Walden University NURS 6501 Advanced
Pathophysiology ACTUAL EXAM
2025/2026 | Midterm Exam Practice Test &
Test Bank | 120 Questions | Correct Answers
| Verified Q&A | Pass Guaranteed - A+
Graded
SECTION 1: CELLULAR ADAPTATION, INJURY, AND DEATH
(15 Questions)
Q1. A 68-year-old male with chronic alcoholism presents with jaundice, ascites, and confusion.
Liver biopsy shows hepatocytes with enlarged nuclei and increased protein synthesis. This cellular
adaptation is best described as:
A. Atrophy
B. Hypertrophy [CORRECT]
C. Hyperplasia
D. Metaplasia
,Correct Answer: B
Rationale: Hypertrophy is an increase in cell size due to increased workload or stimulation. In
chronic liver disease, hepatocytes undergo hypertrophy to compensate for damaged cells. Atrophy
(A) involves decreased cell size (seen in disuse or denervation). Hyperplasia (C) involves
increased cell number (seen in endometrial hyperplasia or BPH). Metaplasia (D) involves
replacement of one differentiated cell type with another (seen in Barrett's esophagus).
Walden Note: Hypertrophy vs. hyperplasia is a frequently tested distinction—remember
hypertrophy increases cell size, hyperplasia increases cell number.
Q2. A 45-year-old female with a 30-pack-year smoking history undergoes bronchoscopy. Biopsy of
the bronchial epithelium shows replacement of pseudostratified ciliated columnar epithelium with
stratified squamous epithelium. This adaptation is termed:
A. Dysplasia
B. Hyperplasia
C. Metaplasia [CORRECT]
D. Anaplasia
Correct Answer: C
Rationale: Metaplasia is the reversible replacement of one differentiated cell type with another,
often in response to chronic irritation. In smokers, columnar epithelium transforms to squamous
epithelium to better withstand irritation. Dysplasia (A) refers to disordered, pre-neoplastic cellular
development. Anaplasia (D) indicates loss of differentiation in malignancy.
Walden Note: Barrett's esophagus (squamous → columnar) and smoker's metaplasia (columnar →
squamous) are classic examples tested on midterms.
,Q3. A patient experiences myocardial infarction followed by percutaneous coronary intervention.
Despite restoration of blood flow, myocardial damage worsens due to free radical generation from
xanthine oxidase and mitochondrial dysfunction. This phenomenon is:
A. Coagulative necrosis
B. Ischemia-reperfusion injury [CORRECT]
C. Apoptosis
D. Liquefactive necrosis
Correct Answer: B
Rationale: Ischemia-reperfusion injury occurs when oxygen restoration paradoxically exacerbates
cellular damage through oxidative stress (free radicals), calcium overload, and inflammatory
neutrophil recruitment. Coagulative necrosis (A) is the ischemic tissue death pattern. Apoptosis (C)
is programmed cell death without inflammation.
Walden Note: The xanthine oxidase pathway generating superoxide radicals is a key mechanism in
reperfusion injury.
Q4. A patient with severe peripheral vascular disease develops dry gangrene in the toes.
Microscopic examination would most likely reveal:
A. Liquefactive necrosis with pus formation
B. Coagulative necrosis with tissue architecture preservation [CORRECT]
C. Caseous necrosis with granulomatous inflammation
, D. Fat necrosis with saponification
Correct Answer: B
Rationale: Dry gangrene results from coagulative necrosis (ischemia without bacterial infection),
characterized by protein denaturation and preservation of tissue architecture, leading to
desiccation and mummification. Wet gangree (A) involves liquefactive necrosis with bacterial
infection. Caseous necrosis (C) is seen in tuberculosis.
Walden Note: Dry gangrene = coagulative necrosis + ischemia; Wet gangrene = liquefactive
necrosis + infection.
Q5. A 55-year-old male with acute pancreatitis presents with severe abdominal pain. CT shows
areas of chalky white deposits in the omentum. This represents:
A. Coagulative necrosis
B. Liquefactive necrosis
C. Fat necrosis [CORRECT]
D. Caseous necrosis
Correct Answer: C
Rationale: Fat necrosis occurs when lipases (activated in pancreatitis) break down triglycerides into
fatty acids, which combine with calcium to form chalky white soap deposits (saponification). This is
specific to traumatic fat injury or pancreatic enzyme release.
Walden Note: Saponification = calcium soaps = fat necrosis = elevated serum calcium
(hypocalcemia can occur as calcium is bound).
