-
Question 1
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: Somatostatin
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.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 2
Incorrect
-
Which of the following has the highest content of triglycerides?
Your Answer: VLDL
Correct Answer: Chylomicron
Explanation:Created by the small intestinal cells, chylomicrons are large lipoprotein molecules which transport lipids to the liver, adipose, cardiac and skeletal tissue. Chylomicrons are mainly composed of triglycerides (,85%) along with some cholesterol and cholesteryl esters. Apo B-48 is the main apolipoprotein content.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 3
Incorrect
-
What is the mostly likely cause of prolonged activated partial thromboplastin time (aPPT) ?
Your Answer: Poor factor VII synthesis
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 4
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.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 5
Incorrect
-
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: Sodium
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.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 6
Correct
-
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: 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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 7
Incorrect
-
Which of the following can occur even in the absence of brainstem co-ordination?
Your Answer: Swallowing
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.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 8
Correct
-
Bile salt reuptake principally occurs where?
Your 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.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 9
Incorrect
-
Which of the following is a true statement regarding the pupil?
Your Answer: Increase in sympathetic activity in fibres innervating the inner eye muscles during darkness results in pupil constriction
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.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 10
Incorrect
-
Which of the following substances is most likely to cause pulmonary vasodilatation?
Your Answer: Catecholamines
Correct 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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 11
Incorrect
-
A 59-year-old woman with hyperaldosteronism is prescribed a diuretic. Which of the following diuretics promotes diuresis by opposing the action of aldosterone?
Your Answer: Thiazide
Correct Answer: Potassium-sparing diuretic
Explanation:The term potassium-sparing refers to an effect rather than a mechanism or location. Potassium-sparing diuretics act by either antagonising the action of aldosterone (spironolactone) or inhibiting Na+ reabsorption in the distal tubules (amiloride). This group of drugs is often used as adjunctive therapy, in combination with other drugs, for the management of chronic heart failure. Spironolactone, the first member of the class, is also used in the management of hyperaldosteronism (including Conn’s syndrome) and female hirsutism (due to additional antiandrogen actions).
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 12
Incorrect
-
Which of the following variables are needed to calculate inspiratory reserve volume of a patient?
Your Answer: Vital capacity and residual volume
Correct 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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 13
Incorrect
-
During pregnancy the uterus enlarges however after delivery it regresses to its original size. Which of the following organelles is responsible for this regression?
Your Answer: Golgi apparatus
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 14
Incorrect
-
A 23 year old woman is Rh -ve and she delivered a baby with a Rh+ blood group. What measure can be performed to prevent Rh incompatibility in the next pregnancy?
Your Answer: Anti IgM
Correct Answer: Immunoglobulin D
Explanation:Rh disease is also known as erythroblastosis fetalis and is a disease of the new-born. In mild states it can cause anaemia with reticulocytosis and in severe forms causes severe anaemia, morbus hemolytcus new-born and hydrops fetalis. RBCs of the Rh+ baby can cross the placenta and enter into the maternal blood. As she is Rh- her body will form antibodies against the D antigen which will pass through the placenta in subsequent pregnancies.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 15
Incorrect
-
C5a (a complement component) is a potent?
Your Answer: Opsonin
Correct Answer: Anaphylotoxin
Explanation:C5a is a strong chemoattractant as well as an anaphylotoxin and is involved in the recruitment of inflammatory cells such as neutrophils, eosinophils, monocytes, and T lymphocytes. It is also involved in activation of phagocytic cells, release of granule-based enzymes and generation of oxidants. All of which contribute to innate immune functions.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 16
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: Fasciculations
Correct Answer: Spasticity
Explanation:An upper motor neuron lesion affects the neural pathway above the anterior horn cell or motor nuclei of the cranial nerves, whereas a lower motor neurone lesion affects nerve fibres travelling from the anterior horn of the spinal cord to the relevant muscles. An upper motor neurone lesions results in the following:
– Spasticity in the extensor muscles (lower limbs) or flexor muscles (upper limbs).
– ‘clasp-knife’ response where initial resistance to movement is followed by relaxation
– Weakness in the flexors (lower limbs) or extensors (upper limbs) with no muscle wasting
– Brisk tendon jerk reflexes
– Positive Babinski sign (on stimulation of the sole of the foot, the big toe is raised rather than curled downwards)
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 17
Incorrect
-
Gastric acid secretion is stimulated by which of the following?
Your Answer: Cholecystokinin
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
-
This question is part of the following fields:
- Endocrine
- Physiology
-
-
Question 18
Incorrect
-
Increased resistance to flow of blood in cerebral vessels is most likely seen in:
Your Answer: Reduction of pa(O2) to < 50 mmHg
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 19
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 20
Incorrect
-
Decreased velocity of impulse conduction through the atrioventricular node (AV node) in the heart will lead to:
Your Answer:
Correct 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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 21
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).
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 22
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.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 23
Incorrect
-
When the pitch of a sound increases, what is the physiological response seen in the listener?
Your Answer:
Correct Answer: The location of maximal basilar membrane displacement moves toward the base of the cochlea
Explanation:An increase in the frequency of sound waves results in a change in the position of maximal displacement of the basilar membrane in the cochlea. Low pitch sound produces maximal displacement towards the cochlear apex and greatest activation of hair cells there. With an increasing pitch, the site of greatest displacement moves towards the cochlear base. However, increased amplitude of displacement, increase in the number of activated hair cells, increased frequency of discharge of units in the auditory nerve and increase in the range of frequencies to which such units respond, are all seen in increases in the intensity or a sound stimulus.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 24
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:
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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 25
Incorrect
-
Carbon dioxide is principally transported in the blood in which form?
Your Answer:
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.
-
This question is part of the following fields:
- Physiology
- Respiratory; Cardiovascular
-
-
Question 26
Incorrect
-
Which of the following muscles aid in inspiration?
Your Answer:
Correct Answer: Diaphragm and external intercostals
Explanation:The diaphragm and external intercostals are muscles of inspiration as they increase the volume of thoracic cavity and reduce the intrathoracic pressure. Muscles of expiration include abdominal muscles and internal intercostals.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 27
Incorrect
-
A 30 year old female suffered from mismatched transfusion induced haemolysis. Which substance will be raised in the plasma of this patient?
Your Answer:
Correct Answer: Bilirubin
Explanation:Bilirubin is a yellow pigment that is formed due to the break down of RBCs. Haemolysis results in haemoglobin that is broken down into a haem portion and globin which is converted into amino acids and used again. Haem is converted into unconjugated bilirubin in the macrophages and shunted to the liver. In the liver it is conjugated with glucuronic acid making it water soluble and thus excreted in the urine. Its normal levels are from 0.2-1 mg/dl. Increased bilirubin causes jaundice and yellowish discoloration of the skin.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 28
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.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 29
Incorrect
-
Which of the following substances brings about a dilatation of the gastrointestinal resistance vessels?
Your Answer:
Correct 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.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 30
Incorrect
-
A 55-year old gentleman presented to the doctor with worsening dysphagia for both solids and liquids over 6 months. This was associated with regurgitation of undigested food and occasional chest pain. Barium swallow revealed distal oesophageal dilatation with lack of peristalsis in the distal two-third oesophagus. The likely diagnosis is:
Your Answer:
Correct Answer: Achalasia
Explanation:Achalasia is an oesophageal motility disorder where inappropriate contractions in the oesophagus lead to reduced peristalsis and failure of the lower oesophageal sphincter to relax properly in response to swallowing. Classical triad of symptoms include dysphagia to fluids followed by solids, chest pain and regurgitation of undigested food. Other symptoms include belching, hiccups, weight loss and cough. Diagnosis is by:
– X-ray with a barium swallow or oesophagography : narrowing at the gastroesophageal junction (‘bird/parrot beak’ or ‘rat tail’ appearance) and various degrees of mega-oesophagus (oesophageal dilatation) as the oesophagus is gradually stretched by retained food. Effectiveness of treatment can be measured with a 5-minute timed barium swallow.
