The top motor is more likely to be kinesin, because kinesins usually move toward the plus end of the microtubules.

17-37 Some lower vertebrates such as fish and amphibians can control their color by regulating specialized pigment cells called melanophores. These cells contain small, pigmented organelles, termed melanosomes, that can be dispersed throughout the cell, making the cell darker, or aggregated in the center of the cell to make the cell lighter. You purify the melanosomes from melanophores that have either aggregated or dispersed melanosomes and find that: 1. aggregated melanosomes co-purify with dynein; 2. dispersed melanosomes co-purify with kinesin. Given this set of data, propose a mechanism for how the aggregation and dispersal of melanosomes occur.

The melanosomes are transported in the cell on microtubules. When it is advantageous for the animal to become lighter, a signal is sent to the pigment cell that causes the melanosomes to associate with dynein. Because dynein is a minus-end directed motor, it will transport the melanosomes toward the center of the cell, causing the melanosomes to aggregate in the center and the cell to take on a lighter appearance. When the animal wants to become darker, a signal is sent to the pigment cell that causes the melanosomes to associate with kinesin. Kinesin is usually a plus-end directed motor and will move the melanosomes away from the center of the cell so that they are more dispersed, making the cell look darker.

17-38 Indicate whether each of the following statements refers to a ciliary microtubule, a microtubule of the mitotic spindle, both types of microtubule, or neither type of microtubule. A. The basal body is the organizing center. B. The monomer is sequestered by profilin. C. It is arranged in a "9 + 2" array. D. It is nucleated at the centrosome. E. It uses dynein motors. F. It is involved in sperm motility. G. It is involved in moving fluid over the surface of cells.

A. ciliary microtubules B. neither C. ciliary microtubules D. microtubules of the mitotic spindle E. both F. neither (this involves flagellar microtubules) G. ciliary microtubules

17-39 Which of the following items is not important for flagellar movement?

sarcoplasmic reticulum

17-40 Figure Q17-40A shows how the movement of dynein causes the flagellum to bend. If instead of the normal situation, the polarity of the adjacent doublet of microtubules were to be reversed (see Figure Q17-40B), what do you predict would happen? Figure Q17-40

Bending would occur, except that the right microtubule doublet would move down relative to the left one.

17-41 Figure Q17-41 shows two isolated outer-doublet microtubules from a eukaryotic flagellum with their associated dynein molecules. Figure Q17-41 A. Sketch what will happen to this structure if it is supplied with ATP. B. Sketch what will happen to this structure if the linking proteins are removed and it is supplied with ATP. C. In a complete flagellum, what would happen if all the dynein molecules were active at the same time? Actin Filaments

A. See Figure A17-41A. B. See Figure A17-41B. [Note to instructor: this question should be marked as correct only if the microtubule is shown bending in the correct direction (in A) and the correct microtubule is shown pushed forward (in B)]. Figure A17-41 C. The flagellum will not bend because there is no significant relative motion of one microtubule doublet to another: each is trying to push its neighbor forward at the same time. For the flagellum to bend, sets of dynein molecules on one side of the flagellum must be selectively activated.

17-42 Which of the following statements about actin is false?

The dynamic instability of actin filaments is important for cell movement.

17-43 Actin can adopt a variety of shapes. Match the name of the actin form with the type of actin structure depicted as black lines within the cells in Q17-43. Figure Q17-43 _____ lamellipodia _____ contractile bundles _____ contractile ring _____ microvilli

__C__ lamellipodia __B__ contractile bundles __D__ contractile ring __A__ microvilli

17-44 Consider the mechanism by which actin and tubulin polymerize. Which of the items below does not describe something similar about the polymerization mechanisms of actin and microtubules?

Depolymerization initiates at the plus ends of filaments.

17-45 For both actin and microtubule polymerization, nucleotide hydrolysis is important for ______.

decreasing the binding strength between subunits on filaments.

17-46 Compared to the normal situation, in which actin monomers carry ATP, what do you predict would happen if actin monomers that bind a nonhydrolyzable form of ATP were incorporated into actin filaments?

Actin filaments would grow longer.

17-47 Cytochalasin is a drug that caps actin filament plus ends, thus preventing actin polymerization. Phalloidin is a drug that binds to and stabilizes actin filaments, preventing actin depolymerization. Even though these drugs have opposite effects on actin polymerization, the addition of either of these drugs instantaneously freezes the cell movements that depends on actin filaments. Explain why drugs that have opposite effects on actin filaments can have a similar effect on cell movements.

