Cellular bio Flashcards

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1
Q

Which of the following is NOT a colligative property?

A. Freezing-point depression

B. Boiling-point elevation

C. Diffusion

D. Vapor pressure

A

C. Diffusion

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2
Q

Secreted and membrane-bound proteins must go through the secretory pathway, targeted there by a signal sequence. The signal sequence must be composed of:

A. hydrophilic amino acids.

B. hydrophobic amino acids.

C. basic amino acids.

D. acidic amino acids.

A

B. hydrophobic amino acids.

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3
Q

What type of cellular junction allows for exchange of cytosol between cells?

A

Gap junctions

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4
Q

At what juncture point is mitosis most heavily regulated?

A

Between G1 and S phases

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5
Q

What is the correct order in the steps of apoptosis?

A

Disassemble cytoskeleton,

break down nuclear membrane,

break down genome,

phagocytic digestion

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6
Q

All of the following are characteristics of G-protein mediated cell signaling EXCEPT:

A. they are the mechanism by which steroid hormones exert their effects.

B. they can result in increased intracellular calcium.

C. they are an example of signal amplification.

D. they are fast and temporary.

A

A. they are the mechanism by which steroid hormones exert their effects.

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7
Q

In Kartagener’s syndrome, defective dynein is produced causing a paralysis of microtubule-based movement of flagellae and cilia. One could expect to find all of the following outcomes EXCEPT:

A. failure to ovulate in women.

B. male infertility.

C. chronic lung infections.

D. ectopic pregnancy in women.

A

A. failure to ovulate in women.

  • A male with Kartagener’s syndrome would be infertile due to immobile sperm, eliminating male infertility.
  • Ova would not enter the Fallopian tubes normally, due to the lack of cilia, causing increased risk of ectopic pregnancy, eliminating ectopic pregnancy in women.
  • The lungs also require cilia to remove bacteria and other particulates, eliminating chronic lung infections.
  • Ovulation, however, is determined by levels of circulating hormones and will not be affected by the lack of dynein.
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8
Q

What organelle would be most closely associated with exocytosis of newly synthesized secretory protein?

A. Lysosomes

B. Golgi apparatus

C. Ribosomes

D. Peroxisomes

A

B. Golgi apparatus

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9
Q

Which types of protein are produced by the ribosomes of the rough endoplasmic reticulum?

I. Membrane-bound

II. Integral

III. Secreted

A. I and II only

B. II and III only

C. I, II, and III

D. I and III only

A

C. I, II, and III

Lysosomal proteins are integral proteins

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10
Q

Compound X binds specifically and irreversibly to the potassium-binding site of sodium-potassium ATPase pumps, inhibiting the activity of this protein. Which of the following would most likely result from poisoning cells with compound X?

A. ATP would still be produced and ion pumping would continue.

B. ATP production would cease.

C. ATP hydrolysis would increase.

D. Secondary active transport would be impaired.

A

D. Secondary active transport would be impaired.

The Na+/K+ ATPase uses the energy from ATP hydrolysis to pump Na+ out of the cell and K+ into the cell, creating concentration gradients for these two ions. The Na+ pumped out can then be used in secondary active transport, where the energy of the Na+ concentration gradient is used to move other molecules into or out of the cell.

If Compound X were to inhibit the activity of this pump, the Na+ concentration gradient would not be formed, and secondary active transport would not be possible. ATPases hydrolyze ATP, they do not synthesize it, and if the pump is inhibited, ATP hydrolysis would be inhibited.

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11
Q

Large proteins, most notably albumin, are dissolved in the plasma and serve an important role in regulation of plasma volume. Reducing the amount of albumin to below-normal levels would most likely have which of the following effects?

A. Movement of water from the tissues into the bloodstream with resulting dehydration, due to increased osmotic pressure.

B. Movement of water from the tissues into the bloodstream with resulting dehydration, due to reduced hydrostatic pressure.

C. Movement of water from the bloodstream into the tissues with resulting swelling, due to increased hydrostatic pressure.

