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For the following questions,match the labelled component of the cell membrane in the figure with its description. For the following questions,match the labelled component of the cell membrane in the figure with its description.    -Which component is the peripheral protein? A) A B) B C) C D) D E) E -Which component is the peripheral protein?


A) A
B) B
C) C
D) D
E) E

F) None of the above
G) A) and E)

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Cell membranes are asymmetrical.Which of the following is the most likely explanation?


A) The cell membrane forms a border between one cell and another in tightly packed tissues such as epithelium.
B) Cell membranes communicate signals from one organism to another.
C) The two sides of a cell membrane face different environments and carry out different functions.
D) The "innerness" and "outerness" of membrane surfaces are predetermined by genes.
E) Proteins can only be associated with the cell membranes on the cytoplasmic side.

F) C) and D)
G) A) and B)

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Which of the following statements correctly describes the normal tonicity conditions for typical plant and animal cells?


A) The animal cell is in a hypotonic solution, and the plant cell is in an isotonic solution.
B) The animal cell is in an isotonic solution, and the plant cell is in a hypertonic solution.
C) The animal cell is in a hypertonic solution, and the plant cell is in an isotonic solution.
D) The animal cell is in an isotonic solution, and the plant cell is in a hypotonic solution.
E) The animal cell is in a hypertonic solution, and the plant cell is in a hypotonic solution.

F) A) and E)
G) A) and B)

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Which of the following is a characteristic feature of a carrier protein in a plasma membrane?


A) It is a peripheral membrane protein.
B) It exhibits a specificity for a particular type of molecule.
C) It requires the expenditure of cellular energy to function.
D) It works against diffusion.
E) It has few, if any, hydrophobic amino acids.

F) A) and C)
G) None of the above

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Use the following information to answer the questions below. Human immunodeficiency virus (HIV) infects cells that have both CD4 and CCR5 cell surface molecules.The viral nucleic acid molecules are enclosed in a protein capsid,and the protein capsid is itself contained inside an envelope consisting of a lipid bilayer membrane and viral glycoproteins.One hypothesis for viral entry into cells is that binding of HIV membrane glycoproteins to CD4 and CCR5 initiates fusion of the HIV membrane with the plasma membrane,releasing the viral capsid into the cytoplasm.An alternative hypothesis is that HIV gains entry into the cell via receptor-mediated endocytosis,and membrane fusion occurs in the endocytotic vesicle.To test these alternative hypotheses for HIV entry,researchers labelled the lipids on the HIV membrane with a red fluorescent dye. Use the following information to answer the questions below. Human immunodeficiency virus (HIV) infects cells that have both CD4 and CCR5 cell surface molecules.The viral nucleic acid molecules are enclosed in a protein capsid,and the protein capsid is itself contained inside an envelope consisting of a lipid bilayer membrane and viral glycoproteins.One hypothesis for viral entry into cells is that binding of HIV membrane glycoproteins to CD4 and CCR5 initiates fusion of the HIV membrane with the plasma membrane,releasing the viral capsid into the cytoplasm.An alternative hypothesis is that HIV gains entry into the cell via receptor-mediated endocytosis,and membrane fusion occurs in the endocytotic vesicle.To test these alternative hypotheses for HIV entry,researchers labelled the lipids on the HIV membrane with a red fluorescent dye.    -If HIV first enters the cell in an endocytotic vesicle,instead of directly fusing with the plasma membrane,then A) HIV infection should be hindered by microtubule polymerization inhibitors such as nocodazole. B) HIV infection should be more efficient at lower temperatures. C) intact cortical actin microfilaments should interfere with HIV infection. D) cells lacking integrins should be resistant to HIV infection. E) addition of ligands for other cell-surface receptors to stimulate their endocytosis should increase the efficiency of HIV infection. -If HIV first enters the cell in an endocytotic vesicle,instead of directly fusing with the plasma membrane,then


A) HIV infection should be hindered by microtubule polymerization inhibitors such as nocodazole.
B) HIV infection should be more efficient at lower temperatures.
C) intact cortical actin microfilaments should interfere with HIV infection.
D) cells lacking integrins should be resistant to HIV infection.
E) addition of ligands for other cell-surface receptors to stimulate their endocytosis should increase the efficiency of HIV infection.

