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  • Question 1 - A 28-year-old man is admitted to the critical care unit. He has been...

    Correct

    • A 28-year-old man is admitted to the critical care unit. He has been diagnosed with adult respiratory distress syndrome and is being ventilated. His haemodynamic condition is improved using a pulmonary artery flotation.

      His readings are listed below:

      Haemoglobin concentration: 10 g/dL
      Mixed venous oxygen saturation: 70%
      Mixed venous oxygen tensions (PvO2): 50 mmHg

      Estimate his mixed venous oxygen content (mL/100mL).

      Your Answer: 9.5

      Explanation:

      Mixed venous oxygen content (CvO2) is the oxygen concentration in 100mL of mixed venous blood taken from the pulmonary artery. It is usually 12-17 mL/dL (70-75%). It is represented mathematically as:

      CvO2 = (1.34 x Hgb x SvO2 x 0.01) + (0.003 x PvO2)

      Where,

      1.34 = Huffner’s constant
      Hgb = Haemoglobin level (g/dL)
      SvO2 = % oxyhaemoglobin saturation of mixed venous blood
      PvO2 = 0.0225 = mL of O2 dissolved per 100mL plasma per kPa, or 0.003 mL per mmHg

      Therefore,

      CvO2 = (1.34 x 10 x 70 x 0.01) + (0.003 x 50)

      CvO2 = 9.38 + 0.15 = 9.53 mL/100mL

    • This question is part of the following fields:

      • Clinical Measurement
      5.6
      Seconds
  • Question 2 - Concerning forced alkaline diuresis, which of the following statements is true? ...

    Correct

    • Concerning forced alkaline diuresis, which of the following statements is true?

      Your Answer: Can be used in a barbiturate overdose

      Explanation:

      In situations of poisoning or drug overdose with acid dugs like salicylates and barbiturates, forced alkaline diuresis may be used.

      With regards to overdose with alkaline drugs, forced acid diuresis is used.

      By changing the pH of the urine, the ionised portion of the drug stays in the urine, and this prevents its diffusion back into the blood. Charged molecules do not readily cross biological membranes.

      The process involves the infusion of specific fluids at a rate of about 500ml per hour. This requires monitoring of the central venous pressure, urine output, plasma electrolytes, especially potassium, and blood gas analysis.

      The fluid regimen recommended is:
      500ml of 1.26% sodium bicarbonate (not 200ml of 8.4%)
      500ml of 5% dextrose and
      500ml of 0.9% sodium chloride.

    • This question is part of the following fields:

      • Physiology
      4.7
      Seconds
  • Question 3 - A single intravenous dose of 100 mg phenytoin was administered to a 70...

    Correct

    • A single intravenous dose of 100 mg phenytoin was administered to a 70 kg patient and plasma concentration monitored.

      The concentration in plasma over time is recorded as follows:

      Time (hours) 1 2 3 4 5
      Concentration (mcg/mL) 100 71 50 35.5 25

      From the data available, the drug is likely eliminated by?

      Your Answer: First-order kinetics with a half-life of 2 hours

      Explanation:

      Elimination of phenytoin from the body follows first-order kinetics. This means that the rate of elimination is proportional to plasma concentration.

      The rate of elimination can be described by the equation:

      C = C0·e-kt

      Where:

      C = drug concentration
      C0 = drug concentration at time zero (extrapolated)
      k = Rate constant
      t = Time

      Enzyme systems become saturated when phenytoin concentrations exceed the normal range and elimination of the drug becomes zero-order. At this point, the drug is metabolised at a fixed rate and metabolism is independent of plasma concentration.

      Aspirin and ethyl alcohol are other drugs that behave this way.

    • This question is part of the following fields:

      • Pharmacology
      2.1
      Seconds
  • Question 4 - When nitrous oxide is stored in cylinders at room temperature, it is a...

    Correct

    • When nitrous oxide is stored in cylinders at room temperature, it is a gas.
      Which of its property is responsible for this?

      Your Answer: Critical temperature

      Explanation:

      The temperature above which a gas cannot be liquefied no matter how much pressure is applied is its critical temperature. The critical temperature of nitrous oxide is 36.5°C

      The minimum pressure that causes liquefaction is the critical pressure of that gas.

      The Poynting effect refers to the phenomenon where mixing of liquid nitrous oxide at low pressure with oxygen at high pressure (in Entonox) leads to formation of gas of nitrous oxide.

      There is no relevance of molecular weight to this question. it does not change with phase of a substance.

    • This question is part of the following fields:

      • Pharmacology
      2.3
      Seconds
  • Question 5 - All of the following statements are true about blood clotting except: ...

