12 sample questions from the 2163 Medicine and Allied Paper 2 questions in FCPS-1 Prep. Choose an answer to see why it is right and why each other option is wrong.
During a total thyroidectomy, the surgeon ligates the inferior thyroid artery close to the gland. After surgery the patient's voice is hoarse and weak. Which nerve has most likely been injured?
Answer: Recurrent laryngeal nerve
The recurrent laryngeal nerve runs in the tracheo-oesophageal groove and crosses the inferior thyroid artery near the lower pole. It supplies all intrinsic laryngeal muscles except cricothyroid, so injury paralyses the vocal cord on that side, causing hoarseness. Bilateral injury can cause stridor and airway obstruction.
Ligate the superior artery close to the gland; the inferior artery away from the gland
Remember Superior = Singer's nerve (pitch). Inferior = the one that makes you Rasp.
Exam trap Many question books swap them. Upper pole = external laryngeal (pitch). Lower pole / inferior thyroid artery = recurrent laryngeal (hoarseness).
Why the other options are wrong
External laryngeal nerve Runs with the superior thyroid artery at the upper pole; injury causes a weak, monotonous voice and loss of high pitch (cricothyroid), not classic hoarseness.
Internal laryngeal nerve Sensory to the larynx above the cords; injury causes loss of the cough reflex and aspiration, not hoarseness.
Ansa cervicalis Supplies infrahyoid strap muscles; cutting it does not affect the voice.
Hypoglossal nerve Supplies tongue muscles; injury causes deviation of the tongue.
Question 2Medium
A kidney transplant patient on ciclosporin has rising creatinine, hypertension and gum hyperplasia. How does ciclosporin work?
Answer: Calcineurin inhibition
Ciclosporin (and tacrolimus) bind immunophilins and inhibit calcineurin, preventing activation of NFAT and IL-2 transcription, so T cells are not activated. Side effects: nephrotoxicity, hypertension, gum hyperplasia, hirsutism, tremor and many CYP3A4 interactions. Levels are monitored.
A patient with a large intrapulmonary shunt has an arterial PO2 that barely rises when given 100% oxygen. Why?
Answer: Shunted blood misses O₂
Hypoxaemia has five mechanisms: low inspired oxygen, hypoventilation (raised PaCO2, normal A–a gradient), diffusion impairment, V/Q mismatch and right-to-left shunt (the last three raise the A–a gradient). In a shunt (V/Q = 0; e.g. pneumonia with consolidated alveoli, atelectasis, ARDS, intracardiac shunts), blood passes unventilated alveoli; blood from normal units is already almost fully saturated on the flat part of the dissociation curve and cannot carry much extra oxygen, so mixed arterial PO2 improves little with 100% oxygen. Hypoxaemia from V/Q mismatch or diffusion impairment corrects well with oxygen. Treating a shunt requires reopening alveoli (PEEP, recruitment) or treating the cause.
Hypoxaemia mechanisms
Normal A–a: low FiO2, hypoventilation
Raised A–a: diffusion, V/Q mismatch, shunt
Shunt: poor response to 100% O2
Treat shunt: PEEP, recruit alveoli
Remember Oxygen can't reach blood that never meets air.
Why the other options are wrong
Fibrosis blocks diffusion Diffusion limitation does improve with oxygen.
Haemoglobin affinity too low Not the mechanism.
The patient hypoventilates Hypoventilation corrects with oxygen.
CO₂ displaces oxygen Not correct.
Question 4Hard
A 60-year-old man with poorly controlled diabetes presents with fever, neck stiffness and confusion. His blood glucose is 20 mmol/L (360 mg/dL). CSF shows 1,200 white cells/µL (90% neutrophils), protein 2.1 g/L and glucose 4.4 mmol/L (80 mg/dL). The junior doctor notes that the CSF glucose is in the usual normal range. Which of the following is the correct interpretation of the CSF glucose?
