NR507 Advanced Pathophysiology - Week 4 Midterm Exam (Version A) Chamberlain University - 100 Verified Questions - A+ Graded
NR507 Midterm Exam - Advanced Pathophysiology
Week 4 Midterm - Version A - Latest 2026/2027 Edition
Chamberlain University College of Nursing - NR-507 Advanced Pathophysiology
Aligned with 2026-2027 Chamberlain NR-507 Curriculum - Brand New Version
25% recall - 55% application - 20%
Total Questions 100 (Q1-Q100) Cognitive Mix
analysis
75% scenario - 20% recall - 5%
Sections 8 sections (see below) Question Style
clinical reasoning
Version Version A (Brand New) Options 4 options (A-D), one correct
Exam Structure Overview - This NR507 Week 4 Midterm (Version A) is organized into eight sections covering
foundational pathophysiology: cellular adaptation and injury, inflammation and immunity, genetics and neoplasia,
fluid/electrolyte and acid-base disorders, cardiovascular, respiratory, renal and endocrine pathophysiology, and integrated
clinical case scenarios. Every question includes the verified correct answer (marked) and a detailed rationale with
pathophysiological reasoning grounded in cellular mechanisms, molecular pathways, clinical manifestations, and Chamberlain
NR-507 competency standards.
Section Topic Area Questions Count
Section 1 Cellular Adaptation, Injury, and Death Q1-Q15 15
Section 2 Inflammation and Immunity Q16-Q30 15
Section 3 Genetics and Neoplasia Q31-Q40 10
Section 4 Fluid, Electrolyte, and Acid-Base Disorders Q41-Q50 10
Section 5 Cardiovascular Pathophysiology Q51-Q65 15
Section 6 Respiratory Pathophysiology Q66-Q75 10
Section 7 Renal and Endocrine Pathophysiology Q76-Q90 15
Section 8 Integrated Clinical Case Scenarios Q91-Q100 10
Instructions: Each question has exactly one correct answer. The correct option is marked with a green check and the label
[CORRECT]; the answer is also restated in the "Correct Answer" line, followed by a detailed rationale with
pathophysiological reasoning. Use this document as a verified study reference for the Chamberlain NR-507 Week 4 Midterm.
Aligned with 2026-2027 Chamberlain NR-507 Curriculum and Chamberlain Competency Standards Page 1
,NR507 Advanced Pathophysiology - Week 4 Midterm Exam (Version A) Chamberlain University - 100 Verified Questions - A+ Graded
Section 1: Cellular Adaptation, Injury, and Death
Atrophy, Hypertrophy, Hyperplasia, Metaplasia, Dysplasia, Necrosis, Apoptosis | Q1-15
Q1. A 68-year-old male with long-standing hypertension has an echocardiogram demonstrating increased left
ventricular wall thickness. Which cellular adaptation best explains this finding, and what is the underlying
mechanism?
A. Dysplasia, because the cardiac cells demonstrate disordered growth and abnormal organization
B. Hypertrophy, because cardiac muscle cells cannot divide and instead increase in cell size in response to increased
workload [✓ CORRECT]
C. Hyperplasia, because cardiac muscle cells divide to increase cell number under increased workload
D. Metaplasia, because cardiac muscle cells transform into a more resistant cell type
Correct Answer: B
Rationale: Hypertrophy is an increase in cell size in response to increased workload. Cardiac muscle cells are terminally
differentiated and cannot divide, so they enlarge rather than proliferate when stressed by chronic pressure overload from
hypertension. Hyperplasia (A) is an increase in cell number, which cardiac myocytes cannot do. Metaplasia (C) is reversible
substitution of one differentiated cell type for another. Dysplasia (D) is disordered cellular growth with abnormal size, shape, and
organization, often premalignant.
Q2. A 55-year-old chronic smoker undergoes bronchoscopy with biopsy of the bronchial epithelium. Histology
reveals that the normal ciliated pseudostratified columnar epithelium has been replaced by stratified
squamous epithelium. Which cellular adaptation is this, and what is its significance?
