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Question 1
Correct
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After having donated a unit of blood. The blood bank will prefer to use which of the following anticoagulants to store the blood?
Your Answer: Citrate
Explanation:Calcium is necessary for coagulation to occur. Citrate being a chelator and combining with calcium ions to form un-ionised compound will prevent coagulation. Following transfusion the citrate is removed by the liver with in a few minutes. Oxalate also works on the same principle but it is toxic to the body.
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This question is part of the following fields:
- General
- Physiology
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Question 2
Correct
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Work of breathing (WOB) is the energy expended to inhale and exhale a breathing gas. Normally, maximal amount of work of breathing is required to overcome:
Your Answer: Elastic lung compliance
Explanation:The forces of elastance (compliance), frictional resistance and inertia have been identified as the forces that oppose lung inflation and deflation. The normal relaxed state of the lung and chest is partially empty. Further exhalation requires muscular work. Inhalation is an active process requiring work. About 60–66% of the total work performed by the respiratory muscles is used to overcome the elastic or compliance characteristics of the lung–chest cage, 30–35% is used to overcome frictional resistance and only 2–5% of the work is used for inertia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 3
Incorrect
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Which statement is correct regarding coagulation?
Your Answer: Warfarin inhibits the intrinsic pathway
Correct Answer: Thrombin converts fibrinogen to fibrin
Explanation:Coagulation of blood is a complex process and an important part of haemostasis. There are two main pathways related to coagulation: the contact activation pathway/intrinsic pathway and tissue factor/extrinsic pathway. The extrinsic pathway is activated by external trauma that causes blood to escape from the vascular system. This pathway is quicker than the intrinsic pathway and involves factor VII. The intrinsic pathway is activated by trauma inside the vascular system, and initiated by platelets, exposed endothelium, chemicals, or collagen. This pathway is slower than the extrinsic pathway, but more important. It involves factors XII, XI, IX, VIII. Both pathways meet to finish the formation of a clot in what is known as the common pathway. The common pathway involves factors I, II, V, and X. They converge on the common pathway in which activation of prothrombin to thrombin leads to conversion of fibrinogen to fibrin and clot formation.
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This question is part of the following fields:
- General
- Physiology
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Question 4
Incorrect
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A lesion involving the suprachiasmatic nucleus of hypothalamus is likely to affect:
Your Answer: Autonomic function
Correct Answer: Regulation of circadian rhythm
Explanation:The suprachiasmatic nucleus (SCN) in the hypothalamus is responsible for controlling endogenous circadian rhythms and destruction of the SCN leads to a loss of circadian rhythm.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 5
Correct
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An 18 year-old with an iron deficient diet was prescribed an iron supplement by her GP. Lack of iron often results in:
Your Answer: Hypochromic anaemia
Explanation:Iron deficiency anaemia is the most common type of anaemia. It can occur due to deficiency of iron from decreased intake, increased loss or inadequate absorption. An MCV less than 80 will indicated iron deficiency anaemia. On the smear the RBCs will be microcytic hypochromic and will also show poikilocytosis. Iron profile tests are important to make a diagnosis. Clinically the patient will be pale and lethargic.
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This question is part of the following fields:
- General
- Physiology
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Question 6
Correct
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A 65-year old patient with altered bowl movement experienced the worsening of shortness of breath and exertional chest pains over the course of 8 weeks. Examination shows pallor and jugular venous distension. Furthermore, a test of the stool for occult blood is positive. Laboratory studies show:
Haemoglobin 7.4 g/dl
Mean corpuscular volume 70 fl Leukocyte count 5400/mm3
Platelet count 580 000/mm3 Erythrocyte sedimentation 33 mm/h
A blood smear shows hypochromic, microcytic RBCs with moderate poikilocytosis. Which of the following is the most likely diagnosis?Your Answer: Iron deficiency anaemia
Explanation:Iron deficiency anaemia is the most common type of anaemia. It can occur due to deficiency of iron due to decreased intake or due to faulty absorption. An MCV less than 80 will indicated iron deficiency anaemia. On the smear the RBC will be microcytic hypochromic and will also show piokilocytosis. iron profiles tests are important to make a diagnosis. Clinically the patient will be pale and lethargic.
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This question is part of the following fields:
- General
- Physiology
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Question 7
Correct
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A 32-year old gentleman came to the emergency department, complaining of progressively increasing weakness in his arms and legs over 5 days. On examination, there is symmetrical weakness on both sides of his face, along with weakness of the proximal and distal muscles of all four limbs. No loss of sensation noted. Deep tendon reflexes could not be elicited and plantar responses were downward. On enquiry, it was revealed that he had an upper respiratory tract infection 10 days ago. The likely diagnosis is:
Your Answer: Guillain–Barré syndrome
Explanation:Guillain–Barré syndrome (GBS) is an acute, autoimmune polyradiculoneuropathy which affects the peripheral nervous system and is usually triggered by an acute infectious process. 75% patients have a history of acute infection within the past 1–4 weeks, usually respiratory or gastrointestinal. immunisations have also been implicated. The most common form is acute inflammatory demyelinating polyneuropathy. It results in an ascending paralysis with complete loss of deep tendon reflexes. Treatment includes immunoglobulins and supportive care. However, the disease may be fatal due to severe pulmonary complications and dysautonomia.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 8
Correct
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Basal Metabolic Rate (BMR) will most likely be reduced by which of the following?
Your Answer: Decrease in body temperature
Explanation:The basal metabolic rate (BMR) is defined as the rate of calorie consumption after an overnight fast, in the absence of any muscular activity, with the patient in a restful state. Various factors affect the BMR including weight, body surface area and age. The BMR is 30 kcal/m2 per hour at birth; at age 2, the rate is 57 kcal/m2 per hour; and at age 20, 41 kcal/m2 per hour. After this, the BMR decreases by 10% between 20-60 years of age. Women are known to have a 10% lower BMR than men (due to higher fat content). A one-degree change in body temperature leads to a 10% change in BMR in the same direction. However, shivering and increasing ambient temperature brings about a rise in BMR, and so does stress, physical activity, caffeine, theophylline and hyperthyroidism. Also, thermogenesis induced by diet results in increased metabolic rate and hence, BMR should be ideally measured after overnight fasting.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 9
Correct
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Chest X-ray of a 45-year old gentleman with a week history of pleurisy showed a small pneumothorax with moderate-sized pleural effusion. Arterial blood gas analysis showed p(CO2) = 23 mmHg, p(O2) = 234.5 mmHg, standard bicarbonate = 16 mmol/l. What are we most likely dealing with?
Your Answer: Compensated respiratory alkalosis
Explanation:Normal pH with low p(CO2) and low standard bicarbonate could indicate either compensated respiratory alkalosis or a compensated metabolic acidosis. However, the history of hyperventilation for 5 days (pleurisy) favours compensated respiratory alkalosis. Compensated metabolic acidosis would have been likely in a diabetic patient with fever, vomiting and high glucose (diabetic ketoacidosis).
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 10
Incorrect
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Calculate the resistance of the artery if the pressure at one end is 60 mmHg, pressure at the other end is 20 mm Hg and the flow rate in the artery is 200 ml/min.
Your Answer:
Correct Answer: 0.2
Explanation:Flow in any vessel = Effective perfusion pressure divided by resistance, where effective perfusion pressure is the mean intraluminal pressure at the arterial end minus the mean pressure at the venous end. Thus, in the given problem, resistance = (60 − 20)/200 = 0.2 mmHg/ml per min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 11
Incorrect
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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:
Correct 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 12
Incorrect
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If a 55-year old gentleman who has suffered a stroke, develops a tremor in his fingers that worsens on reaching for an object, what part of his brain is likely to be involved?
Your Answer:
Correct Answer: Cerebellum
Explanation:The cerebellum plays an important role in the integration of sensory perception and motor output. Multiple neural pathways link the cerebellum with the motor cortex and the spinocerebellar tract. The cerebellum uses the constant feedback on body position to fine-tune the movements and integrates these pathways. The patient described here has a characteristic cerebellar tremor that is a slow, broad tremor of the extremities and occurs at the end of a purposeful movement.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 13
Incorrect
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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:
Correct 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 14
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:
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 15
Incorrect
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Evaluation of a 60-year old gentleman, who has been a coal miner all his life and is suspected to have pulmonary fibrosis reveals the following: FEV1 of 75% (normal > 65%), arterial oxygen saturation 92%, alveolar ventilation 6000 ml/min at a tidal volume of 600 ml and a breathing rate of 12 breaths/min. There are also pathological changes in lung compliance and residual volume. Calculate his anatomical dead space.