– Manometry – probe measures the pressure waves in different parts of oesophagus and stomach while swallowing.
– Endoscopy
– CT scan to exclude other causes like malignancy
– Pathological examination showing defect in the nerves which control oesophageal motility (myenteric plexus).
In Chagas disease, there is destruction of ganglion cells by Trypanosoma cruzi.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 31
Incorrect
-
During a normal respiratory exhalation, what is the recoil alveolar pressure?
Your Answer:
Correct Answer: +10 cmH2O
Explanation:To determine compliance of the respiratory system, changes in transmural pressures (in and out) immediately across the lung or chest cage (or both) are measured simultaneously with changes in lung or thoracic cavity volume. Changes in lung or thoracic cage volume are determined using a spirometer with transmural pressures measured by pressure transducers. For the lung alone, transmural pressure is calculated as the difference between alveolar (pA; inside) and intrapleural (ppl; outside) pressure. To calculate chest cage compliance, transmural pressure is ppl (inside) minus atmospheric pressure (pB; outside). For the combined lung–chest cage, transmural pressure or transpulmonary pressure is computed as pA – pB. pA pressure is determined by having the subject deeply inhale a measured volume of air from a spirometer. Under physiological conditions the transpulmonary or recoil pressure is always positive; intrapleural pressure is always negative and relatively large, while alveolar pressure moves from slightly negative to slightly positive as a person breathes.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 32
Incorrect
-
Which of the following is NOT a nutritional factor involved in wound healing:
Your Answer:
Correct Answer: Vitamin B3
Explanation:Vitamin B6 is required for collagen cross-links.
Vitamin A is required for epithelial cell proliferation.
Zinc is required for RNA and DNA synthesis.
Copper is required for cross-linking of collagen.
-
This question is part of the following fields:
- Cell Injury & Wound Healing
- Physiology
-
-
Question 33
Incorrect
-
Evaluation of a 60-year old gentleman, who has been a coal miner all his life and is suspected to have pulmonary fibrosis reveals the following: normal FEV1, arterial oxygen saturation 92%, alveolar ventilation 6000 ml/min at a tidal volume of 600 ml and a breathing rate of 12 breaths/min. There are also pathological changes in lung compliance and residual volume. Which of the following is most accurate about his residual volume?
Your Answer:
Correct Answer: Cannot be measured directly with a spirometer
Explanation:Residual volume is the air left in the lungs after maximal expiration is done. Thus, this is not a part of vital capacity and cannot be measured with a spirometer directly. It can be measured by the methods such as body plethysmography or inert gas dilution. Expiratory reserve volume is vital capacity minus inspiratory capacity. Resting volume of lungs is he sum of residual volume and expiratory reserve volume. Lungs recoil inward until the recoil pressure becomes zero, which corresponds to a volume significantly lower than residual volume.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 34
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 35
Incorrect
-
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 36
Incorrect
-
Which antibiotic acts by inhibiting protein synthesis?
Your Answer:
Correct 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
-
This question is part of the following fields:
- Pharmacology; Microbiology
- Physiology
-
-
Question 37
Incorrect
-
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:
Correct Answer: 7.28, 55, 81
Explanation:Hypoventilation (or respiratory depression) causes an increase in carbon dioxide (hypercapnia) and respiratory acidosis. It can result due to drugs such as alcohol, benzodiazepines, barbiturates, opiates, mechanical conditions or holding ones breath. Strong opioids such as heroin and fentanyl are commonly implicated and can lead to respiratory arrest. In recreational drug overdose, acute respiratory acidosis occurs with an increase in p(CO2) over 45 mm Hg and acidaemia (pH < 7.35)
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 38
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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 39
Incorrect
-
A patient is diagnosed with Conn’s syndrome. Aldosterone is secreted from where?
Your Answer:
Correct Answer: Zona glomerulosa of the adrenal cortex
Explanation:The adrenal gland comprises an outer cortex and an inner medulla, which represent two developmentally and functionally independent endocrine glands.
The adrenal medulla secretes adrenaline (70%) and noradrenaline (30%)
The adrenal cortex consists of three layers (remembered by the mnemonic GFR):
G = zona glomerulosa – secretes aldosterone
F = zona fasciculata – secretes cortisol and sex steroids
R = zona reticularis – secretes cortisol and sex steroids.
Aldosterone facilitates the reabsorption of sodium and water and the excretion of potassium and hydrogen ions from the distal convoluted tubule and collecting ducts. Conn’s syndrome is characterized by increased aldosterone secretion from the adrenal glands.
-
This question is part of the following fields:
- Endocrine
- Physiology
-
-
Question 40
Incorrect
-
Which of these substances is secreted by pericytes in the juxtaglomerular cells?
Your Answer:
Correct Answer: Renin
Explanation:The juxtaglomerular cells synthesise, store and secrete the enzyme renin in the kidney. They are specialised smooth muscle cells in the wall of the afferent arteriole that delivers blood to the glomerulus and thus play a critical role in the renin– angiotensin system and so in renal autoregulation.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 41
Incorrect
-
A 34-year-old woman is diagnosed with cerebral oedema after suffering a severe head trauma. Which of the following conditions is not likely to be associated with the extracellular oedema?
Your Answer:
Correct Answer: Increased plasma colloid osmotic pressure
Explanation:Cerebral oedema is extracellular fluid accumulation in the brain. Increased capillary permeability, increased capillary pressure, increased interstitial fluid colloid osmotic pressure and lymphatic blockage would increase fluid movement into the interstitial spaces. Increased plasma colloid osmotic pressure, however, would oppose fluid movement from the capillaries into the interstitial compartment.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 42
Incorrect
-
Skeletal muscle fibres are divided into two basic types, type I (slow-twitch fibres) and type II (fast-twitch fibres). Fast muscle fibres do which of the following:
Your Answer:
Correct Answer: Use anaerobic metabolism
Explanation:Skeletal muscles are divided into two types:
1) type I also known as the slow twitch fibres. They use oxygen for their metabolism and as a result they have a high endurance potential. To support this they have abundant mitochondria and myoglobin, so they appear red/dark.
2) type II fibres also called fast twitch fibres, are low endurance fibres used during anaerobic metabolism. They are required for short bursts of strength and cannot sustain contractions for long periods of time.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 43
Incorrect
-
Destruction of the ventromedial nucleus of the hypothalamus will result in:
Your Answer:
Correct Answer: Loss of satiety
Explanation:The ventromedial nucleus of the hypothalamus is divided into an anterior and a superior part. The anterior part controls the female sexual drive, whereas the superior part is responsible for satiety. Destruction of the superior part of the nucleus will result in overeating, as no signal tells the body that it is satisfied.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 44
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 45
Incorrect
-
Which of the following is the cause of flattened (notched) T waves on electrocardiogram (ECG)?
Your Answer:
Correct Answer: Hypokalaemia
Explanation:The T-wave is formed due to ventricular repolarisation. Normally, it is seen as a positive wave. It can be normally inverted (negative) in V1 (occasionally in V2-3 in African-Americans/Afro-Caribbeans). Hyperacute T-waves are the earliest ECG change of acute myocardial infarction. ECG findings of hyperkalaemia include high, tent-shaped T-waves, a small P-wave and a wide QRS complex. Hypokalaemia results in flattened (notched) T-waves, U-waves, ST-segment depression and prolonged QT interval.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 46
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) = 80–100 fl, mean corpuscular haemoglobin concentration (MCHC) = 31–37 g/dl. Which of the following is correct regarding this patient’s erythrocytes:
Your Answer:
Correct Answer: Normal MCV
Explanation:MCV is the mean corpuscular volume and it is calculated from the haematocrit and the RBC count. It is normally 90 fl. Mean corpuscular haemoglobin concentration (MCHC) [g/dl] = haemoglobin [g/dl]/haematocrit = 11/0.27 = 41 g/dl and is higher than normal range (32 to 36 g/dL).
