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Question 1
Incorrect
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Rocuronium is substituted for succinylcholine during induction of anaesthesia for a caesarean section delivery.
Which of the following feature of rocuronium ensures the neonate shows no clinical signs of muscle relaxation?Your Answer: Unaffected by ion trapping
Correct Answer: Highly ionised
Explanation:Drugs cross the placenta by Simple, Ion channel and Facilitated diffusion; Exocytosis and Endocytosis, Osmosis, and Active transport (primary and secondary)
The following factors influence rate of diffusion across the placenta:
Protein binding
Degree of ionisation
Placental blood flow
Maternal and foetal blood pH
Materno-foetal concentration gradient.
Thickness of placental membrane
Molecular weight of drug <600 Daltons cross by diffusion
Lipid solubility (lipid soluble molecules readily diffuse across the placenta)Rocuronium has a F/M ratios of 0.16, a 30% plasma protein binding, low lipid solubility, a low volume of distribution (0.25L/kg), and a high molecular weight (530Da).
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This question is part of the following fields:
- Pharmacology
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Question 2
Correct
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What can an outbreak of flu that has spread globally be termed as?
Your Answer: Pandemic
Explanation:An epidemic is declared when the increase in a give disease is above a certain level in a specific interval of time.
An endemic is the general, usual level of a disease in a population at a particular time.
A pandemic is an epidemic that is spread across many countries and continents.
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This question is part of the following fields:
- Statistical Methods
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Question 3
Incorrect
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A 45-year old male who was involved in a road traffic accident has had to receive a large blood transfusion of whole blood which is two weeks old. Which of these best describes the oxygen carrying capacity of this blood?
Your Answer: The release of oxygen in metabolically active tissues will be the same as fresh blood
Correct Answer: It will have an increased affinity for oxygen
Explanation:With respect to oxygen transport in cells, almost all oxygen is transported within erythrocytes. There is limited solubility and only 1% is carried as solution. Thus, the amount of oxygen transported depends upon haemoglobin concentration and its degree of saturation.
Haemoglobin is a globular protein composed of 4 subunits. Haem is made up of a protoporphyrin ring surrounding an iron atom in its ferrous state. The iron can form two additional bonds – one is with oxygen and the other with a polypeptide chain.
There are two alpha and two beta subunits to this polypeptide chain in an adult and together these form globin. Globin cannot bind oxygen but can bind to CO2 and hydrogen ions.
The beta chains are able to bind to 2,3 diphosphoglycerate. The oxygenation of haemoglobin is a reversible reaction. The molecular shape of haemoglobin is such that binding of one oxygen molecule facilitates the binding of subsequent molecules.The oxygen dissociation curve (ODC) describes the relationship between the percentage of saturated haemoglobin and partial pressure of oxygen in the blood.
Of note, it is not affected by haemoglobin concentration.Chronic anaemia causes 2, 3 DPG levels to increase, hence shifting the curve to the right
Haldane effect – Causes the ODC to shift to the left. For a given oxygen tension there is increased saturation of Hb with oxygen i.e. Decreased oxygen delivery to tissues.
This can be caused by:
-HbF, methaemoglobin, carboxyhaemoglobin
-low [H+] (alkali)
-low pCO2
-ow 2,3-DPG
-ow temperatureBohr effect – causes the ODC to shifts to the right = for given oxygen tension there is reduced saturation of Hb with oxygen i.e. Enhanced oxygen delivery to tissues. This can be caused by:
– raised [H+] (acidic)
– raised pCO2
-raised 2,3-DPG
-raised temperature -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 4
Correct
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Which of the following is the most appropriate first-line pharmacologic treatment for status epilepticus?
Your Answer: Lorazepam
Explanation:Lorazepam is an intermediate-acting benzodiazepine that binds to the GABA-A receptor subunit to increase the frequency of chloride channel opening and cause membrane hyperpolarization.
Lorazepam has emerged as the preferred benzodiazepine for acute management of status epilepticus. Lorazepam differs from diazepam in two important respects. It is less lipid-soluble than diazepam, with a distribution half-life of two to three hours versus 15 minutes for diazepam. Therefore, it should have a longer duration of clinical effect. Lorazepam also binds the GABAergic receptor more tightly than diazepam, resulting in a longer duration of action. The anticonvulsant effects of lorazepam last six to 12 hours, and the typical dose ranges from 4 to 8 mg. This agent also has a broad spectrum of efficacy, terminating seizures in 75-80% of cases. Its adverse effects are identical to those of diazepam. Thus, lorazepam also is an effective choice for acute seizure management, with the added possibility of a longer duration of action than diazepam.
Phenobarbitone is a long-acting barbiturate that binds to GABA-A receptor site and increase the duration of chloride channel opening. It also blocks glutamic acid neurotransmission, and, at high doses, can block sodium channels. It is considered as the drug of choice for seizures in infants.
Phenytoin is an anti-seizure drug that blocks voltage-gated sodium channels. It is preferred in prolonged therapy of status epilepticus because it is less sedating.
In cases wherein airway protection is required, thiopentone and propofol are the preferred drugs.
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This question is part of the following fields:
- Pharmacology
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Question 5
Correct
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Which plasma protein will bind the thyroid hormone triiodothyronine (T3) more readily?
Your Answer: Thyroxine binding globulin
Explanation:Secreted T4 and T3 circulate in the bloodstream almost entirely bound to proteins. Normally only about 0.03% of total plasma T4 and 0.3% of total plasma T3 exist in the free state. Free T3 is biologically active and mediates the effects of thyroid hormone on peripheral tissues in addition to exerting negative feedback on the pituitary and hypothalamus. The major binding protein is thyroxine-binding globulin (TBG), which is synthesized in the liver and binds one molecule of T4 or T3. About 70% of circulating T4 and T3 is bound to TBGl 10% to 15% is bound to another specific thyroid-binding protein called transthyretin (TTR). Albumin binds 15% to 20%, and 3% to lipoproteins. Ordinarily only alterations in TBG concentration significantly affect total plasma T4 and T3 levels.
Two important biological functions have been ascribed to TBG. First, it maintains a large circulating reservoir of T4 that buffers any acute changes in thyroid gland function. Second, binding of plasma T4 and T3 to proteins prevents loss of these relatively small hormone molecules in urine and thereby helps conserve iodide. TTR transports T4 in CSF and provides thyroid hormones to the CNS.
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This question is part of the following fields:
- Physiology
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Question 6
Correct
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A 45-year-old man is being operated on for emergency laparotomy as he presented with bowel perforation. During the surgery, the marginal artery of Drummond is encountered and preserved.
Which of the following two arteries fuse to form the marginal artery of Drummond?Your Answer: Superior mesenteric artery and inferior mesenteric artery
Explanation:The arteries of the midgut (superior mesenteric artery) and hindgut (inferior mesenteric artery) give off terminal branches that form an anastomotic vessel called the marginal artery of Drummond. It runs in the inner margins of the colon and gives off short terminal branches to the bowel wall.
The marginal artery is formed by the main branches and arcades arising from the ileocolic, right colic, middle colic, and left colic arteries. It is most apparent in the ascending, transverse, and descending colons and poorly developed in the sigmoid colon.
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This question is part of the following fields:
- Anatomy
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Question 7
Incorrect
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Which of the following statements best describes adenosine receptors?
Your Answer: The A1 and A2 receptors are present centrally and peripherally
Correct Answer:
Explanation:Adenosine receptors are expressed on the surface of most cells.
Four subtypes are known to exist which are A1, A2A, A2B and A3.Of these, the A1 and A2 receptors are present peripherally and centrally. There are agonists at the A1 receptors which are antinociceptive, which reduce the sensitivity to a painful stimuli for the individual. There are also agonists at the A2 receptors which are algogenic and activation of these results in pain.
The role of adenosine and other A1 receptor agonists is currently under investigation for use in acute and chronic pain states.
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This question is part of the following fields:
- Physiology
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Question 8
Correct
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With regards to the internal carotid artery, which of these statements is correct.
Your Answer: Enters the skull and divides into the anterior and middle cerebral arteries
Explanation:The internal carotid artery passes through the carotid canal in the petrous part of the temporal bone into the cranial cavity. It does NOT groove the sphenoid bone.
The internal carotid artery gives off no branches in the neck and is a terminal branch of the common carotid artery.
These structures pass between the external and internal carotid arteries: the styloglossus and stylopharyngeus muscles, the glossopharyngeal nerve (CN IX), and the pharyngeal branch of the vagus.
Accompanied by its sympathetic plexus, the internal carotid artery, passes through the cavernous sinus and is crossed by the abducent nerve.
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This question is part of the following fields:
- Anatomy
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Question 9
Correct
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A 26-year old male patient was admitted to the surgery department for appendectomy. Medical history revealed that he has major depressive disorder and was on Phenelzine. Aside from abdominal pain, initial assessment was unremarkable. However, thirty minutes after, the patient was referred to you for generalized seizures. He was given an analgesic and it was noted that, during the first 15 minutes of administration, he became anxious, with profuse sweating, which later developed into seizures. Upon physical examination, he was febrile at 38.3°C.
Which of the following statements is the best explanation for the patient's symptoms?Your Answer: Drug interaction with pethidine
Explanation:The clinical picture best describes a probable drug interaction with pethidine.
Phenelzine, a monoamine oxidase (MAO) inhibitor, when given with pethidine, an opioid analgesic, may lead to episodes of hypertension, rigidity, excitation, hyperpyrexia, seizures, coma and death. Studies have shown that pethidine reacts more significantly with MAO inhibitors than morphine.
When pethidine is metabolised to normeperidine, it acts as a serotonin reuptake inhibitor and cause an increase in serotonin levels in the brain. MAO inhibitors can also lead to elevated levels of serotonin because of its mechanism of action by inhibiting the enzyme monoamine oxidase that degrades serotonin.
The excess serotonin levels may lead to serotonin syndrome, of which some of the common precipitating drugs are selective serotonin reuptake inhibitors, MAO inhibitors, tricyclic antidepressants, meperidine, and St. John’s Wort. Onset of symptoms is within hours, which includes fever, agitation, tremor, clonus, hyperreflexia and diaphoresis.
Drug interaction between phenelzine and paracetamol do not commonly precipitate serotonin syndrome.
Neuroleptic malignant syndrome is due to dopamine antagonism, precipitated commonly by antipsychotics. Its onset of symptoms occur in 1 to 3 days, and is characterized by fever, encephalopathy, unstable vitals signs, elevated CPK, and rigidity.
Altered mental status is the most common manifestation of sepsis-associated encephalopathy. Patient also exhibit confusional states and inappropriate behaviour. In some cases, this may lead to coma and death.
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This question is part of the following fields:
- Pharmacology
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Question 10
Correct
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When nitrous oxide is stored in cylinders at room temperature, it is a gas.
Which of its property is responsible for this?
Your Answer: Critical temperature
Explanation:The temperature above which a gas cannot be liquefied no matter how much pressure is applied is its critical temperature. The critical temperature of nitrous oxide is 36.5°C
The minimum pressure that causes liquefaction is the critical pressure of that gas.
The Poynting effect refers to the phenomenon where mixing of liquid nitrous oxide at low pressure with oxygen at high pressure (in Entonox) leads to formation of gas of nitrous oxide.
There is no relevance of molecular weight to this question. it does not change with phase of a substance.
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This question is part of the following fields:
- Pharmacology
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Question 11
Correct
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You draw a patient's blood sample from the median cubital vein in the antecubital fossa.
Which of the following veins also connects to the cephalic vein other than the median cubital vein?Your Answer: Basilic vein
Explanation:The upper limb venous drainage is divided into superficial and deep. The superficial veins are accessible to draw blood for investigations. The cephalic, basilic, and median cubital veins are superficial veins.
The median cubital vein connects the cephalic vein and basilic vein. It is located anteriorly in the antecubital fossa and is preferred for venepuncture due to its palpability and ease of access.
The basilic vein and cephalic vein are the primary veins that drain the upper limb. They begin as the dorsal venous arch. The basilic vein originates from the ulnar side, while the cephalic vein originates from the radial side of the dorsal arch of the upper limb.
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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 ED95 of muscle relaxants is the dose required to reduce twitch height by 95% in half of the target population. The dose of non-depolarizing muscle relaxants used for intubation is 2-3 times the ED95.
