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Question 1
Incorrect
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The chest X-ray of an 10-year-old boy, that presented with low-grade fever and cough, revealed hilar enlargement and parenchymal consolidation in the middle lobes. These X-ray findings are more typical for which of the following diagnoses?
Your Answer: Cryptococcal pneumonia
Correct Answer: Pulmonary tuberculosis
Explanation:Primary pulmonary tuberculosis is seen in patients exposed to Mycobacterium tuberculosis for the firs time. The main radiographic findings in primary pulmonary tuberculosis include homogeneous parenchymal consolidation typically in the lower and middle lobes, lymphadenopathy, miliary opacities and pleural effusion.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 2
Incorrect
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What causes a reduction in pulmonary functional residual capacity?
Your Answer: Asthma
Correct Answer: Pulmonary fibrosis
Explanation:Pulmonary functional residual capacity (FRC) is = volume of air present in the lungs at the end of passive expiration.
Obstructive diseases (e.g. emphysema, chronic bronchitis, asthma) = an increase in FRC due to an increase in lung compliance and air trapping.
Restrictive diseases (e.g. pulmonary fibrosis) result in stiffer, less compliant lungs and a reduction in FRC.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 3
Incorrect
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Which tumour marker is associated with medullary carcinoma of thyroid?
Your Answer: Alpha-fetoprotein
Correct Answer: Calcitonin
Explanation:Medullary carcinoma of thyroid accounts for 3% of thyroid cancers. It arises from the parafollicular cells (C cells) of the thyroid gland that produce calcitonin. It is often familial and caused by mutation of ret proto-oncogene, but can occasionally be sporadic. The familial cases can also occur as part of MEN syndromes IIA and IIB. The high calcitonin leads to down-regulation of the receptors, which does not affect the calcium levels significantly. Medullary carcinoma of thyroid shows characteristic amyloid deposits that stain positively with Congo red. The initial presentation consists of an asymptomatic thyroid nodule. Many cases are diagnosed due to routine screening of relatives of patients with MEN IIA and IIB. Medullary carcinoma can also cause ectopic production of other hormones/peptides such as adrenocorticotrophic hormone, vasoactive intestinal polypeptide, kallikreins and serotonin.
Metastasis from medullary carcinoma spread via the lymphatics to cervical and mediastinal nodes, and can also affect the liver, lungs and bone. Diagnosis is by raised serum calcitonin levels. A provocative test with calcium (15 mg/kg intravenously over 4 hours) also aids in diagnosis by leading to excessive secretion of calcitonin. X-ray might also show dense, conglomerate calcification.
CA-125 is frequently elevated in ovarian carcinomas. CA 15-3 is often associated with breast carcinomas. Alpha-fetoprotein is seen raised in hepatomas and gonadal tumours. Elevated HCG is associated with normal pregnancies, gonadal tumours, and choriocarcinomas. Thyroglobulin is used for surveillance in papillary carcinoma of thyroid. CA 19-9 is used in the management of pancreatic cancer.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 4
Incorrect
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What is the likely course of a pulmonary embolism arising from the leg veins and ending in the apical segmental pulmonary artery that supplies the superior lobe of left lung?.
Your Answer: Inferior vena cava – right atrium – mitral valve – right ventricle – pulmonary trunk – left pulmonary artery – left superior lobar artery – left apical segmental artery
Correct Answer: Inferior vena cava – right atrium – tricuspid valve – right ventricle – pulmonary trunk – left pulmonary artery – left superior lobar artery – left apical segmental artery
Explanation:A clot originating in the leg vein will go to the inferior vena cava, into the right atrium, through the tricuspid valve, into the right ventricle, through the pulmonary trunk, into the left pulmonary artery, into the left superior lobar artery and then finally reach the left apical segmental artery.
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This question is part of the following fields:
- Anatomy
- Thorax
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Question 5
Incorrect
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Post-total gastrectomy, there will be a decreased production of which of the following enzymes?
