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Question 1
Correct
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A 77-year-old woman is scheduled for day case cataract surgery under local anaesthesia. She has no cardiac or respiratory problems. Lisinopril is being used to treat her hypertension, which is under control.
Which of the following preoperative investigations are the most appropriate for this patient?Your Answer: No investigations
Explanation:Because the patient has mild systemic disease, he is ASA 2 and the procedure will be performed under local anaesthesia.
The following factors should be considered when requesting preoperative investigations:
Indications derived from a preliminary clinical examination
Whether or not a general anaesthetic will be used, the possibility of asymptomatic abnormalities, and the scope of the surgery.No special investigations are needed if the patient has no history of significant systemic disease and no abnormal findings on examination during the nurse-led assessment.
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This question is part of the following fields:
- Clinical Measurement
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Question 2
Incorrect
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An 82-year old male has shortness of breath which is made worse when he lies down but investigations have revealed a normal ejection fraction. Why might this be?
Your Answer: He has increased atrial compliance
Correct Answer: He has diastolic dysfunction
Explanation:Decreased stroke volume causes decreased ejection fraction which results in diastolic dysfunction.
Ejection fraction is not a useful measure in someone with diastolic dysfunction because stroke volume may be reduced whilst end-diastolic volume may be reduced.
Diastolic dysfunction may arise with reduced heart compliance.Ejection fraction measures of the proportion of blood leaving the ventricles with each beat and is calculated as follows:
Stroke volume / end-diastolic volume.A healthy ejection fraction is usually taken as 60% (based on a stroke volume of 70ml and end-diastolic volume of 120ml).
Respiratory inspiration causes a decreased pressure in the thoracic cavity, which in turn causes more blood to flow into the atrium.
Sitting up decreases venous because of the action of gravity on blood in the venous system.
Hypotension also decreases venous return.
A less compliant aorta, like in aortic stenosis increases end systolic left ventricular volume which decreases stroke volume.Systemic vascular resistance = mean arterial pressure / cardiac output.
Increased vascular resistance impedes the flow of blood back to the heart.Increased venous return increases end diastolic LV volume as there is more blood returning to the ventricles.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 3
Correct
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A morbidly obese (BMI=48) patient has the following co-morbidities: type II diabetes mellitus and hypertension. It is recommended for the patient to undergo bariatric surgery.
If the patient is laid flat for induction of anaesthesia, what physiologic changes of the respiratory system is the most important to consider?Your Answer: Functional residual capacity will decrease
Explanation:A decrease in the functional residual capacity (FRC) is the most important physiologic change to consider for such patients.
FRC is the sum of the expiratory reserve volume and the residual volume. It is the resting volume of the lung, and is an important marker for lung function. During this time, the alveolar pressure is equal to the atmospheric pressure. When morbidly obese individuals lie supine, the FRC decreases by as much as 40% because the abdominal contents push the diaphragm into the thoracic cavity.
Chest wall compliance is expected to reduce because of fat deposition surrounding adjacent structures.
Inspiratory reserve volume (IRV) is expected to increase, and peak expiratory flow is expected to decrease, however the decrease in FRC is more important to consider because of the risk of hypoxia secondary to premature airway closure and ventilation-perfusion mismatch.
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This question is part of the following fields:
- Physiology
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Question 4
Incorrect
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Which of the following does Lidocaine 1% solution equate to?
Your Answer: 1000 mcg per 100 ml
Correct Answer: 1000 mg per 100 ml
Explanation:Lidocaine 1% is formulated as 1000 mg/100 mL.
% solution is based on (grams of medicine) / 100 ml
% solution ~ (1000 mg) / 100 ml
% solution ~ 10 mg/ml
Examples:
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- Lidocaine 4% = 40 mg/ml of Lidocaine
- Lidocaine 2% = 20 mg/ml of Lidocaine
- Lidocaine 1% = 10 mg/ml of Lidocaine
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This question is part of the following fields:
- Pharmacology
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Question 5
Incorrect
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Which of these anaesthetics has the best chance of preventing HPV (hypoxic pulmonary vasoconstriction)?