    Correct

    • All of the following statements are true about blood clotting except:

      Your Answer: Administration of aprotinin during liver transplantation surgery prolongs survival

      Explanation:

      Even though aprotinin reduces fibrinolysis and therefore bleeding, there is an associated increased risk of death. It was withdrawn in 2007.
      Protein C is dependent upon vitamin K and this may paradoxically increase the risk of thrombosis during the early phases of warfarin treatment.

      The coagulation cascade include two pathways which lead to fibrin formation:
      1. Intrinsic pathway – these components are already present in the blood
      Minor role in clotting
      Subendothelial damage e.g. collagen
      Formation of the primary complex on collagen by high-molecular-weight kininogen (HMWK), prekallikrein, and Factor 12
      Prekallikrein is converted to kallikrein and Factor 12 becomes activated
      Factor 12 activates Factor 11
      Factor 11 activates Factor 9, which with its co-factor Factor 8a form the tenase complex which activates Factor 10

      2. Extrinsic pathway – needs tissue factor that is released by damaged tissue)
      In tissue damage:
      Factor 7 binds to Tissue factor – this complex activates Factor 9
      Activated Factor 9 works with Factor 8 to activate Factor 10

      3. Common pathway
      Activated Factor 10 causes the conversion of prothrombin to thrombin and this hydrolyses fibrinogen peptide bonds to form fibrin. It also activates factor 8 to form links between fibrin molecules.

      4. Fibrinolysis
      Plasminogen is converted to plasmin to facilitate clot resorption

    • This question is part of the following fields:

      • Physiology And Biochemistry
      3.4
      Seconds
  • Question 6 - An experiment is designed to investigate that how three diets having different sugar...

    Correct

    • An experiment is designed to investigate that how three diets having different sugar content affect the body weight to a different level.

      Which one of the following test will determine a statistically significant difference among the diets?

      Your Answer: ANOVA

      Explanation:

      Chi-square test is used to determine the statistically significant different between categorical variables. It also determines the difference between expected frequencies and the observed frequencies.

      Mann Whitney U test is used to determine the statistically significant different between two independent groups.

      Wilcoxon’s test is the test of dependency. it determines the statistically significant difference between two dependent groups.

      Student t-test is one of the most commonly used method to test the hypothesis. It determines the significant difference between the means of two different groups.

      ANOVA (analysis of variance) is similar to student’s t-test.

      ANOVA is a statistical method used to determines the statistically significant difference between the mean of more than two group. In this experiment as we are dealing with three different group, ANOVA is most suitable test to determine the difference between each groups.

    • This question is part of the following fields:

      • Statistical Methods
      2.4
      Seconds
  • Question 7 - A 63-year old man has palpitations and goes to the emergency room. An...

    Correct

    • A 63-year old man has palpitations and goes to the emergency room. An ECG shows tall tented T waves, which corresponds to phase 3 of the cardiac action potential.
      The shape of the T wave is as a result of which of the following?

      Your Answer: Repolarisation due to efflux of potassium

      Explanation:

      Cardiac conduction

      Phase 0 – Rapid depolarization. Opening of fast sodium channels with large influx of sodium

      Phase 1 – Rapid partial depolarization. Opening of potassium channels and efflux of potassium ions. Sodium channels close and influx of sodium ions stop

      Phase 2 – Plateau phase with large influx of calcium ions. Offsets action of potassium channels. The absolute refractory period

      Phase 3 – Repolarization due to potassium efflux after calcium channels close. Relative refractory period

      Phase 4 – Repolarization continues as sodium/potassium pump restores the ionic gradient by pumping out 3 sodium ions in exchange for 2 potassium ions coming into the cell. Relative refractory period

    • This question is part of the following fields:

      • Physiology And Biochemistry
      41.2
      Seconds
  • Question 8 - A 60-year-old male is being reviewed in the peri-operative assessment before total knee...

    Correct

    • A 60-year-old male is being reviewed in the peri-operative assessment before total knee replacement. He had a history of a heart transplant 10 years back. His resting heart rate is 110 beats per minute. On examination, ECG showed sinus tachycardia.

      Which of the following explains this tachycardia?

      Your Answer: Loss of parasympathetic innervation

      Explanation:

      Normally, at rest vagal influence is dominant producing the heart rate of 60-80 beats per minute even if the intrinsic automaticity of Sinoatrial Node is 100-110 beats per minute.

      The transplanted heart has no autonomic nervous supply. So, it will respond to endogenous and exogenous catecholamine. This loss of parasympathetic innervation is responsible for the tachycardia in this patient.

