Cell Physiology

Question 1

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

Answer 1

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.

Question 2

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

Answer 2

D. 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

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

Answer 3

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.

Question 4

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

Answer 4

B. 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

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

Answer 5

C. 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

6. 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)

Answer 6

E. 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 7

7. 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

Answer 7

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).

Question 8

8. 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

Answer 8

B. 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 9

9. 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

Answer 9

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.
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 solutions would cause shrinkage of the RBCs.
150 mM NaCl and 300 mM mannitol are isotonic.
350 mM mannitol and 150 mM CaCl2 are hypertonic. Because the reflection coefficient of urea is <1.0, 300 mM urea is hypotonic.

Question 10

11. 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

Answer 10

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.

Question 11

12. 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

Answer 11

D. 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 = 60mV 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 12

16. 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

Answer 12

D. 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

17. 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

Answer 13

D. 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

18. 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

Answer 14

A. 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

19. 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

Answer 15

B. 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.

Question 16

20. A newly developed local anesthetic blocks Na+ channels in nerves. Which of the following effects on the action potential would it be expected to produce?

(A) Decrease the rate of rise of the upstroke of the action potential
(B) Shorten the absolute refractory period
(C) Abolish the hyperpolarizing afterpotential
(D) Increase the Na+ equilibrium potential
(E) Decrease the Na+ equilibrium potential

Answer 16

A. 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 reduced or 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).

Question 17

21. At the muscle end plate, acetylcholine (ACh) causes the opening of

(A) Na+ channels and depolarization toward the Na+ equilibrium potential
(B) K+ channels and depolarization toward the K+ equilibrium potential
(C) Ca2+ channels and depolarization toward the Ca2+ equilibrium potential
(D) Na+ and K+ channels and depolarization to a value halfway between the Na+ and K+ equilibrium potentials
(E) Na+ and K+ channels and hyperpolarization to a value halfway between the Na+ and K+ equilibrium potentials

Answer 17

D. 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

Question 18

22. An inhibitory postsynaptic potential

(A) depolarizes the postsynaptic membrane by opening Na+ channels
(B) depolarizes the postsynaptic membrane by opening K+ channels
(C) hyperpolarizes the postsynaptic membrane by opening Ca2+ channels
(D) hyperpolarizes the postsynaptic membrane by opening Cl− channels

Answer 18

D. An inhibitory postsynaptic potential hyperpolarizes the postsynaptic membrane, taking it farther from threshold.
Opening Cl− channels would hyperpolarize the postsynaptic membrane by driving the membrane potential toward the Cl− equilibrium potential (about −90 mV).
Opening Ca2+ channels would depolarize the postsynaptic membrane by driving it toward the Ca2+ equilibrium potential.

Question 19

23. Which of the following would occur as a result of the inhibition of Na+, K+-ATPase?

(A) Decreased intracellular Na+ concentration
(B) Increased intracellular K+ concentration
(C) Increased intracellular Ca2+ concentration
(D) Increased Na+–glucose cotransport
(E) Increased Na+–Ca2+ exchange

Answer 19

C. 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

Question 20

24. Which of the following temporal sequences is correct for excitation– contraction coupling in skeletal muscle?

(A) Increased intracellular [Ca2+]; action potential in the muscle membrane; cross-bridge formation
(B) Action potential in the muscle membrane; depolarization of the T tubules; release of Ca2+ from the sarcoplasmic reticulum (SR)
(C) Action potential in the muscle membrane; splitting of adenosine triphosphate (ATP); binding of Ca2+ to troponin C
(D) Release of Ca2+ from the SR; depolarization of the T tubules; binding of Ca2+ to troponin C

Answer 20

B. The correct sequence is action potential in the muscle membrane
1. depolarization of the T tubules;
2. release of Ca2+ from the sarcoplasmic
reticulum (SR);
3. binding of Ca2+ to troponin C;
4. cross-bridge formation; and splitting of
adenosine triphosphate (ATP).

Question 21

25. Which of the following transport processes is involved if transport of glucose from the intestinal lumen into a small intestinal cell is inhibited by abolishing the usual Na+ gradient across the cell membrane?

(A) Simple diffusion
(B) Facilitated diffusion
(C) Primary active transport
(D) Cotransport
(E) Countertransport

Answer 21

D. 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 Na+ gradient—cotransport and countertransport. Because glucose is
moving in the same direction as Na+, one can conclude that it is cotransport.

Question 22

26. In skeletal muscle, which of the following events occurs before depolarization of the T tubules in the mechanism of excitation– contraction coupling?

(A) Depolarization of the sarcolemmal membrane
(B) Opening of Ca2+ release channels on the sarcoplasmic reticulum (SR)
(C) Uptake of Ca2+ into the SR by Ca2+- adenosine triphosphatase (ATPase)
(D) Binding of Ca2+ to troponin C
(E) Binding of actin and myosin

Answer 22

A. 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 Ca2+ concentration, binding of Ca2+ to troponin C, and then contraction.

