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Question 1
Correct
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A 65-year-old female is taking different medications for various medical conditions. Which medication would most likely predispose the patient to develop hyperkalaemia?
Your Answer: Spironolactone
Explanation:The most important potential side effect of spironolactone is hyperkalaemia (high potassium levels), which, in severe cases, can be life-threatening. Hyperkalaemia in these patients can present as a non anion-gap metabolic acidosis.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 2
Correct
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Which condition presents with a positive urine dipstick test for blood, but no blood cells on urine microscopy?
Your Answer: Myoglobinuria
Explanation:Myoglobinuria, or presence of myoglobulin in the urine is seen due to rhabdomyolysis (muscle destruction). Common causes of rhabdomyolysis include trauma, electrical injuries, burns, venom and drugs. Damaged muscle leads to release of myoglobin in the blood. Ideally, the released myoglobin gets filtered and excreted by the kidneys. However, excess myoglobin can occlude the renal filtration system leading to acute tubular necrosis and acute renal dysfunction.
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This question is part of the following fields:
- Pathology
- Renal
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Question 3
Correct
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A suspected recreational drug user was brought to the Emergency department in an unconscious state, and was found to be hypoventilating. Which of the following set of arterial blood gas analysis report is most consistent with hypoventilation as the primary cause? pH, pa(CO2) (mmHg), pa(O2) (mmHg).
Your Answer: 7.28, 55, 81
Explanation:Hypoventilation (or respiratory depression) causes an increase in carbon dioxide (hypercapnia) and respiratory acidosis. It can result due to drugs such as alcohol, benzodiazepines, barbiturates, opiates, mechanical conditions or holding ones breath. Strong opioids such as heroin and fentanyl are commonly implicated and can lead to respiratory arrest. In recreational drug overdose, acute respiratory acidosis occurs with an increase in p(CO2) over 45 mm Hg and acidaemia (pH < 7.35)
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 4
Correct
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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: 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 5
Correct
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A 90-year-old man is prescribed spironolactone after his family notices his legs are swollen. What class of drugs does spironolactone belong to?
Your Answer: Potassium-sparing diuretics
Explanation:Spironolactone is a renal competitive aldosterone antagonist in a class of drugs called ‘potassium-sparing diuretics’, that is primarily used to treat fluid build-up due to heart failure, liver scarring, or kidney disease. It is also used in the treatment of high blood pressure, low blood potassium, early-onset puberty, and acne and excessive hair growth in women. Spironolactone inhibits the effect of aldosterone by competing for intracellular aldosterone receptors in the distal tubule cells. This increases the secretion of water and sodium, while decreasing the excretion of potassium.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 6
Correct
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A 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 7
Correct
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A 65-year-old man complains of headaches, weakness, cramps, and confusion; blood tests reveal he has severe hyponatremia. The most likely cause is:
Your Answer: Severe diarrhoea or vomiting
Explanation:Hyponatraemia occurs when the sodium level in the plasma falls below 135 mmol/l. Hyponatraemia is an abnormality that can occur in isolation or, more commonly as a complication of other medical illnesses. Severe hyponatraemia may cause osmotic shift of water from the plasma into the brain cells. Typical symptoms include nausea, vomiting, headache and malaise. As the hyponatraemia worsens, confusion, diminished reflexes, convulsions, stupor or coma may occur. The cause of hyponatremia is typically classified by a person’s fluid status into low volume, normal volume, and high volume. Low volume hyponatremia can occur from diarrhoea, vomiting, diuretics, and sweating.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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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
Correct
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Which of the following conditions is characterized by generalised oedema due to effusion of fluid into the extracellular space?
Your Answer: Anasarca
Explanation:Anasarca (or ‘generalised oedema’) is a condition characterised by widespread swelling of the skin due to effusion of fluid into the extracellular space. It is usually caused by liver failure (cirrhosis of the liver), renal failure/disease, right-sided heart failure, as well as severe malnutrition/protein deficiency.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 10
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 11
Correct
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A 45-year old gentleman presented to the emergency department at 5.00 AM with pain in his left flank. The pain began suddenly and presented in waves throughout the night. Urine examination was normal except for presence of blood and few white blood cells. The pH and specific gravity of the urine were also found to be within normal range. What is the likely diagnosis?
