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Question 1
Correct
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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: 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 2
Incorrect
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The statement that best describes temperature management is:
Your Answer: Thermocouples use the principle that resistance in metals show a linear increase with temperature
Correct Answer: Gauge thermometers use coils of different metals with different co-efficients of expansion which either tighten or relax with changes in temperature
Explanation:There are different types of temperature measurement. These include:
Thermistor – this is a type of semiconductor, meaning they have greater resistance than conducting materials, but lower resistance than insulating materials. There are small beads of semiconductor material (e.g. metal oxide) which are incorporated into a Wheatstone bridge circuit. As the temperature increases, the resistance of the bead decreases exponentially
Thermocouple – Two different metals make up a thermocouple. Generally, in the form of two wires twisted, welded, or crimped together. Temperature is sensed by measuring the voltage. A potential difference is created that is proportional to the temperature at the junction (Seebeck effect)
Platinum resistance thermometers (PTR) – uses platinum for determining the temperature. The principle used is that the resistance of platinum changes with the change of temperature. The thermometer measures the temperature over the range of 200°C to1200°C. Resistance in metals show a linear increase with temperature
Tympanic thermometers – uses infrared radiation which is emitted by all living beings. It analyses the intensity and wavelength and then transduces the heat energy into a measurable electrical output
Gauge/dial thermometers – Uses coils of different metals with different co-efficient of expansion. These either tighten or relax with changes in temperature, moving a lever on a calibrated dial.
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This question is part of the following fields:
- Clinical Measurement
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Question 3
Correct
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The child-Pugh scoring system can be used, if risk classifying a patient with chronic liver disorder earlier to anaesthesia. Which one is the best combination of clinical signs and examinations used within the Child-Pugh scoring system?
Your Answer: Ascites, grade of encephalopathy, albumin, bilirubin and INR
Explanation:In the Child-Pugh classification system, the following 5 components are determined or calculated in order:
Ascites
Grade of encephalopathy
Serum bilirubin (μmol/L)
Serum Albumin (g/L)
Prothrombin time or INR
Raised liver enzymes are not the component of the classification system.
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This question is part of the following fields:
- Basic Physics
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Question 4
Incorrect
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An elderly man complains of a vague lump near his stomach to his physician. On examination, the lump is visible on coughing and is found within Hesselbach's triangle. Which of the following is true regarding the borders for this triangle?
Your Answer: Inguinal ligament inferiorly, ASIS laterally, lateral border of rectus sheath medially
Correct Answer: Inguinal ligament inferiorly, inferior epigastric vessels laterally, lateral border of rectus sheath medially
Explanation:The inguinal triangle of Hesselbach is an important clinical landmark on the posterior wall of the inguinal canal. It has the following relations:
Inferiorly – medial third of the inguinal ligament
Medially – lower lateral border of the rectus abdominis
Laterally – inferior epigastric vesselsDirect inguinal hernia is when the bowel bulges directly through the abdominal wall. These hernias usually protrude through Hesselbach’s triangle.
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This question is part of the following fields:
- Anatomy
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Question 5
Correct
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Useful diagnostic information can be obtained from measuring the osmolality of biological fluids. Of the following physical principles, which is the most accurate and reliable method of measuring osmolality?
Your Answer: Depression of freezing point
Explanation:Colligative properties are properties of solutions that depend on the number of dissolved particles in solution. They do not depend on the identities of the solutes.
All of the above have colligative properties with the exception of depression of melting point.
The osmolality from the concentration of a substance in a solution is measured by an osmometer. The freezing point of a solution can determines concentration of a solution and this can be measured by using a freezing point osmometer. This is applicable as depression of freezing point is directly correlated to concentration.
Vapour pressure osmometers, which measure vapour pressure, may miss certain volatiles such as CO2, ammonia and alcohol that are in the solution
The use of a freezing point osmometer provides the most accurate and reliable results for the majority of applications.
Colligative properties does not include melting point depression . Mixtures of substances in which the liquid phase components are insoluble, display a melting point depression and a melting range or interval instead of a fixed melting point.
