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  • Question 1 - An orthopaedic surgery is scheduled for a 68-year-old man. He is normally in...

    Correct

    • 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: 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
      20.1
      Seconds
  • Question 2 - 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: Reduced forced vital capacity

      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
      46.8
      Seconds
  • Question 3 - A participant of a metabolism study is to be fed only granulated sugar...

    Correct

    • 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: 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
      11.8
      Seconds
  • Question 4 - 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: Has a sodium concentration of 125-135 mmol/L

      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
      88.9
      Seconds
  • Question 5 - 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:

      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
      0
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  • Question 6 - 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
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  • Question 7 - 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
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  • Question 8 - Which one of the following factor affects the minimal alveolar concentration (MAC)? ...

    Incorrect

    • Which one of the following factor affects the minimal alveolar concentration (MAC)?

      Your Answer:

      Correct Answer: Hypoxaemia

      Explanation:

      The minimal alveolar concentration (MAC) is the concentration of an inhalation anaesthetic agent in the lung alveoli required to stop a response to the surgical stimulus in 50% of the patient.

      Following factors don’t affect the MAC of the inhaled anaesthetic agents:

      Gender, acidosis, alkalosis, hypothyroidism, hyperthyroidism, body weight, serum potassium level, and the duration of the anaesthesia.

      MAC increase in children, elevated temperature, high metabolic rate, sympathetic increase and chronic alcoholism.

      MAC decrease in low temperature, low oxygen level, old age, hypotension (<40 mmHg), depressant drugs e.g. opioids and low level of catecholamines; alpha methyl dopa. Carbon dioxide O2 at the pressure > 120mmHg is being used in anesthetic-Hinkman as an additive effect to decrease MAC, however, increase concentration of CO2 activates the sympathetic system resulting the MAC increases.

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 9 - If a large volume of 0.9% N. saline is administered during resuscitation, it...

    Incorrect

    • If a large volume of 0.9% N. saline is administered during resuscitation, it is most likely to cause?

      Your Answer:

      Correct Answer: Hyperchloremic metabolic acidosis

      Explanation:

      Crystalloids recommended for fluid resuscitation include 0.9% N saline and Hartmann’s solution(a physiological solution). 0.9% N. saline is not a physiological solution for the following reasons:

      Compared with the normal range of 98-102 mmol/L, its chloride concentration is high (154 mmol/L)
      It lacks calcium, magnesium, glucose and potassium
      It does not have bicarbonate or bicarbonate precursor buffer necessary to maintain plasma pH within normal limits

      There is a difference in the activity (concentration) of strong ions at a physiological pH. This imbalance can explain abnormalities of acid base balance. A normal strong ion difference (SID) is in the order of 40.

      SID = ([Na+] + [K+] + [Ca2+] + [Mg2+]) – ([Cl-] + [lactate] + [SO42-])

      This imbalance is made up with the weaker anions to maintain electrical neutrality.
      Administration of a large volume of 0.9% normal saline during resuscitation results in excessive chloride administration and this impairs renal bicarbonate reabsorption. The SID of 0.9% normal saline is 0 (Na+ = 154mmol/L and Cl- = 154mmol/L = 154 – 154 = 0). A large volume of NS will decrease the plasma SID causing an acidosis.

      Other causes of a hyperchloremic acidosis are:

      Diabetic ketoacidosis
      Total Parenteral Nutrition
      Overdose of ammonium chloride and hydrochloric acid
      Gastrointestinal losses of bicarbonate like in diarrhoea and pancreatic fistula
      Proximal renal tubular acidosis with failure of bicarbonate reabsorption

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 10 - Which of the statements below best describe the total cerebral flow (CBF) in...

    Incorrect

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

      Your Answer:

      Correct 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
      0
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  • Question 11 - 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
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  • Question 12 - 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
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  • Question 13 - 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:

      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
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  • Question 14 - Given the following values:

    Expired tidal volume = 800 ml
    Plateau pressure = 50 cmH2O
    PEEP...

    Incorrect

    • Given the following values:

      Expired tidal volume = 800 ml
      Plateau pressure = 50 cmH2O
      PEEP = 10 cmH2O

      Compute for the static pulmonary compliance.

      Your Answer:

      Correct Answer: 20 ml/cmH2O

      Explanation:

      Compliance of the respiratory system describes the expandability of the lungs and chest wall. There are two types of compliance: dynamic and static.

      Dynamic compliance describes the compliance measured during breathing, which involves a combination of lung compliance and airway resistance. Defined as the change in lung volume per unit change in pressure in the presence of flow.

