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Question 1
Incorrect
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A 72-year-old long-term rheumatoid arthritis patient is having shoulder replacement surgery.
He has chronic obstructive pulmonary disease with a limited exercise tolerance. He agrees to the procedure being performed with an interscalene brachial plexus block.
Which of the following neurological complications puts this patient at the greatest risk?Your Answer:
Correct Answer: Phrenic nerve block
Explanation:An ipsilateral phrenic nerve block will result from a successful interscalene block (ISB).
The phrenic nerve is the diaphragm’s sole motor supply, and ipsilateral hemidiaphragmatic paresis affects up to 100% of patients who receive ISBs. Phrenic nerve palsy is usually well tolerated and goes unnoticed by healthy people. However, forced vital capacity decreases by approximately 25%, which can produce ventilatory compromise in patients with limited pulmonary reserve, requiring assisted ventilation.
Vocal cord palsy occurs when the recurrent laryngeal nerve is inadvertently blocked, causing hoarseness and possibly acute respiratory insufficiency. Unless bilateral laryngeal nerve palsy occurs, which can cause severe laryngeal obstruction, this complication is usually of little consequence.
ISB can also cause cranial nerve X and XII palsy (Tapia’s syndrome). One-sided cord paralysis, aphonia, and the patient’s tongue deviating toward the block’s side are all symptoms.
When a local anaesthetic spreads to the stellate ganglion and its cervical sympathetic nerves, Horner’s syndrome can develop. Ptosis of the eyelid, miosis, and anhidrosis of the face are all symptoms. Horner’s syndrome, on the other hand, may not indicate that the brachial plexus is sufficiently blocked.
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This question is part of the following fields:
- Pathophysiology
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Question 2
Incorrect
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All of the following are responses to massive haemorrhage except which of the following?
Your Answer:
Correct Answer: Decreased cardiac output by increased direct parasympathetic stimulation
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 3
Incorrect
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A balanced general anaesthetic including a muscle relaxant is administered at induction. It is observed that the train-of-four count is two after two hours, with no further doses of the muscle relaxant.
What is most likely reason for this?Your Answer:
Correct Answer: Mivacurium with plasma cholinesterase deficiency
Explanation:Mivacurium is metabolised primarily by plasma cholinesterase at an In vitro rate of about 70% that of succinylcholine. Mivacurium is contraindicated in patients with genetic and acquired plasma cholinesterase deficiencies.
The clearance of atracurium is by Hoffman degradation and ester hydrolysis in the plasma and is independent of both hepatic and renal function.
Rocuronium is eliminated primarily by the liver after metabolises to a less active metabolite, 17-desacetyl-rocuronium. Its duration of action is not affected much by renal impairment.
Vecuronium undergoes hepatic metabolism into 3-desacetyl-vecuronium which has 50-80% the activity of the parent drug. It undergoes biliary (40%) and renal excretion (30%). The aminoglycoside antibiotics possess additional neuromuscular blocking activity. The potency of gentamicin > streptomycin > amikacin. Calcium can be used to reverse the muscle weakness produced by gentamicin but not neostigmine. When vecuronium and gentamycin are given together the effect on neuromuscular blockade is synergistic.
Significant residual neuromuscular block 2 hours after the administration of these drugs is unlikely In this scenario.
Any recovery from neuromuscular blockade with suxamethonium in a patient with deficiency of plasma cholinesterase demonstrate four twitches on a train of four count.
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This question is part of the following fields:
- Pharmacology
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Question 4
Incorrect
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A meta analysis has been conducted to see if addition of a new supplement the diet would prevent any further myocardial infraction among the patients who have recently had one. 4 trials (all randomised) were carried out. Which among the following is the most apt interpretation of the data?
Your Answer:
Correct Answer: Taking the supplement increases the chance of a further myocardial infarction
Explanation:Meta analysis performed upon the results, presented by the diamond, is clear from the no effect line. It presents a substantial increase in the probability of another heart attack.
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This question is part of the following fields:
- Statistical Methods
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Question 5
Incorrect
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Campylobacter is which type of bacteria?
Your Answer:
Correct Answer: sdgsdf
Explanation:Campylobacter is the commonest bacterial cause of infectious intestinal disease in the UK. The majority of cases are caused by the Gram-negative bacillus Campylobacter jejuni which is spread by the faecal-oral route. The incubation period is 1-6 days.
Features include a prodrome phase with headaches and malaise, then diarrhoea occurs which is often bloody.
There is often abdominal pain which may mimic appendicitis.It is usually self-limiting but treatment is warranted if the infection is severe or the infection occurs in an immunocompromised patient.
Severe infection comprises of high fever, bloody diarrhoea, or more than eight stools per day or symptoms last for more than one week.
This management would include antibiotics and the first-line antibiotic is clarithromycin.
Ciprofloxacin is an alternative but there are strains with decreased sensitivity to ciprofloxacin which can be frequently isolated.Complications include:
1.Guillain-Barre syndrome may follow Campylobacter
2. Jejuniinfections
3. Reactive arthritis
4. Septicaemia, endocarditis, arthritis -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 6
Incorrect
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Which of the following statements is true regarding sucralfate?
Your Answer:
Correct Answer:
Explanation:Sucralfate is an octasulfate of glucose to which Al(OH)3 has been added. It undergoes extensive cross-linking in an acidic environment and forms a polymer which adheres to the ulcer base for up to 6 hours and protects it from further erosion. Since it is not systemically absorbed it is virtually devoid of side effects. However, it may cause constipation in about 2% of cases due to the Aluminium component in it.
Sucralfate does not have antibacterial action against Helicobacter pylori. However, Bismuth has antibacterial action due to its oligodynamic effect.
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This question is part of the following fields:
- Pharmacology
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Question 7
Incorrect
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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:
Correct 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 8
Incorrect
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An acidic drug with a pKA of 4.3 is injected intravenously into a patient.
At a normal physiological pH, the approximate ratio of ionised to unionised forms of this drug in the plasma is?Your Answer:
Correct Answer: 1000:01:00
Explanation:The pH at which the drug exists in 50 percent ionised and 50 percent unionised forms is known as the pKa.
To calculate the proportion of ionised to unionised form of an ACID, use the Henderson-Hasselbalch equation.
pH = pKa + log ([A-]/[HA])
or
pH = pKa + log [(salt)/(acid)]
pH = pKa + log ([ionised]/[unionised]).Hence, if the pKa − pH = 0, then 50% of drug is ionised and 50% is unionised.
In this example:
7.4 = 4.3 + log ([ionised]/[unionised])
7.4 − 4.3 = log ([ionised]/[unionised])
log 3.1 = log ([ionised]/[unionised])Simply put, the antilog is the inverse log calculation. In other words, if you know the logarithm of a number, you can use the antilog to find the value of the number. The antilogarithm’s definition is as follows:
y = antilog x = 10x
Antilog to the base 10 of 0 = 1, 1 = 10, 2 =100, 3 = 1000, and 4 = 10,000.
If you want to find the antilogarithm of 3.1, for a number between 3 and 4, the antilogarithm will return a value between 1000 and 10,000. The ratio is 1:1 if pKa = pH, that is, pH pKa = log 0. (50 percent ionised and unionised).
According to the above value, there is only one unionised molecule for every approximately 1000 (1259) ionised molecules of this drug in plasma, implying that this drug is largely ionised in plasma (99.99 percent ).
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This question is part of the following fields:
- Pharmacology
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Question 9
Incorrect
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A sevoflurane vaporiser with a 2 percent setting and a 200 kPa ambient pressure is used.
At this pressure, which of the following options best represents vaporiser output?Your Answer:
Correct Answer: The output is 1% because the saturated pressure of sevoflurane is unaffected by ambient pressure
Explanation:Ambient pressure has no effect on a volatile agent’s saturated vapour pressure (SVP). At a temperature of 20°C, the SVP of sevoflurane is approximately 21 kPa, or 21% of atmospheric pressure (100 kPa).
The SVP of sevoflurane remains the same when the ambient pressure is doubled to 200 kPa, but the output of the vaporiser is halved, now 21 percent of 200 kPa, equalling 10.5 percent. The vaporiser’s output has increased to 1%, but the partial pressure output has remained unchanged. The splitting ratio will not change because it is determined by temperature changes.
Calculations can be made as follows:
Vaporizer output % (ambient pressure) = % volatile (calibrated) x 100 kPa calibrated pressure/ambient pressure
2% = 2% (dialled) × 100/100
2% of 100 = 2 kPaAltitude, pressure 50 kPa
4% = 2% (dialled) × 100/50
4% of 50 = 2 kPaHigh pressure at 200 kPa
1% = 2% (dialled) × 100/200
1% of 200 = 2 kPaSevoflurane has a boiling point of 58°C and, unlike desflurane (which has a boiling point of 22.8°C), does not need to be heated and pressurised with a Tec 6 vaporiser.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 10
Incorrect
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The fluids with the highest osmolarity is?
Your Answer:
Correct Answer: 0.45% N. Saline with 5% glucose
Explanation:The concentration of solute particles per litre (mosm/L) = the osmolarity of a solution. Changes in water content, ambient temperature, and pressure affects osmolarity. The osmolarity of any solution can be calculated by adding the concentration of key solutes in it.
Individual manufacturers of crystalloids and colloids may have different absolute values but they are similar to these.
0.45% N. Saline with 5% glucose:
Tonicity – hypertonic
Osmolarity – 405 mosm/L
Kilocalories (kCal) – 1070.9% N. Saline:
Tonicity – isotonic
Osmolarity – 308 mosm/L
Kilocalories (kCal) – 05% Dextrose:
Tonicity – isotonic
Osmolarity – 253 mosm/L
Kilocalories (kCal) – 170Gelofusine (154 mmol/L Na, 120 mmol/L Cl):
Tonicity – isotonic
Osmolarity – 274 mosm/L
Kilocalories (kCal) – 0Hartmann’s solution:
Tonicity – isotonic
Osmolarity – 273 mosm/L
Kilocalories (kCal) – 9 -
This question is part of the following fields:
- Physiology
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Question 11
Incorrect
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In which of the following situations will a regional fall in cerebral blood flow occur, suppose there is no changes in the mean arterial pressure (MAP)?
Your Answer:
Correct Answer: Hyperoxia
Explanation:The response of cerebral blood flow (CBF) to hyperoxia (PaO2 >15 kPa, 113 mmHg), the cerebral oxygen vasoreactivity is less well defined. A study originally described, using a nitrous oxide washout technique, a reduction in CBF of 13% and a moderate increase in cerebrovascular resistance in subjects inhaling 85-100% oxygen. Subsequent human studies, using a variety of differing methods, have also shown CBF reductions with hyperoxia, although the reported extent of this change is variable. Another study assessed how supra-atmospheric pressures influenced CBF, as estimated by changes in middle cerebral artery flow velocity (MCAFV) in healthy individuals. Atmospheric pressure alone had no effect on MCAFV if PaO2 was kept constant. Increases in PaO2 did lead to a significant reduction in MCAFV; however, there were no further reductions in MCAFV when oxygen was increased from 100% at 1 atmosphere of pressure to 100% oxygen at 2 atmospheres of pressure. This suggests that the ability of cerebral vasculature to constrict in response to increasing partial pressure of oxygen is limited.
Increases in arterial blood CO2 tension (PaCO2) elicit marked cerebral vasodilation.
CBF increases with general anaesthesia, ketamine anaesthesia, and hypoviscosity.
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This question is part of the following fields:
- Physiology
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Question 12
Incorrect
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A 77-year-old man, is scheduled for an angiogram to investigate gastro-intestinal bleeding. The radiologist performing the angiogram inserts the catheter into the coeliac axis.
What level of the vertebrae does the coeliac axis normally arise from the aorta?Your Answer:
Correct Answer: T12
Explanation:The coeliac axis refers to one of the splanchnic arteries located within the abdomen.
It arises from the aorta almost horizontally at the level of the T12 vertebrae
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This question is part of the following fields:
- Anatomy
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Question 13
Incorrect
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What is the number of valves between the superior vena cava and the right atrium?
Your Answer:
Correct Answer: None
Explanation:The inflow of blood from the superior vena cava is directed towards the right atrioventricular orifice. It returns deoxygenated blood from all structures superior to the diaphragm, except the lungs and heart.
There are no valves in the superior vena cava which is why it is relatively easy to insert a CVP line from the internal jugular vein into the right atrium. The brachiocephalic vein is similar as it also has no valves.
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This question is part of the following fields:
- Anatomy
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Question 14
Incorrect
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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:
Correct 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 15
Incorrect
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When the volume of gas is measured, its value should be supported by a suitable abbreviation. From the following options, what does the STPD stand for?
Your Answer:
Correct Answer: Standard Temperature and Pressure, Dry
Explanation:Gas is composed of large numbers of molecules moving in random directions, separated by distances. They undergo perfectly elastic collisions with each other and the walls of a container and transfer kinetic energy in form of heat. These assumptions bring the characteristics of gases within the range and reasonable approximation to a real gas, particularly how any change in temperature and pressure affect the behaviour of gas. According to different theories and laws proposed, mathematical equations are derived to calculate the volume of gas, also different abbreviations are being used according to given conditions. The abbreviations used are ATP, BTPS, and STPD.
ATP stands for ambient temperature and barometric pressure, it is used to describe the conditions under which volume of gas is measured.
BTPS stands for body temperature and pressure saturated with water vapor. These are conditions under which volume of gas exist and all results of lung volume determination should be quoted at BTPS.
STPD stands for standard temperature and pressure, dry (0C and 760 mm Hg). These are the conditions that are used to describe quantities of individual gases exchanged in the lungs. -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 16
Incorrect
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All of the following statements about calcium channel antagonists are incorrect except:
Your Answer:
Correct 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 17
Incorrect
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When describing the surface anatomy of the sacrum, which of the following anatomical landmarks refers to the base of an equilateral triangle is formed by the sacral hiatus?
