00
Correct
00
Incorrect
00 : 00 : 00
Session Time
00 : 00
Average Question Time ( Mins)
  • Question 1 - In an experimental study, a healthy subject was given one litre of 5%...

    Correct

    • In an experimental study, a healthy subject was given one litre of 5% dextrose within a 15-minute period. Which of the following mechanisms is expected to affect the urine output?

      Your Answer: Inhibition of arginine vasopressin (AVP) secretion

      Explanation:

      Changes in the osmolality of body fluids (changes as minor as 1% are sufficient) play the most important role in regulating AVP secretion. The receptors that monitor changes in osmolality of body fluids (termed osmoreceptors) are distinct from the cells that synthesize and secrete AVP, and are located in the organum vasculosum of the lamina terminalis (OVLT) of the hypothalamus. The osmoreceptors sense changes in body osmolality by either shrinking or swelling. When the effective osmolality of the plasma increases, the osmoreceptors send signals to the AVP synthesizing/secreting cells located in the supraoptic and paraventricular nuclei of the hypothalamus, and AVP synthesis and secretion are stimulated. Conversely, when the effective osmolality of the plasma is reduced, secretion is inhibited. Because AVP is rapidly degraded in the plasma, circulating levels can be reduced to zero within minutes after secretion is inhibited.

      In this scenario, the osmolality of the plasma will decrease to an estimate of 2.5%, hence inhibition of AVP.

      Stimulation of atrial stretch receptors is incorrect because the increase in plasma volume is still below the threshold for its activation.

      Osmotic diuresis is incorrect because 5% dextrose is isotonic, hence osmotic diuresis is not probable.

      Renin is inhibited when an excess of NaCl in the tubular fluid is sensed by the macula densa.

    • This question is part of the following fields:

      • Physiology
      13.6
      Seconds
  • Question 2 - Regarding bilirubin, which one of the following statement is true? ...

    Correct

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

      Your 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
      13.9
      Seconds
  • Question 3 - You're summoned to the emergency room, where a 39-year-old man has been admitted...

    Correct

    • 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: 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
      26.6
      Seconds
  • Question 4 - Which plasma protein will bind the thyroid hormone triiodothyronine (T3) more readily? ...

    Incorrect

    • Which plasma protein will bind the thyroid hormone triiodothyronine (T3) more readily?

      Your Answer: Immunoglobulins

      Correct Answer: Thyroxine binding globulin

      Explanation:

      Secreted T4 and T3 circulate in the bloodstream almost entirely bound to proteins. Normally only about 0.03% of total plasma T4 and 0.3% of total plasma T3 exist in the free state. Free T3 is biologically active and mediates the effects of thyroid hormone on peripheral tissues in addition to exerting negative feedback on the pituitary and hypothalamus. The major binding protein is thyroxine-binding globulin (TBG), which is synthesized in the liver and binds one molecule of T4 or T3. About 70% of circulating T4 and T3 is bound to TBGl 10% to 15% is bound to another specific thyroid-binding protein called transthyretin (TTR). Albumin binds 15% to 20%, and 3% to lipoproteins. Ordinarily only alterations in TBG concentration significantly affect total plasma T4 and T3 levels.

      Two important biological functions have been ascribed to TBG. First, it maintains a large circulating reservoir of T4 that buffers any acute changes in thyroid gland function. Second, binding of plasma T4 and T3 to proteins prevents loss of these relatively small hormone molecules in urine and thereby helps conserve iodide. TTR transports T4 in CSF and provides thyroid hormones to the CNS.

    • This question is part of the following fields:

      • Physiology
      10.2
      Seconds
  • Question 5 - A mercury barometer can be used to determine absolute pressure. A mercury manometer...

    Incorrect

    • A mercury barometer can be used to determine absolute pressure. A mercury manometer can be used to check blood pressure. The SI units of length(mm) are used to measure pressure.

      Why is pressure expressed in millimetres of mercury (mmHg)?