Your Answer:
Correct Answer: 100 ml
Explanation:Dead space refers to inhaled air that does not take part in gas exchange. Because of this dead space, taking deep breaths slowly is more effective for gas exchange than taking quick, shallow breaths where a large proportion is dead space. Use of a snorkel by a diver increases the dead space marginally. Anatomical dead space refers to the gas in conducting areas such as mouth and trachea, and is roughly 150 ml (2.2 ml/kg body weight). This corresponds to a third of the tidal volume (400-500 ml). It can be measured by Fowler’s method, a nitrogen wash-out technique. It is posture-dependent and increases with increase in tidal volume. Physiological dead space is equal to the anatomical dead space plus the alveolar dead space, where alveolar dead space is the area in the alveoli where no effective exchange takes place due to poor blood flow in capillaries. This physiological dead space is very small normally (< 5 ml) but can increase in lung diseases. Physiological dead space can be measured by Bohr’s method. Total ventilation per minute (minute ventilation) is given by the product of tidal volume and the breathing rate. Here, the total ventilation is 600 ml times 12 breaths/min = 7200 ml/min. The problem mentions alveolar ventilation to be 6000 ml/min. Thus, the difference between the alveolar ventilation and total ventilation is 7200 – 6000 ml/min = 1200 ml/min, or 100 ml per breath at 12 breaths per min. This 100 ml is the dead space volume.
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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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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 17
Incorrect
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A glycogen storage disorder is characterised by increased liver glycogen with a normal structure and no increase in serum glucose after oral intake of a protein-rich diet. Deficiency of which of the following enzymes is responsible for this disorder?
Your Answer:
Correct Answer: Glucose-6-phosphatase
Explanation:The most common glycogen storage disorder is von Gierke’s disease or glycogen storage disease type I. It results from a deficiency of enzyme glucose-6-phosphatase which affects the ability of liver to produce free glucose from glycogen and gluconeogenesis; leading to severe hypoglycaemia. There is also increased glycogen storage in the liver and kidneys causing enlargement and various problems in their functioning. The disease also causes lactic acidosis and hyperlipidaemia. The main treatment includes frequent or continuous feedings of corn-starch or other carbohydrates.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 18
Incorrect
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Which of the following is the cause of flattened (notched) T waves on electrocardiogram (ECG)?
Your Answer:
Correct 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 19
Incorrect
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Calculate the cardiac stroke volume of a patient whose oxygen consumption (measured by analysis of mixed expired gas) is 300 ml/min, arterial O2 content is 20 ml/100 ml blood, pulmonary arterial O2 content is 15 ml/100 ml blood and heart rate is 60/min.
Your Answer:
Correct Answer: 100 ml
Explanation:By Fick’s principle, VO2 = Q × (CA (O2) − CV (O2)) where VO2 = O2 consumption, Q = cardiac output and CA(O2) and CV(O2) are arterial and mixed venous O2 content respectively. Thus, in the given problem, 300 ml O2/min = Q × (20−15) ml O2/100 ml. Thus, Q = 6000 ml blood/min. Then, we can calculate stroke volume by dividing the cardiac output with heart rate. Thus, stroke volume = 6000 ml/min divided by 60/min stroke volume = 100 ml.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 20
Incorrect
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A patient came into the emergency in a state of shock. His blood group is not known, but on testing it clotted when mixed with Type A antibodies. Which blood should be transfused?
Your Answer:
Correct Answer: B +ve
Explanation:There are two stages to determine the blood group, known as ABO typing. The first stage is called forward typing. In this method, RBCs are mixed with two separate solutions of type A or type B antibodies to see if they agglutinate. If this blood clumps, this indicates the presence of antigens within the blood sample. For example, a sample of type B blood will clump when tested with type A antibodies as it contains type B antigens. Group B – has only the B antigen on red cells (and A antibody in the plasma)
Group B – has only the B antigen on red cells (and A antibody in the plasma)
Group AB – has both A and B antigens on red cells (but neither A nor B antibody in the plasma)
Group O – has neither A nor B antigens on red cells (but both A and B antibody are in the plasma). Many people also have a Rh factor on the red blood cell’s surface. This is also an antigen and those who have it are called Rh+. Those who have not are called Rh–. A person with Rh– blood does not have Rh antibodies naturally in the blood plasma (as one can have A or B antibodies, for instance) but they can develop Rh antibodies in the blood plasma if they receive blood from a person with Rh+ blood, whose Rh antigens can trigger the production of Rh antibodies. A person with Rh+ blood can receive blood from a person with Rh– blood without any problems. The patient’s blood group is B positive as he has antigen B, antibody A and Rh antigens.
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This question is part of the following fields:
- General
- Physiology
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Question 21
Incorrect
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Which of the following can occur even in the absence of brainstem co-ordination?
Your Answer:
Correct Answer: Gastric emptying
Explanation:Although gastric emptying is under both neural and hormonal control, it does not require brainstem co-ordination. Increased motility of the orad stomach (decreased distensibility) or of the distal stomach (increased peristalsis), decreased pyloric tone, decreased duodenal motility or a combination of these, all increase the rate of gastric emptying. The major control mechanism for gastric emptying is through duodenal gastric feedback. The duodenum has receptors for the presence of acid, carbohydrate, fat and protein digestion products, osmolarity different from that of plasma, and distension. Activating these receptors decreases the rate of gastric emptying. Neural mechanisms involve both enteric and vagal pathways and a vagotomy impairs the gastric emptying regulation. CCK (cholecystokinin) slows gastric emptying at physiological levels of the hormone. Gastrin, secretin and glucose-1-phosphate also slow gastric emptying, but require higher doses.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 22
Incorrect
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Which of the following is a true statement regarding the pupil?
Your Answer:
Correct Answer: Phentolamine causes pupil constriction
Explanation:A balance between the sympathetic tone to the radial fibres of the iris and parasympathetic tone to the pupillary sphincter muscle determines the pupil size. Phentolamine (α-adrenergic receptor blocker) causes pupillary constriction. Dilatation of the pupil occurs with increased sympathetic activity, decreased parasympathetic activity during darkness or block of muscarinic receptors by atropine.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 23
Incorrect
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A child defecates a few minutes after being fed by the mother. This is most likely due to:
Your Answer:
Correct Answer: Gastrocolic reflex
Explanation:The gastrocolic reflex is a physiological reflex that involves increase in colonic motility in response to stretch in the stomach and by-products of digestion in the small intestine. It is shown to be uneven in its distribution throughout the colon, with the sigmoid colon affected more than the right side of the colon in terms of a phasic response. Various neuropeptides have been proposed as mediators of this reflex, such as serotonin, neurotensin, cholecystokinin and gastrin.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 24
Incorrect
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In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your Answer:
Correct Answer: Exercise
Explanation:The oxygen-haemoglobin dissociation curve plots saturated haemoglobin against the oxygen tension and is usually a sigmoid plot. Each molecule of haemoglobin can bind to four molecules of oxygen reversibly. Factors that can influence the binding include: pH, concentration of 2,3-diphosphoglycerate (2,3-DPG), temperature, type of haemoglobin molecules, and presence of toxins, especially carbon monoxide. Shape of the curve is due to interaction of bound oxygen molecules with the incoming molecules. The binding of first molecule is difficult, with easier binding of the second and third molecule and increase in difficulty with the fourth molecule – partly as a result of crowding and partly as a natural tendency of oxygen to dissociate.
Left shift of curve indicates haemoglobin’s increased affinity for oxygen (seen at lungs). Right shift indicates decreased affinity and is seen with an increase in body temperature, hydrogen ions, 2,3-diphosphoglycerate (DPG), carbon dioxide concentration and exercise. Under normal resting conditions in a healthy individual, the normal position of the curve is at a pH of 7.4. A shift in the position of the curve with a change in pH is called the Bohr effect. Left shift occurs in acute alkalosis, decrease in p(CO2), decrease in temperature and decrease in 2,3-DPG. The fetal haemoglobin curve is to the left of the adult haemoglobin to allow for oxygen diffusion across the placenta. The curve for myoglobin is even further to the left. Carbon monoxide has a much higher affinity for haemoglobin than oxygen does. Thus, carbon monoxide poisoning leads to hypoxia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 25
Incorrect
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A brain tumour causing blockage of the hypophyseal portal system is likely to result in an increased secretion of which of the following hormones?
Your Answer:
Correct Answer: Prolactin
Explanation:The hypophyseal portal system links the hypothalamus and the anterior pituitary. With the help of this system, the anterior pituitary receives releasing and inhibitory hormones from the hypothalamus and regulates the action of other endocrine glands. One of the inhibitory hormones carried by this system is the prolactin-inhibitory hormone. In the absence of this hormone which might occur in case of a blockage of the system, prolactin secretion increases to about three times normal levels.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 26
Incorrect
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A 45-year old gentleman presents with diarrhoea for two weeks. He has no history of fever and the diarrhoea stops on fasting. Which is the most likely type of diarrhoea that he is suffering from?
Your Answer:
Correct Answer: Osmotic
Explanation:The different types of diarrhoea are:
1. Secretory diarrhoea – Due to increased secretion or decreased absorption. There is minimal to no structural damage in this type. The most common cause is cholera toxin which stimulates secretion of anions (especially chloride), with sodium and water.
2. Osmotic diarrhoea – Due to increased osmotic load, there is water loss. This occurs in cases of maldigestion syndromes, such as coeliac or pancreatic disease.
3. Motility-related diarrhoea – Occurs in cases of abnormal gastrointestinal motility. Due to increased motility, there is poor absorption and this leads to diarrhoea. This is seen post-vagotomy or in diabetic neuropathy.