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 47
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.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 48
Incorrect
-
A 56 year old gentleman, who is a chronic smoker presents to the clinic with dyspnoea. His Chest X-ray shows intercostal space widening with increased blackening bilaterally. What is the most likely finding on his pulmonary function test?
Your Answer:
Correct Answer: Increased total lung capacity
Explanation:In patients suspected of having chronic obstructive pulmonary disease, pulmonary function testing (PFT) is useful to confirm airway obstruction, to quantify its severity and reversibility, for following disease progression and monitoring response to treatment. These tests include:
FEV1 – volume of air forcefully expired during the first second after a full breath
Forced vital capacity (FVC) – total volume of air expired with maximal force and flow-volume loops. The hallmark of airway obstruction is reduction of FEV1, FVC and the ratio of FEV1/FVC, with a concave pattern in expiratory tracing on the flow-volume loop. FEV1 and forced vital capacity (FVC) are easily measured with office spirometry and are useful to assess the severity of disease. Other parameters include increased total lung capacity, functional residual capacity and residual volume, which can help distinguish chronic obstructive pulmonary disease (COPD) from restrictive pulmonary disease where these values are lower than normal. Along with these, other tests are decreased vital capacity; and decreased diffusing capacity for carbon monoxide (DLCO). DLCO is non-specific and can be low in other disorders that affect the pulmonary vascular bed, such as interstitial lung disease. DLCO is however useful to distinguish COPD from asthma, in which DLCO is normal or elevated.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 49
Incorrect
-
Extracellular body fluid as compared with intracellular body fluid:
Your Answer:
Correct Answer: Is relatively rich in glucose
Explanation:The percentages of body water contained in various fluid compartments add up to total body water (TBW). This water makes up a significant fraction of the human body, both by weight and by volume. Ensuring the right amount of body water is part of fluid balance, an aspect of homeostasis. The extracellular fluid (ECF) includes all fluids outside the cells. This fluid can be divided into three fluid departments: interstitial (in the tissue spaces) fluid, blood plasma and lymph, and specialised compartments called transcellular fluid. The extracellular fluid surrounds all the cells in the body and is in equilibrium with the intracellular fluid. So, its composition must remain fairly constant even though substances are passing into and out of the cells. The interstitial fluid, though called a fluid, is in a reality a gel-like composition made up of: water, proteoglycan molecules and collagen. The extracellular fluid constitutes 40% of total body water, with intracellular fluid making up the remaining 60%. It is relatively rich in glucose.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 50
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 51
Incorrect
-
A 31-year-old woman is diagnosed with adrenal hyperplasia, and laboratory samples are taken to measure serum aldosterone and another substance. Which is most likely to be the other test that was prescribed to this patient?
Your Answer:
Correct Answer: Plasma renin
Explanation:The evaluation of a patient in whom hyperaldosteronism is first to determine that hyperaldosteronism is present (serum aldosterone) and, if it is present, to differentiate primary from secondary causes of hyperaldosteronism. The aldosterone-to-renin ratio (ARR) is the most sensitive means of differentiating primary from secondary causes of hyperaldosteronism as it is abnormally increased in primary hyperaldosteronism, and decreased or normal but with high renin levels in secondary hyperaldosteronism.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 52
Incorrect
-
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:
Correct 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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 53
Incorrect
-
Calculate the pulmonary vascular resistance in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer:
Correct Answer: 2.0 resistance units (mmHg/l per min)
Explanation:Pulmonary vascular resistance (PVR) = (Mean pulmonary artery pressure – Pulmonary capillary wedge pressure) divided by Cardiac output. To get cardiac output, Fick’s principle needs to be applied which states that VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CA = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24–0.16, CO = 500/0.8, CO = 6.25 l/min. To calculate mean pulmonary artery pressure, we use the formula: Mean pulmonary artery pressure = Diastolic pressure + 1/3(Systolic pressure – Diastolic pressure). Thus, Mean pulmonary artery pressure = 15 + 1/3(25 – 15) = 15 + 3. 33 = 18.33. Substituting these values in the first formula, PVR = 18.3–5/6.25 = 13.5/5.25 = 2.0 resistance units (mmHg/l per min) approximately.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 54
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.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 55
Incorrect
-
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:
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.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 56
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:
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 57
Incorrect
-
Work of breathing (WOB) is the energy expended to inhale and exhale a breathing gas. Normally, maximal amount of work of breathing is required to overcome:
Your Answer:
Correct Answer: Elastic lung compliance
Explanation:The forces of elastance (compliance), frictional resistance and inertia have been identified as the forces that oppose lung inflation and deflation. The normal relaxed state of the lung and chest is partially empty. Further exhalation requires muscular work. Inhalation is an active process requiring work. About 60–66% of the total work performed by the respiratory muscles is used to overcome the elastic or compliance characteristics of the lung–chest cage, 30–35% is used to overcome frictional resistance and only 2–5% of the work is used for inertia.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 58
Incorrect
-
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:
Correct 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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 59
Incorrect
-
A 30 year old female presented in the emergency with an irregular pulse. Her ECG showed absent P-waves with irregular RR interval. What is the most likely diagnosis?
Your Answer:
Correct Answer: Atrial fibrillation
Explanation:Atrial fibrillation is one of the most common cardiac arrhythmias. It is often asymptomatic but may present with symptoms of palpitations, fainting, chest pain and heart failure. Characteristic findings are: absence of P-waves, unorganised electrical activity in their place, irregularity of RR interval due to irregular conduction of impulses to the ventricles and if paroxysmal AF is suspected, episodes may be documented with the use of Holter monitoring
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 60
Incorrect
-
A 60 year old patient with a history of carcinoma of the head of the pancreas, and obstructive jaundice presents with a spontaneous nose bleed and easy bruising. What is the most likely reason for this?
Your Answer:
Correct Answer: Vitamin-K-dependent clotting factors deficiency
Explanation:Vitamin K is a fat soluble vitamin requiring fat metabolism to function properly to allow for its absorption. People with obstructive jaundice develop vitamin k deficiency as fat digestion is impaired. Vit K causes carboxylation of glutamate residue and hence regulates blood coagulation including: prothrombin (factor II), factors VII, IX, X, protein C, protein S and protein Z.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 61
Incorrect
-
Atractyloside is an inhibitor of electron transport chain. It is expected to have little or no effect on the functioning of which of the following cell types?
Your Answer:
Correct Answer: Red blood cells
Explanation:Electron transport chain is a series of electron carriers that are embedded in the mitochondrial membrane. It is the place where ATP is made. Inhibiting the electron transport chain will stop production of ATP. Red blood cells are the only cell in the given option which do not contain ATP.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 62
Incorrect
-
A patient presents with loss of fine touch and sense of proprioception in the lower part of the body (below T6). He is likely to have a lesion involving:
Your Answer:
Correct Answer: Gracile nucleus
Explanation:The gracile nucleus is located in the medulla oblongata and is one of the dorsal column nuclei involved in the sensation of fine touch and proprioception. It contains second-order neurons of the dorsal column–medial lemniscus system, that receive inputs from sensory neurones of the dorsal root ganglia and send axons that synapse in the thalamus.
The gracile nucleus and fasciculus carry epicritic, kinaesthetic and conscious proprioceptive information from the lower part of the body (below the level of T6 in the spinal cord). Similar information from the upper part of body (above T6, except for face and ear) is carried by the cuneate nucleus and fasciculus. The information from face and ear is carried by the primary sensory trigeminal nucleus.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 63
Incorrect
-
Diffusion is the movement of molecules from a region of high concentration to a region of low concentration. Which of these changes will decrease the rate of diffusion of a substance?