Your Answer:
Correct Answer: A line connecting the posterior superior iliac spines
Explanation:The apex of an equilateral triangle completed by the posterior superior iliac spines is where the sacral hiatus or sacrococcygeal membrane can normally located. The failure of posterior fusion of the laminae of the fourth and fifth sacral vertebrae allows the sacral canal to be accessible via the membrane.
In adults, the spine of L4 usually lies on a line drawn between the highest points of the iliac crests (Tuffier’s line). A line connecting each anterior iliac spine, approximates to the L3/4 interspace in the sitting position. Both of these options are incorrect.
A line connecting the greater trochanters is also incorrect.
A line connecting the posterior superior iliac spines is correct, but in adults the presence of a sacral fat pad can still make identification of this landmark less straightforward.
The processes of S5 are remnants only and form the sacral cornua, which are also used to help identify the sacral hiatus.
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This question is part of the following fields:
- Anatomy
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Question 18
Incorrect
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Which of the following statement is not true regarding the effects of Dopamine infusions?
Your Answer:
Correct Answer: Decreasing gastric transit time
Explanation:Moderately high doses of dopamine produce a positive inotropic (direct?1 and D1 action + that due to Noradrenaline 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.
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This question is part of the following fields:
- Pharmacology
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Question 19
Incorrect
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During positive pressure ventilation using positive end-expiratory pressure (PEEP), there is usually an associated reduction in cardiac output
Which of the following is responsible?
Your Answer:
Correct Answer: Reduced venous return to the heart
Explanation:The option that is most responsible is the progressive decrease in venous return of blood to the right atrium. The heart rate does not usually change with PEEP so the fall in cardiac output is due to a reduction in left ventricular (LV) stroke volume (SV).
Note that the interventricular septum does shift toward the left and there is an increased pulmonary vascular resistance (PVR) from overdistention of alveolar air sacs that contribute to the reduction in cardiac output. Any increase in PVR will be associated with reduced pulmonary vascular capacitance.
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This question is part of the following fields:
- Pathophysiology
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Question 20
Incorrect
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A 75-year-old man, visits his general practitioner. He complains of changes to his bowel habit and unexpected weight loss in the last 6 months. He is scheduled for a colonoscopy and biopsy where he is diagnosed with a transverse colon malignancy.
The transverse colon is one of many organs tethered to the posterior wall of the abdominal cavity by a double fold of the peritoneum.
Which of the listed organs is also tethered to the peritoneum in a similar way?Your Answer:
Correct Answer: The stomach
Explanation:The peritoneal cavity is made up of the omentum, the ligaments and the mesentery.
The section of the peritoneum responsible for tethering organs to the posterior abdominal wall is the mesentery.
These tethered organs are classified as intraperitoneal, and these include the stomach, spleen, liver, first and fourth parts of the duodenum, jejunum, ileum, transverse, and sigmoid colon.
Retroperitoneal organs are located posterior to the peritoneum and include: the rest of the duodenum, the ascending colon, the descending colon, the middle third of the rectum, and the remainder of the pancreas
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This question is part of the following fields:
- Anatomy
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Question 21
Incorrect
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A young male is undergoing inguinal hernia repair. During the procedure, the surgeons approach the inguinal canal and expose the superficial inguinal ring.
Which structure forms the lateral edge of the superficial inguinal ring?Your Answer:
Correct Answer: External oblique aponeurosis
Explanation:The superficial inguinal ring is an opening in the aponeurosis of the external oblique muscle, just above and lateral to the pubic crest.
The superficial ring resembles a triangle more than a ring with the base lying on the pubic crest and its apex pointing towards the anterior superior iliac spine. The sides of the triangle are crura of the opening in the external oblique aponeurosis. The lateral crura of the triangle is attached to the pubic tubercle. The medial crura of the triangle is attached to the pubic crest.
The external oblique aponeurosis forms the anterior wall of the inguinal canal and also the lateral edge of the superficial inguinal ring. The rectus abdominis lies posteromedially, and the transversalis posterior to this.
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This question is part of the following fields:
- Anatomy
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Question 22
Incorrect
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If the speed of an ultrasound in soft tissue is 1540 meters per second, what is the estimated wavelength produced if the frequency of a generated ultrasound is 10 megahertz?
Your Answer:
Correct Answer: 0.15 millimetre
Explanation:Wavelength can be computed as follows:
Wavelength = velocity/frequency
In the given problem, the values stated are:
Frequency = 10 x 10^6
Velocity = 1540 meters per secondWavelength = 1540/(10×10^6)
Wavelength = 1540/10,000,000 meters
Wavelength = 0.15 millimetres -
This question is part of the following fields:
- Clinical Measurement
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Question 23
Incorrect
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Which of the following statements is true regarding dopamine?
Your Answer:
Correct Answer: It can increase or decrease cAMP levels
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 DA 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 used in patients with cardiogenic or septic shock and severe CHF wherein it increases BP and urine outflow.
It is administered by i.v. infusion (0.2–1 mg/min) which is regulated by monitoring BP and rate of urine formation
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This question is part of the following fields:
- Pharmacology
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Question 24
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:
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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Question 25
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:
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 26
Incorrect
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You decide to conduct research on the normal rates of gastric emptying in healthy people. The strategy is to give a drug orally and measure plasma concentrations at predetermined intervals.
Which of the following drugs would you choose to use?Your Answer:
Correct Answer: Paracetamol
Explanation:Because of the low pH in the stomach, paracetamol absorption is minimal (pKa value is 9.5). Paracetamol is absorbed quickly and completely in the alkaline environment of the small intestine. Oral bioavailability is approaching 100%. As a result, measuring paracetamol levels in plasma after an oral paracetamol dose has been used as a surrogate marker of gastric emptying. This method has been used to investigate the effects of drugs on gastric emptying. At clinically used doses, paracetamol is ideal because it has very few side effects.
Scintigraphic imaging is the gold standard for determining gastric emptying.
Although aspirin (acetyl salicylic acid) is absorbed primarily in the small intestine, some may also be absorbed in the stomach. The oral bioavailability ranges from 70 to 100 percent, making it less reliable than paracetamol.
Propranolol is a lipophilic drug that is rapidly absorbed after administration. However, it is highly metabolised by the liver in the first pass, and only about 25% of propranolol reaches the systemic circulation. It’s not the best indicator of gastric emptying.
Oral bioavailability of gentamicin and vancomycin is low. Only antibiotic-induced pseudomembranous colitis is treated with oral vancomycin.
Erythromycin is a pro-kinetic agent that acts as a motilin receptor agonist.
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This question is part of the following fields:
- Pharmacology
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Question 27
Incorrect
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A 57-year old lady is admitted to the Emergency Department with signs of a subarachnoid haemorrhage.
On admission, her GCS was 7. She has been intubated, sedated and is being ventilated and is waiting for a CT scan. Her Blood pressure is 140/70mmHg.
The arterial blood gas analysis shows the following:
pH 7.2 (7.35 - 7.45)
PaO2 70 mmHg (80-100)
9.2 kPa (10.5-13.1)
PaCO2 78 mmHg (35-45)
10.2 kPa (4.6-6.0)
BE -3 mEq/L (-3 +/-3)
Standard bic 27 mmol/L (21-27)
SaO2 94%
The most likely cause of an increase in the patient's global cerebral blood flow (CBF) is which of the following?
Your Answer:
Correct Answer: Hypercapnia
Explanation:PaCO2 is one of the most important factors that regulate cerebral vascular tone. CO2 induces cerebral vasodilatation and as a result, it increases CBF. Between 20 mmHg (2.7 kPa) and 80 mmHg (10.7 kPa), there is a linear increase of PaCO2.
Sometimes, there are areas where auto regulation has failed locally but not globally. Similarly, local vs. systemic acidosis will have similar effects. When the PaO2 falls below 50 mmHg (6.5 kPa), the CBF progressively increases.
An increase in the cerebral metabolic rate for oxygen (CMRO2) and therefore CBF can be caused by hyperthermia.
A late feature of cerebral injury is hyperthermia secondary to hypothalamic injury. Therefore this is not the most likely cause of an increased CBF in this scenario. -
This question is part of the following fields:
- Physiology
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Question 28
Incorrect
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About low molecular weight (LMW) heparin, Which of these is true?
Your Answer:
Correct Answer: Is excreted in the urine
Explanation:Low molecular weight heparin (LMWH) creates a complex by binding to antithrombin. This complex binds with and inactivates factor Xa.
There is less risk of bleeding with LMWH because it binds less to platelets, endothelium and von Willebrand factor.
LMW binds Xa more readily. The shorter chains are less likely to bind both antithrombin and thrombin.
There is need for monitoring in renal impairment because LMHW is excreted in the urine (and partly by hepatic metabolism)
LMWH have been shown to be as efficacious as unfractionated heparin. It is also safer and have improved inpatient stay and reduced hospital cost.
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This question is part of the following fields:
- Pharmacology
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Question 29
Incorrect
-
Which one of the following pharmacokinetic models is most suitable for target-controlled infusion (TCI) of propofol in paediatric patients?
Your Answer:
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 30
Incorrect
-
Which of the following may indicate an inadequate reversal of non-depolarising neuromuscular blockade?
Your Answer:
Correct Answer: Post tetanic count of 5
Explanation:A post-tetanic count of 5 denotes a deep neuromuscular blockade.
Post tetanic count (PTC) is a well-established method of evaluating neuromuscular recovery during intense neuromuscular blockade. It cam ne used when there is no response to single twitch, tetanic, or train-of-four (TOF) stimulation to assess the intensity of neuromuscular blockade and to estimate the duration after which the first twitch in the TOF (T1) is likely to reappear.
During a nondepolarizing block, the high frequency of tetanic stimulation will induce a transient increase in the amount of acetylcholine released from the presynaptic nerve ending, such that the intensity of subsequent muscle contractions will be increased (potentiated) briefly (period of post-tetanic potentiation, which may last 2 to 5 min. The neuromuscular response to stimulation during post tetanic potentiation can be used to gauge the depth of block when TOF stimulation otherwise evokes no responses. The number of post tetanic responses is inversely proportional to the depth of block: fewer post tetanic contractions denote a deeper block. When the post tetanic count (PTC) is 6 to 8, recovery to TOF count = 1 is likely imminent from an intermediate-duration blocking agent; when the PTC is 0, the depth of block is profound, and no additional NMBA should be administered.
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This question is part of the following fields:
- Pathophysiology
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Question 31
Incorrect
-
A 72-year-old man complains of severe, central abdominal pain that radiates to the back. He has a past medical history of an abdominal aortic aneurysm.
A focused abdominal ultrasonography test (FAST) is performed, revealing diffuse dilatation of the abdominal aorta. The most prominent dilatation is at the bifurcation site of abdominal aorta into the iliac arteries.
What vertebra level corresponds to the site of the most prominent dilatation as evident on the FAST scan?Your Answer:
Correct Answer: L4
Explanation:The important landmarks of vessels arising from the abdominal aorta at different levels of vertebrae are:
T12 – Coeliac trunk
L1 – Left renal artery
L2 – Testicular or ovarian arteries
L3 – Inferior mesenteric artery
L4 – Bifurcation of the abdominal aorta
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This question is part of the following fields:
- Anatomy
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Question 32
Incorrect
-
Out of the following, which artery is NOT a branch of the hepatic artery?
Your Answer:
Correct Answer: Pancreatic artery
Explanation:The common hepatic artery arises from the celiac artery and has the following branches:
1. hepatic artery proper that branches into –
a. cystic artery to supply the gallbladder
b. left and right hepatic arteries to supply the liver
2. gastroduodenal artery that branches into
a. right gastroepiploic artery
b. superior pancreaticoduodenal artery
3. right gastric arteryThe pancreatic artery is a branch of the splenic artery.
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This question is part of the following fields:
- Anatomy
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Question 33
Incorrect
-
A 70-year old male has diverticular disease and is undergoing a sigmoid colectomy. His risk of developing a post operative would infection can be minimized by which of the following interventions?
Your Answer:
Correct Answer: Administration of single dose of broad spectrum antibiotics prior to the procedure
Explanation:Staphylococcus aureus infection is the most likely cause.
Surgical site infections (SSI) occur when there is a breach in tissue surfaces and allow normal commensals and other pathogens to initiate infection. They are a major cause of morbidity and mortality.
SSI comprise up to 20% of healthcare associated infections and approximately 5% of patients undergoing surgery will develop an SSI as a result.
The organisms are usually derived from the patient’s own body.Measures that may increase the risk of SSI include:
-Shaving the wound using a single use electrical razor with a disposable head
-Using a non iodine impregnated surgical drape if one is needed
-Tissue hypoxia
-Delayed prophylactic antibiotics administration in tourniquet surgery, patients with a prosthesis or valve, in clean-contaminated surgery of in contaminated surgery.Measures that may decrease the risk of SSI include:
1. Intraoperatively
– Prepare the skin with alcoholic chlorhexidine (Lowest incidence of SSI)
-Cover surgical site with dressingIn contrast to previous individual RCT’s, a recent meta analysis has confirmed that administration of supplementary oxygen does not reduce the risk of wound infection and wound edge protectors do not appear to confer benefit.
2. Post operatively
Tissue viability advice for management of surgical wounds healing by secondary intentionUse of diathermy for skin incisions
In the NICE guidelines the use of diathermy for skin incisions is not advocated. Several randomised controlled trials have been undertaken and demonstrated no increase in risk of SSI when diathermy is used. -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 34
Incorrect
-
Which of the following statements is correct regarding hypomagnesaemia?
Your Answer:
Correct Answer: Causes tetany
Explanation:The ECG changes seen in hypomagnesaemia include:
Prolonged PR interval
Prolonged QT interval
Flattening of T waves
ST segment depression
Prominent U wavesThese changes are almost the same as those of hypokalaemia.
