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  • Question 1 - Which of the following drugs would cause the most clinical concern if accidentally...

    Correct

    • Which of the following drugs would cause the most clinical concern if accidentally administered intravenously to a 4-year-old boy?

      Your Answer: 20 mg codeine

      Explanation:

      To begin, one must determine the child’s approximate weight. There are a variety of formulas to choose from. It is acceptable to use the advanced paediatric life support formula:

      (age + 4) 2 = weight

      A 5-year-old child will weigh around 18 kilogrammes.

      The following are the appropriate doses of the drugs listed above:

      Gentamicin (once daily) – 5-7 mg/kg = 90-126 mg and subsequent dose modified according to plasma levels
      Ondansetron – 0.1 mg/kg, but a maximum of 4 mg as a single dose = 1.8 mg
      Codeine should be administered orally at a dose of 1 mg/kg rather than intravenously, as the latter can cause ‘dangerous’ hypotension due to histamine release.
      15 mg/kg paracetamol = 270 mg orally or intravenously (a loading dose of 20 mg/kg, or 360 mg, is sometimes recommended, which is not far short of the doses listed above).
      Cefuroxime – the initial intravenous dose is 20 mg/kg (360 mg) depending on the indication (again, similar to the dose given in the answer options above).

    • This question is part of the following fields:

      • Pharmacology
      5.6
      Seconds
  • Question 2 - Out of the following, which anatomical structure lies within the spiral groove of...

    Correct

    • Out of the following, which anatomical structure lies within the spiral groove of the humerus?

      Your Answer: Radial nerve

      Explanation:

      The shaft of the humerus has two prominent features:
      1. Deltoid tuberosity – attachment for the deltoid muscle
      2. Radial or spiral groove – The radial nerve and profunda brachii artery lie in the groove

      Mid-shaft fractures of the humerus usually occur after a direct blow to the upper arm, which can occur after a fall or RTAs. The most important clinical significance of a mid-shaft humeral fracture is an injury to the radial nerve. The radial nerve originates from the brachial plexus and has roots of C5-T1. It crosses the spiral groove on the posterior side of the shaft of the humerus.
      On examination, the patient may have a wrist drop, loss or weakness of finger extension, and decreased or absent sensation to the posterior forearm, digits 1 to 3, and the radial half of the fourth digit.

      The following parts of the humerus are in direct contact with the indicated
      nerves:
      Surgical neck: axillary nerve.
      Radial groove: radial nerve.
      Distal end of humerus: median nerve.
      Medial epicondyle: ulnar nerve.

    • This question is part of the following fields:

      • Anatomy
      10.1
      Seconds
  • Question 3 - A 70-year-old female presented with a productive cough and is prescribed a bacteriostatic...

    Correct

    • A 70-year-old female presented with a productive cough and is prescribed a bacteriostatic antibiotic?
      Which of the following best explains the mechanism of action of bacteriostatic drugs?

      Your Answer: Protein synthesis inhibition

      Explanation:

      Cell membrane pore formation, Bacterial DNA damage, Peptidoglycan cross-linking inhibition, and peptidoglycan synthesis inhibitor are always lethal and such mechanisms are possible only in bactericidal drugs. But Protein synthesis inhibition would only prevent cell replication or cell growth and is responsible for bacteriostatic effects of the drug.

    • This question is part of the following fields:

      • Pharmacology
      7.7
      Seconds
  • Question 4 - Which of the following is a correctly stated fundamental (base) SI unit? ...

    Correct

    • Which of the following is a correctly stated fundamental (base) SI unit?

      Your Answer: A metre is the unit of length

      Explanation:

      The international system of units, or system international d’unites (SI) is a collection of measurements derived from expanding the metric system.

      There are seven base units, which are:

      Metre (m): a unit of length
      Second (s): a unit of time
      Kilogram (kg): a unit of mass
      Ampere (A): a unit of electrical current
      Kelvin (K): a unit of thermodynamic temperature
      Candela (cd): a unit of luminous intensity
      Mole (mol): a unit of substance.

    • This question is part of the following fields:

      • Clinical Measurement
      13
      Seconds
  • Question 5 - Work is underway to devise a new breast cancer screen testing method that...

    Correct

    • Work is underway to devise a new breast cancer screen testing method that doesn’t involve mammography which is in contrast to the prevailing standard. The initial numbers look promising and are indicating that the new testing method is better able to pick up early stage cancer. There, however, is not substantial difference in survival rates.

