Questions From (1 To 50)

Krok 1 Paper with Explanation

About Lesson

1. After prolonged fasting, the patient developed tissue edema. What is the cause of this phenomenon?

A. Decreased oncotic pressure of blood plasma

B. Increased osmotic pressure of blood plasma

C. Decreased hydrostatic blood pressure

D. Decreased osmotic pressure of blood plasma

E. Increased oncotic blood pressure

Correct Answer: A) Decreased oncotic pressure of blood plasma

Explanation:

Oncotic pressure (colloid osmotic pressure) is primarily generated by plasma proteins, mainly albumin, which help retain water in the intravascular compartment. Prolonged fasting leads to protein malnutrition (hypoalbuminemia), reducing oncotic pressure and causing fluid to shift from blood vessels into the interstitial space, leading to edema.

This phenomenon is commonly seen in kwashiorkor, a condition of severe protein deficiency, where patients develop pitting edema due to low plasma oncotic pressure.

Why Other Options Are Incorrect:

In fasting, protein depletion reduces oncotic pressure, leading to fluid leakage into the interstitial space.

B) Increased osmotic pressure of blood plasma:

If the osmotic pressure of blood plasma increased, it would draw water into the vascular space, preventing edema rather than causing it.

Prolonged fasting does not increase plasma osmotic pressure; instead, it leads to protein loss and decreased oncotic pressure.

C) Decreased hydrostatic blood pressure:

Hydrostatic pressure is the force that pushes fluid out of blood vessels. A decrease in hydrostatic pressure would reduce fluid leakage, making edema less likely, not more.

In fasting, hydrostatic pressure may drop due to dehydration or low blood volume, but edema is mainly due to reduced oncotic pressure.

D) Decreased osmotic pressure of blood plasma:

Osmotic pressure is influenced by solutes like sodium and glucose, while oncotic pressure is influenced by plasma proteins.

Fasting affects oncotic pressure due to protein depletion, not the general osmotic balance of plasma.

E) Increased oncotic blood pressure:

If oncotic pressure increased, fluid would be drawn into the blood vessels, preventing edema.

2. A patient with diabetes mellitus developed ketoacidosis. This acid-base imbalance was caused by increased levels of a certain substance in the blood. Name this substance.

A. β-Hydroxybutyrate

B. Malate

C. Cholesterol

D. Glucose

E. Succinate

Correct Answer: A) β-Hydroxybutyrate

Explanation:

Diabetic ketoacidosis (DKA) is a life-threatening complication of uncontrolled diabetes mellitus, primarily type 1 diabetes. It occurs due to insulin deficiency, leading to increased lipolysis and ketogenesis in the liver. The main ketone bodies produced are:

β-Hydroxybutyrate (major ketone in DKA)
Acetoacetate
Acetone

β-Hydroxybutyrate is the predominant ketone body in DKA and contributes significantly to the metabolic acidosis (low pH, decreased bicarbonate, and increased anion gap) seen in affected patients.

Why Other Options Are Incorrect:

It does not accumulate in significant amounts during DKA or contribute to acid-base imbalance.

B) Malate:

Malate is an intermediate in the TCA (Krebs) cycle and the malate-aspartate shuttle, which helps transfer reducing equivalents across the mitochondrial membrane.

It is not directly involved in ketoacidosis or ketone body formation.

C) Cholesterol:

Cholesterol is a lipid molecule used in membrane synthesis and steroid hormone production.

It does not contribute to ketone body formation or acid-base disturbances like ketoacidosis.

D) Glucose:

Hyperglycemia is present in DKA, but glucose itself does not cause metabolic acidosis.

The acidosis is due to excessive ketone body production, not glucose accumulation.

E) Succinate:

Succinate is another TCA cycle intermediate involved in energy metabolism.

3. Examination of the biopsy materi-al detects a granuloma consisting of lymphocytes, plasma cells, macrophages with a foamy cytoplasm (Mikulicz cells), and a large number of hyaline spheres. What disease can be characterized by such findings?

A. Leprosy

B. Tuberculosis

C. Actinomycosis

D. Rhinoscleroma

E. Syphilis

Correct Answer: D) Rhinoscleroma

Explanation:

Rhinoscleroma is a chronic granulomatous disease caused by Klebsiella rhinoscleromatis, a gram-negative bacillus. It primarily affects the upper respiratory tract (nasal cavity, pharynx, larynx) and is characterized histologically by:

Mikulicz cells → Large foamy macrophages containing Klebsiella rhinoscleromatis
Plasma cells and lymphocytes
Russell bodies (hyaline spheres) → Eosinophilic inclusions derived from plasma cells

The disease progresses in three stages:

Catarrhal Stage – Nonspecific inflammation, nasal congestion
Granulomatous Stage – Formation of granulomas with Mikulicz cells
Sclerotic Stage – Fibrosis leading to airway obstruction

Why Other Options Are Incorrect:

They do not feature Mikulicz cells or Russell bodies.

A) Leprosy:

Leprosy (Hansen’s disease) is caused by Mycobacterium leprae and presents with granulomas containing acid-fast bacilli and foamy macrophages.

The characteristic Virchow cells (foamy macrophages in lepromatous leprosy) resemble Mikulicz cells but do not contain Klebsiella rhinoscleromatis.

B) Tuberculosis:

Tuberculosis (TB) is caused by Mycobacterium tuberculosis and forms caseating granulomas with Langhans giant cells.

Rhinoscleroma does not show caseous necrosis, a hallmark of TB.

C) Actinomycosis:

Actinomycosis is caused by Actinomyces israelii and is characterized by sulfur granules within abscesses and sinus tracts.

It does not involve Mikulicz cells or hyaline bodies.

E) Syphilis:

Syphilitic granulomas (gummas) contain plasma cells, necrotic tissue, and obliterative endarteritis.

4. A 28-year-old patient has accidentally cut himself on a glass. The cut is located on the anterior surface of the forearm, 2 cm below the elbow joint. The patient complains that he cannot rotate his hand inwards. What muscle is most likely to be dysfunctional in this case?

A. M. pronator quadratus

B. M. pronator teres

C. M. supinator

D. M. flexor pollicis longus

E. M. flexor digitorum profundus

Correct Answer: B) M. pronator tere

Explanation:

The pronator teres muscle is primarily responsible for forearm pronation (rotating the palm downward) and assists in elbow flexion. It is innervated by the median nerve.

An injury 2 cm below the elbow on the anterior surface of the forearm is in the region of the median nerve and pronator teres muscle. Damage to this area can impair pronation, which explains the patient’s inability to rotate the hand inwards.

Why Other Options Are Incorrect:

If damaged, the patient would struggle to flex the fingers but not to rotate the forearm.

A) M. pronator quadratus:

This muscle also helps in pronation, but it is located in the distal forearm near the wrist, not near the elbow.

Since the injury is proximal, pronator teres is more likely to be affected.

C) M. supinator:

The supinator muscle is responsible for forearm supination (rotating the palm upward), not pronation.

Damage to the supinator would impair outward rotation, but the patient in this case has difficulty with inward rotation (pronation).

D) M. flexor pollicis longus:

This muscle flexes the thumb and does not contribute to pronation.

Its dysfunction would lead to thumb flexion weakness, not issues with forearm rotation.

E) M. flexor digitorum profundus:

This muscle flexes the fingers (DIP joints) and is not involved in pronation.

5. Autopsy shows that the lung tissue resembles a honeycomb because of bag-like and cylindrical expansions of the bronchi. Microscopically, leukocyte infi-Itration with a predominance of neutrophi-Is is observed in the walls of the affected bronchi. Elastic muscle fibers and carti-laginous plates are partially destroyed and replaced with connective tissue. Adjacent lung tissue has inflammation foci, areas of fibrosis and sclerosis of vessels, and si-gns of emphysema. Hypertrophy of the right ventricle is observed in the heart. What disease can be characterized by these pathological changes?

A. Pulmonary emphysema

B. Multiple bronchiectasis

C. Pneumofibrosis

D. Chronic bronchitis

E. Interstitial pneumonia

Correct Answer: B) Multiple bronchiectasis

Explanation:

The autopsy findings describe honeycomb-like lung tissue with bronchial dilations (bag-like and cylindrical expansions), neutrophilic infiltration, and destruction of bronchial structures, which are characteristic of bronchiectasis.

Bronchiectasis is a chronic, irreversible dilation of the bronchi due to repeated infections, inflammation, and impaired mucociliary clearance. Over time, this leads to:

Chronic neutrophilic inflammation → Persistent infection and mucus accumulation
Bronchial wall destruction → Loss of elastic fibers and cartilage, leading to permanent dilation
Fibrosis and emphysema → Scarring of lung parenchyma
Pulmonary hypertension → Increased resistance in lung circulation
Right ventricular hypertrophy (cor pulmonale) → Due to chronic lung disease

Why Other Options Are Incorrect:

It lacks the honeycomb lung and mucus-filled, dilated bronchi seen in bronchiectasis.

A) Pulmonary emphysema:

Emphysema is characterized by alveolar destruction and enlargement of air spaces, not bronchial dilation.

There is no mention of bronchial wall destruction or chronic neutrophilic infiltration, which are key findings in bronchiectasis.

C) Pneumofibrosis:

Pneumofibrosis involves widespread lung scarring and stiffening, but it does not typically cause dilated bronchi with mucus accumulation (honeycombing seen in bronchiectasis).

Fibrosis is present in bronchiectasis but is not the primary pathology.

D) Chronic bronchitis:

Chronic bronchitis features hypersecretion of mucus and chronic inflammation, but it does not lead to permanent bronchial dilation or the “honeycomb” pattern seen in bronchiectasis.

It is more associated with productive cough for at least 3 months per year for 2 consecutive years.

