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Question 1
Incorrect
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During positive pressure ventilation using positive end-expiratory pressure (PEEP), there is usually an associated reduction in cardiac output
Which of the following is responsible?
Your Answer: Increased pulmonary vascular resistance
Correct Answer: Reduced venous return to the heart
Explanation:The option that is most responsible is the progressive decrease in venous return of blood to the right atrium. The heart rate does not usually change with PEEP so the fall in cardiac output is due to a reduction in left ventricular (LV) stroke volume (SV).
Note that the interventricular septum does shift toward the left and there is an increased pulmonary vascular resistance (PVR) from overdistention of alveolar air sacs that contribute to the reduction in cardiac output. Any increase in PVR will be associated with reduced pulmonary vascular capacitance.
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This question is part of the following fields:
- Pathophysiology
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Question 2
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: Direct stimulation of osteoclasts
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 3
Incorrect
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Iron is one of the most important micronutrients in the body.
Out of the following, which one has the most abundant storage of iron in the body?Your Answer: Ferritin and haemosiderin
Correct Answer: Haemoglobin
Explanation:Iron is a necessary micronutrient for proper erythropoietic function, oxidative metabolism, and cellular immune responses. Although dietary iron absorption (1-2 mg/d) is tightly controlled, it is only just balanced by losses.
The adult body contains 35-45 mg/kg iron (about 4-5 g)
Iron can be found in a variety of forms, including haemoglobin, ferritin, haemosiderin, myoglobin, haem enzymes, and transferrin bound proteins.
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This question is part of the following fields:
- Pathophysiology
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Question 4
Correct
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What is the mechanism of the pupillary reflex arc?
Your Answer: Oculomotor nerve fibres from the Edinger-Westphal nuclei
Explanation:Pupil size is reduced by the pupillary light reflex and during accommodation for near vision. In the pupillary light reflex, light that strikes the retina is processed by retinal circuits that excite W-type retinal ganglion cells. These cells respond to diffuse illumination. The axons of some of the W-type cells project through the optic nerve and tract to the pretectal area, where they synapse in the olivary pretectal nucleus. This nucleus contains neurons that also respond to diffuse illumination. Activity of neurons of the olivary pretectal nucleus causes pupillary constriction by means of bilateral connections with parasympathetic preganglionic neurons in the Edinger-Westphal nuclei. The reflex results in contraction of the pupillary sphincter muscles in both eyes, even when light is shone into only one eye.
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This question is part of the following fields:
- Pathophysiology
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Question 5
Correct
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Clearance techniques are used to assess renal glomerular function.
Which of the following is the most accurate marker for glomerular filtration rate measurement?Your Answer: Inulin
Explanation:The perfect glomerular filtration marker is:
The human body is not harmed by it.
Chemical or physical methods are used to accurately measure
Extracellular fluid (ECF) compartment is freely and evenly diffusible.
Inability to access the intracellular fluid (ICF) compartment
Filtration in the kidney is the only way to remove it from the blood.The ideal marker should not be reabsorbed into the bloodstream by the renal tubules or other urinary system components.
Creatinine is an endogenous substance that is filtered freely by the glomerulus and secreted by the proximal tubule. As a result, creatinine clearance consistently underestimates GFR. In healthy people, this overestimation ranges from 10% to 40%, but it is higher and more unpredictable in patients with chronic kidney disease.
The gold standard method of inulin clearance necessitates an intravenous infusion and several hours of timed urine collection, making it costly and time-consuming. Inulin is hard to come by and is difficult to mix and keep as a solution.
Exogenous filtration markers include the following:
Although plasma clearance of 51chromium EDTA is a widely used method in Europe, tubular reabsorption can occur.
Because 125I-iothalamate can be excreted by renal tubules in the urine, it cannot be used in patients who have an iodine assay.Radioactive substances must be stored, administered, and disposed of according to these methods.
The glomerulus filters para-aminohippuric acid (PAH) freely, and any that remains in the peritubular capillaries is secreted into the proximal convoluted tubules. This marker is used to determine the amount of blood flowing through the kidneys.
