Question 1

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

Accepted Answer

The answer is A [eII 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.

Question 2

During the upstroke of the nerve action potential (A) there is net outward current and the cell interior becomes more negative (B) there is net outward current and the cell interior becomes less negative (C) there is net inward current and the cell interior becomes more negative (D) there is net inward current and the cell interior becomes less negative

Accepted Answer

The answer is D [IV E 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+.

Question 3

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

Accepted Answer

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 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.

Question 4

The correct temporal sequence for events at the neuromuscular junction is (A) action potential in the motor nerve; depolarization of the muscle end plate; uptake of Ca2+ into the presynaptic nerve terminal (B) uptake of Ca2+ into the presynaptic terminal; release of acetylcholine (ACh); depolarization of the muscle end plate (C) release of ACh; action potential in the motor nerve; action potential in the muscle (D) uptake of Ca2+ into the motor end plate; action potential in the motor end plate; action potential in the muscle (E) release of ACh; action potential in the muscle end plate; action potential in the muscle

Accepted Answer

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.

Question 5

Which characteristic or component is shared by skeletal muscle and smooth muscle? (a) Thick and thin filaments arranged in sarcomeres (b) Troponin (C) Elevation of intracellular [Ca2+] for excitation–contraction coupling (D) Spontaneous depolarization of the membrane potential (e) High degree of electrical coupling between cells

Accepted Answer

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, 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.

Question 6

Solutions A and B are separated by a membrane that is permeable to Ca2+ and impermeable to Cl−. Solution A contains 10 mM 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 concentrations of the two solutions are equal ```

Accepted Answer

The answer is D [IV B]. 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).

Question 7

A 42-year-old man with myasthenia gravis notes increased muscle strength when he is treated with an acetylcholinesterase (AChE) inhibitor. The basis for his improvement is increased (A) amount of acetylcholine (ACh) released from motor nerves (B) levels of ACh at the muscle end plates (C) number of ACh receptors on the muscle end plates (D) amount of norepinephrine released from motor nerves (E) synthesis of norepinephrine in motor nerves

Accepted Answer

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.

Question 8

In a hospital error, a 60-year-old woman is infused with large volumes of a solution that causes lysis of her red blood cells (RBCs). The solution was most likely (a) 150 mM NaCl (b) 300 mM mannitol (C) 350 mM mannitol (D) 300 mM urea (e) 150 mM CaCl2

Accepted Answer

The answer is D [III B 2 d]. Lysis of RBCs was caused by entry of water and swelling of cells to point of rupture. Water would flow into RBCs if the ECF became hypotonic (had a lower osmotic pressure) relative to ICF. Isotonic solutions don't cause water to flow into/out of cells because osmotic pressure is the same on both sides of the membrane. Hypertonic solutions would cause RBC shrinkage. Isotonic: 150 mM NaCl and 300 mM mannitol. Hypertonic: 350 mM mannitol and 150 mM CaCl2. Hypotonic: 300 mM urea, reflection coefficient less than 1.0

Question 9

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

Accepted Answer

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.

Question 10

A muscle cell has an intracellular [Na+] of 14 mM and an extracellular [Na+] of 140 mM. Assuming that 2.3 RT/F = 60 mV , what would the membrane potential be if the muscle cell membrane were permeable only to Na+? ``` (A) 80 mV (B) −60 mV (C) 0 mV (D) +60 mV (E) +80 mV ```

Accepted Answer

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/Ci = 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.

Question 11

Repeated stimulation of a skeletal muscle fiber causes a sustained contraction (tetanus). Accumulation of which solute in intracellular fluid is responsible for the tetanus? ``` (A) Na+ (B) K+ (C) Cl− (D) Mg2+ (E) Ca2+ (F) Troponin (G) Calmodulin (H) Adenosine triphosphate (ATP) ```

Accepted Answer

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.

Question 12

The velocity of conduction of action potentials along a nerve will be increased by ``` (A) stimulating the Na+–K+ pump (B) inhibiting the Na+–K+ pump (C) decreasing the diameter of the nerve (D) myelinating the nerve (E) lengthening the nerve fiber ```

Accepted Answer

The answer is D [IV E 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.

Question 13

Solutions A and B are separated by a semipermeable membrane. Solution A contains 1 mM sucrose and 1 mM urea. Solution B contains 1 mM sucrose. The reflection coefficient for sucrose is one, and the reflection coefficient for urea is zero. Which of the following statements about these solutions is correct? (A) Solution A has a higher effective osmotic pressure than solution B (B) Solution A has a lower effective osmotic pressure than solution B (C) Solutions A and B are isosmotic (D) Solution A is hyperosmotic with respect to solution B, and the solutions are isotonic (E) Solution A is hyposmotic with respect to solution B, and the solutions are isotonic

Accepted Answer

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 osmolarity of solution A is 2 mOsm/L, and the calculated osmolarity of solution B is 1 mOsm/L. Therefore, solution A, which has a higher osmolarity, 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.

Question 14

Transport of d- and l-glucose proceeds at the same rate down an electrochemical gradient by which of the following processes? ``` (A) Simple diffusion (B) Facilitated diffusion (C) Primary active transport (D) Cotransport (E) Countertransport ```

Accepted Answer

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.

Question 15

Which of the following will double the permeability of a solute in a lipid bilayer? (A) Doubling the molecular radius of the solute (B) Doubling the oil/water partition coefficient of the solute (C) Doubling the thickness of the bilayer (D) Doubling the concentration difference of the solute across the bilayer

Accepted Answer

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.

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