E) Interstitial pneumonia:

Interstitial pneumonia primarily affects the alveolar walls and interstitium, leading to fibrosis but not bronchial dilation.

6. In a patient with dysentery, colonoscopy revealed that the mucosa of the large intestine was hyperemic and edematous and its surface was covered in gray-green films. What morphological form of dysenteric colitis is it?

A. Purulent

B. Catarrhal

C. Fibrinous

D. Ulcerative

E. Necrotic

Correct Answer: C) Fibrinous

Explanation:

Dysentery, particularly bacillary dysentery caused by Shigella species, leads to dysenteric colitis, which can present in different morphological forms. The presence of gray-green films covering the hyperemic and edematous mucosa suggests fibrinous colitis.

Fibrinous colitis is characterized by:

Formation of pseudomembranes composed of fibrin, necrotic debris, and inflammatory cells
Mucosal hyperemia and edema
Gray-green membranes (pseudomembranes) that adhere to the intestinal wall
Superficial tissue necrosis, but without deep ulceration

This is a hallmark feature of Shigella dysentery, Clostridioides difficile colitis, and other severe inflammatory bowel infections.

Why Other Options Are Incorrect:

The description suggests superficial necrosis with fibrin deposition rather than full-thickness necrosis.

A) Purulent:

Purulent colitis involves massive neutrophilic infiltration and pus formation, but it does not typically result in pseudomembrane formation.

There is no mention of visible pus accumulation in the description.

B) Catarrhal:

Catarrhal inflammation is an early, mild form of colitis characterized by mucosal redness, swelling, and mucus secretion but without fibrinous films.

The presence of gray-green pseudomembranes suggests a more severe, fibrinous form.

D) Ulcerative:

Ulcerative colitis features deep mucosal ulcers, but pseudomembrane formation is more indicative of fibrinous inflammation.

While Shigella dysentery can lead to ulceration, the description focuses on pseudomembranes rather than deep ulcers.

E) Necrotic:

Necrotic colitis involves extensive tissue necrosis leading to black, gangrenous areas rather than gray-green pseudomembranes.

7. A 42-year-old woman has been hospitali-zed with the diagnosis of angina pectoris, ischemic heart disease. Pain attacks occur 1-2 times a day. What drug should he prescribed for treatment in this case?

A. Drotaverine

B. Dipyridamole

C. Theophylline

D. Papaverine

E. Isosorbide mononitrate

Correct Answer: E) Isosorbide mononitrate

Explanation:

The patient is diagnosed with angina pectoris (a type of ischemic heart disease). Angina pectoris is typically treated with nitrates, which are effective in relieving chest pain by reducing myocardial oxygen demand.

Isosorbide mononitrate is a long-acting nitrate that works by dilating blood vessels, thus reducing preload (venous return) and afterload, which decreases the work of the heart and helps to alleviate angina symptoms. Nitrates also improve coronary blood flow, which is essential for patients with ischemic heart disease.

Why Other Options Are Incorrect:

It can relieve smooth muscle spasm but is not the drug of choice for ischemic heart disease.

A) Drotaverine:

Drotaverine is a spasmolytic drug used to relieve smooth muscle spasms, but it does not have the desired effects in the treatment of angina pectoris.

It is more commonly used for conditions like gastrointestinal and genitourinary spasms.

B) Dipyridamole:

Dipyridamole is a vasodilator and antiplatelet agent used in conditions like stroke prevention and peripheral artery disease.

While it has some benefit in improving coronary blood flow, it is not the first-line treatment for angina pectoris.

C) Theophylline:

Theophylline is a bronchodilator used in respiratory conditions like asthma and chronic obstructive pulmonary disease (COPD).

It has little role in treating angina pectoris.

D) Papaverine:

Papaverine is also a spasmolytic drug, but like drotaverine, it does not directly treat the underlying ischemia in angina pectoris.

8. A microspecimen was made from the puncture material obtained from the pati-ent’s regional lymph node and stained according to the Romanowsky-Giemza technique. In this specimen, the doctor detected pale pink thin microorganisms with 12-14 uniform curls and sharp ends. that were 10-13 mem long. What infecti-ous discase can be caused by the detected. pathogen?

A. Leptospirosis

B. Leishmaniasis

C. Syphilis

D. Trypanosomiasis

E. Typhus recurrens

Correct Answer: A) Leptospirosis

Explanation:

The microspecimen described shows thin, pale pink microorganisms with 12-14 uniform curls and sharp ends that are 10-13 micrometers long. These features are characteristic of Leptospira spp., the bacteria responsible for leptospirosis.

Leptospirosis is an infectious disease caused by spirochetes, particularly Leptospira interrogans. The spiral shape with sharp ends and uniform curls described fits the appearance of Leptospira bacteria, which can be observed under a microscope after Romanowsky-Giemsa staining. Leptospirosis is typically transmitted to humans via contact with water or soil contaminated with the urine of infected animals.

Why Other Options Are Incorrect:

Leptospira better matches the microscopic description given in the question.

B) Leishmaniasis:

Leishmaniasis is caused by Leishmania spp., which are protozoa, not spirochetes.

Leishmania species appear as intracellular amastigotes or promastigotes and are not spiral-shaped like Leptospira.

C) Syphilis:

Syphilis is caused by Treponema pallidum, another spirochete, but it typically has more tightly coiled spirals and different staining characteristics.

The description of uniform curls and sharp ends fits better with Leptospira than Treponema.

D) Trypanosomiasis:

Trypanosomiasis (e.g., African sleeping sickness) is caused by Trypanosoma spp., which are flagellated protozoa.

These organisms have elongated, spindle-shaped bodies and are not spirochetes with curled ends.

E) Typhus recurrens:

Typhus recurrens is caused by Borrelia recurrentis, a spirochete, but Borrelia spirochetes tend to be larger and have irregular undulating waves rather than uniform curls.

9. A patient developed motor aphasia as a result of a brain injury. Where in the cerebral cortex is the focus of damage in this case?

A. Postcentral gyrus

B. Wernicke’s arca

C. Heschl’s gyrus

D. Angular gyrus

E. Broca’s arca

Correct Answer: E) Broca’s area

Explanation:

Motor aphasia, also known as expressive aphasia, occurs when there is damage to Broca’s area, which is located in the left frontal lobe, specifically in the posterior part of the frontal gyrus (inferior frontal gyrus).

Broca’s area is responsible for the production of speech and the motor control of speech-related muscles. Damage to this area results in difficulty speaking (motor aspect) while comprehension remains relatively intact. Patients may have difficulty forming grammatically correct sentences or speaking fluently but can often understand language well.

Why Other Options Are Incorrect:

It does not play a role in speech production, which is why it is not the focus of damage in motor aphasia.

A) Postcentral gyrus:

The postcentral gyrus is located in the parietal lobe and is primarily involved in somatosensory processing (touch, temperature, pain).

It does not contribute to speech production and is unrelated to aphasia.

B) Wernicke’s area:

Wernicke’s area is located in the posterior part of the temporal lobe and is responsible for language comprehension.

Damage to Wernicke’s area leads to receptive aphasia (difficulty understanding language), but speech production remains relatively fluent, though nonsensical.

C) Heschl’s gyrus:

Heschl’s gyrus is located in the temporal lobe and is primarily involved in auditory processing, not speech production.

Damage to Heschl’s gyrus would affect hearing, not language production or comprehension.

D) Angular gyrus:

The angular gyrus is located in the parietal lobe and is involved in reading, writing, and language comprehension.

10. A patient complains of impaired memory, rapid fatigability, drowsiness, and chills. What is the most likely cause of this condition?

A. Hyperthyroidism

B. Hypofunction of the neurohypophysis

C. Hypothyroidism

D. Adrenal hypofunction

E. Hyperparathyroidism

Correct Answer: C) Hypothyroidism

Explanation:

The symptoms described—impaired memory, rapid fatigability, drowsiness, and chills—are consistent with hypothyroidism, a condition where the thyroid gland produces insufficient thyroid hormones (T3 and T4).

Thyroid hormones are essential for maintaining normal metabolism, energy levels, and cognitive function. In hypothyroidism, the slowed metabolism leads to symptoms such as:

Fatigue and drowsiness due to low metabolic activity
Memory impairment because of reduced cerebral function
Chills due to decreased heat production from a slower metabolism

Other common symptoms of hypothyroidism include weight gain, constipation, dry skin, and cold intolerance.

Why Other Options Are Incorrect:

Symptoms typically include weakness, bone pain, kidney stones, and constipation, not memory impairment or chills.

A) Hyperthyroidism:

Hyperthyroidism (excess thyroid hormones) often causes symptoms like nervousness, irritability, weight loss, rapid heart rate, and heat intolerance, rather than the fatigue, drowsiness, and chills seen in hypothyroidism.

Memory problems are usually less pronounced in hyperthyroidism.

B) Hypofunction of the neurohypophysis (posterior pituitary):

The posterior pituitary secretes antidiuretic hormone (ADH) and oxytocin. ADH deficiency can lead to diabetes insipidus, causing excessive urination and thirst, but it is not typically associated with impaired memory, fatigue, or chills.

D) Adrenal hypofunction (Addison’s disease):

Addison’s disease involves insufficient cortisol and aldosterone production from the adrenal glands.

Symptoms can include fatigue, weakness, weight loss, low blood pressure, and hyperpigmentation but not typically memory impairment or chills as a primary feature.

E) Hyperparathyroidism:

Hyperparathyroidism results from excessive parathyroid hormone (PTH) and causes elevated calcium levels.

11. Autopsy of the body of a person, who died of renal failure and had been sufferi-ng from bronchicctasis for the past 5 years, revealed dense enlarged kidneys with a thickened white cortical layer and a greasy sheen. What disease was detected in the kidneys?

