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Question 1
Incorrect
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A 30-year-old woman admitted following a tonsillectomy has developed stridor with a respiratory rate of 22 breaths per minute and obstructive movements of the chest and abdomen that is in a see-saw pattern .
Her SpO2 is 92% on 60% oxygen with pulse rate 120 beats per minute while her blood pressure is 180/90mmHg. She is repeatedly trying to remove the oxygen mask and appears anxious.
Her pharynx is suctioned and CPAP applied with 100% oxygen via a Mapleson C circuit.
Which of these is the most appropriate next step in her management?Your Answer:
Correct Answer: Administer intravenous propofol 0.5 mg/kg
Explanation:Continuous closure of the vocal cords resulting in partial or complete airway obstruction is called Laryngospasm. It is a reflex that helps protect against pulmonary aspiration.
Predisposing factors include: Hyperactive airway disease, Insufficient depth of anaesthesia, Inexperience of the anaesthetist, Airway irritation, Smoking, Shared airway surgery and Paediatric patients
Its primary treatment includes checking for blood or stomach aspirate in the pharynx, removing any triggering stimulation, relieving any possible supra-glottic component to airway obstruction and application of CPAP with 100% oxygen.
In this patient, all the above has been done and the next treatment of choice is the administration of a rapidly acting intravenous anaesthetic agent such as propofol (0.5 mg/kg) in increments as it has been reported to relieve laryngospasm in approximately 75% of cases. Administering suxamethonium to an awake patient would be inappropriate at this stage.
Magnesium and lidocaine are used for prevention rather than acute treatment of laryngospasm. Superior laryngeal nerve blocks have been reported to successfully treat recurrent laryngospasm but it is not the next logical step in index patient.
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This question is part of the following fields:
- Pathophysiology
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Question 2
Incorrect
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Which of the following may indicate an inadequate reversal of non-depolarising neuromuscular blockade?
Your Answer:
Correct Answer: Post tetanic count of 5
Explanation:A post-tetanic count of 5 denotes a deep neuromuscular blockade.
Post tetanic count (PTC) is a well-established method of evaluating neuromuscular recovery during intense neuromuscular blockade. It cam ne used when there is no response to single twitch, tetanic, or train-of-four (TOF) stimulation to assess the intensity of neuromuscular blockade and to estimate the duration after which the first twitch in the TOF (T1) is likely to reappear.
During a nondepolarizing block, the high frequency of tetanic stimulation will induce a transient increase in the amount of acetylcholine released from the presynaptic nerve ending, such that the intensity of subsequent muscle contractions will be increased (potentiated) briefly (period of post-tetanic potentiation, which may last 2 to 5 min. The neuromuscular response to stimulation during post tetanic potentiation can be used to gauge the depth of block when TOF stimulation otherwise evokes no responses. The number of post tetanic responses is inversely proportional to the depth of block: fewer post tetanic contractions denote a deeper block. When the post tetanic count (PTC) is 6 to 8, recovery to TOF count = 1 is likely imminent from an intermediate-duration blocking agent; when the PTC is 0, the depth of block is profound, and no additional NMBA should be administered.
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This question is part of the following fields:
- Pathophysiology
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Question 3
Incorrect
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A 54-year-old lady comes in for a right-sided elective bunionectomy with a realignment osteotomy under local anaesthetic on her first (large) toe.
For the operation, which of the following nerve blocks will be most effective?Your Answer:
Correct Answer: Superficial peroneal, deep peroneal and posterior tibial nerves
Explanation:An ankle block is commonly used for anaesthesia and postoperative analgesia when operating on bunions. It results in the selective block of the superficial peroneal, deep peroneal, and posterior tibial nerves.
The deep peroneal nerve supplies sensory input to the web space between the first and second toes (L4-5).
The L2-S1 nerve, often known as the superficial peroneal nerve, is a mixed motor and sensory neuron. It gives sensory supply to the anterolateral region of the leg, the anterior aspect of the 1st, 2nd, 3rd, and 4th toes, and innervates the peroneus longus and brevis muscles (with the exception of the web space between 1st and 2nd toes).
The sensory area of the saphenous nerve (L3-4) in the foot stretches from the proximal portion of the midfoot on the medial side to the proximal part of the midfoot on the lateral side.
The lateral side of the little (fifth) toe is innervated by the sural nerve’s sensory supply (S1-2). The heel, medial (medial plantar nerve), and lateral (lateral plantar nerve) soles of the foot are all served by the posterior tibial nerve.
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This question is part of the following fields:
- Pathophysiology
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Question 4
Incorrect
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In a diagnosis of a compensated respiratory acidosis, which of the following arterial blood gas results is likely to be seen?
Your Answer:
Correct Answer:
Explanation:During normal tissue metabolism, there is production of CO2 (acid) which is then expired by the lungs. If metabolism switches from aerobic to anaerobic due to a lack of oxygen, the tissues are unable to completely oxidise sugars to CO2. As a consequence, the sugars can only be partially oxidised to lactic acid. Since lactic acid cannot be expired by the lungs, it remains in the circulation leading to metabolic acidosis.
Also, normal tissue metabolism leads to the production of some amount of acid from the breakdown of proteins. These acids are excreted from the body by kidney filtration. Renal failure will therefore results in acidosis after several days.