4. Inflammatory diarrhoea – Due to damage to the mucosa or brush border, there is a loss of protein-rich fluids and poor absorption. Features of all the above three types can be seen in this type. Aetiology includes bacterial, viral, parasitic infections or autoimmune problems including inflammatory bowel disease.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 27
Incorrect
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Which of the following associations is correctly matched with the body's defence mechanism in fighting infection?
Your Answer:
Correct Answer: Specific cellular mechanism → cytotoxic T cells
Explanation:The immune system has certain levels of defence against pathogens. First line includes simple barriers such as skin, mucosa and stomach acid that prevent the pathogen from entering into the body. If this barrier is breached then the innate immune system is activated which includes leukocytes (macrophages, neutrophils, mast cells, eosinophils, basophils, natural killer cells). If the pathogens invade the second layer of defence then the third layer, adaptive immunity is activated, which includes B and T lymphocytes. B cells provide a humoral response whereas cytotoxic T cells have specific cellular mechanisms. They maintain a memory of past infections and are activated faster following a recurrence.
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This question is part of the following fields:
- General
- Physiology
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Question 28
Incorrect
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A patient with a long standing lower motor neuron lesion will have:
Your Answer:
Correct Answer: Muscle wasting
Explanation:Lower motor neurons (LMNs) connect the brainstem and spinal cord to muscle fibres. Damage to lower motor neurons is indicated by abnormal electromyographic potentials, fasciculations, paralysis, weakening and wasting of skeletal muscles.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 29
Incorrect
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The primary motor cortex is located in the:
Your Answer:
Correct Answer: Precentral gyrus
Explanation:The primary motor cortex is located in the dorsal part of the precentral gyrus and the anterior bank of the central sulcus. The precentral gyrus lies anterior to the postcentral gyrus and is separated from it by a central sulcus. Its anterior border is the precentral sulcus, while inferiorly it borders to the lateral fissure (Sylvian fissure).
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This question is part of the following fields:
- Neurology
- Physiology
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Question 30
Incorrect
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Calculate the total peripheral resistance for a patient with a blood pressure of 130/70 mm HG and cardiac output of 5 litres / min?
Your Answer:
Correct Answer: 18 mmHg × min/l
Explanation:Total peripheral resistance = Mean arterial pressure/Cardiac output. And the mean arterial pressure = Diastolic pressure + 1/3 (Systolic pressure – Diastolic pressure), i.e., 70 + 1/3 (130-70) = 90 mmHg. Therefore, total peripheral resistance = 90 mmHg/5 l per min = 18 mmHg × min/l.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 31
Incorrect
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The majority of gallstones are mainly composed of:
Your Answer:
Correct Answer: Cholesterol
Explanation:Bile salts are formed out of cholesterol in the liver cells. Occasionally, precipitation of cholesterol occurs resulting into cholesterol stones developing in the gall bladder.
These cholesterol gallstones are the most common type and account for 80% of all gallstones. Another type, accounting for 20% gallstones is pigment stones which are composed of bilirubin and calcium salts. Occasionally, stones of mixed origin are also seen.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 32
Incorrect
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Nephrotic syndrome is a condition that causes proteinuria, hypoalbuminemia and oedema. Which of the following is the cause of the oedema in these patients?
Your Answer:
Correct Answer: Decreased oncotic pressure
Explanation:The glomeruli of the kidneys are the parts that normally filter the blood. They consist of capillaries that are fenestrated and allow fluid, salts and other small solutes to flow through, but normally not proteins. In nephrotic syndrome, the glomeruli become damaged allowing small proteins, such as albumin to pass through the kidneys into urine. Oedema usually occurs due to salt and water retention by the diseased kidneys as well as due to the reduced colloid oncotic pressure (because of reduced albumin in the plasma). Lower serum oncotic pressure causes fluid to accumulate in the interstitial tissues.
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This question is part of the following fields:
- Physiology
- Renal
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Question 33
Incorrect
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What is a major source of fuel being oxidised by the skeletal muscles of a man who has undergone starvation for 7 days?
Your Answer:
Correct Answer: Serum fatty acids
Explanation:Starvation is the most extreme form of malnutrition. Prolonged starvation can lead to permanent organ damage and can be fatal. Starved individuals eventually lose significant fat and muscle mass as the body uses these for energy.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 34
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 35
Incorrect
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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:
Correct 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 36
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 37
Incorrect
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A 56 year old gentleman, who is a chronic smoker presents to the clinic with dyspnoea. His Chest X-ray shows intercostal space widening with increased blackening bilaterally. What is the most likely finding on his pulmonary function test?
Your Answer:
Correct Answer: Increased total lung capacity
Explanation:In patients suspected of having chronic obstructive pulmonary disease, pulmonary function testing (PFT) is useful to confirm airway obstruction, to quantify its severity and reversibility, for following disease progression and monitoring response to treatment. These tests include:
FEV1 – volume of air forcefully expired during the first second after a full breath
Forced vital capacity (FVC) – total volume of air expired with maximal force and flow-volume loops. The hallmark of airway obstruction is reduction of FEV1, FVC and the ratio of FEV1/FVC, with a concave pattern in expiratory tracing on the flow-volume loop. FEV1 and forced vital capacity (FVC) are easily measured with office spirometry and are useful to assess the severity of disease. Other parameters include increased total lung capacity, functional residual capacity and residual volume, which can help distinguish chronic obstructive pulmonary disease (COPD) from restrictive pulmonary disease where these values are lower than normal. Along with these, other tests are decreased vital capacity; and decreased diffusing capacity for carbon monoxide (DLCO). DLCO is non-specific and can be low in other disorders that affect the pulmonary vascular bed, such as interstitial lung disease. DLCO is however useful to distinguish COPD from asthma, in which DLCO is normal or elevated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 38
Incorrect
-
What is the role of factor VII in coagulation?
Your Answer:
Correct Answer: Initiates the process of coagulation in conjunction with tissue factor
Explanation:The main role of factor VII is to initiate the process of coagulation along with tissue factor (TF). TF is found in the blood vessels and is not normally exposed to the bloodstream. When a vessel is injured tissue factor is exposed to blood and circulating factor VII. Factor VII is converted to VIIa by TF.
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This question is part of the following fields:
- General
- Physiology
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Question 39
Incorrect
-
The gradual depolarization in-between action potentials in pacemaker tissue is a result of?
Your Answer:
Correct Answer: A combination of gradual inactivation outward IK along with the presence of an inward ‘funny’ current (If) due to opening of channels permeable to both Na+ and K+ ions
Explanation:One of the characteristic features of the pacemaker cell is the generation of a gradual diastolic depolarization also called the pacemaker potential. In phase 0, the upstroke of the action potential caused by an increase in the Ca2+ conductance, an influx of calcium occurs and a positive membrane potential is generated. The next is phase 3 which is repolarization caused by increased K+ conductance as a result of outwards K+ current. Phase 4 is a slow depolarization which accounts for the pacemaker activity, caused by increased conductance of Na+, inwards Na+ current called IF. it is turned on by repolarization.
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This question is part of the following fields:
- General
- Physiology
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Question 40
Incorrect
-
The mechanism of action of streptokinase involves:
Your Answer:
Correct Answer: Direct conversion of plasminogen to plasmin
Explanation:Streptokinase is an enzyme that is produced by group A beta haemolytic streptococcus and is an effective and cost efficient method for the dissolution of a clot used in cases of MI and pulmonary embolism. It works by directly converting plasminogen to plasmin which breaks down the blood components in the clot and fibrin, dissolving the clot. Streptokinase is a bacterial product and thus the body will develop immunity against it.
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This question is part of the following fields:
- General
- Physiology
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Question 41
Incorrect
-
Which of these substances is secreted by pericytes in the juxtaglomerular cells?
Your Answer:
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 42
Incorrect
-
Myoglobin is released as a result of rhabdomyolysis from damaged skeletal muscles. What function do they perform in the muscle?
Your Answer:
Correct Answer: Acts like haemoglobin and binds with O2
Explanation:Myoglobin is a pigmented globular protein made up of 153 amino acids with a prosthetic group containing haem around which the apoprotein folds. It is the primary oxygen carrying protein of the muscles. The binding of oxygen to myoglobin is unaffected by the oxygen pressure as it has an instant tendency to bind given its hyperbolic oxygen curve. It releases oxygen at very low pO2 levels.
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This question is part of the following fields:
- General
- Physiology
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Question 43
Incorrect
-
A 30 year old man suffered severe blood loss, approx. 20-30% of his blood volume. What changes are most likely seen in the pulmonary vascular resistance (PVR) and pulmonary artery pressure (PAP) respectively following this decrease in cardiac output?
Your Answer:
Correct Answer: Increase Decrease
Explanation:Hypovolemia will result in the activation of the sympathetic adrenal discharge resulting is a decrease pulmonary artery pressure and an elevated pulmonary vascular resistance.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 44
Incorrect
-
Under normal conditions, what is the major source of energy of cardiac muscles?