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This question is part of the following fields:
- Pathophysiology
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Question 6
Correct
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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: 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 7
Incorrect
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A 43-year-old patient was brought to the emergency department with a traumatic amputation of his leg at mid-thigh level. Resuscitation with 1 L gelofusine was done and four units of packed red blood cells were given before theatre. Thirty minutes following blood transfusion, the patient became flushed, breathless, hypotensive, develops haemoglobinuria, and had a fever of 38oC.
Which one of the following correctly explains the patient signs and symptoms?Your Answer: Non-immune haemolytic transfusion reaction
Correct Answer: Activation of classic complement pathway
Explanation:This may be the classical case of blood transfusion reaction due to ABO incompatibility.
Here red cells are destroyed in the bloodstream with the release of haemoglobin in circulation (causing haemoglobinuria). Here, IgM or IgG anti-A or anti-B antibody can cause rapid activation of complement cascade usually the classical pathway. This is called intravascular haemolysis.
There may be extravascular haemolysis by cells of the mononuclear phagocyte system situated in the liver and spleen. Extravascular red cell destruction can increase breakdown products of haemoglobin, such as bilirubin and urobilinogen.
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This question is part of the following fields:
- Pathophysiology
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Question 8
Incorrect
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The arterial-venous oxygen (a-vO2) difference is the difference between the oxygen content of arterial blood (CaO2) and mixed-venous blood (CvO2).
At rest, which organ has the greatest a-vO2 difference?Your Answer: Intestine
Correct Answer: Heart
Explanation:At rest, the heart has the greatest a-vO2 difference, a high capillary to myocyte ratio, short diffusion distances, and a high mitochondrial density. The flow of blood through the coronary arteries is also tightly controlled. At rest, 70-80 percent of the oxygen available to the cardiac muscle is extracted, increasing to 90 percent during exercise.
The a-vO2 difference indicates the body’s or an individual organ’s ability to extract oxygen from the blood.
CaO2 is influenced by a number of factors, including Hb concentration, PaO2 and pulmonary diffusion capacity.
CvO2 is influenced by a number of factors, including capillary density, regional blood flow, heart, resting skeletal muscle, kidney, intestine and skin.
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This question is part of the following fields:
- Pathophysiology
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Question 9
Correct
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A 68-year old female is brought to the Emergency Room for abdominal pain. Medical history revealed that she is on long-term warfarin therapy for deep vein thrombosis.
Upon further investigation, the patient is hypotensive at 80/60 mmHg, and an abdominal mass is palpable on the umbilical area. An initial diagnosis of ruptured abdominal aortic aneurysm (AAA) is made. Moreover, blood tests show an international normalised ratio (INR) of 4.2.
Which of the following products should be initially transfused or administered to the patient to reverse the anticoagulation?Your Answer: Prothrombin complex
Explanation:Warfarin prevents reductive metabolism of the inactive vitamin K epoxide back to its active hydroquinone form. Thus, warfarin inhibits the synthesis of vitamin K dependent clotting factors: X, IX, VII, II (prothrombin), and of the anticoagulants protein C and protein S. The therapeutic range for oral anticoagulant therapy is defined in terms of an international normalized ratio (INR). The INR is the prothrombin time ratio (patient prothrombin time/mean of normal prothrombin time for lab)ISI, where the ISI exponent refers to the International Sensitivity Index and is dependent on the specific reagents and instruments used for the determination. A prolonged INR is widely used as an indication of integrity of the coagulation system in liver disease and other disorders, it has been validated only in patients in steady state on chronic warfarin therapy.
Prothrombin complex concentrate (PCC) is used to replace congenital or acquired vitamin-K deficiency warfarin-induced anticoagulant effect, particularly in the emergent setting.
Intravenous vitamin K has a slower onset of action compared to PCC, but is useful for long term therapy.
Fresh frozen plasma (FFP) prepared from freshly donated blood is the usual source of the vitamin K-dependent factors and is the only source of factor V. The factors needed, however, are found in small quantities compared to PCC.
Cryoprecipitate is indicated for hypofibrinogenemia/dysfibrinogenemia, von Willebrand disease, haemophilia A, factor XIII deficiency, and management of bleeding related to thrombolytic therapy.
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This question is part of the following fields:
- Pathophysiology
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Question 10
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 11
Incorrect
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Concerning calcium metabolism and its control, which of these is correct?
