WGU D236 Pathophysiology - Objective Assessment Exam (2026/2027) Total: 180 Questions
WESTERN GOVERNORS UNIVERSITY | WGU D236 | PATHOPHYSIOLOGY OBJECTIVE ASSESSMENT (OA)
Objective Assessment Exam: Pathophysiology
D236 - Verified Questions and Answers for an A Grade
Latest Update 2026/2027 | Aligned with WGU Course Competencies and OA Format | 100% Verified
180 Questions - Multiple Choice (A-D), One Best Answer - Cognitive Mix: 25% Recall / 50% Application / 25% Analysis
- 75% Scenario-Based
SECTION 1 - CELLULAR ADAPTATION, INJURY, AND NEOPLASIA
Cellular Adaptations, Cell Injury Mechanisms, Necrosis, Apoptosis, Carcinogenesis, Oncogenes, Tumor Suppressor Genes,
Metastasis
Q1: A 62-year-old man with a 15-year history of poorly controlled hypertension presents with
progressive shortness of breath. An echocardiogram reveals a thickened left ventricular wall with
an enlarged overall muscle mass but no increase in the number of cardiac muscle cells. Which
cellular adaptation best explains this finding?
A. Hyperplasia, because increased hormonal stimulation drives an increase in the number of
myocardial cells
B. Hypertrophy, because increased cardiac workload from pressure overload increases the
size of individual myocytes through augmented protein synthesis [CORRECT]
C. Atrophy, because chronic ischemia reduces the size and function of the ventricular
myocytes
D. Dysplasia, because disordered growth of the myocardium produces abnormally sized and
arranged cells
Correct Answer: B
Rationale: Chronic hypertension imposes a pressure (workload) overload on the left ventricle, and
cardiac muscle cells, which cannot divide meaningfully, respond by increasing in individual cell size
through increased synthesis of contractile proteins and organelles; this is hypertrophy. Hyperplasia
requires an increase in cell number and is not a property of mature cardiac myocytes, so it is
mechanistically impossible here. Atrophy is a decrease in cell size from reduced workload,
nutrition, or innervation, the opposite of the pressure-overload state. Dysplasia refers to disordered,
atypical cellular growth seen in premalignant epithelium, not a physiologic response to workload.
1
,WGU D236 Pathophysiology - Objective Assessment Exam (2026/2027) Total: 180 Questions
Q2: A 55-year-old pack-a-day smoker undergoes bronchoscopy, and biopsy of the bronchial
epithelium shows replacement of the normal pseudostratified ciliated columnar epithelium by
stratified squamous epithelium. The nurse practitioner counsels the patient that this change is
potentially reversible. What is this adaptive process, and why is it clinically significant?
A. Hyperplasia; it represents an increase in cell number and is always benign
B. Metaplasia; the squamous epithelium lacks cilia and mucus secretion, and with persistent
irritation such as smoking it can progress to dysplasia and eventually squamous cell
carcinoma
C. Hypertrophy; enlarged epithelial cells thicken the airway wall and reverse completely
after irritant removal
D. Anaplasia; it reflects a malignant transformation that is irreversible regardless of irritant
removal [CORRECT]
Correct Answer: D
Rationale: Metaplasia is the reversible replacement of one adult cell type by another better able to
withstand the adverse environment; in the bronchi, fragile ciliated columnar cells are replaced by
tougher stratified squamous cells. The clinical significance is twofold: the new squamous epithelium
loses the protective mucociliary blanket, and continued exposure to tobacco carcinogens commonly
drives metaplasia into dysplasia and eventually squamous cell carcinoma of the lung. Hyperplasia
and hypertrophy involve increases in cell number and size respectively and do not describe a
cell-type change. Anaplasia denotes loss of differentiation in frank malignancy, which is not
reversible and is not present in this biopsy description.
Q3: A 38-year-old woman has her right leg immobilized in a cast for eight weeks after a tibial
fracture. When the cast is removed, the right calf is noticeably smaller and weaker than the left,
although imaging shows no nerve damage. Which mechanism is primarily responsible for this
adaptive change?