Your Answer: Isoflurane 1 MAC
Correct Answer: Desflurane 2 MAC
Explanation:Resistance pulmonary arteries constrict in response to alveolar and airway hypoxia, diverting blood to better-oxygenated alveoli.
In atelectasis, pneumonia, asthma, and adult respiratory distress syndrome, hypoxic pulmonary vasoconstriction optimises O2 uptake. Hypoxic pulmonary vasoconstriction helps maintain systemic oxygenation during single-lung anaesthesia.
A redox-based O2 sensor within pulmonary artery smooth muscle cells is involved in hypoxic pulmonary vasoconstriction. The production of reactive oxygen species by smooth muscle cells in the pulmonary artery varies in proportion to PaO2. Hypoxic removal of these redox second messengers inhibits voltage-gated potassium channels, depolarizing smooth muscle cells in the pulmonary artery.
L-type calcium channels are activated by depolarization, which raises cytosolic calcium and causes hypoxic pulmonary vasoconstriction. Some anaesthetics suppress this response, increasing the risk of further deterioration in ventilation perfusion mismatch.
Agents that inhibit HPV are ether, halothane, and desflurane (>1.6 MAC).
Agents with no effect on HPV include thiopentone, fentanyl, desflurane (1MAC), isoflurane (<1.5MAC), sevoflurane(1MAC), and propofol. -
This question is part of the following fields:
- Pharmacology
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Question 6
Correct
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Which of the following best explains the association between smoking and lower oxygen delivery to tissues?
Your Answer: Left shift of the oxygen dissociation curve
Explanation:Smoking is a major risk factor associated with perioperative respiratory and cardiovascular complications. Evidence also suggests that cigarette smoking causes imbalance in the prostaglandins and promotes vasoconstriction and excessive platelet aggregation. Two of the constituents of cigarette smoke, nicotine and carbon monoxide, have adverse cardiovascular effects. Carbon monoxide increases the incidence of arrhythmias and has a negative ionotropic effect both in animals and humans.
Smoking causes an increase in carboxyhaemoglobin levels, resulting in a leftward shift in which appears to represent a risk factor for some of these cardiovascular complications.
There are two mechanisms responsible for the leftward shift of oxyhaemoglobin dissociation curve when carbon monoxide is present in the blood. Carbon monoxide has a direct effect on oxyhaemoglobin, causing a leftward shift of the oxygen dissociation curve, and carbon monoxide also reduces the formation of 2,3-DPG by inhibiting glycolysis in the erythrocyte. Nicotine, on the other hand, has a stimulatory effect on the autonomic nervous system. The effects of nicotine on the cardiovascular system last less than 30 min.
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This question is part of the following fields:
- Physiology
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Question 7
Incorrect
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The most abundant intracellular ion is?
Your Answer: Calcium
Correct Answer: Phosphate
Explanation:Phosphate is the principal anion of the intracellular fluid, most of which is bound to either lipids or proteins. They dissociate or associate with different compounds, depending on the enzymatic reaction, thus forming a constantly shifting pool.
Calcium and magnesium are also present intracellularly, however in lesser amounts than phosphate.
Sodium is the most abundant extracellular cation, and Chloride and is the most abundant extracellular anion.
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This question is part of the following fields:
- Physiology
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Question 8
Incorrect
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Which of these statements is true about spirometry?
Your Answer: Residual volume can not be measured using a body plethysmograph
Correct Answer: A capacity is the sum of two or more volumes
Explanation:Functional residual capacity (FRC) is 1.7 to 3.5L/kg
A capacity is the sum of two or more volumes. The total lung capacity (TLC) is total sum of the volume of gas present in all lung compartments upon maximum inspiration. It is represented mathematically as:
Total lung capacity (TLC) = Vital capacity (VC) + Residual volume (RV)
The residual volume (RV) is the volume of gas still present within the lung post maximum exhalation. It cannot be measured by spirometry, but can be using a body plethysmograph and also with the helium dilution technique.
Closing capacity (CC) is the volume of gas within the lungs at which small airways close upon expiration. It increases with age and is especially important when it surpasses the FRC as it causes changes in ventilation/perfusion mismatch and hypoxia.