D. Movement of water from the bloodstream into the tissues with resulting swelling, due to reduced osmotic pressure.

A

D. Movement of water from the bloodstream into the tissues with resulting swelling, due to reduced osmotic pressure

The reduced osmotic pressure of the blood will have the effect of allowing water to leave the bloodstream and enter the tissues, where it will cause the tissues to swell.

(You can also think of it as the tissues’ relative osmotic pressure increasing, thus they have a greater tendency to draw water out of the blood.)

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12
Q

Fluorescence recovery after photobleaching (FRAP) involves the bleaching (a non-reversible process) of a fluorescent compound and observation of the bleached spot for new fluorescent signal. If a circular spot is bleached in a cell membrane uniformly containing high levels of a fluorescent transmembrane protein, what best describes the change in fluorescence over time?

A. Recovery of fluorescence beginning at the periphery of the spot due to lateral diffusion of proteins in the membrane.

B. Recovery of fluorescence beginning at the center of the spot due to the action of flipases.

C. Recovery of fluorescence beginning at the center of the spot due to relaxation of excited electrons.

D. Recovery of fluorescence beginning at the periphery due to diffusion into the membrane of newly translated proteins in the cytoplasm.

A

A. Recovery of fluorescence beginning at the periphery of the spot due to lateral diffusion of proteins in the membrane.

Photobleaching involves the permanent quenching of fluorescence. Given this is an integral membrane protein, it is capable of lateral diffusion within the cell membrane and non-bleached protein will begin to encroach on the bleached spot shortly after bleaching.

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13
Q

Vertebrates have developed various renal structures for osmoregulation based on their habitats. Bony fish that live in seawater drink large amounts of seawater and use cells in gills to pump excess salt out of the body. This is in response to:

A. a need to maintain their tissues in a hypoosmotic state.

B. an influx of water by osmosis into their tissues.

C. a loss of water by active transport to their hypertonic surroundings.

D. a loss of salt to their surroundings.

A

A. a need to maintain their tissues in a hypoosmotic state.

Bony fish living in saltwater maintain hypoosmotic body fluids. The hyperosmotic water they live in tends to draw the water out of their body by osmosis. They can obtain water by ingesting ocean water, but this raises their internal osmolarity. They excrete salt, then, to lower their internal osmolarity.

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14
Q

Which of the following is an example of passive transport?

A. Movement of protons into the mitochondrial matrix in oxidative phosphorylation, with the proton influx driving ATP synthesis.

B. Transport of protons out of the mitochondrial matrix by the electron transport chain.

C. Movement of a bolus of food through the digestive tract.

D. Absorption of glucose from the digestive tract.

A

A. Movement of protons into the mitochondrial matrix in oxidative phosphorylation, with the proton influx driving ATP synthesis.

In oxidative phosphorylation, protons move down their gradient from the intermembrane space of the mitochondria into the mitochondrial matrix. Any time a substance moves across a membrane, down its gradient, the process is a form of passive transport.

Do not let the fact that the proton movement is driving ATP formation confuse you; even though the protons are moving through the ATP synthase, they are still moving down their gradient so the process is passive.

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15
Q

A physiologist observes a dark, dense subcellular structure via electron microscopy that spans the space between two adjacent cell membranes and extends to the underlying cytoskeleton. Upon further investigation, she discovers that two cells with this structure are electrically isolated but that a tracer dye can travel between the cells through the pericellular space. What is the identity of this structure?

A. Communicating junction

B. Tight junction

C. Gap junction

D. Desmosome

A

D. Desmosome

Desmosomes are a type of anchoring junction which connect adjacent cells and extend through the cell membrane to associate with components of the cytoskeleton.

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16
Q

How do deosomes differ from gap junctions and tight junctions?

A

Unlike gap junctions (also known as communicating junctions), desmosomes do not link the cytoplasms of the two cells (the cells are not electrically coupled). Tight junctions form connections with adjacent cells but prevent pericellular diffusion of molecules, such as any kind of dye, where as a deosome would allow a dye through.