      Which of the following term can be used to describe this?

      Your Answer: Lead-time bias

      Explanation:

      Work up bias involves comparing the novel diagnostic test with the current standard test. A portion of the patients undergo the standard test while others undergo the new test as the standard test is costly. The result can be alteration in specify and sensitivity.

      Recall bias is specifically appropriate to the case control studies that is when ever the memories retrieved by the participants differ in accuracy.

      Not publishing the results of a valid study just because they are negative or uninteresting can be termed as publication bias.

      When information gathering is ill suited with respect to time i.e. collecting the data regarding a fatal disease many years after the death of its patients, it is termed as Late – look bias.

      The case in point is an instance of lead time bias when upon comparing two tests, one is able to detect the condition earlier than the other but the overall outcome doesn’t change. There is a possibility that this will make the survival rates for the newer test look more promising.

    • This question is part of the following fields:

      • Statistical Methods
      3.5
      Seconds
  • Question 6 - The structure most likely to be damaged during cannulation of the subclavian vein...

    Correct

    • The structure most likely to be damaged during cannulation of the subclavian vein is?

      Your Answer: Subclavian artery

      Explanation:

      The subclavian artery lies behind and partly above the subclavian vein. 3-4% of the time, it can be inadvertently cannulated during cannulation of the subclavian vein

      Because of its anatomical position, putting pressure on the subclavian artery is impossible so arresting bleeding with pressure when it is punctured is not viable.

      One of the consequences of subclavian vein cannulation (1%) is pleural puncture leading to a pneumothorax. This is because the apical pleura is inferior and caudal to the subclavian vein.

    • This question is part of the following fields:

      • Anatomy
      4.1
      Seconds
  • Question 7 - The outer muscular layer of the oesophagus is covered by? ...

    Correct

    • The outer muscular layer of the oesophagus is covered by?

      Your Answer: Loose connective tissue

      Explanation:

      The oesophagus has four layers namely; 1. the mucosal layer, 2. the submucosal layer, 3. the muscular layer and 4. the layer of loose connective tissue which binds to the outer mucosal layer. The oesophagus lacks the serosal layer and therefore holds sutures poorly.

      The mucosal layer consists of muscularis mucosa and the lamina propria and is made up of non keratinised stratified squamous epithelium. The mucosal layer is the innermost layer of the oesophagus.

      The submucosal layer being the strongest layer of all has mucous glands which are called as the tuboalveolar mucous glands.

      The outer muscular layer has two types of muscle layers of which one is the circular layer and the other the longitudinal layer. The Auerbach’s and Meissner’s nerve plexuses lie in between the longitudinal and circular muscle layers and submucosally. The muscle fibres present in the upper 1/3rd part of the oesophagus are skeletal muscle fibres, the middle 1/3rd layer has both smooth and skeletal muscle fibres, but the lower 1/3rd only has smooth muscle fibres.

      The loose connective tissue layer or the adventitious layer has dense fibrous tissue.

    • This question is part of the following fields:

      • Anatomy
      5.2
      Seconds
  • Question 8 - The thebesian veins contribute to the venous drainage of the heart. Into which...

    Correct

    • The thebesian veins contribute to the venous drainage of the heart. Into which of the following structures do they primarily drain?

      Your Answer: Atrium

      Explanation:

      The heart has two venous drainage systems:
      1. Greater venous system – it parallels the coronary arterial circulation and provides 70% venous drainage to the heart
      2. Lesser venous system – includes the thebasian veins and provides up to 30% of the venous drainage to the heart

      Thebasian veins (also called venae cordis minimae) are the smallest coronary veins and run in the myocardial layer of the heart. They serve to drain the myocardium and are present in all four heart chambers. They are more abundant on the right side of the heart and, more specifically, are most abundant in the right atrium. Thebesian veins drain the subendocardial myocardium either directly, via connecting intramural arteries and veins, or indirectly, via subendocardial sinusoidal spaces.

    • This question is part of the following fields:

      • Anatomy
      1.8
      Seconds
  • Question 9 - The statement that best describes temperature management is: ...