These drugs both stop cell movements because actin polymerization and depolymerization are both required for this process. As cells move forward, the growth of actin filaments near the plasma membrane helps push out the membrane. As this occurs, continuous depolymerization of actin filaments occurs at the actin filaments away from the plasma membrane.

17-48 Actin-You a protein, binding proteins bind to actin and can modify its properties. purify Cap1, that seems to bind and cap one end of an actin filament, although you do not know whether it binds the plus end or the minus end. To determine which end of the actin filament your protein binds to, you decide to examine the effect of Cap1 on actin polymerization by measuring the kinetics of actin filament formation in the presence and the absence of Cap1 protein. You obtain the following results (see Figure Q17-48). Do you think Cap1 binds the plus end or the minus end of actin? Explain your reasoning. Figure Q17-48

You would predict that Cap1 binds the plus end of actin, because it seems to inhibit actin polymerization. Actin filaments grow through the addition of monomers to the plus end of the actin filament. A capping protein that binds the plus end of actin can block monomer addition to the actin filament. Thus, less actin polymerization will be seen in the presence of the Cap1 protein.

17-49 Which of the following statements is false?

Actin filaments are usually excluded from the cell cortex.

17-50 Cell movement involves the coordination of many events in the cell. Which of the following phenomena is not required for cell motility?

Release of Ca from the sarcoplasmic reticulum.

17-51 The following proteins are important for cell movement. Match the following proteins with their function. myosin _____ A. nucleation of new actin filaments at the side of an existing filament ARP proteins _____ B. regulation of the availability of actin monomers profilin _____ C. important for the growth of straight, unbranched actin filaments integrins _____ D. contracting the rear of the cell formins _____ E. involvement in focal contacts

myosin ___D_____ ARP proteins ___A_____ profilin ___B_____ integrins ___E_____ formins ___C_____

17-52 Figure Q17-52 shows the leading edge of a lamellipodium. Which of the following statements is false? Figure Q17-52

Capping proteins bind to the minus end of actin filaments.

17-53 You are examining a cell line in which activation of the Rho family member Rac promotes lamellipodia formation. Which of the following statements is most likely to be true?

Cells carrying a Rac mutation that makes Rac act as if it is always bound to GTP will polymerize more branched actin filaments than normal cells.

17-54 In the budding yeast, activation of the GTP-binding protein Cdc42 occurs on binding of an external signal (pheromone) to a G-protein-coupled receptor. Activation of Cdc42 promotes actin polymerization. Predict what would happen to actin polymerization, in comparison with pheromone-treated cells, in the following cases. A. You add pheromone to an inhibitor of G-protein-coupled receptors. B. You add pheromone to a nonhydrolyzable analog of GTP.

A. Less actin polymerization. Cdc42 will not be able to be activated by the G- protein-coupled receptor. B. More actin polymerization. Cdc42 will be more active, because it will bind the nonhydrolyzable form of GTP and will not be able to be turned off.

17-55 Your friend works in a biotech company that has just discovered a drug that seems to promote lamellipodia formation in cells. Which of the following molecules is unlikely to be involved in the pathway that this drug affects?

myosin Muscle Contraction

17-56 Which of the following structures shorten during muscle contraction?

sarcomeres

17-57 Figure Q17-57shows an electron micrograph of a skeletal muscle fiber, where various points along a fiber and various regions have been labeled. F i g u r e Q 1 7 - 57 Which of the following statements is true about muscle contraction?

Point B will move closer to point C.

17-58 Which of the following conditions is likely to decrease the likelihood of skeletal muscle contraction?

addition of a drug that blocks Ca binding to troponin

17-59 Which of the following statements about skeletal muscle contraction is false?

The changes in voltage across the plasma membrane that occur when a muscle 2+ cell receives a signal from the nervous system cause an influx of Ca into the sarcoplasmic reticulum, triggering a muscle contraction.

17-60 Do you agree or disagree with the following statement? Explain your answer. When skeletal muscle receives a signal from the nervous system to contract, the signal 2+ from the motor neuron triggers the opening of a voltage-sensitive Ca channel in the 2+ muscle cells' plasma membrane, allowing Ca to flow into the cell.

