, TESTBANK FOR Pathophysiology for Advanced Practice Workman
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provided.
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,Workman: Pathophysiology for Advanced Practice, 1e 1
Test Bank
Chapter 1 Cellular Biology, Regulation, and Control Mechanisms
1. Which structure(s) is/are common feature(s) of all cells regardless of cellular function or
degree of maturity?
A. Cilia
B. Nuclear envelope
C. Plasma membranes
D. Smooth endoplasmic reticulum
Correct Answer: C
Rationale:
All human cells have a plasma membrane, although it may have other names in skeletal muscles
and neurons. Other features, such as a nucleus, rough and smooth endoplasmic reticulum,
mitochondria, and others, are not present in every cell type.
2. Why are most positive feedback mechanisms harmful if allowed to continue to function
indefinitely?
A. They amplify the effects of the initiating stimulus, increasing the undesirable response.
B. They require additional energy in the form of ATP and impair normal cellular uptake of
nutrients.
C. They misinterpret the input from homeostatic monitoring centers and delay control response
time.
D. They decrease the range of normal for most parameters, negating the need for homeostasis.
Correct Answer: A
Rationale:
Positive feedback mechanisms have the same responses as the initiating change, which results in
an amplification of the changed parameter, often increasing its rate of change. They can be
helpful in the very short run until other interventions or mechanisms can correct the initiating
problem. However, when they continue uncorrected, they lead to exhaustion of the system.
3. Which property of plasma membranes contributes to a cell’s longevity?
A. Unlimited size
B. Lack of covalent bonds
C. Hydrophilic outer surface
D. High concentration of lipids
Correct Answer: B
Rationale:
An important feature of plasma membranes is flexibility or fluidity, which allows cells to change
shape or “deform” without breaking. This feature is a result of the phospholipids composing
plasma membranes not having covalent bonds that would make them more rigid and more easily
Copyright © 2026 F. A. Davis Company
,Workman: Pathophysiology for Advanced Practice, 1e 2
Test Bank
ruptured when squeezing through small capillaries or when pressure is applied.
4. Which cell type is most likely to have the characteristic of polarity of orientation?
A. Red blood cell (erythrocyte)
B. Bronchial epithelium
C. Skeletal muscle
D. Lysosome
Correct Answer: B
Rationale:
Some cells are symmetrical on all sides, whereas others are asymmetrical and have polarity with
regard to orientation. This is related to differences in plasma membrane surface anatomy for a
cell’s specific function. For example, a respiratory epithelial cell has cilia on the luminal surface
and none on the apical surface that is in contact with the basement membrane. The same is true
about the epithelial lining of the intestinal tract with the luminal surfaces expressing villi.
5. Why does the lack of a nucleus in a mature erythrocyte fail to affect its function?
A. Function of these cells is independent of a constant oxygen supply.
B. Their membranes are more rigid than those of other cell types.
C. These cells no longer produce intracellular proteins.
D. Erythrocytes do not require energy for locomotion.
Correct Answer: C
Rationale:
Without a nucleus, a cell can neither divide nor produce proteins. In the case of mature
erythrocytes, all cell division and protein production, including hemoglobin, are performed at an
earlier, less mature stage. Thus, a mature erythrocyte is already equipped with everything it
needs for function (transport of oxygen, acting as a buffering system) and no longer needs a
nucleus.
6. What is the purpose of “chaperone” proteins?
A. Completing the degradation of intracellular debris
B. Transporting newly synthesized proteins for exocytosis
C. Adding a ligand to the plasma membrane for endocytosis
D. Ensuring the correct folding of newly synthesized proteins
Correct Answer: D
Rationale:
Newly synthesized protein strands are released into the lumen of the RER with their amino acids
in linear order. However, protein function is based on its final structure, which includes how it is
folded. In the smooth ER (SER), “chaperone” proteins serve as guides that help fold the new
protein into its correct conformation (shape) for final function.
Copyright © 2026 F. A. Davis Company
,Workman: Pathophysiology for Advanced Practice, 1e 3
Test Bank
7. Why are hydrolases confined to lysosomes?
A. The pH of lysosomes optimizes hydrolase activity.
B. They ensure degradation products are recycled appropriately.
C. They maintain a constant regeneration of transmembrane proteins.
D. Lysosomes have a direct connection to the nucleus where hydrolase is synthesized.
Correct Answer: A
Rationale:
The hydrolases within a lysosome are most effective at a pH of 4, which is the pH maintained
within lysosomes.