      Your Answer: Pressure is directly proportional to length of the mercury column and is the only constant

      Correct Answer: Pressure is directly proportional to length of the mercury column and is variable

      Explanation:

      A mercury barometer can be used to determine absolute pressure. A glass tube with one closed end serves as the barometer. The open end is inserted into a mercury-filled open vessel. The mercury in the container is pushed into the tube by atmospheric pressure exerted on its surface. Absolute pressure is the distance between the tube’s meniscus and the mercury surface.

      Pressure is defined as force in newtons per unit area (F) (A). 

      Mass of mercury = area (A) × density (ρ) × length (L)
      Pressure = ((A × ρ × L) × 9.8 m/s2)/A
      Pressure = ρ × L x 9.8
      Pressure is proportional to L

      The numerator and denominator of the above equation, area (A), cancel out. The constants are density and the gravitational acceleration value.

      The length is proportional to the applied pressure.

    • This question is part of the following fields:

      • Physiology
      23.6
      Seconds
  • Question 6 - The typical fluid compartments in a normal 70kg male are: ...

    Incorrect

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

      Your Answer: intracellular>interstitial

      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
      91.3
      Seconds
  • Question 7 - 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: Descending loop of Henle

      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
      135.5
      Seconds
  • Question 8 - 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 9 - A 20-year old male was involved in an accident and has presented to...

    Incorrect

    • A 20-year old male was involved in an accident and has presented to the Emergency Department with a pelvic crush injury.

      The clinical exam according to ATLS protocol revealed the following:

      Airway-patent

      Breathing - respiratory rate 25 breaths per minute. Breath sounds are vesicular and there are no added sounds.

      Circulation - Capillary refill time - 4 seconds. Peripheries are cool. Pulse 125 beats/min. BP - 125/95 mmHg.

      Disability - GSC 15, anxious and in pain.

      Secondary survey reveals no other injuries. The patient is administered high flow oxygen and IV access is established.

      The most appropriate IV fluid regimen in this case will be which of the following?

      Your Answer:

      Correct Answer: Judicious infusion of Hartmann's solution to maintain a systolic blood pressure greater than 90mmHg

      Explanation:

      These clinical signs suggest that 15-30% of circulating blood volume has been lost.

      Pelvic fractures are associated with significant haemorrhage (>2000 ml) that can be concealed. This may require aggressive fluid resuscitation which is initially with crystalloids and then blood. What is also important is including stabilisation of the fracture(s) and pain relief.

      The Advanced Trauma Life Support (ATLS) classification of haemorrhagic shock is as follows:

      Class I haemorrhage (blood loss up to 15%):
      <750 ml of blood loss
      Minimal tachycardia
      No changes in blood pressure, RR or pulse pressure
      Patients do not normally not require fluid replacement as will be restored in 24 hours, but in trauma, this needs to be correct.

      Class II haemorrhage (15-30% blood volume loss):
      Uncomplicated haemorrhage requiring crystalloid resuscitation
      Represents about 750 – 1500 ml of blood loss
      Tachycardia, tachypnoea and a decrease in pulse pressure (due to a rise in diastolic component due action of catecholamines).
      There are minimal systolic pressure changes.
      There may be associated anxiety, fright or hostility

      Class III haemorrhage (30-40% blood volume loss):
      Complicated haemorrhagic state – crystalloid and probably blood replacement are required
      There are classical signs of inadequate perfusion, marked tachycardia, tachypnoea, significant changes in mental state and measurable fall in systolic pressure.
      Almost always require blood transfusion, but decision based on patient initial response to fluid resuscitation.

      Class IV haemorrhage (> 40% blood volume loss):
      Preterminal event patient will die in minutes
      Marked tachycardia, significant depression in systolic pressure and very narrow pulse pressure (or unobtainable diastolic pressure)
      Mental state is markedly depressed
      Skin cold and pale.
      Needs rapid transfusion and immediate surgical intervention.

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

      Fluid resuscitation following trauma is a controversial area.

