00
Correct
00
Incorrect
00 : 00 : 00
Session Time
00 : 00
Average Question Time ( Mins)
  • Question 1 - Which of the following statements is about the measurement of glomerular filtration rate...

    Incorrect

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

      Your Answer: One would expect approximately 120 litres to be filtered daily

      Correct 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
      2051.8
      Seconds
  • Question 2 - A common renal adverse effect of non-steroidal anti-inflammatory drugs is? ...

    Correct

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

      Your 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
      13.4
      Seconds
  • Question 3 - The passage of glucose into the brain is facilitated by which transport method?...

    Correct

    • The passage of glucose into the brain is facilitated by which transport method?

      Your Answer: Facilitated diffusion

      Explanation:

      Glucose transport is a highly regulated process accomplished mostly by facilitated diffusion using carrier proteins to cross cell membranes.

      There are many transporters, but the most important are known as glucose transporters (GLUTs).

      Stresses in various form of acute and chronic forms affect the activity of glucose transporters.
      They are responsive to many types of metabolic stress, including hypoxia, injury, hypoglycaemia, numerous metabolic inhibitors, stress hormones, and other influences such as growth factors.

      Numerous signalling pathways appear to be involved in transporter regulation.

      New evidence suggests that stresses regulating GLUTs are not only acute biological stresses. In addition, chronic low-grade inflammation, and their associated chronic diseases also lead to altered glucose transport. These include obesity, type 2 diabetes, cardiovascular disease, and the growth and spread of many tumours that are affected by altered glucose transporters. Some of these glucose transport effects are compensatory, while others are pathogenic.

      Ultimately, deliberate manipulation of GLUTs could be used as treatment for some of these chronic diseases.

    • This question is part of the following fields:

      • Physiology
      12.9
      Seconds
  • Question 4 - 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
      225.7
      Seconds
  • Question 5 - The Fick principle can be used to determine the blood flow to any...

    Incorrect

    • 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: Brain

      Correct 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
      13.3
      Seconds
  • Question 6 - An orthopaedic surgery is scheduled for a 68-year-old man. He is normally in...

    Incorrect

    • An orthopaedic surgery is scheduled for a 68-year-old man. He is normally in good shape. His routine biochemistry results are checked and found to be within normal limits.

      Which of the following pairs has the greatest impact on his plasma osmolarity?

      Your Answer: Calcium and phosphate ions

      Correct Answer: Sodium and potassium cations

      Explanation:

      The number of osmoles (Osm) of solute per litre (L) of solution (Osmol/L) is the unit of measurement for solute concentration. The calculated serum osmolality assumes that the primary solutes in the serum are sodium salts (chloride and bicarbonate), glucose, and urea nitrogen.

      2 (Na + K) + Glucose + Urea (all in mmol/L) = calculated osmolarity

      313 mOsm/L = 2 (144 + 6) + 9.5 + 3.5

      Sodium and potassium ions clearly contribute the most to plasma osmolarity. Glucose and urea, on the other hand, are less so.

      The osmolarity of normal serum is 285-295 mOsm/L. Temperature and pressure affect osmolality, and this calculated variable is less than osmolality for a given solution.

      The number of osmoles (Osm) of solute per kilogramme (Osm/kg) is a measure of osmolality, which is also a measure of solute concentration. Temperature and pressure have no effect on the value. An osmometer is used to measure it in the lab. Osmometers rely on a solution’s colligative properties, such as a decrease in freezing point or a rise in vapour pressure.

      The osmolar gap (OG) is calculated as follows:

      OG = osmolaRity calculated from measured serum osmolaLity

      Excess alcohols, lipids, and proteins in the blood can all contribute to the difference.

    • This question is part of the following fields:

      • Physiology
      43.5
      Seconds
  • Question 7 - A 61-year-old woman with myasthenia gravis is admitted to the ER with type...

    Incorrect

    • A 61-year-old woman with myasthenia gravis is admitted to the ER with type II respiratory failure. There is a suspicion of myasthenic crisis.

      She is in a semiconscious state. Her blood pressure is 160/90 mmHg, pulse is 110 beats per minute, temperature is 37°C, and oxygen saturation is 84 percent.

      With a PaCO2 of 75 mmHg (10 kPa) breathing air, blood gas analysis confirms she is hypoventilating.

      Which of the following values is the most accurate representation of her alveolar oxygen tension (PAO2)?