There is an increased risk of digoxin toxicity and a risk of atrial and ventricular ectopic and ventricular arrhythmias.
There is impaired synthesis and release of parathyroid hormone (PTH) in chronic hypomagnesaemia leading to impaired target organ response to PTH. This produces secondary hypocalcaemia.
The use of potassium ‘wasting’ diuretics (e.g. loop diuretics like furosemide) may lead to Hypomagnesaemia.
A tall T wave is seen in hypermagnesemia.
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This question is part of the following fields:
- Pathophysiology
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Question 35
Incorrect
-
Which of the following is true regarding the dose of propofol?
Your Answer:
Correct 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 36
Incorrect
-
A 54-year-old lady comes in for a right-sided elective bunionectomy with a realignment osteotomy under local anaesthetic on her first (large) toe.
For the operation, which of the following nerve blocks will be most effective?Your Answer:
Correct Answer: Superficial peroneal, deep peroneal and posterior tibial nerves
Explanation:An ankle block is commonly used for anaesthesia and postoperative analgesia when operating on bunions. It results in the selective block of the superficial peroneal, deep peroneal, and posterior tibial nerves.
The deep peroneal nerve supplies sensory input to the web space between the first and second toes (L4-5).
The L2-S1 nerve, often known as the superficial peroneal nerve, is a mixed motor and sensory neuron. It gives sensory supply to the anterolateral region of the leg, the anterior aspect of the 1st, 2nd, 3rd, and 4th toes, and innervates the peroneus longus and brevis muscles (with the exception of the web space between 1st and 2nd toes).
The sensory area of the saphenous nerve (L3-4) in the foot stretches from the proximal portion of the midfoot on the medial side to the proximal part of the midfoot on the lateral side.
The lateral side of the little (fifth) toe is innervated by the sural nerve’s sensory supply (S1-2). The heel, medial (medial plantar nerve), and lateral (lateral plantar nerve) soles of the foot are all served by the posterior tibial nerve.
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This question is part of the following fields:
- Pathophysiology
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Question 37
Incorrect
-
The following statements are about the conjugation of bilirubin. Which is true?
Your Answer:
Correct Answer: Is catalysed by a glucuronyl transferase
Explanation:Bilirubin is formed by metabolizing heme, mostly from haemoglobin in red blood cells.
Bilirubin is conjugated to glucuronic acid in the hepatocytes by the glucuronyl transferase enzyme in order to enable it to become soluble and allow for its secretion across the canalicular membrane and into bile.
The conjugation process is increased by rifampicin and decreased by valproate.
Gilbert’s syndrome is caused by a decrease in glucuronyl transferase in the hepatic system, decreasing the transport of bilirubin into the hepatocyte, causing unconjugated bilirubinaemia.
Crigler-Najjer syndrome is caused by mutations in the genes responsible for hepatic glucuronyl transferase, decreasing the activity of the enzyme, meaning bilirubin cannot be conjugated, causing unconjugated bilirubinaemia.
Dubin-Johnson syndrome does not cause an impairment in the conjugation of bilirubin, but it blocks the transport of bilirubin out of the hepatocyte resulting in conjugated bilirubinaemia.
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This question is part of the following fields:
- Pathophysiology
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Question 38
Incorrect
-
A 40-year old gentleman has palpitations and has gone to the emergency department. He is found to have monomorphic ventricular tachycardia. The resting potential of ventricular monocytes is maintained by which electrolyte?
Your Answer:
Correct Answer: Potassium
Explanation:Potassium maintains the resting potential of cardiac myocytes, with depolarization triggered by a rapid influx of sodium ions, and repolarization due to efflux of potassium. A slow influx of calcium is responsible for the longer duration of a cardiac action potential compared with skeletal muscle.
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 ms.Phase 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 And Biochemistry
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Question 39
Incorrect
-
An elective left colectomy is being performed on a 60-year old male for left-sided colon cancer. The upper and lower parts of the descending colon are supplied by the left colic artery.
Which of the following arteries gives rise to the left colic artery?Your Answer:
Correct Answer: Inferior mesenteric artery
Explanation:The inferior mesenteric artery originates 3-4 cm above the bifurcation of the abdominal aorta. The left colic artery branches off the inferior mesenteric artery, arising close to its origin from the abdominal aorta. Other branches of IMA include the three sigmoid arteries that supply the sigmoid colon.
The left colic artery branches off from IMA to supply the distal 1/3 of the transverse colon and the descending colon. It moves upwards posterior to the left colic mesentery and then travels anteriorly to the psoas major muscle, left ureter, and left internal spermatic vessels, before dividing into ascending and descending branches.
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This question is part of the following fields:
- Anatomy
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Question 40
Incorrect
-
If a patient is to be placed under general anaesthesia using total intravenous technique with target-controlled infusions of propofol and remifentanil, what safety precaution is the most vital in this a scenario?
Your Answer:
Correct Answer: Cannula access site clearly visible and regularly checked
Explanation:According to the Safe Anaesthesia Liaison Group, the most important factor to consider the cannula access, and if the patient is properly receiving the total intravenous anaesthesia. The cannula access must be regularly checked for kinks, leaks and disconnections.
Below are the safety precautions and policies to be followed for total intravenous anaesthesia among children and adults:
When administering TIVA, a non-return valve must be used on any intravenous fluid line;
When using equipment, it is essential that clinical staff know its limitations and uses;
Sites of intravenous infusions should be visible so they may be monitored for disconnection, leaks or perivenous infusion into the subcutaneous tissues; and,
Organisations must give preference to clearly labelled intravenous connectors and valves. -
This question is part of the following fields:
- Pathophysiology
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Question 41
Incorrect
-
Which statement is correct about the Mapleson anaesthetic breathing circuits?
Your Answer:
Correct Answer: Mapleson A is most efficient for spontaneous ventilation
Explanation:Mapleson breathing system (or circuit) analysed five different arrangements of components of the breathing system:
Mapleson A – It is the most efficient for spontaneous respiration. The flow of fresh gas required is 70-85 ml/kg/min, i.e., approximately 5-6 lit./min fresh gas flow for an average adult.
Mapleson B and C – inefficient for both SV and PPV; requires gas flow of two to three times minute volume (100 ml/kg/min). Not commonly used but category C may be used for emergency resuscitation.
Mapleson D – efficient for PPV at gas flow equivalent to patient’s minute volume; the Bain’s circuit is a coaxial version of the Mapleson D
Mapleson E and F – for paediatric use; requires gas flow at two to three times the patient’s minute volume. The Mapleson F consists of an open-ended reservoir bag (Jackson-Rees modification). -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 42
Incorrect
-
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:
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 43
Incorrect
-
An orthopaedic surgery is scheduled for a 68-year-old man. He is normally in good shape. His routine biochemistry results are checked and found to be within normal limits.
Which of the following pairs has the greatest impact on his plasma osmolarity?Your Answer:
Correct Answer: Sodium and potassium cations
Explanation:The number of osmoles (Osm) of solute per litre (L) of solution (Osmol/L) is the unit of measurement for solute concentration. The calculated serum osmolality assumes that the primary solutes in the serum are sodium salts (chloride and bicarbonate), glucose, and urea nitrogen.
2 (Na + K) + Glucose + Urea (all in mmol/L) = calculated osmolarity
313 mOsm/L = 2 (144 + 6) + 9.5 + 3.5
Sodium and potassium ions clearly contribute the most to plasma osmolarity. Glucose and urea, on the other hand, are less so.
The osmolarity of normal serum is 285-295 mOsm/L. Temperature and pressure affect osmolality, and this calculated variable is less than osmolality for a given solution.
The number of osmoles (Osm) of solute per kilogramme (Osm/kg) is a measure of osmolality, which is also a measure of solute concentration. Temperature and pressure have no effect on the value. An osmometer is used to measure it in the lab. Osmometers rely on a solution’s colligative properties, such as a decrease in freezing point or a rise in vapour pressure.
The osmolar gap (OG) is calculated as follows:
OG = osmolaRity calculated from measured serum osmolaLity
Excess alcohols, lipids, and proteins in the blood can all contribute to the difference.
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This question is part of the following fields:
- Physiology
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Question 44
Incorrect
-
An emergency appendicectomy is being performed on a 20 year old man. For maintenance of anaesthesia, he is being ventilated using a circle system with a fresh gas flow (FGF) of 1 L/min (air/oxygen and sevoflurane). The trace on the capnograph shows a normal shape.
The table below demonstrates the changes in the end-tidal and baseline carbon dioxide measurements of the capnograph at 10 and 20 minutes of anaesthesia maintenance. Â
End-tidal CO2: 4.9 kPa vs 8.4kPa (10 minutes vs 20 minutes)
Baseline end-tidal CO2: 0.2 kPa vs 2.4kPa
Pulse 100-107 beats per minute, systolic blood pressure 125-133 mmHg and oxygen saturation 98-99%.Â
Which of the following is the single most important immediate course of action?Your Answer:
Correct Answer: Increase the FGF
Explanation:End-tidal carbon dioxide (ETCO2) monitoring has been an important factor in reducing anaesthesia-related mortality and morbidity. Hypercarbia, or hypercapnia, occurs when levels of CO2 in the blood become abnormally high (Paco2 >45 mm Hg). Hypercarbia is confirmed by arterial blood gas analysis. When using capnography to approximate Paco2, remember that the normal arterial–end-tidal carbon dioxide gradient is roughly 5 mm Hg. Hypercarbia, therefore, occurs when PETco2 is greater than 40 mm Hg.
The most likely explanation for the changes in capnograph is either exhaustion of the soda lime and a progressive rise in circuit dead space.
Inspect the soda lime canister for a change in colour of the granules. To overcome soda lime exhaustion, the first step is to increase the fresh gas flow (FGF) (Option A). Then, if need arises, replace the soda lime granules. Other strategies that can work are changing to another circuit or bypassing the soda lime canister, but remember that both these strategies are employed only after increasing FGF first. Exclude other causes of equipment deadspace too.
There are also other causes for hypercarbia to develop intraoperatively:
1. Hypoventilation is the most common cause of hypercapnia. A. Inadequate ventilation can occur with spontaneous breathing due to drugs like anaesthetic agents, opioids, residual NMDs, chronic respiratory or neuromuscular disease, cerebrovascular accident.
B. In controlled ventilation, hypercapnia due to circuit leaks, disconnection or miscalculation of patient’s minute volume.
2. Rebreathing – Soda lime exhaustion with circle, inadequate fresh gas flow into Mapleson circuits and increased breathing system deadspace.
3. Endogenous source – Tourniquet release, hypermetabolic states (MH or thyroid storm) and release of vascular clamps.
4. Exogenous source – Absorption of CO2 from pneumoperitoneum. -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 45
Incorrect
-
The spinal cord in a neonate terminates at the lower border of:
Your Answer:
Correct Answer: L3
Explanation:The spinal cord and the vertebral canal are as long as each other in early fetal life. The length of the cord increases faster than the growth of the vertebrae during development. By the time of birth, the spinal cord is at the level of the lower border of the 3rd lumbar vertebra, compared to its original position at the level of the 2nd coccygeal vertebra.
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This question is part of the following fields:
- Anatomy
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Question 46
Incorrect
-
At what site would you palpate to assess the posterior tibial pulse?
Your Answer:
Correct Answer: Behind and below the medial ankle
Explanation:The posterior tibial artery originates from the popliteal artery in the popliteal fossa. It passes posterior to the popliteus muscle to pierce the soleus muscle. It descends between the tibialis posterior and flexor digitorum longus muscles.
The posterior tibial artery supplies blood to the posterior compartment of the lower limb. The artery can be palpated posterior to the medial malleolus.
There are 4 main pulse points for the lower limb:
1. Femoral pulse 2-3 cm below the mid-inguinal point
2. Popliteal partially flexed knee to loosen the popliteal fascia
3. Posterior tibial behind and below the medial ankle
4. Dorsal pedis dorsum of the foot over the navicular bone -
This question is part of the following fields:
- Anatomy
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Question 47
Incorrect
-
The pharmacologically inactive precursor of barbiturates is Barbituric acid. Because the molecule is heterocyclic, small structural changes can alter its pharmacological activity (structure function relationship).
Which of the following modifications to the molecule has the greatest impact on the effectiveness of the barbiturate derivative?Your Answer:
Correct Answer: Sulphur at C2
Explanation:Barbituric acid is the barbiturates’ pharmacologically inactive precursor. A pyrimidine heterocyclic nucleus is formed by the condensation of urea and malonic acid. Its pharmacological activity can be influenced by minor structural changes (structure function relationship).
The duration of action and potency as a sedative are influenced by the length of the side chains at C5. Barbiturates with three carbon atoms in their chain last longer than those with two. Anticonvulsant properties are enhanced by branched chains.
The addition of a methyl group at N1 causes a faster onset/offset of action, but it also causes excitatory phenomena (twitching/lower convulsive threshold).
The addition of oxygen and sulphur to C2 increases the molecule’s lipid solubility and thus its potency. Thiopentone (thiobarbiturate) has sulphur groups at C2, making it 20-200 times more lipid soluble than oxybarbiturates.
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This question is part of the following fields:
- Pharmacology
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Question 48
Incorrect
-
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:
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 49
Incorrect
-
The Medical Admissions unit receives a 71-year-old woman. She has type 2 diabetes, which she manages with diet, but she has been feeling ill for the past 48 hours.
Her pulse rate is 110 beats per minute, her blood pressure is 90/50 mmHg, and she is clinically dehydrated. Her respiratory rate is 20 breaths per minute, and chest auscultation reveals no focal signs.