In the supine position, a patient with a normal body mass index and no history of lung pathology, the CC equals the FRC at approximately 44, and at approximately 66 at standing position. -
This question is part of the following fields:
- Clinical Measurement
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Question 9
Correct
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A 71-year-old man, presents with central crushing pain in his chest to the emergency department. On examination, he complains of nausea and is notably sweating. On ECG, elevation in the ST-segment is noted in multiple chest leads, as well as sinus bradycardia. A myocardial infarction can cause a sinus bradycardia.
The sinoatrial (SA) node and the atrioventricular (AV) node receive arterial supply from which vessel?Your Answer: Right coronary artery
Explanation:The left marginal artery comes off the left circumflex artery, and runs alongside the heart.
The left circumflex artery is one of the bifurcations of the left coronary artery, and eventually forms the left marginal artery.
An occlusion in the left circumflex artery often results in a lateral MI.
The right marginal artery originates from the right coronary artery.
The left anterior descending artery (LAD) is another bifurcation of the left coronary artery. An occlusion in the LAD would often result in an anteroseptal MI as is diagnosed on ECG by noting changes in leads V1-V4.
The right coronary artery originates from the right aortic sinus of the ascending aorta, and bifurcates to give rise to many branches, including the sinoatrial artery which supplies the sinoatrial (SA) node in 50-70% of cases, the artery of the atrioventricular (AV) node in 50-60% of cases, the right acute marginal artery which supplies the right ventricle. It also supplies the right atrium, interatrial septum and the posterior inferior third of the interventricular septum.
Arrhythmias and inferior MI often occurs as a result of an occlusion in the right coronary artery, and can be diagnosed by ECG changes in leads II, III and aVF.
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This question is part of the following fields:
- Anatomy
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Question 10
Incorrect
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A 55-year-old male is diagnosed with superior vena cava obstruction. What is the number of collateral circulations that exist for alternate pathways for venous return?
Your Answer: Three
Correct Answer: Four
Explanation:Superior vena cava is the main vein bringing blood back to the heart. It can get partially or completely blocked by various causes, the most common being due to malignant tumours of the mediastinum.
There are collateral pathways that form in long-standing cases with 60% or more stenosis and continue venous drainage in cases of superior vena obstruction. The collaterals are classified into four as follows:
1. The azygos-hemiazygos pathway
Azygos, hemiazygos, intercostal, and lumbar veins.2. The internal and external mammary pathway
internal mammary, superior epigastric, and inferior epigastric veins and superficial veins of the thorax.3. The lateral thoracic pathway
Lateral thoracic, thoracoepigastric, superficial circumflex, long saphenous, and femoral veins to collateralize to the IVC.4. The vertebral pathway
Innominate, vertebral, intercostal, lumbar, and sacral veins to collateralize to the azygos and internal mammary pathways. -
This question is part of the following fields:
- Anatomy
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Question 11
Incorrect
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Health workers are at increased risks of anaesthetic exposure. Therefore, The Control of Substances Hazardous to Health (COSHH) regulations require measures implemented to assess and control the risks related to this exposure.
Among the following control measures, which one is most likely to limit potentially harmful exposure to anaesthetic agents within an operating theatre?Your Answer: Maintenance of 15-20 air changes per hour
Correct Answer: Total intravenous anaesthesia (TIVA)
Explanation:Exposure to anaesthetic hazards is one among the occupational exposures in manipulating toxic agents or inhaling toxic gases during anaesthetic practices.
Toxic gases mainly nitrous oxide, is one of the most gaseous anaesthetic agents that constitutes an important source of pollution. One of the safe and effective technics used in anaesthesia and reducing the amount of pollution is the Total Intravenous Anaesthesia (TIVA) which consists of using opioids in analgesia and propofol for the induction and the maintenance of anaesthesia. It refers to the administration intravenously of an anaesthetic, sedative, and/or tranquilizer. A less polluting but not the best way to get rid of the toxic aesthetic agents is the scavenger system that collects and expels the gas outside the medical environment. Yet, this technique still represents a hazard for the environment and still increase the risk of exposure for the health workers and clinical staff.