For procedures that need a short duration of muscle relaxation and abrupt recovery, the short-acting drug Mivacurium is given at less than 2 times the ED95. What is the explanation for Mivacurium being an exception to this rule?Your Answer: Mivacurium has a high potency
Correct Answer: Dose related histamine release occurs which frequently leads to tachycardia and hypotension
Explanation:Mivacurium, when administered at doses greater than 0.2 mg/kg,increases the risk for hypotension, tachycardia, and erythema. This is due to the ability of mivacurium to release histamine with increasing dose. Contrary to this fact, anaphylaxis is rare for mivacurium because of the short duration of histamine release.
The effective dose 50 (ED50) of mivacurium is between 0.08-0.15 mg/kg. It is administered slowly to prevent and decrease the risk of developing adverse effects.
Mivacurium has a high potency thus a longer duration of action, however this is not the answer that we are looking for.
Although drug metabolism takes longer for mivacurium than succinylcholine, it has no effect on the dose required for intubation.
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This question is part of the following fields:
- Pharmacology
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Question 13
Correct
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All of the following statements about calcium channel antagonists are incorrect except:
Your Answer: May cause potentiation of muscle relaxants
Explanation:Calcium channel blocker (CCB) blocks L-type of voltage-gated calcium channels present in blood vessels and the heart. By inhibiting the calcium channels, these agents decrease the frequency of opening of calcium channels activity of the heart, decrease heart rate, AV conduction, and contractility.
Three groups of CCBs include
1) Phenylalkylamines: Verapamil, Norverapamil
2) Benzothiazepines : Diltiazem
3) Dihydropyridine : Nifedipine, Nicardipine, Nimodipine, Nislodipine, Nitrendipine, Isradipine, Lacidipine, Felodipine and Amlodipine.Even though verapamil as good absorption from GIT, its oral bioavailability is low due to high first-pass metabolism.
Nimodipine is a Cerebro-selective CCB, used to reverse the compensatory vasoconstriction after sub-arachnoid haemorrhage and is more lipid soluble analogue of nifedipine
Calcium channel antagonist can potentiate the effect of non-depolarising muscle relaxants.
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This question is part of the following fields:
- Pharmacology
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Question 14
Incorrect
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The following haemodynamic data is available from a patient with pulmonary artery catheter inserted:
Pulse rate - 100 beats per minute
Blood pressure - 120/70mmHg
Mean central venous pressure (MCVP) - 10mmHg
Right ventricular pressure (RVP) - 30/4 mmHg
Mean pulmonary artery wedge pressure (MPAWP) - 12mmHg
Which value best approximates the patient's coronary perfusion pressure?Your Answer: 60mmHg
Correct Answer: 58mmHg
Explanation:Coronary perfusion pressure(CPP), the difference between aortic diastolic pressure (Pdiastolic) and the left ventricular end-diastolic pressure (LVEDP), is mainly determined by the formula:
CPP = Pdiastolic -LVEDP
where
Pdiastolic is the lowest pressure in the aorta before left ventricular ejection and
LVEDP is measured directly during a cardiac catheterisation or indirectly using a pulmonary artery catheter. The pulmonary artery occlusion or wedge pressure approximates best with LVEDP.Using this patient’s haemodynamic data:
CPP = Pdiastolic – MPAWP
COO = 70 – 12 = 58mmHg -
This question is part of the following fields:
- Clinical Measurement
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Question 15
Correct
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The liver plays a major role in drug metabolism.
Which of the following liver cells is most important in phase I of drug metabolism?Your Answer: Centrilobular cells
Explanation:The metabolism of drugs in the liver occurs in 3 phases
Phase I: This involves functionalization reactions, which are of 3 types, namely hydrolysis, oxidation and reduction reactions catalysed by the cytochrome P450 (CYP) enzymes.
Phase II: This involves conjugation or acetylation reactions. The goal is to create water soluble metabolites that can be excreted from the body.
The liver is the second largest organ. It’s smallest functional unit is the acinus which is divided into 3 zones:
Zone I (periportal): This zone receives the largest amount of oxygen supply as it is the closest to the blood vessels. It is the site of plasma protein synthesis.
Zone II (mediolobular): This is located between the portal triad and central vein.
Zone III (centrilobular): This is closest to the central vein and receives the least amount of oxygen supply.
Kupffer cells are specialized macrophages found in the periportal zone of the liver, and function to remove foreign particles and breakdown red blood cells via phagocytosis.
Ito cells are fat-storing liver cells found in the space of Disse. Their function is to take-uo, store and secrete retinoids, as well as manufacture and release proteins that make up the extracellular matrix.
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This question is part of the following fields:
- Pathophysiology
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Question 16
Incorrect
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A chain smoker is interested in knowing how many years of his life would be lessened by smoking. You tell him explicitly that precise determination is impossible but you can tell him the proportion of people who died due to smoking. Which of the following epidemiological term is apt in this regard?
Your Answer: Relative risk - the risk of an event relative to exposure.
Correct Answer: Attributable risk - the rate in the exposed group minus the rate in the unexposed group
Explanation:Attributable proportion is the proportion of disease that is caused due to exposure. It refers to the proportion of disease that would be eradicated from a particular population if the disease rate was diminished to match that of the unexposed group.
Risk ratio (relative risk) compares the probability of an event in an exposed (experimental) group to that of an event in the unexposed (control) group. Thus two are not the same.
The attributable risk is the rate of a disease in an exposed group to that of a group that has not been exposed to it i.e. how many deaths did the exposure cause.
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This question is part of the following fields:
- Statistical Methods
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Question 17
Incorrect
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A 54-year-old lady comes in for a right-sided elective bunionectomy with a realignment osteotomy under local anaesthetic on her first (large) toe.
For the operation, which of the following nerve blocks will be most effective?Your Answer: Superficial peroneal, posterior tibial and sural nerves
Correct Answer: Superficial peroneal, deep peroneal and posterior tibial nerves
Explanation:An ankle block is commonly used for anaesthesia and postoperative analgesia when operating on bunions. It results in the selective block of the superficial peroneal, deep peroneal, and posterior tibial nerves.
The deep peroneal nerve supplies sensory input to the web space between the first and second toes (L4-5).
The L2-S1 nerve, often known as the superficial peroneal nerve, is a mixed motor and sensory neuron. It gives sensory supply to the anterolateral region of the leg, the anterior aspect of the 1st, 2nd, 3rd, and 4th toes, and innervates the peroneus longus and brevis muscles (with the exception of the web space between 1st and 2nd toes).
The sensory area of the saphenous nerve (L3-4) in the foot stretches from the proximal portion of the midfoot on the medial side to the proximal part of the midfoot on the lateral side.
The lateral side of the little (fifth) toe is innervated by the sural nerve’s sensory supply (S1-2). The heel, medial (medial plantar nerve), and lateral (lateral plantar nerve) soles of the foot are all served by the posterior tibial nerve.
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This question is part of the following fields:
- Pathophysiology
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Question 18
Correct
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Which of the following statement is correct regarding the difference between dabigatran and other anticoagulants?
Your Answer: Competitive thrombin inhibitor blocking both free and bound thrombin
Explanation:Dabigatran template is a prodrug and its active metabolite is a direct thrombin inhibitor. It is a synthetic, reversible, non-peptide thrombin inhibitor. This inhibition of thrombin results in a decrease of fibrin and reduces platelet aggregation.
Drugs like warfarin act by inhibiting the activation of vitamin K-dependent clotting factors. These factors are synthesized by the liver and activated by gamma-carboxylation of glutamate residues with the help of vitamin K. Hydroquinone form of vitamin K is converted to epoxide form in this reaction and regeneration of hydroquinone form by enzyme vitamin K epoxide reductase (VKOR) is required for this activity. Oral anticoagulants prevent this regeneration by inhibiting VKOR, thus vitamin K-dependent factors are not activated. These factors include clotting factors II, VII, IX, and X as well as anti-clotting proteins, protein C and protein S.
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This question is part of the following fields:
- Pharmacology
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Question 19
Incorrect
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Which medical gas cylinders have the correct colour codes?
Your Answer: Carbon dioxide cylinders are brown
Correct Answer: Oxygen cylinders have a black body with white shoulders
Explanation:The following are the colour codes for medical gas cylinders:
Oxygen cylinder has a dark body with white shoulders.
Nitrous oxide is French blue. Air encompasses a grey body with dark and white quarters on the shoulders.
Entonox contains a French blue body with white and blue quarters on the shoulders.
Carbon dioxide barrels are grey in colour.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 20
Incorrect
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A 74-year-old man presents to a hospital for manipulation of Colles fracture. The patient is 50 kg and the anaesthetic plan is to perform an intravenous regional (Bier's) block.
Which of the following is the appropriate dose of local anaesthetic for the procedure?Your Answer: 0.5% bupivacaine (20 ml)
Correct Answer: 0.5% prilocaine (40 ml)
Explanation:Prilocaine is the drug of choice for intravenous regional anaesthesia. 0.5% prilocaine (40 ml) is indicated for this condition.
Lidocaine is another alternative for this condition but volume and dose are likely to be inadequate for the procedure. -
This question is part of the following fields:
- Pharmacology
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Question 21
Correct
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Following are some examples of induction agents. Which one has the longest elimination half-life?
Your Answer: Thiopental
Explanation:Thiopental has the longest elimination half-life of 6-15 hours.
Elimination half-life of other drugs are given as:
– Propofol: 5-12 h
– Methohexitone: 3-5 h
– Ketamine: 2 h
– Etomidate: 1-4 h -
This question is part of the following fields:
- Pharmacology
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Question 22
Correct
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The external urethral sphincter arises from which nerve root?
Your Answer: S2, S3, S4
Explanation:The external urethral sphincter functions to provide voluntary control of urine flow from the bladder to the urethra.
It receives its innervation from the branches of the pudendal nerve which originate from S2, S3 and S4.
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This question is part of the following fields:
- Anatomy
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Question 23
Correct
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The most sensitive indicator of mild obstructive airway disease is?
Your Answer: Forced expiratory flow (FEF25-75%)
Explanation:The volume expired in the first second of maximal expiration after a maximal inspiration is known as forced expiratory volume in one second (FEV1), and it indicates how quickly full lungs can be emptied. It is the most commonly measured parameter for bronchoconstriction assessment.
The maximum volume of air exhaled after a maximal inspiration is known as the ‘slow’ vital capacity (VC). VC is normally equal to FVC after a forced vital capacity (FVC) or slow vital capacity (VC) manoeuvre, unless there is an airflow obstruction, in which case VC is usually higher than FVC.
The FEV1/FVC (Tiffeneau index) is a clinically useful index of airflow restriction that can be used to distinguish between restrictive and obstructive respiratory disorders.
The average expired flow over the middle half (25-75 percent) of the FVC manoeuvre is the forced expiratory volume (FEF25-75). The airflow from the resistance bronchioles corresponds to this. It’s a more sensitive indicator of mild small airway narrowing than FEV1, but it’s difficult to tell if the VC (or FVC) is decreasing or increasing.
The maximum expiratory flow rate achieved is called the peak expiratory flow (PEF), which is usually 8-14 L/second.
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This question is part of the following fields:
- Pathophysiology
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Question 24
Correct
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Regarding the plateau phase of the cardiac potential, which electrolyte is the main determinant?
Your Answer: Ca2+
Explanation:The cardiac action potential has several phases which have different mechanisms of action as seen below:
Phase 0: Rapid depolarisation – caused by a rapid sodium influx.
These channels automatically deactivate after a few msPhase 1: caused by early repolarisation and an efflux of potassium.
Phase 2: Plateau – caused by a slow influx of calcium.
Phase 3 – Final repolarisation – caused by an efflux of potassium.
Phase 4 – Restoration of ionic concentrations – The resting potential is restored by Na+/K+ATPase.
There is slow entry of Na+into the cell which decreases the potential difference until the threshold potential is reached. This then triggers a new action potentialOf note, cardiac muscle remains contracted 10-15 times longer than skeletal muscle.
