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Question 1
Correct
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The proximal tubule is the portion of the ductal system of the nephron of the kidney which leads from Bowman's capsule to the loop of Henle. Which of the following is most likely to be seen in a sample of fluid leaving the proximal tubule?
Your Answer: It will have no amino acids
Explanation:The proximal tubule is the portion of the duct system of the nephron leading from Bowman’s capsule to the loop of Henlé. The most distinctive characteristic of the proximal tubule is its brush border (or ‘striated border’). The luminal surface of the epithelial cells of this segment of the nephron is covered with densely packed microvilli forming a border which greatly increases the luminal surface area of the cells, presumably facilitating their reabsorptive function. Glucose, amino acids, inorganic phosphate, and some other solutes are100% reabsorbed via secondary active transport through co-transporters driven by the sodium gradient out of the nephron.
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This question is part of the following fields:
- Physiology
- Renal
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Question 2
Correct
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Gastric acid secretion is stimulated by which of the following?
Your Answer: Gastrin
Explanation:Gastric acid secretion is stimulated by three factors:
– Acetylcholine, from parasympathetic neurones of the vagus nerve that innervate parietal cells directly
– Gastrin, produced by pyloric G-cells
– Histamine, produced by mast cells.
Gastric acid is inhibited by three factors:
– Somatostatin
– Secretin
– Cholecystokinin
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 3
Correct
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QT interval in the electrocardiogram of a healthy individual is normally:
Your Answer: 0.40 s
Explanation:QT interval extends from beginning of the QRS complex till the end of he T-wave and normally lasts for 0.40 s. It is important in the diagnosis of long-QT and short-QT syndrome. The QT interval varies on the basis of heart rate and may need to be corrected.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 4
Correct
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A 28 years old women presents with a history of chronic cough with fever for the past 2 months. A chest x ray revealed a diffuse bilateral reticulonodular pattern. A transbronchial biopsy was performed and histological examination showed focal areas of inflammation with epithelioid macrophages, Langhans cells and lymphocytes. Which of the immune reaction is responsible for this?
Your Answer: Type IV hypersensitivity
Explanation:A reactivated tuberculosis with granuloma formation is characteristic of type IV reaction. It is also called a delayed type of hypersensitivity reaction and takes around 2-8 days to deliver. It is a cell mediated response with the involvement of CD8 and CD4 cells and the release of IL-1 from macrophages that further activate these CD cells.
Granulomatous reactions are mostly cell-mediated.
Type I reactions are allergic and anaphylactic reactions and type II are complement-mediated immune reactions.
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This question is part of the following fields:
- Inflammation & Immunology; Respiratory
- Pathology
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Question 5
Correct
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A patient is admitted to the ICU, and is prescribed tazobactam, amongst other drugs. What is the mechanism of action of tazobactam?
Your Answer: Inhibits beta-lactamase
Explanation:Tazobactam is a compound which inhibits the action of bacterial beta-lactamases. It is added to the extended-spectrum beta-lactam antibiotic piperacillin to broaden the spectrum of piperacillin by making it effective against organisms that express beta-lactamase and would normally degrade piperacillin.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 6
Correct
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A 42 year old women presents with end stage renal failure and is prepared to receive a kidney from her husband. HLA testing showed that they are not a 100% match and she is given immunosuppressant therapy for this. Three months later when her renal function tests were performed she showed signs of deteriorating renal function, with decreased renal output, proteinuria of +++ and RBCs in the urine. She was given antilymphocyte globulins and her condition reversed. What type of graft did this patient receive?
Your Answer: Allograft
Explanation:Allograft describes a graft between two of the same species. As the donor and the recipients are history-incompatible, rejection of the graft is common and is controlled by immunosuppressive drug therapy. Isograft and syngraft are synonymous and referred to a graft transferred between genetically identical individuals e.g. identical twins. In this case rejection is rare as they are history-compatible.
Autograft refers to transfer of one part of the body to another location.
Xenograft is transfer of tissue from another species e.g. pig to human in valve replacement surgeries and rejection is very high.
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This question is part of the following fields:
- Inflammation & Immunology; Renal
- Pathology
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Question 7
Correct
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A 35-year-old ultra marathon runner becomes severely dehydrated and collapses. This patient most likely has:
Your Answer: Decreased baroreceptor firing rate
Explanation:Baroreceptors are sensors located in the blood vessels of all vertebrate animals. They sense the blood pressure and relay the information to the brain, so that a proper blood pressure can be maintained. Acute dehydration results in decreased plasma volume and increased plasma osmolarity, since more water than salt is lost in sweat. The decrease in plasma volume leads to an inhibition of the baroreceptors and a lower firing rate. The increase in plasma osmolarity leads to increased ADH secretion and high plasma ADH levels, which increases water permeability of collecting duct cells. Therefore more water is reabsorbed by the kidneys and renal water excretion is low.
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This question is part of the following fields:
- Physiology
- Renal
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Question 8
Incorrect
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A 30-year-old man is brought to the emergency department suffering from extreme dehydration, and subsequent hypotension and tachycardia. Which part of the kidney will compensate for this loss?