The magnitude of the melting point depression depends on the mixture composition.
The melting point depression is used to determine the purity and identity of compounds. EMLA (eutectic mixture of local anaesthetics) cream is a mixture of lidocaine and prilocaine and is used as a topical local anaesthetic. The melting point of the combined drugs is lower than that individually and is below room temperature (18°C).
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This question is part of the following fields:
- Physiology
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Question 6
Correct
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Following an acute appendicectomy, a 6-year-old child is admitted to the recovery unit. Your consultant has requested that you prescribe maintenance fluids for the next 12 hours. The child is 21 kg in weight. What is the most suitable fluid volume to be prescribed?
Your Answer: 732 ml
Explanation:After a paediatric case, you’ll frequently have to calculate and prescribe maintenance fluids. The ‘4-2-1 rule’ should be used as a guideline:
1st 10 kg – 4 ml/kg/hr
2nd 10 kg – 2 ml/kg/hr
Subsequent kg – 1 ml/kg/hrHence
1st 10 kg = 4 × 10 = 40 ml
2nd 10 kg = 2 × 10 = 20 ml
Subsequent kg = 1 × 1 = 1 ml
Total = 61 ml/hr61 × 12 = 732 ml over 12 hrs.
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This question is part of the following fields:
- Physiology
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Question 7
Incorrect
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The main site of storage of thyroid hormones in the thyroid gland is?
Your Answer: Bound to colloid
Correct Answer: Thyroglobulin
Explanation:The follicle is the functional unit of the thyroid gland. The follicular cells surround the follicle which is filled with colloid. Suspended within the colloid is the is a pro-hormone complex thyroglobulin.
The synthesis and storage of thyroid hormones is done by follicular cells and the thyroglobulin within the colloid.
Iodide ions (Iˆ’) are actively transported against a concentration gradient into the follicular cell under the influence of thyroid stimulating hormone (TSH). It then undergoes oxidation to active iodine catalysed by thyroid peroxidase (TPO). The synthesis of thyroglobulin is in the follicular cells and it contains up to 140 tyrosine residues. The tyrosine residues of thyroglobulin and active iodine are merged to form mono- and di-iodotyrosines (MIT and DIT). The iodinated thyroglobulin is then taken up into the colloid where it is stored and dimerised. Two DIT molecules are joined to produce thyroxine (T4) while one MIT and one DIT molecule are joined to produce tri-iodotyrosine (T3) by a process catalysed by TPO.
Thyroglobulin droplets are taken up as vesicles into follicular cells by pinocytosis. This process is stimulated by TSH. When these vesicles fuse with lysosomes, hydrolysis of the thyroglobulin molecules and subsequent release of T4 and T3 into the circulation occurs.
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This question is part of the following fields:
- Pathophysiology
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Question 8
Incorrect
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You have always been curious about the effects of statins. While going through a study, something ticks you off and makes you think that they are way more common then the data suggests and are mostly under reported. In search of some concrete evidence, you decide to conduct a study of your own. While doing research, you come across a recent study that highlights the long term effects of statins. Which of the following types of study could that have been?
Your Answer: Case-control study
Correct Answer: Clinical trial, Phase 4
Explanation:In general practice, majority of phase 3 trials and some of the trials conducted in phase 2 are randomized. Because phase 4 trials require a huge sample size, they are not randomized as much. The primal reason behind conducting phase 3 trials is to test the efficiency and safety in a significant sample population. At this point it is assumed that the drug is effective up to a certain extent.
During a case-control study, subjects that exhibit outcomes of interest are compared with those who don’t show the expected outcome. The extent of exposure to a particular risk factor is then matched between cases and controls. If the exposure among cases surpasses controls, it becomes a risk factor for the outcome that is being studied.
Pilot studies are conducted on a lower and much smaller level, to assess if a randomized controlled trial of the crucial components of a study will be plausible.
Phase 4 trials are the ones that are conducted after its established that the drug is effective and is approved by the regulating authority for use. These trials are concerned with the side effects and potential risks associated with the long term usage of the drug.