      Hypokalaemia can cause myocardial excitability and potential for ventricular ectopic and supraventricular arrhythmias. Hypothyroidism is also unlikely to cause tachycardia in this patient.

    • This question is part of the following fields:

      • Pathophysiology
      2.3
      Seconds
  • Question 9 - 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
      3.7
      Seconds
  • Question 10 - 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
      3.4
      Seconds
  • Question 11 - A 78-year-old man with a previous history of ischaemic heart disease is admitted...

    Correct

    • A 78-year-old man with a previous history of ischaemic heart disease is admitted to hospital. He is scheduled for a cardiopulmonary exercise test (CPX) before he undergoes an elective abdominal aneurysm repair.

      What measurement obtained during a CPX test alone provides the best indication for postoperative mortality?

      Your Answer: Anaerobic threshold

      Explanation:

      Cardiopulmonary exercise testing (CPX, CPEX, CPET) is a non-invasive testing method used to determine the performance of the heart, lungs and skeletal muscle. It measures the exercise tolerance of the patient.

      The parameters measured include:

      ECG and ST-segment analysis and blood pressure
      Oxygen consumption (VO2)
      Carbon dioxide production (VCO2)
      Gas flows and volumes
      Respiratory exchange ratio (RER)
      Respiratory rate
      Anaerobic threshold (AT)

      The anaerobic threshold (AT) is an estimate of exercise ability. Any measurement below 11 ml/kg/min is usually related with an increase in mortality, especially when there is a background of myocardial ischaemia occurring during the test.

      Peak VO2 <20 mL/kg with a low AT have a correlation with postoperative complications and a 30 day mortality. The CPX test is used for risk-testing patients prior to surgery to determine the appropriate postoperative care facilities. The V slope measured in CPX testing represents VO2 versus VCO2 relationship. During AT, the ramp of V slope increases, but does not provide a picture of postoperative mortality.

    • This question is part of the following fields:

      • Clinical Measurement
      2
      Seconds
  • Question 12 - Which of the following would most likely explain a failed post-operative analgesia via...

    Correct

    • Which of the following would most likely explain a failed post-operative analgesia via local anaesthesia of a neck abscess?

      Your Answer: pKA

      Explanation:

      For the local anaesthetic base to be stable in solution, it is formulated as a hydrochloride salt. As such, the molecules exist in a quaternary, water-soluble state at the time of injection. However, this form will not penetrate the neuron. The time for onset of local anaesthesia is therefore predicated on the proportion of molecules that convert to the tertiary, lipid-soluble structure when exposed to physiologic pH (7.4).

      The ionization constant (pKa) for the anaesthetic predicts the proportion of molecules that exists in each of these states. By definition, the pKa of a molecule represents the pH at which 50% of the molecules exist in the lipid-soluble tertiary form and 50% in the quaternary, water-soluble form. The pKa of all local anaesthetics is >7.4 (physiologic pH), and therefore a greater proportion the molecules exists in the quaternary, water-soluble form when injected into tissue having normal pH of 7.4.

      Furthermore, the acidic environment associated with inflamed tissues favours the quaternary, water-soluble configuration even further. Presumably, this accounts for difficulty when attempting to anesthetize inflamed or infected tissues; fewer molecules exist as tertiary lipid-soluble forms that can penetrate nerves.

    • This question is part of the following fields:

      • Physiology
      4.6
      Seconds
  • Question 13 - General anaesthesia is administered to a patient in a hospital in Lhasa which...

    Correct

    • General anaesthesia is administered to a patient in a hospital in Lhasa which is one of the highest cities in the world (at 11,975 feet). An Anaesthetic rotameter is normally calibrated at 20 C and 1 bar pressure and is known to be underread at altitude. The temperature of the theatre was 10 C.

      Which one of the following physical properties is responsible for the rotameter inaccuracy in these conditions?

      Your Answer: Density of the gas

      Explanation:

      Since the gas is less dense at higher altitudes, the density of a gas influences flows when passing through the orifice. Due to this reason, for a given flow rate, the bobbin will not be forced as far up the rotameter tube.

      At higher altitudes, the volume of a fixed mass of gas increases, and therefore the molecules of gas are widely spaced resulting in a decrease in density with an increase in altitude.

      Viscosity is simply termed as friction of gas. The viscosity of a gas is important only at low flow rates when the flow characteristic of the gas is laminar.

      Charle’s law stated that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature (T) provided the pressure remains constant.

      Boyle’s law for a fixed amount of gas at constant temperature, the pressure (P) and volume (V) are inversely proportional.

    • This question is part of the following fields:

      • Basic Physics
      3.7
      Seconds
  • Question 14 - Very small SI units are easily expressed using mathematical prefixes.

