1
NUR631 MIDTERM EXAM Actual Exam 2026/2027
Complete Questions and Verified Answers with Detailed
Rationales Advanced Physiology and Pathophysiology 100%
Correct Grade A Pass Guaranteed - A+ Graded
SECTION 1: CELLULAR ADAPTATION, INJURY, AND DEATH (Questions 1-15)
Q1: A 65-year-old male with a history of smoking presents with a chronic cough. A bronchial
biopsy reveals that the normal ciliated pseudostratified columnar epithelial cells have been
replaced by stratified squamous epithelial cells. This change is best described as:
A. Hyperplasia
B. Metaplasia. [CORRECT]
C. Dysplasia
D. Anaplasia
Correct Answer: B
Rationale: Metaplasia is a reversible change in which one differentiated cell type is replaced by
another cell type better able to withstand environmental stress. In smokers, the normal ciliated
columnar epithelium of the airways undergoes metaplasia to stratified squamous epithelium to
better withstand chronic irritation. Hyperplasia is increased cell number. Dysplasia is abnormal
cell growth with loss of uniformity. Anaplasia is loss of cell differentiation characteristic of
malignant tumors.
Q2: A 45-year-old female with breast cancer undergoes radiation therapy to the chest wall. Six
months later, the irradiated skin appears thin, dry, and atrophic with loss of adnexal structures.
This represents:
A. Hypertrophy
B. Hyperplasia
C. Atrophy [CORRECT]
D. Metaplasia
Correct Answer: C
Rationale: Atrophy is a decrease in cell size and number leading to reduced tissue mass.
Radiation-induced atrophy results from DNA damage, reduced cell proliferation, and vascular
injury causing chronic ischemia. The skin demonstrates decreased thickness, loss of hair follicles
,2
and sweat glands, and fibrosis. Hypertrophy and hyperplasia represent increased tissue mass.
Metaplasia involves cell type conversion without size reduction.
Q3: A patient with chronic gastroesophageal reflux disease (GERD) undergoes endoscopy.
Biopsy of the distal esophagus shows columnar epithelium with goblet cells replacing the normal
squamous epithelium. This adaptation is termed:
A. Dysplasia
B. Barrett's esophagus (intestinal metaplasia) [CORRECT]
C. Hyperplasia of squamous cells
D. Anaplasia
Correct Answer: B
Rationale: Barrett's esophagus is intestinal metaplasia where chronic acid exposure causes the
normal stratified squamous epithelium to be replaced by intestinal-type columnar epithelium
with goblet cells. This is a protective adaptation but increases risk for adenocarcinoma. It
represents metaplasia, not dysplasia (which would show nuclear atypia), hyperplasia (increased
cell number), or anaplasia (malignant de-differentiation).
Q4: Myocardial cells in a patient with severe aortic stenosis demonstrate increased cell size with
enlarged nuclei. This cellular adaptation is:
A. Hyperplasia
B. Hypertrophy [CORRECT]
C. Metaplasia
D. Dysplasia
Correct Answer: B
Rationale: Hypertrophy is an increase in cell size (not number) in response to increased workload
or hormonal stimulation. In pressure overload (aortic stenosis), cardiac myocytes cannot divide
(permanent cells), so they undergo hypertrophy to increase contractile force. This results in
concentric left ventricular hypertrophy. Hyperplasia involves cell division (not possible in mature
cardiac myocytes). Metaplasia and dysplasia do not describe this workload response.
Q5: A 28-year-old female presents with cervical dysplasia on Pap smear. Colposcopy shows
disordered squamous epithelium with nuclear hyperchromasia, pleomorphism, and loss of
polarity confined to the lower third of the epithelium. This represents:
A. Carcinoma in situ
B. Cervical intraepithelial neoplasia (CIN) I [CORRECT]
,3
C. Metaplasia
D. Hyperplasia
Correct Answer: B
Rationale: Dysplasia is characterized by abnormal cell growth with variation in cell size and
shape, nuclear hyperchromasia, and disordered tissue architecture. CIN I (mild dysplasia)
involves the lower third of the epithelium; CIN II (moderate) involves lower two-thirds; CIN III
(severe/carcinoma in situ) involves full thickness. This is pre-neoplastic, not invasive cancer.
Metaplasia is reversible cell type change without atypia. Hyperplasia is increased cell number
without atypia.
Q6: A patient suffers an acute myocardial infarction. Twelve hours post-infarction, microscopic
examination of the affected myocardium shows eosinophilic cytoplasm, loss of nuclei, and
preservation of tissue architecture. This type of necrosis is:
A. Liquefactive necrosis
B. Coagulative necrosis [CORRECT]
C. Caseous necrosis
D. Fat necrosis
Correct Answer: B
Rationale: Coagulative necrosis is characteristic of ischemic injury in solid organs (heart, kidney,
spleen) except the brain. Denaturation of structural and enzymatic proteins preserves tissue
architecture, creating "ghost" outlines of cells with eosinophilic cytoplasm and karyolysis.
Liquefactive necrosis occurs in brain and with bacterial infection. Caseous necrosis is seen in
tuberculosis. Fat necrosis occurs in breast and pancreas.