Different sites have different conduction velocities:
1. Atrial conduction – Spreads along ordinary atrial myocardial fibres at 1 m/sec2. AV node conduction – 0.05 m/sec
3. Ventricular conduction – Purkinje fibres are of large diameter and achieve velocities of 2-4 m/sec, the fastest conduction in the heart. This allows a rapid and coordinated contraction of the ventricles
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This question is part of the following fields:
- Physiology
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Question 25
Incorrect
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Which of the following is correct regarding nitric oxide?
Your Answer: Is increased by cyclic AMP activation
Correct Answer: Is produced by both inducible and constitutive forms of nitric oxide synthetase
Explanation:Nitric oxide is generated from L-arginine by nitric oxide synthase. It is produced in response to haemodynamic stress by the vascular endothelium, and it produces both smooth muscle relaxation and reduced vascular resistance.
Nitric oxide may be inactivated through interaction with other oxygen free radicals, (e.g. oxidised low-density lipoprotein (LDL)).
Nitric oxide causes the production of the second messenger, cyclic guanosine monophosphate (cGMP).
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This question is part of the following fields:
- Pathophysiology
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Question 26
Incorrect
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Which of the following closely estimates the interstitial oncotic pressure acting on a pulmonary capillary?
Your Answer: 0 mmHg
Correct Answer: 17 mmHg
Explanation:The starling forces operate to maintain a homeostatic flow across the pulmonary capillary bed.
The outward driving force comprises of the capillary hydrostatic pressure (13 mmHg), negative interstitial fluid pressure (zero to slightly negative), and interstitial colloid osmotic pressure (17 mmHg). The inward driving force is controlled by the plasma colloid osmotic pressure (25 mmHg).
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This question is part of the following fields:
- Basic Physics
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Question 27
Correct
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Which of the following is true regarding the dose of propofol?
Your Answer: 1-2mg/kg
Explanation:Propofol is a short-acting medication used for starting and maintenance of general anaesthesia, sedation for mechanically ventilated adults, and procedural sedation.
The dose of propofol is 1-2 mg/kg.Dose of some other important drugs are listed below:
Thiopental dose: 3-7 mg/kg
Ketamine dose: 1-2 mg/kg
Etomidate dose: 0.3 mg/kg
Methohexitone dose: 1.0-1.5 mg/kg -
This question is part of the following fields:
- Pharmacology
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Question 28
Correct
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These proprietary preparations of local anaesthetic are available in your hospital:
Solution A contains 10 mL 0.5% bupivacaine (plain), and
Solution B contains 10 mL 0.5% bupivacaine with adrenaline 1 in 200,000.
What is the pharmacokinetic difference between the two solutions?Your Answer: The onset of action of solution A is quicker than solution B
Explanation:The reasons for adding adrenaline to a local anaesthetic solution are:
1. To Increase the duration of block
2. To reduce absorption of the local anaesthetic into the circulation
3. To Increase the upper safe limit of local anaesthetic (2.5 mg/kg instead of 2 mg/kg, in this case).The addition of adrenaline to bupivacaine does not affect its potency, lipid solubility, protein binding, or pKa(8.1 with or without adrenaline).
The pH of bupivacaine is between 5-7. Premixed with adrenaline, it is 3.3-5.5.
The onset of a local anaesthetic and its ability to penetrate membranes depends upon degree of ionisation. Compared with the ionised fraction, unionised local anaesthetic readily penetrates tissue membranes to site of action. The onset of action of solution B is slower. this is because the relationship between pKa(8.1) and pH(3.3-5.5) of the solution results in a greater proportion of ionised local anaesthetic molecules compared with solution A. -
This question is part of the following fields:
- Pharmacology
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Question 29
Incorrect
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The muscle that lies behind the first part of the axillary nerve is?
Your Answer: Teres minor
Correct Answer: Subscapularis
Explanation:The axillary nerve lies behind the axillary artery initially, and in front of the subscapularis. It passes downward to the lower border of the subscapularis muscle.
In company with the posterior humeral circumflex artery and vein, it winds backward through a quadrilateral space bounded above by the subscapularis (anterior) and teres minor (posterior), below by the teres major, medially by the long head of the triceps brachii, and laterally by the humerus (surgical neck).
It then divides into an anterior and a posterior part. The anterior division supplies the deltoid (anterior and middle heads) while the posterior division supplies the teres minor and posterior part of deltoid
The posterior division terminates as the superior lateral cutaneous nerve of the arm -
This question is part of the following fields:
- Anatomy
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Question 30
Correct
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The equipment used for patient monitoring in theatre and intensive care settings have electrical safety requirements for the protection of hospital staff and patients.
Of the different classes of electrical equipment listed, which is least likely to cause a patient to suffer a microshock?Your Answer: II (CF)
Explanation:Microshock refers to ventricular fibrillation caused by miniscule amounts of currents or voltages (100-150 microamperes) passing through the myocardial tissue from external cables arising from electrical components within the cardiac muscle, for example, pacemaker electrodes or saline filled venous catheters.
The risk of shock changes with the construction of electrical equipment in question. The main classes of electrical equipment include: I: Appliances have a protective earth connected to an outer casing which prevents live elements from coming in contact with conductive elements. A fault in this equipment class will result in live elements coming in contact with the outer casing and allowing electrical flow into the protective earth. This triggers the protective fuse to disconnect the electric supply to the appliance.
II: These appliances have reinforced insulation. In the event of a fault which causes the first layer of insulation to fail, the second layer is able to prevent contact of live elements with outer casing.
III: These appliances have no insulation to provide safety, and rely solely on the use of separated extra low voltage source (SELV) which limits voltage to 25V AC or 60V DC allowing for a person to come in contact with it without risk of a shock under normal dry conditions. Under wet conditions, voltage supply should be lowered to reduce risk of shock. These devices have no risk of macroshocks, but some risk of microshocks.
Class I and II electrical appliances are further divided into subtypes developed to limit current leakage in the event of a singular fault:
B (body): Upper limit of current leakage is 500 µA. This current can cause skin tingling and microshocks, but is not sufficient to cause injury.
BF (body floating): These appliances have an isolating capacitor or transformer which separate the secondary circuit from the protective earth. The upper limit of current leakage is the same as type B.
CF (cardiac floating): Upper limit of leakage current during a singular fault is 50 microamps. It is least likely to result in a microshock -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 31
Incorrect
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Obeying Boyle's law and Charles's law is a characteristic feature of an ideal gas.
The gas which is most ideal out of the following options is?Your Answer: Oxygen
Correct Answer: Helium
Explanation:The ideal gas equation makes the following assumptions:
The gas particles have a small volume in comparison to the volume occupied by the gas.
Between the gas particles, there are no forces of interaction.
Individual gas particle collisions, as well as gas particle collisions with container walls, are elastic, meaning momentum is conserved.
PV = nRT
Where:P = pressure
V = volume
n = moles of gas
T = temperature
R = universal gas constantHelium is a monoatomic gas with a small helium atom. The attractive forces between helium atoms are small because the helium atom is spherical and has no dipole moment. Because helium atoms are spherical, collisions between them approach the ideal state of elasticity.
Most real gases behave qualitatively like ideal gases at standard temperatures and pressures. When intermolecular forces and molecular size become important, the ideal gas model tends to fail at lower temperatures or higher pressures. It also fails to work with the majority of heavy gases.
Helium, argon, neon, and xenon are noble or inert gases that behave the most like an ideal gas. Xenon is a noble gas with a much larger atomic size than helium.
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This question is part of the following fields:
- Pharmacology
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Question 32
Incorrect
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A 64-year old male has shortness of breath on exertion and presented to the cardiology clinic. He has a transthoracic echo performed to help in assessing the function of his heart.
How can this echo aid in calculating cardiac output?Your Answer: (end diastolic LV volume - end systolic LV volume) / heart rate
Correct Answer: (end diastolic LV volume - end systolic LV volume) x heart rate
Explanation:Cardiac output = stroke volume x heart rate
Left ventricular ejection fraction = (stroke volume / end diastolic LV volume ) x 100%
Stroke volume = end diastolic LV volume – end systolic LV volume
Pulse pressure = Systolic Pressure – Diastolic Pressure
Systemic vascular resistance = mean arterial pressure / cardiac output
Factors that increase pulse pressure include:
-a less compliant aorta (this tends to occur with advancing age)
-increased stroke volume -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 33
Correct
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A 20-year-old boy is undergoing surgery for indirect inguinal hernia repair. The deep inguinal ring is exposed and held with a retractor at its medial aspect during the procedure.
What structure is most likely to lie under the retractor on the medial side?Your Answer: Inferior epigastric artery
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 34
Incorrect
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A 53-year-old-male is being operated on for a right hemicolectomy. In the procedure, the ileocolic artery is ligated. Which vessel does this artery originate from?
Your Answer: Inferior mesenteric artery
Correct Answer: Superior mesenteric artery
Explanation:The ileocolic artery is the terminal branch of the superior mesenteric artery. It supplies:
1. terminal ileum
2. proximal right colon
3. cecum
4. appendix (via its branch of the appendicular artery)As veins accompany arteries in the mesentery and are lined by lymphatics, high ligation is the norm in cancer resections—the ileocolic artery branches off the SMA near the duodenum.
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This question is part of the following fields:
- Anatomy
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Question 35
Incorrect
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After a bariatric surgery, average weight loss observed in patients is 18 kg. The standard deviation was found to be 3 kg. What is the percentage of patients that lie between 9 and 27 kg?
Note: Assume that the curve is normally distributed.Your Answer: 95.00%
Correct Answer: 99.70%
Explanation:9 & 27 can be obtained by subtracting and adding 9 from the mean. 9 is three times the standard deviation and we know that 99.7% values lie within 3 standard deviations from the mean. We can find the interval for 99.7% to verify in the following way:
For 99.7% confidence interval, you can find the range as follows:
1. Multiply the standard error by 3.
2. Subtract the answer from mean value to get the lower limit.
3. Add the answer obtained in step 1 from the mean value to get the upper limit.
4. The range turns out to be 9-27 kg.
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This question is part of the following fields:
- Statistical Methods
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Question 36
Incorrect
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Which of the following statements is TRUE regarding an epidural set?
Your Answer: An 18G Tuohy needle is 8 cm in length
Correct Answer: 19G Tuohy needles have 0.5 cm markings
Explanation:A paediatric 19G Tuohy catheter is available that is 5cm in length and has 0.5cm markings
18G Tuohy catheters are generally 9 to 10cm to hub
Distal end of catheter is angled (15 to 30 degrees) and closed to avoid puncturing the dura
Epidural mesh are usually 0.2 microns and are used to filter bacteria and viruses to ensure sterility of procedure
Transparent catheters are 90cm long with diameters depending on gauge size. It has 1cm graduations from 5 to 20cm to ensure they have been inserted amply and removed completely. Distal end is smooth which can be open or closed (with lateral openings)
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 37
Correct
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A 31-year old Caucasian female came into the emergency department due to difficulty of breathing. History revealed exposure to room odorizes that are rich in alkyl nitrites. Upon physical examination, patient is tachypnoeic at 32 breaths per minute, desaturated at 88% while on a non-rebreather mask at 15 litres per minute oxygen. She was also noted to be cyanotic, however with clear breath sounds.
Considering the history, what is the most probable cause of her difficulty of breathing?Your Answer: Increased affinity of bound oxygen to haemoglobin
Explanation:Amyl nitrate is part of the treatment of cyanide poisoning. The short acting nitrate causes oxidation of Fe2+ in haemoglobin to Fe3+ in methaemoglobin. Methaemoglobin combines with cyanide (cyanmethemoglobin), which reacts with sodium thiosulfate to convert nontoxic thiocyanate and methaemoglobin.
Methaemoglobin is formed when the iron in haemoglobin is converted from the reduced state (Fe2+) to the oxidized state (Fe3+). The oxidized form of haemoglobin (Fe3+) does not bind oxygen as readily as Fe2+, but has high affinity for cyanide. It also results to high affinity of bound oxygen to haemoglobin, thus leading to tissue hypoxia. Arterial oxygen tension is normal despite observations of cyanosis and dyspnoea. Methemoglobinemia can be treated with methylene blue and vitamin C.
Carboxyhaemoglobin can be due to carbon monoxide poisoning. In such cases, patients experience headache and dizziness, but do not develop cyanosis.