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17
Q

If erythrocytes (RBCs) are placed into a hypertonic solution, they will:

A. hemolyze.

B. shrivel.

C. remain the same.

D. swell up.

A

B. shrivel.

Hypertonic = shriveling

hypotonic = swelling

In a hypertonic solution, the extracellular environment has a higher solute concentration. Osmotic pressure would draw water out of the cell, making solutes more concentrated inside the cell, and making the cell shrivel.

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18
Q

As the molality of a solution increases, how are its boiling point and vapor pressure affected?

A. Both boiling point and vapor pressure decrease.

B. Boiling point decreases and vapor pressure increases.

C. Both boiling point and vapor pressure increase.

D. Boiling point increases and vapor pressure decreases.

A

D. Boiling point increases and vapor pressure decreases.

Boiling point increases with an increasing concentration of solutes because more energy is required to separate the water from those solutes.

Vapor pressure decreases with an increasing concentration of solutes; since it takes more energy to trigger evaporation, there will be less solvent evaporating, and thus less pressure due to those evaporated solvent molecules.

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19
Q

A hypertonic solution creates a gradient that will draw fluid through a semi-permeable membrane. Which of the following is a true statement?

A. Osmotic pressure is the force used to allow more solutes to be dissolved in the hypertonic solution.

B. Osmotic pressure is the force exerted to push enough fluid into the solution to make it hypotonic.

C. Osmotic pressure is the force used to push solutes through the semi-permeable membrane to balance the concentrations.

D. Osmotic pressure is the force necessary to counterbalance the pull on fluid exerted by the hypertonic solution.

A

D. Osmotic pressure is the force necessary to counterbalance the pull on fluid exerted by the hypertonic solution.

A hypertonic solution creates a gradient that will draw fluid through a semi-permeable membrane. Osmotic pressure is the pressure used to counteract osmosis or the force needed to keep the solutes in the hypertonic solution from drawing on the available fluid is a true statement.

It does not move or alter solutes and will not change the tonicity of the solution.

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20
Q

How does freezing-point depression work to defy the thermodynamic principle of entropy?

A. During freezing, solvent first forms crystals before solutes are organized into an array; thus higher solute concentrations require lower temperatures to freeze.

B. During freezing, solvent first forms crystals before solutes are organized into an array; thus lower solute concentrations require lower temperatures to freeze.

C. During freezing, solutes are first organized into an array before the solvent forms crystals; thus higher solute concentrations require lower temperatures to freeze.

D. During freezing, solutes are first organized into an array before the solvent forms crystals; thus lower solute concentrations require lower temperatures to freeze.

A

C. During freezing, solutes are first organized into an array before the solvent forms crystals; thus higher solute concentrations require lower temperatures to freeze.

In the process of freezing, solutes are arrayed before the solvent freezes. The more solute present in the solution, the more energy must be removed from that system to organize those solutes into the requisite array. Thus, even more energy needs to be removed to then freeze the solvent.

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21
Q

Clathrin, a substance that aggregates on the cytoplasmic side of cell membranes, is responsible for the coordinated pinching off of membrane in receptor-mediated endocytosis. A lipid-soluble toxin that inactivates clathrin would be associated with:

A. increased protein production on the rough endoplasmic reticulum.

B. increased secretion of hormone into the extracellular fluid.

C. an increase in ATP consumption.

D. reduced delivery of polypeptide hormones to endosomes.

A

D. reduced delivery of polypeptide hormones to endosomes.

Endocytosis is the process by which the cell internalizes receptor–ligand complexes from the cell surface, such as polypeptide hormones bound to their receptor. At the cell surface, the receptor–ligand complexes cluster in clathrin-coated pits and pinch off the vesicles that join acidic compartments known as endosomes.

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22
Q

How does the van’t Hoff factor of NaCl compare to that of NH3?

A. The i of NaCl is 2 while the i of NH3 is 4.

B. The i of both NaCl and NH3 is 1.

C. The i of NaCl is 2 while the i of NH3 is 1.

D. The i of both NaCl and NH3 is 2.

A

C. The i of NaCl is 2 while the i of NH3 is 1.

Since NaCl is ionic, it breaks into Na+ and Cl– giving an i = 2.