      Static compliance describes pulmonary compliance when there is no airflow, like an inspiratory pause. Defined as the change in lung volume per unit change in pressure in the absence of flow.

      For example, if a person was to fill the lung with pressure and then not move it, the pressure would eventually decrease; this is the static compliance measurement. Dynamic compliance is measured by dividing the tidal volume, the average volume of air in one breath cycle, by the difference between the pressure of the lungs at full inspiration and full expiration. Static compliance is always a higher value than dynamic

      Static compliance can be computed using the formula:

      Cstat = Tidal volume/Plateau pressure – PEEP

      Substituting the values given,

      Cstat = 800/50-10
      Cstat = 20 ml/cmH2O

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 15 - 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
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  • Question 16 - A patient on admission is given an infusion of 1000 mL of 10%...

    Incorrect

    • A patient on admission is given an infusion of 1000 mL of 10% glucose and 500 mL of 20% lipid over a 24 hour period.

      Which of these best approximates to the energy input over this time period?

      Your Answer:

      Correct Answer: 1300 kcal

      Explanation:

      1% solution contains 1 g of substance per 100 mL.

      A solution of 10% glucose is 10 g/100mL. Therefore 1000 mL of this glucose solution will contain 100 g.

      1 g of glucose yields about 4 kcal of energy. One litre of 10% glucose will therefore release approximately 4x100g = 400 kcal of energy.

      A solution of 20% fat is 20 g/100mL. Therefore 1000 mL of this fat solution will have 200 g and 500 mL will contain 100 g.

      1 g of fat yields approximately 9 kcal. 500 mL of 20% fat therefore has the potential to yield 900 kcal of energy.

      The total energy input over this 24 hour period is approximately 400kcal + 900kcal = 1300 kcal.

    • This question is part of the following fields:

      • Physiology
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  • Question 17 - 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
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  • 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
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  • Question 19 - Anaesthetic awareness is most probable in general anaesthesia for which surgical operation? ...

    Incorrect

    • Anaesthetic awareness is most probable in general anaesthesia for which surgical operation?

      Your Answer:

      Correct Answer: Emergency surgery for major trauma

      Explanation:

      Awareness during general anaesthesia is a frightening experience, which may result in serious emotional injury and post-traumatic stress disorder.

      The incidence of awareness during general anaesthesia with current anaesthetic agents and techniques has been reported as 0.2-0.4% in nonobstetric and noncardiac surgery, as 0.4% during caesarean section, and as 1.5% in cardiac surgery.

      The incidence during major trauma surgery is higher. Incidence of recall has been reported to be as high as 11-43% in major trauma cases.

    • This question is part of the following fields:

      • Physiology
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  • Question 20 - 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
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  • Question 21 - In an experimental study, a healthy subject was given one litre of 5%...

    Incorrect

    • 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:

      Correct 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
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  • 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
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  • Question 23 - 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
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  • Question 24 - Which of the following statement is true regarding hypoxic pulmonary vasoconstriction (HPV)? ...

    Incorrect

    • Which of the following statement is true regarding hypoxic pulmonary vasoconstriction (HPV)?

      Your Answer:

      Correct Answer: 20 parts per million (ppm) of nitric oxide will reduce hypoxic pulmonary vasoconstriction

      Explanation:

      Hypoxic Pulmonary vasoconstriction (HPV) reflects the constriction of small pulmonary arteries in response to hypoxic alveoli (.i.e.; PO2 below 80-100mmHg or 11-13kPa).

      These blood vessels become independent of the nerve stimulus, when blood with a high PO2 flows through the lung which contains a low alveolar PO2.

      Thus a low PO2 within the alveoli has been shown to impact on hypoxic pulmonary vasoconstriction (HPV) more than a low PO2 within the blood.

      HPV results in the blood flow being directed away from poorly ventilated areas of the lung and helps to reduce the ventilation/perfusion mismatch (not increase).

      In animals, volatile anaesthetic agents can diminish HPV, while in adults, the evidence proves less persuading, in spite of the fact that it certainly doesn’t strengthen the effects.

      HPV response will be suppressed by 20 parts per million (ppm) of nitric oxide.

    • This question is part of the following fields:

      • Physiology
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  • Question 25 - 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
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  • Question 26 - Which of the following, at a given PaO2, increases the oxygen content of...

