Cell Biology-MCB 2210-May Session Final Exam Questions

34. Which of the following are made on RER-bound ribosomes?

. SRP (Signal Recognition Particle): SRP is involved in targeting and guiding newly synthesized proteins to the appropriate cellular compartments, including the endoplasmic reticulum (ER). Loss-of-function mutations in SRP could disrupt the proper targeting of secretory proteins, potentially leading to abnormal localization. b. Translocation channel: The translocation channel, also known as the protein translocation complex, is responsible for transporting proteins across the ER membrane during their synthesis. Loss-of-function mutations in the translocation channel could impair the translocation process, resulting in mislocalization of secretory proteins. c. Clathrin proteins: Clathrin proteins are involved in vesicle formation and trafficking within the cell. They play a role in endocytosis and intracellular transport. While abnormalities in clathrin proteins might impact cellular processes, they are not directly associated with the translocation of secretory proteins into the ER. Based on this information, the most likely correct answer would be: d. Mutations in a and b could result in phenotype 1. Loss-of-function mutations in SRP (option a) and the translocation channel (option b) are both directly involved in the proper targeting and translocation of secretory proteins. Mutations in either of these molecules could lead to abnormal localization of normally secreted proteins, resulting in phenotype 1. Please note that without the visual information from the picture, this is a general explanation based on the knowledge of the functions of the molecules mentioned.

What do you think will be the effect on COPI-coated vesicles if a cell is treated with GTP analogues that can bind to the G-protein involved in vesicle formation, but cannot trigger its active conformation?

. They will accumulate in the nucleus: COPI-coated vesicles are primarily involved in retrograde transport from the Golgi complex to the endoplasmic reticulum (ER). They do not typically have a role in nuclear transport. Therefore, treating the cell with GTP analogues that cannot trigger the active conformation of the G-protein involved in vesicle formation is unlikely to cause the accumulation of COPI-coated vesicles in the nucleus. b. They will accumulate in the cytoplasm: This option is incorrect. Treating the cell with GTP analogues that cannot trigger the active conformation of the G-protein involved in vesicle formation is expected to impair vesicle formation and transport. Consequently, COPI-coated vesicles would be expected to decrease in number rather than accumulate in the cytoplasm. c. They will fuse into one giant vesicle that was seen in the cytoplasm: COPI-coated vesicles are discrete transport carriers responsible for the specific transport of cargo molecules. Treating the cell with GTP analogues that cannot trigger the active conformation of the G-protein involved in vesicle formation would likely disrupt proper vesicle formation and lead to a decrease in the number of vesicles. However, the notion of fusion into one giant vesicle is not expected. d. They will decrease substantially in number in the nucleus: COPI-coated vesicles do not typically have a significant presence in the nucleus. Treating the cell with GTP analogues that cannot trigger the active conformation of the G-protein involved in vesicle formation may affect vesicle formation and transport, but it would not lead to a substantial decrease in the number of COPI-coated vesicles in the nucleus. e. They will decrease substantially in number in the cytoplasm: This is the correct answer. Treating the cell with GTP analogues that cannot trigger the active conformation of the G-protein involved in vesicle formation is expected to impair vesicle formation and transport. As a result, the number of COPI-coated vesicles in the cytoplasm would decrease substantially.

23. Place the following events in the proper order.1 - Cyclic AMP activates one or more cellular signaling proteins.2 - Gα-subunit with its attached GTP activates the effector adenylyl cyclase.3 - Conformational change in the Gα subunit causing a decreased affinity for the Gβγ subunit.4 - Production of a second messenger, like cAMP.5 - Replacement of GDP by GTP on the Gα after interaction with an activated GPCR. 6 - Dissociation of Gα from the G protein complex.

5 - 6 - 3 - 2 - 4 - 1

79. Among these five M-phase events, which happens third? a. nuclear envelope breakdownb. spindle assembly checkpointc. microtubule aster separation d. the contractile ring constricts the cleavage furrowe. chromosome movement to poles and spindle pole separation

Apologies for the confusion. Let's go through each answer choice and explain why they are correct or incorrect: b. Spindle assembly checkpoint: The spindle assembly checkpoint is a regulatory mechanism that ensures proper attachment and alignment of chromosomes to the spindle fibers before the onset of anaphase. It occurs during metaphase, after nuclear envelope breakdown. The spindle assembly checkpoint serves as a surveillance mechanism to prevent premature separation of sister chromatids. Since it occurs after the events of nuclear envelope breakdown and microtubule aster separation, it can be considered the third event in M-phase. Therefore, answer b is correct. a. Nuclear envelope breakdown: This event occurs at the beginning of M-phase, specifically during prophase. The nuclear envelope disassembles, allowing the spindle apparatus to interact with the chromosomes. It happens before the third event, so it is incorrect. c. Microtubule aster separation: Microtubule asters are radial arrays of microtubules that form around the centrosomes during cell division. They are involved in spindle formation and positioning. The separation of microtubule asters occurs during late prophase or prometaphase, which is earlier in M-phase than the third event. Therefore, it is incorrect. d. The contractile ring constricts the cleavage furrow: The contractile ring is responsible for dividing the cytoplasm during cytokinesis, which occurs after the completion of mitosis. It is the final event of cell division, following the separation of chromosomes and the movement of chromosomes to the poles. Therefore, it is not the third event in M-phase, so it is incorrect. e. Chromosome movement to poles and spindle pole separation: This event occurs during anaphase, following metaphase. The spindle fibers pull the sister chromatids apart, moving them toward opposite poles of the cell. It is later in M-phase than the third event, so it is incorrect. To recap, the correct answer is b. Spindle assembly checkpoint because it occurs after nuclear envelope breakdown and microtubule aster separation. The incorrect answers are a. Nuclear envelope breakdown, c. Microtubule aster separation, d. The contractile ring constricts the cleavage furrow, and e. Chromosome movement to poles and spindle pole separation because they either occur earlier or later in M-phase than the third event.

64. A new organism that has only one motor protein was discovered in a water fountain in the Biology/Physics Building. To characterize its activity, you coat glass coverslips with the motor protein so that its tail is stuck to the glass, while its head is exposed. You then treat the coverslips with actin filaments and microtubules and find that the actin filaments move with their (+) ends leading and that microtubules move with their (-) ends leading. This motor behaves like a...

Based on the observations that actin filaments move with their (+) ends leading and microtubules move with their (-) ends leading, we can determine the behavior of the motor protein. Actin filaments are associated with myosin motor proteins, and myosins generally move towards the plus (+) end of actin filaments. Therefore, the motor protein in question behaves as a plus-end directed myosin. On the other hand, microtubules are associated with kinesin motor proteins, and kinesins generally move towards the plus (+) end of microtubules. Therefore, the motor protein in question behaves as a minus-end directed kinesin. a. plus-end directed myosin and a minus-end directed kinesin: This option suggests that the motor protein behaves as both a plus-end directed myosin and a minus-end directed kinesin. However, the given observations indicate that actin filaments move with their (+) ends leading, which is consistent with the behavior of a plus-end directed myosin. It does not support the idea of a minus-end directed kinesin. c. plus-end directed myosin and a plus-end directed kinesin: This option suggests that both the actin filaments and microtubules move with their (+) ends leading, which contradicts the given observations. The actin filaments are observed to move with their (+) ends leading, but the microtubules move with their (-) ends leading. Therefore, this option is incorrect. d. minus-end directed myosin and a minus-end directed kinesin: This option suggests that both the actin filaments and microtubules move with their (-) ends leading. However, the given observations indicate that actin filaments move with their (+) ends leading, which is inconsistent with the behavior of a minus-end directed myosin. Additionally, microtubules move with their (-) ends leading, suggesting the involvement of a plus-end directed kinesin. Therefore, this option is incorrect. e. macromolecule that moves intermediate filaments: This option suggests that the motor protein is involved in the movement of intermediate filaments. However, the given observations describe the movement of actin filaments and microtubules, not intermediate filaments. Therefore, this option is incorrect.

88. What do CDKs that are activated just before the end of G2 do to initiate the next phase of the cell cycle?

Certainly! After CDKs are activated and phosphorylate substrates needed for the cell to enter mitosis, several key events occur: 1. Activation of the mitotic checkpoint: CDK-mediated phosphorylation leads to the activation of the mitotic checkpoint, which ensures that all chromosomes are properly aligned on the mitotic spindle before the onset of anaphase. 2. Chromosome condensation: CDKs phosphorylate specific proteins involved in chromosome condensation, promoting the compaction of chromatin into visible individual chromosomes. 3. Nuclear envelope breakdown: CDKs phosphorylate nuclear envelope proteins, causing the disassembly of the nuclear envelope, which allows the chromosomes to interact with the mitotic spindle. 4. Spindle assembly: CDKs contribute to the assembly and stabilization of the mitotic spindle by phosphorylating proteins involved in microtubule dynamics and organization. 5. Entry into mitosis: Collectively, these phosphorylation events and associated processes initiated by CDKs lead to the cell's entry into mitosis, which is the M-phase of the cell cycle. Now let's discuss why the other answer options are incorrect: b. They de-phosphorylate substrates needed for the cell to enter mitosis: CDKs are primarily involved in phosphorylation events and driving the cell cycle forward. They do not generally have a role in de-phosphorylating substrates required for mitotic entry. c. They ubiquitinate substrates needed for the cell to enter mitosis: Ubiquitination is a process involved in protein degradation and regulation, but it is not the primary mechanism by which CDKs initiate the next phase of the cell cycle. d. They phosphorylate lipids needed for the cell to enter mitosis: CDKs primarily phosphorylate proteins rather than lipids. Their main role is to regulate protein activity and cellular processes. e. They act as proteases to degrade proteins that inhibit mitosis: CDKs themselves are not proteases. However, they can indirectly regulate the degradation of specific proteins through phosphorylation events that mark them for subsequent proteasomal degradation. CDKs do not directly act as proteases to degrade proteins inhibiting mitosis. Therefore, the correct answer is a. They phosphorylate substrates needed for the cell to enter mitosis, as CDKs play a critical role in phosphorylating proteins to initiate mitotic entry and progression.

86. ___ is an enzyme that interacts with growth factor receptors and phosphorylates lipids to trigger signaling cascades that influence cell growth, while ___ is a type of enzyme that phosphorylates protein targets that act to drive cell cycle transitions.

Certainly! Here's an explanation for why option b is correct and why the other options are incorrect: b. PI 3-kinase, CDK: PI 3-kinase is an enzyme that phosphorylates lipids in response to growth factor signaling, triggering downstream signaling cascades involved in cell growth and survival. CDK (cyclin-dependent kinase) is a type of enzyme that phosphorylates protein targets to regulate cell cycle transitions. The statement correctly describes the roles of PI 3-kinase and CDK in cellular processes related to growth factor signaling and cell cycle regulation. a. PI 3-kinase, mTOR: While both PI 3-kinase and mTOR (mechanistic target of rapamycin) are involved in growth factor signaling and cell growth, mTOR is not specifically an enzyme that phosphorylates lipids. Instead, mTOR is a protein kinase that regulates cell growth, protein synthesis, and other cellular processes in response to various signals. c. PI 3-kinase, cyclin: Cyclins are regulatory proteins that bind to and activate CDKs during specific stages of the cell cycle. They are not directly related to PI 3-kinase or lipid phosphorylation. d. PI(3,4,5)P3, PI 3-kinase: This option is incorrect because PI(3,4,5)P3 is a lipid molecule produced by PI 3-kinase through the phosphorylation of phosphoinositides. PI 3-kinase itself is the enzyme responsible for this lipid phosphorylation, rather than being the product of it. e. PI(3,4,5)P3, CDK: PI(3,4,5)P3 is a lipid product generated by PI 3-kinase, as mentioned earlier, and CDK is a protein kinase involved in cell cycle regulation. However, there is no direct link between PI(3,4,5)P3 and CDK in terms of their functional interaction or role in signaling pathways. Therefore, option b (PI 3-kinase, CDK) correctly identifies the enzyme involved in lipid phosphorylation and the enzyme involved in protein phosphorylation for cell growth and cell cycle regulation, respectively.

70. Which specialized attachment(s) allow(s) epithelial cells to act as an occluding barrier layer that can normally prevent penetration of bacteria and other substances into deeper tissue? a. tight junctionsb. desmosomesc. gap junctionsd. adherens junctionse. a, b, and c all act as epithelial barriers

Certainly! Let's go through each option and explain whether it is correct or incorrect in the context of epithelial cells acting as an occluding barrier layer: a. Tight junctions: Tight junctions are specialized attachments between epithelial cells that form a tight seal between adjacent cells. They play a crucial role in creating a barrier that prevents the passage of substances between cells, including bacteria and other substances. Therefore, option a (tight junctions) is correct. b. Desmosomes: Desmosomes are intercellular junctions that provide mechanical strength and adhesion between epithelial cells. While desmosomes contribute to the overall integrity of epithelial tissue, they do not play a direct role in forming a barrier against the penetration of bacteria and other substances. Therefore, option b (desmosomes) is incorrect. c. Gap junctions: Gap junctions are intercellular channels that allow direct communication and passage of small molecules between adjacent cells. They do not contribute significantly to the formation of a physical barrier against the penetration of bacteria and other substances. Therefore, option c (gap junctions) is incorrect. d. Adherens junctions: Adherens junctions are intercellular attachments that provide mechanical support and adhesion between epithelial cells. While they contribute to the overall integrity of epithelial tissue, they do not play a direct role in forming a barrier against the penetration of bacteria and other substances. Therefore, option d (adherens junctions) is incorrect. e. Option e (a, b, and c all act as epithelial barriers): This option is incorrect because, as discussed, only option a (tight junctions) is directly involved in forming the occluding barrier layer that prevents penetration of bacteria and other substances into deeper tissue. Desmosomes, gap junctions, and adherens junctions do not have the same barrier function. In summary, the correct answer is option a (tight junctions). Tight junctions play a vital role in creating an occluding barrier layer that prevents the passage of bacteria and other substances between epithelial cells, ensuring the integrity and protection of deeper tissues.

21. A given protein is regulated by a protein kinase and a protein phosphatase, and is active when phosphorylated. Which would you expect to INCREASE the protein's activity?

Correct: b. Expression of a mutant nonfunctional, dominant-negative phosphatase that blocks the activity of the cell's normal phosphatases -mutated could prevent activity of protein if mutated allele is present, kinase phosphorylates the protein making It active(hyperactive kinases will continually phosphorylate kinases, dephosphorylatini increases activity of the protein Incorrect: a. Expression of a mutant nonfunctional, dominant-negative kinase that blocks the activity of the cell's normal kinases -expression of a mutant non-functional, blocks and inhibit kinase activity that would decrease protein activity instead of increasing it, without phosphyorylaition, the protein will remain in an inactive state c. Expression of a mutant hyperactive phosphatase. -hyperactive phosphatase removes phosphorylation from protein, increased activity of phosphatase results for more dephosphorylation, and decreased protein activity, that is in an inactive state. d. a and b are correct. e. b and c are correct.

65. Specialized cells called margaritacytes respond to changes in ethanol (EtOH) concentration by reorganizing their detoxifying granules along microtubules as shown in the figure below. Imagine that you shift cells from media containing low levels of EtOH to media with high levels of EtOH. Which type of drug would prevent the expected movement of granules in the cell?

Cytoplasmic dynein is a motor protein involved in the movement of cargo along microtubules towards the minus (-) end. In the given scenario, the reorganization of detoxifying granules along microtubules is observed in response to changes in ethanol concentration. This movement is likely mediated by cytoplasmic dynein, which transports the granules towards the minus (-) end of microtubules. If a drug blocks the activity of cytoplasmic dynein, it would prevent the expected movement of granules in the cell. The granules would not be able to move along microtubules towards the minus (-) end as they normally would, resulting in the disruption of their reorganization in response to changes in ethanol concentration. The other options are incorrect because: a. a drug that increases the activity of an unconventional myosin: Unconventional myosins are associated with actin filaments rather than microtubules. They are not directly involved in the movement of cargo along microtubules. b. a drug like taxol, which stabilizes microtubules: Stabilizing microtubules with drugs like taxol would not necessarily prevent the movement of granules along microtubules. It may affect microtubule dynamics but would not specifically inhibit the movement of granules. d. a drug that blocks the activity of conventional kinesin: Conventional kinesin is a motor protein involved in the movement of cargo towards the plus (+) end of microtubules. In the given scenario, the movement of granules is expected towards the minus (-) end of microtubules. Blocking the activity of conventional kinesin would not directly prevent the expected movement of granules. e. a drug that blocks the activity of an unconventional myosin: Unconventional myosins are associated with actin filaments and are not directly involved in the movement of cargo along microtubules. Blocking the activity of an unconventional myosin would not prevent the expected movement of granules along microtubules.

22. Ras is a small G protein that regulates cell proliferation in response to growth factors. When Ras is in its active form, cells proliferate. The activity of Ras is carefully regulated by two other proteins: Ras GEF (guanine nucleotide exchange factor) which stimulates binding of GTP by Ras, and Ras GAP (GTPase activating protein) which stimulates GTP hydrolysis by Ras. The activities of these regulatory proteins are, in turn, also regulated. Ras activity stimulates cell proliferation. Which of the following changes in GAP and GEF proteins might cause a cell to increase proliferation?