Your Answer:
Correct Answer: An increase in the molecular weight of the substance
Explanation:Unless given IV, a drug must cross several semipermeable cell membranes before it reaches the systemic circulation. Drugs may cross cell membranes by diffusion, amongst other mechanisms. The rate of diffusion of a substance is proportional to the difference in the concentration of the diffusing substance between the two sides of the membrane, the temperature of the solution, the permeability of the membrane and, in the case of ions, the electrical potential difference between the two sides of the membrane.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 64
Incorrect
-
Which organ is responsible for the secretion of enzymes that aid in digestion of complex starches?
Your Answer:
Correct Answer: Pancreas
Explanation:α-amylase is secreted by the pancreas, which is responsible for hydrolysis of starch, glycogen and other carbohydrates into simpler compounds.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 65
Incorrect
-
Which type of contractions are responsible for the propulsion of chyme along the small intestine?
Your Answer:
Correct Answer: Segmentation
Explanation:Two major types of intestinal contractions are segmentation and peristalsis:
Segmentation occurs most frequently and primarily involves circular muscle. It is essentially a contraction of 2- or 3-cm long intestinal segments while the muscle on either side of it relaxes. Chyme in the segment is displaced in both directions. As the contracted segment relaxes, the previously relaxed segments on either side may contract. This efficiently mixes the chyme with the digestive secretions and exposes the mucosal absorptive surface to the luminal contents. It also serves a propulsive function and contributes to the movement of chyme.
Peristalsis is a propulsive wave of contraction that is initiated by intestinal distension. It is short lived and travels only a few centimetres before dying out. The combined effects of intestinal peristalsis and segmentation provide for both adequate mixing of the intestinal contents and slow, steady movement of chyme.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 66
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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 67
Incorrect
-
Action potentials are used extensively by the nervous system to communicate between neurones and muscles or glands. What happens during the activation of a nerve cell membrane?
Your Answer:
Correct Answer: Sodium ions flow inward
Explanation:During the generation of an action potential, the membrane gets depolarized which cause the voltage gated sodium channels to open and sodium diffuses inside the neuron, resulting in the membrane potential moving towards a positive value. This positive potential will then open the voltage gated potassium channels and cause more K+ to move out decreasing the membrane potential and restoring the membrane potential to its resting value.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 68
Incorrect
-
What is the normal amount of oxygen that is carried in the blood?
Your Answer:
Correct Answer: 20 ml oxygen/100 ml blood
Explanation:Normally, 100 ml of blood contains 15g haemoglobin and a single gram of haemoglobin can bind to 1.34 ml oxygen when 100% saturated. Thus, 15 × 1.34 = 20 ml O2/100 ml blood. The haemoglobin in venous blood that is leaving the tissues is about 75% saturated with oxygen, and hence it carries about 15 ml O2/100 ml venous blood. This implies that for each 10 ml of blood, 5 ml oxygen is transported to the tissues. With a p(O2) > 100 mm Hg, only 3 ml of oxygen is dissolved in every one litre of plasma. By increasing the pA(O2) by breathing 100% oxygen, one can add an extra amount of oxygen in the plasma, but the amount of oxygen carried by haemoglobin will not increase significantly as it is already > 95% saturated.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 69
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 70
Incorrect
-
Which is the site of action of the drug omeprazole?
Your Answer:
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.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 71
Incorrect
-
Nephrotic syndrome is a condition that causes proteinuria, hypoalbuminemia and oedema. Which of the following is the cause of the oedema in these patients?
Your Answer:
Correct Answer: Decreased oncotic pressure
Explanation:The glomeruli of the kidneys are the parts that normally filter the blood. They consist of capillaries that are fenestrated and allow fluid, salts and other small solutes to flow through, but normally not proteins. In nephrotic syndrome, the glomeruli become damaged allowing small proteins, such as albumin to pass through the kidneys into urine. Oedema usually occurs due to salt and water retention by the diseased kidneys as well as due to the reduced colloid oncotic pressure (because of reduced albumin in the plasma). Lower serum oncotic pressure causes fluid to accumulate in the interstitial tissues.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 72
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:
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 73
Incorrect
-
What is the normal glomerular filtration rate?
Your Answer:
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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 74
Incorrect
-
Post-total gastrectomy, there will be a decreased production of which of the following enzymes?
Your Answer:
Correct Answer: Pepsin
Explanation:Pepsin is a protease that is released from the gastric chief cells and acts to degrade proteins into peptides. Released as pepsinogen, it is activated by hydrochloric acid and into pepsin itself. Gastrin and the vagus nerve trigger the release of pepsinogen and HCl when a meal is ingested. Pepsin functions optimally in an acidic environment, especially at a pH of 2.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 75
Incorrect
-
Arterial blood gas analysis of a man admitted with acute exacerbation of chronic obstructive pulmonary disease (COPD) showed the following: pH = 7.28, p(CO2) = 65.5 mmHg, p(O2)= 60 mmHg and standard bicarbonate = 30.5 mmol/l. This patient had:
Your Answer:
Correct Answer: Respiratory acidosis
Explanation:Acidosis with high p(CO2) and normal standard bicarbonate indicates respiratory acidosis, commonly seen in acute worsening of COPD patients. Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with a low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg).
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 76
Incorrect
-
A 32-year old gentleman came to the emergency department, complaining of progressively increasing weakness in his arms and legs over 5 days. On examination, there is symmetrical weakness on both sides of his face, along with weakness of the proximal and distal muscles of all four limbs. No loss of sensation noted. Deep tendon reflexes could not be elicited and plantar responses were downward. On enquiry, it was revealed that he had an upper respiratory tract infection 10 days ago. The likely diagnosis is:
Your Answer:
Correct Answer: Guillain–Barré syndrome
Explanation:Guillain–Barré syndrome (GBS) is an acute, autoimmune polyradiculoneuropathy which affects the peripheral nervous system and is usually triggered by an acute infectious process. 75% patients have a history of acute infection within the past 1–4 weeks, usually respiratory or gastrointestinal. immunisations have also been implicated. The most common form is acute inflammatory demyelinating polyneuropathy. It results in an ascending paralysis with complete loss of deep tendon reflexes. Treatment includes immunoglobulins and supportive care. However, the disease may be fatal due to severe pulmonary complications and dysautonomia.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 77
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 78
Incorrect
-
After having donated a unit of blood. The blood bank will prefer to use which of the following anticoagulants to store the blood?
Your Answer:
Correct Answer: Citrate
Explanation:Calcium is necessary for coagulation to occur. Citrate being a chelator and combining with calcium ions to form un-ionised compound will prevent coagulation. Following transfusion the citrate is removed by the liver with in a few minutes. Oxalate also works on the same principle but it is toxic to the body.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 79
Incorrect
-
Question 80
Incorrect
-
A medical student is told a substance is freely filtered but is not metabolised, secreted, or stored in the kidney. It has a plasma concentration of 1000 mg/l and its urine excretion rate is 25 mg/min, and the inulin clearance is 100 ml/min. What is the rate of tubular reabsorption of the substance?
Your Answer:
Correct Answer: 75 mg/min
Explanation:Reabsorption or tubular reabsorption is the process by which the nephron removes water and solutes from the tubular fluid (pre-urine) and returns them to the circulating blood. To calculate the reabsorption rate of substance Z we use the following equation: excretion = (filtration + secretion) – reabsorption. As this substance is freely filtered, its filtration rate is equal to that of inulin. So 25 = (100 + 0) – reabsorption. Reabsorption = 100 – 25 therefore reabsorption = 75 mg/min.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 81
Incorrect
-
Renal function is an indication of the state of the kidney, measured by glomerular filtration rate (GFR). In a healthy person, GFR would be greatly increased by:
Your Answer:
Correct Answer: Substantial increases in renal blood flow
Explanation:An increase in the rate of renal blood flow (RBF) greatly increases the glomerular filtration rate (GFR). The more plasma available (from increased RBF), the more filtrate is formed. Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal (kidney) glomerular capillaries into the Bowman’s capsule per unit time. Central to the physiologic maintenance of GFR is the differential basal tone of the afferent and efferent arterioles.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 82
Incorrect
-
Which of the following will increase blood pressure and cause hypokalaemia?