The following are the lab results:
Glucose 27.4 mmol/L (3.5-5.5)
Ketones 2.5 mmol/L (<0.1)
Urinary glucose is zero (dipstick) with ketones
A random blood glucose of 15.3 mmol/L was measured during a visit to the diabetic clinic one month prior to admission, according to her notes, and a urinary dipstick registered a high glucose and ketones++.
The discrepancy between plasma and urinary glucose measurements is best explained by which of the following physiological mechanisms?Your Answer:
Correct Answer: The glomerular filtration rate is abnormally low
Explanation:The glucose molecule enters the Bowman’s capsule freely and becomes part of the filtrate.
All glucose is reabsorbed in the proximal convoluted tubule when blood glucose concentrations are below a certain threshold (approximately 11 mmol/L) (PCT). Active transportation makes this possible. In the proximal tubular cells, sodium/glucose cotransporters (SGLT1 and SGLT2) are the proteins responsible.
Glucose does not normally appear in the urine below the renal threshold.
The renal glucose threshold is not set in stone and is affected by a variety of factors, including GFR, TmG, and the quantity of splay.
The different absorptive and filtering capacities of individual nephrons cause splay, which is the rounding of a glucose reabsorption curve.
The SGLT proteins have a high affinity for glucose, but not an infinite affinity. As a result, some glucose may escape reabsorption before the TmG. A decrease in renal threshold may be caused by an increase in splay.
Because the filtered glucose load is reduced and the PCT can reabsorb all of the filtered glucose despite hyperglycaemia, a low GFR causes an increase in TmG. In contrast, lowering the TmG lowers the threshold because the tubules’ ability to reabsorb glucose is reduced.
A reduction in GFR caused by severe dehydration and reduced perfusion pressure is the most obvious cause of the discrepancy between plasma and urinary glucose in this scenario.
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This question is part of the following fields:
- Pathophysiology
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Question 50
Incorrect
-
A 21-year-old female was brought to the Emergency department with a ruptured ectopic pregnancy.
On clinical examination, the following were the findings:
Pulse: 120 beats per minute
BP: 120/95 mmHg
Respiratory rate: 22 breaths per minute
Capillary refill time: three seconds
Cool peripheries.
Which of the following best describes the cause for this clinical finding?Your Answer:
Correct Answer: Reduction in blood volume of 15-30%
Explanation:Classification of hemorrhagic shock according to Advanced Trauma Life Support is as follows:
– Class I haemorrhage (blood loss up to 15%) in which there is no change in blood pressure, RR, or pulse pressure.
– Class II haemorrhage (15-30% blood volume loss) where there is tachycardia, tachypnoea, and a decrease in pulse pressure.
– Class III haemorrhage (30-40% blood volume loss) where clinical signs of inadequate perfusion, marked tachycardia, tachypnoea, significant changes in mental state, and measurable fall in systolic pressure is seen. It almost always requires a blood transfusion.
– Class IV haemorrhage (> 40% blood volume loss) in which marked tachycardia, significant depression in systolic pressure and very narrow pulse pressure, and markedly depressed mental state with cold and pale skin are seen.
Loss of >50% results in loss of consciousness, pulse, and blood pressure.
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This question is part of the following fields:
- Pathophysiology
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Question 51
Incorrect
-
Which structure passes through the foramen magnum?
Your Answer:
Correct Answer: Spinal roots of the accessory nerve
Explanation:The structures that pass through the foramen magnum are:
Meningeal lymphatics
Spinal cord
Spinal meninges
Sympathetic plexus of vertebral arteries
Vertebral arteries
Vertebral artery spinal branches
The spinal roots of the accessory nerve.The jugular foramen contains the vagus nerve, the accessory nerve and glossopharyngeal nerve.
The vertebral veins does not pass into the skull.
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This question is part of the following fields:
- Anatomy
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Question 52
Incorrect
-
When combined with a general anaesthetic or central neuraxial block, which of the following medications used to treat dementia involves the risk of significant hypotension?
Your Answer:
Correct Answer: Risperidone
Explanation:Atypical antipsychotic drugs include risperidone and quetiapine. They not only inhibit dopamine receptors in the limbic system, but also histamine (H1) and alpha2 adrenoreceptors. When combined with general and/or central neuraxial block, this might result in severe hypotension.
Donepezil (Aricept) is an acetylcholinesterase (AChE) inhibitor that increases the neurotransmitter acetylcholine in the cerebral cortex and hippocampus in a reversible, non-competitive manner. It is used to reduce the advancement of Alzheimer’s disease symptoms (AD). Rivastigmine and galantamine are two more drugs that work in the same way.
Ginkgo Biloba contains anti-oxidant characteristics and is used to treat early-stage Alzheimer’s disease, vascular dementia, and peripheral vascular disease. It lowers platelet adhesiveness and decreases platelet activating factor (PAF) increasing the risk f bleeding, especially in individuals who are also taking anticoagulants and antiplatelet medication.
Memantine is an antagonist of the NMDA receptor. Synaptic plasticity, which is thought to be a critical component of learning and memory, can be inhibited at high doses. The use of ketamine is a relative contraindication since antagonism of this receptor can cause a dissociative state.
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This question is part of the following fields:
- Pharmacology
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Question 53
Incorrect
-
A survey aimed at finding out mean glucose level in individuals that took antipsychotics medicines was conducted. The results were as follows:
Mean Value: 7mmol/L
Standard Deviation: 6mmol/L
Sample Size: 9
Standard Error: 2mmol/L
For a confidence interval of 95%, which of the option presents the correct range up to the nearest value?Your Answer:
Correct Answer: 3-11 mmol/L
Explanation:Key Point: While finding out confidence intervals, standard errors are used. Standard error and Standard deviation are two distinct entities and should not be confused.
For 99.7% confidence interval, you can find the range as follows:
Multiply the standard error by 3.
Subtract the answer from mean value to get the lower limit.
Add the answer obtained in step 1 from the mean value to get the upper limit.
The range turns out to be 1-13 mmol/L.
For a confidence interval of 68%, multiply the standard error with 1 and repeat the process. The range found for this interval is 3-11 mmol/L.
For a 95% confidence interval. Standard Error is multiplied by 1.96 which gives us the limit ranging from 3.08 to 10.92 mmol/L which could be approximated to 3-11 mmol/L.
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This question is part of the following fields:
- Statistical Methods
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Question 54
Incorrect
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What feature is found in the flowmeters of modern anaesthetic machines ?
Your Answer:
Correct Answer: The use of stannic oxide increases the accuracy of flowmeters
Explanation:Flowmeters measure the rate at which a specific gas, that the flowmeter has been calibrated for, passes through. This calibration is done at room temperature and standard atmospheric pressure with an accuracy of +/- 2%.
Reading the flowmeter is done from the top of a bobbin (the midpoint of a ball). Oxygen is the last gas to be added downstream to the mixture delivered to the back bar as a safety feature. This prevents delivery of a hypoxic mixture.
Inaccurate flow measurements occur when the bobbin sticks to the inside wall of the flowmeter. Stannic oxide has been used as a successful antistatic substance thus, reducing the aforementioned risk.
Carbon dioxide being easily delivered is found on some older machines, but those attached flowmeters are limited by a maximum flow of 500 ml /min. Thus avoiding the delivery of a hypercarbic mixture.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 55
Incorrect
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Regarding the anatomical relations of the scalenus muscles, which of these is true?
Your Answer:
Correct Answer: The trunks of the brachial plexus emerge from the lateral border of scalenus anterior
Explanation:The ascending cervical artery lies media the phrenic nerve on scalenus anterior and can easily be mistaken for the phrenic nerve at operation.
The phrenic nerve passes across scalenus anterior and medius inferiorly.
The subclavian artery is separated from the vein by the scalenus anterior.
The brachiocephalic vein is formed at the medial border of scalenus anterior by the subclavian vein and the internal jugular vein.
Emerging from the lateral border of scalenus anterior are the trunks of the brachial plexus .
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This question is part of the following fields:
- Anatomy
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Question 56
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 57
Incorrect
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A study of 1500 participants is designed to determine the normal range for a number of parameters in a certain population.
Analysis has revealed that there is a strong positive correlation between height and forced expiratory volume (FEV1).
Which of the following statistical technique can be utilized to predict FEV1 at any given height in this cohort?Your Answer:
Correct Answer: Linear regression
Explanation:A forest plot is a graphical representation that summarizes the findings of several research, such as a meta-analysis of a series of randomized controlled trials.
The Kaplan-Meier estimate shows survival over time, for example, plotting the number of patients still alive seven years after chemotherapy for lung cancer.
Fisher’s exact test similarly uses contingency tables to assess statistical significance, however, it is typically used when sample sizes are small.
Chi-square test assesses whether an association exists between two categorical variables using the observed and expected frequencies. For instance, is social class (I-V) related to body mass index (BMI) category? Using the observed and anticipated frequencies, the Chi-square test determines whether a connection exists between two categorical variables. For example, is socio-economic status related to BMI category?
Linear regression is a technique which attempts to model the relationship between two variables by fitting a linear equation to observed data. Linear regression uses correlation between two continuous variables. As correlation only indicates the strength of an association only, it cannot be used to forecast the change in one variable when a second variable is altered.
This equation takes the form y = mx + c, where ‘y’ is the dependent variable, ‘x’ is the independent variable, ‘m’ is the slope of the line and ‘c’ is the intercept. In this example, for a range of heights, it would be possible to map a line of best fit to a scatter plot and thus predict the forced expiratory volume (FEV1) for an individual.
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This question is part of the following fields:
- Statistical Methods
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Question 58
Incorrect
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Which of the following statements is the most correct about ketamine?
Your Answer:
Correct Answer: The S (+) isomer is more potent that the R (-) isomer
Explanation:Ketamine, a phencyclidine derivative, is an antagonist at the NMDA receptor. It causes depression of the CNS that is dose dependent and induces a dissociative anaesthetic state with profound analgesia and amnesia.
Ketamine has a chiral centre usually presented as a racemic mixture with two optical isomers, S (+) and R (-) forms. These isomers are in equal proportions. The S (+) isomer is about three times more potent than the R (-) form. The S (+) form is less likely to cause emergence delirium and hallucinations.
Ketamine is extensively metabolised by hepatic microsomal cytochrome P450 enzymes producing norketamine as its main metabolite. Norketamine has a one third to one fifth as potency as its parent compound.
It increases the CMRO2, cerebral blood flow and potentially increase intracranial pressure. -
This question is part of the following fields:
- Pharmacology
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Question 59
Incorrect
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A transport ventilator is powered by an air/oxygen mix using a full oxygen cylinder (class CD) with an internal capacity of 2 litres, and pressure of 23,000 kPa, with a gas flow of 4 litres/minute.
The ventilator also has a control resulting in an additional gas consumption of 1 litre/minute.
How long will it take for the cylinder to empty?Your Answer:
Correct Answer: 92 minutes
Explanation:The Drager Oxylog® 1000 is a pneumatically powered, time-dependent, volume-titrated emergency ventilator with a pressure limit. It is compatible with CD cylinder oxygen. The CD cylinder is a strong and lightweight cylinder usually composed of aluminium or Kevlar. The internal cylinder volume is 2 litres, and the pressure of a full cylinder is 230 bar. The volume of the full cylinder is determined by applying Boyle’s law: P1 × V1 = P2 × V2
Where:
P1= pressure of a full cylinder (230 bar)
V1= volume of oxygen at that pressure (2 litres)
P2= final pressure (1 bar), and
V2= volume of oxygen in the full cylinder.Substituting values into the equation:
230 × 2 = 1 x V2
V2 = 460 litres. The flow of fresh gas is 4 litres/minute + 1 litre/minute required by the control, making a total of 5 litres/minute. The amount of time it takes for the cylinder to empty would be the total volume of oxygen in the full cylinder divided by the amount of oxygen expelled per minute: 460/5 = 92, meaning it would take 92 minutes for the cylinder to empty. -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 60
Incorrect
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A 40-year old female comes to the GP's office with unexplained weight gain, cold intolerance and fatigue. Her thyroid function tests are performed as there is a suspicion of hypothyroidism. A negative feedback mechanism is incorporated in the control of thyroid hormone release. All of choices below are also controlled by a negative feedback loop except:
Your Answer:
Correct Answer: Clotting cascade
Explanation:The correct answer is the clotting cascade, which occurs via a positive feedback mechanism. As clotting factors are attracted to a site, their presence attracts further clotting factors. This continues until a functioning clot is formed.
This patient has presented with symptoms of hypothyroidism and symptoms include weight gain, lethargy, cold intolerance, dry skin, coarse hair and constipation. It can be treated by replacing the missing thyroid hormone with levothyroxine which is a synthetic version of thyroxine (T4).
Serum carbon dioxide (CO2) is controlled via a negative feedback mechanism as well. Chemoreceptors can detect when the serum CO2 is high, and send an impulse to the respiratory centre of the brain to increase the respiratory rate. As a result, more CO2 is exhaled which lowers the serum concentration.
Cortisol is also released according to a negative feedback mechanism. Cortisol acts on both the hypothalamus and the anterior pituitary. Its action serve to decrease the formation of corticotrophin releasing hormone (CRH) and adrenocorticotropic hormone (ACTH), respectively. CRH acts on the anterior pituitary to release ACTH. This then acts on the adrenal gland to cause the release of cortisol. Thus, inhibition of CRH and ACTH formation results in high levels of cortisol which inhibit its further release.
Blood pressure (BP) is controlled via a negative feedback mechanism. Low BP results in renin-angiotensin-aldosterone system (RAAS) activation. This leads to vasoconstriction and retention of salt and water which increased BP.
Blood sugar is controlled via a negative feedback mechanism. A rise in blood sugar causes insulin to be released. Insulin acts to transport glucose into the cell which lowers blood sugar. -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 61
Incorrect
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Transthoracic echocardiogram (TTE) can be used to investigate the function of the heart in patients with suspected heart failure. The aim is to measure the ejection fraction, but to do that, the stroke volume must first be measured. How is stroke volume calculated?