Correct: a. A nonfunctional GAP. -when non-functional GAP is present, fails to simulate GTP hydrolysis by RAS, leading to sustained activation by RAS and cause prolonged RAS signalling for cell proliferation . d. A permanently active GEF. -permanently active GEF present, stimulates binding of GTP by RAS, promoting activation, and enhance raw signalling for cell proliferation. e. a and d are true. -both combined has regulation of RAS activity would be disrupted in favorof sustained activation, in active GTP-bound state for longer duration and increased has signalling and promoting cell proliferation. Incorrect: b. A permanently active GAP. -GAP protein permanently active that stimulates GTP hydrolysis by RAS, Was inactivated the signalling for cell proliferation would be suppressed rather than enhanced, permanently active gap is not expected to increase cell proliferation. c. A nonfunctional GEF. -if GEF protein is non-functional and fails to stimulate binding to GTP by Ras, prevent activation of Raw; without Ras activation, signalling for cell proliferation would not occurs(non-functional GEF is not expected to increase proliferation

11. You express a recombinant membrane protein and put it into an artificial lipid bilayer. You tag it with a fluorescent antibody and measure its mobility using FRAP. You find that it is highly mobile. However, when you do the same experiment on a cell that naturally expresses that protein, you find that it is not mobile. Which explanation could account for these results?

Correct: a. In the cell, the protein binds to the cytoskeleton. -proteins bind to cytoskeleton that is a network of protein filaments inside cells that provides structural support and facilitates cellular movement, if the protein of interest binds to cytoskeleton in natural cell, restrict mobility and prevent from freely diffusing within the membrane. b. In the cell, the protein binds to other immobile membrane proteins. protein binds to other immobile membrane eptieins: membrane proteins often form complexes and interact with other immobile membrane proteins, restrict movement and hinder its mobility within the membrane. d. a and b could. Incorrect: c. In the cell, the protein binds to other mobile membrane proteins. -this does not restrict movement as mentioned in the question above e. a and c could.

46. Which of the models below suggests that the Golgi cisternae are transient structures that form at the cis face and travel through the Golgi complex, moving physically and exiting the organelle at the trans face, while their enzymes change during the journey?

Correct: a. The cisternal maturation model: This model suggests that the Golgi cisternae, which are flattened membrane-bound sacs within the Golgi complex, are transient structures that mature over time. According to this model, new cisternae are formed at the cis face (the entry side) of the Golgi complex, and as they mature, they gradually move through the Golgi stack, with their contents being modified and processed along the way. Eventually, the cisternae reach the trans face (the exit side) of the Golgi complex and shed off as vesicles to transport their modified cargo to their target destinations. In this model, the Golgi cisternae physically move and change their content during the journey through the Golgi complex. Incorrect: b. The cargo carrying model: This is not a widely recognized term or model in the context of Golgi function. It does not describe the dynamics or movement of Golgi cisternae or the changing of enzymes during transit. c. The vesicular transport model: This model describes the transport of cargo within vesicles that bud off from one compartment and fuse with another. In the context of the Golgi complex, it refers to the transport of materials between different cisternae or between the Golgi and other cellular compartments. However, this model does not specifically emphasize the maturation or changing of enzymes within the Golgi cisternae during their journey. d. The secretory transport model: This model primarily focuses on the transport of proteins and other molecules from the endoplasmic reticulum (ER) to the Golgi complex and from the Golgi complex to the plasma membrane or other secretory destinations. It does not specifically address the movement or maturation of Golgi cisternae or the changing of enzymes within them. e. The chemiosmotic model: This model is associated with energy production in mitochondria, specifically in the context of ATP synthesis through the electron transport chain. It does not pertain to the Golgi complex or its cisternae.

19. A neurotransmitter that binds to a postsynaptic cell and opens K+ channels.

Correct: b. inhibits the postsynaptic cell. -neurotransmitter that binds to a postsynaptic cell and opens K+ channels would typically lead to an efflux of potassium ions, is hyperpolarization of post-synaptic cell, K+ channels open and K+ ions are allowed to leave the cell, causing membrane potential to become more negative or depolarised reduces excitability and inhibits its activity Incorrect: a. excites the postsynaptic cell. -opening K+ channels an allowing K+ to leave the cell is hyperpolarization and difficult for post-synaptic cell to generate an action potential to generate an action potential; hyperpolarization inhibits the postsynaptic cell's excitability rather than exciting it. c. depolarises the postsynaptic cell. opening K+ channels and allowing potassium ions to leave the cell leads to hyperpolarization, not depolarisation. Depolarisation refers to a shift in the membrane potential towards a more positive value, opposite to what occurs when K+ channels open. d. must be acetylcholine. -identity of neurotransmitter that K+ channels is not specified or limited to acetylcholine, nervous system and different neurotransmitters can have diverse effects on post-synaptic cells

9. You determine the amino acid sequence of a novel protein. You find that it has a single alpha helical domain. The alpha helix has primarily hydrophobic amino acids on one face of the helix, and primarily polar (hydrophilic) amino acids on the other (an amphipathic helix). Which statement(s) is/ are false?

Correct: c. This protein is almost certainly an isolated single pass transmembrane protein. Incorrect: a.This protein cannot exist as an isolated single pass transmembrane protein. b.This protein cannot be a membrane protein. This protein could be part of a multimeric protein complex that spans the bilayer to allow transport of hydrophilic molecules through the lipid bilayer. b and c are false statements.

10. You engineered a new gene which includes GFP fused to the cytoplasmic domain of a transmembrane protein. You then added a cardiac-specific promoter and incorporate this new gene into the genome of the mouse. When you examine cells from these mice in the fluorescent microscope:

Correct: c. You will see the fluorescence from the protein in the membrane of all cardiac cells -engineered a new gene gene that has gfp in alpha-transmembrane that is transported into the mouses genome 1. cardiac specific motor: promotor used specific to cardiac expression in specific cells, protein fused with GFP, produced in cardiac cells. 2. transmembrane protein: cytoplasmic domain of transmembrane protein used responsible to anchor protein in cell membrane, linked GFP to cytoplasmic domain(effectively linked GFP to cytoplasmic domain linked GFP to transmembrane protein, presence of cell membrane. 3. Fluorescnece visualization: GFP is a fluorescent protein that emits green fluorescence when exposed to specific wavelengths of light, fuse GFP to cytoplasmic domain of a transmembrane protein protein will have to attach to it(GFP fluorescence will be visible in the cell membranes of all cardiac cells) Incorrect: a. You will see the fluorescence throughout the cytoplasm of all the cells of the mouse. b. You will see the fluorescence throughout the cytoplasm of all cardiac cells in the mouse. d. You will see the fluorescence in the membranes of all the cells of the mouse. e. All of the above will be seen.

55. The Critical concentration (Cc) of a newly-discovered cytoskeletal polymer is 2.3μΜ at one end and 7.7μΜ at the other end. These polymers will...

Correct: c. undergo treadmilling when present at 6μΜ. Treadmilling is a phenomenon observed in cytoskeletal polymers where subunits are simultaneously added to one end of the polymer while being removed from the other end, resulting in a net movement of the polymer without a change in overall length. In the given scenario, the critical concentration (Cc) of the cytoskeletal polymer is 2.3μΜ at one end and 7.7μΜ at the other end. This means that at concentrations below 2.3μΜ, the polymer will depolymerize, and at concentrations above 7.7μΜ, the polymer will polymerize. When the polymer is present at 6μΜ, which falls between the critical concentrations of 2.3μΜ and 7.7μΜ, it is within the range where the polymer can exist in both a polymerized and depolymerized state. This condition allows for the simultaneous addition of subunits at one end and removal of subunits at the other end, leading to treadmilling. Therefore, the correct answer is c. undergo treadmilling when present at 6μΜ. incorrect: a. The statement "exhibit dynamic instability when present at 1.5μΜ" is incorrect because the critical concentration of the polymer at one end is 2.3μΜ. Dynamic instability refers to the phenomenon where microtubules, for example, undergo cycles of polymerization and depolymerization. However, in this case, the concentration of 1.5μΜ is below the critical concentration of the polymer at both ends, so it would not exhibit dynamic instability. b. The statement "have a critical concentration of 7.7μΜ at the plus end" is incorrect because the critical concentration is not specific to the plus end of the polymer. The given information states that the critical concentration is 7.7μΜ at one end and 2.3μΜ at the other end, indicating different critical concentrations at the two ends. d. The statement "shrink from both ends when present at 8μΜ" is incorrect because the concentration of 8μΜ is above the critical concentration at both ends. When the concentration is above the critical concentration, polymerization occurs. Therefore, the polymer would not shrink from both ends at this concentration. e. The statement "grow from both ends when present at 5μΜ" is incorrect because the concentration of 5μΜ is below the critical concentration at both ends. When the concentration is below the critical concentration, depolymerization occurs. Therefore, the polymer would not grow from both ends at this concentration.

14. A membrane-associated protein is isolated and found to contain multiple alpha helices consisting of many hydrophobic amino acids. This protein is most likely a:

Correct: d. Transmembrane protein that spans the membrane more than once. -have a hydrophobic amino acid is a transmembrane protein that spans the membrane more than once, also hydrophobic charges within the lipid bilayer of the membrane, while hydrophilic regions are exposed to the aqueous environments on both sides of the membrane. Incorrect: a. Transmembrane protein that spans the membrane once. -the question asks for "multiple" alpha helices, if it was one transmembrane protein that spans the membrane once but miltiple alpha helices suggest that proteins spans the membrane more than once. b. Peripheral membrane protein. -do not penetrate the membrane or span lipid bilayer of membrane, occur through electrostatic interactions or binding other membrane proteins or lipids, multiple alpha helices with hydrophobic amino acids suggests a transmembrane arrangement than a peripheral association c. Lipid-anchored protein. -attached to membrane through a covalently linked lipid molecule, they do not a transmembrane domain or segments, since the given protein has multiple alpha helices that span the membrane. e. None of the above.

48. Which coated vesicles are NOT involved in moving materials in a retrograde direction from the Golgi complex to the ER?

Correct: e. a and c. a. COPII-coated vesicles: COPII-coated vesicles are involved in anterograde transport, meaning they transport materials from the endoplasmic reticulum (ER) to the Golgi complex. These vesicles play a role in exporting newly synthesized proteins and lipids from the ER to the Golgi for further processing. c. Clathrin-coated vesicles: Clathrin-coated vesicles are mainly involved in transporting materials between the trans-Golgi network (TGN) and endosomes, as well as between endosomes and the plasma membrane. They are also involved in receptor-mediated endocytosis, bringing molecules from the extracellular space into the cell. However, they are not primarily responsible for retrograde transport from the Golgi complex to the ER. Incorrect: b. COPI-coated vesicles: COPI-coated vesicles are specifically involved in retrograde transport, meaning they transport materials from the Golgi complex back to the ER. They play a crucial role in retrieving and recycling Golgi resident proteins and enzymes, as well as maintaining the integrity and proper functioning of the Golgi complex.

47. What is the function of the protein coat on budding vesicles?

Correct: e. a and d. The protein coat on budding vesicles serves multiple functions: a. It acts as a mechanical device that helps to form the vesicle: The protein coat plays a crucial role in the initial stages of vesicle formation. It deforms the donor membrane, causing it to curve and bud outwards to form a vesicle. The coat proteins help in shaping the vesicle by providing structural support and aiding in membrane curvature. d. It provides a mechanism for selecting components and cargo to be carried by each vesicle: The protein coat acts as a sorting mechanism, determining which molecules and cargo are incorporated into the vesicle during its formation. The coat proteins interact with specific cargo molecules and membrane receptors, ensuring the selective packaging of specific cargo into the vesicle. Incorrect: b. It protects the forming vesicle from osmotic shock: The protein coat on budding vesicles does not provide protection against osmotic shock. Osmotic shock refers to a sudden change in osmotic pressure, which could cause damage to cellular structures. The primary function of the protein coat is related to vesicle formation and cargo selection, not osmotic protection. c. It prevents the diffusion of harmful materials into the forming vesicle: While the protein coat may selectively incorporate specific cargo into the vesicle, its primary role is not to prevent the diffusion of harmful materials. The selective packaging of cargo is mainly involved in ensuring the proper sorting and transport of specific molecules, rather than providing a general barrier against harmful substances.

13. With regard to the diagram above, which statement(s) is/ are correct?

Correct: e. None are correct. not enough information to specify values or interaction strengths for labeled interaction strengths to compare affinity(Kd) or interaction to strengths for labeled interactions. Incorrect: a. Interaction (i) has the smallest Kd. b. Interaction (i) has the highest affinity.c. Interaction (iii) has the lowest affinity. d. Interaction (iii) has the greatest Kd.

12. You are studying the cellular basis of protein synthesis. You have two proteins- protein A and protein B- and you know the following about them. Protein A adopts different conformations if it is translated in a test tube with purified ribosomes, mRNA, tRNA's, and amino acids than if it is translated in vivo. When protein B is immuno- isolated from cells and denatured by heating in a test-tube, it spontaneously refolds into the right conformation. Which of the statements below is correct?

Correct: a. Protein A but not protein B likely folds with the aid of chaperones and chaperonins in cells. -protein A adopts to different confirmations when translated in a test-tube compared to in vivo suggestions that required involvement of chaperones and chaperonins to assist folding in cells, protein B spontaneously refolds into the right confirmation when denatured, indicating not need these for proper refolding Incorrect: b. Protein B but not protein A likely folds with the aid of chaperones and chaperonins in cells. -protein b is denatured by heating in a test tube, spontaneously refolds into correct confirmation without the aid of chaperones and chaperonins to fold properly. c. Both proteins require chaperones and chaperonins to fold properly. d. Neither protein appears to require chaperones and chaperonins to fold properly.

20. Increasing the activity of vesicular H+-ATPases would:

Correct: b. Decrease pH inside of vesicles. c. Increase the pH of the cytoplasm. e. b and c -passive ions move down electrochemical gradient passive uniporter carriers move one species of molecule -passive move down concentration gradient from low to high Side carrier protein flip back/forth, in.out increases the pH of the lumen. H+ ATPases have the following characteristics: 1. H+ ATPases present in plasma membranes of plants 2. ATP powered pumps, an active transport mechanism 3. pumps protons(H+) against the concentration gradient(active), proton motive force(pmf) is a secondary active transporters(small molecules and ions) 4. primary active transport(direct transport) in plants, these H+ ATPases present on vesicular plasma membrane(functions 1. pH regulation, H+ ATPase optimum pH of 7-7.5(slightly alkaline) of cytoplasm and acidic pH(4.5-5) inside the vesicle 2. salt tolerance, also maintaining the gradient(inside the acidic and outside slightly alkaline) by using ATP Incorrect: a. Increase pH inside of vesicles. d. Decrease the pH of the cytoplasm.

Which is NOT true of phospholipids in a bilayer?

Correct: b. Phospholipids can rotate rapidly around their long axis. flexible tails shaking around, thermal movement occurs rapid rate, spin 180 degrees is favourable and lateral diffusion core leaflet movement moves laterally among neighbours, hydrophobic had to off tails of favorable e. a and c are not true. phosphoglycoglyceraldyhydes(one tail is saturated, carbon from a double bondnhas one hydrogen is unsaturated; has a kink has a wider tail region and overall height -liposome has hydrophobic tail touch water, or inside is in aqueous solution and fatty acid tails of each leaflet face each other transverse: move from hydrophobic region to hydrophilic, not thermodynamically favoured Incorrect: a. Phospholipids are uniform in tail length, making the bilayer a uniform thickness. phospholipid is shorter than sphynglycotion tail(cholesterol= interference tail structure has kinked shorter tails) c. Phospholipids can easily move from one leaflet to the other leaflet by diffusion. d. a and b are not true.

17. Of the steps of the action potential listed, which is the fourth after a cell is stimulated by a current-passing electrode?

Correct: b. the membrane becomes repolarized. Incorrect: 1)a. Voltage-gated sodium channels open. 2)c. the membrane rapidly depolarises near the sodium equilibrium potential. 3)d. V-gated Na+ channels inactivate and V-gated K+ channels begin to open. 5)e. Voltage-gated K+ channels close.

18.Substance X has a large uncharged polar molecule. It is found at higher concentrations inside of the cells than outside. How might substance X enter cells?

Correct: c. Active transport -need energy use, large uncharged polar molecules and is found at higher concentrations inside cells than outside is most likely mechanism than outside, the most likely mechanism and active transport. -active transport mechanisms can utilise energy to transport large polar molecules against the concentration gradient, against the concentration gradient, allowing substance A to enter cels through it is uncharged and polar. Incorrect: a. Diffusion through a channel. -passive use ions, uncharged electrons, is zero left with chemical so is a chemical and active transport needs energy for active transport b. Diffusion directly through a lipid bilayer. d. b and c are correct. e. none of the above.

7. Two cells with different cell-surface markers are fused by the addition of polyethylene glycol (PEG). The cells are then placed at 0oC. What do you expect to observe about the individual cell markers?

Correct: d. The markers will essentially remain where they are, with little migration Incorrect a. The markers will evenly disperse throughout both membranes.b. Only one marker will disperse while the other remains stationary.c. Both sets of markers will not mix, but will migrate to opposite poles from one another..e. The markers will be endocytosed by the fused cell and then redistributed as fused markers.