Your Answer: Chymotrypsin
Correct Answer: Pepsin
Explanation:Pepsin is a protease that is released from the gastric chief cells and acts to degrade proteins into peptides. Released as pepsinogen, it is activated by hydrochloric acid and into pepsin itself. Gastrin and the vagus nerve trigger the release of pepsinogen and HCl when a meal is ingested. Pepsin functions optimally in an acidic environment, especially at a pH of 2.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 6
Incorrect
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When at rest, which of the following will be higher in a marathon runner compared to an untrained individual?
Your Answer: Cardiac output
Correct Answer: Cardiac stroke volume
Explanation:Cardiac muscle hypertrophy is seen in trained athletes as compared to the normal population. This hypertrophy results in higher stroke volume at rest and increased cardiac reserve (maximum cardiac output during exercise). However, the cardiac output at rest is almost the same in both trained and untrained people. This is because in trained athletes, the heart rate is slower, even up to 40-50 beats/min. There is minimal affect of athletic training on oxygen consumption and respiratory rate.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 7
Correct
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A 40-year old lady presented to the hospital with fever and mental confusion for 1 week. On examination, she was found to have multiple petechiae all over her skin and mucosal surfaces. Blood investigations revealed low platelet count and raised urea and creatinine. A platelet transfusion was carried out, following which she succumbed to death. Autopsy revealed pink hyaline thrombi in myocardial arteries. What is the likely diagnosis?
Your Answer: Thrombotic thrombocytopenic purpura
Explanation:Hyaline thrombi are typically associated with thrombotic thrombocytopenic purpura (TTP), which is caused by non-immunological destruction of platelets. Platelet transfusion is contraindicated in TTP. Platelets and red blood cells also get damaged by loose strands of fibrin deposited in small vessels. Multiple organs start developing platelet-fibrin thrombi (bland thrombi with no vasculitis) typically at arteriocapillary junctions. This is known as ‘thrombotic microangiopathy’. Treatment consists of plasma exchange.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 8
Correct
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The following branch of the aorta is unpaired:
Your Answer: Coeliac artery
Explanation:Branches that stem from the abdominal aorta can be divided into three: the visceral branches, parietal branches and terminal branches. Of the visceral branches, the suprarenal, renal, testicular and ovarian arteries are paired while the coeliac artery and superior and inferior mesenteric arteries are unpaired. Of the parietal branches the inferior phrenic and lumbar arteries are paired while the middle sacral artery is unpaired. The terminal branches i.e. the common iliac arteries are paired.
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This question is part of the following fields:
- Abdomen
- Anatomy
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Question 9
Incorrect
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From which branchial (pharyngeal) pouch does the inferior parathyroid gland arise?
Your Answer: 1st
Correct Answer: 3rd
Explanation:The following structures arise from each branchial pouch:
1st pouch – eustachian tube, middle ear, mastoid, and inner layer of the tympanic membrane
2nd pouch – middle ear, palatine tonsils
3rd pouch – inferior parathyroid glands, thymus
4th pouch – superior parathyroid glands, ultimobranchial body which forms the parafollicular C-cells of the thyroid gland, musculature and cartilage of larynx (along with the sixth pharyngeal pouch)
5th pouch – rudimentary structure
6th pouch – along with the fourth pouch, contributes to the formation of the musculature and cartilage of the larynx.
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This question is part of the following fields:
- Anatomy
- Endocrine; Embryology
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Question 10
Incorrect
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A patient presents with loss of pain and temperature sensation in the left leg. He is likely to have a lesion involving:
Your Answer: Left anterior spinothalamic tract
Correct Answer: Right lateral spinothalamic tract
Explanation:The spinothalamic tract is a sensory pathway originating in the spinal cord that transmits information to the thalamus. There are two main parts of the spinothalamic tract: the lateral spinothalamic tract transmits pain and temperature and the anterior spinothalamic tract transmits touch (crude touch). The decussation of this pathway occurs at the level of the spinal cord. Hence, a unilateral lesion of the lateral spinothalamic tract causes contralateral loss of pain and temperature.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 11
Incorrect
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A 47-year old-woman diagnosed with pancreatitis presented to the emergency department complaining of a worsening shortness of breath, fever, agitation and cough. Oxygen saturation was 67% in room air. Her respiratory status continued to deteriorate therefore she was intubated. She was admitted to the intensive care unit for management. Chest X-ray demonstrated bilateral perihilar opacities. The patient failed conventional treatment and died several days later. At autopsy, the lung shows growth of type 2 pneumocytes and thickened alveolar walls. What is the most probable diagnosis?