A. Atrophy through decreased protein synthesis and increased proteasomal degradation of
proteins in disused skeletal muscle fibers [CORRECT]
B. Hypertrophy through compensatory enlargement of the immobilized muscle fibers from
disuse
C. Necrosis of muscle fibers caused by pressure injury from the cast
D. Metaplasia of skeletal muscle into fibrous connective tissue caused by the absence of
loading
Correct Answer: A
Rationale: Disuse atrophy occurs when reduced muscle loading lowers protein synthesis while the
ubiquitin-proteasome pathway accelerates protein degradation, shrinking individual muscle fibers;
the process is reversible with remobilization and resistance exercise. Hypertrophy requires increased
workload, which cannot occur in an immobilized limb, so it cannot explain a smaller calf. Pressure
necrosis would produce pain, skin breakdown, and irreversible cell death rather than a smooth,
symmetric reduction in muscle bulk. Metaplasia is a change of one adult cell type to another and is
not the mechanism of disuse-related muscle loss.
2
,WGU D236 Pathophysiology - Objective Assessment Exam (2026/2027) Total: 180 Questions
Q4: A 30-year-old woman with an elevated estrogen level has a pelvic ultrasound showing a
thickened endometrium; biopsy confirms an increased number of normal-appearing endometrial
glandular cells. Which statement best distinguishes this adaptation from hypertrophy?
A. Both processes require increased cell size, but only hyperplasia occurs in the uterus
B. Hypertrophy increases cell number while hyperplasia increases individual cell size
C. Hyperplasia increases the number of cells driven by hormone stimulation, whereas
hypertrophy increases the size of existing cells driven by increased workload [CORRECT]
D. Hyperplasia is always pathological and precedes cancer, whereas hypertrophy is always
physiological
Correct Answer: C
Rationale: Hyperplasia is an increase in the number of cells in an organ or tissue, typically in
response to hormonal stimulation (as estrogen drives endometrial gland proliferation) or
compensatory demand; hypertrophy is an increase in individual cell size, classically seen in cardiac
and skeletal muscle responding to increased work. The two can coexist, but the defining distinction
is cell number versus cell size. Hyperplasia is not always pathological; physiological examples
include endometrial proliferation in the menstrual cycle and compensatory liver regrowth after
partial hepatectomy.
Q5: A routine Pap smear from a 29-year-old woman is reported as showing 'disordered
epithelial growth with variation in nuclear size and shape, increased mitoses, and loss of normal
maturation orientation.' The colposcopic biopsy is classified as cervical intraepithelial neoplasia
associated with high-risk HPV. Which term describes this finding, and what is its key clinical
implication?
A. Metaplasia; it is a fully benign adaptive change requiring no follow-up
B. Dysplasia; it is a premalignant disordered growth that may be reversible with removal of
the inciting stimulus but can progress to carcinoma in situ and invasive cancer
[CORRECT]
C. Hyperplasia; it is a benign increase in cell number caused by inflammation
D. Apoptosis; it represents programmed cell death induced by HPV oncoproteins
Correct Answer: B
Rationale: Dysplasia is characterized by disordered growth: altered cell size and shape
(pleomorphism), enlarged and hyperchromatic nuclei, increased mitotic activity, and loss of normal
polarity and maturation. In the cervix it is induced by high-risk HPV types such as 16 and 18, whose
E6 and E7 oncoproteins inactivate p53 and Rb; importantly, dysplasia is potentially reversible if the
stimulus is removed but may progress through carcinoma in situ to invasive squamous cell
carcinoma, which is why surveillance and treatment are required. Metaplasia and hyperplasia are
orderly adaptive processes without cytologic atypia. Apoptosis is a mode of cell death, not a growth
disorder.
3
, WGU D236 Pathophysiology - Objective Assessment Exam (2026/2027) Total: 180 Questions
Q6: A patient in septic shock becomes profoundly hypotensive, and tissue perfusion falls.
Within the injured cells, decreased oxygen delivery leads to failure of the sodium-potassium
membrane pump. What intracellular change directly follows this pump failure and initiates acute
cell injury?
