Cell Bio ch.14-20 minus 19

Identify the cytoskeletal structures depicted in the epithelial cells shown in Figure Q17-1.

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

Which of the phylogenetic trees in Figure Q14-53 is the most accurate? (The mitochondria and chloroplasts are from maize, but they are treated as independent "organisms" for the purposes of this question.)

(C). Mitochondria are most closely related to Bacillus, and chloroplasts to cyanobacteria. Maize (a eucaryote) is more closely related to Giardia (a simple eucaryote) than it is to bacteria (procaryotes).

Cells in the G0 state ________________. (a) do not divide (b) cannot re-enter the cell cycle (c) have entered this arrest state from either G1 or G2 (d) have duplicated their DNA

(a)

In stage 1 of photosynthesis, a proton gradient is generated and ATP is synthesized. Where do protons become concentrated in the chloroplast? (a) thylakoid space (b) stroma (c) inner membrane (d) thylakoid membrane

(a)

The principal microtubule organizing center in animal cells is the ____________. (a) centrosome (b) centromere (c) kinetochore (d) cell cortex

(a)

Which of the following statements about kinetochores is true? (a) Kinetochores assemble onto chromosomes during late prophase. (b) Kinetochores contain DNA-binding proteins that recognize sequences at the telomere of the chromosome. (c) Kinetochore proteins bind to the tubulin molecules at the minus end of microtubules. (d) Kinetochores assemble on chromosomes that lack centromeres.

(a)

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

(a)

The ethylene response in plants involves a dimeric transmembrane receptor. When the receptor is not bound to ethylene, the receptor binds to and activates a protein kinase, which activates an intracellular signaling pathway that leads to the degradation of a transcriptional regulator important for transcribing the ethylene response genes (see Fig. Q16-50). You discover a phosphatase that is important for ethylene signaling, and you name it PtpE. Plants lacking PtpE never turn on ethylene-response genes, even in the presence of ethylene. You find that PtpE dephosphorylates serine 121 on the transcriptional regulator. Furthermore, plants lacking PtpE degrade the transcriptional regulator in the presence of ethylene. Which of the following statements below is inconsistent with your data? (a) When the transcriptional regulator is phosphorylated, it activates transcription of the ethylene-response genes. (b) When the transcriptional regulator is not phosphorylated, it binds to DNA. (c) Activation of the protein kinase that binds to the ethylene receptor leads to inactivation of PtpE. (d) Binding of ethylene to its receptor leads to the activation of PtpE.

(a) is inconsistent. Cells lacking PtpE do not transcribe the ethylene-reponse genes, suggesting that dephosphorylation of the transcriptional regulator is required before it can activate transcription. From your data, PtpE is either constitutively active or inactivated by the active protein kinase downstream of the ethylene receptor. The latter scenario would permit maximum repression in the absence of ethylene. All the other choices are consistent with your data.

Which of the following statements about microtubules is true? (a) Motor proteins move in a directional fashion along microtubules by using the inherent structural polarity of a protofilament. (b) The centromere nucleates the microtubules of the mitotic spindle. (c) Because microtubules are subject to dynamic instability, they are used only for transient structures in a cell. (d) ATP hydrolysis by a tubulin heterodimer is important for controlling the growth of a microtubule.

(a) is true. Microtubules are nucleated by the centrosome (not the centromere, choice (b)). Although microtubules are subject to dynamic instability, their interaction with microtubule-binding proteins can stabilize them so that they can be used to form stable structures such as cilia and flagella (choice (c)). GTP (not ATP) hydrolysis is important for controlling the growth of a microtubule (choice (d)).

Which of the following statements about membrane-enclosed organelles is true? (a) In a typical cell, the area of the endoplasmic reticulum membrane far exceeds the area of plasma membrane. (b) The nucleus is the only organelle that is surrounded by a double membrane. (c) Other than the nucleus, most organelles are small and thus, in a typical cell, only about 10% of a cell's volume is occupied by membrane-enclosed organelles; the other 90% of the cell volume is the cytosol. (d) The nucleus is the only organelle that contains DNA.

(a) is true; the area of the endoplasmic reticulum membrane is 20-30 times that of the plasma membrane in a typical cell. Chloroplasts and mitochondria are also surrounded by a double membrane (choice (b)). The cytosol is about half the volume of a typical eucaryotic cell, with membrane-enclosed organelles making up the other half of the volume (choice (c)). Chloroplasts and mitochondria also carry their own genome, whereas the nucleus carries the genome of the organism (choice (d)).

Activated protein kinase C (PKC) can lead to the modification of the membrane lipids in the vicinity of the active PKC. Figure Q16-30 shows how G proteins can indirectly activate PKC. You have discovered the enzyme activated by PKC that mediates the lipid modification. You call the enzyme Rafty and demonstrate that activated PKC directly phosphorylates Rafty, activating it to modify the plasma membrane lipids in the vicinity of the cell where PKC is active; these lipid modifications can be detected by dyes that bind to the modified lipids. Cells lacking Rafty do not have these modifications, even when PKC is active. Which of the following conditions would lead to signal-independent modification of the membrane lipids by Rafty? (a) the expression of a constitutively active phospholipase C (b) a mutation in the GPCR that binds the signal more tightly (c) a Ca2+ channel in the endoplasmic reticulum with an increased affinity for IP3 (d) a mutation in the gene that encodes Rafty such that the enzyme can no longer be phosphorylated by PKC

(a). A constitutively active phospholipase C will lead to the constitutive production of IP3 and diacylglycerol, leading to activation of PKC in a signal- independent manner; thus Rafty activation and the lipid modification will be signal-independent. Choices (b) and (c) will increase activity of the signal transduction pathway in a signal-dependent manner. Choice (d) will prevent PKC from activating Rafty and will thus prevent the lipid modifications.

Which of the situations below will enhance microtubule shrinkage? (a) addition of a drug that inhibits GTP exchange on free tubulin dimers (b) addition of a drug that inhibits GTP hydrolysis of tubulin dimers (c) addition of a drug that increases the affinity of tubulin molecules carrying GDP for other tubulin molecules (d) addition of a drug that blocks the ability of a tubulin dimer to bind to γ- tubulin

(a). A drug that inhibits GTP exchange on free tubulin dimers will effectively decrease the available pool of GTP-bound tubulin dimers available for addition to microtubule ends, thus tipping the balance in favor of microtubule disassembly. A drug that inhibits GTP hydrolysis of tubulin dimers (choice (b)) or that increases the affinity of GDP-bound tubulin dimers for each other (choice (c)) will stabilize growing microtubules. Blocking the ability of a tubulin dimer to bind to γ-tubulin will decrease the rate of new microtubule formation but should not enhance microtubule shrinkage (choice (d)).

Cytokinesis in animal cells ________________. (a) requires ATP (b) leaves a small circular 'scar' of actin filaments on the inner surface of the plasma membrane (c) is often followed by phosphorylation of integrins in the plasma membrane (d) is assisted by motor proteins that pull on microtubules attached to the cell cortex

(a). All cell movement requires ATP, and in cytokinesis actin and myosin molecules are moving relative to one another to cause contraction of the contractile ring. The myosin is an ATPase that hydrolyzes ATP to power this movement. The assembly of the contractile ring requires a general rearrangement of the filaments in the cell cortex. The contractile ring in animal cells disassembles completely after mitosis, leaving no trace (choice (b)). Phosphorylation of integrins, which weakens the hold of these transmembrane proteins on the extracellular matrix, thereby allowing cells to round up, generally precedes cytokinesis and is part of the general rearrangement of cell structure that accompanies cell division (choice (c)). Microtubules do not have an important role in animal cytokinesis (choice (d)).

Which of the following statements about vesicle budding from the Golgi is false? (a) Clathrin molecules are important for binding to and selecting cargos for transport. (b) Adaptins interact with clathrin. (c) Once vesicle budding occurs, clathrin molecules are released from the vesicle. (d) Clathrin molecules act at the cytosolic surface of the Golgi membrane.

(a). Cargo binds to cargo receptors. Adaptin molecules capture cargo receptors, which bind to the appropriate cargo molecules for incorporation into the vesicle.

Which of the following does not occur during M phase in animal cells? (a) growth of the cell (b) condensation of chromosomes (c) breakdown of nuclear envelope (d) attachment of chromosomes to microtubules

(a). Cell size increases throughout interphase and not during M phase. All of the other phenomena are observed in M phase

The link between bond-forming reactions and membrane transport processes in the mitochondria is called __________________. (a) chemiosmotic coupling (b) proton pumping (c) electron transfer (d) ATP synthesis

(a). Choices (b), (c), and (d) are individual parts of the overall process of chemiosmotic coupling.

Modern eucaryotes depend on mitochondria to generate most of the cell's ATP. How many molecules of ATP can a single molecule of glucose generate? (a) 30 (b) 2 (c) 20 (d) 36

(a). Glycolysis of a single glucose molecule generates 2 ATP molecules. Oxidative phosphorylation in the mitochondria generates an additional 28 ATP molecules, making a total of 30 ATP molecules for each glucose molecule.

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). Programmed cell death results from an intracellular suicide program that eliminates unneeded, unwanted, or damaged cells. It occurs frequently and happens even in healthy cells throughout the lifetime of an individual.

Which of the following mechanisms is not directly involved in inactivating an activated RTK? (a) dephosphorylation by serine/threonine phosphatases (b) dephosphorylation by protein tyrosine phosphatases (c) removal of the RTK from the plasma membrane by endocytosis (d) digestion of the RTK in lysosomes

(a). RTKs are phosphorylated on tyrosines by their dimerization partner, which is also a tyrosine kinase, and thus the reversal of these phosphorylations involves protein tyrosine phosphotases, and not protein serine/threonine phosphatases. Endocytosis of the receptor and its ultimate digestion in the lysosome are other methods that the cell uses to downregulate active receptors.

Which of the following precede the re-formation of the nuclear envelope during M phase in animal cells? (a) assembly of the contractile ring (b) decondensation of chromosomes (c) reassembly of the nuclear lamina (d) transcription of nuclear genes

(a). The contractile ring in an animal cell begins to assemble in anaphase. The chromosomes do not decondense (choice (b)), the lamina does not reform (choice (c)), and transcription does not begin (choice (d)) until the formation of the nuclear envelope is complete and nuclear proteins have been imported through the nuclear pores.

Acetylcholine binds to a GPCR on heart muscle, making the heart beat more slowly. The activated receptor stimulates a G protein, which opens a K+ channel in the plasma membrane, as shown in Figure Q16-25. Which of the following would enhance this effect of the acetylcholine? (a) addition of a high concentration of a non-hydrolyzable analog of GTP (b) addition of a drug that prevents the α subunit from exchanging GDP for GTP (c) mutations in the acetylcholine receptor that weaken the interaction between the receptor and acetylcholine (d) mutations in the acetylcholine receptor that weaken the interaction between the receptor and the G protein

(a). The heart is induced to beat more slowly by the binding of acetylcholine to a GPCR, activating a G protein whose βγ complex binds to and opens K+ channels. The addition of high concentrations of a non-hydrolyzable analog of GTP will increase the length of time that the G protein βγ complex remains free of the α subunit and able to activate the K+ channel; this will therefore enhance the effect of acetylcholine (choice (a)). All the other choices will make it more difficult for the signal to proceed from the GPCR to the K+ channel.

A friend declares that chromosomes are held at the metaphase plate by microtubules that push on each chromosome from opposite sides. Which of the following observations does not support your belief that the microtubules are pulling on the chromosomes? (a) the jiggling movement of chromosomes at the metaphase plate (b) the way in which chromosomes behave when the attachment between sister chromatids is severed (c) the way in which chromosomes behave when the attachment to one kinetochore is severed (d) the shape of chromosomes as they move toward the spindle poles at anaphase

(a). The jiggling movement (choice (a)) is simply a sign that the chromosomes are subject to forces from both sides, which could be the microtubules pulling, pushing, or both. All the other observations suggest pulling. When the attachment between sister chromatids is severed (choice (b)), both daughter chromosomes move toward their respective poles, which suggests that they are being pulled. When the attachment to one kinetochore is severed (choice (c)), the whole chromosome moves to the opposite pole, showing that the kinetochore microtubules are pulling on their attached chromatid, not pushing it. Similarly, the shape of the chromosomes as they move toward the pole (choice (d)) suggests that the chromosomes are being pulled.

What is the role of the nuclear localization sequence in a nuclear protein? (a) It is bound by cytoplasmic proteins that direct the nuclear protein to the nuclear pore. (b) It is a hydrophobic sequence that enables the protein to enter the nuclear membranes. (c) It aids in protein unfolding so that the protein can thread through nuclear pores. (d) It prevents the protein from diffusing out of the nucleus through nuclear pores.

(a). The nuclear localization signal typically contains positively charged amino acids, not hydrophobic ones (choice (b)). Proteins are not unfolded as they enter the nucleus (choice (c)). Proteins are actively transported in and out of the nucleus and do not diffuse through the nuclear pores (choice (d)).

Which of the following items below are not important for flagellar movement? (a) sarcoplasmic reticulum (b) ATP (c) dynein (d) microtubules

(a). The sarcoplasmic reticulum is important for muscle contraction. All other items are important for flagellar movement.

The relationship of free-energy change (δG) to the concentrations of reactants and products is important because it predicts the direction of spontaneous chemical reactions. In the hydrolysis of ATP to ADP and inorganic phosphate (Pi), the standard free-energy change (δG°) is -7.3 kcal/mole. The free-energy change depends on concentrations according to the following equation: δG = δG° + 1.42 log10 ([ADP] [Pi]/[ATP]) In a resting muscle, the concentrations of ATP, ADP, and Pi are approximately 0.005 M, 0.001 M, and 0.010 M, respectively. What is the δG for ATP hydrolysis in resting muscle? (a) -11.1 kcal/mole (b) -8.72 kcal/mole (c) 6.01 kcal/mole (d) -5.88 kcal/mole

(a). The δG for hydrolysis is -11.1 kcal/mole. This result is calculated by substituting values into the equation given: δG = -7.3 kcal/mole + 1.42 log10 ([0.001 M] [0.010 M]/[0.005 M]) = -7.3 kcal/mole + 1.42 log10 (0.002) = -11.1 kcal/mole.

Which of the following statements about secretion is true? (a) The membrane of a secretory vesicle will fuse with the plasma membrane when it discharges its contents to the cell's exterior. (b) Vesicles for regulated exocytosis will not bud off the trans Golgi network until the appropriate signal has been received from the cell. (c) The signal sequences of proteins destined for constitutive exocytosis ensure their packaging into the correct vesicles. (d) Proteins destined for constitutive exocytosis aggregate as a result of the acidic pH of the trans Golgi network.

(a). Vesicles for regulated exocytosis bud from the trans Golgi network and accumulate at the plasma membrane until the appropriate signal has been received (choice (b)). There are no signal sequences for proteins destined for exocytosis (choice (c)). Those proteins that are to be secreted by regulated exocytosis aggregate in the trans Golgi as a result of the acidic pH and high Ca2+ concentrations (choice (d)); those proteins that do not aggregate are packed into transport vesicles for constitutive exocytosis.

Condensins ________________. (a) are degraded when cells enter M-phase (b) assemble into complexes on the DNA when phosphorylated by M-Cdk (c) are involved in holding sister chromatids together (d) bind to DNA before DNA replication begins

(b)

Cytochrome oxidase is an enzyme complex that uses metal ions to help coordinate the transfer of four electrons to O2. Which metal atoms are found in the active site of this complex? (a) two iron atoms (b) one iron atom and one copper atom (c) one iron atom and one zinc atom (d) one zinc atom and one copper atom

(b)

In the electron-transport chain in chloroplasts, ________-energy electrons are taken from __________. (a) high; H2O (b) low; H2O (c) high; NADPH (d) low; NADPH

(b)

NADH contains a high-energy bond that, when cleaved, donates a pair of electrons to the electron-transport chain. What are the immediate products of this bond cleavage? (a) NAD+ + OH- (b) NAD+ + H- (c) NAD- + H+ (d) NAD + H

(b)

The F1 portion of the mitochondrial ATP synthase comprises several different protein subunits. Which subunit binds to ADP + Pi and catalyzes the synthesis of ATP as a result of a conformational change? (a) α (b) β (c) δ (d) ε

(b)

The photosystems in chloroplasts contain hundreds of chlorophyll molecules, most of which are part of _______________. (a) plastoquinone (b) the antenna complex (c) the reaction center (d) the ferredoxin complex

(b)

Which of the following events does not usually occur during interphase? (a) Cells grow in size. (b) The nuclear envelope breaks down. (c) DNA is replicated. (d) The centrosomes are duplicated.

(b)

Which of the following statements is false? (a) DNA synthesis begins at origins of replication. (b) The loading of the origin recognition complexes (ORCs) is triggered by S- Cdk. (c) The phosphorylation and degradation of Cdc6 help to ensure that DNA is replicated only once in each cell cycle. (d) DNA synthesis can only begin after pre-replicative complexes assembles on the ORCs.

(b)

Which organelle fragments during mitosis? (a) endoplasmic reticulum (b) Golgi apparatus (c) mitochondrion (d) chloroplast

(b)

You create cells with a version of Cdc6 that cannot be phosphorylated and thus cannot be degraded. Which of the following statements describes the likely consequence of this change in Cdc6? (a) Cells will enter S phase prematurely. (b) Cells will be unable to complete DNA synthesis. (c) The origin recognition complex (ORC) will be unable to bind to DNA. (d) Cdc6 will be produced inappropriately during M phase.

(b)

Which of the following statements about skeletal muscle contraction is false? (a) When a muscle cell receives a signal from the nervous system, voltage- gated channels open in the T-tubule membrane. (b) The changes in voltage across the plasma membrane that occur when a muscle cell receives a signal from the nervous system cause an influx of Ca2+ into the sarcoplasmic reticulum, triggering a muscle contraction. (c) A change in the conformation of troponin leads to changes in tropomyosin such that it no longer blocks the binding of myosin heads to the actin filament. (d) During muscle contraction, the Z discs move closer together as the myosin heads walk toward the plus ends of the actin filaments.

(b) is false. Muscle contraction is triggered by an efflux of Ca2+ from the sarcoplasmic reticulum into the cytosol.

Which of the following statements about the endoplasmic reticulum (ER) is false? (a) The ER is the major site for new membrane synthesis in the cell. (b) Proteins to be delivered to the ER lumen are synthesized on smooth ER. (c) Steroid hormones are synthesized on the smooth ER. (d) The ER membrane is contiguous with the outer nuclear membrane.

(b) is false. Proteins to be delivered to the ER lumen are synthesized on rough ER; these area appear "rough" because ribosomes are attached to the cytosolic surface of these ER regions.

Acetylcholine is a signaling molecule that elicits responses from heart muscle cells, salivary gland cells, and skeletal muscle cells. Which of the following statements is false? (a) Heart muscle cells decrease their rate and force of contraction when they receive acetylcholine, whereas skeletal muscle cells contract. (b) Heart muscle cells, salivary gland cells, and skeletal muscle cells all express an acetylcholine receptor that belongs to the transmitter-gated ion channel family. (c) Active acetylcholine receptors on salivary gland cells and heart muscle cells activate different intracellular signaling pathways. (d) Heart muscle cells, salivary gland cells, and skeletal muscle cells all respond to acetylcholine within minutes of receiving the signal.

(b) is not true. Only skeletal muscle cells express an acetylcholine receptor that belongs to the transmitter-gated ion channel family; salivary gland cells and heart muscle cells express a different receptor. The other choices are all true.

Which of the following statements about organellar movement in the cell is false? (a) Organelles undergo saltatory movement in the cell. (b) Only the microtubule cytoskeleton is involved in organellar movement. (c) Motor proteins involved in organellar movement use ATP hydrolysis for energy. (d) Organelles are attached to the tail domain of motor proteins.

(b) is untrue; both the actin cytoskeleton and the microtubule cytoskeleton are involved in organellar movement.

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 cyclin

(b). A cell with a mutation that prevents ubiquitylation of M cyclin could enter mitosis but could not exit mitosis properly.

Figure Q16-15 shows the pathway through which nitric oxide (NO) triggers smooth muscle relaxation in a blood-vessel wall. Which of the following situations would lead to relaxation of the smooth muscle cells in the absence of acetylcholine? (a) a smooth muscle cell that has a defect in guanylyl cyclase such that it cannot bind NO (b) a muscle cell that has a defect in guanylyl cyclase such that it constitutively converts GTP to cyclic GMP (c) a muscle cell that has cyclic GMP phosphodiesterase constitutively active (d) a drug that blocks an enzyme involved in the metabolic pathway from arginine to NO

(b). A constitutively active guanylyl cyclase will produce cGMP even in the absence of a signal and thus will lead to relaxation of smooth muscle cells in the absence of acetylcholine. Choice (a) would lead to a block in the production of cGMP such that even if NO were to reach the smooth muscle cells, relaxation would not occur. Choice (c) would not lead to muscle cell relaxation independently of acetylcholine, because cyclic GMP phosphodiesterase is involved in the degradation of cGMP. Choice (d) will lead to a block in the production of NO.

An individual transport vesicle (a) contains only one type of protein in its lumen. (b) will fuse with only one type of membrane. (c) is endocytic if it is traveling toward the plasma membrane. (d) is enclosed by a membrane with the same lipid and protein composition as the membrane of the donor organelle.

(b). An individual vesicle may contain more than one type of protein in its lumen (choice (a)), all of which will contain the same sorting signal (or will lack specific sorting signals). Endocytic vesicles (choice (c)) generally move away from the plasma membrane. The vesicle membrane will not necessarily contain the same lipid and protein composition as the donor organelle, because the vesicle is formed from a selected subset of the organelle membrane from which it budded (choice (d)).

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 Q18-3 Which of the following would not explain the results with the mutant? (a) inability to initiate DNA replication (b) inability to begin M phase (c) inability to activate proteins needed to enter S phase (d) inappropriate production of a signal that causes the cells to remain in G1

(b). At 37°C, the cells all have one genome-worth of DNA, meaning that they have not replicated their DNA and therefore have not entered S phase. Cells that are unable to begin M phase should have two genomes-worth of DNA, as they would have completed DNA replication and arrested in G2.