Fume cupboards are also not recommended to use because of their high pollution potency, mainly of the air resulting in a great harm for medical workers.
Supraglottic airways as well as the Air Changes per Hour technics could be harmful for both patients and health workers, increasing the risks of transmitted diseases, namely nosocomial infections.
Therefore, the Total Intravenous Anaesthesia technique (TIVA) is most likely to be safe and recommended to use.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 12
Incorrect
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Heights of 100 individuals(adults) who were administered steroids at any stage during childhood was studied. The mean height was found to be 169cm with the data having a standard deviation of 16cm. What will be the standard error associated with the mean?
Your Answer: 1.69
Correct Answer: 1.6
Explanation:Standard error can be calculated by the following formula:
Standard Error= (Standard Deviation)/√(Sample Size)
= (16) / √(100)
= 16 / 10
= 1.6 -
This question is part of the following fields:
- Statistical Methods
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Question 13
Incorrect
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With a 10-day history of severe vomiting, a 71-year-old man with a gastric outlet obstruction is admitted to the surgical ward.
The serum biochemical results listed below are available:
Sodium 128 mmol/L (137-144)
Potassium 2.6 mmol/L (3.5-4.9)
Chloride 50 mmol/L (95-107)
Urea 12 mmol/L (2.5-7.5)
Creatinine 180 µmol/L (60-110)
Which of the following do you think you are most likely to encounter?Your Answer: Bicarbonate 45-50 mmol/L
Correct Answer: The standard base excess will be higher than actual base excess
Explanation:Hydrochloric acid is lost when you vomit for a long time (HCl). As a result, the following can be expected, in varying degrees of severity:
Hypokalaemia
Hypochloraemia
Increased bicarbonate to compensate for chloride loss and metabolic alkalosisThe alkalosis causes potassium to move from the intracellular to the extracellular compartment at first. Long-term vomiting and dehydration cause potassium to be excreted by the kidneys in order to conserve sodium. Dehydration can cause urea and creatinine levels to rise.
The actual base excess is always greater than the standard base excess.
The actual base excess (BE) is a measurement of a base’s contribution to a blood gas picture’s metabolic component. It’s the amount of base that needs to be added to a blood sample to bring the pH back to 7.4 after the respiratory component of a blood gas picture has been corrected (PaCO2 of 40 mmHg or 5.3 kPa). The BE has a normal range of +2 to 2. A large positive BE indicates a severe metabolic alkalosis, while a large negative BE indicates a severe metabolic acidosis. As a result, the actual BE in vitro is unaffected by CO2.
In vivo, however, standard BE is not independent of pCO2 because blood with haemoglobin acts as a better buffer than total ECF.
As a result, it is impossible to tell the difference between compensating for a respiratory disorder and compensating for the presence of a primary metabolic disorder.
The differences between in vitro and in vivo behaviour can be mostly eliminated if the BE is calculated for a haemoglobin concentration of 50 g/L (the ‘effective’ or virtual value of Hb if it was distributed throughout the extracellular space) rather than the actual haemoglobin. Because haemoglobin has a lower buffering capacity, the standard BE is higher than the actual BE. It reflects the BE better in the extracellular space rather than just the intravascular compartment.
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This question is part of the following fields:
- Pathophysiology
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Question 14
Incorrect
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Gag reflex was assessed as a part of brain stem death in a 22-year-old man with severe traumatic brain injury.
Which of the following nerves forms the afferent limb of this reflex?Your Answer: Vagus nerve
Correct Answer: Glossopharyngeal nerve
Explanation:The gag reflex is a protective mechanism that prevents any foreign material to enter the aerodigestive tract.
This reflex has afferent (sensory) and effect (motor) components.
– Glossopharyngeal nerve form the afferent limb
– Vagus nerve form the efferent limb -
This question is part of the following fields:
- Pathophysiology
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Question 15
Incorrect
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The diaphragm is a muscle that is relatively resistant to non-depolarizing neuromuscular blockade's effects.