Since NH3 is non-ionic, it does not break down and thus has an i = 1.

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23
Q

When it is time to breed, salmon travel from saltwater, in which they are hypotonic, to freshwater, in which they are hypertonic. They maintain solute balance by reversing their osmoregulatory machinery when moving between the two environments. Failure to reverse this machinery when moving to their breeding grounds would most likely result in:

A. improved metabolic activity, as enzyme concentrations increased.

B. no change, because movement from a hypertonic to a hypotonic medium does not present osmotic challenges.

C. death, as cells became too concentrated to carry out normal metabolism.

D. death, as cells underwent lysis due to water influx.

A

D. death, as cells underwent lysis due to water influx.

In a hypotonic environment (freshwater), the cells will have higher osmotic pressure than the surrounding environment. Water would tend to flow into the cells, causing them to swell and eventually burst.

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24
Q

The Na+/K+ ATPase, found in the membranes of all cells, establishes concentration gradients of Na+ and K+ across the cell membrane. Furthermore, K+ can cross the cell membrane though K+ leak channels; the overall effect of the two membrane proteins is to maintain osmotic balance for the cell and to establish an electrical potential across the cell membrane. The movement of K+ through the leak channels is best characterized as:

A. simple diffusion.

B. facilitated diffusion.

C. primary active transport.

D. secondary active transport.

A

B. facilitated diffusion.

if the molecule moves through a channel or a carrier protein, it is specifically called facilitated diffusion.

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25
Q

A researcher discovers that a molecule of interest can travel across the cell membrane. To characterize how this happens, he increases the external concentration and records the rate of entry into the cell. He soon discovers that there appears to be no limit (within physiological conditions) to how quickly the compound can enter the cell. How is the molecule entering the cell?

A. Facilitated diffusion

B. Secondary active transport

C. Primary active transport

D. Simple diffusion

A

D. Simple diffusion

The molecule that can enter the cell membrane with no effective upper limit to its rate is undergoing simple diffusion.

26
Q

Albumin, an extracellular protein found in the plasma, is labeled with green fluorescent protein and translated. In which of the following locations would the fluorescent signal be WEAKEST during production?

A. -Golgi network

B. Endosomes

C. Secretory vesicles

D. Endoplasmic reticulum

A

B. Endosomes

Fluorescent albumin would begin being detected at the rough endoplasmic reticulum following its translation. Next it would travel to the cis-Golgi and progress through the compartments to the trans-Golgi where it would ultimately be shuttled to the cell exterior via a secretory vesicle. Endosomes are formed following endocytosis and it is unlikely that a significant quantity of albumin would be present in endosomes.

27
Q

Secondary active transport can move Molecule A against its gradient by linking that transport to the movement of Molecule B down its gradient. Typically, the gradient of Molecule B is established by a primary active transporter. Which of the following could NOT be Molecule A?

A. vitamin D

B. glucose

C. bicarbonate

D. dipeptide

A

A. vitamin D

In order to be moved across the membrane by secondary active transport, a molecule must not be able to cross on its own. Dipeptides, glucose, and bicarbonate are polar, and as such cannot diffuse across the membrane. However vitamin D is a fat-soluble vitamin derived from cholesterol and can freely cross the cell membrane. It would not require a transporter of any kind.

28
Q

Muscle cells share what structural component with the microfilaments of a cell’s cytoskeleton?

A. Dynein

B. Actin

C. Myosin

D. Sarcomere

A

B. Actin

Microfilaments are built from polymers of actin, as are the thin filaments within muscle cells

29
Q

Which of the following eukaryotic cytoskeleton components is made up of protein filaments that consist of tubulin subunits?

A. Intermediate filaments

B. Microfilaments

C. Actin

D. Microtubules

A

D. Microtubules

30
Q

What class of enzymes is responsible for both the initiation and effector stages of apoptosis?