      This clinical scenario points to a 15-30% blood loss. However, further crystalloid and blood replacement may be required after assessing the clinical situation. There is increasing evidence to suggest that transfusion of large volumes of crystalloid in the hospital setting are likely to be deleterious to the patient and hypotensive resuscitation and judicious blood and blood product resuscitation is a more appropriate option. A ratio of 1 unit of plasma to 1 unit of red blood cells is used to replace fluid volume in adults.

      This patient does not require immediate transfusion of O negative blood and there is time for a formal crossmatch. The argument about colloids versus crystalloids has existed for decades. However, while they have a role in fluid resuscitation, they are not first line.

      There is a risk of anaphylaxis, Hypernatraemia, and acute renal injury with colloidal solutions.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 10 - 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 11 - 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 12 - A 25-year old man needs an emergency appendicectomy and has gone to the...

    Incorrect

    • A 25-year old man needs an emergency appendicectomy and has gone to the operating room. During general anaesthesia, ventilation is achieved using a circle system with a fresh gas flow (FGF) of 1L/min, with and air/oxygen and sevoflurane combination. The capnograph trace is normal.

      Changes to the end tidal and baseline CO2 measurements at 10 and 20 mins respectively are seen on the capnograph below:

      10 minutes 20 minutes
      End-tidal CO2 4.9 kPa 8.4 kPa
      Baseline end-tidal CO2 0.2 kPa 2.4 kPa

      The other vitals were as follows:
      Pulse 100-105 beats per minute
      Systolic blood pressure 120-133 mmHg
      O2 saturation 99%.

      The next most important immediate step is which of the following?

      Your Answer:

      Correct Answer: Increase the FGF

      Explanation:

      This scenario describes rebreathing management.

      Changes is exhaustion of the soda lime and a progressive rise in circuit deadspace is the most likely explanation for the capnograph.

      It is important that the soda lime canister is inspected for a change in colour of the granules. Initially fresh gas flow should be increased and then if necessary, replace the soda lime granules. Other strategies include changing to another circuit or bypassing the soda lime canister after the fresh gas flow is increased.

      Any other causes of increased equipment deadspace should be excluded.

      Intraoperative hypercarbia can be caused by:

      1. Hypoventilation – Breathing spontaneously; drugs which include anaesthetic agents, opioids, residual neuromuscular blockade, pre-existing respiratory or neuromuscular disease and cerebrovascular accident.
      2. Controlled ventilation- circuit leaks, disconnection, miscalculation of patient’s minute volume.
      3. Rebreathing – Soda lime exhaustion with circle, inadequate fresh gas flow into Mapleson circuits, increased breathing system deadspace.
      4. Endogenous source – Tourniquet release, hypermetabolic states (MH or thyroid storm) and release of vascular clamps.
      5. Exogenous source – Absorption of CO2 absorption from the pneumoperitoneum.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 13 - A transport ventilator connected to a size CD oxygen cylinder has a setting...

    Incorrect

    • A transport ventilator connected to a size CD oxygen cylinder has a setting of air/oxygen entrainment ratio of 1:1 and a minute volume set at 10 litres/minute.

      Which value best approximates to the FiO2?

      Your Answer:

      Correct Answer: 0.6

      Explanation:

      A nominal volume of 2 litres is contained in a CD cylinder. It has a pressure of 230 bar when full and contains litres 460 L of useable oxygen at STP.

      For every 1000 mL 100% oxygen there will be an entrainment of 1000 mL or air (20% oxygen) in an air/oxygen mix.

      The average concentration is, therefore, 120/2=60% or 0.6.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 14 - The biochemical assessment of malnutrition can be measured by the amount of plasma...

    Incorrect

    • The biochemical assessment of malnutrition can be measured by the amount of plasma proteins.

      In acute starvation, which of these plasma proteins is the most sensitive indicator?

      Your Answer:

      Correct Answer: Retinol binding globulin

      Explanation:

      The half life of Retinol binding protein (RBP) is 10-12 hours and therefore reflects more acute changes in protein metabolism than any of these proteins. Therefore it is not commonly used as a parameter for nutritional assessment.