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This question is part of the following fields:
- Statistical Methods
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Question 9
Incorrect
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A randomized study aimed at finding out the efficacy of a novel anticoagulant, in preventing stroke in patients suffering from atrial fibrillation, relative to those already available in the market was performed. A 59 year old woman volunteered for it and was randomised to the treatment arm. A year later, following findings were reported: 165 out of 1050 patients who were prescribed the already prevalent medicine had a stroke while the number of patients who had a single stroke after using the new drug was 132 out of 1044. In order to avoid one stroke case, what is the number of patients that need to be treated?
Your Answer: 31
Correct Answer: 32
Explanation:Number needed to treat can be defined as the number of patients who need to be treated to prevent one additional bad outcome.
It can be found as:
NNT=1/Absolute Risk Reduction (rounded to the next integer since number of patients can be integer only).
where ARR= (Risk factor associated with the new drug group) — (Risk factor associated with the currently available drug)
So,
ARR= (165/1050)-(132/1044)
ARR= (0.157-0.126)
ARR= 0.031
NNT= 1/0.031
NNT=32.3
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This question is part of the following fields:
- Statistical Methods
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Question 10
Incorrect
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Pressure volume loop represents the compliance of left ventricle. Considering there is no change in preload and myocardial contractility, which physiological change may result an increase in left ventricular afterload?
Your Answer: Increased stroke volume
Correct Answer: Increased end-systolic volume
Explanation:If there is no change in preload and myocardial contractility, there will be decrease in end-diastolic volume and stroke volume. So there must be increase in end-systolic volume.
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This question is part of the following fields:
- Physiology
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Question 11
Correct
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During a critical liver resection surgery, a 65-year-old man suffers uncontrolled hepatic bleeding. The 'Pringle manoeuvre is performed to stop the bleeding where the hepatic artery, portal vein, and common bile duct are clamped. These structures form the anterior boundary of the epiploic foramen. Which of the following vessels also contributes to the boundary of this region?
Your Answer: Inferior vena cava
Explanation:The epiploic foramen (foramen of Winslow or aditus to the lesser sac) is found behind the free right border of the lesser omentum. A short, 3 cm slit serves as the entrance to the lesser sac from the greater sac.
The epiploic foramen has the following boundaries:
Anteriorly: hepatoduodenal ligament, the bile duct (anteriorly on the right), the hepatic artery (anteriorly on the left), and the portal vein (posteriorly) together with nerves and lymphatics
Superiorly: the peritoneum of the posterior layer of the hepatoduodenal ligament runs over the caudate process of the liver
Posteriorly: inferior vena cava
Floor: upper border of the first part of the duodenum
The anterior and posterior walls of the foramen are normally
apposed, which partly explains why patients can develop large fluid
collections isolated to the greater or lesser sacRapid control of the hepatic artery and portal vein can be obtained by compression of the free edge of the lesser omentum (a €˜Pringle’ manoeuvre), which is a potentially useful technique in liver trauma and surgery.
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This question is part of the following fields:
- Anatomy
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Question 12
Incorrect
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The tissue layer in a patient is infiltrated with local anaesthetic (marcaine 0.125%) with 1 in 120,000 adrenaline as part of an enhanced recovery programme for primary hip replacement surgery. The total volume of solution is 120mL. What is the appropriate combination of constituents in the final solution?
Your Answer: 40mL 0.5% bupivacaine, 2mL 1 in 1,000 adrenaline and 78mL 0.9% N. Saline
Correct Answer: 30mL 0.5% bupivacaine, 1mL 1 in 1,000 adrenaline and 89mL 0.9% N. Saline
Explanation:30mL 0.5% bupivacaine, 1mL 1 in 1,000 adrenaline and 89mL 0.9% N. Saline is the correct answer.
Initial concentration of bupivacaine is 0.5% with a volume of 30mLThe volume is doubled (60mL) by the addition of 0.9% N. saline (30mls) and the concentration of bupivacaine is halved to (0.25%).
