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Question 1
Correct
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The renal cortex and medulla, if seen under the microscope, is lacking one of the following:
Your Answer: Squamous epithelium
Explanation:Capillaries, Henle’s loop, collecting ducts, Bertin columns and type IV collagen in glomerular basement membrane are all structures present in the renal cortex or medulla. The squamous epithelium is the only one that is lacking in both the renal cortex and medulla, because normally it is not found above the outer urethra.
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This question is part of the following fields:
- Pathology
- Renal
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Question 2
Incorrect
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Which of the following muscles aid in inspiration?
Your Answer:
Correct Answer: Diaphragm and external intercostals
Explanation:The diaphragm and external intercostals are muscles of inspiration as they increase the volume of thoracic cavity and reduce the intrathoracic pressure. Muscles of expiration include abdominal muscles and internal intercostals.
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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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What is the percentage of bone calcium that is freely exchangeable with the extracellular fluid that is available for buffering changes in the calcium ion balance?
Your Answer:
Correct Answer: 1%
Explanation:Around 1% of calcium in the body is available for buffering changes in calcium ion balance. These are mainly derived from the bone that are freely exchangeable with extracellular fluid.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 4
Incorrect
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In multiple myeloma, which of these cell types confirms the diagnosis when found in a smear of bone marrow aspirate?
Your Answer:
Correct Answer: Plasma cells
Explanation:A bone marrow aspiration is the diagnostic test for multiple myeloma, which is a malignant bone tumour that usually affects older adults. The smear reveals clusters of plasma cells, while X-rays tend to show circumscribed lytic lesions or diffuse demineralisation.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 5
Incorrect
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Diuretics that act on the ascending limb of the loop of Henle produce:
Your Answer:
Correct Answer: Reduced active transport of sodium
Explanation:The loop of Henlé leads from the proximal convoluted tubule to the distal convoluted tubule. Its primary function uses a counter current multiplier mechanism in the medulla to reabsorb water and ions from the urine. It can be divided into four parts:
1. Descending limb of loop of Henlé – low permeability to ions and urea, while being highly permeable to water
2. Thin ascending limb of loop of Henlé – not permeable to water, but it is permeable to ions
3. Medullary thick ascending limb of loop of Henlé – sodium (Na+), potassium (K+) and chloride (Cl–) ions are reabsorbed by active transport. K+ is passively transported along its concentration gradient through a K+ channel in the basolateral aspect of the cells, back into the lumen of the ascending limb.
4. The cortical thick ascending limb – the site of action where loop diuretics such as furosemide block the K+/Na+/2Cl− co-transporters = reduced active transport.
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This question is part of the following fields:
- Physiology
- Renal
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Question 6
Incorrect
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A 13 year old girl presented with signs of shortness of breath, chest pain, non-productive cough, oedema of the lower extremities and cyanosis of the fingertips. She has a history of a ventricular septal defect not surgically corrected. The most probable cause of these symptoms is:
Your Answer:
Correct Answer: Shunt reversal
Explanation:A ventricular septal defect (VSD) is a common form of congenital heart defects and is characterised by the presence of a hole in the wall that separates the right from the left ventricle. Medium or large defects can cause many complications. One of these complication is Eisenmenger syndrome, characterised by reversal of the shunt (from left-to-right shunt into a right-to-left) ,cyanosis and pulmonary hypertension.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 7
Incorrect
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A patient is suspected to have Blastomyces dermatidis infection. The patient contracted the disease most likely through which port of entry?
Your Answer:
Correct Answer: Respiratory tract
Explanation:Blastomycosis disease is a fungal infection acquired through inhalation of the spores. It caused by the organism Blastomyces dermatitidis and manifests as a primary lung infection in about 70% of cases. The onset is relatively slow and symptoms are suggestive of pneumonia.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 8
Incorrect
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Which of the following diseases is caused by intra-articular and/or extra-articular deposition of calcium pyrophosphate dihydrate (CPPD) crystals, due to unknown causes?