Your Answer:
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 83
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 84
Incorrect
-
A TRUE statement regarding abolition of the cephalic phase of pancreatic secretion is that it:
Your Answer:
Correct Answer: Will result after vagotomy
Explanation:Recognition and integration of the sight, smell and taste of food triggers the cephalic phase of pancreatic secretion, causing an increase in pancreatic HCO3- and enzyme secretion. The degree of enzyme secretion in this phase is about 50% of the maximal response seen with exogenous CCK and secretin. The vagus nerve regulates the secretion through the cholinergic fibres innervating the acinar cells of the pancreas, and through peptidergic nerve fibres, which innervate ductal cells.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 85
Incorrect
-
Acute respiratory distress syndrome (ARDS) is a medical condition that occurs in critically ill patients, and can be triggered by events such as trauma and sepsis. Which of the following variables is most likely to be lower than normal in a patient with ARDS?
Your Answer:
Correct Answer: Lung compliance
Explanation:Acute (or Adult) respiratory distress syndrome (ARDS) is a medical condition occurring in critically ill patients characterized by widespread inflammation in the lungs. The development of acute respiratory distress syndrome (ARDS) starts with damage to the alveolar epithelium and vascular endothelium, resulting in increased permeability to plasma and inflammatory cells. These cells pass into the interstitium and alveolar space, resulting in pulmonary oedema. Damage to the surfactant-producing type II cells and the presence of protein-rich fluid in the alveolar space disrupt the production and function of pulmonary surfactant, leading to micro atelectasis and impaired gas exchange. The pathophysiological consequences of lung oedema in ARDS include a decrease in lung volumes, compliance and large intrapulmonary shunts. ARDS may be seen in the setting of pneumonia, sepsis, following trauma, multiple blood transfusions, severe burns, severe pancreatitis, near-drowning, drug reactions, or inhalation injuries.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 86
Incorrect
-
When does the heart rate decrease?
Your Answer:
Correct Answer: Pressure on the eyeball
Explanation:Various vagotonic manoeuvres (e.g. Valsalva manoeuvre, carotid sinus massage, pressure on eyeballs, ice-water facial immersion, swallowing of ice-cold water) result in increased parasympathetic tone through the vagus nerve which results in a decrease in heart rate. These manoeuvres may be clinically useful in terminating supraventricular arrhythmias.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 87
Incorrect
-
The primary motor cortex is located in the:
Your Answer:
Correct Answer: Precentral gyrus
Explanation:The primary motor cortex is located in the dorsal part of the precentral gyrus and the anterior bank of the central sulcus. The precentral gyrus lies anterior to the postcentral gyrus and is separated from it by a central sulcus. Its anterior border is the precentral sulcus, while inferiorly it borders to the lateral fissure (Sylvian fissure).
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 88
Incorrect
-
A 50 year old man was admitted to the surgical ICU following a hemicolectomy for carcinoma of the caecum. A full blood count revealed: haematocrit = 30%, erythrocytes = 4 × 106/μ, haemoglobin level = 8 g/dl. To determine the likely cause of his anaemia, red blood cell indices were calculated. Which RBC indices are correct?
Your Answer:
Correct Answer: MCHC = haemoglobin concentration/haematocrit
Explanation:Mean corpuscular haemoglobin concentration (MCHC) is calculated simply by dividing the haemoglobin concentration (8 g/dl) by the haematocrit (0.3). The normal range is 31–36 g/dl. This patient has a hypochromic anaemia (MCHC = 8/0.3 = 26.7 g/dl). Dividing the haemoglobin concentration × 10 by erythrocyte number yields mean corpuscular haemoglobin (MCH). Normal range is 25.4–34.6 pg/cell and this patient has a significantly reduced cellular haemoglobin content (MCH = 8 × 10/4 = 20 pg/cell). Mean corpuscular volume (MCV) is calculated by dividing haematocrit × 1000 by erythrocyte number (4 × 106/μl). Normal range is 80–100 fl and this patient has a microcytic anaemia (MCV = 0.3 × 1000/4 = 75 fl). Microcytic, hypochromic anaemia is characteristic for iron-deficiency.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 89
Incorrect
-
A 45-year-old pregnant woman develops high blood pressure at 20 weeks. She complains of headaches and swollen feet, and a test reveals proteinuria (350 mg/day). Which of the following is the most likely diagnosis?
Your Answer:
Correct Answer: Pre-eclampsia
Explanation:Pre-eclampsia (PE) is a disorder of pregnancy characterized by the onset of high blood pressure (two separate readings taken at least 6 h apart of 140/90 or more) and often a significant amount of protein in the urine (>300 mg of protein in a 24-h urine sample). While blood pressure elevation is the most visible sign of the disease, it involves generalised damage to the maternal endothelium of the kidneys and liver, with the release of vasopressive factors only secondary to the original damage. Pre-eclampsia may develop at varying times within pregnancy and its progress differs among patients; most cases present pre-term. It has no known cure apart from ending the pregnancy (induction of labour or abortion). It may also present up to 6 weeks post partum. Risk factors for pre-eclampsia include obesity, prior hypertension, older age, and diabetes mellitus.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 90
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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 91
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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 92
Incorrect
-
During cardiac catheterisation in a 20-year old man, the following data is obtained: Pressure (mmHg), O2 saturation (%) Right atrium 7 (N = 5) 90 (N = 75), Right ventricle 35/7 (N = 25/5) 90 (N = 75), Pulmonary artery 35/8 (N = 25/15), 90 (N = 75), Left atrium 7 (N = 9) 95 (N = 95), Left ventricle 110/7 (N = 110/9) 95 (N = 95), Aorta 110/75 (N = 110/75) 95 (N = 95) where N = Normal value. What is the likely diagnosis?
Your Answer:
Correct Answer: Atrial septal defect
Explanation:A congenital heart disease, ASD or atrial septal defect leads to a communication between the right and left atria due to a defect in the interatrial septum. This leads to mixing of arterial and venous blood from the right and left side of the heart. The hemodynamic significance of this defect depends on the presence of shunting of blood. Normally, the left side of the heart has higher pressure than the right as the left side has to pump blood throughout the body. A large ASD (> 9 mm) will result in a clinically significant left-to-right shunt, causing volume overload of the right atrium and ventricle, eventually leading to heart failure. Cardiac catheterization would reveal very high oxygen saturation in the right atrium, right ventricle and pulmonary artery. Eventually, the left-to-right shunt will lead to pulmonary hypertension and increased afterload in the right ventricle, along with the increased preload due to the shunted blood. This will either cause right ventricular failure, or raise the pressure in the right side of the heart to equal or more than that in the left. Elevation of right atrial pressure to that of left atrial pressure would thus lead to diminishing or complete cessation of the shunt. If left uncorrected, there will be reversal of the shunt, known as Eisenmenger syndrome, resulting in clinical signs of cyanosis as the oxygen-poor blood form right side of the heart will mix with the blood in left side and reach the peripheral vascular system.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 93
Incorrect
-
A 38-year-old woman with end-stage renal disease, is undergoing haemodialysis. She has normocytic normochromic anaemia. What is the best treatment for her?