Your Answer: Distal convoluted tubule
Correct Answer: Collecting ducts
Explanation:The collecting duct system of the kidney consists of a series of tubules and ducts that physically connect nephrons to a minor calyx or directly to the renal pelvis. The collecting duct system is the last component of the kidney to influence the body’s electrolyte and fluid balance. In humans, the system accounts for 4–5% of the kidney’s reabsorption of sodium and 5% of the kidney’s reabsorption of water. At times of extreme dehydration, over 24% of the filtered water may be reabsorbed in the collecting duct system.
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This question is part of the following fields:
- Physiology
- Renal
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Question 9
Correct
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Blood investigations of a patient with vitamin K deficiency revealed a prolonged prothrombin time. This coagulation abnormality is most probably due to:
Your Answer: Factor VII deficiency
Explanation:Factor VII deficiency is a bleeding disorder caused by a deficiency or reduced activity of clotting factor VII. It may be inherited or acquired at some point during life. Inherited factor VII deficiency is an autosomal recessive disorder caused by mutations of the F7 gene. Factor VII is vitamin K-dependent, as are Factors II, IX and X and therefore lack of this vitamin can cause the development of acquired factor VII deficiency. Other causes of acquired deficiency of this factor include liver disease, sepsis and warfarin therapy.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 10
Correct
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Which of the following muscles aid in inspiration?
Your 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 11
Correct
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What is the normal glomerular filtration rate?
Your Answer: 125 mL/min
Explanation:The normal glomerular filtration rate (GFR) in humans is 125 mL/min. After the age of 40, GFR decreases progressively by about 0.4–1.2 mL/min per year.
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This question is part of the following fields:
- Physiology
- Renal
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Question 12
Correct
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Elevated mean corpuscular volume with hypersegmented neutrophils and low reticulocyte index is seen in on the blood count of a middle-aged lady about to undergo elective surgery. On enquiry, she mentions feeling tired for a few months. Which of the following investigations should be carried out in her to reach a diagnosis?
Your Answer: Serum vitamin B12 and folate
Explanation:Elevated levels of MCV indicates megaloblastic anaemia, which are associated with hypersegmented neutrophils. Likely causes include vitamin B12 or folate deficiency. Megaloblastic anaemia results from defective synthesis of DNA. As RNA production continues, the cells enlarge with a large nucleus. The cytoplasmic maturity becomes greater than nuclear maturity. Megaloblasts are produced initially in the marrow, before blood. Dyspoiesis makes erythropoiesis ineffective, causing direct hyperbilirubinemia and hyperuricemia. As all cell lines are affected, reticulocytopenia, thrombocytopenia and leukopenia develop. Large, oval blood cells (macro-ovalocytes) are released in the circulation, along with presence of hypersegmented neutrophils.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 13
Correct
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A 31-year-old woman is diagnosed with adrenal hyperplasia, and laboratory samples are taken to measure serum aldosterone and another substance. Which is most likely to be the other test that was prescribed to this patient?
Your Answer: Plasma renin
Explanation:The evaluation of a patient in whom hyperaldosteronism is first to determine that hyperaldosteronism is present (serum aldosterone) and, if it is present, to differentiate primary from secondary causes of hyperaldosteronism. The aldosterone-to-renin ratio (ARR) is the most sensitive means of differentiating primary from secondary causes of hyperaldosteronism as it is abnormally increased in primary hyperaldosteronism, and decreased or normal but with high renin levels in secondary hyperaldosteronism.
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This question is part of the following fields:
- Physiology
- Renal
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Question 14
Correct
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A 45-year-old pregnant woman develops high blood pressure at 20 weeks. She complains of headaches and swollen feet, and a test reveals proteinuria (350 mg/day). Which of the following is the most likely diagnosis?
Your Answer: Pre-eclampsia
Explanation:Pre-eclampsia (PE) is a disorder of pregnancy characterized by the onset of high blood pressure (two separate readings taken at least 6 h apart of 140/90 or more) and often a significant amount of protein in the urine (>300 mg of protein in a 24-h urine sample). While blood pressure elevation is the most visible sign of the disease, it involves generalised damage to the maternal endothelium of the kidneys and liver, with the release of vasopressive factors only secondary to the original damage. Pre-eclampsia may develop at varying times within pregnancy and its progress differs among patients; most cases present pre-term. It has no known cure apart from ending the pregnancy (induction of labour or abortion). It may also present up to 6 weeks post partum. Risk factors for pre-eclampsia include obesity, prior hypertension, older age, and diabetes mellitus.
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This question is part of the following fields:
- Physiology
- Renal
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Question 15
Correct
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The anatomical dead space in a patient with low oxygen saturation, is 125 ml, with a tidal volume of 500 ml and pa(CO2) of 40 mm Hg. The dead space was determined by Fowler's method. If we assume that the patient's lungs are healthy, what will his mixed expired CO2 tension [pE(CO2)] be?