Your Answer:
Correct Answer: Pseudogout
Explanation:Pseudogout or chondrocalcinosis is a rheumatological disease caused by the accumulation of crystals of calcium pyrophosphate dihydrate (CPPD) in the connective tissues. It is frequently associated with other conditions, such as trauma, amyloidosis, gout, hyperparathyroidism and old age, which suggests that it is secondary to degenerative or metabolic changes in the tissues. The knee is the most commonly affected joint. It causes symptoms similar to those of rheumatoid arthritis or osteoarthritis.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 9
Incorrect
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A 30-year-old female was alarmed when she started to experience hair loss and balding, however, she also noted increased hair on her face and body and developed an acne breakout. Deepening of her voice also became prominent. She was referred to an oncologist and was diagnosed with a hormone-producing tumour. What is the most likely diagnosis ?
Your Answer:
Correct Answer: Arrhenoblastoma
Explanation:Arrhenoblastoma, known as ‘Sertoli–Leydig tumour’ is a rare ovarian stromal neoplasm that secretes testosterone. It is mostly seen in women in the reproductive years. The key clinical features of this tumour is due to excessive production of testosterone which leads to progressive masculinisation in a woman who was typical normal beforehand. The lesion tends to grow slowly and rarely metastasises. Treatment is surgical removal of the tumour and the prognosis is generally good.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 10
Incorrect
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A 15-day old baby was brought to the emergency department with constipation for 4 days. On examination, the abdomen of the baby was found to be distended and tender all over. No bowel sounds were heard. A sigmoid colon biopsy was carried out, which showed absent ganglion cells. What is the diagnosis?
Your Answer:
Correct Answer: Hirschsprung’s disease
Explanation:Hirschsprung’s disease is characterized by congenital absence of the autonomic plexus (Meissner’s and Auerbach’s plexus) in the intestinal wall. Usually limited to the distal colon, it can occasionally involve the entire colon or even the small bowel. There is abnormal or absent peristalsis in the affected segment, resulting in continuous spasm of smooth muscle and partial/complete obstruction. This causes accumulation of intestinal contents and dilatation of proximal segment. Skip lesions are highly uncommon. This disease is seen early in life with 15% patients presenting in first month, 60% by 1 year of age and 85% by the age of 4 years. Symptoms include severe and complete constipation, abdominal distension and vomiting. Patients with involvement of ultra-short segments might have mild constipation with intervening diarrhoea. In older children, symptoms include failure to thrive, anorexia, and lack of an urge to defecate. On examination, an empty rectum is revealed with stool palpable high up in the colon. If not diagnosed in time, it can lead to Hirschsprung’s enterocolitis (toxic megacolon), which can be fulminant and lead to death. Diagnosis involves a barium enema or a rectal suction biopsy. Barium enema shows a transition in diameter between the dilated, normal colon proximal to the narrowed, affected distal segment. It is to be noted that barium enema should be done without prior preparation, which can dilate the abnormal segment, leading to a false-negative result. A 24-hour post-evacuation film can be obtained in the neonatal period – if the colon is still filled with barium, there is a high likelihood of Hirschsprung’s disease. Full-thickness rectal biopsy is diagnostic by showing the absence of ganglion cells. Acetylcholinesterase staining can be done to highlight the enlarged nerve trunks. Abnormal innervation can also be demonstrated by rectal manometry.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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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:
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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A 45-year-old-female is suspected to have a pulmonary mass. Supposing that she has a neoplasm, which of the following are most commonly found to involve the lung:
Your Answer:
Correct Answer: Pulmonary metastases
Explanation:Lung metastases occur when a cancer started in another part of the body (primary site) spreads to the lungs. The lungs are among the most common site where cancer can spread due to its rich systemic venous drainage, almost every type of cancer can spread to the lung. The most common types of cancer that spread to the lung are breast, colorectal, kidney, testicular, bladder, prostate, head and neck cancers.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 13
Incorrect
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Extracellular body fluid as compared with intracellular body fluid:
Your Answer:
Correct Answer: Is relatively rich in glucose
Explanation:The percentages of body water contained in various fluid compartments add up to total body water (TBW). This water makes up a significant fraction of the human body, both by weight and by volume. Ensuring the right amount of body water is part of fluid balance, an aspect of homeostasis. The extracellular fluid (ECF) includes all fluids outside the cells. This fluid can be divided into three fluid departments: interstitial (in the tissue spaces) fluid, blood plasma and lymph, and specialised compartments called transcellular fluid. The extracellular fluid surrounds all the cells in the body and is in equilibrium with the intracellular fluid. So, its composition must remain fairly constant even though substances are passing into and out of the cells. The interstitial fluid, though called a fluid, is in a reality a gel-like composition made up of: water, proteoglycan molecules and collagen. The extracellular fluid constitutes 40% of total body water, with intracellular fluid making up the remaining 60%. It is relatively rich in glucose.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 14
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:
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 15
Incorrect
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Which of the following mediators of inflammation requires arachidonic acid for synthesis?