Your Answer:
Correct Answer: Fatty acids
Explanation:Under basal conditions, most of the energy needed by cardiac muscle for metabolism is derived from fats (60%), 35% by carbohydrates, and 5% by ketones and amino acids. However, after intake of large amounts of glucose, lactate and pyruvate are mainly used. During prolonged starvation, fat acts as the primary source. 50% of the used lipids are sourced from circulating fatty acids.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 45
Incorrect
-
In the kidney, the macula densa is an area of closely packed specialized cells lining the wall of the:
Your Answer:
Correct Answer: Distal convoluted tubule
Explanation:In the kidney, the macula densa is an area of closely packed specialised cells lining the region of the distal convoluted tubule (DCT) lying next to the glomerular vascular pole. The cells of the macula densa are sensitive to the ionic content and water volume of the fluid in the DCT, producing signals that promote renin secretion by other cells of the juxtaglomerular apparatus.
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This question is part of the following fields:
- Physiology
- Renal
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Question 46
Incorrect
-
One sensitive indicator of heavy alcohol dependence is:
Your Answer:
Correct Answer: Elevated serum gamma-glutamyl transpeptidase
Explanation:Elevated serum gamma-glutamyl transpeptidase (GGT) may be the only laboratory abnormality in patients who are dependent on alcohol. Heavy drinkers may also have an increased MCV.
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This question is part of the following fields:
- Hepatobiliary
- Physiology
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Question 47
Incorrect
-
Which of the following will be affected by a lesion in the posterior column-medial lemniscus system?
Your Answer:
Correct Answer: Fine touch
Explanation:The posterior column–medial lemniscus (PCML) pathway is a sensory pathway that transmits fine touch and conscious proprioceptive information from the body to the brain. As the posterior columns are also known as dorsal columns, the pathway is also called the dorsal column–medial lemniscus system or DCML.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 48
Incorrect
-
Atractyloside is an inhibitor of electron transport chain. It is expected to have little or no effect on the functioning of which of the following cell types?
Your Answer:
Correct Answer: Red blood cells
Explanation:Electron transport chain is a series of electron carriers that are embedded in the mitochondrial membrane. It is the place where ATP is made. Inhibiting the electron transport chain will stop production of ATP. Red blood cells are the only cell in the given option which do not contain ATP.
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This question is part of the following fields:
- General
- Physiology
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Question 49
Incorrect
-
A sudden loud sound is more likely to result in cochlear damage than a slowly developing loud sound. This is because:
Your Answer:
Correct Answer: There is a latent period before the attenuation reflex can occur
Explanation:On transmission of a loud sound into the central nervous system, an attenuation reflex occurs after a latent period of 40-80 ms. This reflex contracts the two muscles that pull malleus and stapes closer, developing a high degree of rigidity in the entire ossicular chain. This reduces the ossicular conduction of low frequency sounds to the cochlea by 30-40 decibels. In this way, the cochlea is protected from damage due to loud sounds (these are low frequency sounds) when they develop slowly.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 50
Incorrect
-
Thalamic syndrome will most likely result in:
Your Answer:
Correct Answer: Hyperaesthesia
Explanation:Signs and symptoms of thalamic syndrome include contralateral hemi anaesthesia, burning or aching sensation in one half of a body (hyperaesthesia), often accompanied by mood swings.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 51
Incorrect
-
Which of the following compensatory parameters is responsible for causing an increase in the blood pressure in a 30 year old patient with a BP of 40 mmHg?
Your Answer:
Correct Answer: Baroreceptor reflex
Explanation:The baroreflex or baroreceptor reflex is one of the body’s homeostatic mechanisms for regulating blood pressure. It provides a negative feedback response in which an elevated blood pressure will causes blood pressure to decrease; similarly, decreased blood pressure depresses the baroreflex, causing blood pressure to rise. The system relies on specialised neurones (baroreceptors) in the aortic arch, carotid sinuses and elsewhere to monitor changes in blood pressure and relay them to the brainstem. Subsequent changes in blood pressure are mediated by the autonomic nervous system. Baroreceptors include those in the auricles of the heart and vena cava, but the most sensitive baroreceptors are in the carotid sinuses and aortic arch. The carotid sinus baroreceptors are innervated by the glossopharyngeal nerve (CN IX); the aortic arch baroreceptors are innervated by the vagus nerve (CN X).
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 52
Incorrect
-
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:
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 53
Incorrect
-
During strenuous exercise, what else occurs besides tachycardia?
Your Answer:
Correct Answer: Increased stroke volume
Explanation:During strenuous exercise there is an increase in:
– Heart rate, stroke volume and therefore cardiac output. (CO = HR x SV)
– Respiratory rate (hyperventilation) which will lead to a reduction in Paco2.
– Oxygen demand of skeletal muscle, therefore leading to a reduction in mixed venous blood oxygen concentration.
Renal blood flow is autoregulated, so renal blood flow is preserved and will tend to remain the same. Mean arterial blood pressure is a function of cardiac output and total peripheral resistance and will increase with exercise, mainly as a result of the increase in cardiac output that occurs.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 54
Incorrect
-
A lesion involving the lateral geniculate nucleus of the thalamus is likely to affect:
Your Answer:
Correct Answer: Vision
Explanation:The lateral geniculate nucleus (LGN) of the thalamus is the primary processor of visual information in the central nervous system. The LGN receives information directly from the retina and sends projections directly to the primary visual cortex. The LGN likely helps the visual system focus its attention on the most important information.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 55
Incorrect
-
What is the most likely cause of prolonged bleeding time in a 40 year old women admitted for a laparoscopic cholecystectomy?
Your Answer:
Correct Answer: Thrombocytopaenia
Explanation:Bleeding time is related to platelet function, thus a decrease in platelet function, as seen in thrombocytopenia, DIC and von Willebrand disease in which platelet aggregation is defective, leads to an increase in bleeding time. It is not affected by a decrease or deficiency of any other clotting factors. Aspirin and other COX inhibitors prolong bleeding time along with warfarin and heparin.
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This question is part of the following fields:
- General
- Physiology
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Question 56
Incorrect
-
A patient presents with loss of fine touch and sense of proprioception in the lower part of the body (below T6). He is likely to have a lesion involving:
Your Answer:
Correct Answer: Gracile nucleus
Explanation:The gracile nucleus is located in the medulla oblongata and is one of the dorsal column nuclei involved in the sensation of fine touch and proprioception. It contains second-order neurons of the dorsal column–medial lemniscus system, that receive inputs from sensory neurones of the dorsal root ganglia and send axons that synapse in the thalamus.
The gracile nucleus and fasciculus carry epicritic, kinaesthetic and conscious proprioceptive information from the lower part of the body (below the level of T6 in the spinal cord). Similar information from the upper part of body (above T6, except for face and ear) is carried by the cuneate nucleus and fasciculus. The information from face and ear is carried by the primary sensory trigeminal nucleus.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 57
Incorrect
-
QT interval in the electrocardiogram of a healthy individual is normally:
Your Answer:
Correct 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 58
Incorrect
-
Increased resistance to flow of blood in cerebral vessels is most likely seen in:
Your Answer:
Correct Answer: Elevation in systemic arterial pressure from 100 to 130 mmHg
Explanation:Constant cerebral blood flow is maintained by autoregulation in the brain, which causes an increase in local vascular resistance to offset an increase in blood pressure. There will be an increase in cerebral blood flow (and decrease in resistance to cerebral blood flow) with a decrease in arterial oxygen or an increase in arterial CO2. Similarly, a decrease in viscosity will also increase the blood flow. Due to increased brain metabolism and activity during a seizure, there will also be an increase in the cerebral blood flow.
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This question is part of the following fields:
- Cardiovascular
- Physiology
-
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Question 59
Incorrect
-
The most likely cause of a low p(O2) in arterial blood is:
Your Answer:
Correct Answer: Hypoxic hypoxia
Explanation:Hypoxia is when the whole body or a region is deprived of adequate oxygen supply. Different types of hypoxia include the following:
– Hypoxic hypoxia, which occurs due to poor oxygen supply, as a result of low partial pressure of oxygen in arterial blood. This could be due to low partial pressure of atmospheric oxygen (e.g., at high altitude), sleep apnoea, poor ventilation because of chronic obstructive pulmonary disease or respiratory arrest, or shunts. The other types of hypoxia have a normal partial pressure of oxygen.
– Anaemic hypoxia occurs due to low total oxygen content of the blood, with a normal arterial oxygen pressure.
– Hyperaemic hypoxia occurs due to poor delivery of oxygen to target tissues, such as in carbon monoxide poisoning or methemoglobinemia.
– Histotoxic hypoxia results due to inability of the cells to use the delivered oxygen due to disabled oxidative phosphorylation enzymes.
– Ischaemic (or stagnant) hypoxia occurs due to local flow restriction of well-oxygenated blood, seen in cases like cerebral ischaemia, ischaemic heart disease and intrauterine hypoxia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 60
Incorrect
-
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. What is his alveolar ventilation?
Your Answer:
Correct Answer: 3000 ml/min
Explanation:Alveolar ventilation is the amount of air reaching the alveoli per minute. Alveolar ventilation = respiratory rate × (tidal volume – anatomical dead space volume). Thus, alveolar ventilation = 10 × (550 − 250) = 3000 ml/min.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 61
Incorrect
-
A 25 year old man presented with a history of headache and peripheral cyanosis. He had been living in the Himalayas for 6 months prior to this. What is the reason for his condition?