A. Atrophy; the bronchial cells have decreased in size due to chronic irritation
B. Metaplasia; the change is a reversible substitution of one differentiated cell type for another that better tolerates
stress, though it may predispose to malignant transformation with ongoing injury [✓ CORRECT]
C. Hyperplasia; the bronchial cells have increased in number due to chronic inflammation
D. Dysplasia; the change is disordered growth that is irreversible and already malignant
Correct Answer: B
Rationale: Metaplasia is the reversible change of one differentiated cell type to another, here ciliated columnar epithelium replaced
by stratified squamous epithelium in response to chronic smoke irritation. The squamous epithelium better survives the stress but loses
ciliary clearance function and, with persistent injury, can progress to malignant transformation. Dysplasia (A) is disordered growth,
not a substitution of cell types. Atrophy (B) is a decrease in cell size. Hyperplasia (C) is an increase in cell number, not a change in
cell type.
Q3. A 30-year-old female with menorrhagia has an endometrial biopsy showing increased number of
endometrial glands due to unopposed estrogen stimulation. Which cellular adaptation is occurring?
A. Hyperplasia, because there is an increase in cell number in response to hormonal stimulation [✓ CORRECT]
B. Hypertrophy, because the endometrial cells have increased in size
C. Dysplasia, because the cells show disordered growth and atypia
D. Metaplasia, because one cell type has replaced another
Correct Answer: A
Rationale: Hyperplasia is an increase in cell number in response to increased demand or hormonal stimulation, as in endometrial
hyperplasia from unopposed estrogen. Hypertrophy (A) is an increase in cell size, not number. Metaplasia (C) involves replacement
of one cell type with another. Dysplasia (D) implies disordered, atypical growth that may be premalignant, which is not described
here. Endometrial hyperplasia must be monitored because it can progress to endometrial carcinoma with persistent estrogen
stimulation.
Q4. A cervical biopsy in a 35-year-old female shows disordered cellular growth with abnormal size, shape, and
organization of squamous epithelial cells that has not broken through the basement membrane. Which term
best describes this finding?
A. Carcinoma with metastasis, because the cells have invaded surrounding tissues
B. Hyperplasia, because there is an increased number of normal cells
C. Metaplasia, because one cell type has been replaced by another
D. Dysplasia that may be reversible but can progress to carcinoma in situ if the insult continues [✓ CORRECT]
Aligned with 2026-2027 Chamberlain NR-507 Curriculum and Chamberlain Competency Standards Page 2
, NR507 Advanced Pathophysiology - Week 4 Midterm Exam (Version A) Chamberlain University - 100 Verified Questions - A+ Graded
Correct Answer: D
Rationale: Dysplasia is disordered cellular growth characterized by abnormal size, shape, and organization. It is often reversible if
the inciting stimulus is removed, but with ongoing injury it can progress to carcinoma in situ, in which dysplastic cells remain localized
but have not broken through the basement membrane. Metaplasia (A) is a cell-type substitution. Hyperplasia (C) is an increase in cell
number without atypia. Carcinoma with metastasis (D) implies invasion beyond the basement membrane, which has not occurred
here.
Q5. A 60-year-old male with a cast on his leg for 6 weeks has noticeable decrease in muscle mass of the affected
limb when the cast is removed. Which cellular adaptation explains this finding, and what is its likely
mechanism?
A. Metaplasia due to chronic irritation from the cast material
B. Atrophy due to decreased workload, denervation, ischemia, malnutrition, or aging [✓ CORRECT]
C. Hypertrophy due to increased workload from the cast
D. Apoptosis due to programmed cell death from cast pressure
Correct Answer: B
Rationale: Atrophy is a decrease in cell size due to decreased workload, denervation, ischemia, malnutrition, or aging. Prolonged
immobilization in a cast reduces mechanical workload on the limb muscles, leading to disuse atrophy. Hypertrophy (B) is increased
cell size from increased workload. Metaplasia (C) is cell-type substitution. Apoptosis (D) is programmed cell death of individual cells,
not a uniform decrease in muscle cell size.
Q6. A pathologist examining a tissue specimen from an acutely ischemic myocardium notes cellular swelling and
fatty change. These findings are most characteristic of which stage of cellular injury?