Your Answer: Ureteric calculus
Explanation:A calculus in the ureter, if less than 5mm in diameter is likely to pass spontaneously. However, a larger calculus irritates the ureter and may become lodged, leading to hydroureter and/or hydronephrosis. Likely sites where the calculus might get lodged, include pelviureteric junction, distal ureter at the level of iliac vessels and the vesicoureteric junction. An obstruction can result in reduced glomerular filtration. There can be deterioration in renal function due to hydronephrosis and a raised glomerular pressure, leading to poor renal blood flow. Permanent renal dysfunction usually takes about 4 weeks to occur. Secondary infection can also occur in chronic obstruction.
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This question is part of the following fields:
- Pathology
- Renal
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Question 12
Incorrect
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A 58-year-old woman diagnosed with deep vein thrombosis had been using warfarin for 10 days. When she presented to the doctor she had haemorrhagic bullae and necrotic lesions in her lower limbs and buttocks. Deficiency of which of the following proteins may have caused the necrotic skin lesions?
Your Answer: Protein S
Correct Answer: Protein C
Explanation:Warfarin-induced skin necrosis is a rare complication of anticoagulant therapy that requires immediate drug cessation. The most common cutaneous findings include petechiae that progress to ecchymoses and haemorrhagic bullae. Warfarin inactivates vitamin K-dependent clotting factors II, VII, IX, and X and vitamin K-dependent proteins C and S. The concentration of protein C falls more rapidly than other vitamin K-dependent factors because they have a shorter half-lives. Skin necrosis is seen mainly in patients with prior protein C deficiency.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 13
Correct
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A 23-year-old woman decides to donate a kidney through a kidney chain. Which of the following indices would be expected to be decreased in the donor after full recovery from the operation?
Your Answer: Creatinine clearance
Explanation:Since medication to prevent rejection is so effective, donors do not need to be similar to their recipient. Most donated kidneys come from deceased donors; however, the utilisation of living donors is on the rise. Most problems encountered with live donation are associated with the donor. Firstly, there are the potentially harmful investigative procedures carried out in the assessment phase, the most hazardous being renal angiography, where there is cannulation of the artery and injection of a radio-opaque dye to determine the blood supply to the kidney. Secondly, there are the short-term risks of nephrectomy surgery. According to the literature, there is a mortality rate of between 1 in 1600 and 1 in 3000, but this is no more than is associated with any anaesthetic. In the initial postoperative period creatinine clearance may be decreased but this recovers fully over a few weeks to months. Long-term complications include prolonged wound pain.
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This question is part of the following fields:
- Physiology
- Renal
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Question 14
Correct
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Which of the following is NOT a nutritional factor involved in wound healing:
Your Answer: Vitamin B3
Explanation:Vitamin B6 is required for collagen cross-links.
Vitamin A is required for epithelial cell proliferation.
Zinc is required for RNA and DNA synthesis.
Copper is required for cross-linking of collagen.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Physiology
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Question 15
Incorrect
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A chest X-ray in a healthy, non-smoker, asymptomatic 48-year-old woman reveals a 2cm left lower lobe well-defined round opacity. Which of the following conditions is most probably responsible for this finding?
Your Answer: Mesothelioma
Correct Answer: Pulmonary hamartoma
Explanation:An asymptomatic healthy patient with no history of smoking and a lesion of small size most probably has a benign lung lesion. Hamartomas are one of the most common benign tumours of the lung that accounts for approximately 6% of all solitary pulmonary nodules. Pulmonary hamartomas are usually asymptomatic and therefore are found incidentally when performing an imaging test for other reasons.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 16
Incorrect
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A 27-year-old female was admitted due to severe dehydration. The patient also complained of chest tightness, thus an ECG was requested. The ECG strip showed an isoelectric ST segment, upright T wave, with prominent U waves. What is the most likely electrolyte abnormality responsible for these ECG tracing?
Your Answer: Hyperkalaemia
Correct Answer: Hypokalaemia
Explanation:U waves are prominent if it is >1-2mm or 25% of the height of the T wave. Abnormally prominent U waves are characteristically seen in severe hypokalaemia.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 17
Correct
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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: 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 18
Incorrect
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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: 65 mg/min
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 19
Correct
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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: 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 20
Incorrect
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A 46-year old lady presents with chief complaints of a large mass in the left breast. Histopathology of the mass revealed a stromal component with an epithelial component. What is the likely lesion?
Your Answer: Medullary carcinoma of breast
Correct Answer: Phyllodes tumour
Explanation:Phyllodes tumours are large, quickly growing tumours which arise from the periductal stroma of the breast. These are fibroepithelial tumours and account for less than 1% of breast cancers. These tumours can be benign, borderline or malignant based on the histology. The tumour usually affects adult women, mostly between the age of 40 to 50 years. It can be confused with fibroadenoma, which however affects much younger patients.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 21
Correct
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During strenuous exercise, what else occurs besides tachycardia?