Intermediate filaments are made from elongated fibrous proteins that are assembled into a ropelike structure. Figure Q17-7 shows the structure of an intermediate filament subunit. You are interested in how intermediate filaments are formed, and you create an 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 IF2αd as the subunit, which of the following predictions below describes the most likely outcome? Figure Q17-7 (a) Filaments assembled using IFαd will interact with different cytoskeletal components. (b) Filaments assembled using IFαd will form dimers that are twice as long as dimers assembled from normal intermediate filaments. (c) Sixteen tetramers assembled from IFαd will be needed for a ropelike structure to form. (d) Dimers of IFαd will form by interactions with the N-terminal globular head and the C-terminal globular tail.

(b). Because the α-helical region is twice as long, you would predict that a coiled- coil dimer made up of two IFαd subunits would be about twice as long as a dimer assembled from a normal intermediate filament. Because the globular head and tail regions usually interact with other cellular components, doubling the size of the α-helical region without changes in the globular regions is unlikely to cause changes in protein interactions (choice (a)). Eight tetramers are usually needed to form a ropelike filament, and it is unlikely that 16 will be needed with IFαd, because it is the length of the coiled-coil region that is most affected by a doubling in size of the α-helical region (choice (c)). Interactions in the α-helical region are important for dimerization, and thus choice (d) is untrue.

The growth factor RGF stimulates proliferation of cultured rat cells. The receptor that binds RGF is a receptor tyrosine kinase called RGFR. Which of the following types of alteration to RGF would be most likely to prevent receptor dimerization? (a) a mutation that increases the affinity of RGFR for RGF (b) a mutation that prevents RGFR from binding to RGF (c) changing the tyrosines that are normally phosphorylated on RGFR dimerization to alanines (d) changing the tyrosines that are normally phosphorylated on RGFR dimerization to glutamic acid

(b). Binding of a ligand to RTKs leads to their dimerization, and thus a mutation that prevents RGFR from binding to RGF will prevent dimerization. A mutation that increases the affinity of RGFR for RGF will increase dimerization in the presence of ligand (choice (a)). RTKs become phosphorylated on dimerization. However, changing the relevant tyrosines to alanine will block receptor activation but should not cause dimerization (choice (c)). Because glutamic acid is negatively charged, it can mimic the addition of a phosphate to an amino acid; thus, changing the relevant tyrosines to glutamic acid may mimic receptor activation, but it should not cause or prevent receptor dimerization (choice (d)).

Which of the following conditions below is likely to decrease the likelihood of skeletal muscle contraction? (a) partial depolarization of the T-tubule membrane, such that the resting potential is closer to zero (b) addition of a drug that blocks Ca2+ binding to troponin (c) an increase in the amount of ATP in the cell (d) a mutation in tropomyosin that decreases its affinity for the actin filament

(b). Ca2+ binding to troponin leads to a conformational change that causes a movement in tropomyosin so that myosin can bind to actin to initiate contraction. Thus, if troponin cannot bind Ca2+, the likelihood of contraction decreases. Partial depolarization of the T-tubule membrane will make it easier to depolarize the membrane, increasing the likelihood of muscle contraction (choice (a)). ATP is required for myosin movement, so increasing the amount of ATP in the cell will not decrease contraction (choice (c)). Tropomyosin normally binds to actin and blocks myosin binding, so a mutation in tropomyosin that decreases its affinity for actin should not decrease the likelihood of muscle contraction (choice (d)).

Which of the following reactions have a large enough free-energy change to enable it to be used, in principle, to provide the energy needed to synthesize one molecule of ATP from ADP and Pi under standard conditions? See Table 14-23. Recall that δG° = -n (0.023) δE′0 and δE′0 = E′0 (acceptor) - E′0 (donor). (a) the reduction of a molecule of pyruvate by NADH (b) the reduction of a molecule of cytochrome b by NADH (c) the reduction of a molecule of cytochrome b by reduced ubiquinone (d) the oxidation of a molecule of reduced ubiquinone by cytochrome c

(b). For a reaction to drive ATP synthesis under standard conditions, the δG°′ of the reaction must be less than -7.3 kcal/mol. Because δG°′ = -n (0.023) δE′0, the value of δE′0 must be greater than 317 mV/n, where n is the number of electrons transferred. δE′0 is 130 mV for the reduction of a molecule of pyruvate by NADH, 390 mV for the reduction of a molecule of cytochrome b by NADH, 40 mV for the reduction of a molecule of cytochrome b by ubiquinone, 200 mV for the oxidation of a molecule of ubiquinone by cytochrome c, and 590 mV for the oxidation of cytochrome c by oxygen. The numbers of electrons transferred in each of the above reactions are two, one, one, one, and one, respectively. Thus, only reactions (b) and (e) are sufficient to drive ATP synthesis.

A cell with nuclear lamins that cannot be phosphorylated in M phase will be unable to ________________. (a) reassemble its nuclear envelope at telophase (b) disassemble its nuclear lamina at prometaphase (c) begin to assemble a mitotic spindle (d) condense its chromosomes at prophase

(b). If the lamins cannot be phosphorylated during mitosis, the cells will be unable to disassemble their nuclear lamina, preventing the breakdown of the nuclear envelope at prometaphase. The mitotic spindle begins to form before the nucleus breaks down, forming a sort of cage around the nucleus (choice (c)). Lamins are not involved in chromosome condensation (choice (d)).

Which of the following statements is true? (a) MAP kinase is important for phosphorylating MAP kinase kinase. (b) PI 3-kinase phosphorylates a lipid in the plasma membrane. (c) Ras becomes activated when an RTK phosphorylates its bound GDP to create GTP. (d) STAT proteins phosphorylate JAK proteins, which then enter the nucleus and activate gene transcription.

(b). MAP kinases are phosphorylated by MAP kinase kinases (choice (a)). Ras exchanges its GDP for GTP when activated (choice (c)). JAK proteins are receptor-associated kinases that phosphorylate the transcriptional regulators called STAT proteins (choice (d)).

Which of the following statements is false? (a) Cytochalasins prevent actin polymerization. (b) Actin filaments are usually excluded from the cell cortex. (c) Integrins are transmembrane proteins that can bind to the extracellular matrix. (d) ARPs can promote the formation of branched actin filaments.

(b). Much of the actin in the cell is concentrated in the cell cortex, the region of the cell just beneath the plasma membrane.

Male cockroaches with mutations that strongly decrease the function of an RTK called RTKX are oblivious to the charms of their female comrades. This particular RTK binds to a small molecule secreted by sexually mature females. Most males carrying loss-of-function mutations in the gene for Ras protein are also unable to respond to females. You have just read a paper in which the authors describe how they have screened cockroaches that are mutant in RTKX for additional mutations that partly restore the ability of males to respond to females. These mutations decrease the function of a protein that the authors call Z. Which of the following types of protein could Z be? Explain your answer. (a) a protein that activates the Ras protein by causing Ras to exchange GDP for GTP (b) a protein that stimulates hydrolysis of GTP by the Ras protein (c) an adaptor protein that mediates the binding of the RTKX to the Ras protein (d) a transcriptional regulator required for the expression of the Ras gene

(b). Mutations that increase the activity of Ras should mimic the effect of stimulating RTKX in a receptor-independent fashion. Because the intracellular concentration of GTP is higher than that of GDP, some proportion of the Ras molecules is expected to be GTP-bound and active; ridding the cells of a protein that stimulates GTP hydrolysis will increase this pool of active Ras. Mutants that cannot stimulate Ras to exchange GDP for GTP will have the same phenotype as mutants lacking Ras (choice (a)), as will mutants lacking a transcriptional regulator required for expression of the Ras gene (choice (d)). Defects in an adaptor protein that mediates the binding of receptor X to Ras will have no further effect on a mutant already lacking the receptor (choice (c)).

You are working in a biotech company that has discovered a small-molecule drug called H5434. H5434 binds to LDL receptors when they are bound to cholesterol. H5434 binding does not alter the conformation of the LDL receptor's intracellular domain. Interestingly, in vitro experiments demonstrate that addition of H5434 increases the affinity of LDL for cholesterol and prevents cholesterol from dissociating from the LDL receptor even in acidic conditions. Which of the following is a reasonable prediction of what may happen when you add H5434 to cells? (a) Cytosolic cholesterol levels will remain unchanged relative to normal cells. (b) Cytosolic cholesterol levels will decrease relative to normal cells. (c) The LDL receptor will remain on the plasma membrane. (d) The uncoating of vesicles will not occur.

(b). Normally, cholesterol dissociates from the LDL receptor in the acidic environment of the endosomes and is released into the cytosol. If the drug prevents cholesterol from dissociating from the LDL receptor in acidic conditions, cholesterol may not become released into the cytosol, and thus cytosolic cholesterol levels are likely to decrease relative to those in normal cells. There is no reason to believe that the LDL receptor will remain on the plasma membrane (choice (c)), because the cytosolic region of the receptor is not directly altered by the drug. Vesicle uncoating is also unlikely to be altered (choice (d)), because this occurs after vesicles are pinched off from the membrane.

During the mating process, yeast cells respond to pheromones secreted by other yeast cells. These pheromones bind GPCRs on the surface of the responding cell and lead to the activation of G proteins inside the cell. When a wild-type yeast cell senses the pheromone, its physiology changes in preparation for mating: the cell stops growing until it finds a mating partner. If yeast cells do not undergo the appropriate response after sensing a pheromone, they are considered sterile. Yeast cells that are defective in one or more components of the G protein have characteristic phenotypes in the absence and presence of the pheromone, which are listed in Table 16-27. Table Q16-27 Mating phenotypes of various strains of yeast Which of the following models is consistent with the data from the analysis of these mutants? Explain your answer. (a) α activates the mating response but is inhibited when bound to βγ. (b) βγ activates the mating response but is inhibited when bound to α. (c) The G protein is inactive; either free α or free βγ complex is capable of activating the mating response. (d) The G protein is active; both free α and free βγ complex are required to inhibit the mating response

(b). Single mutations in the β and γ subunits of the G protein permit growth in the absence of pheromone and display a sterile phenotype in the presence of pheromone, whereas loss of the α subunit causes growth arrest in either the presence or the absence of pheromone. Because arrested growth is a normal response that occurs when the cells sense a pheromone, it must be the βγ complex that normally activates the cellular mating response when released from the α subunit. In cells lacking α, the βγ complex causes growth arrest inappropriately because α normally inhibits the action of the complex until it is activated by the binding of pheromone to the yeast cell. This interpretation is consistent with the analysis of the double mutants in which α is deleted together with either β or γ. In these double mutants, normal growth is seen in the absence of pheromone because β and γ act together as a complex to activate target proteins; a lack of either β or γ leads to no response to pheromone, even when α is not present to inhibit the response.

Figure Q15-19 shows the organization of a protein that normally resides in the plasma membrane. The boxes labeled 1 and 2 represent membrane-spanning sequences and the arrow represents a site of action of signal peptidase. Given this diagram, which of the following statements must be true? (a) The N-terminus of this protein is cytoplasmic. (b) The C-terminus of this protein is cytoplasmic. (c) The mature version of this protein will span the membrane twice, with both the N and C-termini in the cytoplasm. (d) None of the above.

(b). The mature version of this protein will span the membrane once, with membrane-spanning segment 2 in the membrane and the C-terminus facing the cytoplasm.

If you shine light on chloroplasts and measure the rate of photosynthesis as a function of light intensity, you get a curve that reaches a plateau at a fixed rate of photosynthesis, x, as shown in Figure Q14-45. Which of the following conditions will increase the value of x? (a) increasing the number of chlorophyll molecules in the antennae complexes (b) increasing the number of reaction centers (c) adding a powerful oxidizing agent (d) decreasing the wavelength of light used

(b). The rate of photosynthesis will increase with increasing light intensity until photons hit all of the reaction centers directly. At saturating levels of light, the number of reaction centers that are still capable of being excited limits the rate of photosynthesis, which can be increased only by increasing the number of reaction centers or by increasing the rate at which the reaction centers are restored to their low-energy state. Increasing the number of chlorophyll molecules in the antennae complexes, the energy per photon of light, or the rate at which chlorophyll molecules are able to transfer energy electrons to one another will have no effect on either of these parameters. Adding a powerful oxidizing agent might, if anything, interfere with the reduction of the reaction center back to its resting state.

14-51 Oxidative phosphorylation, as it occurs in modern eucaryotes, is a complex process that probably arose in simple stages in primitive bacteria. Which mechanism is proposed to have arisen first as this complex system evolved? (a) electron transfers coupled to a proton pump (b) the reaction of oxygen with an ancestor of cytochrome oxidase (c) ATP-driven proton pumps (d) the generation of ATP from the energy of a proton gradient

(c)

Apoptosis differs from necrosis in that necrosis ________________. (a) requires the reception of an extracellular signal (b) causes DNA to fragment (c) causes cells to swell and burst, whereas apoptotic cells shrink and condense (d) involves a caspase cascade

(c)

At the end of DNA replication, the sister chromatids are held together by the ___________. (a) kinetochores (b) securins (c) cohesins (d) histones

(c)

In oxidative phosphorylation, ATP production is coupled to the events in the electron-transport chain. What is accomplished in the final electron transfer event in the electron-transport chain? (a) OH- is oxidized to O2. (b) Pyruvate is oxidized to CO2. (c) O2 is reduced to H2O. (d) NAD+ is reduced to NADH.

(c)

The local mediator nitric oxide stimulates the intracellular enzyme guanylyl cyclase by ________________. (a) activating a G-protein (b) activating a receptor tyrosine kinase (c) diffusing into cells and stimulating the cyclase directly (d) activating an intracellular protein kinase

(c)

Which of the following is not an electron carrier that participates in the electron-transport chain? (a) cytochrome (b) quinone (c) rhodopsin (d) copper ion

(c)

Figure Q16-51 shows that intracellular signaling pathways can be highly interconnected. From the information in Figure Q16-51, which of the following statements is incorrect? (a) The GPCR and the RTK both activate phospholipase C. (b) Activation of either the GPCR or the RTK will lead to activation of transcriptional regulators. (c) CaM-kinase is only activated when the GPCR is active and not when the RTK is active. (d) Ras is activated only when the RTK is active and not when the GPCR is active.

(c) is incorrect. CaM-kinase is activated by calmodulin, which is ultimately activated by phospholipase C. Either the GPCR or the RTK activates phospholipase C. All the other statements are correct.

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? (a) Cells carrying a Rac mutation that makes Rac act as if it is always bound to GTP will polymerize more unbranched actin filaments than normal cells. (b) Cells carrying a Rac mutation that makes Rac unable to exchange GDP for GTP will polymerize more unbranched actin filaments than normal cells. (c) 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. (d) Cells carrying a Rac mutation that makes Rac unable to exchange GDP for GTP will polymerize more branched actin filaments than normal cells.

(c). Activation of Rac promotes lamellipodia formation by enhancing actin nucleation using the ARP complex, which promotes the formation of branched actin filaments. Because lamellipodia formation involves branched actin filaments, a mutation that creates a constitutively active form of Rac (a GTP- bound form of Rac) will promote the formation of a greater number of branched actin filaments. A Rac mutation that makes Rac unable to exchange GDP for GTP will not be active.

Which of the following statements is correct? Kinesins and dyneins ____________________. (a) have tails that bind to the filaments (b) move along both microtubules and actin filaments (c) often move in opposite directions to each other (d) derive their energy from GTP hydrolysis

(c). All other answers are false. The motors heads bind to the filaments (choice (a)). Both motors move along microtubules (choice (b)) and use ATP hydrolysis for energy (choice (d)).

14-23 The relationship of free-energy change (δG) to the concentrations of reactants and products is important because it predicts the direction of spontaneous chemical reactions. Consider, for example, the hydrolysis of ATP to ADP and inorganic phosphate (Pi). The standard free-energy change (δG°) for this reaction is -7.3 kcal/mole. The free-energy change depends on concentrations according to the following equation: δG = δG° + 1.42 log10 ([ADP] [Pi]/[ATP]) In a resting muscle, the concentrations of ATP, ADP, and Pi are approximately 0.005 M, 0.001 M, and 0.010 M, respectively. At [Pi] = 0.010 M, what will be the ratio of [ATP] to [ADP] at equilibrium? (a) 1.38 × 106 (b) 1 (c) 7.2 × 10-8 (d) 5.14

(c). At equilibrium, the δG is equal to zero by definition. The ratio of [ATP] to [ADP] at equilibrium is less than 1:107. This result is calculated by setting δG = 0, so that 1.42 log10 ([ADP] [Pi]/[ATP]) = -δG° = 7.3 kcal/mole. Solving for [ADP]/[ATP], the equation becomes log10 ([ADP] [0.010]/[ATP]) = 7.3/1.42 = 5.14; then [ADP]/[ATP] = (105.14)/(0.010) = 13.8 × 106. Thus, the reciprocal [ATP]/[ADP] is 7.2 × 10-8.

Figure Q17-31A 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 Q17-31B) what do you predict would happen? (a) No bending would occur. (b) Bending would occur exactly as diagrammed in Figure Q17-31A. (c) Bending would occur, except that the right microtubule doublet would move down relative to the left one. (d) The two microtubule doublets would slide away from each other.

(c). Because the polarity of the microtubule bundle is reversed, the dynein motors should walk in the opposite direction from the normal situation diagrammed in Q17-31A. Microtubule sliding would occur if the linking proteins were absent (choice (d)), which is not true here.

Figure Q17-41 shows the leading edge of a lamellipodium. Which of the following statements is false? (a) Nucleation of new filaments near at the leading edge pushes the plasma membrane forward. (b) ARP proteins nucleate the branched actin filaments in the lamellipodia. (c) Capping proteins bind to the minus end of actin filaments. (d) There is more ATP-bound actin at the leading edge than in the actin filaments away from the leading edge.

(c). Capping protein binds to the plus end of actin filaments, preventing further assembly or disassembly from the growing end.

The following happens when a cell-surface receptor activates a G protein. (a) The β subunit exchanges its bound GDP for GTP. (b) The GDP bound to the α subunit is phosphorylated to form bound GTP. (c) The α subunit exchanges its bound GDP for GTP. (d) It activates the α subunit and inactivates the βγ complex.

(c). Choice (d) is incorrect because the βγ complex is often also activated. The other statements are simply untrue.

Stage 1 of oxidative phosphorylation requires the movement of electrons along the electron-transport chain coupled to the pumping of protons into the intermembrane space. What is the final result of these electron transfers? (a) OH- is oxidized to O2. (b) Pyruvate is oxidized to CO2. (c) O2 is reduced to H2O. (d) H- is converted to H2.

(c). Contrary to what the term "oxidative phosphorylation" may imply, the phosphorylation event does not depend on an oxidative reaction but rather on the reduction of molecular oxygen, converting it to water.

Your friend has just joined a lab that studies vesicle budding from the Golgi and has been given a cell line that does not form mature vesicles. He wants to start designing some experiments but wasn't listening carefully when he was told about the molecular defect of this cell line. He's too embarrassed to ask and comes to you for help. He does recall that this cell line forms coated pits but vesicle budding and the removal of coat proteins don't happen. Which of the following proteins might be lacking in this cell line? (a) clathrin (b) Rab (c) dynamin (d) adaptin

(c). Given that coated pits can form but no vesicle budding is seen, dynamin is the most likely answer. Since coated pits are formed, clathrin and adaptin are unlikely to be the answer, because they are involved in the initial shaping of the vesicle into the pit (choice (a) and (d)). Rab proteins are involved in the recognition of the transport vesicle with its target membrane and not with vesicle budding (choice (b)).

Figure Q16-52 shows how normal signaling works with a Ras protein acting downstream of an RTK. You examine a cell line with a constitutively active Ras protein that is always signaling. Which of the following conditions will turn off signaling in this cell line? (a) addition of a drug that prevents protein X from activating Ras (b) addition of a drug that increases the affinity of protein Y and Ras (c) addition of a drug that blocks protein Y from interacting with its target (d) addition of a drug that increases the activity of protein Y

(c). If protein Y cannot interact with its target, signaling will not occur. Increasing the activity of protein Y (choice (c)) or increasing the affinity of protein Y and Ras (choice (b) would not turn off signaling. Preventing protein X from activating Ras (choice (a)) would have no effect in a cell line with a constitutively active Ras protein, because Ras is already active.

Bongkrekic acid is an antibiotic that inhibits the ATP/ADP transport protein in the inner mitochondrial membrane. Which of the following will allow electron transport to occur in mitochondria treated with bongkrekic acid? (a) placing the mitochondria in anaerobic conditions (b) adding FADH2 (c) making the inner membrane permeable to protons (d) inhibiting the ATP synthase

(c). Inhibition of the ATP/ADP translocase prevents the export of ATP generated by oxidative phosphorylation in exchange for an import of ADP into the matrix. The ensuing buildup of ATP at the expense of ADP inhibits the ATP synthase. Because protons are no longer being used to power the ATP synthase, the proton gradient is not dissipated; the increasingly steep proton gradient makes it increasingly difficult for the electron-transport proteins to pump protons out of the matrix, and electron transport quickly stops. Hence, the inner membrane becomes permeable to protons, allowing electron transport to resume (although no ATP will be synthesized).

A protein kinase can act as an integrating device in signaling if it ___________________. (a) phosphorylates more than one substrate (b) catalyzes its own phosphorylation (c) is activated by two or more proteins in different signaling pathways (d) initiates a phosphorylation cascade involving two or more protein kinases

(c). Integrating devices are able to relay signals from more than one signaling pathway. Being activated by two or more proteins in different signalling pathways different allows a kinase (or any other signaling molecule) to be affected by more than one upstream signal. Choices (a), (b), and (d) affect the output signal that a kinase is able to produce, not its ability to integrate upstream signals from more than one signaling pathway.

Which of the following organelles are not part of the endomembrane system? (a) Golgi apparatus (b) the nucleus (c) mitochondria (d) lysosomes

(c). Mitochondria are not part of the endomembrane system, which is thought to have arisen initially through invagination of the plasma membrane. Instead, mitochondria (and chloroplasts) are thought to have evolved from a bacterium that was engulfed by a primitive eucaryotic cell.

