WGU D027 OBJECTIVE ASSESSMENT FINAL EXAM COMPLETE
EXAM QUESTIONS AND VERIFIED ANSWERS | 2026–2027
LATEST UPDATE | GUARANTEED PASS | DETAILED
RATIONALES | FULL STUDY GUIDE | EXAM PREP | PRACTICE
TEST | CERTIFICATION PREPARATION
1. A 58-year-old male with a 20-year history of poorly controlled hypertension presents with
concentric left ventricular hypertrophy on echocardiogram. Which cellular adaptation does this
finding represent?
A. Hyperplasia
B. Metaplasia
C. Hypertrophy
D. Atrophy
Correct Answer: C. Hypertrophy
Rationale: Hypertrophy is an increase in cell size that occurs in tissues incapable of cell division, such
as cardiac muscle. Chronic pressure overload from hypertension triggers concentric hypertrophy as
the myocardium thickens to compensate for increased workload. Hyperplasia (A) involves increased
cell number and does not occur in terminally differentiated cardiac muscle. Metaplasia (B) is the
reversible replacement of one mature cell type with another. Atrophy (D) represents a decrease in cell
size, which is opposite to the finding described.
2. A patient with emphysema demonstrates destruction of alveolar walls with permanent
enlargement of air spaces. Which pathophysiologic mechanism best explains this finding?
A. Increased surfactant production
B. Protease-antiprotease imbalance
C. Excessive mucus gland hyperplasia
D. Smooth muscle hypertrophy of bronchioles
Correct Answer: B. Protease-antiprotease imbalance
Rationale: Emphysema results from destruction of alveolar walls due to an imbalance between
proteases (such as elastase) and antiproteases (such as alpha-1 antitrypsin). This leads to loss of
elastic recoil and permanent enlargement of air spaces. Increased surfactant (A) would not cause
alveolar destruction. Mucus gland hyperplasia (C) is characteristic of chronic bronchitis. Smooth
muscle hypertrophy (D) contributes to bronchospasm in asthma, not alveolar destruction.
3. A chronic smoker undergoes bronchial biopsy revealing stratified squamous epithelium
replacing the normal pseudostratified ciliated columnar epithelium. This cellular change is best
classified as:
A. Dysplasia
B. Anaplasia
C. Hyperplasia
D. Metaplasia
,Correct Answer: D. Metaplasia
Rationale: Metaplasia is the reversible replacement of one differentiated cell type with another that
is better suited to withstand environmental stress. In smokers, chronic irritation causes squamous
metaplasia of the bronchial epithelium. Dysplasia (A) involves disordered growth with nuclear atypia,
which is not described here. Anaplasia (B) refers to loss of differentiation characteristic of
malignancy. Hyperplasia (C) is an increase in cell number of the same cell type.
4. A cervical Pap smear demonstrates disordered squamous epithelium with nuclear
hyperchromasia, pleomorphism, and loss of polarity involving the lower two-thirds of the
epithelium. This finding is most consistent with:
A. Metaplasia
B. Mild dysplasia
C. Moderate dysplasia
D. Carcinoma in situ
Correct Answer: C. Moderate dysplasia
Rationale: Dysplasia is characterized by disordered growth with nuclear atypia, hyperchromasia,
pleomorphism, and loss of polarity. Involvement of the lower two-thirds of the epithelium
corresponds to moderate dysplasia (CIN II). Metaplasia (A) lacks these atypical features. Mild
dysplasia (B) involves only the lower one-third. Carcinoma in situ (D) involves full-thickness dysplasia
without breaching the basement membrane.
5. A myocardial infarction produces tissue that is firm and pale with preserved tissue architecture
on histology. This pattern of necrosis is best described as:
A. Liquefactive necrosis
B. Coagulative necrosis
C. Caseous necrosis
D. Fat necrosis
Correct Answer: B. Coagulative necrosis
Rationale: Coagulative necrosis results from ischemia and is characterized by preservation of tissue
architecture due to denaturation of structural and enzymatic proteins. The heart is a classic site.
Liquefactive necrosis (A) occurs in brain infarcts and abscesses where tissue softens and liquefies.
Caseous necrosis (C) is characteristic of tuberculosis and appears cheese-like. Fat necrosis (D) occurs
in pancreatitis and breast trauma.
6. A patient with a brain infarct develops a softened, liquefied area with complete loss of tissue
architecture. This type of necrosis is:
A. Coagulative necrosis
B. Liquefactive necrosis
C. Caseous necrosis
D. Gangrenous necrosis
Correct Answer: B. Liquefactive necrosis
Rationale: Liquefactive necrosis occurs in the brain due to its high lipid content and the action of
hydrolytic enzymes from microglial cells, resulting in softening and liquefaction of tissue. Coagulative
necrosis (A) preserves architecture and occurs in most other tissues after ischemia. Caseous necrosis
,(C) is associated with tuberculosis. Gangrenous necrosis (D) typically affects extremities and involves
coagulative necrosis with superimposed bacterial infection.