      Your Answer: 9

      Correct Answer: 7.3

      Explanation:

      The following is the alveolar gas equation:

      PAO2 = PiO2 − PaCO2/R

      Where:

      PAO2 is the partial pressure of oxygen in the alveoli.
      PiO2 is the partial pressure of oxygen inhaled.
      PaCO2 stands for partial pressure of carbon dioxide in the arteries.
      The amount of carbon dioxide produced (200 mL/minute) divided by the amount of oxygen consumed (250 mL/minute) equals R = respiratory quotient. With a normal diet, the value is 0.8.

      By subtracting the partial pressure exerted by water vapour at body temperature, the PiO2 can be calculated:

      PiO2 = 0.21 × (100 kPa − 6.3 kPa)
      PiO2 = 19.8

      Substituting:
      PAO2 = 19.8 − 10/0.8
      PAO2 = 19.8 − 12.5
      PAO2 = 7.3k Pa

    • This question is part of the following fields:

      • Physiology
      181.5
      Seconds
  • Question 8 - Which of the statements below best describe the total cerebral flow (CBF) in...

    Correct

    • Which of the statements below best describe the total cerebral flow (CBF) in an adult?

      Your Answer: Accounts for 15% of the cardiac output

      Explanation:

      While the brain only weighs 3% of the body weight, 15% of the cardiac output goes towards the brain.

      Between mean arterial pressures (MAP) of 60-130 mmHg, autoregulation of cerebral blood flow (CBF) occurs. Exceeding this, the CBF is maintained at a constant level. This is controlled mainly by the PaCO2 level, and the autonomic nervous system has minimal role.

      Beyond these limits, the CBF is directly proportional to the MAP, not the systolic blood pressure.

    • This question is part of the following fields:

      • Physiology
      639.9
      Seconds
  • Question 9 - A human's resting oxygen consumption (VO2) is typically 3.5 ml/kg/minute (one metabolic equivalent...

    Correct

    • A human's resting oxygen consumption (VO2) is typically 3.5 ml/kg/minute (one metabolic equivalent or 1 MET).

      Which of the following options is linked to the highest VO2 when a person is at rest?

      Your Answer: Neonate

      Explanation:

      The oxygen consumption rate (VO2) at rest is 3.5 ml/kg/minute (one metabolic equivalent or 1 MET).
      3.86 ml/kg/minute thyrotoxicosis

      Young children consume a lot of oxygen: around 7 ml/kg/min when they are born. The metabolic cost of breathing is higher in children than in adults, and it can account for up to 15% of total oxygen consumption. Similarly, an infant’s metabolic rate is nearly twice that of an adult, resulting in a larger alveolar minute volume and a lower FRC.

      At term, oxygen consumption at rest can increase by as much as 40% (5 ml/kg/minute) and can rise to 60% during labour.

      When compared to normal basal metabolism, sepsis syndrome increases VO2 and resting metabolic rate by 30% (4.55 ml/kg/minute). In septicaemic shock, VO2 decreases.

      Dobutamine hydrochloride was infused into 12 healthy male volunteers at a rate of 2 micrograms per minute per kilogramme, gradually increasing to 4 and 6 micrograms per minute per kilogramme. Dobutamine was infused for 20 minutes for each dose. VO2 increased by 10% to 15%. (3.85-4.0 ml/kg/min).

    • This question is part of the following fields:

      • Physiology
      33.7
      Seconds
  • Question 10 - Of the following, which of these oxygen carrying molecules causes the greatest shift...

    Correct

    • Of the following, which of these oxygen carrying molecules causes the greatest shift of the oxygen-dissociation curve to the left?

      Your Answer: Myoglobin (Mb)

      Explanation:

      Myoglobin is a haemoglobin-like, iron-containing pigment that is found in muscle fibres. It has a high affinity for oxygen and it consists of a single alpha polypeptide chain. It binds only one oxygen molecule, unlike haemoglobin, which binds 4 oxygen molecules.

      The myoglobin ODC is a rectangular hyperbola. There is a very low P50 0.37 kPa (2.75 mmHg). This means that it needs a lower P50 to facilitate oxygen offloading from haemoglobin. It is low enough to be able to offload oxygen onto myoglobin where it is stored. Myoglobin releases its oxygen at the very low PO2 values found inside the mitochondria.

      P50 is defined as the affinity of haemoglobin for oxygen: It is the PO2 at which the haemoglobin becomes 50% saturated with oxygen. Normally, the P50 of adult haemoglobin is 3.47 kPa(26 mmHg).