Pathophysiology ACTUAL EXAM
2025/2026 | Midterm Exam Practice Test &
Test Bank | 120 Questions | Correct Answers
| Verified Q&A | Pass Guaranteed - A+
Graded
SECTION 1: CELLULAR ADAPTATION, INJURY, AND DEATH
(15 Questions)
Q1. A 68-year-old male with chronic alcoholism presents with jaundice, ascites, and confusion.
Liver biopsy shows hepatocytes with enlarged nuclei and increased protein synthesis. This cellular
adaptation is best described as:
A. Atrophy
B. Hypertrophy [CORRECT]
C. Hyperplasia
D. Metaplasia
,Correct Answer: B
Rationale: Hypertrophy is an increase in cell size due to increased workload or stimulation. In
chronic liver disease, hepatocytes undergo hypertrophy to compensate for damaged cells. Atrophy
(A) involves decreased cell size (seen in disuse or denervation). Hyperplasia (C) involves
increased cell number (seen in endometrial hyperplasia or BPH). Metaplasia (D) involves
replacement of one differentiated cell type with another (seen in Barrett's esophagus).
Walden Note: Hypertrophy vs. hyperplasia is a frequently tested distinction—remember
hypertrophy increases cell size, hyperplasia increases cell number.
Q2. A 45-year-old female with a 30-pack-year smoking history undergoes bronchoscopy. Biopsy of
the bronchial epithelium shows replacement of pseudostratified ciliated columnar epithelium with
stratified squamous epithelium. This adaptation is termed:
A. Dysplasia
B. Hyperplasia
C. Metaplasia [CORRECT]
D. Anaplasia
Correct Answer: C
Rationale: Metaplasia is the reversible replacement of one differentiated cell type with another,
often in response to chronic irritation. In smokers, columnar epithelium transforms to squamous
epithelium to better withstand irritation. Dysplasia (A) refers to disordered, pre-neoplastic cellular
development. Anaplasia (D) indicates loss of differentiation in malignancy.
Walden Note: Barrett's esophagus (squamous → columnar) and smoker's metaplasia (columnar →
squamous) are classic examples tested on midterms.
,Q3. A patient experiences myocardial infarction followed by percutaneous coronary intervention.
Despite restoration of blood flow, myocardial damage worsens due to free radical generation from
xanthine oxidase and mitochondrial dysfunction. This phenomenon is:
A. Coagulative necrosis
B. Ischemia-reperfusion injury [CORRECT]
C. Apoptosis
D. Liquefactive necrosis
Correct Answer: B
Rationale: Ischemia-reperfusion injury occurs when oxygen restoration paradoxically exacerbates
cellular damage through oxidative stress (free radicals), calcium overload, and inflammatory
neutrophil recruitment. Coagulative necrosis (A) is the ischemic tissue death pattern. Apoptosis (C)
is programmed cell death without inflammation.
Walden Note: The xanthine oxidase pathway generating superoxide radicals is a key mechanism in
reperfusion injury.
Q4. A patient with severe peripheral vascular disease develops dry gangrene in the toes.
Microscopic examination would most likely reveal:
A. Liquefactive necrosis with pus formation
B. Coagulative necrosis with tissue architecture preservation [CORRECT]
C. Caseous necrosis with granulomatous inflammation
, D. Fat necrosis with saponification
Correct Answer: B
Rationale: Dry gangrene results from coagulative necrosis (ischemia without bacterial infection),
characterized by protein denaturation and preservation of tissue architecture, leading to
desiccation and mummification. Wet gangree (A) involves liquefactive necrosis with bacterial
infection. Caseous necrosis (C) is seen in tuberculosis.
Walden Note: Dry gangrene = coagulative necrosis + ischemia; Wet gangrene = liquefactive
necrosis + infection.
Q5. A 55-year-old male with acute pancreatitis presents with severe abdominal pain. CT shows
areas of chalky white deposits in the omentum. This represents:
A. Coagulative necrosis
B. Liquefactive necrosis
C. Fat necrosis [CORRECT]
D. Caseous necrosis
Correct Answer: C
Rationale: Fat necrosis occurs when lipases (activated in pancreatitis) break down triglycerides into
fatty acids, which combine with calcium to form chalky white soap deposits (saponification). This is
specific to traumatic fat injury or pancreatic enzyme release.
Walden Note: Saponification = calcium soaps = fat necrosis = elevated serum calcium
(hypocalcemia can occur as calcium is bound).
