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Question 1
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 2
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: normal FEV1, arterial oxygen saturation 92%, alveolar ventilation 6000 ml/min at a tidal volume of 600 ml and a breathing rate of 12 breaths/min. There are also pathological changes in lung compliance and residual volume. Which of the following is most accurate about his residual volume?
Your Answer: Cannot be measured directly with a spirometer
Explanation:Residual volume is the air left in the lungs after maximal expiration is done. Thus, this is not a part of vital capacity and cannot be measured with a spirometer directly. It can be measured by the methods such as body plethysmography or inert gas dilution. Expiratory reserve volume is vital capacity minus inspiratory capacity. Resting volume of lungs is he sum of residual volume and expiratory reserve volume. Lungs recoil inward until the recoil pressure becomes zero, which corresponds to a volume significantly lower than residual volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 3
Correct
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Normally, the O2 transfer in the lungs from alveolar to capillary is perfusion-limited. In which of the following situations does it become a diffusion-limited process?
Your Answer: Pulmonary oedema
Explanation:Normally, the transfer of oxygen from air spaces to blood takes place across the alveolar-capillary membrane by simple diffusion and depends entirely on the amount of blood flow (perfusion-limited process). Diseases that affect this diffusion will transform the normal process to a diffusion limited process. Thus, the diseases which cause a thickened barrier (such as pulmonary oedema due to increased extravascular lung water or asbestosis) will limit the diffusion of oxygen. Chronic obstructive lung diseases will have little effect on diffusion. Inhaling hyperbaric gas mixtures might overcome the diffusion limitation in patients with mild asbestosis or interstitial oedema, by increasing the driving force. Strenuous (not mild) exercise might also favour diffusion limitation and decrease passage time. Increasing the rate of ventilation will not have this affect but will only maintain a high oxygen gradient from air to blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 4
Incorrect
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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: 40 ml
Correct 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 5
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 6
Correct
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A 40-year old gentleman, known with a history of peptic ulcer disease, was brought to the clinic in a dehydrated state with persistent vomiting. His blood investigations revealed:
- sodium = 142 mmol/l
- potassium = 2.6 mmol/l
- chloride = 85 mmol/l
- pH = 7.55
- p(CO2) = 50 mmHg
- p(O2) = 107 mmHg
- standard bicarbonate = 40 mmol/l
Your Answer: Metabolic alkalosis
Explanation:High pH with high standard bicarbonate indicates metabolic alkalosis. The pa(CO2) was appropriately low in compensation. This is hypokalaemic hypochloraemic metabolic acidosis due to prolonged vomiting. Treatment includes treating the cause and intravenous sodium chloride with potassium.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 7
Correct
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Acute respiratory distress syndrome (ARDS) is a medical condition that occurs in critically ill patients, and can be triggered by events such as trauma and sepsis. Which of the following variables is most likely to be lower than normal in a patient with ARDS?
Your Answer: Lung compliance
Explanation:Acute (or Adult) respiratory distress syndrome (ARDS) is a medical condition occurring in critically ill patients characterized by widespread inflammation in the lungs. The development of acute respiratory distress syndrome (ARDS) starts with damage to the alveolar epithelium and vascular endothelium, resulting in increased permeability to plasma and inflammatory cells. These cells pass into the interstitium and alveolar space, resulting in pulmonary oedema. Damage to the surfactant-producing type II cells and the presence of protein-rich fluid in the alveolar space disrupt the production and function of pulmonary surfactant, leading to micro atelectasis and impaired gas exchange. The pathophysiological consequences of lung oedema in ARDS include a decrease in lung volumes, compliance and large intrapulmonary shunts. ARDS may be seen in the setting of pneumonia, sepsis, following trauma, multiple blood transfusions, severe burns, severe pancreatitis, near-drowning, drug reactions, or inhalation injuries.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 8
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 9
Incorrect
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During pregnancy the uterus enlarges however after delivery it regresses to its original size. Which of the following organelles is responsible for this regression?
Your Answer: Endoplasmic reticulum
Correct Answer: Lysosomes
Explanation:Lysosomes are formed by budding of the Golgi apparatus and contain enzymes which digest macromolecules. They are found in both plants and animals and are active in autophagic cell death, digestion after phagocytosis and for the cells own recycling process. They fuse with the molecules and release their content resulting in digestion.
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This question is part of the following fields:
- General
- Physiology
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Question 10
Correct
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A 30 year old female suffered from mismatched transfusion induced haemolysis. Which substance will be raised in the plasma of this patient?
Your Answer: Bilirubin
Explanation:Bilirubin is a yellow pigment that is formed due to the break down of RBCs. Haemolysis results in haemoglobin that is broken down into a haem portion and globin which is converted into amino acids and used again. Haem is converted into unconjugated bilirubin in the macrophages and shunted to the liver. In the liver it is conjugated with glucuronic acid making it water soluble and thus excreted in the urine. Its normal levels are from 0.2-1 mg/dl. Increased bilirubin causes jaundice and yellowish discoloration of the skin.
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This question is part of the following fields:
- General
- Physiology
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Question 11
Incorrect
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A 40-year old Caucasian male came to the hospital with complaints of fatigue and lethargy. On examination, he was found to have raised blood pressure. Urine examination showed >300 mg/dl proteinuria (4+) and 24-hour urine protein 3.5g. No glucose, blood, nitrites, urobilinogen or casts were present in urine. What is the most likely diagnosis?
Your Answer: Post-streptococcal glomerulonephritis
Correct Answer: Membranous glomerulonephritis
Explanation:Membranous glomerulonephritis or nephropathy, is a renal disorder with insidious course and usually affects people aged 30-50 years. 85% cases are primary (or idiopathic). The other 15% are secondary to autoimmune conditions like SLE, infections like malaria or hepatitis B, drugs like captopril and NSAIDs, or malignancies (particularly lung or colonic carcinoma). This disease is caused due to circulating immune complexes which are said to form by binding of antibodies to antigens in glomerular basement membrane. This antigens could be endogenous or derived from systemic circulation. This immune complex triggers the complement system, resulting in formation of membrane attack complex (MAC) on glomerular epithelial cells. This further results in release of proteases and oxidants which damage the capillaries making them ‘leaky’. Moreover, the epithelial cells also secrete a mediator to reduce nephron synthesis and distribution.
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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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Calculate the stroke volume in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer: 75 ml
Correct Answer: 90 ml
Explanation:Fick’s principle states that, VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CAO2 = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24–0.16, CO = 500/0.8, CO = 6.25 l/min. Cardiac output is also given by product of stroke volume and heart rate. Thus, stroke volume = cardiac output / heart rate = 6.25/70 × 1000 stroke volume = 90 ml approximately.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 13
Correct
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Which of the following substances brings about a dilatation of the gastrointestinal resistance vessels?
Your Answer: Vasoactive intestinal peptide
Explanation:Gastric vasoconstrictors include catecholamines, angiotensin II and vasopressin. Vasodilators include vasoactive intestinal peptide and the hormones; gastrin, cholecystokinin and glucagon.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 14
Correct
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Which organ is most vulnerable to haemorrhagic shock?
Your Answer: Kidneys
Explanation:At rest, the brain receives 15% cardiac output, muscles 15%, gastrointestinal tract 30% and kidneys receive 20%. However, if normalised by weight, the largest specific blood flow is received by the kidneys at rest (400 ml/min x 100g), making them highly vulnerable in the case of a haemorrhagic shock.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 15
Correct
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Whilst snorkelling, a 30-year old gentleman has the respiratory rate of 10/min, tidal volume of 550 ml and an effective anatomical dead space of 250 ml. Which of the following will bring about a maximum increase in his alveolar ventilation?
Your Answer: A 2x increase in tidal volume and a shorter snorkel
Explanation:Alveolar ventilation = respiratory rate × (tidal volume − anatomical dead space volume). Increase in respiratory rate simply causes movement of air in the anatomical dead space, with no contribution to the alveolar ventilation. By use of a shorter snorkel, the effective anatomical dead space will decrease and will cause a maximum rise in alveolar ventilation along with doubling of tidal volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 16
Correct
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A 35-year-old ultra marathon runner becomes severely dehydrated and collapses. This patient most likely has:
Your Answer: Decreased baroreceptor firing rate
Explanation:Baroreceptors are sensors located in the blood vessels of all vertebrate animals. They sense the blood pressure and relay the information to the brain, so that a proper blood pressure can be maintained. Acute dehydration results in decreased plasma volume and increased plasma osmolarity, since more water than salt is lost in sweat. The decrease in plasma volume leads to an inhibition of the baroreceptors and a lower firing rate. The increase in plasma osmolarity leads to increased ADH secretion and high plasma ADH levels, which increases water permeability of collecting duct cells. Therefore more water is reabsorbed by the kidneys and renal water excretion is low.
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This question is part of the following fields:
- Physiology
- Renal
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Question 17
Correct
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Thalamic syndrome will most likely result in:
Your Answer: Hyperaesthesia
Explanation:Signs and symptoms of thalamic syndrome include contralateral hemi anaesthesia, burning or aching sensation in one half of a body (hyperaesthesia), often accompanied by mood swings.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 18
Correct
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A lesion involving the lateral geniculate nucleus of the thalamus is likely to affect:
Your Answer: Vision
Explanation:The lateral geniculate nucleus (LGN) of the thalamus is the primary processor of visual information in the central nervous system. The LGN receives information directly from the retina and sends projections directly to the primary visual cortex. The LGN likely helps the visual system focus its attention on the most important information.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 19
Correct
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A 45-year-old man presented to the doctor complaining of a flank pain and episodes of haematuria. Abdominal ultrasound revealed a left renal mass and the patient underwent a nephrectomy. Histopathological pattern was triphasic with blastemal, epithelial, and stromal components. The pathologist suggested the tumour resulted from the lack of a tumour suppressor gene on chromosome 11. Which of the following tumours is the pathologist most likely suggesting?
Your Answer: Wilms’ tumour
Explanation:Wilms’ tumour is one of the most common malignant tumours of childhood but it can also rarely be found in adults. In biopsy, classical histopathological findings include the triphasic pattern composed by blastemal, epithelial, and stromal elements. First symptoms in children include an abdominal palpable mass, while in adults pain and haematuria are the most common complaints. Deletions of tumours’ suppressor genes on chromosome 11 are usually associated with Wilms’ tumour.
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This question is part of the following fields:
- Pathology
- Renal
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Question 20
Incorrect
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What is the mostly likely cause of prolonged activated partial thromboplastin time (aPPT) ?
Your Answer: Malabsorption of vitamin K
Correct Answer: Heparin therapy
Explanation:The partial thromboplastin time (PTT) or activated partial thromboplastin time (aPTT) is an indicator for measuring the efficacy of both the intrinsic and common coagulation pathway. Prolonged aPTT may indicate: use of heparin, antiphospholipid antibody and coagulation factor deficiency (e.g., haemophilia). Deficiencies of factors VIII, IX, XI and XII and rarely von Willebrand factor (if causing a low factor VIII level) may lead to a prolonged aPTT correcting on mixing studies.
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This question is part of the following fields:
- General
- Physiology
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Question 21
Incorrect
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Increased resistance to flow of blood in cerebral vessels is most likely seen in:
Your Answer: Inhalation of a gas mixture enriched with CO2
Correct Answer: Elevation in systemic arterial pressure from 100 to 130 mmHg
Explanation:Constant cerebral blood flow is maintained by autoregulation in the brain, which causes an increase in local vascular resistance to offset an increase in blood pressure. There will be an increase in cerebral blood flow (and decrease in resistance to cerebral blood flow) with a decrease in arterial oxygen or an increase in arterial CO2. Similarly, a decrease in viscosity will also increase the blood flow. Due to increased brain metabolism and activity during a seizure, there will also be an increase in the cerebral blood flow.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 22
Correct
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Which of the following muscles aid in inspiration?
Your Answer: Diaphragm and external intercostals
Explanation:The diaphragm and external intercostals are muscles of inspiration as they increase the volume of thoracic cavity and reduce the intrathoracic pressure. Muscles of expiration include abdominal muscles and internal intercostals.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 23
Correct
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A glycogen storage disorder is characterised by increased liver glycogen with a normal structure and no increase in serum glucose after oral intake of a protein-rich diet. Deficiency of which of the following enzymes is responsible for this disorder?
Your Answer: Glucose-6-phosphatase
Explanation:The most common glycogen storage disorder is von Gierke’s disease or glycogen storage disease type I. It results from a deficiency of enzyme glucose-6-phosphatase which affects the ability of liver to produce free glucose from glycogen and gluconeogenesis; leading to severe hypoglycaemia. There is also increased glycogen storage in the liver and kidneys causing enlargement and various problems in their functioning. The disease also causes lactic acidosis and hyperlipidaemia. The main treatment includes frequent or continuous feedings of corn-starch or other carbohydrates.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 24
Correct
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A 40 year old man suffered severe trauma following an MVA. His BP is 70/33 mmhg, heart rate of 140 beats/mins and very feeble pulse. He was transfused 3 units of blood resulting in his BP returning to 100/70 and his heart rate to 90 beats/min. What decreased following transfusion?
Your Answer: Total peripheral resistance
Explanation:The patient is in hypovolemic shock, he is transfused with blood to replace the volume lost. It is important not only to replace fluids but stop active bleeding in resuscitation. Fluid replacement will result in a decreased sympathetic discharge and adequate ventricular filling thus reducing total peripheral resistance and increasing cardiac output and cardiac filling pressures.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 25
Correct
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A 60 year old patient with a history of carcinoma of the head of the pancreas, and obstructive jaundice presents with a spontaneous nose bleed and easy bruising. What is the most likely reason for this?
Your Answer: Vitamin-K-dependent clotting factors deficiency
Explanation:Vitamin K is a fat soluble vitamin requiring fat metabolism to function properly to allow for its absorption. People with obstructive jaundice develop vitamin k deficiency as fat digestion is impaired. Vit K causes carboxylation of glutamate residue and hence regulates blood coagulation including: prothrombin (factor II), factors VII, IX, X, protein C, protein S and protein Z.
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This question is part of the following fields:
- General
- Physiology
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Question 26
Correct
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Decreased velocity of impulse conduction through the atrioventricular node (AV node) in the heart will lead to:
Your Answer: Increased PR interval
Explanation:AV node damage may lead to an increase in the PR interval to as high as 0.25 – 0.40 s (normal = 0.12 – 0.20 s). In the case of severe impairment, there might be a complete failure of passage of impulses leading to complete block. In this case, the atria and ventricles will beat independently of each other.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 27
Incorrect
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Which of the following will be a likely sequelae of complete ileal resection?
Your Answer: Decreased calcium absorption
Correct Answer: Vitamin B12 deficiency
Explanation:The ileum is a part of the small intestine and has a pH of around 7-8 (neutral or slightly alkaline). Its main function is absorption of products of digestion. The ileal wall has multiple villi, which in turn have numerous microvilli. This increases the surface area available for absorption significantly. The cells lining the ileum contain multiple enzymes such as protease and carbohydrase, which aid in the final stages of digestion. Villi contain lacteals which absorb the products of fat digestion, fatty acids and glycerol. Thus, ileal resection will lead to their decreased absorption and increased fat content in the stool. The ileum is also responsible for absorption of vitamin B12.