    Correct

    • The statement that best describes temperature management is:

      Your Answer: Gauge thermometers use coils of different metals with different co-efficients of expansion which either tighten or relax with changes in temperature

      Explanation:

      There are different types of temperature measurement. These include:

      Thermistor – this is a type of semiconductor, meaning they have greater resistance than conducting materials, but lower resistance than insulating materials. There are small beads of semiconductor material (e.g. metal oxide) which are incorporated into a Wheatstone bridge circuit. As the temperature increases, the resistance of the bead decreases exponentially

      Thermocouple – Two different metals make up a thermocouple. Generally, in the form of two wires twisted, welded, or crimped together. Temperature is sensed by measuring the voltage. A potential difference is created that is proportional to the temperature at the junction (Seebeck effect)

      Platinum resistance thermometers (PTR) – uses platinum for determining the temperature. The principle used is that the resistance of platinum changes with the change of temperature. The thermometer measures the temperature over the range of 200°C to1200°C. Resistance in metals show a linear increase with temperature

      Tympanic thermometers – uses infrared radiation which is emitted by all living beings. It analyses the intensity and wavelength and then transduces the heat energy into a measurable electrical output

      Gauge/dial thermometers – Uses coils of different metals with different co-efficient of expansion. These either tighten or relax with changes in temperature, moving a lever on a calibrated dial.

    • This question is part of the following fields:

      • Clinical Measurement
      11.1
      Seconds
  • Question 10 - You've been summoned to the paediatric ward after a 4-year-old child was discovered...

    Correct

    • You've been summoned to the paediatric ward after a 4-year-old child was discovered 'collapsed' in bed.

      The child had been admitted the day before with febrile convulsions and was scheduled to be discharged. It is safe to approach the child.

      What should your first life-saving action be?

      Your Answer: Apply a gentle stimulus and ask the child if they are alright

      Explanation:

      Paediatric life support differs from adult life support in that hypoxia is the primary cause of deterioration.

      After checking for danger, the child should be given a gentle stimulus (such as holding the head and shaking the arm) and asked, Are you alright? according to current advanced paediatric life support (APLS) guidelines. Safety, Stimulate, Shout is a phrase that is frequently remembered. Any airway assessment should be preceded by these actions.

      Although the algorithm includes five rescue breaths, they are performed after the airway assessment.

      It is not recommended to ask parents to leave unless they are obstructing the resuscitation. A team member should be with them at all times to explain what is going on and answer any questions they may have.

      CPR should not begin until the child has been properly assessed and rescue breaths have been administered.

    • This question is part of the following fields:

      • Pathophysiology
      4.3
      Seconds
  • Question 11 - A 52-year old man was placed under general anaesthesia for an emergent open...

    Correct

    • A 52-year old man was placed under general anaesthesia for an emergent open cholecystectomy. As part of the induction, suxamethonium was administered at 1.5mg/kg. Post-operatively, there was failure to restore muscle twitch responses over a course of five hours.

      Clinical chemistry studies were obtained and showed the following results:
      Butrylcholinesterase (BChE) activity: 49 U/L (Reference range: 3300-10,300 U/L)
      Dibucaine number: <4% (Reference range: 83-88%)

      The attending physician gave an initial diagnosis of Suxamethonium Apnoea.

      What is the most probable phenotype of BChE of the patient?

      Your Answer: S (silent)

      Explanation:

      Silent (S) is the most probable phenotype of the patient. In S phenotype, patients have significantly reduced levels of BChE, the lowest among the four phenotypes. Because of this, individuals with S phenotype are subjected to long periods of apnoea. In addition, their dibucaine number is very low.

      Other BChE phenotypes are the following:

      Usual (U)
      Atypical (A)
      Fluoride-resistant (F)

    • This question is part of the following fields:

      • Pathophysiology
      7.9
      Seconds
  • Question 12 - Which of the following, at a given PaO2, increases the oxygen content of...

    Correct

    • Which of the following, at a given PaO2, increases the oxygen content of arterial blood?

      Your Answer: A reduced erythrocyte 2,3-diphosphoglycerate level

      Explanation:

      The oxygen content of arterial blood can be calculated by the following equation:
      (10 x haemoglobin x SaO2 x 1.34) + (PaO2 x 0.0225).
      This is the sum of the oxygen bound to haemoglobin and the oxygen dissolved in the plasma.

      Oxygen content x cardiac output = The amount of oxygen delivered to the tissues in unit time which is known as the oxygen flux.

      Any factor that increases the metabolic demand will encourage oxygen offloading from the haemoglobin in the tissues and this causes the oxygen dissociation curve (ODC) to shift to the right. This subsequently reduced the oxygen content of arterial blood.

      Conditions like fever, metabolic or respiratory acidosis lowers the oxygen content and shifts the ODC to the right.
      A low level of 2,3 diphosphoglycerate (2,3-DPG) is usually related to an increased oxygen content as there is less offloading, and so the ODC is shifted to the left.