Your Answer:
Correct Answer: Erythropoietin
Explanation:E erythropoietin (EPO) is a hormone that is released by the kidney. It is responsible for the regulation of red blood cell production in the body. It can be made using recombinant technology and is used in the treatment of anaemia of chronic renal failure and in patients under going chemotherapy
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 94
Incorrect
-
A sudden loud sound is more likely to result in cochlear damage than a slowly developing loud sound. This is because:
Your Answer:
Correct Answer: There is a latent period before the attenuation reflex can occur
Explanation:On transmission of a loud sound into the central nervous system, an attenuation reflex occurs after a latent period of 40-80 ms. This reflex contracts the two muscles that pull malleus and stapes closer, developing a high degree of rigidity in the entire ossicular chain. This reduces the ossicular conduction of low frequency sounds to the cochlea by 30-40 decibels. In this way, the cochlea is protected from damage due to loud sounds (these are low frequency sounds) when they develop slowly.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 95
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 96
Incorrect
-
Regarding the coagulation cascade, Factor VII:
Your Answer:
Correct Answer: Is a serine protease
Explanation:Factor VII (FVII) is a zymogen for a vitamin K-dependent serine protease essential for the initiation of blood coagulation. It is synthesized primarily in the liver and circulates in plasma. Within the liver, hepatocytes are involved in the synthesis of most blood coagulation factors, such as fibrinogen, prothrombin, factor V, VII, IX, X, XI, XII, as well as protein C and S, and antithrombin, whereas liver sinusoidal endothelial cells produce factor VIII and von Willebrand factor.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 97
Incorrect
-
Which of the following is responsible for the maximum increase in total peripheral resistance on sympathetic stimulation?
Your Answer:
Correct 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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 98
Incorrect
-
Basal Metabolic Rate (BMR) will most likely be reduced by which of the following?
Your Answer:
Correct Answer: Decrease in body temperature
Explanation:The basal metabolic rate (BMR) is defined as the rate of calorie consumption after an overnight fast, in the absence of any muscular activity, with the patient in a restful state. Various factors affect the BMR including weight, body surface area and age. The BMR is 30 kcal/m2 per hour at birth; at age 2, the rate is 57 kcal/m2 per hour; and at age 20, 41 kcal/m2 per hour. After this, the BMR decreases by 10% between 20-60 years of age. Women are known to have a 10% lower BMR than men (due to higher fat content). A one-degree change in body temperature leads to a 10% change in BMR in the same direction. However, shivering and increasing ambient temperature brings about a rise in BMR, and so does stress, physical activity, caffeine, theophylline and hyperthyroidism. Also, thermogenesis induced by diet results in increased metabolic rate and hence, BMR should be ideally measured after overnight fasting.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 99
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.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 100
Incorrect
-
There are several mechanisms involved in the transport of sodium ions from blood to interstitial fluid of the muscle cells. Which of the following mechanisms best describes this phenomenon?
Your Answer:
Correct Answer: Diffusion through channels between endothelial cells
Explanation:Capillaries are the smallest of the body’s blood vessels, measuring 5–10 μm and they help to enable the exchange of water, oxygen, carbon dioxide, and many other nutrients and waste substances between the blood and the tissues surrounding them. The walls of capillaries are composed of only a single layer of cells, the endothelium. Ion channels are pore-forming proteins that help to establish and control the small voltage gradient that exists across the plasma membrane of all living cells by allowing the flow of ions down their electrochemical gradient. An ion channel is an integral membrane protein or more typically an assembly of several proteins. The archetypal channel pore is just one or two atoms wide at its narrowest point. It conducts a specific ion such as sodium or potassium and conveys them through the membrane in single file.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 101
Incorrect
-
Which of the following is likely to induce secretion of glucagon?
Your Answer:
Correct Answer: Low serum concentration of glucose
Explanation:The most potent stimulus for secretion of glucagon is hypoglycaemia whereas hyperglycaemia is a stimulus for insulin release. Glucagon secretion also occurs in response to high levels of amino acids. Somatostatin inhibits glucagon secretion. Parasympathetic stimulation increases pancreatic acinar secretion, but not of α-cells.
-
This question is part of the following fields:
- Endocrinology
- Physiology
-
-
Question 102
Incorrect
-
In a cardiac cycle, what event does the opening of the atrioventricular (AV) valves coincide with?
Your Answer:
Correct Answer: Beginning of diastole
Explanation:Cardiac diastole refers to the time period when the heart is relaxed after contraction and is preparing to refill with blood. Both ventricular and atrial diastole are together known as complete cardiac diastole. At its beginning, the ventricles relax, causing a drop in the ventricular pressure. As soon as the left ventricular pressure drops below that in left atrium, the mitral valve opens and there is ventricular filling of blood. Similarly, the tricuspid valve opens filling the right atrium.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 103
Incorrect
-
What is the reason for a deranged thrombin clotting time?
Your Answer:
Correct Answer: Heparin therapy
Explanation:Thrombic clotting time is also known as thrombin time. It is clinically performed to determine the therapeutic levels of heparin. After plasma is isolated from the blood, bovine thrombin is added to it and the time it takes from the addition to clot is recorded. The reference interval is usually <21s. deranged results are indicative of heparin therapy, hypofibrinogenemia, hyperfibrinogenaemia or lupus anticoagulant.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 104
Incorrect
-
A blood sample from a patient with polycythaemia vera will show which of the following abnormalities?
Your Answer:
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 105
Incorrect
-
A patient admitted for esophagectomy showed low levels of the lightest plasma protein in terms of weight. Which of the following is the lightest plasma protein:
Your Answer:
Correct Answer: Albumin
Explanation:Albumin is the most abundant and the lightest of all the plasma proteins. It maintains osmotic pressure, transports unconjugated bilirubin, thyroid hormones, fatty acids, drugs and acts as a buffer for pH.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 106
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 107
Incorrect
-
Production of pain is most likely associated with:
Your Answer:
Correct 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.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 108
Incorrect
-
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 109
Incorrect
-
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:
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 110
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.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 111
Incorrect
-
The blood-brain barrier is a membrane that separates the circulating blood from the brain extracellular fluid in the central nervous system (CNS). Which of the following statements regarding the blood– brain barrier is CORRECT?
Your Answer:
Correct Answer: It breaks down in areas of brain that are infected
Explanation:The blood–brain barrier is a membrane that controls the passage of substances from the blood into the central nervous system. It is a physical barrier between the local blood vessels and most parts of the central nervous system and stops many substances from travelling across it. During meningitis, the blood–brain barrier may be disrupted. This disruption may increase the penetration of various substances (including either toxins or antibiotics) into the brain. A few regions in the brain, including the circumventricular organs, do not have a blood–brain barrier.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 112
Incorrect
-
Which of the following is a likely consequence of severe diarrhoea?
Your Answer:
Correct 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.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 113
Incorrect
-
Ventricular filling follows a delay caused by?
Your Answer:
Correct 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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 114
Incorrect
-
One sensitive indicator of heavy alcohol dependence is:
Your Answer:
Correct Answer: Elevated serum gamma-glutamyl transpeptidase
Explanation:Elevated serum gamma-glutamyl transpeptidase (GGT) may be the only laboratory abnormality in patients who are dependent on alcohol. Heavy drinkers may also have an increased MCV.
-
This question is part of the following fields:
- Hepatobiliary
- Physiology
-
-
Question 115
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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 116
Incorrect
-
Which statement is correct regarding secretions from the adrenal glands?
Your Answer:
Correct Answer: Aldosterone is producd by the zona glomerulosa
Explanation:The secretions of the adrenal glands by zone are:
Zona glomerulosa – aldosterone
Zona fasciculata – cortisol and testosterone
Zona reticularis – oestradiol and progesterone
Adrenal medulia – adrenaline, noradrenaline and dopamine.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 117
Incorrect
-
In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your Answer:
Correct Answer: Exercise
Explanation:The oxygen-haemoglobin dissociation curve plots saturated haemoglobin against the oxygen tension and is usually a sigmoid plot. Each molecule of haemoglobin can bind to four molecules of oxygen reversibly. Factors that can influence the binding include: pH, concentration of 2,3-diphosphoglycerate (2,3-DPG), temperature, type of haemoglobin molecules, and presence of toxins, especially carbon monoxide. Shape of the curve is due to interaction of bound oxygen molecules with the incoming molecules. The binding of first molecule is difficult, with easier binding of the second and third molecule and increase in difficulty with the fourth molecule – partly as a result of crowding and partly as a natural tendency of oxygen to dissociate.