Your Answer: 30 mmHg
Explanation:According to Bohr’s equation, VD/VT = (pA(CO2) − pE(CO2))/pA(CO2), where pE(CO2) is mixed expired CO2 and pA(CO2) is alveolar CO2pressure. Normally, the pa(CO2) is virtually identical to pA(CO2). Thus, VD/VT = (pa(CO2)) − pE(CO2)/pa(CO2). By Fowler’s method, VD/VT= 0.25. In the given problem, (pa(CO2) − pE(CO2)/pa(CO2) = (40 − pE(CO2)/40 = 0.25. Thus, pE(CO2) = 30 mmHg. If there is a great perfusion/ventilation inequality, pE(CO2) could be significantly lower than 30 mm Hg, and the patient’s physiological dead space would exceed the anatomical dead space.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 16
Incorrect
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Whilst snorkelling, a 30-year old gentleman has the respiratory rate of 10/min, tidal volume of 550 ml and an effective anatomical dead space of 250 ml. Which of the following will bring about a maximum increase in his alveolar ventilation?
Your Answer: A 2x increase in tidal volume and a longer snorkel
Correct Answer: A 2x increase in tidal volume and a shorter snorkel
Explanation:Alveolar ventilation = respiratory rate × (tidal volume − anatomical dead space volume). Increase in respiratory rate simply causes movement of air in the anatomical dead space, with no contribution to the alveolar ventilation. By use of a shorter snorkel, the effective anatomical dead space will decrease and will cause a maximum rise in alveolar ventilation along with doubling of tidal volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 17
Incorrect
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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: Idiopathic thrombocytopenic purpura
Correct 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 18
Correct
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Ventricular filling follows a delay caused by?
Your Answer: AV node
Explanation:The AV node is a conducting tissue found between the atria and the ventricles of the heart. It conducts electrical signal from the atria to the ventricles and acts a delaying mechanism preventing the atria and the ventricles from contracting at the same time. This decremental conduction prevents premature ventricular contraction in cases such as atrial fibrillation. A delay in the AV node is the reason for the PR segment seen on the ECG. In certain types of supraventricular tachycardia, a person could have two AV nodes; this will cause a loop in electrical current and uncontrollably rapid heart beat. When this electricity catches up with itself, it will dissipate and return to a normal heart rate.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 19
Correct
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Hormones of the anterior pituitary include which of the following?
Your Answer: Prolactin
Explanation:The anterior pituitary gland (adenohypophysis or pars distalis) synthesizes and secretes:
1. FSH (follicle-stimulating hormone)
2. LH (luteinizing hormone)
3. Growth hormone
4. Prolactin
5. ACTH (adrenocorticotropic hormone)
6. TSH (thyroid-stimulating hormone).
The posterior pituitary gland (neurohypophysis) stores and secretes 2 hormones produced by the hypothalamus:
1. ADH (antidiuretic hormone or vasopressin)
2. Oxytocin
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 20
Correct
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A 7-year old child from a rural setting complains of recurrent abdominal pain. The child is found to have a heavy parasitic infestation and anaemia. Which type of anaemia is most likely seen in this patient?
Your Answer: Iron deficiency anaemia
Explanation:The most common cause of iron deficiency anaemia in children in developing countries is parasitic infection (hookworm, amoebiasis, schistosomiasis and whipworm).
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 21
Correct
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Arterial blood gas analysis of a man admitted with acute exacerbation of chronic obstructive pulmonary disease (COPD) showed the following: pH = 7.28, p(CO2) = 65.5 mmHg, p(O2)= 60 mmHg and standard bicarbonate = 30.5 mmol/l. This patient had:
Your Answer: Respiratory acidosis
Explanation:Acidosis with high p(CO2) and normal standard bicarbonate indicates respiratory acidosis, commonly seen in acute worsening of COPD patients. Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with a low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg).
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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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Glucose is not secreted by the kidneys, and is filtered without a limit. What is the transport maximum for glucose?
Your Answer: 200 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 23
Correct
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Which of the following is the cause of flattened (notched) T waves on electrocardiogram (ECG)?
Your Answer: Hypokalaemia
Explanation:The T-wave is formed due to ventricular repolarisation. Normally, it is seen as a positive wave. It can be normally inverted (negative) in V1 (occasionally in V2-3 in African-Americans/Afro-Caribbeans). Hyperacute T-waves are the earliest ECG change of acute myocardial infarction. ECG findings of hyperkalaemia include high, tent-shaped T-waves, a small P-wave and a wide QRS complex. Hypokalaemia results in flattened (notched) T-waves, U-waves, ST-segment depression and prolonged QT interval.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 24
Correct
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In a cardiac cycle, what event does the closing of atrioventricular (AV) valves coincide with?
Your Answer: First heart sound
Explanation:In the cardiac cycle, the closing of the atrioventricular (AV) valves coincides with the onset of ventricular systole. This event marks the beginning of the isovolumetric contraction phase, where the ventricles begin to contract, but the volume of blood in the ventricles remains the same because both the AV valves and the semilunar valves (aortic and pulmonary valves) are closed. The closing of the AV valves produces the first heart sound, known as “S1” or “lub.”
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 25
Correct
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The patient who is diagnosed with bladder cancer asked his physician, what could have been the contributing factor in the development of his bladder cancer?