Disagree. The increase in intracellular Ca2+ during muscle contraction comes from an 2+ intracellular source. The Ca is released from the lumen of the sarcoplasmic reticulum, which is a specialized region of endoplasmic reticulum inside a muscle cell. The signal from the nerve terminal triggers an action potential in the muscle cell plasma membrane, which causes a voltage-sensitive transmembrane protein in 2+ the membranous transverse tubules to open a Ca -release channel in the membrane of the sarcoplasmic reticulum.

17-61 You isolate some muscle fibers to examine what regulates muscle contraction. When you 2+ bathe the muscle fibers in a solution containing ATP and Ca , you see muscle 2+ contraction (experiment 3 in Table Q17-61). Ca is necessary, as solutions containing ATP alone or nothing do not stimulate contraction and thus the muscle remains in a relaxed state (experiments 1 and 2 in Table Q17-61). From what you know about the mechanism of muscle contraction, fill in your predictions of whether the muscle will be contracted or relaxed for experiments 4, 5, and 6. Explain your answers. Table Q17-61 2+ Extra credit: In what state would the muscle be if you added Ca but no ATP? How We Know: Pursuing Motor Proteins

See Table A17-61. Table A17-61 In experiment 4, the muscle will be relaxed because troponin will not be able to 2+2+ bind Ca . By preventing troponin from binding to Ca, troponin will not be able to undergo the conformational change that causes tropomyosin to alter its association with actin. This altered association is normally required for myosin to bind actin. In the 2+ absence of troponin regulation by Ca , myosin cannot bind actin and the muscle cannot contract. In experiment 5, the muscle will contract because tropomyosin cannot bind to actin. If tropomyosin cannot bind actin, myosin can. The presence of ATP means that ATP will be available for myosin to hydrolyze, causing muscle contraction. 2+ In experiment 6, the muscle will remain relaxed. The presence of Ca will induce a conformational change in troponin that causes tropomyosin to shift, exposing actin for myosin to bind. However, when myosin binds, the myosin molecule will attach and then release because the myosin will bind the nonhydrolyzable analog of ATP. Because no ATP hydrolysis can occur, the muscle will remain in the relaxed state. 2+ Extra credit: The muscle will be in a rigor state. If Ca is added without ATP, troponin can bind 2+ Ca and undergo a conformational change, which causes tropomyosin to shift and exposes the actin for the myosin to bind. However, myosin will bind actin and remain in the attached state, because a myosin head lacking a bound nucleotide is locked onto the actin until nucleotide binding occurs.

17-62 Kinesins were purified by adding the nonhydrolyzable analog AMP-PNP to cytoplasmic extracts containing microtubules, purifying the microtubules, and then releasing the kinesin proteins, which were still attached to the microtubules, by adding ATP. Would this trick have worked to purify myosin motors attached to actin filaments? Explain.

No, addition of a nonhydrolyzable form of ATP would not increase the affinity of myosin for actin. This is because when the myosin motor binds to ATP, the myosin head undergoes a conformational change that reduces its affinity for actin.

17-63 Consider the in vitro motility assay using purified kinesin and purified polymerized microtubules shown in Figure Q17-63. The three panels are images taken at 1-second intervals. In this figure, three microtubules have been numbered to make it easy to identify them. Which of the following statements about this assay is false? Figure Q17-63

Addition of the nonhydrolyzable ATP analog (AMP-PNP) would cause the microtubules to move faster.

17-64 You are interested in studying kinesin movements. You therefore prepare silica beads and coat them with kinesin molecules so that each bead, on average, has only one kinesin molecule attached to it. You add these kinesin-coated beads to a preparation of microtubules you have polymerized. Using video microscopy, you watch the kinesin [labeled with green fluorescent protein (GFP)] move down the microtubules. A. Kinesin-GFP has been measured to move along microtubules at a rate of 0.3 μm/sec, and single-molecule studies have revealed that kinesin moves along microtubules progressively, with each step being 8 nm. How many steps can the kinesin molecule take in 4 seconds, assuming that the kinesin stays attached to the microtubule for the entire 4 seconds? B. Because each kinesin molecule is thought to take approximately 100 steps before falling off the microtubule, will you see your silica beads detach from the mic