An increased acidosis stimulates the brain’s respiratory centres to increase the respiratory rate. This lowers the CO2 in the blood, leading to a decrease in its acidity. Renal excretion removes the excess acid, resulting in a normal pH, and a reduced PaCO2 and HCO3.
pH PaCO2 (kPa) HCO3
Compensated respiratory acidosis 7.34 7.2 29
Acute respiratory acidosis 7.25 7.3 22
Compensated metabolic acidosis 7.34 3.6 14
Metabolic acidosis 7.21 5.3 15
Metabolic alkalosis 7.51 5.1 30 -
This question is part of the following fields:
- Pathophysiology
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Question 5
Incorrect
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A 5-year-old child is scheduled for squint surgery requiring general anaesthesia.
To begin, she is given sevoflurane for the inhalation induction, then intravenous access is established along with the insertion of a supraglottic airway. Anaesthesia is maintained with fentanyl 1 mcg/kg, with an air/oxygen/sevoflurane mix with spontaneous respirations.
Once the surgery begins, her pulse rate drastically reduces from 120 beats/min to 8 beats/min.
What is the most appropriate next step for this patient?Your Answer:
Correct Answer: Tell surgeon to stop surgical retraction
Explanation:This sudden change in pulse rate is due to the oculocardiac reflex. It is a >20% reduction in pulse rate as a result of placing pressure directly on the eyeball. The reflex arc has an afferent and efferent arm:
The afferent (sensory) arm: The trigeminal nerve (CN V)
The efferent arm: The vagus nerve (CN X)
The most appropriate action is to ask the surgeon to stop retraction of the extraocular muscles, Assess for hypoxia, and give 100% oxygen if indicated.
Atropine of glycopyrrolate can be administered to counteract the reflex, and also prevent any further vagal reflexes.
Administration of fentanyl may increase patient’s risk of bradycardia and sinus arrest in this case.
Adrenaline is not indicated here as other treatment options will provide sufficient relief from arrhythmia.
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This question is part of the following fields:
- Pathophysiology
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Question 6
Incorrect
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The following statements are about the conjugation of bilirubin. Which is true?
Your Answer:
Correct Answer: Is catalysed by a glucuronyl transferase
Explanation:Bilirubin is formed by metabolizing heme, mostly from haemoglobin in red blood cells.
Bilirubin is conjugated to glucuronic acid in the hepatocytes by the glucuronyl transferase enzyme in order to enable it to become soluble and allow for its secretion across the canalicular membrane and into bile.
The conjugation process is increased by rifampicin and decreased by valproate.
Gilbert’s syndrome is caused by a decrease in glucuronyl transferase in the hepatic system, decreasing the transport of bilirubin into the hepatocyte, causing unconjugated bilirubinaemia.
Crigler-Najjer syndrome is caused by mutations in the genes responsible for hepatic glucuronyl transferase, decreasing the activity of the enzyme, meaning bilirubin cannot be conjugated, causing unconjugated bilirubinaemia.
Dubin-Johnson syndrome does not cause an impairment in the conjugation of bilirubin, but it blocks the transport of bilirubin out of the hepatocyte resulting in conjugated bilirubinaemia.
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This question is part of the following fields:
- Pathophysiology
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Question 7
Incorrect
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A 46-year-old woman is listed for clipping of a cerebral aneurysm, following a diagnosis of surgical third nerve palsy.
Which of the following clinical findings correlate with surgical third nerve palsy?Your Answer:
Correct Answer: Ptosis, inferolateral rotation of globe and mydriasis
Explanation:Ptosis and mydriasis are visible in surgical third nerve palsy, and the eye looks ‘down and out.’ The loss of innervation to all of the major structures supplied by the oculomotor nerve is reflected in these characteristics.
Ptosis is caused by the paralysis of the levator palpebrae superioris in oculomotor nerve palsy. Due to the unopposed actions of the superior oblique and lateral rectus muscles, the eye rotates down and out.
Mydriasis is caused by surgical (compressive) causes of third nerve palsy, which disrupt the parasympathetic pupillomotor fibres on the nerve’s periphery.
Medical (ischaemic) causes of a third nerve palsy, on the other hand, leave the superficial parasympathetic fibres relatively unaffected and the pupil unaffected.
Horner’s syndrome is characterised by ptosis, anhidrosis, and miosis, which are caused by a loss of sympathetic innervation to the tarsal muscle of the upper lid, facial skin, and dilator pupillae, respectively.
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This question is part of the following fields:
- Pathophysiology
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Question 8
Incorrect
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Which of the following nerves is responsible for carrying taste sensation from the given part of the tongue?
Your Answer:
Correct Answer: Anterior two thirds of tongue - facial nerve
Explanation:Taste sensation from the anterior two-thirds of the tongue is carried by chorda tympani, a branch of the facial nerve.
The general somatic sensation of the anterior two-third of the tongue is supplied by the lingual nerve, a branch of the mandibular nerve.
Both general somatic sensation and taste from the posterior third of the tongue are carried by the glossopharyngeal nerve.