Your 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 22
Correct
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A 45-year-old pregnant woman develops high blood pressure at 20 weeks. She complains of headaches and swollen feet, and a test reveals proteinuria (350 mg/day). Which of the following is the most likely diagnosis?
Your Answer: Pre-eclampsia
Explanation:Pre-eclampsia (PE) is a disorder of pregnancy characterized by the onset of high blood pressure (two separate readings taken at least 6 h apart of 140/90 or more) and often a significant amount of protein in the urine (>300 mg of protein in a 24-h urine sample). While blood pressure elevation is the most visible sign of the disease, it involves generalised damage to the maternal endothelium of the kidneys and liver, with the release of vasopressive factors only secondary to the original damage. Pre-eclampsia may develop at varying times within pregnancy and its progress differs among patients; most cases present pre-term. It has no known cure apart from ending the pregnancy (induction of labour or abortion). It may also present up to 6 weeks post partum. Risk factors for pre-eclampsia include obesity, prior hypertension, older age, and diabetes mellitus.
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This question is part of the following fields:
- Physiology
- Renal
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Question 23
Correct
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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: 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 24
Correct
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A 45-year-old woman was brought to the emergency department due to fever and chills. She has a history of recurrent UTI and complains of dysuria and urinary frequency. Urinary white blood cell count is >200 cell/high power field. If urine culture is performed, what is the most likely organism that will grow?
Your Answer: Escherichia coli
Explanation:The pathogen that most likely causes recurrent urinary tract infection in young women are E. coli, Enterococcus and Staphylococcus saprophyticus.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 25
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 26
Correct
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Purkinje fibres in the heart conduct action potentials at the rate of:
Your Answer: 1.5–4.0 m/s
Explanation:Purkinje fibres control the heart rate along with the sinoatrial node (SA node) and the atrioventricular node (AV node). The QRS complex is associated with the impulse passing through the Purkinje fibres. These fibres conduct action potential about six times faster than the velocity in normal cardiac muscle.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 27
Correct
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A 45-year-old man complains of shortness of breath, cough and chest pain. Chest X ray revealed a perihilar mass with bronchiectasis in the left mid-lung. Which of the following is most probably associated with these findings?
Your Answer: Bronchial carcinoid
Explanation:Bronchial carcinoids are neuroendocrine tumours that arise from Kulchitsky’s cells of the bronchial epithelium. Kulchitsky’s cells belong to the diffuse endocrine system. Patients affected by this tumour may be asymptomatic or may present with symptoms of airway obstruction, like dyspnoea, wheezing, and cough. Other common findings are recurrent pneumonia, haemoptysis, chest pain and paraneoplastic syndromes. Chest radiographs are abnormal in the majority of cases. Peripheral carcinoids usually present as a solitary pulmonary nodule. For central lesions common findings include hilar or perihilar masses with or without atelectasis, bronchiectasis, or consolidation. Bronchial carcinoids most commonly arise in the large bronchi causing obstruction.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 28
Correct
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A 35-year-old woman is in a comatose state following a traumatic head injury, and is receiving intravenous (IV) antibiotics and IV fluids containing saline and 5% dextrose. A serum biochemistry analysis is performed five days later which shows a low serum potassium level. This is most likely to be due to:
Your Answer: Nothing per oral regimen
Explanation:In this patient the cause for hypokalaemia is insufficient consumption of potassium as she is nil-per mouth with no intravenous supplementation. Parenteral nutrition has been used for comatose patients, although enteral feeding is usually preferable, and less prone to complications.
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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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Multiple, non-tender lymphadenopathy with biopsy showing several crowded follicles of small, monomorphic lymphocytes and the absence of Reed-Sternberg cells is seen in which of the following?
Your Answer: Chronic lymphocytic leukaemia
Correct Answer: Poorly differentiated lymphocytic lymphoma
Explanation:Malignant lymphoma usually causes non-tender lymphadenopathy, unlike the tender lymphadenopathy caused by infections (including infectious mononucleosis caused by Epstein-Barr virus). Also, the lymphoid hyperplasia seen in infectious mononucleosis is benign and polyclonal.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 30
Correct
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Different substances have different renal clearance rates. Which of the following substances should have the lowest renal clearance rate in a healthy patient?