Your Answer: The major stimulant to parathyroid hormone secretion is a fall in the plasma unionised calcium concentration
Correct Answer: Cholecalciferol is 25-hydroxylated in the liver
Explanation:When there is a fall in ionised plasma calcium levels, the chief cells of the parathyroid glands are stimulated to secrete parathyroid hormone (PTH).
50% of extracellular calcium occurs as non-ionised, protein- (albumin-)bound calcium.
The degree of ionisation increases with low ph and decreases with high pH.
There is increased renal calcium excretion with secretion of calcitonin.
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This question is part of the following fields:
- Pathophysiology
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Question 12
Incorrect
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In endurance athletes, which of the following physiological adaptations to exercise is the best predictor of performance?
Your Answer: Increased percentage of slow twitch muscles
Correct Answer: Velocity of blood lactate accumulation
Explanation:Multiple regression analysis revealed that velocity of lactate accumulation (VOBLA) accounted for 92 percent of the variation in marathon running velocity (VM), and VOBLA plus training volume prior to the marathon accounted for 96 percent of the variation. Percent ST muscle fibre distribution (r = 0.55-0.69) and capillary density (r = 052-0.63) were found to be positively correlated with all performance variables. As a result, marathon running performance was linked to VOBLA and the ability to run at a pace close to it during the race. The percent ST, capillary density, and training volume were all related to these properties.
Another metabolic adaptation compared to normal people is the early selection of fat for oxidation by muscle, especially when glucose availability is limited during high-intensity exercise. This helps to delay the onset of muscle fatigue, but it does not prevent VOBLA.
For a given level of exercise, training can also result in cardiovascular adaptation, such as increased heart size, increased contractility, and a slower heart rate. All of these factors contribute to an increase in maximal oxygen consumption (VO2 max), but genetic factors, despite intensive training, play a large role in an athlete’s performance.
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This question is part of the following fields:
- Pathophysiology
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Question 13
Correct
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Following a traumatic head injury, you are summoned to the emergency department to transfer a patient to the CT scanner. With a Glasgow coma score of 3, he has already been intubated and ventilated.
It is important to ensure that cerebral protection strategies are implemented during the transfer.
Which of the following methods is the most effective for reducing venous obstruction?Your Answer: Position with a head-up tilt of 30°
Explanation:ICP is significant because it influences cerebral perfusion pressure and cerebral blood flow. The normal ICP ranges from 5 to 13 mmHg.
The components within the skull include the brain (80%/1400 ml), blood (10%/150 ml), and cerebrospinal fluid (CSF) (10%/150 ml).
Because the skull is a rigid box, if one of the three components increases in volume, one or more of the remaining components must decrease in volume to compensate, or the ICP will rise (Monroe-Kellie hypothesis).
Primary brain injury occurs as a result of a head injury and is unavoidable unless precautions are taken to reduce the risk of head injury. A reduction in oxygen delivery due to hypoxemia (low arterial PaO2) or anaemia, a reduction in cerebral blood flow due to hypotension or reduced cardiac output, and factors that cause a raised ICP and reduced CPP are all causes of secondary brain injury. Secondary brain injury can be avoided with proper management.
The most important initial management task is to make certain that:
There is protection of the airway and the cervical spine
There is proper ventilation and oxygenation
Blood pressure and cerebral perfusion pressure are both adequate (CPP).Following the implementation of these management principles, additional strategies to reduce ICP and preserve cerebral perfusion are required. The volume of one or more of the contents of the skull can be reduced using techniques that can be used to reduce ICP.
Reduce the volume of brain tissue
Blood volume should be reduced.
CSF volume should be reduced.The following are some methods for reducing the volume of brain tissue:
Abscess removal or tumour resection
Steroids (especially dexamethasone) are used to treat oedema in the brain.
To reduce intracellular volume, use mannitol/furosemide or hypertonic saline.
To increase intracranial volume, a decompressive craniectomy is performed.The following are some methods for reducing blood volume:
Haematomas must be evacuated.
Barbiturate coma reduces cerebral metabolic rate and oxygen consumption, lowering cerebral blood volume as a result.
Hypoxemia, hypercarbia, hyperthermia, vasodilator drugs, and hypotension should all be avoided in the arterial system.
