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  • Question 1 - Pressure volume loop represents the compliance of left ventricle.

    Considering there...

    Incorrect

    • Pressure volume loop represents the compliance of left ventricle.

      Considering there is no change in preload and myocardial contractility, which physiological change may result an increase in left ventricular afterload?

      Your Answer: Reduced cardiac work

      Correct Answer: Increased end-systolic volume

      Explanation:

      If there is no change in preload and myocardial contractility, there will be decrease in end-diastolic volume and stroke volume. So there must be increase in end-systolic volume.

    • This question is part of the following fields:

      • Physiology
      67.9
      Seconds
  • Question 2 - Following a near drowning accident, a 5-year-old child is admitted to the emergency...

    Incorrect

    • 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: 15-19 kg

      Correct 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
      16.3
      Seconds
  • Question 3 - The resistance to flow in a blood vessel is affected by the following...

    Correct

    • The resistance to flow in a blood vessel is affected by the following except?

      Your Answer: Thickness of the vessel wall

      Explanation:

    • This question is part of the following fields:

      • Physiology
      25.6
      Seconds
  • Question 4 - Metabolization of many drugs used in anaesthesia involves the cytochrome P450 (CYP) isoenzymes.

    The...

    Correct

    • Metabolization of many drugs used in anaesthesia involves the cytochrome P450 (CYP) isoenzymes.

      The CYP enzyme most likely to be subject to genetic variability and thus cause adverse drug reactions is which of these?

      Your Answer: CYP2D6

      Explanation:

      Approximately 25% of phase-1 drug reactions is made responsible by CYP2D6.

      As much as a 1,000-fold difference in the ability to metabolise drugs by CYP2D6 can happen between phenotypes, and this may result in adverse drug reactions (ADRs).

      The metabolism of antiemetics, beta-blockers, codeine, tramadol, oxycodone, hydrocodone, tamoxifen, antidepressants, neuroleptics, and antiarrhythmics is also as a result of CYP2D6.

      Patients who take drugs that are metabolised by CYP2D6 but have poor CYP2D6 metabolism are more likely to have ADRs. People with ultra-rapid CYP2D6 metabolism may have a decreased drug effect due to low plasma concentrations of these drugs.

      All the other CYP enzymes are subject to genetic polymorphism. Variants are less likely to lead to adverse drug reactions.

    • This question is part of the following fields:

      • Physiology
      11.6
      Seconds
  • Question 5 - Which statement best describes the bispectral index (BIS)? ...

    Correct

    • Which statement best describes the bispectral index (BIS)?

      Your Answer: It decreases during normal sleep

      Explanation:

      The bispectral index (BIS) is one of several systems used in anaesthesiology as of 2003 to measure the effects of specific anaesthetic drugs on the brain and to track changes in the patient’s level of sedation or hypnosis. It is a complex mathematical algorithm that allows a computer inside an anaesthesia monitor to analyse data from a patient’s electroencephalogram (EEG) during surgery. It is a dimensionless number (0-100) that is a summative measurement of time domain, frequency domain and high order spectral parameters derived from electroencephalogram (EEG) signals.

      Sleep and anaesthesia have similar behavioural characteristics but are physiologically different but BIS monitors can be used to measure sleep depth. With increasing sleep depth during slow-wave sleep, BIS levels decrease. This correlates with changes in regional cerebral blood flow when measured using positron emission tomography (PET).

      BIS shows a dose-response relationship with the intravenous and volatile anaesthetic agents. Opioids produce a clinical change in the depth of sedation or analgesia but fail to produce significant changes in the BIS. Ketamine increases CMRO2 and EEG activity.

      BIS is unable to predict movement in response to a surgical stimulus. Some of these are spinal reflexes and not perceived by the cerebral cortex.

      BIS is used during cardiopulmonary bypass to measure depth of anaesthesia and an index of cerebral perfusion. However, it cannot predict subtle or significant cerebral damage.

    • This question is part of the following fields:

      • Physiology
      12.8
      Seconds
  • Question 6 - Useful diagnostic information can be obtained from measuring the osmolality of biological fluids....

    Correct

    • Useful diagnostic information can be obtained from measuring the osmolality of biological fluids.

      Of the following physical principles, which is the most accurate and reliable method of measuring osmolality?

      Your Answer: Depression of freezing point

      Explanation:

      Colligative properties are properties of solutions that depend on the number of dissolved particles in solution. They do not depend on the identities of the solutes.

      All of the above have colligative properties with the exception of depression of melting point.