Question 23

27. Which of the following is an inhibitory neurotransmitter in the central nervous system (CNS)?

(A) Norepinephrine
(B) Glutamate
(C) γ-Aminobutyric acid (GABA)
(D) Serotonin
(E) Histamine

Answer 23

C. γ-Aminobutyric acid (GABA) is an inhibitory neurotransmitter.

Norepinephrine, glutamate, serotonin, and histamine are excitatory neurotransmitters.

Question 24

28. Adenosine triphosphate (ATP) is used indirectly for which of the following processes?

(A) Accumulation of Ca2+ by the sarcoplasmic reticulum (SR)
(B) Transport of Na+ from intracellular to extracellular fluid
(C) Transport of K+ from extracellular to intracellular fluid
D) Transport of H+ from parietal cells into the lumen of the stomach
(E) Absorption of glucose by intestinal epithelial cells

Answer 24

E. 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).

Question 25

29. Which of the following causes rigor in skeletal muscle?

(A) Lack of action potentials in motoneurons
(B) An increase in intracellular Ca2+ level
(C) A decrease in intracellular Ca2+ level
(D) An increase in adenosine triphosphate (ATP) level
(E) A decrease in ATP level

Answer 25

E. Rigor is a state of permanent contraction that occurs in skeletal
muscle when adenosine triphosphate (ATP) levels are depleted. With no ATP bound, myosin remains attached to actin and the cross-bridge cycle cannot continue. If there were no action potentials in motoneurons, the muscle fibers they innervate would not contract at all, since action potentials are required for release of Ca2+ from the sarcoplasmic reticulum (SR).
When intracellular Ca2+ concentration increases, Ca2+ binds troponin C, permitting the cross-bridge cycle to occur.
Decreases in intracellular Ca2+ concentration cause relaxation.

Question 26

30. Degeneration of dopaminergic neurons has been implicated in

(A) schizophrenia
(B) Parkinson disease
(C) myasthenia gravis
(D) curare poisoning

Answer 26

B. Dopaminergic neurons and D2 receptors are deficient in people with Parkinson disease.
Schizophrenia involves increased levels of D2 receptors.
Myasthenia gravis and curare poisoning involve the neuromuscular junction, which uses acetylcholine (ACh) as a neurotransmitter.

Question 27

31. Assuming complete dissociation of all solutes, which of the following solutions would be hyperosmotic to 1 mM NaCl?

(A) 1 mM glucose
(B) 1.5 mM glucose
(C) 1 mM CaCl2
(D) 1 mM sucrose
(E) 1 mM KCl

Answer 27

C. Osmolarity is the concentration of particles (osmolarity = g × C).
When two solutions are compared, that with the higher osmolarity is hyperosmotic. The 1 mM CaCl2 solution (osmolarity = 3 mOsm/L) is hyperosmotic to 1 mM NaCl (osmolarity= 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 KCl is isosmotic.

Question 28

32. A new drug is developed that blocks the transporter for H+ secretion in gastric parietal cells. Which of the following transport processes is being inhibited?

(A) Simple diffusion
(B) Facilitated diffusion
(C) Primary active transport
(D) Cotransport
(E) Countertransport

Answer 28

C. H+ secretion by gastric parietal cells occurs by H+–K+ adenosine
riphosphatase (ATPase), a primary active transporter.

Question 29

33. A 56-year-old woman with severe muscle weakness is hospitalized. The only abnormality in her laboratory values is an elevated serum K+ concentration. The elevated serum K+ causes muscle weakness because

(A) the resting membrane potential is hyperpolarized
(B) the K+ equilibrium potential is hyperpolarized
(C) the Na+ equilibrium potential is hyperpolarized
(D) K+ channels are closed by depolarization
(E) K+ channels are opened by depolarization
(F) Na+ channels are closed by depolarization
(G) Na+ channels are opened by depolarization

Answer 29

F. Elevated serum K+ concentration causes depolarization of the K+ equilibrium potential and therefore depolarization of the resting membrane potential in skeletal muscle. Sustained depolarization closes the inactivation gates on Na+ channels
and prevents the occurrence of action potentials in the muscle.

Question 30

34. In contraction of gastrointestinal smooth muscle, which of the following events occurs after binding of Ca2+ to calmodulin?

(A) Depolarization of the sarcolemmal membrane
(B) Ca2+-induced Ca2+ release
(C) Increased myosin-light-chain kinase
(D) Increased intracellular Ca2+ concentration
(E) Opening of ligand-gated Ca2+ channels

Answer 30

C. The steps that produce contraction in smooth muscle occur in the following order:
1. various mechanisms that raise intracellular Ca2+ concentration, including
depolarization of the sarcolemmal membrane, which opens voltage-gated Ca2+ channels, and opening of ligand-gated Ca2+ channels;
2. Ca2+-induced Ca2+ released from SR; 3. increased intracellular Ca2+ concentration;
4. binding of Ca2+ to calmodulin;
5. increased myosin-light chain kinase;
6. phosphorylation of myosin;
7. binding of myosin to actin;
8. cross-bridge cycling, which produces contraction