Primary FRCA Sample Questions

Correct answer: C. Velocity increases and pressure decreases

• The Bernoulli principle states total energy per unit volume remains constant along a streamline
• At a constriction, the same mass flow requires higher velocity in the narrower section
• Increased kinetic energy from higher velocity requires decreased pressure to conserve total energy
• Option B would require both kinetic energy and static-pressure energy to increase, contrary to Bernoulli’s equation at constant height
• This pressure drop at the constriction allows room air entrainment in Venturi masks to deliver fixed oxygen concentrations

Correct answer: Remains constant until the liquid phase is exhausted

• Nitrous oxide has a critical temperature of 36.4°C, so at room temperature (20°C) it exists as a liquid in equilibrium with its vapour.
• The pressure reading of 52 bar represents the saturated vapour pressure of nitrous oxide at 20°C.
• As vapour is withdrawn, liquid evaporates to maintain equilibrium, keeping the pressure constant at approximately 52 bar.
• Option E is incorrect because Boyle's law assumes a fixed mass of gas at a constant temperature. In a nitrous oxide cylinder, the mass of the vapour phase changes as liquid evaporates. Furthermore, vapours near their saturation point deviate significantly from ideal gas behaviour.
• The cylinder must be weighed to determine the remaining contents, as the pressure gauge will only begin to fall once all the liquid is exhausted.

Correct answer: D. Decrease to one-sixteenth

• The Hagen-Poiseuille equation shows flow is proportional to the fourth power of radius (Q = πΔPr⁴/(8ηL))
• Halving radius means the new radius is 0.5 times the original, so (0.5)⁴ = 1/16
• Flow decreases to one-sixteenth of the original value when radius is halved
• Option B incorrectly applies a square relationship rather than fourth power
• This principle explains why small changes in airway diameter dramatically affect resistance and flow. For steady laminar flow through a rigid cylindrical tube, small changes in radius produce large changes in resistance and flow.

Correct answer: E. Increase by 7.5 mm Hg

• A 10 cm change in transducer height alters the pressure reading by 7.5 mm Hg
• Lowering the transducer below the reference point increases the displayed pressure
• This occurs because the hydrostatic pressure of the fluid column adds to the measured pressure
• The transducer must be kept horizontally level with the right atrium for accurate readings
• Transducer levelling is specific to invasive monitoring. In non-invasive measurement, cuff height relative to the heart can similarly affect the recorded pressure

Correct answer: A. 660 nm and 940 nm

• Conventional two-wavelength pulse oximeters typically use approximately 660 nm and 940 nm. Exact LED wavelengths may vary slightly between devices.
• Measurement depends on differential absorption by oxyhaemoglobin and deoxyhaemoglobin.
• These wavelengths are chosen because oxyhaemoglobin and deoxyhaemoglobin have sufficiently different absorption characteristics at these points
• At 660 nm deoxyhaemoglobin absorbs more light than oxyhaemoglobin; at 940 nm the reverse is true
• This two-wavelength approach allows simple and cost-effective fingertip devices compared to laboratory co-oximeters using multiple wavelengths

Correct answer: B. Ketamine blocks the channel pore while glutamate binds to the extracellular surface

• Non-competitive antagonists bind at a different site from the agonist, so increasing agonist concentration cannot restore the original response.
• Ketamine antagonises NMDA receptors by blocking the channel pore, while the agonist (glutamate) binding site is on the extracellular surface.
• Option B correctly describes the non-competitive mechanism with different binding sites for agonist and antagonist.
• Option A describes competitive antagonism, which is surmountable by increasing agonist concentration.
• The spatial separation of binding sites explains why ketamine's blockade cannot be overcome simply by increasing glutamate concentration.

Correct answer: C. Time for 50% reduction in concentration

• Half-life is the time taken for a 50% reduction in drug concentration
• This value remains constant regardless of starting concentration in first-order kinetics
• Half-life describes exponential elimination where the rate of elimination is proportional to concentration
• A 63% reduction (concentration falling to 36.8% of its initial value) corresponds to one time constant, not the half-life
• Understanding half-life is essential for predicting drug accumulation and dosing intervals

Correct answer: B. Nitrous oxide uptake increases alveolar volatile concentration

• Nitrous oxide is 30 times more soluble than nitrogen despite both being relatively insoluble
• Initial high uptake of nitrous oxide from the alveolus causes loss of alveolar volume
• This volume loss concentrates simultaneously administered agents and increases their alveolar partial pressures
• The higher FA/FI ratio for the volatile agent leads to faster induction
• Option C is incorrect as the mechanism is concentration rather than delivery-based

Correct answer: B. Hepatic and extrahepatic via glucuronide and sulphate conjugation

• Propofol is metabolised mainly in the liver, with extrahepatic metabolism also contributing.
• Major pathways include direct glucuronidation and oxidation followed by glucuronide or sulphate conjugation.
• The metabolites are inactive and predominantly excreted in urine.
• Although its clearance is high, propofol should still be titrated carefully in patients with hepatic impairment.