NADH and FADH2 carry high-energy electrons that are used to power the production of ATP in the mitochondria. These cofactors are generated during glycolysis, the citric acid cycle, and the fatty acid oxidation cycle. Which molecuale below can produce the most ATP? Explain your answer. (a) NADH from glycolysis (b) FADH2 from the fatty acid cycle (c) NADH from the citric acid cycle (d) FADH2 from the citric acid cycle

(c). NADH produced in glycolysis does not contribute directly to ATP production in the mitochondria because it cannot be imported into the matrix. If the energy is transferred to a different carrier, some of the stored energy is lost. FADH2, from either the fatty acid cycle or the citric acid cycle, contributes less energy than NADH from the citric acid cycle because the electrons are donated further down the chain. Fewer electron transfers mean that fewer protons are pumped across the membrane.

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? (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). Normal cells arrest at the G2 checkpoint if DNA replication is incomplete or DNA is damaged. Cells without this mechanism may enter M phase with unreplicated or damaged DNA, whereas normal cells would not.

Cells have oligosaccharides displayed on their cell surface that are important for cell-cell recognition. Your friend discovered a transmembrane glycoprotein, GP1, on a pathogenic yeast cell that is recognized by human immune cells. He decides to purify large amounts of GP1 by expressing it in bacteria. To his purified protein he then adds a branched 14-sugar oligosaccharide to the asparagine of the only Asn-X-Ser sequence found on GP1 (Figure Q15-33). Unfortunately, immune cells do not seem to recognize this synthesized glycoprotein. Which of the following statements is a likely explanation for this problem? (a) The oligosaccharide should have been added to the serine instead of the asparagine. (b) The oligosaccharide should have been added one sugar at a time. (c) The oligosaccharide needs to be further modified before it is mature. (d) The oligosaccharide needs a disulfide bond.

(c). Oligosaccharides are usually further modified by enzymes in the ER and the Golgi before the glycoprotein is inserted into the plasma membrane. The other choices are untrue, and thus are not good explanations. Oligosaccharides are added to the Asn and not the Ser (choice (a)) and are added as a branched 14- sugar oligosaccharide (choice (b)). Disulfide bonding occurs between cysteines of proteins and not in sugars (choice (d)).

The microtubules in a cell form a structural framework that can have all the following functions except which of the following? (a) holding internal organelles such as the Golgi apparatus in particular positions in the cell (b) creating long thin cytoplasmic extensions that protrude from one side of the cell (c) strengthening the plasma membrane (d) moving materials from one place to another inside a cell

(c). One function of actin filaments, but not microtubules, is to provide a meshwork beneath the plasma membrane that helps to form and strengthen this membrane. Microtubules have all of the other functions that are listed.

Most proteins destined to enter the endoplasmic reticulum (a) are transported across the membrane after their synthesis is complete. (b) are synthesized on free ribosomes in the cytosol. (c) begin to cross the membrane while still being synthesized. (d) remain within the endoplasmic reticulum.

(c). Proteins destined to enter the endoplasmic reticulum have an N-terminal signal sequence that leads to the docking of the ribosome synthesizing the protein onto the ER and the entry of the protein across the ER membrane as the polypeptide chain is being synthesized.

The growth factor Superchick stimulates the proliferation of cultured chicken cells. The receptor that binds Superchick is a receptor tyrosine kinase (RTK), and many chicken tumor cell lines have mutations in the gene that encodes this receptor. Which of the following types of mutation would be expected to promote uncontrolled cell proliferation? (a) a mutation that prevents dimerization of the receptor (b) a mutation that destroys the kinase activity of the receptor (c) a mutation that inactivates the protein tyrosine phosphatase that normally removes the phosphates from tyrosines on the activated receptor (d) a mutation that prevents the binding of the normal extracellular signal to the receptor

(c). RTKs are usually activated by signal-induced dimerization, which allows the receptors to phosphorylate themselves and activate intracellular signaling proteins that are stimulated by the phosphorylated receptor. After it is activated, the receptor is dephosphorylated, and thereby inactivated, by a protein tyrosine phosphatase. Therefore, a mutation in the gene that encodes the protein tyrosine phosphatase will inappropriately increase the activity of the receptor and promote uncontrolled cell proliferation. Mutations that prevent dimerization of the receptor (including mutations that prevent ligand binding) or autophosphorylation (which requires the kinase activity of the receptor) will inactivate the receptor.

In which cellular location would you expect to find ribosomes translating mRNAs that encode ribosomal proteins? (a) the nucleus (b) on the rough ER (c) in the cytosol (d) in the lumen of the ER

(c). Ribosomes are cytoplasmic proteins and thus their protein components are translated in the cytosol.

Which of the following structures shorten during muscle contraction? (a) myosin filaments (b) flagella (c) sarcomeres (d) actin filaments

(c). Sarcomeres contain actin filaments and myosin filaments that slide past each other during muscle contraction, leading to shortening of the sarcomere; the actin filaments and myosin filaments do not change in length. Flagella are microtubule- based structures that are not present on muscle cells.

Which of the following statements about the anaphase-promoting complex (APC) is false? (a) It promotes the degradation of proteins that regulate M phase. (b) It inhibits M-Cdk activity. (c) It is continuously active throughout the cell cycle. (d) M-Cdk stimulates its activity.

(c). The APC becomes activated in mid to late M phase.

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). The concentration of mitotic cyclin rises gradually during G2 and is ubiquitylated and degraded during late M phase.

Which of the following statements about the cytoskeleton is false? (a) The cytoskeleton is made up of three types of protein filament. (b) The bacterial cytoskeleton is important for cell division and DNA segregation. (c) Protein monomers that are held together with covalent bonds form cytoskeletal filaments. (d) The cytoskeleton of a cell can change in response to the environment.

(c). The protein monomers of the cytoskeleton are held together by noncovalent interactions between the protein monomers. All other statements are true.

Which of the following statements about the unfolded protein response (UPR) is false? (a) Activation of the UPR results in the production of more ER membrane. (b) Activation of the UPR results in the production of more chaperone proteins. (c) Activation of the UPR occurs when receptors in the cytoplasm sense misfolded proteins. (d) Activation of the UPR results in the cytoplasmic activation of gene regulatory proteins.

(c). The receptors for the unfolded proteins are on the ER membrane, and they sense the misfolded proteins using their luminal domains.

Which of the following statements is true? (a) The signal sequences on mitochondrial proteins are usually C-terminal. (b) Most mitochondrial proteins are not imported from the cytosol but are synthesized inside the mitochondria. (c) Chaperone proteins in the mitochondria facilitate the movement of proteins across the outer and inner mitochondrial membranes. (d) Mitochondrial proteins cross the membrane in their native, folded state.

(c). The signal sequences on a protein destined for the mitochondria are on its N-terminus (choice (a)). Although some mitochondrial proteins are synthesized inside the mitochondria from the mitochondrial genome, most mitochondrial proteins are encoded by genes in the nucleus and imported into the mitochondria after synthesis in the cytosol (choice (b)). Mitochondrial proteins are unfolded as they enter the mitochondria through protein translocators (choice (d)).

The enzyme ribulose bisphosphate carboxylase (rubisco) normally adds carbon dioxide to ribulose 1,5-bisphosphate. However, it will also catalyze a competing reaction in which O2 is added to ribulose 1,5-bisphosphate to form 3-phosphoglycerate and phosphoglycolate. Assume that phosphoglycolate is a compound that cannot be used in any further reactions. If O2 and CO2 have the same affinity for rubisco, which of the following is the lowest ratio of CO2 to O2 at which a net synthesis of sugar can occur? (a) 1:3 (b) 1:2 (c) 3:1 (d) 2:1

(c). Three molecules of O2 are required to form three molecules of 3-phosphoglycerate and three molecules of phosphoglycolate. To break even (i.e., simply to keep the Calvin cycle going with no net sugar produced), you need to have enough 3-phosphoglycerate to synthesize ribulose 1,5-bisphosphate again. Therefore, for every three molecules of O2 that react with ribulose 1,5-bisphosphate, you need to generate two additional molecules of 3-phosphoglycerate. For every three molecules of CO2 that go into the Calvin cycle, one molecule of 3-phosphoglycerate is formed. So if you have at least six molecules of CO2 per three molecules of O2 going through the Calvin cycle, you will break even. Only if you have a ratio of CO2 to O2 higher than 6:3 (2:1) can you have a net synthesis of carbohydrate.

Your friend works in a biotechnology company and has discovered a drug that blocks the ability of Ran to exchange GDP for GTP. What is the most likely effect of this drug on nuclear transport? (a) Nuclear transport receptors would be unable to bind cargo. (b) Nuclear transport receptors would be unable to enter the nucleus. (c) Nuclear transport receptors would be unable to release their cargo in the nucleus. (d) Nuclear transport receptors would interact irreversibly with the nuclear pore fibrils.

(c). When Ran-GTP binds to the nuclear transport receptor, cargo is released. If Ran could not exchange its GTP for GDP, this would not happen. Ran-GTP is not needed for cargo binding, for nuclear entry, or for interactions with the nuclear pore fibrils during nuclear import

You are interested in how cyclic-AMP-dependent protein kinase A (PKA) functions to affect learning and memory, and you decide to study its function in the brain. It is known that, in the cells you are studying, PKA works via a signal transduction pathway like the one depicted in Figure Q16-28. Furthermore, it is also known that activated PKA phosphorylates the transcriptional regulator called Nerd that then activates transcription of the gene Brainy. Which situation described below will lead to an increase in Brainy transcription? (a) a mutation in the Nerd gene that produces a protein that cannot be phosphorylated by PKA (b) a mutation in the nuclear import sequence of PKA from PPKKKRKV to PPAAAAAV (c) a mutation in the gene that encodes cAMP phosphodiesterase that makes the enzyme inactive (d) a mutation in the gene that encodes adenylyl cyclase that renders the enzyme unable to interact with the α subunit of the G protein

(c). cAMP phosphodiesterase is important for converting cAMP into AMP and thus downregulating the activity of PKA. Without cAMP phosphodiesterase, transcription of Brainy will be increased. All the other choices will lead to inactivation of the signaling pathway and a decrease in Brainy transcription. A mutant form of Nerd that cannot be phosphorylated will not be active (choice (a)). If PKA cannot be imported into the nucleus, it will be unable to phosphorylate Nerd (choice (b)). Adenylyl cyclase interaction with the α subunit of the G protein is important for the G protein's activation (choice (d)).

Below is a list of breakthroughs in energy metabolism in living systems. Which is the correct order in which they are thought to have evolved? A. H2O-splitting enzyme activity B. light-dependent transfer of electrons from H2S to NADPH C. the consumption of fermentable organic acids D. oxygen-dependent ATP synthesis (a) A, C, D, B (b) C, A, B, D (c) B, C, A, D (d) C, B, A, D

(d)

N-linked oligosaccharides on secreted glycoproteins are attached to (a) nitrogen atoms in the polypeptide backbone. (b) the serine or threonine in the sequence Asn-X-Ser/Thr. (c) the N-terminus of the protein. (d) the asparagine in the sequence Asn-X-Ser/Thr.

(d)

Photosynthesis is a process that takes place in chloroplasts and uses light energy to generate high-energy electrons, which are passed along an electron-transport chain. Where are the proteins of the electron-transport chain located in chloroplasts? (a) thylakoid space (b) stroma (c) inner membrane (d) thylakoid membrane

(d)

Stage 2 of photosynthesis, sometimes referred to as the dark reactions, involves the reduction of CO2 to produce organic compounds such as sucrose. What cofactor is the electron donor for carbon fixation? (a) H2O (b) NADH (c) FADH2 (d) NADPH

(d)

The ATP synthase found in chloroplasts is structurally similar to the ATP synthase in mitochondria. Given that ATP is being synthesized in the stroma, where will the F0 portion of the ATP synthase be located? (a) thylakoid space (b) stroma (c) inner membrane (d) thylakoid membrane

(d)

The G1 DNA damage checkpoint ________________. (a) causes cells to proceed through S phase more quickly (b) involves the degradation of p53 (c) is activated by errors caused during DNA replication (d) involves the inhibition of cyclin-Cdk complexes by p21

(d)

Which component of the electron-transport chain is required to combine the pair of electrons with molecular oxygen? (a) cytochrome c (b) cytochrome b-c1 complex (c) ubiquinone (d) cytochrome c oxidase

(d)

Which of the following choices reflects the appropriate order through which a protein destined for the plasma membrane travels? (a) lysosome → endosome → plasma membrane (b) ER → lysosome → plasma membrane (c) Golgi → lysosome → plasma membrane (d) ER → Golgi → plasma membrane

(d)

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)

Which of the following statements is true? (a) The mitotic spindle is largely made of intermediate filaments. (b) The contractile ring is made largely of microtubules and actin filaments. (c) The contractile ring divides the nucleus in two. (d) The mitotic spindle helps segregate the chromosomes to the two daughter cells.

(d)

Consider the in vitro motility assay using purified kinesin and purified polymerized microtubules shown in Figure Q17-50. 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? (a) Kinesin molecules are attached by their tails to a glass slide. (b) The microtubules used in this assay must be polymerized using conditions that stabilize tubule formation or else they would undergo dynamic instability. (c) ATP must be added for this assay to work. (d) Addition of the nonhydrolyzable ATP analog (AMP-PNP) would cause the microtubules to move faster.

(d) is false. Addition of AMP-PNP would block movement, because ATP- hydrolysis is required for the kinesin to step along a microtubule. Addition of AMP-PNP would cause the microtubules to attach to the kinesin heads without being released. Kinesin molecules are attached to the slide by their tails (the cargo-binding domain) so that the heads are available to move the microtubules along the slides. If they were in solution with the microtubules, there would be no force and thus no movement (choice (a)). The microtubules used in this assay are stabilized with a nonhydrolyzable form of GTP, because otherwise they might shrink during the course of the assay (choice (b)). ATP is required for kinesin movement, and thus must be added for this assay to work (choice (c)).

Akt promotes the survival of many cells. It is activated by an intracellular signaling pathway that is triggered by an RTK that activates PI 3-kinase, as diagrammed in Figure Q16-48. Figure Q16-48 Which of the following statements is false? (a) In the presence of a survival signal, Akt localizes to the plasma membrane by binding to PIP3. (b) In the absence of survival signal, Bad inhibits the cell death inhibitor protein Bcl2. (c) In the presence of survival signal, the cell death inhibitory protein Bcl2 is active. (d) In the absence of survival signal, Bad is phosphorylated.

(d) is incorrect. Bad is phosphorylated in the presence of survival signal. When the survival signal is not present, Bad binds to the cell death inhibitor protein Bcl2, promoting cell death. All the other statements are correct.

Which of the following statements is false? (a) Nucleotides and amino acids can act as extracellular signal molecules. (b) Some signal molecules can bind directly to intracellular proteins that bind DNA and regulate gene transcription. (c) Some signal molecules are transmembrane proteins. (d) Dissolved gases such as nitric oxide (NO) can act as signal molecules, but because they cannot interact with proteins they must act by affecting membrane lipids.

(d) is not true. NO can diffuse across the plasma membrane and directly activate intracellular proteins such as the enzyme guanylyl cyclase.

Which of the following statements about the cytoskeleton is true? (a) All eucaryotic cells have actin, microtubules, and intermediate filaments in their cytoplasm. (b) The cytoskeleton provides a rigid and unchangeable structure important for the shape of the cell. (c) The three cytoskeletal filaments perform distinct tasks in the cell and act completely independently of one another. (d) Actin filaments and microtubules have an inherent polarity, with a plus end that grows more quickly than the minus end.

(d) is true. Not all eucaryotic cells have cytoplasmic intermediate filaments (choice (a)). The cytoskeleton is not rigid and unchangeable; in fact, it can be quite dynamic (choice (b)). Each of the three cytoskeletal systems is not completely independent. For example, proteins such as plectin are known to link intermediate filaments to the actin and microtubule cytoskeleton (choice (c)).

Microtubules are important for transporting cargo in nerve cell axons, as diagrammed in Figure Q17-24. Notice that the two types of cargo are traveling in opposite directions. Which of the following statements is likely to be false? (a) The gray cargo is attached to dynein. (b) The black cargo and the gray cargo require ATP hydrolysis for their motion. (c) The black cargo moving toward the axon terminal contains a domain that specifically interacts with the tail domain of a particular kind of motor. (d) The black cargo and the gray cargo are moving along microtubules of opposite polarity.

(d) is unlikely to be the case. Microtubules in nerve axons are generally organized such that their plus ends are facing the axon terminal while the minus ends reside in the cell body. Thus, the gray cargo is likely to be attached to a dynein motor, because it is moving toward the cell body (choice (a)). Because cargo attaches to the tail domains of both dynein and kinesin motors, the attachment of a cargo to its tail is unlikely to affect directionality (choice (c)). Both dynein and kinesis require ATP hydrolysis for their movement (choice (b)).

Which of the following statements regarding dynamic instability is false? (a) Each microtubule filament grows and shrinks independently of its neighbors. (b) The GTP cap helps protect a growing microtubule from depolymerization. (c) GTP hydrolysis by the tubulin dimer promotes microtubule shrinking. (d) The newly freed tubulin dimers from a shrinking microtubule can be immediately captured by growing microtubules and added to their plus end.

(d) is untrue. A newly dissociated tubulin dimer will be bound to GDP; this GDP will need to be exchanged for GTP before it can be added to a newly growing microtubule.

Which of the statements below about intermediate filaments is false? (a) They can stay intact in cells treated with concentrated salt solutions. (b) They can be found in the cytoplasm and the nucleus. (c) They can be anchored to the plasma membrane at cell-cell junction. (d) Each filament is about 10 μm in diameter.

(d) is untrue. Intermediate filaments are about 10 nm (not μm) in diameter. All the other statements are true.

Keratins, neurofilaments, and vimentins are all categories of intermediate filaments. Which of the following properties below is not true of these types of intermediate filaments? (a) They strengthen cells against mechanical stress. (b) Dimers associate by noncovalent bonding to form a tetramer. (c) They are found in the cytoplasm. (d) Phosphorylation causes disassembly during every mitotic cycle.

(d) is untrue. Keratins, neurofilaments, and vimentins are cytoplasmic intermediate filaments (choice (c)), which tend to be very stable once formed. The nuclear intermediate filaments are disassembled and reformed during mitosis; this process is regulated by phosphorylation.

Adrenaline stimulates glycogen breakdown in skeletal muscle cells by ultimately activating glycogen phosphorylase, the enzyme that breaks down glycogen, as depicted in Figure Q16-29. Which of the following statements below is false? (a) A constitutively active mutant form of PKA in skeletal muscle cells would lead to a decrease in the amount of unphosphorylated phosphorylase kinase. (b) A constitutively active mutant form of PKA in skeletal muscle cells would not increase the affinity of adrenaline for the adrenergic receptor. (c) A constitutively active mutant form of PKA in skeletal muscle cells would lead to an excess in the amount of glucose available. (d) A constitutively active mutant form of PKA in skeletal muscle cells would lead to an excess in the amount of glycogen available.

(d). A constitutively active mutant form of PKA in skeletal muscle cells would lead to a decrease in the amount of glycogen available, because active PKA stimulates enzymes that are responsible for the breakdown of glycogen so that glucose can be produced. All the other statements are true.

Which of the following protein families are not involved in directing transport vesicles to the target membrane? (a) SNAREs (b) Rabs (c) tethering proteins (d) adaptins

(d). Adaptins are involved in vesicle budding and are removed during the uncoating process and thus should not be present when the vesicle reaches its target.

Electron transport is coupled to ATP synthesis in mitochondria, in chloroplasts, and in the thermophilic bacterium Methanococcus. Which of the following is likely to affect the coupling of electron transport to ATP synthesis in all of these systems? (a) a potent inhibitor of cytochrome oxidase (b) the removal of oxygen (c) the absence of light (d) an ADP analogue that inhibits ATP synthase

(d). All chemiosmotic coupling systems involve a proton gradient that ATP synthase uses to bind to ADP and phosphorylate it. Hence agents that prevent ADP from binding the synthase or that dissipate the proton gradient affect all chemiosmotic systems. Cytochrome oxidase and oxygen are required only for mitochondria and aerobic bacteria (not Methanococcus); light is required only for chloroplasts and photosynthetic bacteria (not Methanococcus).

All members of the steroid hormone receptor family __________________. (a) are cell-surface receptors 16-9 (b) do not undergo conformational changes (c) are found only in the cytoplasm (d) interact with signal molecules that diffuse through the plasma membrane

(d). All members of the steroid hormone receptor family are intracellular proteins (thus choice (a) is not correct) that interact with signal molecules that can diffuse through the plasma membrane. Once activated, steroid hormone receptors regulate gene transcription in the nucleus (choice (c)). The binding of the signal molecule induces a large conformational change in the receptor protein (choice (b)). This conformational change activates the steroid hormone receptors, allowing them to promote or inhibit the transcription of the appropriate genes.

Which of the following is not good direct evidence that the cell-cycle control system is conserved through billions of 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). Although it is true that yeast cells have one Cdk and human cells have many, this statement does not provide any evidence for the conservation of function across evolutionary time. All the other statements do provide such evidence.

Which of the following statements is false? (a) The cleavage furrow is a puckering of the plasma membrane caused by the constriction of a ring of filaments attached to the plasma membrane. (b) The cleavage furrow will not begin to form in the absence of a mitotic spindle. (c) The cleavage furrow always forms perpendicular to the interpolar microtubules. (d) The cleavage furrow always forms in the middle of the cell.

(d). Although the furrow always forms perpendicular to the interpolar microtubules about midway between the spindle poles (choice (c)), if the spindle were in an asymmetrical position (which can occur normally during development), cell division would not always occur in the middle of the cell.

Which of the following statements about disulfide bond formation is false? (a) Disulfide bonds do not form under reducing environments. (b) Disulfide bonding occurs by the oxidation of pairs of cysteine side chains on the protein. (c) Disulfide bonding stabilizes the structure of proteins. (d) Disulfide bonds form spontaneously within the ER because the lumen of the ER is oxidizing.

(d). An enzyme in the ER lumen catalyzes disulfide bond formation.

