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Question 1
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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: 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%).
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This question is part of the following fields:
- Physiology
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Question 2
Correct
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A 30-year old female was anaesthetically induced for an elective open cholecystectomy. Upon mask ventilation, patient's oxygen saturation level dropped to 90% despite maximal head extension, jaw thrust and two handed mask seal. Intubation was performed twice but failed. Use of bougie also failed to localize the trachea. Oxygen levels continued to drop, but was maintained between 80 and 88% with mask ventilation.
Which of the following options is the best action to take for this patient?Your Answer: Insert a supraglottic airway
Explanation:A preplanned preinduction strategy includes the consideration of various interventions designed to facilitate intubation should a difficult airway occur. Non-invasive interventions intended to manage a difficult airway include, but are not limited to: (1) awake intubation, (2) video-assisted laryngoscopy, (3) intubating stylets or tube-changers, (4) SGA for ventilation (e.g., LMA, laryngeal tube), (5) SGA for intubation (e.g., ILMA), (6) rigid laryngoscopic blades of varying design and size, (7) fibreoptic-guided intubation, and (8) lighted stylets or light wands.
Most supraglottic airway devices (SADs) are designed for use during routine anaesthesia, but there are other roles such as airway rescue after failed tracheal intubation, use as a conduit to facilitate tracheal intubation and use by primary responders at cardiac arrest or other out-of-hospital emergencies. Supraglottic airway devices are intrinsically more invasive than use of a facemask for anaesthesia, but less invasive than tracheal intubation. Supraglottic airway devices can usefully be classified as first and second generation SADs and also according to whether they are specifically designed to facilitate tracheal intubation. First generation devices are simply ‘airway tubes’, whereas second generation devices incorporate specific design features to improve safety by protecting against regurgitation and aspiration.
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This question is part of the following fields:
- Physiology
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Question 3
Correct
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Of the following, which option best describes the muscle type that has the fastest twitch response to stimulation?
Your 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.
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This question is part of the following fields:
- Physiology
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Question 4
Correct
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The passage of glucose into the brain is facilitated by which transport method?
Your Answer: Facilitated diffusion
Explanation:Glucose transport is a highly regulated process accomplished mostly by facilitated diffusion using carrier proteins to cross cell membranes.
There are many transporters, but the most important are known as glucose transporters (GLUTs).
Stresses in various form of acute and chronic forms affect the activity of glucose transporters.
They are responsive to many types of metabolic stress, including hypoxia, injury, hypoglycaemia, numerous metabolic inhibitors, stress hormones, and other influences such as growth factors.Numerous signalling pathways appear to be involved in transporter regulation.
New evidence suggests that stresses regulating GLUTs are not only acute biological stresses. In addition, chronic low-grade inflammation, and their associated chronic diseases also lead to altered glucose transport. These include obesity, type 2 diabetes, cardiovascular disease, and the growth and spread of many tumours that are affected by altered glucose transporters. Some of these glucose transport effects are compensatory, while others are pathogenic.
Ultimately, deliberate manipulation of GLUTs could be used as treatment for some of these chronic diseases.
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This question is part of the following fields:
- Physiology
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Question 5
Correct
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Which of the following, at a given PaO2, increases the oxygen content of arterial blood?
Your 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.
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This question is part of the following fields:
- Physiology
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Question 6
Incorrect
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Concerning forced alkaline diuresis, which of the following statements is true?
Your Answer: Increases the excretion of unionised drugs in the urine
Correct Answer: Can be used in a barbiturate overdose
Explanation:In situations of poisoning or drug overdose with acid dugs like salicylates and barbiturates, forced alkaline diuresis may be used.
With regards to overdose with alkaline drugs, forced acid diuresis is used.
By changing the pH of the urine, the ionised portion of the drug stays in the urine, and this prevents its diffusion back into the blood. Charged molecules do not readily cross biological membranes.