2,3-diphosphoglycerate causes a shift in the oxygen dissociation curve to the right, decreasing haemoglobin’s affinity to oxygen to facilitate unloading of oxygen to the tissues.
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This question is part of the following fields:
- Pathophysiology
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Question 38
Correct
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In a study lasting over a period of two years, in which the mean age of 800 patients was 82 years, the efficacy of hip protectors in reducing femoral neck fractures was discussed.
Both experimental and control group had 400 members. Instances of fractures reported over the two year time duration were 10 for the control group (that were prescribed hip protector) and 20 for the control group.
What is the value of Absolute Risk Reduction?Your Answer: 0.025
Explanation:ARR= (Risk factor associated with the new drug group) — (Risk factor associated with the currently available drug)
So,
ARR= (10/400)-(20/400)
ARR= 0.025-0.05
ARR= 0.025 (Numerical Value)
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This question is part of the following fields:
- Statistical Methods
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Question 39
Correct
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A 30-year-old woman admitted following a tonsillectomy has developed stridor with a respiratory rate of 22 breaths per minute and obstructive movements of the chest and abdomen that is in a see-saw pattern .
Her SpO2 is 92% on 60% oxygen with pulse rate 120 beats per minute while her blood pressure is 180/90mmHg. She is repeatedly trying to remove the oxygen mask and appears anxious.
Her pharynx is suctioned and CPAP applied with 100% oxygen via a Mapleson C circuit.
Which of these is the most appropriate next step in her management?Your Answer: Administer intravenous propofol 0.5 mg/kg
Explanation:Continuous closure of the vocal cords resulting in partial or complete airway obstruction is called Laryngospasm. It is a reflex that helps protect against pulmonary aspiration.
Predisposing factors include: Hyperactive airway disease, Insufficient depth of anaesthesia, Inexperience of the anaesthetist, Airway irritation, Smoking, Shared airway surgery and Paediatric patients
Its primary treatment includes checking for blood or stomach aspirate in the pharynx, removing any triggering stimulation, relieving any possible supra-glottic component to airway obstruction and application of CPAP with 100% oxygen.
In this patient, all the above has been done and the next treatment of choice is the administration of a rapidly acting intravenous anaesthetic agent such as propofol (0.5 mg/kg) in increments as it has been reported to relieve laryngospasm in approximately 75% of cases. Administering suxamethonium to an awake patient would be inappropriate at this stage.
Magnesium and lidocaine are used for prevention rather than acute treatment of laryngospasm. Superior laryngeal nerve blocks have been reported to successfully treat recurrent laryngospasm but it is not the next logical step in index patient.
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This question is part of the following fields:
- Pathophysiology
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Question 40
Correct
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Which of the following hormones is secreted by the posterior pituitary?
Your Answer: Oxytocin
Explanation:The posterior pituitary is made up mostly of neural tissue. It is responsible for the storage and release of 2 hormones:
– antidiuretic hormone (ADH)
– oxytocin.These two hormones are synthesised in the supraoptic and paraventricular nuclei of the hypothalamus.
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This question is part of the following fields:
- Pathophysiology
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Question 41
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: French gauge (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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Question 42
Correct
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Which one of the following statement is true regarding United Kingdom gas cylinders?
Your Answer: Tensile tests are performed on sections of one cylinder in every hundred
Explanation:Medical gas cylinders are made up of molybdenum steel but not cast iron. They are checked and assessed at a regular interval.
At least one cylinder in each hundred are tested for tensile, pressure, smash, twist and straightening.
Nitrous Oxide cylinders contain a mixture of liquid and vapour at a pressure of approx. 4500 kPa or 45 Bar. Carbon dioxide cylinder contain gas at the pressure of 5000kPa.
The filling ratio is the ratio of mass of liquified gas in the cylinder to the mass of water required to fill the cylinder at the temperature of 15ºC. In the united kingdom, filling ratio of liquid nitrous oxide is 0.75. The cylinders are usually attached to the anaesthetic machine. As nitrous oxide is an N-methyl-d-aspartate receptor antagonist that may reduce the incidence of chronic post-surgical pain.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 43
Correct
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With regards to oxygen delivery in the body, which of these statements is true?
Your Answer: Anaemia will reduce oxygen delivery
Explanation:Oxygen delivery depends on 2 variables.
1) Content of oxygen in blood
2) Cardiac outputOxygen content (arterial) = (Hb (g/dL) x 1.39 x SaO2 (%) ) + (0.023 x PaO2 (kPa))
Oxygen content (mixed venous) = (Hb (g/dL) x 1.39 x mixed venous saturation) + (0.023 x mixed venous partial pressure of oxygen in kPA)
Huffner’s constant = 1.39 = 1g of Hb binds to 1.39 ml of O2
Oxygen delivery DO2 (ml/min) = 10 x Cardiac output (L/min) x Oxygen content
Normally 1000ml/minOxygen consumption VO2 (ml/min) = 10 x Cardiac output (L/min) x Difference in arterial and mixed venous oxygen content
Normally 250 ml/minOxygen extraction ratio (OER) = VO2/DO2
Normally approximately 25% -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 44
Correct
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What statement about endotoxins is true?
Your Answer: Can often survive autoclaving
Explanation:Endotoxins are the lipopolysaccharides found in the outer cell wall of Gram-negative bacteria. They are responsible for providing the structure and stability of the cell wall.
They cannot be destroyed by normal sterilisation as they are heat stable molecules. They require the use of certain sterilant such as superoxide, peroxide and hypochlorite to be neutralised.
They stimulate strong immune responses, but can only be destroyed partially by specific antibodies. Repeat infections occur as memory T cells cannot be formed.
It can cause septicaemia and associated symptoms such as fever, shock, hypotension and nausea.
It activates the alternative complement pathway and the coagulation pathway using secreted cytokines.
It is not involved in botulism as clostridium botulinum, the responsible organism, secretes a neurotoxic exotoxin.
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This question is part of the following fields:
- Pathophysiology
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Question 45
Correct
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Which among the given choices can be used to describe a persistent and expected level of disease in a particular population?
Your Answer: Endemic
Explanation:Phase 0 trials assist the scientists in studying the behaviour of drugs in humans by micro dosing patients. They are used to speed up the developmental process. They have no measurable therapeutic effect and efficiency.
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This question is part of the following fields:
- Statistical Methods
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Question 46
Correct
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Which of the following is true in the Kreb's cycle?
Your Answer: Alpha-ketoglutarate is a five carbon molecule
Explanation:Krebs’ cycle (tricarboxylic acid cycle or citric acid cycle) is a sequence of reactions to release stored energy through oxidation of acetyl coenzyme A (acetyl-CoA). Some of the products are carbon dioxide and hydrogen atoms.
The sequence of reactions, known collectively as oxidative phosphorylation, only occurs in the mitochondria (not cytoplasm).
The Krebs cycle can only take place when oxygen is present, though it does not require oxygen directly, because it relies on the by-products from the electron transport chain, which requires oxygen. It is therefore considered an aerobic process. It is the common pathway for the oxidation of carbohydrate, fat and some amino acids, required for the formation of adenosine triphosphate (ATP).
Pyruvate enters the mitochondria and is converted into acetyl-CoA. Acetyl-CoA is then condensed with oxaloacetate, to form citrate which is a six carbon molecule. Citrate is subsequently converted into isocitrate, alpha-ketoglutarate, succinyl-CoA, succinate, fumarate, malate and finally oxaloacetate.
The only five carbon molecule in the cycle is Alpha-ketoglutarate.
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This question is part of the following fields:
- Physiology
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Question 47
Correct
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Concerning calcium metabolism and its control, which of these is correct?
Your 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 48
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: Reaction resulting from hereditary C1-inhibitor deficiency
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 49
Incorrect
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Which of the following options will best reflect the adequacy of preoxygenation prior to rapid sequence induction of a patient?
Your Answer: Arterial partial pressure of oxygen (PaO2)
Correct Answer: Expired fraction of oxygen (FEO2)
Explanation:The most important determinant of preoxygenation adequacy is expired fraction of oxygen. Denitrogenating of the functional residual capacity is the purpose of preoxygenation. This is dependent on three vital factors: (1) respiratory rate; (2) inspired volume, and; (3) inspired oxygen concentration (FiO2).
Arterial oxygen saturation does not efficiently determine adequacy of preoxygenation because of its inability to measure tissue reserves. Arterial partial pressure of oxygen is also unsuitable for determining preoxygenation adequacy. Moreover, the absence of central cyanosis is a very crude sign of low tissue oxygenation.
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This question is part of the following fields:
- Pathophysiology
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Question 50
Correct
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At sea level, Sevoflurane is administered via a plenum vaporiser. 100 mL of the fresh gas flow is bypassed into the vaporising chamber. Temperature within the vaporising chamber is maintained at 20°C.
The following fresh gas flows approximates best for the delivery of 1% sevoflurane.Your Answer: 2.7 L/minute
Explanation:The equation for calculating vaporiser output is:
Vaporiser output (VO) mL = Carrier gas flow (mL/minute) × SVP of agent (kPa)
Ambient pressure (kPa) − SVP of agent (kPa)The saturated vapour pressure of sevoflurane at 1 atm (100 kPa) and 20°C is 21 kPa.
VO = (100 mL × 21 kPa)/(100 kPa − 21kPa) for sevoflurane,
VO = 26.6 mL26.6 mL of 100% sevoflurane and 100 mL bypass carrier gas is being added to the fresh gas flow per minute.
2660 mL of 1% sevoflurane and 100 mL bypass carrier gas is approximately 2.7 L/minute.
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This question is part of the following fields:
- Pharmacology
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Question 51
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 52
Correct
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Concerning the intercostal nerves, which one of the following is true?
Your Answer: Each is connected to a ganglion of the sympathetic trunk
Explanation:The intercostal nerves arise from the ventral rami of the first 11 thoracic spinal nerves. they course along the costal groove on the lower margin of the rib.
The twelfth intercoastal nerve is called the subcostal nerve. This is because it is below the 12th rib.
Each intercostal nerve is connected to a ganglion of the sympathetic trunk from which it carries preganglionic and postganglionic fibres that innervate blood vessels, sweat glands, and muscles.
The lateral and medial pectoral nerves innervates pectoralis major muscle.
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This question is part of the following fields:
- Anatomy
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Question 53
Correct
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A 72-year-old woman with a medical history of ischaemic heart disease, hypertension, and hypothyroidism was brought to ER with a change in her mental state over the past few hours. Medications used by her were hydrochlorothiazide, aspirin, ramipril, and levothyroxine.
On physical examination, decreased skin turgor, orthostatic hypotension, and disorientation of time and place were found. There were no significant neurological signs.
Initial biochemical tests are as follows:
Na: 111 mmol/L (135-145)
K: 4.1 mmol/L (3.5-5.1)
Cl: 105 mmol/L (99-101)
Bic: 29 mmol/L (22-29)
Urea: 16.4 mmol/L (1.7-8.3)
Creatinine: 320µmol/L (44-80)
Glucose: 13.5mmol/L (3.5-5.5)
Plasma osmolality: 278mOsm/kg
Urinary osmolality: 450mOsm/kg
TSH: 6.2 miu/L (0.1-6.0)
Free T4: 10.1 pmol/L (10-25)
Free T3: 1.4nm/L (1.0-2.5)
Which of the following is most likely cause for this condition of the patient?Your Answer: Drug idiosyncrasy
Explanation:Based on the laboratory reports, the patient is suffering from significant hyponatremia. The symptoms of hyponatremia are mainly neurological and depend on the severity and rapidity of onset of hyponatremia.
Patient symptom according to the hyponatremia level is correlated below:
125 – 130mmol/L – Nausea and malaise
115 – 125mmol/L – Headache, lethargy, seizures, and coma
<120mmol/L - Up to 11% present with coma. -
This question is part of the following fields:
- Pathophysiology
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Question 54
Correct
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The following foetal anatomical features functionally closes earliest at birth?
Your Answer: Foramen ovale
Explanation:Foramen ovale, ductus arteriosus (DA) and ductus venosus (DV) are the three important cardiac shunts in-utero.
At birth the umbilical vessels constrict in response to stretch as they are clamped. Blood flow through the ductus venosus (DV) decreases but the DV closes passively in 3-10 days.