Maximum water absorption occurs in the colon followed by the jejunum. Hence, ileal resection is less likely to lead to fluid volume deficiency. Also, most minerals (like calcium, iron etc.) are absorbed in the duodenum, and thus will not be affected by ileal resection.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 28
Incorrect
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For calculation of cardiac output by Fick's principle, which of the following vessels is the best source of venous blood to determine the arterial-to-venous oxygen tension difference?
Your Answer: Pulmonary vein
Correct Answer: Pulmonary artery
Explanation:Fick’s principle states that the total uptake (or release) of a substance by peripheral tissues is equal to the product of the blood flow to the peripheral tissues and the arterial– venous concentration difference (gradient) of the substance. It is used to measure the cardiac output, and the formula is Cardiac output = oxygen consumption divided by arteriovenous oxygen difference. Assuming there are no shunts across the pulmonary system, the pulmonary blood flow equals the systemic blood flow. The arterial and venous blood oxygen content is measured by sampling from the pulmonary artery (low oxygen content) and pulmonary vein (high oxygen content). Peripheral arterial blood is used as a surrogate for the pulmonary vein.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 29
Incorrect
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A 77-year-old woman's renal function is tested. The following results were obtained during a 24-h period:
Urine flow rate: 2. 0 ml/min
Urine inulin: 0.5 mg/ml
Plasma inulin: 0.02 mg/ml
Urine urea: 220 mmol/l
Plasma urea: 5 mmol/l.
What is the urea clearance?Your Answer: 4.4 ml/min
Correct Answer: 88 ml/min
Explanation:Urea is reabsorbed in the inner medullary collecting ducts of the nephrons. The clearance (C) of any substance can be calculated as follows: C = (U × V)/P, where U and P are the urine and plasma concentrations of the substance, respectively and V is the urine flow rate. So, glomerular filtration rate = (0.220 × 2. 0)/0.005 = 88 ml/min.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 30
Incorrect
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Which of the following is involved in vitamin B12 absorption?
Your Answer: Ca2+
Correct Answer: Intrinsic factor
Explanation:Absorption of vitamin B12 is by an active transport process and occurs in the ileum. Most cobalamins are bound to proteins and are released in the stomach due to low pH and pepsin. The cobalamins then bind to R proteins, i.e. haptocorrin (HC) secreted from salivary glands and gastric juice. Another cobalamin binding protein is Intrinsic factor (IF) secreted from the gastric parietal cells. The cobalamin-HC complex is digested by pancreatic proteases in the intestinal lumen, and the free cobalamin then binds to IF. The complex then reaches a transmembrane receptor in the ileum and undergoes endocytosis. Cobalamin is then released intracellularly and binds to transcobalamin II (TC II). The newly formed complex then exits the ileal cell and enters the blood circulation.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 31
Correct
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A 7-year-old girl is given cephalexin to treat an infection and develops hives, with localised facial oedema. Which of the following conditions will cause localised oedema?
Your Answer: Angio-oedema
Explanation:Angio-oedema, is the rapid swelling of the skin, mucosa and submucosal tissues. The underlying mechanism typically involves histamine or bradykinin. The version related to histamine is to due an allergic reaction to agents such as insect bites, food, or medications. The version related to bradykinin may occur due to an inherited C1 esterase inhibitor deficiency, medications e.g. angiotensin converting enzyme inhibitors, or a lymphoproliferative disorder.
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This question is part of the following fields:
- Physiology
- Renal
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Question 32
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 33
Correct
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Nephrotic syndrome is a condition that causes proteinuria, hypoalbuminemia and oedema. Which of the following is the cause of the oedema in these patients?
Your Answer: Decreased oncotic pressure
Explanation:The glomeruli of the kidneys are the parts that normally filter the blood. They consist of capillaries that are fenestrated and allow fluid, salts and other small solutes to flow through, but normally not proteins. In nephrotic syndrome, the glomeruli become damaged allowing small proteins, such as albumin to pass through the kidneys into urine. Oedema usually occurs due to salt and water retention by the diseased kidneys as well as due to the reduced colloid oncotic pressure (because of reduced albumin in the plasma). Lower serum oncotic pressure causes fluid to accumulate in the interstitial tissues.
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This question is part of the following fields:
- Physiology
- Renal
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Question 34
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: The fluid pressure in the scala tympani decreases as a sound becomes louder
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 35
Incorrect
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Which statement is correct regarding coagulation?
Your Answer: Vitamin K is an essential co-factor for hepatic synthesis of factor VIII
Correct Answer: Thrombin converts fibrinogen to fibrin
Explanation:Coagulation of blood is a complex process and an important part of haemostasis. There are two main pathways related to coagulation: the contact activation pathway/intrinsic pathway and tissue factor/extrinsic pathway. The extrinsic pathway is activated by external trauma that causes blood to escape from the vascular system. This pathway is quicker than the intrinsic pathway and involves factor VII. The intrinsic pathway is activated by trauma inside the vascular system, and initiated by platelets, exposed endothelium, chemicals, or collagen. This pathway is slower than the extrinsic pathway, but more important. It involves factors XII, XI, IX, VIII. Both pathways meet to finish the formation of a clot in what is known as the common pathway. The common pathway involves factors I, II, V, and X. They converge on the common pathway in which activation of prothrombin to thrombin leads to conversion of fibrinogen to fibrin and clot formation.
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This question is part of the following fields:
- General
- Physiology
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Question 36
Correct
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The majority of gallstones are mainly composed of:
Your Answer: Cholesterol
Explanation:Bile salts are formed out of cholesterol in the liver cells. Occasionally, precipitation of cholesterol occurs resulting into cholesterol stones developing in the gall bladder.
These cholesterol gallstones are the most common type and account for 80% of all gallstones. Another type, accounting for 20% gallstones is pigment stones which are composed of bilirubin and calcium salts. Occasionally, stones of mixed origin are also seen.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 37
Incorrect
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The transmembrane proteins responsible for resting membrane potential of vascular smooth muscle cells was blocked by a drug. Which of the following transmembrane proteins were blocked by this drug?
Your Answer: Na+/K+ pump
Correct Answer: K+ channels
Explanation:The resting membrane potential is due to selective permeability of the membrane to potassium ions. The Na/K pump is responsible for the generation of a gradient across the membrane and it is due to the inherent ability of the K channels to allow diffusion back into the nerve at rest which charges the cells. In reality, the resting membrane potential is more positive because of small contributions by Na+ channels, Cl− channels and non-selective cation channels.
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This question is part of the following fields:
- General
- Physiology
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Question 38
Incorrect
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What will the destruction of endoplasmic reticulum stop?
Your Answer: Glycosylation of protein and carbohydrate moieties
Correct Answer: Synthesis of proteins
Explanation:The rough endoplasmic reticulum is the factory for the manufacturing of proteins. It contains ribosomes attached to it and transports proteins that are destined for membranes and secretions. The rough ER is connected to the nuclear envelope and to the cisternae of the Golgi apparatus by vesicles that shuttle between the two compartments.
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This question is part of the following fields:
- General
- Physiology
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Question 39
Correct
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Which of the following is a potential cause of a positive D-dimer assay?
Your Answer: Deep venous thrombosis
Explanation:A D-dimer test is performed to detect and diagnose thrombotic conditions and thrombosis. A negative result would rule out thrombosis and a positive result although not diagnostic, is highly suspicious of thrombotic conditions like a deep vein thrombosis, pulmonary embolism as well as DIC.
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This question is part of the following fields:
- General
- Physiology
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Question 40
Correct
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Urine specific gravity allows the assessment of which of the following renal functions?
Your Answer: Concentration
Explanation:Concentrating ability of kidneys is assessed by measuring the urine specific gravity. Normal values of urine specific gravity fall between 1.002 and 1.030 g/ml.
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This question is part of the following fields:
- Pathology
- Renal
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Question 41
Incorrect
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A medical student is asked to calculate the net pressure difference in a capillary wall, considering: Interstitial fluid hydrostatic pressure = –3 mmHg, Plasma colloid osmotic pressure = 28 mmHg, Capillary hydrostatic pressure = 17 mmHg, Interstitial fluid colloid osmotic pressure = 8 mmHg, and Filtration coefficient = 1. Which is the correct answer?
Your Answer: -3 mmHg
Correct Answer: 0 mmHg
Explanation:The rate of filtration at any point along a capillary depends on a balance of forces sometimes called Starling’s forces after the physiologist who first described their operation in detail. The Starling principle of fluid exchange is key to understanding how plasma fluid (solvent) within the bloodstream (intravascular fluid) moves to the space outside the bloodstream (extravascular space). Fluid movement = k[(pc– pi)–(Πc– Πi)] where k = capillary filtration coefficient, pc = capillary hydrostatic pressure, pi= interstitial hydrostatic pressure, Πc = capillary colloid osmotic pressure, Πi = interstitial colloid osmotic pressure. Therefore: 1 × [capillary hydrostatic pressure (17) – interstitial fluid hydrostatic pressure (–3)] – [plasma colloid osmotic pressure (28) – interstitial fluid colloid osmotic pressure (8)] = 0 mmHg
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 42
Correct
-
The renal cortex and medulla, if seen under the microscope, is lacking one of the following:
Your Answer: Squamous epithelium
Explanation:Capillaries, Henle’s loop, collecting ducts, Bertin columns and type IV collagen in glomerular basement membrane are all structures present in the renal cortex or medulla. The squamous epithelium is the only one that is lacking in both the renal cortex and medulla, because normally it is not found above the outer urethra.
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This question is part of the following fields:
- Pathology
- Renal
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Question 43
Incorrect
-
A 76-year-old man with a urinary tract obstruction due to prostatic hyperplasia develops acute renal failure. Which of the following physiological abnormalities of acute renal failure will be most life threatening for this patient?
Your Answer: Fluid overload
Correct Answer: Acidosis
Explanation:Acute renal failure (ARF) is a rapid loss of renal function due to damage to the kidneys, resulting in retention of nitrogenous (urea and creatinine) and non-nitrogenous waste products that are normally excreted by the kidney. This accumulation may be accompanied by metabolic disturbances, such as metabolic acidosis and hyperkalaemia, changes in body fluid balance and effects on many other organ systems. Metabolic acidosis and hyperkalaemia are the two most serious biochemical manifestations of acute renal failure and may require medical treatment with sodium bicarbonate administration and antihyperkalaemic measures. If not appropriately treated these can be life-threatening. ARF is diagnosed on the basis of characteristic laboratory findings, such as elevated blood urea nitrogen and creatinine, or inability of the kidneys to produce sufficient amounts of urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 44
Correct
-
In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your Answer: Exercise
Explanation:The oxygen-haemoglobin dissociation curve plots saturated haemoglobin against the oxygen tension and is usually a sigmoid plot. Each molecule of haemoglobin can bind to four molecules of oxygen reversibly. Factors that can influence the binding include: pH, concentration of 2,3-diphosphoglycerate (2,3-DPG), temperature, type of haemoglobin molecules, and presence of toxins, especially carbon monoxide. Shape of the curve is due to interaction of bound oxygen molecules with the incoming molecules. The binding of first molecule is difficult, with easier binding of the second and third molecule and increase in difficulty with the fourth molecule – partly as a result of crowding and partly as a natural tendency of oxygen to dissociate.
Left shift of curve indicates haemoglobin’s increased affinity for oxygen (seen at lungs). Right shift indicates decreased affinity and is seen with an increase in body temperature, hydrogen ions, 2,3-diphosphoglycerate (DPG), carbon dioxide concentration and exercise. Under normal resting conditions in a healthy individual, the normal position of the curve is at a pH of 7.4. A shift in the position of the curve with a change in pH is called the Bohr effect. Left shift occurs in acute alkalosis, decrease in p(CO2), decrease in temperature and decrease in 2,3-DPG. The fetal haemoglobin curve is to the left of the adult haemoglobin to allow for oxygen diffusion across the placenta. The curve for myoglobin is even further to the left. Carbon monoxide has a much higher affinity for haemoglobin than oxygen does. Thus, carbon monoxide poisoning leads to hypoxia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 45
Incorrect
-
Glucose is the most important source of energy for cellular respiration. The transport of glucose in the renal tubular cells occurs via:
Your Answer: Active transport
Correct Answer: Secondary active transport with sodium
Explanation:In 1960, Robert K. Crane presented for the first time his discovery of the sodium-glucose cotransport as the mechanism for glucose absorption. Glucose transport through biological membranes requires specific transport proteins. Transport of glucose through the apical membrane of renal tubular as well as intestinal epithelial cells depends on the presence of secondary active Na+–glucose symporters, SGLT-1 and SGLT-2, which concentrate glucose inside the cells, using the energy provided by co-transport of Na+ ions down their electrochemical gradient.
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This question is part of the following fields:
- Physiology
- Renal
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Question 46
Correct
-
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 47
Correct
-
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 48
Correct
-
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: 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 49
Correct
-
Post-total gastrectomy, there will be a decreased production of which of the following enzymes?
Your 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 50
Incorrect
-
Some substances, such as Chromium-51 and Technetium-99, are freely filtered but not secreted or absorbed by the kidney. In these cases, their clearance rate is equal to:
Your Answer: Urinary excretion rate of the substance
Correct Answer: Glomerular filtration rate
Explanation:If a substance passes through the glomerular membrane with perfect ease, the glomerular filtrate contains virtually the same concentration of the substance as does the plasma and if the substance is neither secreted nor reabsorbed by the tubules, all of the filtered substance continues on into the urine. Glomerular filtration rate (GFR) describes the flow rate of filtered fluid through the kidney.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 51
Correct
-
Which of the following substances is most likely to cause pulmonary vasodilatation?
Your Answer: Nitric oxide
Explanation:In the body, nitric oxide is synthesised from arginine and oxygen by various nitric oxide synthase (NOS) enzymes and by sequential reduction of inorganic nitrate. The endothelium of blood vessels uses nitric oxide to signal the surrounding smooth muscle to relax, so dilating the artery and increasing blood flow. Nitric oxide/oxygen blends are used in critical care to promote capillary and pulmonary dilation to treat primary pulmonary hypertension in neonatal patients post-meconium aspiration and related to birth defects.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 52
Correct
-
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 53
Incorrect
-
A lesion involving the lateral portion of the dorsal columns at the level of the nape of the neck will most likely affect:
Your Answer: Sweating of the ipsilateral face
Correct Answer: Vibratory sensations from the ipsilateral arm
Explanation:At the level mentioned in the question, the lateral portion of dorsal columns comprises of the fasciculus cuneatus. Axons carrying the sensations of touch, vibration and proprioception from the ipsilateral arm enter the spinal cord and ascend in the fasciculus cuneatus, synapsing in the nucleus cuneatus of the caudal medulla. Secondary neurons from this nucleus give rise to internal arcuate fibres, which decussate and ascend to the thalamus as the medial lemniscus. Tertiary neurons from there project to the ipsilateral somatosensory cortex. Thus, any damage to the fasciculus cuneatus will result in a deficit in tactile, proprioceptive and vibratory sensations in the ipsilateral arm, and not the contralateral arm.