      So for a given PaO2, a high blood oxygen content is related to any factors that can shift the ODC to the left and not to the right.

      A low haematocrit usually means that there is a decreased haemoglobin concentration, and therefore is associated with decreased oxygen binding to haemoglobin.

    • This question is part of the following fields:

      • Physiology
      13.3
      Seconds
  • Question 13 - The main site of storage of thyroid hormones in the thyroid gland is?...

    Correct

    • The main site of storage of thyroid hormones in the thyroid gland is?

      Your Answer: Thyroglobulin

      Explanation:

      The follicle is the functional unit of the thyroid gland. The follicular cells surround the follicle which is filled with colloid. Suspended within the colloid is the is a pro-hormone complex thyroglobulin.

      The synthesis and storage of thyroid hormones is done by follicular cells and the thyroglobulin within the colloid.

      Iodide ions (I−) are actively transported against a concentration gradient into the follicular cell under the influence of thyroid stimulating hormone (TSH). It then undergoes oxidation to active iodine catalysed by thyroid peroxidase (TPO). The synthesis of thyroglobulin is in the follicular cells and it contains up to 140 tyrosine residues. The tyrosine residues of thyroglobulin and active iodine are merged to form mono- and di-iodotyrosines (MIT and DIT). The iodinated thyroglobulin is then taken up into the colloid where it is stored and dimerised. Two DIT molecules are joined to produce thyroxine (T4) while one MIT and one DIT molecule are joined to produce tri-iodotyrosine (T3) by a process catalysed by TPO.

      Thyroglobulin droplets are taken up as vesicles into follicular cells by pinocytosis. This process is stimulated by TSH. When these vesicles fuse with lysosomes, hydrolysis of the thyroglobulin molecules and subsequent release of T4 and T3 into the circulation occurs.

    • This question is part of the following fields:

      • Pathophysiology
      3.5
      Seconds
  • Question 14 - A 30-year-old man has been diagnosed with a direct inguinal hernia.

    A direct...

    Correct

    • A 30-year-old man has been diagnosed with a direct inguinal hernia.

      A direct inguinal hernia passes through Hesselbach's triangle. What forms the medial edge of this triangle?

      Your Answer: Rectus abdominis muscle

      Explanation:

      The inguinal triangle of Hesselbach’s is an important clinical landmark on the posterior wall of the inguinal canal. It has the following relations:
      Inferiorly – medial third of the inguinal ligament
      Medially – lower lateral border of the rectus abdominis
      Laterally – inferior epigastric vessels

      Direct inguinal hernia is when the bowel bulges directly through the abdominal wall. These hernias usually protrude through Hesselbach’s triangle

    • This question is part of the following fields:

      • Anatomy
      20.1
      Seconds
  • Question 15 - Which of the following statements is true regarding drug dose and response? ...

    Correct

    • Which of the following statements is true regarding drug dose and response?

      Your Answer: Intrinsic activity determines maximal response

      Explanation:

      There are two types of drug dose-response relationships, namely, the graded dose-response and the quantal dose-response relationships.

      Drug response curves are plotted as percentage response again LOG drug concentration. This graph is sigmoid in shape.

      Agonists are drugs with high affinity and high intrinsic activity. Meanwhile, the antagonist is a drug with high affinity but no intrinsic activity. Intrinsic activity determines the maximal response. The maximal response can be achieved even by activation of a small proportion of receptor sites.

    • This question is part of the following fields:

      • Pharmacology
      8.2
      Seconds
  • Question 16 - Which statement is correct concerning breathing systems? ...

    Correct

    • Which statement is correct concerning breathing systems?

      Your Answer: The reservoir bag can limit the pressure in the breathing system to about 40 cm of water

      Explanation:

      Mapleson classified breathing systems into A, B, C, D and E. Jackson-Rees subsequently modified the Mapleson E by adding a double-ended bag to the end of the reservoir tubing, creating the Mapleson F. A Mapleson E or T-piece does not have a reservoir bag.

      A Mapleson A system is a very efficient system for use during spontaneous ventilation. However, it is not suitable for use with patients less than 25 kg, due to the increased dead space at the distal / patient end. This system can be modified into a Lack system or coaxial Mapleson A, where the fresh gas flows through an outer tube (30 mm) and exhaled gases flow through the inner tube (14 mm).