Your Answer: Smoking
Explanation:Tobacco smoking is the main known contributor to urinary bladder cancer. In most populations, smoking is associated with over half of bladder cancer cases in men and one-third of cases among women.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 26
Correct
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An ECG of a 30 year old woman revealed low voltage QRS complexes. This patient is most probably suffering from?
Your Answer: Pericardial effusion
Explanation:The QRS complex is associated with current that results in the contraction of both the ventricles. As ventricles have more muscle mass than the atria, they result in a greater deflection on the ECG. The normal duration of a QRS complex is 10s. A wide and deep Q wave depicts myocardial infarction. Abnormalities in the QRS complex maybe indicative of a bundle block, ventricular tachycardia or hypertrophy of the ventricles. Low voltage QRS complexes are characteristic of pericarditis or a pericardial effusion.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 27
Correct
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A 43-year-old diabetic man complains of headaches, palpitations, anxiety, abdominal pain and weakness. He is administered sodium bicarbonate used to treat:
Your Answer: Metabolic acidosis
Explanation:Sodium bicarbonate is indicated in the management of metabolic acidosis, which may occur in severe renal disease, uncontrolled diabetes, circulatory insufficiency due to shock or severe dehydration, extracorporeal circulation of blood, cardiac arrest and severe primary lactic acidosis. Bicarbonate is given at 50-100 mmol at a time under scrupulous monitoring of the arterial blood gas readings. This intervention, however, has some serious complications including lactic acidosis, and in those cases, should be used with great care.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 28
Correct
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Lung compliance is increased by:
Your Answer: Emphysema
Explanation:Lung compliance is increased by emphysema, acute asthma and increasing age and decreased by alveolar oedema, pulmonary hypertension, atelectasis and pulmonary fibrosis.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 29
Incorrect
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Causes of metabolic acidosis with a normal anion gap include:
Your Answer: Lactic acidosis
Correct Answer: Diarrhoea
Explanation:Excess acid intake and excess bicarbonate loss as in diarrhoea, are causes of metabolic acidosis with a normal anion gap. The other conditions all result in an increased anion gap.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 30
Correct
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A 60-year-old woman has had persistent diarrhoea for a week. A stool test reveals an infection by Clostridium difficile. Which of the following antibiotics could be used to treat the infection?
Your Answer: Oral vancomycin
Explanation:Three antibiotics are effective against Clostridium difficile:
Metronidazole 500 mg orally three times daily is the drug of choice, because of superior tolerability, lower price and comparable efficacy.
Oral vancomycin 125 mg four times daily is second-line therapy in particular cases of relapse or where the infection is unresponsive to metronidazole treatment.
Thirdly, the use of linezolid might also be considered.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 31
Correct
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A 30 year old female presented in the emergency with an irregular pulse. Her ECG showed absent P-waves with irregular RR interval. What is the most likely diagnosis?
Your Answer: Atrial fibrillation
Explanation:Atrial fibrillation is one of the most common cardiac arrhythmias. It is often asymptomatic but may present with symptoms of palpitations, fainting, chest pain and heart failure. Characteristic findings are: absence of P-waves, unorganised electrical activity in their place, irregularity of RR interval due to irregular conduction of impulses to the ventricles and if paroxysmal AF is suspected, episodes may be documented with the use of Holter monitoring
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 32
Correct
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Which of the following physiological changes will you see in a young man who has been trekking in the Himalayas for 3 years?
Your Answer: Increased renal excretion of HCO3 –
Explanation:The atmospheric pressure is lower at high altitudes as compared with sea level. This leads to a decrease in the partial pressure of oxygen. Once 2100 m (7000 feet) of altitude is reached, there is a drop in saturation of oxyhaemoglobin. The oxygen saturation of haemoglobin determines the oxygen content in the blood. The body physiological tries to adapt to high altitude by acclimatization. Immediate effects include hyperventilation, fluid loss (due to a decreased thirst drive), increase in heart rate and slightly lowered stroke volume. Long term effects include lower lactate production, compensatory alkali loss in urine, decrease in plasma volume, increased erythropoietin release and red cell mass, increased haematocrit, higher concentration of capillaries in striated muscle tissue, increase in myoglobin, increase in mitochondria, increase in aerobic enzyme concentration such as 2,3-DPG and pulmonary vasoconstriction.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 33
Correct
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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: 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 34
Correct
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A 40 year old man suffered severe trauma following an MVA. His BP is 70/33 mmhg, heart rate of 140 beats/mins and very feeble pulse. He was transfused 3 units of blood resulting in his BP returning to 100/70 and his heart rate to 90 beats/min. What decreased following transfusion?
Your Answer: Total peripheral resistance
Explanation:The patient is in hypovolemic shock, he is transfused with blood to replace the volume lost. It is important not only to replace fluids but stop active bleeding in resuscitation. Fluid replacement will result in a decreased sympathetic discharge and adequate ventricular filling thus reducing total peripheral resistance and increasing cardiac output and cardiac filling pressures.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 35
Correct
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During cardiac catheterisation in a 20-year old man, the following data is obtained: Pressure (mmHg), O2 saturation (%) Right atrium 7 (N = 5) 90 (N = 75), Right ventricle 35/7 (N = 25/5) 90 (N = 75), Pulmonary artery 35/8 (N = 25/15), 90 (N = 75), Left atrium 7 (N = 9) 95 (N = 95), Left ventricle 110/7 (N = 110/9) 95 (N = 95), Aorta 110/75 (N = 110/75) 95 (N = 95) where N = Normal value. What is the likely diagnosis?