Q7: A patient with acute pancreatitis develops chalky white deposits in the peripancreatic fat.
Microscopically, these show shadowy outlines of necrotic adipocytes with calcium deposits. This
represents:
A. Coagulative necrosis
B. Liquefactive necrosis
C. Fat necrosis [CORRECT]
D. Fibrinoid necrosis
Correct Answer: C
Rationale: Fat necrosis occurs when lipases (activated in pancreatitis or released from breast
trauma) hydrolyze triglycerides into fatty acids, which combine with calcium to form chalky
, 4
white soaps (saponification). Microscopically, shadowy outlines of necrotic fat cells are seen
with basophilic calcium deposits. This is distinct from other necrosis types and specific to
adipose tissue injury with enzymatic lipolysis.
Q8: A 70-year-old male with atherosclerotic peripheral vascular disease develops dry gangrene
of the toes. The affected tissue appears black, dry, and shrunken. The pathophysiological
mechanism is:
A. Liquefactive necrosis with bacterial infection
B. Coagulative necrosis with desiccation [CORRECT]
C. Caseous necrosis with granuloma formation
D. Apoptosis with tissue shrinkage
Correct Answer: B
Rationale: Dry gangrene is coagulative necrosis of distal extremities due to ischemia without
bacterial infection. The tissue becomes dry, shrunken, and black (mummified) due to desiccation.
The lack of blood flow prevents bacterial growth and liquefaction. Wet gangrene involves
bacterial infection with liquefactive necrosis, edema, and putrefaction. Caseous necrosis is
specific to TB. Apoptosis is single-cell death, not gross tissue necrosis.
Q9: A patient undergoing chemotherapy develops programmed cell death characterized by cell
shrinkage, chromatin condensation, nuclear fragmentation, and formation of apoptotic bodies
without inflammation. This process is:
A. Necrosis
B. Apoptosis [CORRECT]
C. Autophagy
D. Pyroptosis
Correct Answer: B
Rationale: Apoptosis is programmed cell death characterized by cell shrinkage, membrane
blebbing, chromatin condensation (pyknosis), nuclear fragmentation (karyorrhexis), and
formation of membrane-bound apoptotic bodies phagocytosed by macrophages without
inflammation. Chemotherapy-induced DNA damage triggers intrinsic mitochondrial pathway of
apoptosis. Necrosis is accidental cell death with inflammation. Autophagy is cellular recycling.
Pyroptosis is inflammatory programmed death.
Q10: Reperfusion injury following myocardial ischemia is mediated primarily by:
A. ATP depletion
NUR631 MIDTERM EXAM Actual Exam 2026/2027
Complete Questions and Verified Answers with Detailed
Rationales Advanced Physiology and Pathophysiology 100%
Correct Grade A Pass Guaranteed - A+ Graded
SECTION 1: CELLULAR ADAPTATION, INJURY, AND DEATH (Questions 1-15)
Q1: A 65-year-old male with a history of smoking presents with a chronic cough. A bronchial
biopsy reveals that the normal ciliated pseudostratified columnar epithelial cells have been
replaced by stratified squamous epithelial cells. This change is best described as:
A. Hyperplasia
B. Metaplasia. [CORRECT]
C. Dysplasia
D. Anaplasia
Correct Answer: B
Rationale: Metaplasia is a reversible change in which one differentiated cell type is replaced by
another cell type better able to withstand environmental stress. In smokers, the normal ciliated
columnar epithelium of the airways undergoes metaplasia to stratified squamous epithelium to
better withstand chronic irritation. Hyperplasia is increased cell number. Dysplasia is abnormal
cell growth with loss of uniformity. Anaplasia is loss of cell differentiation characteristic of
malignant tumors.
Q2: A 45-year-old female with breast cancer undergoes radiation therapy to the chest wall. Six
months later, the irradiated skin appears thin, dry, and atrophic with loss of adnexal structures.
This represents:
A. Hypertrophy
B. Hyperplasia
C. Atrophy [CORRECT]
D. Metaplasia
Correct Answer: C
Rationale: Atrophy is a decrease in cell size and number leading to reduced tissue mass.
Radiation-induced atrophy results from DNA damage, reduced cell proliferation, and vascular
injury causing chronic ischemia. The skin demonstrates decreased thickness, loss of hair follicles
,2
and sweat glands, and fibrosis. Hypertrophy and hyperplasia represent increased tissue mass.
Metaplasia involves cell type conversion without size reduction.
Q3: A patient with chronic gastroesophageal reflux disease (GERD) undergoes endoscopy.
Biopsy of the distal esophagus shows columnar epithelium with goblet cells replacing the normal
squamous epithelium. This adaptation is termed:
A. Dysplasia
B. Barrett's esophagus (intestinal metaplasia) [CORRECT]
C. Hyperplasia of squamous cells
D. Anaplasia
Correct Answer: B
Rationale: Barrett's esophagus is intestinal metaplasia where chronic acid exposure causes the
normal stratified squamous epithelium to be replaced by intestinal-type columnar epithelium
with goblet cells. This is a protective adaptation but increases risk for adenocarcinoma. It
represents metaplasia, not dysplasia (which would show nuclear atypia), hyperplasia (increased
cell number), or anaplasia (malignant de-differentiation).