When these muscle relaxants are used, which of the following peripheral nerve stimulator twitch patterns is best for monitoring the return of diaphragmatic function?Your Answer: Single-twitch stimulation
Correct Answer: Post-tetanic count stimulation
Explanation:Certain skeletal muscles are more resistant to the effects of neuromuscular blocking agents, both non-depolarizing and depolarizing. The diaphragm is the most resistant. The muscles of the larynx and the corrugator supercilii are less resistant. The abdominal, orbicularis oris, and limb peripheral muscles are the most sensitive muscles.
Twitch stimulation patterns:
Supramaximal single stimulus:
The frequency ranges from 1 Hz to 0.1 Hz (one every second to one every 10 seconds)
The response is proportional to the frequency of the event.
It has limited clinical utility because it only tells you whether or not a patient is paralysed (no information on degree of paralysis).Over the course of 0.5 seconds (2 Hz), four supramaximal stimulate were applied:
It is possible to see ‘fade’ and use it as a basis for evaluation.
This stimulation pattern is used to determine the degree of blockade (1-2 twitches is appropriate for abdominal surgery)
If the train of four (TOF) count is 1-2, reversal agents can be used in conjunction with medium-acting neuromuscular blocking agents.Ratio of TOF:
This is the ratio of the 4th twitch amplitude to the 1st twitch amplitude.
The ratio decreases with non-depolarising block and is inversely proportional to the degree of block, allowing objective measurement of residual neuromuscular blockade.
To achieve adequate reversal, the ratio (as measured by accelerography) must be between 0.7 and 0.9.Count of twitches after a tetanic experience(PTC):
50 Hz for 5 seconds, then a 3 second pause, followed by a single 1 Hz twitch stimulus.
When the TOF count is zero, this stimulation pattern is used to assess deep blockade (that is, in neurosurgery, microsurgery or ophthalmic surgery when even small movements of a patient will disturb the surgical field)
It gives an estimate of how long it will take for the response to return to single twitches, allowing assessment of blocks that are too deep for any other technique.
A palpable post-tetanic count (PTC) of 2 indicates no twitch response for about 20-30 minutes, and a PTC of 5 indicates no twitch response for about 10-15 minutes.This is without a doubt the best way to keep track of paralysis in patients who need to avoid diaphragmatic movement. It’s best to use drug infusions and aim for a PTC of 2. After a tetanic stimulus, acetylcholine is mobilised, causing post-tetanic potentiation.
Stimulation in Two Bursts:
750 milliseconds between two short bursts of 50 Hz
This stimulation pattern is used to assess small amounts of residual blockade manually (tactile). -
This question is part of the following fields:
- Clinical Measurement
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Question 16
Correct
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Which nerve does NOT pass through the posterior triangle of the neck?
Your Answer: Ansa cervicalis
Explanation:The sternocleidomastoid muscle divides the neck into anterior and posterior triangles on both sides of the neck.
The posterior triangle has the following boundaries:
anteriorly – sternocleidomastoid muscle
posteriorly – trapezius
roof – investing layer of deep cervical fascia
floor – prevertebral fascia overlying splenius capitis, levator scapulae, and the scalene musclesThe contents of the posterior triangle are:
1. fat
2. lymph nodes (level V)
3. accessory nerve
4. cutaneous branches of the cervical plexus – greater auricular nerve, transverse cervical nerve, lesser occipital nerve, supraclavicular nerve (A major branch of this plexus is the phrenic nerve, which arises from the anterior divisions of spinal nerves C3-C5)
5. inferior belly of omohyoid
6. branches of the thyrocervical trunk (transverse cervical and suprascapular arteries)
7. third part of the subclavian artery
8. external jugular vein -
This question is part of the following fields:
- Anatomy
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Question 17
Incorrect
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Regarding gas laws, which one best explains the ability of a Bourdon gauge to measure temperature?
Your Answer: Henry's Law
Correct Answer: Charles's law
Explanation:Charles’ Law states that there is a direct correlation between temperature and volume, where pressure and amount gas are constant. As temperature increases, volume also increases.