All the muscles of the tongue except palatoglossus are supplied by the hypoglossal nerve whereas palatoglossus is supplied by the vagus nerve. (This is because palatoglossus is the only tongue muscle derived from the fourth branchial arch)
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This question is part of the following fields:
- Pathophysiology
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Question 9
Incorrect
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With a cervical dilation of 9 cm, a 23-year-old term primigravida is in labour. She is otherwise in good health. She's been in labour for 14 hours and counting.
Early foetal pulse decelerations can be seen on the cardiotocograph, and a recent foetal scalp blood sample revealed a pH of 7.25.
Which of the following is true about this patient's care and management?Your Answer:
Correct Answer: Monitor for downward trend in fetal scalp blood pH as caesarean section is not indicated at the present time
Explanation:Once the decision to deliver a baby by caesarean section has been made, it should be carried out with a level of urgency commensurate with the risk to the baby and the mother’s safety.
There are four types of caesarean section urgency:
Category 1 – Endangering the life of the mother or the foetus
Category 2 – Maternal or foetal compromise that is not immediately life threatening
Category 3 – Early delivery is required, but there is no risk to the mother or the foetus.
Category 4: Elective delivery at a time that is convenient for both the mother and the maternity staff.Caesarean sections for categories 1 and 2 should be performed as soon as possible after the decision is made, especially for category 1. For category 1 caesarean sections, a decision to deliver time of 30 minutes is currently used.
In most cases, Category 2 caesarean sections should be performed within 75 minutes of making the decision.
The condition of the woman and the unborn baby should be considered when making a decision for a quick delivery, as it may be harmful in some cases.
There is no evidence of foetal compromise in the example above (early foetal pulse decelerations and a pH of less than 7.25). Early foetal pulse decelerations are most likely caused by the uterus compressing the foetal head. The foetus is not harmed by these. A spinal anaesthetic is preferred over a general anaesthetic whenever possible.
If the foetal scalp blood pH is greater than 7.25, it’s a good idea to repeat the test later and look for any changes. When a foetus decelerates, the mother should be given oxygen, kept in a left lateral position, and kept hydrated to avoid the need for a caesarean section.
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This question is part of the following fields:
- Pathophysiology
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Question 10
Incorrect
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Following an uneventful laparoscopic right hemicolectomy, a previously fit and well 75-year-old male is admitted to the critical care unit.
You've been summoned to examine the patient because he's become oliguric.
Which of the following is most likely to indicate that acute kidney injury is caused by a prerenal cause?Your Answer:
Correct Answer: Serum urea: creatinine ratio 200
Explanation:Prerenal failure has a serum urea: creatinine ratio of >100, while acute kidney injury has a ratio of 40.
In prerenal failure, ADH levels are typically high, resulting in water, urea, and sodium resorption. The fractional sodium excretion is less than 1%, but it is greater than 2% in acute tubular necrosis.
Prerenal azotaemia has higher serum urea nitrogen/serum creatinine ratios (>20), whereas acute tubular necrosis has lower ratios (10-15). The normal range is between 12 and 20.
Urinary sodium is less than 20 in prerenal failure and greater than 40 in acute tubular necrosis.
Prerenal failure has a urine osmolality of >500, while acute tubular necrosis has an osmolality of 350.
Prerenal failure has a urine/serum creatinine ratio of >40, while acute tubular necrosis has a urine/serum creatinine ratio of 20.The concentrations of serum urea or creatinine change in inverse proportion to glomerular filtration. Changes in serum creatinine concentrations are more reliable than changes in serum urea concentrations in predicting GFR. Creatinine is produced at a constant rate from creatine, and blood concentrations are almost entirely determined by GFR.
A number of factors influence urea formation, including liver function, protein intake, and protein catabolism rate. Urea excretion is also influenced by hydration status, the amount of water reabsorption, and GFR.
A high serum creatinine level, as well as a urine output of less than 10 mL/hour and the production of concentrated looking urine, do not necessarily indicate a specific cause of oliguria.
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This question is part of the following fields:
- Pathophysiology
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Question 11
Incorrect
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Which of the following statements is true about an acute pulmonary embolism?
Your Answer:
Correct Answer: Thrombolysis administered through a peripheral vein is as effective as through a pulmonary artery catheter
Explanation:Acute pulmonary embolism occurs when a blood clot becomes embedded in a pulmonary artery and restricts lung blood flow.
Thrombolysis is recommended in patients with extremely compromised circulation rather than reduced oxygen in the blood. It is effective when administered via a peripheral vein or a pulmonary artery catheter.
Anticoagulant therapy (heparin use) decreases the risk of further embolic evens and decreases constriction of pulmonary vessels.
An ECG may be normal in patients with an acute pulmonary embolism.
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This question is part of the following fields:
- Pathophysiology
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Question 12
Incorrect
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Calcium homeostasis is regulated by parathormone (PTH).
Which of the following PTH actions is most likely to cause calcium to be released from bone?Your Answer:
Correct Answer: Indirect stimulation of osteoclasts
Explanation:The hormone parathyroid hormone (PTH) and the receptor parathyroid hormone type 1 (PTH1-Rc) are important regulators of blood calcium homeostasis.
PTH can cause a rapid release of calcium from the matrix in bone, but it also affects long-term calcium metabolism by acting directly on bone-forming osteoblasts (by binding to PTH1-Rc) and indirectly on bone-resorbing osteoclasts.