Which of the following statements is true? (a) Anaphase A must be completed before anaphase B can take place. (b) In cells in which anaphase B predominates, the spindle will elongate much less than in cells in which anaphase A dominates. (c) In anaphase A, both kinetochore and interpolar microtubules shorten. (d) In anaphase B, microtubules associated with the cell cortex shorten.

(d). Anaphase A and anaphase B generally occur at the same time (choice (a)). In cells in which anaphase B predominates, the spindle will elongate more than in cells in which anaphase A predominates (choice (b)). In anaphase A, only the kinetochore microtubules shorten (choice (c)).

Experimental evidence supporting the chemiosmotic hypothesis was gathered by using artificial vesicles containing a protein that can pump protons in one direction across the vesicle membrane to create a proton gradient. Which protein was used to generate the gradient in a highly controlled manner? (a) cytochrome c oxidase (b) NADH dehydrogenase (c) cytochrome c (d) bacteriorhodopsin

(d). Bacteriorhodopsin is a transmembrane protein that pumps protons across the membrane when exposed to light. The other proteins pump protons but are part of the electron-transport chain, so they would not be good options to show that it is only the proton gradient that is required in this system rather than any specific intermediate in the electron-transport chain.

Which of the following statements about apoptosis is true? (a) Cells that constitutively express Bcl2 will be more prone to undergo apoptosis. (b) The prodomain of procaspases contains the catalytic activity necessary for procaspase activation. (c) Bax and Bak promote apoptosis by binding to procaspases in the apoptosome. (d) Apoptosis is promoted by the release of cytochrome c into the cytosol from mitochondria.

(d). Bcl2 tends to inhibit rather than promote apoptosis (choice (a)). The prodomain of procaspases does not contain the catalytic activity needed for procaspase activation and is usually discarded from the active caspase (choice (b)). When activated, Bax and Bak promote apoptosis by stimulating the release of cytochrome c from mitochondria into the cytosol, not by binding to procaspases in the apoptosome (choice (c)).

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). Cdks do not phosphorylate each other (choice (a)). The Cdks do not activate the cyclins (choice (b)), and Cdks are not degraded during specific phases of the cycle (choice (c)). The cyclins, however, are degraded in a cell-cycle-dependent fashion, and they are required for Cdk activity.

Which of the following statements about actin is false? (a) ATP hydrolysis decreases actin filament stability. (b) Actin at the cell cortex helps govern the shape of the plasma membrane. (c) Actin filaments are nucleated at the side of existing actin filaments in lamellipodia. (d) The dynamic instability of actin filaments is important for cell movement.

(d). Dynamic instability is a phenomenon associated with microtubules and not actin. Actin disassembly and assembly are both important for cell movement. However, this differs from dynamic instability in that the growth of actin filaments occurs at the leading edge; this growth occurs in a directed fashion because of actin-binding proteins that promote the formation of new filaments at the leading edge. Actin-binding proteins that destabilize actin filaments promote actin disassembly away from the leading edge. The actin assembly and disassembly in moving cells differs from the stochastic growth and disassembly of the microtubules.

You are interested in Fuzzy, a soluble protein that functions within the ER lumen. Given that information, which of the following statement must be true? (a) Fuzzy has a C-terminal signal sequence that binds to SRP. (b) Only one ribosome can be bound to the mRNA encoding Fuzzy during translation. (c) Fuzzy must contain a hydrophobic stop-transfer sequence. (d) Once the signal sequence from Fuzzy has been cleaved, the signal peptide will be ejected into the ER membrane and degraded.

(d). ER signal sequences are typically at the N-terminus (choice (a)). More than one ribosome can bind to an mRNA molecule (choice (b)). Hydrophobic stop-transfer sequences are found on membrane-inserted proteins and not on soluble proteins (choice (c)).

Disassembly of the nuclear envelope ________________. (a) causes the inner nuclear membrane to separate from the outer nuclear membrane (b) results in the conversion of the nuclear envelope into protein-free membrane vesicles (c) is triggered by the phosphorylation of integrins (d) must occur for kinetochore microtubules to form in animal cells

(d). In animal cells, kinetochore microtubules cannot form if the chromosomes in the nucleus are separated from the microtubules in the cytoplasm because of the presence of the nuclear envelope. (But in some other cells, such as the yeast S. cerevisiae, the nuclear envelope never breaks down and yet chromosomes can attach to microtubules emanating from the spindle poles within the nucleus.) The nuclear envelope disassembles by breaking up into vesicles containing lipids from both the outer and inner envelopes (choice (a)). Integral membrane proteins of the nuclear envelope and some of the nuclear lamins remain associated with the vesicles (choice (b)). Phosphorylation of lamins (not integrins) triggers the breakdown of the nuclear lamina (choice (c)).

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 below is unlikely to be involved in the pathway that this drug affects? (a) Rac (b) ARP (c) actin (d) myosin

(d). Myosins are not directly involved in lamellipodia formation. Lamellipodia formation involves branched actin structures that use ARP for their formation, and thus ARP and actin are likely to be involved (choices (b) and (c)). The Rho family member Rac triggers lamellipodia formation when activated, and thus may be involved (choice (a)).

Which ratio of NADH to NAD+ in solution will generate the largest, positive redox potential? (a) 1:10 (b) 10:1 (c) 1:1 (d) 5:1

(d). NAD+ is the electron acceptor; NADH is the electron donor. If there is an excess of NAD+ in solution, there is less capacity to donate electrons (which would reflect a negative redox potential) and more capacity to accept electrons (which would reflect a positive redox potential).

Which of the following statements is false? (a) Cytokinesis in plant cells is mediated by the microtubule cytoskeleton. (b) Small membrane vesicles derived from the Golgi apparatus deliver new cell wall material for the new wall of the dividing cell. (c) The phragmoplast forms from the remains of interpolar microtubules of the mitotic spindle. (d) Motor proteins walking along the cytoskeleton are important for the contractile ring that guides formation of the new cell wall.

(d). No contractile ring is formed during plant cytokinesis.

Where are proteins in the chloroplast synthesized? (a) in the cytosol (b) in the chloroplast (c) on the endoplasmic reticulum (d) in both the cytosol and the chloroplast

(d). Proteins in the chloroplast are synthesized in the cytosol and in the chloroplast. The chloroplast proteins that are encoded by the nuclear DNA are synthesized in the cytosol, and the sorting signals on the protein direct them to the chloroplast. The chloroplast proteins encoded by the chloroplast DNA are synthesized on ribosomes inside the chloroplast.

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.

(d). The cells produced would get smaller and smaller, as they would not have sufficient time to double their mass before dividing.

The hydrolysis of GTP to GDP carried out by tubulin molecules ________________. (a) provides the energy needed for tubulin to polymerize (b) occurs because the pool of free GDP has run out (c) tips the balance in favor of microtubule assembly (d) allows the behavior of microtubules called dynamic instability

(d). The hydrolysis of GTP to GDP occurs after a GTP-bound tubulin molecule is incorporated into a microtubule, and it makes the microtubule more susceptible to disassembly. It is the resulting switch in microtubule stability that gives rise to the phenomenon known as dynamic instability.

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? (a) The piercing of the oocyte membrane by a needle is insufficient to cause mitotic spindle formation. (b) An increased volume of cytoplasm is insufficient to cause mitotic spindle formation. (c) Injection of extra RNA molecules is insufficient to cause mitotic spindle formation. (d) Components of an interphase nucleus are insufficient to cause mitotic spindle formation.

(d). The interphase cytoplasmic extract used as the control would not have contained nuclear components (because the nuclear membrane was intact) and so one cannot conclude that components of an interphase nucleus are insufficient to cause mitotic spindle formation. The other statements are all true.

When introduced into mitotic cells, which of the following is expected to impair anaphase B but not anaphase A? (a) an antibody against myosin (b) ATPγS, a nonhydrolyzable ATP analog that binds to and inhibits ATPases (c) an antibody against the motor proteins that move from the plus end of microtubules to the minus end (d) an antibody against the motor proteins that move from the minus end of microtubules toward the plus end

(d). The motor proteins used in anaphase B to push the interpolar microtubules apart move toward the plus end of microtubules, whereas the motor proteins used in anaphase A move toward the minus end. Myosin (choice (a)) is not involved in either anaphase A or anaphase B. Motor proteins require ATP hydrolysis (they are ATPases) to function and are used in both anaphase A and anaphase B, so both types of anaphase will be affected by ATPγS (choice (b)). The motor proteins used in anaphase A move toward the minus end of microtubules, as do the motor proteins attached to the cell cortex used in anaphase B (choice (c)).

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 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? (a) addition of phosphatase inhibitors (b) addition of ATP (c) addition of a concentrated salt solution that is 10 times the concentration normally found in the nucleoplasm (d) addition of protein kinase inhibitors

(d). The phosphorylation of nuclear lamina by protein kinases induces conformational changes that weaken the binding between nuclear lamina tetramers; thus, inhibiting protein kinases may enhance assembly of the nuclear lamina. Adding phosphatase inhibitors (choice (a)) or ATP (choice (b)) will enhance the activity of any co-purifying protein kinases and enhance disassembly. Because noncovalent protein-protein interactions hold the nuclear lamina together, the addition of a very concentrated salt solution will inhibit proper nuclear lamina assembly (choice (c)).

Cell movement involves the coordination of many events in the cell. Which of the following phenomena are not required for cell motility? (a) Myosin-mediated contraction at the rear of the moving cell. (b) Integrin association with the extracellular environment. (c) Nucleation of new actin filaments. (d) Release of Ca2+ from the sarcoplasmic reticulum.

(d). The release of Ca2+ from the sarcoplasmic reticulum is important for muscle contraction, not cell motility.

After isolating the rough endoplasmic reticulum from the rest of the cytoplasm, you purify the RNAs attached to it. Which of the following proteins do you expect the RNA from the rough endoplasmic reticulum to encode? (a) soluble secreted proteins (b) ER membrane proteins (c) plasma membrane proteins (d) all of the above

(d). The rough ER consists of ER membranes and polyribosomes that are in the process of translating and translocating proteins into the ER membrane and lumen. Thus all proteins that end up in the lysosome, Golgi apparatus, or plasma membrane, or are secreted, will be encoded by the RNAs associated with the rough ER.

The relationship of free-energy change (δG) to the concentrations of reactants and products is important because it predicts the direction of spontaneous chemical reactions. In the hydrolysis of ATP to ADP and inorganic phosphate (Pi), the standard free-energy change (δG°) is -7.3 kcal/mole. The free-energy change depends on concentrations according to the following equation: δG = δG° + 1.42 log10 ([ADP] [Pi]/[ATP]) In a resting muscle, the concentrations of ATP, ADP, and Pi are approximately 0.005 M, 0.001 M, and 0.010 M, respectively. What is the δG for ATP synthesis in resting muscle? (a) -6.01 kcal/mole (b) 5.88 kcal/mole (c) 8.72 kcal/mole (d) 11 kcal/mole

(d). The δG for hydrolysis is -11.1 kcal/mole. This result is calculated by substituting values into the equation given: δG = -7.3 kcal/mole + 1.42 log10 ([0.001 M] [0.010 M]/[0.005 M]) = -7.3 kcal/mole + 1.42 log10 (0.002) = -11.1 kcal/mole. The δG for synthesis is +11.1 kcal/mole because the forward and reverse reactions always have the same numerical value for δG, but with the sign reversed.

Name two types of protein modification that can occur in the ER but not in the cytosol.

1. Proteins in the ER can undergo disulfide bond formation. (This does not occur in the cytosol because of its reducing environment.) 2. Proteins in the ER can undergo glycosylation. (Glycosylating enzymes are not found in the cytosol.) (Signal-sequence cleavage is also an acceptable answer, although not really what this question is referring to.)

Name three possible fates for an endocytosed molecule that has reached the endosome.

1. recycled to the original membrane 2. destroyed in the lysosome 3. transcytosed across the cell to a different membrane

Describe how a standard flashlight battery can convert energy into useful work and explain how this is similar to the energy conversions in the mitochondria.

A battery contains chemicals that generate negatively charged ions at one pole, and it is able to cause the continuous transfer of electrons along a metal wire if that pole is connected to the other end of the battery. The energy released by the electron transfer process driven by the battery can be harnessed to do useful work, as when it is used to run an electric motor.. Likewise, the energy released by the electron transfers that occur between the protein complexes in the electron transport chain does useful work when it drives the movement of protons to one side of the membrane, since the resulting proton gradient is then used to generate chemical energy in the form of ATP.

You discover a fungus that contains a strange star-shaped organelle not found in any other eucaryotic cell you have seen. On further investigation you find the following. 1. The organelle possesses a small genome in its interior. 2. The organelle is surrounded by two membranes. 3. Vesicles do not pinch off from the organelle membrane. 4. The interior of the organelle contains proteins similar to those of many bacteria. 5. The interior of the organelle contains ribosomes. How might this organelle have arisen?

A genome, a double membrane, ribosomes, and proteins similar to those found in bacteria are evidence for an organelle's having evolved from an engulfed bacterium.

Name the membrane-enclosed compartments in a eucaryotic cell where each of the functions listed below takes place. A. photosynthesis B. transcription C. oxidative phosphorylation D. modification of secreted proteins E. steroid hormone synthesis F. degradation of worn-out organelles G. new membrane synthesis H. breakdown of lipids and toxic molecules

A photosynthesis = chloroplast B. transcription = nucleus C. oxidative phosphorylation = mitochondrion D. modification of secreted proteins = Golgi apparatus and rough endoplasmic reticulum (ER) E. steroid hormone synthesis = smooth ER F. degradation of worn-out organelles = lysosome G. new membrane synthesis = ER H. breakdown of lipids and toxic molecules = peroxisome

You have discovered a new protein that regulates microtubule dynamics. First, you isolated proteins from a cellular extract that bound to a tubulin affinity column. You then separated the proteins from each other by loading the mixture of proteins on an ion-exchange column, eluting the column with increasing salt concentration and collecting small "fractions" of protein as they dripped from the column. To test whether each fraction contained microtubule regulators, you mixed it with fluorescent tubulin and purified centrosomes, and then analyzed the reaction microscopically to measure the size of the aster microtubules formed. You found that fractions 8, 9, and 10 promoted the formation of unusually long aster microtubules. Because electrophoretic separation of the fractions on a gel revealed a plentiful protein with an apparent molecular weight of 98 kD, you named the protein p98. A. Does p98 behave like a MAP or a catastrophin? B. Propose two ways in which p98 might change the dynamic behavior of microtubules to account for the observed change in microtubule length. Hint: There are four simple possible mechanisms. C. Video microscopy of fluorescent tubulin in reactions with purified centrosomes allowed you to follow the behavior of individual microtubules over time. You graphed the changes in microtubule length in the absence (Figure Q18-51A) and presence (Figure Q18-51B) of p98. Five representative microtubules are shown for each condition. Does p98 alter the rate of microtubule growth or shrinkage? Does p98 alter the frequency of catastrophes (a sudden and rapid decline in microtubule length) or rescues (when a microtubule switches from shrinking to growing)? Explain your answers. D. After demonstrating the consequences of p98 on microtubule dynamics in vitro with the use of purified components, you want to determine whether the protein has the same effects in a complex cellular extract that naturally contains p98. Briefly describe an experiment that will allow you to determine what p98 normally does in mitotic extracts.

A. B. The four possible mechanisms by which a protein can promote p98 behaves like a MAP, because it promotes microtubule growth. microtubule growth are (1) increasing the rate of polymerization, (2) decreasing the rate of depolymerization, (3) inhibiting the occurrence of catastrophes (when a microtubule shifts abruptly from growing to shrinking), and (4) stimulating the occurrence of rescues (when a microtubule switches from shrinking to growing). C. p98 increased the rate of microtubule growth, as seen by the steeper slopes of the lines when p98 was added. The data do not allow a conclusion to be made about the rate of microtubule shrinkage. The protein decreased the rate of catastrophes: two of the five microtubules in the absence of p98 underwent catastrophe, whereas none did so in the presence of p98. No rescues were observed in either the presence or the absence of p98, so it is uncertain whether the protein alters the rescue rate. D. The p98 protein can be removed from an extract of Xenopus eggs in mitosis by using antibodies that specifically recognize p98. The depleted extract with little to no p98 protein can then be mixed with sperm nuclei, centrosomes, and fluorescent tubulin. On the basis of the previous experiments, the p98-depleted extract would be expected to have shorter and more dynamic microtubules. Ideally, this kind of experiment would be supplemented with the production and examination of intact cells in which p98 function has been abolished by mutation or another means, such as RNA-mediated interference.

A calmodulin-regulated kinase (CaM-kinase) is involved in spatial learning and memory. This kinase is able to phosphorylate itself such that its kinase activity is now independent of the intracellular concentration of Ca2+. Thus the kinase stays active after Ca2+ levels have dropped. Mice completely lacking this CaM-kinase have severe spatial learning defects but are otherwise normal. A. Each of the following mutations also leads to similar learning defects. For each case explain why. (1) a mutation that prevents the kinase from binding ATP (2) a mutation that deletes the calmodulin-binding part of the kinase (3) a mutation that destroys the site of autophosphorylation B. What would be the effect on the activity of CaM-kinase if there were a mutation that reduced its interaction with the protein phosphatase responsible for inactivating the kinase?

A. Because a complete lack of the CaM-kinase causes a learning defect, we can assume that mutations leading to inactivation of the kinase would also have a similar effect. (1) Protein kinases have a binding site for ATP, which is the source of the phosphate used for phosphorylating their target proteins; if the kinase cannot bind ATP, it will be inactive. (2) Because binding to calmodulin in the presence of Ca2+ activates CaM-kinases, deletion of the calmodulin-binding portion would inactivate the kinase. (3) A mutation that destroys the site of autophosphorylation will also impair the normal function of the kinase, because the kinase will become inactive as soon as Ca2+ levels decrease. B. The kinase would stay active for longer after a transient increase in intracellular Ca2+ concentration.

The graph in Figure Q17-22 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. 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.

Your friend is studying a segment of a newly discovered virus that carries an enhancer of gene expression that confers responsiveness to glucocorticoid (a hormone) on genes that are linked to it. He constructs two versions of a reporter gene: one has only a minimal promoter linked to it (which contains sites for RNA polymerase binding); the other reporter gene has both this minimal promoter plus the viral enhancer attached to it. The reporter gene allows him to measure the amount of transcription that occurs from each construct. Your friend puts each of these constructs into two different cell lines and examines the expression of the reporter gene in each cell line, as shown in Figure Q16-21. He is puzzled by these findings and asks for your help in interpreting them. A. From these data, can you tell whether both cell lines contain glucocorticoid receptors? Why? B. What might account for the difference in the transcription of the reporter gene in cell lines 1 and 2 after introduction of the construct containing the viral enhancer in the presence of glucocorticoid?

A. Both cell lines seem to contain the glucocorticoid receptor, because both cell lines demonstrate increases in reporter gene expression in response to the addition of glucocorticoid. With the introduction of the construct containing the glucocorticoid-responsive viral enhancer, cell line 1 showed a 1000-fold increase and cell line 2 showed an 80-fold increase. B. There are rour reasonable explanations for this difference in glucocorticoid responses in the two cell lines are (any one would be correct). 1. Cell line 1 contains a protein that does not exist in cell line 2. This protein cooperates with the glucocorticoid receptor to activate the transcription of the reporter gene, leading to higher levels of the reporter gene in cell line 1 than in cell line 2. 2. Cell line 2 contains a slightly defective glucocorticoid receptor that does not bind DNA as strongly as the normal glucocorticoid receptor in cell line 1, thus leading to lower levels of reporter gene activity in cell line 2 than in cell line 1. 3. Cell line 1 contains a mutant glucocorticoid receptor that binds DNA much more strongly than the normal glucocorticoid receptor found in cell line 2. Therefore, higher levels of reporter gene activity are seen in cell line 1 than in cell line 2. 4. Cell line 2 contains a protein that does not exist in cell line 1. This protein acts as an antagonist to the glucocorticoid receptor to decrease the level of transcription from the reporter gene, leading to lower levels of reporter gene expression in cell line 2 than in cell line 1.

Acetylcholine acts at a GPCR on heart muscle to make the heart beat more slowly. It does so by ultimately opening K+ channels in the plasma membrane (as diagrammed in Figure Q16-25), which decreases the cell's excitability by making it harder to depolarize the plasma membrane. Indicate whether each of the following conditions would increase or decrease the effect of acetylcholine. A. addition of a drug that stimulates the GTPase activity of the Gα subunit B. mutations in the K+ channel that keep it closed all the time C. modification of the Gα subunit by cholera toxin D. a mutation that decreases the affinity of the βγ complex of the G protein for the K+ channel E. a mutation in the acetylcholine receptor that prevents its localization on the cell surface F. adding acetylcholinesterase to the external environment of the cell

A. Decrease. An increase in the GTPase activity of the Gα subunit will decrease the length of time that the G protein is active. B. Decrease. If the K+ channel remains closed, acetylcholine will not slow the heart. C. Increase. Cholera toxin inhibits the GTPase activity of the Gα subunit, keeping the subunit in an active state for a longer time. D. Decrease. The activated βγ complex binds to and activates the K+ channel; decreasing their affinity for each other will decrease the time that the K+ channel is open, effectively decreasing the effect of acetylcholine. E. Decrease. If there is no receptor on the cell surface, cells will be unable to respond to acetylcholine. F. Decrease. Acetylcholinesterase degrades acetylcholine and thus will decrease the effect of acetylcholine.

Before chromosomes segregate in M phase, they and the segregation machinery must be appropriately prepared. Indicate whether the following statements are true or false. If false, change a single noun to make the statement true. A. Sister chromatids are held together by condensins from the time they arise by DNA replication until the time they separate at anaphase. B. Cohesins are required to make the chromosomes more compact and thus to prevent tangling between different chromosomes. C. The mitotic spindle is composed of actin filaments and myosin filaments. D. Microtubule-dependent motor proteins and microtubule polymerization and depolymerization are mainly responsible for the organized movements of chromosomes during mitosis. E. The centromere nucleates a radial array of microtubules called an aster, and its duplication is triggered by S-Cdk. F. Each centrosome contains a pair of centrioles and hundreds of γ-tubulin rings that nucleate the growth of microtubules.