      The adjustable pressure limiting valve (APL) or expiratory valve allows exhaled gas and excess fresh gas to leave the breathing system. It is a one-way, adjustable spring-loaded valve, and gases escape when the pressure in the system exceeds the valve opening pressure. During spontaneous ventilation a pressure of less than 1 cm of water (0.1 kPa) is needed when the valve is in the open position (not 2 cm of H2O).

      The reservoir bag is highly compliant and when over inflated, the rubber bag can limit the pressure in the system to about 40 cm of H2O.

      This is due to the law of Laplace, which states that the pressure will fall as the radius of the bag increases:

      Pressure = 2 x tension/radius.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      7.1
      Seconds
  • Question 17 - Which among the given options can be used to find out the number...

    Correct

    • Which among the given options can be used to find out the number needed to treat?

      Your Answer: 1 / (Absolute risk reduction)

      Explanation:

      Number needed to treat can be defined as the number of patients who need to be treated to prevent one additional bad outcome.

      It can be found as:

      NNT=1/Absolute Risk Reduction (rounded to the next integer since number of patients can be integer only).

    • This question is part of the following fields:

      • Statistical Methods
      12.5
      Seconds
  • Question 18 - When an inotrope is given to the body, it has the following effects...

    Correct

    • When an inotrope is given to the body, it has the following effects on the cardiovascular system:

      The automaticity of the sino-atrial node increases
      Lusitropy is accelerated
      Dromotropy is increased
      Chronotropy is increased
      Inotropy increases
      There is increased excitability of the conducting system

      The most probably mechanism of action of this compound is?

      Your Answer: Increase in intracellular calcium influenced by a conformational change of a Gs protein

      Explanation:

      A beta-1 adrenoreceptor agonist is most likely the ligand that causes increased automaticity, increased chronotropy, increased excitability, and increased inotropy on the sino-atrial node. However, alpha-1 adrenoreceptor effects may cause an increase in systemic vascular resistance. Noradrenaline, adrenaline, dopamine, and ephedrine are examples of drugs with mixed alpha and beta effects.

      Adrenaline, noradrenaline, dopamine, dopexamine, dobutamine, ephedrine, and isoprenaline are examples of drugs that have some beta-1 activity. The beta-1 receptor is a G protein-coupled metabotropic receptor. When the beta-1 agonist binds to the cell surface membrane, it causes a conformational change in the Gs unit, which triggers a cAMP-dependent pathway and a calcium influx into the cell.

      Catecholamines also help to relax the heart muscle (positive lusitropy). Dromotropy is the ability to increase the atrioventricular (AV) node’s conduction velocity.

      Inodilators cause an increase in intracellular calcium as a result of phosphodiesterase III (PDIII) inhibition. Milrinone, enoximone, and amrinone are some examples. Positive inotropy is caused by increased calcium entry into the myocytes. Lusitropy is also increased by phosphodiesterase inhibitors. Increased cAMP inhibits myosin light chain kinase, resulting in reduced phosphorylation of vascular smooth muscle myosin, lowering systemic and pulmonary vascular resistance.

      The mechanism of action of alpha-1 adrenoreceptor agonists is an increase in intracellular calcium caused by an increase in inositol triphosphate (IP3). IP3 is a second messenger that causes an increase in systemic vascular resistance by stimulating the influx of Ca2+ into smooth muscle cells. Reflex bradycardia can occur as a result of the subsequent increase in blood pressure. Phenylephrine and metaraminol are examples of pure alpha-1 agonists.

      Levosimendin is a novel inotrope that makes myocytes more sensitive to intracellular Ca2+. It causes a positive inotropy without changing heart rate or oxygen consumption significantly.

      The Na-K-ATPase membrane pump in the myocardium is inhibited by digoxin. This inhibition promotes sodium-calcium exchange, resulting in an increase in intracellular Ca2+ and increased contraction force. The parasympathetic effects of digoxin on the AV node result in bradycardia. Systemic vascular resistance will not be affected by it.

    • This question is part of the following fields:

      • Pathophysiology
      9.2
      Seconds
  • Question 19 - One of the following neuromuscular blocking agents is the most potent: ...

    Correct

    • One of the following neuromuscular blocking agents is the most potent:

      Your Answer: Vecuronium

      Explanation:

      The measure of drug potency or therapeutic response is the ED95. This is defined as the dose of a neuromuscular blocking drug required to produce a 95% depression of muscle twitch height. The ED50 and ED90 describe a depression of twitch height by 50% and 90% respectively.