PEEP/airway obstruction/CVP lines in neck: patient positioning with 30° head up, avoid neck compression with ties/excessive rotation, avoid PEEP/airway obstruction/CVP lines in neckThe following are some methods for reducing CSF volume:
To reduce CSF volume, an external ventricular drain or a ventriculoperitoneal shunt is inserted (although more a long term measure).
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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: Abducens nerve
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
Correct
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A 66-year-old man with a haemorrhagic stroke is admitted to the medical admissions unit.
He has been taking warfarin for a long time because of atrial fibrillation. His INR at the time of admission was 9.1.
Which of the following treatment options is the most effective in managing his condition?Your Answer: Prothrombin complex concentrate
Explanation:Haemorrhage, including intracranial bleeding, is a common and potentially fatal side effect of warfarin therapy, and reversing anticoagulation quickly and completely can save lives. When complete and immediate correction of the coagulation defect is required in orally anticoagulated patients with life-threatening haemorrhage, clotting factor concentrates are the only viable option.
For rapid reversal of vitamin K anticoagulants, prothrombin complex concentrates (PCC) are recommended. They contain the vitamin K-dependent clotting factors II, VII, IX, and X and are derived from human plasma. They can be used as an adjunctive therapy in patients with major bleeding because they normalise vitamin K dependent clotting factors and restore haemostasis.
The most common treatments are fresh frozen plasma (FFP) and vitamin K. The efficacy of this approach is questioned due to the variable content of vitamin K-dependent clotting factors in FFP and the effects of dilution. Significant intravascular volume challenge, as well as the possibility of rare complications like transfusion-associated lung injury or blood-borne infection, are all potential issues.
To avoid anaphylactic reactions, vitamin K should be given as a slow intravenous infusion over 30 minutes. Regardless of the route of administration, the reversal of INRs with vitamin K can take up to 24 hours to reach its maximum effect.
Reversal of anticoagulation in patients with warfarin-associated intracranial haemorrhage may be considered with factor VIIa (recombinant), but its use is controversial. There are concerns about thromboembolic events following treatment, as well as questions about assessing efficacy in changes in the INR. If the drug is to be administered, patients should be screened for an increased risk of thrombosis before the drug is given.
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This question is part of the following fields:
- Pathophysiology
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Question 16
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: Bicarbonate 45-50 mmol/L
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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Question 17
Correct
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Which of the following statements is correct regarding hypomagnesaemia?
Your Answer: Causes tetany
Explanation:The ECG changes seen in hypomagnesaemia include:
Prolonged PR interval
Prolonged QT interval
Flattening of T waves
ST segment depression
Prominent U wavesThese changes are almost the same as those of hypokalaemia.
There is an increased risk of digoxin toxicity and a risk of atrial and ventricular ectopic and ventricular arrhythmias.
There is impaired synthesis and release of parathyroid hormone (PTH) in chronic hypomagnesaemia leading to impaired target organ response to PTH. This produces secondary hypocalcaemia.
The use of potassium ‘wasting’ diuretics (e.g. loop diuretics like furosemide) may lead to Hypomagnesaemia.
A tall T wave is seen in hypermagnesemia.
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This question is part of the following fields:
- Pathophysiology
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Question 18
Incorrect
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Which of the following ionic changes is associated with the ventricular myocyte action potential's initial repolarization phase?
Your Answer: Decreased Ca2+ and increased K+ conductance
Correct Answer: Ceased Na+ and increase K+ conductances
Explanation:The Purkinje system, as well as the action potentials of ventricular and atrial myocytes, have the same ionic changes. It lasts about 200 milliseconds and has a resting membrane potential, as well as fast depolarisation and plateau phases.
There are five stages to the process:
Increased Na+ and decreased K+ conductance in Phase 0 (rapid depolarisation).
1st phase (initial repolarisation) : Na+ conductance decreased, while K+ conductance increased.
Phase two (plateau phase) : Ca2+ conductance increased
Phase three (repolarisation phase) : Lower Ca2+ conductance and higher K+ conductance
4th Phase (resting membrane potential) : K+ conductance increased, Na+ conductance decreased, and Ca2+ conductance decreased. -
This question is part of the following fields:
- Pathophysiology
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Question 19
Correct
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A 57-year old woman, presents to her general practitioner. She has a 2 week history of a vaginal hysterectomy for which she was placed under general anaesthesia.