    One femtolitre is equal...

    Incorrect

    • Very small SI units are easily expressed using mathematical prefixes.

      One femtolitre is equal to which of the following volumes?

      Your Answer: 0.000, 000, 000, 001 L

      Correct Answer: 0.000, 000, 000, 000, 001 L

      Explanation:

      Small measurement units are denoted by the following SI mathematical prefixes:

      1 deci = 0.1
      1 milli = 0.001
      1 micro = 0.000001
      1 nano = 0.000000001
      1 pico = 0.000000000001
      1 femto = 0.000000000000001 (used to measure red blood cell volume)
      1 atto = 0.000000000000000001

    • This question is part of the following fields:

      • Basic Physics
      3.2
      Seconds
  • Question 15 - A 79-year-old female complains of painful legs, especially in her thigh region. The...

    Correct

    • A 79-year-old female complains of painful legs, especially in her thigh region. The pain starts after walking and settles with rest. She occasionally has to take paracetamol to relieve the pain. She is a known case of hyperlipidaemia, type 2 diabetes mellitus, hypertension, and depression.

      Her physician makes a provisional diagnosis of claudication of the femoral artery, which is a continuation of the external iliac artery.
      Which of the following anatomical landmarks does the external iliac artery cross to become the femoral artery?

      Your Answer: Inguinal ligament

      Explanation:

      The external iliac artery is the larger of the two branches of the common iliac artery. It forms the main blood supply to the lower limbs. The common iliac bifurcates into the internal and external iliac artery anterior to the sacroiliac joint.

      The external iliac artery courses on the medial border of the psoas major muscles and exits the pelvic girdle posterior to the inguinal ligament. Here, midway between the anterior superior iliac spine and the pubic symphysis, the external iliac artery becomes the femoral artery and descends along the anteromedial part of the thigh in the femoral triangle.

      The pectineus forms the posterior border of the femoral canal.
      The femoral vein forms the lateral border of the femoral canal.
      The medial border of the adductor longus muscle forms the medial wall of the femoral triangle.
      The medial border of the sartorius muscle forms the lateral wall of the femoral triangle.

    • This question is part of the following fields:

      • Anatomy
      3
      Seconds
  • Question 16 - Radical prostatectomy is being performed on a 60-year-old man for carcinoma of the...

    Correct

    • Radical prostatectomy is being performed on a 60-year-old man for carcinoma of the prostate gland.

      What is the direct blood supply of the prostate?

      Your Answer: Inferior vesical artery

      Explanation:

      The prostate gland is primarily supplied by the inferior vesical artery, which branches off from the anterior division of the internal iliac artery. The inferior vesical artery supplies the base of the bladder, the distal ureters, and the prostate. The branches to the prostate communicate with the corresponding vessels of the opposite side.

      The inferior vesical artery branches into two main arteries:
      1. Urethral artery – supplies the transition zone and is the main arterial supply for the adenomas in BPH
      2. Capsular artery – supplies the glandular tissue

      The venous drainage of the prostate is from the prostatic venous plexus, which drains into the paravertebral veins.

    • This question is part of the following fields:

      • Anatomy
      7.4
      Seconds
  • Question 17 - What statement about endotoxins is true? ...

    Incorrect

    • What statement about endotoxins is true?

      Your Answer: Are produced mainly by Gram positive bacteria

      Correct Answer: Can often survive autoclaving

      Explanation:

      Endotoxins are the lipopolysaccharides found in the outer cell wall of Gram-negative bacteria. They are responsible for providing the structure and stability of the cell wall.

      They cannot be destroyed by normal sterilisation as they are heat stable molecules. They require the use of certain sterilant such as superoxide, peroxide and hypochlorite to be neutralised.

      They stimulate strong immune responses, but can only be destroyed partially by specific antibodies. Repeat infections occur as memory T cells cannot be formed.

      It can cause septicaemia and associated symptoms such as fever, shock, hypotension and nausea.

      It activates the alternative complement pathway and the coagulation pathway using secreted cytokines.

      It is not involved in botulism as clostridium botulinum, the responsible organism, secretes a neurotoxic exotoxin.

    • This question is part of the following fields:

      • Pathophysiology
      27.4
      Seconds
  • Question 18 - Which of the following statements is true with regards to the Krebs' cycle...

    Incorrect

    • Which of the following statements is true with regards to the Krebs' cycle (also known as the tricarboxylic acid cycle or citric acid cycle)?