      The ED95 (mg/kg) of the commonly used neuromuscular blocking agents are:

      suxamethonium: 0.27
      rocuronium: 0.31
      vecuronium: 0.04
      pancuronium: 0.07
      cisatracurium: 0.04
      mivacurium: 0.08

    • This question is part of the following fields:

      • Pharmacology
      4.8
      Seconds
  • Question 20 - A 21-year-old female was brought to the Emergency department with a ruptured ectopic...

    Correct

    • A 21-year-old female was brought to the Emergency department with a ruptured ectopic pregnancy.

      On clinical examination, the following were the findings:
      Pulse: 120 beats per minute
      BP: 120/95 mmHg
      Respiratory rate: 22 breaths per minute
      Capillary refill time: three seconds
      Cool peripheries.

      Which of the following best describes the cause for this clinical finding?

      Your Answer: Reduction in blood volume of 15-30%

      Explanation:

      Classification of hemorrhagic shock according to Advanced Trauma Life Support is as follows:

      – Class I haemorrhage (blood loss up to 15%) in which there is no change in blood pressure, RR, or pulse pressure.

      – Class II haemorrhage (15-30% blood volume loss) where there is tachycardia, tachypnoea, and a decrease in pulse pressure.

      – Class III haemorrhage (30-40% blood volume loss) where clinical signs of inadequate perfusion, marked tachycardia, tachypnoea, significant changes in mental state, and measurable fall in systolic pressure is seen. It almost always requires a blood transfusion.

      – Class IV haemorrhage (> 40% blood volume loss) in which marked tachycardia, significant depression in systolic pressure and very narrow pulse pressure, and markedly depressed mental state with cold and pale skin are seen.

      Loss of >50% results in loss of consciousness, pulse, and blood pressure.

    • This question is part of the following fields:

      • Pathophysiology
      18.6
      Seconds
  • Question 21 - A chain smoker is interested in knowing how many years of his life...

    Incorrect

    • A chain smoker is interested in knowing how many years of his life would be lessened by smoking. You tell him explicitly that precise determination is impossible but you can tell him the proportion of people who died due to smoking. Which of the following epidemiological term is apt in this regard?

      Your Answer: Attributable proportion - the rate in the exposed group minus the rate in the unexposed group.

      Correct Answer: Attributable risk - the rate in the exposed group minus the rate in the unexposed group

      Explanation:

      Attributable proportion is the proportion of disease that is caused due to exposure. It refers to the proportion of disease that would be eradicated from a particular population if the disease rate was diminished to match that of the unexposed group.

      Risk ratio (relative risk) compares the probability of an event in an exposed (experimental) group to that of an event in the unexposed (control) group. Thus two are not the same.

      The attributable risk is the rate of a disease in an exposed group to that of a group that has not been exposed to it i.e. how many deaths did the exposure cause.

    • This question is part of the following fields:

      • Statistical Methods
      50.4
      Seconds
  • Question 22 - Which of the following antihypertensive drugs is an alpha-blocker? ...

    Correct

    • Which of the following antihypertensive drugs is an alpha-blocker?

      Your Answer: Doxazosin

      Explanation:

      Doxazosin is selective alpha 1 blocker (it causes less tachycardia than a non-selective alpha-blocker) and is the drug of choice for a patient with hypertension and benign hyperplasia of the prostate (BHP).

      The major adverse effect of an alpha-blocker is first-dose hypotension.

      Atenolol and Labetalol are beta blockers. It works by relaxing blood vessels and slowing heart rate to improve blood flow and decrease blood pressure.

      Clonidine is an α2A-adrenergic agonist used to treat high blood pressure, ADHD, drug withdrawal (alcohol, opioids, or nicotine), menopausal flushing, diarrhea, spasticity, and certain pain conditions.

      Methyldopa is a centrally-acting alpha-2 adrenergic agonist used to manage hypertension alone or in combination with hydrochlorothiazide, and to treat hypertensive crises.

    • This question is part of the following fields:

      • Pharmacology
      4.9
      Seconds
  • Question 23 - Which of the following correctly explains the mechanism of sevoflurane preconditioning? ...

    Correct

    • Which of the following correctly explains the mechanism of sevoflurane preconditioning?

      Your Answer: Opening of mitochondrial KATP channels

      Explanation:

      Sevoflurane is highly fluorinated methyl isopropyl ether widely used as an inhalational anaesthetic. It is suggested that sevoflurane preconditioning occurs via the opening of mitochondrial Potassium ATP dependent channel similar to that of Ischemic Preconditioning protection.