Your Answer:
Correct Answer: Prostaglandins
Explanation:Arachidonic acid is normally present in the phospholipids that make up the cell membrane and is cleaved by phospholipase A2 from the phospholipid. Arachidonic acid is a precursor for the production of eicosanoids which include: 1) prostaglandins, prostacyclins and thromboxane, 2) leukotrienes and 3) anandamides. The production of these products along with their action on the body is called the arachidonic acid cascade.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 16
Incorrect
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A 47-year-old male smoker, who had been self-medicating with oral steroids for the last two years due to persistent breathlessness presented to the doctor complaining of a productive cough, fever and chest pain. A chest X-ray revealed bilateral patchy opacities. He was diagnosed with bilateral bronchopneumonia. Which of these organisms is most probably causing these findings?
Your Answer:
Correct Answer: Nocardia asteroides
Explanation:Nocardia is a Gram-positive aerobic actinomycete. Several species have been identified but the most common human pathogen is Nocardia asteroides. The predominant clinical finding in the majority of patients affected by nocardiosis is pulmonary disease. Predisposing factors for pulmonary nocardiosis include leukaemia, human immunodeficiency virus (HIV) infection, organ transplantation, diabetes and receiving prolonged corticosteroids.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 17
Incorrect
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A 47 -year-old male was admitted due to a bleeding peptic ulcer. On his 3rd hospital day, he developed a cardiac arrhythmia. His serum potassium was markedly elevated. What is the most likely cause of hyperkalaemia in this patient?
Your Answer:
Correct Answer: Multiple blood transfusions
Explanation:Patients with gastrointestinal bleeding often require blood transfusion. Among the various side effects of blood transfusions, is the increase of potassium levels. The use of stored blood for transfusions is followed by an increase of serum potassium levels. Potassium level increases are more pronounced in patients who receive blood stored for more than 12 d. Furthermore, the lysis and destruction of red blood cells, especially in the transfusion of older PRBCs, can further increase potassium levels. Excessive use of a PPi has been associated with hyperkaelemia however would be less likely in this acute setting.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 18
Incorrect
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Which portion of the renal tubule absorbs amino acids and glucose?
Your Answer:
Correct Answer: Proximal convoluted tubule
Explanation:In relation to the morphology of the kidney as a whole, the convoluted segments of the proximal tubules are confined entirely to the renal cortex. Glucose, amino acids, inorganic phosphate and some other solutes are reabsorbed via secondary active transport in the proximal renal tubule through co-transport channels driven by the sodium gradient.
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This question is part of the following fields:
- Physiology
- Renal
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Question 19
Incorrect
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A 27-year-old woman, who had been taking a combined oral contraceptive for 6 months, presented with inguinal pain and oedema of the left leg. Which of the following investigations would you recommend to help confirm the diagnosis?
Your Answer:
Correct Answer: Duplex scan
Explanation:Oral combined contraceptive pill (OCCP) is a drug used for birth control and treating a number of other conditions. Women who take the OCP have a higher risk of developing deep vein thrombosis (DVT), usually in the legs. Duplex ultrasonography is a safe and non-invasive technique which is used for diagnosing the presence of lower extremity thrombi.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 20
Incorrect
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A 59-year old gentleman admitted for elective cholecystectomy was found to have a haemoglobin 12.5 g/dl, haematocrit 37%, mean corpuscular volume 90 fl, platelet count 185 × 109/l, and white blood cell count 32 × 109/l; along with multiple, small mature lymphocytes on peripheral smear. The likely diagnosis is:
Your Answer:
Correct Answer: Chronic lymphocytic leukaemia
Explanation:CLL or chronic lymphocytic leukaemia is the most common leukaemia seen in the Western world. Twice more common in men than women, the incidence of CLL increases with age. About 75% cases are seen in patients aged more than 60 years. The blood, marrow, spleen and lymph nodes all undergo infiltration, eventually leading to haematopoiesis (anaemia, neutropenia, thrombocytopenia), hepatomegaly, splenomegaly and decreased production of immunoglobulin. In 98% cases, CD+5 B cells undergo malignant transformation.