A. Nephroblastoma

B. Glomerulonephritis

C. Chronic pyelonephritis

D. Necrotic nephrosis

E. Secondary amyloidosis

Correct Answer: E. Secondary amyloidosis

Explanation:

Secondary amyloidosis occurs when there is chronic inflammation or infection, leading to the deposition of amyloid proteins in various organs, including the kidneys. In this case, the patient has had bronchiectasis, a chronic respiratory condition that can lead to secondary amyloidosis due to long-standing infection and inflammation. The dense, enlarged kidneys with a thickened white cortical layer and greasy sheen are characteristic of amyloid deposits, which give the kidneys a distinctive appearance.

3. Why Other Options Are Incorrect:

D. Necrotic nephrosis:
Necrotic nephrosis refers to acute kidney damage and necrosis of the renal tubules, typically caused by ischemia or toxins. The appearance described in the autopsy, with a thickened white cortical layer and greasy sheen, is more consistent with amyloid deposits, not necrotic tissue.

A. Nephroblastoma:
Nephroblastoma, also known as Wilms tumor, is a kidney cancer primarily found in children, not associated with chronic conditions like bronchiectasis or renal failure in adults. It typically presents as a solid mass in the kidney, not with the described appearance of amyloid deposits.

B. Glomerulonephritis:
While glomerulonephritis can cause renal failure and kidney enlargement, it is typically associated with inflammation of the glomeruli, not the amyloid-like thickening and greasy sheen described here. The kidneys in glomerulonephritis usually show a more diffuse, non-specific pattern of injury, not the distinct appearance of amyloidosis.

C. Chronic pyelonephritis:
Chronic pyelonephritis is a result of long-term kidney infections and can cause kidney scarring, but it does not typically lead to the thick, greasy sheen of amyloid deposition. The kidneys may be scarred or deformed, but the description of a greasy sheen suggests amyloidosis, not pyelonephritis.

12.During the surgical treatment of a femoral hernia, the doctor takes note of the external opening of the femoral canal. What anatomical structure forms it?

A. Septum femorale

B. Hiatus saphenus

C. Fascia pectinea

D. Arcus iliopectineus

E. Fossa femoralis

Correct Answer: A. Septum femorale

Explanation:

The external opening of the femoral canal is formed by the septum femorale. This is a thin connective tissue structure that separates the femoral vein from the femoral canal. It is located at the femoral ring, which is the entrance to the femoral canal, and it plays a critical role in the development and management of femoral hernias, as it is through this opening that abdominal contents may herniate into the femoral canal.

3. Why Other Options Are Incorrect:

E. Fossa femoralis:
The fossa femoralis is a triangular space located in the upper thigh, bounded by the inguinal ligament, the adductor longus, and the sartorius muscle. While it is related to the femoral canal, it does not directly form the external opening of the femoral canal.

B. Hiatus saphenus:
The hiatus saphenus is the opening in the fascia lata where the great saphenous vein enters the thigh. It does not form the external opening of the femoral canal and is not directly related to femoral hernia formation.

C. Fascia pectinea:
The fascia pectinea is a continuation of the inguinal ligament and covers the pectineus muscle. It contributes to the structure of the femoral triangle, but it does not form the external opening of the femoral canal.

D. Arcus iliopectineus:
The arcus iliopectineus is a fibrous band that forms the boundary of the femoral canal but does not directly form the external opening. The septum femorale is the actual structure that forms the external opening of the femoral canal.

13. Teturam (Disulfiram) that is an aldehyde dehydrogenase inhibitor is wi-dely used in medical practice for alcoholi-sm prevention. What metabolite causes aversion to alcohol, if its levels are high in the blood?

A. Ethanol

B. Methanol

C. Malonaldehyde

D. Acetaldehyde

E. Propionaldehyde

Correct Answer: D. Acetaldehyde

Explanation:

Disulfiram inhibits the enzyme aldehyde dehydrogenase, which normally breaks down acetaldehyde, a toxic intermediate product of ethanol metabolism. As a result, when an individual consumes alcohol while taking disulfiram, acetaldehyde accumulates in the blood, leading to unpleasant symptoms such as nausea, vomiting, and flushing. This aversion to alcohol helps in the treatment of alcoholism.

3. Why Other Options Are Incorrect:

E. Propionaldehyde:
Propionaldehyde is another aldehyde, but it is not involved in the disulfiram-induced aversion to alcohol.

A. Ethanol:
Ethanol is the alcohol consumed, and while it is the substance that is metabolized, it is not the metabolite that causes the aversion. The issue is the buildup of acetaldehyde when ethanol is metabolized in the presence of disulfiram.

B. Methanol:
Methanol is a different type of alcohol that is toxic when ingested and is metabolized to formaldehyde and formic acid. It is not involved in the action of disulfiram.

C. Malonaldehyde:
Malonaldehyde is a product of lipid peroxidation and is not directly related to the effects of disulfiram or alcohol metabolism.

14. Autopsy of the body of a 33-year-old woman revealed thickening of the stomach wall in the pyloric region (the layers of the wall can be distinguished on section) wi-th the growth of a dense whitish tissue in the submucosal layer and its small strands. located in the muscular layer. The relief of the mucosa is preserved, the folds are rigid and immobile. What is the nature of the tumor growth in this case?

A. Endophytic

B. Exophytic

C. Node

D. Ulceroinfiltrative form

E. Ulcer

Correct Answer: D. Ulceroinfiltrative form

Explanation:

The described features of a thickened stomach wall, dense whitish tissue in the submucosal layer, and rigidity of the mucosal folds are consistent with an ulceroinfiltrative form of tumor growth, commonly seen in gastric cancer. This type of tumor involves both the mucosal layer and the deeper layers, causing thickening and rigidity of the stomach wall, while preserving the mucosal relief.

3. Why Other Options Are Incorrect:

E. Ulcer:
An ulcer alone would typically show a breach in the mucosal lining and an ulcerated surface, which is not described in this case. The presence of dense tissue and submucosal involvement points more to an ulceroinfiltrative tumor.

A. Endophytic:
Endophytic tumors grow inwardly, invading the tissues. Although this is similar to ulceroinfiltrative growth, it does not fully capture the description of the tumor’s effect on the mucosa and the deeper layers as described in the case.

B. Exophytic:
Exophytic tumors grow outward and typically form a mass projecting into the lumen. This is not consistent with the description of thickening and immobility of the mucosal folds.

C. Node:
A “node” is typically a mass that is discrete and encapsulated, which does not match the description of a widespread thickening and infiltrative growth pattern in the stomach.

15. The patient’s complex therapy includes sulfonamides, which has caused the development of hemolytic anemia. This complication can be caused by insuffici-ent activity of a certain enzyme. Name this enzyme.

A. Pyruvate kinase

B. Glucose-6-phosphate dehydrogenase

C. Glucokinase

D. Aldolase

E. Glucose-6-phosphatase

Correct Answer: B. Glucose-6-phosphate dehydrogenase

Explanation:

Sulfonamides can induce hemolytic anemia in patients with a deficiency in glucose-6-phosphate dehydrogenase (G6PD), an enzyme involved in protecting red blood cells from oxidative damage. A deficiency in G6PD can lead to hemolysis when exposed to oxidative agents, such as those in sulfonamides.

3. Why Other Options Are Incorrect:

E. Glucose-6-phosphatase:
Glucose-6-phosphatase deficiency causes glycogen storage disease type 1, which presents with metabolic issues, not hemolytic anemia due to sulfonamides.

A. Pyruvate kinase:
Pyruvate kinase deficiency can cause hemolytic anemia, but it is not specifically linked to sulfonamide use.

C. Glucokinase:
Glucokinase deficiency typically leads to diabetes mellitus, not hemolytic anemia caused by sulfonamides.

D. Aldolase:
Aldolase is involved in glycolysis, but a deficiency is more related to muscle weakness and not hemolytic anemia caused by sulfonamides.

16. A patient developed hemolytic jaundi-ce as a result of transfusion of Rh-incompatible blood. What blood test value can confirm this type of jaundice?

A. Decreased stercobilin levels

B. Accumulation of urobilinogen

C. Decreased levels of conjugated bilirubin

D. Decreased levels of unconjugated bili- rubin E. Accumulation of unconjugated bilirubin

Correct Answer: E. Accumulation of unconjugated bilirubin

Explanation:

In hemolytic jaundice, such as that caused by Rh incompatibility, there is an increased breakdown of red blood cells, leading to the accumulation of unconjugated bilirubin in the blood. This type of jaundice occurs due to the overwhelming destruction of red blood cells, which exceeds the liver’s ability to conjugate and excrete bilirubin.

3. Why Other Options Are Incorrect:

D. Decreased levels of unconjugated bilirubin:
This is incorrect because in hemolytic jaundice, unconjugated bilirubin levels would actually be elevated due to increased red blood cell breakdown.

A. Decreased stercobilin levels:
Stercobilin is a product of the excretion of conjugated bilirubin in the intestine. Decreased stercobilin levels are not directly associated with hemolytic jaundice.

B. Accumulation of urobilinogen:
While urobilinogen levels may increase in hemolytic jaundice, the primary finding is the accumulation of unconjugated bilirubin in the blood.

C. Decreased levels of conjugated bilirubin:
In hemolytic jaundice, there is typically an increase in unconjugated bilirubin, not a decrease in conjugated bilirubin. Decreased conjugated bilirubin would be more indicative of hepatic or obstructive jaundice.

17. What hormone stimulates water reabsorption in the distal tubules of the kidneys and maintains blood pressure by directly stimulating the vascular wall?