A. False. Sister chromatids are held together by cohesins from the time they arise by DNA replication until the time they separate at anaphase. B. False. Condensins are required to make the chromosomes more compact and thus to prevent tangling between different chromosomes. C. False. The contractile ring is composed of actin filaments and myosin filaments. D. True. E. False. The centrosome nucleates a radial array of microtubules called an aster, and its duplication is triggered by S-Cdk. F. True.

Two protein kinases, PK1 and PK2, work sequentially in an intracellular signaling pathway. You create cells that contain inactivating mutations in the genes that encode either PK1 or PK2 and find that these cells no longer respond to a particular extracellular signal. You also create cells containing a version of PK1 that is permanently active and find that the cells behave as though they are receiving the signal even when the signal is not present. When you introduce the permanently active version of PK1 into cells that have an inactivating mutation in PK2, you find that these cells also behave as though they are receiving the signal even when no signal is present. A. From these results, does PK1 activate PK2 or does PK2 activate PK1? Explain your answer. B. You now create a permanently active version of PK2 and find that cells containing this version behave as though they are receiving the signal even when the signal is not present. What do you predict will happen if you introduce the permanently active version of PK2 into cells that have an inactivating mutation in PK1?

A. Normally, PK2 activates PK1. We are told that PK1 and PK2 normally work sequentially in an intracellular signaling pathway. If PK1 is permanently activated, a response is seen independently of whether or not PK2 is present. If PK1 activated PK2, no response should be seen if PK1 were activated in the absence of PK2. B. Protein B is an inhibitor of DNA synthesis. Consequently, receptors with binding sites for B and D (see experiment 7) stimulate a lower DNA synthesis rate than do receptors that bind only D (experiment 5). C. Protein C has no detectable role in DNA synthesis. Receptors that can bind only C (experiment 4) activate DNA synthesis about as much as receptors that do not bind any of the four proteins (experiment 9; our negative control). Furthermore, the binding of protein C does not affect the response mediated by protein D when the receptor can bind both C and D (experiment 8). B. You would predict that no response to signal would be observed. This is because PK2 normally needs to activate PK1 for the cells to respond to the signal. When PK2 is permanently activated in the absence of PK1, PK1 is not there to relay the signal.

When activated by the extracellular signal protein platelet-derived growth factor (PDGF), the PDGF receptor phosphorylates itself on multiple tyrosines (as indicated in Figure 16-46A 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, and include the proteins A, B, C, and D. One of the cell's responses to PDGF is an increase in DNA synthesis, which can be measured by the incorporation of radioactive thymidine into the DNA. To determine which protein or proteins, A, B, C, or D, are responsible for the 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 their own PDGF receptor. In these cells, the various mutant 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 16-46B. A. From these data, which, if any, of proteins A, B, C, and D are involved in the stimulation of DNA synthesis by PDGF? Explain your answer. B. Which, if any, of these proteins inhibit DNA synthesis? Explain your answer. C. Which, if any, of these proteins seem to have no detectable role in DNA synthesis? Explain your answer.

A. Proteins A and D stimulate DNA synthesis. PDGF receptors that can bind to only A or D (see experiments 2 and 5) can stimulate DNA synthesis to about 50% of normal amounts (which is represented by experiment 1). Proteins A and D are both needed and are used in an additive fashion; this is evident from experiment 6: when a PDGF receptor can bind both A and D, DNA synthesis levels are close to that obtained with the normal receptor. 16-47 A. Normally, PK2 activates PK1. We are told that PK1 and PK2 normally work sequentially in an intracellular signaling pathway. If PK1 is permanently activated, a response is seen independently of whether or not PK2 is present. If PK1 activated PK2, no response should be seen if PK1 were activated in the absence of PK2. B. Protein B is an inhibitor of DNA synthesis. Consequently, receptors with binding sites for B and D (see experiment 7) stimulate a lower DNA synthesis rate than do receptors that bind only D (experiment 5). C. Protein C has no detectable role in DNA synthesis. Receptors that can bind only C (experiment 4) activate DNA synthesis about as much as receptors that do not bind any of the four proteins (experiment 9; our negative control). Furthermore, the binding of protein C does not affect the response mediated by protein D when the receptor can bind both C and D (experiment 8)

You are interested in cell size regulation and discover that signaling through an enzyme-coupled receptor is important for the growth (enlargement) of mouse liver cells. Activation of the receptor activates adenylyl cyclase, which ultimately leads to the activation of PKA, which then phosphorylates a transcription factor called TFS on threonine 42. This phosphorylation is necessary for the binding of TFS to its specific sites on DNA, where it then activates the transcription of Sze2, a gene that encodes a protein important for liver cell growth. You find that liver cells lacking the receptor are 15% smaller than normal cells, whereas cells that express a constitutively activated version of PKA are 15% larger than normal liver cells. Given these results, predict whether you would expect the cell's size to be bigger or smaller than normal cells if cells were treated in the following fashion. A. You change threonine 42 on TFS to an alanine residue. B. You create a version of the receptor that is constitutively active. C. You add a drug that inhibits adenylyl cyclase. D. You add a drug that increases the activity of cyclic AMP phosphodiesterase. E. You mutate the cAMP-binding sites in the regulatory subunits of PKA, so that the complex binds cAMP more tightly.

A. Smaller. This mutation will make a TFS that cannot be phosphorylated by PKA. B. Bigger. C. Smaller. D. Smaller. cAMP phosphodiesterase is involved in converting cAMP to AMP and will down-regulate this signaling pathway. E. Bigger. Higher affinity of the PKA complex for cAMP will increase its activity, and thus cells will be bigger.

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

A. True. In nearly all cells in an organism, the S, G2, and M phases of the cell cycle take the same amount of time. The different timing of cell division in different cell types is due to variable lengths of the G1 phase or to withdrawal into the G0 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.

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 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 microtubule during your 4 seconds of observation? C. What would you predict would happen to the kinesin-coated silica beads if you were to add AMP-PNP (a nonhydrolyzable ATP analog)?

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 observation, because each bead will take more than 100 steps in that 4- second time frame (see 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.

A. Match each equation in column A with the corresponding standard redox potential in column B. Column A Column B 1. H2O ↔ ½O2 + 2H+ + 2 e- A) +30 mV 2. reduced ubiquinone ↔ oxidized ubiquinone + 2H+ + 2 e- B) +820 mV 3. NADH ↔ NAD+ + H+ + 2 e- C) +230 mV 4. reduced cytochrome c ↔ oxidized cytochrome c + e- D) -320 mV B. How do these standard redox potentials support our understanding of the stepwise electron transfers that occur in the electron-transport chain? C. Why would it not be advantageous for living systems to evolve a mechanism for the direct transfer of electrons from NADH to O2?

A. 1—B; 2—A; 3—D; 4—C B. Each successive member of the electron-transport chain is a better electron acceptor, which permits a unidirectional series of electron transfers until reaching O2, which is the best electron acceptor and the final destination of the electrons, forming water as oxygen is consumed. C. If NADH directly donated electrons to O2, a large amount of energy would be released as heat and lost as a way for the cell to generate chemical energy in the form of ATP.

Name the stage of M phase in which the following events occur. Place the numbers 1-8 next to the letter headings to indicate the normal order of events. A. alignment of the chromosomes at the spindle equator B. attachment of spindle microtubules to chromosomes C. breakdown of nuclear envelope D. pinching of cell in two E. separation of two centrosomes and initiation of mitotic spindle assembly F. re-formation of the nuclear envelope G. condensation of the chromosomes H. separation of sister chromatids

A. 5, metaphase B. 4, prometaphase C. 3, prometaphase D. 8, cytokinesis E. 2, prophase F. 7, telophase G. 1, prophase H. 6, anaphase

The rod photoreceptors in the eye are extremely sensitive to light. The cells sense light through a signal transduction cascade involving light activation of a GPCR that activates a G protein that activates cyclic GMP phosphodiesterase. How would you expect the addition of the following drugs to affect the light-sensing ability of the rod cells? Explain your answers. A. a drug that inhibits cyclic GMP phosphodiesterase B. a drug that is a nonhydrolyzable analog of GTP

A. A drug that inhibits cyclic GMP phosphodiesterase would decrease any light response in the rod cell. Normally, cyclic GMP is continuously being produced in the eye. The perception of light by a rod cell normally leads to the activation of cyclic GMP phosphodiesterase, which then hydrolyzes cyclic GMP molecules. This causes Na+ channels to close, which changes the membrane potential and alters the signal sent to the brain. If cyclic GMP phosphodiesterase were blocked, levels of cyclic GMP would remain high and there would be no cellular response to light. B. A drug that is a nonhydrolyzable analog of GTP would lead to a prolonged response to light. This is because a nonhydrolyzable analog of GTP would prevent the G protein from turning itself off by hydrolyzing its bound GTP to GDP. Continued activation of the G protein would keep cyclic GMP phosphodiesterase levels higher than normal, leading to a prolonged period of lowered levels of cyclic GMP. This in turn would cause Na+ channels to be closed for longer than normal, leading to a prolonged change in the membrane potential and an extended light response.

Receipt of extracellular signals can change cell behavior quickly (e.g., in seconds or less) or much more slowly (e.g., in hours). A. What kind of molecular changes could cause quick changes in cell behaviour? B. What kind of molecular changes could cause slow changes in cell behaviour? C. Explain why the response you named in A results in a quick change, whereas the response you named in B results in a slow change.

A. Any answer that involves the modification of existing cell components is correct. Protein phosphorylation, protein dephosphorylation, protein ubiquitylation, lipid phosphorylation, and lipid cleavage are all examples of correct answers. B. Responses that involve alterations in gene expression occur slowly. C. Modification of existing cellular components can happen quickly, whereas responses that depend on changes in gene expression take much longer, because the genes will need to be transcribed, the mRNAs will need to be translated, and the proteins need to accumulate to high enough levels to instigate change.

Figure Q15-22 shows the orientation of a multipass transmembrane protein after it has completed its entry into the ER membrane (part A) and after it gets delivered to the plasma membrane (part B). This protein has an N-terminal signal sequence (depicted as the dark gray membrane-spanning box), which signal peptidase cleaves off in the endoplasmic reticulum. The other membrane-spanning domains in the protein are represented as open boxes. Given that any hydrophobic membrane-spanning domain can act as either a start-transfer region or a stop-transfer region, draw the final consequences of the actions described below on the orientation of the protein in the plasma membrane. Indicate on your drawing the extracellular space, the cytosolic face, and the plasma membrane, as well as the N- and C-termini of the protein. A. deleting the first signal sequence B. changing the hydrophobic amino acids in the first, cleaved, sequence to charged amino acids C. changing the hydrophobic residues in every other transmembrane sequence to charged residues, starting with the first, cleaved, signal sequence

A. Deleting the first signal sequence completely would convert the next membrane-spanning domain into an internal start-transfer signal and would invert the orientation of the protein (see Figure A15-22A). B. Changing the hydrophobic amino acids to charged amino acids destroys the ability of the sequence both to act as a signal sequence and to become a membrane-spanning sequence. Therefore, the adjacent membrane-spanning domain will now become an internal start-transfer sequence and the protein will be inverted, as seen above in part A. The mutated signal sequence would not get cleaved off, because it would remain on the cytoplasmic side of the membrane and signal peptidase is found only inside the ER (see Figure A15-22B). C. Mutating every other membrane-spanning region so that they are now charged (and thus cannot span the membrane) would decrease the number of transmembrane regions and increase the size of the internal loops between membrane-spanning regions (see Figure 15-22C).

Indicate whether the following statements are true or false. If a statement is false, explain why it is false. A. The driving force that pulls protons into the matrix is called the proton-motive force, which is a combination of the large force due to the pH gradient and the smaller force that results from the voltage gradient across the inner mitochondrial membrane. B. Under anaerobic conditions, the ATP synthase can hydrolyze ATP instead of synthesizing it. C. ATP is moved out of the matrix, across the inner mitochondrial membrane, in a co-transporter that also brings ADP into the matrix. D. Brown fat cells make less ATP because they have an inefficient ATP synthase.

A. False. Although it is true that both the pH gradient and the voltage gradient are components of the proton-motive force, it is the voltage gradient (also referred to as the membrane potential) that is the greater of the two. B. True. C. True. D. False. The inner mitochondrial membranes in brown fat cells contain a transport protein that allows protons to move down their gradient without passing through the ATP synthase. As a result, less ATP is made and most of the energy from the proton gradient is released as heat.

Indicate whether the following statements are true or false. If a statement is false, explain why it is false. A. The dark reactions of photosynthesis occur only in the absence of light. B. Much of the glyceraldehyde 3-phosphate made in the chloroplast ends up producing the molecules needed by the mitochondria to produce ATP. C. Ribulose 1,5-bisphosphate is similar to oxaloacetate in the Krebs cycle in that they are both regenerated at the end of their respective cycles. D. Each round of the Calvin cycle uses five molecules of CO2 to produce one molecule of glyceraldehyde 3-phosphate and one of pyruvate.

A. False. The dark reactions are those involved in carbon fixation and are named as such because they do not require light. B. True. C. True. D. False. Three molecules of CO2 are required for each round of the Calvin cycle, and the product is one molecule of glyceraldehyde 3-phosphate and the recycling of the ribulose 1,5-bisphosphate molecule.

Indicate whether the following statements are true or false. If a statement is false, explain why it is false. A. Ubiquinone is associated with the inner mitochondrial membrane as a protein-bound electron carrier molecule. B. Ubiquinone can transfer only one electron in each cycle. C. The iron-sulfur centers in NADH dehydrogenase are relatively poor electron acceptors. D. Cytochrome oxidase binds O2 using an iron-heme group, where four electrons are shuttled one at a time.

A. False. Ubiquinone is an aromatic compound that uses its long hydrocarbon tail to associate with the inner mitochondrial membrane. B. False. Ubiquinone can transfer one or two electrons. In the case in which only one electron is transferred, the molecule contains an unpaired electron, which is highly reactive. C. True. D. True.

In the budding yeast, activation of the GTP-binding protein Cdc42 occurs on binding of an external signal (pheromone) to a G-protein-linked receptor. Activation of Cdc42 promotes actin polymerization. Predict 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-linked receptors. B. You add pheromone to a nonhydrolyzable analogue of GTP.

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

Indicate by writing "yes" or "no" whether amplification of a signal could occur at the particular steps described below. Explain your answers. A. An extracellular signaling molecule binds and activates a GPCR. B. The activated GPCRs cause Gα to separate from Gβ and Gγ. C. Adenylyl cyclase produces cyclic AMP. D. cAMP activates protein kinase A. E. Protein kinase A phosphorylates target proteins.

A. No. Each signaling molecule activates only one receptor molecule. B. Yes. Each activated GPCR activates many G-protein molecules. C. Yes. Each activated adenylyl cyclase molecule can generate many molecules of cAMP. D. No. In unstimulated cells, protein kinase A is held inactive in a protein complex. Binding of cAMP to the complex induces a conformational change, releasing the active protein kinase A. Therefore, one cAMP cannot activate more than one molecule of protein kinase A. E. Yes. Each activated protein kinase A molecule can phosphorylate many molecules of each type of target protein.

In which of the four compartments of a mitochondrion are each of the following located? A. porin-outer membrane B. the mitochondrial genome-matrix C. citric acid cycle enzymes-matrix D. proteins of the electron-transport chain-inner membrane E. ATP synthase-inner membrane F. membrane transport protein for pyruvate-intermembrane space

A. Porin is in the outer membrane. B. The mitochondrial genome is in the matrix. C. The citric acid cycle enzymes are in the matrix. D. The proteins in the electron-transport chain are in the inner membrane. E. ATP synthase is in the inner membrane. F. The transport protein for pyruvate is in the inner membrane.

Figure Q17-32 shows two isolated outer doublet microtubules from a eucaryotic flagellum with their associated dynein molecules. 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?

A. See Figure A17-32A. B. See Figure A17-32B. (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 A17-32 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.

The F0 portion of the ATP synthase is a multisubunit complex that spans the inner mitochondrial membrane. A. What are the designations for the subunits, and which are present in multiple copies in the assembled complex? B. Explain how the F0 complex harnesses the proton-motive force to help synthesize ATP. What would happen if the proton gradient were reversed?

A. The individual subunits are a, b, and c. The a and b subunits are present in the complex as single polypeptides. However, there are between 10 and 14 polypeptides of subunit c that form off a ring in the F0 complex. B. Protons flow through a channel that exists between each of the c subunits in the ring: these form part of the "stalk." The flow of protons makes the entire stalk rotate, which causes a conformational change in each of the three β subunits in the F1 portion of the synthase. If the proton gradient is reversed, the flow of protons is reversed and the stalk rotates in the opposite direction, causing ATP hydrolysis rather than ATP synthesis.

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 hydrolyzed, and you then severed the microtubules in the middle with a laser 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.

What would happen in each of the following cases? Assume in each case that the protein involved is a soluble protein, not a membrane protein. A. You add a signal sequence (for the ER) to the N-terminal end of a normally cytosolic protein. B. You change the hydrophobic amino acids in an ER signal sequence into charged amino acids. C. You change the hydrophobic amino acids in an ER signal sequence into other, hydrophobic, amino acids. D. You move the N-terminal ER signal sequence to the C-terminal end of the protein.

A. The protein will now be transported into the ER lumen. B. The altered signal sequence will not be recognized and the protein will remain in the cytosol. C. The protein will still be delivered into the ER. It is the distribution of hydrophobic amino acids that is important, not the actual sequence. D. The protein will not enter the ER. Because the C-terminus of the protein is the last part to be made, the ribosomes synthesizing this protein will not be recognized by the SRP and carried to the ER.

Using genetic engineering techniques, you have created a set of proteins that contain two (and only two) conflicting signal sequences that specify different compartments. Predict which signal would win out for the following combinations. Explain your answers. A. Signals for import into the nucleus and import into the ER. B. Signals for export from the nucleus and import into the mitochondria. C. Signals for import into mitochondria and retention in the ER.

A. The protein would enter the ER. The signal for a protein to enter the ER is recognized as the protein is being synthesized and will end up either in the ER or on the ER membrane. Cytosolic nuclear transport proteins recognize proteins destined for the nucleus once those proteins are fully synthesized and fully folded. B. The protein would enter the mitochondria. For a nuclear export signal to work, the protein would have to end up in the nucleus first and thus would need a nuclear import signal for the nuclear export signal to be used. C. The protein would enter the mitochondria. To be retained in the ER, the protein needs to enter the ER. Because there is no signal for ER import, the ER retention signal would not function.

The respiratory chain is relatively inaccessible to the experimental manipulation of intact mitochondria. After disrupting mitochondria with ultrasound, however, it is possible to isolate functional submitochondrial particles, which consist of broken cristae that have resealed inside-out into small closed vesicles. In these vesicles the components that originally faced the matrix are now exposed to the surrounding medium. A. How might such an arrangement aid in the study of electron transport and ATP synthesis? B. Consider an anaerobic preparation of such submitochondrial particles. If a small amount of oxygen is added, do you predict that the preparation will consume oxygen in respiration reactions? Will the medium outside the particles become more acidic or more basic? What, if anything, will change if the flow of protons through ATP synthase is blocked by an inhibitor? Explain your answer.

A. This arrangement of components within the vesicles allows the experimental manipulation of the medium surrounding the vesicles to permit the consequences of different conditions in the mitochondrial matrix to be examined. The medium can be altered by changing pH, adding electron carriers and oxygen, and providing ADP and Pi, for example. The oxidation of electron carriers, the consumption of oxygen, and the production of ATP can be measured in the medium. By changing the composition of the medium, it should be possible, for example, to identify the electron carriers that can donate electrons from the matrix to the transport chain (the side of the membrane that normally faces the matrix is now on the outside), to assess the redox potentials of various components of the transport chain, and to determine the dependence of ATP synthesis on the pH gradient across the membrane and on the ATP:ADP ratio. B. Respiration reactions will rapidly consume at least some of the added oxygen. During the anaerobic conditions, the electron carriers in the electron-transport chain were reduced; on the addition of oxygen, electrons will be transferred to oxygen, thereby reducing the oxygen and oxidizing the carriers. Concomitantly with the electron flow, protons will be pumped from the medium into the vesicles, thereby making the medium slightly more basic and the inside of the vesicles acidic. Inhibition of the ATP synthase will not have an immediate effect on oxygen consumption or proton pumping. However, the proton concentration inside the vesicles will quickly become too high to continue the activity of the electron-transport-coupled proton pumping, and thus electron transport and oxygen consumption will cease.

Indicate whether the following statements are true or false. If a statement is false, explain why it is false. A. The number and location of mitochondria within a cell can change, depending on the both the cell type and the amount of energy required. B. The inner mitochondrial membrane contains porins, which allow pyruvate to enter for use in the citric acid cycle. C. The inner mitochondrial membrane is actually a series of discrete flattened membrane-enclosed compartments called cristae, similar to what is seen in the Golgi apparatus. D. The intermembrane space of the mitochondria is chemically equivalent to the cytosol with respect to pH and the small molecules present.

A. True. B. False. The outer mitochondrial membrane contains porins, allowing the passage of all molecules with a mass of less than 5000 daltons. Although pyruvate must pass through the inner membrane, it does so in a highly regulated manner via specific transporter channels. C. False. Although the cristae do look like individual compartments on the basis of the images of the inner structure of the mitochondria, the inner membrane is a single, albeit highly convoluted, membrane. D. True.