      The half life of Transthyretin (thyroxine binding pre-albumin) is only one to two days and so levels are less sensitive and this protein is not an albumin precursor. 15 mg/dL represents early malnutrition and a need for nutritional support.

      Albumin levels have been frequently as a marker of nutrition but this is not a very sensitive marker. It’s half life more than 30 days and significant change takes some time to be noticed. Also, synthesis of albumin is decreased with the onset of the stress response after burns. Unrelated to nutritional status, the synthesis of acute phase proteins increases and that of albumin decreases.

      A more accurate indicator of protein stores is transferrin. It’s response to acute changes in protein status is much faster. The half life of serum transferrin is shorter (8-10 days) and there are smaller body stores than albumin. A low serum transferrin level is below 200 mg/dL and below 100 mg/dL is considered severe. Serum transferrin levels can also affect serum transferrin level.

      Fibronectin is used a nutritional marker but levels decrease after seven days of starvation. It is a glycoprotein which plays a role in enhancing the phagocytosis of foreign particles.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 15 - Which of the following best explains the association between smoking and lower oxygen...

    Incorrect

    • Which of the following best explains the association between smoking and lower oxygen delivery to tissues?

      Your Answer:

      Correct Answer: Left shift of the oxygen dissociation curve

      Explanation:

      Smoking is a major risk factor associated with perioperative respiratory and cardiovascular complications. Evidence also suggests that cigarette smoking causes imbalance in the prostaglandins and promotes vasoconstriction and excessive platelet aggregation. Two of the constituents of cigarette smoke, nicotine and carbon monoxide, have adverse cardiovascular effects. Carbon monoxide increases the incidence of arrhythmias and has a negative ionotropic effect both in animals and humans.

      Smoking causes an increase in carboxyhaemoglobin levels, resulting in a leftward shift in which appears to represent a risk factor for some of these cardiovascular complications.

      There are two mechanisms responsible for the leftward shift of oxyhaemoglobin dissociation curve when carbon monoxide is present in the blood. Carbon monoxide has a direct effect on oxyhaemoglobin, causing a leftward shift of the oxygen dissociation curve, and carbon monoxide also reduces the formation of 2,3-DPG by inhibiting glycolysis in the erythrocyte. Nicotine, on the other hand, has a stimulatory effect on the autonomic nervous system. The effects of nicotine on the cardiovascular system last less than 30 min.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 16 - 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 17 - The following statement is true with regards to the Nernst equation: ...

    Incorrect

    • The following statement is true with regards to the Nernst equation:

      Your Answer:

      Correct Answer: It is used to calculate the potential difference across a membrane when the individual ions are in equilibrium

      Explanation:

      The Nernst equation is used to calculate the membrane potential at which the ions are in equilibrium across the cell membrane.

      The normal resting membrane potential is -70 mV (not + 70 mV).

      The equation is:
      E = RT/FZ ln {[X]o
      /[X]i}

      Where:
      E is the equilibrium potential
      R is the universal gas constant
      T is the absolute temperature
      F is the Faraday constant
      Z is the valency of the ion
      [X]o is the extracellular concentration of ion X
      [X]i is the intracellular concentration of ion X.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 18 - The following is normally higher in concentration extracellularly than intracellularly ...

    Incorrect

    • The following is normally higher in concentration extracellularly than intracellularly

      Your Answer:

      Correct Answer: Sodium

      Explanation:

      The ions found in higher concentrations intracellularly than outside the cells are:

      ATP
      AMP
      Potassium
      Phosphate, and
      Magnesium Adenosine diphosphate (ADP)

      Sodium is a primarily extracellular ion.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 19 - 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 20 - The single most important prerequisite for accuracy in measuring basal metabolic rate (BMR)...