Your Answer: Atrial septal defect
Explanation:A congenital heart disease, ASD or atrial septal defect leads to a communication between the right and left atria due to a defect in the interatrial septum. This leads to mixing of arterial and venous blood from the right and left side of the heart. The hemodynamic significance of this defect depends on the presence of shunting of blood. Normally, the left side of the heart has higher pressure than the right as the left side has to pump blood throughout the body. A large ASD (> 9 mm) will result in a clinically significant left-to-right shunt, causing volume overload of the right atrium and ventricle, eventually leading to heart failure. Cardiac catheterization would reveal very high oxygen saturation in the right atrium, right ventricle and pulmonary artery. Eventually, the left-to-right shunt will lead to pulmonary hypertension and increased afterload in the right ventricle, along with the increased preload due to the shunted blood. This will either cause right ventricular failure, or raise the pressure in the right side of the heart to equal or more than that in the left. Elevation of right atrial pressure to that of left atrial pressure would thus lead to diminishing or complete cessation of the shunt. If left uncorrected, there will be reversal of the shunt, known as Eisenmenger syndrome, resulting in clinical signs of cyanosis as the oxygen-poor blood form right side of the heart will mix with the blood in left side and reach the peripheral vascular system.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 36
Correct
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A medical student is asked to calculate the net pressure difference in a capillary wall, considering: Interstitial fluid hydrostatic pressure = –3 mmHg, Plasma colloid osmotic pressure = 28 mmHg, Capillary hydrostatic pressure = 17 mmHg, Interstitial fluid colloid osmotic pressure = 8 mmHg, and Filtration coefficient = 1. Which is the correct answer?
Your Answer: 0 mmHg
Explanation:The rate of filtration at any point along a capillary depends on a balance of forces sometimes called Starling’s forces after the physiologist who first described their operation in detail. The Starling principle of fluid exchange is key to understanding how plasma fluid (solvent) within the bloodstream (intravascular fluid) moves to the space outside the bloodstream (extravascular space). Fluid movement = k[(pc– pi)–(Πc– Πi)] where k = capillary filtration coefficient, pc = capillary hydrostatic pressure, pi= interstitial hydrostatic pressure, Πc = capillary colloid osmotic pressure, Πi = interstitial colloid osmotic pressure. Therefore: 1 × [capillary hydrostatic pressure (17) – interstitial fluid hydrostatic pressure (–3)] – [plasma colloid osmotic pressure (28) – interstitial fluid colloid osmotic pressure (8)] = 0 mmHg
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 37
Correct
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Urine specific gravity allows the assessment of which of the following renal functions?
Your Answer: Concentration
Explanation:Concentrating ability of kidneys is assessed by measuring the urine specific gravity. Normal values of urine specific gravity fall between 1.002 and 1.030 g/ml.
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This question is part of the following fields:
- Pathology
- Renal
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Question 38
Correct
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A 54-year-old woman is re-admitted to the hospital with shortness of breath and sharp chest pain 2 weeks after surgical cholecystectomy. The most probable cause of these clinical findings is:
Your Answer: Pulmonary embolus
Explanation:Pulmonary embolism is caused by the sudden blockage of a major lung blood vessel, usually by a blood clot. Symptoms include sudden sharp chest pain, cough, dyspnoea, palpitations, tachycardia or loss of consciousness. Risk factors for developing pulmonary embolism include long periods of inactivity, recent surgery, trauma, pregnancy, oral contraceptives, oestrogen replacement, malignancies and venous stasis.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 39
Correct
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Streptokinase is used to break down clots in some cases of myocardial infarction, pulmonary embolism, and arterial thromboembolism; however, it is not recommended to use it again after 4 days from the first administration. Which complication could arise from repeated use?
Your Answer: Allergic reaction
Explanation:Streptokinase belongs to a group of medications known as ‘fibrinolytics’ and is an extracellular metallo-enzyme produced by beta-haemolytic streptococci, used as an effective clot-dissolving medication in patients with myocardial infarction and pulmonary embolism. As Streptokinase is a bacterial product, the body has the ability to build up an immunity to it. Therefore, it is recommended that this medication should not be used again after four days from the first administration, as it may not be as effective and may also cause an allergic reaction.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 40
Incorrect
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A 30-year-old woman feels thirsty. This thirst is probably due to:
Your Answer: Hypokalaemia
Correct Answer: Increased level of angiotensin II
Explanation:Thirst is the basic need or instinct to drink. It arises from a lack of fluids and/or an increase in the concentration of certain osmolites such as salt. If the water volume of the body falls below a certain threshold or the osmolite concentration becomes too high, the brain signals thirst. Excessive thirst, known as polydipsia, along with excessive urination, known as polyuria, may be an indication of diabetes. Angiotensin II is a hormone that is a powerful dipsogen (i.e. it stimulates thirst) that acts via the subfornical organ. It increases secretion of ADH in the posterior pituitary and secretion of ACTH in the anterior pituitary.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 41
Correct
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A 79-year-old has been bedridden for 2 months after suffering from a stroke. She suddenly developed shortness of breath and chest pain, and was diagnosed with a pulmonary embolism. Which of the following is most likely to increase in this case?
