Membrane Transport: Clinical Case
*The answer is A.* Water moves from high concentration of solute to low concentration of solute. Therefore, more water will leave the tube to go to the less concentrated area and the tube will decrease in volume.
A thistle tube containing a 5% sugar solution is covered at one end by a differentially permeable membrane and is placed in a beaker containing a 10% sugar solution. What will happen to the volume in the thistle tube? A. It will decrease in volume. B. It will increase in volume. C. It will remain the same. D. It will initially decrease in volume and then increase. E. It will initially increase in volume and then decrease.
*The answer is A.* The term "hyperosmotic" refers to a solution that has a higher osmolarity relative to another solution. The osmolarity of a 1-millimolar NaCl solution is 2 mOsm/L. The osmolarity of a 1-millimolar solution of either glucose or sucrose is only 1 mOsm/L. The osmolarity of a 1.5-millimolar glucose solution is 1.5 mOsm/L. These solutions are all "hypo-osmotic" relative to 1 millimolar NaCl. The osmolarity of a 1-millimolar KCl solution is 2 mOsm/L. It is "iso-osmotic" relative to 1 millimolar NaCl. Only 1 millimolar CaCl2, with an osmolarity of 3 mOsm/L is hyperosmotic relative to 1 millimolar NaCl.
Assuming complete dissociation of all solutes, which of the following solutions would be hyperosmotic relative to 1 millimolar NaCl? A) 1 millimolar CaCl2 B) 1 millimolar glucose C) 1 millimolar KCl D) 1 millimolar sucrose E) 1.5 millimolar glucose
*The answer is A.* Net diffusion of a substance across a permeable membrane is proportional to the concentration difference of the substance on either side of the membrane. Initially, the concentration difference is 5 millimolar (10 millimolar − 5 millimolar). When the intracellular concentration doubles to 20 millimolar, the concentration difference becomes 15 millimolar (20 millimolar − 5 millimolar). The concentration difference has tripled; therefore, the rate of diffusion would also increase by a factor of 3.
If the intracellular concentration of a membrane-permeant substance doubles from 10 to 20 millimolar and the extracellular concentration remains at 5 millimolar, the rate of diffusion of that substance across the plasma membrane will increase by a factor of how much? A) 2 B) 3 C) 4 D) 5 E) 6
*The answer is B.* The redistribution of fluid volume shown in diagram B reflects the net diffusion of water, or osmosis, due to differences in the osmolarity of the solutions on either side of the semipermeable membrane. Osmosis occurs from solutions of high water concentration to low water concentration or from low osmolarity to high osmolarity. In diagram B, osmosis has occurred from X to Y and from Y to Z. Therefore, the osmolarity of solution Z is higher than that of solution Y, and the osmolarity of solution Y is higher than that of solution X.
In the experiment illustrated in diagram A, equal volumes of solutions X, Y, and Z are placed into the compartments of the two U-shaped vessels shown. The two compartments of each vessel are separated by semipermeable membranes (i.e., impermeable to ions and large polar molecules). Diagram B illustrates the fluid distribution across the membranes at equilibration. Assuming complete dissociation, identify each of the solutions shown.
*The answer is B.* In contrast to primary and secondary active transport, neither facilitated diffusion nor simple diffusion requires additional energy and, therefore, can work in the absence of ATP. Only facilitated diffusion displays saturation kinetics and involves a carrier protein. By definition, neither simple nor facilitated diffusion can move molecules from low to high concentration. The concept of specific inhibitors is not applicable to simple diffusion that occurs through a lipid bilayer without the aid of protein.
Simple diffusion and facilitated diffusion share which of the following characteristics? A. Can be blocked by specific inhibitors B. Do not require adenosine triphosphate (ATP) C. Require transport protein D. Saturation kinetics E. Transport solute against concentration gradient
*The answer is B.* Because the membrane is only permeable to water, the solute molecules cannot cross through. Water will move from an area of low solute concentration to an area of high solute concentration (in other words, an area of high water concentration to an area of low water concentration). Since the solute concentration in B is lower than the solute concentration in A, water will move from side B to side A to balance out the concentration gradient.