    Incorrect

    • The single most important prerequisite for accuracy in measuring basal metabolic rate (BMR) using indirect calorimetry is performing the test:

      Your Answer:

      Correct Answer: In a neutral thermal environment

      Explanation:

      The basal metabolic rate (BMR) is the amount of energy required to maintain basic bodily functions in the resting state. The unit is Watt (Joule/second) or calories per unit time.

      Indirect calorimetry measures O2 consumption and CO2 production where gases are collected in a canopy which is the gold standard, Douglas bag, face-mask dilution technique or interfaced with a ventilator.

      The BMR can be calculated using the Weir formula:

      Metabolic rate (kcal per day) = 1.44 (3.94 VO2 + 1.11 VCO2)

      The BMR should be measured while lying down and at rest with the following conditions met:

      It should follow a 12 -hour fast
      No stimulants ingested within a 12-hour period
      It should be done in a neutral thermal environment (between 20°C-25°C)

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 21 - The SI unit of measurement is kgm2s-2 in the System international d'unités (SI).

    Which...

    Incorrect

    • The SI unit of measurement is kgm2s-2 in the System international d'unités (SI).

      Which of the following derived units of measurement has this format?

      Your Answer:

      Correct Answer: Energy

      Explanation:

      The derived SI unit of force is Newton.
      F = m·a (where a is acceleration)
      F = 1 kg·m/s2

      The joule (J) is a converted unit of energy, work, or heat. When a force of one newton (N) is applied over a distance of one metre (Nm), the following amount of energy is expended:

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

      The unit of velocity is metres per second (m/s or ms-1).

      The watt (W), or number of joules expended per second, is the SI unit of power:

      J/s = kg·m2·s-2/s
      J/s = kg·m2·s-3

      Pressure is measured in pascal (Pa) and is defined as force (N) per unit area (m2):
      Pa = kg·m·s-2/m2
      Pa = kg·m-1·s-2

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 22 - Of the following, which of these oxygen carrying molecules causes the greatest shift...

    Incorrect

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

      Your Answer:

      Correct 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
      0
      Seconds
  • Question 23 - In which of the following situations will a regional fall in cerebral blood...

    Incorrect

    • In which of the following situations will a regional fall in cerebral blood flow occur, suppose there is no changes in the mean arterial pressure (MAP)?

      Your Answer:

      Correct Answer: Hyperoxia

      Explanation:

      The response of cerebral blood flow (CBF) to hyperoxia (PaO2 >15 kPa, 113 mmHg), the cerebral oxygen vasoreactivity is less well defined. A study originally described, using a nitrous oxide washout technique, a reduction in CBF of 13% and a moderate increase in cerebrovascular resistance in subjects inhaling 85-100% oxygen. Subsequent human studies, using a variety of differing methods, have also shown CBF reductions with hyperoxia, although the reported extent of this change is variable. Another study assessed how supra-atmospheric pressures influenced CBF, as estimated by changes in middle cerebral artery flow velocity (MCAFV) in healthy individuals. Atmospheric pressure alone had no effect on MCAFV if PaO2 was kept constant. Increases in PaO2 did lead to a significant reduction in MCAFV; however, there were no further reductions in MCAFV when oxygen was increased from 100% at 1 atmosphere of pressure to 100% oxygen at 2 atmospheres of pressure. This suggests that the ability of cerebral vasculature to constrict in response to increasing partial pressure of oxygen is limited.

      Increases in arterial blood CO2 tension (PaCO2) elicit marked cerebral vasodilation.

      CBF increases with general anaesthesia, ketamine anaesthesia, and hypoviscosity.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 24 - Which statement is true when describing carbonic anhydrase? ...

    Incorrect

    • Which statement is true when describing carbonic anhydrase?

      Your Answer:

      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
      0
      Seconds
  • Question 25 - Which of the following statements best describes adenosine receptors? ...

    Incorrect

    • Which of the following statements best describes adenosine receptors?

      Your Answer:

      Correct Answer:

      Explanation:

      Adenosine receptors are expressed on the surface of most cells.
      Four subtypes are known to exist which are A1, A2A, A2B and A3.