Your Answer:
Correct Answer: Physiological polycythaemia
Explanation:Polycythaemia is a condition that results in an increase in the total number of red blood cells (RBCs) in the blood. It can be due to myeloproliferative syndrome or due to chronically low oxygen levels or rarely malignancy. In primary polycythaemia/polycythaemia vera the increase is due to an abnormality in the bone marrow, resulting in increases RBCs, white blood cells (WBCs) and platelets. In secondary polycythaemia the increase occurs due to high levels of erythropoietin either artificially or naturally. The increase is about 6-8 million/cm3 of blood. A type of secondary polycythaemia is physiological polycythaemia where people living in high altitudes who are exposed to hypoxic conditions produce more erythropoietin as a compensatory mechanism for thin oxygen and low oxygen partial pressure.
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This question is part of the following fields:
- General
- Physiology
-
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Question 62
Incorrect
-
Question 63
Incorrect
-
A 23 year old woman is Rh -ve and she delivered a baby with a Rh+ blood group. What measure can be performed to prevent Rh incompatibility in the next pregnancy?
Your Answer:
Correct Answer: Immunoglobulin D
Explanation:Rh disease is also known as erythroblastosis fetalis and is a disease of the new-born. In mild states it can cause anaemia with reticulocytosis and in severe forms causes severe anaemia, morbus hemolytcus new-born and hydrops fetalis. RBCs of the Rh+ baby can cross the placenta and enter into the maternal blood. As she is Rh- her body will form antibodies against the D antigen which will pass through the placenta in subsequent pregnancies.
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This question is part of the following fields:
- General
- Physiology
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Question 64
Incorrect
-
Which of the following is responsible for the activation of pepsinogen released in the stomach?
Your Answer:
Correct Answer: Acid pH and pepsin
Explanation:Pepsinogen is the inactive precursor of pepsin. Once secreted, it comes in contact with hydrochloric acid and pepsin, previously formed, and undergoes cleavage to form active pepsin.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 65
Incorrect
-
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:
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 66
Incorrect
-
What is the reason for a deranged thrombin clotting time?
Your Answer:
Correct Answer: Heparin therapy
Explanation:Thrombic clotting time is also known as thrombin time. It is clinically performed to determine the therapeutic levels of heparin. After plasma is isolated from the blood, bovine thrombin is added to it and the time it takes from the addition to clot is recorded. The reference interval is usually <21s. deranged results are indicative of heparin therapy, hypofibrinogenemia, hyperfibrinogenaemia or lupus anticoagulant.
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This question is part of the following fields:
- General
- Physiology
-
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Question 67
Incorrect
-
Different portions of the renal tubule have varying degrees of water permeability. Which of the following renal sites is characterised by low water permeability under normal circumstances?
Your Answer:
Correct Answer: Thick ascending limb of the loop of Henlé
Explanation:Within the nephron of the kidney, the ascending limb of the loop of Henle is a segment of the loop of Henle downstream of the descending limb, after the sharp bend of the loop. Both the thin and the thick ascending limbs of the loop of Henlé have very low permeability to water. Since there are no regulatory mechanisms to alter its permeability, it remains poorly permeable to water under all circumstances. Sodium and chloride are transported out of the luminal fluid into the surrounding interstitial spaces, where they are reabsorbed. Water must remain behind because it is not reabsorbed, so the solute concentration becomes less and less (the luminal fluid becomes more dilute). This is one of the principal mechanisms (along with diminution of ADH secretion) for the production of a dilute, hypo-osmotic urine (water diuresis).
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This question is part of the following fields:
- Physiology
- Renal
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Question 68
Incorrect
-
Which of the following is involved in vitamin B12 absorption?
Your Answer:
Correct Answer: Intrinsic factor
Explanation:Absorption of vitamin B12 is by an active transport process and occurs in the ileum. Most cobalamins are bound to proteins and are released in the stomach due to low pH and pepsin. The cobalamins then bind to R proteins, i.e. haptocorrin (HC) secreted from salivary glands and gastric juice. Another cobalamin binding protein is Intrinsic factor (IF) secreted from the gastric parietal cells. The cobalamin-HC complex is digested by pancreatic proteases in the intestinal lumen, and the free cobalamin then binds to IF. The complex then reaches a transmembrane receptor in the ileum and undergoes endocytosis. Cobalamin is then released intracellularly and binds to transcobalamin II (TC II). The newly formed complex then exits the ileal cell and enters the blood circulation.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 69
Incorrect
-
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:
Correct 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 70
Incorrect
-
The transmembrane proteins responsible for resting membrane potential of vascular smooth muscle cells was blocked by a drug. Which of the following transmembrane proteins were blocked by this drug?
Your Answer:
Correct Answer: K+ channels
Explanation:The resting membrane potential is due to selective permeability of the membrane to potassium ions. The Na/K pump is responsible for the generation of a gradient across the membrane and it is due to the inherent ability of the K channels to allow diffusion back into the nerve at rest which charges the cells. In reality, the resting membrane potential is more positive because of small contributions by Na+ channels, Cl− channels and non-selective cation channels.
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This question is part of the following fields:
- General
- Physiology
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Question 71
Incorrect
-
Calculate the stroke volume 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:
Correct Answer: 90 ml
Explanation:Fick’s principle states that, VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CAO2 = 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. Cardiac output is also given by product of stroke volume and heart rate. Thus, stroke volume = cardiac output / heart rate = 6.25/70 × 1000 stroke volume = 90 ml approximately.
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This question is part of the following fields:
- Cardiovascular
- Physiology
-
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Question 72
Incorrect
-
There are several mechanisms involved in the transport of sodium ions from blood to interstitial fluid of the muscle cells. Which of the following mechanisms best describes this phenomenon?
Your Answer:
Correct Answer: Diffusion through channels between endothelial cells
Explanation:Capillaries are the smallest of the body’s blood vessels, measuring 5–10 μm and they help to enable the exchange of water, oxygen, carbon dioxide, and many other nutrients and waste substances between the blood and the tissues surrounding them. The walls of capillaries are composed of only a single layer of cells, the endothelium. Ion channels are pore-forming proteins that help to establish and control the small voltage gradient that exists across the plasma membrane of all living cells by allowing the flow of ions down their electrochemical gradient. An ion channel is an integral membrane protein or more typically an assembly of several proteins. The archetypal channel pore is just one or two atoms wide at its narrowest point. It conducts a specific ion such as sodium or potassium and conveys them through the membrane in single file.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
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Question 73
Incorrect
-
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 74
Incorrect
-
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:
Correct 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 75
Incorrect
-
Which organ is most vulnerable to haemorrhagic shock?
Your Answer:
Correct Answer: Kidneys
Explanation:At rest, the brain receives 15% cardiac output, muscles 15%, gastrointestinal tract 30% and kidneys receive 20%. However, if normalised by weight, the largest specific blood flow is received by the kidneys at rest (400 ml/min x 100g), making them highly vulnerable in the case of a haemorrhagic shock.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 76
Incorrect
-
Renin is secreted by pericytes in the vicinity of the afferent arterioles of the kidney from the juxtaglomerular cells. Plasma renin levels are decreased in patients with:
Your Answer:
Correct Answer: Primary aldosteronism
Explanation:Primary aldosteronism, also known as primary hyperaldosteronism or Conn’s syndrome, is excess production of the hormone aldosterone by the adrenal glands resulting in low renin levels. Most patients with primary aldosteronism (Conn’s syndrome) have an adrenal adenoma. The increased plasma aldosterone concentration leads to increased renal Na+ reabsorption, which results in plasma volume expansion. The increase in plasma volume suppresses renin release from the juxtaglomerular apparatus and these patients usually have low plasma renin levels. Salt restriction and upright posture decrease renal perfusion pressure and therefore increases renin release from the juxtaglomerular apparatus. Secondary aldosteronism is due to elevated renin levels and may be caused by heart failure or renal artery stenosis.
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This question is part of the following fields:
- Physiology
- Renal
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Question 77
Incorrect
-
A young women following a road traffic accident suffered heavy blood loss and developed subsequent anaemia. Which of the following is a consequence of this?
Your Answer:
Correct Answer: A high reticulocyte count
Explanation:Anaemia refers to a decrease in the circulating levels of haemoglobin in the blood resulting in a reduced ability of the body to transport oxygen effectively. Anaemia from blood loss results in the body further compensating by releasing stored RBCs and immature RBCs from the bone marrow. Thus resulting in a high reticulocyte count.
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This question is part of the following fields:
- General
- Physiology
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Question 78
Incorrect
-
How are amino acids transported across the luminal surface of the small intestinal epithelium?