7. A 32-year-old male presents with recurrent hemolytic episodes after eating fava beans. Genetic
testing confirms G6PD deficiency. Which enzyme pathway is directly impaired in this condition?
A. Glycolysis
B. Pentose phosphate pathway
C. Krebs cycle
D. Electron transport chain
Correct Answer: B. Pentose phosphate pathway
Rationale: G6PD (glucose-6-phosphate dehydrogenase) is the rate-limiting enzyme of the pentose
phosphate pathway, which produces NADPH to maintain reduced glutathione and protect red blood
cells from oxidative damage. Glycolysis (A) generates ATP but does not directly involve G6PD. The
Krebs cycle (C) and electron transport chain (D) are mitochondrial pathways unaffected by G6PD
deficiency.
8. A patient with celiac disease is being evaluated. Which antibody is most specific for this
condition?
A. Anti-Saccharomyces cerevisiae antibody
B. Anti-tissue transglutaminase IgA
C. Antinuclear antibody
D. Rheumatoid factor
Correct Answer: B. Anti-tissue transglutaminase IgA
Rationale: Anti-tissue transglutaminase IgA (anti-tTG IgA) is the most sensitive and specific serologic
test for celiac disease. Anti-Saccharomyces cerevisiae antibody (A) is associated with Crohn's disease.
Antinuclear antibody (C) is seen in autoimmune conditions like lupus. Rheumatoid factor (D) is
associated with rheumatoid arthritis.
9. A 28-year-old female presents with fatigue, weight gain, cold intolerance, and constipation.
Laboratory findings reveal elevated TSH and low free T4. What is the most likely diagnosis?
A. Graves disease
B. Hashimoto thyroiditis
C. Subclinical hyperthyroidism
D. Thyroid storm
Correct Answer: B. Hashimoto thyroiditis
Rationale: Hashimoto thyroiditis is the most common cause of hypothyroidism and presents with
elevated TSH, low free T4, and positive thyroid peroxidase antibodies. Symptoms reflect decreased
metabolic rate. Graves disease (A) causes hyperthyroidism with low TSH and elevated free T4.
Subclinical hyperthyroidism (C) presents with low TSH and normal free T4. Thyroid storm (D) is a
severe hyperthyroid state.
10. Which medication is considered first-line therapy for a patient with uncomplicated
hypertension?
, A. Hydrochlorothiazide
B. Clonidine
C. Hydralazine
D. Minoxidil
Correct Answer: A. Hydrochlorothiazide
Rationale: Thiazide diuretics such as hydrochlorothiazide are recommended as first-line therapy for
uncomplicated hypertension due to proven efficacy in reducing cardiovascular events and favorable
cost profile. Clonidine (B) is a central alpha-2 agonist typically reserved for resistant hypertension.
Hydralazine (C) and minoxidil (D) are direct vasodilators used in specific clinical situations such as
hypertensive emergencies or pregnancy.
11. A patient with type 2 diabetes and hypertension requires antihypertensive therapy. Which
class is most appropriate given the need for renal protection?
A. Beta-blockers
B. ACE inhibitors
C. Thiazide diuretics
D. Alpha-blockers
Correct Answer: B. ACE inhibitors
Rationale: ACE inhibitors are preferred in diabetic patients with hypertension because they reduce
intraglomerular pressure and slow progression of diabetic nephropathy. Beta-blockers (A) may mask
hypoglycemia symptoms and are not first-line in diabetes. Thiazide diuretics (C) can worsen glucose
tolerance. Alpha-blockers (D) are not first-line for hypertension and lack renal protective effects.
12. A 45-year-old male on warfarin for atrial fibrillation presents with an INR of 6.5 and no active
bleeding. Which intervention is most appropriate?
A. Administer vitamin K intravenously
B. Hold warfarin and monitor INR
C. Administer fresh frozen plasma
D. Administer protamine sulfate
Correct Answer: B. Hold warfarin and monitor INR
Rationale: For an elevated INR without bleeding, the appropriate action is to hold warfarin and
monitor until the INR decreases. Vitamin K (A) is reserved for serious bleeding or very high INR with
bleeding risk. Fresh frozen plasma (C) is used for active bleeding with coagulopathy. Protamine
sulfate (D) reverses heparin, not warfarin.
13. A patient with heart failure is prescribed digoxin. Which assessment finding would require
holding the medication?