A. You would expect the kinesin molecule to travel 150 steps. The calculation is as follows: 0.3 μm = 300 nm. Therefore, in 4 seconds, the kinesin molecule could travel 1200 nm if it were to move at a rate of 0.3 μm/sec. Because the step size is 8 nm, 1200 nm/(8 nm per step) = 150 steps. B. Yes, you should see silica beads detach some time during the 4 seconds of that 4-second observation, because each kinesin will take more than 100 steps in time frame (see part A above). C. If you were to add AMP-PNP, you would no longer see the silica beads moving down the microtubule. It is thought that one molecule of ATP is hydrolyzed per step that kinesin takes; without ATP hydrolysis, translocation of the beads will be inhibited. However, you may still see the beads associated with the microtubules, because AMP-PNP does not inhibit the association of kinesin with the microtubule.

18-1 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase may be used more than once. The four phases of the cell cycle, in order, are G, __________________, 1 __________________, and __________________. A cell contains the most DNA after __________________ phase of the cell cycle. A cell is smallest in size after __________________ phase of the cell cycle. Growth occurs in __________________, __________________, and __________________ phases of the cell cycle. A cell does not enter mitosis until it has completed __________________ synthesis. DNA M protein Gnucleotide S 1 Gorganelle 2

The four phases of the cell cycle, in order, are G, S, G, and M. A cell contains the most 12 DNA after S phase of the cell cycle. A cell is smallest in size after M phase of the cell cycle. Growth occurs in G, S, and G phases of the cell cycle. A cell does not enter 12 mitosis until it has completed DNA synthesis.

18-2 What would be the most obvious outcome of repeated cell cycles consisting of S phase and M phase only?

The cells produced would get smaller and smaller.

18-3 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 Q18-3 are obtained. Figure 18-3 Which of the following would not explain the results with the mutant?

inability to begin M phase

18-4 Which of the following events does not usually occur during interphase?

The nuclear envelope breaks down.

18-5 In which phase of the cell cycle do cells check to determine whether the DNA is fully and correctly replicated?

at the end of G 2

18-6 What would happen to the progeny of a cell that proceeded to mitosis and cell division after entering S phase but had not completed S phase? Keep in mind that highly condensed chromatin, including the centromere region, is replicated late in S phase. Explain your answer.

The daughter cells would probably die. Those chromosomes that had not completed replication in S phase would have only one centromere, because the centromere is the last part of the chromosome to be replicated; the chromosome would therefore be segregated to only one of the two daughter cells at random. At least one, and probably both, of the daughter cells would thus receive an incomplete set of chromosomes and would be unlikely to be viable. Even if one daughter cell, by chance, received a full set of chromosomes, some of these chromosomes would be incompletely replicated and the cell would probably still not be viable.

18-7 Are the statements below true or false? Explain your answer. A. Statement 1: Generally, in a given organism, the S, G, and M phases of the cell 2 cycle take a defined and stereotyped amount of time in most cells. B. Statement 2: Therefore, the cell-cycle control system operates primarily by a timing mechanism, in which the entry into one phase starts a timer set for sufficient time to complete the required tasks. After a given amount of time has elapsed, a molecular "alarm" triggers movement to the next phase.

A. True. In nearly all cells in an organism, the S, G, and M phases of the cell cycle 2 take the same amount of time. The different timing of cell division in different cell types is due to variable lengths of the G phase or to withdrawal into the G 10 state. B. False. The cell-cycle control system does use a timing mechanism of sorts, but it also employs various surveillance and feedback mechanisms (checkpoints). Cells will not embark on later events or phases until the earlier events or phases have been completed successfully. In response to a defect or a delay in a cell-cycle event, cells engage molecular brakes to arrest the progression of the cell cycle at various checkpoints, to allow time for completion or repair.

18-8 Which of the following statements about the cell cycle is false?

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

18-9 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 G? 2

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

18-10 Match the following labels to the numbered label lines on Figure Q18-10. Figure Q18-10 A. G phase 1 B. mitotic cyclin C. mitotic Cdk D. S phase E. S-phase cyclin F. S-phase Cdk G. MPF H. G phase 2

A—5; B—6; C—3; D—2; E—4; F—1; G—7; H—8

18-11 Progression through the cell cycle requires a cyclin to bind to a Cdk because _________.

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

18-12 Levels of Cdk activity change during the cell cycle, in part because ________________.

cyclin levels change during the cycle.