      Of these, the A1 and A2 receptors are present peripherally and centrally. There are agonists at the A1 receptors which are antinociceptive, which reduce the sensitivity to a painful stimuli for the individual. There are also agonists at the A2 receptors which are algogenic and activation of these results in pain.

      The role of adenosine and other A1 receptor agonists is currently under investigation for use in acute and chronic pain states.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 26 - A 30-year old female athlete was brought to the Emergency Room for complaints...

    Incorrect

    • A 30-year old female athlete was brought to the Emergency Room for complaints of light-headedness and nausea. Clinical chemistry studies were done and the results were the following:

      Na: 144 mmol/L (Reference: 137-144 mmol/L)
      K: 6 mmol/L (Reference: 3.5-4.9 mmol/L)
      Cl: 115 mmol/L (Reference: 95-107 mmol/L)
      HCO3: 24 mmol/L (Reference: 20-28 mmol/L)
      BUN: 9.5 mmol/L (Reference: 2.5-7.5 mmol/L)
      Crea: 301 µmol/l (Reference: 60 - 110 µmol/L)
      Glucose: 3.5 mmol/L (Reference: 3.0-6.0 mmol/L)

      Taking into consideration the values above, in which of the following ranges will his osmolarity fall into?

      Your Answer:

      Correct Answer: 300-313

      Explanation:

      Osmolarity refers to the osmotic pressure generated by the dissolved solute molecules in 1 L of solvent. Measurements of osmolarity are temperature dependent because the volume of the solvent varies with temperature. The higher the osmolarity of a solution, the more it attracts water from an opposite compartment.

      Osmolarity can be computed using the following formulas:

      Osmolarity = Concentration x number of dissociable particles; OR
      Plasma osmolarity (Posm) = 2([Na+]) + (glucose in mmol/L) + (BUN in mmol/L)

      Posm = 2 (144) + 3.5 + 9.5 = 301 mOsm/L

      Suppose there is electrical neutrality, the formula will double the cation activity to account for the anions.

      Plasma osmolarity (Posm) = 2([Na+] + [K+]) + (glucose in mmol/L) + (BUN in mmol/L)

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

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 27 - 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 28 - 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 29 - In the fetal circulation, the cerebral and coronary circulations are preferentially supplied by...

    Incorrect

    • In the fetal circulation, the cerebral and coronary circulations are preferentially supplied by oxygen-rich blood over other organs. This is possible because of which phenomenon?

      Your Answer:

      Correct Answer: Well oxygenated blood from the inferior vena cava is preferentially streamed across the patent foramen ovale

      Explanation:

      During fetal development, blood oxygenated by the placenta flows to the foetus through the umbilical vein, bypasses the fetal liver through the ductus venosus, and returns to the fetal heart through the inferior vena cava.

      Blood returning from the inferior vena cava then enters the right atrium and is preferentially shunted to the left atrium through the patent foramen ovale. Blood in the left atrium is then pumped from the left ventricle to the aorta. The oxygenated blood ejected through the ascending aorta is preferentially directed to the fetal coronary and cerebral circulations.

      Deoxygenated blood returns from the superior vena cava to the right atrium and ventricle to be pumped into the pulmonary artery. Fetal pulmonary vascular resistance (PVR), however, is higher than fetal systemic vascular resistance (SVR); this forces deoxygenated blood to mostly bypass the fetal lungs. This poorly oxygenated blood enters the aorta through the patent ductus arteriosus and mixes with the well-oxygenated blood in the descending aorta. The mixed blood in the descending aorta then returns to the placenta for oxygenation through the two umbilical arteries.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 30 - Regarding the plateau phase of the cardiac potential, which electrolyte is the main...

    Incorrect

    • Regarding the plateau phase of the cardiac potential, which electrolyte is the main determinant?

      Your Answer:

      Correct Answer: Ca2+

      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

SESSION STATS - PERFORMANCE PER SPECIALTY

Physiology (6/7) 86%
Passmed