      Your Answer: Krebs' cycle can function under anaerobic conditions

      Correct Answer: Alpha-ketoglutarate is a five carbon molecule

      Explanation:

      Krebs’ cycle (tricarboxylic acid cycle or citric acid cycle) is a sequence of reactions in which acetyl coenzyme A (acetyl-CoA) is metabolised and this results in carbon dioxide and hydrogen atoms production.

      This series of reactions occur in the mitochondria of eukaryotic cells, not the cytoplasm. The cycle requires oxygen and so, cannot function under anaerobic conditions.

      It is the common pathway for carbohydrate, fat and some amino acids oxidation and is required for high energy phosphate bond formation in adenosine triphosphate (ATP).

      When pyruvate enters the mitochondria, it is converted into acetyl-CoA. This represents the formation of a 2 carbon molecule from a 3 carbon molecule. There is loss of one CO2 but formation of one NADH molecule. Acetyl-CoA is condensed with oxaloacetate, the anion of a 4 carbon acid, to form citrate which is a 6 carbon molecule.

      Citrate is then converted into isocitrate, alpha-ketoglutarate, succinyl-CoA, succinate, fumarate, malate and finally oxaloacetate.

      The only 5 carbon molecule in the cycle is alpha-ketoglutarate.

    • This question is part of the following fields:

      • Physiology
      38.3
      Seconds
  • Question 19 - Following a near drowning accident, a 5-year-old child is admitted to the emergency...

    Correct

    • Following a near drowning accident, a 5-year-old child is admitted to the emergency department and advanced paediatric life support is started.

      What is the child's approximate weight, according to the preferred formulae of the Resuscitation Council (UK), the European Resuscitation Council, and the Royal College of Anaesthetists?

      Your Answer: 20-25kg

      Explanation:

      For estimating a child’s weight, the Resuscitation Council (UK) and European Resuscitation Council teach the following formula:

      Weight = (age + 4) × 2

      The weight of the child will be around 20 kg.

      This formula is used in the Primary FRCA exam by the Royal College of Anaesthetists.

      In ‘developed’ countries, the traditional ‘APLS formula’ for estimating weight in children based on age (wt in kg = [age+4] x 2) is acknowledged as underestimating weight by 33.4 percent on average, with the degree of underestimation increasing with increasing age.

      However, more recently, the APLS formula ‘Weight=3(age)+7’ has been found to provide a mean underestimate of only 6.9%. This formula is applicable to children aged 1 to 13 years.

      The estimated weight based on age using this formula is 25 kg.

    • This question is part of the following fields:

      • Physiology
      270.9
      Seconds
  • Question 20 - Which of the following is a correctly stated fundamental (base) SI unit? ...

    Incorrect

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

      Your Answer: A watt is the unit of energy

      Correct 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
      5.2
      Seconds
  • Question 21 - In endurance athletes, which of the following physiological adaptations to exercise is the...

    Incorrect

    • In endurance athletes, which of the following physiological adaptations to exercise is the best predictor of performance?

      Your Answer: Increased capillary density in muscle

      Correct Answer: Velocity of blood lactate accumulation

      Explanation:

      Multiple regression analysis revealed that velocity of lactate accumulation (VOBLA) accounted for 92 percent of the variation in marathon running velocity (VM), and VOBLA plus training volume prior to the marathon accounted for 96 percent of the variation. Percent ST muscle fibre distribution (r = 0.55-0.69) and capillary density (r = 052-0.63) were found to be positively correlated with all performance variables. As a result, marathon running performance was linked to VOBLA and the ability to run at a pace close to it during the race. The percent ST, capillary density, and training volume were all related to these properties.

      Another metabolic adaptation compared to normal people is the early selection of fat for oxidation by muscle, especially when glucose availability is limited during high-intensity exercise. This helps to delay the onset of muscle fatigue, but it does not prevent VOBLA.

      For a given level of exercise, training can also result in cardiovascular adaptation, such as increased heart size, increased contractility, and a slower heart rate. All of these factors contribute to an increase in maximal oxygen consumption (VO2 max), but genetic factors, despite intensive training, play a large role in an athlete’s performance.

    • This question is part of the following fields:

      • Pathophysiology
      213.9
      Seconds
  • Question 22 - Which of the following statement is not true regarding Adrenaline or Epinephrine? ...

    Incorrect

    • Which of the following statement is not true regarding Adrenaline or Epinephrine?

      Your Answer: Inhibits Insulin secretion

      Correct Answer: Inhibits Glucagon secretion in the pancreas

      Explanation:

      Adrenaline acts on ?1, ?2,?1, and ?2 receptors and also on dopamine receptors (D1, D2) and have sympathomimetic effects.