Fine motor control of the fingers is mainly carried by the ipsilateral lateral corticospinal tract in the lateral funiculus of the cord. Motor control of the contralateral foot is carried by the ipsilateral corticospinal tract in the lateral funiculus of the cord. Lack of sweating of the face could be produced by interruption of sympathetic innervation. Proprioception from the ipsilateral leg is carried by the fasciculus gracilis in the medial part of the dorsal columns.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 54
Incorrect
-
An 18 year-old with an iron deficient diet was prescribed an iron supplement by her GP. Lack of iron often results in:
Your Answer: Megaloblastic anaemia
Correct Answer: Hypochromic anaemia
Explanation:Iron deficiency anaemia is the most common type of anaemia. It can occur due to deficiency of iron from decreased intake, increased loss or inadequate absorption. An MCV less than 80 will indicated iron deficiency anaemia. On the smear the RBCs will be microcytic hypochromic and will also show poikilocytosis. Iron profile tests are important to make a diagnosis. Clinically the patient will be pale and lethargic.
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This question is part of the following fields:
- General
- Physiology
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Question 55
Incorrect
-
A medical student is told a substance is freely filtered but is not metabolised, secreted, or stored in the kidney. It has a plasma concentration of 1000 mg/l and its urine excretion rate is 25 mg/min, and the inulin clearance is 100 ml/min. What is the rate of tubular reabsorption of the substance?
Your Answer: 25 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 56
Incorrect
-
A child defecates a few minutes after being fed by the mother. This is most likely due to:
Your Answer: Enterocolic reflex
Correct Answer: Gastrocolic reflex
Explanation:The gastrocolic reflex is a physiological reflex that involves increase in colonic motility in response to stretch in the stomach and by-products of digestion in the small intestine. It is shown to be uneven in its distribution throughout the colon, with the sigmoid colon affected more than the right side of the colon in terms of a phasic response. Various neuropeptides have been proposed as mediators of this reflex, such as serotonin, neurotensin, cholecystokinin and gastrin.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 57
Correct
-
The most important difference between interstitial fluid and plasma is the:
Your Answer: Protein concentration
Explanation:Interstitial fluid (or tissue fluid or intercellular fluid) is a solution that surrounds the cells of multicellular animals. It is the main component of the extracellular fluid, which also includes plasma, lymph and transcellular fluid. Plasma, the major component in blood, communicates freely with interstitial fluid through pores and intercellular clefts in capillary endothelium. Interstitial fluid consists of a water solvent containing amino acids, sugars, fatty acids, coenzymes, hormones, neurotransmitters, salts, as well as waste products from the cells. Red blood cells, platelets and plasma proteins cannot pass through the walls of the capillaries. The resulting mixture that does pass through is essentially blood plasma without the plasma proteins. Tissue fluid also contains certain types of white blood cells. Once the extracellular fluid collects into small vessels it is considered to be lymph, and the vessels that carry it back to the blood are called the lymphatic vessels. The lymphatic system returns protein and excess interstitial fluid to the circulation.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 58
Correct
-
Which of these conditions causes haematuria, hypertension and proteinuria in children, usually after a streptococcal infection?
Your Answer: Acute nephritic syndrome
Explanation:Nephritic syndrome (or acute nephritic syndrome) is a syndrome comprising of signs of nephritis. Children between 2 and 12 are most commonly affected, but it may occur at any age. Predisposing factors/causes include:
Infections with group A streptococcal bacteria (acute post-streptococcal glomerulonephritis).
Primary renal diseases: immunoglobulin A nephropathy, membranoproliferative glomerulonephritis, idiopathic rapidly progressive crescentic glomerulonephritis.
Secondary renal diseases: subacute bacterial endocarditis, infected ventriculo–peritoneal shunt, glomerulonephritis with visceral abscess, glomerulonephritis with bacterial, viral or parasitic infections.
Multisystem diseases.
By contrast, nephrotic syndrome is characterized by only proteins moving into the urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 59
Correct
-
A 7-year-old boy with facial oedema was brought to the hospital by his parents. Renal function is normal and urinalysis revealed the presence of a profound proteinuria. Which of the following is the most probable cause of these findings?
Your Answer: Minimal-change disease
Explanation:Minimal-change disease (MCD) refers to a histopathologic glomerular lesion, typically found in children, that is almost always associated with nephrotic syndrome. The most noticeable symptom of MCD is oedema, which can develop very rapidly. Due to the renal loss of proteins muscle wasting and growth failure may be seen in children. Renal function is usually not affected and a proteinuria of more than 40 mg/h/m2 is the only abnormal finding in urinalysis.
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This question is part of the following fields:
- Pathology
- Renal
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Question 60
Incorrect
-
One sensitive indicator of heavy alcohol dependence is:
Your Answer: Decreased serum alkaline phosphatase
Correct Answer: Elevated serum gamma-glutamyl transpeptidase
Explanation:Elevated serum gamma-glutamyl transpeptidase (GGT) may be the only laboratory abnormality in patients who are dependent on alcohol. Heavy drinkers may also have an increased MCV.
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This question is part of the following fields:
- Hepatobiliary
- Physiology
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Question 61
Incorrect
-
A patient with a long standing lower motor neuron lesion will have:
Your Answer: Spasticity
Correct Answer: Muscle wasting
Explanation:Lower motor neurons (LMNs) connect the brainstem and spinal cord to muscle fibres. Damage to lower motor neurons is indicated by abnormal electromyographic potentials, fasciculations, paralysis, weakening and wasting of skeletal muscles.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 62
Incorrect
-
As per the Poiseuille-Hagen formula, doubling the diameter of a vessel will change the resistance of the vessel from 16 peripheral resistance units (PRU) to:
Your Answer: 32 PRU
Correct Answer: 1 PRU
Explanation:Poiseuille-Hagen formula for flow in along narrow tube states that F = (PA– PB) × (Π/8) × (1/η) × (r4/l) where F = flow, PA– PB = pressure difference between the two ends of the tube, η = viscosity, r = radius of tube and L = length of tube. Also, flow is given by pressure difference divided by resistance. Hence, R = 8ηL ÷ Πr4. Hence, the resistance of the vessel changes in inverse proportion to the fourth power of the diameter. So, if the diameter of the vessel is increased to twice the original, it will lead to decrease in resistance to one-sixteenth its initial value.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 63
Correct
-
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 64
Correct
-
Calculate the resistance of the artery if the pressure at one end is 60 mmHg, pressure at the other end is 20 mm Hg and the flow rate in the artery is 200 ml/min.
Your Answer: 0.2
Explanation:Flow in any vessel = Effective perfusion pressure divided by resistance, where effective perfusion pressure is the mean intraluminal pressure at the arterial end minus the mean pressure at the venous end. Thus, in the given problem, resistance = (60 − 20)/200 = 0.2 mmHg/ml per min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 65
Correct
-
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 66
Incorrect
-
Which of the following proteins prevents red blood cells (RBCs) from bursting when they pass through capillaries?
Your Answer: Integrin
Correct Answer: Spectrin
Explanation:Spectrin is a structural protein found in the cytoskeleton that lines the intercellular side of the membrane of cells which include RBCs. They maintain the integrity and structure of the cell. It is arranged into a hexagonal arrangement formed from tetramers of spectrin and associated with short actin filaments that form junctions allowing the RBC to distort its shape.
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This question is part of the following fields:
- General
- Physiology
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Question 67
Correct
-
Ventricular filling follows a delay caused by?
Your Answer: AV node
Explanation:The AV node is a conducting tissue found between the atria and the ventricles of the heart. It conducts electrical signal from the atria to the ventricles and acts a delaying mechanism preventing the atria and the ventricles from contracting at the same time. This decremental conduction prevents premature ventricular contraction in cases such as atrial fibrillation. A delay in the AV node is the reason for the PR segment seen on the ECG. In certain types of supraventricular tachycardia, a person could have two AV nodes; this will cause a loop in electrical current and uncontrollably rapid heart beat. When this electricity catches up with itself, it will dissipate and return to a normal heart rate.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 68
Incorrect
-
During strenuous exercise, what else occurs besides tachycardia?
Your Answer: Rise in Paco2
Correct Answer: Increased stroke volume
Explanation:During strenuous exercise there is an increase in:
– Heart rate, stroke volume and therefore cardiac output. (CO = HR x SV)
– Respiratory rate (hyperventilation) which will lead to a reduction in Paco2.
– Oxygen demand of skeletal muscle, therefore leading to a reduction in mixed venous blood oxygen concentration.
Renal blood flow is autoregulated, so renal blood flow is preserved and will tend to remain the same. Mean arterial blood pressure is a function of cardiac output and total peripheral resistance and will increase with exercise, mainly as a result of the increase in cardiac output that occurs.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 69
Correct
-
A 40 year old man suffered severe trauma following a MVA. His BP is 72/30 mmhg, heart rate of 142 beats/mins and very feeble pulse. He was transfused 3 units of blood and his BP returned to 100/70 and his heart rate slowed to 90 beats/min. What decreased after transfusion?
Your Answer: Total peripheral resistance
Explanation:The patient is in hypovolemic shock, he is transfused with blood, this fluid resuscitation will result in a decreased sympathetic discharge and adequate ventricular filling which will result in the decreases TPR with an increased CO and cardiac filling pressures
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 70
Incorrect
-
Carbon dioxide is principally transported in the blood in which form?
Your Answer: Carboxyhaemoglobin
Correct Answer: Bicarbonate
Explanation:Carbon dioxide is transported in the blood in various forms:
– Bicarbonate (80–90%)
– Carbamino compounds (5–10%)
– Physically dissolved in solution (5%).
Carbon dioxide is carried on the haemoglobin molecule as carbamino-haemoglobin; carboxyhaemoglobin is the combination of haemoglobin with carbon monoxide.
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This question is part of the following fields:
- Physiology
- Respiratory; Cardiovascular
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Question 71
Incorrect
-
The renal tubule is the portion of the nephron that contains the fluid that has been filtered by the glomerulus. Which of the following substances is actively secreted into the renal tubules?
Your Answer: Glucose
Correct Answer: Potassium
Explanation:The renal corpuscle filters out solutes from the blood, delivering water and small solutes to the renal tubule for modification. In normal circumstances more than 90% of the filtered load of K is reabsorbed by the proximal tubules and loops of Henlé and almost all K appearing in the urine has been secreted by the late distal tubules and collecting tubules. So the rate of excretion is usually independent of the rate of filtration, but is closely tied to the rate of secretion and control of K excretion, largely accomplished by control of the secretion rate. Around 65–70% of the filtered potassium is reabsorbed along with water in the proximal tubule and the concentration of potassium in the tubular fluid varies little from that of the plasma.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 72
Correct
-
A 76-year-old woman is diagnosed with diabetes mellitus after a urine test revealed she has glucosuria. Glucosuria may occur due to inadequate glucose reabsorption at:
Your Answer: Proximal convoluted tubule
Explanation:Glucose is reabsorbed almost 100% via sodium–glucose transport proteins (apical) and GLUT (basolateral) in the proximal convoluted tubule. Glycosuria or glucosuria is a condition of osmotic diuresis typical in those suffering from diabetes mellitus. Due to a lack of insulin, plasma glucose levels are above normal. This leads to saturation of receptors in the kidneys and glycosuria usually at plasma glucose levels above 11 mmol/l. Rarely, glycosuria is due to an intrinsic problem with glucose reabsorption within the kidneys (such as Fanconi syndrome), producing a condition termed renal glycosuria.
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This question is part of the following fields:
- Physiology
- Renal
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Question 73
Incorrect
-
Which of the following substances will enhance the activity of antithrombin III?
Your Answer: Citrate
Correct Answer: Heparin
Explanation:Antithrombin III is a glycoprotein that inactivates multiple enzymes involved in the coagulation system. It inactivates factor X, factor IX, factor II, factor VII, factor XI and factor XII. Its activity is greatly increased by the action of heparin.
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This question is part of the following fields:
- General
- Physiology
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Question 74
Incorrect
-
Evaluation of a 60-year old gentleman, who has been a coal miner all his life and is suspected to have pulmonary fibrosis reveals the following: 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: 120 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 75
Correct
-
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 76
Correct
-
A 15-day old male baby was brought to the emergency department with sweating and his lips turning blue while feeding. He was born full term. On examination, his temperature was 37.9°C, blood pressure 75/45 mmHg, pulse was 175/min, and respiratory rate was 42/min. A harsh systolic ejection murmur could be heard at the left upper sternal border. X-ray chest showed small, boot-shaped heart with decreased pulmonary vascular markings. He most likely has:
Your Answer: Tetralogy of Fallot
Explanation:The most common congenital cyanotic heart disease and the most common cause of blue baby syndrome, Tetralogy of Fallot shows four cardiac malformations occurring together. These are ventricular septal defect (VSD), pulmonary stenosis (right ventricular outflow obstruction), overriding aorta (degree of which is variable), and right ventricular hypertrophy. The primary determinant of severity of disease is the degree of pulmonary stenosis. Tetralogy of Fallot is seen in 3-6 per 10,000 births and is responsible for 5-7% congenital heart defects, with slightly higher incidence in males. It has also been associated with chromosome 22 deletions and DiGeorge syndrome. It gives rise to right-to-left shunt leading to poor oxygenation of blood. Primary symptom is low oxygen saturation in the blood with or without cyanosis at birth of within first year of life. Affected children ay develop acute severe cyanosis or ‘tet spells’ (sudden, marked increase in cyanosis, with syncope, and may result in hypoxic brain injury and death). Other symptoms include heart murmur, failure to gain weight, poor development, clubbing, dyspnoea on exertion and polycythaemia. Chest X-ray reveals characteristic coeur-en-sabot (boot-shaped) appearance of the heart. Treatment consists of immediate care for cyanotic spells and Blalock–Taussig shunt (BT shunt) followed by corrective surgery.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 77
Correct
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Which of the following is a likely consequence of severe diarrhoea?
Your Answer: A decrease in the sodium content of the body
Explanation:Diarrhoea can occur due to any of the numerous aetiologies, which include infectious, drug-induced, food related, surgical, inflammatory, transit-related or malabsorption. Four mechanisms have been implicated in diarrhoea: increased osmotic load, increased secretion, inflammation and decreased absorption time. Diarrhoea can result in fluid loss with consequent dehydration, electrolyte loss (Na+, K+, Mg2+, Cl–) and even vascular collapse. Loss of bicarbonate ions can lead to a metabolic acidosis.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 78
Incorrect
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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: Decrease Increase
Correct Answer: Increase Decrease
Explanation:Hypovolemia will result in the activation of the sympathetic adrenal discharge resulting is a decrease pulmonary artery pressure and an elevated pulmonary vascular resistance.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 79
Incorrect
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During cardiac catheterisation, if the blood sample from the catheter shows an oxygen saturation of 70%, and the pressure ranging from 12 to 24 mm Hg, it implies that the catheter tip is located in the:
Your Answer: Right atrium
Correct Answer: Pulmonary artery
Explanation:Normal values for various parameters are as follows:
Systolic arterial blood pressure (SBP): 90–140 mmHg.
Diastolic arterial blood pressure: 60–90 mmHg.