63. All of the following statements about cytoskeletal filaments and motor proteins are true except for one. Which statement is false? a. all cytoskeletal motor proteins use ATP hydrolysis to power movementb. actin filaments are polar and associate with motor proteinsc. microtubules are polar and associate with motor proteinsd. intermediate filaments are non-polar and do not associate with motor proteins e. some cytoskeletal motor proteins use GTP hydrolysis to power movement

Correct; . some cytoskeletal motor proteins use GTP hydrolysis to power movement. Explanation: Cytoplasmic motor proteins, which include kinesins and dyneins, utilize ATP hydrolysis, not GTP hydrolysis, to power their movement along cytoskeletal filaments. ATP hydrolysis provides the energy required for the conformational changes in motor proteins that enable their movement along actin filaments or microtubules. Therefore, statement e is false. Incorrect: a. all cytoskeletal motor proteins use ATP hydrolysis to power movement: This statement is true. Cytoskeletal motor proteins, including kinesins, dyneins, and myosins, use ATP hydrolysis to power their movement along cytoskeletal filaments. ATP hydrolysis provides the energy needed for the conformational changes in motor proteins required for their movement. b. actin filaments are polar and associate with motor proteins: This statement is true. Actin filaments are polar structures, with distinct plus (+) and minus (-) ends. Motor proteins such as myosins associate with actin filaments and move along them, utilizing ATP hydrolysis to generate force and perform various cellular functions. c. microtubules are polar and associate with motor proteins: This statement is true. Microtubules are polar structures, with a distinct plus (+) end and a minus (-) end. Motor proteins such as kinesins and dyneins associate with microtubules and utilize ATP hydrolysis to generate force and perform functions such as intracellular transport and cell division. d. intermediate filaments are non-polar and do not associate with motor proteins: This statement is false. Intermediate filaments are indeed non-polar structures, lacking intrinsic polarity like actin filaments and microtubules. However, they can associate with motor proteins such as dynein, which can interact with intermediate filaments to perform functions such as positioning organelles and facilitating cellular mechanics.

91. A graduate student wants to perform a genetic screen to identify cellular factors that promote cancer-like properties (transformation) in a cultured mammalian cell line. She designs a screen in which each cell receives a different small interfering RNA (siRNA) that will prevent that gene from being expressed in the cell. This screen will likely identify which type of gene(s)?

I apologize for the confusion caused by my previous response. Let's now address question 91. The correct answer is a. tumor-suppressor genes. Explanation: In the described genetic screen using siRNA to prevent gene expression, the aim is to identify cellular factors that promote cancer-like properties or transformation in a cultured mammalian cell line. Transformation refers to the acquisition of characteristics associated with cancer cells, such as uncontrolled proliferation, evasion of cell death, and enhanced migration. To identify genes that promote transformation, the screen typically focuses on genes whose loss of function leads to an increase in cancer-like properties. These genes are known as tumor-suppressor genes. Tumor suppressors normally function to regulate cell growth, prevent mutations, and inhibit the development of tumors. Loss or inactivation of tumor suppressor genes can contribute to cellular transformation and cancer development. By using siRNA to prevent the expression of different genes in the cell line, the screen will likely identify tumor-suppressor genes whose inhibition promotes cancer-like properties. The loss of these genes would lead to the acquisition of transformative characteristics. Now let's discuss the incorrect options: b. caretaker genes: Caretaker genes are involved in maintaining genomic stability and DNA repair mechanisms. While mutations in caretaker genes can increase the risk of developing cancer, they are not specifically associated with promoting cancer-like properties or transformation. Therefore, they are not the primary targets for identifying factors promoting transformation in this context. c. proto-oncogenes: Proto-oncogenes are genes that, when mutated or overexpressed, can promote cell proliferation and contribute to the development of cancer. However, in the given scenario, the aim is to identify factors that promote cancer-like properties, not those that are already known to have oncogenic potential. Therefore, the most appropriate choice for the type of gene(s) likely to be identified in this genetic screen is a. tumor-suppressor genes.

96. What has happened to your telomeres since May 12th, when you began taking Cell Biology?

I apologize for the incorrect response. Let me provide an explanation for why the other options are incorrect: a. They have gotten longer in my proliferating somatic cells: Telomeres typically shorten during each round of cell division in somatic cells, so they are unlikely to get longer. c. They have gotten longer in my senescing cells: Senescing cells generally exhibit telomere shortening, not lengthening. d. They have gotten longer in my necrotic cells: Necrotic cells undergo cell death and cellular breakdown, so changes in telomere length are not relevant in this context. e. They are the same length in all of my cells: While telomeres are usually maintained at a relatively stable length in most cells, they can undergo gradual shortening during cell divisions, so this option is not entirely accurate. Therefore, option b, "they have gotten shorter in my proliferating somatic cells," is the correct answer as telomeres tend to shorten with each round of cell division in somatic cells.

42. If you took a gene that is normally present in the mitochondrial genome and codes for an enzyme that works in the matrix, and then inserted the gene into the nucleus of a cell, what would you expect the final destination of the protein to be? a. ER lumen b. Nucleus. c. Secreted. d. Mitochondrial matrix e. None of the above

I apologize for the incorrect response. You are correct. If a gene that is normally present in the mitochondrial genome and codes for an enzyme in the mitochondrial matrix is inserted into the nucleus of a cell, the protein's final destination would not be any of the options provided. When a mitochondrial gene is transferred to the nucleus, the resulting protein product would typically not be targeted to the mitochondria. Mitochondrial protein targeting requires specific mitochondrial targeting signals that are typically encoded within the mitochondrial genome. Without these targeting signals, the protein would not be recognized and imported by the mitochondrial import machinery. Instead, the protein would likely be synthesized in the cytoplasm and would remain there or be directed to other cellular compartments based on other signals or sorting mechanisms present in the protein. Therefore, the correct answer is e. None of the above.

90. In the scenario depicted below, what would happen if the p16 protein was lost? a. Rb would not be phosphorylatedb. Rb would repress DNA synthesisc. CDK4 would be an inactive kinased. Cells would be stuck in G1e. Cells would enter S-phase prematurely

If the p16 protein is lost, the correct answer is e. Cells would enter S-phase prematurely. Let's understand why this is the case and why the other options are incorrect: a. Rb would not be phosphorylated: The loss of p16 would lead to increased activity of cyclin D-CDK4/6 complexes, which phosphorylate and inactivate Rb. Therefore, Rb would still be phosphorylated in the absence of p16. b. Rb would repress DNA synthesis: The loss of p16 would result in increased cyclin D-CDK4/6 activity, leading to phosphorylation and inactivation of Rb. Inactivated Rb cannot repress DNA synthesis, so this option is incorrect. c. CDK4 would be an inactive kinase: The loss of p16 would result in increased cyclin D-CDK4 activity, so CDK4 would not be inactive. This option is incorrect. d. Cells would be stuck in G1: The loss of p16 would relieve the inhibition on cyclin D-CDK4/6 complexes, allowing them to phosphorylate Rb and promote cell cycle progression. Cells would not be stuck in G1, so this option is incorrect. e. Cells would enter S-phase prematurely: The loss of p16 would result in increased cyclin D-CDK4/6 activity, leading to early phosphorylation and inactivation of Rb. Inactivated Rb releases E2F transcription factors, promoting the expression of genes required for S-phase entry. Therefore, cells would enter S-phase prematurely if p16 is lost. Therefore, the correct answer is e. Cells would enter S-phase prematurely.

60. If you microinject red-fluorescent tubulin subunits into a cell during interphase and fix them after a few minutes, how would the microtubules be labeled?

If you microinject red-fluorescent tubulin subunits into a cell during interphase and fix them after a few minutes, the microtubules would be labeled as follows: c. Only the (+) ends will be labeled red. During interphase, microtubules undergo dynamic instability, which involves cycles of polymerization and depolymerization. The growing ends of microtubules, known as the (+) ends, are more dynamic and exhibit higher rates of polymerization. When red-fluorescent tubulin subunits are microinjected, they preferentially incorporate into the (+) ends of the microtubules. Therefore, after fixing the cells, only the (+) ends of the microtubules will be labeled with red fluorescence. The rest of the microtubule structure, including the (-) ends and the microtubule lattice, will not be labeled with red fluorescence. Incorrect: a. None of the microtubules will be labeled red: This option is incorrect because when red-fluorescent tubulin subunits are microinjected, they will incorporate into the microtubules, specifically at the (+) ends. b. Only the (-) ends will be labeled red: This option is incorrect because the (-) ends of microtubules are typically associated with the microtubule organizing center (MTOC) or the centrosome, and not the site of active growth. Therefore, the (-) ends are less likely to be labeled with red fluorescence after microinjection of red-fluorescent tubulin subunits. d. The (-) and (+) ends will be labeled red, and the label at each end will be equivalent: This option is incorrect because the (-) ends are less likely to be labeled with red fluorescence. The (+) ends, on the other hand, are the sites of active growth and are more likely to incorporate the injected red-fluorescent tubulin subunits. e. The entire length of all the microtubules will be labeled red: This option is incorrect because the injected red-fluorescent tubulin subunits will preferentially incorporate into the (+) ends of the microtubules, where active polymerization occurs. The rest of the microtubule structure, including the (-) ends and the microtubule lattice, will not be labeled with red fluorescence.

71. Which of the following statements about cilia is true?

Let's evaluate each statement and determine which one is true regarding cilia: a. All cilia and flagella are sensory organelles: This statement is incorrect. While some cilia and flagella can have sensory functions, not all of them serve as sensory organelles. Cilia and flagella can have various functions, including sensory reception, locomotion, and fluid movement. b. Cilia and flagella both lack microtubule-based axonemes: This statement is incorrect. Cilia and flagella are both composed of microtubule-based structures called axonemes, which are responsible for their movement and motility. c. Intraflagellar transport (IFT) of vesicles can be driven by ciliary dynein: This statement is true. Intraflagellar transport (IFT) is a process by which vesicles and other components are transported along the length of cilia and flagella. This transport is facilitated by motor proteins, including ciliary dynein, which generates the force necessary for vesicle movement. d. Primary cilia lack ciliary dynein and are non-motile: This statement is incorrect. Primary cilia do possess ciliary dynein, although their main function is not motility. Primary cilia are involved in sensing environmental cues and playing important roles in signal transduction and cell signaling. e. Flagella lack cytoplasmic dynein and kinesin: This statement is incorrect. Flagella, like cilia, can utilize motor proteins such as cytoplasmic dynein and kinesin for their movement. These motor proteins play essential roles in generating the force required for flagellar movement. In summary, the correct statement is: c. Intraflagellar transport (IFT) of vesicles can be driven by ciliary dynein.

80. Which is the second event of those listed?a. de-phosphorylation of lamin Ab. centrosome migration and aster assembly c. separation of sister chromatidsd. chromosome capture and congressione. centrosome duplication

Let's go through each answer choice and determine the second event: a. De-phosphorylation of lamin A: This event refers to the removal of phosphate groups from lamin A, a component of the nuclear lamina. However, the de-phosphorylation of lamin A is not specifically associated with a particular phase or event in the cell cycle. Therefore, it is not the second event among the options provided. b. Centrosome migration and aster assembly: Centrosome migration and aster assembly occur during the early stages of cell division, specifically in prophase. The centrosomes move to opposite poles of the cell and establish the spindle apparatus. This event typically occurs before the separation of sister chromatids and can be considered the second event. Therefore, answer b is correct. c. Separation of sister chromatids: The separation of sister chromatids occurs during anaphase, which is a later stage of cell division. It follows the alignment of chromosomes at the metaphase plate. Therefore, it is not the second event among the options provided. d. Chromosome capture and congression: Chromosome capture and congression refer to the process of chromosomes aligning at the metaphase plate, facilitated by the attachment of spindle fibers to the kinetochores. This event occurs during metaphase and is typically preceded by centrosome migration and aster assembly. Therefore, it is not the second event. e. Centrosome duplication: Centrosome duplication is a process that occurs during the interphase of the cell cycle, specifically during the S phase. It involves the replication of the centrosome, which is crucial for spindle formation during cell division. Since it occurs prior to the events of cell division, it is not the second event. In summary, among the provided options, the second event is b. Centrosome migration and aster assembly. The incorrect answers are a. de-phosphorylation of lamin A, c. separation of sister chromatids, d. chromosome capture and congression, and e. centrosome duplication.

29. If a nuclear localization signal were added to the gene encoding a lysosomal protein, where would the protein likely be found?

Lysosomes are membrane-bound organelles responsible for the degradation and recycling of cellular waste material. Lysosomal proteins are specifically targeted to lysosomes through a process involving signals known as lysosomal targeting signals or sorting signals. When a protein is synthesized, it contains various signals or motifs that direct it to its appropriate subcellular location. In the case of lysosomal proteins, they typically possess specific lysosomal targeting signals. These signals can be amino acid sequences or structural motifs that are recognized by sorting machinery within the cell. The addition of a nuclear localization signal (NLS) to the gene encoding a lysosomal protein would introduce a sequence that is recognized by the nuclear import machinery, directing proteins to the nucleus. However, it does not alter the existing lysosomal targeting signals within the protein. In a situation where a lysosomal protein has both an NLS and the original lysosomal targeting signal, the protein would still primarily follow its original sorting pathway. The lysosomal targeting signal would guide the protein to the lysosomes, overriding the nuclear import signal. Therefore, despite the presence of the added NLS, the lysosomal protein would still be targeted to lysosomes due to the dominance of its native lysosomal targeting signal. As a result, the protein would likely be found in lysosomes rather than the nucleus. Incorrect: a. Nucleus: Adding a nuclear localization signal (NLS) to the gene encoding a lysosomal protein would not redirect the protein to the nucleus. The NLS would be recognized by importin proteins, which facilitate nuclear import, but the protein's native lysosomal targeting signals would override the NLS, guiding the protein to lysosomes instead of the nucleus. b. Cytoplasm: The cytoplasm is a general cellular compartment where various cellular processes occur. While the protein would pass through the cytoplasm during its journey to the lysosomes, it would not specifically localize there. The lysosomal targeting signals within the protein would direct it to lysosomes rather than the cytoplasm. c. Mitochondria: Lysosomal proteins are not typically targeted to mitochondria. Mitochondrial proteins possess their own specific targeting signals that direct them to the mitochondria. Adding an NLS to the gene encoding a lysosomal protein would not alter its targeting to lysosomes or redirect it to the mitochondria. e. Secreted from the cell: Lysosomal proteins are not secreted from the cell. They are primarily involved in intracellular degradation and recycling processes within lysosomes. The addition of an NLS to the gene encoding a lysosomal protein would not cause it to be secreted from the cell but rather guide it to lysosomes. In summary, the addition of an NLS to the gene encoding a lysosomal protein would not change its targeting to lysosomes. The protein would still be directed to lysosomes due to the presence of its native lysosomal targeting signals, overriding the NLS.

81. A graduate student wants to test whether beads coated with various biomolecules can generate microtubule spindles in Xenopus egg extracts. Which of the following beads would be likely to form spindles in extracts? a. beads coated with chromatin b. beads coated with RanGEF c. either a or bd. beads coated with RanGAP e. either a or d

Option c states that beads coated with either chromatin (option a) or RanGEF (option b) would be likely to form spindles in Xenopus egg extracts. Chromatin plays a critical role in spindle formation as it provides the necessary structure and organization for proper chromosome segregation. Additionally, RanGEF is involved in regulating various aspects of spindle assembly and dynamics. Therefore, beads coated with either chromatin or RanGEF have the potential to induce spindle formation in Xenopus egg extracts.

45.PTS1- and PTS2-bearing matrix proteins use:a. a common cytosolic receptor.b. a common translocation machinery on the peroxisomal membrane.c. a common receptor on the nuclear pore that catalyzes entry into the nucleus via pore targeting sequences.d. a common receptor protein within the peroxisomal matrix that activates protein processing for PTS1- and PTS2-bearing proteins.

PTS1- and PTS2-bearing matrix proteins use a common translocation machinery on the peroxisomal membrane. This machinery is responsible for recognizing and importing proteins containing either PTS1 or PTS2 targeting signals into the peroxisomal matrix. Once the PTS1- or PTS2-bearing proteins reach the peroxisomal membrane, a receptor protein on the membrane recognizes the specific targeting signal and facilitates the translocation of the protein into the peroxisome. This translocation machinery is shared by both types of matrix proteins, allowing them to be imported into the peroxisome. Therefore, the correct answer is b. a common translocation machinery on the peroxisomal membrane.

101. Which is NOT a direct advantage of having apoptotic cell death pathways in multicellular organisms?

The answer is d. elimination of functional neuronal cells. Explanation: Apoptotic cell death pathways play important roles in multicellular organisms by eliminating unwanted or potentially harmful cells. However, the elimination of functional neuronal cells is not a direct advantage of apoptosis. The other options are all direct advantages of having apoptotic cell death pathways in multicellular organisms: a. elimination of cells with the potential to cause cancer: Apoptosis helps to eliminate cells that have undergone genetic mutations or abnormalities, reducing the risk of cancer development. b. elimination of misplaced cells by developmental quality-control processes: During development, apoptosis helps remove cells that are in the wrong location or have not properly integrated into the tissue or organ, ensuring proper tissue organization. c. elimination of self-reactive immune cells that could cause autoimmunity: Apoptosis helps eliminate immune cells that recognize and attack the body's own tissues, preventing the development of autoimmune diseases. e. elimination of cells that are infected with a virus: Apoptosis can be triggered as a defense mechanism against viral infections, eliminating infected cells to limit the spread of the virus within the organism. In summary, while apoptosis serves several crucial roles in multicellular organisms, the elimination of functional neuronal cells is not one of its direct advantages.

