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Question 1
Correct
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After a total colectomy and ileotomy, a 50-year old diabetic man who was a known case of diabetic nephropathy had persistent metabolic acidosis. The patient appeared well perfused, with normal vital signs and normal fluid balance. Investigations revealed:
Sodium = 132 mmol/l
Potassium = 6.6 mmol/l
Creatinine = 185 μmol/l (2.16 mg/dl)
Chloride = 109 μmol/l
8am cortisol = 500 nmol/l (18 μg/dl)
pH = 7.29, p(CO2) = 27 mmHg
p(O2) = 107 mmHg
standard bicarbonate = 12 mmol/l.
What is the likely causes of his acidosis?Your Answer: 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 2
Incorrect
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A 42-year old woman presents to the doctor with jaundice. Her investigations show conjugated hyperbilirubinemia, raised urine bilirubin levels and low urine urobilinogen levels. What is the likely cause of her jaundice?
Your Answer: Deficiency of glucuronyl transferase
Correct Answer: Blockage of the common bile duct
Explanation:The description of the patient here fits the diagnosis of obstructive jaundice or cholestasis, which results in conjugated hyperbilirubinemia. Cholestasis occurs due to impairment of bile flow, which can be anywhere from the liver cell canaliculus to the ampulla of Vater. Causes can be divided into intrahepatic and extrahepatic.
– Intrahepatic causes include hepatitis, drug toxicity, alcoholic liver disease, primary biliary cirrhosis, cholestasis of pregnancy and metastatic cancer.
– Extrahepatic causes include common duct stone, pancreatic cancer, benign stricture of the common duct, ductal carcinoma, pancreatitis and sclerosing cholangitis.
There is absence of bile constituents in the intestine, which causes spillage in the systemic circulation. Symptoms include pale stools, dark urine, pruritus, malabsorption leading to steatorrhea and deficiency of fat-soluble vitamins. Chronic cases can result in osteoporosis or osteomalacia due to vitamin D deficiency and Ca2+ malabsorption. Cholesterol and phospholipid retention produces hyperlipidaemia despite fat malabsorption (although increased liver synthesis and decreased plasma esterification of cholesterol also contribute); triglyceride levels are largely unaffected. The lipids circulate as a unique, low-density lipoprotein called lipoprotein X.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 3
Correct
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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: 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 4
Correct
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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: 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 5
Correct
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The pleural cavity is the space between the two pulmonary pleurae which cover the lungs. What is the normal amount of pleural fluid?
Your Answer: 10 ml
Explanation:Pleural fluid is a serous fluid produced by the serous membrane covering normal pleurae. Most fluid is produced by the parietal circulation (intercostal arteries) via bulk flow and reabsorbed by the lymphatic system. The total volume of fluid present in the intrapleural space is estimated to be only 2–10 ml. A small amount of protein is present in intrapleural fluid. Normally, the rate of reabsorption increases as a physiological response to accumulating fluid.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 6
Correct
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Which of the following is a true statement regarding the pupil?
Your 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 7
Incorrect
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A 55 year old lady underwent an uneventful appendicectomy. Two hours later, her arterial blood gas analysis on room revealed pH: 7.30, p(CO2): 53 mmHg and p(O2): 79 mmHg. What is the most likely cause of these findings?
Your Answer: Pulmonary embolus
Correct Answer: Alveolar hypoventilation
Explanation:In the given problem, there is respiratory acidosis due to hypercapnia from a low respiratory rate and/or volume (hypoventilation). Causes of hypoventilation include conditions impairing the central nervous system (CNS) respiratory drive, impaired neuromuscular transmission and other causes of muscular weakness (drugs and sedatives), along with obstructive, restrictive and parenchymal pulmonary disorders. Hypoventilation leads to hypoxia and hypercapnia reduces the arterial pH. Severe acidosis leads to pulmonary arteriolar vasoconstriction, systemic vascular dilatation, reduced myocardial contractility, hyperkalaemia, hypotension and cardiac irritability resulting in arrhythmias. Raised carbon dioxide concentration also causes cerebral vasodilatation and raised intracranial pressure. Over time, buffering and renal compensation occurs. However, this might not be seen in acute scenarios where the rise in p(CO2) occurs rapidly.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 8
Correct
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After a prolonged coronary artery bypass surgery, a 60-year old gentleman was transfused 3 units of fresh-frozen plasma and 2 units of packed red cells. Two days later, the nurse noticed that he was tachypnoeic and chest X-ray showed signs consistent with adult respiratory distress syndrome. Which of the following variables will be low in this patient?