A. Apoptosis with chromatin condensation and apoptotic body formation
B. Coagulative necrosis with preserved tissue architecture
C. Irreversible injury with nuclear pyknosis and karyorrhexis
D. Reversible injury characterized by impaired Na+/K+ ATPase activity and accumulation of lipids [✓
CORRECT]
Correct Answer: D
Rationale: Cellular swelling and fatty change are hallmarks of reversible cellular injury. Cellular swelling results from failure of
the Na+/K+ ATPase, allowing sodium and water to enter the cell. Fatty change occurs in lipid-rich organs like the liver due to
impaired lipid metabolism. These changes are reversible if perfusion is restored. Irreversible injury (A) shows nuclear changes such
as pyknosis, karyorrhexis, and karyolysis. Apoptosis (C) is programmed cell death. Coagulative necrosis (D) is a later, irreversible
pattern seen in ischemic solid organs.
Q7. Histologic examination of a myocardial infarction at 24 hours demonstrates preserved cellular outlines with
loss of nuclei and eosinophilic cytoplasm. Which type of necrosis is this?
A. Caseous necrosis, characteristic of tuberculosis
B. Gangrenous necrosis, characteristic of lower extremity ischemia
C. Coagulative necrosis, characteristic of ischemic injury in solid organs such as the heart, kidney, and liver [✓
CORRECT]
D. Liquefactive necrosis, characteristic of brain infarction
Correct Answer: C
Rationale: Coagulative necrosis is characteristic of ischemic injury in solid organs (heart, kidney, liver). The cell outlines are
preserved because cellular proteins are denatured and coagulated, but the nuclei are lost. This is the classic pattern of myocardial
infarction. Liquefactive necrosis (A) is enzymatic digestion seen in brain infarcts and abscesses. Caseous necrosis (B) is the
cheese-like necrosis of tuberculosis. Gangrenous necrosis (D) affects lower extremities via ischemia (dry) or bacterial infection (wet).
Q8. A 72-year-old male suffers an ischemic stroke. CT scan days later shows an area of tissue liquefaction in the
cerebral cortex. Which type of necrosis explains this finding?
A. Fat necrosis, characteristic of acute pancreatic enzyme release
B. Coagulative necrosis, because neurons lack connective tissue
C. Caseous necrosis, characteristic of central nervous system tuberculosis
D. Liquefactive necrosis, characteristic of brain infarction due to enzymatic digestion of tissue by lysosomal
enzymes [✓ CORRECT]
Aligned with 2026-2027 Chamberlain NR-507 Curriculum and Chamberlain Competency Standards Page 3
NR507 Midterm Exam - Advanced Pathophysiology
Week 4 Midterm - Version A - Latest 2026/2027 Edition
Chamberlain University College of Nursing - NR-507 Advanced Pathophysiology
Aligned with 2026-2027 Chamberlain NR-507 Curriculum - Brand New Version
25% recall - 55% application - 20%
Total Questions 100 (Q1-Q100) Cognitive Mix
analysis
75% scenario - 20% recall - 5%
Sections 8 sections (see below) Question Style
clinical reasoning
Version Version A (Brand New) Options 4 options (A-D), one correct
Exam Structure Overview - This NR507 Week 4 Midterm (Version A) is organized into eight sections covering
foundational pathophysiology: cellular adaptation and injury, inflammation and immunity, genetics and neoplasia,
fluid/electrolyte and acid-base disorders, cardiovascular, respiratory, renal and endocrine pathophysiology, and integrated
clinical case scenarios. Every question includes the verified correct answer (marked) and a detailed rationale with
pathophysiological reasoning grounded in cellular mechanisms, molecular pathways, clinical manifestations, and Chamberlain
NR-507 competency standards.
Section Topic Area Questions Count
Section 1 Cellular Adaptation, Injury, and Death Q1-Q15 15
Section 2 Inflammation and Immunity Q16-Q30 15
Section 3 Genetics and Neoplasia Q31-Q40 10
Section 4 Fluid, Electrolyte, and Acid-Base Disorders Q41-Q50 10
Section 5 Cardiovascular Pathophysiology Q51-Q65 15
Section 6 Respiratory Pathophysiology Q66-Q75 10
Section 7 Renal and Endocrine Pathophysiology Q76-Q90 15
Section 8 Integrated Clinical Case Scenarios Q91-Q100 10
Instructions: Each question has exactly one correct answer. The correct option is marked with a green check and the label
[CORRECT]; the answer is also restated in the "Correct Answer" line, followed by a detailed rationale with
pathophysiological reasoning. Use this document as a verified study reference for the Chamberlain NR-507 Week 4 Midterm.