A. Caspases

B. Hydrolases

C. Carboxylases

D. Lipases

A

A. Caspases

Caspases carry out the work of apoptosis. Initiators are drawn to the cell due to internal or external triggers and self-activate to create a cascade that then allows effectors to carry out the steps of programmed cell death.

31
Q

Kinesin plays a significant role in mitosis and helps in the spindle apparatus assemble. In which direction does kinesin travel and in which phase of mitosis does its activity begin?

A. Away from the MTOC; prophase

B. Toward the MTOC; anaphase

C. Away from the MTOC; anaphase

D. Toward the MTOC; prophase

A

A. Away from the MTOC; prophase

Kinesin would travel away from the microtubule organizing center (MTOC) and begin activity in the prophase when microtubule assembly begins. The spindle apparatus, composed of microtubules, assembles with the minus ends toward the MTOC and the plus ends away from the MTOC. Kinesin travels in the plus direction and would be visible in the prophase as this is when the spindle apparatus begins forming.

32
Q

If tumor suppressor activity was induced exogenously to exceed normal levels by several-fold, which cellular process would most likely be blocked?

A. Mitosis

B. Transcription

C. Meiosis

D. Translation

A

A. Mitosis

33
Q

Tumor suppressor genes produce proteins that either prevent a cell’s conversion to a cancerous state or stop a cell from multiplying if the conversion occurs. Which of the following would be the best indicator of the need for tumor suppressor activity?

A. Intracellular transport via microtubules.

B. Heavy regulation between the G1 and S phases of the cell cycle.

C. Eukaryotic transcripts produced without splicing or other post-processing.

D. DNA damage and unusual levels of replication activity.

A

D. DNA damage and unusual levels of replication activity.

Tumor suppressor activity is typically triggered by unusual occurrences with genes rather than with transcriptional activity

34
Q

Why is the first step of apoptosis to disassemble the cytoskeleton?

A. Disassembling the cytoskeleton provides a channel for cytokines to enter the cell.

B. Disassembling the cytoskeleton allows for the filaments to be recycled as part of programmed cell death.

C. Disassembling the cytoskeleton provides a source of ATP to power the process of programmed cell death.

D. Disassembling the cytoskeleton allows the cell to shrink, thus minimizing damage to neighboring cells as it is destroyed.

A

D. Disassembling the cytoskeleton allows the cell to shrink, thus minimizing damage to neighboring cells as it is destroyed.

35
Q

How could an oncogene impact a fully differentiated, non-dividing cell?

A. Oncogenes could trigger the cell to leave G1 and begin cycling through S phase.

B. Oncogenes could trigger the cell to leave G0 and begin cycling through S phase.

C. Oncogenes could trigger the cell to leave M phase and begin cycling through G0 phase.

D. Oncogenes could trigger the cell to leave G2 and begin cycling through M phase.

A

B. Oncogenes could trigger the cell to leave G0 and begin cycling through S phase.

Fully differentiated, non-dividing cells reside in G0. Since oncogenes deregulate cell growth by increasing replication and division, the oncogene would need to shift the cell from G0 into the start of an active cell cycle and the replication events of S phase.

36
Q

Where in the adult human is both mitosis and meiosis occurring in parallel?

A. Seminiferous tubules

B. Ovaries

C. Bone marrow

D. Hypodermis

A

A. Seminiferous tubules

37
Q

Since proto-oncogenes have the ability to become cancerous, why are they maintained as part of the genome?

A. Proto-oncogenes stimulate multipotent stem cells and are needed for hematopoietic production.

B. Proto-oncogenes prevent transposons from creating further cancer-causing mutations.

C. Proto-oncogenes only become cancerous after rare recombinational events and thus the cancer risk is low.

D. Proto-oncogenes often encode gene products, like growth factors, that are necessary for the cell when expressed at normal levels.

A

D. Proto-oncogenes often encode gene products, like growth factors, that are necessary for the cell when expressed at normal levels.

Proto-oncogenes are often “normal” versions of a gene that encode a needed cellular product. Gene mutation or exposure to a mutagen can convert them to their oncogenic state, but the cell cannot rid itself of the gene due to what it makes.