If the volume is doubled again (120mL) by the addition of further 0.9% N. saline (59mls) the final concentration of bupivacaine is halved again to 0.125%. Total N. saline = 89mls
The 1 mL of 1 in 1000 adrenaline has also been diluted into the final volume of 120 mL making it a 1 in 120000 concentration.
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This question is part of the following fields:
- Pharmacology
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Question 13
Correct
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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: 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 14
Correct
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A new proton pump inhibitor (PPI) is being evaluated in elderly patients who are taking aspiring. Study designed has 120 patients receiving the PPI, while a control group of 240 individuals is given the standard PPI. Over a span of 6 years, 24 of the group receiving the new PPI had an upper GI bleed compared to 60 individuals who received the standard PPI. How would you calculate the absolute risk reduction?
Your Answer: 5%
Explanation:Absolute risk reduction = (Control event rate) – (Experimental event rate)
Experimental event rate = 24 / 120 = 0.2
Control event rate = 60 / 240 = 0.25
Absolute risk reduction = 0.25 – 0.2 = 0.05 = 5% reduction
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This question is part of the following fields:
- Statistical Methods
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Question 15
Correct
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An inguinal hernia repair under general anaesthesia is scheduled for a fit 36-year-old man (75 kg). For perioperative and postoperative analgesia, you decide to perform an inguinal field block. Which of the following local anaesthetic solutions is the most appropriate?
Your Answer: 30 mL bupivacaine 0.5%
Explanation:Perioperative and postoperative analgesia can both be provided by an inguinal hernia field block. The Iliohypogastric and ilioinguinal nerves, as well as the skin, superficial fascia, and deeper structures, must be blocked for maximum effectiveness. The local anaesthetic should ideally have a long duration of action, be highly concentrated, and have a volume of at least 30 mL.
Plain bupivacaine has a maximum safe dose of 2 mg/kg body weight.
Because the patient weighs 75 kg, 150 mg bupivacaine can be safely administered. Both 30 mL 0.5 percent bupivacaine (150 mg) and 60 mL 0.25 percent bupivacaine (150 mg) are acceptable doses, but 30 mL 0.5 percent bupivacaine represents the optimal volume and strength, potentially providing a denser and longer block.
The maximum safe dose of plain lidocaine has been estimated to be between 3.5 and 5 mg/kg. The patient weighs 75 kg and can receive a maximum of 375 mg using the higher dosage regimen:
There are 200 mg of lidocaine in 10 mL of 2% lidocaine (and therefore 11 mL contains 220 mg)
200 mg of lidocaine is contained in 20 mL of 1% lidocaine.While alternatives are available, Although the doses of 11 mL lidocaine 2% and 20 mL lidocaine 1% are well within the dose limit, the volumes used are insufficient for effective field block for this surgery.
With 1 in 200,000 epinephrine, the maximum safe dose of lidocaine is 7 mg/kg. The patient can be given 525 mg in this case. Even with epinephrine, 60 mL of 1% lidocaine is 600 mg, which could be considered an overdose.
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This question is part of the following fields:
- Pharmacology
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Question 16
Correct
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Which among the following is summed up by F statistic?
Your Answer: ANOVA
Explanation:ANOVA is based upon within group variance (i.e. the variance of the mean of a sample) and between group variance (i.e. the variance between means of different samples). The test works by finding out the ratio of the two variances mentioned above. (Commonly known as F statistic).
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This question is part of the following fields:
- Statistical Methods
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Question 17
Correct
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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: 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 18
Correct
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A pre-operative evaluation for a trans-sphenoidal pituitary adenectomy is being performed on a 57-year-old woman. Her vision is causing her problems. A macroadenoma compressing the optic chiasm is visible on MRI. What is the most likely visual field defect to be discovered during an examination?
Your Answer: Bitemporal hemianopia
Explanation:The pituitary gland plays a crucial role in the neuro-endocrine axis. It is located at the base of the skull in the sella turcica of the sphenoid bone. It is connected superiorly to the hypothalamus, third ventricle, and visual pathways, and laterally to the cavernous sinuses, internal carotid arteries, and cranial nerves III, IV, V, and VI.