PTH causes changes in the synthesis and/or activity of several proteins, including osteoclast-differentiating factor, also known as TRANCE or RANKL, when it acts on osteoblasts.
RANK receptors are found on the cell surfaces of osteoclast precursors. The osteoclasts are activated when RANKL binds to the RANK receptors. Osteoclasts lack PTH receptors, whereas osteoblasts do. Osteoclasts are activated indirectly when the RANK receptor binds to the RANKL secreted by osteoblasts, resulting in bone resorption. PTH1 receptors are found in osteoclasts, but they are few.
PTH activates G-protein coupled receptors in all target cells via adenylate cyclase.
The PTH2 receptor is most abundant in the nervous system and pancreas, but it is not a calcium metabolism regulator. It is abundant in the septum, midline thalamic nuclei, several hypothalamic nuclei, and the dorsal horn of the spinal cord, as well as the cerebral cortex and basal ganglia. Expression in pancreatic islet somatostatin cells is the most prominent on the periphery.
The distribution of the receptor is being used to test functional hypotheses. It may play a role in pain modulation and hypothalamic releasing-factor secretion control.
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This question is part of the following fields:
- Pathophysiology
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Question 13
Incorrect
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A previously fit 26-year-old is undergoing surgery to repair an inguinal hernia. He is breathing on his own, and a supraglottic airway is being maintained via a circle system with air/oxygen and sevoflurane.
With a fresh gas flow of 14 L/min, the end-tidal CO2 reading is 8.1 kPa. CO2 pressure is 1.9 kPa. The percentages of oxygen inhaled and exhaled are 38 and 33 percent, respectively.
What do you think is the most likely source for these readings?Your Answer:
Correct Answer: Incompetent expiratory valve
Explanation:The patient is rebreathing carbon dioxide that has been exhaled.
Exhaustion of the soda lime and failure of the expiratory valve are the two most likely causes. A leak in the inspiratory limb is a less likely cause. Increased inhaled and exhaled carbon dioxide levels may appear with a normal-looking capnogram if the expiratory valve is ineffective.
The patient will exhale into both the inspiratory and expiratory limbs if the inspiratory valve is inoperable. A slanted downstroke inspiratory phase (as the patient inhales carbon dioxide-containing gas from the inspiratory limb) and increased end-tidal carbon dioxide can be seen on the capnogram.
Even if the soda lime were exhausted, a high fresh gas flow would be enough to prevent rebreathing. The difference in oxygen concentrations in inspired and expired breaths would be less pronounced.
Hypercapnia is caused by respiratory obstruction and malignant hyperthermia, but not by rebreathing.
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This question is part of the following fields:
- Pathophysiology
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Question 14
Incorrect
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A 43-year old woman, presented to the emergency department. She has suffered trauma to her right orbital floor.
On examination, it is noted that her right eye is deviated upwards when compared to her left. She also has a deliberate tilt in her head to the left in an attempt to compensate for loss of intorsion.
This clinical sign is caused by damage to which of the following cranial nerves?Your Answer:
Correct Answer: Trochlear nerve
Explanation:The trochlear nerve (CN IV) is the fourth and smallest cranial nerve. It’s role is to provide somatic motor innervation of the superior oblique muscle which is responsible for oculomotion.
Injury to the trochlear nerve will result in vertical diplopia, which worsens when looking downwards or inwards. This diplopia presents as an upward deviation of the eye with a head tilt away from the site of the lesion.
The abducens nerve (CN VI) provides somatic motor innervation for the lateral rectus muscle which functions to abduct the eye. Injury to this nerve will cause diplopia and an inability to abduct the eye, causing the patient to have to rotate their head to look sideways.
The facial nerve (CN VII) provides sensory, motor and parasympathetic innervations. It’s motor aspect controls the muscles of facial expression. Damage will cause paralysis of facial expression.
The oculomotor nerve (CN III) provides motor and parasympathetic innervations. Its motor component controls most of the other extraocular muscles. Damage to it will result in ptosis, dilatation of the pupil and a down and out eye position.
The ophthalmic division of the trigeminal nerve (CN VI) is responsible for sensory innervation of skin, mucous membranes and sinuses of the upper face and scalp.
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This question is part of the following fields:
- Pathophysiology
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Question 15
Incorrect
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Concerning the anterior pituitary gland, one of following is true.
Your Answer:
Correct Answer: Produces glycoproteins
Explanation:The posterior pituitary and the hypothalamus are connected by the pituitary stalk. It contains in the pituitary sella and has the optic chiasm and hypothalamus as superior relations.
The anterior pituitary produces thyroid-stimulating hormone (TSH), luteinising hormone (LH) and follicle-stimulating hormone (FSH) . These hormones are Glycoproteins and share a common alpha subunit with unique beta subunits.
The secretion of pituitary hormones are pulsatile. Examples are LH, adrenocorticotropic hormone (ACTH) and growth hormone (GH).
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This question is part of the following fields:
- Pathophysiology
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Question 16
Incorrect
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A 42-year-old man presented with a bitemporal hemianopia with enlarged hands and feet. On examination, he was found to be hypertensive.
Which of the following correctly explains the cause of his visual field defect?Your Answer:
Correct Answer: Pituitary macroadenoma secreting growth hormone (GH)
Explanation:Pituitary macroadenoma is a benign tumour with growth larger than 10mm (those under 10mm are called microadenoma)
Compression of optic chiasm by pituitary adenoma is responsible for causing visual field defects like bitemporal hemianopia, optic neuropathy.