Boyle’s Law states that Pressure is inversely proportional to volume, assuming that temperature and amount of gas are constant. As volume increases, pressure decreases. In Dalton’s law of partial pressure, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressure of the gases in mixture.
According to Henry’s Law for concentration of dissolved gases, at a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid. An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Gay-Lussac’s Law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. This law is very similar to Charles’ Law, with the only difference being the type of container. Whereas the container in a Charles’ Law experiment is flexible, it is rigid in a Gay-Lussac’s Law experiment.
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This question is part of the following fields:
- Basic Physics
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Question 18
Incorrect
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All of the following options describes a thermistor for the measurement of temperature except:
Your Answer:
Correct Answer: Resistance of the bead increases exponentially as the temperature increases
Explanation:There are different types of temperature measurement. These include:
Thermistor – this is a type of semiconductor, meaning they have greater resistance than conducting materials, but lower resistance than insulating materials. There are small beads of semiconductor material (e.g. metal oxide) which are incorporated into a Wheatstone bridge circuit. As the temperature increases, the resistance of the bead decreases exponentially
Thermocouple – Two different metals make up a thermocouple. Generally, in the form of two wires twisted, welded, or crimped together. Temperature is sensed by measuring the voltage. A potential difference is created that is proportional to the temperature at the junction (Seebeck effect)
Platinum resistance thermometers (PTR) – uses platinum for determining the temperature. The principle used is that the resistance of platinum changes with the change of temperature. The thermometer measures the temperature over the range of 200°C to1200°C. Resistance in metals show a linear increase with temperature
Tympanic thermometers – uses infrared radiation which is emitted by all living beings. It analyses the intensity and wavelength and then transduces the heat energy into a measurable electrical output
Gauge/dial thermometers – Uses coils of different metals with different co-efficient of expansion. These either tighten or relax with changes in temperature, moving a lever on a calibrated dial.
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This question is part of the following fields:
- Clinical Measurement
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Question 19
Incorrect
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Out of the following, which anatomical structure lies within the spiral groove of the humerus?
Your Answer:
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 20
Incorrect
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Which is the most appropriate statement describing the function of flowmeters?
Your Answer:
Correct Answer: Constant pressure, variable orifice - Heidbrink flowmeters
Explanation:There are different models of flowmeters determined by the applied pressure and its orifice. For instance, the watersight flowmeter functions through applying variable pressure, and it has a variable orifice. In contrast, the bubble flowmeter is operated using a constant pressure and orifice. Flowmeters such as rotameters, Heidbrink and Peak have a constant pressure but variable orifice. On the other hand, flowmeters including a simple pressure gauge, water depression, and pneumotachograph have a constant orifice but variable pressure.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 21
Incorrect
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It is safe to perform a central neuraxial block without an increased risk of developing a vertebral canal haematoma in?
Your Answer:
Correct Answer: A patient on clopidogrel 7 days after last dose
Explanation:The incidence of vertebral canal haematoma following neuraxial blockade was reported (third National Audit Project (NAP3)) as 0.85 per 100 000 (95% CI 0-1.8 per 100 000). The incidence following neuraxial blockade in coagulopathic patients is likely to be higher hence coagulopathy remains a relative contraindication for conducting a spinal or epidural. When indicate, risk and benefits are weighed, and it is only performed by experienced personnel in this case.
Acceptable time to perform a block after the last dose of rivaroxaban in a patient with a creatinine clearance of greater than 30mL/minute is 18 hours.
Acceptable time to perform a block after the last dose of subcutaneous LMWH as prophylaxis is 12 hours.
Acceptable time to perform a block after the last dose of subcutaneous UFH as prophylaxis is 4 hours.
Acceptable time to perform a block after the last dose of thrombolytic therapy (streptokinase or alteplase) is 10 days.
Clopidogrel should be stopped 7 days prior to surgery, particularly if a central neuraxial procedure is considered.
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This question is part of the following fields:
- Pathophysiology
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Question 22
Incorrect
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A 28-year male patient presents to the GP with a 2-day history of abdominal pain and bloody diarrhoea. He reports that he was completely fine until one week ago when headache and general tiredness appeared. After further questioning, he revealed eating at a dodgy takeaway 3 days before the start of his symptoms.