As the pulmonary circulation is established, there is a drastic fall in pulmonary vascular resistance and an increased pulmonary blood flow. This increases flow and pressure in the Left Atrium that exceeds that of the right atrium. The difference in pressure usually leads to the IMMEDIATE closure of the foramen ovale.
The DA is functionally closed within the first 36-hours of birth in a healthy full-term newborn. Subsequent endothelial and fibroblast proliferation leads to permanent anatomical closure within 2 – 3 weeks.
Oxygenated blood from the placenta passes via the umbilical vein to the liver. Blood also bypasses the liver via the ductus venosus into the inferior vena cava (IVC). The Crista dividens is a tissue flap situated at the junction of the IVC and the right atrium (RA). This flap directs the oxygen-rich blood, along the posterior aspect of the IVC, through the foramen ovale into the left atrium (LA).
The Eustachian valve also known as the valve of The IVC is a remnant of the crista dividens.
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This question is part of the following fields:
- Anatomy
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Question 55
Correct
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A 65-year-old man got operated on for carotid endarterectomy for his carotid artery disease. He is recovering well post-surgery. However, on follow-up in the ward, he has hoarseness of his voice.
Which of the following explains the hoarseness?Your Answer: Damage to the vagus
Explanation:During carotid endarterectomy, injury to the vagus nerve or its branches can cause hoarseness. Injury to the vagus nerve can result in adductor vocal cord paralysis. It can also cause other symptoms like dysphagia or even vocal cord immobility.
Carotid endarterectomy is the procedure to relieve an obstruction in the carotid artery by opening the artery at its origin and stripping off the atherosclerotic plaque with the intima. Because of the internal carotid artery relations, there is a risk of cranial nerve injury during the procedure involving one or more of the following nerves: CN IX, CN X (or its branch, the superior laryngeal nerve), CN XI, or CN XII.
However, only damage to the vagus would account for speech difficulties.
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This question is part of the following fields:
- Anatomy
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Question 56
Correct
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A laceration to the upper lateral margin of the popliteal fossa will pose the greatest risk of injury for which nerve?
Your Answer: Common peroneal nerve
Explanation:The common peroneal (fibular) nerve descends obliquely along the lateral side of the popliteal fossa to the fibular head, medial to biceps femoris.
The sural nerve exits at the fossa’s lower inferolateral aspect and is more at risk in short saphenous vein surgery.
The tibial nerve lies more medially and is even less likely to be injured in this location.
The boundaries of the popliteal fossa are:
Superolateral – the biceps femoris tendon
Superomedial – semimembranosus reinforced by semitendinosus
Inferomedial and inferolateral – medial and lateral heads of gastrocnemiusThe contents of the Popliteal fossa are:
1. The popliteal artery
2. The popliteal vein
3. The Tibial nerve and common Fibular nerve
4. Posterior femoral cutaneous nerve: descends and pierces the roof
5. Small saphenous vein
6. popliteal lymph nodes
7. fat -
This question is part of the following fields:
- Anatomy
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Question 57
Incorrect
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A 70-year-old man presents with bilateral buttock claudication that spreads down the thigh and erectile dysfunction in a vascular clinic.
The left femoral pulse is not palpable on examination, and the right is weakly palpable. Leriche syndrome is diagnosed as the blood flow at the abdominal aortic bifurcation is blocked due to atherosclerosis. He is prepared for aortoiliac bypass surgery.
Which vertebral level will you find the affected artery that requires bypassing?Your Answer: L2
Correct Answer: L4
Explanation:The bifurcation of the abdominal aorta into common iliac arteries occurs at the level of L4. The bifurcation is a common site for atherosclerotic plaques as it is an area of high turbulence.
Leriche Syndrome is an aortoiliac occlusive disease and affects the distal abdominal aorta, iliac arteries, and femoropopliteal vessels. It has a triad of symptoms:
1. Claudication (cramping lower extremities pain that is reproducible by exercise)
2. Impotence (reduced penile arterial flow)
3. Absent/weak femoral pulses (hallmark)T12 – aorta enters the diaphragm with the thoracic duct and azygous veins
L2 – testicular or ovarian arteries branch off the aorta
L3 – inferior mesenteric artery
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This question is part of the following fields:
- Anatomy
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Question 58
Correct
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The immediate physiological response to massive perioperative blood loss is:
Your Answer: Stimulation of baroreceptors in carotid sinus and aortic arch
Explanation:With regards to compensatory response to blood loss, the following sequence of events take place:
1. Decrease in venous return, right atrial pressure and cardiac output
2. Baroreceptor reflexes (carotid sinus and aortic arch) are immediately activated
3. There is decreased afferent input to the cardiovascular centre in medulla. This inhibits parasympathetic reflexes and increases sympathetic response
4. This results in an increased cardiac output and increased SVR by direct sympathetic stimulation. There is increased circulating catecholamines and local tissue mediators (adenosine, potassium, NO2)
5. Fluid moves into the intravascular space as a result of decreased capillary hydrostatic pressure absorbing interstitial fluid.A slower response is mounted by the hypothalamus-pituitary-adrenal axis.
6. Reduced renal blood flow is sensed by the intra renal baroreceptors and this stimulates release of renin by the juxta-glomerular apparatus.
7. There is cleavage of circulating Angiotensinogen to Angiotensin I, which is converted to Angiotensin II in the lungs (by Angiotensin Converting Enzyme ACE)Angiotensin II is a powerful vasoconstrictor that sets off other endocrine pathways.
8. The adrenal cortex releases Aldosterone
9. There is antidiuretic hormone release from posterior pituitary (also in response to hypovolaemia being sensed by atrial stretch receptors)
10. This leads to sodium and water retention in the distal convoluted renal tubule to conserve fluid
Fluid conservation is also aided by an increased amount of cortisol which is secreted in response to the increase in circulating catecholamines and sympathetic stimulation. -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 59
Incorrect
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A new intravenous neuromuscular blocking agent has been developed. It has a hepatic extraction ratio of 0.25 and three quaternary nitrogen atoms in its structure. It has been discovered that it has a half-life of fifteen minutes in healthy volunteers.
Which of the following elimination mechanisms is the most likely to explain this pharmacological behaviour?Your Answer: It is an ester metabolised in the plasma and tissues
Correct Answer: It is filtered and not reabsorbed by the renal tubules
Explanation:The neuromuscular blocking agent is likely to be filtered and not reabsorbed by the renal tubules due to an exclusion process.
Neuromuscular blocking agents that contain one or more quaternary nitrogen atoms are polar and ionised. As a result, the molecules have low lipid solubility, low membrane diffusion capacity, and low distribution volume.
It’s unlikely that a compound with three quaternary nitrogen atoms is an ester. Its high polarity would prevent molecules from moving quickly into tissues.
When drugs have a low hepatic extraction ratio (0.3), the venous and arterial drug concentrations are nearly identical. The liver is not the primary site of drug metabolism.
Therefore:
Changes in liver blood flow have no effect on clearance.
Protein binding, intrinsic metabolism, and excretion are all very sensitive to changes in clearance.
When taken orally, there is no first-pass metabolism.There is no reason for the lungs to eliminate any neuromuscular blocking agent.
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This question is part of the following fields:
- Pharmacology
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Question 60
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 61
Correct
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One litre of water at 0°C and a pressure of 1 bar is in a water-bath. A 1 kW element is used in heating it.
Given that the specific heat capacity of water is 4181 J/(kg°C) or J/(kg K), how long will it take to raise the temperature of the water by 10°C?Your Answer: 42 seconds
Explanation: -
This question is part of the following fields:
- Physiology
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Question 62
Incorrect
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Out of the following, which anatomical structure lies within the spiral groove of the humerus?
Your Answer: Axillary nerve
Correct Answer: Radial nerve
Explanation:The shaft of the humerus has two prominent features:
1. Deltoid tuberosity – attachment for the deltoid muscle
2. Radial or spiral groove – The radial nerve and profunda brachii artery lie in the grooveMid-shaft fractures of the humerus usually occur after a direct blow to the upper arm, which can occur after a fall or RTAs. The most important clinical significance of a mid-shaft humeral fracture is an injury to the radial nerve. The radial nerve originates from the brachial plexus and has roots of C5-T1. It crosses the spiral groove on the posterior side of the shaft of the humerus.
On examination, the patient may have a wrist drop, loss or weakness of finger extension, and decreased or absent sensation to the posterior forearm, digits 1 to 3, and the radial half of the fourth digit.The following parts of the humerus are in direct contact with the indicated
nerves:
Surgical neck: axillary nerve.
Radial groove: radial nerve.
Distal end of humerus: median nerve.
Medial epicondyle: ulnar nerve. -
This question is part of the following fields:
- Anatomy
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Question 63
Incorrect
-
The SI unit of energy is the joule. Energy can be kinetic, potential, electrical or chemical energy.
Which of these correlates with the most energy?Your Answer: Raising the temperature of 1 kg water from 0°C to 100°C (the specific heat capacity of water is 4.2 kJ per kg/°C)
Correct Answer: Energy released when 1 kg fat is metabolised to CO2 and water (the energy content of fat is 37 kJ/g)
Explanation:The derived unit of energy, work or amount of heat is joule (J). It is defined as the amount of energy expended if a force of one newton (N) is applied through a distance of one metre (N·m)
J = 1 kg·m/s2·m = 1 kg·m2/s2 or 1 kg·m2·s-2
Kinetic energy (KE) = ½ MV2
An object with a mass of 1500 kg moving at 30 m/s correlates to 675 kJ:
KE = ½ (1500) × (30)2 = 750 × 900 = 675 kJ
Total energy released when 1 kg fat is metabolised to CO2 and water is 37 MJ. 1 g fat produces 37 kJ/g, therefore 1 kg fat produces 37,000 × 1000 = 37 MJ.
Raising the temperature of 1 kg water from 0°C to 100°C correlates to 420 kJ. The amount of energy needed to change the temperature of 1 kg of the substance by 1°C is the specific heat capacity. We have 1 kg water therefore:
4,200 J × 100 = 420,000 J = 420 kJ
In order to calculate the energy involved in raising a 100 kg mass to a height of 1 km against gravity, we need to calculate the potential energy (PE) of the mass:
PE = mass × height attained × acceleration due to gravity
PE = 100 kg × 1000 m × 10 m/s2 = 1 MJThe heat generated when a direct current of 10 amps flows through a heating element for 10 seconds when the potential difference across the element is 1000 volts can be calculated by applying Joule’s law of heating:
Work done (WD) = V (potential difference) × I (current) × t (time)
WD = 10 × 10 × 1000 = 100 kJ -
This question is part of the following fields:
- Physiology
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Question 64
Correct
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Conclusive evidence suggests that rate for the prevalence of schizophrenia in United Kingdom is around 1%.
Which term can be used to describe that?Your Answer: Endemic
Explanation:An epidemic is declared when the increase in a give disease is above a certain level in a specific interval of time.
An endemic is the general, usual level of a disease in a population at a particular time.
A pandemic is an epidemic that is spread across many countries and continents.
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This question is part of the following fields:
- Statistical Methods
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Question 65
Incorrect
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In medical testing, there are true negative, true positive, false positive and false negative results for some test.
How are the sensitivity of these predictive tests calculated?Your Answer: True negatives / (true negatives + false negatives)
Correct Answer: True positives / (true positives + false negatives)
Explanation:The following terms are used in medical testing:
True negative – The test is negative and the patient does not have the disease.
True positive – The test is positive and the patient has the disease.
False positive – The test is positive but the patient does not have the disease.
False negative – The test is negative but the patient has the disease.The sensitivity of a predictive test = true positives / (true positives + false negatives).
The specificity of a test = true negatives / (false positives + true negatives).
The negative predictive value of a test = true negatives / (false negatives + true negatives).
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This question is part of the following fields:
- Statistical Methods
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Question 66
Incorrect
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A 25 year-old female came to the out-patient department with complaints of vaginal discharge with a distinct fishy odour. She was later diagnosed with bacterial vaginosis and was prescribed to take metronidazole.