Pituitary tumours make up about 10-15% of all intracranial tumours. The majority of adenomas are benign. Over-secretion of pituitary hormones (most commonly prolactin, growth hormone, or ACTH), under-secretion of hormones, or localised or generalised pressure effects can all cause symptoms.
Compression of the optic chiasm can result in visual field defects, the most common of which is bitemporal hemianopia. This is caused by compression of the nasal retinal fibres, which carry visual impulses from temporal vision across the optic chiasm to the contralateral sides before continuing to the optic tracts.
The interruption of the visual pathways distal to the optic chiasm causes a homonymous visual field defect. The loss of the right or left halves of each eye’s visual field is referred to as homonymous hemianopia. It’s usually caused by a middle or posterior cerebral artery territory stroke that affects the occipital lobe’s optic radiation or visual cortex.
Binasal hemianopia is a condition in which vision is lost in the inner half of both eyes (nasal or medial). It’s caused by compression of the temporal visual pathways, which don’t cross at the optic chiasm and instead continue to the ipsilateral optic tracts. Binasal hemianopia is a rare complication caused by the internal carotid artery impinging on the temporal (lateral) visual fibres.
A monocular visual loss (that is, loss of vision in only one eye) can be caused by a variety of factors, but if caused by nerve damage, the damage would be proximal to the optic chiasm on the ipsilateral side.
A central scotoma is another name for central visual field loss. Every normal mammalian eye has a scotoma, also known as a blind spot, in its field of vision. The optic disc is a region of the retina that lacks photoreceptor cells and is where the retinal ganglion cell axons that make up the optic nerve exit the retina. When both eyes are open, visual signals that are absent in one eye’s blind spot are provided for the other eye by the opposite visual cortex, even if the other eye is closed.
Scotomata can be caused by a variety of factors, including demyelinating disease such as multiple sclerosis, damage to nerve fibre layer in the retina, methyl alcohol, ethambutol, quinine, nutritional deficiencies, and vascular blockages either in the retina or in the optic nerve.
Bilateral scotoma can occur when a pituitary tumour compresses the optic chiasm, causing a bitemporal paracentral scotoma, which then spreads out to the periphery, causing bitemporal hemianopsia. A central scotoma in a pregnant woman could be a sign of severe pre-eclampsia.
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This question is part of the following fields:
- Pathophysiology
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Question 19
Correct
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The thyroid gland:
Your Answer: Internalises iodine through active transport
Explanation:The thyroid gland is a gland shaped like a butterfly which lies at the base of the anterior neck. It controls metabolism using hormone secretion.
Iodine is extremely important for the synthesis of hormones within the thyroid. It is internalised into the thyroid follicular cells via the sodium/iodide symporter (NIS).
The parathyroid glands are found posterior to the thyroid gland, with the recurrent laryngeal nerves running posteromedially.
The expected weight of a normal thyroid gland is about 30 grams.
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This question is part of the following fields:
- Pathophysiology
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Question 20
Correct
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A 40-year-old woman was diagnosed with hyperparathyroidism. She is undergoing a parathyroidectomy, and during the surgery, the inferior parathyroid gland is found to be enlarged. There is a vessel adjacent to this gland on its lateral side. What is this vessel most likely to be?
Your Answer: Common carotid artery
Explanation:There are four parathyroid glands that lie on the medial half of the posterior surface of each lobe of the thyroid gland, inside its sheath. There are two superior and two inferior parathyroid glands.
The common carotid artery is a lateral relation of the inferior parathyroid.
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This question is part of the following fields:
- Anatomy
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Question 21
Incorrect
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A 61-year-old woman with myasthenia gravis is admitted to the ER with type II respiratory failure. There is a suspicion of myasthenic crisis. She is in a semiconscious state. Her blood pressure is 160/90 mmHg, pulse is 110 beats per minute, temperature is 37°C, and oxygen saturation is 84 percent. With a PaCO2 of 75 mmHg (10 kPa) breathing air, blood gas analysis confirms she is hypoventilating. Which of the following values is the most accurate representation of her alveolar oxygen tension (PAO2)?