On examination, she has notable weakness of dorsiflexion of her left foot and a high stepping gait.
Which nerve was most likely injured during her surgery?Your Answer: Common peroneal nerve
Explanation:The common peroneal (fibular) nerve is a peripheral nerve in the lower limb. It arises of the L4-S2 nerve roots and has sensory and motor innervations:
Sensory: Provides innervation of the lateral leg and foot dorsum.
Motor: Provides innervation of the short head of the biceps femoris, as well as muscles of the anterior and lateral leg compartments.
It is the most commonly damaged nerve in the lower extremity, as it is easily compressed by a plaster cast or injured when the fibula is fractured.
Damage to the common peroneal nerve will result in loss of dorsiflexion at ankle (footdrop, as feet are permanently plantarflexed), with the accompanying high stepping gait.
The saphenous and sural nerve only provide sensory innervation.
The tibial nerve arises from the sciatic nerve (like the common peroneal), but it provides motor innervation to the posterior leg compartments and intrinsic foot muscles. Injury to the tibial nerve will cause loss of plantar flexion, toe flexion and weakened foot inversion.
Extreme hip flexion into the lithotomy or Lloyd-Davies position can result in stretch damage to the neurones (sciatic and obturator nerves) or by applying direct pressure (femoral nerve compression).
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This question is part of the following fields:
- Pathophysiology
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Question 20
Correct
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Which of the following statements is true about monoamine oxidase (MOA) enzymes?
Your 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 21
Correct
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A 40-year old farmer came into the emergency room with a chief complaint of 4 episodes of non-bloody diarrhoea. This was associated with frequent urination, vomiting and salivation. History also revealed frequent use of insecticides. Upon physical examination, there was miosis and bradycardia.
Given the different types of bonds, which is the most likely bond formed between insecticide poisoning and receptors?Your Answer: Covalent
Explanation:Organophosphate poisoning occurs most often due to accidental exposure to toxic amounts of pesticides. Signs and symptoms include diarrhoea, urination, miosis, bradycardia, emesis, lacrimation, lethargy and salivation.
Organophosphates are classified as indirect acting cholinomimetics, and their mode of action involves: (1) the inhibition of acetylcholinesterase (AChE) by forming a stable covalent bond on the active site serine; and, (2) amplification of endogenously release acetylcholine (ACh), hence the clinical manifestation.
There are 4 types of bonds or interactions: ionic, covalent, hydrogen bonds, and van der Waals interactions. Ionic and covalent bonds are strong interactions that require a larger energy input to break apart. When an element donates an electron from its outer shell, a positive ion is formed. The element accepting the electron is now negatively charged. Because positive and negative charges attract, these ions stay together and form an ionic bond. Covalent bonds form when an electron is shared between two elements and are the strongest and most common form of chemical bond in living organisms. Covalent bonds form between the elements that make up the biological molecules in our cells. Unlike ionic bonds, covalent bonds do not dissociate in water.
When polar covalent bonds containing a hydrogen atom form, the hydrogen atom in that bond has a slightly positive charge. This is because the shared electron is pulled more strongly toward the other element and away from the hydrogen nucleus. Because the hydrogen atom is slightly positive, it will be attracted to neighbouring negative partial charges. When this happens, a weak interaction occurs between the slightly positive charge of the hydrogen atom of one molecule and the slightly negative charge of the other molecule. This interaction is called a hydrogen bond.
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This question is part of the following fields:
- Pathophysiology
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Question 22
Incorrect
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One of the following sets of arterial blood gases best demonstrates compensated respiratory acidosis?
Your Answer: pH=7.24, PaCO2=10.1kPa, PaO2=7.0kPa, HCO3- =27mmol/L
Correct Answer: pH=7.36, PaCO2=8.5kPa, PaO2=7.5kPa, HCO3- = 43mmol/L
Explanation:pH=7.36, PaCO2=8.5kPa, PaO2=7.5kPa, HCO3- = 43mmol/L is the correct answer.