8. Why is it necessary to limit the intracellular presence of reactive oxygen species (ROS)?
A. High levels of ROS slow cellular locomotion.
B. ROS inhibit the activity of lysosomal catalase and hydrolase.
C. ROS contribute to an overly acidic intracellular environment.
D. They can damage/inhibit the function of intracellular components.
Correct Answer: D
Rationale:
Because ROS can also oxidize almost anything and cause oxidative stress and damage to other
organelles, tissues, and organs, preventing ROS from coming into contact with other normal cell
structures is critical.
9. Under normal physiologic conditions, which cell type has the highest concentration of
mitochondria?
A. Dermal skin cells
B. Intradermal skin cells
C. Mature erythrocytes
D. Skeletal muscle cells
Correct Answer: D
Rationale:
Mitochondria are the “powerhouses” for ATP production under aerobic conditions. They are
highly concentrated in tissues that are most metabolically active, such as skeletal muscle, heart
muscle, and liver cells, and they are only present in minimal or basal concentrations in less
metabolically active cells, such as skin cells and erythrocytes.
10. Why is glucose inhibited from entering most cells through simple passive diffusion even
though it is a nonpolar molecule?
A. The molecule expresses an overall negative charge.
B. Glucose is highly hydrophilic.
C. The molecule is too large for unaided diffusion.
Copyright © 2026 F. A. Davis Company
,Workman: Pathophysiology for Advanced Practice, 1e 4
Test Bank
D. Glucose is devoid of lipid-soluble components.
Correct Answer: C
Rationale:
Simple passive diffusion allows small, hydrophobic, uncharged (nonpolar) molecules to diffuse
across plasma membranes easily. Although glucose is a nonpolar molecule, it is composed of 24
molecules linked together covalently, making it too large to move across most plasma
membranes by simple diffusion.
11. Which factor regarding the generation of action potentials (AP) is true?
A. The speed of depolarization is determined by the number of Na+-K+ ATPase pumps present in
the membrane.
B. Action potentials can only be generated in cells that normally express a high degree of
polarity when the membrane is at rest.
C. Although an AP may be initiated at any area of an excitable membrane, propagation of the AP
is unidirectional within one cell.
D. Excessive amounts of extracellular calcium enhance the capacity of an excitable membrane to
generate an action potential.
Correct Answer: B
Rationale:
All cells have an ICF that is more negative than the ECF, with nonexcitable membranes having
an electrical resting membrane potential of between -5 and -10 mv, and the ICF is only slightly
more negative than the ECF. Cells with excitable membranes have a much greater charge
difference between the ICF and the ECF, usually between -70 and -85 mv, which is required to
generate an action potential that can be transmitted to other cells within the tissue.
12. In any excitable membrane, what is the trigger for opening of the potassium channels during
the process of depolarization?
A. Rising intracellular ATP concentration
B. Loss of intracellular fluid protein content
C. Increased flexibility of plasma membranes
D. Closure of the voltage-regulated sodium channels
Correct Answer: D
Rationale:
A stimulus for depolarization causes a small portion of the excitable membrane to be more
permeable to Na+, which allows Na+ to influx down its concentration gradient into the cell.
Movement of Na+ into the cell increases ICF positivity. When ICF positivity reaches threshold
level (about +55 mv), all voltage-gated Na+ channels along the plasma membrane open at the
same time, allowing Na+ to rapidly enter the cell until the Na+ concentrations in the ICF and the
ECF surrounding the cell are the same and depolarized with regard to Na+. At this time, the
voltage-regulated Na+ channels close. Closure of these channels triggers K+ channels to open,
allowing intracellular K+ molecules to rapidly leave the cell down its concentration gradient until
Copyright © 2026 F. A. Davis Company
,Workman: Pathophysiology for Advanced Practice, 1e 5
Test Bank
the K+ gradient disappears. At this time, the membrane is completely depolarized for both ions,
and an action potential (AP) has been generated.
13. A new drug being tested for use as an antihypertensive exerts its effects by moving
extracellular chloride ions (Cl–) into nerve cells and all other tissues with excitable
membranes. What changes would you expect to see as a result of this action?