Answer: Low, as the CSF:blood ratio is about 0.2
Glucose enters the CSF by facilitated transport (GLUT1) across the choroid plexus and blood–brain barrier, so CSF glucose is normally about 60% of the blood level. It must always be interpreted against a simultaneous blood glucose. Here the ratio is 4.4/20, about 0.22; a ratio below 0.4 is low and, with a neutrophil pleocytosis and high protein, supports bacterial meningitis, where bacteria and neutrophils consume glucose and transport is impaired.
CSF glucose
Normal CSF:blood ratio ~0.6
Ratio < 0.4: low
Always take a simultaneous blood glucose
Low in bacterial, TB, fungal meningitis
Remember Judge CSF glucose by the ratio, not the number.
Exam trap Hyperglycaemia can make a low CSF glucose look normal.
Why the other options are wrong
Normal, as it lies within the reference range A 'normal' value is low when blood glucose is very high.
High, from glucose leaking into CSF CSF glucose follows blood glucose but is not raised by leakage.
Unreliable unless measured fasting A simultaneous blood sample is what is needed, not fasting.
Normal, which excludes bacterial meningitis The ratio is low and the cell picture is bacterial.
Eight days after a 15-month-old receives MMR vaccine, he develops fever and a faint rash, and has a brief febrile seizure. Why do these effects appear about a week after vaccination rather than on the same day?
Answer: Vaccine measles virus replicates first
MMR contains live attenuated viruses. Fever, rash and the small risk of febrile seizures occur around 6–11 days after vaccination, when the attenuated measles virus is replicating and stimulating the immune response, mimicking a very mild form of the natural infection. Mumps components can cause mild parotid swelling at about 3 weeks, and rubella can cause transient arthralgia. Reactions to inactivated vaccines, by contrast, occur within the first 48 hours. The febrile seizures have no long-term consequence.
Timing of vaccine reactions
Inactivated: reactions within 48 h
MMR: fever, rash day 6–11
Mumps parotitis ~3 weeks
Due to vaccine virus replication
Remember Live vaccines cause a mini-illness about a week later.
Exam trap A febrile seizure after MMR is not a contraindication to the second dose.
Why the other options are wrong
The adjuvant takes a week to dissolve MMR does not contain an aluminium adjuvant.
Delayed IgE-mediated allergy to egg Egg allergy is not a contraindication, and IgE reactions occur within minutes.
Contaminating bacteria multiply slowly Vaccines are sterile; this is viral replication.
Maternal antibodies are released late Maternal antibodies are already present; they are not 'released'.
Question 6Hard
A 30-year-old man treated with intravenous artesunate for severe falciparum malaria with 12% parasitaemia recovers and goes home. Two weeks later he returns tired and pale: Hb has fallen from 11 to 7.2 g/dL, LDH is high and haptoglobin undetectable. The blood film shows no parasites and the direct Coombs test is negative. What is the most likely mechanism?
Answer: Early loss of once-infected red cells
Artesunate kills ring-stage parasites inside red cells so fast that the spleen removes the dead parasites by pitting and returns the red cells to the circulation. These once-infected cells have a shortened lifespan and are destroyed together 1–3 weeks later, giving post-artesunate delayed haemolysis. It is commonest after high parasitaemia, and Hb should be checked about 2 weeks after treatment. Most cases need only monitoring, with transfusion if severe.
Post-artesunate haemolysis
Spleen pits dead parasites
Red cells returned, short-lived
Haemolysis 1–3 weeks later
Check Hb at 2 weeks after high parasitaemia
Remember Fast cure today, falling Hb in two weeks.
Exam trap Do not assume treatment failure; check the film for parasites first.
Why the other options are wrong
Recrudescence of falciparum infection The film shows no parasites.
G6PD-related oxidant haemolysis Artesunate is not an oxidant drug like primaquine.
Blackwater fever from immune haemolysis Blackwater fever occurs during acute malaria, often with quinine.
Hypersplenism following acute malaria Hypersplenism does not cause sudden haemolysis with absent haptoglobin.