93. Somatic cells can be reprogrammed to a stem cell like state by forcing the expression of 4 pluripotency genes in these cells. The resulting induced-pluripotent stem cells (iPS cells) resemble cancer cells in several ways. Why might iPS cells have cancer-like properties? a. one of the reprogramming factors is c-myc, a proto-oncogeneb. one of the reprogramming factors is c-src, a tyrosine kinasec. one of the reprogramming factors is Ras, a proto-oncogened. one of the reprogramming factors is Rb, a tumor-suppressor genee. one of the reprogramming factors nucleates actin in the somatic nucleus

The correct answer is a. One of the reprogramming factors is c-myc, a proto-oncogene. Explanation: Induced pluripotent stem cells (iPS cells) are generated by reprogramming somatic cells to a pluripotent state by expressing certain transcription factors. While iPS cells hold great potential for regenerative medicine, they do share some similarities with cancer cells. One of the reprogramming factors commonly used is c-myc, which is a proto-oncogene. Proto-oncogenes are genes that, when overexpressed or mutated, have the potential to become oncogenes and promote cell proliferation. c-myc is a transcription factor involved in cell cycle regulation, cellular growth, and differentiation. Its overexpression can disrupt normal cellular processes, leading to uncontrolled growth and potential transformation. The other options (b, c, d, e) are not directly associated with the cancer-like properties of iPS cells: b. c-src is a tyrosine kinase involved in cellular signaling, but its presence alone does not confer cancer-like properties to iPS cells. c. Ras is another proto-oncogene involved in cell signaling, but its presence alone does not explain the cancer-like properties of iPS cells. d. Rb (Retinoblastoma) is a tumor suppressor gene that regulates cell cycle progression. Its inclusion in reprogramming factors does not directly contribute to cancer-like properties. e. Nucleation of actin by one of the reprogramming factors is not directly linked to cancer-like properties of iPS cells. Therefore, the most significant factor contributing to the cancer-like properties of iPS cells is the presence of c-myc, a proto-oncogene, among the reprogramming factors.

104. Suppose you treat a culture of human cells with mutagenic ultraviolet (UV) radiation and you want to determine how many cells have initiated apoptosis and how many have not. Which of the following features would be present in the normal (non- apoptotic) cells?

The correct answer is a. cytochrome c will be found in mitochondria. Explanation: In normal (non-apoptotic) cells, cytochrome c is predominantly located within the mitochondria. Cytochrome c plays a crucial role in the electron transport chain and energy production within mitochondria. During apoptosis, however, there is a disruption of the mitochondrial membrane integrity, leading to the release of cytochrome c from the mitochondria into the cytoplasm. This release of cytochrome c triggers a series of events in the apoptotic pathway. Let's examine the other options and explain why they are incorrect: b. cytochrome c will be found in the cytoplasm: This option is incorrect because in normal cells, cytochrome c is primarily localized within the mitochondria and not in the cytoplasm. The presence of cytochrome c in the cytoplasm is indicative of mitochondrial membrane permeabilization and initiation of apoptosis. c. phosphatidylserine will be found in mitochondria: This option is incorrect because phosphatidylserine is a phospholipid primarily located in the inner leaflet of the plasma membrane. Translocation of phosphatidylserine to the outer leaflet of the plasma membrane is one of the hallmarks of apoptosis, not a feature of normal cells. d. phosphatidylserine will be found in the cytoplasm: This option is incorrect because phosphatidylserine is primarily located in the inner leaflet of the plasma membrane. During apoptosis, phosphatidylserine is exposed on the outer leaflet of the plasma membrane, which serves as an "eat me" signal for phagocytic cells. e. cytochrome c will be found in the outer leaflet of the plasma membrane: This option is incorrect because cytochrome c is normally localized within the mitochondria and not in the outer leaflet of the plasma membrane. The presence of cytochrome c in the outer leaflet of the plasma membrane is not a characteristic of normal cells. In summary, in normal (non-apoptotic) cells, cytochrome c is found in the mitochondria. During apoptosis, cytochrome c is released from the mitochondria into the cytoplasm, which is a critical step in initiating the apoptotic process.

100. ____ are the ____ enzymes that drive the process of apoptosis.

The correct answer is b. Caspases, protease. Caspases are a group of protease enzymes that play a central role in the process of apoptosis, which is programmed cell death. They are responsible for cleaving various cellular proteins and initiating the biochemical and morphological changes associated with apoptosis. Caspases are key regulators of the apoptotic pathway and are considered the primary executioners of apoptosis. a. Separases, protease: Separases are enzymes involved in the separation of sister chromatids during cell division (mitosis and meiosis), not in apoptosis. They cleave the protein cohesion, which holds the sister chromatids together. c. Cdc25 & Cdc14, phosphatase: Cdc25 and Cdc14 are phosphatases that regulate the progression of the cell cycle, not apoptosis. They are involved in controlling the activation of cyclin-dependent kinases (CDKs) by dephosphorylating them. d. Caspases, protein kinase: Caspases are proteases, not protein kinases. Protein kinases are enzymes that catalyze the phosphorylation of proteins, while caspases are responsible for proteolysis (cleaving proteins). e. Telomerases, reverse transcriptase: Telomerases are enzymes that add repetitive DNA sequences (telomeres) to the ends of chromosomes. They are involved in maintaining the length of telomeres and preventing chromosome shortening. Telomerases are not directly involved in apoptosis. Reverse transcriptase is an enzyme found in retroviruses that synthesizes DNA from an RNA template, but it is not associated with apoptosis.

56. Suppose you have purified a population of actin filaments, treated them with red- fluorescent phalloidin, and added them to a test tube containing green-fluorescent actin subunits under conditions that strongly favor polymerization. After a few minutes, you fix the filaments to a slide and view the fluorescence. What would be the pattern of green labeling?

The correct answer is b. both ends will be labeled green, but the label will be longer at the plus end. When actin filaments are treated with fluorescent phalloidin and added to a test tube containing green-fluorescent actin subunits under conditions favoring polymerization, the green-fluorescent actin subunits will incorporate into the ends of the existing actin filaments. Actin polymerization occurs by the addition of new subunits to the plus (barbed) end of the filament, while the minus (pointed) end remains relatively stable. As a result, both ends of the actin filaments will be labeled green due to the incorporation of the green-fluorescent actin subunits. However, the label will be longer at the plus end because this is where the polymerization is occurring and new subunits are being added. Therefore, the correct pattern of green labeling would be b. both ends will be labeled green, but the label will be longer at the plus end.

103. Eukaryotic cells contain _____ autophagosomes that fuse with _____ to promote organelle degradation in a process called _____. a. single-membrane, lysosomes, autophagyb. single-membrane, autophagosomes, apoptosis c. double-membrane, lysosomes, autophagyd. double-membrane, autophagosomes, autophagy e. double-membrane, lysosomes, apoptosis

The correct answer is c. double-membrane, lysosomes, autophagy. Explanation: Autophagy is a cellular process that involves the degradation and recycling of cellular components, such as organelles and proteins, to maintain cellular homeostasis. It plays a vital role in various physiological and pathological processes. In the context of the given statement, eukaryotic cells contain double-membrane structures called autophagosomes, which sequester and enclose the cellular components targeted for degradation. These autophagosomes then fuse with lysosomes, which are membrane-bound organelles containing various hydrolytic enzymes. The fusion of autophagosomes with lysosomes forms autolysosomes, where the enclosed cellular components are degraded by lysosomal enzymes. This degradation process within autolysosomes is called autophagy. Let's examine the other options and explain why they are incorrect: a. single-membrane, lysosomes, autophagy: This option is incorrect because autophagosomes have a double membrane, not a single membrane. b. single-membrane, autophagosomes, apoptosis: This option is incorrect because autophagosomes are involved in autophagy, not apoptosis. Apoptosis is a programmed cell death process, and it typically involves distinct cellular mechanisms compared to autophagy. d. double-membrane, autophagosomes, autophagy: This option is incorrect because it redundantly states that autophagosomes are involved in autophagy, which is already known. e. double-membrane, lysosomes, apoptosis: This option is incorrect because lysosomes are not directly involved in apoptosis. Apoptosis is mainly regulated by caspases, which are proteases involved in the apoptotic signaling cascade. Lysosomes are primarily responsible for the degradation of cellular materials through processes like autophagy. In summary, the correct option is c because eukaryotic cells contain double-membrane autophagosomes that fuse with lysosomes to promote organelle degradation in a process called autophagy.

92. A graduate student wants to perform a genetic screen to identify cellular factors that promote cancer-like properties (transformation) in a cultured mammalian cell line. She designs a screen in which each cell receives a plasmid that will express a different gene at abnormally high levels in the cell. This screen will likely identify which type of gene(s)? a. caretaker genesb. tumor-suppressor genes c. proto-oncogenesd. a & be. a, b, & c

The correct answer is c. proto-oncogenes. Explanation: In the given scenario, the graduate student is designing a genetic screen where each cell receives a plasmid that expresses a different gene at abnormally high levels. This approach aims to identify genes that, when overexpressed, promote cancer-like properties or transformation in the cultured mammalian cell line. a. Caretaker genes: Caretaker genes are involved in maintaining genomic stability and DNA repair mechanisms. Although mutations or dysregulation of caretaker genes can contribute to genomic instability and increase the risk of cancer, overexpression of caretaker genes is less likely to directly promote cancer-like properties. Therefore, it is less likely to identify caretaker genes through the described overexpression screen. b. Tumor-suppressor genes: Tumor-suppressor genes normally function to regulate cell growth, prevent mutations, and inhibit tumor formation. However, their overexpression is unlikely to directly promote cancer-like properties. In fact, overexpression of tumor-suppressor genes can often have a suppressive effect on cancer development. Therefore, it is less likely to identify tumor-suppressor genes through the described overexpression screen. c. Proto-oncogenes: Proto-oncogenes are genes that, when mutated or overexpressed, have the potential to become oncogenes and drive cell proliferation. Overexpression of proto-oncogenes can directly promote cancer-like properties in cells. By overexpressing different genes, the genetic screen is more likely to identify proto-oncogenes that drive transformation and contribute to cancer development. Therefore, the correct answer is c. proto-oncogenes, as the overexpression screen is specifically designed to identify genes that, when overexpressed, promote cancer-like properties.

94. Stem cells are defined by their capacity to... a. self-renew b. differentiate into more specialized cell types c. remain in a terminally differentiated stated. both a & b are truee. a, b, & c are all true

The correct answer is d. both a & b are true. Explanation: Stem cells possess two defining characteristics: self-renewal and the ability to differentiate into more specialized cell types. a. Self-renewal refers to the ability of stem cells to divide and produce identical copies of themselves, maintaining an undifferentiated state. This process ensures the long-term maintenance of the stem cell population. b. Differentiation refers to the ability of stem cells to give rise to various specialized cell types. Stem cells have the potential to differentiate into specific cell lineages and contribute to the development and repair of different tissues and organs in the body. c. Stem cells do not remain in a terminally differentiated state. Unlike terminally differentiated cells, which have lost the ability to divide and differentiate further, stem cells retain their capacity for self-renewal and differentiation. Therefore, the correct statement is that stem cells have the capacity for both self-renewal (a) and differentiation into more specialized cell types (b).

87. how do cells regulate the concentrations of cyclins and other key cell cycle proteins?

The correct answer is e. both a & b are true. Cells regulate the concentrations of cyclins and other key cell cycle proteins through both the adjustment of their rates of synthesis and degradation at different points in the cell cycle. a. Cells adjust the rates of synthesis of these proteins at different points in the cell cycle. For example, certain cyclins are synthesized during specific stages of the cell cycle to ensure their availability when needed for progression to the next phase. b. Cells also adjust the rates of degradation of these proteins at different points in the cell cycle. The degradation of cyclins and other cell cycle proteins is typically mediated by specific ubiquitin ligases, such as the anaphase-promoting complex/cyclosome (APC/C). The timely degradation of cyclins helps regulate the progression of the cell cycle by ensuring that they are present only when required. Therefore, both mechanisms of adjusting synthesis and degradation rates contribute to the precise regulation of cyclin and other cell cycle protein concentrations during different stages of the cell cycle.

77. Which of the following statements about the immunological synapse is false?

The false statement about the immunological synapse is: e. its formation involves internalization of autophagy vacuoles. The immunological synapse is a specialized structure formed between an antigen-presenting cell (APC) and a T cell during immune responses. It plays a crucial role in facilitating communication and signal exchange between these cells. Let's examine the other statements to understand why they are true: a. Its formation involves rearrangements of the actin and microtubule cytoskeletons: This statement is true. During the formation of the immunological synapse, there are rearrangements of the actin and microtubule cytoskeletons. These changes help in the spatial organization of signaling molecules and facilitate cellular movement and communication. b. It is a focal point for tyrosine kinase signaling: This statement is true. The immunological synapse is a critical site for tyrosine kinase signaling. It involves the activation and recruitment of various signaling molecules, including tyrosine kinases, which play important roles in immune cell activation and response. c. It is a focal point for endocytosis and exocytosis: This statement is true. The immunological synapse serves as a site for both endocytosis (internalization of molecules into the cell) and exocytosis (release of molecules from the cell). These processes facilitate the exchange of membrane receptors, signaling molecules, and other components between the interacting cells. d. Its formation involves cell adhesion molecules (CAMs): This statement is true. The formation of the immunological synapse involves the interaction of cell adhesion molecules (CAMs) present on the surface of the interacting cells. CAMs play a crucial role in mediating cell-cell adhesion and stabilizing the synapse structure. In summary, the false statement is: e. its formation involves internalization of autophagy vacuoles. Autophagy vacuoles are not involved in the formation of the immunological synapse.

89. Consider the following figure describing a mitotic cyclin-CDK complex in yeast: In this scheme, the "UConn" kinase can add an inhibitory phosphate to a specific amino acid in the CDK, while a "Husky" phosphatase can de-phosphorylate the same amino acid in the CDK. Given this information, which of these statements would be true? a. if UConn kinase is overactive, the cell will enter mitosis too soonb. if Husky phosphatase is overactive, the cell will grow unusually large in G2c. if UConn kinase is inactivated, the cell will grow unusually large in G2d. if Husky phosphatase is blocked by a drug, the daughter cells will be unusually smalle. if UConn kinase is blocked by a drug, the cell will enter mitosis too soon

The correct answer is e. if UConn kinase is blocked by a drug, the cell will enter mitosis too soon. Let's go through each option and explain why the other options are incorrect: a. if UConn kinase is overactive, the cell will enter mitosis too soon: This statement is incorrect because an overactive UConn kinase would add an inhibitory phosphate to the CDK, which would prevent the cell from entering mitosis. It would delay mitotic entry rather than causing it to happen too soon. b. if Husky phosphatase is overactive, the cell will grow unusually large in G2: This statement is incorrect because an overactive Husky phosphatase would de-phosphorylate the inhibitory phosphate on the CDK, allowing the cell to enter mitosis. It would not affect cell growth in G2 phase. c. if UConn kinase is inactivated, the cell will grow unusually large in G2: This statement is incorrect because if UConn kinase is inactivated, it would remove the inhibitory phosphate from the CDK, promoting mitotic entry rather than affecting cell growth in G2 phase. d. if Husky phosphatase is blocked by a drug, the daughter cells will be unusually small: This statement is incorrect because blocking Husky phosphatase would lead to the persistence of the inhibitory phosphate on the CDK, preventing mitotic entry. It would not directly affect the size of the daughter cells. e. if UConn kinase is blocked by a drug, the cell will enter mitosis too soon: This statement is correct. Blocking UConn kinase would prevent the addition of the inhibitory phosphate on the CDK, resulting in the activation of the cyclin-CDK complex and premature entry into mitosis. Therefore, the correct answer is e. if UConn kinase is blocked by a drug, the cell will enter mitosis too soon.

61. A protein complex called the ____ nucleates cytoplasmic microtubules, while motile cilia and flagella originate at a ____.

The correct answer is: b. γ-TURC, basal body The γ-TURC (γ-tubulin ring complex) is a protein complex that nucleates the formation of cytoplasmic microtubules. It serves as a template for the assembly of microtubules by providing a platform for the addition of α- and β-tubulin subunits. Motile cilia and flagella, on the other hand, originate at a structure called the basal body. The basal body is derived from the centriole and serves as the anchor point for the assembly of cilia and flagella. a. centriole, centromere: The centriole is indeed involved in nucleating cytoplasmic microtubules, but it is not directly responsible for the origin of motile cilia and flagella. The centromere, on the other hand, is a region of DNA involved in chromosome attachment during cell division and is not related to microtubule nucleation or cilia/flagella formation. c. centromere, centriole: Similar to option (a), the centromere is unrelated to microtubule nucleation or cilia/flagella formation. The centriole is involved in nucleating cytoplasmic microtubules, but it does not serve as the origin for motile cilia and flagella. d. basal body, γ-TURC: This option incorrectly associates the basal body with the γ-TURC. While the basal body is indeed responsible for the origin of motile cilia and flagella, it is not directly associated with the γ-TURC, which is involved in nucleating cytoplasmic microtubules. e. basal body, centrosome: This option correctly associates the basal body with the origin of motile cilia and flagella. However, it incorrectly identifies the centrosome as the nucleating complex for cytoplasmic microtubules, whereas the correct nucleating complex is the γ-TURC.