A. Intracellular accumulation of sodium and water producing cellular swelling, with
potassium diffusing out of the cell [CORRECT]
B. Loss of intracellular water and shrinkage of the cell caused by hyperosmolarity of the
cytoplasm
C. Immediate rupture of the nuclear membrane with digestion of DNA by endonucleases
D. Conversion of the cell to anaerobic carbohydrate synthesis with depletion of intracellular
chloride
Correct Answer: A
Rationale: Hypoxia forces cells into anaerobic glycolysis, depleting ATP; without ATP the
energy-dependent sodium-potassium pump fails, so sodium accumulates intracellularly while
potassium leaks out, and water follows sodium osmotically, causing acute cellular swelling - the
earliest reversible manifestation of cell injury. The shrinkage option reverses the water movement;
injured hypoxic cells swell rather than shrink. Nuclear membrane rupture and DNA digestion relate
to later, irreversible stages or to apoptosis, not the initial pump failure. Anaerobic metabolism
produces lactate and drops intracellular pH; it does not synthesize carbohydrate or alter chloride as
the primary injurious event.
Q7: A patient with a completely occluded coronary artery undergoes successful percutaneous
reperfusion, but afterward myocardial dysfunction paradoxically worsens and infarct size
enlarges. Blood studies show elevated markers of oxidative damage to cell membranes. What is
the mechanism of this phenomenon?
A. Excess calcium efflux that stabilizes mitochondrial membranes and halts ATP
production
B. Depletion of intracellular sodium that causes the cell to take up excess bicarbonate
C. Sudden increase in anaerobic glycolysis that overproduces ATP and injures
mitochondrial DNA
D. Restoration of blood flow delivering neutrophils that generate more damaging reactive
oxygen species than the tissue can neutralize [CORRECT]
Correct Answer: D
Rationale: Reperfusion injury occurs when restored blood flow reoxygenates previously ischemic
tissue; damaged mitochondria and infiltrating neutrophils generate a burst of reactive oxygen
species (superoxide, hydroxyl radicals, hydrogen peroxide) that overwhelms antioxidant defenses
and extends membrane lipid peroxidation and cell death. Ischemic cells accumulate intracellular
calcium rather than losing it, and calcium overload further damages mitochondria, so any option
describing calcium efflux states the opposite direction. Sodium depletion is not a feature of
reperfusion; sodium overload accompanies pump failure. Anaerobic glycolysis ceases upon
reoxygenation; the injury comes from oxidative free radicals, not from ATP overproduction.
4
WESTERN GOVERNORS UNIVERSITY | WGU D236 | PATHOPHYSIOLOGY OBJECTIVE ASSESSMENT (OA)
Objective Assessment Exam: Pathophysiology
D236 - Verified Questions and Answers for an A Grade
Latest Update 2026/2027 | Aligned with WGU Course Competencies and OA Format | 100% Verified
180 Questions - Multiple Choice (A-D), One Best Answer - Cognitive Mix: 25% Recall / 50% Application / 25% Analysis
- 75% Scenario-Based
SECTION 1 - CELLULAR ADAPTATION, INJURY, AND NEOPLASIA
Cellular Adaptations, Cell Injury Mechanisms, Necrosis, Apoptosis, Carcinogenesis, Oncogenes, Tumor Suppressor Genes,
Metastasis
Q1: A 62-year-old man with a 15-year history of poorly controlled hypertension presents with
progressive shortness of breath. An echocardiogram reveals a thickened left ventricular wall with
an enlarged overall muscle mass but no increase in the number of cardiac muscle cells. Which
cellular adaptation best explains this finding?
A. Hyperplasia, because increased hormonal stimulation drives an increase in the number of
myocardial cells
B. Hypertrophy, because increased cardiac workload from pressure overload increases the
size of individual myocytes through augmented protein synthesis [CORRECT]
C. Atrophy, because chronic ischemia reduces the size and function of the ventricular
myocytes
D. Dysplasia, because disordered growth of the myocardium produces abnormally sized and
arranged cells
Correct Answer: B
Rationale: Chronic hypertension imposes a pressure (workload) overload on the left ventricle, and
cardiac muscle cells, which cannot divide meaningfully, respond by increasing in individual cell size
through increased synthesis of contractile proteins and organelles; this is hypertrophy. Hyperplasia
requires an increase in cell number and is not a property of mature cardiac myocytes, so it is
mechanistically impossible here. Atrophy is a decrease in cell size from reduced workload,
nutrition, or innervation, the opposite of the pressure-overload state. Dysplasia refers to disordered,
atypical cellular growth seen in premalignant epithelium, not a physiologic response to workload.