A. Thyroxine

B. Oxytocin

C. Aldosteronc

D. Adrenaline

E. Vasopressin

Correct Answer: E. Vasopressin

Explanation:

Vasopressin (also known as antidiuretic hormone, ADH) regulates water balance by stimulating water reabsorption in the distal tubules and collecting ducts of the kidneys. It also constricts blood vessels, contributing to blood pressure regulation.

3. Why Other Options Are Incorrect:

D. Adrenaline:
Adrenaline increases heart rate and causes vasoconstriction, which raises blood pressure. However, it does not significantly influence water reabsorption in the kidneys.

A. Thyroxine:
Thyroxine is a thyroid hormone that regulates metabolism but does not directly affect water reabsorption in the kidneys or blood pressure through vascular constriction.

B. Oxytocin:
Oxytocin primarily regulates uterine contractions during labor and milk ejection during breastfeeding. It does not play a role in water reabsorption or blood pressure regulation.

C. Aldosterone:
Aldosterone stimulates sodium reabsorption in the distal tubules and collecting ducts, which indirectly helps to maintain blood volume and blood pressure. However, it does not directly stimulate water reabsorption; vasopressin has a more direct effect.

18. A transplanted kidney responds to pai-nful stimuli by stopping urination. What causes this response?

A. Decreased secretion of ADH

B. Decreased secretion of ACTH

C. Effect of the parasympathetic nervous system

D. Effect of the sympathetic nervous system

E. Increased secretion of ADH

Correct Answer: D. Effect of the sympathetic nervous system

Explanation:

Painful stimuli activate the sympathetic nervous system, which leads to vasoconstriction of renal blood vessels and a reduction in glomerular filtration rate (GFR). This results in decreased urine production, a phenomenon called “stress-induced oliguria.”

3. Why Other Options Are Incorrect:

E. Increased secretion of ADH:
While ADH secretion may increase in response to stress, the immediate cessation of urination in this case is due to sympathetic vasoconstriction, not ADH.

A. Decreased secretion of ADH:
ADH (vasopressin) is primarily involved in water reabsorption in the kidneys. Pain activates ADH secretion rather than decreasing it.

B. Decreased secretion of ACTH:
ACTH controls adrenal cortisol release, which is not directly responsible for the acute suppression of urination due to pain.

C. Effect of the parasympathetic nervous system:
The parasympathetic nervous system generally promotes relaxation and increased urinary output, the opposite of the observed effect.

19. A patient has inflammation of the oral mucosa, accompanied by unbearable pain. What nerve is affected in this case, causing this condition?

A. Facial

B. Vagus

C. Trigeminal

D. Chorda tympani

E. Glossopharyngeal

Correct Answer: C. Trigeminal

Explanation:

The trigeminal nerve (CN V) is responsible for sensory innervation of the face, including the oral mucosa. Inflammation and pain in the oral cavity suggest trigeminal nerve involvement, particularly the maxillary (V2) or mandibular (V3) divisions.

Why Other Options Are Incorrect:

E) Glossopharyngeal nerve (CN IX) – Provides sensory innervation to the posterior one-third of the tongue and pharynx but does not innervate the oral mucosa extensively..

A) Facial nerve (CN VII) – Primarily responsible for facial muscle movement and taste but does not provide significant pain sensation to the oral mucosa.

B) Vagus nerve (CN X) – Involved in autonomic functions and sensation in the pharynx and larynx, not the oral mucosa.

D) Chorda tympani – A branch of CN VII that carries taste sensation from the anterior two-thirds of the tongue, not pain from the oral mucosa.

20. Autopsy of the body of a woman who died of uremia revealed that the ki-dneys differed in size, their surface had large tubercles, and there were dense commissures between the renal surface. and the renal capsule. Microscopically, the following is observed in the renal tissues: encapsulated abscesses, proliferation of connective tissue with lymphohistiocytic infiltration, metaplastic foci where transi-tional epithelium transforms into stratified epithelium, dystrophy and atrophy of the tubules. What is the most likely diagnosis in this case?

A. Tubulointerstitial nephritis

B. Chronic pyelonephritis

C. Chronic glomerulonephritis

D. Acute glomerulonephritis

E. Acute pyelonephritis

Correct Answer: B) Chronic pyelonephritis

Explanation:

Chronic pyelonephritis is a chronic inflammatory condition of the renal parenchyma and pelvis due to recurrent infections. It causes kidney scarring, abscesses, and fibrosis, leading to dystrophic changes, tubular atrophy, and uremia.

Why Other Options Are Incorrect:

E) Acute pyelonephritis – Features abscesses but lacks chronic fibrosis and scarring.

A) Tubulointerstitial nephritis – Involves inflammation of interstitial tissue without large abscesses.

C) Chronic glomerulonephritis – Primarily affects glomeruli, not tubules and pelvis.

D) Acute glomerulonephritis – An acute disease without fibrosis or chronic abscesses.

21. A pathogen has been isolated from a patient with acute gastroenteritis. The pathogen must be identified based on its antigenic structure. What serological reaction should be used for this purpose?

A. Opsonization

B. Precipitation

C. Agglutination

D. Neutralization

E. Complement fixation

Correct Answer: C) Agglutination

Explanation:

Agglutination tests detect specific antigens using antibodies, forming visible clumps, making it useful for pathogen identification.

Why Other Options Are Incorrect:

E) Complement fixation – Detects antibodies rather than pathogen antigens.

A) Opsonization – Enhances phagocytosis, not antigen detection.

B) Precipitation – Identifies soluble antigens, not bacterial surface antigens.

D) Neutralization – Blocks pathogen activity but does not identify it.

22. After taking trimethoprim/sulfametho-xazole, a patient with an acute respiratory disease developed redness and edema of the skin of her face and hands and intense itching. What type of hyperemia is it?

A. Venous hyperemia

B. Vacate arterial hyperemia

C. Neurotonic arterial hyperemia

D. Neuroparalytic arterial hyperemia

E. Metabolic arterial hyperemia

Correct Answer: C) Neurotonic arterial hyperemia

Explanation:

Neurotonic arterial hyperemia occurs due to increased blood flow from nervous system activation, commonly seen in allergic reactions (redness, swelling, itching).

Why Other Options Are Incorrect:

E) Metabolic arterial hyperemia – Related to increased tissue metabolism, not allergies.

A) Venous hyperemia – Caused by impaired venous outflow, not an allergic reaction.

B) Vacate arterial hyperemia – Not a recognized term.

D) Neuroparalytic arterial hyperemia – Due to nerve paralysis, not an immune response.

23. A patient has varicose veins and thrombophlebitis on the posterolateral surface of the lower leg. What vein is affected in this case?

A. Fibular vein

B. Large saphenous vein

C. Anterior tibial vein

D. Small saphenous vein

E. Posterior tibial vein

Correct Answer: D) Small saphenous vein

Explanation:

The small saphenous vein, located on the posterolateral aspect of the leg, is prone to varicose veins and thrombophlebitis.

Why Other Options Are Incorrect:

E) Posterior tibial vein – A deep vein, less prone to varicosities..

A) Fibular vein – Deep vein, less commonly affected.

B) Large saphenous vein – Runs along the medial side, not posteriorly.

C) Anterior tibial vein – Located anteriorly, not posterolateral.

24. A patient presents with a decreased excitation conduction velocity in the atri-oventricular node. What ECG component will have an increased duration in this case?

A. RR interval

B. ST segment

C. P wave

D. QRS complex

E. PQ interval

Correct Answer: E) PQ interval

Explanation:

The PQ interval represents the time from atrial depolarization to ventricular depolarization, which increases if AV node conduction is delayed.

Why Other Options Are Incorrect:

D) QRS complex – Represents ventricular depolarization.

A) RR interval – Represents heart rate variability, not conduction delay.

B) ST segment – Represents ventricular repolarization.

C) P wave – Represents atrial depolarization, not AV conduction.

25. Dystrophic changes in the cardiac muscle are accompanied by dilation of the heart chambers, decreased force of cardi-ac contractions, increased volume of the blood that remains in the heart chambers during systole, and venous overflow. What medical condition can be characterized by these phenomena?

A. Tonogenic dilatation

B. Stage of cardiosclerosis

C. Cardiac tamponade

D. Myogenic dilatation

E. Developing stage of myocardial hypert-rophy

Correct Answer: D) Myogenic dilatation

Explanation:

Myogenic dilatation occurs due to myocardial weakness, reducing contractility and causing chamber dilation.

Why Other Options Are Incorrect:

E) Developing stage of myocardial hypertrophy – Hypertrophy involves thickened walls, not dilation.

A) Tonogenic dilatation – Caused by increased preload, not myocardial weakness.

B) Stage of cardiosclerosis – Refers to fibrotic changes, not acute dilation.

C) Cardiac tamponade – Due to fluid accumulation, not dystrophy.

26. Before a neurosurgical operation, it became necessary to asses the electrical activity of the subcortical brain structures. What method can be used for this assessment?

A. Stereotactic surgery

B. X-ray

C. Electrocardiography

D. Tomography

E. Ultrasound

Correct Answer: A) Stereotactic surgery

Explanation:

Stereotactic surgery uses electrodes to assess deep brain electrical activity.

Why Other Options Are Incorrect:

E) Ultrasound – Not used for brain electrical activity.

B) X-ray – Provides structural images, not electrical activity.

C) Electrocardiography – Measures heart, not brain activity.

D) Tomography – Provides imaging but not electrical readings.

E) Ultrasound – Not used for brain electrical activity.

27. A patient takes choleretic drugs. What process, besides bile secretion, do they sti-mulate?

A. Secretion of gastric juice

B. Gastric motility

C. Water absorption

D. Secretion of pancreatic juice

E. Intestinal motility

Correct Answer: D) Secretion of pancreatic juice

Explanation:

Choleretic drugs stimulate pancreatic juice secretion via reflex pathways.