Your Answer: 8.5
Correct Answer: 7.3
Explanation:The following is the alveolar gas equation:
PAO2 = PiO2 ˆ’ PaCO2/R
Where:
PAO2 is the partial pressure of oxygen in the alveoli.
PiO2 is the partial pressure of oxygen inhaled.
PaCO2 stands for partial pressure of carbon dioxide in the arteries.
The amount of carbon dioxide produced (200 mL/minute) divided by the amount of oxygen consumed (250 mL/minute) equals R = respiratory quotient. With a normal diet, the value is 0.8.By subtracting the partial pressure exerted by water vapour at body temperature, the PiO2 can be calculated:
PiO2 = 0.21 × (100 kPa ˆ’ 6.3 kPa)
PiO2 = 19.8Substituting:
PAO2 = 19.8 ˆ’ 10/0.8
PAO2 = 19.8 ˆ’ 12.5
PAO2 = 7.3k Pa -
This question is part of the following fields:
- Physiology
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Question 22
Incorrect
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A young male is operated on for an open inguinal hernia repair. During the procedure, the cord is mobilized, and the deep inguinal ring is located. What structure forms the lateral wall of the deep inguinal ring?
Your Answer: Inferior epigastric artery
Correct Answer: Transversalis fascia
Explanation:The deep inguinal ring is the entrance of the inguinal canal. It is an opening in the transversalis fascia around 1 cm above the inguinal ligament. Therefore, the superolateral wall is made by the transervalis fascia.
The inferior epigastric vessels run medially to the deep inguinal ring forming its inferomedial border.
The inguinal canal extends obliquely from the deep inguinal ring to the superficial inguinal ring.
An indirect inguinal hernia arises through the deep inguinal ring lateral to the inferior epigastric vessels. -
This question is part of the following fields:
- Anatomy
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Question 23
Incorrect
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In the erect position, the partial pressure of oxygen in the alveoli (PAO2) is higher in the apical lung units than in the basal lung units. What is the most significant reason for this?
Your Answer: The apical units are better ventilated
Correct Answer: The V/Q ratio of apical units is greater than that of basal units
Explanation:In any alveolar unit, the V/Q ratio affects alveolar oxygen (PAO2) and carbon dioxide tension (PACO2).
The partial pressure of alveolar carbon dioxide (PACO2) is plotted against the partial pressure of alveolar oxygen in a Ventilation-Perfusion (V/Q) ratio graph (PAO2). Given a set of model assumptions, the curve represents all of the possible values for PACO2 and PAO2 that an individual alveolus could have.
In the case of an infinity V/Q ratio (ventilation but no perfusion or dead space), the PACO2 of the alveolus will equal zero, while the PAO2 will approach that of external air (150mmmHg). At the apex of the lung, the V/Q ratio is 3.3, compared to 0.67 at the base.
PACO2 and PAO2 approach the partial pressures for these gases in the venous blood when the V/Q ratio is zero (no ventilation but perfusion). At the base of the lung, the V/Q ratio is 0.67, whereas at the apex, it is 3.3.
PAO2 at the apex is typically 132mmHg, and PACO2 is typically 28mmHg.
The average PAO2 at the base is 89 mmHg, while the average PACO2 is 42 mmHg.
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This question is part of the following fields:
- Physiology
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Question 24
Incorrect
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A normal woman at term, not in labour, has her arterial blood gas analysed. Which set of results is most likely her own? Option - pH - PaCO2 - HCO3 - PaO2. A: 7.35 - 28 mmHg (3.73 kPa) - 27 mmol/L - 104 mmHg (13.8kPa), B: 7.43 - 32 mmHg (4.27 kPa) - 21 mmol/L - 104 mmHg (13.8kPa), C: 7.44 - 36 mmHg (4.8 kPa) - 27 mmol/L - 104 mmHg (13.8kPa), D: 7.45 - 40 mmHg (5.33 kPa) - 21 mmol/L - 104 mmHg (13.8kPa), E - 7.46 - 44 mmHg (5.87kPa) - 21 mmol/L - 104 mmHg (13.8kPa).