1
,WGU D236 Pathophysiology - Objective Assessment Exam (2026/2027) Total: 180 Questions
Q2: A 55-year-old pack-a-day smoker undergoes bronchoscopy, and biopsy of the bronchial
epithelium shows replacement of the normal pseudostratified ciliated columnar epithelium by
stratified squamous epithelium. The nurse practitioner counsels the patient that this change is
potentially reversible. What is this adaptive process, and why is it clinically significant?
A. Hyperplasia; it represents an increase in cell number and is always benign
B. Metaplasia; the squamous epithelium lacks cilia and mucus secretion, and with persistent
irritation such as smoking it can progress to dysplasia and eventually squamous cell
carcinoma
C. Hypertrophy; enlarged epithelial cells thicken the airway wall and reverse completely
after irritant removal
D. Anaplasia; it reflects a malignant transformation that is irreversible regardless of irritant
removal [CORRECT]
Correct Answer: D
Rationale: Metaplasia is the reversible replacement of one adult cell type by another better able to
withstand the adverse environment; in the bronchi, fragile ciliated columnar cells are replaced by
tougher stratified squamous cells. The clinical significance is twofold: the new squamous epithelium
loses the protective mucociliary blanket, and continued exposure to tobacco carcinogens commonly
drives metaplasia into dysplasia and eventually squamous cell carcinoma of the lung. Hyperplasia
and hypertrophy involve increases in cell number and size respectively and do not describe a
cell-type change. Anaplasia denotes loss of differentiation in frank malignancy, which is not
reversible and is not present in this biopsy description.
Q3: A 38-year-old woman has her right leg immobilized in a cast for eight weeks after a tibial
fracture. When the cast is removed, the right calf is noticeably smaller and weaker than the left,
although imaging shows no nerve damage. Which mechanism is primarily responsible for this
adaptive change?
A. Atrophy through decreased protein synthesis and increased proteasomal degradation of
proteins in disused skeletal muscle fibers [CORRECT]
B. Hypertrophy through compensatory enlargement of the immobilized muscle fibers from
disuse
C. Necrosis of muscle fibers caused by pressure injury from the cast
D. Metaplasia of skeletal muscle into fibrous connective tissue caused by the absence of
loading
Correct Answer: A
Rationale: Disuse atrophy occurs when reduced muscle loading lowers protein synthesis while the
ubiquitin-proteasome pathway accelerates protein degradation, shrinking individual muscle fibers;
the process is reversible with remobilization and resistance exercise. Hypertrophy requires increased
workload, which cannot occur in an immobilized limb, so it cannot explain a smaller calf. Pressure
necrosis would produce pain, skin breakdown, and irreversible cell death rather than a smooth,
symmetric reduction in muscle bulk. Metaplasia is a change of one adult cell type to another and is
not the mechanism of disuse-related muscle loss.
2
,WGU D236 Pathophysiology - Objective Assessment Exam (2026/2027) Total: 180 Questions
Q4: A 30-year-old woman with an elevated estrogen level has a pelvic ultrasound showing a
thickened endometrium; biopsy confirms an increased number of normal-appearing endometrial
glandular cells. Which statement best distinguishes this adaptation from hypertrophy?
A. Both processes require increased cell size, but only hyperplasia occurs in the uterus
B. Hypertrophy increases cell number while hyperplasia increases individual cell size
C. Hyperplasia increases the number of cells driven by hormone stimulation, whereas
hypertrophy increases the size of existing cells driven by increased workload [CORRECT]
D. Hyperplasia is always pathological and precedes cancer, whereas hypertrophy is always
physiological
Correct Answer: C
Rationale: Hyperplasia is an increase in the number of cells in an organ or tissue, typically in
response to hormonal stimulation (as estrogen drives endometrial gland proliferation) or
compensatory demand; hypertrophy is an increase in individual cell size, classically seen in cardiac
and skeletal muscle responding to increased work. The two can coexist, but the defining distinction
is cell number versus cell size. Hyperplasia is not always pathological; physiological examples
include endometrial proliferation in the menstrual cycle and compensatory liver regrowth after
partial hepatectomy.
