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WGU D236 PATHOPHYSIOLOGY EXAM REVIEW 2026/2027 |
Complete Guide with Questions & Verified Answers 100%
Correct | Pass Guaranteed - A+ Graded
Section 1: Cellular Adaptation, Injury & Death (Q1-18)
Question 1
A 72-year-old man has his right leg immobilized in a cast for 6 weeks after a tibial
fracture. After cast removal, the calf circumference is visibly smaller than the left.
Which cellular adaptation has occurred?
A. Hypertrophy
B. Hyperplasia
C. Atrophy
D. Metaplasia
Correct Answer: C. Atrophy [CORRECT]
Rationale: Atrophy is the decrease in cell size caused by decreased workload, blood
supply, innervation, or nutrition. In this case, disuse from immobilization leads to
reduced protein synthesis and increased protein degradation in muscle cells,
resulting in smaller cell size. Hypertrophy involves increased cell size from increased
workload, which is the opposite process. Hyperplasia involves increased cell number,
not relevant here as muscle cells are permanent cells that do not divide. Metaplasia is
the reversible replacement of one differentiated cell type with another due to chronic
irritation, which does not occur from simple disuse.
Question 2
,2
A 58-year-old man with uncontrolled hypertension (BP 168/102 mmHg) for 10 years
develops left ventricular hypertrophy on echocardiogram. Which statement best
explains this adaptation?
A. Hyperplasia of cardiac myocytes due to chronic volume overload
B. Hypertrophy of cardiac myocytes due to increased workload
C. Metaplasia of cardiac myocytes into skeletal muscle cells
D. Dysplasia of cardiac myocytes indicating pre-malignant change
Correct Answer: B. Hypertrophy of cardiac myocytes due to increased workload
[CORRECT]
Rationale: Pathologic hypertrophy occurs when cells increase in size in response to
increased demand or stress; here, chronic pressure overload from hypertension
causes cardiac myocytes to enlarge (concentric LVH) without an increase in cell
number. Hyperplasia is incorrect because cardiac myocytes are permanent cells that
cannot divide and increase in number. Metaplasia does not occur in cardiac muscle
under hemodynamic stress. Dysplasia represents disordered growth and is not a
response to mechanical overload.
Question 3
A 45-year-old woman with polycystic ovary syndrome and chronic anovulation
develops endometrial hyperplasia on biopsy. Which cellular process explains this
finding?
A. Atrophy from estrogen deficiency
B. Metaplasia of endometrial cells
C. Hyperplasia from unopposed estrogen stimulation
D. Hypertrophy of endometrial stromal cells
Correct Answer: C. Hyperplasia from unopposed estrogen stimulation [CORRECT]
Rationale: Hyperplasia is an increase in cell number due to hormonal stimulation; in
PCOS, chronic anovulation leads to unopposed estrogen acting on the endometrium,
causing proliferation of glandular cells. Atrophy would result from estrogen
deficiency, not excess. Metaplasia involves transformation to a different cell type,
,3
which is not the primary process in endometrial hyperplasia. Hypertrophy refers to
increased cell size, whereas endometrial hyperplasia is characterized by increased
glandular cell number.
Question 4
A 55-year-old man with long-standing GERD has a distal esophageal biopsy showing
columnar epithelium with goblet cells replacing the normal squamous epithelium.
Which cellular adaptation is present?
A. Dysplasia
B. Metaplasia
C. Hyperplasia
D. Anaplasia
Correct Answer: B. Metaplasia [CORRECT]
Rationale: Metaplasia is the reversible replacement of one differentiated cell type
with another due to chronic irritation; here, chronic acid exposure causes squamous
epithelium to transform into columnar epithelium (Barrett's esophagus). Dysplasia
involves disordered growth with abnormal size, shape, and organization, which may
develop later in Barrett's but is not the initial adaptation shown. Hyperplasia involves
increased cell number of the same type, not replacement. Anaplasia is a feature of
malignant cells showing loss of differentiation, not a reversible adaptation.
Question 5
A 30-year-old woman has a cervical Pap smear showing disordered growth with
nuclear enlargement and hyperchromasia in the lower third of the epithelium. Which
term describes this finding?
A. Carcinoma in situ
B. Dysplasia
, 4
C. Metaplasia
D. Hyperplasia
Correct Answer: B. Dysplasia [CORRECT]
Rationale: Dysplasia is characterized by disordered epithelial growth with abnormal
cell size, shape, and organization, often pre-neoplastic and reversible if the stimulus
(e.g., HPV) is removed; this biopsy describes classic cervical dysplasia. Carcinoma in
situ involves full-thickness dysplasia without basement membrane invasion, more
severe than lower-third involvement. Metaplasia is replacement of one cell type with
another, not disordered growth of existing cells. Hyperplasia is increased cell number
without the cytologic atypia seen here.
Question 6
A 65-year-old man suffers an acute MI. The infarcted myocardial tissue shows
preserved cellular outlines with loss of nuclei and increased eosinophilia. Which type
of necrosis is present?
A. Liquefactive necrosis
B. Caseous necrosis
C. Coagulative necrosis
D. Fat necrosis
Correct Answer: C. Coagulative necrosis [CORRECT]
Rationale: Coagulative necrosis is characteristic of ischemic injury in most tissues
except the brain; the architecture is preserved for several days due to denaturation of
structural and enzymatic proteins, leaving ghost outlines of cells. Liquefactive
necrosis occurs in the brain and with bacterial infections due to enzymatic digestion.
Caseous necrosis is associated with tuberculosis and has a cheese-like appearance
without preserved architecture. Fat necrosis occurs in adipose tissue, particularly in
pancreatitis or breast trauma, with saponification.