They become oncogenic through many mechanisms, are not inherently involved in hematopoiesis, and do not block transposon activity.

38
Q

An inhibitor of β-tubulin polymerization is applied to actively replicating cells. Which of the following processes would be most impacted by this inhibitor?

A. Flagellar motion

B. Cytokinesis

C. Mitosis

D. Transport of vesicles down a neuronal axon

A

C. Mitosis

β-tubulin polymerization is important in the formation of microtubules. All other answers, except cytokinesis, depend on microtubules, but mitosis would be affected most because formation of the mitotic spindle and movement and separation of replicated chromosomes rely on polymerization and depolymerization of microtubules.

39
Q

Treatment of a cell with a membrane-impermeable protease is most likely to impact the receptor for which of the following?

A. Progesterone

B. Cholecalciferol (vitamin D)

C. Diethylstilbestrol

D. Prolactin

A

D. Prolactin

The protease, given that it cannot enter the cell, will destroy extracellular receptors which would generally bind peptide hormones. Prolactin, a peptide hormone produced in the anterior pituitary, binds to an extracellular receptor.

40
Q

Which of the following describes an amplification step in the G-protein signaling cascade?

A. Activation of multiple receptors by the ligand

B. Dissociation of G from the beta and gamma subunits

C. Binding of multiple ligands to the receptor

D. Activation of multiple G-proteins by the bound receptor

A

D. Activation of multiple G-proteins by the bound receptor

41
Q

Hyperpolarization-activated cyclic nucleotide-gated (HCN) channels are ion channels which have activity modulated by cyclic nucleotides. A given HCN channel is inhibited by high levels of cAMP. Activating which G-protein cascade would increase activity of this channel?

A. Gq, which would stimulate phospholipase

B. Gi, which would inhibit adenylyl cyclase

C. G, which interacts with adenylyl cyclase or phospholipase

D. Gs, which would stimulate adenylyl cyclase

A

B. Gi, which would inhibit adenylyl cyclase

Given that high levels of cAMP inhibit this channel, we are looking for a cascade that will activate the HCN channel by decreasing cAMP levels. Gi accomplishes this by inhibiting adenylyl cyclase; this would lead to a drop in cAMP levels.

42
Q

What is the Gs phase of the G-protein cascade?

A

Gs activatesadenylate cyclase, leading to an increase in cAMP, and further inhibit the channel.

43
Q

What is the Gq phase of the G-protein cascade?

A

Gq activation results in the production of DAG and IP3 which result in activation of protein kinase C and increased intracellular calcium levels.

44
Q

Colchicine is a compound which interferes with the formation of microtubules. Which of the following would be affected LEAST by the administration of colchicine?

A. Organelle movement

B. Flagellae

C. Amoeboid motility of cells

D. Mitotic spindles

A

C. Amoeboid motility of cells

Actin microfilaments, not microtubules, are responsible for amoeboid movement of cells.

45
Q

Before a gene targeted mouse embryonic stem cell line can be used to generate a transgenic animal, chromosome quantification is performed to ensure the cell line has the appropriate number of chromosomes. To do this, cells are grown in the presence of colcemid for four hours, cells are then harvested and nuclear clusters are collected and stained. Colcemid treatment most likely:

A. stops the cell cycle in prophase I.

B. arrests the cells in metaphase.

C. causes cell cycle arrest in G1.

D. increases cell proliferation by activating cell cycle progression

A

B. arrests the cells in metaphase.

If the harvested cells are being used for nuclear harvesting and chromosome counting, it would be best if chromosomes were condensed. Chromosomes are condensed by metaphase, so arresting cells in this mitotic phase would be ideal.

46
Q

The cell signaling master regulator Akt binds phosphatidylinositol (4,5)-bisphosphate (PIP2) only when it is phosphorylated to form phosphatidylinositol (3,4,5)-trisphosphate (PIP3). This occurs via a pleckstrin homology (PH) domain, which is about 120 amino acids in length. Akt itself is then phosphorylated on amino acids 308 and 473, which allows it to positively regulate cellular survival and metabolism. Which of the following is most supported by the information above?