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 eucaryotic 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

Circle the phrase in each pair that is likely to occur more rapidly in response to an extracellular signal. A. changes in cell secretion / increased cell division B. changes in protein phosphorylation / changes in proteins being synthesized C. changes in mRNA levels / changes in membrane potential

A. changes in cell secretion B. changes in protein phosphorylation C. changes in membrane potential

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

When adrenaline binds to adrenergic receptors on the surface of a muscle cell, it activates a G protein, initiating an intracellular signaling pathway in which the activated α subunit activates adenylyl cyclase, thereby increasing cAMP levels in the cell. The cAMP molecules then activate a cAMP-dependent kinase (PKA) that, in turn, activates enzymes that result in the breakdown of muscle glycogen, thus lowering glycogen levels. You obtain muscle cells that are defective in various components of the signaling pathway. Referring to Figure Q16-29, indicate how glycogen levels would be affected in the presence of adrenaline in the following cells. Would they be higher or lower than in normal cells treated with adrenaline? A. cells that lack adenylyl cyclase B. cells that lack the GPCR C. cells that lack cAMP phosphodiesterase D. cells that have an α subunit that cannot hydrolyze GTP but can interact properly with the β and γ subunits

A. higher B. higher C. lower D. lower

In the cell, the concentration of actin monomer is higher than the concentration required for purified actin monomers to polymerize in vitro. Thymosin is a protein that can bind actin monomers. If you were to add a drug that inhibits the ability of thymosin to bind actin monomers, what effect would this have on actin polymerization? Explain your answer.

Addition of a drug that keeps thymosin from binding actin monomer will increase the rate of actin polymerization in the cell. The reason that actin monomers do not spontaneously form filaments in the cell, despite their high concentration, is that the monomers are normally bound by proteins (such as thymosin) and are thereby prevented from adding to the end of an actin filament.

Do you agree or disagree with the following statement? Explain your answer. Nucleotide hydrolysis has a similar role in actin polymerization and in tubulin polymerization.

Agree. ATP-bound actin monomers are added to actin filaments; GTP-bound tubulin subunits are added to the growing end of a microtubule. Nucleotide hydrolysis occurs after addition of the subunit to the filament in both actin and microtubules. ATP hydrolysis in actin polymerization decreases the binding strength between monomers in the actin filaments; GTP hydrolysis during tubulin polymerization decreases the binding strength between the tubulin subunits in the microtubule. Because the nucleotide hydrolysis decreases the binding strength between subunits, depolymerization is enhanced in both actin filaments and microtubules.

When the neurotransmitter acetylcholine is applied to skeletal muscle cells, it binds the acetylcholine receptor and causes the muscle cells to contract. Succinylcholine, which is a chemical analog of acetylcholine, binds to the acetylcholine receptor on skeletal muscle cells but causes the muscle to relax; it is therefore often used by surgeons as a muscle relaxant. Propose a model for why succinylcholine causes muscle relaxation. What might be the mechanism to explain the different activities of acetylcholine and succinylcholine on the acetylcholine receptor?

Although succinylcholine can bind to the acetylcholine receptor, it does not activate the receptor and therefore does not cause the muscle to contract. Instead, succinylcholine blocks the ability of acetylcholine to bind to the receptor and thereby prevents acetylcholine from stimulating muscle contraction.

For each of the following sentences, choose one of the options enclosed in square brackets to make a correct statement. "An electron bound to a molecule with low affinity for electrons is a [high/low]-energy electron. Transfer of an electron from a molecule with low affinity to one with higher affinity has a [positive/negative] δG° and is thus [favorable/unfavorable] under standard conditions. If the reduced form of a redox pair is a strong electron donor with a [high/low] affinity for electrons, it is easily oxidized; the oxidized member of such a redox pair is a [weak/strong] electron acceptor."

An electron bound to a molecule with low affinity for electrons is a high-energy electron. Transfer of an electron from a molecule with low affinity to one with higher affinity has a negative δG° and is thus favorable under standard conditions. If the reduced form of a redox pair is a strong electron donor with a low affinity for electrons, it is easily oxidized; the oxidized member of such a redox pair is a weak electron acceptor.

For each of the following sentences, select the best word or phrase from the list below to fill in the blanks. Not all words or phrases will be used; each word or phrase should be used only once. An extracellular signal molecule can act to change a cell's behavior by acting through cell-surface __________________ that control intracellular signaling proteins. These intracellular signaling proteins ultimately change the activity of __________________ proteins that bring about cell responses. Intracellular signaling proteins can __________________ the signal received to evoke a strong response from just a few extracellular signal molecules. A cell that receives more than one extracellular signal at the same time can __________________ this information using intracellular signaling proteins. __________________ proteins can act as molecular switches, letting a cell know that a signal has been received. Enzymes that phosphorylate proteins, termed ___________, can also serve as molecular switches; the actions of these enzymes are countered by the activity of __________________.

An extracellular signal molecule can act to change a cell's behavior by acting through cell-surface receptors that control intracellular signaling proteins. These intracellular signaling proteins ultimately change the activity of effector proteins that bring about cell responses. Intracellular signaling proteins can amplify the signal received to evoke a strong response from just a few extracellular signal molecules. A cell that receives more than one extracellular signal at the same time can integrate this information using intracellular signaling proteins. GTP- binding proteins can act as molecular switches, letting a cell know that a signal has been received. Enzymes that phosphorylate proteins, termed protein kinases, can also serve as molecular switches; the actions of these enzymes are countered by the activity of protein phosphatases.

Consider an animal cell that has eight chromosomes (four pairs of homologous chromosomes) in G1 phase. How many of each of the following structures will the cell have at mitotic prophase? A. sister chromatids B. centromeres C. kinetochores D. centrosomes E. centrioles

A—16; B—16; C—16; D—2; E—4

Match each of the main classes of spindle microtubules from list 1 with their functions and features from list 2.

A—1; B—2, 5; C—3, 4

A plasma membrane protein carries an oligosaccharide containing mannose (Man), galactose (Gal), sialic acid (SA), and N-acetylglucosamine (GlcNAc). These sugars are added to the protein as it proceeds through the secretory pathway. First, a core oligosaccharide containing Man and GlcNAc is added, followed by Gal, Man, SA, and GlcNAc in a particular order. Each addition is catalyzed by a different transferase acting at a different stage as the protein proceeds through the secretory pathway. You have isolated mutants defective for each of the transferases, purified the membrane protein from each of the mutants, and identified which sugars are present in each mutant protein. Table Q15-37 summarizes the results. From these results, match each of the transferases (A, B, C, D) to its subcellular location selected from the list below. (Assume that each location contains only one enzyme.) 1. central Golgi cisternae 2. cis Golgi network 3. ER 4. trans Golgi network

A—3 (oligosaccharide protein transferase = ER) B—1 (galactose transferase = central Golgi cisternae) C—4 (SA transferase = trans Golgi network) D—2 (GlcNAc transferase = cis Golgi network Proteins are modified in a stepwise fashion in the Golgi apparatus, with early steps taking place in the cis Golgi, intermediate steps taking place in the central Golgi cisternae, and late steps occurring in the trans Golgi network. If each enzyme produces the substrate for the next step, then a mutant lacking the enzyme that catalyzes the addition of the first sugar will be missing all of the sugars, a mutant lacking the enzyme that catalyzes the addition of the second sugar will contain the first sugar but will lack the other three, and so on. By this logic, mannose and GlcNAc must be the first sugars added, additional GlcNAc the second, galactose the third, and SA the last. Hence, the oligosaccharide protein transferase must be in the ER, the GlcNAc transferase in the cis Golgi, the galactose transferase in the central Golgi, and the SA transferase in the trans Golgi.

Match the set of labels below with the numbered label lines on Figure Q15-36. A. cisterna B. Golgi stack C. secretory vesicle D. trans Golgi network E. cis Golgi network

A—3; B—1; C—5; D—4; E—2

Match the following labels to the numbered lines on Figure Q17-27. 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 Q17-27 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.

Match the following labels to the numbered label lines on Figure Q18-15. Figure Q18-15 A. G1 phase B. mitotic cyclin C. mitotic Cdk D. S phase E. S-phase cyclin F. S-phase Cdk G. MPF H. G2 phase

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

Which stage of mitosis in an animal cell does each part of Figure Q18-29 represent?

A—telophase; B—prophase; C—anaphase; D—prometaphase

Of the following mutations, which are likely to cause cell-cycle arrest? If you predict a cell-cycle arrest, indicate whether the cell will arrest in early G1, late G1, or G2. Explain your answers. A. a mutation in a gene encoding a cell-surface mitogen receptor that makes the receptor active even in the absence of the mitogen B. a mutation that destroyed the kinase activity of S-Cdk C. a mutation that allowed G1-Cdk to be active independently of its phosphorylation status D. a mutation that removed the phosphorylation sites on the Rb protein E. a mutation that inhibited the activity of Rb

B and D are likely to cause cell-cycle arrest. A. Because ligand-independent activation of a mitogen receptor will probably make the cell divide when it otherwise might not, this mutation is unlikely to cause a cell-cycle arrest. B. This mutation is likely to cause cell-cycle arrest in late G1. Without S-Cdk activity, the cells will probably be unable to enter S phase. C. Phosphorylation-independent activity of G1-Cdk is unlikely to lead to cell- cycle arrest; the cells should progress through the cycle, although the kinetics and fine regulation of the cycle may be altered. D. This mutation is likely to cause cell-cycle arrest in early G1. Unphosphorylated Rb will inhibit the transcription of genes required for progression through G1 and into S phase. This inhibition is normally released on phosphorylation of Rb. If Rb cannot be phosphorylated, it will always inhibit transcription of these genes, leading to arrest in early G1. E. If Rb is inactivated by a mutation, cells will be more likely to divide in the absence of extracellular mitogens, which is the opposite of a cell-cycle arrest.

Irradiated mammalian cells usually stop dividing and arrest at a G1 checkpoint. Place the 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

Can signaling via a steroid hormone receptor lead to amplification of the original signal? If so, how?

Because the interactions of the signal molecule with its receptor and of the activated receptor with its gene are both one-to-one, there is no amplification in this part of the signaling pathway. The signal can, however, be amplified when the target genes are transcribed, because each activated gene produces multiple copies of mRNA, each of which is used to make multiple copies of the protein that the gene encodes.

The cytoskeleton of an animal cell changes markedly between G1 and early M phase (prophase) of the cell cycle. For each of the following sentences, choose one of the options enclosed in square brackets that best describes the changes to the cytoskeleton and its components. Before mitosis, the number of centrosomes must [increase/decrease]. At the beginning of [anaphase/prophase] in animal cells, the centrosomes separate in a process driven partly by interactions between the [plus/minus] ends of microtubules arising from the two centrosomes. Centrosome separation initiates the assembly of the bipolar mitotic spindle and is associated with a sudden [increase/decrease] in the dynamic instability of microtubules. In comparison with an interphase microtubule array, a mitotic aster contains a [smaller/larger] number of [longer/shorter] microtubules. Extracts from M-phase cells exhibit [increased/decreased/unchanged] rates of microtubule polymerization and increased frequencies of microtubule [shrinkage/growth]. The changes in microtubule dynamics are largely due to [enhanced/reduced] activity of microtubule- associated proteins and [increased/decreased] activity of catastrophins. The new balance between polymerization and depolymerization of microtubules is necessary for the mitotic spindle to move the [replicated chromosomes/daughter chromosomes] to the metaphase plate.

Before mitosis, the number of centrosomes must increase. At the beginning of prophase in animal cells, the centrosomes separate in a process driven partly by interactions between the plus ends of microtubules arising from the two centrosomes. Centrosome separation initiates the assembly of the bipolar mitotic spindle and is associated with a sudden increase in the dynamic instability of microtubules. In comparison with an interphase microtubule array, a mitotic aster contains a larger number of shorter microtubules. Extracts from M-phase cells exhibit unchanged rates of microtubule polymerization and increased frequencies of microtubule shrinkage. The changes in microtubule dynamics are largely due to reduced activity of microtubule-associated proteins and increased activity of catastrophins. The new balance between polymerization and depolymerization of microtubules is necessary for the mitotic spindle to move the replicated chromosomes to the metaphase plate.

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

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. Ca2+ can trigger biological effects in cells because an unstimulated cell has an extremely __________________ concentration of free Ca2+ in the cytosol, compared with its concentration in the __________________ space and in the __________________, creating a steep electrochemical gradient. When Ca2+ enters the cytosol, it interacts with Ca2+-responsive proteins such as __________________, which also binds diacylglycerol, and __________________, which activates CaM-kinases.

Ca2+ can trigger biological effects in cells because an unstimulated cell has an extremely low concentration of free Ca2+ in the cytosol, compared with its concentration in the extracellular space and in the endoplasmic reticulum, creating a steep electrochemical gradient. When Ca2+ enters the cytosol, it interacts with Ca2+-responsive proteins such as protein kinase C, which also binds diacylglycerol, and calmodulin, which activates CaM-kinases.

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.

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. Cells signal to one another in various ways. Some use extracellular signal molecules that are dissolved gases, such as __________________, which can diffuse easily into cells. Others use cytokines, which bind to cytokine receptors. Cytokine receptors have no intrinsic enzyme activity but are associated with cytoplasmic tyrosine kinases called __________________s, which become activated on the binding of cytokine to its receptor and go on to phosphorylate and activate cytoplasmic transcriptional regulators called __________________s. Some intracellular signaling pathways involve chains of protein kinases that phosphorylate each other, as seen in the __________________ signaling module. Lipids can also relay signals in the cell, as we observe when phospholipase C cleaves the sugar-phosphate head off a lipid molecule to generate the two small messenger molecules __________________ (which remains embedded in the plasma membrane) and __________________ (which diffuses into the cytosol).

Cells signal to one another in various ways. Some use extracellular signal molecules that are dissolved gases, such as NO, which can diffuse easily into cells. Others use cytokines, which bind to cytokine receptors. Cytokine receptors have no intrinsic enzyme activity but are associated with cytoplasmic tyrosine kinases called JAKs, which become activated on the binding of cytokine to its receptor and go on to phosphorylate and activate cytoplasmic transcriptional regulators called STATs. Some intracellular signaling pathways involve chains of protein kinases that phosphorylate each other, as seen in the MAP kinase signalling module. Lipids can also partake in relaying signals in the cell, as seen when phospholipase C cleaves the sugar-phosphate head off a lipid molecule to generate the two small messenger molecules diacylglycerol (which remains embedded in the plasma membrane) and IP3 (which diffuses into the cytosol).

Examine the schematic representation of centrosome duplication in Figure Q18- 33. By analogy with DNA replication, would you classify centrosome duplication as conservative or semi-conservative? Explain your answer.

Centrosome duplication is semi-conservative. The paired centrioles in the centrosome separate, and each serves to nucleate the assembly of a new centriole. As a consequence, each new centrosome consists of one old and one new centriole. Thus, centrosome duplication is analogous to DNA replication, in which the new double helix consists of one old DNA strand and one newly replicated DNA strand.

Which of the following statements about molecular switches is false? (a) Phosphatases remove the phosphate from GTP on GTP-binding proteins, turning them off. (b) Protein kinases transfer the terminal phosphate from ATP onto a protein. (c) Serine/threonine kinases are the most common types of protein kinase. (d) A GTP-binding protein exchanges its bound GDP for GTP to become activated.

Choice (a) is false. GTP-binding proteins themselves hydrolyze their bound GTP to GDP, using their own intrinsic GTPase activity.

The last common ancestor to plants and animals was a unicellular eucaryote. Thus, it is thought that multicellularity and the attendant demands for cell communication arose independently in these two lineages. This evolutionary viewpoint accounts nicely for the vastly different mechanisms that plants and animals use for cell communication. Fungi use signaling mechanisms and components that are very similar to those used in animals. Which of the phylogenetic trees shown in Figure 16-49 does this observation support?

Choice (b) is the correct answer (see Figure Q16-49). The similarities in signaling mechanisms between animals and fungi support the phylogenetic tree in which fungi branched from the animal lineage after plants and animals separated. This branching order is supported by a wide variety of other data, including genomic sequence comparisons.

Which of the following types of ion movement might be expected to require co-transport of protons from the intermembrane space to the matrix, inasmuch as it could not be driven by the membrane potential across the inner membrane? (Assume that each ion being moved is moving against its concentration gradient.) (a) import of Ca2+ into the matrix from the intermembrane space (b) import of acetate ions into the matrix from the intermembrane space (c) exchange of Fe2+ in the matrix for Fe3+ in the intermembrane space (d) exchange of ATP from the matrix for ADP in the intermembrane space

Choice (b) is the correct answer. Because the inside of the membrane (the mitochondrial matrix) is more negative than the outside, in principle any traffic resulting in an increase in the positive charge in the matrix can be driven by the membrane potential. Hence, exchange of Fe2+ (or ATP) in the matrix for Fe3+ (or ADP) in the intermembrane space can be driven by the membrane potential and need not require the co-transport of protons down the pH gradient. The same applies to the import of Ca2+. But import of acetate ions into the matrix and exchange of Ca2+ in the matrix for Na+ in the intermembrane space result in an increase in the amount of negative charge in the matrix and it therefore cannot be driven by the charge difference between the two mitochondrial compartments.

Which of the following statements is true? (a) Because the electrons in NADH are at a higher energy than the electrons in reduced ubiquinone, the NADH dehydrogenase complex can pump more protons than can the cytochrome b-c1 complex. (b) The pH in the mitochondrial matrix is higher than the pH in the intermembrane space. (c) The proton concentration gradient and the membrane potential across the inner mitochondrial membrane tend to work against each other in driving protons from the intermembrane space into the matrix. (d) The difference in proton concentration across the inner mitochondrial membrane has a much larger effect than the membrane potential on the total proton-motive force.

Choice (b) is the correct answer. The pumping of protons out of the matrix into the intermembrane space creates a difference in proton concentration between the two sides of the membrane, with the matrix at a higher pH (i.e., more alkaline) than the intermembrane space, which tends to equilibrate with the cytosol (which has a neutral pH). The electrons in NADH are at a higher energy than the electrons in reduced ubiquinone, but proton pumping is not determined simply by the energy of the electron donors (choice (a)). Instead, the number of protons that can be pumped by each complex is determined by the difference in energy between the electrons in each substrate/product pair (i.e., the difference between the electrons in NADH and reduced ubiquinone, compared with that between reduced ubiquinone and reduced cytochrome c). The proton concentration gradient and the membrane potential generated by the electron-transport chain work in the same direction (choice (c)), creating a steep electrochemical gradient for protons across the membrane. For choice (d), the difference in proton concentration has a smaller effect than the membrane potential on the total proton-motive force.

The lab you work in has discovered a previously unidentified extracellular signal molecule called QGF, a 75,000-dalton protein. You add purified QGF to different types of cells to determine its effect on these cells. When you add QGF to heart muscle cells, you observe an increase in cell contraction. When you add it to fibroblasts, they undergo cell division. When you add it to nerve cells, they die. When you add it to glial cells, you do not see any effect on cell division or survival. Given these observations, which of the following statements is most likely to be true? (a) Because it acts on so many diverse cell types, QGF probably diffuses across the plasma membrane into the cytoplasm of these cells. (b) Glial cells do not have a receptor for QGF. (c) QGF activates different intracellular signaling pathways in heart muscles, fibroblasts, and nerve cells to produce the different responses observed. (d) Heart muscle cells, fibroblasts, and nerve cells must all have the same receptor for GQF.

Choice (c) is most likely to be true. Because heart muscles, fibroblasts, and nerve cells all respond to QGF with different outcomes, it is likely that QGF activates different effector proteins in these different cell types, leading to the diversity of outcomes observed in the experiment. QGF is unlikely to diffuse across the cell membrane, given that it is a large protein (choice (a)). Although glial cells do not die or divide in response to QGF, they could have a receptor for QGF, as receptor activation could lead to some other response (choice (b)). A signal molecule can often bind to different types of receptor on different cell types, so choice (d) may or may not be correct.

Which of the following statements is true? (a) Only compounds with negative redox potentials can donate electrons to other compounds under standard conditions. (b) Compounds that donate one electron have higher redox potentials than those of compounds that donate two electrons. (c) The δE′0 of a redox pair does not depend on the concentration of each member of the pair. (d) The free-energy change, δG, for an electron transfer reaction does not depend on the concentration of each member of a redox pair.

Choice (c) is the correct answer. By definition, E′0 refers to the standard state of equal concentrations of each member of the redox pair. Therefore δE′0 does not vary with the actual concentrations. Compounds with positive redox potentials can donate electrons to other compounds under standard conditions, so long as the electron acceptor has a higher (more positive) redox potential; thus option (a) is incorrect. Compounds that are able to donate only one electron do not necessarily have higher redox potentials than compounds that are able to donate two electrons; thus option (b) is incorrect. (Water, for example, has a very high redox potential.) Although the δE′0 of a reaction is directly proportional to the δG°′ of a reaction and both are independent of the concentrations of substrates and products, the δG depends on these concentrations; thus option (d) is incorrect

Which of the following statements is true? (a) Ubiquinone is a small hydrophobic protein containing a metal group that acts as an electron carrier. (b) A 2Fe2S iron-sulfur center carries one electron, whereas a 4Fe4S center carries two. (c) Iron-sulfur centers generally have a higher redox potential than do cytochromes. (d) Mitochondrial electron carriers with the highest redox potential generally contain copper ions and/or heme groups.

Choice (d) is the correct answer. Cytochrome oxidase, which is the last carrier in the mitochondrial electron-transport chain and therefore has the highest redox potential, contains copper ions and a heme group. Ubiquinone is not a protein and does not contain a metal group (choice (a)). Both 2Fe2S and 4Fe4S centers carry one electron (choice (b)). Iron-sulfur centers generally have a lower redox potential than do cytochromes (choice (c)). The heme group in cytochrome c contains a charged iron ion. The interiors of proteins are often hydrophobic, favoring a relatively high redox potential, because reduction of the iron ion decreases its charge, and charges are energetically unfavorable in a hydrophobic environment.

Bacteria undergo chemotaxis toward amino acids, which usually indicates the presence of a food source. Chemotaxis receptors bind a particular amino acid and cause changes in the bacterial cell that induce the cell to move toward the source of the amino acid. Four types of chemotaxis receptor that mediate responses to different amino acids have been identified in a bacterium. The receptors are called ChrA, ChrB, ChrC, and ChrD. Each receptor specifically senses serine, aspartate, glutamate, or glycine, although you do not know which receptor senses which amino acid. You have been given a wild-type bacterial strain that contains all four receptors, as well as various mutant bacterial strains that are lacking one or more of the receptors. To figure out which receptor senses which amino acid, you conduct experiments in which you fill a capillary tube with an amino acid to attract the bacteria, dip the capillary tube into a solution containing bacteria, remove the capillary tube after 5 minutes, and count the number of bacteria in the capillary tube. Your results are shown in Table Q16-54. Table Q16-54 Chemotaxis in wild-type and mutant strains of bacteria From these results, indicate which receptor is used for which amino acid.