Mean arterial blood pressure (MAP): SBP + (2 × DBP)/3 (normal range: 70-105 mmHg).
Right atrial pressure (RAP): 2–6 mmHg.
Systolic right ventricular pressure (RVSP): 15–25 mmHg.
Diastolic right ventricular pressure (RVDP): 0–8 mmHg.
Pulmonary artery pressure (PAP): Systolic (PASP) is 15-25 mmHg and Diastolic (PADP) is 8–15 mmHg.
Pulmonary artery wedge pressure (PAWP): 6–12 mmHg.
Left atrial pressure (LAP): 6–12 mmHg.
Thus, the given value indicates that the position of catheter tip is likely to be in the pulmonary artery.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 80
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 I
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 81
Correct
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What is the most likely cause of prolonged bleeding time in a 40 year old women admitted for a laparoscopic cholecystectomy?
Your Answer: Thrombocytopaenia
Explanation:Bleeding time is related to platelet function, thus a decrease in platelet function, as seen in thrombocytopenia, DIC and von Willebrand disease in which platelet aggregation is defective, leads to an increase in bleeding time. It is not affected by a decrease or deficiency of any other clotting factors. Aspirin and other COX inhibitors prolong bleeding time along with warfarin and heparin.
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This question is part of the following fields:
- General
- Physiology
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Question 82
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 83
Incorrect
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Atractyloside is an inhibitor of electron transport chain. It is expected to have little or no effect on the functioning of which of the following cell types?
Your Answer: Cardiac muscle cells
Correct Answer: Red blood cells
Explanation:Electron transport chain is a series of electron carriers that are embedded in the mitochondrial membrane. It is the place where ATP is made. Inhibiting the electron transport chain will stop production of ATP. Red blood cells are the only cell in the given option which do not contain ATP.
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This question is part of the following fields:
- General
- Physiology
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Question 84
Incorrect
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Which of the following will increase blood pressure and cause hypokalaemia?
Your Answer: Desmopressin
Correct Answer: Angiotensin II
Explanation:Angiotensin is a peptide that is released in response to a decrease in blood volume and blood pressure. It has multiple functions but mainly acts to cause vasoconstriction, increase BP and release aldosterone from the adrenal cortex. It is a powerful vasoconstrictor and release of aldosterone causes increased retention of sodium and excretion of potassium.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 85
Incorrect
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A 45-year-old-female is suspected to have a pulmonary mass. Supposing that she has a neoplasm, which of the following are most commonly found to involve the lung:
Your Answer: Small-cell anaplastic carcinomas in smokers
Correct Answer: Pulmonary metastases
Explanation:Lung metastases occur when a cancer started in another part of the body (primary site) spreads to the lungs. The lungs are among the most common site where cancer can spread due to its rich systemic venous drainage, almost every type of cancer can spread to the lung. The most common types of cancer that spread to the lung are breast, colorectal, kidney, testicular, bladder, prostate, head and neck cancers.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 86
Incorrect
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Under normal conditions, what is the major source of energy of cardiac muscles?
Your Answer: Glucose
Correct Answer: Fatty acids
Explanation:Under basal conditions, most of the energy needed by cardiac muscle for metabolism is derived from fats (60%), 35% by carbohydrates, and 5% by ketones and amino acids. However, after intake of large amounts of glucose, lactate and pyruvate are mainly used. During prolonged starvation, fat acts as the primary source. 50% of the used lipids are sourced from circulating fatty acids.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 87
Incorrect
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What is the normal glomerular filtration rate?
Your Answer: 100 mL/min
Correct Answer: 125 mL/min
Explanation:The normal glomerular filtration rate (GFR) in humans is 125 mL/min. After the age of 40, GFR decreases progressively by about 0.4–1.2 mL/min per year.
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This question is part of the following fields:
- Physiology
- Renal
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Question 88
Incorrect
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Which of the following is a true statement regarding the pupil?
Your Answer: Pupil diameter is determined by the balance between parasympathetic tone to the radial fibres of the iris and sympathetic tone to the pupillary sphincter muscle.
Correct Answer: Phentolamine causes pupil constriction
Explanation:A balance between the sympathetic tone to the radial fibres of the iris and parasympathetic tone to the pupillary sphincter muscle determines the pupil size. Phentolamine (α-adrenergic receptor blocker) causes pupillary constriction. Dilatation of the pupil occurs with increased sympathetic activity, decreased parasympathetic activity during darkness or block of muscarinic receptors by atropine.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 89
Incorrect
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A 45-year-old man with short bowel syndrome requires parenteral nutrition. The solution of choice for parenteral nutrition is:
Your Answer: Darrow’s solution
Correct Answer: Crystalline amino acids
Explanation:Total parenteral nutrition (TPN), is the practice of feeding a person intravenously, circumventing the gut. It is normally used in the following situations: surgery, when feeding by mouth is not possible, when a person’s digestive system cannot absorb nutrients due to chronic disease or if a person’s nutrient requirement cannot be met by enteral feeding and supplementation. A sterile bag of nutrient solution, between 500 ml and 4L, is provided. The pump infuses a small amount (0.1–10 ml/h) continuously to keep the vein open. The nutrient solution consists of water, glucose, salts, amino acids, vitamins and sometimes emulsified fats. Ideally each patient is assessed individually before commencing on parenteral nutrition, and a team consisting of doctors, nurses, clinical pharmacists and dietitians evaluate the patient’s individual data and decide what formula to use and at what rate.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 90
Incorrect
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A 56-year-old man undergoes tests to determine his renal function. His results over a period of 24 hours were:
Urine flow rate: 2. 0 ml/min
Urine inulin: 1.0 mg/ml
Plasma inulin: 0.01 mg/ml
Urine urea: 260 mmol/l
Plasma urea: 7 mmol/l
What is the glomerular filtration rate?Your Answer: 250 ml/min
Correct Answer: 200 ml/min
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal (kidney) glomerular capillaries into the Bowman’s capsule per unit time. GFR is equal to the inulin clearance because inulin is freely filtered into Bowman’s capsule but is not reabsorbed or secreted. The clearance (C) of any substance can be calculated as follows: C = (U × V)/P, where U and P are the urine and plasma concentrations of the substance, respectively and V is the urine flow rate. Thus, glomerular filtration rate = (1.0 × 2. 0)/0.01 = 200 ml/min.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 91
Incorrect
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A blood sample from a patient with polycythaemia vera will show which of the following abnormalities?
Your Answer: Decreased MCV
Correct Answer: High platelet count
Explanation:Polycythaemia is a condition that results in an increase in the total number of red blood cells (RBCs) in the blood. It can be due to a myeloproliferative syndrome, chronically low oxygen levels or rarely malignancy. In primary polycythaemia/ polycythaemia vera the increase is due to an abnormality in the bone marrow, resulting in increased RBCs, white blood cells (WBCs) and platelets. In secondary polycythaemia the increase occurs due to high levels of erythropoietin either artificially or naturally. The increase is about 6-8 million/cm3 of blood. A type of secondary polycythaemia is physiological polycythaemia where people living in high altitudes who are exposed to hypoxic conditions produce more erythropoietin as a compensatory mechanism for thin oxygen and low oxygen partial pressure.
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This question is part of the following fields:
- General
- Physiology
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Question 92
Correct
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What is the result of maltase deficiency in the brush border of the small intestine?
Your Answer: Results in increased passage of maltose in stool
Explanation:Maltase is an enzyme produced from the surface cells of the villi, lining the small intestine and aids in hydrolysing the disaccharide maltose, which splits into two molecules of α-glucose. It is done by breaking the glycosidic bond between the ‘first’ carbon of one glucose and the ‘fourth’ carbon of the other (a 1–4 bond). Hence, a deficiency of enzyme maltase will result in the increased passage of maltose in the stool.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 93
Correct
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Chronic obstructive pulmonary disease (COPD) is likely to result in:
Your Answer: Respiratory acidosis
Explanation:COPD leads to respiratory acidosis (chronic). This occurs due to hypoventilation which involves multiple causes, such as poor responsiveness to hypoxia and hypercapnia, increased ventilation/perfusion mismatch leading to increased dead space ventilation and decreased diaphragm function.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 94
Correct
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Production of pain is most likely associated with:
Your Answer: Substance P
Explanation:Substance P is a short-chain polypeptide that functions as a neurotransmitter and as a neuromodulator, and is thus, a neuropeptide. It has been linked with pain regulation, mood disorders, stress, reinforcement, neurogenesis, respiratory rhythm, neurotoxicity, nausea and emesis. It is also a potent vasodilator as it brings about release of nitric oxide from the endothelium. Its release can also cause hypotension.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 95
Incorrect
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Extracellular body fluid as compared with intracellular body fluid:
Your Answer: Contains more amino acids
Correct Answer: Is relatively rich in glucose
Explanation:The percentages of body water contained in various fluid compartments add up to total body water (TBW). This water makes up a significant fraction of the human body, both by weight and by volume. Ensuring the right amount of body water is part of fluid balance, an aspect of homeostasis. The extracellular fluid (ECF) includes all fluids outside the cells. This fluid can be divided into three fluid departments: interstitial (in the tissue spaces) fluid, blood plasma and lymph, and specialised compartments called transcellular fluid. The extracellular fluid surrounds all the cells in the body and is in equilibrium with the intracellular fluid. So, its composition must remain fairly constant even though substances are passing into and out of the cells. The interstitial fluid, though called a fluid, is in a reality a gel-like composition made up of: water, proteoglycan molecules and collagen. The extracellular fluid constitutes 40% of total body water, with intracellular fluid making up the remaining 60%. It is relatively rich in glucose.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 96
Incorrect
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The primary area involved in the pathology of Parkinson's disease is:
Your Answer: Motor cortex
Correct Answer: Substantia nigra
Explanation:Parkinson’s disease is a degenerative, movement disorder of the central nervous system, and is typically characterized by muscle rigidity, tremor and bradykinesia (in extreme cases, akinesia). Secondary symptoms include high-level cognitive dysfunction and subtle language problems.
Parkinson’s disease is also called ‘primary Parkinsonism’ or ‘idiopathic Parkinson’s disease and is the most common cause of Parkinsonism, a group of similar symptoms. The disorder is caused due to loss of pigmented dopaminergic cells in the pars compacta region of the substantia nigra.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 97
Incorrect
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A 50-year-old woman goes to the doctor complaining of myalgia, muscle cramps, and weakness; she is diagnosed with severe hypokalaemia. Which of the following is the most common cause of hypokalaemia?
Your Answer: Chronic renal failure
Correct Answer: Prolonged vomiting
Explanation:Potassium is one of the body’s major ions. Nearly 98% of the body’s potassium is intracellular. The ratio of intracellular to extracellular potassium is important in determining the cellular membrane potential. Small changes in the extracellular potassium level can have profound effects on the function of the cardiovascular and neuromuscular systems. Hypokalaemia may result from conditions as varied as renal or gastrointestinal (GI) losses, inadequate diet, transcellular shift (movement of potassium from serum into cells) and medications. The important causes of hypokalaemia are:
Renal losses: renal tubular acidosis, hyperaldosteronism, magnesium depletion, leukaemia (mechanism uncertain).
GI losses: vomiting or nasogastric suctioning, diarrhoea, enemas or laxative use, ileal loop.
Medication effects: diuretics (most common cause), β-adrenergic agonists, steroids, theophylline, aminoglycosides.
Transcellular shift: insulin, alkalosis.
Severe hypokalaemia, with serum potassium concentrations of 2.5–3 meq/l, may cause muscle weakness, myalgia, tremor, muscle cramps and constipation.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 98
Incorrect
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A 35-year-old woman in her 37th week of pregnancy complains of urinary incontinence. She is most likely to have:
Your Answer: Overflow incontinence
Correct Answer: Stress incontinence
Explanation:Urinary incontinence is the involuntary excretion of urine from one’s body. It is often temporary and it almost always results from an underlying medical condition. Several types include:
– Stress incontinence is the voiding of urine following increased abdominal pressure e.g. laughing, coughing, pregnancy etc. It is the most common form of incontinence in women, most commonly due to pelvic floor muscle weakness, physical changes from pregnancy, childbirth and menopause. In men it is a common problem following a prostatectomy. Most lab results such as urine analysis, cystometry and postvoid residual volume are normal.
– Urge incontinence is involuntary loss of urine occurring for no apparent reason while suddenly feeling the need or urge to urinate. The most common cause of urge incontinence are involuntary and inappropriate detrusor muscle contractions.
– Functional incontinence – occurs when a person does not recognise the need to go to the toilet, recognise where the toilet is or get to the toilet in time. The urine loss may be large. Causes of functional incontinence include confusion, dementia, poor eyesight, poor mobility, poor dexterity or unwillingness. t
– Overflow incontinence – sometimes people find that they cannot stop their bladders from constantly dribbling or continuing to dribble for some time after they have passed urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 99
Incorrect
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A 29-year-old woman presents to the doctor complaining of cough, shortness of breath, fever and weight loss. Chest X-ray revealed bilateral hilar and mediastinal lymph node enlargement and bilateral pulmonary opacities. Non-caseating granulomas were found on histological examination. The most likely diagnosis is:
Your Answer: Asbestosis
Correct Answer: Sarcoidosis
Explanation:Sarcoidosis is an inflammatory disease of unknown aetiology that affects multiple organs but predominantly the lungs and intrathoracic lymph nodes. Systemic and pulmonary symptoms may both be present. Pulmonary involvement is confirmed by a chest X-ray and other imaging studies. The main histological finding is the presence of non-caseating granulomas.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 100
Correct
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What is the pH of freshly formed saliva at ultimate stimulation?
Your Answer: 8
Explanation:Saliva has four major components: mucus (lubricant), α-amylase (enzyme that initiates digestion of starch), lingual lipase (enzyme that begins fat digestion), and a slightly alkaline electrolyte solution for moistening food. As the secretion rate of saliva increases, its osmolality increases. Moreover, the pH changes from slightly acidic (at rest) to basic (pH 8) at ultimate stimulation. This occurs due to increase of HCO3-. Amylase and mucus also increase in concentration after stimulation.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 101
Incorrect
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Which of the following associations is correctly matched with the body's defence mechanism in fighting infection?
Your Answer: Non-specific cellular mechanism → B lymphocytes
Correct Answer: Specific cellular mechanism → cytotoxic T cells
Explanation:The immune system has certain levels of defence against pathogens. First line includes simple barriers such as skin, mucosa and stomach acid that prevent the pathogen from entering into the body. If this barrier is breached then the innate immune system is activated which includes leukocytes (macrophages, neutrophils, mast cells, eosinophils, basophils, natural killer cells). If the pathogens invade the second layer of defence then the third layer, adaptive immunity is activated, which includes B and T lymphocytes. B cells provide a humoral response whereas cytotoxic T cells have specific cellular mechanisms. They maintain a memory of past infections and are activated faster following a recurrence.
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This question is part of the following fields:
- General
- Physiology
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Question 102
Correct
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Depression of the normal coagulation system and excessive bleeding after surgery can occur in which of the following medical conditions?