Your Answer: Bronchiectasis
Correct Answer: Adult respiratory distress syndrome
Explanation:Acute (or adult) respiratory distress syndrome (ARDS) is a life-threatening lung condition characterised by a non-cardiogenic pulmonary oedema that leads to acute respiratory failure. The most common risk factors for ARDS include trauma with direct lung injury, sepsis, pneumonia, pancreatitis, burns, drug overdose, massive blood transfusion and shock. Acute onset of dyspnoea with hypoxemia, anxiety and agitation is typical. Chest X ray most commonly demonstrates bilateral pulmonary infiltrates. Histological changes include the exudative, proliferative and fibrotic phase. ARDS is mainly a clinical diagnosis.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 12
Incorrect
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Glucose is not secreted by the kidneys, and is filtered without a limit. What is the transport maximum for glucose?
Your Answer: 850 mg/dl
Correct Answer: 300 mg/dl
Explanation:Transport maximum (or Tm) refers to the point at which increases in concentration do not result in an increase in movement of a substance across a membrane. Glucose is not secreted, thus excretion = filtration – reabsorption. Both filtration and reabsorption are directly proportional to the concentration of glucose in the plasma. However, reabsorption has a transport maximum of about 300 mg/dl in healthy nephrons, while filtration has effectively no limit (within reasonable physiological ranges). So, if the concentration rises above 300 mg/dl, the body cannot retain all the glucose, leading to glucosuria. Glucosuria is nearly always caused by elevated blood glucose levels, most commonly due to untreated diabetes mellitus.
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This question is part of the following fields:
- Physiology
- Renal
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Question 13
Incorrect
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A 49-year-old woman with acute renal failure has a total plasma [Ca2+] = 2. 5 mmol/l and a glomerular filtration rate of 160 l/day. What is the estimated daily filtered load of calcium?
Your Answer: 550 mmol/day
Correct Answer: 240 mmol/day
Explanation:Calcium is the most abundant mineral in the human body. The average adult body contains in total approximately 1 kg of calcium of which 99% is in the skeleton in the form of calcium phosphate salts. The extracellular fluid (ECF) contains approximately 22 mmol, of which about 9 mmol is in the plasma. About 40% of total plasma Ca2+ is bound to proteins and not filtered at the glomerular basement membrane. Therefore, the estimated daily filtered load is 1.5 mmol/l × 160 l/day = 240 mmol/day. The exact amount of free versus total Ca2+ depends on the blood pH: free Ca2+ increases during acidosis and decreases during alkalosis.
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This question is part of the following fields:
- Physiology
- Renal
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Question 14
Incorrect
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In the kidney, the macula densa is an area of closely packed specialized cells lining the wall of the:
Your Answer: Loop of Henlé
Correct Answer: Distal convoluted tubule
Explanation:In the kidney, the macula densa is an area of closely packed specialised cells lining the region of the distal convoluted tubule (DCT) lying next to the glomerular vascular pole. The cells of the macula densa are sensitive to the ionic content and water volume of the fluid in the DCT, producing signals that promote renin secretion by other cells of the juxtaglomerular apparatus.
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This question is part of the following fields:
- Physiology
- Renal
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Question 15
Incorrect
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Selective destruction of which of the following cells will affect antibody synthesis?
Your Answer: T lymphocytes
Correct Answer: Plasma cells
Explanation:Plasma cell are memory cells. After the antigen Is engulfed by the B cells it is presented to the CD4+ helper cells via the MCH II receptor and this leads to their activation which in turn stimulates the B cells to form antibodies against that specific antigen. Some B cells differentiate into plasma cells also called memory cells that get activated after subsequent infection.
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This question is part of the following fields:
- General
- Physiology
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Question 16
Incorrect
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Which of the following is a likely cause of jaundice?