The process involves the infusion of specific fluids at a rate of about 500ml per hour. This requires monitoring of the central venous pressure, urine output, plasma electrolytes, especially potassium, and blood gas analysis.
The fluid regimen recommended is:
500ml of 1.26% sodium bicarbonate (not 200ml of 8.4%)
500ml of 5% dextrose and
500ml of 0.9% sodium chloride. -
This question is part of the following fields:
- Physiology
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Question 7
Correct
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Which of the following would most likely explain a failed post-operative analgesia via local anaesthesia of a neck abscess?
Your Answer: pKA
Explanation:For the local anaesthetic base to be stable in solution, it is formulated as a hydrochloride salt. As such, the molecules exist in a quaternary, water-soluble state at the time of injection. However, this form will not penetrate the neuron. The time for onset of local anaesthesia is therefore predicated on the proportion of molecules that convert to the tertiary, lipid-soluble structure when exposed to physiologic pH (7.4).
The ionization constant (pKa) for the anaesthetic predicts the proportion of molecules that exists in each of these states. By definition, the pKa of a molecule represents the pH at which 50% of the molecules exist in the lipid-soluble tertiary form and 50% in the quaternary, water-soluble form. The pKa of all local anaesthetics is >7.4 (physiologic pH), and therefore a greater proportion the molecules exists in the quaternary, water-soluble form when injected into tissue having normal pH of 7.4.
Furthermore, the acidic environment associated with inflamed tissues favours the quaternary, water-soluble configuration even further. Presumably, this accounts for difficulty when attempting to anesthetize inflamed or infected tissues; fewer molecules exist as tertiary lipid-soluble forms that can penetrate nerves.
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This question is part of the following fields:
- Physiology
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Question 8
Incorrect
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A 25-year old lady is in the operating room and has had general anaesthesia for a knee arthroscopy.
Induction was done with fentanyl 1mcg/kg and propofol 2mg/kg. A supra-glottic airway was inserted and using and air oxygen mixture with 2.5% sevoflurane, her anaesthesia was maintained. The patient is allowed to spontaneously breathe using a Bain circuit, and the fresh gas flow is 9L/min. Over the next 30 minutes, the end-tidal Co2 rises from 4.5kPa to 8.4kPa, and the baseline reading on the capnograph is 0kPa.
The most appropriate initial action is which of the following?Your Answer: Malignant hyperthermia
Correct Answer: Hypoventilation
Explanation:The commonest and most likely cause of a gradual rise in end-tidal CO2 (EtCO2) occurring during anaesthesia in a spontaneously breathing patient is hypoventilation. This occurs from the respiratory depressant effects of the opioid and sevoflurane.
Malignant hyperthermia should be sought if the EtCO2 shows further progressive rise.
Causes of rebreathing and a rise in the baseline of the capnograph can be caused by exhausted soda lime and inadequate fresh gas flow into the Bain circuit.
A sudden rise in EtCO2 can be caused deflation of the tourniquet.
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This question is part of the following fields:
- Physiology
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Question 9
Incorrect
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A human's resting oxygen consumption (VO2) is typically 3.5 ml/kg/minute (one metabolic equivalent or 1 MET).
Which of the following options is linked to the highest VO2 when a person is at rest?Your Answer: Sepsis syndrome
Correct Answer: Neonate
Explanation:The oxygen consumption rate (VO2) at rest is 3.5 ml/kg/minute (one metabolic equivalent or 1 MET).
3.86 ml/kg/minute thyrotoxicosisYoung children consume a lot of oxygen: around 7 ml/kg/min when they are born. The metabolic cost of breathing is higher in children than in adults, and it can account for up to 15% of total oxygen consumption. Similarly, an infant’s metabolic rate is nearly twice that of an adult, resulting in a larger alveolar minute volume and a lower FRC.
At term, oxygen consumption at rest can increase by as much as 40% (5 ml/kg/minute) and can rise to 60% during labour.