The mechanism of action of metronidazole is?Your Answer: Interferes with bacterial RNA synthesis
Correct Answer: Interferes with bacterial DNA synthesis
Explanation:Metronidazole is a nitroimidazole antiprotozoal drug that is selectively absorbed by anaerobic bacteria and sensitive protozoa. Once taken up be anaerobes, it is nonenzymatically reduced by reacting with reduced ferredoxin. This reduction results in products that accumulate in and are toxic to anaerobic cells. The metabolites of metronidazole are taken up into bacterial DNA, forming unstable molecules. This action occurs only when metronidazole is partially reduced, and, because this reduction usually happens only in anaerobic cells, it has relatively little effect on human cells or aerobic bacteria.
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This question is part of the following fields:
- Pharmacology
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Question 67
Correct
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Which compound is secreted only from the adrenal medulla?
Your Answer: Adrenaline
Explanation:The adrenal medulla comprises chromaffin cells (pheochromocytes), which are functionally equivalent to postganglionic sympathetic neurons. They synthesize, store and release the catecholamines noradrenaline (norepinephrine) and adrenaline (epinephrine) into the venous sinusoids.
The majority of the chromaffin cells synthesize adrenaline. -
This question is part of the following fields:
- Anatomy
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Question 68
Incorrect
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Systemic vascular resistance (multiplied by 80) to produce the units of dynes.s.cm-5 is represented by?
Your Answer: Mean pulmonary artery pressure (MPAP) - pulmonary artery occlusion pressure (PAOP)/ cardiac output (CO)
Correct Answer: Mean arterial pressure (MAP) - central venous pressure (CVP)/cardiac output (CO)
Explanation:Systemic vascular resistance (SVR) is a derived value based on:
SVR = (MAP-CVP)/CO x 80
= (60 -10)/5 x 80 = 800 dynes.s.cm-5
A correction factor of 80 is needed in converting mmHg to dynes.s.cm-5
Normal values is between 700 -1600 dynes.s.cm-5Pulmonary resistance (PVR) = (MPAP-PCWP)/CO x 80
= (10 – 5)/5 x 80 = 80 dynes.s.cm-5
To account for body size, cardiac index (CI) can be used instead of CO. CI = CO/body surface area (m2) or mL/minute/m2.
N/B: either MAP and CVP, or MPAP and PCWP are used in calculation to get dynes.s.cm-5 -
This question is part of the following fields:
- Clinical Measurement
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Question 69
Incorrect
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In a diagnosis of a compensated respiratory acidosis, which of the following arterial blood gas results is likely to be seen?
Your Answer: pH = 7.34
PaCO2 = 7.2 kPa
HCO3 = 29Correct Answer:
Explanation:During normal tissue metabolism, there is production of CO2 (acid) which is then expired by the lungs. If metabolism switches from aerobic to anaerobic due to a lack of oxygen, the tissues are unable to completely oxidise sugars to CO2. As a consequence, the sugars can only be partially oxidised to lactic acid. Since lactic acid cannot be expired by the lungs, it remains in the circulation leading to metabolic acidosis.
Also, normal tissue metabolism leads to the production of some amount of acid from the breakdown of proteins. These acids are excreted from the body by kidney filtration. Renal failure will therefore results in acidosis after several days.
An increased acidosis stimulates the brain’s respiratory centres to increase the respiratory rate. This lowers the CO2 in the blood, leading to a decrease in its acidity. Renal excretion removes the excess acid, resulting in a normal pH, and a reduced PaCO2 and HCO3.
pH PaCO2 (kPa) HCO3
Compensated respiratory acidosis 7.34 7.2 29
Acute respiratory acidosis 7.25 7.3 22
Compensated metabolic acidosis 7.34 3.6 14
Metabolic acidosis 7.21 5.3 15
Metabolic alkalosis 7.51 5.1 30 -
This question is part of the following fields:
- Pathophysiology
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Question 70
Incorrect
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Which one of the following pharmacokinetic models is most suitable for target-controlled infusion (TCI) of propofol in paediatric patients?
Your Answer: Minto
Correct Answer: Kataria
Explanation:Marsh (adult) model, when used with children caused over-estimation of plasma concentration. To address this issue Kataria et al developed a three-compartmental model for propofol in children. The pharmacokinetic models used by Target controlled infusion (TCI) systems are used to calculate the relative sizes of the central (vascular), vessel-rich peripheral, and vessel-poor peripheral compartments. The relative volumes of these compartments are different in young children when compared to adults.
Kataria, therefore, is the correct option as described above.
The Maitre model is a three-compartmental model for alfentanil TCI.
The Marsh model describes a propofol TCI model for adults
The Minto model applies to TCI remifentanil.
The Schnider model is also an adult model for propofol that incorporates age and lean body mass as covariates.
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This question is part of the following fields:
- Pharmacology
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Question 71
Correct
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The statement that best describes lactic acidosis is:
Your Answer: It can be precipitated by intravenous fructose
Explanation:An elevated arterial blood lactate level and an increase anion gap ([Na + K] – [Cl + HCO3]) of >20mmol gives rise to lactic acidosis. It can also be a result of overproduction and/or reduced metabolism of lactic acid.
The liver and kidney are the main sites of lactate metabolism, not skeletal muscle.
The two types of lactic acidosis that are known are:
Type A – due to tissue hypoxia, inadequate tissue perfusion and anaerobic glycolysis. These may be seen in cardiac arrest, shock, hypoxaemia and anaemia. The management of type A lactic acidosis involves reversing the underlying cause of the tissue hypoxia.
Type B – occurs in the absence of tissue hypoxia. Some of the causes of this include hepatic failure, renal failure, diabetes mellitus, pancreatitis and infection. Some drugs can also cause this lie aspirin, ethanol, methanol, biguanides and intravenous fructose.
The mainstay of treatment involves:
1. Optimising tissue oxygen delivery
2. Correcting the cause
3. Intravenous sodium bicarbonateIn resistant cases, peritoneal dialysis can be performed.
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This question is part of the following fields:
- Physiology
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Question 72
Incorrect
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Fixed performance devices like high air flow oxygen enrichment (HAFOE) masks have large volumes of air entrained into a flow of 100% oxygen.
The term that best describes the physics behind air entrainment is?Your Answer: Venturi effect
Correct Answer: Bernoulli's principle
Explanation:Bernoulli’s principle states that as the speed of a moving fluid increases, there is a simultaneously decrease in static pressure or a decrease in the fluid’s potential energy.
This is seen in the simultaneous increase in speed and kinetic energy and fall in pressure that causes entrainment of large volumes of air into a flow of 100% oxygen in the nozzle of HAFOE masks.The reduction in fluid pressure that happens when a fluid flows through a constriction in a tube is the Venturi effect.
When a flow of gas or liquid attaches itself to a nearby surface and remains attached even when the surface curves away from the initial direction of flow, this is the Coanda effect.
The branch of engineering and technology that is concerned with the building of devices that use the flow and pressure of a fluid for functions usually performed by electronic devices is Fluidics . Fluidic logic is used to power some ventilators.
The branch of engineering that utilises pressurised gases is Pneumatics.
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This question is part of the following fields:
- Basic Physics
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Question 73
Correct
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A 50-year-old male is planned for elective parotidectomy for pleomorphic adenoma. The surgeon intends to use a nerve integrity monitor thus avoiding neuromuscular blockade. Which of the following nerves is liable to injury in parotidectomy?
Your Answer: Facial nerve
Explanation:Parotidectomy is basically an anatomical dissection. Identification of the facial nerve trunk is essential during parotid gland surgery because facial nerve injury is the most daunting potential complication of parotid gland surgery owing to the close relation between the gland and the extratemporal course of the facial nerve. After exiting the stylomastoid foramen, the facial nerve enters the substance of the parotid gland and then gives off five terminal branches:
From superior to inferior, these are the:
– Temporal branch supplying the extrinsic ear muscles, occipitofrontalis and orbicularis oculi
– Zygomatic branch supplying orbicularis oculi
– Buccal branch supplying buccinator and the lip muscles
– Mandibular branch supplying the muscles of the lower lip and chin
– Cervical branch supplying platysma.There are two approaches to identify the facial nerve trunk during parotidectomy—conventional antegrade dissection of the facial nerve, and retrograde dissection. Numerous soft tissue and bony landmarks have been proposed to assist the surgeon in the early identification of this nerve. Most commonly used anatomical landmarks to identify facial nerve trunk are stylomastoid foramen, tympanomastoid suture (TMS), posterior belly of digastric (PBD), tragal pointer (TP), mastoid process and peripheral branches of the facial nerve.
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This question is part of the following fields:
- Anatomy
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Question 74
Incorrect
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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: 61 ml
Correct 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 75
Incorrect
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A 39-year old man came to the Out-Patient department for symptoms of gastroesophageal reflux disease. Medical history revealed he is on anti-epileptic medication Phenytoin. His plasma phenytoin levels are maintained between 10-12 mcg/mL (Therapeutic range: 10-20 mcg/mL). He is given a H2 antagonist receptor agent (Cimetidine) for his GERD symptoms.
Upon follow-up, his plasma phenytoin levels increased to 38 mcg/mL.
Regarding metabolism and elimination, which of the following best explains the pharmacokinetics of phenytoin at higher plasma levels?Your Answer: Plasma concentration plotted against time is exponential
Correct Answer: Plasma concentration plotted against time is linear
Explanation:Drug elimination is the termination of drug action, and may involve metabolism into inactive state and excretion out of the body. Duration of drug action is determined by the dose administered and the rate of elimination following the last dose.
There are two types of elimination: first-order and zero-order elimination.
In first-order elimination, the rate of elimination is proportionate to the concentration; the concentration decreases exponentially over time. It observes the characteristic half-life elimination, where the concentration decreases by 50% for every half-life.
In zero-order elimination, the rate of elimination is constant regardless of concentration; the concentration decreases linearly over time. A constant amount of the drug being excreted over time, and it occurs when drugs have saturated their elimination mechanisms.
Since phenytoin is observed in elevated levels, the elimination mechanisms for it has been saturated and, thus, will have to undergo zero-order elimination.
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This question is part of the following fields:
- Pharmacology
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Question 76
Correct
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During exercise, muscle blood flow can increase by 20 to 50 times.
Which mechanism is the most important for increased blood flow?Your Answer: Local autoregulation
Explanation:Skeletal muscle blood flow is in the range of 1-4 ml/min per 100 g when at rest. Blood flow can reach 50-100 ml/min per 100 g during exercise. With maximal vasodilation, blood flow can increase 20 to 50 times.
The adrenal medulla releases catecholamines and increases neural sympathetic activity during exercise. Normally, alpha-1 and alpha-2 would cause vasoconstriction in the muscle groups being used, but vasodilatory metabolites override these effects, resulting in a so-called functional sympathectomy. Local hypoxia and hypercarbia, nitric oxide, K+ ions, adenosine, and lactate are some of the stimuli that cause vasodilation.
However, the splanchnic and cutaneous circulations, which supply inactive muscles, vasoconstrict.
Sympathetic cholinergic innervation of skeletal muscle arteries is found in some species (such as cats and dogs, but not humans). Vasodilation is induced by stimulating smooth muscle beta-2 adrenoreceptors, but at rest, the alpha-adrenoreceptor effects of adrenaline and noradrenaline predominate. During exercise, the skeletal muscle pump promotes venous emptying, but it does not necessarily increase blood flow.
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This question is part of the following fields:
- Physiology
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Question 77
Incorrect
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Which measure of central tendency is most useful for a continuous, non-skewed data?
Your Answer: Median
Correct Answer: Mean
Explanation:Mean, also known as the average, is the most common measure of central tendency. It is the sum of all observed values divided by the number of observation. It is not useful for skewed data, which has an abnormal distribution. It is useful, instead, for numerical data that have symmetric distribution. It reflects the contributions of each data in the group, and are sensitive to outliers.
The median is the value that falls in the middle position when the observations are ranked in order from the smallest to the largest. If the number of observations is odd, the median is the middle number. If it is even, the median is the average of the two middle numbers. Unlike the mean, the median is useful on skewed data, and can be used for ordinal or numerical data if skewed.
The mode is the value that occurs with the greatest frequency in a set of observations, and is utilized for bimodal distribution.
The variance and the standard deviation are not measures of central tendency, but of dispersion.
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This question is part of the following fields:
- Statistical Methods
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Question 78
Incorrect
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Metabolization of many drugs used in anaesthesia involves the cytochrome P450 (CYP) isoenzymes.