Since the pH is the lower limit of normal, it is compensated despite a raised PaCO2. Retention of bicarbonate ions by the renal system suggests this process is chronic.
pH=7.24, PaCO2=3.5kPa, PaO2=8.5kPa, HCO3- =18mmol/L represents an acute uncompensated metabolic acidosis
The remaining stems are degrees of uncompensated respiratory acidosis and therefore incorrect.
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This question is part of the following fields:
- Pathophysiology
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Question 23
Incorrect
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A 52-year-old patient is brought to ER with a chief complaint of chest pain for two hours. Chest pain was tightness in nature, located in the centre of the chest and radiate into the neck and left arm. The patient otherwise looks fit and well.
Just after admitting the patient, he suffered VF cardiac arrest and is immediately defibrillated with the return of spontaneous circulation (ROSC).
On clinical examination following was the finding:
BP: 82/45 mmHg
Heart rate: 120 beats/min
Oxygen saturation on air: 25%
Heart sounds: Normal
There is no sign of pulmonary oedema. The patient is anxious, cold, and clammy.
A 12 lead ECG was done which revealed a sinus rhythm of 120 with ST-segment depression and T wave inversion in leads II, III, and aVF. Which of the following is considered best for the initial treatment of the patient?Your Answer: Inhaled high flow oxygen
Correct Answer: Oral aspirin
Explanation:This is a classical case of unstable angina or NSTEMI (Non-ST-elevation myocardial infarction). As soon as the diagnosis of unstable angina or NSTEMI is made the initial treatment is Aspirin and antithrombin therapy.
Betablocker is known to reduce mortality from acute myocardial infarction by reducing oxygen demand. If there is no contraindication (heart block, bradycardia, hypotension, severe left ventricular dysfunction, and asthma), a beta-blocker should be given early. This patient has hypotension and therefore metoprolol is contraindicated.
If three doses of nitroglycerine tablets or Nitrolingual sprays and intravenous beta-blockers too cannot relieve the symptoms intravenous Glyceryl Trinitrate (GTN) should be considered provided that there is no hypotension. But in this case, the patient is hypotensive, and therefore, it is contraindicated.
If the symptoms are not relieved after three serial doses of nitroglycerine or if symptoms recur despite adequate anti-anginal treatment morphine sulphate is indicated.
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This question is part of the following fields:
- Pathophysiology
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Question 24
Correct
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What is the most sensitive method of detecting an intra-operative air embolism?
Your Answer: Transoesophageal echocardiogram
Explanation:An intra-operative air embolism occurs when air becomes trapped in the blood vessels during surgery.
A transoesophageal echocardiography (OE) uses invasive echocardiography to monitor the integrity and performance of the heart. It is the gold standard as it provides real-time imaging of the heart to enable early diagnosis and treatment.
Precordial doppler ultrasonography can also be used to detect into-operative air emboli. It is non-invasive and more practical, but is less sensitive.
A change in end-tidal CO2 could be indicative of and increase in physiological dead-space, but could also be indicative of any processes that reduces the excretion or increases the production of CO2, making it non-specific.
A transoesophageal stethoscope can be used to listen for the classic mill-wheel murmur produced by a large air embolus.
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This question is part of the following fields:
- Pathophysiology
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Question 25
Incorrect
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A 33-year-old woman known to be hypothyroid and taking 150 mcg l-thyroxine daily is reviewed in the preoperative assessment clinic prior to a laparoscopic cholecystectomy.
She has required three increases in her thyroid replacement therapy in the last six months.
Her thyroid function tests are as follows:
TSH 11 (normal range 0.4-4mU/L)
T3 20 (normal range 9-25mU/L)
T4 6.2 (normal range 3.5-7.8mU/L)
What will explain this biochemical picture?Your Answer: Generalised resistance to thyroid hormone (GrTH)
Correct Answer: Poor compliance with medication
Explanation:In patients with an intact hypothalamic-pituitary axis, serial TSH measurements are used to determine the adequacy of treatment with thyroid hormones . changes in TSH levels becoming apparent after approximately eight weeks of therapy with thyroid hormone replacement. Change in T3/T4 levels are seen before changes in TSH .
In patients taking thyroid replacement therapy, the most frequent reason for persistent elevation of serum TSH is poor compliance. Patients who do not regularly take their L-thyroxine try and catch up just before a visit to a clinician for blood test.