    • This question is part of the following fields:

      • Pharmacology
      6.6
      Seconds
  • Question 24 - The immediate physiological response to massive perioperative blood loss is: ...

    Correct

    • The immediate physiological response to massive perioperative blood loss is:

      Your Answer: Stimulation of baroreceptors in carotid sinus and aortic arch

      Explanation:

      With regards to compensatory response to blood loss, the following sequence of events take place:

      1. Decrease in venous return, right atrial pressure and cardiac output
      2. Baroreceptor reflexes (carotid sinus and aortic arch) are immediately activated
      3. There is decreased afferent input to the cardiovascular centre in medulla. This inhibits parasympathetic reflexes and increases sympathetic response
      4. This results in an increased cardiac output and increased SVR by direct sympathetic stimulation. There is increased circulating catecholamines and local tissue mediators (adenosine, potassium, NO2)
      5. Fluid moves into the intravascular space as a result of decreased capillary hydrostatic pressure absorbing interstitial fluid.

      A slower response is mounted by the hypothalamus-pituitary-adrenal axis.
      6. Reduced renal blood flow is sensed by the intra renal baroreceptors and this stimulates release of renin by the juxta-glomerular apparatus.
      7. There is cleavage of circulating Angiotensinogen to Angiotensin I, which is converted to Angiotensin II in the lungs (by Angiotensin Converting Enzyme ACE)

      Angiotensin II is a powerful vasoconstrictor that sets off other endocrine pathways.
      8. The adrenal cortex releases Aldosterone
      9. There is antidiuretic hormone release from posterior pituitary (also in response to hypovolaemia being sensed by atrial stretch receptors)
      10. This leads to sodium and water retention in the distal convoluted renal tubule to conserve fluid
      Fluid conservation is also aided by an increased amount of cortisol which is secreted in response to the increase in circulating catecholamines and sympathetic stimulation.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      11.5
      Seconds
  • Question 25 - Regarding gas laws, which one best explains the ability of a Bourdon gauge...

    Correct

    • Regarding gas laws, which one best explains the ability of a Bourdon gauge to measure temperature?

      Your Answer: Charles's law

      Explanation:

      Charles’ Law states that there is a direct correlation between temperature and volume, where pressure and amount gas are constant. As temperature increases, volume also increases.

      Boyle’s Law states that Pressure is inversely proportional to volume, assuming that temperature and amount of gas are constant. As volume increases, pressure decreases. In Dalton’s law of partial pressure, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressure of the gases in mixture.

      According to Henry’s Law for concentration of dissolved gases, at a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid. An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

      Gay-Lussac’s Law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. This law is very similar to Charles’ Law, with the only difference being the type of container. Whereas the container in a Charles’ Law experiment is flexible, it is rigid in a Gay-Lussac’s Law experiment.

    • This question is part of the following fields:

      • Basic Physics
      9.4
      Seconds
  • Question 26 - Which of the following drug can be the first-line drug for both broad...

    Correct

    • Which of the following drug can be the first-line drug for both broad and narrow complex tachyarrhythmia?

      Your Answer: Amiodarone

      Explanation:

      Amiodarone is the longest-acting anti-arrhythmic drug. It possesses the action of all classes of antiarrhythmic drugs (Sodium channel blockade, Beta blockade, Potassium channel blockade, and Calcium channel blockade). Due to this property, it has the widest anti-arrhythmic spectrum and thus can be used in both broad and narrow complex tachyarrhythmia.

      Adenosine is shortest acting anti-arrhythmic drug.

    • This question is part of the following fields:

      • Pharmacology
      6.7
      Seconds
  • Question 27 - A 60-year-old man is having his central venous pressure (CVP) measured. A long...

    Correct

    • A 60-year-old man is having his central venous pressure (CVP) measured. A long venous line was inserted via the femoral vein into the right atrium for this purpose. The catheter is advanced through the IVC.

      Which level of the vertebra is the site where this vessel enters the thorax?

      Your Answer: T8

      Explanation:

      The diaphragm divides the thoracic cavity from the abdominal cavity. Structures penetrate the diaphragm at different vertebral levels through openings in the diaphragm to communicate between the two cavities. The diaphragm has openings at three vertebral levels:

      T8: vena cava, terminal branches of the right phrenic nerve
      T10: oesophagus, vagal trunks, left anterior phrenic vessels, oesophageal branches of the left gastric vessels
      T12: descending aorta, thoracic duct, azygous and hemi-azygous vein

    • This question is part of the following fields:

      • Anatomy
      20.7
      Seconds
  • Question 28 - A young male is operated on for an open inguinal hernia repair. During...