Q5: A routine Pap smear from a 29-year-old woman is reported as showing 'disordered
epithelial growth with variation in nuclear size and shape, increased mitoses, and loss of normal
maturation orientation.' The colposcopic biopsy is classified as cervical intraepithelial neoplasia
associated with high-risk HPV. Which term describes this finding, and what is its key clinical
implication?
A. Metaplasia; it is a fully benign adaptive change requiring no follow-up
B. Dysplasia; it is a premalignant disordered growth that may be reversible with removal of
the inciting stimulus but can progress to carcinoma in situ and invasive cancer
[CORRECT]
C. Hyperplasia; it is a benign increase in cell number caused by inflammation
D. Apoptosis; it represents programmed cell death induced by HPV oncoproteins
Correct Answer: B
Rationale: Dysplasia is characterized by disordered growth: altered cell size and shape
(pleomorphism), enlarged and hyperchromatic nuclei, increased mitotic activity, and loss of normal
polarity and maturation. In the cervix it is induced by high-risk HPV types such as 16 and 18, whose
E6 and E7 oncoproteins inactivate p53 and Rb; importantly, dysplasia is potentially reversible if the
stimulus is removed but may progress through carcinoma in situ to invasive squamous cell
carcinoma, which is why surveillance and treatment are required. Metaplasia and hyperplasia are
orderly adaptive processes without cytologic atypia. Apoptosis is a mode of cell death, not a growth
disorder.
3
, WGU D236 Pathophysiology - Objective Assessment Exam (2026/2027) Total: 180 Questions
Q6: A patient in septic shock becomes profoundly hypotensive, and tissue perfusion falls.
Within the injured cells, decreased oxygen delivery leads to failure of the sodium-potassium
membrane pump. What intracellular change directly follows this pump failure and initiates acute
cell injury?
A. Intracellular accumulation of sodium and water producing cellular swelling, with
potassium diffusing out of the cell [CORRECT]
B. Loss of intracellular water and shrinkage of the cell caused by hyperosmolarity of the
cytoplasm
C. Immediate rupture of the nuclear membrane with digestion of DNA by endonucleases
D. Conversion of the cell to anaerobic carbohydrate synthesis with depletion of intracellular
chloride
Correct Answer: A
Rationale: Hypoxia forces cells into anaerobic glycolysis, depleting ATP; without ATP the
energy-dependent sodium-potassium pump fails, so sodium accumulates intracellularly while
potassium leaks out, and water follows sodium osmotically, causing acute cellular swelling - the
earliest reversible manifestation of cell injury. The shrinkage option reverses the water movement;
injured hypoxic cells swell rather than shrink. Nuclear membrane rupture and DNA digestion relate
to later, irreversible stages or to apoptosis, not the initial pump failure. Anaerobic metabolism
produces lactate and drops intracellular pH; it does not synthesize carbohydrate or alter chloride as
the primary injurious event.
Q7: A patient with a completely occluded coronary artery undergoes successful percutaneous
reperfusion, but afterward myocardial dysfunction paradoxically worsens and infarct size
enlarges. Blood studies show elevated markers of oxidative damage to cell membranes. What is
the mechanism of this phenomenon?
A. Excess calcium efflux that stabilizes mitochondrial membranes and halts ATP
production
B. Depletion of intracellular sodium that causes the cell to take up excess bicarbonate
C. Sudden increase in anaerobic glycolysis that overproduces ATP and injures
mitochondrial DNA
D. Restoration of blood flow delivering neutrophils that generate more damaging reactive
oxygen species than the tissue can neutralize [CORRECT]
Correct Answer: D
Rationale: Reperfusion injury occurs when restored blood flow reoxygenates previously ischemic
tissue; damaged mitochondria and infiltrating neutrophils generate a burst of reactive oxygen
species (superoxide, hydroxyl radicals, hydrogen peroxide) that overwhelms antioxidant defenses
and extends membrane lipid peroxidation and cell death. Ischemic cells accumulate intracellular
calcium rather than losing it, and calcium overload further damages mitochondria, so any option
describing calcium efflux states the opposite direction. Sodium depletion is not a feature of
reperfusion; sodium overload accompanies pump failure. Anaerobic glycolysis ceases upon
reoxygenation; the injury comes from oxidative free radicals, not from ATP overproduction.
4