Correct answer: E. pKa

• The pKa determines the proportion of un-ionised drug available at physiological pH
• Un-ionised drug crosses the nerve sheath and membrane more rapidly
• Local anaesthetics with low pKa have faster onset because more drug exists in the un-ionised form at pH 7.4
• Lipid solubility correlates mainly with potency, whereas protein binding and tissue retention contribute to duration.
• Lidocaine (pKa 7.8) has faster onset than bupivacaine (pKa 8.1) due to this principle

Correct answer: Decreased oxygen affinity of haemoglobin due to low pH or high carbon dioxide

• The Bohr effect is defined as the decrease in oxygen affinity of haemoglobin in the presence of low pH or high carbon dioxide
• This facilitates oxygen release in peripheral tissues where pH is lower and carbon dioxide is higher
• Option D correctly states both the direction (decreased affinity) and the conditions (low pH or high carbon dioxide)
• Option A incorrectly suggests increased affinity in acidic conditions, which is the opposite of the Bohr effect
• The related Haldane effect describes increased carbon dioxide carriage by deoxygenated haemoglobin, and should not be confused with the Bohr effect
• This effect is clinically important in ensuring oxygen delivery matches metabolic demand in working tissues

Correct answer: D. Residual volume

• Basic spirometry measures volumes that can be actively inspired or expired by the patient
• Residual volume is the gas remaining in the lungs after maximal forced expiration and cannot be exhaled, therefore cannot be measured by spirometry
• Total lung capacity and functional residual capacity also cannot be measured by spirometry as they both contain the residual volume component
• Tidal volume, inspiratory reserve volume, expiratory reserve volume, and vital capacity can all be measured directly from a spirometry trace
• Helium dilution or body plethysmography is used to measure functional residual capacity (FRC); residual volume (RV) and total lung capacity (TLC) are then calculated from FRC using spirometric volumes (e.g., RV = FRC − ERV; TLC = RV + VC)

Correct answer: A. 200 ml/litre

• Arterial oxygen content (CaO₂) equals haemoglobin-bound oxygen plus dissolved oxygen.
• The equation is: CaO₂ = (1.34 × Hb × SaO₂) + (0.23 × PaO₂).
• Haemoglobin-bound oxygen: 1.34 × 150 × 0.98 ≈ 197 ml O₂/L.
• Dissolved oxygen: 0.23 × 12 = 2.8 ml O₂/L.
• Total arterial oxygen content ≈ 200 ml O₂/L, therefore the closest answer is 200 ml/L.
• In normal physiology, over 98% of oxygen is transported bound to haemoglobin, with dissolved oxygen contributing only a small proportion.

Correct answer: Freely filtered, neither secreted nor reabsorbed

• GFR measurement requires a marker that accurately reflects glomerular filtration without tubular modification.
• The ideal substance must be freely filtered at the glomerulus and neither secreted nor reabsorbed by the renal tubules.
• Inulin is the classic gold-standard exogenous marker that perfectly meets these criteria.
• Creatinine is commonly used clinically as an endogenous marker, but it is actively secreted to a small extent (approximately 10–20%) by the proximal tubule, leading to a slight overestimation of true GFR.
• Substances with a high renal extraction ratio (e.g. para-aminohippuric acid) are used to measure renal plasma flow, not GFR.
• A marker for GFR should also be freely soluble in plasma, non-toxic, easily measured, and not protein-bound or metabolised, so that its clearance equals the volume of plasma filtered per unit time.

Correct answer: B. 5–15 mmHg

• Normal ICP in a supine adult is conventionally quoted as 5–15 mmHg
• ICP is posture-dependent and is lowest in the upright position
• Increases in ICP above critical levels decrease cerebral perfusion pressure and risk cerebral ischaemia
• Sustained ICP above 22 mmHg is a treatment threshold in severe traumatic brain injury, although its effect on cerebral perfusion also depends on arterial pressure
• ICP monitoring is essential in neurocritical care to guide management and prevent secondary brain injury