33. Which of the following could prevent a secretory protein from being secreted normally?

The correct answer is: e. All of the above. Each of the listed scenarios can prevent a secretory protein from being secreted normally. Let's examine each option: a. A mutation that causes addition of mannose-6-phosphate to the oligosaccharide structure: Mannose-6-phosphate addition to the oligosaccharide structure is a signal for targeting proteins to the lysosomes. If this modification occurs on a secretory protein instead of a lysosomal protein, it can redirect the protein to the lysosomes for degradation rather than being properly secreted. b. A mutation that prevents addition of N-linked oligosaccharides in the ER: N-linked oligosaccharides are important for proper protein folding and quality control in the endoplasmic reticulum (ER). If the addition of N-linked oligosaccharides is prevented due to a mutation, it can lead to protein misfolding and ER retention, preventing the normal secretion of the protein. c. A mutation that results in deletion of the start transfer sequence: The start transfer sequence is responsible for guiding a nascent polypeptide chain to the ER membrane during protein synthesis. Deletion of the start transfer sequence would prevent the proper targeting and translocation of the protein into the ER, leading to its mislocalization and impaired secretion. d. Addition of an ER-retrieval sequence: An ER-retrieval sequence is a signal that directs proteins back to the ER from other compartments. If such a sequence is added to a secretory protein, it would cause the protein to be continually retrieved from its intended destination and returned to the ER, preventing its normal secretion. Therefore, all of the listed scenarios can interfere with the normal secretion of a secretory protein, and the correct answer is e. All of the above.

102. Multiple signaling pathways can lead to apoptosis. Which of the following statements is false? a. extrinsic pathways of programmed cell death involve binding of extracellular signaling proteins to cell surface death receptorsb. intrinsic pathways of programmed cell death involve a response to damaged DNA, a lack of oxygen, or a lack of nutrients c. extrinsic pathways of programmed cell death can be triggered in virus-infected target cells when these cells bind to cytotoxic T-lymphocytes (CTLs)d. extrinsic pathways of programmed cell death can be triggered by an epinephrine-related hormone called the Gronkowski particle e. intrinsic pathways of programmed cell death rely on the release of a protein called cytochrome c from mitochondria

The false statement is d. extrinsic pathways of programmed cell death can be triggered by an epinephrine-related hormone called the Gronkowski particle. Explanation: Extrinsic pathways of programmed cell death, also known as extrinsic apoptosis, involve the binding of extracellular signaling proteins (such as death ligands) to cell surface death receptors. This binding activates a cascade of events that ultimately leads to apoptosis. Examples of death receptors include Fas receptor and tumor necrosis factor receptor (TNFR). Intrinsic pathways of programmed cell death, also known as intrinsic apoptosis, are triggered by internal cellular signals. These signals can be a response to various conditions such as damaged DNA, lack of oxygen (hypoxia), or lack of nutrients. In intrinsic apoptosis, the release of cytochrome c from mitochondria is a crucial event that activates downstream caspases and initiates the apoptotic process. The statement c. extrinsic pathways of programmed cell death can be triggered in virus-infected target cells when these cells bind to cytotoxic T-lymphocytes (CTLs) is true. This is known as the cytotoxic T-cell-mediated killing of virus-infected cells, where CTLs recognize virus-infected cells through specific interactions and induce apoptosis in those cells. However, the statement d. extrinsic pathways of programmed cell death can be triggered by an epinephrine-related hormone called the Gronkowski particle is false. There is no known hormone or particle called the Gronkowski particle that triggers extrinsic apoptosis.

99. Which statement is false?a. viruses can cause cancer by inactivating p53 and Rbb. viruses cannot cause cancerc. viruses can transit through the secretory machinery on their way to exiting cells d. viruses can cause cancer by hijacking proto-oncogenes and altering theme. viruses can be transported by microtubule-associated motor proteins in cells

The false statement is option b: "viruses cannot cause cancer." Viruses can indeed cause cancer through various mechanisms, including inactivating tumor suppressor genes like p53 and Rb (option a), hijacking proto-oncogenes and altering them (option d), and using microtubule-associated motor proteins for intracellular transport (option e). Additionally, some viruses can transit through the secretory machinery during their life cycle (option c). Therefore, option b is incorrect. a. Viruses can cause cancer by inactivating p53 and Rb: This statement is true. Some viruses have the ability to disrupt the normal functions of tumor suppressor genes, such as p53 and Rb, which play crucial roles in regulating cell growth and preventing cancer. c. Viruses can transit through the secretory machinery on their way to exiting cells: This statement is true. During their replication cycle, some viruses can utilize the host cell's secretory machinery to exit the infected cell and spread to neighboring cells or tissues. d. Viruses can cause cancer by hijacking proto-oncogenes and altering them: This statement is true. Certain viruses can integrate their genetic material into the host cell's DNA, which may include proto-oncogenes. By altering the expression or function of these genes, viruses can promote uncontrolled cell growth and contribute to the development of cancer. e. Viruses can be transported by microtubule-associated motor proteins in cells: This statement is true. Microtubule-associated motor proteins, such as dynein and kinesin, are responsible for intracellular transport within cells. Some viruses can exploit these motor proteins to move within the host cell and reach specific cellular locations for replication or release.

69. Which of the following statements about cell connections is false?

The false statement is: b. Some adhesions result from protein-carbohydrate interactions between different cell types. Adhesions between cells and the extracellular matrix (ECM) are primarily mediated by protein-protein interactions, not protein-carbohydrate interactions. Proteins such as integrins are responsible for anchoring cells to the ECM through direct interactions with ECM proteins. The other statements are true: a. Some adhesions result from protein-protein interactions between cells and ECM. c. Some adhesion proteins are linked to intermediate filaments. d. Some adhesion proteins are linked to F-actin. e. Some adhesion proteins are linked to microtubules. These statements accurately describe the various types of adhesion proteins and their associations with different cytoskeletal components. Apologies for the confusion in my previous responses. Let's explain why options b and c are incorrect: b. A permanently active GAP: If the GAP protein is permanently active and continuously stimulates GTP hydrolysis by Ras, it would lead to rapid and excessive inactivation of Ras. With Ras constantly inactivated, its signaling for cell proliferation would be suppressed rather than enhanced. Therefore, a permanently active GAP is not expected to increase cell proliferation. Option b is incorrect. c. A nonfunctional GEF: If the GEF protein is nonfunctional and fails to stimulate the binding of GTP by Ras, it would prevent the activation of Ras. Without Ras activation, its signaling for cell proliferation would not occur. Therefore, a nonfunctional GEF is not expected to increase cell proliferation. Option c is also incorrect. In summary, both options b and c are incorrect. A permanently active GAP (option b) leads to excessive inactivation of Ras, while a nonfunctional GEF (option c) prevents Ras activation. Neither of these conditions would promote increased cell proliferation. The correct answer remains option e, which includes a nonfunctional GAP and a permanently active GEF, disrupting Ras regulation in favor of sustained activation and increased cell proliferation.

78. The structure that contains two barrel-shaped centrioles surrounded by an amorphous area of electron-dense pericentriolar material is called a...

The structure that contains two barrel-shaped centrioles surrounded by an amorphous area of electron-dense pericentriolar material is called a: d. centrosome The centrosome is an organelle found in animal cells and serves as the main microtubule organizing center (MTOC). It consists of a pair of centrioles, which are cylindrical structures composed of microtubules, and the pericentriolar material that surrounds them. The centrosome plays a crucial role in cell division, spindle formation, and organization of the cytoskeleton. It is involved in processes such as cell motility, intracellular transport, and establishment of cell polarity. Option c, "centromere," refers to the region of a chromosome that plays a role in the attachment of spindle fibers during cell division and is not associated with centrioles or the pericentriolar material. Therefore, the correct answer is d. centrosome. Sure! Let's go through the wrong answer choices and explain why they are incorrect: a. pericentromere: This term refers to the region of a chromosome located near the centromere, which is involved in the attachment of spindle fibers during cell division. It is not the correct term for the structure described in the question, which is the centrosome. b. kinetochore: The kinetochore is a structure located at the centromere of a chromosome. It serves as the attachment site for spindle fibers during cell division. While it is related to chromosome organization and spindle attachment, it is not the structure described in the question. c. centromere: The centromere is a specific region of a chromosome that plays a crucial role in the separation of sister chromatids during cell division. It is not the correct term for the structure described in the question, which is the centrosome. e. both c & d: This option combines the incorrect answer choices of centromere and centrosome. As explained above, the centromere and centrosome are distinct structures with different functions. Therefore, this option is incorrect. In summary, the correct answer is d. centrosome, as it accurately describes the structure containing centrioles and pericentriolar material. The other options, pericentromere, kinetochore, centromere, and both c & d, are incorrect as they either refer to different structures or combine unrelated terms.

85. Assume that the total length of the cell cycle of one of your skin cells is 24 h when grown in a culture flask. If M-phase is 2 h, S-phase is 8 h, and G1 is 4 h: How long is G2? Cytokinesis?

subtract the durations of M-phase, S-phase, and G1 from the total length of the cell cycle. Given: Total length of the cell cycle = 24 hours M-phase = 2 hours S-phase = 8 hours G1 = 4 hours To find G2, we subtract the durations of M-phase, S-phase, and G1 from the total length of the cell cycle: G2 = Total length of the cell cycle - (M-phase + S-phase + G1) G2 = 24 hours - (2 hours + 8 hours + 4 hours) G2 = 24 hours - 14 hours G2 = 10 hours Therefore, G2 is 10 hours. To find cytokinesis, we need to consider that cytokinesis typically occurs during the late stages of M-phase. Since M-phase is given as 2 hours, we can assume that cytokinesis is included within this duration.

73. Fibroblasts generally have a perinuclear Golgi and peripheral ER. Integrins are expressed by fibroblasts and are found in transport vesicles of the conventional secretory pathway. After vesicle budding from the ER, what would allow normal transport of integrins to the plasma membrane?

To determine what allows the normal transport of integrins from the ER to the plasma membrane after vesicle budding, let's evaluate each option: a. Vesicles would be carried by kinesin-I only: Kinesin-I is a motor protein that generally moves cargo towards the plus end of microtubules. However, relying solely on kinesin-I for transport would limit the ability of vesicles to reach the plasma membrane, as it would not facilitate movement towards the cell periphery. Therefore, option a is unlikely to be the correct answer. b. Vesicles would be carried by cytoplasmic dynein only: Cytoplasmic dynein is a motor protein that generally moves cargo towards the minus end of microtubules. Similarly to option a, relying solely on cytoplasmic dynein would hinder the transport of vesicles to the plasma membrane. Thus, option b is unlikely to be correct. c. Integrins in transport vesicles would not reach the plasma membrane: This option suggests that transport vesicles containing integrins would fail to reach the plasma membrane. However, if that were the case, integrins would not be properly delivered to their destination, leading to functional impairment. Given that integrins are expressed by fibroblasts and are involved in various cellular processes, this option is not likely to be accurate. d. Vesicles would be carried by cytoplasmic dynein followed by vesicles being carried by kinesin-I: This option proposes a sequential mechanism involving both cytoplasmic dynein and kinesin-I for vesicle transport. Cytoplasmic dynein could initially move vesicles from the perinuclear region towards the cell periphery, followed by kinesin-I taking over to transport the vesicles towards the plasma membrane. This bidirectional transport system allows efficient and regulated delivery of cargo to its destination. Therefore, option d is a plausible answer. e. Vesicles would be carried by kinesin-I followed by vesicles being carried by cytoplasmic dynein: This option suggests the opposite direction of transport, with kinesin-I moving vesicles towards the plasma membrane first, followed by cytoplasmic dynein bringing them back towards the perinuclear region. While bidirectional transport can occur in certain cellular contexts, it is less likely to be the primary mechanism for delivering integrins to the plasma membrane. Therefore, option e is less likely to be correct. In summary, the answer is likely: d. Vesicles would be carried by cytoplasmic dynein followed by vesicles being carried by kinesin-I. This sequential mechanism involving both cytoplasmic dynein and kinesin-I allows proper transport of integrin-containing vesicles from the ER to the plasma membrane in fibroblasts.

75. Which of the following events happens fourth during neutrophil extravasation?

To determine which event happens fourth during neutrophil extravasation, let's evaluate each option: a. The neutrophil crawls between endothelial cells towards the underlying tissue: This event, known as transmigration or diapedesis, typically occurs after the neutrophil has firmly adhered to the endothelium. It happens later in the extravasation process and is not the fourth event. b. The neutrophil binds to selectins on endothelial cells: This event, known as rolling, is one of the initial steps in neutrophil extravasation. Selectins on the endothelial cells interact with corresponding ligands on the neutrophil, allowing for transient binding and rolling along the vessel wall. However, this event typically occurs before the neutrophil firmly binds to the endothelium, so it is not the fourth event. c. Integrins are activated in the neutrophil: This event involves the activation of integrin adhesion molecules on the neutrophil's surface, allowing them to bind tightly to the endothelial cells. While integrin activation is crucial for firm adhesion, it generally occurs after the rolling of the neutrophil. Therefore, it is unlikely to be the fourth event. d. The neutrophil binds tightly to the endothelium and stops rolling: This event, known as firm adhesion, follows the rolling phase. Once integrins are activated, the neutrophil undergoes a transition from rolling to firmly adhering to the endothelial cells. This event involves stronger interactions between integrins and endothelial adhesion molecules, leading to the cessation of rolling. Considering the chronological order of events, this is a likely candidate for the fourth event. e. The neutrophil begins to roll on the endothelial surface: This option corresponds to the initial rolling of the neutrophil on the endothelial surface mediated by selectin interactions. As discussed earlier, rolling typically occurs before the neutrophil firmly adheres to the endothelium. Therefore, it is unlikely to be the fourth event. In summary, the fourth event during neutrophil extravasation is likely: d. The neutrophil binds tightly to the endothelium and stops rolling. This event follows the initial rolling phase and involves the firm adhesion of neutrophils to the endothelial cells.

76. Which of the following events happens third during muscle contraction?

To determine which event happens third during muscle contraction, let's evaluate each option: a. Calcium is released from the sarcoplasmic reticulum: Calcium release from the sarcoplasmic reticulum is a crucial step in muscle contraction, but it typically occurs before the other events listed. It triggers the activation of the contractile apparatus. Therefore, it is unlikely to be the third event. b. The myosin II power stroke: The myosin II power stroke is a key step in muscle contraction, where the myosin heads undergo a conformational change, pulling the actin filaments. However, the power stroke typically occurs after the myosin heads bind to the actin filament. Therefore, it is unlikely to be the third event. c. Tropomyosin shifts position on the actin helix: Tropomyosin shifting position on the actin helix is an important event in muscle contraction. It occurs in response to the binding of calcium ions to troponin, which exposes the myosin-binding sites on the actin filament. This event happens before the myosin heads bind to the actin filament. Therefore, it is a plausible candidate for the third event. d. Myosin II heads bind to the actin filament: The binding of myosin II heads to the actin filament is a crucial step in muscle contraction, allowing the formation of cross-bridges between myosin and actin. However, this event typically occurs after tropomyosin shifts position and the myosin-binding sites on the actin filament are exposed. Therefore, it is unlikely to be the third event. e. Secretory vesicles are exocytosed from a motor neuron: Exocytosis of secretory vesicles from a motor neuron is not directly involved in muscle contraction. It is a process associated with the release of neurotransmitters at the neuromuscular junction, which initiates muscle contraction. Therefore, it is unlikely to be the third event. In summary, the third event during muscle contraction is likely: c. Tropomyosin shifts position on the actin helix. This event occurs after calcium release from the sarcoplasmic reticulum and precedes the binding of myosin II heads to the actin filament.

72. Which of the following events happens third in the current model for fibroblast motility?

To determine which event happens third in the current model for fibroblast motility, let's evaluate each option: a. Loss of integrin adhesion at the rear of the cell: This event typically occurs after the cell has moved forward and retraction of the rear part of the cell begins. It happens later in the motility process and is not the third event. b. Protrusion of lamellipodia and filopodia at the leading edge: This event involves the extension of thin, sheet-like lamellipodia and slender filopodia at the leading edge of the cell. It is one of the initial steps in cell migration and occurs early in the motility process. Therefore, it is unlikely to be the third event. c. Forward translocation of the cell body: This event refers to the movement of the cell body in the direction of cell migration. It involves the coordinated action of various cellular components and occurs as the cell progresses in its motility. While it is an important aspect of cell movement, it typically follows the initial protrusion of lamellipodia and filopodia. Thus, it is unlikely to be the third event. d. Adhesion of integrins to the ECM beneath the lamellipodium: This event involves the attachment of integrin proteins to the extracellular matrix (ECM) beneath the lamellipodium. It is necessary for the cell to anchor and exert traction during migration. While integrin adhesion is crucial for cell motility, it generally occurs after the protrusion of lamellipodia and filopodia. Therefore, it is unlikely to be the third event. e. Actin nucleation by the Arp2/3 complex and formins: This event involves the formation of new actin filaments through the action of the Arp2/3 complex and formin proteins. Actin nucleation is crucial for lamellipodia and filopodia formation and occurs early in the process of cell migration. Therefore, it is a potential candidate for the third event. Considering the explanations above, the third event in the current model for fibroblast motility is likely to be: e. Actin nucleation by the Arp2/3 complex and formins.