Your Answer: Liver disease
Explanation:As most of the coagulation factors are synthesized in the liver, liver diseases like hepatitis or cirrhosis will depress the coagulation system. Vitamin K deficiency can also decrease the production of vitamin K dependent coagulation factors VII, XI, X and prothrombin.
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This question is part of the following fields:
- General
- Physiology
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Question 103
Incorrect
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There are several mechanisms involved in the transport of sodium ions from blood to interstitial fluid of the muscle cells. Which of the following mechanisms best describes this phenomenon?
Your Answer: Active transport through endothelial cell membranes
Correct Answer: Diffusion through channels between endothelial cells
Explanation:Capillaries are the smallest of the body’s blood vessels, measuring 5–10 μm and they help to enable the exchange of water, oxygen, carbon dioxide, and many other nutrients and waste substances between the blood and the tissues surrounding them. The walls of capillaries are composed of only a single layer of cells, the endothelium. Ion channels are pore-forming proteins that help to establish and control the small voltage gradient that exists across the plasma membrane of all living cells by allowing the flow of ions down their electrochemical gradient. An ion channel is an integral membrane protein or more typically an assembly of several proteins. The archetypal channel pore is just one or two atoms wide at its narrowest point. It conducts a specific ion such as sodium or potassium and conveys them through the membrane in single file.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 104
Correct
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Which antibiotic acts by inhibiting protein synthesis?
Your Answer: Erythromycin
Explanation:Penicillins and cephalosporins (e.g. cefuroxime, cefotaxime, ceftriaxone) inhibit bacterial cell wall synthesis through the inhibition of peptidoglycan cross-linking.
Macrolides (e.g. erythromycin), tetracyclines, aminoglycosides and chloramphenicol act by interfering with bacterial protein synthesis.
Sulphonamides (e.g. trimethoprim, co-trimoxazole) work by inhibiting the synthesis of nucleic acid
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This question is part of the following fields:
- Pharmacology; Microbiology
- Physiology
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Question 105
Correct
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When the pitch of a sound increases, what is the physiological response seen in the listener?
Your Answer: The location of maximal basilar membrane displacement moves toward the base of the cochlea
Explanation:An increase in the frequency of sound waves results in a change in the position of maximal displacement of the basilar membrane in the cochlea. Low pitch sound produces maximal displacement towards the cochlear apex and greatest activation of hair cells there. With an increasing pitch, the site of greatest displacement moves towards the cochlear base. However, increased amplitude of displacement, increase in the number of activated hair cells, increased frequency of discharge of units in the auditory nerve and increase in the range of frequencies to which such units respond, are all seen in increases in the intensity or a sound stimulus.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 106
Incorrect
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After a severe asthma attack, a 26-year-old woman is left in a markedly hypoxic state. In which of the following organs are the arterial beds most likely to be vasoconstricted due to the hypoxia?
Your Answer: Skeletal muscle
Correct Answer: Lungs
Explanation:Hypoxic pulmonary vasoconstriction is a local response to hypoxia resulting primarily from constriction of small muscular pulmonary arteries in response to reduced alveolar oxygen tension. This unique response of pulmonary arterioles results in a local adjustment of perfusion to ventilation. This means that if a bronchiole is obstructed, the lack of oxygen causes contraction of the pulmonary vascular smooth muscle in the corresponding area, shunting blood away from the hypoxic region to better-ventilated regions. The purpose of hypoxic pulmonary vasoconstriction is to distribute blood flow regionally to increase the overall efficiency of gas exchange between air and blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 107
Correct
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A 49-year-old man, smoker, complains of a persisting and worsening cough over the past few months. He also has noted blood in his sputum. The patient has no other major health conditions. Which of the following investigative procedures should be done first?
Your Answer: Sputum cytology
Explanation:Sputum cytology is a diagnostic test used for the examination of sputum under a microscope to determine if abnormal cells are present. It may be used as the first diagnostic procedure to help detect a suspected lung cancer or certain non-cancerous lung conditions.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 108
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 109
Incorrect
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A diabetic 58-year-old man, after the injection of radiographic contrast, has a decreased urine output and decreased level of consciousness. Which of the following conditions has he most likely developed
Your Answer: Rapidly progressive glomerulonephritis
Correct Answer: Acute tubular necrosis
Explanation:Acute tubular necrosis (ATN) involves damage to the tubule cells of the kidneys and is the most common cause of acute kidney injury. ATN in the majority of the cases is caused by ischaemia of the kidneys due to lack of perfusion and oxygenation but it may also occur due to poison or harmful substance. Contrast used for radiology may cause ATN in patients with several risk factors e.g. diabetic nephropathy. Symptoms may include oliguria, nausea, fluid retention, fatigue and decreased consciousness.
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This question is part of the following fields:
- Pathology
- Renal
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Question 110
Correct
-
A 47-year old-woman diagnosed with pancreatitis presented to the emergency department complaining of a worsening shortness of breath, fever, agitation and cough. Oxygen saturation was 67% in room air. Her respiratory status continued to deteriorate therefore she was intubated. She was admitted to the intensive care unit for management. Chest X-ray demonstrated bilateral perihilar opacities. The patient failed conventional treatment and died several days later. At autopsy, the lung shows growth of type 2 pneumocytes and thickened alveolar walls. What is the most probable diagnosis?
Your Answer: Adult respiratory distress syndrome
Explanation:Acute (or adult) respiratory distress syndrome (ARDS) is a life-threatening lung condition characterised by a non-cardiogenic pulmonary oedema that leads to acute respiratory failure. The most common risk factors for ARDS include trauma with direct lung injury, sepsis, pneumonia, pancreatitis, burns, drug overdose, massive blood transfusion and shock. Acute onset of dyspnoea with hypoxemia, anxiety and agitation is typical. Chest X ray most commonly demonstrates bilateral pulmonary infiltrates. Histological changes include the exudative, proliferative and fibrotic phase. ARDS is mainly a clinical diagnosis.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 111
Correct
-
A woman that presented with dyspnoea, chest pain and cough was found to have a serous pleural effusion. This finding is most likely to be associated with which of the following conditions?
Your Answer: Congestive heart failure
Explanation:A pleural effusion is defined as an abnormal collection of fluid in the pleural space. Pleural effusion can result from excess fluid production or decreased absorption or both. Thoracentesis and laboratory testing help determine the origin of the accumulated fluid. Serous fluid accumulation in the pleural space indicates the presence of a hydrothorax and is most likely to develop secondary to congestive heart failure.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 112
Incorrect
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When at rest, which of the following will be higher in a marathon runner compared to an untrained individual?
Your Answer: Oxygen consumption
Correct Answer: Cardiac stroke volume
Explanation:Cardiac muscle hypertrophy is seen in trained athletes as compared to the normal population. This hypertrophy results in higher stroke volume at rest and increased cardiac reserve (maximum cardiac output during exercise). However, the cardiac output at rest is almost the same in both trained and untrained people. This is because in trained athletes, the heart rate is slower, even up to 40-50 beats/min. There is minimal affect of athletic training on oxygen consumption and respiratory rate.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 113
Correct
-
What percentage of the cardiac output is delivered to the brain?
Your Answer: 15%
Explanation:Among all body organs, the brain is most susceptible to ischaemia. Comprising of only 2.5% of total body weight, the brain receives 15% of the cardiac output. Oxygen extraction is also higher with venous oxygen levels approximating 13 vol%, and arteriovenous oxygen difference of 7 vol%.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 114
Incorrect
-
Which of the following brings about a reduction in gastric blood flow?
Your Answer: Acetylcholine
Correct Answer: Vasopressin
Explanation:Gastric blood flow is increased by vagal stimulation, gastrin, histamine and acetylcholine as they stimulate gastric section and the production of vasodilator metabolites. Acetylcholine and histamine also have a direct action on the gastric arterioles. Similarly, gastric blood flow is reduced by inhibitors of secretion – catecholamines, secretin and vasopressin.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 115
Correct
-
Selective destruction of which of the following cells will affect antibody synthesis?
Your Answer: Plasma cells
Explanation:Plasma cell are memory cells. After the antigen Is engulfed by the B cells it is presented to the CD4+ helper cells via the MCH II receptor and this leads to their activation which in turn stimulates the B cells to form antibodies against that specific antigen. Some B cells differentiate into plasma cells also called memory cells that get activated after subsequent infection.
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This question is part of the following fields:
- General
- Physiology
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Question 116
Correct
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A 47-year-old male smoker, who had been self-medicating with oral steroids for the last two years due to persistent breathlessness presented to the doctor complaining of a productive cough, fever and chest pain. A chest X-ray revealed bilateral patchy opacities. He was diagnosed with bilateral bronchopneumonia. Which of these organisms is most probably causing these findings?
Your Answer: Nocardia asteroides
Explanation:Nocardia is a Gram-positive aerobic actinomycete. Several species have been identified but the most common human pathogen is Nocardia asteroides. The predominant clinical finding in the majority of patients affected by nocardiosis is pulmonary disease. Predisposing factors for pulmonary nocardiosis include leukaemia, human immunodeficiency virus (HIV) infection, organ transplantation, diabetes and receiving prolonged corticosteroids.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 117
Incorrect
-
If the blood flow is constant, oxygen extraction by tissues will show the greatest decrease due to which of the following interventions?
Your Answer: Exercise
Correct 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 118
Correct
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The bronchial circulation is a part of the circulatory system that supplies nutrients and oxygen to the pulmonary parenchyma. What percentage of cardiac output is received by bronchial circulation?
Your Answer: 2%
Explanation:The bronchial circulation is part of the systemic circulation and receives about 2% of the cardiac output from the left heart. Bronchial arteries arise from branches of the aorta, intercostal, subclavian or internal mammary arteries. The bronchial arteries supply the tracheobronchial tree with both nutrients and O2. It is complementary to the pulmonary circulation that brings deoxygenated blood to the lungs and carries oxygenated blood away from them in order to oxygenate the rest of the body.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 119
Incorrect
-
Chest X-ray of a 45-year old gentleman with a week history of pleurisy showed a small pneumothorax with moderate-sized pleural effusion. Arterial blood gas analysis showed p(CO2) = 23 mmHg, p(O2) = 234.5 mmHg, standard bicarbonate = 16 mmol/l. What are we most likely dealing with?
Your Answer: Mixed acidosis
Correct Answer: Compensated respiratory alkalosis
Explanation:Normal pH with low p(CO2) and low standard bicarbonate could indicate either compensated respiratory alkalosis or a compensated metabolic acidosis. However, the history of hyperventilation for 5 days (pleurisy) favours compensated respiratory alkalosis. Compensated metabolic acidosis would have been likely in a diabetic patient with fever, vomiting and high glucose (diabetic ketoacidosis).
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 120
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 121
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 122
Correct
-
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 123
Incorrect
-
Which of the following enzymes is secreted by the small intestinal mucosa?
Your Answer: Enterokinase
Correct Answer: Lactase
Explanation:Lactase, an enzyme belonging to β-galactosidase family of enzymes, brings about the hydrolysis of the disaccharide lactose into galactose and glucose. In humans, it is present along the brush border membrane of the cells lining the small intestinal villi. Deficiency of lactase causes lactose intolerance.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 124
Correct
-
A 65-year-old man with no history of smoking complains of shortness of breath and persistent cough over the past 8 months. He reveals that in the 1960s he worked for several years as a boiler operator. Chest X-ray shows diffuse lung infiltrates. Which of the following is the most probable cause of these findings?
Your Answer: Asbestosis
Explanation:Asbestosis is a chronic lung disease which leads to long-term respiratory complications and is caused by the inhalation of asbestos fibres. Symptoms due to long exposure to asbestos usually appear 10 to 40 years after initial exposure and include shortness of breath, cough, weight loss, clubbing of the fingers and chest pain. Typical chest X-ray findings include diffuse lung infiltrates that cause the appearance of shaggy heart borders.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 125
Correct
-
Which of the following physiological changes will you see in a young man who has been trekking in the Himalayas for 3 years?
Your Answer: Increased renal excretion of HCO3 –
Explanation:The atmospheric pressure is lower at high altitudes as compared with sea level. This leads to a decrease in the partial pressure of oxygen. Once 2100 m (7000 feet) of altitude is reached, there is a drop in saturation of oxyhaemoglobin. The oxygen saturation of haemoglobin determines the oxygen content in the blood. The body physiological tries to adapt to high altitude by acclimatization. Immediate effects include hyperventilation, fluid loss (due to a decreased thirst drive), increase in heart rate and slightly lowered stroke volume. Long term effects include lower lactate production, compensatory alkali loss in urine, decrease in plasma volume, increased erythropoietin release and red cell mass, increased haematocrit, higher concentration of capillaries in striated muscle tissue, increase in myoglobin, increase in mitochondria, increase in aerobic enzyme concentration such as 2,3-DPG and pulmonary vasoconstriction.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 126
Incorrect
-
Which of the following will be a seen in a patient with a plasma thyroid-stimulating hormone (TSH) level of 14 mU/l (normal < 5 mU/l) and a low T3 resin uptake of 19% (normal 25–35%)?
Your Answer: Palpitations
Correct Answer: Periorbital swelling and lethargy
Explanation:Low T3 resin uptake combined with raised TSH is indicative of hypothyroidism. Signs and symptoms include dull expression, facial puffiness, lethargy, periorbital swelling due to infiltration with mucopolysaccharides, bradycardia and cold intolerance. Anxiety, palpitations, tachycardia, raised body temperature, heat intolerance and weight loss are all seen in hyperthyroidism.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 127
Incorrect
-
Which of the following clinical signs will be demonstrated in a case of Brown-Séquard syndrome due to hemisection of the spinal cord at mid-thoracic level?
Your Answer: Ipsilateral spastic paralysis, ipsilateral loss of vibration and proprioception (position sense) and ipsilateral loss of pain and temperature sensation beginning one or two segments below the lesion
Correct Answer: Ipsilateral spastic paralysis, ipsilateral loss of vibration and proprioception (position sense) and contralateral loss of pain and temperature sensation beginning one or two segments below the lesion
Explanation:Brown–Séquard syndrome results due to lateral hemisection of the spinal cord and results in a loss of motricity (paralysis and ataxia) and sensation. The hemisection of the cord results in a lesion of each of the three main neural systems: the principal upper motor neurone pathway of the corticospinal tract, one or both dorsal columns and the spinothalamic tract. As a result of the injury to these three main brain pathways the patient will present with three lesions. The corticospinal lesion produces spastic paralysis on the same side of the body (the loss of moderation by the upper motor neurons). The lesion to fasciculus gracilis or fasciculus cuneatus results in ipsilateral loss of vibration and proprioception (position sense). The loss of the spinothalamic tract leads to pain and temperature sensation being lost from the contralateral side beginning one or two segments below the lesion. At the lesion site, all sensory modalities are lost on the same side, and an ipsilateral flaccid paralysis.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 128
Incorrect
-
The blood-brain barrier is a membrane that separates the circulating blood from the brain extracellular fluid in the central nervous system (CNS). Which of the following statements regarding the blood– brain barrier is CORRECT?
