Question From (1 To 50 )
1. A man was hospitalized with an injury to the psoas major muscle. He has lost the ability to extend the lower leg in the knee joint. What nerve is damaged in this case?
A. Femoral
B. Genitofemoral
C. Ilioinguinal
D. Obturator
E. Iliohypogastric
Answer: Femoral
Explanation
| The correct answer is A. Femoral nerve. The femoral nerve is a nerve that arises from the lumbar plexus (L2-L4) and innervates the muscles of the anterior compartment of the thigh, including the quadriceps muscle. It also innervates the skin on the anterior and medial aspects of the thigh and the medial aspect of the leg and foot. The psoas major muscle is a hip flexor that originates from the lumbar vertebrae and inserts onto the lesser trochanter of the femur. The femoral nerve supplies the psoas major muscle, and damage to this nerve can result in weakness or paralysis of the hip flexors, as well as weakness or paralysis of the knee extensors, which can result in an inability to extend the lower leg in the knee joint. Therefore, in this case, the loss of ability to extend the lower leg in the knee joint suggests that the femoral nerve is damaged. |
2. A 64-year-old woman has suffered a pathological fracture of the humerus. Biopsy detects atypical plasma cells. X-ray shows tumor-like formations at the fracture site. What disease is likely in this case?
A. Adenocarcinoma metastasis
B. Chondrosarcoma
C. Myeloma disease
D. Fibrous dysplasia of bone
E. Chronic osteomyelitis
Answer:Myeloma disease
Explanation
| The correct answer is C. Myeloma disease. Multiple myeloma is a type of cancer that arises from plasma cells, which are a type of white blood cell that produce antibodies. It is characterized by the proliferation of abnormal plasma cells in the bone marrow, which can lead to the formation of tumors in the bone and the destruction of normal bone tissue. In this case, the presence of atypical plasma cells on biopsy and tumor-like formations at the fracture site suggest that the patient may have multiple myeloma. Pathological fractures, which are fractures that occur through weakened bone due to an underlying disease process, are a common complication of multiple myeloma. Adenocarcinoma metastasis and chondrosarcoma can also cause pathological fractures, but the presence of atypical plasma cells on biopsy is more suggestive of multiple myeloma. Fibrous dysplasia of bone and chronic osteomyelitis are not typically associated with the presence of atypical plasma cells and are therefore less likely diagnoses in this case. |
3. A person entered a room with increased levels of carbon dioxide in the air. How will the breathing of this person change?
A. Respiration rate will decrease
B. Respiration depth will increase
C. Respiration rate will increase
D. Respiration depth will decrease
E. Respiration rate and depth will increase
Answer: Respiration rate and depth will increase
Explanation
| The correct answer is C. Respiration rate will increase. When a person enters a room with increased levels of carbon dioxide in the air, the concentration of carbon dioxide in their bloodstream will increase. High levels of carbon dioxide in the blood can stimulate the respiratory center in the brainstem, which can lead to an increase in the rate and depth of breathing. The respiratory center responds to the increase in carbon dioxide by sending signals to the respiratory muscles, which include the diaphragm and the intercostal muscles. The diaphragm contracts more forcefully, and the intercostal muscles work harder to expand the chest, resulting in an increase in the rate and depth of breathing. Therefore, in this case, the breathing of the person will change by an increase in the respiration rate. The increase in respiration rate is an adaptive response to maintain the proper balance of oxygen and carbon dioxide in the bloodstream |
4. A 10-year-old child underwent a Mantoux test (with tuberculin). After 48 hours, a papule up to 8 mm in diameter appeared at the site of tuberculin injection. What type of hypersensitivity reaction developed after administration of tuberculin?
A. Serum sickness
B. Atopic reaction
C. Arthus reaction
D. B. Type II hypersensitivity reaction
E. Type IV hypersensitivity reaction
Answer: Type IV hypersensitivity reaction
Explanation
| The correct answer is E. Type IV hypersensitivity reaction. The Mantoux test is a type of tuberculosis (TB) skin test that involves the injection of purified protein derivative (PPD) tuberculin into the skin of the forearm. The test is used to detect whether a person has been exposed to the bacteria that cause TB. After the injection of tuberculin, a delayed-type hypersensitivity reaction occurs in individuals who have been exposed to TB in the past. This reaction is mediated by T cells and is classified as a type IV hypersensitivity reaction. In this case, the development of a papule at the site of injection after 48 hours suggests that the child has developed a positive reaction to the Mantoux test. The size of the papule (up to 8 mm) is considered a moderate reaction. A positive reaction indicates exposure to TB in the past, but it does not necessarily mean that the person has active TB disease. Serum sickness (A), atopic reactions (B), and type II hypersensitivity reactions (D) are not associated with the Mantoux test. The Arthus reaction (C) is a type III hypersensitivity reaction that occurs when there is an excess of antigen in the body, leading to immune complex formation and tissue damage. This reaction is not typically associated with the Mantoux test. |
5. A patient was hospitalized with the provisional diagnosis of typhoid fever. The disease onset was three days ago. The temperature is 39°C. What method of laboratory diagnostics must be used to confirm this diagnosis?
A. Serology
B. Obtaining a coproculture
C. Obtaining a biliculture
D. Obtaining a blood culture
E. Obtaining a urinoculture
Answer: Obtaining a blood culture
Explanation
| The correct answer is D. Obtaining a blood culture. Typhoid fever is a bacterial infection caused by Salmonella typhi, which is transmitted through contaminated food and water. The diagnosis of typhoid fever is based on the presence of clinical symptoms such as fever, abdominal pain, and diarrhea, as well as the results of laboratory tests. Blood culture is the most reliable method for diagnosing typhoid fever. The blood culture should be obtained during the first week of illness when the bacterial load is highest. The culture should be incubated for 7-10 days to increase the sensitivity of the test. Serology (A) can also be used to diagnose typhoid fever, but it is less reliable than blood culture, particularly during the early stages of the disease. Obtaining a coproculture (B) or a biliculture (C) can be useful for detecting other bacterial pathogens that cause gastroenteritis, but they are not specific for typhoid fever. Obtaining a urinoculture (E) is not useful for diagnosing typhoid fever as S. typhi is not typically found in the urine of infected patients |
6. In adipocytes of adipose tissue, the pentose-phosphate pathway has the nature of a cycle. What is the main function of this cycle in adipose tissue?
A. Neutralization of xenobiotics
B. Production of ribose phosphates
C. Generation of NADPH2
D. Oxidation of glucose to end products
E. Energy generation
Answer: Generation of NADPH2
Explanation
| The correct answer is C. Generation of NADPH2. The pentose-phosphate pathway, also known as the hexose monophosphate shunt, is a metabolic pathway that occurs in the cytoplasm of cells. The pathway generates NADPH2 and ribose-5-phosphate, which are important for various biosynthetic processes in the cell. In adipocytes of adipose tissue, the pentose-phosphate pathway has the nature of a cycle. The main function of this cycle in adipose tissue is to generate NADPH2, which is required for the biosynthesis of fatty acids and other lipids. NADPH2 is also important for the maintenance of the redox balance in the cell and for the detoxification of reactive oxygen species (ROS). Therefore, in adipose tissue, the pentose-phosphate pathway is essential for the synthesis of fatty acids and other lipids, which are stored as triglycerides in adipocytes. The pathway also plays a crucial role in protecting the cell from oxidative stress by generating NADPH2, which is an antioxidant cofactor. |
7. A man was hospitalized with abdominal pain and profuse salivation, sweating, and tears. Examination detects miosis. The day before, he was treating plants with a solution of an insecticidal substance without wearing personal protective equipment. The substance that has caused the poisoning in this case belongs to:
A. Copper salts
B. Organochlorine compounds
C. Nitrates
D. Nicotinic cholinomimetics
E. Anticholinesterase agents
Answer:. Anticholinesterase agents
Explanation
| The correct answer is E. Anticholinesterase agents. The symptoms described in the case (abdominal pain, profuse salivation, sweating, tears, and miosis) are consistent with cholinergic toxicity, which can occur as a result of exposure to certain insecticides and nerve agents. These compounds act by inhibiting the activity of the enzyme acetylcholinesterase, which breaks down the neurotransmitter acetylcholine in the nervous system. Anticholinesterase agents, such as organophosphates and carbamates, are widely used as insecticides and can cause cholinergic toxicity in humans if exposure occurs through inhalation or skin contact. The symptoms of cholinergic toxicity can range from mild to severe and can include gastrointestinal symptoms, respiratory distress, muscle weakness, and seizures. In this case, the history of exposure to an insecticidal substance without wearing personal protective equipment, along with the symptoms of cholinergic toxicity, suggests that the man has been poisoned with an anticholinesterase agent. Copper salts (A) and nitrates (C) are not typically associated with the symptoms described in the case. Organochlorine compounds (B) are no longer widely used as insecticides due to their persistence in the environment and toxicity to non-target species. Nicotinic cholinomimetics (D) are not commonly used as insecticides and are not typically associated with the symptoms described. |
8. After the treatment with antitubercular agent, 40-year-old woman developed optic neuritis, memory impairment, and seizures. What medicine was she taking?