Left shift of curve indicates haemoglobin’s increased affinity for oxygen (seen at lungs). Right shift indicates decreased affinity and is seen with an increase in body temperature, hydrogen ions, 2,3-diphosphoglycerate (DPG), carbon dioxide concentration and exercise. Under normal resting conditions in a healthy individual, the normal position of the curve is at a pH of 7.4. A shift in the position of the curve with a change in pH is called the Bohr effect. Left shift occurs in acute alkalosis, decrease in p(CO2), decrease in temperature and decrease in 2,3-DPG. The fetal haemoglobin curve is to the left of the adult haemoglobin to allow for oxygen diffusion across the placenta. The curve for myoglobin is even further to the left. Carbon monoxide has a much higher affinity for haemoglobin than oxygen does. Thus, carbon monoxide poisoning leads to hypoxia.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 118
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 119
Incorrect
-
During strenuous exercise, what else occurs besides tachycardia?
Your Answer:
Correct Answer: Increased stroke volume
Explanation:During strenuous exercise there is an increase in:
– Heart rate, stroke volume and therefore cardiac output. (CO = HR x SV)
– Respiratory rate (hyperventilation) which will lead to a reduction in Paco2.
– Oxygen demand of skeletal muscle, therefore leading to a reduction in mixed venous blood oxygen concentration.
Renal blood flow is autoregulated, so renal blood flow is preserved and will tend to remain the same. Mean arterial blood pressure is a function of cardiac output and total peripheral resistance and will increase with exercise, mainly as a result of the increase in cardiac output that occurs.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 120
Incorrect
-
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:
Correct 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
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 121
Incorrect
-
Multiple cells were labelled using a fluorescent dye that doesn’t cross the cell membrane. One cell in the middle was bleached with a light that destroys the dye, but the cell soon recovers its stain. The presence of which structures best explains this?
Your Answer:
Correct Answer: Gap junctions
Explanation:Gap junctions are attachments between cells that permit intercellular communication e.g. they permit current flow and electrical coupling between myocardial cells. They allow direct electrical transmission among cells and also permit certain substance to pass through as well. They are either homotypic, formed by two identical hemichannels or heterotypic, formed by different hemichannels.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 122
Incorrect
-
A 78-year-old diabetic man undergoes renal function tests. Which of the following substances will be the most accurate for measuring glomerular filtration rate (GFR)?
Your Answer:
Correct Answer: Inulin
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal glomerular capillaries into the Bowman’s capsule per unit time. Clinically, this is often measured to determine renal function. Inulin was originally used as it is not reabsorbed by the kidney after glomerular filtration, therefore its rate of excretion is directly proportional to the rate of filtration of water and solutes across the glomerular filter. However, in clinical practice, creatinine clearance is used to measure GFR. Creatinine is an endogenous molecule, synthesised in the body, that is freely filtered by the glomerulus (but also secreted by the renal tubules in very small amounts). Creatinine clearance exceeds GFR due to creatinine secretion, and is therefore a close approximation of the GFR.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 123
Incorrect
-
Which of the following is a true statement regarding secretion of gastric acid?
Your Answer:
Correct Answer: Acetylcholine increases gastric acid secretion
Explanation:Gastric acid secretion is increased by acetylcholine, histamine and gastrin, with the help of cAMP as a secondary messenger. They increase H+ and Cl- secretion by increasing the number of H+/K+ ATPase molecules and Cl- channels. In contrast, gastric acid secretion is decreased by somatostatin, epidermal growth factor and prostaglandins.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 124
Incorrect
-
A young women following a road traffic accident suffered heavy blood loss and developed subsequent anaemia. Which of the following is a consequence of this?
Your Answer:
Correct Answer: A high reticulocyte count
Explanation:Anaemia refers to a decrease in the circulating levels of haemoglobin in the blood resulting in a reduced ability of the body to transport oxygen effectively. Anaemia from blood loss results in the body further compensating by releasing stored RBCs and immature RBCs from the bone marrow. Thus resulting in a high reticulocyte count.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 125
Incorrect
-
The most likely cause of prominent U waves on the electrocardiogram (ECG) of a patient is:
Your Answer:
Correct Answer: Hypokalaemia
Explanation:The U-wave, not always visible in ECGs, is thought to represent repolarisation of papillary muscles or Purkinje fibres. When seen, it is very small and occurs after the T-wave. Inverted U-waves indicate myocardial ischaemia or left ventricular volume overload. Prominent U-waves are most commonly seen in hypokalaemia. Other causes include hypercalcaemia, thyrotoxicosis, digitalis exposure, adrenaline and class 1A and 3 anti-arrhythmic agents. It can also be seen in congenital long-QT syndrome and in intracranial haemorrhage.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 126
Incorrect
-
Which portion of the renal tubule absorbs amino acids and glucose?
Your Answer:
Correct Answer: Proximal convoluted tubule
Explanation:In relation to the morphology of the kidney as a whole, the convoluted segments of the proximal tubules are confined entirely to the renal cortex. Glucose, amino acids, inorganic phosphate and some other solutes are reabsorbed via secondary active transport in the proximal renal tubule through co-transport channels driven by the sodium gradient.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 127
Incorrect
-
Which of the following proteins prevents red blood cells (RBCs) from bursting when they pass through capillaries?
Your Answer:
Correct Answer: Spectrin
Explanation:Spectrin is a structural protein found in the cytoskeleton that lines the intercellular side of the membrane of cells which include RBCs. They maintain the integrity and structure of the cell. It is arranged into a hexagonal arrangement formed from tetramers of spectrin and associated with short actin filaments that form junctions allowing the RBC to distort its shape.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 128
Incorrect
-
An ECG of a 30 year old woman revealed low voltage QRS complexes. This patient is most probably suffering from?
Your Answer:
Correct Answer: Pericardial effusion
Explanation:The QRS complex is associated with current that results in the contraction of both the ventricles. As ventricles have more muscle mass than the atria, they result in a greater deflection on the ECG. The normal duration of a QRS complex is 10s. A wide and deep Q wave depicts myocardial infarction. Abnormalities in the QRS complex maybe indicative of a bundle block, ventricular tachycardia or hypertrophy of the ventricles. Low voltage QRS complexes are characteristic of pericarditis or a pericardial effusion.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 129
Incorrect
-
Which of the following substances will enhance the activity of antithrombin III?
Your Answer:
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 130
Incorrect
-
A 50 year old woman presented with excessive bleeding after an inguinal hernia repair. Labs are suggestive of a primary haemostasis defect. Deficiency of which of the following is most likely to cause it?
Your Answer:
Correct Answer: Platelets
Explanation:Primary haemostatic control means the first line of defence against immediate bleeding. This is carried out by the platelets. They immediately form a haemostatic plug at the site of injury. Coagulation starts within 20s after an injury to the blood vessel which damage the endothelial cells. Secondary haemostasis follows which includes activation of the coagulation factors to form fibrin strands which mesh together forming the platelet plug. Platelets interact with platelet collagen receptor, glycoprotein Ia/IIa and to collagen fibres in the vascular endothelium. This adhesion is mediated by von Willebrand factor (vWF), which forms links between the platelet glycoprotein Ib/IX/V and collagen fibrils. The platelets are then activated and release the contents of their granules into the plasma, in turn activating other platelets and white blood cells.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 131
Incorrect
-
Calculate the cardiac output of a patient with the following measurements: oxygen uptake 200 ml/min, oxygen concentration in the peripheral vein 7 vol%, oxygen concentration in the pulmonary artery 10 vol% and oxygen concentration in the aorta 15 vol%.