The U shaped pipe shown in the figure below is divided with a membrane that is only permeable to water. Which of the following best describes how water will flow through the membrane? A. Water will flow from side A into side B B. Water will flow from side B into side A C. The water levels are already equal to water will not flow through the semipermeable membrane
*The answer is D.* The membrane is permeable to Ca2+, but impermeable to Cl-. Although there is a concentration gradient across the membrane for both ions, only Ca2+ can diffuse down this gradient. Ca2+ will diffuse from solution A to solution B, leaving negative charge behind in solution A. The magnitude of this voltage can be calculated for electrochemical equilibrium with the Nernst equation as follows: ECa2+ = 2.3 RT/zF log CA/CB = 60 mV/+2 log 10 mM/1 mM = 30 mV log 10 = 30 mV. The sign is determined with an intuitive approach—Ca2+ diffuses from solution A to solution B, so solution A develops a negative voltage (-30 mV). Net diffusion of Ca2+ will cease when this voltage is achieved, that is, when the chemical driving force is exactly balanced by the electrical driving force (not when the Ca2+ concentrations of the solutions become equal).
Solutions A and B are separated by a membrane that is permeable to Ca2+ and impermeable to Cl-. Solution A contains 10mM CaCl2, and solution B contains 1 mM CaCl2. Assuming that 2.3 RT/F = 60 mV, Ca2+ will be at electrochemical equilibrium when (A) solution A is +60 mV (B) solution A is +30 mV (C) solution A is -60 mV (D) solution A is -30 mV (E) solution A is +120 mV (F) solution A is -120 mV (G) the Ca2+ concentrations of the two solutions are equal (H) the Cl- concentrations of the two solutions are equal
*The answer is B.* Flux is proportional to the concentration difference across the membrane, J = -PA (CA - CB). Originally, CA - CB = 10 mM - 5 mM = 5 mM. When the urea concentration was doubled in solution A, the concentration difference became 20 mM - 5 mM = 15 mM, or three times the original difference. Therefore, the flux would also triple. Note that the negative sign preceding the equation is ignored if the lower concentration is subtracted from the higher concentration.
Solutions A and B are separated by a membrane that is permeable to urea. Solution A is 10 mM urea, and solution B is 5 mM urea. If the concentration of urea in solution A is doubled, the flux of urea across the membrane will (A) double (B) triple (C) be unchanged (D) decrease to one-half (E) decrease to one-third
*The answer is D.* Because the membrane is permeable only to K+ ions, K+ will diffuse down its concentration gradient from solution A to solution B, leaving some Cl- ions behind in solution A. A diffusion potential will be created, with solution A negative with respect to solution B. Generation of a diffusion potential involves movement of only a few ions and, therefore, does not cause a change in the concentration of the bulk solutions.
Solutions A and B are separated by a semipermeable membrane that is permeable to K+, but not to Cl-. Solution A is 100 mM KCl, and solution B is 1 mM KCl. Which of the following statements about solution A and solution B is true? (A) K+ ions will diffuse from solution A to solution B until the [K+] of both solutions is 50.5 mM (B) K+ ions will diffuse from solution B to solution A until the [K+ ] of both solutions is 50.5 mM (C) KCl will diffuse from solution A to solution B until the [KCl] of both solutions is 50.5 mM (D) K+ will diffuse from solution A to solution B until a membrane potential develops with solution A negative with respect to solution B (E) K+ will diffuse from solution A to solution B until a membrane potential develops with solution A positive with respect to solution B
*The answer is C.* Because starch will turn turn blue-black in the presence of iodine, and the dialysis tubing changed color, we know that iodine must have gotten into the dialysis tubing to combine with the starch solution. This means that the tubing must be permeable to iodine. Similarly, if the dialysis tubing was permeable to starch, starch would have diffused across the tubing into the beaker, changing the color of the iodine solution. We know from the problem that this did not occur. This means that the tubing was not permeable to starch.