Your Answer: Common bile duct obstruction if the serum amino transferases are elevated and alkaline phosphatase is low
Correct Answer: Hepatic disease if plasma albumin is low and serum aminotransferase elevations > 500 units
Explanation:Jaundice can occur due to any of the possible causes and treatment depends upon diagnosing the correct condition. Mild hyperbilirubinemia with normal levels of aminotransferase and alkaline phosphatase is often unconjugated (e.g., due to haemolysis or Gilbert’s syndrome rather than hepatobiliary disease). Moderate or severe hyperbilirubinemia along with increased urinary bilirubin (bilirubinuria), high alkaline phosphatase or aminotransferase levels suggest hepatobiliary disease. Hyperbilirubinemia produced by any hepatobiliary disease is largely conjugated. In this case, other blood tests include hepatitis serology for suspected hepatitis, prothrombin time (PT) or international normalised ratio (INR), albumin and globulin levels, and antimitochondrial antibody levels (suspected primary biliary cirrhosis). Low albumin and high globulin levels suggest chronic rather than acute liver disease. In cases where there is only a an elevation of alkaline phosphatase, γ-glutamyl transpeptidase (GGT) levels should be checked – the levels of which will be found high in hepatobiliary disease, but not in bone disorder which can also lead to elevated alkaline phosphatase levels. In diseases of hepatobiliary origin, aminotransferase elevations > 500 units suggest a hepatocellular cause, whereas disproportionate increases of alkaline phosphatase (e.g., alkaline phosphatase > 3 times normal and aminotransferase < 200 units) suggest cholestasis. Because hepatobiliary disease alone rarely causes bilirubin levels > 30 mg/dl, higher levels are suggestive of a combination of severe hepatobiliary disease and haemolysis or renal dysfunction. Imaging is best for diagnosing infiltrative and cholestatic causes of jaundice. Liver biopsy is rarely needed, but can be of use in intrahepatic cholestasis and in some types of hepatitis.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 17
Incorrect
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A 60-year-old male who was admitted due to cerebrovascular disease on his 5th hospital stay developed pneumonia. The most likely organism that causes hospital acquired pneumonia is pseudomonas aeruginosa. What is the most likely mechanism for the pathogenesis on pseudomonas infection?
Your Answer: Endotoxin
Correct Answer: Exotoxin
Explanation:Pseudomonas aeruginosa is a common Gram-negative, rod-shaped bacterium that can cause disease in plants and animals, including humans. It is citrate, catalase, and oxidase positive. P. aeruginosa uses the virulence factor exotoxin A to inactivate eukaryotic elongation factor 2 via ADP-ribosylation in the host cell, much the same as the diphtheria toxin does. Without elongation factor 2, eukaryotic cells cannot synthesize proteins and necrotise. The release of intracellular contents induces an immunologic response in immunocompetent patients.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 18
Incorrect
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Following an accident, a man was unable to extend the wrist and metacarpophalangeal joints, despite sensation being intact. Which nerve was likely damaged?
Your Answer: Median nerve
Correct Answer: Posterior interosseous nerve
Explanation:The posterior interosseous nerve which causes, what is known as the posterior interosseous nerve syndrome. The nerve is compressed before it bifurcates to form the medial and lateral branches. The compression is beyond the origin of the branches to the radial wrist extensors and the radial sensory nerve. The result of such a case is paralysis of the digital extensors and the extensor carpi ulnaris, resulting in dorsoradial deviation of the wrist.
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This question is part of the following fields:
- Anatomy
- Upper Limb
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Question 19
Incorrect
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A 42 year old man with an abdominal aortic aneurysm (AAA) underwent an abdominal aortic angiography which revealed that his inferior mesenteric artery was occluded. If this patient showed no symptoms, the most likely reason is that the area the inferior mesenteric artery supplies, must be supplied by collateral blood flow from which arteries?
Your Answer: Sigmoid and superior rectal
Correct Answer: Left and middle colic
Explanation:The arterial branches that form an anastomosis between the superior mesenteric artery and the inferior mesenteric artery are the left colic artery and the middle colic artery. The middle colic artery is the most distal branch of the superior mesenteric artery while the left colic forms the most proximal branch of the inferior mesenteric artery. These two arteries will give collateral blood flow in the case that the inferior mesenteric artery gets occluded. The superior mesenteric artery gives off the following branches; ileocolic, appendicular, ileal artery, right colic and middle colic arteries. The left colic, sigmoid and superior rectal arteries are branches of the inferior mesenteric artery. The marginal artery branches off directly from the abdominal aorta.