Your Answer: C
Correct Answer: B
Explanation:Due to an increased tidal volume with little change or slight increase in respiratory rate, Minute ventilation at term is increased by about 50%. Hypothalamic function are thought to influence by Progesterone, oestradiol and prostaglandins. This causes a mild compensated respiratory alkalosis.
Maternal PaCO2 is usually decreased to about 32 mmHg (4.27 kPa) as a result of this increased alveolar ventilation at term . A compensatory decrease in serum bicarbonate from 27 to 21 mmol/L by renal excretion lessens the impact of maternal alkalosis.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 25
Correct
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A 72-year old man is experiencing a cardiac risk evaluation for the management of obstructive umbilical hernia. Echocardiogram demonstrates an aortic valve area=0.59cm with a pressure of 70mmHg. Five years ago, he had mild myocardial infarction complicated with pulmonary oedema. Now he encounters angina with little exertion. Which of the following factor is the foremost profoundly weighted using Deysky's cardiac risk scoring system in this case?
Your Answer: Aortic stenosis
Explanation:Detsky’s Modified cardiac risk classification system in patients undergoing non-cardiac surgery:
Age more than 70: 05 points
History of myocardial infarction:
Less than 6 months: 10 points
More than 6 months: 5 pointsAngina Pectoris:
Angina with minimal exertion: 10 points
Angina at any level of exertion: 20 points
Pulmonary Oedema:
Within 7 days: 10 points
At any time: 5 pointsSuspected aortic valve stenosis with valve area 30 points = high risk.
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This question is part of the following fields:
- Pathophysiology
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Question 26
Incorrect
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A 68-year-old man with nausea and vomiting is admitted to the hospital. For temporal arteritis, he takes 40 mg prednisolone orally in divided doses. His prescription chart will need to be adjusted to reflect his inability to take oral medications. What is the equivalent dose of intravenous hydrocortisone to 40 mg oral prednisolone?
Your Answer: 80 mg
Correct Answer: 160 mg
Explanation:Prednisolone 5 mg is the same as 20 mg hydrocortisone.
Prednisolone 40 mg is the same as 8 x 20 mg or 160 mg of prednisolone.
Mineralocorticoid effects and variations in action duration are not taken into account in these comparisons.
5 mg of prednisolone is the same as Dexamethasone 750 mcg, Hydrocortisone 20 mg, Methylprednisolone 4 mg, and Cortisone acetate 25 mg.
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This question is part of the following fields:
- Pharmacology
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Question 27
Correct
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Which of the following statements is true regarding sucralfate?
Your Answer: Has very few side effects
Explanation:Sucralfate is an octasulfate of glucose to which Al(OH)3 has been added. It undergoes extensive cross-linking in an acidic environment and forms a polymer which adheres to the ulcer base for up to 6 hours and protects it from further erosion. Since it is not systemically absorbed it is virtually devoid of side effects. However, it may cause constipation in about 2% of cases due to the Aluminium component in it.
Sucralfate does not have antibacterial action against Helicobacter pylori. However, Bismuth has antibacterial action due to its oligodynamic effect.
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This question is part of the following fields:
- Pharmacology
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Question 28
Incorrect
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Which peripheral nerve of the foot is often utilized to evaluate for neuromuscular blockade?
Your Answer: Superficial peroneal nerve
Correct Answer: Posterior tibial nerve
Explanation:The posterior tibial nerve lies on the posterior surface of the tibialis posterior and, lower down the leg, on the posterior surface of the tibia. The nerve accompanies the posterior tibial artery and lies at first on its medial side, then crosses posterior to it, and finally lies on its lateral side. The nerve, with the artery, passes behind the medial malleolus, between the tendons of the flexor digitorum longus and the flexor hallucis longus.
It gives off muscular branches to the soleus, flexor digitorum longus, flexor hallucis longus, and tibialis posterior. A medial calcaneal branches off to supply the skin over the medial surface of the heel, and an articular nerve to supply the ankle joint. Finally, it terminates to become the medial and lateral plantar nerves.