A. Para-aminosalicylic acid
B. Isoniazid
C. Thioacetazone
D. Kanamycin
E. Rifampicin
Answer: Isoniazid
Explanation
| The correct answer is B. Isoniazid. Isoniazid is a first-line antitubercular agent used to treat tuberculosis (TB). It works by inhibiting the synthesis of mycolic acids, which are essential components of the cell wall of the TB bacterium. One of the potential side effects of isoniazid is peripheral neuropathy, which can manifest as optic neuritis (inflammation of the optic nerve), memory impairment, and seizures. This side effect is thought to be related to the depletion of vitamin B6 (pyridoxine), which is a cofactor required for the metabolism of isoniazid. Para-aminosalicylic acid (A) and thioacetazone (C) are second-line antitubercular agents that are not commonly used due to their potential side effects. Kanamycin (D) is an aminoglycoside antibiotic that can be used as a second-line agent for the treatment of multidrug-resistant TB, but it is not typically associated with the side effects described in the case. Rifampicin (E) is a first-line antitubercular agent that is not typically associated with optic neuritis, memory impairment, or seizures. |
9. What internal organ plays the largest role in humoral regulation of erythropoiesis?
A. Kidneys
B. Pancreas
C. Lungs
D. B. Liver
E. Gastrointestinal tract
Answer: Kidneys
Explanation
| The correct answer is A. Kidneys. Erythropoiesis is the process of red blood cell production in the bone marrow. The regulation of erythropoiesis is controlled by the hormone erythropoietin (EPO), which is produced primarily in the kidneys in response to hypoxia (low oxygen levels) in the body. When the oxygen-carrying capacity of the blood decreases, as in conditions such as anemia or high-altitude exposure, the kidneys respond by increasing the production and secretion of EPO. EPO stimulates the production of red blood cells in the bone marrow, which increases the oxygen-carrying capacity of the blood and helps to restore normal oxygen levels in the body. Therefore, the kidneys play the largest role in the humoral regulation of erythropoiesis. Other organs, such as the liver and the spleen, can also produce small amounts of EPO, but the kidneys are the primary source of this hormone. The pancreas (B), lungs (C), and gastrointestinal tract (E) are not directly involved in the regulation of erythropoiesis. The liver (D) is involved in the breakdown of old red blood cells and the recycling of iron, but it does not play a major role in the humoral regulation of erythropoiesis. |
10. A 47-year-old woman has interphalangeal and metacarpophalangeal joints that can be easily dislocated or subluxated and a characteristic deviation of the fingers that resembles “walrus flippers”. Microscopy reveals proliferation of synovial villi, cartilage destruction, and pannus formation. What disease causes these pathological changes?
A. Ankylosing spondylitis (Bekhterev’s disease)
B. Osteoarthrosis
C. Rheumatoid arthritis
D. Systemic lupus erythematosus
E. Rheumatic arthritis
Answer: Rheumatoid arthritis
Explanation
| The correct answer is C. Rheumatoid arthritis. Rheumatoid arthritis (RA) is a chronic autoimmune disorder characterized by inflammation of the synovial membrane, which lines the joints and tendons. The inflammation leads to the proliferation of synovial villi, cartilage destruction, and pannus formation, which can result in joint deformities and subluxations. RA commonly affects the small joints of the hands and feet, including the interphalangeal and metacarpophalangeal joints. The characteristic deviation of the fingers, which resembles “walrus flippers,” is caused by the subluxation of the joints. Ankylosing spondylitis (A) is a chronic inflammatory disorder that primarily affects the spine and sacroiliac joints. Osteoarthrosis (B) is a degenerative joint disease characterized by the breakdown of articular cartilage in the joints. Systemic lupus erythematosus (D) is a multisystem autoimmune disorder that can affect the skin, joints, kidneys, and other organs. Rheumatic arthritis (E) is not a recognized medical condition and may be a misspelling of rheumatoid arthritis. |
11. Problems with the processes of lipid breakdown in small intestine are caused by disturbed lipase activity. What factor activates lipase?
A. Pepsin
B. Bile acids
C. Enterokinase
D. Hydrochloric acid inhibitors
E. Na+ salts
Answer: Bile acids
Explanation
| The correct answer is B. Bile acids. Lipase is an enzyme that breaks down triglycerides into fatty acids and glycerol in the small intestine. The activity of lipase is activated by bile acids, which are produced by the liver and stored in the gallbladder. Bile acids are released into the small intestine in response to the presence of dietary fats. Bile acids emulsify the dietary fats, breaking them down into smaller droplets, which increases the surface area available for lipase to act on. Bile acids also activate lipase by binding to it and inducing a conformational change that enhances its activity. Pepsin (A) is a digestive enzyme produced in the stomach that breaks down proteins into smaller peptides. Enterokinase (C) is an enzyme produced by the duodenum that activates trypsinogen, a precursor of the enzyme trypsin. Hydrochloric acid inhibitors (D) are drugs that reduce the production of stomach acid and are used to treat conditions such as gastroesophageal reflux disease (GERD) and peptic ulcers. Na+ salts (E) are not directly involved in the activation of lipase or the process of lipid breakdown in the small intestine. |
12. A 35-year-old has been man hospitalized with complaints of a runny nose and headache that last for 5 days already. After examination, he was diagnosed with maxillary sinusitis (inflammation of the maxillary sinus). Through what nasal passage did the infection reach this sinus?
A. Inferior nasal meatus
B. Nasopharyngeal meatus
C. Middle nasal meatus
D. Common nasal meatus
E. Superior nasal meatus
Answer: Middle nasal meatus
Explanation
| The correct answer is C. Middle nasal meatus. The nasal cavity is divided into several passages called meatuses. The maxillary sinus is located in the cheekbones and is one of the four pairs of paranasal sinuses that are connected to the nasal cavity. The maxillary sinus drains into the middle nasal meatus through the maxillary sinus ostium. Inflammation of the maxillary sinus, known as maxillary sinusitis, can occur as a result of infection, allergies, or other factors. The infection can spread to the maxillary sinus through the middle nasal meatus, which is located below the middle concha (a bony structure in the nasal cavity). The inferior nasal meatus (A) is the lowest meatus in the nasal cavity and is located below the inferior concha. The nasopharyngeal meatus (B) is not a recognized anatomical structure. The common nasal meatus (D) and superior nasal meatus (E) are also not recognized anatomical structures. |
13. A 47-year-old woman complains of protracted vomiting. She has lost a large amount of gastric juice. What acid-base imbalance can be suspected in this case?
A. Non-gaseous acidosis
B. Gaseous acidosis
C. Gaseous alkalosis
D. Metabolic acidosis
E. Non-gaseous alkalosis
Answer: Non-gaseous alkalosis
Explanation
| The correct answer is E. Non-gaseous alkalosis. Vomiting, particularly if it is protracted and severe, can result in the loss of a large amount of gastric juice, which contains hydrochloric acid (HCl). The loss of HCl can lead to a decrease in the concentration of hydrogen ions (H+) in the body, which can cause alkalosis. Alkalosis is a condition in which the pH of the blood is elevated above the normal range of 7.35-7.45. Alkalosis can be classified as respiratory or metabolic, depending on the underlying cause. Respiratory alkalosis results from hyperventilation, which leads to a decrease in the partial pressure of carbon dioxide (CO2) in the blood. Metabolic alkalosis results from a primary increase in the concentration of bicarbonate (HCO3-) in the blood. In this case, the vomiting has resulted in the loss of HCl from the body, which would lead to a decrease in the concentration of H+ ions and an increase in the concentration of bicarbonate ions. This would result in an increase in the pH of the blood, indicating non-gaseous alkalosis. Non-gaseous acidosis (A), gaseous acidosis (B), gaseous alkalosis (C), and metabolic acidosis (D) are not likely to occur as a result of vomiting-induced loss of gastric juice. Non-gaseous acidosis and metabolic acidosis, in particular, are caused by an excess of acid in the body, which is not the case in this scenario. |
14. A surgeon performs an operation on the sigmoid colon, stopping the bleeding from aa. Sigmoideae. They are the branches of the following artery:
A. A. mesenterica inferion
B. A. mesenterica superior
C. Truncus coeliacus
D. A. colica sinistra
E. A. colica dextra
Answer: A. mesenterica inferion
Explanation
| The correct answer is A. A. mesenterica inferior. The sigmoid colon is part of the large intestine and is located in the left lower quadrant of the abdomen. The sigmoid colon receives its blood supply from the sigmoid arteries, which are branches of the inferior mesenteric artery. The inferior mesenteric artery (IMA) is a branch of the abdominal aorta and supplies blood to the distal part of the large intestine, including the left colic flexure, descending colon, sigmoid colon, and rectum. The sigmoid arteries (aa. sigmoideae) are branches of the IMA that supply blood to the sigmoid colon. The superior mesenteric artery (SMA) is another major branch of the abdominal aorta that supplies blood to the small intestine, cecum, ascending colon, and transverse colon. The celiac trunk (truncus coeliacus) is a major branch of the abdominal aorta that supplies blood to the foregut, including the stomach, duodenum, liver, pancreas, and spleen. The left colic artery (a. colica sinistra) is a branch of the inferior mesenteric artery that supplies blood to the descending colon. The right colic artery (a. colica dextra) is a branch of the superior mesenteric artery that supplies blood to the ascending colon. |
15. A patient was hospitalized with complaints of periodic attacks of palpitations that pass on their own. ECG detected an episode of contractions with the rate of 200/min. and the following characteristics:regular rhythm, no P wave, unchanged QRS complex, deformed T wave. What type of arrhythmia is it?
A. Ventricular extrasystole
B. Paroxysmal supraventricular
tachycardia
C. Atrial extrasystole
D. First-degree AV block
E. Complete AV block
Answer: Paroxysmal supraventricular
tachycardia
Explanation
| The correct answer is B. Paroxysmal supraventricular tachycardia (PSVT). PSVT is a type of arrhythmia that occurs when there is an abnormal electrical pathway in the heart that causes the heart to beat faster than normal. The rapid heart rate in PSVT is usually between 150 and 220 beats per minute and starts and stops suddenly, often without any obvious trigger. The ECG findings in PSVT typically show a regular rhythm with no P wave, as the atria are not contributing to the rapid heart rate. The QRS complex is usually unchanged, as the ventricles are still being activated normally. The T wave may be deformed due to the rapid heart rate and altered repolarization. Ventricular extrasystole (A) is a premature contraction of the ventricles that occurs before the next expected normal heartbeat. Atrial extrasystole (C) is a premature contraction of the atria that occurs before the next expected normal heartbeat. First-degree AV block (D) is a delay in the conduction of electrical impulses from the atria to the ventricles. Complete AV block (E) is a complete blockage of the conduction of electrical impulses from the atria to the ventricles, resulting in an independent pacemaker rhythm in the ventricles. |
16. A patient diagnosed with chronic glomerulonephritis developed persistent arterial hypertension. What group of drugs should be used for the treatment of this patient?