Your Answer:
Correct Answer: End diastolic volume - end systolic volume
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 62
Incorrect
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A 27-year-old woman takes part in a study looking into the effects of different dietary substrates on metabolism. She receives a 24-hour ethyl alcohol infusion.
A constant volume, closed system respirometer is used to measure CO2 production and consumption. The production of carbon dioxide is found to be 200 mL/minute.
Which of the following values most closely resembles her anticipated O2 consumption at the conclusion of the trial?Your Answer:
Correct Answer: 300 mL/minute
Explanation:The respiratory quotient (RQ) is the ratio of CO2 produced by the body to O2 consumed in a given amount of time.
CO2 produced / O2 consumed = RQ
CO2 is produced at a rate of 200 mL per minute, while O2 is consumed at a rate of 250 mL per minute. An RQ of around 0.8 is typical for a mixed diet.
The RQ will change depending on the energy substrates consumed in the diet. Granulated sugar is a refined carbohydrate that contains 99.999 percent carbohydrate and no lipids, proteins, minerals, or vitamins.
Glucose and other hexose sugars (glucose and other hexose sugars):
RQ=1Fats:
RQ = 0.7Proteins:
Approximately 0.9 RQEthyl alcohol is a type of alcohol.
200/300 = 0.67 RQ
For complete oxidation, lipids and alcohol require more oxygen than carbohydrates.
When carbohydrate is converted to fat, the RQ can rise above 1.0. Fat deposition and weight gain are likely to occur in these circumstances.
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This question is part of the following fields:
- Physiology
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Question 63
Incorrect
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The population incidence of a disease is best described by?
Your Answer:
Correct Answer: The number of new cases of a disease per population in a defined period
Explanation:The incidence of a disease is the number of new cases of the disease in a population over a defined time period.
The prevalence of a disease is the number of cases of the disease in a population over a defined time period describes. It is NOT the number of new cases.
The number of new cases of a disease only, has no denominator (time period or population) from which to derive an incidence.
The number of new cases of a disease seeking medical treatment is the incidence of patients seeking medical treatment NOT the incidence of the disease in a population.
The death rate from a disease is the number of patients dying from the disease in a population.
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This question is part of the following fields:
- Statistical Methods
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Question 64
Incorrect
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During the design phase of a study, which among the given is aimed at addressing confounding factors?
Your Answer:
Correct Answer: Randomisation
Explanation:Randomisation allows for performance of experimental trials in a random order. Using this method gives us control over the confounding variables that are not supposed to be held constant.
For an instance, by employing randomisation we get to control biological differences among individual human beings during experimental trials.
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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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The tip of a pulmonary artery flotation catheter becomes wedged when threaded through the chambers of the heart and the pulmonary artery.
Which of the following options best describes the sequence of pressures measured at the catheter's tip during its passage through a normal patient's pulmonary artery?Your Answer:
Correct Answer: 0-12 mmHg, 2-25 mmHg, 12-25 mmHg and 8-12 mmHg
Explanation:The tricuspid valve allows the tip of a pulmonary artery catheter to pass through the right atrium and into the right ventricle.
The balloon will be inflated before crossing the pulmonary valve and entering the pulmonary artery, where it will eventually wedge or occlude the artery, providing an indirect measure of left atrial pressure.
0-12 mmHg in the right atrium
2-25 mmHg in the right ventricle
12-25 mmHg in the pulmonary artery
8-12 mmHg is the occlusion pressure -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 66
Incorrect
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The structure most likely to be damaged during cannulation of the subclavian vein is?
Your Answer:
Correct Answer: Subclavian artery
Explanation:The subclavian artery lies behind and partly above the subclavian vein. 3-4% of the time, it can be inadvertently cannulated during cannulation of the subclavian vein
Because of its anatomical position, putting pressure on the subclavian artery is impossible so arresting bleeding with pressure when it is punctured is not viable.
One of the consequences of subclavian vein cannulation (1%) is pleural puncture leading to a pneumothorax. This is because the apical pleura is inferior and caudal to the subclavian vein.
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This question is part of the following fields:
- Anatomy
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Question 67
Incorrect
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Which of the following is a true statement about invasive arterial pressure monitoring?
Your Answer:
Correct Answer: Increased resonance elevates the systolic and lower the diastolic pressures
Explanation:The arterial cannula inserted should have parallel walls in order to reduce the risk of interruption of blood flow to distal limbs.
It is essential that the monitor used to display the arterial pressure waves has a frequency capacity of 0.5-40Hz. This is because the pressure waves are a combination of different sine waves of varying frequencies and amplitudes.
The diameter of the catheter is directly proportional to the natural frequency which is the frequency at which the system responsible for monitoring the waves resonates and amplifies the signals. This should be at least ten fold in comparison to the fundamental frequency. The diameter of the catheter is also inversely proportional to the square root of the system compliance, the tubing length and the fluid density within the system.
The presence of an air bubble, a clot or an easily malleable diaphragm and tube can result in wave damping. Increased damping will cause a reduction in the systolic pressure, and an increase in diastolic pressure. The maximum damping value of an appropriate monitoring system would be 0.64.
A rigid, non-malleable diaphragm and tubing can cause a resonance within the system. This resonance will result in an increase in the systolic pressure and a reduction in the diastolic pressure
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This question is part of the following fields:
- Clinical Measurement
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Question 68
Incorrect
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Which of the following statements about a cervical rib is correct?
Your Answer:
Correct Answer: It originates from the 7th cervical vertebra
Explanation:The costal elements of the seventh cervical vertebrae form projections known as cervical ribs, which are present in approximately 0.5% – 1 % individuals.
A cervical rib commonly comprises of a head, neck and tubercle. The body of the rib varies from person to person. It extends into the posterior triangle of the neck, where it is either free anteriorly, or attached to the first rib / sternum.
Patients with a cervical rib mostly are asymptomatic and it is usually diagnosed as an incidental finding on chest x-ray. However, in some cases, the subclavian artery and the lower trunk of the brachial plexus are compressed where they pass over the cervical rib. This leads to a condition known as ”neurovascular compression syndrome,” in which these neurovascular structures are compressed between the cervical rib and scalenus anterior.
The most common cause of neurogenic symptoms in approximately 80% of the patients with cervical rib is neck trauma.
In most cases, the tingling, numbness and impaired circulation to the upper limb appears only after puberty. This is because the neck elongates, and the shoulders droop slightly.
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This question is part of the following fields:
- Anatomy
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Question 69
Incorrect
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All of the following statements are true regarding calcium except:
Your Answer:
Correct Answer: Serum calcium accounts for 10% of total body calcium stores
Explanation:Calcium is a very important ion and is involved in:
-cell homeostasis
-coagulation
-muscle contraction
-neuronal impulse transmission/membrane stabilization
-bone formation and skeletal strength
-secretion processes99% is found in bone and 1% in the plasma. Of the 1% that is in the plasma
-45% is free ionized calcium
-45% is bound to proteins, mainly Albumin
-10% is present as an anion complexReduced levels of IONIZED calcium give rise to features of hypocalcaemia , resulting in increased excitability of membranes. This results when the total calcium concentration goes below 2 mmol/L.
Features of mild to moderate hypocalcaemia are:
-paraesthesia (peri-oral, fingers)
-tetany
-spasm
-muscle cramps
-ECG changes (prolonged QT)
-Trousseau’s sign (inflation of tourniquet induces carpopedal spasm)
-Chvostek’s sign (tapping the facial nerve – cranial nerve VII – causes facial muscle twitch/spasm)Features of severe hypocalcaemia are:
-cardiogenic shock and congestive cardiac failure due to reduced myocardial contractility
respiratory distress due to bronchospasm, agitation, confusion, seizuresFeatures of hypercalcaemia (remember ‘bones, stones, groans and psychic moans’):
-Abdominal pain
-Vomiting
-Constipation
-Polyuria
-Polydipsia
-Depression
-Lethargy
-Anorexia
-Weight loss
-Hypertension
-Confusion
-Pyrexia
-Calcification in the cornea
-Renal stones
-Renal failure
-Decreased Q-T interval
-Cardiac shock/collapse -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 70
Incorrect
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Drug X, a new intravenous induction drug, is being administered as a bolus at regular time intervals, and the following data were observed:
Time following injection (hours) vs Plasma concentration of drug X (mcg/mL)
2 / 400
6 / 100
10 / 25
14 / 6.25
Which of the following values estimate the plasma half-life (T½) of drug X?
Your Answer:
Correct Answer: 2 hours
Explanation:Half life (T½) is the time required to change the amount of drug in the body by one-half (or 50%) during elimination. The time course of a drug in the body will depend on both the volume of distribution and the clearance.
Extrapolating the values from the plasma concentration vs time:
Plasma concentration at 0 hours = 800 mcg/mL
Plasma concentration at 2 hours = 400 mcg/mL
Plasma concentration at 4 hours = 200 mcg/mL
Plasma concentration at 6 hours = 100 mcg/mL
Plasma concentration at 8 hours = 50 mcg/mL
Plasma concentration at 10 hours = 25 mcg/mL
Plasma concentration at 12 hours = 12.5 mcg/mL
Plasma concentration at 14 hours = 6.25 mcg/mL -
This question is part of the following fields:
- Statistical Methods
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Question 71
Incorrect
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A log-dose response curve is plotted after drug A is given. The shape of this curve is sigmoid, with a maximum response of 100%.
The log-dose response curve of drug A shifts to the right with a maximum response of 100 percent when drug B is administered.
What does this mean in terms of drug B?Your Answer:
Correct Answer: Drug B has affinity for the receptor but has no intrinsic efficacy
Explanation:Drug A is a pure agonist for the receptor, with high intrinsic efficacy and affinity, according to the log-dose response curve.
Drug B, on the other hand, works as a competitive antagonist. It binds to the receptor but has no inherent efficacy. Drug A’s efficacy will not change, but its potency will be reduced.
A partial agonist is a drug with partial intrinsic efficacy and affinity for the receptor. Giving a partial agonist after a pure agonist will not increase receptor occupancy or decrease receptor activity, and thus will not affect drug A’s efficacy. The inverse agonist flumazenil can reverse all benzodiazepines.
An inverse agonist is a drug that binds to the receptor but has the opposite pharmacological effect.
A non-competitive antagonist is a drug that has affinity for a receptor but has different pharmacological effects and reduces the efficacy of an agonist for that receptor.
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This question is part of the following fields:
- Pharmacology
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Question 72
Incorrect
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Which statement is the most accurate when describing electrical equipment and shock?
Your Answer:
Correct Answer: Type CF is considered to safe for direct connection with the heart
Explanation:There are different classes of electrical equipment that can be classified in the table below:
Class 1 – provides basic protection only. It must be connected to earth and insulated from the mains supply
Class II – provides double insulation for all equipment. It does not require an earth.
Class III – uses safety extra low voltage (SELV) which does not exceed 24 V AC. There is no risk of gross electrocution but risk of microshock exists.
Type B – All of above with low leakage currents (0.5mA for Class IB, 0.1 mA for Class IIB)
Type BF – Same as with other equipment but has ‘floating circuit’ which means that the equipment applied to patient is isolated from all its other parts.
Type CF – Class I or II equipment with ‘floating circuits’ that is considered to be safe for direct connection with the heart. There are extremely low leakage currents (0.05mA for Class I CF and 0.01mA for Class II CF)
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This question is part of the following fields:
- Clinical Measurement
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Question 73
Incorrect
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A 57-year old woman, presents to her general practitioner. She has a 2 week history of a vaginal hysterectomy for which she was placed under general anaesthesia.
On examination, she has notable weakness of dorsiflexion of her left foot and a high stepping gait.
Which nerve was most likely injured during her surgery?Your Answer:
Correct Answer: Common peroneal nerve
Explanation:The common peroneal (fibular) nerve is a peripheral nerve in the lower limb. It arises of the L4-S2 nerve roots and has sensory and motor innervations:
Sensory: Provides innervation of the lateral leg and foot dorsum.
Motor: Provides innervation of the short head of the biceps femoris, as well as muscles of the anterior and lateral leg compartments.
It is the most commonly damaged nerve in the lower extremity, as it is easily compressed by a plaster cast or injured when the fibula is fractured.
Damage to the common peroneal nerve will result in loss of dorsiflexion at ankle (footdrop, as feet are permanently plantarflexed), with the accompanying high stepping gait.
The saphenous and sural nerve only provide sensory innervation.
The tibial nerve arises from the sciatic nerve (like the common peroneal), but it provides motor innervation to the posterior leg compartments and intrinsic foot muscles. Injury to the tibial nerve will cause loss of plantar flexion, toe flexion and weakened foot inversion.
Extreme hip flexion into the lithotomy or Lloyd-Davies position can result in stretch damage to the neurones (sciatic and obturator nerves) or by applying direct pressure (femoral nerve compression).
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This question is part of the following fields:
- Pathophysiology
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Question 74
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 75
Incorrect
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Which of the following best explains the statement Epinephrine is formulated as 1 in 1000 solution
Your Answer:
Correct Answer: 1000 mg per 1000 ml solution
Explanation:The statement Epinephrine is formulated as 1 in 1000 solution means 1 gm epinephrine is present in 1000 ml of solution.
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This question is part of the following fields:
- Pharmacology
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Question 76
Incorrect
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You've been summoned to help resuscitate a 6-year-old child who has suffered a cardiac arrest. The ECG monitor shows electrical activity that isn't pulsed.
Which of the following statements is the most appropriate during resuscitation?Your Answer:
Correct Answer: The dose of intravenous adrenaline is 180 mcg
Explanation:To begin, one must determine the child’s approximate weight. There are a variety of formulas to choose from. It is acceptable to use the advanced paediatric life support formula:
(Age + 4) 2 = Weight
A 5-year-old child will weigh around 18 kilogrammes.