Your Answer: Ventilation/perfusion ratio
Explanation:Pulmonary embolism (PE) is a blockage of an artery in the lungs by an embolus that has travelled from elsewhere in the body through the bloodstream. The change in cardiopulmonary function is proportional to the extent of the obstruction, which varies with the size and number of emboli obstructing the pulmonary arteries. The resulting physiological changes may include pulmonary hypertension with right ventricular failure and shock, dyspnoea with tachypnoea and hyperventilation, arterial hypoxaemia and pulmonary infarction. Consequent alveolar hyperventilation is manifested by a lowered pa(CO2). After occlusion of the pulmonary artery, areas of the lung are ventilated but not perfused, resulting in wasted ventilation with an increased ventilation/perfusion ratio – the physiological hallmark of PE – contributing to a further hyperventilatory state. The risk of blood clots is increased by cancer, prolonged bed rest, smoking, stroke, certain genetic conditions, oestrogen-based medication, pregnancy, obesity, and post surgery.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 42
Correct
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Normally, the O2 transfer in the lungs from alveolar to capillary is perfusion-limited. In which of the following situations does it become a diffusion-limited process?
Your Answer: Pulmonary oedema
Explanation:Normally, the transfer of oxygen from air spaces to blood takes place across the alveolar-capillary membrane by simple diffusion and depends entirely on the amount of blood flow (perfusion-limited process). Diseases that affect this diffusion will transform the normal process to a diffusion limited process. Thus, the diseases which cause a thickened barrier (such as pulmonary oedema due to increased extravascular lung water or asbestosis) will limit the diffusion of oxygen. Chronic obstructive lung diseases will have little effect on diffusion. Inhaling hyperbaric gas mixtures might overcome the diffusion limitation in patients with mild asbestosis or interstitial oedema, by increasing the driving force. Strenuous (not mild) exercise might also favour diffusion limitation and decrease passage time. Increasing the rate of ventilation will not have this affect but will only maintain a high oxygen gradient from air to blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 43
Incorrect
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A suspected recreational drug user was brought to the Emergency department in an unconscious state, and was found to be hypoventilating. Which of the following set of arterial blood gas analysis report is most consistent with hypoventilation as the primary cause? pH, pa(CO2) (mmHg), pa(O2) (mmHg).
Your Answer: 7.35, 50, 69
Correct Answer: 7.28, 55, 81
Explanation:Hypoventilation (or respiratory depression) causes an increase in carbon dioxide (hypercapnia) and respiratory acidosis. It can result due to drugs such as alcohol, benzodiazepines, barbiturates, opiates, mechanical conditions or holding ones breath. Strong opioids such as heroin and fentanyl are commonly implicated and can lead to respiratory arrest. In recreational drug overdose, acute respiratory acidosis occurs with an increase in p(CO2) over 45 mm Hg and acidaemia (pH < 7.35)
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 44
Incorrect
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When does the heart rate decrease?
Your Answer: Exercise
Correct Answer: Pressure on the eyeball
Explanation:Various vagotonic manoeuvres (e.g. Valsalva manoeuvre, carotid sinus massage, pressure on eyeballs, ice-water facial immersion, swallowing of ice-cold water) result in increased parasympathetic tone through the vagus nerve which results in a decrease in heart rate. These manoeuvres may be clinically useful in terminating supraventricular arrhythmias.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 45
Incorrect
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A 62-year-old woman presented to the doctor complaining of spine pain, fatigue and oliguria. She is diagnosed with chronic renal failure. Dipstick testing shows no protein, glucose, nitrite or ketones but a semi-quantitative sulphosalicylic acid test for urine protein is positive. Which of the following is the most probable cause of chronic renal failure in this patient.
Your Answer: Systemic lupus erythematosus
Correct Answer: Multiple myeloma
Explanation:Dipstick results are negative because the proteins found in the urine of this patient are not albumin but Bence Jones proteins. A Bence Jones protein is a monoclonal globulin protein commonly detected in patients affected by multiple myeloma. Multiple myeloma is a malignancy of plasma cells characterised by the production of monoclonal immunoglobulin. Symptoms include bone pain, bone fractures, bleeding, neurologic symptoms, fatigue, frequent infections and weight loss.