F) A) and C)
G) B) and D)

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According to the fluid mosaic model of membrane structure,proteins of the membrane are mostly


A) spread in a continuous layer over the inner and outer surfaces of the membrane.
B) confined to the hydrophobic interior of the membrane.
C) embedded in a lipid bilayer.
D) randomly oriented in the membrane, with no fixed inside-outside polarity.
E) free to depart from the fluid membrane and dissolve in the surrounding solution.

F) None of the above
G) B) and D)

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Water passes quickly through cell membranes because


A) the bilayer is hydrophilic.
B) it moves through hydrophobic channels.
C) water movement is tied to ATP hydrolysis.
D) it is a small, polar, charged molecule.
E) it moves through aquaporins in the membrane.

F) A) and D)
G) A) and E)

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When a membrane is freeze-fractured,the bilayer splits down the middle between the two layers of phospholipids.In an electron micrograph of a freeze-fractured membrane,the bumps seen on the fractured surface of the membrane are


A) peripheral proteins.
B) phospholipids.
C) carbohydrates.
D) integral proteins.
E) cholesterol molecules.

F) A) and C)
G) B) and D)

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Use the following information to answer the questions below. The solutions in the two arms of this U-tube are separated by a membrane that is permeable to water and glucose but not to sucrose.Side A is half-filled with a solution of 2 M sucrose and 1 M glucose.Side B is half-filled with 1 M sucrose and 2 M glucose.Initially,the liquid levels on both sides are equal. Use the following information to answer the questions below. The solutions in the two arms of this U-tube are separated by a membrane that is permeable to water and glucose but not to sucrose.Side A is half-filled with a solution of 2 M sucrose and 1 M glucose.Side B is half-filled with 1 M sucrose and 2 M glucose.Initially,the liquid levels on both sides are equal.    -After the system reaches equilibrium,what changes are observed? A) The molarity of sucrose and glucose are equal on both sides. B) The molarity of glucose is higher in side A than in side B. C) The water level is higher in side A than in side B. D) The water level is unchanged. E) The water level is higher in side B than in side A. -After the system reaches equilibrium,what changes are observed?


A) The molarity of sucrose and glucose are equal on both sides.
B) The molarity of glucose is higher in side A than in side B.
C) The water level is higher in side A than in side B.
D) The water level is unchanged.
E) The water level is higher in side B than in side A.

F) A) and D)
G) A) and E)

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Which of the following is a reasonable explanation for why unsaturated fatty acids help keep any membrane more fluid at lower temperatures?


A) The double bonds form kinks in the fatty acid tails, preventing adjacent lipids from packing tightly.
B) Unsaturated fatty acids have a higher cholesterol content and therefore more cholesterol in membranes.
C) Unsaturated fatty acids are more polar than saturated fatty acids.
D) The double bonds block interaction among the hydrophilic head groups of the lipids.
E) The double bonds result in shorter fatty acid tails and thinner membranes.

F) None of the above
G) C) and D)

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For the following questions,match the labelled component of the cell membrane in the figure with its description. For the following questions,match the labelled component of the cell membrane in the figure with its description.    -Which component is cholesterol? A) A B) B C) C D) D E) E -Which component is cholesterol?


A) A
B) B
C) C
D) D
E) E

F) B) and C)
G) C) and D)

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Which of the following would likely move through the lipid bilayer of a plasma membrane most rapidly?


A) CO₂
B) an amino acid
C) glucose
D) K⁺
E) starch

F) None of the above
G) All of the above

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Fibronectin binds to integrin,an integral plasma membrane protein.Binding of fibronectin to the extracellular domain of integrin changes integrin's conformation.The change in protein conformation is transmitted to the cytoplasm resulting in a cascade of reactions.What does this description describe?