A. Calcium antagonists
B. Myotropic antispasmodics
C. Angiotensin-converting enzyme inhibitors
D. α-blockers
E. Ganglionic blockers
Answer:Angiotensin-converting enzyme inhibitors
Explanation
| The correct answer is C. Angiotensin-converting enzyme (ACE) inhibitors. Chronic glomerulonephritis is a type of kidney disease that can result in kidney damage and impaired kidney function, which can lead to hypertension (high blood pressure). Hypertension can further damage the kidneys, leading to a vicious cycle of kidney damage and hypertension. ACE inhibitors are a group of drugs commonly used to treat hypertension in patients with chronic glomerulonephritis. ACE inhibitors work by blocking the conversion of angiotensin I to angiotensin II, a potent vasoconstrictor that can increase blood pressure. By blocking the production of angiotensin II, ACE inhibitors can lower blood pressure and reduce the workload on the kidneys, potentially slowing the progression of kidney damage. Calcium antagonists (A) are another group of drugs used to treat hypertension by blocking the influx of calcium into smooth muscle cells, causing vasodilation and lowering blood pressure. Myotropic antispasmodics (B) are drugs that act on smooth muscle cells to relax them, but they are not commonly used to treat hypertension. α-blockers (D) are drugs that block the action of α-adrenergic receptors, leading to vasodilation and lowering of blood pressure, but they are not typically used as first-line therapy for hypertension in patients with chronic glomerulonephritis. Ganglionic blockers (E) are drugs that block the transmission of impulses in the autonomic ganglia, leading to a decrease in sympathetic tone and lowering of blood pressure, but they are not commonly used to treat hypertension due to their side effects. |
17. A repeated Widal agglutination test shows an increase from 1:100 to 1:400 in the titers of antibodies to S. typhi O-antigens in the patient’s serum. How can the obtained results be interpreted?
A. The patient is a chronic carrier of typhoid microbes
B. The patient was previously vaccinated against typhoid fever
C. The patient has typhoid fever
D. The patient has a past history of typhoid fever
E. The patient is an acute carrier of typhoid microbes
Answer: The patient has typhoid fever
Explanation
| The correct answer is C. The patient has typhoid fever. The Widal agglutination test is a serological test used to diagnose typhoid fever, a systemic infection caused by the bacterium Salmonella enterica serovar Typhi. The test measures the levels of antibodies (IgM and IgG) against the O and H antigens of S. Typhi in the patient’s serum. In a patient with typhoid fever, the Widal test typically shows a rise in the titer of antibodies against the O and H antigens of S. Typhi after 7-14 days of infection. A four-fold or greater increase in the titer of antibodies against the O antigens of S. Typhi is considered to be a significant rise and is indicative of a recent infection. In this case, the repeated Widal test showed a four-fold increase in the titer of antibodies against the S. Typhi O-antigens from 1:100 to 1:400. This indicates that the patient has typhoid fever, as there has been a significant rise in the titer of antibodies against the O-antigens of S. Typhi. Chronic carriers of typhoid microbes (A) are individuals who continue to shed S. Typhi in their feces or urine for more than a year after an acute infection. Vaccination against typhoid fever (B) can result in the production of antibodies against the O and H antigens of S. Typhi, but the titers of antibodies do not typically increase to the levels seen in acute infections. A past history of typhoid fever (D) may result in the presence of antibodies against the O and H antigens of S. Typhi, but a four-fold increase in the titer of antibodies is indicative of a recent infection. Acute carriers of typhoid microbes (E) are individuals who are currently infected with S. Typhi and are shedding the bacteria in their feces or urine. |
18. Autopsy of the body of a man who died of ethylene glycol poisoning revealed slightly enlarged edematous kidneys with the capsule that could be very easily removed. The cortical substance is pale gray and wide. The medullary substance is dark red. What kidney pathology developed in the patient?
A. Acute glomerulonephritis
B. Acute tubulointerstitial nephritis
C. Necrotic nephrosis
D. Acute pyelonephritis
E. Lipoid nephrosis X
Answer:Necrotic nephrosis
Explanation
| The correct answer is C. Necrotic nephrosis. Ethylene glycol is a toxic substance that is commonly found in antifreeze, windshield washer fluid, and other industrial solvents. It can cause toxic effects on the kidneys, leading to acute kidney injury (AKI) and kidney failure. The autopsy findings of slightly enlarged edematous kidneys with a pale gray cortical substance and a dark red medullary substance are consistent with necrotic nephrosis, also known as acute tubular necrosis (ATN). ATN is a type of AKI that is characterized by the death of tubular cells in the kidneys, resulting in the formation of casts and obstruction of the tubules. This can lead to a decrease in kidney function and the development of kidney failure. The easy removal of the kidney capsule is also a characteristic feature of ATN, as the tubular cells are sloughed off and the interstitial tissue becomes edematous, leading to a separation of the capsule from the underlying tissue. Acute glomerulonephritis (A) is a type of kidney disease that is characterized by inflammation of the glomeruli in the kidneys, leading to hematuria, proteinuria, and decreased kidney function. Acute tubulointerstitial nephritis (B) is a type of kidney disease that is characterized by inflammation of the tubules and interstitium in the kidneys, leading to acute kidney injury and renal failure. Acute pyelonephritis (D) is a type of kidney infection that is characterized by inflammation of the renal pelvis and parenchyma, leading to fever, flank pain, and urinary symptoms. Lipoid nephrosis (E), also known as minimal change disease, is a type of kidney disease that is characterized by proteinuria and edema, but the renal pathology is not consistent with the findings described in this case. |
19. A doctor suspects diphtheria in a patient. Bacterioscopy of a throat swab detected rod-shaped bacteria with volutin granules. What etiotropic drug would be the drug of choice in this case?
A. K.Bacteriophage
B. Eubiotic
C. Antidiphtheric antitoxic serum
D. Interferon
E. Diphtheria toxoid
Answer: Antidiphtheric antitoxic serum
Explanation
| The correct answer is C. Antidiphtheric antitoxic serum. Diphtheria is a serious bacterial infection caused by the bacterium Corynebacterium diphtheriae. The bacteria produce a toxin that can cause severe damage to the respiratory tract, heart, and nervous system. Bacterioscopy of a throat swab that detects rod-shaped bacteria with volutin granules is consistent with the diagnosis of diphtheria. The definitive diagnosis of diphtheria is usually confirmed by culture and identification of the bacterium. The treatment of diphtheria involves the administration of antitoxin to neutralize the toxin produced by the bacteria. Antidiphtheric antitoxic serum is the drug of choice for the treatment of diphtheria. It contains antibodies that can neutralize the diphtheria toxin and prevent further damage to the body. K.Bacteriophage (A) is a type of virus that can infect and kill bacteria, but it is not typically used to treat diphtheria. Eubiotic (B) is a general term that refers to drugs or substances that promote the growth of beneficial bacteria in the body, but it is not specific to the treatment of diphtheria. Interferon (D) is a type of protein that is produced by the body’s immune system in response to viral infections and other pathogens, but it is not typically used to treat bacterial infections such as diphtheria. Diphtheria toxoid (E) is a vaccine that can be used to prevent diphtheria, but it is not effective in treating an active infection. |
20. Vitamin A deficiency causes impaired twilight vision. What cells have this receptor function?
A. Neurosensory rod cells
B. Neurosensory cone cells
C. Retinal horizontal cells
D. Ganglionic neurons
E. Bipolar neurons
Answer: Neurosensory rod cells
Explanation
| The correct answer is A. Neurosensory rod cells. Vitamin A is an essential nutrient that is required for the proper functioning of the visual system. One of the most well-known symptoms of vitamin A deficiency is impaired twilight vision, also known as night blindness. The retina of the eye contains two types of photoreceptor cells: rods and cones. Rods are responsible for vision in low-light conditions, such as at night or in dimly lit environments, whereas cones are responsible for color vision and vision in bright light. The cells in the retina that are responsible for twilight vision are the neurosensory rod cells. These cells contain a photopigment called rhodopsin, which is composed of a protein called opsin and a molecule of vitamin A called retinal. When light enters the eye and is absorbed by rhodopsin, retinal undergoes a conformational change that triggers a series of chemical reactions that ultimately lead to the generation of an electrical signal that is transmitted to the brain. Vitamin A is required for the synthesis of retinal, and a deficiency of vitamin A can lead to a decrease in the amount of rhodopsin in the neurosensory rod cells, resulting in impaired twilight vision. Retinal horizontal cells (C), ganglionic neurons (D), and bipolar neurons (E) are other types of cells found in the retina, but they are not directly involved in the process of vision in low-light conditions. Neurosensory cone cells (B) are responsible for color vision and vision in bright light, but they are not involved in twilight vision. |
21. During a regular examination, blood was taken from a vein of a pregnant woman was positive. The patient and her husband deny extramarital sexual intercourse. What must be done to confirm or refute the diagnosis of syphilis?