ChrA senses glycine, ChrB senses aspartate, ChrC senses glutamate, and ChrD senses serine. To figure this out, you must match the pattern of intact receptors with the pattern of responses to the various amino acids. For example, ChrD is missing in strain 2, which does not sense serine. Therefore, ChrD is the receptor used to sense serine. Since we know that one of the receptors senses glutamate, but all strains respond to glutamate, ChrC must be the sensor for glutamate because it is present in all strains.

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 from the motor neuron triggers the opening of a voltage-sensitive Ca2+ channel in the muscle cells' plasma membrane, allowing Ca2+ to flow into the cell.

Disagree. The increase in intracellular Ca2+ during muscle contraction comes from an intracellular source. The Ca2+ 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 the membranous transverse tubules to open a Ca2+ release channel in the membrane of the sarcoplasmic reticulum.

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.

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. Eucaryotic cells are continually taking up materials from the extracellular space by the process of endocytosis. One type of endocytosis is __________________, which uses __________________ proteins to form small vesicles containing fluids and molecules. After these vesicles have pinched off from the plasma membrane, they will fuse with the __________________, where materials that are taken into the vesicle are sorted. A second type of endocytosis is __________________, which is used to take up large vesicles that can contain microorganisms and cellular debris. Macrophages are especially suited for this process, as they extend __________________ (sheetlike projections of their plasma membrane) to surround the invading microorganisms.

Eucaryotic cells are continually taking up materials from the extracellular space by the process of endocytosis. One type of endocytosis is pinocytosis, which uses clathrin proteins to form small vesicles containing fluids and molecules. After these vesicles have pinched off from the plasma membrane, they will fuse with the endosome, where materials that are taken into the vesicle are sorted. A second type of endocytosis is phagocytosis, which is used to take up large vesicles that can contain microorganisms and cellular debris. Macrophages are especially suited for this process, as they extend pseudopods (sheetlike projections of their plasma membrane) to surround the invading microorganisms.

For each of the following sentences, select the best word or phrase from the list below to fill in the blanks. Not all words or phrases will be used; each word or phrase should be used only once. G-protein-coupled receptors (GPCRs) all have a similar structure with __________________ transmembrane domains. When a GPCR binds an extracellular signal, an intracellular G protein, composed of __________________ subunits, becomes activated. __________________ of the G-protein subunits are tethered to the plasma membrane by short lipid tails. When unstimulated, the α subunit is bound to __________________, which is exchanged for __________________ on stimulation. The intrinsic __________________ activity of the α subunit is important for inactivating the G protein. __________________ inhibits this activity of the α subunit, thereby keeping the subunit in an active state.

G-protein-coupled receptors (GPCRs) all have a similar structure with seven transmembrane domains. When a GPCR binds an extracellular signal, an intracellular G protein, composed of three subunits, becomes activated. Two of the G-protein subunits are tethered to the plasma membrane by short lipid tails. When unstimulated, the α subunit is bound to GDP, which is exchanged for GTP on stimulation. The intrinsic GTPase activity of the α subunit is important for inactivating the G protein. Cholera toxin inhibits this activity of the α subunit, thereby keeping the subunit in an active state.

What is the cause of the massive amount of programmed cell death of nerve cells (neurons) that occurs in the developing vertebrate nervous system, and what purpose does it serve?

Immature neurons are produced in excess of the number that will eventually be required. They compete for the limited amount of survival factors secreted by the target cells they contact. Those cells that fail to get enough survival factor undergo programmed cell death. Up to half or more of the original nerve cells die in this way. This competitive mechanism helps match the number of developing nerve cells to the number of target cells they contact.

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

In G0 the cell-cycle control system is partly dismantled, so that some of the Cdks and cyclins are not present. Cells paused in G1, by contrast, still contain all the components of the cell-cycle control system. Whereas the latter cells can rapidly progress through the cycle when conditions are right, G0 cells need to synthesize the missing cell-cycle control proteins so as to re-enter the cycle, which usually takes 8 hours or more.

The lengths of microtubules in various stages of mitosis depend on the balance between the activities of catastrophins, which destabilize microtubules, and microtubule-associated proteins (MAPs), which stabilize them. If you created cells with an increased number of catastrophin molecules, do you predict the length of the mitotic spindle will be longer, shorter, or unchanged, relative to the corresponding stage of mitosis in wild-type cells? What do you predict for a cell with increased numbers of MAPs? Explain your reasoning.

In a cell with excessive catastrophin molecules, the balance between catastrophins and MAPs will be disrupted. Increased catastrophin activity relative to MAP activity will lead to an increased frequency of microtubule catastrophes, the sudden shift from growth to shrinkage. Thus, microtubules will be shorter on average because they spend less time growing slowly and more time shrinking rapidly. Shorter microtubules would probably result in a shorter mitotic spindle. Conversely, increased MAP activity relative to catastrophin activity will stabilize microtubules by enhancing polymerization or inhibiting depolymerization, thereby promoting the formation of longer microtubules and probably a longer mitotic spindle. However, in some cases the normal balance between catastrophins and MAPs has been shown to be necessary for the formation of a bipolar spindle. It is possible that increasing the amount of MAP or catastrophin proteins will result in microtubules that are so long or so short that they cannot form a spindle at all.

Explain how scientists used artificial vesicles to prove that the generation of ATP by the ATP synthase was not powered by a single high-energy intermediate but rather by a proton gradient. Be sure to describe the two experiments that were negative controls (no ATP generated), the positive control (ATP generated as expected), and a fourth experiment proving that the gradient is the required energy source.

In all the experiments, artificial liposomes were generated and exposed to light, and the surrounding solution was checked for an increase in ATP. In the first experiment, the liposomal membranes contained only bacteriorhodopsin, a bacterial protein that pumps protons and is activated by light. In this negative control, ATP was not expected to be produced, and it was not. In the second experiment, also a negative control, the liposomes contained only ATP synthase. Again, if the chemiosmotic hypothesis was correct, ATP should not have been generated, which was what was observed. In the third experiment, both bacteriorhodopsin and ATP synthase were present in the liposomal membrane. When exposed to light, protons were pumped into the vesicle and ATP was generated. In the fourth experiment, to show that the ATP production was solely a result of the proton gradient, an uncoupling agent was added to the solution containing liposomes with bacteriorhodopsin and ATP synthase. In this case, even though the protons were being pumped into the liposomes, a gradient did not build up; this was because of the presence of the uncoupling agent, which made the membrane permeable to protons. No ATP was generated, proving that it was the proton gradient that was the energy source for ATP synthesis.

Some bacteria can live both aerobically and anaerobically. How does the ATP synthase in the plasma membrane of the bacterium help such bacteria to keep functioning in the absence of oxygen?

In the absence of oxygen, the respiratory chain no longer pumps protons, and thus no proton electrochemical gradient is generated across the bacterial membrane. In these conditions the ATP synthase uses some of the ATP generated by glycolysis in the cytosol to pump protons out of the bacterium, thus forming the proton gradient across the membrane that the bacterium requires for importing vital nutrients by coupled transport.

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. In the carbon fixation process in chloroplasts, carbon dioxide is initially added to the sugar __________________. The final product of carbon fixation in chloroplasts is the three-carbon compound __________________. This is converted into __________________ (which can be used directly by the mitochondria), into __________________ (which is exported to other cells), and into __________________ (which is stored in the stroma). The carbon fixation cycle requires energy in the form of __________________ and reducing power in the form of __________________.

In the carbon fixation process in chloroplasts, carbon dioxide is initially added to the sugar ribulose 1,5-bisphosphate. The final product of carbon fixation in chloroplasts is the three-carbon compound glyceraldehyde 3-phosphate. This is converted into pyruvate (which can be used directly by the mitochondria), into sucrose (which is exported to other cells), and into starch (which is stored in the stroma). The carbon fixation cycle requires energy in the form of ATP and reducing power in the form of NADPH.

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. Mutations in genes that disrupt intermediate filaments cause some rare human 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, 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, __________________s 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 __________________, an important protein for cross-linking intermediate filaments, have a disease that combines symptoms of epidermolysis bullosa simplex, muscular dystrophy, and neurodegeneration.

Intermediate filaments are found mainly in cells that are subject to mechanical stress. Mutations in genes 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.

Name the three main classes of cell-surface receptor.

Ion-channel-coupled receptors; G-protein-coupled receptors; enzyme-coupled receptors.

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 __________________ G1 and G2 phases and 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 __________________, has a __________________ effect on MPF activity and plays a part in regulating interactions with its __________________s. The components of MPF are evolutionarily __________________ from yeast to humans, so that the corresponding human genes are __________________ to function in yeast. able hexokinase short asynchronously inhibitory sperm Cdk conserved cyclin divergent egg fibroblast G1 G2 long M maturation promoting factor oscillate PI 3-kinase regulin S steady stimulatory substrate synchronously ubiquitin unable uniform

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 G2-like phase. When the cells are triggered to resume cycling, the cell divisions have especially short G1 and G2 phases and occur synchronously. Studies with Xenopus eggs identified a 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.

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 __________________ and involve a "__________________ plus two" array of microtubules. The motor protein __________________ generates the bending motion in cilia; the lack of this protein can cause Kartagener's syndrome in humans.

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 of a microtubule. γ-Tubulin rings are important for microtubule nucleation and 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.

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. Mitochondria can use both __________________ and __________________ directly as fuel. __________________ produced in the citric acid cycle donates electrons to the electron-transport chain. The citric acid cycle oxidizes __________________ and produces __________________ as a waste product. __________________ acts as the final electron acceptor in the electron-transport chain. The synthesis of ATP in mitochondria is also known as __________________.

Mitochondria can use both pyruvate and fatty acids directly as fuel. NADH produced in the citric acid cycle donates electrons to the electron-transport chain. The citric acid cycle oxidizes acetyl groups and produces carbon dioxide as a waste product. Oxygen acts as the final electron acceptor in the electron-transport chain. The synthesis of ATP in mitochondria is also known as oxidative phosphorylation.

Why should it be that drugs such as colchicine that inhibit microtubule polymerization and drugs such as Taxol that stabilize microtubules both inhibit mitosis?

Mitosis requires that the spindle microtubules behave dynamically—continuously polymerizing and depolymerising—to probe the cell cortex, to seek attachments to kinetochores, and to segregate the chromosomes.. Static microtubules are unable to do any of these things.

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. NADH donates electrons to the __________________ of the three respiratory enzyme complexes in the mitochondrial electron-transport chain. __________________ is a small protein that acts as a mobile electron carrier in the respiratory chain. __________________ transfers electrons to oxygen. Electron transfer in the chain occurs in a series of __________________ reactions. The first mobile electron carrier in the respiratory chain is __________________.

NADH donates electrons to the first of the three respiratory enzyme complexes in the mitochondrial electron-transport chain. Cytochrome c is a small protein that acts as a mobile electron carrier in the respiratory chain. Cytochrome oxidase transfers electrons to oxygen. Electron transfer in the chain occurs in a series of oxidation-reduction reactions. The first mobile electron carrier in the respiratory chain is ubiquinone.

For each of the following sentences, choose one of the two options enclosed in square brackets to make a correct statement. New plasma membrane reaches the plasma membrane by the [regulated/constitutive] exocytosis pathway. New plasma membrane proteins reach the plasma membrane by the [regulated/constitutive] exocytosis pathway. Insulin is secreted from pancreatic cells by the [regulated/constitutive] exocytosis pathway. The interior of the trans Golgi network is [acidic/alkaline]. Proteins that are constitutively secreted [aggregate/do not aggregate] in the trans Golgi network.

New plasma membrane reaches the plasma membrane by the constitutive exocytosis pathway. New plasma membrane proteins reach the plasma membrane by the constitutive exocytosis pathway. Insulin is secreted from pancreatic cells by the regulated exocytosis pathway. The interior of the trans Golgi network is acidic. Proteins that are constitutively secreted do not aggregate in the trans Golgi network.

Which of the following statements is true? (a) Extracellular signal molecules that are hydrophilic must bind to a cell- surface receptor so as to signal a target cell to change its behavior. (b) To function, all extracellular signal molecules must be transported by their receptor across the plasma membrane into the cytosol. (c) A cell-surface receptor capable of binding only one type of signal molecule can mediate only one kind of cell response. (d) Any foreign substance that binds to a receptor for a normal signal molecule will always induce the same response that is produced by that signal molecule on the same cell type.

Only choice (a) is true. A hydrophilic molecule cannot diffuse across the membrane and it can therefore only affect a cell if it binds to a cell-surface receptor (choice (a)). Most signal molecules remain bound to the extracellular domain of the receptor, whereas the intracellular domain mediates signal transduction; although many signal molecules are endocytosed with their receptor, they remain inside membrane-bounded compartments and are therefore not transported into the cytosol (choice (b)). A cell-surface receptor capable of binding only one type of signal molecule can stimulate more than one kind of cell response, depending on the types of intracellular signaling pathway it activates (choice (c)). Foreign substances that bind to a receptor for a normal signal molecule can sometimes induce the same response as the natural signal molecule, but in other cases they can block the binding of the natural signal molecule without activating the receptor (choice (d)).

Use the terms provided below to fill in the blanks. Not all words or phrases will be used; each word or phrase may be used more than once. Photons from sunlight that are in the ______________ wavelength range are preferentially absorbed by chlorophyll molecules to raise the energy levels of electrons in the __________ ring. The __________ emitted are lower in energy, which is reflected in the ________, green wavelengths detected by the human eye. red benzene heme blue shorter electrons longer photons porphyrin orange

Photons from sunlight that are in the red wavelength range are preferentially absorbed by chlorophyll molecules to raise the energy levels of electrons in the porphyrin ring. The photons emitted are lower in energy, which is reflected in the longer, green wavelengths detected by the human eye.

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. Plasma membrane proteins are inserted into the membrane in the __________________. The address information for protein sorting in a eucaryotic cell is contained in the __________________ of the proteins. Proteins enter the nucleus in their __________________ form. Proteins that remain in the cytosol do not contain a __________________. Proteins are transported into the Golgi apparatus via __________________. The proteins transported into the endoplasmic reticulum by __________________ are in their __________________ form.

Plasma membrane proteins are inserted into the membrane in the endoplasmic reticulum. The address information for protein sorting in a eucaryotic cell is contained in the amino acid sequence of the proteins. Proteins enter the nucleus in their folded form. Proteins that remain in the cytosol do not contain a sorting signal. Proteins are transported into the Golgi apparatus via transport vesicles. The proteins transported into the endoplasmic reticulum by protein translocators are in their unfolded form.

The number of cells in an adult tissue or animal depends on cell proliferation. What else does it depend on?

Programmed cell death also influences cell numbers. Most animal cells require survival signals from other cells to avoid programmed cell death, so that the levels of such signals can help determine how many cells live and how many die.

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. Proteins are transported out of a cell via the __________________ or __________________ pathway. Fluids and macromolecules are transported into the cell via the __________________ pathway. All proteins being transported out of the cell pass through the __________________ and the __________________. Transport vesicles link organelles of the __________________ system. The formation of __________________ in the endoplasmic reticulum stabilizes protein structure.

Proteins are transported out of a cell via the secretory or exocytic pathway. Fluid and macromolecules are transported into the cell via the endocytic pathway. All proteins being transported out of the cell pass through the endoplasmic reticulum and the Golgi apparatus. Transport vesicles link organelles of the endomembrane system. The formation of disulfide bonds in the endoplasmic reticulum stabilizes protein structure.

You isolate some muscle fibers to examine what regulates muscle contraction. When you bathe the muscle fibers in a solution containing ATP and Ca2+, you see muscle contraction (experiment 3 in Table Q17-49). Ca2+ 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 Q17-49). 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 Q17-49 Extra credit: In what state would the muscle be if you added Ca2+ but no ATP?

See Table A17-49. Table A17-49 In experiment 4, the muscle will be relaxed because troponin will not be able to bind Ca2+. By preventing troponin from binding to Ca2+, 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 absence of troponin regulation by Ca2+, 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 it will be available for myosin to hydrolyze, causing muscle contraction. In experiment 6, the muscle will remain relaxed. The presence of Ca2+ 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. Extra credit: The muscle will be in a rigor state. If Ca2+ is added without ATP, troponin can bind Ca2+, undergoing 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.

You have created a GFP fusion to a protein that is normally secreted from yeast cells. Because you have learned about the use of temperature-sensitive mutations in yeast to study protein and vesicle transport, you obtain three mutant yeast strains, each defective in some aspect of the protein secretory process. Being a good scientist, you of course also obtain a wild-type control strain. You decide to examine the fate of your GFP fusion protein in these various yeast strains and engineer the mutant strains to express your GFP fusion protein. However, in your excitement to do the experiment, you realize that you did not label any of the mutant yeast strains and no longer know which strain is defective in what process. You end up numbering your strains with the numbers 1 to 4, and then you carry out the experiment anyway, obtaining the results shown in Figure Q15-46 (the black dots represent your GFP fusion protein). Figure Q15-46 Name the process that is defective in each of these strains. Remember that one of these strains is your wild-type control.

Strain A has protein accumulating in the ER, which means that this cell has a mutation that blocks transport from the ER to the Golgi apparatus. Strain B has secreted protein, and therefore is the wild-type control. Strain C has protein accumulating in the Golgi apparatus, and thus has a mutation that blocks exit of proteins from the Golgi apparatus. Strain D has protein accumulating in the cis-Golgi network, and thus has a mutation that blocks the travel of proteins through the Golgi cisternae.

Briefly describe the mechanism by which an internal stop-transfer sequence in a protein causes the protein to become embedded in the lipid bilayer as a transmembrane protein with a single membrane-spanning region. Assume that the protein has an N-terminal signal sequence and just one internal hydrophobic stop-transfer sequence.

The N-terminal signal sequence initiates translocation and the protein chain starts to thread through the translocation channel. When the stop-transfer sequence enters the translocation channel, the channel discharges both the signal sequence and the stop-transfer sequence sideways into the lipid bilayer. The signal sequence is then cleaved, so that the protein remains held in the membrane by the hydrophobic stop-transfer sequence.

Antibodies are Y-shaped molecules that have two identical binding sites. Suppose that you have obtained an antibody that is specific for the extracellular domain of an RTK. When the antibody binds to the RTK, it brings together two RTK molecules. If cells containing the RTK were exposed to the antibody, would you expect the kinase to be activated, inactivated, or unaffected? Explain your reasoning.

The RTK will probably become activated on binding of the antibody molecule. This is because signal-induced dimerization usually activates RTKs. When RTK molecules are brought together, their cytoplasmic kinase domains become activated and each receptor phosphorylates the other.

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 protein called __________________, or __________________, to trigger S and M phases. Inactivation of __________________ is required to exit from M phase after chromosome segregation. APC Cdks condensation cyclin-dependent kinases cytokinesis G1 phase G1-Cdk G2 phase interphase intraphase kinesins M phase M-Cdk meiosis metaphase microtubules mitosis myosins S phase S-Cdk

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 G1 phase, S phase, and G2 phase. During M phase, 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 and M phases. Inactivation of M-Cdk is required to exit from M phase after chromosome segregation.

The citric acid cycle generates NADH and FADH2, which are then used in the process of oxidative phosphorylation to make ATP. If the citric acid cycle (which does not use oxygen) and oxidative phosphorylation are separate processes, as they are, then why is it that the citric acid cycle stops almost immediately when O2 is removed?

The citric acid cycle stops almost immediately when oxygen is removed, because several steps in the cycle require the oxidized forms of NAD+ and FAD. In the absence of oxygen, these electron carriers can be reduced by the reactions of the citric acid cycle but cannot be reoxidized by the electron-transport chain that participates in oxidative phosphorylation.

Figure Q18-30 shows a living cell from the lung epithelium of a newt at different stages in M phase. Order these light micrographs into the correct sequence and identify the stage in M phase that each represents.

The correct order is listed as follows, with stages in parenthesis: E (prophase), D (prometaphase), C (metaphase), A (anaphase), F (telophase), and B (cytokinesis)

A gene regulatory protein, A, contains a typical nuclear localization signal but surprisingly is usually found in the cytosol. When the cell is exposed to hormones, protein A moves from the cytosol into the nucleus, where it turns on genes involved in cell division. When you purify protein A from cells that have not been treated with hormones, you find that protein B is always complexed with it. To determine the function of protein B, you engineer cells lacking the gene for protein B. You compare normal and defective cells by using differential centrifugation to separate the nuclear fraction from the cytoplasmic fraction and then separate the proteins in these fractions by gel electrophoresis. You identify the presence of protein A and protein B by looking for their characteristic bands on the gel. The gel you run is shown in Figure Q15-16. Figure Q15-16 On the basis of these results, what is the function of protein B? Explain your conclusion and propose a mechanism for how protein B works.

The data on the gel show that protein A is always found in the nucleus in the absence of protein B. Therefore, any mechanism that is proposed must explain this result. One possible answer is that protein B binds protein A and masks the nuclear localization signal. In the presence of hormone, protein B interacts with the hormone, which changes its conformation so that it can no longer bind protein A. When protein B no longer binds to protein A, the nuclear localization signal on protein A is now exposed and protein A can enter the nucleus. Therefore, in the absence of protein B, the nuclear localization signal on protein A is always exposed and protein A resides in the nucleus. Another possible answer is that protein B binds protein A and sequesters it by keeping protein A in some subcellular compartment, away from the nucleus. In the presence of hormone, protein B interacts with the hormone, changing its conformation so that it can no longer bind to protein A. When protein B is not present, protein A can enter the nucleus in the presence or absence of hormone.

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, and it 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.

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. The four phases of the cell cycle, in order, are G1, __________________, __________________, 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 G1 nucleotide S G2 organelle

The four phases of the cell cycle, in order, are G1, S, G2, and M. A cell contains the most DNA after S phase of the cell cycle. A cell is smallest in size after M phase of the cell cycle. Growth occurs in G1, S, and G2 phases of the cell cycle. A cell does not enter mitosis until it has completed DNA synthesis.

If a lysosome breaks, what protects the rest of the cell from lysosomal enzymes?