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This question is part of the following fields:
- Pathophysiology
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Question 17
Incorrect
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A 55-year-old woman presents for transsphenoidal surgery following a diagnosis of pituitary macroadenoma.
Which of the following is the most common visual field defect caused by such lesions?Your Answer:
Correct Answer: Bitemporal hemianopia
Explanation:Pituitary tumours that compress the optic chiasma primarily affect the neurones that decussate at this location. Bitemporal hemianopia is caused by neurones that emerge from the nasal half of the retina and transmit the temporal half of the visual field.
The axons of ganglion cells in the retina form the optic nerve.
It exits the orbit through the optic foramen and projects to the thalamic lateral geniculate body. The optic chiasma forms above the sella turcica as the nasal fibres decussate along the way. The optic radiation travels from the lateral geniculate body to the occipital cortex.
Lesions at various points along this pathway cause the following visual field defects:
Scotoma implies partial retinal or optic nerve damage.
Monocular vision loss occurs when the optic nerve is completely damaged.
Pathology at the optic chiasma causes bitemporal hemianopia.
Cortical blindness with occipital cortex pathology and homonymous hemianopia with lesions compromising the optic radiation. -
This question is part of the following fields:
- Pathophysiology
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Question 18
Incorrect
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You've been summoned to the paediatric ward after a 4-year-old child was discovered 'collapsed' in bed.
The child had been admitted the day before with febrile convulsions and was scheduled to be discharged. It is safe to approach the child.
What should your first life-saving action be?Your Answer:
Correct Answer: Apply a gentle stimulus and ask the child if they are alright
Explanation:Paediatric life support differs from adult life support in that hypoxia is the primary cause of deterioration.
After checking for danger, the child should be given a gentle stimulus (such as holding the head and shaking the arm) and asked, Are you alright? according to current advanced paediatric life support (APLS) guidelines. Safety, Stimulate, Shout is a phrase that is frequently remembered. Any airway assessment should be preceded by these actions.
Although the algorithm includes five rescue breaths, they are performed after the airway assessment.
It is not recommended to ask parents to leave unless they are obstructing the resuscitation. A team member should be with them at all times to explain what is going on and answer any questions they may have.
CPR should not begin until the child has been properly assessed and rescue breaths have been administered.
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This question is part of the following fields:
- Pathophysiology
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Question 19
Incorrect
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A 4-year-old boy with status epilepticus was brought to ER and has already received two doses of intravenous lorazepam but is still continuing to have seizures.
Which of the following drug would be best for his treatment?Your Answer:
Correct Answer: Phenytoin 20 mg/kg IV
Explanation:When the convulsion lasts for five or more than five minutes, or if there are recurrent episodes of convulsions in a 5 minute period without returning to the baseline, it is termed as Status Epilepticus.
The first priority in the patient with seizures is maintaining the airway, breathing, and circulation.Guideline for the management of Status Epilepticus in children by Advanced Life Support Group is as follow:
Step 1 (Five minutes after the start of seizures):
If intravascular access is available start treatment with lorazepam 0.1 mg/kg IV
If no intravascular access then give buccal midazolam 0.5 mg/kg or rectal diazepam 0.5 mg/kg.Step 2 (Ten minutes after the start of seizure):
If the convulsions continue then a second dose of benzodiazepine should be given. Senior should be called on-site and phenytoin should be prepared.
No more than two doses or benzodiazepines should be given (including any doses given before arrival at the hospital)
If still no IV access then obtain intraosseous access (IO).Step 3 (Ten minutes after step 2)
Senior help along with anaesthetic/ICU help should be sought
Phenytoin 20 mg/kg IV over 20 minutes
If the seizure stops before the full dose of phenytoin is given then the infusion should be completed as this provides up to 24 hours of anticonvulsant effect
In children already receiving phenytoin as treatment for epilepsy then an alternative is phenobarbitone 20 mg/kg IV over five minutes
Once the phenytoin is started, senior staff may wish to give rectal paraldehyde 0.4 mg/kg although this is no longer included in the routine algorithm recommended by APLS.Step 4 (20 minutes after step 3)
If 20 minutes after starting phenytoin the child remains in status epilepticus then rapid sequence induction of anaesthesia with thiopentone and a short acting paralysing agent is needed and the child transferred to paediatric intensive care.
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This question is part of the following fields:
- Pathophysiology
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Question 20
Incorrect
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A 52-year old man was placed under general anaesthesia for an emergent open cholecystectomy. As part of the induction, suxamethonium was administered at 1.5mg/kg. Post-operatively, there was failure to restore muscle twitch responses over a course of five hours.
Clinical chemistry studies were obtained and showed the following results:
Butrylcholinesterase (BChE) activity: 49 U/L (Reference range: 3300-10,300 U/L)
Dibucaine number: <4% (Reference range: 83-88%)
The attending physician gave an initial diagnosis of Suxamethonium Apnoea.
What is the most probable phenotype of BChE of the patient?Your Answer:
Correct Answer: S (silent)
Explanation:Silent (S) is the most probable phenotype of the patient. In S phenotype, patients have significantly reduced levels of BChE, the lowest among the four phenotypes. Because of this, individuals with S phenotype are subjected to long periods of apnoea. In addition, their dibucaine number is very low.