      The osmolality from the concentration of a substance in a solution is measured by an osmometer. The freezing point of a solution can determines concentration of a solution and this can be measured by using a freezing point osmometer. This is applicable as depression of freezing point is directly correlated to concentration.

      Vapour pressure osmometers, which measure vapour pressure, may miss certain volatiles such as CO2, ammonia and alcohol that are in the solution

      The use of a freezing point osmometer provides the most accurate and reliable results for the majority of applications.

      Colligative properties does not include melting point depression . Mixtures of substances in which the liquid phase components are insoluble, display a melting point depression and a melting range or interval instead of a fixed melting point.

      The magnitude of the melting point depression depends on the mixture composition.

      The melting point depression is used to determine the purity and identity of compounds. EMLA (eutectic mixture of local anaesthetics) cream is a mixture of lidocaine and prilocaine and is used as a topical local anaesthetic. The melting point of the combined drugs is lower than that individually and is below room temperature (18°C).

    • This question is part of the following fields:

      • Physiology
      84.3
      Seconds
  • Question 7 - Regarding amide local anaesthetics, which one factor has the most significant effect on...

    Incorrect

    • Regarding amide local anaesthetics, which one factor has the most significant effect on its duration of action?

      Your Answer: Lipid solubility

      Correct Answer: Protein binding

      Explanation:

      When drugs are bound to proteins, drugs cannot cross membranes and exert their effect. Only the free (unbound) drug can be absorbed, distributed, metabolized, excreted and exert pharmacologic effect. Thus, when amide local anaesthetics are bound to ?1-glycoproteins, their duration of action are reduced.

      The potency of local anaesthetics are affected by lipid solubility. Solubility influences the concentration of the drug in the extracellular fluid surrounding blood vessels. The brain, which is high in lipid content, will dissolve high concentration of lipid soluble drugs. When drugs are non-ionized and non-polarized, they are more lipid-soluble and undergo more extensive distribution. Hence allowing these drugs to penetrate the membrane of the target cells and exert their effect.

      Tissue pKa and pH will determine the degree of ionization.

    • This question is part of the following fields:

      • Physiology
      33.7
      Seconds
  • Question 8 - The Fick principle can be used to determine the blood flow to any...

    Correct

    • The Fick principle can be used to determine the blood flow to any organ of the body.

      At rest, which one of these organs has the highest blood flow (ml/min/100g)?

      Your Answer: Thyroid gland

      Explanation:

      After the carotid body, the thyroid gland is the second most richly vascular organ in the body.

      The global blood flow to the thyroid gland can be measured using:
      1. Colour ultrasound sonography
      2. Quantitative perfusion maps using MRI of the thyroid gland using an arterial spin labelling (ASL) method.

      This table shows the blood flow to various organs of the body at rest:
      Organ Blood Flow(ml/minute/100g)
      Hepatoportal 58
      Kidney 420
      Brain 54
      Skin 13
      Skeletal muscle 2.7
      Heart 87
      Carotid body 2000
      Thyroid gland 560

    • This question is part of the following fields:

      • Physiology
      11.4
      Seconds
  • Question 9 - A common renal adverse effect of non-steroidal anti-inflammatory drugs is? ...

    Incorrect

    • A common renal adverse effect of non-steroidal anti-inflammatory drugs is?

      Your Answer: Renal papillary necrosis

      Correct Answer: Haemodynamic renal insufficiency

      Explanation:

      Prostaglandins do not play a major role in regulating RBF in healthy resting individuals. However, during pathophysiological conditions such as haemorrhage and reduced extracellular fluid volume (ECVF), prostaglandins (PGI2, PGE1, and PGE2) are produced locally within the kidneys and serve to increase RBF without changing GFR. Prostaglandins increase RBF by dampening the vasoconstrictor effects of both sympathetic activation and angiotensin II. These effects are important because they prevent severe and potentially harmful vasoconstriction and renal ischemia. Synthesis of prostaglandins is stimulated by ECVF depletion and stress (e.g. surgery, anaesthesia), angiotensin II, and sympathetic nerves.

      Non-steroidal anti-inflammatory drugs (NSAIDs), such as ibuprofen and naproxen, potently inhibit prostaglandin synthesis. Thus administration of these drugs during renal ischemia and hemorrhagic shock is contraindicated because, by blocking the production of prostaglandins, they decrease RBF and increase renal ischemia. Prostaglandins also play an increasingly important role in maintaining RBF and GFR as individuals age. Accordingly, NSAIDs can significantly reduce RBF and GFR in the elderly.