Which of the following diagnosis is most likely?Your Answer:
Correct Answer: Campylobacter
Explanation:Giardiasis is known to have a longer incubation time and doesn’t cause bloody diarrhoea.
Cholera usually doesn’t cause bloody diarrhoea.
Generally, most of the E.coli strains do not cause bloody diarrhoea.
Diverticulitis can be a cause of bloody stool but the history here points out to an infectious cause.
Campylobacter infection is the most probable cause as it is characterized by a prodrome, abdominal pain and bloody diarrhoea
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 23
Incorrect
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During a squint surgery, a 5-year-old child developed severe bradycardia as a result of the oculocardiac reflex.
The afferent limb of this reflex is formed by which nerve?Your Answer:
Correct Answer: Trigeminal nerve
Explanation:When the eye is compressed or the extra-ocular muscles are tractioned, the oculocardiac reflex causes a decrease in heart rate.
The ophthalmic division of the trigeminal nerve provides the afferent limb. This synapses with the vagus nerve’s visceral motor nucleus in the brainstem. The efferent signal is carried by the vagus nerve to the heart, where increased parasympathetic tone reduces sinoatrial node output and slows heart rate.
The most common symptom is sinus bradycardia, but junctional rhythm and asystole can also occur.
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This question is part of the following fields:
- Pathophysiology
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Question 24
Incorrect
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The average diastolic blood pressure of a control group was found out to be 80 with a standard deviation of 5 in a study aimed at exploring the efficiency of a novel anti-hypertensive drug. The trial was randomised.
Making an assumption that the data is normally distributed, find out the number of patients that had diastolic blood pressure over 90.Your Answer:
Correct Answer: 3%
Explanation:Since the data is normally distributed, 95% of the values lie with in the interval 70 to 90. This can be calculated as follows:
Interval= Mean ± ( 2 times standard deviation)
= 80 ± 2(5)
= 80 ± 10
= 70 & 90The rest of the 5% are distributed symmetrically beyond 90 and below 70 which means 2.5% of the values lie above 90.
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This question is part of the following fields:
- Statistical Methods
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Question 25
Incorrect
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A 30-year old male has Von Willebrand's disease and attends the hospital to get an infusion of desmopressin acetate. The way this works is by stimulating the release of von Willebrand factor from cells, which in turn increases factor VIII and platelet plug formation in clotting.
In patients that have no clotting abnormalities, the substance that keeps the blood soluble and prevents platelet activation normally is which of these?Your Answer:
Correct Answer: Prostacyclin
Explanation:Even though aprotinin reduces fibrinolysis and therefore bleeding, there is an associated increased risk of death. It was withdrawn in 2007.
Protein C is dependent upon vitamin K and this may paradoxically increase the risk of thrombosis during the early phases of warfarin treatment.The coagulation cascade include two pathways which lead to fibrin formation:
1. Intrinsic pathway – these components are already present in the blood
Minor role in clotting
Subendothelial damage e.g. collagen
Formation of the primary complex on collagen by high-molecular-weight kininogen (HMWK), prekallikrein, and Factor 12
Prekallikrein is converted to kallikrein and Factor 12 becomes activated
Factor 12 activates Factor 11
Factor 11 activates Factor 9, which with its co-factor Factor 8a form the tenase complex which activates Factor 102. Extrinsic pathway – needs tissue factor that is released by damaged tissue)
In tissue damage:
Factor 7 binds to Tissue factor – this complex activates Factor 9
Activated Factor 9 works with Factor 8 to activate Factor 103. Common pathway
Activated Factor 10 causes the conversion of prothrombin to thrombin and this hydrolyses fibrinogen peptide bonds to form fibrin. It also activates factor 8 to form links between fibrin molecules.4. Fibrinolysis
Plasminogen is converted to plasmin to facilitate clot resorption -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 26
Incorrect
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Left ventricular afterload is mostly calculated from systemic vascular resistance.