Your Answer:
Correct Answer: 4000 ml/min
Explanation:The Fick’s principle states that the uptake of a substance by an organ equals the arteriovenous difference of the substance multiplied by the blood flowing through the organ. We can thus calculate the pulmonary blood flow with pulmonary arterial (i.e., mixed venous) oxygen content, aortic oxygen content and oxygen uptake. The pulmonary blood flow, systemic blood flow and cardiac output can be considered the same assuming there are no intracardiac shunts. Thus, we can calculate the cardiac output. Cardiac output = oxygen uptake/(aortic − mixed venous oxygen content) = 200 ml/min/(15 ml O2/100 ml − 10 ml O2/100 ml) = 200 ml/min/(5 ml O2/100 ml) = 200 ml/min/0.05 = 4000 ml/min.
It is crucial to remember to use pulmonary arterial oxygen content and not peripheral vein oxygen content, when calculating the cardiac output by Fick’s method.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 132
Incorrect
-
Which of the following organs is most likely to have dendritic cells?
Your Answer:
Correct Answer: Skin
Explanation:Dendritic cells are part of the immune system and they function mainly as antigen presenting cells. They are present in small quantities in tissues which are in contact in the external environment. Mainly in the skin and to a lesser extent in the lining of the nose, lungs, stomach and intestines. In the skin they are known as Langerhans cells.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 133
Incorrect
-
The neurotransmitters adrenaline, noradrenaline and dopamine are derived from which amino acid?
Your Answer:
Correct Answer: Tyrosine
Explanation:Tyrosine is the precursor to adrenaline, noradrenaline and dopamine. Tyrosine hydroxylase converts tyrosine to DOPA, which is in turn converted to dopamine, then to noradrenaline and finally adrenaline.
-
This question is part of the following fields:
- Endocrine
- Physiology
-
-
Question 134
Incorrect
-
The Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity. According to this equation, the buffering capacity of the system is at maximum when the number of free anions compared with undissociated acid is:
Your Answer:
Correct Answer: Equal
Explanation:In 1908, Lawrence Joseph Henderson wrote an equation describing the use of carbonic acid as a buffer solution. Later, Karl Albert Hasselbalch re-expressed that formula in logarithmic terms, resulting in the Henderson–Hasselbalch equation. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid–base reactions. Two equivalent forms of the equation are: pH = pKa + log10 [A–]/[HA] or pH = pKa + log10 [base]/[acid]. Here, pKa is − log10(Ka) where Ka is the acid dissociation constant, that is: pKa = –log10(Ka) = –log10 ([H3 O+][A–]/[HA]) for the reaction: HA + H2 O ≈ A– + H3 O+ In these equations, A– denotes the ionic form of the relevant acid. Bracketed quantities such as [base] and [acid] denote the molar concentration of the quantity enclosed. Maximum buffering capacity is found when pH = pKa or when the number of free anions to undissociated acid is equal and buffer range is considered to be at a pH = pKa ± 1.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 135
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.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 136
Incorrect
-
When one is silently counting, what part of his brain will show increased regional cerebral blood flow (rCBF)?
Your Answer:
Correct Answer: Supplementary motor area
Explanation:Regional cerebral blood flow (rCBF) increases in the superior speech cortex (supplementary motor area) during periods of silent counting, whereas speaking aloud will do so in the motor cortex and medial temporal lobe, along with the superior speech cortex.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 137
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 138
Incorrect
-
A 76-year-old woman is diagnosed with diabetes mellitus after a urine test revealed she has glucosuria. Glucosuria may occur due to inadequate glucose reabsorption at:
Your Answer:
Correct Answer: Proximal convoluted tubule
Explanation:Glucose is reabsorbed almost 100% via sodium–glucose transport proteins (apical) and GLUT (basolateral) in the proximal convoluted tubule. Glycosuria or glucosuria is a condition of osmotic diuresis typical in those suffering from diabetes mellitus. Due to a lack of insulin, plasma glucose levels are above normal. This leads to saturation of receptors in the kidneys and glycosuria usually at plasma glucose levels above 11 mmol/l. Rarely, glycosuria is due to an intrinsic problem with glucose reabsorption within the kidneys (such as Fanconi syndrome), producing a condition termed renal glycosuria.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 139
Incorrect
-
Hormones of the anterior pituitary include which of the following?
Your Answer:
Correct Answer: Prolactin
Explanation:The anterior pituitary gland (adenohypophysis or pars distalis) synthesizes and secretes:
1. FSH (follicle-stimulating hormone)
2. LH (luteinizing hormone)
3. Growth hormone
4. Prolactin
5. ACTH (adrenocorticotropic hormone)
6. TSH (thyroid-stimulating hormone).
The posterior pituitary gland (neurohypophysis) stores and secretes 2 hormones produced by the hypothalamus:
1. ADH (antidiuretic hormone or vasopressin)
2. Oxytocin
-
This question is part of the following fields:
- Endocrine
- Physiology
-
-
Question 140
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 141
Incorrect
-
A 25-year-old woman complains of generalised swelling and particularly puffiness around the eyes which is worst in the morning. Laboratory studies showed:
Blood urea nitrogen (BUN) = 30 mg/dl
Creatinine = 2. 8 mg/dl
Albumin = 2. 0 mg/dl
Alanine transaminase (ALT) = 25 U/l
Bilirubin = 1 mg/dl
Urine analysis shows 3+ albumin and no cells.
Which of the following is the most likely diagnosis?Your Answer:
Correct Answer: Nephrotic syndrome
Explanation:Nephrotic syndrome is a disorder in which the glomeruli have been damaged, characterized by:
– Proteinuria (>3.5 g per 1.73 m2 body surface area per day, or > 40 mg per square meter body surface area per hour in children)
– Hypoalbuminemia (< 2,5 g/dl) – Hyperlipidaemia, and oedema (generalized anasarca).
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 142
Incorrect
-
Which of the following will be a seen in a patient with a plasma thyroid-stimulating hormone (TSH) level of 14 mU/l (normal < 5 mU/l) and a low T3 resin uptake of 19% (normal 25–35%)?
Your Answer:
Correct Answer: Periorbital swelling and lethargy
Explanation:Low T3 resin uptake combined with raised TSH is indicative of hypothyroidism. Signs and symptoms include dull expression, facial puffiness, lethargy, periorbital swelling due to infiltration with mucopolysaccharides, bradycardia and cold intolerance. Anxiety, palpitations, tachycardia, raised body temperature, heat intolerance and weight loss are all seen in hyperthyroidism.
-
This question is part of the following fields:
- Endocrinology
- Physiology
-
-
Question 143
Incorrect
-
Which statement is correct regarding coagulation?
Your Answer:
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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 144
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).
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 145
Incorrect
-
Depression of the normal coagulation system and excessive bleeding after surgery can occur in which of the following medical conditions?
Your Answer:
Correct 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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 146
Incorrect
-
In what form are fats primarily transported in the body?
Your Answer:
Correct Answer: Free fatty acids
Explanation:Fat is mainly transported in the body as free fatty acids. Once out of the adipose cell, the free fatty acids get ionized and combine with albumin.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 147
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:
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 148
Incorrect
-
A lesion involving the suprachiasmatic nucleus of hypothalamus is likely to affect:
Your Answer:
Correct Answer: Regulation of circadian rhythm
Explanation:The suprachiasmatic nucleus (SCN) in the hypothalamus is responsible for controlling endogenous circadian rhythms and destruction of the SCN leads to a loss of circadian rhythm.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 149
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:
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 150
Incorrect
-
Calculate the total peripheral resistance for a patient with a blood pressure of 130/70 mm HG and cardiac output of 5 litres / min?
Your Answer:
Correct Answer: 18 mmHg × min/l
Explanation:Total peripheral resistance = Mean arterial pressure/Cardiac output. And the mean arterial pressure = Diastolic pressure + 1/3 (Systolic pressure – Diastolic pressure), i.e., 70 + 1/3 (130-70) = 90 mmHg. Therefore, total peripheral resistance = 90 mmHg/5 l per min = 18 mmHg × min/l.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
SESSION STATS - PERFORMANCE PER SPECIALTY