A. Perform sedimentation tests
B. Perform the Treponema pallidum
immobilization test
C. Obtain a smear from the urethra
D. Perform the complement fixation
test
E. Repeat the Wasserman test
Answer: Perform the Treponema pallidum
immobilization test
Explanation
| The correct answer is B. Perform the Treponema pallidum immobilization test. Syphilis is a sexually transmitted infection caused by the bacterium Treponema pallidum. The diagnosis of syphilis is usually made by serological tests that detect antibodies against T. pallidum in the blood. The Wasserman test is a type of serological test that was historically used to diagnose syphilis, but it has largely been replaced by more specific and sensitive tests. However, a positive Wasserman test can be an indication of a possible syphilis infection. To confirm or refute the diagnosis of syphilis, a more specific test for T. pallidum should be performed. The Treponema pallidum immobilization test (TPI) is a highly specific test that detects antibodies against T. pallidum. It is considered to be one of the most reliable tests for the diagnosis of syphilis. Other serological tests that are used to diagnose syphilis include the fluorescent treponemal antibody absorption (FTA-ABS) test, the enzyme-linked immunosorbent assay (ELISA), and the chemiluminescence immunoassay (CLIA). These tests detect antibodies against T. pallidum and can be used to confirm the diagnosis of syphilis. Sedimentation tests (A), such as the erythrocyte sedimentation rate (ESR) test, are not specific to the diagnosis of syphilis and are not commonly used for this purpose. Obtaining a smear from the urethra (C) is not a reliable test for the diagnosis of syphilis, as T. pallidum is not typically found in the urethra. The complement fixation test (D) is a non-specific test that detects antibodies against a variety of antigens, and it is not commonly used for the diagnosis of syphilis. Repeat testing with the Wasserman test (E) may be useful to confirm the initial positive result, but a more specific test for T. pallidum should also be performed. |
22. A patient has been prescribed pyridoxal phosphate. This drug is recommended for correction of the following processes:
A. Deamination of purine nucleotides
B. Synthesis of purine and pyrimidine bases
C. Protein synthesis
D. Oxidative decarboxylation of keto acids
E. Transamination and decarboxylation of amino acids
Answer: Transamination and decarboxylation of amino acids
Explanation
| The correct answer is E. Transamination and decarboxylation of amino acids. Pyridoxal phosphate (PLP) is the active form of vitamin B6 and is an important cofactor for many enzymes involved in amino acid metabolism. PLP is required for the transamination of amino acids, which is the process by which amino groups are transferred from one amino acid to another. PLP is also required for the decarboxylation of amino acids, which is a process that removes a carboxyl group from an amino acid, resulting in the formation of a new molecule. PLP is particularly important for the metabolism of the essential amino acid tryptophan, which is converted to niacin (vitamin B3) in the body. PLP is also required for the metabolism of other amino acids, including histidine, cysteine, and glycine. Deamination of purine nucleotides (A) is a process that involves the removal of an amino group from a purine base, and it is not directly related to the function of PLP. Synthesis of purine and pyrimidine bases (B) is a complex process that involves multiple enzymes and is not directly related to the function of PLP. Protein synthesis (C) is a process that involves the ribosome and transfer RNA and does not directly require PLP. Oxidative decarboxylation of keto acids (D) is a process that involves the conversion of a keto acid to an acyl-CoA molecule and is not directly related to the function of PLP. |
23. A 32-year-old woman was diagnosed with myocarditis. ECG detects disturbance of the cardiac rhythm (a non-sinus rhythm). What cardiomyocytes are dysfunctional in this case?
A. Conducting cardiomyocytes of the bundle of His
B. Conducting cardiomyocytes of the His bundle branches
C. Contractile cardiomyocytes
D. Transitional conducting cardiomyocytes
E. Pacemaker cells
Answer: Pacemaker cells
Explanation
| The correct answer is E. Pacemaker cells. Myocarditis is an inflammation of the heart muscle (myocardium) that can affect the structure and function of the heart. One of the possible complications of myocarditis is the development of arrhythmias, which are abnormal heart rhythms that can interfere with the pumping action of the heart. The electrical conduction system of the heart is responsible for generating and transmitting electrical impulses that coordinate the contraction of the heart muscle. The system includes specialized cells called pacemaker cells, which are located in the sinoatrial (SA) node of the heart. Pacemaker cells are responsible for initiating the electrical impulses that regulate the heartbeat. Disturbance of the cardiac rhythm (a non-sinus rhythm) detected on an electrocardiogram (ECG) suggests dysfunction of the pacemaker cells. A non-sinus rhythm means that the electrical impulses that regulate the heartbeat are not originating from the SA node, but from another part of the heart. This can result in an irregular or abnormal heart rhythm. Conducting cardiomyocytes of the bundle of His (A), conducting cardiomyocytes of the His bundle branches (B), and transitional conducting cardiomyocytes (D) are all involved in the transmission of electrical impulses through the heart, but they are not directly responsible for initiating the impulses. Contractile cardiomyocytes (C) are responsible for the contraction of the heart muscle, but they do not have a direct role in the generation or transmission of electrical impulses. |
24. A was hospitalized with provisional diagnosis of acute pancreatitis. What enzyme activity must be measured in the patient’s blood and urine to confirm this diagnosis?
A. Lactate dehydrogenase
B. α-amylase
C. Cholinesterase
D. AST
E. ALT
Answer: α-amylase
Explanation
| The correct answer is B. α-amylase. Acute pancreatitis is a condition characterized by inflammation of the pancreas, which can be caused by a number of factors including alcohol consumption, gallstones, and high levels of triglycerides in the blood. One of the hallmarks of acute pancreatitis is an increase in the activity of enzymes produced by the pancreas, such as α-amylase and lipase. Measurement of α-amylase activity in the blood and urine is a commonly used test to confirm the diagnosis of acute pancreatitis. The pancreas produces α-amylase, an enzyme that breaks down carbohydrates, and increased levels of α-amylase in the blood and urine can indicate damage to the pancreas. Other tests that may be used to diagnose acute pancreatitis include measurement of lipase activity in the blood, which is another enzyme produced by the pancreas, as well as imaging tests such as ultrasound, computed tomography (CT) scan, or magnetic resonance imaging (MRI). Lactate dehydrogenase (A), cholinesterase (C), AST (D), and ALT (E) are enzymes that are produced by various organs and tissues in the body, but they are not specific to the pancreas and are not typically used to diagnose acute pancreatitis. |
25. Against the background of ionizing radiation exposure, decrease in the granulocyte count was detected in the patient’s blood. What causes agranulocytosis in this case?
A. Increased leukocyte destruction
B. Disturbed release of mature leukocytes from the bone marrow
C. Increased migration of granulocytes into tissues
D. Autoimmune process development
E. Leukopoiesis inhibition
Answer: Leukopoiesis inhibition
Explanation
| The correct answer is E. Leukopoiesis inhibition. Agranulocytosis is a condition characterized by a decrease in the number of granulocytes (a type of white blood cell) in the blood. It can be caused by a number of factors, including exposure to ionizing radiation. Ionizing radiation can damage the bone marrow, which is the site of production of blood cells including granulocytes. The damage to the bone marrow can result in a decrease in the production of granulocytes, a condition known as leukopenia or neutropenia. If the granulocyte count falls below a certain threshold, agranulocytosis can occur. Therefore, in this case, the decrease in granulocyte count is likely due to inhibition of leukopoiesis, which is the process of producing white blood cells in the bone marrow. Ionizing radiation can damage the DNA of cells in the bone marrow, including the stem cells that give rise to blood cells, leading to a decrease in their production. Increased leukocyte destruction (A) is not typically associated with agranulocytosis. Disturbed release of mature leukocytes from the bone marrow (B) can result in a decrease in the number of circulating white blood cells, but it is not specific to granulocytes. Increased migration of granulocytes into tissues (C) is not typically associated with agranulocytosis. Autoimmune process development (D) can lead to the destruction of white blood cells, including granulocytes, but it is not typically associated with exposure to ionizing radiation. |
26. There are several stages in the process of translation. At one of these stages, a complex forms that consists of a ribosome, mRNA, and aminoacyltRNA-methionine. What is the name of this stage?
A. Termination
B. Transcription
C. Initiation
D. Elongation
E. Repair
Answer: Initiation
Explanation
| The correct answer is C. Initiation. Translation is the process by which the genetic information encoded in mRNA is used to synthesize proteins. It involves the formation of a complex between the mRNA and a ribosome, as well as the binding of aminoacyl-tRNA molecules to the ribosome in the correct sequence. Initiation is the first stage of translation, during which the ribosome binds to the mRNA and identifies the start codon that signals the beginning of the protein-coding sequence. The initiator aminoacyl-tRNA molecule, which carries a methionine residue, then binds to the start codon, forming a complex consisting of the ribosome, mRNA, and aminoacyl-tRNA-methionine. This complex is known as the initiation complex. Once the initiation complex has formed, the ribosome moves along the mRNA molecule in a process called elongation, during which additional aminoacyl-tRNA molecules are added to the growing protein chain. Finally, termination occurs when the ribosome reaches a stop codon, and the newly synthesized protein is released from the ribosome. Transcription (B) is the process by which the genetic information encoded in DNA is used to synthesize mRNA. Elongation (D) is a stage of translation that follows initiation, during which additional amino acids are added to the growing protein chain. Repair (E) is a general term that refers to the process of fixing damaged or mutated DNA. |
27. Laboratory diagnostics of hepatitis B has determined the presence of viral DNA in the patient’s blood. What reaction is usually used for this purpose?