Aligned with 2026-2027 Chamberlain NR-507 Curriculum and Chamberlain Competency Standards Page 1
,NR507 Advanced Pathophysiology - Week 4 Midterm Exam (Version A) Chamberlain University - 100 Verified Questions - A+ Graded
Section 1: Cellular Adaptation, Injury, and Death
Atrophy, Hypertrophy, Hyperplasia, Metaplasia, Dysplasia, Necrosis, Apoptosis | Q1-15
Q1. A 68-year-old male with long-standing hypertension has an echocardiogram demonstrating increased left
ventricular wall thickness. Which cellular adaptation best explains this finding, and what is the underlying
mechanism?
A. Dysplasia, because the cardiac cells demonstrate disordered growth and abnormal organization
B. Hypertrophy, because cardiac muscle cells cannot divide and instead increase in cell size in response to increased
workload [✓ CORRECT]
C. Hyperplasia, because cardiac muscle cells divide to increase cell number under increased workload
D. Metaplasia, because cardiac muscle cells transform into a more resistant cell type
Correct Answer: B
Rationale: Hypertrophy is an increase in cell size in response to increased workload. Cardiac muscle cells are terminally
differentiated and cannot divide, so they enlarge rather than proliferate when stressed by chronic pressure overload from
hypertension. Hyperplasia (A) is an increase in cell number, which cardiac myocytes cannot do. Metaplasia (C) is reversible
substitution of one differentiated cell type for another. Dysplasia (D) is disordered cellular growth with abnormal size, shape, and
organization, often premalignant.
Q2. A 55-year-old chronic smoker undergoes bronchoscopy with biopsy of the bronchial epithelium. Histology
reveals that the normal ciliated pseudostratified columnar epithelium has been replaced by stratified
squamous epithelium. Which cellular adaptation is this, and what is its significance?
A. Atrophy; the bronchial cells have decreased in size due to chronic irritation
B. Metaplasia; the change is a reversible substitution of one differentiated cell type for another that better tolerates
stress, though it may predispose to malignant transformation with ongoing injury [✓ CORRECT]
C. Hyperplasia; the bronchial cells have increased in number due to chronic inflammation
D. Dysplasia; the change is disordered growth that is irreversible and already malignant
Correct Answer: B
Rationale: Metaplasia is the reversible change of one differentiated cell type to another, here ciliated columnar epithelium replaced
by stratified squamous epithelium in response to chronic smoke irritation. The squamous epithelium better survives the stress but loses
ciliary clearance function and, with persistent injury, can progress to malignant transformation. Dysplasia (A) is disordered growth,
not a substitution of cell types. Atrophy (B) is a decrease in cell size. Hyperplasia (C) is an increase in cell number, not a change in
cell type.
Q3. A 30-year-old female with menorrhagia has an endometrial biopsy showing increased number of
endometrial glands due to unopposed estrogen stimulation. Which cellular adaptation is occurring?
A. Hyperplasia, because there is an increase in cell number in response to hormonal stimulation [✓ CORRECT]
B. Hypertrophy, because the endometrial cells have increased in size
C. Dysplasia, because the cells show disordered growth and atypia
D. Metaplasia, because one cell type has replaced another
Correct Answer: A
Rationale: Hyperplasia is an increase in cell number in response to increased demand or hormonal stimulation, as in endometrial
hyperplasia from unopposed estrogen. Hypertrophy (A) is an increase in cell size, not number. Metaplasia (C) involves replacement
of one cell type with another. Dysplasia (D) implies disordered, atypical growth that may be premalignant, which is not described
here. Endometrial hyperplasia must be monitored because it can progress to endometrial carcinoma with persistent estrogen
stimulation.
Q4. A cervical biopsy in a 35-year-old female shows disordered cellular growth with abnormal size, shape, and
organization of squamous epithelial cells that has not broken through the basement membrane. Which term
best describes this finding?