Starch turns a blue-black color in the presence of iodine. Julian decides to use this information to test the permeability of some dialysis tubing. Julian fills the tubing with a starch solution and places it in a beaker of iodine and water. After two hours, the starch solution in the dialysis tubing turns blue-black, and the iodine solution is the same color. What do Julian's results show about the permeability of the tubing? A. The membrane is permeable to both starch and iodine B. The dialysis tubing is permeable to starch but not to iodine C. The dialysis tubing is permeable to iodine but not to starch D. The dialysis tubing is not permeable to starch or iodine
*The answer is D.* Water moves from high concentration of solute to low concentration of solute. Therefore, more water will enter the thistle tube, which has a high concentration of sugar, than leave the thistle tube. Choice A is incorrect because although water will enter the tube, it will also leave the tube. But the net water movement will be inside the tube.
Suppose a thistle tube containing a 10% sugar solution is covered at one end by a differentially permeable membrane and is placed in a beaker containing a 5% sugar solution. What will happen to the water? A. Water will only enter the thistle tube. B. Water will only leave the thistle tube. C. There will be no net water movement. D. More water will enter the thistle tube than leave the thistle tube. E. More water will leave the thistle tube than enter the thistle tube.
*The answer is B.* The sodium-potassium pump usually pumps 3 sodium out of the cell and 2 potassium into the cell.
The sodium potassium pump usually pumps A. Potassium out of the cell B. Sodium out of the cell C. Sodium into the cell D. Both potassium and sodium into the cell
*The answer is D.* Trace A reflects the kinetics of a process that is limited by an intrinsic Vmax. Of the choices provided, only the transport of K+, which occurs through the activity of the Na+, K+-ATPase, is the result of an active transport event. The movement of CO2 and O2 through a biological membrane and the movement of Ca++ and Na+ through ion channels are all examples of simple diffusion.
Trace A best describes the kinetics of which of the following events? A) Movement of CO2 across the plasma membrane B) Movement of O2 across a lipid bilayer C) Na+ flux through an open nicotinic acetylcholine receptor channel D) Transport of K+ into a muscle cell E) Voltage-dependent movement of Ca++ into the terminal of a motor neuron
*The answer is E.* Trace B is indicative of a process not limited by an intrinsic Vmax. This excludes active transport and facilitated diffusion. Therefore, of the choices provided, only the rate of transport of O2 across an artificial lipid bilayer via simple diffusion would be accurately reflected by trace B.
Trace B best describes the kinetics which of the following events? A) Na+-dependent transport of glucose into an epithelial cell B) Transport of Ca++ into the sarcoplasmic reticulum of a smooth muscle cell C) Transport of K+ into a muscle cell D) Transport of Na+ out of a nerve cell E) Transport of O2 across an artificial lipid bilayer
*The answer is E.* A 1 millimolar solution has an osmolarity of 1 milliosmole when the solute molecule does not dissociate. However, NaCl and KCl both dissociate into two molecules, and CaCl2 dissociates into three molecules. Therefore, 12 millimolar NaCl has an osmolarity of 24 milliosmoles, 4 millimolar KCl has an osmolarity of 8 milliosmoles, and 2 millimolar CaCl2 has an osmolarity of 6 milliosmoles. These add up to 38 milliosmoles.
What is the calculated osmolarity of a solution containing 12 millimolar NaCl, 4 millimolar KCl, and 2 millimolar CaCl2 (in mOsm/L)? A) 16 B) 26 C) 29 D) 32 E) 38 F) 42
*The answer is B.* A solution of 280 milliosmoles, is iso osmotic relative to "normal" intracellular osmolarity. If red blood cells were placed in of 280 milliosmolar NaCl alone, no change in cell volume would occur because intracellular and extracellular osmolarities would be equal. The presence of 20-millimolar urea, however, increases the solution's osmolarity and makes it hypertonic relative to the intracellular solution at the beginning of the experiment. Water will initially move out of the cell, but because the plasma membrane is permeable to urea, urea will diffuse into the cell and equilibrate across the plasma membrane. As a result, water will re-enter the cell, and the cell will return to its original volume.