Your Answer: Compliance of the lung
Explanation:Acute or adult respiratory distress syndrome (ARDS) is a reaction to several forms of lung injuries and is commonly associated with sepsis and SIRS (systemic inflammatory response syndrome), severe traumatic injury, severe head injury, narcotics overdose, drowning, pulmonary contusion, and multiple blood transfusions. There is an increase in risk due to pre-existing liver disease or coagulation abnormalities. It results due to indirect toxic effects of neutrophil-derived inflammatory mediators in the lungs. ARDS is defined by the 1994 American–European Consensus Committee as the acute onset of bilateral infiltrates on chest X-ray, a partial pressure of arterial oxygen (pa(O2)) to fraction of inspired oxygen Fi(O2) ratio of less than 200 mmHg and a pulmonary artery occlusion pressure of less than 18 or the absence of clinical evidence of left arterial hypertension. ARDS is basically pulmonary oedema in the absence of volume overload or poor left ventricular function. This is different from acute lung injury, which shows a pa(O2)/Fi(O2) ratio of less than 300 mmHg. Pathogenesis of ARDS starts from damage to alveolar epithelium and vascular endothelium, causing increased permeability. Damage to surfactant-producing type II cells disrupts the production and function of pulmonary surfactant, causing micro atelectasis and poor gas exchange. There is a decrease in lung compliance and increase in work of breathing. Eventually, there is resorption of alveolar oedema, regeneration of epithelial cells, proliferation and differentiation of type II alveolar cells and alveolar remodelling. Some show resolution and some progress to fibrosing alveolitis, which involves the deposition of collagen in alveolar, vascular and interstitial spaces.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 9
Incorrect
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Which of the following is true about myasthenia gravis?
Your Answer: Response of skeletal muscles to direct electrical stimulation is weakened
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 10
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: Motility related
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 11
Incorrect
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A sudden loud sound is more likely to result in cochlear damage than a slowly developing loud sound. This is because:
Your Answer: A sudden sound carries more energy
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 12
Incorrect
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Which of the following can lead to haemolytic anaemia?
Your Answer: Living at high altitude
Correct Answer: Presence of haemoglobin S
Explanation:Haemoglobin S is an abnormal type of haemoglobin seen in sickle cell anaemia. This allows for the haemoglobin to crystalize within the RBC upon exposure to low partial pressures of oxygen. This results in rupture of the RBCs as they pass through microcirculation, especially in the spleen. This can cause blockage of the vessel down stream and ischaemic death of tissues, accompanied by severe pain.
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This question is part of the following fields:
- General
- Physiology
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Question 13
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: A-ve
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 14
Incorrect
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Which of the following conditions causes an elevation of the pH in the tissues with elevated arterial CO2 content?
Your Answer: Chronic renal failure
Correct Answer: Metabolic alkalosis
Explanation:Metabolic alkalosis is a metabolic condition in which the pH of tissue is elevated beyond the normal range (7.35-7.45). This is the result of decreased hydrogen ion concentration, leading to increased bicarbonate, or alternatively a direct result of increased bicarbonate concentrations. Normally, arterial pa(CO2) increases by 0.5–0.7 mmHg for every 1 mEq/l increase in plasma bicarbonate concentration, a compensatory response that is very quick. If the change in pa(CO2) is not within this range, then a mixed acid–base disturbance occurs.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 15
Correct
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Calculate the pulmonary vascular resistance in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer: 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 16
Correct
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Which of the following variables are needed to calculate inspiratory reserve volume of a patient?
Your Answer: Tidal volume, vital capacity and expiratory reserve volume
Explanation:Vital capacity = inspiratory reserve volume + tidal volume + expiratory reserve volume. Thus, inspiratory reserve volume can be calculated if tidal volume, vital capacity and expiratory reserve volume are known.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 17
Incorrect
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Most of the coagulation factors are serine proteases. Which of the following is not one of them?
Your Answer: Factor XII
Correct Answer: Factor XIII
Explanation:Serine protease coagulation factors include: thrombin, plasmin, Factors X, XI and XII. Factor VIII and factor V are glycoproteins and factor XIII is a transglutaminase.
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This question is part of the following fields:
- General
- Physiology
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Question 18
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: Follicle-stimulating hormone
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 19
Correct
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A 56-year-old woman weighs 75 kg. In this patient, total body water, intracellular fluid and extracellular fluid are respectively:
Your Answer: 45 l, 30 l, 15 l
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. The total body water (TBW) content of humans is approximately 60% of body weight. Two-thirds is located in the intracellular and one-third in the extracellular compartment. So, in a 75-kg individual, TBW = 60 × 75/100 = 45 l. Intracellular content = 2/3 × 45 = 30 l and extracellular content = 1/3 × 45 = 15 l.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 20
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: 50 ml
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 21
Correct
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Which of the following is the cause of flattened (notched) T waves on electrocardiogram (ECG)?
Your Answer: Hypokalaemia
Explanation:The T-wave is formed due to ventricular repolarisation. Normally, it is seen as a positive wave. It can be normally inverted (negative) in V1 (occasionally in V2-3 in African-Americans/Afro-Caribbeans). Hyperacute T-waves are the earliest ECG change of acute myocardial infarction. ECG findings of hyperkalaemia include high, tent-shaped T-waves, a small P-wave and a wide QRS complex. Hypokalaemia results in flattened (notched) T-waves, U-waves, ST-segment depression and prolonged QT interval.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 22
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: Decreased total lung capacity
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 23
Correct
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Which of the following is responsible for the maximum increase in total peripheral resistance on sympathetic stimulation?