Why Other Options Are Incorrect:

E) Intestinal motility – Not significantly affected.

A) Secretion of gastric juice – Not directly affected.

B) Gastric motility – Not primarily targeted.

C) Water absorption – Not a direct effect.

28. The causative agent of hepatitis D (delta agent) is a defective virus that can only replicate in the cells that are already infected with:

A. Epstein-Barr virus

B. Human immunodeficiency virus

C. Hepatitis A virus

D. Hepatitis E virus

E. Hepatitis B virus

Correct Answer: E) Hepatitis B virus

Explanation:

Hepatitis D virus (HDV) is a defective RNA virus that requires Hepatitis B virus (HBV) for replication because it lacks its own envelope proteins. HDV uses HBV’s surface antigen (HBsAg) to enter hepatocytes and complete its life cycle.

Why Other Options Are Incorrect:

D) Hepatitis E virus – Causes acute hepatitis, not involved in HDV replication.

A) Epstein-Barr virus – Causes infectious mononucleosis, not involved in HDV replication.

B) Human immunodeficiency virus (HIV) – Affects immune cells but does not provide the necessary proteins for HDV.

C) Hepatitis A virus – A non-chronic virus with no relation to HDV.

29. A man complains of increased diuresis (up to 5-7 liters of urine per 24 hours). Examination detects reduced vasopressin secretion. What cells exhibit insufficient secretory activity in this case?

A. Neurosecretory cells of the hypotha-lamus

B. Cells of the pars tuberalis

C. Pituicytes

D. Endocrinocytes of the intermediate pi- tuitary

E. Endocrinocytes of the anterior pituitary

Correct Answer: A) Neurosecretory cells of the hypothalamus

Explanation:

Vasopressin (antidiuretic hormone, ADH) is synthesized by neurosecretory cells in the hypothalamus (supraoptic and paraventricular nuclei) and stored in the posterior pituitary. Deficiency leads to diabetes insipidus, characterized by polyuria and polydipsia due to the kidneys’ inability to retain water.

Why Other Options Are Incorrect:

E) Endocrinocytes of the anterior pituitary – Secrete hormones like ACTH, TSH, GH, but not vasopressin.

B) Cells of the pars tuberalis – These do not produce vasopressin.

C) Pituicytes – Support cells in the posterior pituitary; they do not secrete hormones.

D) Endocrinocytes of the intermediate pituitary – Secrete melanocyte-stimulating hormone (MSH), not vasopressin.

30. A 1.5-year-old boy constantly suffers from pyoderma and had three cases of pneumonia. In his blood, there are reduced levels of immunoglobulins G and A and no plasma cells. What type of immunodefici- ency has developed in the child?

A. Swiss-type immunodeficiency

B. Louis-Bar syndrome

C. Bruton’s hypogammaglobulinemia

D. Wiskott-Aldrich syndrome

E. Thymic hypoplasia

Correct Answer: C) Bruton’s hypogammaglobulinemia

Explanation:

Bruton’s X-linked agammaglobulinemia (XLA) is a primary immunodeficiency caused by a mutation in the BTK (Bruton’s tyrosine kinase) gene, which is essential for B-cell maturation. Without functional BTK, B cells cannot develop into plasma cells, leading to:

Absent or very low levels of immunoglobulins (IgG, IgA, and IgM).
Recurrent bacterial infections such as pyoderma and pneumonia due to a lack of humoral immunity.
Absent or severely reduced plasma cells in blood and lymphoid tissues.
Symptoms usually appear after 6 months of age, when maternal IgG levels decline.

Why Other Options Are Incorrect:

E) Thymic hypoplasia (DiGeorge syndrome) – Affects T-cell development, leading to viral and fungal infections rather than the bacterial infections seen in Bruton’s disease.

A) Swiss-type immunodeficiency – Refers to severe combined immunodeficiency (SCID), which affects both B and T cells, leading to more severe infections, including fungal and viral infections, not just recurrent bacterial infections.

B) Louis-Bar syndrome (Ataxia-Telangiectasia) – Affects both immunity and the nervous system, with characteristic ataxia, telangiectasia, and immune dysfunction (low IgA, not IgG).

D) Wiskott-Aldrich syndrome – Features thrombocytopenia (low platelets), eczema, and recurrent infections, but does not present with absent plasma cells.

31. A 30-year-old patient complains of intense thirst and dry mouth that appeared after a severe nervous breakdown. Laboratory testing detects elevated blood glucose of 10 mmol/L. What endocrine gland is affected in this patient?

B. Pancreas

C. Thyroid gland

D. Pineal gland

E. Gonads

Correct Answer: B) Pancreas

Explanation:

The pancreas is responsible for insulin production via the beta cells of the islets of Langerhans. This patient presents with symptoms of polydipsia (intense thirst) and dry mouth, along with an elevated blood glucose level of 10 mmol/L, which suggests a disturbance in glucose metabolism.

A severe nervous breakdown (psychological stress) can trigger stress-induced hyperglycemia or latent diabetes mellitus, revealing an underlying pancreatic dysfunction, most commonly Type 2 Diabetes Mellitus.

Why Other Options Are Incorrect:

E) Gonads – The gonads (testes/ovaries) primarily produce sex hormones like testosterone, estrogen, and progesterone, which do not directly regulate blood glucose levels. They are unrelated to this patient’s symptoms.

A) Adrenal glands – The adrenal glands produce cortisol and catecholamines, which can contribute to transient hyperglycemia in response to stress. However, adrenal dysfunction typically presents with other symptoms like hypertension (Cushing’s syndrome) or hypotension (Addison’s disease), rather than persistent hyperglycemia.

C) Thyroid gland – Hyperthyroidism can cause increased metabolism and sometimes mild hyperglycemia, but it is not the primary endocrine gland responsible for glucose regulation. The symptoms of hyperthyroidism (weight loss, palpitations, tremors) are absent in this case.

D) Pineal gland – The pineal gland primarily regulates circadian rhythms via melatonin production. It has no direct role in glucose metabolism or endocrine regulation of blood sugar.

32. A large cell with slightly basophilic cytoplasm and a bean-shaped nucleus was detected in a smear prepared from peri-pheral blood. This cell is the largest one among the cells in sight. What cell is it?

A. Monocyte

B. Medium size lymphocyte

C. Plasma cell

D. Small lymphocyte

E. Macrophage

Correct Answer: A) Monocyte

Explanation:

Monocytes are the largest white blood cells found in peripheral blood. They have a characteristic bean-shaped (kidney-shaped) nucleus and slightly basophilic cytoplasm. Monocytes play a crucial role in the immune system by differentiating into macrophages or dendritic cells when they migrate into tissues.

Why Other Options Are Incorrect:

E) Macrophage – While macrophages are large cells, they are not found in peripheral blood but rather in tissues, where they function as phagocytes. The large cell described in this case is circulating in peripheral blood, making it a monocyte rather than a macrophage.

B) Medium-size lymphocyte – Lymphocytes have round nuclei rather than bean-shaped nuclei. They are generally smaller than monocytes and have scant cytoplasm.

C) Plasma cell – Plasma cells are differentiated B cells responsible for antibody production. They have eccentric nuclei (off-center) and a perinuclear halo, which is not described in the given question.

D) Small lymphocyte – Small lymphocytes have a compact, round nucleus with very little cytoplasm, making them significantly smaller than monocytes.

33. The pregnant woman’s condition was complicated by gestosis. Laboratory testi-ng detects ketonuria. What substance has appeared in the patient’s urine?

A. Lactate

B. Acetoacetate

C. Creatinine

D. Pyruvate

E. Uratcs

Correct Answer: B) Acetoacetate

Explanation:

Ketonuria refers to the presence of ketone bodies in the urine, which occurs due to increased fat metabolism and ketogenesis. Acetoacetate, along with β-hydroxybutyrate and acetone, are the primary ketone bodies produced when there is an energy imbalance, such as gestational toxemia (gestosis), starvation, or uncontrolled diabetes mellitus.

In pregnancy-related complications like gestosis (preeclampsia/eclampsia), metabolic disturbances can lead to increased fat breakdown and ketone body production, resulting in ketonuria.

Why Other Options Are Incorrect:

E) Urates – Urates (uric acid salts) are associated with gout and renal dysfunction but have no relation to ketosis or ketonuria.

A) Lactate – Lactate is produced during anaerobic metabolism but does not contribute to ketonuria. It is elevated in conditions like lactic acidosis (e.g., hypoxia, shock).

C) Creatinine – Creatinine is a waste product of muscle metabolism and is used as a marker of kidney function. It is not a ketone body and does not cause ketonuria.

D) Pyruvate – Pyruvate is an intermediate in glucose metabolism, converting into lactate or entering the Krebs cycle. It is not a ketone body and does not appear in urine due to ketosis.

34. A patient with heart failure developed arrhythmia, where ECG shows the atrial rate of 70/min and the ventricular rate of 35/min. What cardiac function is impaired in the patient?

A. Excitability

B. Conductivity

C. Automatism

D. Contractility

E. Excitability and conductivity

Correct Answer: B) Conductivity

Explanation:

The ECG shows an atrial rate of 70/min and a ventricular rate of 35/min, which suggests heart block or a disruption in the conduction pathway between the atria and ventricles. The ventricular rate is slower than the atrial rate, indicating that the impulse from the atria is not being efficiently conducted to the ventricles. This type of conduction issue is commonly seen in second- or third-degree atrioventricular (AV) block, where there is impaired conduction of electrical impulses from the atria to the ventricles.

In this case, the problem lies in the conductivity of the electrical system, specifically in the AV node or His-Purkinje system, which normally conducts the impulse from the atria to the ventricles.