18-13 The concentration of mitotic cyclin (M cyclin) ________________.

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

18-14 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?

inactivation of an enzyme that ubiquitylates M cyclin

18-15 You engineer yeast cells that express the M cyclin during S phase by replacing the promoter sequence of the M cyclin gene with that of S cyclin. Keeping in mind that yeast cells have one common Cdk that binds to all cyclins, which of the following outcomes is least likely during this experiment?

G cyclins will be expressed during S phase. 1

18-16 You have isolated a mutant in which a fraction of the new cells die soon after cell division and a fraction of the living cells have an extra copy of one or more chromosomes. When you grow the cells under conditions in which they transit the cell cycle more slowly, the defect disappears, suggesting that the mitotic spindle and segregation machinery are normal. Propose a basis for the defect.

This mutant may be lacking the checkpoint mechanism that delays the onset of anaphase and chromosome segregation until all chromosomes have attached properly to the mitotic spindle. If cells attempt chromosome segregation before all chromosomes have attached properly, some of the daughter cells will receive too few chromosomes (and thus will probably die) and other cells will receive additional chromosomes. Normally, cells use such a surveillance system to monitor the spindle attachment of each chromosome and engage molecular brakes until all chromosomes have attached properly. The molecular brakes will be dispensable in some dividing cells if all of the chromosomes rapidly become properly attached to the spindle. In other dividing cells, it will take longer for some of the chromosomes to find their appropriate attachments, and thus the molecular brakes will be essential to ensure faithful segregation of the duplicated copies of each chromosome to the two daughter cells. Slowing the cycle will allow more cells to segregate their chromosomes properly even in the absence of this "spindle attachment" checkpoint.

18-17 Which of the following statements is false?

The activating phosphatase (Cdc25) removes all phosphates from mitotic Cdk so that M-Cdk will be active. How We Know: Discovery of Cyclins and CDKs

18-18 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. Many features of __________________ cells make them suitable for biochemical studies of the cell-cycle control system. For example, the cells are unusually large and are arrested in a __________________-like phase. When the cells are triggered to resume cycling, the cell divisions have especially __________________ G and G phases and 12 occur __________________. Studies with Xenopus eggs identified a partly purified activity called __________________ that drives a resting Xenopus oocyte into M phase. MPF activity was found to __________________ during the cell cycle, although the amount of its kinase component, called __________________, remained constant. The regulatory component of MPF, called __________________,

Many features of egg cells make them suitable for biochemical studies of the cell-cycle control system. For example, the cells are unusually large and are arrested in a G-like 2phase. When the cells are triggered to resume cycling, the cell divisions have especially short G and G phases and occur synchronously. Studies with Xenopus eggs identified a 12 partly purified activity called maturation promoting factor that drives a resting Xenopus oocyte into M phase. MPF activity was found to be oscillate during the cell cycle, although the amount of its kinase component, called Cdk, remained constant. The regulatory component of MPF, called cyclin, has a stimulatory effect on MPF activity and plays a part in regulating interactions with its substrates. The components of MPF are evolutionarily conserved from yeast to humans, so that the corresponding human genes are able to function in yeast.

18-19 MPF activity was discovered when cytoplasm from a Xenopus M-phase cell was injected into Xenopus oocytes, inducing the oocytes to form a mitotic spindle. In a control experiment, Xenopus interphase cytoplasm was injected into oocytes and shown not to induce the formation of a mitotic spindle. Which of the following statements is not a legitimate conclusion from the control experiment?

Components of an interphase nucleus are insufficient to cause mitotic spindle formation.

18-20 Which of the following is not good direct evidence that the cell-cycle control system is conserved through billions of years of divergent evolution?

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

18-21 Your friend is intrigued by Cdks and purifies Cdk from the daisy plant. She is able to determine the sequence of 18 amino acids from the daisy protein. Using these data, she aligns the daisy sequence with one human Cdk protein and the Cdk proteins from two different kinds of yeast, as shown in Figure Q18-21. Sequences identical between the human and fungal proteins have been boxed. Figure Q18-21 Such conserved amino acid sequences are often involved in protein-protein interactions. Indeed, the threonine (T) in the central "YTHE/K" block is known to be phosphorylated by akinase that activates Cdks in human and yeast. The surrounding conserved sequences could thus be important for the interaction of this Cdk-activating kinase with Cdk. In the daisy Cdk, however, not all of the amino acids in these conserved bloacks match the sequences from yeast and human, as indicated by the arrows marked #1 and #2. A. When you