Question 7Hard
Three months after a kidney transplant, a woman's creatinine rises. Biopsy shows neutrophils in peritubular capillaries, and donor-specific anti-HLA antibodies are found. Which of the following stains would confirm antibody-mediated rejection?
Answer: C4d in peritubular capillaries
In acute antibody-mediated (humoral) rejection, donor-specific antibodies bind HLA on graft endothelium and activate the classical complement pathway. C4d, a split product of C4, binds covalently to the endothelium and remains after the antibody has gone. Linear C4d staining of peritubular capillaries is therefore a marker of antibody-mediated injury. T-cell tubulitis is the hallmark of acute cellular rejection.
Humoral rejection
Donor-specific anti-HLA antibodies
Classical complement activated
C4d stays on capillaries
Cellular rejection: tubulitis
Remember C4d marks antibody damage.
Exam trap Steroids alone often fail in humoral rejection; antibody removal may be needed.
Why the other options are wrong
Congo red staining of glomeruli Stains amyloid.
CD8 T cells invading the tubules Acute cellular rejection.
IgA deposits in the mesangium IgA nephropathy.
Viral inclusions in tubular cells BK polyomavirus nephropathy.
Question 8Hard
A child with the inherited form of retinoblastoma develops tumours in both eyes at a young age, while the sporadic form gives a single later tumour. Which idea best explains this difference?
Answer: Two hits are needed; inherited cases start with one
The retinoblastoma gene RB1 is a tumour suppressor, and both copies must be lost for a tumour to form (Knudson's two-hit hypothesis). In the inherited form, one defective allele is present in every cell from birth, so a single further mutation in any retinal cell causes a tumour, giving early, bilateral and multiple tumours. In the sporadic form, both hits must occur in the same cell, which is rare, so tumours are single and later.
Two-hit hypothesis
RB1 is a tumour suppressor
Both alleles must be lost
Inherited: one hit already present
Bilateral, early tumours
Remember Inherited cancer starts one hit ahead.
Exam trap Tumour suppressors need both alleles lost; oncogenes need only one activated.
Why the other options are wrong
The inherited gene is an oncogene RB1 is a tumour suppressor, not an oncogene.
Inherited tumours need only one normal allele Both alleles must be lost, not one left normal.
The retinoblastoma gene promotes growth Loss of RB removes a brake; it does not promote growth.
Inherited cases lack any functional tumour gene from birth One allele is still functional until the second hit.
Question 9Medium
Two weeks after a sore throat, an 8-year-old boy has puffy eyelids, cola-coloured urine and BP 140/90 mmHg. Urine shows red cell casts and 1+ protein, and serum albumin is 38 g/L. What is the main mechanism of his oedema?
Answer: Renal salt and water retention
This is post-streptococcal glomerulonephritis, a nephritic syndrome. Immune-complex inflammation reduces glomerular filtration, so the kidney retains sodium and water, expanding plasma and extracellular volume. Capillary hydrostatic pressure rises, causing oedema and hypertension, with a normal serum albumin. In nephrotic syndrome, by contrast, heavy proteinuria and low albumin are central, although renal sodium retention contributes there too.
Nephritic oedema
Low GFR: salt and water retained
Volume expansion: hypertension
Albumin normal
Red cell casts, mild proteinuria
Remember Nephritic oedema is volume overload; nephrotic oedema starts with low albumin.
Exam trap Periorbital oedema is common to both nephritic and nephrotic syndromes; look at albumin and urine.
Why the other options are wrong
Low plasma oncotic pressure Albumin is normal at 38 g/L.
Increased capillary permeability Not the main mechanism in nephritic syndrome.
Lymphatic obstruction Not involved here.
Raised capillary oncotic leakage Not a recognised mechanism.
Question 10Easy
A 50-year-old woman started on amitriptyline for neuropathic pain complains of a very dry mouth. Blockade of which receptors on salivary acinar cells causes this?