A. Heart rate of 72 beats per minute
B. Heart rate of 58 beats per minute
C. Blood pressure of 128/78 mm Hg
D. Respiratory rate of 18 breaths per minute
Correct Answer: B. Heart rate of 58 beats per minute
EXAM QUESTIONS AND VERIFIED ANSWERS | 2026–2027
LATEST UPDATE | GUARANTEED PASS | DETAILED
RATIONALES | FULL STUDY GUIDE | EXAM PREP | PRACTICE
TEST | CERTIFICATION PREPARATION
1. A 58-year-old male with a 20-year history of poorly controlled hypertension presents with
concentric left ventricular hypertrophy on echocardiogram. Which cellular adaptation does this
finding represent?
A. Hyperplasia
B. Metaplasia
C. Hypertrophy
D. Atrophy
Correct Answer: C. Hypertrophy
Rationale: Hypertrophy is an increase in cell size that occurs in tissues incapable of cell division, such
as cardiac muscle. Chronic pressure overload from hypertension triggers concentric hypertrophy as
the myocardium thickens to compensate for increased workload. Hyperplasia (A) involves increased
cell number and does not occur in terminally differentiated cardiac muscle. Metaplasia (B) is the
reversible replacement of one mature cell type with another. Atrophy (D) represents a decrease in cell
size, which is opposite to the finding described.
2. A patient with emphysema demonstrates destruction of alveolar walls with permanent
enlargement of air spaces. Which pathophysiologic mechanism best explains this finding?
A. Increased surfactant production
B. Protease-antiprotease imbalance
C. Excessive mucus gland hyperplasia
D. Smooth muscle hypertrophy of bronchioles
Correct Answer: B. Protease-antiprotease imbalance
Rationale: Emphysema results from destruction of alveolar walls due to an imbalance between
proteases (such as elastase) and antiproteases (such as alpha-1 antitrypsin). This leads to loss of
elastic recoil and permanent enlargement of air spaces. Increased surfactant (A) would not cause
alveolar destruction. Mucus gland hyperplasia (C) is characteristic of chronic bronchitis. Smooth
muscle hypertrophy (D) contributes to bronchospasm in asthma, not alveolar destruction.
3. A chronic smoker undergoes bronchial biopsy revealing stratified squamous epithelium
replacing the normal pseudostratified ciliated columnar epithelium. This cellular change is best
classified as:
A. Dysplasia
B. Anaplasia
C. Hyperplasia
D. Metaplasia
,Correct Answer: D. Metaplasia
Rationale: Metaplasia is the reversible replacement of one differentiated cell type with another that
is better suited to withstand environmental stress. In smokers, chronic irritation causes squamous
metaplasia of the bronchial epithelium. Dysplasia (A) involves disordered growth with nuclear atypia,
which is not described here. Anaplasia (B) refers to loss of differentiation characteristic of
malignancy. Hyperplasia (C) is an increase in cell number of the same cell type.
4. A cervical Pap smear demonstrates disordered squamous epithelium with nuclear
hyperchromasia, pleomorphism, and loss of polarity involving the lower two-thirds of the
epithelium. This finding is most consistent with:
A. Metaplasia
B. Mild dysplasia
C. Moderate dysplasia
D. Carcinoma in situ
Correct Answer: C. Moderate dysplasia
Rationale: Dysplasia is characterized by disordered growth with nuclear atypia, hyperchromasia,
pleomorphism, and loss of polarity. Involvement of the lower two-thirds of the epithelium
corresponds to moderate dysplasia (CIN II). Metaplasia (A) lacks these atypical features. Mild
dysplasia (B) involves only the lower one-third. Carcinoma in situ (D) involves full-thickness dysplasia
without breaching the basement membrane.
5. A myocardial infarction produces tissue that is firm and pale with preserved tissue architecture
on histology. This pattern of necrosis is best described as:
A. Liquefactive necrosis
B. Coagulative necrosis
C. Caseous necrosis
D. Fat necrosis
Correct Answer: B. Coagulative necrosis
Rationale: Coagulative necrosis results from ischemia and is characterized by preservation of tissue
architecture due to denaturation of structural and enzymatic proteins. The heart is a classic site.
Liquefactive necrosis (A) occurs in brain infarcts and abscesses where tissue softens and liquefies.
Caseous necrosis (C) is characteristic of tuberculosis and appears cheese-like. Fat necrosis (D) occurs
in pancreatitis and breast trauma.
6. A patient with a brain infarct develops a softened, liquefied area with complete loss of tissue
architecture. This type of necrosis is:
A. Coagulative necrosis
B. Liquefactive necrosis
C. Caseous necrosis
D. Gangrenous necrosis
Correct Answer: B. Liquefactive necrosis
Rationale: Liquefactive necrosis occurs in the brain due to its high lipid content and the action of
hydrolytic enzymes from microglial cells, resulting in softening and liquefaction of tissue. Coagulative
necrosis (A) preserves architecture and occurs in most other tissues after ischemia. Caseous necrosis
,(C) is associated with tuberculosis. Gangrenous necrosis (D) typically affects extremities and involves
coagulative necrosis with superimposed bacterial infection.