Your Answer: It allows 5-hydroxytryptophan (5HT) to cross to a very limited degree
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 129
Incorrect
-
upon stroking the plantar surface of a patient's foot, extension of toes was noted. This is likely to be accompanied with:
Your Answer: Hypotonia
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 130
Incorrect
-
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: 20 ml
Correct Answer: 100 ml
Explanation:By Fick’s principle, VO2 = Q × (CA (O2) − CV (O2)) where VO2 = O2 consumption, Q = cardiac output and CA(O2) and CV(O2) are arterial and mixed venous O2 content respectively. Thus, in the given problem, 300 ml O2/min = Q × (20−15) ml O2/100 ml. Thus, Q = 6000 ml blood/min. Then, we can calculate stroke volume by dividing the cardiac output with heart rate. Thus, stroke volume = 6000 ml/min divided by 60/min stroke volume = 100 ml.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 131
Incorrect
-
The anatomical dead space in a patient with low oxygen saturation, is 125 ml, with a tidal volume of 500 ml and pa(CO2) of 40 mm Hg. The dead space was determined by Fowler's method. If we assume that the patient's lungs are healthy, what will his mixed expired CO2 tension [pE(CO2)] be?
Your Answer: 0 mmHg
Correct Answer: 30 mmHg
Explanation:According to Bohr’s equation, VD/VT = (pA(CO2) − pE(CO2))/pA(CO2), where pE(CO2) is mixed expired CO2 and pA(CO2) is alveolar CO2pressure. Normally, the pa(CO2) is virtually identical to pA(CO2). Thus, VD/VT = (pa(CO2)) − pE(CO2)/pa(CO2). By Fowler’s method, VD/VT= 0.25. In the given problem, (pa(CO2) − pE(CO2)/pa(CO2) = (40 − pE(CO2)/40 = 0.25. Thus, pE(CO2) = 30 mmHg. If there is a great perfusion/ventilation inequality, pE(CO2) could be significantly lower than 30 mm Hg, and the patient’s physiological dead space would exceed the anatomical dead space.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 132
Correct
-
Which portion of the renal tubule absorbs amino acids and glucose?
Your Answer: Proximal convoluted tubule
Explanation:In relation to the morphology of the kidney as a whole, the convoluted segments of the proximal tubules are confined entirely to the renal cortex. Glucose, amino acids, inorganic phosphate and some other solutes are reabsorbed via secondary active transport in the proximal renal tubule through co-transport channels driven by the sodium gradient.
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This question is part of the following fields:
- Physiology
- Renal
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Question 133
Incorrect
-
Which of the following can occur even in the absence of brainstem co-ordination?
Your Answer: Defecation
Correct Answer: Gastric emptying
Explanation:Although gastric emptying is under both neural and hormonal control, it does not require brainstem co-ordination. Increased motility of the orad stomach (decreased distensibility) or of the distal stomach (increased peristalsis), decreased pyloric tone, decreased duodenal motility or a combination of these, all increase the rate of gastric emptying. The major control mechanism for gastric emptying is through duodenal gastric feedback. The duodenum has receptors for the presence of acid, carbohydrate, fat and protein digestion products, osmolarity different from that of plasma, and distension. Activating these receptors decreases the rate of gastric emptying. Neural mechanisms involve both enteric and vagal pathways and a vagotomy impairs the gastric emptying regulation. CCK (cholecystokinin) slows gastric emptying at physiological levels of the hormone. Gastrin, secretin and glucose-1-phosphate also slow gastric emptying, but require higher doses.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 134
Incorrect
-
Which is the site of action of the drug omeprazole?
Your Answer: Cl–/HCO3 – exchange
Correct Answer: H+/K+ ATPase
Explanation:H+/K+-ATPase or ‘proton pump’ located in the canalicular membrane plays a major role in acid secretion. The ATPase here is magnesium-dependent. Omeprazole is a proton pump inhibitor and blocks H+/K+- ATPase.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 135
Incorrect
-
QT interval in the electrocardiogram of a healthy individual is normally:
Your Answer:
Correct Answer: 0.40 s
Explanation:QT interval extends from beginning of the QRS complex till the end of he T-wave and normally lasts for 0.40 s. It is important in the diagnosis of long-QT and short-QT syndrome. The QT interval varies on the basis of heart rate and may need to be corrected.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 136
Incorrect
-
In which situation is a stretch reflex such as knee jerk likely to be exaggerated?
Your Answer:
Correct Answer: In upper motor neuron lesion
Explanation:A stretch reflex is a monosynaptic reflex that causes muscle contraction in response to stretching within that muscle. The sensory apparatus in a muscle that are sensitive to stretch are the muscle spindles. The patellar (knee jerk) reflex is an example. In upper motor neuron lesions, the stretch reflexes tend to be brisk due to loss of inhibitory signals on gamma neurons through the lateral reticulospinal tract.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 137
Incorrect
-
Bile salt reuptake principally occurs where?
Your Answer:
Correct Answer: In the ileum
Explanation:90 – 95% of the bile salts are absorbed from the small intestine (mostly terminal ileum and then excreted again from the liver. This is known as the enterohepatic circulation. The entire pool recycles twice per meal and approximately 6 to 8 times per day.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 138
Incorrect
-
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:
Correct 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 139
Incorrect
-
Renin is secreted by pericytes in the vicinity of the afferent arterioles of the kidney from the juxtaglomerular cells. Plasma renin levels are decreased in patients with:
Your Answer:
Correct Answer: Primary aldosteronism
Explanation:Primary aldosteronism, also known as primary hyperaldosteronism or Conn’s syndrome, is excess production of the hormone aldosterone by the adrenal glands resulting in low renin levels. Most patients with primary aldosteronism (Conn’s syndrome) have an adrenal adenoma. The increased plasma aldosterone concentration leads to increased renal Na+ reabsorption, which results in plasma volume expansion. The increase in plasma volume suppresses renin release from the juxtaglomerular apparatus and these patients usually have low plasma renin levels. Salt restriction and upright posture decrease renal perfusion pressure and therefore increases renin release from the juxtaglomerular apparatus. Secondary aldosteronism is due to elevated renin levels and may be caused by heart failure or renal artery stenosis.
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This question is part of the following fields:
- Physiology
- Renal
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Question 140
Incorrect
-
The chest X-ray of an 10-year-old boy, that presented with low-grade fever and cough, revealed hilar enlargement and parenchymal consolidation in the middle lobes. These X-ray findings are more typical for which of the following diagnoses?
Your Answer:
Correct Answer: Pulmonary tuberculosis
Explanation:Primary pulmonary tuberculosis is seen in patients exposed to Mycobacterium tuberculosis for the firs time. The main radiographic findings in primary pulmonary tuberculosis include homogeneous parenchymal consolidation typically in the lower and middle lobes, lymphadenopathy, miliary opacities and pleural effusion.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 141
Incorrect
-
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:
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 142
Incorrect
-
A 62-year-old man presented with a persistent cough and weight loss. Chest x-ray demonstrated widespread nodular opacities. After a bronchoalveolar lavage, atypical cells were detected. Which is the most probable diagnosis?
Your Answer:
Correct Answer: Bronchioalveolar carcinoma
Explanation:Bronchioloalveolar carcinoma (BAC) is a term used to define a particular subtype of adenocarcinoma which develops in cells near the alveoli, in the outer regions of the lungs. On a chest X-ray it can appear as a single peripheral spot or as scattered spots throughout the lungs. Symptoms include cough, haemoptysis, chest pain, dyspnoea and loss of weight.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 143
Incorrect
-
Causes of metabolic acidosis with a normal anion gap include:
Your Answer:
Correct Answer: Diarrhoea
Explanation:Excess acid intake and excess bicarbonate loss as in diarrhoea, are causes of metabolic acidosis with a normal anion gap. The other conditions all result in an increased anion gap.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 144
Incorrect
-
A 30-year-old man is brought to the emergency department suffering from extreme dehydration, and subsequent hypotension and tachycardia. Which part of the kidney will compensate for this loss?
Your Answer:
Correct Answer: Collecting ducts
Explanation:The collecting duct system of the kidney consists of a series of tubules and ducts that physically connect nephrons to a minor calyx or directly to the renal pelvis. The collecting duct system is the last component of the kidney to influence the body’s electrolyte and fluid balance. In humans, the system accounts for 4–5% of the kidney’s reabsorption of sodium and 5% of the kidney’s reabsorption of water. At times of extreme dehydration, over 24% of the filtered water may be reabsorbed in the collecting duct system.
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This question is part of the following fields:
- Physiology
- Renal
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Question 145
Incorrect
-
A 62-year-old woman presented to the doctor complaining of spine pain, fatigue and oliguria. She is diagnosed with chronic renal failure. Dipstick testing shows no protein, glucose, nitrite or ketones but a semi-quantitative sulphosalicylic acid test for urine protein is positive. Which of the following is the most probable cause of chronic renal failure in this patient.
Your Answer:
Correct Answer: Multiple myeloma
Explanation:Dipstick results are negative because the proteins found in the urine of this patient are not albumin but Bence Jones proteins. A Bence Jones protein is a monoclonal globulin protein commonly detected in patients affected by multiple myeloma. Multiple myeloma is a malignancy of plasma cells characterised by the production of monoclonal immunoglobulin. Symptoms include bone pain, bone fractures, bleeding, neurologic symptoms, fatigue, frequent infections and weight loss.
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This question is part of the following fields:
- Pathology
- Renal
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Question 146
Incorrect
-
The presence of oval fat bodies in the urine is most likely to be seen in which of the following conditions?
Your Answer:
Correct Answer: Nephrotic syndrome
Explanation:Nephrotic syndrome is associated with the presence of oval fat bodies on urinalysis due to increased proteinuria and lipiduria.
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This question is part of the following fields:
- Pathology
- Renal
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Question 147
Incorrect
-
Which condition presents with a positive urine dipstick test for blood, but no blood cells on urine microscopy?
Your Answer:
Correct 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 148
Incorrect
-
A 26-year-old female patient had the following blood report: RBC count = 4. 0 × 106/μl, haematocrit = 27% and haemoglobin = 11 g/dl, mean corpuscular volume (MCV) = 90 fl, mean corpuscular haemoglobin concentration (MCHC) = 41 g/dl. Further examination of blood sample revealed increased osmotic fragility of the erythrocytes. Which of the following is the most likely cause of this patient’s findings?
Your Answer:
Correct Answer: Spherocytosis
Explanation:Spherocytes are small rounded RBCs. It is due to an inherited defect of the RBC cytoskeleton membrane tethering proteins. Membrane blebs form that are lost over time and cells become round instead of biconcave. As it is a normochromic anaemia, the MCV is normal. it is diagnosed by osmotic fragility test which reveals increased fragility in a hypotonic solution.
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This question is part of the following fields:
- General
- Physiology
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Question 149
Incorrect
-
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:
Correct 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 150
Incorrect
-
After surgery, a patient developed a stitch granuloma . Which leukocyte in the peripheral blood will become an activated macrophage in this granuloma?
Your Answer:
Correct Answer: Monocyte
Explanation:Monocytes are leukocytes that protect the body against infections and move to the site of infection within 8-12 hours to deal with it. They are produced in the bone marrow and shortly after being produced are released into the blood stream where they circulate until an infection is detected. When called upon they leave the circulation and transform into macrophages within the tissue fluid and thus gain the capability to phagocytose the offending substance. Monocyte count is part of a complete blood picture. Monocytosis is the state of excess monocytes in the peripheral blood and may be indicative of various disease states. Examples of processes that can increase a monocyte count include: • chronic inflammation • stress response • hyperadrenocorticism • immune-mediated disease • pyogranulomatous disease • necrosis • red cell regeneration.
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This question is part of the following fields:
- General
- Physiology
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Question 151
Incorrect
-
A 40-year old lady with a flail chest due to trauma was breathing with the help of a mechanical ventilator in the ICU, and was heavily sedated on muscle relaxants. Due to sudden power failure, a nurse began to hand-ventilate the patient with a Ambu bag. What change will occur in the following parameters: (Arterial p(CO2), pH) in the intervening period between power failure and hand ventilation?
Your Answer:
Correct Answer: Increase, Decrease
Explanation:Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. Respiratory acidosis can be acute or chronic. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg). The given problem represents acute respiratory acidosis and thus, will show a increase in arterial p(CO2) and decrease in pH.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 152
Incorrect
-
What is the role of factor VII in coagulation?
Your Answer:
Correct Answer: Initiates the process of coagulation in conjunction with tissue factor
Explanation:The main role of factor VII is to initiate the process of coagulation along with tissue factor (TF). TF is found in the blood vessels and is not normally exposed to the bloodstream. When a vessel is injured tissue factor is exposed to blood and circulating factor VII. Factor VII is converted to VIIa by TF.
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This question is part of the following fields:
- General
- Physiology
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Question 153
Incorrect
-
The proximal tubule is the portion of the ductal system of the nephron of the kidney which leads from Bowman's capsule to the loop of Henle. Which of the following is most likely to be seen in a sample of fluid leaving the proximal tubule?
Your Answer:
Correct Answer: It will have no amino acids
Explanation:The proximal tubule is the portion of the duct system of the nephron leading from Bowman’s capsule to the loop of Henlé. The most distinctive characteristic of the proximal tubule is its brush border (or ‘striated border’). The luminal surface of the epithelial cells of this segment of the nephron is covered with densely packed microvilli forming a border which greatly increases the luminal surface area of the cells, presumably facilitating their reabsorptive function. Glucose, amino acids, inorganic phosphate, and some other solutes are100% reabsorbed via secondary active transport through co-transporters driven by the sodium gradient out of the nephron.
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This question is part of the following fields:
- Physiology
- Renal
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Question 154
Incorrect
-
What is the normal duration of the ST segment?
Your Answer:
Correct Answer: 0.08 s
Explanation:The ST segment lies between the QRS complex and the T-wave. The normal duration of the ST segment is 0.08 s. ST-segment elevation or depression may indicate myocardial ischaemia or infarction.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 155
Incorrect
-
Rapid eye movement (REM) sleep is likely to be affected by a lesion in the:
Your Answer:
Correct Answer: Pons
Explanation:Rapid eye movement (REM) sleep is also known as paradoxical sleep, as the summed activity of the brain’s neurons is quite similar to that during waking hours. Characterised by rapid movements of the eyes, most of the vividly recalled dreams occur during this stage of sleep. The total time of REM sleep for an adult is about 90–120 min per night.
Certain neurones in the brainstem, known as REM sleep-on cells, which are located in the pontine tegmentum, are particularly active during REM sleep and are probably responsible for its occurrence. The eye movements associated with REM are generated by the pontine nucleus with projections to the superior colliculus and are associated with PGO (pons, geniculate, occipital) waves.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 156
Incorrect
-
The mechanism of action of streptokinase involves:
Your Answer:
Correct Answer: Direct conversion of plasminogen to plasmin
Explanation:Streptokinase is an enzyme that is produced by group A beta haemolytic streptococcus and is an effective and cost efficient method for the dissolution of a clot used in cases of MI and pulmonary embolism. It works by directly converting plasminogen to plasmin which breaks down the blood components in the clot and fibrin, dissolving the clot. Streptokinase is a bacterial product and thus the body will develop immunity against it.