Often diagnosed on blood tests while being evaluated for lymphadenopathy, CLL causes symptoms like fatigue, anorexia, weight loss, pallor, dyspnoea on exertion, abdominal fullness or distension. Findings include multiple lymphadenopathy with minimal-to- moderate hepatomegaly and splenomegaly. Increased susceptibility to infections is seen. Herpes Zoster is common. Diffuse or maculopapular skin infiltration can also be seen in T-cell CLL.
Diagnosis is by examination of peripheral blood smear and marrow: hallmark being a sustained, absolute leucocytosis (>5 ×109/l) and increased lymphocytes in the marrow (>30%). Other findings can include hypogammaglobulinemia (<15% of cases) and, rarely, raised lactate dehydrogenase (LDH). Only 10% cases demonstrate moderate anaemia and/or thrombocytopenia.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 21
Incorrect
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After a total colectomy and ileotomy, a 50-year old diabetic man who was a known case of diabetic nephropathy had persistent metabolic acidosis. The patient appeared well perfused, with normal vital signs and normal fluid balance. Investigations revealed:
Sodium = 132 mmol/l
Potassium = 6.6 mmol/l
Creatinine = 185 μmol/l (2.16 mg/dl)
Chloride = 109 μmol/l
8am cortisol = 500 nmol/l (18 μg/dl)
pH = 7.29, p(CO2) = 27 mmHg
p(O2) = 107 mmHg
standard bicarbonate = 12 mmol/l.
What is the likely causes of his acidosis?Your Answer:
Correct Answer: Renal tubular acidosis
Explanation:Acidosis here is due to low bicarbonate. The low p(CO2) is seen in compensation. The anion gap is normal, ruling out intra-abdominal ischaemia (which leads to metabolic acidosis). If it was a gastrointestinal aetiology, low potassium would be seen. The history of diabetic nephropathy predisposes to renal tubular acidosis. Type 4 (hyporeninaemic hypoaldosteronism) is associated with high potassium and is found in diabetic and hypertensive renal disease.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 22
Incorrect
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A 56-year-old man undergoes tests to determine his renal function. His results over a period of 24 hours were:
Urine flow rate: 2. 0 ml/min
Urine inulin: 1.0 mg/ml
Plasma inulin: 0.01 mg/ml
Urine urea: 260 mmol/l
Plasma urea: 7 mmol/l
What is the glomerular filtration rate?Your Answer:
Correct Answer: 200 ml/min
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal (kidney) glomerular capillaries into the Bowman’s capsule per unit time. GFR is equal to the inulin clearance because inulin is freely filtered into Bowman’s capsule but is not reabsorbed or secreted. The clearance (C) of any substance can be calculated as follows: C = (U × V)/P, where U and P are the urine and plasma concentrations of the substance, respectively and V is the urine flow rate. Thus, glomerular filtration rate = (1.0 × 2. 0)/0.01 = 200 ml/min.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 23
Incorrect
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Which of the following is the most abundant WBC seen in a smear from a healthy person.
Your Answer:
Correct Answer: Neutrophils
Explanation:neutrophils are the most abundant cell type of the WBC. These phagocytes are found normally in the blood and increase in number are seen during an acute inflammation. These the percentages of WBC in blood Neutrophils: 40 to 60%
Lymphocytes: 20 to 40%
Monocytes: 2 to 8%
Eosinophils: 1 to 4%
Basophils: 0.5 to 1%
Band (young neutrophil): 0 to 3%. eosinophils, basophils, neutrophils are known as granulocytes and monocytes and lymphocytes as agranulocytes.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 24
Incorrect
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Pseudomonas aeruginosa is a multidrug resistant pathogen that causes hospital-acquired infections. It is usually treated with piperacillin or another antibiotic. Which of the following is the other antibiotic?