    Incorrect

    • A young male is operated on for an open inguinal hernia repair. During the procedure, the cord is mobilized, and the deep inguinal ring is located.

      What structure forms the lateral wall of the deep inguinal ring?

      Your Answer: External oblique aponeurosis

      Correct Answer: Transversalis fascia

      Explanation:

      The deep inguinal ring is the entrance of the inguinal canal. It is an opening in the transversalis fascia around 1 cm above the inguinal ligament. Therefore, the superolateral wall is made by the transervalis fascia.

      The inferior epigastric vessels run medially to the deep inguinal ring forming its inferomedial border.

      The inguinal canal extends obliquely from the deep inguinal ring to the superficial inguinal ring.
      An indirect inguinal hernia arises through the deep inguinal ring lateral to the inferior epigastric vessels.

    • This question is part of the following fields:

      • Anatomy
      64.2
      Seconds
  • Question 29 - A 58-year-old man, visits his general practitioner complaining of a lump in his...

    Correct

    • A 58-year-old man, visits his general practitioner complaining of a lump in his groin. He explains he is otherwise well and reports no other symptoms. The lump is examined and is found to be soft, and can be reduced without causing the patient pain. The GP diagnoses an inguinal hernia. To determine the nature of the hernia, the GP reduced the lump and applies pressure on the deep inguinal ring.

      The deep inguinal ring has what anatomical landmark?

      Your Answer: Superior to the midpoint of the inguinal ligament

      Explanation:

      The deep inguinal ring lies approximately 1.5-2cm above the midpoint of the inguinal ligament, the halfway point between the anterior superior iliac spine and the pubic tubercle, next to the epigastric vessels.

      It is an important point in determining the nature of an inguinal hernia (direct or indirect). The patient is asked to cough after the hernia is reduced, with pressure applied to the deep inguinal ring. The hernia reappearing indicates it is direct, moving through the posterior wall of the inguinal canal.

      Inferior and lateral to the pubic tubercle is the normal anatomical position of the neck of a femoral hernia.

      Superior and medial to the pubic tubercle is the site of the superficial inguinal ring, and the normal anatomical position of the neck of an inguinal hernia.

      The mid-inguinal point is located halways between the pubic symphysis and the anterior superior iliac spine. It is the surface marking for taking the femoral pulse.

    • This question is part of the following fields:

      • Anatomy
      47.8
      Seconds
  • Question 30 - The fluids with the highest osmolarity is? ...

    Correct

    • The fluids with the highest osmolarity is?

      Your Answer: 0.45% N. Saline with 5% glucose

      Explanation:

      The concentration of solute particles per litre (mosm/L) = the osmolarity of a solution. Changes in water content, ambient temperature, and pressure affects osmolarity. The osmolarity of any solution can be calculated by adding the concentration of key solutes in it.

      Individual manufacturers of crystalloids and colloids may have different absolute values but they are similar to these.

      0.45% N. Saline with 5% glucose:
      Tonicity – hypertonic
      Osmolarity – 405 mosm/L
      Kilocalories (kCal) – 107

      0.9% N. Saline:
      Tonicity – isotonic
      Osmolarity – 308 mosm/L
      Kilocalories (kCal) – 0

      5% Dextrose:
      Tonicity – isotonic
      Osmolarity – 253 mosm/L
      Kilocalories (kCal) – 170

      Gelofusine (154 mmol/L Na, 120 mmol/L Cl):
      Tonicity – isotonic
      Osmolarity – 274 mosm/L
      Kilocalories (kCal) – 0

      Hartmann’s solution:
      Tonicity – isotonic
      Osmolarity – 273 mosm/L
      Kilocalories (kCal) – 9

    • This question is part of the following fields:

      • Physiology
      4.3
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Pharmacology (7/7) 100%
Anatomy (7/8) 88%
Clinical Measurement (2/2) 100%
Statistical Methods (2/3) 67%
Pathophysiology (5/5) 100%
Physiology (2/2) 100%
Anaesthesia Related Apparatus (1/1) 100%
Physiology And Biochemistry (1/1) 100%
Basic Physics (1/1) 100%
Passmed