Advanced Pathopharmacological Foundations | Questions &
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Graded
Section 1: Cellular & Genetic Foundations (Questions 1-15)
Question 1
A 65-year-old male with chronic heart failure has an enlarged heart on chest X-ray.
Myocardial biopsy shows increased cell size with normal DNA content. Which cellular
adaptation is MOST likely occurring?
A. Hyperplasia
B. Hypertrophy
C. Hyperplasia with hypertrophy
D. Metaplasia
Rationale: Hypertrophy is an increase in cell SIZE (not number) due to increased
functional demand, resulting in enlarged cells with normal DNA content per cell. Cardiac
myocytes are terminally differentiated cells incapable of division; therefore, they
undergo hypertrophy (not hyperplasia) in response to increased workload. Option A
(hyperplasia) involves increased cell NUMBER, which requires cell division and is not
possible in cardiac myocytes. Option C is incorrect because cardiac cells cannot
hyperplasia. Option D (metaplasia) is the replacement of one differentiated cell type
with another, not relevant here. WGU D027 Competency: Cellular adaptation
mechanisms. Pathophysiology Principle: Permanent cells (neurons, cardiac myocytes,
,skeletal muscle) undergo hypertrophy only; labile and stable cells can undergo
hyperplasia.
Correct Answer: B
Question 2
A 55-year-old female smoker develops squamous cell carcinoma of the bronchus. Prior
biopsy showed ciliated pseudostratified columnar epithelium replaced by stratified
squamous epithelium in the bronchial mucosa. Which cellular adaptation preceded the
malignancy?
A. Dysplasia
B. Metaplasia
C. Atrophy
D. Anaplasia
Rationale: Metaplasia is the reversible replacement of one differentiated cell type with
another, often in response to chronic irritation. In smokers, chronic irritation of bronchial
mucosa causes columnar epithelium to transform to squamous epithelium (squamous
metaplasia). This metaplastic epithelium can then progress to dysplasia and
carcinoma. Option A (dysplasia) is disordered growth with nuclear atypia, which follows
metaplasia in carcinogenesis. Option C (atrophy) is decreased cell size/number. Option
D (anaplasia) is loss of differentiation in malignancy. WGU D027 Competency: Cellular
adaptation and carcinogenesis sequence. Pathophysiology Principle: Metaplasia →
Dysplasia → Carcinoma in situ → Invasive carcinoma.
Correct Answer: B
Question 3
,A 70-year-old male with peripheral arterial disease develops dry gangrene of the toes.
Which type of necrosis is MOST likely present?
A. Coagulative necrosis
B. Liquefactive necrosis
C. Caseous necrosis
D. Fat necrosis
Rationale: Dry gangrene results from ischemia (arterial occlusion) causing coagulative
necrosis in solid organs and extremities. Coagulative necrosis preserves tissue
architecture initially due to denaturation of structural proteins and enzymes, resulting in
firm, dry tissue. Option B (liquefactive necrosis) occurs in brain infarcts and bacterial
infections due to enzymatic digestion. Option C (caseous necrosis) is characteristic of
tuberculosis. Option D (fat necrosis) occurs in acute pancreatitis and breast trauma.
WGU D027 Competency: Types of necrosis and clinical correlations. Pathophysiology
Principle: Ischemia in solid organs → coagulative necrosis; ischemia in brain →
liquefactive necrosis.
Correct Answer: A
Question 4
A 45-year-old male presents with acute severe epigastric pain radiating to his back.
Serum amylase and lipase are markedly elevated. CT shows pancreatic inflammation
with areas of chalky white deposits. Which type of necrosis is present?
A. Coagulative necrosis
B. Liquefactive necrosis
C. Caseous necrosis
D. Fat necrosis
, Rationale: Acute pancreatitis causes fat necrosis due to release of pancreatic lipases
that hydrolyze triglycerides in peripancreatic fat, producing free fatty acids that combine
with calcium to form chalky white calcium soaps (saponification). Option A
(coagulative) is ischemic. Option B (liquefactive) is enzymatic digestion in brain or
infection. Option C (caseous) is TB. WGU D027 Competency: Necrosis types and clinical
correlations. Pathophysiology Principle: Pancreatic lipase release → triglyceride
hydrolysis → fatty acid release → calcium soap formation (saponification).
Correct Answer: D
Question 5
A 25-year-old female presents with cervical dysplasia on Pap smear. Biopsy shows
disordered epithelial maturation, nuclear hyperchromasia, and increased mitotic figures
extending above the basal third. Which cellular adaptation is described?
A. Metaplasia
B. Dysplasia
C. Hyperplasia
D. Neoplasia
Rationale: Dysplasia is disordered epithelial growth characterized by loss of normal
maturation, nuclear pleomorphism, hyperchromasia, and increased mitotic activity. In
cervical intraepithelial neoplasia (CIN), dysplasia is graded by extent of epithelial
involvement (CIN 1: lower third; CIN 2: lower two-thirds; CIN 3: full thickness). Option A
(metaplasia) is reversible cell type replacement without atypia. Option C (hyperplasia) is
increased cell number with normal morphology. Option D (neoplasia) implies
autonomous growth; dysplasia is potentially reversible. WGU D027 Competency:
Dysplasia grading and premalignant changes. Pathophysiology Principle: Dysplasia is
potentially reversible; carcinoma in situ is non-invasive but irreversible.