1. The answer is A [II A 1, C]. 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.
2. The answer is D [IV D 1 a, b, 2 b]. During the upstroke of the action potential, the cell depolarizes, or becomes less negative. The depolarization is caused by inward current, which is, by definition, the movement of positive charge into the cell. In nerve and in most types of muscle, this inward current is carried by Na\
3. The answer is D [IV B]. Because the membrane is permeable only to K+ ions, K+ will diffuse down its concentration gradient from solution A to solution B, leaving some CI 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.
4. The answer is B [V B 1-6]. Acetylcholine (ACh) is stored in vesicles and is released when an action potential in the motor nerve opens Ca2+ channels in the presynaptic terminal. ACh diffuses across the synaptic cleft and opens Na+ and K+ channels in the muscle end plate, depolarizing it (but not producing an action potential). Depolarization of the muscle end plate causes local currents in adjacent muscle membrane, depolarizing the membrane to threshold and producing action potentials.
5. The answer is C [VI A, B 1^4; VII B 1-4]. An elevation of intracellular [Ca2+] is common to the mechanism of excitation-contraction coupling in skeletal and smooth muscle. In skeletal muscle, the Ca2+ binds to troponin C, initiating the cross-bridge cycle. In smooth muscle, Ca2+ binds to calmodulin. The Ca2+-calmodulin complex activates myosin light-chain kinase, which phosphorylates myosin so that shortening can occur. The striated appearance of the sarcomeres and the presence of troponin are characteristic of skeletal, not smooth, muscle. Spontaneous depolarizations and gap junctions are characteristics of unitary smooth muscle but not skeletal muscle.
6. The answer is E [VI B 6]. During repeated stimulation of a muscle fiber, Ca2+ is released from the sarcoplasmic reticulum (SR) more quickly than it can be reaccumulated; therefore, the intracellular [Ca2+ ] does not return to resting levels as it would after a single twitch. The increased [Ca2+] allows more cross-bridges to form and, therefore, produces increased tension (tetanus). Intracellular Na+ and K+ concentrations do not change during the action potential. Very few Na+ or K+ ions move into or out of the muscle cell, so bulk concentrations are unaffected. Adenosine triphosphate (ATP) levels would, if anything, decrease during tetanus.
7. The answer is D [IV B], The membrane is permeable to Ca2+, but impermeable to CI. 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).
8. The answer is B [V B 8]. Myasthenia gravis is characterized by a decreased density of acetylcholine (ACh) receptors at the muscle end plate. An acetylcholinesterase (AChE) inhibitor blocks degradation of ACh in the neuromuscular junction, so levels at the muscle end plate remain high, partially compensating for the deficiency of receptors.
9. The answer is D [III B 2 d]. Lysis of the patient's red blood cells (RBCs) was caused by entry of water and swelling of the cells to the point of rupture. Water would flow into the RBCs if the extracellular fluid became hypotonic (had a lower osmotic pressure) relative to the intracellular fluid-hypotonic urea. By definition, isotonic solutions do not cause water to flow into or out of cells because the osmotic pressure is the same on both sides of the cell membrane. Hypertonic mannitol would cause shrinkage of the RBCs.
10. The answer is E [IV D 3 a]. Because the stimulus was delivered during the absolute refractory period, no action potential occurs. The inactivation gates of the Na+ channel were closed by depolarization and remain closed until the membrane is repolarized. As long as the inactivation gates are closed, the Na+ channels cannot be opened to allow for another action potential.
11. The answer is B [II A], 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.
12. The answer is D [IV B 3 a, b]. The Nernst equation is used to calculate the equilibrium potential for a single ion. In applying the Nernst equation, we assume that the membrane is freely permeable to that ion alone. ENa+ = 2.3 RT/zF log Ce/C1 = 60 mV log 140/14 = 60 mV log 10 = 60 mV. Notice that the signs were ignored and that the higher concentration was simply placed in the numerator to simplify the log calculation. To determine whether ENa+ is +60 mV or -60 mV, use the intuitive approach-Na+ will diffuse from extracellular to intracellular fluid down its concentration gradient, making the cell interior positive.
13. The answer is E [IV D 2 d]. The hyperpolarizing afterpotential represents the period during which K+ permeability is highest, and the membrane potential is closest to the K+ equilibrium potential. At that point, K+ is closest to electrochemical equilibrium. The force driving K+ movement out of the cell down its chemical gradient is balanced by the force driving K* into the cell down its electrical gradient.
14. The answer is A [IV D 2 b (l)-(3)]. The upstroke of the nerve action potential is caused by opening of the Na+ channels (once the membrane is depolarized to threshold). When the Na+ channels open, Na+ moves into the cell down its electrochemical gradient, driving the membrane potential toward the Na+ equilibrium potential.
15. The answer is D [IV D 2 c]. The process responsible for repolarization is the opening of K+ channels. The K* permeability becomes very high and drives the membrane potential toward the K+ equilibrium potential by flow of K+ out of the cell.
16. The answer is D [IV D 4 b]. Myelin insulates the nerve, thereby increasing conduction velocity; action potentials can be generated only at the nodes of Ranvier, where there are breaks in the insulation. Activity of the Na+-K+ pump does not directly affect the formation or conduction of action potentials. Decreasing nerve diameter would increase internal resistance and therefore slow the conduction velocity.