The CYP enzyme most likely to be subject to genetic variability and thus cause adverse drug reactions is which of these?Your Answer: CYP3A4
Correct Answer: CYP2D6
Explanation:Approximately 25% of phase-1 drug reactions is made responsible by CYP2D6.
As much as a 1,000-fold difference in the ability to metabolise drugs by CYP2D6 can happen between phenotypes, and this may result in adverse drug reactions (ADRs).
The metabolism of antiemetics, beta-blockers, codeine, tramadol, oxycodone, hydrocodone, tamoxifen, antidepressants, neuroleptics, and antiarrhythmics is also as a result of CYP2D6.
Patients who take drugs that are metabolised by CYP2D6 but have poor CYP2D6 metabolism are more likely to have ADRs. People with ultra-rapid CYP2D6 metabolism may have a decreased drug effect due to low plasma concentrations of these drugs.
All the other CYP enzymes are subject to genetic polymorphism. Variants are less likely to lead to adverse drug reactions.
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This question is part of the following fields:
- Physiology
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Question 79
Correct
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One of the causes of increased pulse pressure is when the aorta becomes less compliant because of age-related changes. Another cause of increased pulse pressure is which of the following?
Your Answer: Increased stroke volume
Explanation:Impaired ventricular relaxation reduces diastolic filling and therefore preload.
Decreased blood volume decreases preload due to reduced venous return.
Heart failure is characterized by reduced ejection fraction and therefore stroke volume.
Cardiac output = stroke volume x heart rate
Left ventricular ejection fraction = (stroke volume / end diastolic LV volume ) x 100%
Stroke volume = end diastolic LV volume – end systolic LV volume
Pulse pressure (is increased by stroke volume) = Systolic Pressure – Diastolic Pressure
Systemic vascular resistance = mean arterial pressure / cardiac output
Factors that increase pulse pressure include:
-a less compliant aorta (this tends to occur with advancing age)
-increased stroke volume
Aortic stenosis would decrease stroke volume as end systolic volume would increase.
This is because of an increase in afterload, an increase in resistance that the heart must pump against due to a hard stenotic valve. -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 80
Correct
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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: 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 81
Incorrect
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All of the following statements about pH electrode are incorrect except:
Your Answer: Produces a non-linear output of 60 millivolts per unit pH
Correct Answer: A semi-permeable membrane reduces protein contamination
Explanation:Pulse oximeters combine the principles of oximetry and plethysmography to noninvasively measure oxygen saturation in arterial blood. A sensor containing two or three light emitting diodes and a photodiode is placed across a perfused body part, commonly a finger, to be transilluminated. Oximetry depends on oxyhaemoglobin and deoxyhaemoglobin, and their ability to absorb the beams of light produced by the light emitting diodes: red light at 660 nm and infrared light at 960 nm.
The isosbestic point is the point wherein two different substances absorb light to the same extent. For oxyhaemoglobin and deoxyhaemoglobin, the points are at 590 nm and 805 nm. These are considered reference points where light absorption is independent of the degree of saturation.
Non-constant absorption of light is often due to the presence of an arterial pulsation, whilst constant absorption of light is seen in non-pulsatile tissues.
Most pulse oximeters are inaccurate at low SpO2, but is accurate at +/- 2% within the range of 70% to 100% SpO2. All pulse oximeters demonstrate a delay in between changes in SaO2 and SpO2, and display average readings every 10 to 20 seconds, hence they are unable to detect acute desaturation episodes.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 82
Correct
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Regarding the treatment of bladder cancer, a study concerned with the usage of a combined or monotherapy was conducted. A forest plot was used for the visual representation of the data.
Which of the following is true regarding forest plots?Your Answer: Forest plots can present data from multiple studies
Explanation:Being the part of a meta analysis, forest plots are more valued as evidence then randomised control trials.
The notion that forest plots can only be used if the results are substantial is not true. They are good indicators of the significance of the data. If the diamond intersects the central line, the data is rendered significant. It also aggregates means and confidence intervals from studies conducted in the past which makes the study much more reliable as errors associated with individual studies tend to have less of an impact in this way.
The suggestion that forest plots are primarily used for qualitative data is factually incorrect. Forest plots require numerical values to function.
All in all, forest plots help us in determining whether or not there is a significant trend in that particular field of study.
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This question is part of the following fields:
- Statistical Methods
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Question 83
Incorrect
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A 50-year-old woman's blood pressure readings in the clinic are 170/109 mmHg, 162/100 mmHg and 175/107 mmHg and her routine haematology, biochemistry, and 12-lead ECG are normal.
She is assessed on the day of surgery prior to laparoscopic inguinal hernia repair and is found to be normally fit and well. Documentation of previous blood pressure measurements from her general practitioner in the primary healthcare setting are not available.
What is your next course of action?Your Answer: Postpone surgery until further biochemical tests to exclude secondary hypertension
Correct Answer: Proceed with scheduled surgery without treatment
Explanation:The AAGBI and the British Hypertension Society has published guidelines for the measurement of adult blood pressure and management of hypertension before elective surgery.
The objective is to ensure that patients admitted for elective surgery have a known systolic blood pressure below 160 mmHg and diastolic blood pressures below 100 mmHg. The primary health care teams, if possible, should ensure that this is the case and provide evidence to the pre-assessment clinic staff or on admission.
Avoiding cancellation on the day of surgery because of white coat hypertension is a secondary objective.
Patients with blood pressures below 180 mmHg systolic and 110 mmHg diastolic (measured in the preop assessment clinic), who present to pre-operative assessment clinics without documented evidence of primary care blood pressures should proceed to elective surgery.
In this question, the history/assessment does not appear to point to obvious end-organ damage so there is no indication for further investigation for secondary causes of hypertension or an echocardiogram at this point. Further review and treatment at this point is not required.
However, you should write to the patient’s GP and encourage serial blood pressure measurements in the primary health care setting.
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This question is part of the following fields:
- Pathophysiology
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Question 84
Correct
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The passage of glucose into the brain is facilitated by which transport method?
Your Answer: Facilitated diffusion
Explanation:Glucose transport is a highly regulated process accomplished mostly by facilitated diffusion using carrier proteins to cross cell membranes.
There are many transporters, but the most important are known as glucose transporters (GLUTs).
Stresses in various form of acute and chronic forms affect the activity of glucose transporters.
They are responsive to many types of metabolic stress, including hypoxia, injury, hypoglycaemia, numerous metabolic inhibitors, stress hormones, and other influences such as growth factors.Numerous signalling pathways appear to be involved in transporter regulation.
New evidence suggests that stresses regulating GLUTs are not only acute biological stresses. In addition, chronic low-grade inflammation, and their associated chronic diseases also lead to altered glucose transport. These include obesity, type 2 diabetes, cardiovascular disease, and the growth and spread of many tumours that are affected by altered glucose transporters. Some of these glucose transport effects are compensatory, while others are pathogenic.
Ultimately, deliberate manipulation of GLUTs could be used as treatment for some of these chronic diseases.
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This question is part of the following fields:
- Physiology
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Question 85
Correct
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Which of the following is true regarding correlation coefficient?
Your Answer: It can assume any value between -1 and 1
Explanation:The degree of correlation is summarised by the correlation coefficient (r). This indicates how closely the points lie to a line drawn through the plotted data. In parametric data this is called Pearson’s correlation coefficient and can take any value between -1 to +1. A correlation of -1.0 indicates a perfect negative correlation, and a correlation of 1.0 indicates a perfect positive correlation.
For example
r = 1 – strong positive correlation (e.g. systolic blood pressure always increases with age)
r = 0 – no correlation (e.g. there is no correlation between systolic blood pressure and age)
r = – 1 – strong negative correlation (e.g. systolic blood pressure always decreases with age)
Whilst correlation coefficients give information about how one variable may increase or decrease as another variable increases they do not give information about how much the variable will change. They also do not provide information on cause and effect.
In contrast to the correlation coefficient, linear regression may be used to predict how much one variable changes when a second variable is changed.
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This question is part of the following fields:
- Statistical Methods
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Question 86
Correct
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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: 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 87
Incorrect
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An 84-year-old woman has a fall. She fractures the neck of her femur and requires emergency surgery.
On history and examination, she appears to also have a possible heart failure for which an echocardiogram is scheduled.
Her measurements are:
End-diastolic volume: 40mL (70-240)
End-systolic volume: 30mL (16-140)
Calculate her approximate ejection fraction.Your Answer: 65%
Correct Answer: 25%
Explanation:An echocardiogram provides real-time visualisation of cardiac structures. The ejection fraction (EF) is normally measured using this system.
The ejection fraction (EF) can be deduced mathematically if the patient’s end-diastolic volume (EDV), end-systolic volume (ESV) and stroke volume (SV) are known, as:
SV = EDV – ESV, and
EF = SV/EDV x 100
The normal range for EF is >55-70%.
For this patient,
SV= 40 – 30 = 10 mL, therefore
EF = 10/40 x 100 = 25%
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This question is part of the following fields:
- Clinical Measurement
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Question 88
Incorrect
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All of the following statements about that parasympathetic nervous system (PNS) are true except:
Your Answer: Facial nerve supplies the submandibular, sublingual and lacrimal glands via the pterygopalatine and submandibular ganglions
Correct Answer: The PNS has nicotinic receptors throughout the system
Explanation:With regards to the autonomic nervous system (ANS)
1. It is not under voluntary control
2. It uses reflex pathways and different to the somatic nervous system.
3. The hypothalamus is the central point of integration of the ANS. However, the gut can coordinate some secretions and information from the baroreceptors which are processed in the medulla.With regards to the central nervous system (CNS)
1. There are myelinated preganglionic fibres which lead to the
ganglion where the nerve cell bodies of the non-myelinated post ganglionic nerves are organised.
2. From the ganglion, the post ganglionic nerves then lead on to the innervated organ.Most organs are under control of both systems although one system normally predominates.
The nerves of the sympathetic nervous system (SNS) originate from the lateral horns of the spinal cord, pass into the anterior primary rami and then pass via the white rami communicates into the ganglia from T1-L2.
There are short pre-ganglionic and long post ganglionic fibres.
Pre-ganglionic synapses use acetylcholine (ACh) as a neurotransmitter on nicotinic receptors.
Post ganglionic synapses uses adrenoceptors with norepinephrine / epinephrine as the neurotransmitter.
However, in sweat glands, piloerector muscles and few blood vessels, ACh is still used as a neurotransmitter with nicotinic receptors.The ganglia form the sympathetic trunk – this is a collection of nerves that begin at the base of the skull and travel 2-3 cm lateral to the vertebrae, extending to the coccyx.
There are cervical, thoracic, lumbar and sacral ganglia and visceral sympathetic innervation is by cardiac, coeliac and hypogastric plexi.
Juxta glomerular apparatus, piloerector muscles and adipose tissue are all organs under sole sympathetic control.
The PNS has a craniosacral outflow. It causes reduced arousal and cardiovascular stimulation and increases visceral activity.
The cranial outflow consists of
1. The oculomotor nerve (CN III) to the eye via the ciliary ganglion,
2. Facial nerve (CN VII) to the submandibular, sublingual and lacrimal glands via the pterygopalatine and submandibular ganglions
3. Glossopharyngeal (CN IX) to lungs, larynx and tracheobronchial tree via otic ganglion
4. The vagus nerve (CN X), the largest contributor and carries ¾ of fibres covering innervation of the heart, lungs, larynx, tracheobronchial tree parotid gland and proximal gut to the splenic flexure, liver and pancreasThe sacral outflow (S2 to S4) innervates the bladder, distal gut and genitalia.
The PNS has long preganglionic and short post ganglionic fibres.
Preganglionic synapses, like in the SNS, use ACh as the neuro transmitter with nicotinic receptors.
Post ganglionic synapses also use ACh as the neurotransmitter but have muscarinic receptors.Different types of these muscarinic receptors are present in different organs:
There are:
M1 = pupillary constriction, gastric acid secretion stimulation
M2 = inhibition of cardiac stimulation
M3 = visceral vasodilation, coronary artery constriction, increased secretions in salivary, lacrimal glands and pancreas
M4 = brain and adrenal medulla
M5 = brainThe lacrimal glands are solely under parasympathetic control.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 89
Incorrect
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Which of the following statement is false regarding dopamine?