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This question is part of the following fields:
- Abdomen
- Anatomy
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Question 20
Correct
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A victim of mob justice was brought to the A & E with a stab wound in the anterior chest 2 cm lateral to the left sternal border. He underwent an emergency thoracotomy that revealed clots in the pericardium, with a puncture wound in the right ventricle. To evacuate the clots from the pericardial cavity the surgeon slipped his hand behind the heart at its apex. He extended his finger upwards until its tip was stopped by a line of pericardial reflection which forms the:
Your Answer: Oblique pericardial sinus
Explanation:Transverse sinus: part of pericardial cavity that is behind the aorta and pulmonary trunk and in front of the superior vena cava separating the outflow vessels from the inflow vessels.
Oblique pericardial sinus: is behind the left atrium where the visceral pericardium reflects onto the pulmonary veins and the inferior vena cava. Sliding a finger under the heart will take you to this space.
Cardiac notch: indentation of the ‘of the heart’ on the superior lobe of the left lung.
Hilar reflection: the reflection of the pleura onto the root of the lung to continue as mediastinal pleura.
Costomediastinal recess: part of the pleural sac where the costal pleura transitions to become the mediastinal pleura.
Sulcus terminalis: a groove between the right atrium and the vena cava
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This question is part of the following fields:
- Anatomy
- Thorax
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Question 21
Correct
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The wound healing process is documented in patients undergoing laparoscopic procedures. The port incisions are sutured closed and the wounds observed every few weeks for re-epithelialisation and tensile strength. Which substance is mostly likely to be found at a cellular level involved in wound healing?
Your Answer: Tyrosine kinase
Explanation:Cell surface growth factor receptors require intercellular proteins such as tyrosine kinase which are necessary to initiate a series of events that eventually lead to cell division and growth. Tyrosine kinase is an enzyme that transfers a phosphate group to the tyrosine residue in a protein. This phosphorylation will lead to an up regulation of the enzyme activity.
Fibronectin acts in the extracellular matrix to bind macromolecules (such as proteoglycans) via integrin receptors to aid attachment and migration of cells.
Laminin is an extracellular matrix component that is abundant in basement membranes.
Hyaluronic acid is one of the proteoglycans in the extracellular matrix.
Collagen fibres are part of the extracellular matrix that gives strength and stability to connective tissues.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 22
Incorrect
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The gradual depolarization in-between action potentials in pacemaker tissue is a result of?
Your Answer: An increase in the ‘delayed rectifier’ current due to outward movement of inward K+ (IK)
Correct Answer: A combination of gradual inactivation outward IK along with the presence of an inward ‘funny’ current (If) due to opening of channels permeable to both Na+ and K+ ions
Explanation:One of the characteristic features of the pacemaker cell is the generation of a gradual diastolic depolarization also called the pacemaker potential. In phase 0, the upstroke of the action potential caused by an increase in the Ca2+ conductance, an influx of calcium occurs and a positive membrane potential is generated. The next is phase 3 which is repolarization caused by increased K+ conductance as a result of outwards K+ current. Phase 4 is a slow depolarization which accounts for the pacemaker activity, caused by increased conductance of Na+, inwards Na+ current called IF. it is turned on by repolarization.
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This question is part of the following fields:
- General
- Physiology
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Question 23
Incorrect
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Which lymph nodes are most likely to enlarge due to the spread of infection through the lymphatic channels in a patient with a boil on his scrotum?
Your Answer: Internal iliac nodes
Correct Answer: Superficial inguinal nodes
Explanation:The superficial inguinal nodes drain the perineum and the external genitalia which include the scrotum and the labia majora. The testes, however, drain to the lumbar nodes.
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This question is part of the following fields:
- Anatomy
- Pelvis
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Question 24
Correct
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A patient presented with continuous bleeding several hours after dental extraction. Which of the following findings is most often associated with clinical bleeding?