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This question is part of the following fields:
- General
- Physiology
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Question 157
Incorrect
-
A neonate with failure to pass meconium is being evaluated. His abdomen is distended and X-ray films of the abdomen show markedly dilated small bowel and colon loops. The likely diagnosis is:
Your Answer:
Correct Answer: Aganglionosis in the rectum
Explanation:Hirschsprung’s disease (also known as aganglionic megacolon) leads to colon enlargement due to bowel obstruction by an aganglionic section of bowel that starts at the anus. A blockage is created by a lack of ganglion cells needed for peristalsis that move the stool. 1 in 5000 children suffer from this disease, with boys affected four times more commonly than girls. It develops in the fetus in early stages of pregnancy. Symptoms include not having a first bowel movement (meconium) within 48 hours of birth, repeated vomiting and a swollen abdomen. Two-third of cases are diagnosed within 3 months of birth. Some children may present with delayed toilet training and some might not show symptoms till early childhood. Diagnosis is by barium enema and rectal biopsy (showing lack of ganglion cells).
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 158
Incorrect
-
A 62-year-old male smoker, presented with shortness of breath, chronic cough and haemoptysis over the last three months. He has developed a fat pad in the base of his neck, rounded face, acne and osteoporosis. Which of the following is the most likely pulmonary disease that is causing these symptoms and findings?
Your Answer:
Correct Answer: Small-cell anaplastic carcinoma
Explanation:Small cell lung cancer is a highly aggressive form of lung cancer. It is thought to originate from neuroendocrine cells in the bronchus called Feyrter cells and is often associated to ectopic production of hormones like ADH and ACTH that result in paraneoplastic syndromes and Cushing’s syndrome.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 159
Incorrect
-
Which of the following will show decreased hearing when tested by air conduction but normal hearing when tested by bone conduction?
Your Answer:
Correct Answer: Fibrosis causing fixation of the ossicles
Explanation:As the cochlea is embedded into bone, the vibrations from the bone are transmitted directly to the fluid in the cochlea. Hence, any damage to the ossicles or tympanic membrane will not show an abnormal result on bone conduction test.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 160
Incorrect
-
Which of the following is a likely cause of jaundice?
Your Answer:
Correct Answer: Hepatic disease if plasma albumin is low and serum aminotransferase elevations > 500 units
Explanation:Jaundice can occur due to any of the possible causes and treatment depends upon diagnosing the correct condition. Mild hyperbilirubinemia with normal levels of aminotransferase and alkaline phosphatase is often unconjugated (e.g., due to haemolysis or Gilbert’s syndrome rather than hepatobiliary disease). Moderate or severe hyperbilirubinemia along with increased urinary bilirubin (bilirubinuria), high alkaline phosphatase or aminotransferase levels suggest hepatobiliary disease. Hyperbilirubinemia produced by any hepatobiliary disease is largely conjugated. In this case, other blood tests include hepatitis serology for suspected hepatitis, prothrombin time (PT) or international normalised ratio (INR), albumin and globulin levels, and antimitochondrial antibody levels (suspected primary biliary cirrhosis). Low albumin and high globulin levels suggest chronic rather than acute liver disease. In cases where there is only a an elevation of alkaline phosphatase, γ-glutamyl transpeptidase (GGT) levels should be checked – the levels of which will be found high in hepatobiliary disease, but not in bone disorder which can also lead to elevated alkaline phosphatase levels. In diseases of hepatobiliary origin, aminotransferase elevations > 500 units suggest a hepatocellular cause, whereas disproportionate increases of alkaline phosphatase (e.g., alkaline phosphatase > 3 times normal and aminotransferase < 200 units) suggest cholestasis. Because hepatobiliary disease alone rarely causes bilirubin levels > 30 mg/dl, higher levels are suggestive of a combination of severe hepatobiliary disease and haemolysis or renal dysfunction. Imaging is best for diagnosing infiltrative and cholestatic causes of jaundice. Liver biopsy is rarely needed, but can be of use in intrahepatic cholestasis and in some types of hepatitis.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 161
Incorrect
-
If a 55-year old gentleman who has suffered a stroke, develops a tremor in his fingers that worsens on reaching for an object, what part of his brain is likely to be involved?
Your Answer:
Correct Answer: Cerebellum
Explanation:The cerebellum plays an important role in the integration of sensory perception and motor output. Multiple neural pathways link the cerebellum with the motor cortex and the spinocerebellar tract. The cerebellum uses the constant feedback on body position to fine-tune the movements and integrates these pathways. The patient described here has a characteristic cerebellar tremor that is a slow, broad tremor of the extremities and occurs at the end of a purposeful movement.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 162
Incorrect
-
Where is factor VIII predominantly synthesised?
Your Answer:
Correct Answer: Vascular endothelium
Explanation:Factor VIII is an important part of the coagulation cascade. Deficiency causes haemophilia A. It is synthesised predominantly by the vascular endothelium and is not affected by liver disease. In the circulation it is bound to von Willebrand factor and it forms a stable complex with it. It is activated by thrombin or factor Xa and acts as a co factor to factor IXa to activate factor X which is a co factor to factor Va. Thrombin cleaves fibrinogen in fibrin and forms a meshwork to trap RBC and platelets to form a clot.
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This question is part of the following fields:
- General
- Physiology
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Question 163
Incorrect
-
Different portions of the renal tubule have varying degrees of water permeability. Which of the following renal sites is characterised by low water permeability under normal circumstances?
Your Answer:
Correct Answer: Thick ascending limb of the loop of Henlé
Explanation:Within the nephron of the kidney, the ascending limb of the loop of Henle is a segment of the loop of Henle downstream of the descending limb, after the sharp bend of the loop. Both the thin and the thick ascending limbs of the loop of Henlé have very low permeability to water. Since there are no regulatory mechanisms to alter its permeability, it remains poorly permeable to water under all circumstances. Sodium and chloride are transported out of the luminal fluid into the surrounding interstitial spaces, where they are reabsorbed. Water must remain behind because it is not reabsorbed, so the solute concentration becomes less and less (the luminal fluid becomes more dilute). This is one of the principal mechanisms (along with diminution of ADH secretion) for the production of a dilute, hypo-osmotic urine (water diuresis).
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This question is part of the following fields:
- Physiology
- Renal
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Question 164
Incorrect
-
Gastric acid secretion is stimulated by which of the following?
Your Answer:
Correct Answer: Gastrin
Explanation:Gastric acid secretion is stimulated by three factors:
– Acetylcholine, from parasympathetic neurones of the vagus nerve that innervate parietal cells directly
– Gastrin, produced by pyloric G-cells
– Histamine, produced by mast cells.
Gastric acid is inhibited by three factors:
– Somatostatin
– Secretin
– Cholecystokinin
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 165
Incorrect
-
A 65-year old patient with altered bowl movement experienced the worsening of shortness of breath and exertional chest pains over the course of 8 weeks. Examination shows pallor and jugular venous distension. Furthermore, a test of the stool for occult blood is positive. Laboratory studies show:
Haemoglobin 7.4 g/dl
Mean corpuscular volume 70 fl Leukocyte count 5400/mm3
Platelet count 580 000/mm3 Erythrocyte sedimentation 33 mm/h
A blood smear shows hypochromic, microcytic RBCs with moderate poikilocytosis. Which of the following is the most likely diagnosis?Your Answer:
Correct Answer: Iron deficiency anaemia
Explanation:Iron deficiency anaemia is the most common type of anaemia. It can occur due to deficiency of iron due to decreased intake or due to faulty absorption. An MCV less than 80 will indicated iron deficiency anaemia. On the smear the RBC will be microcytic hypochromic and will also show piokilocytosis. iron profiles tests are important to make a diagnosis. Clinically the patient will be pale and lethargic.
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This question is part of the following fields:
- General
- Physiology
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Question 166
Incorrect
-
What causes a reduction in pulmonary functional residual capacity?
Your Answer:
Correct Answer: Pulmonary fibrosis
Explanation:Pulmonary functional residual capacity (FRC) is = volume of air present in the lungs at the end of passive expiration.
Obstructive diseases (e.g. emphysema, chronic bronchitis, asthma) = an increase in FRC due to an increase in lung compliance and air trapping.
Restrictive diseases (e.g. pulmonary fibrosis) result in stiffer, less compliant lungs and a reduction in FRC.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 167
Incorrect
-
Lung compliance is increased by:
Your Answer:
Correct Answer: Emphysema
Explanation:Lung compliance is increased by emphysema, acute asthma and increasing age and decreased by alveolar oedema, pulmonary hypertension, atelectasis and pulmonary fibrosis.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 168
Incorrect
-
Hyperplastic arteriosclerosis with fibrinoid necrosis, petechial haemorrhages, microinfarcts in the kidneys and elevated plasma renin are common findings in which of the following patients?
Your Answer:
Correct Answer: A 45-year-old woman with scleroderma
Explanation:Scleroderma, also known as systemic sclerosis, is a chronic disease of the connective tissue. Involvement of the kidneys occurs in patients with diffuse scleroderma, causing rapid onset of high blood pressure with hyperreninemia, thrombotic microangiopathy, and progressive renal failure.
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This question is part of the following fields:
- Pathology
- Renal
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Question 169
Incorrect
-
Fine-needle aspiration is a type of biopsy procedure. When performing a fine-needle aspiration of the lungs, which is the most common complication of the procedure?
Your Answer:
Correct Answer: Pneumothorax
Explanation:Pneumothorax is the most common complication of a fine-needle aspiration procedure. Various factors, such as lesion size, have been associated with increased risk of pneumothorax .
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 170
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:
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 171
Incorrect
-
A 45-year old gentleman presents with diarrhoea for two weeks. He has no history of fever and the diarrhoea stops on fasting. Which is the most likely type of diarrhoea that he is suffering from?
Your Answer:
Correct Answer: Osmotic
Explanation:The different types of diarrhoea are:
1. Secretory diarrhoea – Due to increased secretion or decreased absorption. There is minimal to no structural damage in this type. The most common cause is cholera toxin which stimulates secretion of anions (especially chloride), with sodium and water.
2. Osmotic diarrhoea – Due to increased osmotic load, there is water loss. This occurs in cases of maldigestion syndromes, such as coeliac or pancreatic disease.
3. Motility-related diarrhoea – Occurs in cases of abnormal gastrointestinal motility. Due to increased motility, there is poor absorption and this leads to diarrhoea. This is seen post-vagotomy or in diabetic neuropathy.
4. Inflammatory diarrhoea – Due to damage to the mucosa or brush border, there is a loss of protein-rich fluids and poor absorption. Features of all the above three types can be seen in this type. Aetiology includes bacterial, viral, parasitic infections or autoimmune problems including inflammatory bowel disease.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 172
Incorrect
-
Mechanical distortion, and not K+ channels are responsible for distortion of which of the following structures?
Your Answer:
Correct Answer: Pacinian corpuscle
Explanation:Pacinian corpuscles are a type of mechanoreceptor, sensitive to deep pressure, touch and high-frequency vibration. The Pacinian corpuscles are ovoid and about 1 mm long. In the centre of the corpuscle is the inner bulb, which is a fluid-filled cavity with a single afferent unmyelinated nerve ending. Any deformation in the corpuscle causes the generation of action potentials by opening of pressure-sensitive sodium ion channels in the axon membrane. This allows influx of sodium ions, creating a receptor potential (independent of potassium channels).
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This question is part of the following fields:
- Neurology
- Physiology
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Question 173
Incorrect
-
In a hypertensive patient with secondary hyperaldosteronism, aldosterone is released mainly in response to:
Your Answer:
Correct Answer: Angiotensin II
Explanation:Secondary hyperaldosteronism in hypertension is either due to primary renin overproduction by the kidneys or renin overproduction secondary to decreased renal blood flow. The main stimulus for aldosterone release are adrenocorticotrophic hormone (ACTH), angiotensin II and high plasma K+ levels. Low plasma Na+ might also stimulate the adrenal cortex. Fluid overload will reduce aldosterone secretion. Atrial natriuretic peptide is secreted under conditions of expanded extracellular volume and will not lead to aldosterone secretion.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 174
Incorrect
-
Pain in the right upper quadrant of the abdomen on ingestion of a fatty meal is seen in a condition which involves which of the following substances?
Your Answer:
Correct Answer: Cholecystokinin
Explanation:The clinical scenario described here favours the presence of gallstones. During food ingestion, vagal discharges stimulate gallbladder contraction. Moreover, presence of fat and amino acids in the intestinal lumen stimulates the release of cholecystokinin (CCK) in the duodenum. This causes sustained gallbladder contraction and relaxation of the sphincter of Oddi. If gallstones are present, there will be inflammation in the gallbladder and CCK will aggravate it due to contractions.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 175
Incorrect
-
Which part of the nephron would have to be damaged to stop the reabsorption of the majority of salt and water?
Your Answer:
Correct Answer: Proximal tubule
Explanation:The proximal tubule is the portion of the duct system of the nephron of the kidney which leads from Bowman’s capsule to the loop of Henle. It is conventionally divided into the proximal convoluted tubule (PCT) and the proximal straight tubule (PST). The proximal tubule reabsorbs the majority (about two-thirds) of filtered salt and water. This is done in an essentially iso-osmotic manner. Both the luminal salt concentration and the luminal osmolality remain constant (and equal to plasma values) along the entire length of the proximal tubule. Water and salt are reabsorbed proportionally because the water is dependent on and coupled with the active reabsorption of Na+. The water permeability of the proximal tubule is high and therefore a significant transepithelial osmotic gradient is not possible.
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This question is part of the following fields:
- Physiology
- Renal
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Question 176
Incorrect
-
An experiment was conducted in which the skeletal muscle protein (not smooth muscle) involved in contraction was selectively inhibited. Which protein was inhibited?
Your Answer:
Correct Answer: Troponin
Explanation:The mechanism of contraction of smooth muscles is different from that of skeletal muscles in which the contractile protein is troponin whilst in smooth muscle contraction is a protein called calmodulin. Calmodulin reacts with calcium ions and stimulates the formation of myosin crossbridges.
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This question is part of the following fields:
- General
- Physiology
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Question 177
Incorrect
-
A patient came into the emergency in a state of shock. His blood group is not known, but on testing it clotted when mixed with Type A antibodies. Which blood should be transfused?
Your Answer:
Correct Answer: B +ve
Explanation:There are two stages to determine the blood group, known as ABO typing. The first stage is called forward typing. In this method, RBCs are mixed with two separate solutions of type A or type B antibodies to see if they agglutinate. If this blood clumps, this indicates the presence of antigens within the blood sample. For example, a sample of type B blood will clump when tested with type A antibodies as it contains type B antigens. Group B – has only the B antigen on red cells (and A antibody in the plasma)
Group B – has only the B antigen on red cells (and A antibody in the plasma)
Group AB – has both A and B antigens on red cells (but neither A nor B antibody in the plasma)
Group O – has neither A nor B antigens on red cells (but both A and B antibody are in the plasma). Many people also have a Rh factor on the red blood cell’s surface. This is also an antigen and those who have it are called Rh+. Those who have not are called Rh–. A person with Rh– blood does not have Rh antibodies naturally in the blood plasma (as one can have A or B antibodies, for instance) but they can develop Rh antibodies in the blood plasma if they receive blood from a person with Rh+ blood, whose Rh antigens can trigger the production of Rh antibodies. A person with Rh+ blood can receive blood from a person with Rh– blood without any problems. The patient’s blood group is B positive as he has antigen B, antibody A and Rh antigens.