10 mcg/kg (0.1 ml/kg of 1 in 10 000 adrenaline) = 180 mcg is the appropriate dose of intravenous or intraosseous adrenaline.
The correct energy level to deliver is 4 J/kg, which equals 72 joules.
The pad size that is appropriate for this patient is 8-12 cm. For an infant, a 4.5 cm pad is appropriate.
To allow adequate separation in infants and small children, the pads should be placed anteriorly and posteriorly on the chest.
When using a bag and mask to ventilate, take two breaths for every 15 chest compressions. If chest compressions are being applied intubated and without interruption, a ventilation rate of 10-20 breaths per minute should be given.
Chest compressions should be done at a rate of 100-120 per minute, the same as an adult.
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This question is part of the following fields:
- Pharmacology
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Question 77
Incorrect
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Which of the following best explains the association between smoking and lower oxygen delivery to tissues?
Your Answer:
Correct 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 78
Incorrect
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Which of the following correctly explains the mechanism of lowering blood pressure by nitroglycerine?
Your Answer:
Correct Answer: Nitric oxide
Explanation:Nitroglycerine is rapidly denitrated enzymatically in the smooth muscle cell to release the free radical nitric oxide (NO).
Released NO activated cytosolic guanylyl cyclase which increases cGMP (cyclin guanosine monophosphate) which causes dephosphorylation of myosin light chain kinase (MLCK) through a cGMP-dependent protein kinase.
Reduced availability of phosphorylated (active) MLCK interferes with activation of myosin and in turn, it fails to interact with actin to cause contraction. Consequently, relaxation occurs.
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This question is part of the following fields:
- Pharmacology
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Question 79
Incorrect
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Which of the following lung parameters can be measured directly using spirometry?
Your Answer:
Correct Answer: Vital capacity
Explanation:Spirometry measures the total volume of air that can be forced out in one maximum breath, that is the total lung capacity (TLC), to maximal expiration, that is the residual volume (RV).
It is conducted using a spirometer which is capable of measuring lung volumes using techniques of dilution.
During spirometry, the following measurements can be determined:
Forced vital capacity (FVC)/vital capacity (VC): The maximum volume of air exhaled in one single forced breathe.
Forced expiratory volume in one second (FEV1)
FEV1/FVC ratio
Peak expiratory flow (PEF): the maximum amount of air flow exhaled in one blow.
Forced expiratory flow (mid expiratory flow): the flow at 25%, 50% and 75% of FVC
Inspiratory vital capacity (IVC): The maximum volume of air inhaled after a full total expiration.Anatomical dead space is measured using a single breath nitrogen washout called the Fowler’s method.
Residual volume and total lung capacity are both measured using the body plethysmograph or helium dilution
The functional residual capacity is usually measured using a nitrogen washout or the helium dilution technique.
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This question is part of the following fields:
- Clinical Measurement
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Question 80
Incorrect
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Which of the following is the smallest value of pressure?
Your Answer:
Correct Answer: 14.69 psi
Explanation:The SI unit of pressure is the pascal (Pa) and it is equal to one newton (N) per square meter (m2) or N/m2.
1 atmosphere (atm) is the equivalent of:
101325 Pa760 mmHg
1.01325 bar
1033.23 cmH2O.
14.69 pounds per square inch (psi)
1013.25 millibar (mbar) or hectopascals (hPa), and14.69 psi is equal to one atmosphere. The other values are equal to two atmospheres of pressure.
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This question is part of the following fields:
- Basic Physics
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Question 81
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 82
Incorrect
-
Out of the following, which therapy for acute severe asthma or life-threatening asthma has been NOT been approved in recent guidelines?
Your Answer:
Correct Answer: Heliox
Explanation:Recommendations from the British Thoracic Society for acute severe asthma or life-threatening asthma are:
1. Give controlled supplementary oxygen to all hypoxemic patients with acute severe asthma titrated to maintain a SpOâ‚‚ level of 94 98%.
2. Use high-dose inhaled ?â‚‚ agonists as first-line agents in patients with acute asthma and administer them as early as possible. Reserve
intravenous ?â‚‚ agonists for those patients in whom inhaled therapy cannot be used reliably.
3. Give steroids in adequate doses to all patients with an acute asthma attack.
4. Add nebulized ipratropium bromide (0.5 mg 4–6 hourly) to ?₂ agonist treatment for acute severe or life-threatening asthma or those with a poor initial response to ?₂ agonist therapy.
5. Consider aminophylline for children with severe or life-threatening asthma unresponsive to maximal doses of bronchodilators and steroids.A review (including 12 case reports, three RCTs, and five other observational studies) of ketamine use in adults and children in status asthmaticus reported that ketamine is a potential bronchodilator. Still, prospective trials are needed before conclusions about effectiveness can be drawn.
Heliox has no place in the current guidelines issued by the British Thoracic Society.
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This question is part of the following fields:
- Anatomy
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Question 83
Incorrect
-
After establishing a cardiopulmonary bypass, the right atrium is opened to repair the tricuspid valve.
Out of the following, which is NOT a part of the right atrium?Your Answer:
Correct Answer: Trabeculae carnae
Explanation:The right atrium receives blood supply from the SVC, IVC, and coronary sinus. It forms the right border of the heart.
The interior of the right atrium has 5 distinct features:
1. Sinus venarum – smooth, thin-walled posterior part of the right atrium where the SVC, IVC, and coronary sinus open
2. Musculi pectinati – an anterior rough, wall of pectinate muscles
3. Tricuspid valve orifice – the opening through which the right atrium empties blood into the right ventricle
4. Crista terminalis – separates the rough (musculi pectinati) from the smooth (sinus venarum) internally
5. Fossa ovalis – a thumbprint size depression in the interatrial septum which is a remnant of the oval foramen and its valve in the foetusThe trabeculae carneae are irregular muscular elevations that form the interior of the right ventricle.
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This question is part of the following fields:
- Anatomy
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Question 84
Incorrect
-
Given the following values:
Expired tidal volume = 800 ml
Plateau pressure = 50 cmH2O
PEEP = 10 cmH2O
Compute for the static pulmonary compliance.Your Answer:
Correct Answer: 20 ml/cmH2O
Explanation:Compliance of the respiratory system describes the expandability of the lungs and chest wall. There are two types of compliance: dynamic and static.
Dynamic compliance describes the compliance measured during breathing, which involves a combination of lung compliance and airway resistance. Defined as the change in lung volume per unit change in pressure in the presence of flow.
Static compliance describes pulmonary compliance when there is no airflow, like an inspiratory pause. Defined as the change in lung volume per unit change in pressure in the absence of flow.
For example, if a person was to fill the lung with pressure and then not move it, the pressure would eventually decrease; this is the static compliance measurement. Dynamic compliance is measured by dividing the tidal volume, the average volume of air in one breath cycle, by the difference between the pressure of the lungs at full inspiration and full expiration. Static compliance is always a higher value than dynamic
Static compliance can be computed using the formula:
Cstat = Tidal volume/Plateau pressure – PEEP
Substituting the values given,
Cstat = 800/50-10
Cstat = 20 ml/cmH2O -
This question is part of the following fields:
- Physiology
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Question 85
Incorrect
-
A 59-year-old smoker booked for an emergency laparotomy is in the anaesthetic room prior to intubation. He is breathing room air and an arterial blood gas is obtained on insertion of an arterial cannula and sent for analysis.
The following results are available:
Haemoglobin 75 g/L
PaO2 10.7 kPa
PaCO2 5.2 kPa
After intravenous induction, intubation is difficult and he rapidly begins to de-saturate.
Which of the following is most effective in prolonging the oxygen de-saturation time?Your Answer:
Correct Answer: Pre-oxygenation with 100% O2 for three minutes
Explanation:Breathing 100% oxygen for three minutes will provide the best reservoir of oxygen during apnoea by oxygenating the functional residual capacity (FRC).
Sitting at 45 degrees might increase the FRC and improve oxygen reserve but not compared with 100% oxygenation.
The following table compares the oxygen reserves in the body following pre-oxygenation with room air and 100% oxygen:
Compartment Factors Room air (mL) 100% O2 (mL)
Lung FAO2, FRC 630 2850
Plasma PaO2, DF, PV 7 45
Red blood cells Hb, TGV, SaO2 788 805
Myoglobin – 200 200
Interstitial space – 25 160FAO2 = alveolar fraction of oxygen.
FRC = Functional residual capacity.
PaO2 = partial pressure of oxygen dissolved in arterial blood
DF = dissolved form.
PV = plasma volume.
TG = total globular volume .
Hb = haemoglobin concentration.
SaO2 = arterial oxygen saturationStopping smoking one month prior to surgery will not be more effective than pre-oxygenation with 100% oxygen though it may reduce postoperative pulmonary complications. Note that both long term and short term abstinence reduces pulse rate and blood pressure thus reducing oxygen consumption and also reduce carboxyhaemoglobin levels.
Blood transfusion will not make a big difference in oxygen reserve, particularly if a blood transfusion is administered within 12-24-hours before surgery.
Heliox (79% helium and 21% oxygen) despite its lower viscosity is unlikely to be more effective than 100% oxygen .
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This question is part of the following fields:
- Pathophysiology
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Question 86
Incorrect
-
Which of the following statements is true regarding oxytocin?
Your Answer:
Correct Answer: Reduces the threshold for depolarisation of the uterine smooth muscle
Explanation:Oxytocin is secreted by the posterior pituitary along with Antidiuretic Hormone (ADH). It increases the contraction of the upper segment (fundus and body) of the uterus whereas the lower segment is relaxed facilitating the expulsion of the foetus.
Oxytocin acts through G protein-coupled receptor and phosphoinositide-calcium second messenger system to contract uterine smooth muscle.
It has 0.5 to 1 % ADH activity introducing possibilities of water intoxication when used in high doses.
The sensitivity of the uterus to oxytocin increases as the pregnancy progresses.
It is used for induction of labour in post maturity and uterine inertia.
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This question is part of the following fields:
- Pharmacology
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Question 87
Incorrect
-
At a pH of 7, pure water has an hydrogen ion concentration of?
Your Answer:
Correct Answer: 100 nanomol/L
Explanation:pH is the negative log to the base 10 of hydrogen ion concentration:
So, what power produces the answer?
pH = – log10 [H+]
Making [H+] the subject:
[H+] = 10-pH
Substituting, [H+] = 10-7
One nanomole = 1 x 10-9 or 0.000000001
10-7 = 1x 0.0000001 or 10 x 0.00000001 or 100 x 0.000000001
100 nanomole
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This question is part of the following fields:
- Basic Physics
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Question 88
Incorrect
-
A caudal epidural block is planned for a 6-year-old child scheduled for an inguinoscrotal hernia repair under general anaesthesia. The weight of the child is 20kg.
The most important safety aspect the anaesthetist must keep in mind while performing the block is?Your Answer:
Correct Answer: Limiting the bupivacaine dose to no more than 40 mg
Explanation:Choosing an appropriate dose of local anaesthetic to reduce the chance of toxicity is the most important safety aspect in performing a caudal block.
The caudal will have to be inserted following induction of anaesthesia as performing it in an awake child is not a viable option.
The patient is placed in the lateral position and the sacral hiatus is identified. Under strict asepsis, a needle ( usually a 21-23FG needle) is advanced at an angle of approximately 55-65° to the coronal plane at the apex of the sacrococcygeal membrane. When there is loss of resistance, thats the endpoint. The needle must first be aspirated before anaesthetic agent is injected because there is a risk (1 in 2000) of perforating the dura or vascular puncture.
Alternatively, a 22-gauge plastic cannula can be used. Following perforation of the sacrococcygeal membrane, the stilette is removed and only the blunter plastic cannula is advanced. This reduces the risk of intravascular perforation.
Eliciting an appropriate end motor response at an appropriate current strength when the caudal and epidural spaces are stimulated helps in improving the efficacy and safety of neural blockade. A 22G insulated needle is advanced in the caudal canal until a pop is felt. If the needle is placed correctly, an anal sphincter contractions (S2 to S4) is seen when an electrical stimulation of 1-10 mA is applied.
The application of ultrasound guidance in identification of the caudal epidural space has been shown to prevent inadvertent dural puncture and to increase the safety and efficacy of the block in children.
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This question is part of the following fields:
- Pharmacology
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Question 89
Incorrect
-
All the following statements are false regarding carbamazepine except
Your Answer:
Correct Answer: Has neurotoxic side effects
Explanation:Phenytoin, Carbamazepine, and Valproate act by inhibiting the sodium channels when these are open. These drugs also prolong the inactivated stage of these channels (Sodium channels are refractory to stimulation till these reach the closed/ resting phase from inactivated phase)
Carbamazepine is the drug of choice for partial seizures and trigeminal neuralgia
It can have neurotoxic side effects. Major neurotoxic effects include dizziness, headache, ataxia, vertigo, and diplopia
After single oral doses of carbamazepine, the absorption is fairly complete and the elimination half-life is about 35 hours (range 18 to 65 hours). During multiple dosing, the half-life is decreased to 10-20 hours, probably due to autoinduction of the oxidative metabolism of the drug.
It is metabolized in liver into active metabolite, carbamazepine-10,11-epoxide.
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This question is part of the following fields:
- Pharmacology
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Question 90
Incorrect
-
Which of the following options are correct regarding epidemiology in public health?
Your Answer:
Correct Answer: Cross-sectional surveys can be used to estimate the prevalence of a condition in the population
Explanation:The incidence is the number of new cases per population in a given time period whereas prevalence is the total number of cases per population at a particular point in time.