      Natural catecholamines are Adrenaline, Noradrenaline, and Dopamine

      Adrenaline is a sympathomimetic amine with both alpha and beta-adrenergic stimulating properties.
      Adrenaline is the drug of choice for anaphylactic shock
      Adrenaline is also used in patients with cardiac arrest. The preferred route is i.v. followed by the intra-osseous and endotracheal route.

      Adrenaline is released by the adrenal glands, acts on ? 1 and 2, ? 1 and 2 receptors, and is responsible for fight or flight response.

      It acts on ? 2 receptors in skeletal muscle vessels-causing vasodilation.

      It acts on ? adrenergic receptors to inhibit insulin secretion by the pancreas. It also stimulates glycogenolysis in the liver and muscle, stimulates glycolysis in muscle.

      It acts on ? adrenergic receptors to stimulate glucagon secretion in the pancreas. It also stimulates Adrenocorticotrophic Hormone (ACTH) and stimulates lipolysis by adipose tissue

    • This question is part of the following fields:

      • Pharmacology
      23.2
      Seconds
  • Question 23 - Which of the following statements is the most correct about ketamine? ...

    Correct

    • Which of the following statements is the most correct about ketamine?

      Your Answer: The S (+) isomer is more potent that the R (-) isomer

      Explanation:

      Ketamine, a phencyclidine derivative, is an antagonist at the NMDA receptor. It causes depression of the CNS that is dose dependent and induces a dissociative anaesthetic state with profound analgesia and amnesia.

      Ketamine has a chiral centre usually presented as a racemic mixture with two optical isomers, S (+) and R (-) forms. These isomers are in equal proportions. The S (+) isomer is about three times more potent than the R (-) form. The S (+) form is less likely to cause emergence delirium and hallucinations.

      Ketamine is extensively metabolised by hepatic microsomal cytochrome P450 enzymes producing norketamine as its main metabolite. Norketamine has a one third to one fifth as potency as its parent compound.
      It increases the CMRO2, cerebral blood flow and potentially increase intracranial pressure.

    • This question is part of the following fields:

      • Pharmacology
      19.9
      Seconds
  • Question 24 - A 40-year old farmer came into the emergency room with a chief complaint...

    Incorrect

    • A 40-year old farmer came into the emergency room with a chief complaint of 4 episodes of non-bloody diarrhoea. This was associated with frequent urination, vomiting and salivation. History also revealed frequent use of insecticides. Upon physical examination, there was miosis and bradycardia.

      Given the different types of bonds, which is the most likely bond formed between insecticide poisoning and receptors?

      Your Answer: Hydrophobic

      Correct Answer: Covalent

      Explanation:

      Organophosphate poisoning occurs most often due to accidental exposure to toxic amounts of pesticides. Signs and symptoms include diarrhoea, urination, miosis, bradycardia, emesis, lacrimation, lethargy and salivation.

      Organophosphates are classified as indirect acting cholinomimetics, and their mode of action involves: (1) the inhibition of acetylcholinesterase (AChE) by forming a stable covalent bond on the active site serine; and, (2) amplification of endogenously release acetylcholine (ACh), hence the clinical manifestation.

      There are 4 types of bonds or interactions: ionic, covalent, hydrogen bonds, and van der Waals interactions. Ionic and covalent bonds are strong interactions that require a larger energy input to break apart. When an element donates an electron from its outer shell, a positive ion is formed. The element accepting the electron is now negatively charged. Because positive and negative charges attract, these ions stay together and form an ionic bond. Covalent bonds form when an electron is shared between two elements and are the strongest and most common form of chemical bond in living organisms. Covalent bonds form between the elements that make up the biological molecules in our cells. Unlike ionic bonds, covalent bonds do not dissociate in water.

      When polar covalent bonds containing a hydrogen atom form, the hydrogen atom in that bond has a slightly positive charge. This is because the shared electron is pulled more strongly toward the other element and away from the hydrogen nucleus. Because the hydrogen atom is slightly positive, it will be attracted to neighbouring negative partial charges. When this happens, a weak interaction occurs between the slightly positive charge of the hydrogen atom of one molecule and the slightly negative charge of the other molecule. This interaction is called a hydrogen bond.

    • This question is part of the following fields:

      • Pathophysiology
      26.5
      Seconds
  • Question 25 - The half-empty cylinder weighs 4.44 kg.
    The tare weight of a nitrous...

    Incorrect

    • The half-empty cylinder weighs 4.44 kg.
      The tare weight of a nitrous oxide cylinder is 4 kg.
      The molecular weight of nitrous oxide is 44gm.

      Based on the data, how many litres of nitrous oxide remains in the cylinder for use?