A) transport
B) enzymatic activity
C) intracellular joining
D) cell-cell recognition
E) signal transduction

F) A) and D)
G) A) and E)

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Use the following information to answer the questions below. Cystic fibrosis is a genetic disease in humans in which the CFTR protein,which functions as a chloride ion channel,is missing or nonfunctional in cell membranes. -In the small airways of the lung,a thin layer of liquid is needed between the epithelial cells and the mucus layer in order for cilia to beat and move the mucus and trapped particles out of the lung.One hypothesis is that the volume of this airway surface liquid is regulated osmotically by transport of sodium and chloride ions across the epithelial cell membrane.How would the lack of a functional chloride channel in cystic fibrosis patients affect sodium ion transport and the volume of the airway surface liquid?


A) Sodium ion transport will increase; higher osmotic potential will increase airway surface liquid volume.
B) Sodium ion transport will increase; higher osmotic potential will decrease airway surface liquid volume.
C) Sodium ion transport will decrease; lower osmotic potential will decrease airway surface liquid volume.
D) Sodium ion transport will decrease; lower osmotic potential will increase the airway surface liquid volume.
E) Sodium ion transport will be unaffected; lack of chloride transport still reduces osmotic potential and decreases the airway surface liquid volume.

F) All of the above
G) C) and D)

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Mammalian blood contains the equivalent of 0.15 M NaCl.Seawater contains the equivalent of 0.45 M NaCl.What will happen if red blood cells are transferred to seawater?


A) Water will leave the cells, causing them to shrivel and collapse.
B) NaCl will be exported from the red blood cells by facilitated diffusion.
C) The blood cells will take up water, swell, and eventually burst.
D) NaCl will passively diffuse into the red blood cells.
E) The blood cells will expend ATP for active transport of NaCl into the cytoplasm.

F) B) and C)
G) D) and E)

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Which of the following is true of the evolution of cell membranes?


A) Cell membranes have stopped evolving now that they are fluid mosaics.
B) Cell membranes cannot evolve if the membrane proteins do not.
C) The evolution of cell membranes is driven by the evolution of glycoproteins and glycolipids.
D) All components of membranes evolve in response to natural selection.
E) An individual organism selects its preferred type of cell membrane for particular functions.

F) C) and D)
G) A) and B)

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Which of the following types of molecules are the major structural components of the cell membrane?


A) phospholipids and cellulose
B) nucleic acids and proteins
C) phospholipids and proteins
D) proteins and cellulose
E) glycoproteins and cholesterol

F) A) and B)
G) A) and C)

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  -What will happen to a red blood cell (RBC) ,which has an internal ion concentration of about 0.9 percent,if it is placed into a beaker of pure water? A) The cell would shrink because the water in the beaker is hypotonic relative to the cytoplasm of the RBC. B) The cell would shrink because the water in the beaker is hypertonic relative to the cytoplasm of the RBC. C) The cell would swell because the water in the beaker is hypotonic relative to the cytoplasm of the RBC. D) The cell will remain the same size because the solution outside the cell is isotonic. E) The cell will remain the same size because the solution outside the cell is hypotonic. -What will happen to a red blood cell (RBC) ,which has an internal ion concentration of about 0.9 percent,if it is placed into a beaker of pure water?


A) The cell would shrink because the water in the beaker is hypotonic relative to the cytoplasm of the RBC.
B) The cell would shrink because the water in the beaker is hypertonic relative to the cytoplasm of the RBC.
C) The cell would swell because the water in the beaker is hypotonic relative to the cytoplasm of the RBC.
D) The cell will remain the same size because the solution outside the cell is isotonic.
E) The cell will remain the same size because the solution outside the cell is hypotonic.

F) A) and B)
G) C) and D)

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An organism non-selectively takes in droplets of water and small molecules.What is this process called?


A) receptor-mediated endocytosis
B) pinocytosis
C) phagocytosis
D) cotransport
E) facilitated diffusion

F) None of the above
G) A) and D)

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In most cells,there are electrochemical gradients of many ions across the plasma membrane even though there are usually only one or two electrogenic pumps present in the membrane.The gradients of the other ions are most likely accounted for by


A) cotransport proteins.
B) ion channels.
C) carrier proteins.
D) passive diffusion across the plasma membrane.
E) cellular metabolic reactions that create or destroy ions.

F) C) and D)
G) A) and E)

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