Other BChE phenotypes are the following:
Usual (U)
Atypical (A)
Fluoride-resistant (F) -
This question is part of the following fields:
- Pathophysiology
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Question 21
Incorrect
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A 70-year-old man presents to hospital complaining of dysphagia. He is scheduled for a rigid oesophagoscopy.
On examination, He is noted to have severe osteoarthritis in his cervical spine resulting in limited rotation and flexion-extension. He has no other neurological signs or symptoms.
He is given anaesthesia for the procedure, which is complicated by a difficult intubation (Cormack-Lehane 3), but was eventually achieved using a gum elastic bougie.
After recovering from anaesthesia, he is examined and found to have severe motor weakness of upper limbs, and mild motor weakness of lower limbs, bladder dysfunction and sensory loss of varying degrees below the level of C5.
What incomplete spinal cord lesion is most likely to be responsible for his symptoms?Your Answer:
Correct Answer: Central cord syndrome
Explanation:Central cord syndrome is the most commonly occurring type of partial spinal cord lesion. It is more likely to occur in older patients with cervical spondylosis and a hyperextension injury. The injury to the spinal cord occurs in the grey matter causing the following symptoms:
Disproportionally higher motor function weakness in the upper limbs than in lower limbs
Dysfunction of the bladder
Degrees of sensory loss below the level of the lesionAn anterior spinal artery infarction will interrupt the corticospinal tract resulting in paralysis of motor function, loss of pain and temperature sensation, all occurring below the level of the injury.
Brown-Sequard syndrome occurs as a result of the hemisection of the spinal cord. Its symptoms include ipsilateral upper motor neurone paralysis and loss of proprioception, with contralateral loss of pain and temperature sensation.
Spinal cord infarctions rarely occur in the posterior spinal artery.
Cauda equina syndrome occurs as a result of compression of the lumbosacral spinal nerve roots below the level of the conus medullaris. Injury to these nerves will cause partial or complete loss of movement and sensation in this distribution.
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This question is part of the following fields:
- Pathophysiology
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Question 22
Incorrect
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When an inotrope is given to the body, it has the following effects on the cardiovascular system:
The automaticity of the sino-atrial node increases
Lusitropy is accelerated
Dromotropy is increased
Chronotropy is increased
Inotropy increases
There is increased excitability of the conducting system
The most probably mechanism of action of this compound is?Your Answer:
Correct Answer: Increase in intracellular calcium influenced by a conformational change of a Gs protein
Explanation:A beta-1 adrenoreceptor agonist is most likely the ligand that causes increased automaticity, increased chronotropy, increased excitability, and increased inotropy on the sino-atrial node. However, alpha-1 adrenoreceptor effects may cause an increase in systemic vascular resistance. Noradrenaline, adrenaline, dopamine, and ephedrine are examples of drugs with mixed alpha and beta effects.
Adrenaline, noradrenaline, dopamine, dopexamine, dobutamine, ephedrine, and isoprenaline are examples of drugs that have some beta-1 activity. The beta-1 receptor is a G protein-coupled metabotropic receptor. When the beta-1 agonist binds to the cell surface membrane, it causes a conformational change in the Gs unit, which triggers a cAMP-dependent pathway and a calcium influx into the cell.
Catecholamines also help to relax the heart muscle (positive lusitropy). Dromotropy is the ability to increase the atrioventricular (AV) node’s conduction velocity.
Inodilators cause an increase in intracellular calcium as a result of phosphodiesterase III (PDIII) inhibition. Milrinone, enoximone, and amrinone are some examples. Positive inotropy is caused by increased calcium entry into the myocytes. Lusitropy is also increased by phosphodiesterase inhibitors. Increased cAMP inhibits myosin light chain kinase, resulting in reduced phosphorylation of vascular smooth muscle myosin, lowering systemic and pulmonary vascular resistance.
The mechanism of action of alpha-1 adrenoreceptor agonists is an increase in intracellular calcium caused by an increase in inositol triphosphate (IP3). IP3 is a second messenger that causes an increase in systemic vascular resistance by stimulating the influx of Ca2+ into smooth muscle cells. Reflex bradycardia can occur as a result of the subsequent increase in blood pressure. Phenylephrine and metaraminol are examples of pure alpha-1 agonists.
Levosimendin is a novel inotrope that makes myocytes more sensitive to intracellular Ca2+. It causes a positive inotropy without changing heart rate or oxygen consumption significantly.
The Na-K-ATPase membrane pump in the myocardium is inhibited by digoxin. This inhibition promotes sodium-calcium exchange, resulting in an increase in intracellular Ca2+ and increased contraction force. The parasympathetic effects of digoxin on the AV node result in bradycardia. Systemic vascular resistance will not be affected by it.
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This question is part of the following fields:
- Pathophysiology
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Question 23
Incorrect
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Regarding the Valsalva manoeuvre, which of the following describes the cardiovascular changes in phase III in a normal patient?
Your Answer:
Correct Answer: Normal intrathoracic pressure, decrease in blood pressure, and increase in heart rate
Explanation:When a person forcefully expires against a closed glottis, changes occur in intrathoracic pressure that dramatically affect venous return, cardiac output, arterial pressure, and heart rate. This forced expiratory effort is called a Valsalva maneuver.