Your Answer: Pulmonary vascular resistance is increased
Correct Answer: Urine output decreases due to inhibition of proximal tubule Na+ reabsorption
Explanation:Dopamine (DA) is a dopaminergic (D1 and D2) as well as adrenergic ? and?1 (but not ?2 ) agonist.
The D1 receptors in renal and mesenteric blood vessels are the most sensitive: i.v. infusion of a low dose of Dopamine dilates these vessels (by raising intracellular cAMP). This increases g.f.r. In addition, DA exerts a natriuretic effect by D1 receptors on proximal tubular cells.
Moderately high doses produce a positive inotropic (direct?1 and D1 action + that due to NA release), but the little chronotropic effect on the heart.
Vasoconstriction (?1 action) occurs only when large doses are infused.
At doses normally employed, it raises cardiac output and systolic BP with little effect on diastolic BP. It has practically no effect on nonvascular ? and ? receptors; does not penetrate the blood-brain barrier—no CNS effects.
Dopamine is less arrhythmogenic than adrenaline
Regarding dopamine part of the dose is converted to Noradrenaline in sympathetic nerve terminals.
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This question is part of the following fields:
- Pharmacology
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Question 90
Correct
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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, 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 91
Correct
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Which of these thyroid hormones is considered the most potent and most physiologically active?
Your Answer: T3
Explanation:Triiodothyronine (T3) is more potent than thyroxine (T4). It is able to bind to more receptors (90%) compared to T4 (10%), and the onset of action is more immediate (within 12 hours) than T4 (2 days).
Ninety-three percent of thyroid hormones synthesized is T4, and the remaining 7% is T3. The half-life of T3 is shorter (1 day), and its affinity for thyroxine-binding globulin is lower than T4.
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This question is part of the following fields:
- Pathophysiology
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Question 92
Incorrect
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Drug A has a 1 L/kg volume of distribution and a 0.1 elimination rate constant (k).
Drug B has a 2 L/kg volume of distribution and a 0.2 elimination rate constant (k).
Which of the following statements best describes the pharmacokinetics of drug A in a single compartment?Your Answer: Drug A has the same clearance as drug B
Correct Answer: Drug A has a lower clearance than drug B
Explanation:The fall in plasma concentration of a drug with time decreases exponentially in a single compartment pharmacokinetic model (wash-out curve).
A straight line is produced when the logarithm (ln) of a drug’s plasma concentration is plotted against time because a constant proportion of the drug is removed from the plasma per unit time. The line’s gradient or slope can be expressed mathematically as k. (the rate constant). The gradient is related to the half life (T1/2) because it can be used to predict a drug’s plasma concentration at any time.
According to the following formula, clearance (CL), volume of distribution (Vd), and elimination rate constant (k) are mathematically related.
CL = Vd x k
For drug A, CL = 1 x 0.1 = 0.1units per minute
For drug B, Cl = 2 x 0.2 = 0.4 units per minute
Hence, it is proved that Drug A has a lower clearance than drug B.
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This question is part of the following fields:
- Pharmacology
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Question 93
Correct
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A 25-year old man needs an emergency appendicectomy and has gone to the operating room. During general anaesthesia, ventilation is achieved using a circle system with a fresh gas flow (FGF) of 1L/min, with and air/oxygen and sevoflurane combination. The capnograph trace is normal.
Changes to the end tidal and baseline CO2 measurements at 10 and 20 mins respectively are seen on the capnograph below:
10 minutes 20 minutes
End-tidal CO2 4.9 kPa 8.4 kPa
Baseline end-tidal CO2 0.2 kPa 2.4 kPa
The other vitals were as follows:
Pulse 100-105 beats per minute
Systolic blood pressure 120-133 mmHg
O2 saturation 99%.
The next most important immediate step is which of the following?Your Answer: Increase the FGF
Explanation:This scenario describes rebreathing management.
Changes is exhaustion of the soda lime and a progressive rise in circuit deadspace is the most likely explanation for the capnograph.
It is important that the soda lime canister is inspected for a change in colour of the granules. Initially fresh gas flow should be increased and then if necessary, replace the soda lime granules. Other strategies include changing to another circuit or bypassing the soda lime canister after the fresh gas flow is increased.
Any other causes of increased equipment deadspace should be excluded.
Intraoperative hypercarbia can be caused by:
1. Hypoventilation – Breathing spontaneously; drugs which include anaesthetic agents, opioids, residual neuromuscular blockade, pre-existing respiratory or neuromuscular disease and cerebrovascular accident.
2. Controlled ventilation- circuit leaks, disconnection, miscalculation of patient’s minute volume.
3. Rebreathing – Soda lime exhaustion with circle, inadequate fresh gas flow into Mapleson circuits, increased breathing system deadspace.
4. Endogenous source – Tourniquet release, hypermetabolic states (MH or thyroid storm) and release of vascular clamps.
5. Exogenous source – Absorption of CO2 absorption from the pneumoperitoneum. -
This question is part of the following fields:
- Physiology
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Question 94
Incorrect
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You are given an intravenous induction agent. The following are its characteristics:
A racemic mixture of cyclohexanone rings with one chiral centre
Local anaesthetic properties.
Which of the following statements about its primary mechanism of action is most accurate?Your Answer: Reversible competitive antagonist affecting Na+ channels
Correct Answer: Non-competitive antagonist affecting Ca2+ channels
Explanation:Ketamine is the substance in question. Its structure and pharmacodynamic effects make it a one-of-a-kind intravenous induction agent. The molecule is made up of two cyclohexanone rings (2-(O-chlorophenyl)-2-methylamino cyclohexanone and 2-(O-chlorophenyl)-2-methylamino cyclohexanone). Ketamine has local anaesthetic properties and acts primarily on the brain and spinal cord.
It affects Ca2+ channels as a non-competitive antagonist for the N-D-methyl-aspartate (NMDA) receptor. It also acts as a local anaesthetic by interfering with neuronal Na+ channels.
Ketamine causes profound dissociative anaesthesia (profound amnesia and analgesia) as well as sedation.
Phenoxybenzamine, an alpha-1 adrenoreceptor antagonist, is an example of an irreversible competitive antagonist. It forms a covalent bond with the calcium influx receptor.
Benzodiazepines are GABAA receptor agonists that affect chloride influx.
Flumazenil is an inverse agonist that affects GABAA receptor chloride influx.
Ketamine is a cyclohexanone derivative that acts as a non-competitive Ca2+ channel antagonist.
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This question is part of the following fields:
- Pharmacology
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Question 95
Incorrect
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General anaesthesia is administered to a patient in a hospital in Lhasa which is one of the highest cities in the world (at 11,975 feet). An Anaesthetic rotameter is normally calibrated at 20 C and 1 bar pressure and is known to be underread at altitude. The temperature of the theatre was 10 C.
Which one of the following physical properties is responsible for the rotameter inaccuracy in these conditions?Your Answer: Density of the bobbin
Correct Answer: Density of the gas
Explanation:Since the gas is less dense at higher altitudes, the density of a gas influences flows when passing through the orifice. Due to this reason, for a given flow rate, the bobbin will not be forced as far up the rotameter tube.
At higher altitudes, the volume of a fixed mass of gas increases, and therefore the molecules of gas are widely spaced resulting in a decrease in density with an increase in altitude.
Viscosity is simply termed as friction of gas. The viscosity of a gas is important only at low flow rates when the flow characteristic of the gas is laminar.
Charle’s law stated that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature (T) provided the pressure remains constant.
Boyle’s law for a fixed amount of gas at constant temperature, the pressure (P) and volume (V) are inversely proportional.
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This question is part of the following fields:
- Basic Physics
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Question 96
Correct
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A study of blood pressure measurements is being performed in patients with chronic kidney disease.
Considering that the results are normally distributed, what percentage of values lie within two standard deviations of the mean blood pressure reading?Your Answer: 95.40%
Explanation:Normal distribution, also called Gaussian distribution, the most common distribution function for independent, randomly generated variables, and describes the spread for many biological and clinical measurements.
Properties of the Normal distribution
symmetrical i.e. Mean = mode = median
68.3% of values lie within 1 SD of the mean
95.4% of values lie within 2 SD of the mean
99.7% of values lie within 3 SD of the mean
The empirical rule, or the 68-95-99.7 rule, tells you where most of the values lie in a normal distribution: Around 68% of values are within 1 standard deviation of the mean.
Around 95% of values are within 2 standard deviations of the mean. Around 99.7% of values are within 3 standard deviations of the mean.
the standard deviation (SD) is a measure of how much dispersion exists from the mean.SD = square root (variance)
The empirical rule, or the 68-95-99.7 rule states where most of the values lie in a normal distribution. Around 68% of values fall within 1 S.D of the mean, about 95% within 2 S.D of the mean, and about 99.7% of values within 3 S.D of the mean. Therefore, 95.4% is the most reasonable answer if results are normally distributed.
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This question is part of the following fields:
- Statistical Methods
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Question 97
Incorrect
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A graph is created to show the exponential relationship between bacterial growth (y-axis) and time (x-axis).
Which of the following statements is most true about this kind of exponential relationship?Your Answer: The negative x axis is a horizontal asymptote
Correct Answer: y = ex
Explanation:The relationship between bacterial growth and time is a tear-away exponential. The mathematical relationship between y and x in this case is:
y = ex
Where: the power is x, and the base is e.
Euler’s number (e) is a mathematical constant that is the base for all logarithms occurring naturally. Its value is 2.718.
The statement X increasing with an increase in Y is proportional to Y refers to the change in y in terms of x when considering any exponential relationship.
This is not a build-up exponential, and that is mathematically stated as y = 1-e-kt.
The negative x axis being a horizontal asymptote and the y intercept being 0, 1 are examples of tearaway exponentials , but do not describe an exponential process.
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This question is part of the following fields:
- Statistical Methods
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Question 98
Correct
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This vertebrae can be easily differentiated from the rest because of its prominent spinous process.
Your Answer: C7
Explanation:The spinous process is the part of a vertebrae that is directed posteriorly.
Typical cervical vertebra have spinous processes that are small and bifid, except for C7, which has a long and prominent spinous process.
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This question is part of the following fields:
- Anatomy
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Question 99
Correct
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A 64-year-old man is admitted to the critical care unit. He has a recent medical history of faecal peritonitis for which a laparotomy was performed. His vitals have been monitored using an invasive pulmonary artery flotation catheter.
His vital readings are:
Temperature: 38.1°C
Blood pressure: 79/51 mmHg (mean 58 mmHg)
Pulmonary artery pressure: 19/6 mmHg (mean 10 mmHg)
Pulmonary capillary occlusion pressure: 5 mmHg
Central venous pressure: 12 mmHg
Cardiac output: 5 L/min
Mixed venous oxygen saturation: 82%
Calculate his approximate pulmonary vascular resistance.
Note: A correction factor of 80 is require to convert mmHg to dynes·s·cm-5Your Answer: 80 dynes·s·cm-5
Explanation:Pulmonary vascular resistance (PVR) refers to the resistance to blood flow to the left atrium from the pulmonary artery.
It is derived mathematically by:PVR = MPAP – PCWP
CO
where,
MPAP: Mean pulmonary artery pressure
PCWP: Pulmonary capillary occlusion pressure
CO: Cardiac outputFor this patient:
PVR = 10 – 5 = 1mmHg
5Remember, multiply by correction factor 80 to change units:
PVR = 1mmHg x 80 = 80 dynes·s·cm-5
Normal values range between 20-130 dynes·s·cm-5
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This question is part of the following fields:
- Clinical Measurement
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Question 100
Incorrect
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An arterial pressure transducer is supposedly in direct correlation to change, thus it is dependent on zero gradient drift and zero offset. Which of the following values will best compensate for the gradient drift?
Your Answer: 200 mmHg and 760 mmHg
Correct Answer: 0 mmHg and 200 mmHg
Explanation:Since an arterial pressure transducer, and every other measuring apparatus, is prone to errors due to offset and gradient drifts, regular calibration is required to maintain accuracy of the instrument. The two-point calibration pressure values of 0 mmHg and 200 mmHg are within the physiologic range and can best compensate for the gradient drift.
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This question is part of the following fields:
- Clinical Measurement
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