The lysosomal enzymes are all acid hydrolases, which have optimal activity at the low pH (about 5.0) found in the interior of lysosomes. If a lysosome were to break, the acid hydrolases would find themselves at pH 7.2, the pH of the cytosol, and would therefore do little damage to cellular constituents.

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. A. minus end of microtubule B. tail of motor protein C. cargo of motor protein D. head of motor protein 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.

Activated GPCRs activate G proteins by reducing the strength of binding of GDP to the α subunit of the G protein, allowing GDP to dissociate and GTP (which is present at much higher concentrations in the cell than GDP) to bind in its place. How would the activity of a G protein be affected by a mutation that reduces the affinity of the α subunit for GDP without significantly changing its affinity for GTP?

The mutant G protein would be constantly active. Each time the α subunit hydrolyzed GTP to GDP, the GDP would dissociate spontaneously, allowing GTP to bind and reactivate the α subunit, especially because the intracellular concentration of GTP is higher than that of GDP. Normally, GDP is tightly bound by the α subunit, which keeps the G protein in its inactive state until interaction with an appropriate activated GPCR stimulates the release of GDP.

Explain why the signal molecules used in neuronal signaling work at a longer range than those used in contact-dependent signaling.

The neurotransmitter released from a neuron in neuronal signaling must diffuse across the synaptic cleft to reach receptors on the target cell. In contrast, in contact-dependent signaling, the signal molecule is attached to the plasma membrane of the signaling cell and interacts with receptors located on the plasma membrane of the receiving cell; thus, the cells must be in direct contact for this type of signaling to occur.

You are trying to identify the peroxisome-targeting sequence in the thiolase enzyme in yeast. The thiolase enzyme normally resides in the peroxisome and therefore must contain amino acid sequences that are used to target the enzyme for import into the peroxisome. To identify the targeting sequences, you create a set of hybrid genes that encode fusion proteins containing part of the thiolase protein fused to another protein, histidinol dehydrogenase (HDH). HDH is a cytosolic enzyme required for the synthesis of the amino acid histidine and cannot function if it is localized in the peroxisome. You genetically engineer a series of yeast cells to express these fusion proteins instead of their own versions of these enzymes. If the fusion proteins are imported into the peroxisome, the HDH portion of the protein cannot function and the yeast cells cannot grow on a medium lacking histidine. You obtain the results shown in Figure Q15-14. Figure Q15-14 What region of the thiolase protein contains the peroxisomal targeting sequence? Explain your answer.

The peroxisomal targeting sequence lies between amino acids number 100 and number 125. Any fusion protein containing this sequence can be targeted for import into the peroxisome (because the yeast cannot grow on a medium lacking histidine), whereas the fusion proteins lacking this region do not target the fusion protein for import into the peroxisome (because the yeast do grow on medium lacking histidine). The most important pieces of data are the fusion proteins containing amino acids 100-200 of the thiolase protein fused to HDH and the fusion protein containing amino acids 1-125 of the thiolase protein fused to HDH. Neither of these fusion proteins not allow growth on medium lacking histidine and can be used to define the minimal region necessary for targeting thiolase for import into the peroxisome. (Note that although these experiments show that amino acids 100-125 are necessary, these experiments do not show that this region is sufficient for peroxisomal targeting. It is possible that the region consisting of amino acids 100-125 is sufficient, or it could be that this region collaborates with redundant signals between amino acids 1 and 100 or between amino acids 125 and 200.)

Imagine that you could microinject cytochrome c into the cytosol of both wild- type cells and cells that were lacking both Bax and Bak, which are apoptosis- promoting members of the Bcl-2 family of proteins. Would you expect one, both, or neither of the cell lines to undergo apoptosis? Explain your reasoning.

The presence or absence of Bak and Bax would not affect whether a microinjection of cytochrome c would promote apoptosis, because Bax and Bak act upstream of cytochrome c by promoting its release from mitochondria. By promoting the formation of the apoptosome and the activation of procaspases, microinjection of cytochrome c bypasses the need for Bax or Bak in promoting apoptosis.

If you remove the ER-retention signal from a protein that normally resides in the ER lumen, where do you predict the protein will ultimately end up? Explain your reasoning.

The protein would end up in the extracellular space. Normally, the protein would go from the ER to the Golgi apparatus, get captured because of its ER-retrieval signal, and return to the ER. However, without the ER-retrieval signal, the protein would evade capture, ultimately leave the Golgi via the default pathway, and become secreted into the extracellular space. The protein would not be retained anywhere else along the secretory pathway: it presumably has no signals to promote such localization because it normally resides in the ER lumen.

Intracellular steroid hormone receptors have binding sites for a signaling molecule and a DNA sequence. How is it that the same steroid hormone receptor, which binds to a specific DNA sequence, can regulate different genes in different cell types?

The specific genes regulated in response to an activated steroid hormone receptor depend not only on the genes' having the appropriate DNA sequence for binding the receptor but also on a variety of other nuclear proteins that influence gene expression, some of which vary between different cell types.

Is the following statement true or false? After the nuclear envelope breaks down, microtubules gain access to the chromosomes and, every so often, a randomly probing microtubule captures a chromosome and ultimately connects to the kinetochore to become a kinetochore microtubule of the spindle.

The statement is true.

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 survival, __________________, and size of each cell in an animal are controlled by extracellular signal molecules secreted by neighboring and distant cells. Many of these signal molecules bind to a cell-surface __________________ and trigger various intracellular signaling pathways. One class of signal molecules, called __________________, stimulates cell division by releasing the molecular brakes that keep cells in the __________________ or __________________ phase of the cell cycle. Members of a second class of signal molecules are called __________________, because they stimulate cell growth and an increase in cell mass. The third class of signal molecules, called __________________, inhibits __________________ by regulating members of the __________________ family of proteins. In addition to such stimulatory factors, some signal proteins such as __________________ act negatively on other cells, inhibiting their survival, growth, or proliferation. anaphase annihilation apoptosis arrestase Bcl-2 biosynthetic cascades caspase Cdk cyclin differentiation G0 G1 G2 growth factors interphase ligand M mitogens myostatin nourishment nutrition phosphatases proliferation receptor S survival factors transcription

The survival, proliferation, and size of each cell in an animal are controlled by extracellular signal molecules secreted by neighboring and distant cells. Many of these signal molecules bind to a cell-surface receptor and trigger various intracellular signaling pathways. One class of signal molecules, called mitogens, stimulates cell division by releasing the molecular brakes that keep cells in the G0 or G1 phase of the cell cycle. Members of a second class of signal molecules are called growth factors because they stimulate cell growth and an increase in cell mass. The third class of signal molecules, called survival factors, inhibits apoptosis by regulating members of the Bcl-2 family of proteins. In addition to such stimulatory factors, some signal proteins such as myostatin act negatively on other cells, inhibiting their survival, growth, or proliferation.

In a cell capable of regulated secretion, what are the three main classes of proteins that must be separated before they leave the trans Golgi network?

The three main classes of protein that must be sorted before they leave the trans Golgi network in a cell capable of regulated secretion are (1) those destined for lysosomes, (2) those destined for secretory vesicles, and (3) those destined for immediate delivery to the cell surface.

Consider a redox reaction between molecules A and B. Molecule A has a redox potential of -100 mV and molecule B has a redox potential of +100 mV. For the transfer of electrons from A to B, is the δG° positive or negative or zero? Under what conditions will the reverse reaction, transfer of electrons from B to A, occur?

The δG° is negative. The sign of δG° is the opposite of that of δE′0 = E′0 (acceptor) - E′0 (donor). The acceptance of electrons by B from A has a δE′0 = 100 + 100 = 200 . The reverse reaction, the donation of electrons from B to A, has a positive δG° and is therefore unfavorable under standard conditions. Remember that, by definition, the concentrations of A and its redox pair A′ are equal under standard conditions; similarly, the concentration of B is equal to the concentration of its redox pair B′. B will be able to donate electrons to A only when [B] > [B′] and/or [A] < [A′] to such an extent that the δG for electron transfer becomes negative.

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. There are many actin-binding proteins in cells that can bind to actin and modify its activity. Some proteins such as __________________ bind to actin monomers, sequestering them until needed for filament formation. __________________ proteins bind to the end of actin filaments. The __________________ proteins are important for nucleation of the branched actin structures commonly found in the __________________ of moving cells, whereas __________________ proteins are important for the formation of capping integrin filopodia Rho severing lamellipodia thymosin ARP formin bundling GAP gelsolin contracting cytochalasin lamin unbranched actin filaments commonly found in __________________. Proteins belonging to the __________________ family of GTPases regulate changes in the actin cytoskeleton in response to extracellular signals.

There are many actin-binding proteins in cells that can bind to actin and modify its activity. Some proteins such as thymosin bind to actin monomers, sequestering them until needed for filament formation. Capping proteins bind to the end of actin filaments. The ARP proteins are important for nucleation of the branched actin structures commonly found in the lamellipodia of moving cells, whereas formin proteins are important for the formation of unbranched actin filaments commonly found in filopodia. Proteins belonging to the Rho family of GTPases regulate changes in the actin cytoskeleton in response to extracellular signals.

Your friend is studying mouse fur color and has isolated the GPCR responsible for determining its color, as well as the extracellular signal that activates the receptor. She finds that, on addition of the signal to pigment cells (cells that produce the pigment determining fur color), cAMP levels rise in the cell. She starts a biotech company, and the company isolates more components of the signaling pathway responsible for fur color. Using transgenic mouse technology, the company genetically engineers mice that are defective in various proteins involved in determining fur color. The company obtains the following results. Normal mice have beige (very light brown) fur color. Mice lacking the extracellular signal have white fur. Mice lacking the GPCR have white fur. Mice lacking cAMP phosphodiesterase have dark brown fur. Your friend has also made mice that are defective in the α subunit of the G protein in this signaling pathway. The defective α subunit works normally except that, once it binds GTP, it cannot hydrolyze GTP to GDP. What color do you predict that the fur of these mice will be? Why?

These mice will have dark brown fur. The inability to hydrolyze GTP to GDP will lead to inappropriate activation of the signaling pathway that makes pigment. Too much pigment will be produced, as seen in the mice lacking cAMP phosphodiesterase (which lack the ability to damp the signal), and the mice will end up with dark brown fur.

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.

In 1925, David Keilin used a simple spectroscope to observe the characteristic absorption bands of the cytochromes that participate in the electron-transport chain in mitochondria. A spectroscope passes a very bright light through the sample of interest and then through a prism to display the spectrum from red to blue. If molecules in the sample absorb light of particular wavelengths, dark bands will interrupt the colors of the rainbow. His key discovery was that the absorption bands disappeared when oxygen was introduced and then reappeared when the samples became anoxic. Subsequent findings demonstrated that different cytochromes absorb light of different frequencies. When light of a characteristic wavelength shines on a mitochondrial sample, the amount of light absorbed is proportional to the amount of a particular cytochrome present in its reduced form. Thus, spectrophotometric methods can be used to measure how the amounts of reduced cytochromes change over time in response to various treatments. If isolated mitochondria are incubated with a source of electrons such as succinate, but without oxygen, electrons enter the respiratory chain, reducing each of the electron carriers almost completely. When oxygen is then introduced, the carriers oxidize at different rates, as can be seen from the decline in the amount of reduced cytochrome (see Figure Q14-49). Note that cytochromes a and a3 cannot be distinguished and thus are listed as cytochrome (a + a3). How does this result allow you to order the electron carriers in the respiratory chain? What is their order?

This result allows you to order the electron carriers in the respiratory chain because when oxygen is added, the last carrier in the chain will be oxidized first. This is because oxygen is the final sink for the electrons that flow through the chain, and it participates directly in a redox reaction with the last electron carrier. The wave of oxidation will then proceed backward through the chain toward the first electron carrier in the chain; this is because the oxidation of each carrier will convert it to a form that can accept electrons from the "upstream" carrier in the chain, thereby oxidizing each upstream carrier sequentially. The order of cytochromes in the respiratory chain is the reverse of the order in which they are oxidized (i.e., the order in which the reduced form is lost). Listed from first to last, the cytochromes in the chain are b, c1, c, (a + a3).

v-SNAREs and t-SNARES mediate the recognition of a vesicle with its target membrane so that a vesicle displaying a particular type of v-SNARE will only fuse with a target membrane containing a complementary type of t-SNARE. In some cases, v-SNAREs and t-SNAREs may also mediate the fusion of identical membranes. In yeast cells, right before the formation of a new cell, vesicles derived from the vacuole will come together and fuse to form a new vacuole destined for the new cell. Unlike the situation we have discussed in class, the vacuolar vesicles contain both v-SNAREs and t-SNAREs. Your friend is trying to understand the role of these SNAREs in the formation of the new vacuole and consults with you regarding the interpretation of his data. Your friend has designed an ingenious assay for the fusion of vacuolar vesicles by using alkaline phosphatase. The protein alkaline phosphatase is made in a "pro" form that must be cleaved for the protein to be active. Your friend has designed two different strains of yeast: strain A produces the "pro" form of alkaline phosphatase (pro-Pase), whereas strain B produces the protease that can cleave pro-Pase into the active form (Pase). Neither strain has the active form of the alkaline phosphatase, but when vacuolar vesicles from the strains A and B are mixed, fusion of vesicles generates active alkaline phosphatase, whose activity can be measured and quantified. Your friend has taken each of these yeast strains and further engineered them so that they express only the v-SNAREs, only the t-SNAREs, both SNAREs (the normal situation), or neither SNARE. He then isolates vacuolar vesicles from all strains and tests the ability of each variant form of strain A to fuse with each variant form of strain B, by using the alkaline phosphatase assay. The data are shown in the graph in Figure Q15-29B. On this graph, the SNARE present on the vesicle of the particular yeast strain is indicated as "v" (for the presence of the v-SNARE) and "t" (for the presence of the t-SNARE). What do his data say about the requirements for v-SNAREs and t-SNAREs in the vacuolar vesicles? Is it important to have a specific type of SNARE (that is, v-SNARE or t-SNARE) on each vesicle?

To get maximal levels of vacuolar vesicle fusion, vesicles from each strain must carry both v-SNAREs and t-SNARES. Experiment 1, which represents the normal scenario, is the only experiment in which 100% alkaline phosphatase activity is measured. However, as long as complementary SNAREs are present on the vesicles, some vesicle fusion does occur (see experiments 3, 4, 6, 7, 8, and 9). If both vesicles are missing either v-SNAREs (experiment 2) or t-SNAREs (experiment 5) or both SNAREs (experiment 10 and 11), the level of fusion is very low. It does not matter whether a t-SNARE or a v-SNARE is on the vesicle of a particular strain, as long as the vesicle from the other strain contains a complementary SNARE (compare experiments 3 and 4, 6 and 7, and 8 and 9).

Fibroblast cells from patients W, X, Y, and Z, each of whom has a different inherited defect, all contain "inclusion bodies," which are lysosomes filled with undigested material. You wish to identify the cellular basis of these defects. The possibilities are: 1. a defect in one of the lysosomal hydrolases 2. a defect in the phosphotransferase that is required for mannose-6-phosphate tagging of the lysosomal hydrolases 3. a defect in the mannose-6-phosphate receptor, which binds mannose-6-phosphate-tagged lysosomal proteins in the trans Golgi network and delivers them to lysosomes When you incubate some of these mutant fibroblasts in a medium in which normal cells have been grown, you find that the inclusion bodies disappear. Because of these results, you suspect that the constitutive exocytic pathway in normal cells is secreting lysosomal hydrolases that are being taken up by the mutant cells. (It is known that some mannose-6-phosphate receptor molecules are found in the plasma membrane and can take up and deliver lysosomal proteins via the endocytic pathway.) You incubate cells from each patient with medium from normal cells and medium from each of the other mutant cell cultures, and get the results summarized in Table Q15-44. Indicate which defect (1, 2, 3) each patient (W, X, Y, Z) is most likely to have.

W—3 (defect in mannose-6-phosphate receptor) X—2 (defect in phosphotransferase) Y—1; Z—1 (defect in lysosomal hydrolases); these will be defects in two different lysosomal acid hydrolases. A cell that has no mannose-6-phosphate receptor will be able to make all the lysosomal hydrolases properly but will not be able to send them to the lysosome and will also not be able to scavenge hydrolases from the external media. Hence, this cell line cannot be rescued by a culture medium that has had lysosomal hydrolases secreted into it and thus will not be rescued by any of the media tested here. A cell line that has no phosphotransferase will be able to scavenge hydrolases from the external medium, but because all of the cell's own hydrolases will lack the mannose-6-phosphate tag, it will be rescued only by medium from a cell line that is able to make all of the hydrolases. Cell lines lacking one hydrolase will be rescued by medium from any cell line that is able to secrete that hydrolase in a mannose-6-phosphate-tagged form; in addition, media from cultures of cells lacking a hydrolase will rescue any cell line with another type of defect.

Actin-binding proteins bind to actin and can modify its properties. You purify a protein, 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 Q17-36). Do you think Cap1 binds the plus end or the minus end of actin? Explain your reasoning.

You would predict that Cap1 binds the plus ends 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 ends of actin can block monomer addition to the actin filament. Thus, less actin polymerization will be seen in the presence of the Cap1 protein.

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

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

Given the generic signaling pathway in Figure Q16-9, write the number corresponding to the item on the line next to the descriptor below. _________ receptor protein _________ effectorproteins _________ intracellular signaling proteins _________ ligand

____2____ receptorprotein ____4____ ____3____ ____1___ effector proteins intracellular signaling proteins ligand

Rank the following types of cell signaling from 1 to 4, with 1 representing the type of signaling in which the signal molecule travels the least distance and 4 the type of signaling in which the signal molecule travels the largest distance. ______ paracrine signaling ______ contact-dependent signaling ______ neuronal signaling ______ endocrine signaling

____3__ paracrine signaling ____1__ contact-dependent signaling ____2__ neuronal signaling ____4__ endocrine signaling

Match the target of the G protein with the appropriate signaling outcome. adenylyl cyclase ________ ion channels _________ phospholipase C _________ A. cleavage of inositol phospholipids B. increase in cAMP levels C. changes in membrane potential

adenylyl cyclase ion channels phospholipase C ___B_____ ___C_____ ___A_____

Cell lines A and B both survive in tissue culture containing serum but do not proliferate. Factor F is known to stimulate proliferation in cell line A. Cell line A produces a receptor protein (R) that cell line B does not produce. To test the role of receptor R, you introduce this receptor protein into cell line B, using recombinant DNA techniques. You then test all of your various cell lines in the presence of serum for their response to factor F, with the results summarized in Table Q16-1. Table Q16-1 Which of the following cannot be concluded from your results above? (a) Binding of Factor F to its receptor is required for proliferation of cell line A. (b) Receptor R binds to Factor F to induce cell proliferation in cell line A. (c) Cell line A expresses a receptor for Factor F. (d) Factor F is not required for proliferation in cell line B.

b) is false. Expressing receptor R in cell line B causes proliferation in the absence of F, suggesting that something else in the serum can bind to R in cell line B. We do not know the effect of eliminating receptor R from cell line A, and thus we cannot say whether binding of receptor R to its ligand is required for cell line A to proliferate. The data support all the other answers. Because cell line A proliferates only in the presence of F, binding of factor F must be required for cell line A to proliferate (choice (a)). Because cell line A proliferates in response to factor F, it must express a receptor for factor F since it responds to factor F (choice (c)). Because cell line B can proliferate in the absence of F once R is expressed, factor F is not required for cell line B to proliferate (choice (d)).

If you add a compound to illuminated chloroplasts that inhibits the NADP+ reductase, NADPH generation ceases, as expected. However, ferredoxin does not accumulate in the reduced form because it is able to donate its electrons not only to NADP+ (via NADP+ reductase) but also back to the cytochrome b6-f complex. Thus, in the presence of the compound, a "cyclic" form of photosynthesis occurs in which electrons flow in a circle from ferredoxin, to the cytochrome b6-f complex, to plastocyanin, to photosystem I, to ferredoxin. What will happen if you now also inhibit photosystem II? (a) Less ATP will be generated per photon absorbed. (b) ATP synthesis will cease. (c) Plastoquinone will accumulate in the oxidized form. (d) Plastocyanin will accumulate in the oxidized form.

c). If you now inhibit photosystem II you will deprive plastoquinone, which can still donate its electrons to the cytochrome b6-f complex, of an electron source. Hence, plastoquinone will accumulate in its oxidized form. In contrast, all of the other components downstream of plastoquinone will be able to cycle between their oxidized and reduced states. ATP synthesis will continue, because electrons are still being fed through the cytochrome b6-f complex, and the same amount of ATP will be generated.

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. Cells can signal to each other in various ways. A signal that must be relayed to the entire body is most efficiently sent by __________________ cells, which produce hormones that are carried throughout the body through the bloodstream. On the other hand, __________________ methods of cell signaling do not require the release of a secreted molecule and are used for very localized signaling events. During __________________ signaling, the signal remains in the neighborhood of the secreting cell and thus acts as a local mediator on nearby cells. Finally, __________________ signaling converts electrical impulses into a chemical signal. Cells receive signals through a __________________, which can be an integral membrane protein or can reside inside the cell.

ells can signal to each other in various ways. A signal that must be relayed to the entire body is most efficiently sent by endocrine cells, which produce hormones that are carried throughout the body through the bloodstream. On the other hand, contact-dependent methods of cell signaling do not require the release of a secreted molecule and are used for very localized signaling events. During paracrine signaling, the signal remains in the neighborhood of the secreting cell and thus acts as a local mediator on nearby cells. Finally, neuronal signaling involves the conversion of electrical impulses into a chemical signal. Cells receive signals through a receptor, which can be an integral membrane protein or can reside inside the cell.

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

lamins ____C____ neurofilaments ____A____ vimentins ____D____ keratins ____B____

The following proteins are important for cell movement. Match the following proteins with their function. myosin II _________ ARP proteins _________ profilin _________ integrins _________ A. nucleation of new actin filaments B. regulation of the availability of actin monomers C. involvement in focal contacts D. contracting the rear of the cell

myosin II ___D_____ ARPproteins ___A_____ profilin___B_____ integrins ___C_____