    Incorrect

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

      Your Answer:

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

      Explanation:

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

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

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

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

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

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

    • This question is part of the following fields:

      • Physiology
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  • Question 27 - 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
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  • Question 28 - Left ventricular afterload is mostly calculated from systemic vascular resistance.

    Which...

    Incorrect

    • Left ventricular afterload is mostly calculated from systemic vascular resistance.

      Which one of the following factors has most impact on systemic vascular resistance?

      Your Answer:

      Correct Answer: Small arterioles

      Explanation:

      Systemic vascular resistance (SVR), also known as total peripheral resistance (TPR), is the amount of force exerted on circulating blood by the vasculature of the body. Three factors determine the force: the length of the blood vessels in the body, the diameter of the vessels, and the viscosity of the blood within them. The most important factor that determines the systemic vascular resistance (SVR) is the tone of the small arterioles.

      These are otherwise known as resistance arterioles. Their diameter ranges between 100 and 450 µm. Smaller resistance vessels, less than 100 µm in diameter (pre-capillary arterioles), play a less significant role in determining SVR. They are subject to autoregulation.

      Any change in the viscosity of blood and therefore flow (such as due to a change in haematocrit) might also have a small effect on the measured vascular resistance.

      Changes of blood temperature can also affect blood rheology and therefore flow through resistance vessels.

      Systemic vascular resistance (SVR) is measured in dynes·s·cm-5

      It can be calculated from the following equation:

      SVR = (mean arterial pressure − mean right atrial pressure) × 80 cardiac output

    • This question is part of the following fields:

      • Physiology
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  • Question 29 - A 57-year old lady is admitted to the Emergency Department with signs of...

    Incorrect

    • A 57-year old lady is admitted to the Emergency Department with signs of a subarachnoid haemorrhage.

      On admission, her GCS was 7. She has been intubated, sedated and is being ventilated and is waiting for a CT scan. Her Blood pressure is 140/70mmHg.

      The arterial blood gas analysis shows the following:

      pH 7.2 (7.35 - 7.45)
      PaO2 70 mmHg (80-100)
      9.2 kPa (10.5-13.1)
      PaCO2 78 mmHg (35-45)
      10.2 kPa (4.6-6.0)
      BE -3 mEq/L (-3 +/-3)
      Standard bic 27 mmol/L (21-27)
      SaO2 94%

      The most likely cause of an increase in the patient's global cerebral blood flow (CBF) is which of the following?


      Your Answer:

      Correct Answer: Hypercapnia

      Explanation:

      PaCO2 is one of the most important factors that regulate cerebral vascular tone. CO2 induces cerebral vasodilatation and as a result, it increases CBF. Between 20 mmHg (2.7 kPa) and 80 mmHg (10.7 kPa), there is a linear increase of PaCO2.

      Sometimes, there are areas where auto regulation has failed locally but not globally. Similarly, local vs. systemic acidosis will have similar effects. When the PaO2 falls below 50 mmHg (6.5 kPa), the CBF progressively increases.

      An increase in the cerebral metabolic rate for oxygen (CMRO2) and therefore CBF can be caused by hyperthermia.
      A late feature of cerebral injury is hyperthermia secondary to hypothalamic injury. Therefore this is not the most likely cause of an increased CBF in this scenario.

    • This question is part of the following fields:

      • Physiology
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  • Question 30 - Which of the following statement is true or false regarding to the respiratory...

    Incorrect

    • Which of the following statement is true or false regarding to the respiratory tract?

      Your Answer:

      Correct Answer: The sympathetic innervation of the bronchi is derived from T2 - T4

      Explanation:

      The diaphragm has three opening through which different structures pass from the thoracic cavity to the abdominal cavity:

      Inferior vena cava passes at the level of T8.

      Oesophagus, oesophageal vessels and vagi at T10.

      Aorta, thoracic duct and azygous vein through T12.

      Sympathetic trunk and pulmonary branches of vagus nerve form a posterior pulmonary plexus at the root of the lung. Fibres continue posteriorly from superficial cardiac plexus to form Anterior pulmonary plexus. It contains vagi nerves and superficial cardiac plexus. These fibres then follow the blood vessel and bronchi into the lungs.

      The lower border of the pleura is at the level of:

      8th rib in the midclavicular line

      10th rib in the lower level of midaxillary line

      T12 at its termination.

      Both lungs have oblique fissure while right lung has transverse fissure too.

      The trachea expands from the lower edge of the cricoid cartilage (at the level of the 6th cervical vertebra) to the carina.

    • This question is part of the following fields:

      • Physiology
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