Your Answer: Glucose
Explanation:Under normal conditions the renal clearance of glucose is zero, since glucose is completely reabsorbed in the renal tubules and not excreted. Glycosuria – the excretion of glucose into the urine- is nearly always caused by elevated blood glucose levels, most commonly due to untreated diabetes mellitus.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 31
Correct
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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: 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 32
Correct
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A 40-year old lady presented to the hospital with fever and mental confusion for 1 week. On examination, she was found to have multiple petechiae all over her skin and mucosal surfaces. Blood investigations revealed low platelet count and raised urea and creatinine. A platelet transfusion was carried out, following which she succumbed to death. Autopsy revealed pink hyaline thrombi in myocardial arteries. What is the likely diagnosis?
Your Answer: Thrombotic thrombocytopenic purpura
Explanation:Hyaline thrombi are typically associated with thrombotic thrombocytopenic purpura (TTP), which is caused by non-immunological destruction of platelets. Platelet transfusion is contraindicated in TTP. Platelets and red blood cells also get damaged by loose strands of fibrin deposited in small vessels. Multiple organs start developing platelet-fibrin thrombi (bland thrombi with no vasculitis) typically at arteriocapillary junctions. This is known as ‘thrombotic microangiopathy’. Treatment consists of plasma exchange.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 33
Correct
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Glucose is the most important source of energy for cellular respiration. The transport of glucose in the renal tubular cells occurs via:
Your Answer: Secondary active transport with sodium
Explanation:In 1960, Robert K. Crane presented for the first time his discovery of the sodium-glucose cotransport as the mechanism for glucose absorption. Glucose transport through biological membranes requires specific transport proteins. Transport of glucose through the apical membrane of renal tubular as well as intestinal epithelial cells depends on the presence of secondary active Na+–glucose symporters, SGLT-1 and SGLT-2, which concentrate glucose inside the cells, using the energy provided by co-transport of Na+ ions down their electrochemical gradient.
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This question is part of the following fields:
- Physiology
- Renal
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Question 34
Correct
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Lack of findings in the bladder but presence of atypical epithelial cells in urinalysis is most often associated with which of the following conditions?
Your Answer: Transitional cell carcinoma of renal pelvis
Explanation:The presence of atypical cells in urinalysis without findings in the bladder suggests a lesion located higher up, most probably in ureters or renal pelvis. Transitional cell cancer of the renal pelvis is a disease in which malignant cells form in the renal pelvis and is characterised by the presence of abnormal cells in urine cytology.
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This question is part of the following fields:
- Pathology
- Renal
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Question 35
Correct
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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: 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 36
Correct
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Which of the following is most likely to cause hypovolaemic hypernatremia:
Your Answer: Hyperalimentation
Explanation:Hypernatremia, characterised by a high serum sodium concentration, is rarely associated with volume overload (hypervolemia). A hypovolaemic hypernatremia may be seen during excessive administration of hypertonic sodium bicarbonate, hypertonic saline or hyperalimentation.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 37
Incorrect
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A 26-year-old female sought consultation due to excessive vaginal discharge. Vaginal smear showed numerous bacilli under the microscope. The organism was non-pathogenic. What is the most likely organism:
Your Answer: Gardnerella vaginalis
Correct Answer: Lactobacillus species
Explanation:Lactobacillus is a Gram-positive facultative bacteria. It is commonly present in the vagina and the gastrointestinal tract. Colonization of Lactobacillus is usually benign and it makes up a small portion of the gastrointestinal flora.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 38
Correct
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A 49-year-old woman with acute renal failure has a total plasma [Ca2+] = 2. 5 mmol/l and a glomerular filtration rate of 160 l/day. What is the estimated daily filtered load of calcium?
Your Answer: 240 mmol/day
Explanation:Calcium is the most abundant mineral in the human body. The average adult body contains in total approximately 1 kg of calcium of which 99% is in the skeleton in the form of calcium phosphate salts. The extracellular fluid (ECF) contains approximately 22 mmol, of which about 9 mmol is in the plasma. About 40% of total plasma Ca2+ is bound to proteins and not filtered at the glomerular basement membrane. Therefore, the estimated daily filtered load is 1.5 mmol/l × 160 l/day = 240 mmol/day. The exact amount of free versus total Ca2+ depends on the blood pH: free Ca2+ increases during acidosis and decreases during alkalosis.