A. Akt is phosphorylated by a phosphatase; activating mutations in Akt would likely drive cell proliferation.

B. Akt is capable of binding phospholipids very specifically.

C. Akt is phosphorylated on two amino acids, which are likely serine, threonine or tryptophan.

D. The PH domain of Akt is a very specific domain that is hydrophobic in nature; loss of this domain would mean Akt is unable to control downstream cell signaling events.

A

B. Akt is capable of binding phospholipids very specifically.

Akt is capable of binding PIP3 but not PIP2, indicating a fairly high level of specificity.

47
Q

Alkylating agents are highly reactive compounds that can be used as anti-cancer drugs. They covalently link an alkyl group (R-CH2) to a chemical species in nucleic acids; most are bipolar, meaning that they contain two groups capable of reacting with DNA. They can thus form cross bridges between a single strand of DNA or between two separate strands. Alkylating agents would most likely interfere with which stage of the cell cycle?

A. Anaphase

B. Metaphase

C. S phase

D. Prophase

A

C. S phase

If DNA is being covalently crosslinked, either to itself or to another DNA strand, DNA polymerase will be incapable of replicating the strands. DNA replication occurs in S phase

48
Q

By shutting off their apoptotic machinery, cancer cells are most likely:

A. conserving energy for faster reproduction and growth associated with cancer cells.

B. protecting themselves from the suicide process that their abnormal behavior would otherwise activate.

C. able to activate Smac and the caspase cascade.

D. preparing for the cleavage of key cellular proteins.

A

B. protecting themselves from the suicide process that their abnormal behavior would otherwise activate.

For a cancer cell, the primary advantage to shutting off the apoptotic pathway is to prevent cell death in response to abnormal cellular behavior (extremely rapid cell division, etc.).

49
Q

All of the following statements about G proteins are true EXCEPT:

A. activation of G proteins can result in an increase or decrease in cAMP levels.

B. G protein-mediated responses are slow.

C. G proteins can trigger an increase in intracellular calcium.

D. G proteins often work by activating kinases that phosphorylate, and thus change the activity of, intracellular enzymes.

A

B. G protein-mediated responses are slow.

Cellular changes brought about by G proteins are typically rapid and temporary. These signal cascades ultimately operate as switches to turn enzymes on or off; since no new proteins are being made, the response is typically fast.

50
Q

The properties relevant to the mimic’s ability to penetrate the cellular membrane are:

I. its polarity.

II. its size.

III. its ability to interact with IAP.

A. I only

B. I and II only

C. II and III only

D. I, II, and III

A

B. I and II only

Item I is true: the polarity of a molecule is the single biggest factor affecting its ability to cross the cell membrane.

Item II is true: the size of a molecule is also important in determining its ability to cross the cell membrane.

51
Q

A zymogen of a caspase protein is present within the cytosol of a cell. Which of the following is true?

A. The zymogen must be cleaved in a certain way before the caspase protein is produced.

B. The zymogen acts as an enzyme to produce the caspase.

C. The zymogen binds to the caspase protein, forming a zymogen-caspase complex.

D. Once activated, the zymogen will transform itself into the caspase.

A

A. The zymogen must be cleaved in a certain way before the caspase protein is produced.

A zymogen is an inactive pro-form of a cellular protein. Most zymogens are activated by cleavage at a specific location within the amino acid sequence of the protein.

52
Q

Which of the following amino acids of the Na+/K+ ATPase are most likely to be found within the interior portion of the lipid bilayer?

A. Glycine, alanine, leucine, and isoleucine

B. Lysine, leucine, isoleucine, and tyrosine

C. Arginine, lysine, and histidine

D. Aspartate, glutamate, glutamine, and glycine

A

A. Glycine, alanine, leucine, and isoleucine

Since the interior portion of the lipid bilayer is hydrophobic, amino acids found there must also be hydrophobic. Glycine, alanine, leucine, and isoleucine are all nonpolar amino acids.

53
Q

Through which mechanism does the Na+/K+ ATPase move ions across the plasma membrane?