74. A student identifies a new cell type that migrates very fast when the cells are grown in culture dishes. She is an expert in wound-healing assays and wants to test a role for specific GTPases in this process using RNA interference (RNAi). She finds that wounds close completely after 8h when using control cells, but that depletion of RhoA, Rac1, or Cdc42 gives the surprising results shown below. What conclusion can be drawn from these data?

To draw a conclusion from the provided data, let's examine the results for depletion of RhoA, Rac1, and Cdc42: - Control cells: Wounds close completely after 8 hours. - Depletion of RhoA: Wounds do not close completely after 8 hours. - Depletion of Rac1: Wounds do not close completely after 8 hours. - Depletion of Cdc42: Wounds close completely after 8 hours. Based on these observations, we can make the following conclusions: a. RhoA, Rac1, and Cdc42 are each important for migration by this cell line: This conclusion is not supported by the data, as depletion of Cdc42 did not impair wound closure. Therefore, option a is not correct. b. Only RhoA and Rac1 are important for migration by this cell line: This conclusion is also not supported by the data, as depletion of RhoA or Rac1 individually resulted in impaired wound closure. Therefore, option b is not correct. c. Only RhoA is important for cell migration by this cell line: This conclusion is not supported by the data, as depletion of Rac1 also resulted in impaired wound closure. Therefore, option c is not correct. d. RhoA, Rac1, and Cdc42 are important for migration by all cell lines: The data provided does not allow us to conclude whether these GTPases are important for migration in all cell lines. Therefore, option d is not correct. e. Only Cdc42 is important for migration by this cell line: This conclusion is supported by the data since depletion of Cdc42 did not affect wound closure, while depletion of RhoA or Rac1 impaired it. Therefore, option e is the most accurate conclusion that can be drawn from the data. In summary, based on the provided data, the conclusion is: e. Only Cdc42 is important for migration by this cell line.

32. You isolated cellular components to study the transport of proteins in a cell. After a lot of hard work, you take a vacation, but you forget to write down how far along in the purification you progressed. If you put some of your preparation in a test tube, and then add mRNA for a secretory protein, you find that the mRNA is completely translated and the new protein is sensitive to protease. Your preparation has which of the following?

c. Ribosomes and microsomes, but no SRP. In the given scenario, the mRNA for a secretory protein is completely translated, indicating that protein synthesis occurred. The newly synthesized protein is sensitive to protease, suggesting that it is located within microsomes. However, since the mRNA was completely translated, it suggests the absence of the SRP (signal recognition particle) since the SRP would normally target the ribosomes to the endoplasmic reticulum (ER) membrane. Option c correctly reflects the presence of ribosomes and microsomes, which are responsible for protein synthesis and compartmentalization, respectively. However, the absence of SRP indicates that it is not present in the preparation. Therefore, the correct answer is c. Ribosomes and microsomes, but no SRP.

95. Suppose you discover a protein that you believe promotes pluripotency. How could you show this experimentally? a. in a somatic cell nuclear transfer experiment, deplete it from eggs to see if nuclear reprogramming becomes less efficientb. express it in somatic cells with the four primary pluripotency factors to see if you can generate induced pluripotent stem (iPS) cells more efficientlyc. in a somatic cell nuclear transfer experiment, deplete it from eggs to see if nuclear reprogramming becomes more efficientd. both a & b would be a good way to show thise. both b & c would be a good way to show this

You're correct, both options a and b are valid ways to experimentally demonstrate that a protein promotes pluripotency. My apologies for the oversight. Let's examine each option in more detail: a. In a somatic cell nuclear transfer experiment, depleting the protein from eggs to see if nuclear reprogramming becomes less efficient: This approach involves removing or depleting the protein of interest from the recipient eggs used in somatic cell nuclear transfer (SCNT) experiments. If the protein is crucial for pluripotency promotion, its depletion would likely lead to reduced efficiency or success in reprogramming the somatic cell nucleus, as the eggs would lack the necessary factors to support efficient reprogramming. b. Expressing the protein in somatic cells with the four primary pluripotency factors to see if you can generate induced pluripotent stem (iPS) cells more efficiently: This approach involves introducing the protein of interest into somatic cells along with the four primary pluripotency factors. If the protein indeed promotes pluripotency, its presence should enhance the efficiency or effectiveness of generating induced pluripotent stem (iPS) cells. This can be assessed by evaluating the number and quality of iPS cell colonies formed or measuring the reprogramming efficiency. c. In a somatic cell nuclear transfer experiment, depleting the protein from eggs to see if nuclear reprogramming becomes more efficient: This option is incorrect because depleting the protein from eggs in a somatic cell nuclear transfer experiment is not expected to make nuclear reprogramming more efficient. Depleting essential factors or proteins from the recipient eggs would likely have a negative impact on their developmental potential and their ability to support successful reprogramming. In summary, option a and b are both valid approaches to experimentally demonstrate the role of a protein in promoting pluripotency, while option c is incorrect as depleting the protein from eggs would not be expected to increase the efficiency of nuclear reprogramming.

27. Of these five steps listed, what is the FOURTH step in making a nuclear protein?

a. An importin binds to the NLS and moves the protein through the nuclear pore into the nucleus. Here's the correct sequence of steps in making a nuclear protein: The mRNA encoding for the nuclear protein is transcribed in the nucleus. The mRNA is transported through the nuclear pore to the cytoplasm. The mRNA is translated by ribosomes in the cytoplasm, resulting in the synthesis of the protein. An importin protein, which acts as a receptor for nuclear localization signals (NLS) present on the protein, binds to the NLS on the protein in the cytoplasm. The importin/protein complex is recognized by the nuclear pore complex and moves through the nuclear pore into the nucleus. Once inside the nucleus, the importin/protein complex encounters Ran-GTP (Guanosine Triphosphate-bound form of Ran), causing the complex to dissociate, releasing the protein into the nucleus. Therefore, the correct answer is a. An importin binds to the NLS and moves the protein through the nuclear pore into the nucleus.

57. The pyrene-actin assembly assay below shows the amount of F-actin formed from G- actin over time in the presence of the Arp2/3 complex (solid line). When this experiment is repeated with G-actin, Arp2/3 complex, and unknown additional proteins, a different curve is obtained (dashed line). Which proteins are likely responsible for this change?

a. Cdc42 and Rac are small GTPases that can regulate actin polymerization by activating the Arp2/3 complex and promoting branched actin filament formation. b. CapZ and alpha-actinin are actin-binding proteins involved in actin filament capping and crosslinking, respectively. They can modulate actin dynamics by regulating filament length and stability. c. Cofilin and gelsolin are actin-severing proteins that promote filament disassembly and turnover. They can enhance actin depolymerization and regulate filament length. d. Auto-inhibited formins are a family of actin nucleation-promoting factors that can be regulated by intramolecular interactions. When activated, formins can initiate linear actin filament assembly. e. WASP (Wiskott-Aldrich syndrome protein) and Cdc42 are proteins involved in the regulation of actin polymerization through the Arp2/3 complex. WASP can bind to and activate the Arp2/3 complex in response to Cdc42 signaling.

66. Structures such as ______ contain actin bundling proteins, while ______ are associated with proteins involved in actin filament branching.

a. Lamellipodia, stress fibers: Lamellipodia are flat, sheet-like membrane protrusions at the leading edge of a migrating cell. They are primarily composed of a network of actin filaments, but they do not contain actin bundling proteins. On the other hand, stress fibers are contractile bundles of actin filaments associated with myosin motors. They are not directly involved in actin filament branching. b. Endosomes, lamellipodia: Endosomes are membrane-bound vesicles involved in intracellular transport and sorting of molecules. They are not associated with actin filament branching. Lamellipodia, as mentioned earlier, are flat membrane protrusions composed of a network of actin filaments, but they do not contain actin bundling proteins. c. Lamellipodia, filopodia: This option is incorrect because both lamellipodia and filopodia are associated with actin filament branching. Lamellipodia are not specifically associated with actin bundling proteins. d. Stress fibers, microvilli: Stress fibers are contractile bundles of actin filaments associated with myosin motors. They are not directly involved in actin filament branching. Microvilli, on the other hand, are finger-like protrusions on the surface of cells that contain actin bundles, but they do not have a direct association with actin filament branching. Option e, filopodia and endosomes, is the correct answer as filopodia are associated with actin bundling proteins, while endosomes are not directly associated with actin filament branching.

26. What would you expect to happen if you massively overexpressed a very active RanGEF in the cytoplasm, such that the normal activity of RanGAP was completely overridden?

a. Ran would not be recycled by Ntf2: This statement is correct. If a very active RanGEF is massively overexpressed, overriding the normal activity of RanGAP, the hydrolysis of RanGTP to RanGDP would be impaired. As a result, Ran would not be efficiently recycled by Ntf2, which is involved in transporting RanGDP back to the nucleus. b. Proteins destined for export to the cytoplasm would probably not function properly, as the cargo/exportin/RanGTP complex would not dissociate: This statement is correct. The dissociation of the cargo/exportin/RanGTP complex is mediated by the hydrolysis of RanGTP to RanGDP by RanGAP. If the normal activity of RanGAP is completely overridden by the overexpressed RanGEF, the complex would not dissociate properly, potentially leading to improper functioning of the exported proteins. d. The importin returning to the cytoplasm would never bind cargo proteins because it would remain bound to the excess RanGTP present there: This statement is correct. Importins bind to cargo proteins in the cytoplasm and form a complex with RanGTP. The subsequent dissociation of the importin from the cargo proteins requires the hydrolysis of RanGTP to RanGDP by RanGAP. With the normal activity of RanGAP overridden, the importin would remain bound to RanGTP and not be available to bind new cargo proteins. Therefore, option e, which includes options a, b, and d, is the correct answer.

25. Ran's role in regulating nucleocytoplasmic transport is based on a mechanism in which the cell maintains a ______ nuclear concentration and a very low cytoplasmic concentration of______.

a. Ran would not be recycled by Ntf2b. Proteins destined for export to the cytoplasm would probably not function properly, as the cargo/exportin/RanGTP complex would not dissociate.c. Import of proteins would be accelerated.d. The importin returning to the cytoplasm would never bind cargo proteins because it would remain bound to the excess RanGTP present there.e. a, b and d are correct. The correct option is: e. a, b, and d are correct. If a very active RanGEF is massively overexpressed in the cytoplasm, overriding the normal activity of RanGAP, several consequences would occur: a. Ran would not be recycled by Ntf2: RanGAP normally promotes the hydrolysis of RanGTP to RanGDP in the cytoplasm, which is then transported back to the nucleus by the nuclear transport factor 2 (Ntf2). However, if RanGAP is completely overridden by the overexpressed RanGEF, the recycling of Ran by Ntf2 would be impaired, leading to an accumulation of RanGTP in the cytoplasm. b. Proteins destined for export to the cytoplasm would probably not function properly, as the cargo/exportin/RanGTP complex would not dissociate: The export of proteins from the nucleus to the cytoplasm is facilitated by the binding of cargo proteins to exportins, forming a complex with RanGTP. Normally, this complex dissociates in the cytoplasm due to the hydrolysis of RanGTP to RanGDP by RanGAP. However, with the excess RanGTP present in the cytoplasm due to the overexpressed RanGEF, the cargo/exportin/RanGTP complex would remain intact, potentially impairing the proper functioning of exported proteins. d. The importin returning to the cytoplasm would never bind cargo proteins because it would remain bound to the excess RanGTP present there: Importin proteins are responsible for facilitating the import of cargo proteins into the nucleus. They bind to cargo proteins in the cytoplasm and form a complex with RanGTP. Normally, in the nucleus, the RanGTP is hydrolyzed to RanGDP by RanGAP, causing the importin to release the cargo protein. However, in this scenario, with excess RanGTP present in the cytoplasm, the importin would remain bound to RanGTP and not be available to bind new cargo proteins, thus preventing the import of proteins into the nucleus. Therefore, options a, b, and d are all correct, and the correct answer is e. a, b, and d.

36. The following are involved with synthesis of a secreted protein. Which is the FOURTH step of the five listed? a. The SRP dissociates from the ribosome and start transfer sequence b. The SRP binds to the start transfer sequence.c. The signal (start transfer) sequence is removed by signal peptidase. d. The SRP binds to SRP receptor e. Protein synthesis halts temporarily

a. The SRP dissociates from the ribosome and start transfer sequence. In the synthesis of a secreted protein, the fourth step is when the SRP (Signal Recognition Particle) dissociates from the ribosome and the start transfer sequence. Here's an explanation of the steps: a. The SRP binds to the start transfer sequence: This is the initial step where the SRP recognizes the signal or start transfer sequence on the growing polypeptide chain and binds to it. This binding halts further protein synthesis temporarily. b. The SRP binds to the SRP receptor: After the SRP binds to the start transfer sequence, the SRP-ribosome complex interacts with the SRP receptor on the target membrane (such as the endoplasmic reticulum or the membrane of the secretory pathway). c. Protein synthesis halts temporarily: The binding of the SRP to the start transfer sequence causes a pause in protein synthesis. d. The SRP dissociates from the ribosome and start transfer sequence: At this point, the SRP dissociates from both the ribosome and the start transfer sequence, allowing protein synthesis to resume and the ribosome to continue translating the polypeptide chain. e. The signal (start transfer) sequence is removed by signal peptidase: After the protein has been translocated into the appropriate cellular compartment, the signal sequence is cleaved by signal peptidase. Therefore, the correct answer is a. The SRP dissociates from the ribosome and start transfer sequence.

31. If free ribosomes in the presence of RER vesicles with a mutant, nonfunctional SRP receptor that cannot bind the SRP, are placed in a test tube with mRNA for a protein with an n-terminal signal sequence, and everything else needed synthesize proteins, along with the addition of SRP, where are the proteins found after their production?

a. There are no full-length proteins made. In the given scenario, the mutant, nonfunctional SRP receptor on the RER vesicles cannot bind the SRP. The SRP is responsible for targeting the ribosome-nascent chain complex to the endoplasmic reticulum (ER) membrane. Without a functional SRP receptor, the ribosomes cannot properly target and dock onto the RER. As a result, the ribosomes will continue to synthesize proteins in the presence of mRNA and other necessary components, but the synthesized proteins will not be successfully inserted into the RER. Without proper targeting and insertion into the RER, no full-length proteins will be made. Therefore, the correct answer is a. There are no full-length proteins made. incorrect: b. Inside the RER-derived vesicles: The mutant, nonfunctional SRP receptor prevents the binding of the SRP to the RER vesicles. Without functional SRP receptor-mediated targeting, the ribosomes cannot properly dock onto the RER and insert the synthesized proteins into the vesicles. As a result, the proteins will not be found inside the RER-derived vesicles. c. Floating free in the aqueous solution in the test tube: This is the correct answer. Without a functional SRP receptor, the ribosomes will remain free in the aqueous solution of the test tube. They will continue to synthesize proteins, but without proper targeting to the RER, the synthesized proteins will not be inserted into any specific organelle or compartment. d. Inside microsomes: Microsomes are small vesicles derived from the endoplasmic reticulum. In the given scenario, the mutant, nonfunctional SRP receptor prevents proper targeting of the ribosomes to the RER. Without successful targeting to the RER, the synthesized proteins will not be inserted into microsomes. e. None of the above: This option is not correct because, in this particular scenario, the synthesized proteins will not be found inside the RER-derived vesicles, microsomes, or any specific organelle. They will be free in the aqueous solution in the test tube, as stated in option c. Therefore, the correct answer is a. There are no full-length proteins made. The mutant SRP receptor prevents proper targeting and insertion of the synthesized proteins into the RER, resulting in the absence of full-length proteins.

5. Which of the following statements about lipid rafts is false?

a. They can be regions enriched in lipids with saturated fatty acid chains. b. Because of the different lipid properties, lipid raft formation is an energetically favorable situation. c. They can have a different complement of membrane proteins than other regions of thebilayer. d. Lipid rafts can be cholesterol rich regions. e. All are correct. hydrophobic to grou sphynomosines and proteins are membrane proteins crossed hydrophobic us alpha alpha-helices

15. The permeability to _____ is the most important for generating peak positive potential during an action potential.

a. sodium ions -inward flow, increases the concentration of positively charged cations in the cell and depolarises -cell is higher than the cell's resting potential, sodium channels closest peak action potential, the same as Na+ equilibrium potential that is +40 mV -Na+ channels are open and dominant membrane conductance is of sodium ion, sodium channels are opened and other channels not open

Antibodies can be used to

a. visualize proteins in fixed and permeabilized cells b. identify specific proteins on an immunoblot (western blot) grind cells, determine that the large and small protein location c. precipitate proteins from cellular extracts immunoprecipitation/immunoisolation d. identify and isolate organelle fractions increasing density of sucrose with elements get only mitochondria In top fractions. Correct: e. all are correct.