A. Hemagglutination inhibition test
B. Enzyme-linked immunosorbent
assay
C. Complement fixation test
D. Indirect hemagglutination test
E. Polymerase chain reaction amino
acids
Answer: Polymerase chain reaction amino
acids
Explanation
| The correct answer is E. Polymerase chain reaction (PCR). Hepatitis B is a viral infection that can cause liver inflammation and damage. It is caused by the hepatitis B virus (HBV), which is a DNA virus that replicates in the liver. The presence of viral DNA in the blood is a strong indicator of active infection. PCR is a sensitive and specific laboratory technique that is commonly used to detect the presence of viral DNA in the blood. It involves amplifying a specific segment of the viral DNA using primers that are complementary to the target sequence. The amplified DNA can then be detected using various methods, such as gel electrophoresis or fluorescent dyes. PCR is a highly sensitive technique that can detect very low levels of viral DNA in the blood, making it a useful tool for diagnosing hepatitis B infection. It can also be used to monitor the response to treatment and to detect the emergence of drug-resistant strains of the virus. Hemagglutination inhibition test (A), enzyme-linked immunosorbent assay (B), complement fixation test (C), and indirect hemagglutination test (D) are all serological tests that detect antibodies produced by the immune system in response to the hepatitis B virus. These tests are useful for diagnosing past infection or determining immunity to the virus, but they do not directly detect the presence of viral DNA in the blood. |
28. A patient presents with a purulent inflammatory process in the thigh region (a post-injection abscess). What lymph nodes become enlarged because of this process?
A. Popliteal
B. Inguinal
C. Submandibular
D. Paratracheal
E. Posterior cervical
Answer: Inguinal
Explanation
| The correct answer is B. Inguinal. Lymph nodes are small, bean-shaped structures that are part of the lymphatic system, which is an important component of the immune system. Lymph nodes filter lymphatic fluid and help to identify and fight infections. Infections and inflammatory processes in certain parts of the body can cause enlargement of nearby lymph nodes, as immune cells are activated and travel to the lymph nodes to fight the infection. In the case of a purulent inflammatory process in the thigh region, the lymph nodes that are most likely to become enlarged are the inguinal lymph nodes, which are located in the groin area. The inguinal lymph nodes receive lymphatic drainage from the lower limbs, the external genitalia, and the lower abdominal wall. Infections or inflammatory processes in these areas can cause the inguinal lymph nodes to become enlarged and tender. Popliteal lymph nodes (A) are located behind the knee and receive lymphatic drainage from the lower leg and foot. Infections or inflammatory processes in these areas can cause the popliteal lymph nodes to become enlarged, but they are less likely to be involved in a purulent inflammatory process in the thigh region. Submandibular lymph nodes (C) are located in the neck and receive lymphatic drainage from the tongue, submandibular gland, and teeth. Paratracheal (D) and posterior cervical (E) lymph nodes are also located in the neck and receive lymphatic drainage from the upper respiratory tract and head and neck region. These lymph nodes are less likely to be involved in a purulent inflammatory process in the thigh region. |
29. A patient died of secondary bacterial pneumonia. Autopsy revealed pale yellow muscles with numerous foci of calcinosis. In the muscles, microscopy shows dystrophic changes, absence of striations, and reduced glycogen levels. Edema and inflammation were detected in the stroma. The cellular infiltrate is represented by lymphocytes, macrophages, and plasma cells. Sclerotic changes were detected in the heart, lungs, and liver. These pathological changes are characteristic of the following disease:
A. Zenker’s degeneration of muscles in typhoid fever
B. Systemic scleroderma
C. Myopathy
D. Myositis
E. Dermatomyositis (WagnerUnverricht Hepp disease)
Answer: Dermatomyositis (WagnerUnverricht Hepp disease)
Explanation
| The correct answer is E. Dermatomyositis (also known as Wagner-Unverricht-Hepp syndrome). Dermatomyositis is an autoimmune disease that affects the muscles and skin. It is characterized by muscle weakness and inflammation, as well as skin rashes. The exact cause of dermatomyositis is not known, but it is thought to involve an abnormal immune response that targets the muscles and skin. The pathological changes described in the question are characteristic of dermatomyositis. The presence of pale yellow muscles with numerous foci of calcinosis suggests the deposition of calcium salts in the muscle tissue, which can occur in patients with dermatomyositis. Microscopic examination of the muscle tissue typically shows dystrophic changes, such as the absence of striations and reduced glycogen levels, as well as edema and inflammation in the stroma. The cellular infiltrate is typically composed of lymphocytes, macrophages, and plasma cells. In addition to muscle involvement, patients with dermatomyositis can also experience skin rashes, such as a characteristic rash on the eyelids and over the knuckles known as a Gottron’s rash. Sclerotic changes, or the formation of scar tissue, can also occur in the heart, lungs, and liver in some cases. Zenker’s degeneration of muscles in typhoid fever (A) is a rare complication of typhoid fever that can lead to muscle weakness and atrophy. Systemic scleroderma (B) is a connective tissue disorder that can cause thickening and hardening of the skin and internal organs. Myopathy (C) and myositis (D) are general terms that refer to diseases or conditions that affect the muscles, but they do not specifically describe the characteristic pathological changes seen in dermatomyositis. |
30. Autopsy of the body of a patient who died with signs of cardiopulmonary failure shows deformed bronchi with sack- like protrusions of the bronchial wall and purulent inflammation. Hypertrophy of the right ventricle was detected in the heart. Amyloidosis can be observed in the kidneys. The patient’s history indicates that for the last 8 years the patient complained of asphyxia and cough with purulent sputum, the patient’s fingers resembled drumsticks. What disease can be characterized by these pathological changes?
A. Acute bronchitis
B. Abscess
C. Tuberculosis
D. Bronchiectasis
E. Chronic bronchitis
Answer: Bronchiectasis
Explanation
| The correct answer is D. Bronchiectasis. Bronchiectasis is a chronic respiratory condition characterized by the permanent dilation and distortion of the bronchi and bronchioles, which are the airways that lead to the lungs. The pathological changes described in the question, including the deformed bronchi with sack-like protrusions of the bronchial wall and purulent inflammation, are typical of bronchiectasis. Bronchiectasis can be caused by a variety of factors, including infections, immunodeficiencies, congenital conditions, and obstructive lung diseases. Patients with bronchiectasis often experience chronic cough, sputum production, and recurrent respiratory infections. The hypertrophy of the right ventricle that was detected in the heart is a common complication of chronic respiratory diseases, including bronchiectasis, and is a result of increased resistance to blood flow in the lungs. Amyloidosis, which was observed in the kidneys, is a deposition of abnormal protein called amyloid in various organs and tissues throughout the body, including the kidneys. It can be associated with chronic respiratory diseases such as bronchiectasis. Acute bronchitis (A) is a temporary inflammation of the bronchi that is usually caused by a viral infection and typically resolves within a few weeks. Abscess (B) is a localized collection of pus within tissues, which can occur in various parts of the body but is not specific to the respiratory system. Tuberculosis (C) is a bacterial infection that primarily affects the lungs and can cause chronic cough, fever, weight loss, and night sweats, but it typically does not cause the characteristic deformities of the bronchi seen in bronchiectasis. Chronic bronchitis (E) is a type of obstructive lung disease that is characterized by chronic cough and sputum production, but it does not typically cause the characteristic deformities of the bronchi seen in bronchiectasis. |
31. An anti-inflammatory drug that blocks the treatment of a patient. What anti- cyclooxygenase activity was used in inflammatory drug is it?
A. Creatine
B. Aspirin (Acetylsalicylic acid)
C. Thiamine
D. Analgin (Metamizole sodium) evaporation
E. Allopurinol evaporation
Answer: Aspirin (Acetylsalicylic acid)
Explanation
| The correct answer is B. Aspirin (Acetylsalicylic acid). Aspirin is a nonsteroidal anti-inflammatory drug (NSAID) that is commonly used to relieve pain, reduce fever, and decrease inflammation. As an NSAID, aspirin works by blocking the activity of cyclooxygenase (COX), an enzyme that is involved in the production of prostaglandins, which are chemical messengers that promote inflammation, pain, and fever. There are two isoforms of COX, COX-1 and COX-2. COX-1 is expressed in many tissues throughout the body and is involved in the production of prostaglandins that protect the stomach lining, promote blood clotting, and regulate kidney function, among other functions. COX-2, on the other hand, is primarily expressed in response to inflammation and is involved in the production of prostaglandins that promote inflammation, pain, and fever. Aspirin inhibits both COX-1 and COX-2, although it has a greater effect on COX-1. This means that aspirin not only reduces inflammation and pain but also has effects on blood clotting and other physiological processes that are mediated by prostaglandins. Creatine (A) is a compound that is involved in energy metabolism in muscle cells and is often used as a dietary supplement to improve athletic performance. Thiamine (C) is a B-vitamin that is involved in energy metabolism and is necessary for the proper functioning of the nervous system. Analgin (D) is a pain reliever that works by blocking pain signals in the central nervous system. Allopurinol (E) is a medication that is used to treat gout by reducing the production of uric acid in the body. None of these drugs have anti-cyclooxygenase activity. |
32. Medical examination detected angina pectoris in a patient. The doctor prescribed the patient metoprolol that reduces the strength and frequency of cardiac contractions and, as a result, reduces the myocardial oxygen demand. What is the mechanism of the therapeutic action of this drug?