A. Membrane hypopolarization; increased rate of depolarization
B. Membrane hypopolarization; decreased rate of depolarization
C. Membrane hyperpolarization; increased rate of depolarization
D. Membrane hyperpolarization; decreased rate of depolarization
Correct Answer: D
Rationale:
When polarity is increased, the membrane is hyperpolarized, with a greater charge difference
between the two fluid compartments and reduced excitability. Such a condition requires a larger,
stronger, or more sustained stimulus to cause a threshold level Na+ influx and a responding
depolarization. Classic ways that true hyperpolarization occurs most often are a reduction of
positively charged ions, especially K+ ions in the ICF, or an increase in ICF concentration of
negatively charged ions, especially chloride ion levels.
14. Why does increasing the voltage difference between the ECF and the ICF of excitable tissues
slow the rate of depolarization?
A. A greater number of cations must enter the cell to reach the depolarization threshold.
B. The sodium-potassium ATPase pump requires additional ATP molecules for activation.
C. Affected membranes are more rigid and suppress movement of voltage-regulated gates.
D. Increased intracellular negativity causes intracellular proteins to inhibit membrane activity.
Correct Answer: A
Rationale:
When polarity is increased, the membrane is hyperpolarized, with a greater charge difference
between the two fluid compartments and reduced excitability. Such a condition requires a larger,
stronger, or more sustained stimulus to cause a threshold level Na+ influx and a responding
depolarization. Classic ways that true hyperpolarization occurs most often are a reduction of
positively charged ions, especially K+ ions in the ICF, or an increase in ICF concentration of
negatively charged ions, especially chloride ion levels.
15. What can be expected for the activity of a drug when the receptor sites that usually bind the
drug in its target tissue have “down-regulated”?
A. More drug will be required to achieve the normal effect (response).
B. Less drug will be required to achieve the normal effect (response).
C. Drug clearance by enzymatic degradation will be enhanced.
D. Drug clearance by enzymatic degradation will be reduced.
Copyright © 2026 F. A. Davis Company
,Workman: Pathophysiology for Advanced Practice, 1e 6
Test Bank
Correct Answer: A
Rationale:
For a drug that affects tissue function by binding to specific cellular receptors, when the number
of receptors in/on the cells within that tissue have down-regulated, fewer are available for
binding, which is a chance event. Thus, more drug will be needed to ensure drug molecules
actually reach enough receptors to change activity.
16. How does hypercalcemia decrease action potential generation in excitable membranes?
A. Inhibiting the binding of adenosine triphosphate
B. Competing with sodium ions at sodium fast channels
C. Hyperpolarizing the intracellular side of the plasma membrane
D. Enhancing intracellular potassium diffusion into the extracellular fluid environment
Correct Answer: B
Rationale:
Calcium is a divalent cation that competes with sodium at sodium fast channels on excitable
membranes, inhibiting inward flow of sodium and reducing or delaying depolarization even
when extracellular sodium channel levels are normal.
17. What is the expected response to autocrine signaling?
A. Stimulation of receptor down-regulation
B. Blunting of signaling cell’s response
C. Stimulation of receptor up-regulation
D. Enhancement of the signaling cell’s response
Correct Answer: D
Rationale:
Autocrine signaling generates a more widespread and rapid response that is self-perpetuating
because not only is the signal transmitted to other cells, the signaling cell also has receptors for
the signal that can trigger the signaling cell to enhance its own response.
18. Which condition most represents the “all or none” principle of excitable membranes?
A. Direction of action potential propagation varies by cell type.
B. Subthreshold level stimuli have weaker propagation of the action potential.
C. Threshold level stimuli result in whole membrane depolarization with conduction.
D. Threshold level stimuli result in whole membrane depolarization with propagation.
Correct Answer: D
Rationale:
When a stimulus is strong enough to result in depolarization of an action potential in the
stimulated cell, the action potential is propagated at the same strength to the next cell in the line.
A weaker stimulus only results in local depolarization that does not generate a full action
potential even though it may be conducted throughout the membrane of the stimulated cell. It is
not propagated to the next cell in the line.