17. The answer is D [III A, B 4]. Solution A contains both sucrose and urea at concentrations of 1 mM, whereas solution B contains only sucrose at a concentration of 1 mM. The calculated osmolality of solution A is 2 mOsm/L, and the calculated osmolality of solution B is 1 mOsm/L. Therefore, solution A, which has a higher osmolality, is hyperosmotic with respect to solution B. Actually, solutions A and B have the same effective osmotic pressure (i.e., they are isotonic) because the only "effective" solute is sucrose, which has the same concentration in both solutions. Urea is not an effective solute because its reflection coefficient is zero.
18. The answer is B [V C 4 b (3)]. Dopaminergic neurons and D2 receptors are deficient in people with Parkinson's disease. Schizophrenia involves increased levels of D2 receptors. Myasthenia gravis and curare poisoning involve the neuromuscular junction, which uses acetylcholine (ACh) as a neurotransmitter.
19. The answer is B [II A 4 a-c]. Increasing oil/water partition coefficient increases solubility in a lipid bilayer and therefore increases permeability. Increasing molecular radius and increased membrane thickness decrease permeability. The concentration difference of the solute has no effect on permeability.
20. The answer is A [IV D 2 b (2), (3), d, 3 a]. Complete blockade of the Na+ channels would prevent action potentials. The upstroke of the action potential depends on the entry of Na+ into the cell through these channels and therefore would also be abolished. The absolute refractory period would be lengthened because it is based on the availability of the Na+ channels. The hyperpolarizing afterpotential is related to increased K+ permeability. The Na+ equilibrium potential is calculated from the Nernst equation and is the theoretical potential at electrochemical equilibrium (and does not depend on whether the Na+ channels are open or closed).
21. The answer is D [V B 5]. Binding of acetylcholine (ACh) to receptors in the muscle end plate opens channels that allow passage of both Na+ and K+ ions. Na+ ions will flow into the cell down its electrochemical gradient, and K+ ions will flow out of the cell down its electrochemical gradient. The resulting membrane potential will be depolarized to a value that is approximately halfway between their respective equilibrium potentials.
22. The answer is D [V C 2 b]. An inhibitory postsynaptic potential hyperpolarizes the postsynaptic membrane, taking it farther from threshold. Opening CI" channels would hyperpolarize the postsynaptic membrane by driving the membrane potential toward the CI' equilibrium potential (about -90 mV). Opening Ca2+ channels would depolarize the postsynaptic membrane by driving it toward the Ca2+ equilibrium potential.
23. The answer is C [II D 2 a]. Inhibition of Na+,K+-adenosine triphosphatase (ATPase) leads to an increase in intracellular Na+ concentration. Increased intracellular Na+ concentration decreases the Na+ gradient across the cell membrane, thereby inhibiting Na+-Ca2+ exchange and causing an increase in intracellular Ca2+ concentration. Increased intracellular Na+ concentration also inhibits Na+-glucose cotransport.
24. The answer is B [VIB 1-4]. The correct sequence is action potential in the muscle membrane; depolarization of the T tubules; release of Ca2t from the sarcoplasmic reticulum (SR); binding of Ca2+ to troponin C; cross-bridge formation; and splitting of adenosine triphosphate (ATP).
25. The answer is C [III A], Osmolality is the concentration of particles (osmolality = g x C). When two solutions are compared, that with the higher osmolality is hyperosmotic. The 1 mM CaCl2 solution (osmolality = 3 mOsm/L) is hyperosmotic to 1 mM NaCl (osmolality = 2 mOsm/L). The 1 mM glucose, 1.5 mM glucose, and 1 mM sucrose solutions are hyposmotic to 1 mM NaCl, whereas 1 mM KC1 is isosmotic.
26. The answer is A [VI A3]. In the mechanism of excitation-contraction coupling, excitation always precedes contraction. Excitation refers to the electrical activation of the muscle cell, which begins with an action potential (depolarization) in the sarcolemmal membrane that spreads to the T tubules. Depolarization of the T tubules then leads to the release of Ca2+ from the nearby sarcoplasmic reticulum (SR), followed by an increase in intracellular Ca2t concentration, binding of Ca2+ to troponin C, and then contraction.
27. The answer is C [V C 2 a-b]. Gamma-aminobutyric acid (GABA) is an inhibitory neurotransmitter. Norepinephrine, glutamate, serotonin, and histamine are excitatory neurotransmitters.
28. The answer is E [IID 2]. All of the processes listed are examples of primary active transport [and therefore use adenosine triphosphate (ATP) directly], except for absorption of glucose by intestinal epithelial cells, which occurs by secondary active transport (i.e., cotransport). Secondary active transport uses the Na+ gradient as an energy source and, therefore, uses ATP indirectly (to maintain the Na+ gradient).
29. The answer is A [II A 1, C 1]. Only two types of transport occur "downhill"—simple and facilitated diffusion. If there is no stereospecificity for the d- or l-isomer, one can conclude that the transport is not carrier-mediated and, therefore, must be simple diffusion.
30. The answer is D [IID 2 a, E 1], In the "usual" Na+ gradient, the [Na+] is higher in extracellular than in intracellular fluid (maintained by the Na+-K+ pump). Two forms of transport are energized by this type of Na+ gradient-cotransport and countertransport. Because glucose is moving in the same direction as Na+, one can conclude that it is cotransport.
31. The answer is A [IIA1]. Depolarization opens Na+ channels in the nerve membrane, allowing Na+ ions to diffuse into the cell down an electrochemical gradient. This process is purely passive.
32. The answer is C [II D 1, 2 a]. Transport of K* from extracellular fluid to intracellular fluid occurs against an electrochemical gradient and is an example of primary active transport. The carrier is the Na*-K+ pump, which is inhibited by cardiac glycosides, such as digitalis.
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