A. Carcinoma with metastasis, because the cells have invaded surrounding tissues
B. Hyperplasia, because there is an increased number of normal cells
C. Metaplasia, because one cell type has been replaced by another
D. Dysplasia that may be reversible but can progress to carcinoma in situ if the insult continues [✓ CORRECT]
Aligned with 2026-2027 Chamberlain NR-507 Curriculum and Chamberlain Competency Standards Page 2
, NR507 Advanced Pathophysiology - Week 4 Midterm Exam (Version A) Chamberlain University - 100 Verified Questions - A+ Graded
Correct Answer: D
Rationale: Dysplasia is disordered cellular growth characterized by abnormal size, shape, and organization. It is often reversible if
the inciting stimulus is removed, but with ongoing injury it can progress to carcinoma in situ, in which dysplastic cells remain localized
but have not broken through the basement membrane. Metaplasia (A) is a cell-type substitution. Hyperplasia (C) is an increase in cell
number without atypia. Carcinoma with metastasis (D) implies invasion beyond the basement membrane, which has not occurred
here.
Q5. A 60-year-old male with a cast on his leg for 6 weeks has noticeable decrease in muscle mass of the affected
limb when the cast is removed. Which cellular adaptation explains this finding, and what is its likely
mechanism?
A. Metaplasia due to chronic irritation from the cast material
B. Atrophy due to decreased workload, denervation, ischemia, malnutrition, or aging [✓ CORRECT]
C. Hypertrophy due to increased workload from the cast
D. Apoptosis due to programmed cell death from cast pressure
Correct Answer: B
Rationale: Atrophy is a decrease in cell size due to decreased workload, denervation, ischemia, malnutrition, or aging. Prolonged
immobilization in a cast reduces mechanical workload on the limb muscles, leading to disuse atrophy. Hypertrophy (B) is increased
cell size from increased workload. Metaplasia (C) is cell-type substitution. Apoptosis (D) is programmed cell death of individual cells,
not a uniform decrease in muscle cell size.
Q6. A pathologist examining a tissue specimen from an acutely ischemic myocardium notes cellular swelling and
fatty change. These findings are most characteristic of which stage of cellular injury?
A. Apoptosis with chromatin condensation and apoptotic body formation
B. Coagulative necrosis with preserved tissue architecture
C. Irreversible injury with nuclear pyknosis and karyorrhexis
D. Reversible injury characterized by impaired Na+/K+ ATPase activity and accumulation of lipids [✓
CORRECT]
Correct Answer: D
Rationale: Cellular swelling and fatty change are hallmarks of reversible cellular injury. Cellular swelling results from failure of
the Na+/K+ ATPase, allowing sodium and water to enter the cell. Fatty change occurs in lipid-rich organs like the liver due to
impaired lipid metabolism. These changes are reversible if perfusion is restored. Irreversible injury (A) shows nuclear changes such
as pyknosis, karyorrhexis, and karyolysis. Apoptosis (C) is programmed cell death. Coagulative necrosis (D) is a later, irreversible
pattern seen in ischemic solid organs.
Q7. Histologic examination of a myocardial infarction at 24 hours demonstrates preserved cellular outlines with
loss of nuclei and eosinophilic cytoplasm. Which type of necrosis is this?
A. Caseous necrosis, characteristic of tuberculosis
B. Gangrenous necrosis, characteristic of lower extremity ischemia
C. Coagulative necrosis, characteristic of ischemic injury in solid organs such as the heart, kidney, and liver [✓
CORRECT]
D. Liquefactive necrosis, characteristic of brain infarction
Correct Answer: C
Rationale: Coagulative necrosis is characteristic of ischemic injury in solid organs (heart, kidney, liver). The cell outlines are
preserved because cellular proteins are denatured and coagulated, but the nuclei are lost. This is the classic pattern of myocardial
infarction. Liquefactive necrosis (A) is enzymatic digestion seen in brain infarcts and abscesses. Caseous necrosis (B) is the
cheese-like necrosis of tuberculosis. Gangrenous necrosis (D) affects lower extremities via ischemia (dry) or bacterial infection (wet).
Q8. A 72-year-old male suffers an ischemic stroke. CT scan days later shows an area of tissue liquefaction in the
cerebral cortex. Which type of necrosis explains this finding?
A. Fat necrosis, characteristic of acute pancreatic enzyme release
B. Coagulative necrosis, because neurons lack connective tissue
C. Caseous necrosis, characteristic of central nervous system tuberculosis
D. Liquefactive necrosis, characteristic of brain infarction due to enzymatic digestion of tissue by lysosomal
enzymes [✓ CORRECT]
Aligned with 2026-2027 Chamberlain NR-507 Curriculum and Chamberlain Competency Standards Page 3