The saphenous nerve is a branch of the femoral nerve that gives off branches that supply the skin on the posteromedial surface of the leg.
The sural nerve is a branch of the tibial nerve that supplies the skin on the lower part of the posterolateral surface of the leg.
The superficial peroneal nerve is one of the terminal branches of the common peroneal nerve. It arises in the substance of the peroneus longus muscle on the lateral side of the neck of the fibular. It ascends between the peroneus longus and brevis muscles, and in the lower part of the leg it becomes cutaneous. Muscular branches of the superficial peroneal nerve supply the peroneus longus and brevis muscles, while medial and lateral cutaneous branches are distributed to the skin on the lower part of the leg and dorsum of the foot. In addition, the cutaneous branches supply the dorsal surfaces of the skin of all the toes, except the adjacent sides of the first and second toes and the lateral side of the little toe.
The superficial peroneal, sural and saphenous nerves cannot be used to assess neuromuscular blocks since they are sensory nerves.
The deep peroneal nerve enters the dorsum of the foot by passing deep to the extensor retinacula on the lateral side of the dorsalis pedis artery. It divides into terminal, medial, and lateral branches. The medial branch supplies the skin of the adjacent sides of the big and second toes. The lateral branch supplies the extensor digitorum brevis muscle. Both terminal branches give articular branches to the joints of the foot. This nerve is too deep to use for neuromuscular blockade assessment.
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This question is part of the following fields:
- Anatomy
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Question 29
Incorrect
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A meta analysis has been conducted to see if addition of a new supplement the diet would prevent any further myocardial infraction among the patients who have recently had one. 4 trials (all randomised) were carried out. Which among the following is the most apt interpretation of the data?
Your Answer: There is a non-significant trend that taking the supplement reduces the chance of a further myocardial infarctions
Correct Answer: Taking the supplement increases the chance of a further myocardial infarction
Explanation:Meta analysis performed upon the results, presented by the diamond, is clear from the no effect line. It presents a substantial increase in the probability of another heart attack.
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This question is part of the following fields:
- Statistical Methods
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Question 30
Correct
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While on the ward, you notice a patient that is lying down supine and attached to a monitor is hypotensive with a blood pressure of 90/70mmHg and he is also tachycardic with a pulse of 120 beats/minute. After adjusting the bed with the patient's legs raised by 45 degrees, you reassess the blood pressure after 1 minute and his blood pressure has increased to 100/75mmHg. You then prescribe IV fluids and ask for 500ml of normal saline to be given intravenously over 15 minutes. The increase in the blood pressure can be explained by which physiological association?
Your Answer: Venous return is proportional to stroke volume
Explanation:Cardiac muscle contraction strength is dependent on the action of adrenaline and noradrenaline, but these hormones contribute to cardiac contractility, not to Starling’s law.
Stroke volume (via a cardiac monitor) and/or pulse pressure (via an arterial line) should be measured to assess the effects of a passive leg raise. An increase in stroke volume by 9% or in pulse pressure by 10% are considered indicative of fluid responsiveness.
A passive leg raise can lead to transient increases in blood pressure and stroke volume as it increased the amount of venous return to the heart. Venous return increases in this situation as it transfers a larger volume of blood from the lower limbs to the right heart. It therefore mimics a fluid challenge. However its effects are short lasting and often lead to minimal increases in blood pressure. It therefore should not be used to treat shock in isolation. The passive leg raise is useful in determining the likelihood that a patient with shock will respond to fluid resuscitation.
Sarcomeres, which can be in cardiac, smooth or skeletal muscle, function optimally when stretched to a specific point.
Blood that enters the ventricles during diastole causing stretching of sarcomeres within cardiac muscle. The extent to which they stretch is proportional to the strength of ventricular muscle contraction. Therefore, the venous return (amount of blood returned to the heart) is proportional to stroke volume. The end diastolic volume is determined by venous return and is also proportional to stroke volume. -
This question is part of the following fields:
- Physiology And Biochemistry
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