    • This question is part of the following fields:

      • Physiology
      24
      Seconds
  • Question 10 - Which of the following is true in the Kreb's cycle? ...

    Correct

    • Which of the following is true in the Kreb's cycle?

      Your Answer: Alpha-ketoglutarate is a five carbon molecule

      Explanation:

      Krebs’ cycle (tricarboxylic acid cycle or citric acid cycle) is a sequence of reactions to release stored energy through oxidation of acetyl coenzyme A (acetyl-CoA). Some of the products are carbon dioxide and hydrogen atoms.

      The sequence of reactions, known collectively as oxidative phosphorylation, only occurs in the mitochondria (not cytoplasm).

      The Krebs cycle can only take place when oxygen is present, though it does not require oxygen directly, because it relies on the by-products from the electron transport chain, which requires oxygen. It is therefore considered an aerobic process. It is the common pathway for the oxidation of carbohydrate, fat and some amino acids, required for the formation of adenosine triphosphate (ATP).

      Pyruvate enters the mitochondria and is converted into acetyl-CoA. Acetyl-CoA is then condensed with oxaloacetate, to form citrate which is a six carbon molecule. Citrate is subsequently converted into isocitrate, alpha-ketoglutarate, succinyl-CoA, succinate, fumarate, malate and finally oxaloacetate.

      The only five carbon molecule in the cycle is Alpha-ketoglutarate.

    • This question is part of the following fields:

      • Physiology
      40
      Seconds
  • Question 11 - You're summoned to the emergency room, where a 39-year-old man has been admitted...

    Incorrect

    • You're summoned to the emergency room, where a 39-year-old man has been admitted following a cardiac arrest. He was rescued from a river, but little else is known about him.

      CPR is being performed on the patient, who has been intubated. He's received three DC shocks and is still in VF. A rectal temperature of 29.5°C is taken with a low-reading thermometer.

      Which of the following statements about his resuscitation is correct?

      Your Answer: 1 mg IV adrenaline and 300 mg IV amiodarone should be administered

      Correct Answer: No further DC shocks and no drugs should be given until his core temperature is greater than 30°C

      Explanation:

      The guidelines for the management of cardiac arrest in hypothermic patients published by the UK Resuscitation Council differ slightly from the standard algorithm.

      In a patient with a core temperature of less than 30°C, do the following:

      If you’re on the shockable side of the algorithm (VF/VT), you should give three DC shocks.
      Further shocks are not recommended until the patient has been rewarmed to a temperature of more than 30°C because the rhythm is refractory and unlikely to change.
      There should be no drugs given because they will be ineffective.

      In a patient with a core temperature of 30°C to 35°C, do the following:

      DC shocks are used as usual.
      Because they are metabolised much more slowly, the time between drug doses should be doubled.

      Active rewarming and protection against hyperthermia should be given to the patient.

      Option e is false because there is insufficient information to determine whether resuscitation should be stopped.

    • This question is part of the following fields:

      • Physiology
      44.7
      Seconds
  • Question 12 - Comparing pressure-volume curves in patients during an asthma attack with that of healthy...

    Incorrect

    • Comparing pressure-volume curves in patients during an asthma attack with that of healthy subjects.

      The increased resistive work of breathing in the patients with asthma is best indicated by?

      Your Answer: Longer expiratory time

      Correct Answer: Larger hysteresis loop

      Explanation:

      A major source of caloric expenditure and oxygen consumption in the body is work of breathing (WOB) and 70% of this is to overcome elastic forces. The remaining 30% is for flow-resistive work

      In a normal patient breathing normally, the total area of hysteresis pressure volume curve represents the flow-resistive WOB.

      The area of the expiratory resistive work increases during an asthma attack making the compliance curve larger in area. The larger the area the greater the work required to breathe.

    • This question is part of the following fields:

      • Physiology
      84.2
      Seconds
  • Question 13 - Which of the following statements is about the measurement of glomerular filtration rate...

    Correct

    • Which of the following statements is about the measurement of glomerular filtration rate (GFR) is correct?

      Your Answer: The result matches clearance of the indicator if it is renally inert

      Explanation:

      The measurements of GFR are done using renally inert indicators like inulin, where passive rate of filtration at the glomerulus = rate of excretion. Normal value is about 180 litres per day.

      GFR is altered by renal blood flow but blood flow does not need to be measured.