      Your Answer: 112 litres

      Correct Answer: 224 litres

      Explanation:

      The Tare weight of a cylinder is the weight when it is empty. So,

      Weight of cylinder – tare weight = weight of remaining N2O (g).
      4.44 kg – 4 kg = 0.44 kg
      Here,
      0.44 kg of nitrous oxide remains in the cylinder

      Since the molecular weight of nitrous oxide is 44 g and one mole of an ideal gas will occupy a volume of 22.4 litres at STP
      Therefore amount left in the cylinder is several (gN2O/44) x 22.4 litres of N2O.

      (440/44) x 22.4 = 224 litres.

    • This question is part of the following fields:

      • Basic Physics
      80.4
      Seconds
  • Question 26 - A peripheral nerve stimulator is used to stimulate the ulnar nerve at the...

    Incorrect

    • A peripheral nerve stimulator is used to stimulate the ulnar nerve at the wrist to indicate the degree of neuromuscular blockade.

      Which single muscle or group of muscles of the hand supplied by the ulnar nerve is best for monitoring the twitch function during neuromuscular blockade?

      Your Answer: Abductor digiti minimi

      Correct Answer: Adductor pollicis

      Explanation:

      In anaesthesia, adductor pollicis neuromuscular monitoring with ulnar nerve stimulation is commonly used. It is the gold standard for measuring the degree of block and comparing neuromuscular blocking drugs and their effects on other muscles.

      Electrodes are usually placed over the ulnar nerve at the wrist to monitor the adductor pollicis.

      Neuromuscular blocking drugs have different sensitivity levels in different muscle groups.

      To achieve the same level of blockade, the diaphragm requires 1.4 to 2 times the amount of neuromuscular blocking agent as the adductor pollicis muscle. The small muscles of the larynx and the ocular muscles are two other respiratory muscles that are less resistant than the diaphragm (especially corrugator supercilii).

      The abdominal muscles, Orbicularis oculi, peripheral muscles of the limbs, Geniohyoid, Masseter, and Upper airway muscles are the most sensitive to neuromuscular blocking agents.

      The C8-T1 nerve roots, which are part of the medial cord of the brachial plexus, form the ulnar nerve. It enters the hand via the ulnar canal, superficial to the flexor retinaculum, after following the ulnar artery at the wrist.

      The nerve then splits into two branches: superficial and deep. The palmaris brevis is supplied by the superficial branch, which also provides palmar digital nerves to one and a half fingers. The dorsal surface of the medial/ulnar 1.5 fingers, as well as the corresponding skin over the hand, are also supplied by it (as well as the palmar surface).

      The ulnar nerve’s deep branch runs between the abductor and flexor digiti minimi, which it supplies. It also innervates the opponens, and with the deep palmar arch, it curves around the hook of the hamate and laterally across the palm. All of the interossei, the medial two lumbricals, the adductor pollicis, and, in most cases, the flexor pollicis brevis are supplied there.

    • This question is part of the following fields:

      • Anatomy
      75.2
      Seconds
  • Question 27 - A 70-year-old man presents to hospital complaining of dysphagia. He is scheduled for...

    Incorrect

    • A 70-year-old man presents to hospital complaining of dysphagia. He is scheduled for a rigid oesophagoscopy.

      On examination, He is noted to have severe osteoarthritis in his cervical spine resulting in limited rotation and flexion-extension. He has no other neurological signs or symptoms.

      He is given anaesthesia for the procedure, which is complicated by a difficult intubation (Cormack-Lehane 3), but was eventually achieved using a gum elastic bougie.

      After recovering from anaesthesia, he is examined and found to have severe motor weakness of upper limbs, and mild motor weakness of lower limbs, bladder dysfunction and sensory loss of varying degrees below the level of C5.

      What incomplete spinal cord lesion is most likely to be responsible for his symptoms?

      Your Answer: Anterior spinal artery thrombosis

      Correct Answer: Central cord syndrome

      Explanation:

      Central cord syndrome is the most commonly occurring type of partial spinal cord lesion. It is more likely to occur in older patients with cervical spondylosis and a hyperextension injury. The injury to the spinal cord occurs in the grey matter causing the following symptoms:

      Disproportionally higher motor function weakness in the upper limbs than in lower limbs
      Dysfunction of the bladder
      Degrees of sensory loss below the level of the lesion

      An anterior spinal artery infarction will interrupt the corticospinal tract resulting in paralysis of motor function, loss of pain and temperature sensation, all occurring below the level of the injury.

      Brown-Sequard syndrome occurs as a result of the hemisection of the spinal cord. Its symptoms include ipsilateral upper motor neurone paralysis and loss of proprioception, with contralateral loss of pain and temperature sensation.