A. Primary active transport, since it is powered by ATP hydrolysis.

B. Secondary active transport, since it is powered by ATP synthesis.

C. Facilitated diffusion, since it is powered by ATP hydrolysis.

D. Simple diffusion, since it involves the movement of very small but charged molecules.

A

A. Primary active transport, since it is powered by ATP hydrolysis.

Since the Na+/K+ ATPase hydrolyzes ATP to transport ions against their concentration gradients, it is an example of primary active transport.

54
Q

Which of the following proteins is LEAST like a Na+/K+ ATPase?

A. Hexokinase

B. Sodium-glucose cotransporter

C. Kinesin motor proteins

D. Myosin

A

B. Sodium-glucose cotransporter

The Na+/K+ ATPase hydrolyzes ATP to move sodium and potassium across the cell membrane. Hexokinase hydrolyzes ATP to phosphorylate glucose in the first step of glycolysis. While the sodium gradient is established by the Na+/K+ ATPase, the cotransporter itself does not hydrolyze ATP . Sodium-glucose cotransporters rely on a sodium gradient to move glucose across the cell membrane.

55
Q

How do kinesins use ATP?

A

Kinesins hydrolyze ATP to “walk” along microtubules

56
Q

How does myosin use ATP?

A

Myosin hydrolyzes ATP to reset to a high-energy conformation during muscle contraction

57
Q
A
58
Q

The inheritable disease myoclonic epilepsy and ragged red fiber disease (MERRF) is caused by a mutation in one of the mitochondrial transfer RNA genes. The disease is characterized by a decrease in synthesis of the mitochondrial proteins required for electron transport and ATP production. A cell containing many mitochondria afflicted by MERRF would demonstrate:

I. decreased intracellular Na+ concentrations.

II. decreased negativity of the interior of the cell.

III. decreased activity of secondary active transport proteins.

A. II only

B. III only

C. I and II only

D. II and III only

A

D. II and III only

Decreased negativity of the interior of the cell is true: normally 3 Na+ ions are pumped out the cell while 2 K+ ions are pumped in. With the inhibition of the pump, MERRF would lead to less positive charge leaving the cell and more remaining inside. Thus, the cell would become less negative on the inside.

Decreased activity of secondary active transport proteins is true: since not as much Na+ is pumped out of the cell, secondary active transporters will have less of a Na+ gradient to work with, and thus will decrease in activity.

59
Q

A cell that is upregulating expression of the Na+/K+ ATPase would perform which of the following?

A. Nuclear transcription, cytoplasmic translation, transport to the plasma membrane via a chaperone protein.

B. Cytoplasmic transcription, splicing and translation, transport to the plasma membrane via a vesicle.

C. Nuclear transcription, translation at the smooth ER, protein modification, transport to the plasma membrane via a vesicle.

D. Nuclear transcription and splicing, translation at the rough ER, protein modification, transport to the plasma membrane via a vesicle.

A

D. Nuclear transcription and splicing, translation at the rough ER, protein modification, transport to the plasma membrane via a vesicle.

All eukaryotic cells perform transcription and splicing in the nucleus. Proteins that end up on the plasma membrane are translated by ribosomes on the rough ER and enter the secretory pathway.

60
Q

Which of the following best explains how K+ transport across the plasma membrane differs from K+ transport across the outer mitochondrial membrane (OMM)?

A. K+ moves across both membranes via facilitated diffusion.

B. K+ moves across the OMM through porins and across the plasma membrane via active transport.

C. K+ moves across both membranes via active transport.

D. K+ moves across the OMM through porins and across the plasma membrane via active transport or facilitated diffusion.

A

D. K+ moves across the OMM through porins and across the plasma membrane via active transport or facilitated diffusion.

The outer mitochondrial membrane contains pores (made of porin proteins) that allow the cytoplasm and the intermembrane space of the mitochondrial to be contiguous. Potassium can move across the plasma membrane via active transport, through the Na+/K+ ATPase, or via facilitated diffusion through the K+ leak channels.