Tissue-level unresponsiveness to thyroid hormone is caused by mutation in the gene controlling a receptor for T3 and is rare.
Reduced responsiveness of target tissues to thyroid hormone aka resistance to thyroid hormones (rTH) occurs when there is a mutation in the thyroid hormone receptor ? gene. It is a rare autosomal dominant inherited syndrome of reduced end-organ responsiveness to thyroid hormone and has two types:
Generalised resistance (GrTH)
Pituitary resistance (PrTH)Patients with rTH have normal or slightly elevated serum thyroid stimulating hormone (TSH) level, elevated serum free thyroxine (FT4) and free triiodothyronine (FT3) concentrations.
Drugs that increase metabolism of thyroxine include:
Warfarin
Rifampin
Phenytoin
Phenobarbital
St John’s Wort
CarbamazepineThese drugs lower circulating thyroid hormones and would be associated with a raised TSH but low T3/T4.
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This question is part of the following fields:
- Pathophysiology
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Question 26
Correct
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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: 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 27
Incorrect
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Regarding thyroid hormones, one of the following is true.
Your Answer: Starvation causes plasma T3 to rise
Correct Answer: Thyroid binding globulin (TBG) is increased in pregnancy
Explanation:T3 is produced by peripheral de-iodination of T4. It is more active than T4.
TBG, like most binding proteins, is increased in pregnancy. Because of this, measurement of free thyroid hormone concentration is more important than total.
T4 and T3 concentrations are decreased in Illness and starvation.
L-T4 that is the active molecule while D-T4 is inactive.
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This question is part of the following fields:
- Pathophysiology
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Question 28
Correct
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Which of the following statements is true about the in-hospital management of ventricular fibrillation?
Your Answer: Amiodarone may be administered following a third DC shock
Explanation:Ventricular fibrillation (VT) is an arrhythmia caused by a distortion in the organized contraction of the ventricles leading to an inability to pump blood out into the body.
Amiodarone is an anti arrhythmic drug used for the treatment of ventricular and atrial fibrillations. It is the gold standard of treatment for refractory pulseless ventricular tachycardia (VT) and ventricular fibrillation (VF).
Guidelines for emergency treatment state that only the rescuer carrying out chest compressions on the patient may stand near the defibrillator as it charges.
Cardio-pulmonary resuscitation (CPR) during cardiac arrest is required for 2 minute cycles.
Hypovolaemia is as a cause of pulseless electrical activity (PEA) can be reversed using fluid resuscitation, whereas hypotension during cardiac arrest is either persistent or undetectable and is therefore irreversible.
Hyperkalaemia and hypocalcaemia are treated using calcium salts, but calcium chloride is often preferred over calcium gluconate.
During a pulseless VT or VF, a single precordial thump will be effective if administered within the first seconds of the occurrence of a shockable rhythm.
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This question is part of the following fields:
- Pathophysiology
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Question 29
Incorrect
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Which of the following statements is true about an acute pulmonary embolism?
Your Answer: Embolectomy is more effective than thrombolysis in improving survival
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 30
Incorrect
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Tubes for vascular access and body cavity drainage are available in a variety of sizes.
When choosing an intravenous or intra-arterial cannula, which of the following measurements is used?Your Answer: Internal diameter (mm)
Correct Answer: Standard wire gauge (SWG)
Explanation:Standard wire gauge cannulas for intravenous and intraarterial use are available (SWG or G). The SWG is a former imperial unit (which requires metric conversion). The cross sectional area of wires is becoming more popular as a size measurement.
The number of wires that will fit into a standard hole template is referred to as SWG.
This standard sized hole can accommodate 22 thin wires side by side (each wire the diameter of a 22 gauge cannula)
In the same hole, 14 thicker wires would fit (each wire the diameter of a 14 gauge cannula)While the diameter and thus radius of a parallel sided tube are the most important determinants of fluid flow rate, they are not commonly used to compare cannula sizes.
The circumference of French gauge (FG) catheters (urinary or chest drains) is measured. Sizes of double lumen tracheal tubes are FG. Internal diameter is used to measure single lumen tubes.
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This question is part of the following fields:
- Pathophysiology
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