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This question is part of the following fields:
- Physiology
- Renal
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Question 39
Correct
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A 45-year old man presents with sclerosing cholangitis, blood in his stools and apparent iron deficiency anaemia. What will be the most likely finding on his colonic biopsy?
Your Answer: Pseudopolyps
Explanation:Sclerosing cholangitis along with the passage of blood in stools suggests ulcerative colitis affecting the mucosa and submucosa of rectum and colon, with a sharp demarcation with the normal tissue. The musclaris layer is involved in severe cases. Initially, the mucosa is erythematous, friable with scattered haemorrhagic areas and loss of normal vascular pattern. Severe disease is indicated by presence of large mucosal ulcers with purulent exudate. There can be islands of normal mucosa between the ulcerated mucosa, along with few hyperplastic inflammatory mucosal lesions (pseudopolyps). Ulcerative colitis does not lead to development of fistulas or abscesses.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 40
Correct
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What is the normal duration of PR interval on an electrocardiogram of a healthy individual?
Your Answer: 0.12–0.20 s
Explanation:PR interval extends from the beginning of the P-wave until the beginning of the QRS complex. The normal duration of the PR interval is 0.12-0.20 s. It can be prolonged in first degree heart block, and reduced in Wolff-Parkinson-White syndrome.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 41
Correct
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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: 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 42
Correct
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In a cardiac cycle, what event does the closing of atrioventricular (AV) valves coincide with?
Your Answer: First heart sound
Explanation:In the cardiac cycle, the closing of the atrioventricular (AV) valves coincides with the onset of ventricular systole. This event marks the beginning of the isovolumetric contraction phase, where the ventricles begin to contract, but the volume of blood in the ventricles remains the same because both the AV valves and the semilunar valves (aortic and pulmonary valves) are closed. The closing of the AV valves produces the first heart sound, known as “S1” or “lub.”
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 43
Correct
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A 70-year-old male who has smoked since his teens complains of progressive shortness of breath and a persistent cough. He is diagnosed with COPD. Which of the following abnormalities is most likely to be present in his pulmonary function tests?
Your Answer: Increased residual volume
Explanation:Chronic obstructive pulmonary disease (COPD) is a type of obstructive lung disease characterized by long-term poor airflow. The main symptoms include shortness of breath and cough with sputum production. The best diagnostic test for evaluating patients with suspected chronic obstructive pulmonary disease (COPD) is lung function measured with spirometry. Key spirometrical measures may be obtained with a portable office spirometer and should include forced vital capacity (FVC) and the normal forced expiratory volume in the first second of expiration (FEV1). The ratio of FEV1 to forced vital capacity (FEV1/FVC) normally exceeds 0.75. Patients with COPD typically present with obstructive airflow. Complete pulmonary function testing may show increased total lung capacity, functional residual capacity and residual volume. A substantial loss of lung surface area available for effective oxygen exchange causes diminished carbon monoxide diffusion in the lung (DLco) in patients with emphysema. Tobacco smoking is the most common cause of COPD, with factors such as air pollution and genetics playing a smaller role.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 44
Correct
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Gastric acid secretion is stimulated by which of the following?
Your Answer: Gastrin
Explanation:Gastric acid secretion is stimulated by three factors:
– Acetylcholine, from parasympathetic neurones of the vagus nerve that innervate parietal cells directly
– Gastrin, produced by pyloric G-cells
– Histamine, produced by mast cells.
Gastric acid is inhibited by three factors:
– Somatostatin
– Secretin
– Cholecystokinin
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 45
Correct
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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: 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 46
Correct
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Which of the following is the most accurate test for the diagnosis of primary syphilis?
Your Answer: Dark-field microscopy
Explanation:Primary syphilis is transmitted via sexual contact. Lesions on genitalia, called a chancre occur after an asymptomatic incubation period of 10-90 days (average 21 days) after exposure. This chancre is a typically solitary (can be multiple), firm, painless, ulceration over the skin at the point of exposure to spirochete, seen on penis, vagina or rectum. It heals spontaneously after 4-6 weeks. Local lymphadenopathy can be seen.
Diagnosis is made by microscopy of fluid from lesion using dark-field illumination, taking care to not confuse with other treponemal disease. Screening tests include rapid plasma regain (RPR) and Venereal Diseases Research Laboratory (VDRL) tests. False positives are known to occur with these tests and can be seen in viral infections like hepatitis, varicella, Epstein-Barr virus, tuberculosis, lymphoma, pregnancy and IV drug use. More specific tests should therefore be carried out in case these screening tests are positive.