For chronic diseases, prevalence is used rather than incidence. Prevalence differs from incidence proportion as prevalence includes all cases (new and pre-existing cases) in the population at the specified time whereas incidence is limited to new cases only. In acute diseases the prevalence and incidence are similar. For conditions such as the common cold the incidence may be greater than the prevalence
Prevalence is usually greater than the incidence rate for a chronic disease as prevalence includes all cases (new and pre-existing cases) in the population at a specified time whereas incidence is only limited to new cases.
Prevalence = incidence * duration of condition
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This question is part of the following fields:
- Statistical Methods
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Question 91
Incorrect
-
The following is normally higher in concentration extracellularly than intracellularly
Your Answer:
Correct Answer: Sodium
Explanation:The ions found in higher concentrations intracellularly than outside the cells are:
ATP
AMP
Potassium
Phosphate, and
Magnesium Adenosine diphosphate (ADP)Sodium is a primarily extracellular ion.
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This question is part of the following fields:
- Physiology
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Question 92
Incorrect
-
A new clinical trial evaluates the effect of a new drug Z on all-cause mortality. The rate of death in the group receiving this drug is 8%, compared with 16% in the control group.
What is the number needed to treat with drug Z to prevent death?Your Answer:
Correct Answer: 13
Explanation:Number needed to treat is a measure of the impact of a treatment or intervention that is often used to communicate results to patients, clinicians, the public and policymakers. It states how many patients need to be treated for one additional patient to experience an adverse outcome (e.g. a death).
It is calculated as the inverse of the absolute risk reduction and is rounded to the next highest whole number.
The absolute risk reduction is 8% (16% – 8%). 100/8 = 12.5, so rounding up the next integer this gives at NNT of 13. i.e. you would need to give the new drug to 13 people to ensure that you prevented one death.
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This question is part of the following fields:
- Statistical Methods
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Question 93
Incorrect
-
Which of the following statements is true regarding Entonox?
Your Answer:
Correct Answer: Exists only in gaseous form in a cylinder (under normal working conditions)
Explanation:Entonox is a gas that consists of 50% oxygen and 50% Nitrous oxide. Nitrous oxide is sometimes used for anaesthetics but in this combination, it works as a short-acting painkiller.
Under normal working conditions, it exists only in gaseous form in a cylinder. The gauge pressure of a full Entonox cylinder is 137 bar.
Entonox cylinders should be stored horizontally at a temperature above 0 C. At temperatures below this the nitrous oxide component may separate.
Pseudocritical temperature and pseudocritical pressure can be defined as the molal average critical temperature and pressure of mixture components. In other words, the pseudo-critical temperature is the temperature at which the two gases separate. The pseudo-critical temperature of Entonox is approximately -5.50 C
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This question is part of the following fields:
- Pharmacology
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Question 94
Incorrect
-
Which medical gas cylinders have the correct colour codes?
Your Answer:
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 95
Incorrect
-
Suppose the afterload and myocardial contractility remain unchanged, which of the following factors in the pressure-volume loop indicates an increase in the preload of the left ventricle?
Your Answer:
Correct Answer: Increased end-diastolic volume
Explanation:If the afterload and myocardiac contractility remains unchanged, an increase in the preload can be attributed to an increase in end-diastolic volume.
Preload can be defined as the initial stretching of the cardiac myocytes prior to contraction. Preload, therefore, is related to muscle sarcomere length. Because sarcomere length cannot be determined in the intact heart, other indices of preload are used such as ventricular end-diastolic volume or pressure. When venous return to the heart is increased, the end-diastolic pressure and volume of the ventricles are increased, which stretches the sarcomeres, thereby increasing their preload.
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This question is part of the following fields:
- Basic Physics
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Question 96
Incorrect
-
A 62-year-old woman, presents to emergency department with an ischaemic left colon.
Multiple arteries arise from the aorta at the level of the L3 vertebrae, which is most likely to be involved in this pathology?Your Answer:
Correct Answer: Inferior mesenteric artery
Explanation:The inferior mesenteric artery arises from the abdominal aorta at the level of the L3 vertebrae and supplies blood to the final third of the transverse colon, the descending colon, the sigmoid colon and the uppermost part of the rectum.
It is the artery most likely to affect the left colon.
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This question is part of the following fields:
- Anatomy
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Question 97
Incorrect
-
An 80-year-old female presents to the emergency department with symptoms consistent with mesenteric ischemia. She is quickly shifted to the operation theatre for an emergency laparotomy.
On exploration, the segment of the colon from the splenic flexure down to the rectum is ischemic.
The artery blocked in this scenario arises at what vertebral level from the aorta?Your Answer:
Correct Answer: L3
Explanation:The hindgut is from the distal third of the transverse colon down to the upper one-third of the anal canal. The inferior mesenteric artery supplies the hindgut.
The inferior mesenteric artery arises from the aorta behind the inferior border of the third part of the duodenum 3–4 cm above the aortic bifurcation, at the third lumbar vertebra level.
The important landmarks of vessels arising from the abdominal aorta at different levels of vertebrae are:
T12 – Coeliac trunk
L1 – Left renal artery
L2 – Testicular or ovarian arteries
L3 – Inferior mesenteric artery
L4 – Bifurcation of the abdominal aorta
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This question is part of the following fields:
- Anatomy
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Question 98
Incorrect
-
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:
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 99
Incorrect
-
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:
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 100
Incorrect
-
A 63-year old man has palpitations and goes to the emergency room. An ECG shows tall tented T waves, which corresponds to phase 3 of the cardiac action potential.
The shape of the T wave is as a result of which of the following?Your Answer:
Correct Answer: Repolarisation due to efflux of potassium
Explanation:Cardiac conduction
Phase 0 – Rapid depolarization. Opening of fast sodium channels with large influx of sodium
Phase 1 – Rapid partial depolarization. Opening of potassium channels and efflux of potassium ions. Sodium channels close and influx of sodium ions stop
Phase 2 – Plateau phase with large influx of calcium ions. Offsets action of potassium channels. The absolute refractory period
Phase 3 – Repolarization due to potassium efflux after calcium channels close. Relative refractory period
Phase 4 – Repolarization continues as sodium/potassium pump restores the ionic gradient by pumping out 3 sodium ions in exchange for 2 potassium ions coming into the cell. Relative refractory period
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 101
Incorrect
-
Arterial pressure waveforms give an indication of the operation of the heart and the patient's clinical state.
Which of the following listed characteristics of arterial waveforms is most indicative of myocardial contractility?Your Answer:
Correct Answer: Slope of the upstroke of the curve.
Explanation:Arterial pressure waveforms is an invasive form of monitoring cardiac parameters. It provides a lot of information on the performance of the heart from different sections, including:
Cardiac measurements:
Heart rate
Systolic pressure
Diastolic pressure
Mean arterial pressure
Pulse pressure
Change in pulse amplitude corresponding to respiratory changes
Slope of anacrotic limb associated with aortic stenosisFrom the shape of the arterial waveform displayed:
Slope of anacrotic limb represents aortic valve and LVOT flow
Indications of aortic stenosis (AS): Slurred wave, collapsing wave
Rapid systolic decline in LVOTO
Bisferiens wave in HOCM
Low dicrotic notch in states with poor peripheral resistance
Position and quality of dicrotic notch as a reflection of the damping coefficientFor this question, the upstroke slope of the pressure wave is indicative of myocardial contractility and is mathematically represented as:
dP/dt, which represents a change of pressure with regards to time.
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This question is part of the following fields:
- Clinical Measurement
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Question 102
Incorrect
-
Sugammadex binds to certain drugs that affect neuromuscular function during anaesthesia in a stereospecific, non-covalent, and irreversible manner.
It has the greatest impact on the activity of which of the following drugs?Your Answer:
Correct Answer: Vecuronium
Explanation:Sugammadex is a modified cyclodextrin that works as an aminosteroid neuromuscular blocking (nmb) reversal agent. By encapsulating each molecule in the plasma, it rapidly reverses rocuronium and, to a lesser extent, vecuronium-induced neuromuscular blockade. Consequently, a  concentration gradient favours the movement of these nmb agents away from the neuromuscular junction. Pancuronium-induced neuromuscular blockade at low levels has also been reversed.
By inhibiting voltage-dependent calcium channels at the neuromuscular junction, antibiotics in the aminoglycoside group potentiate neuromuscular blocking agents. This can be reversed by giving calcium but not neostigmine or sugammadex.
Sugammadex will not reverse the effects of mivacurium, which belongs to the benzylisoquinolinium class of drugs.
A phase II or desensitisation block occurs when the motor end-plate becomes less sensitive to acetylcholine as a result of an overdose or repeated administration of suxamethonium. The use of neostigmine has been shown to be effective in reversing this weakness.
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This question is part of the following fields:
- Pharmacology
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Question 103
Incorrect
-
What is the most sensitive method of detecting an intra-operative air embolism?
Your Answer:
Correct Answer: Transoesophageal echocardiogram
Explanation:An intra-operative air embolism occurs when air becomes trapped in the blood vessels during surgery.
A transoesophageal echocardiography (OE) uses invasive echocardiography to monitor the integrity and performance of the heart. It is the gold standard as it provides real-time imaging of the heart to enable early diagnosis and treatment.
Precordial doppler ultrasonography can also be used to detect into-operative air emboli. It is non-invasive and more practical, but is less sensitive.
A change in end-tidal CO2 could be indicative of and increase in physiological dead-space, but could also be indicative of any processes that reduces the excretion or increases the production of CO2, making it non-specific.
A transoesophageal stethoscope can be used to listen for the classic mill-wheel murmur produced by a large air embolus.
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This question is part of the following fields:
- Pathophysiology
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Question 104
Incorrect
-
All of the following statements are false regarding salmeterol except:
Your Answer:
Correct Answer: Is more potent than salbutamol at the beta-2 receptor
Explanation:Salmeterol is a long-acting Beta 2 selective agonist. Therefore it is only used for prophylaxis whereas salbutamol is a short-acting Beta 2 agonist and is thus used for the treatment of acute attacks of asthma.
Salmeterol is 15 times more potent than salbutamol at the Beta 2 receptor but 4 times less potent at the Beta 1 receptor.
Tachyphylaxis to the unwanted side effects commonly occurs, but not to bronchodilation.
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This question is part of the following fields:
- Pharmacology
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Question 105
Incorrect
-
What separates the tunica media from the tunica adventitia in a blood vessel?
Your Answer:
Correct Answer: External elastic lamina
Explanation:Blood vessels (except capillaries and venules) have three distinctive layers (innermost to outermost):
1. Tunica intima
2. Tunica media
3. Tunica adventitiaThe tunica media contains smooth muscle cells and is separated from the intima layer by the internal elastic lamina and the adventitia by the external elastic lamina.
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This question is part of the following fields:
- Anatomy
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Question 106
Incorrect
-
A patient on admission is given an infusion of 1000 mL of 10% glucose and 500 mL of 20% lipid over a 24 hour period.
Which of these best approximates to the energy input over this time period?Your Answer:
Correct Answer: 1300 kcal
Explanation:1% solution contains 1 g of substance per 100 mL.
A solution of 10% glucose is 10 g/100mL. Therefore 1000 mL of this glucose solution will contain 100 g.
1 g of glucose yields about 4 kcal of energy. One litre of 10% glucose will therefore release approximately 4x100g = 400 kcal of energy.
A solution of 20% fat is 20 g/100mL. Therefore 1000 mL of this fat solution will have 200 g and 500 mL will contain 100 g.
1 g of fat yields approximately 9 kcal. 500 mL of 20% fat therefore has the potential to yield 900 kcal of energy.
The total energy input over this 24 hour period is approximately 400kcal + 900kcal = 1300 kcal.
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This question is part of the following fields:
- Physiology
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Question 107
Incorrect
-
A 78-year-old man with a previous history of ischaemic heart disease is admitted to hospital. He is scheduled for a cardiopulmonary exercise test (CPX) before he undergoes an elective abdominal aneurysm repair.
What measurement obtained during a CPX test alone provides the best indication for postoperative mortality?Your Answer:
Correct Answer: Anaerobic threshold
Explanation:Cardiopulmonary exercise testing (CPX, CPEX, CPET) is a non-invasive testing method used to determine the performance of the heart, lungs and skeletal muscle. It measures the exercise tolerance of the patient.
The parameters measured include:
ECG and ST-segment analysis and blood pressure
Oxygen consumption (VO2)
Carbon dioxide production (VCO2)
Gas flows and volumes
Respiratory exchange ratio (RER)
Respiratory rate
Anaerobic threshold (AT)The anaerobic threshold (AT) is an estimate of exercise ability. Any measurement below 11 ml/kg/min is usually related with an increase in mortality, especially when there is a background of myocardial ischaemia occurring during the test.
Peak VO2 <20 mL/kg with a low AT have a correlation with postoperative complications and a 30 day mortality. The CPX test is used for risk-testing patients prior to surgery to determine the appropriate postoperative care facilities. The V slope measured in CPX testing represents VO2 versus VCO2 relationship. During AT, the ramp of V slope increases, but does not provide a picture of postoperative mortality.
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This question is part of the following fields:
- Clinical Measurement
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Question 108
Incorrect
-
Anaesthetic awareness is most probable in general anaesthesia for which surgical operation?
Your Answer:
Correct Answer: Emergency surgery for major trauma
Explanation:Awareness during general anaesthesia is a frightening experience, which may result in serious emotional injury and post-traumatic stress disorder.
The incidence of awareness during general anaesthesia with current anaesthetic agents and techniques has been reported as 0.2-0.4% in nonobstetric and noncardiac surgery, as 0.4% during caesarean section, and as 1.5% in cardiac surgery.
The incidence during major trauma surgery is higher. Incidence of recall has been reported to be as high as 11-43% in major trauma cases.