      Foetal haemoglobin has 2 ? and 2 ?chains. The ODC is left shifted – this means that P50 lies between 2.34-2.67 kPa [18-20 mmHg]) compared with the adult curve and it has a higher affinity for oxygen. Foetal haemoglobin has no ? chains so this means that there is less binding of 2.3 diphosphoglycerate (2,3 DPG).

      Carbon monoxide binds to haemoglobin with an affinity more than 200-fold higher than that of oxygen. This therefore decreases the amount of haemoglobin that is available for oxygen transport. Carbon monoxide binding also increases the affinity of haemoglobin for oxygen, which shifts the oxygen-haemoglobin dissociation curve to the left and thus impedes oxygen unloading in the tissues.

      In sickle cell disease, (HbSS) has a P50 of 4.53 kPa(34 mmHg).

    • This question is part of the following fields:

      • Physiology
      48.7
      Seconds
  • Question 11 - 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: Forms a greater proportion of the total body weight in the obese than in the lean woman

      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
      53.7
      Seconds
  • Question 12 - 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: The resuscitation should be stopped

      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
      57.8
      Seconds
  • Question 13 - Which statement is true when describing carbonic anhydrase? ...

    Incorrect

    • Which statement is true when describing carbonic anhydrase?

      Your Answer: Contains iron

      Correct Answer: Isoenzyme IV is found in the brush border of the proximal convoluted tubule

      Explanation:

      Carbonic anhydrase is an enzyme which contains zinc and can be found in:
      1. Erythrocytes
      2. Pulmonary endothelium
      3. The intestine
      4. Pancreas
      5. Cardiac muscle and skeletal muscle.

      To date, there have been seven isoenzymes identified. Of note, isoenzyme IV is found in the brush border of the proximal convoluted tubule and isoenzyme II is found within the luminal cells.

      Acetazolamides a carbonic anhydrase inhibitor and is used as prophylaxis against mountain sickness and in glaucoma management.

      Spironolactone is a potassium diuretic and is an aldosterone antagonist.

    • This question is part of the following fields:

      • Physiology
      54.8
      Seconds
  • Question 14 - 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
      Seconds
  • Question 15 - 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 16 - The statement that best describes lactic acidosis is: ...

    Incorrect

    • The statement that best describes lactic acidosis is:

      Your Answer:

      Correct Answer: It can be precipitated by intravenous fructose

      Explanation:

      An elevated arterial blood lactate level and an increase anion gap ([Na + K] – [Cl + HCO3]) of >20mmol gives rise to lactic acidosis. It can also be a result of overproduction and/or reduced metabolism of lactic acid.

      The liver and kidney are the main sites of lactate metabolism, not skeletal muscle.

      The two types of lactic acidosis that are known are:

      Type A – due to tissue hypoxia, inadequate tissue perfusion and anaerobic glycolysis. These may be seen in cardiac arrest, shock, hypoxaemia and anaemia. The management of type A lactic acidosis involves reversing the underlying cause of the tissue hypoxia.

      Type B – occurs in the absence of tissue hypoxia. Some of the causes of this include hepatic failure, renal failure, diabetes mellitus, pancreatitis and infection. Some drugs can also cause this lie aspirin, ethanol, methanol, biguanides and intravenous fructose.

      The mainstay of treatment involves:
      1. Optimising tissue oxygen delivery
      2. Correcting the cause
      3. Intravenous sodium bicarbonate

      In resistant cases, peritoneal dialysis can be performed.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 17 - Of the following, which option best describes the muscle type that has the...

    Incorrect

    • Of the following, which option best describes the muscle type that has the fastest twitch response to stimulation?

      Your Answer:

      Correct Answer: Type IIb skeletal muscle

      Explanation:

      Human skeletal muscle is composed of a heterogeneous collection of muscle fibre types which differ histologically, biochemically and physiologically.

      It can be biochemically classified into 2 groups. This is based on muscle fibre myosin ATPase histochemistry. These are:

      Type 1 (slow twitch): Muscle fibres depend upon aerobic glycolytic metabolism and aerobic oxidative metabolism. They are rich in mitochondria, have a good blood supply, rich in myoglobin and are resistant to fatigue.

      Type II (fast twitch): Muscle fibres are sub-divided into:
      Type IIa – relies on aerobic/oxidative metabolism
      Type IIb – relies on anaerobic/glycolytic metabolism.

      Fast twitch muscle fibres produce short bursts of power but are more easily fatigued.