A. Blockade of nicotinic acetylcholine receptors
B. Blockade of β2-adrenergic receptors
C. Stimulation of β1-adrenergic receptors
D. Blockade of β1-adrenergic receptors
E. Blockade of muscarinic acetylcholine receptors
Answer: Blockade of β1-adrenergic receptors
Explanation
| The correct answer is D. Blockade of β1-adrenergic receptors. Angina pectoris is a condition characterized by chest pain or discomfort that occurs when the heart muscle does not receive enough oxygen-rich blood. Metoprolol is a beta-blocker medication that is commonly used to treat angina pectoris. It works by blocking the activity of beta-adrenergic receptors, specifically the β1-adrenergic receptors found in the heart. Beta-adrenergic receptors are proteins found on the surface of cells in various tissues throughout the body, including the heart. These receptors bind to the neurotransmitter molecules norepinephrine and epinephrine (also known as adrenaline), which are released by the sympathetic nervous system during times of stress or excitement. When norepinephrine or epinephrine binds to the β1-adrenergic receptors in the heart, it increases the strength and frequency of cardiac contractions, which can increase the myocardial oxygen demand and worsen symptoms of angina. Metoprolol works by blocking the β1-adrenergic receptors in the heart, which reduces the strength and frequency of cardiac contractions. This decreases the myocardial oxygen demand and can relieve symptoms of angina. Metoprolol also has other effects, such as reducing heart rate and blood pressure, which can also be beneficial in treating angina. Blockade of nicotinic acetylcholine receptors (A) and muscarinic acetylcholine receptors (E) are not mechanisms of action of metoprolol. Blockade of β2-adrenergic receptors (B) is not a primary mechanism of action of metoprolol, although it can have some effect on these receptors at higher doses. Stimulation of β1-adrenergic receptors (C) is not a mechanism of action of metoprolol, as this would have the opposite effect of increasing the myocardial oxygen demand and worsening symptoms of angina. |
33. A patient diagnosed with diabetes mellitus presents with increased levels of ketone bodies in the blood. From what compound are ketone bodies synthesized?
A. Oxyacyl-CoA
B. Acetyl-CoA
C. Acyl-CoA
D. Succinyl-CoA
E. Butyryl-CoA
Answer: Acetyl-CoA
Explanation
| The correct answer is B. Acetyl-CoA. Ketone bodies are water-soluble molecules that are produced during the metabolism of fatty acids in the liver. They are an important source of energy for many tissues in the body, including the brain, during periods of fasting, starvation, or prolonged exercise. In diabetes mellitus, the body has difficulty producing or using insulin, a hormone that regulates blood glucose levels. This can lead to high levels of glucose in the blood, which can cause the body to switch to using fat as an alternative source of energy. When fatty acids are metabolized in the liver, they are converted into acetyl-CoA, which can then be used to synthesize ketone bodies. Therefore, in a patient with diabetes mellitus, increased levels of ketone bodies in the blood can be a sign of increased fatty acid metabolism and can indicate poor glycemic control. Oxyacyl-CoA (A), acyl-CoA (C), succinyl-CoA (D), and butyryl-CoA (E) are also involved in fatty acid metabolism but are not directly involved in the synthesis of ketone bodies. |
34. Copper deficiency has an effect on energy metabolism in the human body. What substance becomes deficient as a result of this process?
A. A Succinate dehydrogenase
B. Lactate dehydrogenase
C. Pyruvate carboxylase
D. Cytochrome oxidase
E. Arginase
Answer: Cytochrome oxidase
Explanation
| Copper is an essential trace element that is required for the proper functioning of many enzymes in the body, including cytochrome oxidase, which is a key enzyme involved in the electron transport chain that generates ATP, the primary source of energy for the body. Cytochrome oxidase is found in the mitochondria of cells and is responsible for transferring electrons from cytochrome c to oxygen, which results in the production of water and the release of energy that is used to generate ATP. Copper is a critical component of cytochrome oxidase, as it is required for the proper folding and function of the enzyme. Therefore, a deficiency of copper can lead to decreased activity of cytochrome oxidase, which can impair the ability of cells to generate ATP and can affect energy metabolism in the body. Symptoms of copper deficiency can include fatigue, weakness, and anemia, among other symptoms. Succinate dehydrogenase (A), lactate dehydrogenase (B), pyruvate carboxylase (C), and arginase (E) are enzymes involved in other metabolic pathways and are not directly affected by copper deficiency. |
35. Blood test revealed the total leukocyte count of 11*109 /L in the patient’s blood, with neutrophils making up 80% of all leukocytes. Among them, banded neutrophils make up 9%. What is the nature of the changes in the cellular composition of WBC in this patient?
A. Neutrophilic right shift
B. Neutropenia
C. Leukopenia
D. Neutrophilic left shift
E. Lymphocytosis
Answer: Neutrophilic left shift
Explanation
| The correct answer is D. Neutrophilic left shift. Leukocytes, also known as white blood cells (WBCs), are a key component of the immune system and are involved in defending the body against infection and disease. The total leukocyte count measures the number of all types of WBCs in the blood. Neutrophils are a type of WBC that are involved in the initial response to infection and are often the first cells to arrive at the site of an infection. Banded neutrophils, also known as band cells or immature neutrophils, are a type of neutrophil that are released from the bone marrow in response to an infection or inflammation. A neutrophilic left shift refers to an increase in the number of banded neutrophils in the blood, which indicates that the bone marrow is producing more neutrophils than usual in response to an infection or inflammation. In this case, the blood test revealed a total leukocyte count of 11*10^9/L, which is within the normal range, but with neutrophils making up 80% of all leukocytes and banded neutrophils making up 9%, indicating a neutrophilic left shift. Neutropenia (B) and leukopenia (C) refer to a decrease in the total number of WBCs in the blood. Lymphocytosis (E) refers to an increase in the number of lymphocytes, another type of WBC, in the blood. None of these conditions are present in this patient. Neutrophilic right shift (A) is not a recognized term and does not describe any known changes in the cellular composition of WBCs. |
36. Despite profuse sweating, a person feels stuffy and hot in a tropical forest at a relatively low air temperature (26-27°C). Why is profuse sweating not an effective method of heat transfer in this case?
A. High air humidity increases sweat evaporation
B. High air humidity reduces sweat evaporation
C. Air temperature reduces sweat evaporation
D. High air humidity reduces radiation
E. Air temperature increases sweat evaporation evaporation
Answer: High air humidity reduces sweat evaporation
Explanation
| The correct answer is B. High air humidity reduces sweat evaporation. Sweating is a natural mechanism that helps regulate body temperature by dissipating heat as sweat evaporates from the skin. However, the effectiveness of sweating as a method of heat transfer is influenced by several factors, including air temperature and humidity. In a tropical forest with a high air temperature of 26-27°C, sweating can be an effective method of heat transfer if the air humidity is low. However, in this case, the question mentions that the person feels stuffy and hot despite profuse sweating, which suggests that the air humidity is high. When the air is humid, it already contains a high level of moisture, which can reduce the rate of evaporation of sweat from the skin. This means that even if the person is sweating profusely, the sweat may not be evaporating quickly enough to effectively cool the body. As a result, the person may feel stuffy and hot despite sweating. Therefore, in a high-humidity environment, sweating may not be an effective method of heat transfer, and other methods, such as seeking shade or using fans or air conditioning, may be necessary to cool the body. High air humidity (B) reduces sweat evaporation, while high air temperature (C) and air temperature (E) increase sweat evaporation. High air humidity (D) does not directly affect radiation, which is a different mechanism of heat transfer. |
37. A 45-year-old man with acute pneumonia was prescribed a penicillin antibiotic. However, when tested for personal tolerance to this antibiotic, he developed an allergic response. What drug should be prescribed for treatment in this case?
A. Erythromycin
B. Benzylpenicillin
C. Ciprofloxacin
D. Phenoxymethylpenicillin
E. Bicillin-5
Answer: Ciprofloxacin
Explanation
| The correct answer is C. Ciprofloxacin. Penicillin is a type of antibiotic that is commonly used to treat bacterial infections. However, some people may be allergic to penicillin, which can cause an allergic reaction that ranges from mild to severe. In this case, the patient developed an allergic response to penicillin, so an alternative antibiotic should be prescribed. Ciprofloxacin is a broad-spectrum antibiotic that is effective against a wide range of bacteria. It belongs to the fluoroquinolone class of antibiotics and works by interfering with bacterial DNA synthesis, which leads to the death of the bacteria. Ciprofloxacin is not related to penicillin and does not contain any penicillin derivatives, so it is a suitable alternative for patients who are allergic to penicillin. Erythromycin (A), benzylpenicillin (B), phenoxymethylpenicillin (D), and bicillin-5 (E) are all types of antibiotics that contain penicillin or penicillin derivatives. Therefore, they are not suitable alternatives for a patient who is allergic to penicillin. |
38. Residents of areas with a cold a certain hormone that has an adaptive thermoregulatory value. What hormone is it?
A. Insulin
B. Glucagon
C. Cortisol
D. Somatotropin
E. Thyroxine
Answer: Thyroxine
Explanation
| The correct answer is E. Thyroxine. Thyroxine, also known as T4, is a hormone that is produced by the thyroid gland and is involved in regulating metabolism and body temperature. In areas with cold climates, the body needs to generate more heat to maintain a constant body temperature, and thyroxine plays an important role in this process. Thyroxine increases the metabolic rate of cells, which generates heat as a byproduct of metabolism. This increases the body’s overall heat production, which can help to offset the effects of cold temperatures. In addition, thyroxine can also increase the body’s sensitivity to other hormones that regulate body temperature, such as adrenaline. Therefore, in areas with cold climates, the body may produce more thyroxine in order to adapt to the colder temperatures and maintain a constant body temperature. Insulin (A) and glucagon (B) are hormones that are involved in regulating blood glucose levels and are not directly involved in thermoregulation. Cortisol (C) is a hormone that is involved in regulating stress and inflammation, among other functions, but is not directly involved in thermoregulation. Somatotropin (D), also known as growth hormone, is involved in regulating growth and metabolism but is not directly involved in thermoregulation. |
39. What biogenic amine normalizes circadian rhythms?
A. Melatonin
B. Adrenaline
C. Noradrenaline
D. Histamine
E. Dopamine
Answer: Melatonin
Explanation
| The correct answer is A. Melatonin. Melatonin is a biogenic amine that is produced by the pineal gland in the brain and is involved in regulating circadian rhythms, which are the natural 24-hour cycles of the body that control sleep-wake cycles, hormone levels, and other physiological processes. Melatonin is released in response to darkness and helps to induce sleep by promoting relaxation and reducing alertness. It also helps to regulate the timing of the sleep-wake cycle and other circadian rhythms by interacting with receptors in the brain that are involved in regulating these processes. Therefore, melatonin is a key hormone involved in regulating circadian rhythms, and it is often used as a supplement to help treat sleep disorders or to help adjust to changes in time zones, such as during long-distance travel. Adrenaline (B), noradrenaline (C), histamine (D), and dopamine (E) are biogenic amines that are involved in other physiological processes, such as the regulation of the stress response (adrenaline and noradrenaline), the regulation of inflammation and allergic reactions (histamine), and the regulation of mood, motivation, and movement (dopamine). |
40. In an experiment, certain nuclei of the hypothalamus were destroyed in homeothermic animals, which resulted in them being unable to maintain their body temperature. What structure has been destroyed?