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This question is part of the following fields:
- Pathology
- Renal
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Question 46
Incorrect
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A victim of road traffic accident presented to the emergency department with a blood pressure of 120/90 mm Hg, with a drop in systolic pressure to 100 mm Hg on inhalation. This is known as:
Your Answer: Pulsus parvus
Correct Answer: Pulsus paradoxus
Explanation:Weakening of pulse with inhalation and strengthening with exhalation is known as pulsus paradoxus. This represents an exaggeration of the normal variation of the pulse in relation to respiration. It indicates conditions such as cardiac tamponade and lung disease. The paradox refers to the auscultation of extra cardiac beats on inspiration, as compared to the pulse. Due to a decrease in blood pressure, the radial pulse becomes impalpable along with an increase in jugular venous pressure height (Kussmaul sign). Normal systolic blood pressure variation (with respiration) is considered to be >10 mmHg. It is >100 mmHg in Pulsus paradoxus. It is also predictive of the severity of cardiac tamponade.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 47
Incorrect
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A 76-year-old woman is diagnosed with diabetes mellitus after a urine test revealed she has glucosuria. Glucosuria may occur due to inadequate glucose reabsorption at:
Your Answer: Loop of Henlé
Correct Answer: Proximal convoluted tubule
Explanation:Glucose is reabsorbed almost 100% via sodium–glucose transport proteins (apical) and GLUT (basolateral) in the proximal convoluted tubule. Glycosuria or glucosuria is a condition of osmotic diuresis typical in those suffering from diabetes mellitus. Due to a lack of insulin, plasma glucose levels are above normal. This leads to saturation of receptors in the kidneys and glycosuria usually at plasma glucose levels above 11 mmol/l. Rarely, glycosuria is due to an intrinsic problem with glucose reabsorption within the kidneys (such as Fanconi syndrome), producing a condition termed renal glycosuria.
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This question is part of the following fields:
- Physiology
- Renal
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Question 48
Incorrect
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A 7-year-old boy is diagnosed with metabolic acidosis as a result of severe dehydration. Which of the following conditions is most likely to cause severe dehydration and metabolic acidosis?
Your Answer: Severe persistent vomiting
Correct Answer: Severe diarrhoea
Explanation:Diarrhoea is defined as having three or more loose or liquid stools per day, or as having more stools than is normal for that person. Severe diarrhoea, causing fluid loss and loss of bicarbonate, will result in marked dehydration and metabolic acidosis.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 49
Correct
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The blood-brain barrier is a membrane that separates the circulating blood from the brain extracellular fluid in the central nervous system (CNS). Which of the following statements regarding the blood– brain barrier is CORRECT?
Your Answer: It breaks down in areas of brain that are infected
Explanation:The blood–brain barrier is a membrane that controls the passage of substances from the blood into the central nervous system. It is a physical barrier between the local blood vessels and most parts of the central nervous system and stops many substances from travelling across it. During meningitis, the blood–brain barrier may be disrupted. This disruption may increase the penetration of various substances (including either toxins or antibiotics) into the brain. A few regions in the brain, including the circumventricular organs, do not have a blood–brain barrier.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 50
Correct
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Glucose is the most important source of energy for cellular respiration. The transport of glucose in the renal tubular cells occurs via:
Your Answer: Secondary active transport with sodium
Explanation:In 1960, Robert K. Crane presented for the first time his discovery of the sodium-glucose cotransport as the mechanism for glucose absorption. Glucose transport through biological membranes requires specific transport proteins. Transport of glucose through the apical membrane of renal tubular as well as intestinal epithelial cells depends on the presence of secondary active Na+–glucose symporters, SGLT-1 and SGLT-2, which concentrate glucose inside the cells, using the energy provided by co-transport of Na+ ions down their electrochemical gradient.
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This question is part of the following fields:
- Physiology
- Renal
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Question 51
Incorrect
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Calculate the pulmonary vascular resistance in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer: 1.5 resistance units (mmHg/l per min)
Correct Answer: 2.0 resistance units (mmHg/l per min)
Explanation:Pulmonary vascular resistance (PVR) = (Mean pulmonary artery pressure – Pulmonary capillary wedge pressure) divided by Cardiac output. To get cardiac output, Fick’s principle needs to be applied which states that VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CA = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24–0.16, CO = 500/0.8, CO = 6.25 l/min. To calculate mean pulmonary artery pressure, we use the formula: Mean pulmonary artery pressure = Diastolic pressure + 1/3(Systolic pressure – Diastolic pressure). Thus, Mean pulmonary artery pressure = 15 + 1/3(25 – 15) = 15 + 3. 33 = 18.33. Substituting these values in the first formula, PVR = 18.3–5/6.25 = 13.5/5.25 = 2.0 resistance units (mmHg/l per min) approximately.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 52
Incorrect
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Purkinje fibres in the heart conduct action potentials at the rate of:
Your Answer: 5.0–8.5 m/s
Correct Answer: 1.5–4.0 m/s
Explanation:Purkinje fibres control the heart rate along with the sinoatrial node (SA node) and the atrioventricular node (AV node). The QRS complex is associated with the impulse passing through the Purkinje fibres. These fibres conduct action potential about six times faster than the velocity in normal cardiac muscle.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 53
Incorrect
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A 60-year old gentleman visited his general practitioner complaining of high grade fever for 7 days and a dull, aching pain in his left lumbar region. On enquiry, he admitted to having a burning sensation while passing urine. His blood results showed an elevated white blood cell count with a left shift. In his condition, which is the most characteristic finding on urine examination?