Which of the following best describes the changes in cell volume that will occur when red blood cells (previously equilibrated in a 280-milliosmolar solution of NaCl) are placed in a solution of 280-milliosmolar NaCl containing 20-millimolar urea, a relatively large but permeant molecule? A. Cells will shrink, then swell and lyse B. Cells will shrink, then return to original volume C. Cells will swell and lyse D. Cells will swell, then return to original volume E. Cells will not change in cell volume
*The answer is A.* Both types of transport occur down an electrochemical gradient ("downhill"), and do not require metabolic energy. Saturability and inhibition by other sugars are characteristic only of carrier-mediated glucose transport; thus, facilitated diffusion is saturable and inhibited by galactose, whereas simple diffusion is not.
Which of the following characteristics is shared by simple and facilitated diffusion of glucose? (A) Occurs down an electrochemical gradient (B) Is saturable (C) Requires metabolic energy (D) Is inhibited by the presence of galactose (E) Requires a Na+ gradient
*The answer is C.* Facilitated transport does not require the expenditure of energy. It requires specific carrier proteins, transports molecules down the concentration gradient, transports molecules from one side of the membrane to the other side, and transports molecules through the membrane much faster than simple diffusion.
Which of the following conditions does NOT apply to facilitated transport? A. requires specific carrier proteins B. transports molecules down the concentration gradient C. requires the expenditure of energy D. transports molecules from one side of the membrane to the other side E. transports molecules through the membrane much faster than simple diffusion
*The answer is A.* Carrier proteins do not bind to molecules irreversibly, but release them at their destination site.
Which of the following conditions does NOT apply to facilitated transport? A. carrier proteins bind to molecules irreversibly B. transports molecules down the concentration gradient C. does not require ATP D. transports molecules from one side of the membrane to the other side E. transports molecules through the membrane much faster than simple diffusion
*The answer is C.* The membrane channel facilitates the diffusion of the molecule from high concentration to low concentration.
Which of the following responses best describes what this animation is demonstrating? A. Osmosis B. Diffusion C. Facilitated Diffusion D. Active Transport
*The answer is D. *In passive transport, molecules always move along the concentration gradient, from an area of high concentration to an area of low concentration.
Which of the following statements about passive transport is true? A. Membrane proteins are always needed in order for it to take place B. Passive transport refers only to the movement of water molecules C. The input of ATP is required in order to facilitate transportation D. Substances move from areas of high concentration to areas of low concentration
*The answer is E.* Facilitated diffusion and both primary and secondary active transport all involve protein transporters or carriers that must undergo some rate-limited conformational change. The rate of simple diffusion is linear with solute concentration.
Which of the following transport mechanisms is not rate limited by an intrinsic Vmax? A) Facilitated diffusion via carrier proteins B) Primary active transport via carrier proteins C) Secondary co-transport D) Secondary counter-transport E) Simple diffusion through protein channels
*The answer is A.* Hemolysis describes the condition of a red blood cell when placed in a hypotonic solution. This is because there is a higher concentration of solute inside of the cell so there is a net water movement inside of the cell, which causes it to lyse.
Which term best describes the condition of red blood cells when placed in a hypotonic solution? A. Hemolysis B. Plasmolysis C. Crenation D. Turgor pressure E. Osmotic pressure
*The answer is C.* Crenation describes the condition of red blood cells when placed in a hypertonic solution. This is because there is a higher concentration of solute outside of the cell so there is a net water movement outside of the cell, which causes it to shrink or crenate.
Which term describes the condition of red blood cells when placed in a hypertonic solution? A. Hemolysis B. Plasmolysis C. Crenation D. Turgor pressure E. Osmotic pressure