Initially during a Valsalva, intrathoracic (intrapleural) pressure becomes very positive due to compression of the thoracic organs by the contracting rib cage. This increased external pressure on the heart and thoracic blood vessels compresses the vessels and cardiac chambers by decreasing the transmural pressure across their walls. Venous compression, and the accompanying large increase in right atrial pressure, impedes venous return into the thorax. This reduced venous return, and along with compression of the cardiac chambers, reduces cardiac filling and preload despite a large increase in intrachamber pressures. Reduced filling and preload leads to a fall in cardiac output by the Frank-Starling mechanism. At the same time, compression of the thoracic aorta transiently increases aortic pressure (phase I); however, aortic pressure begins to fall (phase II) after a few seconds because cardiac output falls. Changes in heart rate are reciprocal to the changes in aortic pressure due to the operation of the baroreceptor reflex. During phase I, heart rate decreases because aortic pressure is elevated; during phase II, heart rate increases as the aortic pressure falls.
When the person starts to breathe normally again, the intrathoracic pressure declines to normal levels, the aortic pressure briefly decreases as the external compression on the aorta is removed, and heart rate briefly increases reflexively (phase III). This is followed by an increase in aortic pressure (and reflex decrease in heart rate) as the cardiac output suddenly increases in response to a rapid increase in cardiac filling (phase IV). Aortic pressure also rises above normal because of a baroreceptor, sympathetic-mediated increase in systemic vascular resistance that occurred during the Valsava.
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This question is part of the following fields:
- Pathophysiology
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Question 24
Incorrect
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Which of the following facts about IgE is true?
Your Answer:
Correct Answer: Is increased in the serum of atopic individuals
Explanation:Immunoglobulin E (IgE) are an antibody subtype produced by the immune system. They are the least abundant type and function in parasitic infections and allergy responses.
The most predominant type of immunoglobulin is IgG. It is able to be transmitted across the placenta to provide immunity to the foetus.
IgE is involved in the type I hypersensitivity reaction as it stimulates mast cells to release histamine. It has no role in type 2 hypersensitivity.
Its concentration in the serum is normally the least abundant, however certain reactions cause a rise in its concentration, such as atopy, but not in acute asthma.
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This question is part of the following fields:
- Pathophysiology
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Question 25
Incorrect
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An individual who recently moved his residence from the plains to the mountains will experience immediately which physiologic response to high altitude and hypoxia?
Your Answer:
Correct Answer: Increased cardiac output
Explanation:A person remaining at high altitudes for days, weeks, or years becomes more and more acclimatized to the low PO2, so it causes fewer deleterious effects on the body.
After acclimatization, it becomes possible for the person to work harder without hypoxic effects or to ascend to still higher altitudes. The principal means by which acclimatization comes about are (1) a great increase in pulmonary ventilation, (2) increased numbers of red blood cells, (3) diffusing capacity of the lungs, (4) increased vascularity of the peripheral tissues, and (5) increased ability of the tissue cells to use oxygen despite low PO2.
The cardiac output often increases as much as 30% immediately after a person ascends to high altitude but then decreases back toward normal over a period of weeks as the blood haematocrit increases, so the amount of oxygen transported to the peripheral body tissues remains about normal.
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This question is part of the following fields:
- Pathophysiology
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Question 26
Incorrect
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Regarding oxygen consumption, which of these organs has the highest consumption at rest?
Your Answer:
Correct Answer: Kidney
Explanation:Oxygen delivery is related to blood flow as most of the oxygen binds to haemoglobin in red blood cells, although a small amount is dissolved in the plasma. Blood flow per 100 g of tissue is greatest in the kidneys.
The following is the oxygen consumption rate of different organs in ml/minute/100g
Hepatoportal = 2.2
Kidney = 6.8
Brain = 3.7
Skin = 0.38
Skeletal muscle = 0.18
Heart = 11 -
This question is part of the following fields:
- Pathophysiology
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Question 27
Incorrect
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Which of the following statements is true about monoamine oxidase (MOA) enzymes?
Your Answer:
Correct Answer: Type A and type B are found in the liver and brain
Explanation:Monoamine oxidase (MOA) enzymes are responsible for the catalyses of monoamine oxidative deamination. It assists the degradation of serotonin, norepinephrine (NE) and dopamine.
They are found in the mitochondria of most central and peripheral nerve tissues.
There are 2 different types:
Type A: Whose main function it to inactivate dopamine, tyramine, norepinephrine and 5-hydroxytryptamine. In addition to the nervous system, it is also found in the liver, brain gastrointestinal tract, pulmonary endothelium and placenta
Type B: Whose main function is to inactivate dopamine, tyramine, tryptamine and phenylethylamine. In addition to the nervous system, it is also found in the liver, brain (especially in the basal ganglia) and blood platelets. -
This question is part of the following fields:
- Pathophysiology
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Question 28
Incorrect
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The following statements are about chronic obstructive pulmonary disease (COPD). Which is true?
Your Answer:
Correct Answer: Inhaled corticosteroid usage slows the decline in health status
Explanation:Chronic obstructive pulmonary disease (COPD) is an obstructive, inflammatory lung condition. It encompasses symptoms of emphysema, chronic bronchitis and asthma.