A. Medial hypothalamic nuclei
B. Lateral hypothalamic nuclei
C. Posterior hypothalamic nuclei
D. Supraoptic nuclei
E. Ventral hypothalamic nuclei
Answer: Ventral hypothalamic nuclei
Explanation
| The correct answer is E. Ventral hypothalamic nuclei. The hypothalamus is a small region of the brain that is involved in regulating a wide range of physiological processes, including body temperature, hunger and thirst, and the sleep-wake cycle. The hypothalamus is composed of several nuclei, each of which is involved in regulating different aspects of these processes. In particular, the ventral hypothalamic nuclei (VHN) are involved in regulating body temperature by controlling the activity of the autonomic nervous system, which is responsible for regulating many involuntary physiological processes, including body temperature. When the VHN are destroyed, as in the experiment described in the question, the animals are unable to maintain their body temperature and may develop hypothermia or hyperthermia, depending on the conditions. This is because the VHN normally regulate the activity of the autonomic nervous system to help maintain a constant body temperature in response to changes in the environment. The other hypothalamic nuclei listed in the answer choices are involved in other physiological processes. The medial hypothalamic nuclei (A) are involved in regulating hunger and satiety, while the lateral hypothalamic nuclei (B) are involved in regulating the sleep-wake cycle and feeding behavior. The posterior hypothalamic nuclei (C) are involved in regulating body temperature and other autonomic functions, such as blood pressure and heart rate. The supraoptic nuclei (D) are involved in regulating water balance and the release of hormones from the pituitary gland. |
41. A 9-year-old child developed a severe case of purulent destructive pneumonia, for which the child was receiving a massive antibacterial therapy. The disease: was rapidly progressing. Against the background of marked intoxication, a sharp drop in blood pressure was registered, the child went into a state of shock that resulted in the child’s death. What etiopathogenetic type of shock developed in the child?
A. Toxic shock syndrome
B. Hypovolemic shock
C. Hemolytic shock
D. Anaphylactic shock
E. Cardiogenic shock
Answer: Toxic shock syndrome
Explanation
| The correct answer is A. Toxic shock syndrome. Toxic shock syndrome (TSS) is a rare but serious condition that can occur when certain types of bacteria produce toxins that enter the bloodstream and cause a systemic inflammatory response. TSS can be caused by several types of bacteria, including Staphylococcus aureus and Streptococcus pyogenes. In this case, the child developed a severe case of purulent destructive pneumonia, which suggests that the infection was caused by bacteria. The child was receiving a massive antibacterial therapy, which indicates that the infection was severe and required aggressive treatment. The sudden drop in blood pressure and the development of shock suggest that the child’s immune system had a strong response to the infection, leading to the release of a large amount of inflammatory cytokines and other mediators. This systemic inflammatory response can cause damage to tissues and organs throughout the body, leading to shock and organ failure. Therefore, based on the symptoms described in the question, it is likely that the child developed toxic shock syndrome as a result of the severe bacterial infection. Hypovolemic shock (B) is caused by a decrease in blood volume, such as from severe bleeding or dehydration. Hemolytic shock (C) is caused by the breakdown of red blood cells, such as in severe cases of blood transfusion reactions or in certain types of infections. Anaphylactic shock (D) is caused by an extreme allergic reaction, often to a medication or food. Cardiogenic shock (E) is caused by a failure of the heart to pump blood effectively, such as in a heart attack or severe heart failure. None of these conditions are consistent with the symptoms described in the question. |
42. A patient diagnosed with essential hypertension was undergoing hypothiazide (hydrochlorothiazide) treatment. He complains of general weakness, loss of appetite, and palpitations. Examination reveals muscle hypotonia, flaccid paralyses, and decreased intestinal peristalsis. What is the likely cause of the patient’s condition?
A. Hyponatremia
B. Hypercalcemia
C. Hypokalemia
D. Hyperuricemia
E. Hyperkalemia
Answer: Hypokalemia
Explanation
| The correct answer is C. Hypokalemia. Hydrochlorothiazide is a diuretic medication that is commonly used to treat hypertension by increasing the excretion of water and sodium from the body. However, diuretics can also increase the excretion of other electrolytes, including potassium, which can lead to a condition called hypokalemia. Hypokalemia is characterized by low levels of potassium in the blood, which can cause a wide range of symptoms, including weakness, fatigue, loss of appetite, palpitations, muscle hypotonia, flaccid paralysis, and decreased intestinal peristalsis, as described in the question. Potassium is an important electrolyte that is involved in many physiological processes, including muscle and nerve function, fluid balance, and the regulation of heart rhythm. When potassium levels in the blood become too low, it can interfere with these processes and lead to a variety of symptoms. Therefore, in this case, it is likely that the patient’s symptoms are due to hypokalemia caused by the hydrochlorothiazide treatment. Treatment for hypokalemia may include potassium supplements or a change in medication. Hyponatremia (A) is a condition characterized by low levels of sodium in the blood, which can cause symptoms such as nausea, headache, confusion, and seizures. Hypercalcemia (B) is a condition characterized by high levels of calcium in the blood, which can cause symptoms such as fatigue, constipation, and kidney stones. Hyperuricemia (D) is a condition characterized by high levels of uric acid in the blood, which can cause gout and kidney stones. Hyperkalemia (E) is a condition characterized by high levels of potassium in the blood, which can cause symptoms such as muscle weakness, fatigue, and heart rhythm disturbances. None of these conditions are consistent with the symptoms described in the question. |
43. Numerous glucose oxidation metabolites are dissolved in the cytoplasm of myocytes. Which one of those metabolites directly converts into lactate?
A. Oxaloacetate
B. Fructose-6-phosphate
C. Glucose-6-phosphate
D. Glycerophosphate
E. Pyruvate
Answer: Pyruvate
Explanation
| The correct answer is E. Pyruvate. Glucose is a major source of energy for cells, including myocytes (muscle cells), and is metabolized through a series of reactions known as glycolysis. During glycolysis, glucose is converted into pyruvate through a series of enzymatic reactions. Pyruvate can then be further metabolized through aerobic respiration (in the presence of oxygen) or anaerobic respiration (in the absence of oxygen). In the absence of oxygen, pyruvate is converted into lactate through a process called lactate fermentation. This process is important for providing energy to cells in situations where oxygen is limited, such as during intense exercise. Therefore, of the answer choices listed, pyruvate is the metabolite that directly converts into lactate. Oxaloacetate (A) is a metabolic intermediate in the citric acid cycle, which is involved in aerobic respiration. Fructose-6-phosphate (B) and glucose-6-phosphate (C) are intermediate metabolites in glycolysis but do not directly convert into lactate. Glycerophosphate (D) is a metabolic intermediate in the breakdown of fats and is not directly involved in glucose metabolism. |
44. Laboratory testing detects glucose in the urine of an 18-year-old patient, while glucose levels in the patient’s blood plasma are normal. What is the likely cause of this disorder?
A. Tubular reabsorption
B. Tubular secretion
C. Secretion of glucocorticoids
D. Insulin secretion
E. Glomerular filtration
Answer: Tubular reabsorption
Explanation
| The correct answer is A. Tubular reabsorption. Normally, glucose is filtered out of the blood plasma by the glomeruli in the kidneys and then reabsorbed back into the bloodstream by the tubules in the kidneys. This process is known as tubular reabsorption and is important for maintaining normal glucose levels in the blood. When blood glucose levels are high, such as after a meal, the excess glucose is excreted in the urine. However, if the tubules in the kidneys are not able to reabsorb all of the filtered glucose, some of it can remain in the urine, resulting in a condition called glucosuria (glucose in the urine). Therefore, in this case, the likely cause of the glucosuria is a problem with tubular reabsorption in the kidneys. This could be due to a variety of factors, including kidney damage, certain medications, or a genetic disorder that affects tubular function. Tubular secretion (B) is the process by which substances are actively transported from the blood into the tubules in the kidneys for excretion in the urine. This process is not directly related to glucosuria. Secretion of glucocorticoids (C) is the release of hormones such as cortisol from the adrenal glands, which can affect glucose metabolism, but is not directly related to glucosuria. Insulin secretion (D) is the release of the hormone insulin from the pancreas, which helps to regulate blood glucose levels, but is not directly related to glucosuria. Glomerular filtration (E) is the process by which substances are filtered out of the blood by the glomeruli in the kidneys, including glucose. While glomerular filtration can contribute to the development of glucosuria, the primary cause in this case is likely to be a problem with tubular reabsorption. |
45. In a scientific experiment, a structure in one of the cell components has been destroyed, impairing the cell’s ability to. What nuclei were destroyed type of inflammation is observed on the visceral pleura?