7. A 32-year-old male presents with recurrent hemolytic episodes after eating fava beans. Genetic
testing confirms G6PD deficiency. Which enzyme pathway is directly impaired in this condition?
A. Glycolysis
B. Pentose phosphate pathway
C. Krebs cycle
D. Electron transport chain
Correct Answer: B. Pentose phosphate pathway
Rationale: G6PD (glucose-6-phosphate dehydrogenase) is the rate-limiting enzyme of the pentose
phosphate pathway, which produces NADPH to maintain reduced glutathione and protect red blood
cells from oxidative damage. Glycolysis (A) generates ATP but does not directly involve G6PD. The
Krebs cycle (C) and electron transport chain (D) are mitochondrial pathways unaffected by G6PD
deficiency.
8. A patient with celiac disease is being evaluated. Which antibody is most specific for this
condition?
A. Anti-Saccharomyces cerevisiae antibody
B. Anti-tissue transglutaminase IgA
C. Antinuclear antibody
D. Rheumatoid factor
Correct Answer: B. Anti-tissue transglutaminase IgA
Rationale: Anti-tissue transglutaminase IgA (anti-tTG IgA) is the most sensitive and specific serologic
test for celiac disease. Anti-Saccharomyces cerevisiae antibody (A) is associated with Crohn's disease.
Antinuclear antibody (C) is seen in autoimmune conditions like lupus. Rheumatoid factor (D) is
associated with rheumatoid arthritis.
9. A 28-year-old female presents with fatigue, weight gain, cold intolerance, and constipation.
Laboratory findings reveal elevated TSH and low free T4. What is the most likely diagnosis?
A. Graves disease
B. Hashimoto thyroiditis
C. Subclinical hyperthyroidism
D. Thyroid storm
Correct Answer: B. Hashimoto thyroiditis
Rationale: Hashimoto thyroiditis is the most common cause of hypothyroidism and presents with
elevated TSH, low free T4, and positive thyroid peroxidase antibodies. Symptoms reflect decreased
metabolic rate. Graves disease (A) causes hyperthyroidism with low TSH and elevated free T4.
Subclinical hyperthyroidism (C) presents with low TSH and normal free T4. Thyroid storm (D) is a
severe hyperthyroid state.
10. Which medication is considered first-line therapy for a patient with uncomplicated
hypertension?
, A. Hydrochlorothiazide
B. Clonidine
C. Hydralazine
D. Minoxidil
Correct Answer: A. Hydrochlorothiazide
Rationale: Thiazide diuretics such as hydrochlorothiazide are recommended as first-line therapy for
uncomplicated hypertension due to proven efficacy in reducing cardiovascular events and favorable
cost profile. Clonidine (B) is a central alpha-2 agonist typically reserved for resistant hypertension.
Hydralazine (C) and minoxidil (D) are direct vasodilators used in specific clinical situations such as
hypertensive emergencies or pregnancy.
11. A patient with type 2 diabetes and hypertension requires antihypertensive therapy. Which
class is most appropriate given the need for renal protection?
A. Beta-blockers
B. ACE inhibitors
C. Thiazide diuretics
D. Alpha-blockers
Correct Answer: B. ACE inhibitors
Rationale: ACE inhibitors are preferred in diabetic patients with hypertension because they reduce
intraglomerular pressure and slow progression of diabetic nephropathy. Beta-blockers (A) may mask
hypoglycemia symptoms and are not first-line in diabetes. Thiazide diuretics (C) can worsen glucose
tolerance. Alpha-blockers (D) are not first-line for hypertension and lack renal protective effects.
12. A 45-year-old male on warfarin for atrial fibrillation presents with an INR of 6.5 and no active
bleeding. Which intervention is most appropriate?
A. Administer vitamin K intravenously
B. Hold warfarin and monitor INR
C. Administer fresh frozen plasma
D. Administer protamine sulfate
Correct Answer: B. Hold warfarin and monitor INR
Rationale: For an elevated INR without bleeding, the appropriate action is to hold warfarin and
monitor until the INR decreases. Vitamin K (A) is reserved for serious bleeding or very high INR with
bleeding risk. Fresh frozen plasma (C) is used for active bleeding with coagulopathy. Protamine
sulfate (D) reverses heparin, not warfarin.
13. A patient with heart failure is prescribed digoxin. Which assessment finding would require
holding the medication?
A. Heart rate of 72 beats per minute
B. Heart rate of 58 beats per minute
C. Blood pressure of 128/78 mm Hg
D. Respiratory rate of 18 breaths per minute
Correct Answer: B. Heart rate of 58 beats per minute