84. Which of the following forces does NOT occur during M-phase of the cell cycle?

all of the listed forces play important roles: a. A balance of poleward microtubule flux and a polar ejection force: These forces contribute to the movement and positioning of chromosomes during mitosis. The poleward microtubule flux assists in pulling chromosomes toward the spindle poles, while the polar ejection force helps to separate and align the chromosomes. b. Pulling or pushing forces from dynein and kinesin motor proteins: Dynein and kinesin motor proteins are involved in the movement of various cellular components, including chromosomes and microtubules, during mitosis. They generate pulling or pushing forces to contribute to spindle assembly, chromosome movement, and spindle pole separation. c. A contractile force from F-actin and myosin II beneath the cleavage furrow: During cytokinesis, the final stage of M-phase, a contractile ring composed of actin filaments and myosin II forms beneath the cleavage furrow. Contraction of this ring generates a contractile force that drives the physical separation of the dividing cell. d. Forces from polymerization and depolymerization of microtubules: Dynamic changes in microtubule polymerization and depolymerization contribute to spindle assembly, chromosome movement, and spindle elongation during M-phase. Therefore, all of the listed forces (a, b, c, and d)

Which of the following polypeptides would be expected to be glycosylated?

b. A lysosomal enzyme. Glycosylation is a post-translational modification process in which sugar molecules (oligosaccharides) are attached to proteins or lipids. It is a common modification that can have various functions, including protein stability, cell signaling, and protein trafficking. In the given options, lysosomal enzymes (option b) are known to undergo glycosylation. Lysosomal enzymes are synthesized in the rough endoplasmic reticulum (ER) and then transported to the Golgi complex, where they undergo glycosylation before being sorted and delivered to lysosomes. Glycosylation plays a role in targeting lysosomal enzymes to the lysosomes and ensuring their proper function within these organelles. The other options (a, c, d, and e) do not typically undergo glycosylation: a. A histone protein bound to nuclear DNA: Histones are small, highly basic proteins that are involved in DNA packaging in the nucleus. They do not undergo glycosylation. c. COPI vesicle coat protein: COPI (coat protein complex I) is involved in retrograde transport between the Golgi complex and the endoplasmic reticulum (ER). While COPI proteins can be post-translationally modified, glycosylation is not a common modification for these proteins. d. The SRP (signal recognition particle): The signal recognition particle is involved in targeting and directing newly synthesized proteins to the endoplasmic reticulum (ER) for insertion or secretion. The SRP itself is an RNA-protein complex and does not undergo glycosylation. e. A cytoplasmic chaperone: Chaperone proteins assist in protein folding and stability. Cytoplasmic chaperones, such as heat shock proteins (HSPs), are not typically glycosylated. Therefore, the lysosomal enzyme (option b) is the most likely candidate to be glycosylated among the given options.

8. You are examining the subcellular localization of a putative membrane protein using immunofluorescence microscopy. Which is a true statement about what you might expect?

b. Although the lipid bilayer is ~ 4nm across, you would expect to observe an apparent structure ~ 200 nm across due to the resolution limit of light microscopy. incorrect: a. If the antibody you are using was raised to a portion of the protein that is expected to be exposed to the cytoplasm, you will not need to permeabilize the membrane with detergent. c. If the proteins are located 50 nm apart from each other, you will observe a dotted appearance of the membrane due to the separation of each labeled protein. small enough proteins can overlap as an overall pattern d. None are true.

43. If a patient has a lysosomal storage disease in which all lysosomal proteins are found secreted from the cell, and those proteins have no Man-6-phosphate attached to their oligosaccharides, then the defect is most likely:

b. In the enzyme that adds Man-6-P to lysosomal proteins. The addition of mannose-6-phosphate to lysosomal proteins is an important post-translational modification that targets these proteins to the lysosomes. The enzyme responsible for adding Man-6-P to lysosomal proteins is called N-acetylglucosamine-1-phosphotransferase (NAGT), also known as the enzyme that generates the Man-6-P signal. If this enzyme is defective or absent, lysosomal proteins would not receive the Man-6-P modification. As a result, these proteins would not be recognized and targeted to the lysosomes. Instead, they would be secreted from the cell, leading to a lysosomal storage disease. Therefore, the most likely defect in this case is a deficiency in the enzyme that adds mannose-6-phosphate to lysosomal proteins. The correct answer is b. In the enzyme that adds Man-6-P to lysosomal proteins.

38. If a nuclear localization signal were added to the gene encoding a plasma membrane protein, where would the protein likely be found?

b. Plasma membrane. If a nuclear localization signal (NLS) were added to the gene encoding a plasma membrane protein, it would not change the protein's normal destination, which is the plasma membrane. The NLS functions as a signal for nuclear import, directing proteins into the nucleus, but it does not influence the normal localization of a plasma membrane protein. The plasma membrane proteins are synthesized on the endoplasmic reticulum (ER) and then targeted to the plasma membrane through specific sorting and trafficking mechanisms. Adding an NLS to the gene encoding a plasma membrane protein would not alter these normal processes, and the protein would still be directed to the plasma membrane. Therefore, option b is the correct answer: The protein would likely be found in the plasma membrane.

83. In anaphase A, kinetochore MTs ______. In anaphase B, polar microtubules ______.

c is the correct answer and why the other options are incorrect: a. Slide, slide: This option suggests that both kinetochore microtubules and polar microtubules slide during both anaphase A and anaphase B. However, during anaphase A, kinetochore microtubules do not slide; instead, they shorten as the sister chromatids separate. In anaphase B, polar microtubules do slide as the spindle poles move apart. b. Shrink, shrink: This option suggests that both kinetochore microtubules and polar microtubules shrink during both anaphase A and anaphase B. While kinetochore microtubules do shrink in anaphase A, polar microtubules do not shrink in anaphase B; instead, they elongate and push the poles apart. c. Shrink, slide: This is the correct answer. During anaphase A, kinetochore microtubules shrink, resulting in the separation of sister chromatids. In anaphase B, polar microtubules slide past each other, helping to elongate the cell and move the poles apart. This option accurately describes the movements of microtubules during both phases. d. Slide, shrink: This option suggests that kinetochore microtubules slide during anaphase A and shrink during anaphase B. However, it is the opposite. Kinetochore microtubules shrink during anaphase A, facilitating chromosome separation, and polar microtubules shrink during anaphase B. e. Grow, stabilize: This option suggests that kinetochore microtubules grow and polar microtubules stabilize during both anaphase A and anaphase B. However, kinetochore microtubules do not grow during anaphase A; instead, they depolymerize as the sister chromatids separate. Polar microtubules do not stabilize during anaphase B; they undergo elongation to facilitate cell elongation.

What likely determines the specificity of vesicle fusion to a target membrane?

c) interactions between specific combinations of interacting proteins, including tethering proteins, Rabs, and SNAREs. Vesicle fusion, the process by which a vesicle membrane and a target membrane merge, requires precise molecular interactions between various proteins to ensure specificity and accuracy. The fusion process involves multiple steps and a series of protein-protein interactions. Tethering proteins play a role in the initial recognition and docking of vesicles to their target membrane. They help bring the vesicle and target membrane in close proximity, facilitating subsequent interactions. Rab proteins are small GTPases that act as molecular switches and are involved in vesicle trafficking. They participate in vesicle tethering and facilitate the attachment of vesicles to their target membranes by interacting with tethering proteins. SNAREs (soluble N-ethylmaleimide-sensitive factor attachment protein receptors) are a group of proteins involved in membrane fusion. They are present on both the vesicle and target membranes and form specific protein complexes, called SNARE complexes. The assembly of complementary SNARE proteins between the vesicle and target membranes provides the driving force for membrane fusion. Therefore, the specificity of vesicle fusion to a target membrane is likely determined by the interactions between specific combinations of interacting proteins, including tethering proteins, Rabs, and SNAREs. These proteins work together to ensure the accurate and specific fusion of vesicles with their intended target membranes.

What does UGGT do if it binds to a misfolded glycoprotein? It interacts with Calnexin and ERp57 and transports the protein to the proteasome. It adds a single mannose back to one of the glucose residues at the exposed end of the recently trimmed oligosaccharide. It adds a single glucose back to one end of the recently trimmed oligosaccharide. It degrades the protein. a and b are correct.

c. It adds a single glucose back to one end of the recently trimmed oligosaccharide. UGGT (UDP-glucose:glycoprotein glucosyltransferase) is an enzyme involved in the quality control of glycoproteins in the endoplasmic reticulum (ER). When UGGT recognizes a misfolded glycoprotein, it acts to add a single glucose residue back to the recently trimmed oligosaccharide chains of the protein. This reglucosylation process serves as a signal for the protein to undergo further rounds of folding attempts with the assistance of chaperone proteins such as Calnexin and Calreticulin. By adding a glucose residue, UGGT helps to promote proper folding and prevent premature degradation of the misfolded glycoprotein. Therefore, the correct answer is c. It adds a single glucose back to one end of the recently trimmed oligosaccharide.

59. Suppose you have just polymerized microtubules from a 1ml solution of tubulin subunits in vitro and the reaction has reached steady state. What do you expect would happen if you then diluted the microtubule solution into a 10ml pool of buffer (that does not contain any additional protein)?

c. Microtubule shrinkage. Diluting the microtubule solution into a larger volume of buffer reduces the concentration of tubulin subunits available for polymerization. As a result, the concentration of free tubulin subunits in the solution decreases, which disrupts the equilibrium between polymerized and free tubulin. The microtubules will tend to depolymerize or shrink since the concentration of tubulin subunits available for polymerization is reduced. Therefore, the expected outcome is microtubule shrinkage in response to the dilution into the larger buffer pool without any additional protein present. incorrect: a. Microtubule stabilization: Diluting the microtubule solution into a larger buffer would not lead to microtubule stabilization. In fact, the decrease in tubulin concentration would likely destabilize the microtubules and promote depolymerization. b. Microtubule growth: Similarly, diluting the microtubule solution would result in a decrease in tubulin concentration, making it less likely for microtubule growth to occur. Microtubule growth requires the addition of tubulin subunits to the growing ends, which would be limited in a diluted solution. c. Microtubule shrinkage: This option is correct. Diluting the microtubule solution into a larger buffer would decrease the concentration of tubulin subunits available for polymerization. With a reduced tubulin concentration, microtubules are more likely to undergo depolymerization or shrinkage. d. Microtubule severing: Dilution into a larger buffer does not introduce any factors or proteins that would promote microtubule severing. This option is incorrect in this scenario. e. Microtubule degradation by proteases: Dilution into a buffer without additional proteins would not introduce proteases or factors that would lead to microtubule degradation. This option is incorrect.

How are proteins imported into mitochondria and chloroplasts?

c. Post-translationally. Proteins are imported into mitochondria and chloroplasts post-translationally, meaning that the import process occurs after the synthesis of the protein is complete. These organelles have specific protein import machinery that recognizes targeting signals on the precursor proteins and facilitates their transport across the organelle membranes. For mitochondria, proteins are generally synthesized in the cytoplasm as precursor proteins with an N-terminal mitochondrial targeting sequence, often referred to as a mitochondrial targeting peptide. The precursor proteins are then recognized by the import machinery at the outer mitochondrial membrane, which facilitates their translocation across the outer and inner membranes into the mitochondrial matrix or specific subcompartments. Similarly, chloroplasts have their own protein import systems. Proteins destined for the chloroplast are synthesized in the cytoplasm with an N-terminal chloroplast transit peptide. The chloroplast import machinery recognizes this transit peptide and guides the proteins across the chloroplast membranes, allowing them to reach their specific destinations within the organelle. Therefore, the correct answer is c. Post-translationally.

You are tracing the path of the LDL receptor for cholesterol transport from its synthesis to its final destination on the plasma membrane. The order of listed compartments in which you would find it is: I. medial Golgi II. Trans Golgi network III. Secretory vesicles IV. cis-Golgi V. Rough ER

c. V, IV, I, II, III The correct order is as follows: V. Rough ER: The LDL receptor is synthesized in the rough endoplasmic reticulum (ER), where it undergoes initial processing and folding. IV. cis-Golgi: From the rough ER, the LDL receptor moves to the Golgi apparatus, specifically the cis-Golgi compartment, where further modifications and processing take place. I. Medial Golgi: The LDL receptor then progresses to the medial Golgi region of the Golgi apparatus, where additional modifications occur. II. Trans Golgi network: Next, the LDL receptor reaches the trans Golgi network, which serves as a sorting station for proteins destined for different cellular compartments. III. Secretory vesicles: Finally, the LDL receptor is packaged into secretory vesicles, which transport it from the Golgi apparatus to the plasma membrane for secretion and incorporation into the cell surface. Therefore, the correct answer is c. V, IV, I, II, III.

67. An undergraduate student observes the directed motion of fluorescent COPI- and COPII-coated vesicles within a cell under different conditions. She notices that vesicle movement stops when nocodazole is added to the culture medium, but resumes when nocodazole is removed from the medium. Which conclusion makes the most sense? a. normal movement was driven by diffusion

c. normal movement was driven by processive kinesin or dynein motor proteins. Nocodazole is a drug that disrupts microtubules by depolymerizing them. Microtubules are essential for intracellular transport processes mediated by motor proteins such as kinesins and dyneins. Kinesins are generally associated with plus-end-directed movement, while dyneins are typically involved in minus-end-directed movement along microtubules. Based on the observation that vesicle movement stops when nocodazole is added and resumes when it is removed, it suggests that the normal movement of COPI- and COPII-coated vesicles is dependent on functional microtubules and motor proteins. Therefore, the most likely explanation is that the directed motion of the vesicles is driven by processive kinesin or dynein motor proteins along the microtubules in the cell. a. normal movement was driven by diffusion: This option is unlikely because the directed motion of vesicles suggests an active transport process rather than passive diffusion. b. normal movement was driven by a processive plus-end directed dynein motor protein: This option is not supported by the observation that vesicle movement stops when nocodazole is added. Nocodazole disrupts microtubules, which are necessary for dynein-mediated transport. Therefore, the absence of vesicle movement in the presence of nocodazole indicates that dynein alone is not responsible for the normal movement. d. normal movement was driven by a non-processive myosin II motor protein: This option is not supported by the observation that vesicle movement stops when nocodazole is added. Myosin II is primarily associated with actin-based movement rather than microtubule-based movement. e. normal movement stopped because nocodazole killed the cell: This option is unlikely because the observation is about the cessation and resumption of vesicle movement, not cell viability. Additionally, nocodazole does not necessarily cause cell death but rather disrupts microtubule dynamics. Overall, the most reasonable conclusion is that normal movement of COPI- and COPII-coated vesicles in the cell is driven by processive kinesin or dynein motor proteins along microtubules, as indicated by the cessation and resumption of vesicle movement upon the addition and removal of nocodazole, respectively.

Overall, cell surface area would DECREASE if:

correct answer is e) a and d are true. a. If the rate of clathrin-coated pit formation increased: Clathrin-coated pits are involved in endocytosis, the process of bringing substances into the cell. When clathrin-coated pits form more rapidly, it leads to increased endocytosis and internalization of materials from the cell surface. This results in a decrease in cell surface area. d. If the rate of constitutive exocytosis decreased: Constitutive exocytosis is the process by which materials are continuously secreted from the cell. When the rate of constitutive exocytosis decreases, fewer materials are being released from the cell, resulting in a decrease in the overall cell surface area. Both increased clathrin-coated pit formation (option a) and decreased constitutive exocytosis (option d) contribute to a decrease in cell surface area. Therefore, option e) a and d are true. incorrect: b. If the rate of clathrin-coated pit formation decreased: This option suggests that a decrease in the rate of clathrin-coated pit formation would lead to a decrease in cell surface area. However, clathrin-coated pits are involved in endocytosis, the process of bringing substances into the cell. If the rate of clathrin-coated pit formation decreases, there would be less internalization of materials from the cell surface, which would result in an increase in cell surface area, not a decrease. c. If the rate of constitutive exocytosis increased: Constitutive exocytosis is the process by which materials are continuously secreted from the cell. If the rate of constitutive exocytosis increased, more materials would be released from the cell, which would lead to an increase in the overall cell surface area. Therefore, an increased rate of constitutive exocytosis would not cause a decrease in cell surface area. In summary, the correct answer is e) a and d are true, as an increased rate of clathrin-coated pit formation (option a) and a decreased rate of constitutive exocytosis (option d) would both contribute to a decrease in cell surface area. The other options (b and c) are incorrect because they suggest scenarios that would actually result in an increase in cell surface area.

58. A research lab at UConn is testing a new drug that blocks the activity of formin proteins. When this drug is added to human osteoblasts under conditions that favor collagen secretion (i.e., in the presence of vitamin C), the collagen folds into a triple helix and accumulates in the Golgi apparatus. How would you interpret these results?

correct interpretation of the results is: b. Linear actin assembly is important for collagen trafficking out of the Golgi. The given information states that when the drug that blocks the activity of formin proteins is added to human osteoblasts under conditions that favor collagen secretion, the collagen folds into a triple helix and accumulates in the Golgi apparatus. This suggests that the drug's effect on formin proteins, which are involved in linear actin assembly, disrupts the normal trafficking of collagen out of the Golgi. Linear actin assembly has been implicated in the movement and trafficking of cargo within cells, including from the Golgi apparatus to other cellular compartments. In this case, the inhibition of formin activity by the drug likely hinders the formation of linear actin structures necessary for collagen trafficking out of the Golgi. As a result, collagen accumulates within the Golgi apparatus. incorrect: a. The statement "collagen does not fold properly in the presence of the drug" is incorrect because the information provided states that collagen folds into a triple helix in the presence of the drug. This suggests that the drug does not affect the folding of collagen but rather impacts its trafficking. c. The statement "linear actin assembly is important for Golgi-to-ER collagen trafficking" is incorrect because the information provided indicates that collagen accumulates in the Golgi apparatus in the presence of the drug. This suggests that the drug's effect on formin proteins disrupts the trafficking of collagen out of the Golgi rather than the Golgi-to-ER trafficking. d. The statement "branched actin assembly is important for ER-to-Golgi collagen trafficking" is incorrect because there is no mention of the drug affecting the ER-to-Golgi trafficking of collagen in the given information. The drug specifically blocks the activity of formin proteins, which are primarily involved in linear actin assembly. e. The statement "branched actin assembly is important for collagen trafficking out of the Golgi" is incorrect because the given information does not provide any indication that branched actin assembly is involved in collagen trafficking out of the Golgi. The drug's effect on formin proteins suggests that linear actin assembly is disrupted, leading to the accumulation of collagen in the Golgi.