Copyright © 2026 F. A. Davis Company
,Workman: Pathophysiology for Advanced Practice, 1e 7
Test Bank
19. By which process do tyrosine kinases stimulate an increase in cellular activity?
A. Antagonizing membrane receptors
B. Agonizing ligand-associated receptors
C. Causing molecular phosphorylation
D. Dephosphorylating cyclic guanosine monophosphate
Correct Answer: C
Rationale:
Tyrosine kinases are intracellular stimulatory enzymes that activate other molecules by attaching
a phosphate group to them (phosphorylation).
20. By which mechanism does a G protein-coupled receptor affect cellular responses in a
specific tissue?
A. Increasing intracellular concentration of a second messenger; altering nuclear gene expression
B. Increasing intracellular concentration of a second messenger; maintaining nuclear gene
expression
C. Decreasing intracellular concentration of a second messenger; maintaining nuclear gene
expression
D. Decreasing intracellular concentration of a second messenger; altering nuclear gene
expression
Correct Answer: A
Rationale:
A ligand (first messenger), first binds to a membrane-bound G protein-coupled receptor (GPCR).
Binding of the ligand to this receptor results in activation of a G protein that then binds with and
activates the adenylate cyclase enzyme system present on the inner aspect of the plasma
membrane, resulting in a greatly increased intracellular concentration of cAMP, which then
alters gene expression.
21. Why are lipid-based signaling actions independent of a “first messenger”?
A. They are synthesized as active molecules and do not need further activation.
B. Lipid-based substances enter the ICF without a membrane receptor.
C. Lipid-based structures express an overall positive charge.
D. Lipid-based structures express an overall negative charge.
Correct Answer: B
Rationale:
Lipid structures are soluble in the plasma membrane and do not require the use of a membrane-
bound first messenger.
22. What is the most important characteristic or feature of “gap junctions” for intercellular
communication?
A. Unidirectional
B. Open to larger molecules
Copyright © 2026 F. A. Davis Company
, Workman: Pathophysiology for Advanced Practice, 1e 8
Test Bank
C. Rapid signal transmission
D. Confines signals to a single cell
Correct Answer: C
Rationale:
Gap junctions are very small proteins assembled into tiny tunnel-like channels through the
plasma membrane connecting adjacent cells for the purpose of allowing very rapid transfer or
exchange of small molecules for signaling from the cytosol of one cell to its neighboring cells.
23. How do lower-than-normal concentrations of sodium ions in the extracellular fluid lead to
slower rates of depolarization?
A. Decreasing the positivity of the ICF negates the RMP
B. Increasing the negativity of the ICF reduces its osmolarity
C. Inward flow of sodium ions requires more time to reach threshold levels
D. Outward flow of potassium and calcium ions hyperpolarizes excitable membranes
Correct Answer: C
Rationale:
Depolarization requires sufficient inward flow of ECF sodium to the ICF to raise the ICF to +55
mv. With less ECF sodium available, more time or a much stronger stimulus is needed to allow
sufficient sodium ions to enter the cell and raise the positivity to threshold levels.
24. What is the major pathologic mechanism responsible for the clinical manifestation of
hepatomegaly of Gaucher disease?
A. Excessive accumulation of large precursor molecules in liver macrophages
B. Failure of liver cells to respond to growth-inhibiting signals, resulting in organ hyperplasia
C. Failure to activate intracellular proteins critical to essential liver function
D. Inhibition of hepatocyte mitochondrial activity leading to intracellular swelling
Correct Answer: A
Rationale:
Gaucher disease is an inborn error of metabolism of lysosomal storage in which the enzyme
beta-glucocerebrosidase is deficient. As a result, the lysosomes cannot break down beta-
glucocerebroside and it accumulates in liver lysosomes, greatly enlarging the lysosomes and
leading to hepatomegaly.
Copyright © 2026 F. A. Davis Company
Important Notes
The file includes the complete test bank, organized chapter by chapter.
A sample of selected pages has been provided for preview.
All available appendices and Excel files (if included in the original resources) are
provided.
We continuously update our files to ensure you receive the latest and most accurate
editions.
New editions are added regularly – stay connected for updates!
⚠️Note on Answer Keys: If the answer key is not included within the chapter
questions, you will find the complete answers and solutions at the end of each
chapter.
✅ Why Buy From Us?