      Cardiac and smooth muscle twitches are relatively slow compared with skeletal muscle.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 18 - The renal glomerulus is able to filter 180 litres of blood per day,...

    Incorrect

    • The renal glomerulus is able to filter 180 litres of blood per day, as determined by the starling forces present in the glomerulus. Ninety-nine percent of which is reabsorbed thereafter.

      Water is reabsorbed in the highest proportion in which segment of the nephron?

      Your Answer:

      Correct Answer: Proximal convoluted tubule

      Explanation:

      Sixty-seven percent of filtered water is reabsorbed in the proximal tubule. The driving force for water reabsorption is a transtubular osmotic gradient established by reabsorption of solutes (e.g., NaCl, Na+-glucose).

      Henle’s loop reabsorbs approximately 25% of filtered NaCl and 15% of filtered water. The thin ascending limb reabsorbs NaCl by a passive mechanism, and is impermeable to water. Reabsorption of water, but not NaCl, in the descending thin limb increases the concentration of NaCl in the tubule fluid entering the ascending thin limb. As the NaCl-rich fluid moves toward the cortex, NaCl diffuses out of the tubule lumen across the ascending thin limb and into the medullary interstitial fluid, down a concentration gradient as directed from the tubule fluid to the interstitium. This mechanism is known as the counter current multiplier.

      The distal tubule and collecting duct reabsorb approximately 8% of filtered NaCl, secrete variable amounts of K+ and H+, and reabsorb a variable amount of water (approximately 8%-17%).

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 19 - The action potential in a muscle fibre is initiated by which of these...

    Incorrect

    • The action potential in a muscle fibre is initiated by which of these ions?

      Your Answer:

      Correct Answer: Sodium ions

      Explanation:

      The cardiac action potential has several phases which have different mechanisms of action as seen below:
      Phase 0: Rapid depolarisation – caused by a rapid sodium influx.
      These channels automatically deactivate after a few ms

      Phase 1: caused by early repolarisation and an efflux of potassium.

      Phase 2: Plateau – caused by a slow influx of calcium.

      Phase 3 – Final repolarisation – caused by an efflux of potassium.

      Phase 4 – Restoration of ionic concentrations – The resting potential is restored by Na+/K+ATPase.
      There is slow entry of Na+into the cell which decreases the potential difference until the threshold potential is reached. This then triggers a new action potential

      Of note, cardiac muscle remains contracted 10-15 times longer than skeletal muscle.

      Different sites have different conduction velocities:
      1. Atrial conduction – Spreads along ordinary atrial myocardial fibres at 1 m/sec

      2. AV node conduction – 0.05 m/sec

      3. Ventricular conduction – Purkinje fibres are of large diameter and achieve velocities of 2-4 m/sec, the fastest conduction in the heart. This allows a rapid and coordinated contraction of the ventricles

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 20 - 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 21 - Which of the following would most likely explain a failed post-operative analgesia via...

    Incorrect

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

      Your Answer:

      Correct 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
      0
      Seconds
  • Question 22 - 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:

      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
      0
      Seconds
  • Question 23 - 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:

      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
      0
      Seconds
  • Question 24 - 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
      0
      Seconds
  • Question 25 - In a normal healthy adult breathing 100 percent oxygen, which of the following...

    Incorrect

    • In a normal healthy adult breathing 100 percent oxygen, which of the following is the most likely cause of an alveolar-arterial (A-a) oxygen difference of 30 kPa?

      Your Answer:

      Correct Answer: Atelectasis

      Explanation:

      The ‘ideal’ alveolar PO2 minus arterial PO2 is the alveolar-arterial (A-a) oxygen difference.

      The ‘ideal’ alveolar PO2 is derived from the alveolar air equation and is the PO2 that the lung would have if there was no ventilation-perfusion (V/Q) inequality and it was exchanging gas at the same respiratory exchange ratio as real lung.

      The amount of oxygen in the blood is measured directly in the arteries.

      The A-a oxygen difference (or gradient) is a useful measure of shunt and V/Q mismatch, and it is less than 2 kPa in normal adults breathing air (15 mmHg). Because the shunt component is not corrected, the A-a difference increases when breathing 100 percent oxygen, and it can be up to 15 kPa (115 mmHg).

      An abnormally low or abnormally high V/Q ratio within the lung can cause an increased A-a difference, though the former is more common. Atelectasis, which results in a low V/Q ratio, is the most likely cause of an A-a difference in a healthy adult breathing 100 percent oxygen.