      The reabsorption of Na leads to a low excretion rate and low urine concentration and therefore its use as an indicator would lead to an erroneously LOW GFR.

      If there is tubular secretion of any solute, the clearance value will be higher than that of inulin. This will be either due to tubular reabsorption or the solute not being freely filtered at the glomerulus.

    • This question is part of the following fields:

      • Physiology
      50.7
      Seconds
  • Question 14 - The typical fluid compartments in a normal 70kg male are: ...

    Incorrect

    • The typical fluid compartments in a normal 70kg male are:

      Your Answer:

      Correct Answer: intracellular>extracellular

      Explanation:

      Body fluid compartments in a 70kg male:
      Total volume=42L (60% body weight)
      Intracellular fluid compartment (ICF) =28L
      Extracellular fluid compartment (ECF) = 14L

      ECF comprises:
      Intravascular fluid (plasma) = 3L
      Extravascular fluid = 11L

      Extravascular fluids comprises:
      Interstitial fluid = 10.5L
      Transcellular fluid = 0.5L

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 15 - Cells use adenosine-5-triphosphate (ATP) as a coenzyme and is a source of energy.

    Glucose...

    Incorrect

    • Cells use adenosine-5-triphosphate (ATP) as a coenzyme and is a source of energy.

      Glucose metabolism produces the most ATP from which of the following biochemical processes?



      Your Answer:

      Correct Answer: Electron transport phosphorylation in the mitochondria

      Explanation:

      Glycolysis occurs in the cytoplasm of the cell. It converts 1 glucose molecule (6-carbon) to pyruvate (two 3-carbon molecules) and produces 4 ATP molecules and 2NADH but uses 2 ATP in the process with an overall net energy production of 2 ATP.

      Pyruvate is then oxidised to acetyl coenzyme A (generating 2 NADH per pyruvate molecule). This takes place in the mitochondria and then enters the Krebs cycle (citric acid cycle). It produces 2 ATP, 8 NADH and 2 FADH2 per glucose molecule.

      Electron transport phosphorylation takes place in the mitochondria. The aim of this process is to break down NADH and FADH2 and also to pump H+ into the outer compartment of the mitochondria. It produces 32 ATP with an overall net production of 36ATP.

      In anaerobic respiration which occurs in the cytoplasm, pyruvate is reduced to NAD producing 2 ATP.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 16 - The SI unit of energy is the joule. Energy can be kinetic, potential,...

    Incorrect

    • The SI unit of energy is the joule. Energy can be kinetic, potential, electrical or chemical energy.

      Which of these correlates with the most energy?

      Your Answer:

      Correct Answer: Energy released when 1 kg fat is metabolised to CO2 and water (the energy content of fat is 37 kJ/g)

      Explanation:

      The derived unit of energy, work or amount of heat is joule (J). It is defined as the amount of energy expended if a force of one newton (N) is applied through a distance of one metre (N·m)

      J = 1 kg·m/s2·m = 1 kg·m2/s2 or 1 kg·m2·s-2

      Kinetic energy (KE) = ½ MV2

      An object with a mass of 1500 kg moving at 30 m/s correlates to 675 kJ:

      KE = ½ (1500) × (30)2 = 750 × 900 = 675 kJ

      Total energy released when 1 kg fat is metabolised to CO2 and water is 37 MJ. 1 g fat produces 37 kJ/g, therefore 1 kg fat produces 37,000 × 1000 = 37 MJ.

      Raising the temperature of 1 kg water from 0°C to 100°C correlates to 420 kJ. The amount of energy needed to change the temperature of 1 kg of the substance by 1°C is the specific heat capacity. We have 1 kg water therefore:

      4,200 J × 100 = 420,000 J = 420 kJ

      In order to calculate the energy involved in raising a 100 kg mass to a height of 1 km against gravity, we need to calculate the potential energy (PE) of the mass:

      PE = mass × height attained × acceleration due to gravity
      PE = 100 kg × 1000 m × 10 m/s2 = 1 MJ

      The heat generated when a direct current of 10 amps flows through a heating element for 10 seconds when the potential difference across the element is 1000 volts can be calculated by applying Joule’s law of heating:

      Work done (WD) = V (potential difference) × I (current) × t (time)
      WD = 10 × 10 × 1000 = 100 kJ

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 17 - In the erect position, the partial pressure of oxygen in the alveoli (PAO2)...

    Incorrect

    • In the erect position, the partial pressure of oxygen in the alveoli (PAO2) is higher in the apical lung units than in the basal lung units.