Your Answer: Factor IX deficiency
Explanation:Factor IX deficiency, also called Haemophilia B or Christmas disease, is a disorder caused by missing or defective clotting factor IX. Deficiency of the factor IX causes irregular bleeding that can happen spontaneously or after mild trauma, surgery and dental extractions.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 25
Incorrect
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What is the normal duration of the ST segment?
Your Answer: 0.2 s
Correct Answer: 0.08 s
Explanation:The ST segment lies between the QRS complex and the T-wave. The normal duration of the ST segment is 0.08 s. ST-segment elevation or depression may indicate myocardial ischaemia or infarction.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 26
Incorrect
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Which of the following can occur even in the absence of brainstem co-ordination?
Your Answer: Defecation
Correct Answer: Gastric emptying
Explanation:Although gastric emptying is under both neural and hormonal control, it does not require brainstem co-ordination. Increased motility of the orad stomach (decreased distensibility) or of the distal stomach (increased peristalsis), decreased pyloric tone, decreased duodenal motility or a combination of these, all increase the rate of gastric emptying. The major control mechanism for gastric emptying is through duodenal gastric feedback. The duodenum has receptors for the presence of acid, carbohydrate, fat and protein digestion products, osmolarity different from that of plasma, and distension. Activating these receptors decreases the rate of gastric emptying. Neural mechanisms involve both enteric and vagal pathways and a vagotomy impairs the gastric emptying regulation. CCK (cholecystokinin) slows gastric emptying at physiological levels of the hormone. Gastrin, secretin and glucose-1-phosphate also slow gastric emptying, but require higher doses.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 27
Incorrect
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Which of the following is NOT true regarding malignant hyperpyrexia
Your Answer: Dantrolene is the treatment
Correct Answer: It can be caused by nitrous oxide
Explanation:Malignant hyerpyrexia occurs in 1 in 150,000. All inhalational anaesthetic agents and suxamethonium, except nitrous oxide can cause malignant hyperpyrexia.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 28
Incorrect
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An experiment is carried out to observe engulfment and phagocytosis of microbes. Strep pneumoniae are added to a solution of leukocytes with a substance added to enhance the process of phagocytosis. What is this substance?
Your Answer: NADPH oxidase
Correct Answer: Complement C3b
Explanation:C3b is cleaved from C3 complement with the help of the enzyme C3- convertase. It binds to the cell surface of the offending substance and opsonizes it. This makes it easy for the phagocytes to detect and eliminate them.
IgM does not act as an opsonin but igG does.
Selectins aid leukocytes to bind to the endothelial surfaces.
C5a is a chemo-attractant and histamine a vasodilator.
NADPH oxidises offending substance after phagocytosis within the macrophage.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 29
Correct
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Which antibiotic acts by inhibiting protein synthesis?
Your Answer: Erythromycin
Explanation:Penicillins and cephalosporins (e.g. cefuroxime, cefotaxime, ceftriaxone) inhibit bacterial cell wall synthesis through the inhibition of peptidoglycan cross-linking.
Macrolides (e.g. erythromycin), tetracyclines, aminoglycosides and chloramphenicol act by interfering with bacterial protein synthesis.
Sulphonamides (e.g. trimethoprim, co-trimoxazole) work by inhibiting the synthesis of nucleic acid
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This question is part of the following fields:
- Pharmacology; Microbiology
- Physiology
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Question 30
Incorrect
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An 80 year-old quadriplegic man has been lying supine for 7 weeks in a critical care ward. He develops a right lung abscess that is draining by gravity to a particular region of the lung. Which is the most likely site of pus collection?
Your Answer: Middle lobe
Correct Answer: Superior segment of the lower lobe
Explanation:The superior segmental bronchus of the lower lobe of the right lung branches posteriorly off the intermediate bronchus or the inferior lobe bronchus. It is therefore more likely to receive fluid or foreign bodies that enter the right main bronchus especially when the patient is supine.
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This question is part of the following fields:
- Anatomy
- Thorax
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SESSION STATS - PERFORMANCE PER SPECIALTY