📚 Complete & organized chapter-by-chapter – no missing content, no guessing.
⚡ Instant digital delivery – get your file the moment you pay, no waiting.
📅 Always up to date – we track new editions so you always get the latest version.
💬 Friendly support – real humans ready to help, anytime you need us.
🔒 Safe & secure – thousands of satisfied students trust us every semester.
🛡️Our Guarantees
💰 Money-Back Guarantee: Not satisfied? We offer a full refund – no questions asked.
🔄 Wrong File? No Problem: Contact us and we will replace it immediately with the
correct version, free of charge.
⏰ 24/7 Support: We are always here – reach out anytime and expect a fast response.
Contact Email:
,Workman: Pathophysiology for Advanced Practice, 1e 1
Test Bank
Chapter 1 Cellular Biology, Regulation, and Control Mechanisms
1. Which structure(s) is/are common feature(s) of all cells regardless of cellular function or
degree of maturity?
A. Cilia
B. Nuclear envelope
C. Plasma membranes
D. Smooth endoplasmic reticulum
Correct Answer: C
Rationale:
All human cells have a plasma membrane, although it may have other names in skeletal muscles
and neurons. Other features, such as a nucleus, rough and smooth endoplasmic reticulum,
mitochondria, and others, are not present in every cell type.
2. Why are most positive feedback mechanisms harmful if allowed to continue to function
indefinitely?
A. They amplify the effects of the initiating stimulus, increasing the undesirable response.
B. They require additional energy in the form of ATP and impair normal cellular uptake of
nutrients.
C. They misinterpret the input from homeostatic monitoring centers and delay control response
time.
D. They decrease the range of normal for most parameters, negating the need for homeostasis.
Correct Answer: A
Rationale:
Positive feedback mechanisms have the same responses as the initiating change, which results in
an amplification of the changed parameter, often increasing its rate of change. They can be
helpful in the very short run until other interventions or mechanisms can correct the initiating
problem. However, when they continue uncorrected, they lead to exhaustion of the system.
3. Which property of plasma membranes contributes to a cell’s longevity?
A. Unlimited size
B. Lack of covalent bonds
C. Hydrophilic outer surface
D. High concentration of lipids
Correct Answer: B
Rationale:
An important feature of plasma membranes is flexibility or fluidity, which allows cells to change
shape or “deform” without breaking. This feature is a result of the phospholipids composing
plasma membranes not having covalent bonds that would make them more rigid and more easily
Copyright © 2026 F. A. Davis Company
,Workman: Pathophysiology for Advanced Practice, 1e 2
Test Bank
ruptured when squeezing through small capillaries or when pressure is applied.
4. Which cell type is most likely to have the characteristic of polarity of orientation?
A. Red blood cell (erythrocyte)
B. Bronchial epithelium
C. Skeletal muscle
D. Lysosome
Correct Answer: B
Rationale:
Some cells are symmetrical on all sides, whereas others are asymmetrical and have polarity with
regard to orientation. This is related to differences in plasma membrane surface anatomy for a
cell’s specific function. For example, a respiratory epithelial cell has cilia on the luminal surface
and none on the apical surface that is in contact with the basement membrane. The same is true
about the epithelial lining of the intestinal tract with the luminal surfaces expressing villi.
5. Why does the lack of a nucleus in a mature erythrocyte fail to affect its function?
A. Function of these cells is independent of a constant oxygen supply.
B. Their membranes are more rigid than those of other cell types.
C. These cells no longer produce intracellular proteins.
D. Erythrocytes do not require energy for locomotion.
Correct Answer: C
Rationale:
Without a nucleus, a cell can neither divide nor produce proteins. In the case of mature
erythrocytes, all cell division and protein production, including hemoglobin, are performed at an
earlier, less mature stage. Thus, a mature erythrocyte is already equipped with everything it
needs for function (transport of oxygen, acting as a buffering system) and no longer needs a
nucleus.
6. What is the purpose of “chaperone” proteins?
A. Completing the degradation of intracellular debris
B. Transporting newly synthesized proteins for exocytosis
C. Adding a ligand to the plasma membrane for endocytosis
D. Ensuring the correct folding of newly synthesized proteins
Correct Answer: D
Rationale:
Newly synthesized protein strands are released into the lumen of the RER with their amino acids
in linear order. However, protein function is based on its final structure, which includes how it is
folded. In the smooth ER (SER), “chaperone” proteins serve as guides that help fold the new
protein into its correct conformation (shape) for final function.