      Spinal cord infarctions rarely occur in the posterior spinal artery.

      Cauda equina syndrome occurs as a result of compression of the lumbosacral spinal nerve roots below the level of the conus medullaris. Injury to these nerves will cause partial or complete loss of movement and sensation in this distribution.

    • This question is part of the following fields:

      • Pathophysiology
      474.5
      Seconds
  • Question 28 - Regarding the use of soda lime as part of a modern circle system...

    Incorrect

    • Regarding the use of soda lime as part of a modern circle system with a vaporiser outside the circuit (VOC), which of the following is its most deleterious consequence?

      Your Answer: Compound A formation

      Correct Answer: Carbon monoxide formation

      Explanation:

      When using dry soda lime for VOCs, very high amounts of carbon monoxide may be produced, regardless of the inhalational anaesthetic agent used. The carbon monoxide produced is sufficient enough to cause cytotoxic and anaemic hypoxia. To prevent this, soda lime canisters are shaken well to even out the packing of granules. This can help to evenly distribute gas flow for proper CO2 absorption and ventilation.

      Compound A is formed when dry soda lime, or soda lime in high temperature, reacts with the inhalational anaesthetic Sevoflurane. Animal studies have shown renal toxicity in rats, but renal adverse effects in humans are yet to be observed.

      When monitors are not employed with VOCs, deleterious effects are not for certain. However, monitors not employed with vaporiser inside the circuit (VIC) can lead to significant adverse events.

    • This question is part of the following fields:

      • Pathophysiology
      23.5
      Seconds
  • Question 29 - A 65-year-old man got operated on for carotid endarterectomy for his carotid artery...

    Incorrect

    • A 65-year-old man got operated on for carotid endarterectomy for his carotid artery disease. He is recovering well post-surgery. However, on follow-up in the ward, he has hoarseness of his voice.

      Which of the following explains the hoarseness?

      Your Answer: Damage to the hypoglossal nerve

      Correct Answer: Damage to the vagus

      Explanation:

      During carotid endarterectomy, injury to the vagus nerve or its branches can cause hoarseness. Injury to the vagus nerve can result in adductor vocal cord paralysis. It can also cause other symptoms like dysphagia or even vocal cord immobility.

      Carotid endarterectomy is the procedure to relieve an obstruction in the carotid artery by opening the artery at its origin and stripping off the atherosclerotic plaque with the intima. Because of the internal carotid artery relations, there is a risk of cranial nerve injury during the procedure involving one or more of the following nerves: CN IX, CN X (or its branch, the superior laryngeal nerve), CN XI, or CN XII.

      However, only damage to the vagus would account for speech difficulties.

    • This question is part of the following fields:

      • Anatomy
      36.5
      Seconds
  • Question 30 - A 5-year-old child is scheduled for squint surgery requiring general anaesthesia.

    To begin, she...

    Incorrect

    • A 5-year-old child is scheduled for squint surgery requiring general anaesthesia.

      To begin, she is given sevoflurane for the inhalation induction, then intravenous access is established along with the insertion of a supraglottic airway. Anaesthesia is maintained with fentanyl 1 mcg/kg, with an air/oxygen/sevoflurane mix with spontaneous respirations.

      Once the surgery begins, her pulse rate drastically reduces from 120 beats/min to 8 beats/min.

      What is the most appropriate next step for this patient?

      Your Answer: Intravenous adrenaline 10 mcg/kg

      Correct Answer: Tell surgeon to stop surgical retraction

      Explanation:

      This sudden change in pulse rate is due to the oculocardiac reflex. It is a >20% reduction in pulse rate as a result of placing pressure directly on the eyeball. The reflex arc has an afferent and efferent arm:

      The afferent (sensory) arm: The trigeminal nerve (CN V)

      The efferent arm: The vagus nerve (CN X)

      The most appropriate action is to ask the surgeon to stop retraction of the extraocular muscles, Assess for hypoxia, and give 100% oxygen if indicated.

      Atropine of glycopyrrolate can be administered to counteract the reflex, and also prevent any further vagal reflexes.

      Administration of fentanyl may increase patient’s risk of bradycardia and sinus arrest in this case.

      Adrenaline is not indicated here as other treatment options will provide sufficient relief from arrhythmia.

    • This question is part of the following fields:

      • Pathophysiology
      486.8
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Clinical Measurement (2/3) 67%
Physiology (3/4) 75%
Pharmacology (3/4) 75%
Physiology And Biochemistry (2/2) 100%
Statistical Methods (2/2) 100%
Pathophysiology (1/7) 14%
Anatomy (3/5) 60%
Basic Physics (1/3) 33%
Passmed