Your Answer: Proteinuria
Correct Answer: White blood cell casts
Explanation:Tubulointerstitial nephritis is the term given to primary injury to renal tubules and the renal interstitium, which ultimately results in a decline in renal function. Acute tubulointerstitial nephritis (acute pyelonephritis) is often seen as a result of infection or drug reactions. The most characteristic feature of this condition on urine analysis is the presence of white blood cell casts.
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This question is part of the following fields:
- Pathology
- Renal
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Question 54
Correct
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Gram positive bacteria differ from gram negative bacteria due to the presence of which of the following structures?
Your Answer: Outer membrane
Explanation:The reason bacteria are either Gram-positive or Gram-negative is due to the structure of their cell envelope (the cell envelope is defined as the cell membrane and cell wall plus an outer membrane, if one is present.) Gram-positive bacteria, for example, retain the crystal violet due to the amount of peptidoglycan in the cell wall. It can be said therefore that the Gram-stain procedure separates bacteria into two broad categories based on structural differences in the cell envelope.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 55
Incorrect
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A 15 year old girl presented to the emergency with a history of chronic cough, fever and weight loss. Her chest X-ray showed multiple nodules 1-4 cm in size and some of them with cavitation especially in the upper lobe. A sputum sample was positive for acid fast bacilli. Which of the following cells played a part in the development of the lung lesions?
Your Answer: Mast cell
Correct Answer: Macrophage
Explanation:The characteristic cells in granulomatous inflammation are giant cells, formed from merging macrophages and epithelioid cells elongated with granular eosinophilic cytoplasm. Granulomatous reactions are seen in patients with tuberculosis. A tuberculous/caseating granuloma is characterised by a zone of central necrosis lined with giant multinucleated giant cells (Langhans cells) and surrounded by epithelioid cells, lymphocytes and fibroblasts. The caseous zone is present due to the damaged and dead giant cells and epithelioid cells.
Mast cells are only few in number and fibroblasts lay down collagen.
Basophils are not present.
The giant cell made up of macrophages are the most abundant cells in this inflammatory process.
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This question is part of the following fields:
- Inflammation & Immunology; Respiratory
- Pathology
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Question 56
Incorrect
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Lack of findings in the bladder but presence of atypical epithelial cells in urinalysis is most often associated with which of the following conditions?
Your Answer: Squamous cell carcinoma of penis
Correct Answer: Transitional cell carcinoma of renal pelvis
Explanation:The presence of atypical cells in urinalysis without findings in the bladder suggests a lesion located higher up, most probably in ureters or renal pelvis. Transitional cell cancer of the renal pelvis is a disease in which malignant cells form in the renal pelvis and is characterised by the presence of abnormal cells in urine cytology.
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This question is part of the following fields:
- Pathology
- Renal
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Question 57
Incorrect
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The Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity. According to this equation, the buffering capacity of the system is at maximum when the number of free anions compared with undissociated acid is:
Your Answer: Three times more
Correct Answer: Equal
Explanation:In 1908, Lawrence Joseph Henderson wrote an equation describing the use of carbonic acid as a buffer solution. Later, Karl Albert Hasselbalch re-expressed that formula in logarithmic terms, resulting in the Henderson–Hasselbalch equation. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid–base reactions. Two equivalent forms of the equation are: pH = pKa + log10 [A–]/[HA] or pH = pKa + log10 [base]/[acid]. Here, pKa is − log10(Ka) where Ka is the acid dissociation constant, that is: pKa = –log10(Ka) = –log10 ([H3 O+][A–]/[HA]) for the reaction: HA + H2 O ≈ A– + H3 O+ In these equations, A– denotes the ionic form of the relevant acid. Bracketed quantities such as [base] and [acid] denote the molar concentration of the quantity enclosed. Maximum buffering capacity is found when pH = pKa or when the number of free anions to undissociated acid is equal and buffer range is considered to be at a pH = pKa ± 1.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 58
Incorrect
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Which of these substances is secreted by pericytes in the juxtaglomerular cells?
Your Answer: Erythropoietin
Correct Answer: Renin
Explanation:The juxtaglomerular cells synthesise, store and secrete the enzyme renin in the kidney. They are specialised smooth muscle cells in the wall of the afferent arteriole that delivers blood to the glomerulus and thus play a critical role in the renin– angiotensin system and so in renal autoregulation.
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This question is part of the following fields:
- Physiology
- Renal
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Question 59
Correct
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What is the normal duration of the ST segment?
Your 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 60
Incorrect
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The bronchial circulation is a part of the circulatory system that supplies nutrients and oxygen to the pulmonary parenchyma. What percentage of cardiac output is received by bronchial circulation?
Your Answer:
Correct Answer: 2%
Explanation:The bronchial circulation is part of the systemic circulation and receives about 2% of the cardiac output from the left heart. Bronchial arteries arise from branches of the aorta, intercostal, subclavian or internal mammary arteries. The bronchial arteries supply the tracheobronchial tree with both nutrients and O2. It is complementary to the pulmonary circulation that brings deoxygenated blood to the lungs and carries oxygenated blood away from them in order to oxygenate the rest of the body.
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This question is part of the following fields:
- Physiology
- Respiratory
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SESSION STATS - PERFORMANCE PER SPECIALTY