      What is the most significant reason for this?

      Your Answer:

      Correct Answer: The V/Q ratio of apical units is greater than that of basal units

      Explanation:

      In any alveolar unit, the V/Q ratio affects alveolar oxygen (PAO2) and carbon dioxide tension (PACO2).

      The partial pressure of alveolar carbon dioxide (PACO2) is plotted against the partial pressure of alveolar oxygen in a Ventilation-Perfusion (V/Q) ratio graph (PAO2). Given a set of model assumptions, the curve represents all of the possible values for PACO2 and PAO2 that an individual alveolus could have.

      In the case of an infinity V/Q ratio (ventilation but no perfusion or dead space), the PACO2 of the alveolus will equal zero, while the PAO2 will approach that of external air (150mmmHg). At the apex of the lung, the V/Q ratio is 3.3, compared to 0.67 at the base.

      PACO2 and PAO2 approach the partial pressures for these gases in the venous blood when the V/Q ratio is zero (no ventilation but perfusion). At the base of the lung, the V/Q ratio is 0.67, whereas at the apex, it is 3.3.

      PAO2 at the apex is typically 132mmHg, and PACO2 is typically 28mmHg.

      The average PAO2 at the base is 89 mmHg, while the average PACO2 is 42 mmHg.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 18 - The following is true about the extracellular fluid (ECF) in a normal adult...

    Incorrect

    • The following is true about the extracellular fluid (ECF) in a normal adult woman weighing 60 kg.

      Your Answer:

      Correct Answer: Has a total volume of about 12 litres

      Explanation:

      Total body water (TBW) is about 50% to 70% in adults depending on how much fat is present. ECF is relatively contracted in an obese person.

      The simple rule is 60-40-20. (60% of weight = total body water, 40% of body weight is ICF and 20% is ECF)

      For this woman, the total body water is 36 litres (0.6 × 60). ECF is 12 litres (1/3 of TBW) and 24 litres (2/3 of TBW) is intracellular fluid .

      Sodium concentration is approximately 135-145 mmol/L in the ECF.

      The ECF is made up of both intravascular and extravascular fluid and plasma proteins is found in both.

    • This question is part of the following fields:

      • Physiology
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  • Question 19 - Regarding bilirubin, which one of the following statement is true? ...

    Incorrect

    • Regarding bilirubin, which one of the following statement is true?

      Your Answer:

      Correct Answer: Conjugated bilirubin is stored in the gall bladder

      Explanation:

      Bilirubin is the tetrapyrrole and a catabolic product of heme. 70-90% of bilirubin is end product of haemoglobin degradation in the liver.

      Bilirubin circulates in the blood in 2 forms; unconjugated and conjugated bilirubin.

      Unconjugated bilirubin is insoluble in water. It travels through the bloodstream to the liver, where it changes from insoluble into a soluble form (i.e.; unconjugated into conjugated form).

      This conjugated bilirubin travels from the liver into the small intestine and the gut bacteria convert bilirubin into urobilinogen and then into urobilin (not urobilin to urobilinogen). A very small amount passes into the kidneys and is excreted in urine.

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 20 - A participant of a metabolism study is to be fed only granulated sugar...

    Incorrect

    • A participant of a metabolism study is to be fed only granulated sugar and water for 48 hours. What would be his expected respiratory quotient at the end of the study?

      Your Answer:

      Correct Answer: 1

      Explanation:

      The respiratory quotient is the ratio of CO2 produced to O2 consumed while food is being metabolized:

      RQ = CO2 eliminated/O2 consumed

      Most energy sources are food containing carbon, hydrogen and oxygen. Examples include fat, carbohydrates, protein, and ethanol. The normal range of respiratory coefficients for organisms in metabolic balance usually ranges from 1.0-0.7.

      Granulated sugar is a refined carbohydrate with no significant fat, protein or ethanol content.

      The RQ for carbohydrates is = 1.0

      The RQ for the rest of the compounds are:

      Fats RQ = 0.7
      The chemical composition of fats differs from that of carbohydrates in that fats contain considerably fewer oxygen atoms in proportion to atoms of carbon and hydrogen.

      Protein RQ = 0.8
      Due to the complexity of various ways in which different amino acids can be metabolized, no single RQ can be assigned to the oxidation of protein in the diet; however, 0.8 is a frequently utilized estimate.

    • This question is part of the following fields:

      • Physiology
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SESSION STATS - PERFORMANCE PER SPECIALTY

Physiology (7/13) 54%
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