Your Answer:
Correct Answer: Co-transport with sodium ions
Explanation:Once complex peptides are broken down into amino acids by the peptidases present in the brush border of small intestine, they are ready for absorption by at least four sodium-dependent amino acid co-transporters – one each for acidic, basic, neutral and amino acids, present on the luminal plasma membrane. These transporters first bind sodium and can then bind the amino acids. Thus, amino acid absorption is totally dependent on the electrochemical gradient of sodium across the epithelium. The basolateral membrane in contrast, possesses additional transporters to carry amino acids from the cell into the blood, but these are sodium-independent.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 79
Incorrect
-
A 47 year-old woman was admitted for elective cholecystectomy, with a past history of easy bruising and heavy menstrual periods. The patient was also diagnosed with Willebrand's disease. Willebrand's disease is:
Your Answer:
Correct Answer: Autosomal dominant
Explanation:von Willebrand disease is an autosomal dominant disorder marked by the deficiency of vWF, a large protein synthesized by the endothelial cells and megakaryocytes. It mediates adhesion of platelets to the subendothelium at site of vascular injury. Disease characteristics include impaired platelet adhesion, prolonged bleeding time and a functional deficiency of factor VIII (vWF is its carrier protein).
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This question is part of the following fields:
- General
- Physiology
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Question 80
Incorrect
-
The most likely cause of prominent U waves on the electrocardiogram (ECG) of a patient is:
Your Answer:
Correct Answer: Hypokalaemia
Explanation:The U-wave, not always visible in ECGs, is thought to represent repolarisation of papillary muscles or Purkinje fibres. When seen, it is very small and occurs after the T-wave. Inverted U-waves indicate myocardial ischaemia or left ventricular volume overload. Prominent U-waves are most commonly seen in hypokalaemia. Other causes include hypercalcaemia, thyrotoxicosis, digitalis exposure, adrenaline and class 1A and 3 anti-arrhythmic agents. It can also be seen in congenital long-QT syndrome and in intracranial haemorrhage.
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This question is part of the following fields:
- Cardiovascular
- Physiology
-
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Question 81
Incorrect
-
A 25-year-old woman complains of generalised swelling and particularly puffiness around the eyes which is worst in the morning. Laboratory studies showed:
Blood urea nitrogen (BUN) = 30 mg/dl
Creatinine = 2. 8 mg/dl
Albumin = 2. 0 mg/dl
Alanine transaminase (ALT) = 25 U/l
Bilirubin = 1 mg/dl
Urine analysis shows 3+ albumin and no cells.
Which of the following is the most likely diagnosis?Your Answer:
Correct Answer: Nephrotic syndrome
Explanation:Nephrotic syndrome is a disorder in which the glomeruli have been damaged, characterized by:
– Proteinuria (>3.5 g per 1.73 m2 body surface area per day, or > 40 mg per square meter body surface area per hour in children)
– Hypoalbuminemia (< 2,5 g/dl) – Hyperlipidaemia, and oedema (generalized anasarca).
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This question is part of the following fields:
- Physiology
- Renal
-
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Question 82
Incorrect
-
Acute respiratory distress syndrome (ARDS) is a medical condition that occurs in critically ill patients, and can be triggered by events such as trauma and sepsis. Which of the following variables is most likely to be lower than normal in a patient with ARDS?
Your Answer:
Correct Answer: Lung compliance
Explanation:Acute (or Adult) respiratory distress syndrome (ARDS) is a medical condition occurring in critically ill patients characterized by widespread inflammation in the lungs. The development of acute respiratory distress syndrome (ARDS) starts with damage to the alveolar epithelium and vascular endothelium, resulting in increased permeability to plasma and inflammatory cells. These cells pass into the interstitium and alveolar space, resulting in pulmonary oedema. Damage to the surfactant-producing type II cells and the presence of protein-rich fluid in the alveolar space disrupt the production and function of pulmonary surfactant, leading to micro atelectasis and impaired gas exchange. The pathophysiological consequences of lung oedema in ARDS include a decrease in lung volumes, compliance and large intrapulmonary shunts. ARDS may be seen in the setting of pneumonia, sepsis, following trauma, multiple blood transfusions, severe burns, severe pancreatitis, near-drowning, drug reactions, or inhalation injuries.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 83
Incorrect
-
Evaluation of a 60-year old gentleman, who has been a coal miner all his life and is suspected to have pulmonary fibrosis reveals the following: normal FEV1, arterial oxygen saturation 92%, alveolar ventilation 6000 ml/min at a tidal volume of 600 ml and a breathing rate of 12 breaths/min. There are also pathological changes in lung compliance and residual volume. Which of the following is most accurate about his residual volume?
Your Answer:
Correct Answer: Cannot be measured directly with a spirometer
Explanation:Residual volume is the air left in the lungs after maximal expiration is done. Thus, this is not a part of vital capacity and cannot be measured with a spirometer directly. It can be measured by the methods such as body plethysmography or inert gas dilution. Expiratory reserve volume is vital capacity minus inspiratory capacity. Resting volume of lungs is he sum of residual volume and expiratory reserve volume. Lungs recoil inward until the recoil pressure becomes zero, which corresponds to a volume significantly lower than residual volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 84
Incorrect
-
When the pitch of a sound increases, what is the physiological response seen in the listener?
Your Answer:
Correct Answer: The location of maximal basilar membrane displacement moves toward the base of the cochlea
Explanation:An increase in the frequency of sound waves results in a change in the position of maximal displacement of the basilar membrane in the cochlea. Low pitch sound produces maximal displacement towards the cochlear apex and greatest activation of hair cells there. With an increasing pitch, the site of greatest displacement moves towards the cochlear base. However, increased amplitude of displacement, increase in the number of activated hair cells, increased frequency of discharge of units in the auditory nerve and increase in the range of frequencies to which such units respond, are all seen in increases in the intensity or a sound stimulus.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 85
Incorrect
-
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 86
Incorrect
-
A 38-year-old woman with end-stage renal disease, is undergoing haemodialysis. She has normocytic normochromic anaemia. What is the best treatment for her?
Your Answer:
Correct Answer: Erythropoietin
Explanation:E erythropoietin (EPO) is a hormone that is released by the kidney. It is responsible for the regulation of red blood cell production in the body. It can be made using recombinant technology and is used in the treatment of anaemia of chronic renal failure and in patients under going chemotherapy
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This question is part of the following fields:
- General
- Physiology
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Question 87
Incorrect
-
A medical student is told a substance is freely filtered but is not metabolised, secreted, or stored in the kidney. It has a plasma concentration of 1000 mg/l and its urine excretion rate is 25 mg/min, and the inulin clearance is 100 ml/min. What is the rate of tubular reabsorption of the substance?
Your Answer:
Correct Answer: 75 mg/min
Explanation:Reabsorption or tubular reabsorption is the process by which the nephron removes water and solutes from the tubular fluid (pre-urine) and returns them to the circulating blood. To calculate the reabsorption rate of substance Z we use the following equation: excretion = (filtration + secretion) – reabsorption. As this substance is freely filtered, its filtration rate is equal to that of inulin. So 25 = (100 + 0) – reabsorption. Reabsorption = 100 – 25 therefore reabsorption = 75 mg/min.
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This question is part of the following fields:
- Physiology
- Renal
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Question 88
Incorrect
-
What is the basic chemical reaction that takes place in the breakdown of complex foodstuffs?
Your Answer:
Correct Answer: Hydrolysis
Explanation:Breakdown of complex food into simpler compounds is achieved by hydrolysis, with the help of different enzymes specific for different compounds.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 89
Incorrect
-
An experiment was conducted in which the skeletal muscle protein (not smooth muscle) involved in contraction was selectively inhibited. Which protein was inhibited?
Your Answer:
Correct Answer: Troponin
Explanation:The mechanism of contraction of smooth muscles is different from that of skeletal muscles in which the contractile protein is troponin whilst in smooth muscle contraction is a protein called calmodulin. Calmodulin reacts with calcium ions and stimulates the formation of myosin crossbridges.
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This question is part of the following fields:
- General
- Physiology
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Question 90
Incorrect
-
During pregnancy the uterus enlarges however after delivery it regresses to its original size. Which of the following organelles is responsible for this regression?
Your Answer:
Correct Answer: Lysosomes
Explanation:Lysosomes are formed by budding of the Golgi apparatus and contain enzymes which digest macromolecules. They are found in both plants and animals and are active in autophagic cell death, digestion after phagocytosis and for the cells own recycling process. They fuse with the molecules and release their content resulting in digestion.
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This question is part of the following fields:
- General
- Physiology
-
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Question 91
Incorrect
-
Which of the following proteins prevents red blood cells (RBCs) from bursting when they pass through capillaries?
Your Answer:
Correct Answer: Spectrin
Explanation:Spectrin is a structural protein found in the cytoskeleton that lines the intercellular side of the membrane of cells which include RBCs. They maintain the integrity and structure of the cell. It is arranged into a hexagonal arrangement formed from tetramers of spectrin and associated with short actin filaments that form junctions allowing the RBC to distort its shape.
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This question is part of the following fields:
- General
- Physiology
-
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Question 92
Incorrect
-
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:
Correct 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 93
Incorrect
-
A 60 year old patient with a history of carcinoma of the head of the pancreas, and obstructive jaundice presents with a spontaneous nose bleed and easy bruising. What is the most likely reason for this?