The Treponema pallidum hemagglutination assay (TPHA) and the fluorescent treponemal antibody absorption (FTAABS) test are based on monoclonal antibodies and immunofluorescence and are more specific. However, they can too show false positives with other treponemal diseases like yaws or pinta. Other confirmatory tests include those based on enzyme-linked immunoassays.
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This question is part of the following fields:
- Pathology
- Urology
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Question 47
Correct
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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: 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 48
Correct
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Driving pressure is considered to be a strong predictor of mortality in patients with ARDS. What is the normal mean intravascular driving pressure for the respiratory circulation?
Your Answer: 10 mmHg
Explanation:Driving pressure is the difference between inflow and outflow pressure. For the pulmonary circulation, this is the difference between pulmonary arterial (pa) and left atrial pressure (pLA). Normally, mean driving pressure is about 10 mmHg, computed by subtracting pLA (5 mmHg) from pA (15 mmHg). This is in contrast to a mean driving pressure of nearly 100 mmHg in the systemic circulation.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 49
Correct
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Which of the following malignancies is likely to have the best prognosis?
Your Answer: Papillary carcinoma of the thyroid
Explanation:Papillary carcinoma accounts for 70-80% of all thyroid cancers and is seen commonly in people aged 30-60 years. It is more aggressive in elderly patients. 10-20% cases may have recurrence or persistent disease. More common in females with a female to male ratio of 3:1. Papillary carcinomas can also contain follicular carcinomas. The common route of spread is through lymphatics to regional nodes in one-third cases and pulmonary metastasis can also occur. Papillary carcinomas of the thyroid have the best prognosis, especially in patients less than 45 years of age with small tumours confined to the thyroid gland.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 50
Correct
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Which of the following conditions may cause hypervolaemic hyponatraemia?
Your Answer: Cirrhosis
Explanation:Hypovolaemic hyponatraemia: reduced extracellular fluid
Renal loss of sodium and water; urine Na >20 mmol/day
Causes:
Diuretic use
Salt wasting nephropathy
Cerebral salt wasting
Mineralocorticoid deficiency/adrenal insufficiency
Renal tubular acidosis
Extrarenal loss of sodium and water with renal conservation; urine Na <20 mmol/day
Causes:
Burns
Gastrointestinal loss
Pancreatitis
Blood loss
3rd space loss (bowel obstruction, peritonitis)Hypervolaemic hyponatraemia: expanded intracellular fluid and extracellular fluid but reduced effective arterial blood volume
Causes:
Congestive cardiac failure
Cirrhosis
Nephrotic syndromeEuvolaemic hyponatraemia: expanded intracellular and extracellular fluid but oedema absent
Causes:
Thiazide diuretics (can be euvolaemic or hypovolaemic)
Hypothyroidism
Adrenal insufficiency (can be euvolaemic or hypovolaemic)
SIADH (cancer, central nervous system disorders, drugs, pulmonary disease, nausea, postoperative pain, HIV, infection, Guillain‐Barre syndrome, acute intermittent porphyria)
Decreased solute ingestion (beer potomania/tea and toast diet) -
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 51
Correct
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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: 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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Question 52
Correct
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Arterial blood gas analysis of a man admitted with acute exacerbation of chronic obstructive pulmonary disease (COPD) showed the following: pH = 7.28, p(CO2) = 65.5 mmHg, p(O2)= 60 mmHg and standard bicarbonate = 30.5 mmol/l. This patient had:
Your Answer: Respiratory acidosis
Explanation:Acidosis with high p(CO2) and normal standard bicarbonate indicates respiratory acidosis, commonly seen in acute worsening of COPD patients. Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with a low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg).