Copyright © 2026 F. A. Davis Company
,Workman: Pathophysiology for Advanced Practice, 1e 3
Test Bank
7. Why are hydrolases confined to lysosomes?
A. The pH of lysosomes optimizes hydrolase activity.
B. They ensure degradation products are recycled appropriately.
C. They maintain a constant regeneration of transmembrane proteins.
D. Lysosomes have a direct connection to the nucleus where hydrolase is synthesized.
Correct Answer: A
Rationale:
The hydrolases within a lysosome are most effective at a pH of 4, which is the pH maintained
within lysosomes.
8. Why is it necessary to limit the intracellular presence of reactive oxygen species (ROS)?
A. High levels of ROS slow cellular locomotion.
B. ROS inhibit the activity of lysosomal catalase and hydrolase.
C. ROS contribute to an overly acidic intracellular environment.
D. They can damage/inhibit the function of intracellular components.
Correct Answer: D
Rationale:
Because ROS can also oxidize almost anything and cause oxidative stress and damage to other
organelles, tissues, and organs, preventing ROS from coming into contact with other normal cell
structures is critical.
9. Under normal physiologic conditions, which cell type has the highest concentration of
mitochondria?
A. Dermal skin cells
B. Intradermal skin cells
C. Mature erythrocytes
D. Skeletal muscle cells
Correct Answer: D
Rationale:
Mitochondria are the “powerhouses” for ATP production under aerobic conditions. They are
highly concentrated in tissues that are most metabolically active, such as skeletal muscle, heart
muscle, and liver cells, and they are only present in minimal or basal concentrations in less
metabolically active cells, such as skin cells and erythrocytes.
10. Why is glucose inhibited from entering most cells through simple passive diffusion even
though it is a nonpolar molecule?
A. The molecule expresses an overall negative charge.
B. Glucose is highly hydrophilic.
C. The molecule is too large for unaided diffusion.
Copyright © 2026 F. A. Davis Company
,Workman: Pathophysiology for Advanced Practice, 1e 4
Test Bank
D. Glucose is devoid of lipid-soluble components.
Correct Answer: C
Rationale:
Simple passive diffusion allows small, hydrophobic, uncharged (nonpolar) molecules to diffuse
across plasma membranes easily. Although glucose is a nonpolar molecule, it is composed of 24
molecules linked together covalently, making it too large to move across most plasma
membranes by simple diffusion.
11. Which factor regarding the generation of action potentials (AP) is true?
A. The speed of depolarization is determined by the number of Na+-K+ ATPase pumps present in
the membrane.
B. Action potentials can only be generated in cells that normally express a high degree of
polarity when the membrane is at rest.
C. Although an AP may be initiated at any area of an excitable membrane, propagation of the AP
is unidirectional within one cell.
D. Excessive amounts of extracellular calcium enhance the capacity of an excitable membrane to
generate an action potential.
Correct Answer: B
Rationale:
All cells have an ICF that is more negative than the ECF, with nonexcitable membranes having
an electrical resting membrane potential of between -5 and -10 mv, and the ICF is only slightly
more negative than the ECF. Cells with excitable membranes have a much greater charge
difference between the ICF and the ECF, usually between -70 and -85 mv, which is required to
generate an action potential that can be transmitted to other cells within the tissue.
12. In any excitable membrane, what is the trigger for opening of the potassium channels during
the process of depolarization?
A. Rising intracellular ATP concentration
B. Loss of intracellular fluid protein content
C. Increased flexibility of plasma membranes
D. Closure of the voltage-regulated sodium channels
Correct Answer: D
Rationale:
A stimulus for depolarization causes a small portion of the excitable membrane to be more
permeable to Na+, which allows Na+ to influx down its concentration gradient into the cell.