Why Other Options Are Incorrect:

E) Excitability and conductivity – While conductivity is the correct answer, excitability is not impaired in this case, so this answer is not accurate.

A) Excitability – Excitability refers to the ability of cardiac cells to respond to electrical stimuli. This is not the primary issue in this case since the atria are still firing at a normal rate of 70/min, indicating that excitability is not impaired.

C) Automatism – Automatism is the ability of cardiac cells, particularly in the sinoatrial (SA) node, to generate spontaneous electrical impulses. This is not impaired in this patient because the atria are still generating impulses at a normal rate (70/min).

D) Contractility – Contractility refers to the heart muscle’s ability to contract and pump blood. Although this patient has heart failure, the issue in this scenario is related to the conduction of impulses, not the strength of heart muscle contraction.

35. A 40-ycar-old woman, who was systematically taking acetylsalicylic acid, developed hemorrhages. Impaired functi-onal activity of platelets was detected. This phenomenon is associated with inhibition of a certain enzyme. Name this enzyme.

A. Cyclooxygenase

B. Cytochrome oxidase

C. Na+, K+-ATPase

D. Glucose-6-phosphate dehydrogenase

E. Cholinesterase

Correct Answer: A) Cyclooxygenase

Explanation:

Acetylsalicylic acid (ASA), commonly known as aspirin, inhibits the enzyme cyclooxygenase (COX), which plays a crucial role in the synthesis of prostaglandins and thromboxanes from arachidonic acid. In particular, thromboxane A2 is important for platelet aggregation and vasoconstriction. By inhibiting COX, aspirin prevents the formation of thromboxane A2, leading to impaired platelet aggregation and a reduced ability to form blood clots, which can result in hemorrhages or easy bruising. This is why aspirin is commonly used as an anticoagulant for cardiovascular protection.

Why Other Options Are Incorrect:

E) Cholinesterase – Cholinesterase is involved in the breakdown of acetylcholine at synaptic junctions in the nervous system. It does not play a role in platelet aggregation or bleeding disorders.

B) Cytochrome oxidase – Cytochrome oxidase is part of the electron transport chain in mitochondria, responsible for cellular respiration and energy production. It is not involved in platelet function or blood clotting.

C) Na+, K+-ATPase – Na+, K+-ATPase is an enzyme that pumps sodium and potassium ions across the cell membrane to maintain the resting membrane potential and cellular function. It does not directly affect platelet function or the clotting cascade.

D) Glucose-6-phosphate dehydrogenase – This enzyme is part of the pentose phosphate pathway and is involved in the production of NADPH, essential for redox reactions. A deficiency in this enzyme can lead to hemolysis but does not directly affect platelet function.

36. In a patient with cardiac arrhythmia, ECG shows the following: heart rate 60/min, prolonged PQ interval, periodic loss of ORS complex. What heart rhythm disorder is observed in the patient?

A. Incomplete second-degree AV block

B. Right bundle branch block

C. Complete AV block

D. Sick sinus syndrome

E. Incomplete first-degree AV block

Correct Answer: A) Incomplete second-degree AV block

Explanation:

The ECG findings suggest a prolonged PQ interval and periodic loss of the QRS complex. This pattern is characteristic of second-degree AV block. Specifically, incomplete second-degree AV block (Mobitz Type I or Wenckebach block) involves progressive prolongation of the PR interval until a QRS complex is dropped (i.e., a missed beat). This pattern repeats itself periodically, indicating the type of block, where conduction through the AV node is delayed and occasionally fails.

A heart rate of 60/min is typical, as the conduction delay slows the overall rhythm.
The periodic loss of the QRS complex (dropped beats) occurs after progressively longer PR intervals, which is the hallmark of Mobitz Type I.

Why Other Options Are Incorrect:

E) Incomplete first-degree AV block – In first-degree AV block, the PR interval is prolonged but regular, and every atrial impulse is conducted to the ventricles. There is no dropped beat as seen in second-degree AV block, so this diagnosis is not consistent with the described findings.

B) Right bundle branch block – Right bundle branch block (RBBB) would show a characteristic wide QRS complex and a delay in the right ventricle activation. This does not match the findings described, as the problem in the scenario is related to AV conduction, not a bundle branch issue.

C) Complete AV block – In complete (third-degree) AV block, there is no communication between the atria and ventricles, and the atrial and ventricular rhythms are independent. The ventricles usually pace themselves at a slow rate. The described ECG findings, however, indicate some degree of atrioventricular conduction, not complete dissociation.

D) Sick sinus syndrome – Sick sinus syndrome is typically characterized by sinus node dysfunction, such as bradycardia, sinus pauses, or alternating tachycardia and bradycardia. The described findings do not fit this syndrome, as the issue here is related to AV conduction, not sinus node dysfunction.

37. During examination by a pediatrician, multiple petechiac were detected on the skin of a 10-year-old child. The child has bleeding gums and low levels of vitamin C in urine. What process is impaired in this case?

As Collagen synthesis

B. Activation of hyaluronidase

C. Protcoglycan synthesis

D. Proteoglycan breakdown

E. Collagen breakdown

Correct Answer: A) Collagen synthesis

Explanation:

The symptoms described—multiple petechiae, bleeding gums, and low levels of vitamin C in urine—are characteristic of scurvy, which is caused by a deficiency of vitamin C. Vitamin C is essential for the enzyme prolyl hydroxylase, which is involved in the hydroxylation of proline residues during the synthesis of collagen. Without sufficient vitamin C, the collagen produced is defective, leading to weakened blood vessel walls and other connective tissue issues. This results in petechiae, bruising, and gum bleeding, as seen in this child.

Why Other Options Are Incorrect:

E) Collagen breakdown – Collagen breakdown is a normal process that happens during tissue remodeling, but scurvy is a disorder of collagen synthesis, not breakdown. The primary issue in scurvy is defective collagen formation due to a vitamin C deficiency.

B) Activation of hyaluronidase – Hyaluronidase is an enzyme involved in the breakdown of hyaluronic acid, a major component of the extracellular matrix. While hyaluronidase is important for tissue remodeling, it is not related to vitamin C deficiency or the symptoms of scurvy.

C) Proteoglycan synthesis – Proteoglycans are complex molecules that contribute to the structure of the extracellular matrix, but their synthesis is not directly affected by vitamin C deficiency. This is not the primary process impaired in scurvy.

D) Proteoglycan breakdown – Proteoglycan breakdown does not directly cause the symptoms seen in scurvy, which are more closely linked to the synthesis and stability of collagen.

38. Cellular mediators play various roles in the pathogenesis of inflammation. What exactly is the effect of histamine in the pathogenesis of the inflammatory process?

A. Inhibition of leukocyte proliferation

B. Inhibition of phagocytosis

C. Increase of venular permeability

D. Intensification of platelet aggregation

E. Stimulation of leukocyte proliferation

Correct Answer: C) Increase of venular permeability

Explanation:

Histamine is one of the key mediators in the inflammatory response. It is primarily released by mast cells and basophils in response to injury or infection. The main effect of histamine in inflammation is the increase in venular permeability, which allows for the movement of immune cells (like neutrophils and monocytes) and plasma proteins (including antibodies and clotting factors) from the bloodstream into the affected tissues. This leads to edema (swelling) and facilitates the immune response by allowing immune cells to reach the site of injury or infection more effectively.

Why Other Options Are Incorrect:

E) Stimulation of leukocyte proliferation – Histamine does not directly stimulate leukocyte proliferation. It plays a role in the movement of existing leukocytes to the site of inflammation rather than their production or proliferation.

A) Inhibition of leukocyte proliferation – Histamine does not inhibit leukocyte proliferation. In fact, histamine can contribute to the recruitment of leukocytes to the site of inflammation by increasing vascular permeability and facilitating their movement out of the bloodstream.

B) Inhibition of phagocytosis – Histamine does not inhibit phagocytosis. Inflammation actually promotes phagocytosis by immune cells (e.g., macrophages and neutrophils) to remove pathogens and debris. Histamine plays a role in increasing vascular permeability to allow phagocytes to reach the site.

D) Intensification of platelet aggregation – While histamine has a minor effect on platelet aggregation, it is not its primary function in inflammation. Platelet aggregation is more prominently affected by other mediators like thromboxane A2.

39. Examination of a patient at a clinical diagnostic laboratory detects that the acti-vity of LDH-1 isoenzyme is high in the patient’s blood serum. Such clinical and laboratory findings are characteristic of the pathology of the following internal organ:

A. Liver

B. Skeletal muscles

C. Pancreas

D. Heart

E. Kidneys

Correct Answer: D) Heart

Explanation:

Lactate dehydrogenase (LDH) is an enzyme involved in the conversion of lactate to pyruvate in various tissues. The enzyme exists in several isoenzymatic forms (LDH-1 to LDH-5), each predominantly found in different organs. The LDH-1 isoenzyme is primarily found in the heart muscle, and its elevated activity in the blood serum is typically associated with myocardial injury or acute myocardial infarction (heart attack). When the heart muscle is damaged, LDH-1 is released into the bloodstream, causing elevated levels of this isoenzyme.

Why Other Options Are Incorrect:

E) Kidneys – The kidneys express LDH-3 and LDH-4, not LDH-1. Kidney dysfunction can elevate overall LDH levels, but it is not specifically associated with LDH-1.

A) Liver – The liver primarily expresses LDH-5, not LDH-1. While liver dysfunction can lead to elevated LDH levels in general, LDH-1 is more specific to heart tissue.

B) Skeletal muscles – Skeletal muscles have a higher concentration of LDH-5, not LDH-1. Skeletal muscle damage typically results in elevated LDH-5 levels, not LDH-1.