A. The lysine (K) at the position of arrow #2 in the daisy Cdk is likely the culprit for the disrupted interaction between Cdk and Cdk-activating kinase (see Figure A18- 21). This changes a conserved glutamic acid (E) to a lysine, which results in a change from a negatively charged amino acid to a positively charged amino acid. Although there is also a difference in the daisy sequence at the position of arrow #1, this change from an arginine (R) to a lysine is less dramatic, as both are positively charged amino acids. Figure A18-21 B. You would predict that the human Cdk will interact with the yeast Cdk-activating kinase, because the amino acids at the positions of arrows #1 and #2 are identical between human and yeast.

18-22 Mitogens are _____.

extracellular signals that stimulate cell division.

18-23 The Retinoblastoma (Rb) protein blocks cells from entering the cell cycle by ______.

inhibiting cyclin transcription.

18-24 Irradiated mammalian cells usually stop dividing and arrest at a G checkpoint. Place the 1 following events in the order in which they occur. A. production of p21 B. DNA damage C. inhibition of cyclin-Cdk complexes D. accumulation and activation of p53

B, D, A, C

18-25 The G DNA damage checkpoint ________________. 1

involves the inhibition of cyclin-Cdk complexes by p21.

18-26 Cells in the G state ________________. 0

do not divide.

18-27 What is the main molecular difference between cells in a G state and cells that have 0 simply paused in G? 1 S Phase

In G, the cell-cycle control system is partly dismantled, so that some of the Cdks and 0 cyclins are not present. Cells paused in G, by contrast, still contain all the components of 1 the cell-cycle control system. Whereas the latter cells can rapidly progress through the cycle when conditions are right, G cells need to synthesize the missing cell-cycle control 0 proteins so as to re-enter the cycle, which usually takes 8 hours or more.

18-28 Which of the following statements is false?

The loading of the origin recognition complexes (ORCs) is triggered by S-Cdk.

18-29 How does S-Cdk help guarantee that replication occurs only once during each cell cycle?

It phosphorylates the Cdc6 protein, marking it for destruction.

18-30 You create cells with a version of Cdc6 that cannot be phosphorylated and thus cannot be degraded. likely of this Which of the following statements describes the consequence change in Cdc6?

Cells will be unable to complete DNA synthesis.

18-31 Cells can pause in G when DNA is damaged, and can pause in S when there are 1 replication errors. Indicate whether the mechanism below applies to a G arrest, an S- 1 phase arrest, both types of arrest, or neither. A. p53 activates the transcription of a Cdk inhibitor. B. Cyclins are phosphorylated and destroyed. C. Cdk is unable to phosphorylate its substrates. D. The Cdc25 phosphatase is inhibited. M Phase

A. G arrest 1 B. neither C. both D. S-phase arrest

18-32 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. The cell cycle consists of an alternation between __________________, which appears as a period of dramatic activity under the microscope, and a preparative period called __________________, which consists of three phases called __________________, __________________, and __________________. During M phase, the nucleus divides in a process called __________________, and the cytoplasm splits in two in a process called __________________. The cell-cycle control system relies on sharp increases in the activities of regulatory proteins called __________________, or __________________, to trigger S phase and M phase. Inactivation of __________________ is required to exit from M phase after chromosome segregation. APC G

The cell cycle consists of an alternation between M phase, which appears as a period of dramatic activity under the microscope, and a preparative period called interphase, which consists of three phases called Gphase, S phase, and Gphase. During M phase, 12 the nucleus divides in a process called mitosis, and the cytoplasm splits in two in a process called cytokinesis. The cell-cycle control system relies on sharp increases in the activities of regulatory proteins called cyclin-dependent kinases, or Cdks, to trigger S phase and M phase. Inactivation of M-Cdk is required to exit from M phase after chromosome segregation.

18-33 Which of the following does not occur during M phase in animal cells?

growth of the cell

18-34 Condensins ________________.

assemble into complexes on the DNA when phosphorylated by M-Cdk.

18-35 At the end of DNA replication, the sister chromatids are held together by the ___________.

cohesins.

18-36 Which of the following statements is true?

The mitotic spindle helps segregate the chromosomes to the two daughter cells. Mitosis