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This question is part of the following fields:
- General
- Physiology
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Question 178
Incorrect
-
The gradual depolarization in-between action potentials in pacemaker tissue is a result of?
Your Answer:
Correct Answer: A combination of gradual inactivation outward IK along with the presence of an inward ‘funny’ current (If) due to opening of channels permeable to both Na+ and K+ ions
Explanation:One of the characteristic features of the pacemaker cell is the generation of a gradual diastolic depolarization also called the pacemaker potential. In phase 0, the upstroke of the action potential caused by an increase in the Ca2+ conductance, an influx of calcium occurs and a positive membrane potential is generated. The next is phase 3 which is repolarization caused by increased K+ conductance as a result of outwards K+ current. Phase 4 is a slow depolarization which accounts for the pacemaker activity, caused by increased conductance of Na+, inwards Na+ current called IF. it is turned on by repolarization.
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This question is part of the following fields:
- General
- Physiology
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Question 179
Incorrect
-
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:
Correct 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 180
Incorrect
-
A neurotransmitter of the nigrostriatal pathway is:
Your Answer:
Correct Answer: Dopamine
Explanation:Dopamine acts as a neurotransmitter in the brain, activating dopamine receptors. It is also a neurohormone released from the hypothalamus. It plays an important role in the reward system. It is believed that dopamine provides a teaching signal to parts of the brain responsible for acquiring new motor sequences (behaviours), by activation of dopamine neurons when an unexpected reward is presented. Loss of dopamine neurones in the nigrostriatal pathway causes Parkinson’s disease. In the frontal lobes, dopamine controls the flow of information from other areas of the brain, and thus, dopamine disorders in this region can cause a decline in neurocognitive functions, especially memory, attention and problem solving. Reduced dopamine concentrations in the prefrontal cortex are thought to contribute to attention-deficit disorder and some symptoms of schizophrenia. Dopamine is also the primary neuroendocrine regulator of the secretion of prolactin from the anterior pituitary gland. Dopamine is also commonly associated with the pleasure system of the brain. This plays a key role in understanding the mechanism of action of drugs (such as cocaine and the amphetamines), which seem to be directly or indirectly related to the increase of dopamine.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 181
Incorrect
-
what is the cause of a prolonged PT(prothrombin time)?
Your Answer:
Correct Answer: Liver disease
Explanation:PT measure the intrinsic pathway of coagulation. It determines the measure of the warfarin dose regime, liver disease and vit K deficiency status along with the clotting tendency of blood. PT measured factors are II,V,VII,X and fibrinogen. It is used along with aPTT which measure the intrinsic pathway.
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This question is part of the following fields:
- General
- Physiology
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Question 182
Incorrect
-
Abnormal breathing is noticed in a of victim of a road traffic accident, who sustained a head injury. The breathing pattern is characterised by alternate periods of waxing and waning tidal volumes with interspersed periods of apnoea. This breathing pattern is known as:
Your Answer:
Correct Answer: Cheyne–Stokes breathing
Explanation:Cheyne-Stokes breathing is an abnormal breathing pattern with breathing periods of gradually waxing and waning tidal volumes, with apnoeic periods interspersed. It is usually the first breathing pattern to be seen with a rise in intracranial pressure and is caused by failure of the respiratory centre in the brain to compensate quickly enough to changes in serum partial pressure of oxygen and carbon dioxide. The aetiology includes strokes, head injuries, brain tumours and congestive heart failure. It is also a sign of altitude sickness in normal people, a symptom of carbon monoxide poisoning or post-morphine administration. Biot’s respiration (cluster breathing) is characterized by cluster of quick, shallow inspirations followed by regular or irregular periods of apnoea. It is different from ataxic respiration, which has completely irregular breaths and pauses. It results due to damage to the medulla oblongata by any reason (stroke, uncal herniation, trauma) and is a poor prognostic indicator. Kussmaul breathing, also known as ‘air hunger’, is basically respiratory compensation for metabolic acidosis and is characterized by quick, deep and laboured breathing. It is most often seen in in diabetic ketoacidosis. Due to forced inspiratory rate, the patients will show a low p(CO2). Ondine’s curse is congenital central hypoventilation syndrome or primary alveolar hypoventilation, which can be fatal and leads to sleep apnoea. It involves an inborn failure to control breathing autonomically during sleep and in severe cases, can affect patients even while awake. It is known to occur in 1 in 200000 liveborn children. Treatment includes tracheostomies and life long mechanical ventilator support.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 183
Incorrect
-
Dysarthria, nystagmus and a tremor worsening with directed movement are likely to be seen in:
Your Answer:
Correct Answer: Cerebellar disease
Explanation:The given symptoms are seen in diseases affecting the cerebellum. A cerebellar tremor is a slow tremor that occurs at the end of a purposeful movement. It is seen in cerebellar disease, such as multiple sclerosis or some inherited degenerative disorders and chronic alcoholism. Classically, tremors are produced in the same side of the body as a one-sided lesion. Cerebellar disease can also result in a wing-beating’ type of tremor called rubral or Holmes’ tremor – a combination of rest, action and postural tremors. Other signs of cerebellar disease include dysarthria (speech problems), nystagmus (rapid, involuntary rolling of the eyes), gait problems and postural tremor of the trunk and neck.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 184
Incorrect
-
A teenage Somalian boy presents with a complaint of an enlarged lower jaw. His blood film shows blast cells and macrophages. Which virus is responsible for this?
Your Answer:
Correct Answer: Epstein–Barr virus
Explanation:Burkitt’s lymphoma is a type of non-Hodgkin’s lymphoma. Histologically it is characterised by a starry sky appearance due to numerous neoplastic macrophages which are required to clear the rapidly dividing tumour cells/blast cells. Burkitt’s lymphoma commonly affects the jaw bone, forming a huge tumour mass. It is associated with translocation of c-myc gene and has three types: 1) endemic/African type, 2)sporadic and 3)immunodeficiency-associated. The first type is strongly associated with EBV.
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This question is part of the following fields:
- General
- Physiology
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Question 185
Incorrect
-
A patient presents with loss of pain and temperature sensation in the left leg. He is likely to have a lesion involving:
Your Answer:
Correct Answer: Right lateral spinothalamic tract
Explanation:The spinothalamic tract is a sensory pathway originating in the spinal cord that transmits information to the thalamus. There are two main parts of the spinothalamic tract: the lateral spinothalamic tract transmits pain and temperature and the anterior spinothalamic tract transmits touch (crude touch). The decussation of this pathway occurs at the level of the spinal cord. Hence, a unilateral lesion of the lateral spinothalamic tract causes contralateral loss of pain and temperature.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 186
Incorrect
-
A 50-year old gentleman who suffered a stroke was brought to the emergency department by his relatives. The patient however denied the presence of paralysis of his left upper and lower limbs. What is the most likely site of the lesion in this patient?
Your Answer:
Correct Answer: Right posterior parietal cortex
Explanation:A large injury to the non-dominant parietal cortex can make the patient neglect or refuse to acknowledge the presence of paralysis on the contralateral side. This can also involve the perception of the external world. Smaller injuries in this area which involve the precentral gyrus (primary motor cortex) or postcentral gyrus (primary sensory cortex) cause contralateral spastic paralysis or contralateral loss of tactile sensation respectively. A lesion in posterior inferior gyrus of the dominant frontal lobe results in motor aphasia. Involvement of the posterior superior gyrus of the dominant frontal lobe produces sensory aphasia.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 187
Incorrect
-
A syndrome responsible for failure to absorb vitamin B12 from the GIT is called?
Your Answer:
Correct Answer: Pernicious anaemia
Explanation:Pernicious anaemia is a type of autoimmune disease in which antibodies form against the parietal cells or intrinsic factor. Intrinsic factor is required for the absorption of vitamin B12. Blood testing typically shows a macrocytic, normochromic anaemia and low levels of serum vitamin B12. A Schilling test can then be used to distinguish between pernicious anaemia, vitamin B12 malabsorption and vitamin B12 deficiency. Symptoms include shortness of breath, pallor and diarrhoea etc.
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This question is part of the following fields:
- General
- Physiology
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Question 188
Incorrect
-
Which of the following is true about myasthenia gravis?
Your Answer:
Correct Answer: Response of skeletal muscle to nerve stimulation is weakened
Explanation:An autoimmune disorder, myasthenia gravis leads to progressive muscle weakness. It occurs due to formation of antibodies against the nicotinic acetylcholine (ACh) receptor of the motor endplate, which leads to impaired neuromuscular transmission. Thus, nerve stimulation will lead to a weakened muscle response, but direct electrical stimulation will bring about a normal response. Diagnostic test includes improvement of muscle weakness by small doses of acetylcholinesterase inhibitors (physostigmine or edrophonium). However, a large dose of physostigmine worsens the weakness due to desensitisation of the endplate to persistent Ach. One of the investigative tools includes radiolabelled snake venom α-bungarotoxin. It is an in vitro study performed on muscle biopsy specimens and used to quantify the number of ACh receptors at the motor endplate.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 189
Incorrect
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Which of the following will be affected by a lesion in the posterior column-medial lemniscus system?
Your Answer:
Correct Answer: Fine touch
Explanation:The posterior column–medial lemniscus (PCML) pathway is a sensory pathway that transmits fine touch and conscious proprioceptive information from the body to the brain. As the posterior columns are also known as dorsal columns, the pathway is also called the dorsal column–medial lemniscus system or DCML.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 190
Incorrect
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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:
Correct 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 191
Incorrect
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A 54-year-old woman is re-admitted to the hospital with shortness of breath and sharp chest pain 2 weeks after surgical cholecystectomy. The most probable cause of these clinical findings is:
Your Answer:
Correct Answer: Pulmonary embolus
Explanation:Pulmonary embolism is caused by the sudden blockage of a major lung blood vessel, usually by a blood clot. Symptoms include sudden sharp chest pain, cough, dyspnoea, palpitations, tachycardia or loss of consciousness. Risk factors for developing pulmonary embolism include long periods of inactivity, recent surgery, trauma, pregnancy, oral contraceptives, oestrogen replacement, malignancies and venous stasis.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 192
Incorrect
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How much blood can the pulmonary vessels of a 45-year-old healthy man accommodate when he is at rest?
Your Answer:
Correct Answer: 500 ml
Explanation:Pulmonary circulation is the portion of the cardiovascular system which carries deoxygenated blood away from the heart, to the lungs, and returns oxygenated blood back to the heart. The vessels of the pulmonary circulation are very compliant (easily distensible) and so typically accommodate about 500 ml of blood in an adult man. This large lung blood volume can serve as a reservoir for the left ventricle, particularly during periods when left ventricular output momentarily exceeds venous return.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 193
Incorrect
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A 50 year old woman presented with excessive bleeding after an inguinal hernia repair. Labs are suggestive of a primary haemostasis defect. Deficiency of which of the following is most likely to cause it?
Your Answer:
Correct Answer: Platelets
Explanation:Primary haemostatic control means the first line of defence against immediate bleeding. This is carried out by the platelets. They immediately form a haemostatic plug at the site of injury. Coagulation starts within 20s after an injury to the blood vessel which damage the endothelial cells. Secondary haemostasis follows which includes activation of the coagulation factors to form fibrin strands which mesh together forming the platelet plug. Platelets interact with platelet collagen receptor, glycoprotein Ia/IIa and to collagen fibres in the vascular endothelium. This adhesion is mediated by von Willebrand factor (vWF), which forms links between the platelet glycoprotein Ib/IX/V and collagen fibrils. The platelets are then activated and release the contents of their granules into the plasma, in turn activating other platelets and white blood cells.
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This question is part of the following fields:
- General
- Physiology
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Question 194
Incorrect
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A 49-year-old woman with acute renal failure has a total plasma [Ca2+] = 2. 5 mmol/l and a glomerular filtration rate of 160 l/day. What is the estimated daily filtered load of calcium?
Your Answer:
Correct Answer: 240 mmol/day
Explanation:Calcium is the most abundant mineral in the human body. The average adult body contains in total approximately 1 kg of calcium of which 99% is in the skeleton in the form of calcium phosphate salts. The extracellular fluid (ECF) contains approximately 22 mmol, of which about 9 mmol is in the plasma. About 40% of total plasma Ca2+ is bound to proteins and not filtered at the glomerular basement membrane. Therefore, the estimated daily filtered load is 1.5 mmol/l × 160 l/day = 240 mmol/day. The exact amount of free versus total Ca2+ depends on the blood pH: free Ca2+ increases during acidosis and decreases during alkalosis.
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This question is part of the following fields:
- Physiology
- Renal
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Question 195
Incorrect
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With respect to far accommodation, which of the following is a TRUE statement?
Your Answer:
Correct Answer: The ciliary muscles are relaxed
Explanation:Myopia or near-sightedness is a disease due to elongated eyeballs or too strong a lens. For far accommodation (focus of a distant object onto the retina), the lens needs to decrease its refractive power, or in other words, increase its focal length. This is done by relaxation of ciliary muscles which tightens the zonular fibres and flattening of the lens. Relaxation of the zonular fibres, rounding of the lens, shortening of the focal length and constriction of the pupil occurs during near accommodation.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 196
Incorrect
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If your EEG shows waves with a frequency range of 8-12 Hz, the waves most likely to be seen are:
Your Answer:
Correct Answer: Alpha
Explanation:Electroencephalography (EEG) is the neurophysiological measurement of the electrical activity of the brain. It is done by placing electrodes on the scalp or subdurally. In reality, the electrical currents are not measured, but rather the voltage differences between different parts of the brain. Four major types of EEG activity are recognized, which are alpha, beta, delta and theta.
Alpha waves, also known as Berger’s waves ranges in frequency from 8-12 Hz. Best detected with eyes closed, alpha waves are characteristic of a relaxed, alert state of consciousness. An alpha-like normal variant called mu is sometimes seen over the motor cortex (central scalp) and attenuates with movement or, rather, with the intention to move.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 197
Incorrect
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Question 198
Incorrect
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Which of the following is true regarding factor XI?
Your Answer:
Correct Answer: Deficiency causes haemophilia C
Explanation:Factor XI is also known as plasma thromboplastin and is one of the enzymes of the coagulation cascade. It is produced in the liver and is a serine protease. It is activated by factor XIIa, thrombin and by itself. Deficiency of factor XI causes the rare type of haemophilia C. Low levels of factor XI also occur in other disease states, including Noonan syndrome. High levels of factor XI have been seen in thrombosis.
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This question is part of the following fields:
- General
- Physiology
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Question 199
Incorrect
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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:
Correct 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 200
Incorrect
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What is a major source of fuel being oxidised by the skeletal muscles of a man who has undergone starvation for 7 days?
Your Answer:
Correct Answer: Serum fatty acids
Explanation:Starvation is the most extreme form of malnutrition. Prolonged starvation can lead to permanent organ damage and can be fatal. Starved individuals eventually lose significant fat and muscle mass as the body uses these for energy.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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SESSION STATS - PERFORMANCE PER SPECIALTY