C) Pancreas – The pancreas also contains a mix of LDH isoenzymes, but LDH-1 is not predominantly found here. Pancreatitis would typically involve other enzymes, like amylase and lipase, being elevated.

40. A patient developed signs of mucosal inflammation in the anterior and middle ethmoidal cells. Through what structure of the nasal cavity was the infection able to spread in this case?

A. Choanac

B. Superior nasal meatus

C. Middle nasal meatus

D. Common nasal meatus

E. Inferior nasal meatus

Correct Answer: C) Middle nasal meatus

Explanation:

The ethmoidal sinuses are located within the ethmoid bone and are divided into anterior and posterior ethmoidal cells. The middle ethmoidal cells drain into the middle nasal meatus, which is the passage within the nasal cavity situated below the middle turbinate. If an infection or mucosal inflammation occurs in the anterior or middle ethmoidal cells, it can spread to the middle nasal meatus, which provides the drainage pathway from these sinuses. This is a typical route for the spread of sinus infections, specifically from the ethmoidal cells.

Why Other Options Are Incorrect:

E) Inferior nasal meatus – The inferior nasal meatus drains the lacrimal sac and nasolacrimal duct but not the ethmoidal sinuses.

A) Choana – The choanae are the openings at the back of the nasal cavity leading into the nasopharynx. It does not serve as a drainage pathway for the ethmoidal sinuses.

B) Superior nasal meatus – The superior nasal meatus is located beneath the superior turbinate and drains the posterior ethmoidal cells and the sphenoid sinus, not the anterior or middle ethmoidal cells.

D) Common nasal meatus – There is no such anatomical structure as the “common nasal meatus.” The terms refer to the specific superior, middle, and inferior nasal meatuses, which are distinct anatomical spaces in the nasal cavity.

41. Bacteria have spread to the alveolar space of the acinus. During their interacti-on with the surfactant, the cells localized in the alveolar walls and on the alveolar surface became activated. Name these cells.

A. Endothelial cells

B. Alveolar macrophages

C. Type II alveolar cells

D. Clara cells

E. Type I alveolar cells

Correct Answer: B) Alveolar macrophages

Explanation:

The cells described in the question are alveolar macrophages, which are specialized immune cells located in the alveolar space. These macrophages play a critical role in the lung’s defense against infections, including bacterial invasions. When bacteria spread to the alveolar space, the alveolar macrophages interact with them and become activated. They help in the process of phagocytosis (engulfing and digesting the bacteria), as well as secreting pro-inflammatory cytokines and mediators that promote the immune response. These cells are particularly important in clearing pathogens and debris from the lungs.

Why Other Options Are Incorrect:

E) Type I alveolar cells – Type I alveolar cells are the main cells involved in the gas exchange process in the lungs, covering the majority of the alveolar surface. They do not directly participate in immune responses to infection.

A) Endothelial cells – Endothelial cells line the blood vessels and are involved in vascular function but are not involved in the defense against respiratory infections in the alveoli.

C) Type II alveolar cells – Type II alveolar cells are responsible for producing surfactant, a substance that reduces surface tension in the alveoli and prevents lung collapse. They do not primarily function in the immune response to bacterial infections.

D) Clara cells – Clara cells, or club cells, are found in the bronchioles and are involved in detoxification and the production of protective proteins. They are not primarily involved in immune responses within the alveoli.

42. Autopsy of the body of a 70-year-old man who died of cardiovascular fai-lure revealed chronic venous hyperemia of organs, left ventricular hypertrophy wi-th small-focal cardiosclerosis, and three- dimensional yellowish-white plaques in the tunica intima of the aorta with pasty masses in their center that are embedded into the wall. What pathological process is most likely observed in the aorta?

A. Lipoidosis

B. Liposclerosis

C. Arteriolosclerosis

D. Calcinosis

E. Atheromatosis

Correct Answer: E) Atheromatosis

Explanation:

The autopsy findings suggest atheromatosis, which is a term referring to the presence of atherosclerotic plaques in the arteries. The description of yellowish-white plaques in the tunica intima of the aorta, along with pasty masses in their center, is characteristic of atherosclerotic plaques. These plaques are made up of lipid deposits, fibrous tissue, and sometimes calcium, forming a thickened and narrowed artery wall. Over time, the plaques may undergo necrosis, leading to the formation of the pasty masses (which could be lipid-rich debris) embedded into the arterial wall.

Why Other Options Are Incorrect:

D) Calcinosis – Calcinosis refers to the deposition of calcium salts in tissues, which is sometimes seen in arteries as part of arteriosclerosis. However, the described plaques are primarily lipid-rich and do not primarily involve calcium deposits, making atheromatosis a more accurate diagnosis.

A) Lipoidosis – Lipoidosis refers to a condition in which lipids accumulate abnormally in tissues, but it is not typically associated with the formation of atherosclerotic plaques in the arteries. The description here is more consistent with atherosclerosis.

B) Liposclerosis – Liposclerosis involves the accumulation of lipids in the arterial walls but is not a widely recognized term in modern pathology. The process described in the question is more consistent with atherosclerosis rather than liposclerosis.

C) Arteriolosclerosis – Arteriolosclerosis is the thickening and hardening of small arteries (arterioles), often due to high blood pressure or diabetes. It does not describe the large plaques found in the aorta in this case.

43. The infectious disease doctor has di-agnosed the patient with acute enterocoli-tis syndrome with impaired digestion and absorption of breakdown products. What cells are damaged in the intestinal epitheli-um in this case, causing these disorders?

A. Cells with apical granules

B. Goblet cells

C. Columnar cells without the border

D. Endocrine cells

E. Columnar cells with the border

Correct Answer: E) Columnar cells with the border

Explanation:

The columnar cells with the brush border (also known as enterocytes) are primarily responsible for digestion and absorption in the intestine. These cells have microvilli on their apical surface, forming the brush border that is essential for nutrient absorption (carbohydrates, proteins, fats, and vitamins). In acute enterocolitis, the inflammation and damage to these enterocytes can lead to impaired digestion and malabsorption of the breakdown products, which explains the patient’s symptoms of impaired digestion and absorption.

Why Other Options Are Incorrect:

D) Endocrine cells – Enteroendocrine cells are involved in hormone secretion, such as gastrin, cholecystokinin, and secretin, which regulate digestion and motility. However, damage to these cells does not directly cause impaired digestion and absorption of nutrients..

A) Cells with apical granules – These cells are typically involved in secretion (like paneth cells, which have antimicrobial properties), but they are not primarily responsible for nutrient absorption. Damage to these cells would not directly cause the impaired digestion and absorption seen in enterocolitis.

B) Goblet cells – Goblet cells secrete mucus that protects the intestinal lining and aids in lubrication, but they do not play a central role in the absorption of nutrients. Although goblet cell dysfunction can contribute to the inflammatory process, they are not responsible for digestion and absorption.

C) Columnar cells without the border – These cells are not typically found in the normal intestinal epithelium. The columnar cells responsible for nutrient absorption are those with a brush border. Cells without a border do not perform absorption functions.

44. A person, provisionally diagnosed with acute pancreatitis, has been hospitalized. What enzyme activity must be measured in the patient’s blood and urine to confirm this diagnosis?

A. Cholinesterase

B. o-amylase

C. AST

D. Lactate dehydrogenase

E. ALT

Correct Answer: B) O-amylase

Explanation:

In the case of acute pancreatitis, the enzyme amylase (specifically α-amylase) is typically elevated in the blood and urine. Amylase is produced primarily by the pancreas and salivary glands, and when the pancreas becomes inflamed, such as in acute pancreatitis, it releases amylase into the bloodstream. The levels of α-amylase in the blood and urine are commonly measured to help confirm the diagnosis. While lipase is also a more specific marker for pancreatic inflammation, amylase remains a traditional and commonly used enzyme for diagnosis.

Why Other Options Are Incorrect:

E) ALT (Alanine aminotransferase) – ALT is another enzyme primarily used to evaluate liver function and is elevated in liver conditions like hepatitis or liver injury. It is not directly linked to the diagnosis of acute pancreatitis.

A) Cholinesterase – Cholinesterase is an enzyme involved in the breakdown of acetylcholine and is primarily used to assess liver function or exposure to certain toxins or organophosphates. It is not relevant for diagnosing pancreatitis.

C) AST (Aspartate aminotransferase) – AST is a liver enzyme that is elevated in liver diseases and muscle damage, but it is not specific for acute pancreatitis. It is typically used to assess liver function rather than pancreatic function.

D) Lactate dehydrogenase (LDH) – LDH is a nonspecific enzyme that is elevated in a variety of conditions, including liver disease, myocardial infarction, and hemolysis. It is not specific for acute pancreatitis and therefore is not the best choice for confirming this diagnosis.

45. A 22-year-old woman came to a dermatologist with complaints of a purulent rash on her face and back. Her medical record indicates a H. pylori infecti-on. Taking into account this concomitant pathology, the doctor prescribed her an antibacterial drug that will be effective both against the pathogens of soft tissue infections and against H. pylori. What anti-bacterial drug did the doctor prescribe?

A. Fluconazole

B. Clarithromycin

C. Isoniazid

D. Oseltamivir

E. Rifampicin

Correct Answer: B) Clarithromycin

Explanation:

Clarithromycin is an antibiotic that is effective against both Helicobacter pylori (H. pylori) and a wide range of bacterial pathogens that can cause soft tissue infections (such as Staphylococcus aureus and Streptococcus species). It is a macrolide antibiotic, commonly used in the treatment of H. pylori infection as part of a combination therapy (typically with amoxicillin or metronidazole) to eradicate the bacteria from the stomach. It also has activity against gram-positive cocci and some gram-negative organisms, making it suitable for treating purulent skin and soft tissue infections.