Q4: Myocardial cells in a patient with severe aortic stenosis demonstrate increased cell size with
enlarged nuclei. This cellular adaptation is:
A. Hyperplasia
B. Hypertrophy [CORRECT]
C. Metaplasia
D. Dysplasia
Correct Answer: B
Rationale: Hypertrophy is an increase in cell size (not number) in response to increased workload
or hormonal stimulation. In pressure overload (aortic stenosis), cardiac myocytes cannot divide
(permanent cells), so they undergo hypertrophy to increase contractile force. This results in
concentric left ventricular hypertrophy. Hyperplasia involves cell division (not possible in mature
cardiac myocytes). Metaplasia and dysplasia do not describe this workload response.
Q5: A 28-year-old female presents with cervical dysplasia on Pap smear. Colposcopy shows
disordered squamous epithelium with nuclear hyperchromasia, pleomorphism, and loss of
polarity confined to the lower third of the epithelium. This represents:
A. Carcinoma in situ
B. Cervical intraepithelial neoplasia (CIN) I [CORRECT]
,3
C. Metaplasia
D. Hyperplasia
Correct Answer: B
Rationale: Dysplasia is characterized by abnormal cell growth with variation in cell size and
shape, nuclear hyperchromasia, and disordered tissue architecture. CIN I (mild dysplasia)
involves the lower third of the epithelium; CIN II (moderate) involves lower two-thirds; CIN III
(severe/carcinoma in situ) involves full thickness. This is pre-neoplastic, not invasive cancer.
Metaplasia is reversible cell type change without atypia. Hyperplasia is increased cell number
without atypia.
Q6: A patient suffers an acute myocardial infarction. Twelve hours post-infarction, microscopic
examination of the affected myocardium shows eosinophilic cytoplasm, loss of nuclei, and
preservation of tissue architecture. This type of necrosis is:
A. Liquefactive necrosis
B. Coagulative necrosis [CORRECT]
C. Caseous necrosis
D. Fat necrosis
Correct Answer: B
Rationale: Coagulative necrosis is characteristic of ischemic injury in solid organs (heart, kidney,
spleen) except the brain. Denaturation of structural and enzymatic proteins preserves tissue
architecture, creating "ghost" outlines of cells with eosinophilic cytoplasm and karyolysis.
Liquefactive necrosis occurs in brain and with bacterial infection. Caseous necrosis is seen in
tuberculosis. Fat necrosis occurs in breast and pancreas.
Q7: A patient with acute pancreatitis develops chalky white deposits in the peripancreatic fat.
Microscopically, these show shadowy outlines of necrotic adipocytes with calcium deposits. This
represents:
A. Coagulative necrosis
B. Liquefactive necrosis
C. Fat necrosis [CORRECT]
D. Fibrinoid necrosis
Correct Answer: C
Rationale: Fat necrosis occurs when lipases (activated in pancreatitis or released from breast
trauma) hydrolyze triglycerides into fatty acids, which combine with calcium to form chalky
, 4
white soaps (saponification). Microscopically, shadowy outlines of necrotic fat cells are seen
with basophilic calcium deposits. This is distinct from other necrosis types and specific to
adipose tissue injury with enzymatic lipolysis.
Q8: A 70-year-old male with atherosclerotic peripheral vascular disease develops dry gangrene
of the toes. The affected tissue appears black, dry, and shrunken. The pathophysiological
mechanism is:
A. Liquefactive necrosis with bacterial infection
B. Coagulative necrosis with desiccation [CORRECT]
C. Caseous necrosis with granuloma formation
D. Apoptosis with tissue shrinkage
Correct Answer: B
Rationale: Dry gangrene is coagulative necrosis of distal extremities due to ischemia without
bacterial infection. The tissue becomes dry, shrunken, and black (mummified) due to desiccation.
The lack of blood flow prevents bacterial growth and liquefaction. Wet gangrene involves
bacterial infection with liquefactive necrosis, edema, and putrefaction. Caseous necrosis is
specific to TB. Apoptosis is single-cell death, not gross tissue necrosis.
Q9: A patient undergoing chemotherapy develops programmed cell death characterized by cell
shrinkage, chromatin condensation, nuclear fragmentation, and formation of apoptotic bodies
without inflammation. This process is:
A. Necrosis
B. Apoptosis [CORRECT]
C. Autophagy
D. Pyroptosis
Correct Answer: B
Rationale: Apoptosis is programmed cell death characterized by cell shrinkage, membrane
blebbing, chromatin condensation (pyknosis), nuclear fragmentation (karyorrhexis), and
formation of membrane-bound apoptotic bodies phagocytosed by macrophages without
inflammation. Chemotherapy-induced DNA damage triggers intrinsic mitochondrial pathway of
apoptosis. Necrosis is accidental cell death with inflammation. Autophagy is cellular recycling.
Pyroptosis is inflammatory programmed death.
Q10: Reperfusion injury following myocardial ischemia is mediated primarily by:
A. ATP depletion