Answer: Muscarinic M3 receptors
Parasympathetic fibres reach the parotid through the glossopharyngeal nerve and otic ganglion, and the submandibular and sublingual glands through the facial nerve, chorda tympani and submandibular ganglion. They release acetylcholine onto M3 receptors on acinar cells, which act through Gq and calcium to produce large volumes of watery saliva. Tricyclics, antihistamines with anticholinergic action, antipsychotics and bladder antimuscarinics block these receptors and cause dry mouth, raising the risk of caries. Pilocarpine, a muscarinic agonist, increases saliva.
Salivary secretion control
Main drive: parasympathetic
ACh on M3 receptors (Gq, Ca²⁺)
Anticholinergics: dry mouth
Pilocarpine increases saliva
Remember Anticholinergic drugs dry the mouth by blocking M3.
Exam trap Both autonomic divisions stimulate saliva, but the parasympathetic effect dominates.
Why the other options are wrong
Alpha-1 adrenergic receptors Sympathetic; gives scanty viscous saliva, not the main flow.
Histamine H1 receptors H1 block causes sedation, not dry mouth.
Beta-1 adrenergic receptors Sympathetic; adds protein to saliva.
Nicotinic acetylcholine receptors Found at ganglia, not on acinar cells.
Question 11Easy
A 26-year-old woman with episodic wheeze has an FEV₁ of 2.0 L that rises to 2.5 L 15 minutes after inhaled salbutamol. What does this show?
Answer: Reversible airflow obstruction
A rise in FEV₁ of at least 12% and 200 mL after a bronchodilator shows that the airway narrowing is reversible and supports asthma; here FEV₁ rose by 500 mL, or 25%. In asthma the obstruction comes from bronchial smooth muscle contraction, mucosal oedema and mucus, and β₂ agonists relax the smooth muscle. COPD typically shows little reversibility. Restrictive disease reduces FEV₁ and FVC together with a normal or high ratio and does not respond to bronchodilators.
Bronchodilator reversibility
FEV₁ rise ≥12% and ≥200 mL
Supports asthma
COPD: little reversibility
Here: +500 mL, +25%
Remember A big FEV₁ rise after salbutamol suggests asthma.
Exam trap A negative reversibility test does not exclude asthma; testing when well or on treatment may show little change.
Why the other options are wrong
Fixed airflow obstruction Fixed obstruction would not improve this much.
Restrictive lung defect Restriction is not reversed by bronchodilators.
Upper airway obstruction Upper airway obstruction is shown by a flattened flow–volume loop.
Normal day-to-day variation A 25% rise is well beyond normal variation.
Question 12Hard
In a 65-year-old man with suspected autonomic failure, blood pressure is recorded during a Valsalva manoeuvre. When he releases the strain, his pressure slowly returns to baseline but does not rise above it, and his heart rate does not slow. What does the missing overshoot indicate?
Answer: No sympathetic vasoconstriction
During the strain phase, raised intrathoracic pressure reduces venous return and cardiac output, so pressure falls and baroreceptors increase sympathetic vasoconstriction and heart rate. On release, venous return and output recover while the arterioles are still constricted, so pressure normally overshoots above baseline (phase IV), and the baroreflex then slows the heart through the vagus. Without sympathetic vasoconstriction there is no overshoot and no reflex bradycardia. An absent overshoot is a sensitive sign of sympathetic adrenergic failure.
Valsalva manoeuvre
Strain: venous return falls
Sympathetic vasoconstriction builds
Release: pressure overshoots (phase IV)
No overshoot: sympathetic failure
Remember No overshoot after the strain means the sympathetic arm is gone.
Exam trap Patients with heart failure can also lose the overshoot, so interpret the test with cardiac status in mind.
Why the other options are wrong
Normal autonomic reflex function A normal response shows an overshoot with reflex slowing.
Excess resting vagal tone Excess vagal tone would cause bradycardia, not this pattern.
Hypovolaemia without nerve damage Hypovolaemia alone does not abolish the overshoot after strain.
Stiffness of the aortic valve Valve disease does not explain the missing overshoot.