Inhaling high dose steroids are prescribed to treat COPD. They are effective at reducing symptoms and improving lung function. They also work to reduce the number of hospitalisations by decreasing the number of acute exacerbation events. Despite providing effective symptom relief, it cannot slow down the decline of FEV1 as COPD is an irreversible condition.
COPD reduces the FEV1 measurements, as well as the FEV1/FVC ratio.
Breathlessness is a major COPD symptom and can occur at any point in the disease progression, including at an FEV1 >50%.
FEV1 is used in COPD staging, and it is classed as follows:
>80%: Mild or stage I
50 – 79%: Moderate or stage II
30 – 49%: Severe or stage III
<30%: Very severe or stage IV
Patients with mild COPD are usually able to manage their condition on their own, however once the disease progresses to moderate, more GP visits are required, with those in the severe category requiring frequent hospitalisation.Asthma is correlated to an increase in transfer factor. COPD (emphysema) is correlated to a decreased transfer factor.
COPD predisposes to eventual pulmonary hypertension as a result of an increase in pulmonary vascular resistance.
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This question is part of the following fields:
- Pathophysiology
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Question 29
Incorrect
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In North America, there have been reports of paediatric patients dying after undergoing adenotonsillectomy for obstructive sleep apnoea.
Respiratory depression/obstruction is thought to be the cause of death. The codeine dose was 0.5-1 mg/kg, given every 4-6 hours.
In this group of patients, which of the following is the most likely cause of respiratory depression and obstruction?Your Answer:
Correct Answer: Exaggerated metabolism of codeine
Explanation:Codeine is easily absorbed from the gastrointestinal tract and converted to morphine and norcodeine in the liver via O- and N-demethylation. Morphine and norcodeine are excreted almost entirely by the kidney, primarily as conjugates with glucuronic acid.
By glucuronidation, phase II metabolism enzyme UDP-glucuronosyl transferase-2B7 converts morphine to morphine-3-glucuronide (M3G) and morphine-6-glucuronide (M6G) (UGT2B7).
Approximately 60% of morphine is converted to M3G, with the remaining 6-10% converted to M6G. M3G is inactive, but M6G is said to be 4 to 650 times more potent on the MOP receptor than morphine.
When codeine is consumed, cytochrome P450 2D6 in the liver converts it to morphine (CYP2D6).
Some people have DNA variations that increase the activity of this enzyme, causing codeine to be converted to morphine more quickly and completely than in others. After taking codeine, these ultra-rapid metabolisers are more likely to have higher than normal levels of morphine in their blood.
Respiratory depression/obstruction can be caused by high levels of morphine and M6G, especially in people who have a history of obstructive sleep apnoea. The estimated number of ultra-rapid metabolisers ranges from 1 to 7 per 100 people, but some ethnic groups may have as many as 28 per 100 people.
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This question is part of the following fields:
- Pathophysiology
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Question 30
Incorrect
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With a 10-day history of severe vomiting, a 71-year-old man with a gastric outlet obstruction is admitted to the surgical ward.
The serum biochemical results listed below are available:
Sodium 128 mmol/L (137-144)
Potassium 2.6 mmol/L (3.5-4.9)
Chloride 50 mmol/L (95-107)
Urea 12 mmol/L (2.5-7.5)
Creatinine 180 µmol/L (60-110)
Which of the following do you think you are most likely to encounter?Your Answer:
Correct Answer: The standard base excess will be higher than actual base excess
Explanation:Hydrochloric acid is lost when you vomit for a long time (HCl). As a result, the following can be expected, in varying degrees of severity:
Hypokalaemia
Hypochloraemia
Increased bicarbonate to compensate for chloride loss and metabolic alkalosisThe alkalosis causes potassium to move from the intracellular to the extracellular compartment at first. Long-term vomiting and dehydration cause potassium to be excreted by the kidneys in order to conserve sodium. Dehydration can cause urea and creatinine levels to rise.
The actual base excess is always greater than the standard base excess.
The actual base excess (BE) is a measurement of a base’s contribution to a blood gas picture’s metabolic component. It’s the amount of base that needs to be added to a blood sample to bring the pH back to 7.4 after the respiratory component of a blood gas picture has been corrected (PaCO2 of 40 mmHg or 5.3 kPa). The BE has a normal range of +2 to 2. A large positive BE indicates a severe metabolic alkalosis, while a large negative BE indicates a severe metabolic acidosis. As a result, the actual BE in vitro is unaffected by CO2.
In vivo, however, standard BE is not independent of pCO2 because blood with haemoglobin acts as a better buffer than total ECF.
As a result, it is impossible to tell the difference between compensating for a respiratory disorder and compensating for the presence of a primary metabolic disorder.
The differences between in vitro and in vivo behaviour can be mostly eliminated if the BE is calculated for a haemoglobin concentration of 50 g/L (the ‘effective’ or virtual value of Hb if it was distributed throughout the extracellular space) rather than the actual haemoglobin. Because haemoglobin has a lower buffering capacity, the standard BE is higher than the actual BE. It reflects the BE better in the extracellular space rather than just the intravascular compartment.
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This question is part of the following fields:
- Pathophysiology
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