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 53
Correct
-
Intravenous diazepam was administered to a man who was brought to the emergency department with status epilepticus. He was administered 15 l/min oxygen via a reservoir bag mask. Blood investigations showed sodium = 140 mmol/l, potassium = 4 mmol/l and chloride = 98 mmol/l. His arterial blood gas analysis revealed pH 7.08, p(CO2)= 61.5 mmHg, p(O2) = 111 mmHg and standard bicarbonate = 17 mmol/l. This patient had:
Your Answer: Mixed acidosis
Explanation:Acidosis with high p(CO2) and low standard bicarbonate indicates mixed acidosis. Lower p(O2) is due to breathing of 70% oxygen. The prolonged seizures lead to lactic acidosis and the intravenous diazepam is responsible for the respiratory acidosis. Treatment includes airway manoeuvres and oxygen, assisted ventilation if needed, and treatment with fluids.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 54
Correct
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Calculate the stroke volume in a patient admitted for coronary bypass surgery, with the following parameters pre-operatively:
Oxygen consumption = 300 ml/min
Arterial oxygen content = 20 ml/100 ml blood
Pulmonary arterial oxygen content = 15 ml/100 ml blood and Heart rate = 100 beats/min.Your Answer: 60 ml
Explanation:By Fick’s principle, cardiac output can be calculated as follows: VO2 = CO × (CAO2– CVO2) where VO2= oxygen consumption, CO = cardiac output, CAO2 = arterial oxygen content and CvO2 = mixed venous oxygen content. Thus, in the given problem, 300 ml/min = CO × (20 – 15) ml/100 ml CO = 300 × 100/5 ml/min CO = 6000 ml/min. Also, cardiac output = stroke volume × heart rate. Thus, 6000 ml/min = stroke volume × 100 beats/min. Hence, stroke volume = 6000/100 ml/min which is 60 ml/min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 55
Correct
-
During cardiac catheterisation in a 20-year old man, the following data is obtained: Pressure (mmHg), O2 saturation (%) Right atrium 7 (N = 5) 90 (N = 75), Right ventricle 35/7 (N = 25/5) 90 (N = 75), Pulmonary artery 35/8 (N = 25/15), 90 (N = 75), Left atrium 7 (N = 9) 95 (N = 95), Left ventricle 110/7 (N = 110/9) 95 (N = 95), Aorta 110/75 (N = 110/75) 95 (N = 95) where N = Normal value. What is the likely diagnosis?
Your Answer: Atrial septal defect
Explanation:A congenital heart disease, ASD or atrial septal defect leads to a communication between the right and left atria due to a defect in the interatrial septum. This leads to mixing of arterial and venous blood from the right and left side of the heart. The hemodynamic significance of this defect depends on the presence of shunting of blood. Normally, the left side of the heart has higher pressure than the right as the left side has to pump blood throughout the body. A large ASD (> 9 mm) will result in a clinically significant left-to-right shunt, causing volume overload of the right atrium and ventricle, eventually leading to heart failure. Cardiac catheterization would reveal very high oxygen saturation in the right atrium, right ventricle and pulmonary artery. Eventually, the left-to-right shunt will lead to pulmonary hypertension and increased afterload in the right ventricle, along with the increased preload due to the shunted blood. This will either cause right ventricular failure, or raise the pressure in the right side of the heart to equal or more than that in the left. Elevation of right atrial pressure to that of left atrial pressure would thus lead to diminishing or complete cessation of the shunt. If left uncorrected, there will be reversal of the shunt, known as Eisenmenger syndrome, resulting in clinical signs of cyanosis as the oxygen-poor blood form right side of the heart will mix with the blood in left side and reach the peripheral vascular system.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 56
Correct
-
A victim of road traffic accident presented to the emergency department with a blood pressure of 120/90 mm Hg, with a drop in systolic pressure to 100 mm Hg on inhalation. This is known as:
Your Answer: Pulsus 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 57
Correct
-
A 40-year old woman presents with tightening of the skin over her fingers which makes movement of her fingers difficult.. She also gives a history of her fingers turning blue on exposure to low temperatures. She admits to gradual weight loss. Investigations reveal negative rheumatoid factor, negative antinuclear antibody and a positive anticentromere body. Which of the following conditions is she likely to have?
Your Answer: Oesophageal stricture
Explanation:Scleroderma is a connective tissue disorder that ranges in severity and progression. The disease could show generalised skin thickening with rapid, fatal, visceral involvement; or only cutaneous involvement (typically fingers and face). The slow progressive form is also known as ‘limited cutaneous scleroderma’ or CREST syndrome (calcinosis cutis, Raynaud’s phenomenon, (o)oesophageal dysmotility, sclerodactyly, and telangiectasia).