      Hypoventilation may cause an increase in alveolar (and thus arterial) CO2, lowering alveolar PO2 according to the alveolar air equation.

      The alveolar PO2 is also reduced at high altitude.

      Healthy people are unlikely to have a right-to-left shunt or an oxygen transport diffusion defect.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 26 - An intravenous infusion is started with a 500 mL bag of 0.18 percent...

    Incorrect

    • An intravenous infusion is started with a 500 mL bag of 0.18 percent N. saline and 4% dextrose.

      Which of the following best describes its make-up?

      Your Answer:

      Correct Answer: Osmolarity 284 mOsmol/L, sodium 15 mequivalents and glucose 20 g

      Explanation:

      30 mmol Na+ and 30 mmol Cl- are found in 1 litre of 0.18 percent N. saline with 4% dextrose. Percent (percent) refers to the number of grammes of a compound per 100 mL, so a litre of 4 percent dextrose solution contains 40 grammes.

      As a result, a 500 mL bag of 1/5th N. saline and 4% dextrose contains approximately 15 mequivalents of sodium and 20 g of glucose. It is hypotonic due to its osmolarity of 284.

      Because of the risk of hyponatraemia, it is no longer considered the crystalloid of choice for fluid maintenance in children.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 27 - Which of the following statements is true about fluid balance? ...

    Incorrect

    • Which of the following statements is true about fluid balance?

      Your Answer:

      Correct Answer: After intravenous administration of crystalloids, the distribution of these fluids throughout the body depends on its osmotic activity

      Explanation:

      When there is capillary leakage as seen in dependent oedema or ascites, oncotic pressure becomes a problem.

      The intracellular sodium concentration is very sensitive to the extracellular sodium concentrations. When there is an imbalance, osmosis occurs resulting in shifts in water between the two compartments.

      The microvascular endothelium relies upon osmosis and other processes as it is not freely permeable to water.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 28 - One of the non-pharmacologic management of COPD is smoking cessation. Given a case...

    Incorrect

    • One of the non-pharmacologic management of COPD is smoking cessation. Given a case of a 60-year old patient with history of smoking for 30 years and a FEV1 of 70%, what would be the most probable five-year course of his FEV1 if he ceases to smoke?

      Your Answer:

      Correct Answer: The FEV1 will decrease at the same rate as a non-smoker

      Explanation:

      For this patient, his forced expiratory volume in 1 second (FEV1) will decrease at the same rate as a non-smoker.

      There is a notable, but slow, decline in FEV1 when an individual reaches the age of 26. An average reduction of 30 mls every year in non-smokers, while a more significant reduction of 50-70 mls is observed in approximately 20% of smokers.

      Considering the age of the patient, individuals who begin smoking cessation by the age of 60 are far less likely to achieve normal FEV1 levels, even in the next five years. It is expected that their FEV1 will be approximately 14% less than their peers of the same age.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 29 - Using a negative feedback loop, Haem production is controlled by which of these...

    Incorrect

    • Using a negative feedback loop, Haem production is controlled by which of these enzymes?

      Your Answer:

      Correct Answer: ALA synthetase

      Explanation:

      Heme a exists in cytochrome a and heme c in cytochrome c; they are both involved in the process of oxidative phosphorylation. 5′-Aminolevulinic acid synthase (ALA-S) is the regulated enzyme for heme synthesis in the liver and erythroid cells.

      There are two forms of ALA Synthase, ALAS1, and ALAS2.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 30 - Following an acute appendicectomy, a 6-year-old child is admitted to the recovery unit.

    Your...

    Incorrect

    • Following an acute appendicectomy, a 6-year-old child is admitted to the recovery unit.

      Your consultant has requested that you prescribe maintenance fluids for the next 12 hours. The child is 21 kg in weight.

      What is the most suitable fluid volume to be prescribed?

      Your Answer:

      Correct Answer: 732 ml

      Explanation:

      After a paediatric case, you’ll frequently have to calculate and prescribe maintenance fluids. The ‘4-2-1 rule’ should be used as a guideline:

      1st 10 kg – 4 ml/kg/hr
      2nd 10 kg – 2 ml/kg/hr
      Subsequent kg – 1 ml/kg/hr

      Hence

      1st 10 kg = 4 × 10 = 40 ml
      2nd 10 kg = 2 × 10 = 20 ml
      Subsequent kg = 1 × 1 = 1 ml
      Total = 61 ml/hr

      61 × 12 = 732 ml over 12 hrs.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Physiology (4/13) 31%
Passmed