Why Other Options Are Incorrect:

E) Rifampicin – Rifampicin is an antibiotic used primarily to treat tuberculosis and gram-positive infections like staphylococcal infections. However, it is not typically used for H. pylori infections and is not a first-line choice for common soft tissue infections.

A) Fluconazole – Fluconazole is an antifungal drug used to treat fungal infections, such as candidiasis. It is not effective against bacterial infections or H. pylori.

C) Isoniazid – Isoniazid is an antitubercular drug used to treat tuberculosis. It is not effective against H. pylori or soft tissue bacterial infections.

D) Oseltamivir – Oseltamivir (Tamiflu) is an antiviral medication used to treat influenza (the flu). It is not effective against bacterial infections or H. pylori.

46. Autopsy of the body of a person, who died after an abdominal surgery, revealed numerous thrombi in the vei ns of the lesser pelvis. Clinical di-agnosis of thromboembolic syndrome was made. Where in the body can the thromboembolas be found in this case?

A. Pulmonary arteries

B. Left ventricle of the heart

C. Portal vein Dr Leg veins

E. Brain

Correct Answer: A) Pulmonary arteries

Explanation:

In thromboembolic syndrome, a thrombus (blood clot) that forms in the veins can travel through the bloodstream to other parts of the body. When the thrombus originates in the pelvic veins, it is most likely to embolize (travel) to the pulmonary arteries, leading to a pulmonary embolism. This occurs because the blood from the pelvic veins drains into the inferior vena cava, which empties into the right atrium of the heart. From there, the clot can move to the right ventricle, then into the pulmonary arteries, where it may cause a blockage in the lungs.

Why Other Options Are Incorrect:

E) Brain – A thromboembolism can affect the brain (leading to ischemic stroke), but this is much more likely if the thrombus originates from the heart (such as in atrial fibrillation) or from the arteries (such as in carotid artery disease). Pelvic vein thrombi typically travel to the lungs, not the brain.

B) Left ventricle of the heart – The left ventricle typically does not receive blood from the veins (such as the pelvic veins). The left ventricle pumps oxygenated blood to the body, so a clot from the pelvic veins would not enter the left ventricle.

C) Portal vein – The portal vein carries blood from the gastrointestinal tract and spleen to the liver. Although thrombi can form in the portal vein (leading to portal vein thrombosis), thromboembolic events originating from the pelvic veins are more likely to travel to the lungs, not the portal vein.

D) Leg veins – While thrombi can form in the deep veins of the legs (leading to deep vein thrombosis or DVT), in this case, the thrombi originated in the pelvic veins, not the leg veins. The pelvic veins can lead to thromboembolic events traveling to the lungs.

47. A woman periodically has arterial hypertension attacks, accompanied by headaches, palpitations, excessive sweati-ng, sharp pain in the epigastric regi-on, and clevated glucose levels in blood plasma. High levels of metanephrines were detected in blood plasma and urine. What neoplastic disorder is most likely to be associated with such symptoms?

A. Ovarian tumor

B. Pheochromocytoma

C. Parathyroid adenoma

D. Thyroid adenoma

E. Stomach cancer

Correct Answer: B) Pheochromocytoma

Explanation:

The symptoms described—periodic attacks of arterial hypertension, headaches, palpitations, excessive sweating, sharp epigastric pain, and elevated blood glucose levels—are suggestive of pheochromocytoma, a neuroendocrine tumor typically located in the adrenal medulla. Pheochromocytomas secrete excessive amounts of catecholamines (such as adrenaline and noradrenaline), which cause the symptoms of hypertension, tachycardia, and sweating. The detection of high levels of metanephrines (metabolites of catecholamines) in the blood and urine is a key diagnostic marker for this condition.

Why Other Options Are Incorrect:

E) Stomach cancer – Stomach cancer can lead to symptoms like abdominal pain, nausea, vomiting, and weight loss, but it does not cause hypertension, palpitations, sweating, or elevated metanephrines.

A) Ovarian tumor – While ovarian tumors can sometimes cause hormonal imbalances, they are not typically associated with the classic symptoms of hypertension, palpitations, and sweating. Ovarian tumors may cause elevated estrogen or androgens, but they are not linked to metanephrine elevation in the blood and urine.

C) Parathyroid adenoma – Parathyroid adenomas are typically associated with hyperparathyroidism, which leads to elevated calcium levels. This condition does not explain the symptoms of hypertension, headaches, sweating, and high glucose levels, nor does it cause elevated metanephrines.

D) Thyroid adenoma – A thyroid adenoma (benign tumor of the thyroid) might affect thyroid hormone levels, potentially causing symptoms like weight changes, heat intolerance, and palpitations, but it is not typically associated with the symptoms described here, and it would not cause elevated metanephrine levels.

48. In a 40-year-old man, testicular inflammation was complicated by hydrocele testis. A surgery is necessary. What testicular tunic would be the last to be dissected by the surgeon during the operation?

A. Tunica dartos

B. Parietal layer of the tunica vaginalis of the testicle

C. Internal spermatic fascia

D. External spermatic fascia

E. Cremaster muscle

Correct Answer: B) Parietal layer of the tunica vaginalis of the testicle

Explanation:

The tunica vaginalis is a serous membrane that surrounds the testicle. It consists of two layers: the parietal layer (outer layer) and the visceral layer (inner layer, which is closely adherent to the testicle). When performing a surgery to correct a hydrocele testis (fluid accumulation between the layers of the tunica vaginalis), the surgeon will typically dissect through the tunica vaginalis, starting from the more superficial layers and proceeding deeper. The parietal layer of the tunica vaginalis would be the last to be dissected, as it covers the testicle externally, and the surgeon would first access and handle the other tunics and structures before exposing this layer.

Why Other Options Are Incorrect:

E) Cremaster muscle – The cremaster muscle lies between the internal spermatic fascia and the tunica vaginalis. It would also be dissected prior to the parietal layer.

A) Tunica dartos – The tunica dartos is a smooth muscle layer under the skin of the scrotum, and it is one of the outermost layers. It would be dissected before the parietal layer of the tunica vaginalis.

C) Internal spermatic fascia – The internal spermatic fascia is deeper than the external layers but still superficial to the tunica vaginalis. It would be dissected before reaching the parietal layer.

D) External spermatic fascia – The external spermatic fascia is the most superficial layer and would be the first to be dissected, so it is not the last.

49. Microscopy of a woman’s vaginal swab detects cells with cytoplasmic inclusions. The doctor has provisionally diagnosed the patient with chlamydiosis. What test should be used to detect antibodies and confirm this diagnosis?

A. Precipitation reaction

B. Agglutination reaction

C. Enzyme immunoassay

D. Vidal’s reaction

E. Reverse indirect hemagglutination

Correct Answer: C) Enzyme immunoassay

Explanation:
+

Chlamydia infection (chlamydiosis) is commonly diagnosed using methods that detect either the bacteria directly or the immune response to the infection. An enzyme immunoassay (EIA) is a highly effective method for detecting antibodies to Chlamydia trachomatis in blood samples. This assay uses an enzyme-linked antibody to detect the presence of specific IgM or IgG antibodies, which are produced in response to the infection. EIA is widely used due to its high sensitivity and specificity for detecting chlamydial infections.

Why Other Options Are Incorrect:

E) Reverse indirect hemagglutination – Reverse indirect hemagglutination is a test used for detecting antibodies to various infections, but it is not the standard method for diagnosing Chlamydia. An enzyme immunoassay is more specific and commonly used.

A) Precipitation reaction – A precipitation reaction involves the formation of visible precipitates when antibodies and antigens react. While useful in some contexts, it is not commonly used for detecting Chlamydia antibodies.

B) Agglutination reaction – Agglutination tests can detect antibodies to various pathogens by observing the clumping of particles (such as latex beads) coated with antigen. However, this test is not the primary diagnostic method for Chlamydia antibodies.

D) Vidal’s reaction – Vidal’s reaction is used for diagnosing typhoid fever (caused by Salmonella typhi) by detecting O and H antigens. It is not relevant for Chlamydia infection.

50. The act of chewing is impaired in a patient, because a pathological process has affected the structures that form the afferent pathway of the corresponding reflex arc. What nerve is damaged in this patient?

A. N. hypoglossus

B. N. vagus

C. N. trigeminus

D. N. glossopharyngeus and n. vagus

E. N. glossopharyngeus

Correct Answer: C) N. trigeminus

Explanation:

The act of chewing (mastication) is primarily controlled by the mandibular branch of the trigeminal nerve (V3), which innervates the muscles of mastication (such as the masseter, temporalis, and pterygoid muscles). The trigeminal nerve (CN V) is responsible for the afferent sensory pathway in the reflex arc related to chewing, transmitting sensory information from the muscles and oral cavity. If this nerve is damaged, it can impair the ability to chew effectively.

Why Other Options Are Incorrect:

E) N. glossopharyngeus – The glossopharyngeal nerve is responsible for taste and sensation from the posterior tongue and part of the pharynx but is not involved in the reflex arc of chewing.

A) N. hypoglossus – The hypoglossal nerve (CN XII) primarily innervates the muscles of the tongue. While it plays a role in speech and swallowing, it is not directly involved in the sensory or motor control of chewing.

B) N. vagus – The vagus nerve (CN X) is involved in a variety of functions, including parasympathetic control of the heart and gut, but it is not a major player in the afferent pathway for chewing.

D) N. glossopharyngeus and n. vagus – The glossopharyngeal nerve (CN IX) is involved in taste and sensation from the posterior third of the tongue and the vagus nerve (CN X) is involved in parasympathetic functions, but neither of these is directly involved in the afferent pathway of the chewing reflex.

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