Movement of Na+ into the cell increases ICF positivity. When ICF positivity reaches threshold
level (about +55 mv), all voltage-gated Na+ channels along the plasma membrane open at the
same time, allowing Na+ to rapidly enter the cell until the Na+ concentrations in the ICF and the
ECF surrounding the cell are the same and depolarized with regard to Na+. At this time, the
voltage-regulated Na+ channels close. Closure of these channels triggers K+ channels to open,
allowing intracellular K+ molecules to rapidly leave the cell down its concentration gradient until
Copyright © 2026 F. A. Davis Company
,Workman: Pathophysiology for Advanced Practice, 1e 5
Test Bank
the K+ gradient disappears. At this time, the membrane is completely depolarized for both ions,
and an action potential (AP) has been generated.
13. A new drug being tested for use as an antihypertensive exerts its effects by moving
extracellular chloride ions (Cl–) into nerve cells and all other tissues with excitable
membranes. What changes would you expect to see as a result of this action?
A. Membrane hypopolarization; increased rate of depolarization
B. Membrane hypopolarization; decreased rate of depolarization
C. Membrane hyperpolarization; increased rate of depolarization
D. Membrane hyperpolarization; decreased rate of depolarization
Correct Answer: D
Rationale:
When polarity is increased, the membrane is hyperpolarized, with a greater charge difference
between the two fluid compartments and reduced excitability. Such a condition requires a larger,
stronger, or more sustained stimulus to cause a threshold level Na+ influx and a responding
depolarization. Classic ways that true hyperpolarization occurs most often are a reduction of
positively charged ions, especially K+ ions in the ICF, or an increase in ICF concentration of
negatively charged ions, especially chloride ion levels.
14. Why does increasing the voltage difference between the ECF and the ICF of excitable tissues
slow the rate of depolarization?
A. A greater number of cations must enter the cell to reach the depolarization threshold.
B. The sodium-potassium ATPase pump requires additional ATP molecules for activation.
C. Affected membranes are more rigid and suppress movement of voltage-regulated gates.
D. Increased intracellular negativity causes intracellular proteins to inhibit membrane activity.
Correct Answer: A
Rationale:
When polarity is increased, the membrane is hyperpolarized, with a greater charge difference
between the two fluid compartments and reduced excitability. Such a condition requires a larger,
stronger, or more sustained stimulus to cause a threshold level Na+ influx and a responding
depolarization. Classic ways that true hyperpolarization occurs most often are a reduction of
positively charged ions, especially K+ ions in the ICF, or an increase in ICF concentration of
negatively charged ions, especially chloride ion levels.
15. What can be expected for the activity of a drug when the receptor sites that usually bind the
drug in its target tissue have “down-regulated”?
A. More drug will be required to achieve the normal effect (response).
B. Less drug will be required to achieve the normal effect (response).
C. Drug clearance by enzymatic degradation will be enhanced.
D. Drug clearance by enzymatic degradation will be reduced.
Copyright © 2026 F. A. Davis Company
,Workman: Pathophysiology for Advanced Practice, 1e 6
Test Bank
Correct Answer: A
Rationale:
For a drug that affects tissue function by binding to specific cellular receptors, when the number
of receptors in/on the cells within that tissue have down-regulated, fewer are available for
binding, which is a chance event. Thus, more drug will be needed to ensure drug molecules
actually reach enough receptors to change activity.
16. How does hypercalcemia decrease action potential generation in excitable membranes?
A. Inhibiting the binding of adenosine triphosphate
B. Competing with sodium ions at sodium fast channels
C. Hyperpolarizing the intracellular side of the plasma membrane
D. Enhancing intracellular potassium diffusion into the extracellular fluid environment
Correct Answer: B
Rationale:
Calcium is a divalent cation that competes with sodium at sodium fast channels on excitable
membranes, inhibiting inward flow of sodium and reducing or delaying depolarization even
when extracellular sodium channel levels are normal.
17. What is the expected response to autocrine signaling?
A. Stimulation of receptor down-regulation
B. Blunting of signaling cell’s response
C. Stimulation of receptor up-regulation
D. Enhancement of the signaling cell’s response
Correct Answer: D
Rationale:
Autocrine signaling generates a more widespread and rapid response that is self-perpetuating
because not only is the signal transmitted to other cells, the signaling cell also has receptors for
the signal that can trigger the signaling cell to enhance its own response.