When compared to normal basal metabolism, sepsis syndrome increases VO2 and resting metabolic rate by 30% (4.55 ml/kg/minute). In septicaemic shock, VO2 decreases.
Dobutamine hydrochloride was infused into 12 healthy male volunteers at a rate of 2 micrograms per minute per kilogramme, gradually increasing to 4 and 6 micrograms per minute per kilogramme. Dobutamine was infused for 20 minutes for each dose. VO2 increased by 10% to 15%. (3.85-4.0 ml/kg/min).
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This question is part of the following fields:
- Physiology
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Question 10
Incorrect
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The fluids with the highest osmolarity is?
Your Answer: Gelofusine
Correct Answer: 0.45% N. Saline with 5% glucose
Explanation:The concentration of solute particles per litre (mosm/L) = the osmolarity of a solution. Changes in water content, ambient temperature, and pressure affects osmolarity. The osmolarity of any solution can be calculated by adding the concentration of key solutes in it.
Individual manufacturers of crystalloids and colloids may have different absolute values but they are similar to these.
0.45% N. Saline with 5% glucose:
Tonicity – hypertonic
Osmolarity – 405 mosm/L
Kilocalories (kCal) – 1070.9% N. Saline:
Tonicity – isotonic
Osmolarity – 308 mosm/L
Kilocalories (kCal) – 05% Dextrose:
Tonicity – isotonic
Osmolarity – 253 mosm/L
Kilocalories (kCal) – 170Gelofusine (154 mmol/L Na, 120 mmol/L Cl):
Tonicity – isotonic
Osmolarity – 274 mosm/L
Kilocalories (kCal) – 0Hartmann’s solution:
Tonicity – isotonic
Osmolarity – 273 mosm/L
Kilocalories (kCal) – 9 -
This question is part of the following fields:
- Physiology
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Question 11
Correct
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Which of the following best explains the association between smoking and lower oxygen delivery to tissues?
Your 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.
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This question is part of the following fields:
- Physiology
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Question 12
Correct
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All of the following statements about cerebrospinal fluid are incorrect except:
Your Answer: Has a glucose concentration 2/3 that of the plasma glucose
Explanation:The pH of CSF is 7.31 which is lower than plasma.
Compared to plasma, it has a lower concentration of potassium, calcium, and protein and a higher concentration of sodium, chloride, bicarbonate and magnesium.
CSF usually has no cells present but if white cells are present, there should be no more than 4/ml.
The pressure of CSF should be less than 20 cm of water.
The concentration of glucose is approximately two-thirds of that of plasma, and it has a concentration of approximately 3.3-4 mmol/L.
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This question is part of the following fields:
- Physiology
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Question 13
Incorrect
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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: 295-299
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
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This question is part of the following fields:
- Physiology
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Question 14
Correct
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Given the following values:
Expired tidal volume = 800 ml
Plateau pressure = 50 cmH2O
PEEP = 10 cmH2O
Compute for the static pulmonary compliance.Your 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
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Question 15
Incorrect
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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: Velocity
Correct Answer: Energy
Explanation:The derived SI unit of force is Newton.
F = m·a (where a is acceleration)
F = 1 kg·m/s2The 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-2The 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-3Pressure 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
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Question 16
Incorrect
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Which statement is true when describing carbonic anhydrase?
Your Answer: Is present in the plasma
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.
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This question is part of the following fields:
- Physiology
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Question 17
Correct
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The single most important prerequisite for accuracy in measuring basal metabolic rate (BMR) using indirect calorimetry is performing the test:
Your 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
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Question 18
Correct
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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: 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.
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This question is part of the following fields:
- Physiology
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Question 19
Correct
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Regarding the plateau phase of the cardiac potential, which electrolyte is the main determinant?
Your 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 msPhase 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 potentialOf 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/sec2. 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
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This question is part of the following fields:
- Physiology
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Question 20
Correct
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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: 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
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This question is part of the following fields:
- Physiology
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