Your Answer: Arterioles
Explanation:Arterioles are also known as the resistance vessels as they are responsible for approximately half the resistance of the entire systemic circulation. They are richly innervated by the autonomic nervous system and hence, will bring about the maximum increase in peripheral resistance on sympathetic stimulation.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 24
Incorrect
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A patient is diagnosed with Conn’s syndrome. Aldosterone is secreted from where?
Your Answer: Adrenal medulla
Correct Answer: Zona glomerulosa of the adrenal cortex
Explanation:The adrenal gland comprises an outer cortex and an inner medulla, which represent two developmentally and functionally independent endocrine glands.
The adrenal medulla secretes adrenaline (70%) and noradrenaline (30%)
The adrenal cortex consists of three layers (remembered by the mnemonic GFR):
G = zona glomerulosa – secretes aldosterone
F = zona fasciculata – secretes cortisol and sex steroids
R = zona reticularis – secretes cortisol and sex steroids.
Aldosterone facilitates the reabsorption of sodium and water and the excretion of potassium and hydrogen ions from the distal convoluted tubule and collecting ducts. Conn’s syndrome is characterized by increased aldosterone secretion from the adrenal glands.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 25
Incorrect
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upon stroking the plantar surface of a patient's foot, extension of toes was noted. This is likely to be accompanied with:
Your Answer: Fasciculations
Correct Answer: Spasticity
Explanation:An upper motor neuron lesion affects the neural pathway above the anterior horn cell or motor nuclei of the cranial nerves, whereas a lower motor neurone lesion affects nerve fibres travelling from the anterior horn of the spinal cord to the relevant muscles. An upper motor neurone lesions results in the following:
– Spasticity in the extensor muscles (lower limbs) or flexor muscles (upper limbs).
– ‘clasp-knife’ response where initial resistance to movement is followed by relaxation
– Weakness in the flexors (lower limbs) or extensors (upper limbs) with no muscle wasting
– Brisk tendon jerk reflexes
– Positive Babinski sign (on stimulation of the sole of the foot, the big toe is raised rather than curled downwards)
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This question is part of the following fields:
- Neurology
- Physiology
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Question 26
Incorrect
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Post-total gastrectomy, there will be a decreased production of which of the following enzymes?
Your Answer: Cholecystokinin
Correct Answer: Pepsin
Explanation:Pepsin is a protease that is released from the gastric chief cells and acts to degrade proteins into peptides. Released as pepsinogen, it is activated by hydrochloric acid and into pepsin itself. Gastrin and the vagus nerve trigger the release of pepsinogen and HCl when a meal is ingested. Pepsin functions optimally in an acidic environment, especially at a pH of 2.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 27
Incorrect
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The blood-brain barrier is a membrane that separates the circulating blood from the brain extracellular fluid in the central nervous system (CNS). Which of the following statements regarding the blood– brain barrier is CORRECT?
Your Answer: It permits carbon dioxide to pass via facilitated diffusion
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 28
Incorrect
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A 43-year-old diabetic man complains of headaches, palpitations, anxiety, abdominal pain and weakness. He is administered sodium bicarbonate used to treat:
Your Answer: Respiratory acidosis
Correct Answer: Metabolic acidosis
Explanation:Sodium bicarbonate is indicated in the management of metabolic acidosis, which may occur in severe renal disease, uncontrolled diabetes, circulatory insufficiency due to shock or severe dehydration, extracorporeal circulation of blood, cardiac arrest and severe primary lactic acidosis. Bicarbonate is given at 50-100 mmol at a time under scrupulous monitoring of the arterial blood gas readings. This intervention, however, has some serious complications including lactic acidosis, and in those cases, should be used with great care.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 29
Incorrect
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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: All answers are correct
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 30
Correct
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A 45-year-old pregnant woman develops high blood pressure at 20 weeks. She complains of headaches and swollen feet, and a test reveals proteinuria (350 mg/day). Which of the following is the most likely diagnosis?
Your Answer: Pre-eclampsia
Explanation:Pre-eclampsia (PE) is a disorder of pregnancy characterized by the onset of high blood pressure (two separate readings taken at least 6 h apart of 140/90 or more) and often a significant amount of protein in the urine (>300 mg of protein in a 24-h urine sample). While blood pressure elevation is the most visible sign of the disease, it involves generalised damage to the maternal endothelium of the kidneys and liver, with the release of vasopressive factors only secondary to the original damage. Pre-eclampsia may develop at varying times within pregnancy and its progress differs among patients; most cases present pre-term. It has no known cure apart from ending the pregnancy (induction of labour or abortion). It may also present up to 6 weeks post partum. Risk factors for pre-eclampsia include obesity, prior hypertension, older age, and diabetes mellitus.
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This question is part of the following fields:
- Physiology
- Renal
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SESSION STATS - PERFORMANCE PER SPECIALTY