6. Which statement(s) about the non-uniformity of lipid bilayers is/are correct?

correct: b. A given leaflet of the bilayer of a plasma membrane can vary in its composition from location to location. c. The two leaflets of the lipid bilayer making up the plasma membrane do not have identical composition d. b and c are correct heterogeneous gives to membrane to different regions and different inner regions, different lipids inner and outer leaflet Incorrect: a. All membranes in a particular eukaryotic cell type have the same composition. e. None is correct.

37. If you engineered a new protein by taking the gene for a single pass transmembrane protein that has an N-terminal signal sequence and remove the stop transfer sequence from the middle of the gene, what is a possible destination of the new protein?

e. None is a possible destination. By removing the stop transfer sequence from the middle of the gene encoding a single-pass transmembrane protein with an N-terminal signal sequence, the normal mechanism for anchoring the protein within the membrane would be disrupted. The stop transfer sequence is essential for correctly positioning the protein within the membrane. Without the stop transfer sequence, the protein would not be able to properly integrate into any membrane or organelle. It would likely result in a misfolded or non-functional protein that might be targeted for degradation or undergo abnormal cellular localization. Therefore, none of the options provided would be a possible destination for the new protein. Once again, I apologize for the confusion, and thank you for pointing out the error. The correct answer is e. None is a possible destination.

44.Which is the FOURTH step of the five listed in lysosomal enzyme sorting? Co-translational addition of an oligosaccharide to the protein. Binding of Man-6-Phosphate to its receptor in the TGN. Removal of the man-6-phosphate signal Transport of the enzyme from the TGN Transport from the ER to the Golgi

d. Transport of the enzyme from the trans-Golgi network (TGN). After lysosomal enzymes are synthesized in the endoplasmic reticulum (ER) and undergo co-translational addition of an oligosaccharide (option a), they are transported from the ER to the Golgi apparatus (option e). In the Golgi apparatus, the enzymes receive mannose-6-phosphate (Man-6-P) residues on their oligosaccharides through the action of the enzyme N-acetylglucosamine-1-phosphotransferase (option b). The Man-6-P serves as a targeting signal for the lysosomes. Next, the enzymes are transported from the Golgi apparatus to the TGN. In the TGN, the Man-6-P receptors recognize and bind to the Man-6-P residues on the lysosomal enzymes (option b). This binding ensures the sorting and packaging of the enzymes into vesicles for transport to lysosomes. The removal of the Man-6-P signal occurs after the enzymes have been delivered to the lysosomes (option c). This removal allows the enzymes to function properly within the lysosomes. Therefore, the correct answer is d. Transport of the enzyme from the TGN.

41. Put the following in the correct order:I. Protein is imported through the TIM complexII. Protein is imported through the TOM complexIII. Fully-folded, mature protein is located in the mitochondrial matrix IV. Protein is folded by mitochondrial chaperone using ATPV. Import signal is recognized by receptor protein a. V, I, II, IV, IIIb. II, V, III, I, IVc. II, III, V, IV, Id. V, II, I, IV, IIIe. none of the above is correct.

d. V, II, I, IV, III The correct sequence is as follows: Import signal is recognized by a receptor protein (V). Protein is imported through the TOM complex (Translocase of the Outer Mitochondrial Membrane) (II). Protein is imported through the TIM complex (Translocase of the Inner Mitochondrial Membrane) (I). Protein is folded by mitochondrial chaperones using ATP (IV). Fully-folded, mature protein is located in the mitochondrial matrix (III). Therefore, the correct answer is d. V, II, I, IV, III.

24. Which of the following mechanisms could cause an increase in cytosolic calcium ion levels in response to an appropriate stimulus?

d. a and c (Interaction of an extracellular messenger with a G protein coupled receptor causing the activation of phospholipase C, and extracellular messengers signaling through receptor tyrosine kinases to activate phospholipase C). Both of these mechanisms can lead to the activation of phospholipase C, which generates inositol trisphosphate (IP3) as a second messenger. IP3 then binds to IP3 receptors on the endoplasmic reticulum, triggering the release of calcium ions from intracellular stores into the cytosol, resulting in an increase in cytosolic calcium ion levels. Certainly! Let's go through the other options and explain why they are false: a. Interaction of an extracellular messenger with a G protein-coupled receptor causing the activation of phospholipase C. This option is true and can cause an increase in cytosolic calcium ion levels. Activation of phospholipase C leads to the cleavage of phosphatidylinositol 4,5-bisphosphate (PIP2) into inositol trisphosphate (IP3) and diacylglycerol (DAG). IP3, as mentioned earlier, triggers the release of calcium ions from intracellular stores. b. Extracellular messengers signaling through receptor tyrosine kinases to activate members of the MAPK subfamily. This option is false. Activation of members of the MAPK (Mitogen-Activated Protein Kinase) subfamily does not directly result in an increase in cytosolic calcium ion levels. MAPK signaling pathways typically regulate cellular processes such as cell growth, differentiation, and proliferation, but they do not directly impact calcium ion levels. c. Extracellular messengers signaling through receptor tyrosine kinases to activate phospholipase C. This option is true and can cause an increase in cytosolic calcium ion levels. Activation of phospholipase C through receptor tyrosine kinases can lead to the production of IP3, which, as mentioned earlier, triggers the release of calcium ions from intracellular stores.

82. Which of these is an abnormality that can delay the metaphase-to-anaphase transition?

e. All of these abnormalities could delay the metaphase-to-anaphase transition. a. attachment of both kinetochores to microtubules from the same spindle pole b. a failure of one kinetochore to attach to spindle microtubulesc. a mutation that inhibits the activity of the APCd. a lack of tension at the kinetochore e. all of these could delay progress Let's break down each option to understand why they can lead to a delay in the metaphase-to-anaphase transition: a. Attachment of both kinetochores to microtubules from the same spindle pole: During metaphase, the kinetochores, which are protein structures on the chromosomes, are supposed to attach to microtubules emanating from opposite spindle poles. This balanced attachment ensures proper alignment and tension. If both kinetochores attach to microtubules from the same spindle pole, it leads to an unbalanced force distribution and can delay the metaphase-to-anaphase transition. b. Failure of one kinetochore to attach to spindle microtubules: Proper attachment of kinetochores to spindle microtubules is essential for accurate chromosome segregation. If one kinetochore fails to attach to microtubules, it can prevent the chromosomes from aligning properly at the metaphase plate, thereby delaying the transition to anaphase. c. Mutation inhibiting the activity of the APC (Anaphase-Promoting Complex): The APC is a protein complex responsible for regulating the metaphase-to-anaphase transition by initiating the degradation of specific proteins, including securin and cyclin. If there is a mutation that inhibits the activity of the APC, it can prevent the degradation of securin and cyclin, leading to a delay in the metaphase-to-anaphase transition. d. Lack of tension at the kinetochore: Tension at the kinetochore, generated by opposing forces from microtubules attached to sister chromatids, is crucial for maintaining chromosome alignment and ensuring proper segregation. If there is a lack of tension due to improper attachment or other factors, it can activate the spindle assembly checkpoint, which halts the metaphase-to-anaphase transition until tension is properly established. Therefore, any of these abnormalities can disrupt the normal progression from metaphase to anaphase, leading to a delay in this transition.

51. All of the following could prevent a secretory protein from being secreted normally EXCEPT:

e. Glycosylation of the protein in the ER. Glycosylation of a secretory protein in the endoplasmic reticulum (ER) is a normal and essential post-translational modification. It involves the addition of carbohydrate chains (oligosaccharides) to specific amino acid residues on the protein. Glycosylation serves various functions, including protein folding, stability, and proper trafficking within the secretory pathway. The other options listed can all prevent a secretory protein from being secreted normally: a. A mutation that results in the deletion of the start transfer sequence: The start transfer sequence is a signal sequence at the N-terminus of a protein that directs it to the ER for translocation and secretion. If this sequence is deleted due to a mutation, the protein may fail to enter the ER correctly and consequently be prevented from normal secretion. b. The addition of an ER-retrieval sequence: An ER-retrieval sequence is a specific signal that targets proteins from the Golgi complex back to the ER. If this sequence is added to a secretory protein, it will be recognized as an ER-resident protein and retrieved back to the ER instead of being transported to its intended destination. c. A mutation that causes the addition of mannose-6-phosphate to the carbohydrate tree structure: Mannose-6-phosphate is a specific modification that targets proteins to the lysosomes. If a secretory protein acquires this modification due to a mutation, it will be diverted from the normal secretory pathway and targeted for degradation within the lysosomes. d. A mutation that prevents the addition of N-linked oligosaccharides in the ER: N-linked glycosylation is a common type of protein glycosylation that occurs in the ER. If a mutation affects the process of adding N-linked oligosaccharides to a secretory protein, it can lead to improper folding, trafficking, or recognition of the protein, ultimately preventing its normal secretion. In summary, the option that does not prevent normal secretion of a secretory protein is e. Glycosylation of the protein in the ER. Glycosylation is a normal and necessary process for proper protein function and secretion within the secretory pathway.

30. What effect does the docking of the SRP/ribosome/growing polypeptide chain complex to the SRP receptor have on protein synthesis? a. The SRP will release from the ribosome.b. Protein synthesis ceases temporarily.c. Protein synthesis resumes.d. The ribosome releases the growing polypeptide chain. e. a and c are correct.

e. a and c are correct. When the SRP/ribosome/growing polypeptide chain complex docks to the SRP receptor, both of the following effects occur: a. The SRP will release from the ribosome: The docking of the SRP/ribosome/growing polypeptide chain complex to the SRP receptor triggers the release of the SRP (signal recognition particle) from the ribosome. The SRP is responsible for targeting the ribosome-nascent chain complex to the endoplasmic reticulum (ER) membrane. c. Protein synthesis resumes: After the SRP releases from the ribosome, protein synthesis resumes. The ribosome continues the process of translating the mRNA and synthesizing the polypeptide chain. Therefore, options a and c are both correct. The docking of the SRP/ribosome/growing polypeptide chain complex to the SRP receptor leads to the release of the SRP from the ribosome and allows protein synthesis to resume.

28. Competition experiments have been used to determine that there are different kinds of exportins for each species of RNA. You are studying importins for proteins. You translate in vitro two different nls-containing proteins. Protein A is fused to GFP, while the other (protein B) is left unlabeled. Which statement below correctly describes results and interpretations that you might obtain from observing the behavior of these proteins injected into the cytoplasm of oocytes?

e. b and c. Here's an explanation of the options: a. Injecting an excess of protein B would not affect import of protein A into the nucleus if both use the same importin: This statement is not necessarily true. If protein A and protein B use the same importin for nuclear import, injecting an excess of protein B could potentially compete with protein A for binding to the importin, thereby affecting the import of protein A into the nucleus. b. Injecting an excess of protein B would decrease import of Protein A into the nucleus if both use the same importin: This statement is likely true. If protein A and protein B use the same importin for nuclear import and an excess of protein B is injected, it would likely compete with protein A for binding to the importin, leading to a decrease in the import of protein A into the nucleus. c. Injecting an excess of Protein B would not alter import of Protein A into the nucleus if both use different importins: This statement is likely true. If protein A and protein B use different importins for nuclear import and an excess of protein B is injected, it would not directly affect the import of protein A, as they utilize different importin pathways. Therefore, the correct answer is e. b and c. Injecting an excess of protein B would likely decrease the import of protein A if they use the same importin, but it would not alter the import of protein A if they use different importins.

62. ParM is a cytoskeletal protein expressed in some E.coli bacteria. (1) ParM is non- polar, (2) ParM has ATPase activity, and (3) ParM forms polymers that exhibit dynamic instability. How do these three properties compare to those found in eukaryotic cytoskeletal proteins?

e. it is 1- like intermediate filament proteins, 2- like actin, 3- like tubulin Explanation: Statement 1 states that ParM is non-polar. This property is similar to intermediate filament proteins found in eukaryotic cytoskeletal proteins. Intermediate filament proteins, such as keratins, vimentin, and neurofilaments, are also non-polar and form a structural network in the cytoplasm. Statement 2 states that ParM has ATPase activity. This property is similar to actin, which is a major component of the eukaryotic cytoskeleton. Actin also has ATPase activity, which is important for regulating actin filament dynamics and functions. Statement 3 states that ParM forms polymers that exhibit dynamic instability. This property is similar to tubulin, which is a component of microtubules in eukaryotic cells. Microtubules also exhibit dynamic instability, constantly undergoing assembly and disassembly. Therefore, ParM exhibits properties similar to intermediate filament proteins (1), actin (2), and tubulin (3) found in eukaryotic cytoskeletal proteins. a. it is 1- like tubulin, 2- like actin, 3- like tubulin: This option suggests that ParM is similar to tubulin in terms of polarity (1), similar to actin in terms of ATPase activity (2), and similar to tubulin in terms of forming polymers with dynamic instability (3). However, ParM is non-polar, which makes it different from tubulin. Additionally, ParM does not exhibit actin-like ATPase activity. Therefore, this option is incorrect. b. it is 1- like intermediate filament proteins, 2- like tubulin, 3- like tubulin: This option suggests that ParM is similar to intermediate filament proteins in terms of polarity (1), similar to tubulin in terms of ATPase activity (2), and similar to tubulin in terms of forming polymers with dynamic instability (3). However, ParM does not exhibit tubulin-like ATPase activity. Therefore, this option is incorrect. d. it is 1- like actin, 2- like actin, 3- like actin: This option suggests that ParM is similar to actin in terms of polarity (1), similar to actin in terms of ATPase activity (2), and similar to actin in terms of forming polymers with dynamic instability (3). While ParM shares some similarities with actin, it is non-polar, which makes it different from actin. Therefore, this option is incorrect. Based on the provided properties of ParM (non-polar, ATPase activity, dynamic instability), the correct option is e. it is 1- like intermediate filament proteins, 2- like actin, 3- like tubulin.

16. the permeability to ____ is most responsible for generating the negative value of the resting membrane potential in mammalian cells.

e. none of the above ions are most responsible ion. -potassium ions are most responsible, resting potential determined by bye concentration gradients of ins across by the membrane permeability for Na+ and K+, ions move down gradients via channels has a separation of charges of the resting potential closest to equilibrium potential of potassium ion(potential would be generated by K+ if this is the only ion in the system). -resting membrane potential is slightly positive or depolarized in the equilibrium potential for resting membrane potential inward a potassium channel. Resting calcium conductance is exceedingly small, calcium does not contribute too resting membrane potential.

68. During actin assembly in membrane protrusions, which of these five steps occurs second? a. a Rho-family GTPase is activatedb. the Arp2/3 complex becomes activatedc. a WASP-family protein is activatedd. actin filament growth is stopped by CapZ e. an actin filament elongates

the second step is typically the activation of the Arp2/3 complex. a. Activating a Rho-family GTPase, such as Cdc42 or Rac, is an upstream event that occurs before the activation of the Arp2/3 complex. c. Activating a WASP-family protein is also an upstream event that occurs before the activation of the Arp2/3 complex. d. Stopping actin filament growth by CapZ is a regulatory event that typically occurs after the activation of the Arp2/3 complex. e. Actin filament elongation is an ongoing process that occurs after the activation of the Arp2/3 complex. Therefore, the correct answer is b. The activation of the Arp2/3 complex occurs as the second step during actin assembly in membrane protrusions.

The resolution limit of a typical fluorescent microscope is ~200 nm. If two ribosomes that are 50 nm in diameter and 500 nm apart are labeled with a fluorescent antibody, what will you see?

will be able to detect at 500 nm distance apart the two diameters, the two spots will be able to view both as two separate 50 nm diameter spots. Higher resolution means wider and smaller diffraction, lower numerical apenditire has lower lens and higher diffraction. It means it is differentiation. If this was below the resolution limit is below 200 nm, it will always look at 200 nm.

Which would be the best way to visualize dynamics of an intracellular protein in a living cell?

wrong: a. using a fluorescently labeled antibody raised against the protein. can occur on the inside immunofluorescence: fixable(formaldehyde) is frozen to place to permebeclize to the membrane and add detergent to poke wholes into the membrane that will kill the cell in order to access proteins. c. using a gold-labeled antibody in combination with transmission electron microscopy sticks to the elements in the cell wall and causes it to model the cell, but the internal components of the cell is killed d. immunoprecipitating the protein from a cytosolic extract homogenising the cells -antibodies, precipitating a protein and antibody to isolate the final to "pull down complexes" in the crushed cell correct: b. expressing a fusion of the protein to GFP promoter(RNA) add to protein to make it green, CMV promoter is strong genetic RNA transferase, all cell types express this