WGU D236 PATHOPHYSIOLOGY EXAM REVIEW 2026/2027 |
Complete Guide with Questions & Verified Answers 100%
Correct | Pass Guaranteed - A+ Graded
Section 1: Cellular Adaptation, Injury & Death (Q1-18)
Question 1
A 72-year-old man has his right leg immobilized in a cast for 6 weeks after a tibial
fracture. After cast removal, the calf circumference is visibly smaller than the left.
Which cellular adaptation has occurred?
A. Hypertrophy
B. Hyperplasia
C. Atrophy
D. Metaplasia
Correct Answer: C. Atrophy [CORRECT]
Rationale: Atrophy is the decrease in cell size caused by decreased workload, blood
supply, innervation, or nutrition. In this case, disuse from immobilization leads to
reduced protein synthesis and increased protein degradation in muscle cells,
resulting in smaller cell size. Hypertrophy involves increased cell size from increased
workload, which is the opposite process. Hyperplasia involves increased cell number,
not relevant here as muscle cells are permanent cells that do not divide. Metaplasia is
the reversible replacement of one differentiated cell type with another due to chronic
irritation, which does not occur from simple disuse.
Question 2
,2
A 58-year-old man with uncontrolled hypertension (BP 168/102 mmHg) for 10 years
develops left ventricular hypertrophy on echocardiogram. Which statement best
explains this adaptation?
A. Hyperplasia of cardiac myocytes due to chronic volume overload
B. Hypertrophy of cardiac myocytes due to increased workload
C. Metaplasia of cardiac myocytes into skeletal muscle cells
D. Dysplasia of cardiac myocytes indicating pre-malignant change
Correct Answer: B. Hypertrophy of cardiac myocytes due to increased workload
[CORRECT]
Rationale: Pathologic hypertrophy occurs when cells increase in size in response to
increased demand or stress; here, chronic pressure overload from hypertension
causes cardiac myocytes to enlarge (concentric LVH) without an increase in cell
number. Hyperplasia is incorrect because cardiac myocytes are permanent cells that
cannot divide and increase in number. Metaplasia does not occur in cardiac muscle
under hemodynamic stress. Dysplasia represents disordered growth and is not a
response to mechanical overload.
Question 3
A 45-year-old woman with polycystic ovary syndrome and chronic anovulation
develops endometrial hyperplasia on biopsy. Which cellular process explains this
finding?
A. Atrophy from estrogen deficiency
B. Metaplasia of endometrial cells
C. Hyperplasia from unopposed estrogen stimulation
D. Hypertrophy of endometrial stromal cells
Correct Answer: C. Hyperplasia from unopposed estrogen stimulation [CORRECT]
Rationale: Hyperplasia is an increase in cell number due to hormonal stimulation; in
PCOS, chronic anovulation leads to unopposed estrogen acting on the endometrium,
causing proliferation of glandular cells. Atrophy would result from estrogen
deficiency, not excess. Metaplasia involves transformation to a different cell type,
,3
which is not the primary process in endometrial hyperplasia. Hypertrophy refers to
increased cell size, whereas endometrial hyperplasia is characterized by increased
glandular cell number.
Question 4
A 55-year-old man with long-standing GERD has a distal esophageal biopsy showing
columnar epithelium with goblet cells replacing the normal squamous epithelium.
Which cellular adaptation is present?
A. Dysplasia
B. Metaplasia
C. Hyperplasia
D. Anaplasia
Correct Answer: B. Metaplasia [CORRECT]
Rationale: Metaplasia is the reversible replacement of one differentiated cell type
with another due to chronic irritation; here, chronic acid exposure causes squamous
epithelium to transform into columnar epithelium (Barrett's esophagus). Dysplasia
involves disordered growth with abnormal size, shape, and organization, which may
develop later in Barrett's but is not the initial adaptation shown. Hyperplasia involves
increased cell number of the same type, not replacement. Anaplasia is a feature of
malignant cells showing loss of differentiation, not a reversible adaptation.
Question 5
A 30-year-old woman has a cervical Pap smear showing disordered growth with
nuclear enlargement and hyperchromasia in the lower third of the epithelium. Which
term describes this finding?
A. Carcinoma in situ
B. Dysplasia
, 4
C. Metaplasia
D. Hyperplasia
Correct Answer: B. Dysplasia [CORRECT]
Rationale: Dysplasia is characterized by disordered epithelial growth with abnormal
cell size, shape, and organization, often pre-neoplastic and reversible if the stimulus
(e.g., HPV) is removed; this biopsy describes classic cervical dysplasia. Carcinoma in
situ involves full-thickness dysplasia without basement membrane invasion, more
severe than lower-third involvement. Metaplasia is replacement of one cell type with
another, not disordered growth of existing cells. Hyperplasia is increased cell number
without the cytologic atypia seen here.
Question 6
A 65-year-old man suffers an acute MI. The infarcted myocardial tissue shows
preserved cellular outlines with loss of nuclei and increased eosinophilia. Which type
of necrosis is present?
A. Liquefactive necrosis
B. Caseous necrosis
C. Coagulative necrosis
D. Fat necrosis
Correct Answer: C. Coagulative necrosis [CORRECT]
Rationale: Coagulative necrosis is characteristic of ischemic injury in most tissues
except the brain; the architecture is preserved for several days due to denaturation of
structural and enzymatic proteins, leaving ghost outlines of cells. Liquefactive
necrosis occurs in the brain and with bacterial infections due to enzymatic digestion.
Caseous necrosis is associated with tuberculosis and has a cheese-like appearance
without preserved architecture. Fat necrosis occurs in adipose tissue, particularly in
pancreatitis or breast trauma, with saponification.