A. Centrosome
B. Glycocalyx
C. Microfibrils
D. Ribosomes
E. Mitochondria
Answer: Centrosome
Explanation
| The destruction of the centrosome in a cell can lead to a variety of cellular abnormalities, including problems with cell division, cilia and flagella function, and the organization of the cytoskeleton. These abnormalities can result in a wide range of cellular dysfunctions, including impaired cell growth, motility, and signaling. It is not clear from the question what type of inflammation is observed on the visceral pleura. The visceral pleura is the inner layer of the pleural membrane that surrounds the lungs, and inflammation of this membrane can be caused by a variety of factors, including infection, injury, and autoimmune disorders. Therefore, while the destruction of the centrosome can have significant effects on cellular function, it is not directly related to the type of inflammation observed on the visceral pleura. More information would be needed to determine the cause of the inflammation. |
46. All nonsteroidal antiinflammatory drugs can damage the gastric mucosa. To find the substances that do not cause this complication, it is necessary to know what it is associated with. To reduce the severity of this complication, the drug’s effect on a certain molecular substrate must be reduced. Name this molecular substrate.
A. Lysosomal enzymes
B. Cyclooxygenase-1
C. Adenylate cyclase
D. Cyclooxygenase-2
E. Kallikrein
Answer:Cyclooxygenase-1
Explanation
| The correct answer is B. Cyclooxygenase-1. Nonsteroidal anti-inflammatory drugs (NSAIDs) are a class of drugs that are commonly used to relieve pain, inflammation, and fever. However, they can also cause damage to the gastric mucosa, leading to gastrointestinal complications such as ulcers and bleeding. NSAIDs exert their therapeutic effects by inhibiting the activity of cyclooxygenase (COX), an enzyme that is involved in the production of prostaglandins, which are important mediators of inflammation and pain. There are two isoforms of COX: COX-1 and COX-2. COX-1 is constitutively expressed in many tissues, including the gastric mucosa, where it plays a role in maintaining the integrity of the mucosal lining. Inhibition of COX-1 by NSAIDs can reduce the production of protective prostaglandins in the gastric mucosa, leading to damage and ulceration. COX-2, on the other hand, is induced in response to inflammation and is responsible for the production of prostaglandins that mediate pain and inflammation. Selective COX-2 inhibitors were developed as a class of NSAIDs that would have fewer gastrointestinal side effects than traditional NSAIDs by targeting COX-2 while sparing COX-1. Therefore, to reduce the severity of gastrointestinal complications associated with NSAIDs, the drug’s effect on COX-1 must be reduced. This can be achieved through the use of selective COX-2 inhibitors or by adding gastroprotective agents such as proton pump inhibitors or misoprostol to the treatment regimen. |
47. A child has 3 copies of chromosome 18, which resulted in characteristic cranial elongation from front to back, maldevelopments of the musculoskeletal system, fused fingers, and maldevelopments of skeletal muscles. What hereditary pathology is observed in this child?
A. Edwards syndrome
B. Down syndrome
C. Klinefelter syndrome
D. Turner syndrome
E. Patau syndrome
Answer: Edwards syndrome
Explanation
| he correct answer is A. Edwards syndrome. Edwards syndrome, also known as trisomy 18, is a genetic disorder caused by the presence of three copies of chromosome 18 instead of the normal two copies. This occurs due to a random error in cell division during embryonic development and is not inherited from the parents. The extra chromosome 18 leads to a wide range of developmental abnormalities, including cranial elongation from front to back, maldevelopments of the musculoskeletal system, fused fingers, and maldevelopments of skeletal muscles, as described in the question. Other common features of Edwards syndrome include growth retardation, heart defects, and intellectual disability. The condition is typically diagnosed prenatally through genetic testing or ultrasound and is associated with a high rate of fetal loss, with only a small percentage of affected individuals surviving to birth. Those who do survive often have significant medical and developmental challenges and require ongoing support and care. Down syndrome (B) is another genetic disorder caused by the presence of an extra chromosome, specifically an extra copy of chromosome 21. This condition is characterized by intellectual disability, distinctive facial features, and a variety of medical complications, but does not typically involve the cranial or musculoskeletal abnormalities described in the question. Klinefelter syndrome (C) is a genetic disorder that affects males and is caused by the presence of an extra X chromosome. This condition can lead to a variety of physical and developmental abnormalities, including infertility, but does not typically involve the cranial or musculoskeletal abnormalities described in the question. Turner syndrome (D) is a genetic disorder that affects females and is caused by the absence of one of the X chromosomes. This condition can lead to a variety of physical and developmental abnormalities, including short stature, but does not typically involve the cranial or musculoskeletal abnormalities described in the question. Patau syndrome (E) is another genetic disorder caused by the presence of an extra chromosome, specifically an extra copy of chromosome 13. This condition is associated with a wide range of physical and developmental abnormalities, including intellectual disability, heart defects, and cleft lip and palate, but does not typically involve the cranial or musculoskeletal abnormalities described in the question. |
48. During surgery for gallstones in bile ducts, the surgeon must find the common hepatic duct. It is located between the layers of the following ligament:
A. Round ligament of the liver
B. Hepatogastric ligament
C. Hepatoduodenal ligament
D. Ligamentum venosum
E. Hepatorenal ligament
Answer: Hepatoduodenal ligament
Explanation
| The correct answer is C. Hepatoduodenal ligament. The hepatoduodenal ligament is a double-layered fold of the peritoneum that attaches the liver to the duodenum and contains several important structures, including the common hepatic duct, the hepatic artery, and the portal vein. It is located on the right side of the abdomen, just to the right of the midline. During a surgery for gallstones in bile ducts, the surgeon must locate the common hepatic duct, which is the duct that carries bile from the liver to the gallbladder and small intestine. The common hepatic duct is located within the hepatoduodenal ligament, between its two layers of peritoneum. The other ligaments listed in the answer choices are also associated with the liver: The round ligament of the liver (A) is a fibrous remnant of a fetal blood vessel called the umbilical vein.The hepatogastric ligament (B) is a portion of the lesser omentum that attaches the liver to the stomach.The ligamentum venosum (D) is a fibrous remnant of a fetal blood vessel called the ductus venosus. The hepatorenal ligament (E) is a portion of the lesser omentum that attaches the liver to the right kidney.However, none of these ligaments contain the common hepatic duct. Patau syndrome (E) is another genetic disorder caused by the presence of an extra chromosome, specifically an extra copy of chromosome 13. This condition is associated with a wide range of physical and developmental abnormalities, including intellectual disability, heart defects, and cleft lip and palate, but does not typically involve the cranial or musculoskeletal abnormalities described in the question. |
49. X-ray detects a shadow in the area of the patient’s dural sinus that runs from the crista galli of the ethmoid bone of the skull to the internal occipital protuberance. In this case, pathological changes can be detected in the area of the following sinus:
A. Sigmoid sinus
B. Straight sinus
C. Superior sagittal sinus
D. Inferior sagittal sinus
E. Transverse sinus
Answer: Superior sagittal sinus
Explanation
| The correct answer is C. Superior sagittal sinus. The superior sagittal sinus is a large dural venous sinus that runs along the midline of the cranial vault, from the crista galli of the ethmoid bone at the front of the skull to the internal occipital protuberance at the back of the skull. It runs within the upper margin of the falx cerebri, a fold of dura mater that separates the two hemispheres of the brain. A shadow detected by X-ray in the area of the superior sagittal sinus may indicate the presence of pathological changes in this region. These changes could include thrombosis (the formation of a blood clot), stenosis (narrowing of the sinus), or other abnormalities that affect blood flow through the sinus. The other sinuses listed in the answer choices are also dural venous sinuses that are located within the skull: The sigmoid sinus (A) is a continuation of the transverse sinus that runs through the petrous part of the temporal bone.The straight sinus (B) is a short, wide sinus that runs along the junction of the tentorium cerebelli and falx cerebri.The inferior sagittal sinus (D) is a small midline sinus that runs within the lower margin of the falx cerebri.The transverse sinus (E) is a paired sinus that runs laterally along the base of the occipital bone and connects to the sigmoid sinus.However, none of these sinuses run from the crista galli to the internal occipital protuberance, as described in the question. |
50. Autopsy of the body of a man who died of croupous pneumonia revealed an opaque liquid in the pleural cavity and a grayish film on the visceral pleura. What type of inflammation is observed on the visceral pleura?
A. Purulent
B. Fibrinous
C. Hemorrhagic
D. Catarrhal
E. Granulomatous
Answer: Fibrinous
Explanation
| The correct answer is B. Fibrinous. Croupous pneumonia, also known as lobar pneumonia, is a type of pneumonia that is characterized by the consolidation of a single lobe of the lung due to inflammation and accumulation of fluid and cellular debris within the alveoli. In some cases, this inflammation can also spread to the pleural cavity, which is the space between the visceral and parietal pleura that surrounds the lungs. The opaque liquid observed in the pleural cavity is likely an accumulation of fluid resulting from the inflammation. The grayish film observed on the visceral pleura is also a result of the inflammation and is likely a fibrinous exudate, which is a type of inflammatory fluid that contains high levels of fibrinogen. Fibrinogen is a protein that is involved in blood clotting, and its presence in the inflammatory fluid can lead to the formation of a fibrin network that covers the surface of the pleural membrane, creating a film-like appearance. Therefore, the type of inflammation observed on the visceral pleura in this case is fibrinous. This type of inflammation is characterized by the exudation of inflammatory fluid that contains high levels of fibrinogen, leading to the formation of a fibrin network that can cover the surface of the affected tissue. The other types of inflammation listed in the answer choices are: Purulent (A): characterized by the presence of pus, which is a thick, yellowish fluid composed of dead white blood cells, bacteria, and cellular debris.Hemorrhagic (C): characterized by the presence of bleeding and blood vessels within the inflamed tissue.Catarrhal (D): characterized by the exudation of a mucous-like fluid from the inflamed tissue.Granulomatous (E): characterized by the formation of granulomas, which are small nodules of inflammatory cells that can form in response to chronic inflammation. |