Which one of the following factors has most impact on systemic vascular resistance?Your Answer:
Correct Answer: Small arterioles
Explanation:Systemic vascular resistance (SVR), also known as total peripheral resistance (TPR), is the amount of force exerted on circulating blood by the vasculature of the body. Three factors determine the force: the length of the blood vessels in the body, the diameter of the vessels, and the viscosity of the blood within them. The most important factor that determines the systemic vascular resistance (SVR) is the tone of the small arterioles.
These are otherwise known as resistance arterioles. Their diameter ranges between 100 and 450 µm. Smaller resistance vessels, less than 100 µm in diameter (pre-capillary arterioles), play a less significant role in determining SVR. They are subject to autoregulation.
Any change in the viscosity of blood and therefore flow (such as due to a change in haematocrit) might also have a small effect on the measured vascular resistance.
Changes of blood temperature can also affect blood rheology and therefore flow through resistance vessels.
Systemic vascular resistance (SVR) is measured in dynes·s·cm-5
It can be calculated from the following equation:
SVR = (mean arterial pressure − mean right atrial pressure) × 80 cardiac output
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This question is part of the following fields:
- Physiology
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Question 27
Incorrect
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The child-Pugh scoring system can be used, if risk classifying a patient with chronic liver disorder earlier to anaesthesia.
Which one is the best combination of clinical signs and examinations used within the Child-Pugh scoring system?Your Answer:
Correct Answer: Ascites, grade of encephalopathy, albumin, bilirubin and INR
Explanation:In the Child-Pugh classification system, the following 5 components are determined or calculated in order:
Ascites
Grade of encephalopathy
Serum bilirubin (?mol/L)
Serum Albumin (g/L)
Prothrombin time or INR
Raised liver enzymes are not the component of the classification system.
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This question is part of the following fields:
- Basic Physics
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Question 28
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:
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 29
Incorrect
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While administering a general anaesthetic to a 65-year-old man booked for a hip hemiarthroplasty, with a weight 70 kg, and an ASA 1 score, you give 1 g of paracetamol IV but notice that he had received the same dose on the ward one hour prior.
What is the most appropriate subsequent management of this patient?Your Answer:
Correct Answer: Do nothing and give the next doses of paracetamol at standard 6 hour intervals
Explanation:After ingestion of more than 150 mg/kg paracetamol within 24 hours, hepatotoxicity can occur but can also develop rarely after ingestion of doses as low as 75 mg/kg within 24 hours. Hepatocellular damage will not occur in this patient and therefore no need to engage management pathway for paracetamol overdose. If his weight was <33 kg or he already had a history of impaired liver function, then the management would bde different. Subsequent post-operative doses will be a standard dose of 1 g 6 hourly. This is a drug administration error and should be reported as an incident even though the patient will not be harmed.
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This question is part of the following fields:
- Pharmacology
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Question 30
Incorrect
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A 28-year-old man is admitted to the critical care unit. He has been diagnosed with adult respiratory distress syndrome and is being ventilated. His haemodynamic condition is improved using a pulmonary artery flotation.
His readings are listed below:
Haemoglobin concentration: 10 g/dL
Mixed venous oxygen saturation: 70%
Mixed venous oxygen tensions (PvO2): 50 mmHg
Estimate his mixed venous oxygen content (mL/100mL).Your Answer:
Correct Answer: 9.5
Explanation:Mixed venous oxygen content (CvO2) is the oxygen concentration in 100mL of mixed venous blood taken from the pulmonary artery. It is usually 12-17 mL/dL (70-75%). It is represented mathematically as:
CvO2 = (1.34 x Hgb x SvO2 x 0.01) + (0.003 x PvO2)
Where,
1.34 = Huffner’s constant
Hgb = Haemoglobin level (g/dL)
SvO2 = % oxyhaemoglobin saturation of mixed venous blood
PvO2 = 0.0225 = mL of O2 dissolved per 100mL plasma per kPa, or 0.003 mL per mmHgTherefore,
CvO2 = (1.34 x 10 x 70 x 0.01) + (0.003 x 50)
CvO2 = 9.38 + 0.15 = 9.53 mL/100mL
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This question is part of the following fields:
- Clinical Measurement
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