Your Answer:
Correct Answer: Vitamin-K-dependent clotting factors deficiency
Explanation:Vitamin K is a fat soluble vitamin requiring fat metabolism to function properly to allow for its absorption. People with obstructive jaundice develop vitamin k deficiency as fat digestion is impaired. Vit K causes carboxylation of glutamate residue and hence regulates blood coagulation including: prothrombin (factor II), factors VII, IX, X, protein C, protein S and protein Z.
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This question is part of the following fields:
- General
- Physiology
-
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Question 94
Incorrect
-
A patient admitted for esophagectomy showed low levels of the lightest plasma protein in terms of weight. Which of the following is the lightest plasma protein:
Your Answer:
Correct Answer: Albumin
Explanation:Albumin is the most abundant and the lightest of all the plasma proteins. It maintains osmotic pressure, transports unconjugated bilirubin, thyroid hormones, fatty acids, drugs and acts as a buffer for pH.
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This question is part of the following fields:
- General
- Physiology
-
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Question 95
Incorrect
-
A 30-year-old woman feels thirsty. This thirst is probably due to:
Your Answer:
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 96
Incorrect
-
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:
Correct 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 97
Incorrect
-
When does the heart rate decrease?
Your Answer:
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 98
Incorrect
-
With respect to far accommodation, which of the following is a TRUE statement?
Your Answer:
Correct Answer: The ciliary muscles are relaxed
Explanation:Myopia or near-sightedness is a disease due to elongated eyeballs or too strong a lens. For far accommodation (focus of a distant object onto the retina), the lens needs to decrease its refractive power, or in other words, increase its focal length. This is done by relaxation of ciliary muscles which tightens the zonular fibres and flattening of the lens. Relaxation of the zonular fibres, rounding of the lens, shortening of the focal length and constriction of the pupil occurs during near accommodation.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 99
Incorrect
-
Regarding the coagulation cascade, Factor VII:
Your Answer:
Correct Answer: Is a serine protease
Explanation:Factor VII (FVII) is a zymogen for a vitamin K-dependent serine protease essential for the initiation of blood coagulation. It is synthesized primarily in the liver and circulates in plasma. Within the liver, hepatocytes are involved in the synthesis of most blood coagulation factors, such as fibrinogen, prothrombin, factor V, VII, IX, X, XI, XII, as well as protein C and S, and antithrombin, whereas liver sinusoidal endothelial cells produce factor VIII and von Willebrand factor.
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This question is part of the following fields:
- General
- Physiology
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Question 100
Incorrect
-
Which of the following has the highest content of triglycerides?
Your Answer:
Correct Answer: Chylomicron
Explanation:Created by the small intestinal cells, chylomicrons are large lipoprotein molecules which transport lipids to the liver, adipose, cardiac and skeletal tissue. Chylomicrons are mainly composed of triglycerides (,85%) along with some cholesterol and cholesteryl esters. Apo B-48 is the main apolipoprotein content.
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This question is part of the following fields:
- Gastroenterology
- Physiology
-
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Question 101
Incorrect
-
A blood sample of a 58 year old male patient, who underwent an abdominal aortic aneurysm repair, was sent to the laboratory. The laboratory technician said that the patient’s blood agglutinates with antisera anti-A and anti-D, while the patient’s serum agglutinates cells of blood group B. What is the blood group of this patient?
Your Answer:
Correct Answer: A positive
Explanation:Group A – has only the A antigen on red cells (and B antibody in the plasma)
Group B – has only the B antigen on red cells (and A antibody in the plasma)
Group AB – has both A and B antigens on red cells (but neither A nor B antibody in the plasma)
Group O – has neither A nor B antigens on red cells (but both A and B antibody are in the plasma). Many people also have a so-called Rh factor on the red blood cell’s surface. This is also an antigen and those who have it are called Rh+. A person with Rh– blood does not have Rh antibodies naturally in the blood plasma (as one can have A or B antibodies, for instance) but they can develop Rh antibodies in the blood plasma if they receive blood from a person with Rh+ blood, whose Rh antigens can trigger the production of Rh antibodies. A person with Rh+ blood can receive blood from a person with Rh– blood without any problems. In this scenario the person has blood group A+ as he has A antigen, anti B antibody and Rh antigen
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This question is part of the following fields:
- General
- Physiology
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Question 102
Incorrect
-
A 20-year old gentleman was brought to the emergency department with headache and nausea for 2 days. He also complained of intolerance to bright light and loud sounds. Lumbar puncture showed glucose < 45 mg/dl, protein > 5 mg/dl and neutrophil leucocytosis. The likely diagnosis is:
Your Answer:
Correct Answer: Meningitis
Explanation:Diagnosis of meningitis can be carried out with examination of cerebrospinal fluid (CSF) with a lumbar puncture (LP). In a case of bacterial meningitis, the CSF analysis will show:
– Opening pressure: > 180 mmH2O
– White blood cell count: 10–10 000/μl with neutrophil predominance
– Glucose: < 40 mg/dl – CSF glucose to serum glucose ratio: < 0.4 – Protein: > 4.5 mg/dl
– Gram stain: positive in > 60%
– Culture: positive in > 80%
– Latex agglutination: may be positive in meningitis due to Streptococcus pneumoniae, Neisseria meningitidis, Haemophilus influenzae, Escherichia coli and group B streptococci
– Limulus, lysates: positive in Gram-negative meningitis
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This question is part of the following fields:
- Neurology
- Physiology
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Question 103
Incorrect
-
A 34-year-old woman is diagnosed with cerebral oedema after suffering a severe head trauma. Which of the following conditions is not likely to be associated with the extracellular oedema?
Your Answer:
Correct Answer: Increased plasma colloid osmotic pressure
Explanation:Cerebral oedema is extracellular fluid accumulation in the brain. Increased capillary permeability, increased capillary pressure, increased interstitial fluid colloid osmotic pressure and lymphatic blockage would increase fluid movement into the interstitial spaces. Increased plasma colloid osmotic pressure, however, would oppose fluid movement from the capillaries into the interstitial compartment.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
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Question 104
Incorrect
-
After surgery, a patient developed a stitch granuloma . Which leukocyte in the peripheral blood will become an activated macrophage in this granuloma?
Your Answer:
Correct Answer: Monocyte
Explanation:Monocytes are leukocytes that protect the body against infections and move to the site of infection within 8-12 hours to deal with it. They are produced in the bone marrow and shortly after being produced are released into the blood stream where they circulate until an infection is detected. When called upon they leave the circulation and transform into macrophages within the tissue fluid and thus gain the capability to phagocytose the offending substance. Monocyte count is part of a complete blood picture. Monocytosis is the state of excess monocytes in the peripheral blood and may be indicative of various disease states. Examples of processes that can increase a monocyte count include: • chronic inflammation • stress response • hyperadrenocorticism • immune-mediated disease • pyogranulomatous disease • necrosis • red cell regeneration.
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This question is part of the following fields:
- General
- Physiology
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Question 105
Incorrect
-
After a severe asthma attack, a 26-year-old woman is left in a markedly hypoxic state. In which of the following organs are the arterial beds most likely to be vasoconstricted due to the hypoxia?
Your Answer:
Correct Answer: Lungs
Explanation:Hypoxic pulmonary vasoconstriction is a local response to hypoxia resulting primarily from constriction of small muscular pulmonary arteries in response to reduced alveolar oxygen tension. This unique response of pulmonary arterioles results in a local adjustment of perfusion to ventilation. This means that if a bronchiole is obstructed, the lack of oxygen causes contraction of the pulmonary vascular smooth muscle in the corresponding area, shunting blood away from the hypoxic region to better-ventilated regions. The purpose of hypoxic pulmonary vasoconstriction is to distribute blood flow regionally to increase the overall efficiency of gas exchange between air and blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 106
Incorrect
-
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:
Correct 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 107
Incorrect
-
Gastric acid secretion is stimulated by which of the following?
Your Answer:
Correct 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 108
Incorrect
-
In a hypertensive patient with secondary hyperaldosteronism, aldosterone is released mainly in response to:
Your Answer:
Correct Answer: Angiotensin II
Explanation:Secondary hyperaldosteronism in hypertension is either due to primary renin overproduction by the kidneys or renin overproduction secondary to decreased renal blood flow. The main stimulus for aldosterone release are adrenocorticotrophic hormone (ACTH), angiotensin II and high plasma K+ levels. Low plasma Na+ might also stimulate the adrenal cortex. Fluid overload will reduce aldosterone secretion. Atrial natriuretic peptide is secreted under conditions of expanded extracellular volume and will not lead to aldosterone secretion.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 109
Incorrect
-
Where is factor VIII predominantly synthesised?
Your Answer:
Correct Answer: Vascular endothelium
Explanation:Factor VIII is an important part of the coagulation cascade. Deficiency causes haemophilia A. It is synthesised predominantly by the vascular endothelium and is not affected by liver disease. In the circulation it is bound to von Willebrand factor and it forms a stable complex with it. It is activated by thrombin or factor Xa and acts as a co factor to factor IXa to activate factor X which is a co factor to factor Va. Thrombin cleaves fibrinogen in fibrin and forms a meshwork to trap RBC and platelets to form a clot.