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 58
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 59
Correct
-
A 30-year-old woman feels thirsty. This thirst is probably due to:
Your 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 60
Correct
-
The most likely cause of a low p(O2) in arterial blood is:
Your 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 61
Incorrect
-
A 54-year-old woman with amyotrophic lateral sclerosis is diagnosed with respiratory acidosis. The patient’s renal excretion of potassium would be expected to:
Your Answer: Rise, since acid and potassium excretion are coupled
Correct Answer: Fall, since tubular secretion of potassium is inversely coupled to acid secretion
Explanation:Respiratory acidosis is a medical emergency in which decreased ventilation (hypoventilation) increases the concentration of carbon dioxide in the blood and decreases the blood’s pH (a condition generally called acidosis). Secretion of acid and potassium by the renal tubule are inversely related. So, increased excretion of H+ during renal compensation for respiratory acidosis will result in decreased secretion (or increased retention) of potassium ions, with the result that the body’s potassium store rises. An increase in K+ excretion would be associated with renal compensation for respiratory alkalosis. The filtered load of K+depends only on K+ plasma concentration and glomerular filtration rate, not on plasma pH.
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This question is part of the following fields:
- Physiology
- Renal
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Question 62
Correct
-
If the blood flow is constant, oxygen extraction by tissues will show the greatest decrease due to which of the following interventions?
Your Answer: Tissue cooling
Explanation:With a constant blood flow to a given tissue bed, there will be an increase in oxygen extraction by the tissue with the following; an increase in tissue metabolism and oxygen requirements: warming (or fever), exercise, catecholamines and thyroxine. With cooling, the demand for oxygen decreases, leading to decreased oxygen extraction.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 63
Correct
-
Which of the following muscles aid in inspiration?
Your Answer: Diaphragm and external intercostals
Explanation:The diaphragm and external intercostals are muscles of inspiration as they increase the volume of thoracic cavity and reduce the intrathoracic pressure. Muscles of expiration include abdominal muscles and internal intercostals.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 64
Correct
-
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 65
Correct
-
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: 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 66
Correct
-
A chloride sweat test was performed on a 13-year-old boy. Results indicated a high likelihood of cystic fibrosis. This diagnosis is associated with a higher risk of developing which of the following?
Your Answer: Bronchiectasis
Explanation:Cystic fibrosis is a life-threatening disorder that causes the build up of thick mucus in the lungs, digestive tract, and other areas of the body. It is a hereditary autosomal-recessive disease caused by mutations of the CFTR gene. Cystic fibrosis eventually results in bronchiectasis which is defined as a permanent dilatation and obstruction of bronchi or bronchioles.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 67
Correct
-
If a catheter is placed in the main pulmonary artery of a healthy 30-year-old woman, which of the following will be its mean pulmonary arterial pressure?
Your Answer: 15 mmHg
Explanation:The pulmonary artery pressure (PA pressure) is a measure of the blood pressure found in the main pulmonary artery. The hydrostatic pressure of the pulmonary circulation refers to the actual pressure inside pulmonary vessels relative to atmospheric pressure. Hydrostatic (blood pressure) in the pulmonary vascular bed is low compared with that of similar systemic vessels. The mean pulmonary arterial pressure is about 15 mmHg (ranging from about 13 to 19 mmHg) and is much lower than the average systemic arterial pressure of 90 mmHg.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 68
Correct
-
Calculate the cardiac output 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: 6.25 l/min
Explanation:As per Fick’s principle, 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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 69
Correct
-
The proximal tubule is the portion of the ductal system of the nephron of the kidney which leads from Bowman's capsule to the loop of Henle. Which of the following is most likely to be seen in a sample of fluid leaving the proximal tubule?
Your Answer: It will have no amino acids
Explanation:The proximal tubule is the portion of the duct system of the nephron leading from Bowman’s capsule to the loop of Henlé. The most distinctive characteristic of the proximal tubule is its brush border (or ‘striated border’). The luminal surface of the epithelial cells of this segment of the nephron is covered with densely packed microvilli forming a border which greatly increases the luminal surface area of the cells, presumably facilitating their reabsorptive function. Glucose, amino acids, inorganic phosphate, and some other solutes are100% reabsorbed via secondary active transport through co-transporters driven by the sodium gradient out of the nephron.
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This question is part of the following fields:
- Physiology
- Renal
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Question 70
Correct
-
Which is a feature of the action of insulin?
Your Answer: Promotes protein synthesis
Explanation:Insulin is produced by the beta-cells of the islets of Langerhans in the pancreas. Its actions include:
– promoting uptake of glucose into cells
– glycogen synthesis (glycogenesis)
– protein synthesis
– stimulation of lipogenesis (fat formation).
– driving potassium into cells – used to treat hyperkaelamia.
Parathyroid hormone and activated vitamin D are the principal hormones involved in calcium/phosphate metabolism, rather than insulin.
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This question is part of the following fields:
- Endocrine
- Physiology
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SESSION STATS - PERFORMANCE PER SPECIALTY