Your Answer:
Correct Answer: Azlocillin
Explanation:Azlocillin, like piperacillin, is an acylampicillin antibiotic with an extended spectrum of activity and greater in vitro potency than the carboxypenicillins. Azlocillin is similar to mezlocillin and piperacillin. It demonstrates antibacterial activity against a broad spectrum of bacteria, including Pseudomonas aeruginosa.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 25
Incorrect
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A young girl who presented with a clinical picture of type I hypersensitivity reaction with eosinophilia is most likely to have?
Your Answer:
Correct Answer: Liver flukes
Explanation:Usually a parasitic infection will be associated with a type I hypersensitivity reaction.
Amyloid deposition will not cause an immune reaction.
Organic dust will lead to a type III hypersensitivity reaction.
Cell mediated as well as humoral immune mechanism play a part in syphilis, but they are do not specifically cause a type I reaction.
Malaria is cause by plasmodium and is not cause of a hypersensitivity reaction.
Atopic dermatitis will not be accompanied by eosinophilia.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 26
Incorrect
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Decreased velocity of impulse conduction through the atrioventricular node (AV node) in the heart will lead to:
Your Answer:
Correct Answer: Increased PR interval
Explanation:AV node damage may lead to an increase in the PR interval to as high as 0.25 – 0.40 s (normal = 0.12 – 0.20 s). In the case of severe impairment, there might be a complete failure of passage of impulses leading to complete block. In this case, the atria and ventricles will beat independently of each other.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 27
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:
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 28
Incorrect
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Which part of the nephron would have to be damaged to stop the reabsorption of the majority of salt and water?
Your Answer:
Correct Answer: Proximal tubule
Explanation:The proximal tubule is the portion of the duct system of the nephron of the kidney which leads from Bowman’s capsule to the loop of Henle. It is conventionally divided into the proximal convoluted tubule (PCT) and the proximal straight tubule (PST). The proximal tubule reabsorbs the majority (about two-thirds) of filtered salt and water. This is done in an essentially iso-osmotic manner. Both the luminal salt concentration and the luminal osmolality remain constant (and equal to plasma values) along the entire length of the proximal tubule. Water and salt are reabsorbed proportionally because the water is dependent on and coupled with the active reabsorption of Na+. The water permeability of the proximal tubule is high and therefore a significant transepithelial osmotic gradient is not possible.
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This question is part of the following fields:
- Physiology
- Renal
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Question 29
Incorrect
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A 45-year-old man with short bowel syndrome requires parenteral nutrition. The solution of choice for parenteral nutrition is:
Your Answer:
Correct Answer: Crystalline amino acids
Explanation:Total parenteral nutrition (TPN), is the practice of feeding a person intravenously, circumventing the gut. It is normally used in the following situations: surgery, when feeding by mouth is not possible, when a person’s digestive system cannot absorb nutrients due to chronic disease or if a person’s nutrient requirement cannot be met by enteral feeding and supplementation. A sterile bag of nutrient solution, between 500 ml and 4L, is provided. The pump infuses a small amount (0.1–10 ml/h) continuously to keep the vein open. The nutrient solution consists of water, glucose, salts, amino acids, vitamins and sometimes emulsified fats. Ideally each patient is assessed individually before commencing on parenteral nutrition, and a team consisting of doctors, nurses, clinical pharmacists and dietitians evaluate the patient’s individual data and decide what formula to use and at what rate.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 30
Incorrect
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A Jewish man was diagnosed with haemophilia C. Which of the following factors is deficient in this form of haemophilia?
Your Answer:
Correct Answer: Factor XI
Explanation:Haemophilia C, also known as plasma thromboplastin antecedent (PTA) deficiency or Rosenthal syndrome, is a condition caused by the deficiency of the coagulation factor XI. The condition is rare and it is usually found in Ashkenazi Jews.
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This question is part of the following fields:
- Haematology
- Pathology
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SESSION STATS - PERFORMANCE PER SPECIALTY