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This question is part of the following fields:
- General
- Physiology
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Question 110
Incorrect
-
Skeletal muscle fibres are divided into two basic types, type I (slow-twitch fibres) and type II (fast-twitch fibres). Fast muscle fibres do which of the following:
Your Answer:
Correct Answer: Use anaerobic metabolism
Explanation:Skeletal muscles are divided into two types:
1) type I also known as the slow twitch fibres. They use oxygen for their metabolism and as a result they have a high endurance potential. To support this they have abundant mitochondria and myoglobin, so they appear red/dark.
2) type II fibres also called fast twitch fibres, are low endurance fibres used during anaerobic metabolism. They are required for short bursts of strength and cannot sustain contractions for long periods of time.
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This question is part of the following fields:
- General
- Physiology
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Question 111
Incorrect
-
Which of the following is true about myasthenia gravis?
Your Answer:
Correct Answer: Response of skeletal muscle to nerve stimulation is weakened
Explanation:An autoimmune disorder, myasthenia gravis leads to progressive muscle weakness. It occurs due to formation of antibodies against the nicotinic acetylcholine (ACh) receptor of the motor endplate, which leads to impaired neuromuscular transmission. Thus, nerve stimulation will lead to a weakened muscle response, but direct electrical stimulation will bring about a normal response. Diagnostic test includes improvement of muscle weakness by small doses of acetylcholinesterase inhibitors (physostigmine or edrophonium). However, a large dose of physostigmine worsens the weakness due to desensitisation of the endplate to persistent Ach. One of the investigative tools includes radiolabelled snake venom α-bungarotoxin. It is an in vitro study performed on muscle biopsy specimens and used to quantify the number of ACh receptors at the motor endplate.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 112
Incorrect
-
A 7-year-old girl is given cephalexin to treat an infection and develops hives, with localised facial oedema. Which of the following conditions will cause localised oedema?
Your Answer:
Correct Answer: Angio-oedema
Explanation:Angio-oedema, is the rapid swelling of the skin, mucosa and submucosal tissues. The underlying mechanism typically involves histamine or bradykinin. The version related to histamine is to due an allergic reaction to agents such as insect bites, food, or medications. The version related to bradykinin may occur due to an inherited C1 esterase inhibitor deficiency, medications e.g. angiotensin converting enzyme inhibitors, or a lymphoproliferative disorder.
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This question is part of the following fields:
- Physiology
- Renal
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Question 113
Incorrect
-
A 50 year old woman presented with excessive bleeding after an inguinal hernia repair. Labs are suggestive of a primary haemostasis defect. Deficiency of which of the following is most likely to cause it?
Your Answer:
Correct Answer: Platelets
Explanation:Primary haemostatic control means the first line of defence against immediate bleeding. This is carried out by the platelets. They immediately form a haemostatic plug at the site of injury. Coagulation starts within 20s after an injury to the blood vessel which damage the endothelial cells. Secondary haemostasis follows which includes activation of the coagulation factors to form fibrin strands which mesh together forming the platelet plug. Platelets interact with platelet collagen receptor, glycoprotein Ia/IIa and to collagen fibres in the vascular endothelium. This adhesion is mediated by von Willebrand factor (vWF), which forms links between the platelet glycoprotein Ib/IX/V and collagen fibrils. The platelets are then activated and release the contents of their granules into the plasma, in turn activating other platelets and white blood cells.
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This question is part of the following fields:
- General
- Physiology
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Question 114
Incorrect
-
A 30 year old female suffered from mismatched transfusion induced haemolysis. Which substance will be raised in the plasma of this patient?
Your Answer:
Correct Answer: Bilirubin
Explanation:Bilirubin is a yellow pigment that is formed due to the break down of RBCs. Haemolysis results in haemoglobin that is broken down into a haem portion and globin which is converted into amino acids and used again. Haem is converted into unconjugated bilirubin in the macrophages and shunted to the liver. In the liver it is conjugated with glucuronic acid making it water soluble and thus excreted in the urine. Its normal levels are from 0.2-1 mg/dl. Increased bilirubin causes jaundice and yellowish discoloration of the skin.
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This question is part of the following fields:
- General
- Physiology
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Question 115
Incorrect
-
Which of the following will be a seen in a patient with a plasma thyroid-stimulating hormone (TSH) level of 14 mU/l (normal < 5 mU/l) and a low T3 resin uptake of 19% (normal 25–35%)?
Your Answer:
Correct Answer: Periorbital swelling and lethargy
Explanation:Low T3 resin uptake combined with raised TSH is indicative of hypothyroidism. Signs and symptoms include dull expression, facial puffiness, lethargy, periorbital swelling due to infiltration with mucopolysaccharides, bradycardia and cold intolerance. Anxiety, palpitations, tachycardia, raised body temperature, heat intolerance and weight loss are all seen in hyperthyroidism.
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This question is part of the following fields:
- Endocrinology
- Physiology
-
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Question 116
Incorrect
-
In which situation is a stretch reflex such as knee jerk likely to be exaggerated?
Your Answer:
Correct Answer: In upper motor neuron lesion
Explanation:A stretch reflex is a monosynaptic reflex that causes muscle contraction in response to stretching within that muscle. The sensory apparatus in a muscle that are sensitive to stretch are the muscle spindles. The patellar (knee jerk) reflex is an example. In upper motor neuron lesions, the stretch reflexes tend to be brisk due to loss of inhibitory signals on gamma neurons through the lateral reticulospinal tract.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 117
Incorrect
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A 50 year old man was admitted to the surgical ICU following a hemicolectomy for carcinoma of the caecum. A full blood count revealed: haematocrit = 30%, erythrocytes = 4 × 106/μ, haemoglobin level = 8 g/dl. To determine the likely cause of his anaemia, red blood cell indices were calculated. Which RBC indices are correct?
Your Answer:
Correct Answer: MCHC = haemoglobin concentration/haematocrit
Explanation:Mean corpuscular haemoglobin concentration (MCHC) is calculated simply by dividing the haemoglobin concentration (8 g/dl) by the haematocrit (0.3). The normal range is 31–36 g/dl. This patient has a hypochromic anaemia (MCHC = 8/0.3 = 26.7 g/dl). Dividing the haemoglobin concentration × 10 by erythrocyte number yields mean corpuscular haemoglobin (MCH). Normal range is 25.4–34.6 pg/cell and this patient has a significantly reduced cellular haemoglobin content (MCH = 8 × 10/4 = 20 pg/cell). Mean corpuscular volume (MCV) is calculated by dividing haematocrit × 1000 by erythrocyte number (4 × 106/μl). Normal range is 80–100 fl and this patient has a microcytic anaemia (MCV = 0.3 × 1000/4 = 75 fl). Microcytic, hypochromic anaemia is characteristic for iron-deficiency.
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This question is part of the following fields:
- General
- Physiology
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Question 118
Incorrect
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A 77-year-old woman's renal function is tested. The following results were obtained during a 24-h period:
Urine flow rate: 2. 0 ml/min
Urine inulin: 0.5 mg/ml
Plasma inulin: 0.02 mg/ml
Urine urea: 220 mmol/l
Plasma urea: 5 mmol/l.
What is the urea clearance?Your Answer:
Correct Answer: 88 ml/min
Explanation:Urea is reabsorbed in the inner medullary collecting ducts of the nephrons. 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. So, glomerular filtration rate = (0.220 × 2. 0)/0.005 = 88 ml/min.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 119
Incorrect
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What is the normal amount of oxygen that is carried in the blood?
Your Answer:
Correct Answer: 20 ml oxygen/100 ml blood
Explanation:Normally, 100 ml of blood contains 15g haemoglobin and a single gram of haemoglobin can bind to 1.34 ml oxygen when 100% saturated. Thus, 15 × 1.34 = 20 ml O2/100 ml blood. The haemoglobin in venous blood that is leaving the tissues is about 75% saturated with oxygen, and hence it carries about 15 ml O2/100 ml venous blood. This implies that for each 10 ml of blood, 5 ml oxygen is transported to the tissues. With a p(O2) > 100 mm Hg, only 3 ml of oxygen is dissolved in every one litre of plasma. By increasing the pA(O2) by breathing 100% oxygen, one can add an extra amount of oxygen in the plasma, but the amount of oxygen carried by haemoglobin will not increase significantly as it is already > 95% saturated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 120
Incorrect
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A 40-year old lady with a flail chest due to trauma was breathing with the help of a mechanical ventilator in the ICU, and was heavily sedated on muscle relaxants. Due to sudden power failure, a nurse began to hand-ventilate the patient with a Ambu bag. What change will occur in the following parameters: (Arterial p(CO2), pH) in the intervening period between power failure and hand ventilation?
Your Answer:
Correct Answer: Increase, Decrease
Explanation: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. Respiratory acidosis can be acute or chronic. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with 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). The given problem represents acute respiratory acidosis and thus, will show a increase in arterial p(CO2) and decrease in pH.
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This question is part of the following fields:
- Physiology
- Respiratory
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SESSION STATS - PERFORMANCE PER SPECIALTY