18. Which condition most represents the “all or none” principle of excitable membranes?
A. Direction of action potential propagation varies by cell type.
B. Subthreshold level stimuli have weaker propagation of the action potential.
C. Threshold level stimuli result in whole membrane depolarization with conduction.
D. Threshold level stimuli result in whole membrane depolarization with propagation.
Correct Answer: D
Rationale:
When a stimulus is strong enough to result in depolarization of an action potential in the
stimulated cell, the action potential is propagated at the same strength to the next cell in the line.
A weaker stimulus only results in local depolarization that does not generate a full action
potential even though it may be conducted throughout the membrane of the stimulated cell. It is
not propagated to the next cell in the line.
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,Workman: Pathophysiology for Advanced Practice, 1e 7
Test Bank
19. By which process do tyrosine kinases stimulate an increase in cellular activity?
A. Antagonizing membrane receptors
B. Agonizing ligand-associated receptors
C. Causing molecular phosphorylation
D. Dephosphorylating cyclic guanosine monophosphate
Correct Answer: C
Rationale:
Tyrosine kinases are intracellular stimulatory enzymes that activate other molecules by attaching
a phosphate group to them (phosphorylation).
20. By which mechanism does a G protein-coupled receptor affect cellular responses in a
specific tissue?
A. Increasing intracellular concentration of a second messenger; altering nuclear gene expression
B. Increasing intracellular concentration of a second messenger; maintaining nuclear gene
expression
C. Decreasing intracellular concentration of a second messenger; maintaining nuclear gene
expression
D. Decreasing intracellular concentration of a second messenger; altering nuclear gene
expression
Correct Answer: A
Rationale:
A ligand (first messenger), first binds to a membrane-bound G protein-coupled receptor (GPCR).
Binding of the ligand to this receptor results in activation of a G protein that then binds with and
activates the adenylate cyclase enzyme system present on the inner aspect of the plasma
membrane, resulting in a greatly increased intracellular concentration of cAMP, which then
alters gene expression.
21. Why are lipid-based signaling actions independent of a “first messenger”?
A. They are synthesized as active molecules and do not need further activation.
B. Lipid-based substances enter the ICF without a membrane receptor.
C. Lipid-based structures express an overall positive charge.
D. Lipid-based structures express an overall negative charge.
Correct Answer: B
Rationale:
Lipid structures are soluble in the plasma membrane and do not require the use of a membrane-
bound first messenger.
22. What is the most important characteristic or feature of “gap junctions” for intercellular
communication?
A. Unidirectional
B. Open to larger molecules
Copyright © 2026 F. A. Davis Company
, Workman: Pathophysiology for Advanced Practice, 1e 8
Test Bank
C. Rapid signal transmission
D. Confines signals to a single cell
Correct Answer: C
Rationale:
Gap junctions are very small proteins assembled into tiny tunnel-like channels through the
plasma membrane connecting adjacent cells for the purpose of allowing very rapid transfer or
exchange of small molecules for signaling from the cytosol of one cell to its neighboring cells.
23. How do lower-than-normal concentrations of sodium ions in the extracellular fluid lead to
slower rates of depolarization?
A. Decreasing the positivity of the ICF negates the RMP
B. Increasing the negativity of the ICF reduces its osmolarity
C. Inward flow of sodium ions requires more time to reach threshold levels
D. Outward flow of potassium and calcium ions hyperpolarizes excitable membranes
Correct Answer: C
Rationale:
Depolarization requires sufficient inward flow of ECF sodium to the ICF to raise the ICF to +55
mv. With less ECF sodium available, more time or a much stronger stimulus is needed to allow
sufficient sodium ions to enter the cell and raise the positivity to threshold levels.
24. What is the major pathologic mechanism responsible for the clinical manifestation of
hepatomegaly of Gaucher disease?
A. Excessive accumulation of large precursor molecules in liver macrophages
B. Failure of liver cells to respond to growth-inhibiting signals, resulting in organ hyperplasia
C. Failure to activate intracellular proteins critical to essential liver function
D. Inhibition of hepatocyte mitochondrial activity leading to intracellular swelling
Correct Answer: A
Rationale:
Gaucher disease is an inborn error of metabolism of lysosomal storage in which the enzyme
beta-glucocerebrosidase is deficient. As a result, the lysosomes cannot break down beta-
glucocerebroside and it accumulates in liver lysosomes, greatly enlarging the lysosomes and
leading to hepatomegaly.
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