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This question is part of the following fields:
- Physiology
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Question 109
Incorrect
-
The typical fluid compartments in a normal 70kg male are:
Your Answer:
Correct Answer: intracellular>extracellular
Explanation:Body fluid compartments in a 70kg male:
Total volume=42L (60% body weight)
Intracellular fluid compartment (ICF) =28L
Extracellular fluid compartment (ECF) = 14LECF comprises:
Intravascular fluid (plasma) = 3L
Extravascular fluid = 11LExtravascular fluids comprises:
Interstitial fluid = 10.5L
Transcellular fluid = 0.5L -
This question is part of the following fields:
- Physiology
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Question 110
Incorrect
-
Pressure volume loop represents the compliance of left ventricle.
Considering there is no change in preload and myocardial contractility, which physiological change may result an increase in left ventricular afterload?Your Answer:
Correct Answer: Increased end-systolic volume
Explanation:If there is no change in preload and myocardial contractility, there will be decrease in end-diastolic volume and stroke volume. So there must be increase in end-systolic volume.
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This question is part of the following fields:
- Physiology
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Question 111
Incorrect
-
Seven days after undergoing an inguinal hernia repair, a 24-year old male presents with a wound that is erythematous, tender and has a purulent discharge. The most likely cause of this is which of the following?
Your Answer:
Correct Answer: Infection with Staphylococcus aureus
Explanation:Staphylococcus aureus infection is the most likely cause.
Surgical site infections (SSI) occur when there is a breach in tissue surfaces and allow normal commensals and other pathogens to initiate infection. They are a major cause of morbidity and mortality.
SSI comprise up to 20% of healthcare associated infections and approximately 5% of patients undergoing surgery will develop an SSI as a result.
The organisms are usually derived from the patient’s own body.Measures that may increase the risk of SSI include:
-Shaving the wound using a single use electrical razor with a disposable head
-Using a non iodine impregnated surgical drape if one is needed
-Tissue hypoxia
-Delayed prophylactic antibiotics administration in tourniquet surgery, patients with a prosthesis or valve, in clean-contaminated surgery of in contaminated surgery.Measures that may decrease the risk of SSI include:
1. Intraoperatively
– Prepare the skin with alcoholic chlorhexidine (Lowest incidence of SSI)
-Cover surgical site with dressingIn contrast to previous individual RCT’s, a recent meta analysis has confirmed that administration of supplementary oxygen does not reduce the risk of wound infection and wound edge protectors do not appear to confer benefit.
2. Post operatively
Tissue viability advice for management of surgical wounds healing by secondary intentionUse of diathermy for skin incisions
In the NICE guidelines the use of diathermy for skin incisions is not advocated. Several randomised controlled trials have been undertaken and demonstrated no increase in risk of SSI when diathermy is used. -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 112
Incorrect
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Which type of epithelium lines the luminal surface of the oesophagus?
Your Answer:
Correct Answer: Non keratinised stratified squamous epithelium
Explanation:Normally, the oesophagus is lined by non-keratinized stratified squamous epithelium. This epithelium can undergo metaplasia and convert to the columnar epithelium (stomach’s lining) in long-standing GERD that leads to Barret’s oesophagus.
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This question is part of the following fields:
- Anatomy
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Question 113
Incorrect
-
Regarding anaesthetic breathing circuits, which one would be the best for spontaneous breathing?
Your Answer:
Correct Answer: Lack circuit
Explanation:Among the breathing circuits, the Lack circuit is the most efficient for spontaneous breathing.
An outer coaxial tube is present to deliver fresh air; exhaust air is routed to an inner tube, which is then delivered to a scavenging system. An expiratory valve is seen at the patient end, which is an advantage over other circuits. Moreover, the Lack circuit prevents rebreathing slightly greater than the alveolar minute ventilation at 4-5 litres per minute.
The Bain circuit prevents rebreathing at 160-200ml/kg per minute, and is a co-axial version of the Mapleson D circuit.
The Mapleson E circuit prevent rebreathing at a fresh gas flow (FGF) of approximately twice the patient’s normal minute volume. A modification of this, the Mapleson F, has a reservoir bag at the opposite end for the FGF. This circuit is appropriate for paediatric patients with a body weight less than 20 kg.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 114
Incorrect
-
Among the following options which one compares variance within the group and variance between groups?
Your Answer:
Correct Answer: ANOVA
Explanation:ANOVA is based upon within group variance (i.e. the variance of the mean of a sample) and between group variance (i.e. the variance between means of different samples). The test works by finding out the ratio of the two variances mentioned above. (Commonly known as F statistic).
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This question is part of the following fields:
- Statistical Methods
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Question 115
Incorrect
-
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:
Correct 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 116
Incorrect
-
Which plasma protein will bind the thyroid hormone triiodothyronine (T3) more readily?
Your Answer:
Correct 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 117
Incorrect
-
What vessel is NOT considered a major branch of the descending thoracic aorta?
Your Answer:
Correct Answer: Inferior thyroid artery
Explanation:The descending thoracic aorta begins at the lower border of T4 near the midline as a continuation of the arch of the aorta. It descends and ends at the level of T12 at the aortic hiatus in the diaphragm, where it becomes the abdominal aorta.
The aorta gives off the following branches: (descending order)
1. Bronchial arteries
2. Mediastinal arteries
3. Oesophageal arteries
4. Pericardial arteries
5. Superior phrenic arteriesThe posterior intercostal arteries are branches that originate throughout the length of the posterior aspect of the descending thoracic aorta.
The inferior thyroid artery is usually derived from the thyrocervical trunk, a branch of the subclavian artery.
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This question is part of the following fields:
- Anatomy
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Question 118
Incorrect
-
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:
Correct 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 119
Incorrect
-
Regarding gas laws, which one best explains the ability of a Bourdon gauge to measure temperature?
Your Answer:
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 120
Incorrect
-
Considering research studies, which of the following is considered as a limitation of the Delphi method?
Your Answer:
Correct Answer: Potential low response rates
Explanation:The Delphi technique was developed in the 1950s and is a widely used and accepted method for achieving convergence of opinion concerning real-world knowledge solicited from experts within certain topic areas. Choosing the appropriate subjects is the most important step in the entire process because it directly relates to the quality of the results generated, despite this, there is no exact criterion currently listed in the literature concerning the selection of Delphi participants.
Therefore, due to the potential scarcity of qualified participants and the relatively small number of subjects used in a Delphi study, the ability to achieve and maintain an ideal response rate can either ensure or jeopardize the validity of a Delphi study.
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This question is part of the following fields:
- Statistical Methods
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Question 121
Incorrect
-
The rapid depolarisation phase of the myocardial action potential is caused by:
Your Answer:
Correct Answer: Rapid sodium influx
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 And Biochemistry
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Question 122
Incorrect
-
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:
Correct 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 123
Incorrect
-
Which of the following statements is true with regards to 2,3-diphosphoglycerate (2,3-DPG)?
Your Answer:
Correct Answer: Production is increased in heart failure
Explanation:During glycolysis, 2,3-diphosphoglycerate (2,3-DPG) is
created in erythrocytes by the Rapoport-Luebering shunt.The production of 2,3-DPG increases for several conditions
in the presence of decreased peripheral tissue O2 availability.
Some of these conditions include hypoxaemia, chronic lung
disease anaemia, and congestive heart failure. Thus,
2,3-DPG production is likely an important adaptive mechanism.High levels of 2,3-DPG cause a shift of the curve to the right.
Low levels of 2,3-DPG cause a shift of the curve to the left,
as seen in states such as septic shock and hypophosphatemia. -
This question is part of the following fields:
- Physiology
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Question 124
Incorrect
-
Which one of the following lies above the cephalic vein?
Your Answer:
Correct Answer: None of the above
Explanation:The cephalic vein is one of the primary superficial veins of the upper limb. It overlies most of the fascial planes as it is located in the superficial fascia along the anterolateral surface of the biceps.
It originates in the anatomical snuffbox from the radial side of the superficial venous network of the dorsum of the hand. It travels laterally up the arm to join the basilic vein via the median cubital vein at the elbow.
Near the shoulder, it passes between the deltoid and pectoralis major muscles. It pierces the coracoid membrane (continuation of the clavipectoral fascia) to terminate in the axillary vein’s first part.
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This question is part of the following fields:
- Anatomy
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Question 125
Incorrect
-
A 27-year-old woman presents to emergency department. She is experiencing generalised seizures.
She is given emergency management of her symptoms before being referred to the neurologist who diagnoses her with new onset of tonic-clonic epilepsy.
What is the most appropriate first line of treatment?Your Answer:
Correct Answer: Lamotrigine should be offered as first line of treatment
Explanation:Tonic-clonic (Grand mal) epilepsy is characterised by a general loss of consciousness with violent involuntary muscle contractions.
The NICE guidelines for treatment indicates the use of sodium valproate and lamotrigine, but sodium valproate unsuitable in this case and she is a woman of reproductive age and it is known to have teratogenic effects. Lamotrigine is a more suitable choice, prescribed as 800mg daily.
NICE guidelines also advice an additional prescription of 5mg of folic acid daily for women on anticonvulsant therapy looking to get pregnant. It also warns of the need for extra contraceptive precaution as there is a possibly that the anticonvulsant agent can reduce levels of contraceptive agents.
Stimulation of the vagal nerve stimulation is only necessary in patients who are refractory to medical treatment and not candidates for surgical resection.
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This question is part of the following fields:
- Pathophysiology
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Question 126
Incorrect
-
With regards to devices for temperature management, all of these are used EXCEPT:
Your Answer:
Correct Answer: Thermistors use the resistance of a semiconductor bead which 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 127
Incorrect
-
A 21-year-old woman presents to ER following the deliberate ingestion of 2 g of amitriptyline. On clinical examination:
Glasgow coma score: 10
Pulse rate: 140 beats per minute
Blood pressure: 80/50 mmHg.
ECG showed a QRS duration of 233 Ms.
Which of the following statement describes the most important initial course of action?Your Answer:
Correct Answer: Give fluid boluses
Explanation:The first line of treatment in case of hypotension is fluid resuscitation.
Activated charcoal can be used within one hour of tricyclic antidepressant ingestion but an intact and secure airway must be checked before intervention. The risk of aspiration should be assessed.
Vasopressors are indicated for the treatment of hypotension following (Tricyclic Antidepressant) TCA overdose when patients fail to respond to fluids and bicarbonate.
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This question is part of the following fields:
- Pharmacology
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Question 128
Incorrect
-
Regarding a paramagnetic oxygen analyser, the following statements are TRUE:
Your Answer:
Correct Answer: Utilises null deflection
Explanation:The electrons in the outer shell of an oxygen molecule are unpaired, thus it has paramagnetic properties and is attracted into a magnetic field.
It utilizes null deflection -True
Null deflection is a crucial principle in paramagnetic analysers (reflected beam of light on two photocells) which gives very accurate results (typically 0.1%).It can be used to measure the concentration of diamagnetic gases – False
Since most other gases are weakly diamagnetic they are repelled by a magnetic field (nitric oxide is also paramagnetic).Can measure gases dissolved in the blood – False
For accurate analysis the sample gas must be dried before passing into the analysis cell, for example, by passage through silica gel. Therefore, they are unsuitable to measure gases dissolved in blood.Does not require calibration – False
As with most measurement instruments paramagnetic analysers must be calibrated before use.E) The readings are unaffected by water vapour – False
Water vapour affects the readings hence for accurate analysis the sample gas must be dried before passing into the analysis cell, for example, by passage through silica gel. That is why they are unsuitable to measure dissolved blood gases. -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 129
Incorrect
-
A 73-year-old woman is admitted to the coronary care unit. She has been diagnosed with an acute myocardial infarction and has gone into a cardiogenic shock. As part of her treatment, she is prescribed a dobutamine infusion and placed on full haemodynamic monitoring.
Over the next four days, her measurement are:
Day1 Day2 Day3 Day4
Infusion rate of dobutamine (mcg/kg/min): 5 10 15 25
Cardiac output (L/min): 3.4 3.2 3.1 3.2
Pulse rate (beats/min): 110 112 105 100
Blood pressure (mmHg): 80/60 75/61 83/60 81/56
Haemodynamic changes are noted as a result of response to treatment. What is the most probable cause?Your Answer:
Correct Answer: Tolerance by down-regulation of beta1 receptors
Explanation:Tachyphylaxis is the swiftly declining response to successive drug doses which vastly reduces its effectiveness in a short space of time, mostly as a result of an acute consumption of neurotransmitters.
Tolerance or desensitisation is the slow decline in a person’s reaction to a drug due to continued use. It requires a longer time span than tachyphylaxis, usually over days or weeks.
Down- regulation is a reduction in the amount of receptors available on target cells which decreases the affinity of the agent to the cells. For this to occur, the down-regulation of receptors must occur at a rate faster than receptor synthesis. This down-regulation often occurs with beta1 receptors due to:
1) The transportation or receptors from the cell surface to the interior of the cell
2) Degradation of receptors occurring over time.
In this case, dobutamine is prescribed to treat cardiogenic shock. It is able to function by binding to beta1-adrenergic receptors to increase the contraction of the heart, thereby improving cardiac output. It also binds to beta2- and alpha1-adrenergic receptors to balance out the effects produced by binding to beta1 receptors and reduce the risk of system vasculature responses.
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This question is part of the following fields:
- Clinical Measurement
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Question 130
Incorrect
-
A randomized study aimed at finding out the efficacy of a novel anticoagulant, in preventing stroke in patients suffering from atrial fibrillation, relative to those already available in the market was performed. A 59 year old woman volunteered for it and was randomised to the treatment arm. A year later, following findings were reported:
165 out of 1050 patients who were prescribed the already prevalent medicine had a stroke while the number of patients who had a single stroke after using the new drug was 132 out of 1044.
In order to avoid one stroke case, what is the number of patients that need to be treated?Your Answer:
Correct Answer: 32
Explanation:Number needed to treat can be defined as the number of patients who need to be treated to prevent one additional bad outcome.
It can be found as:
NNT=1/Absolute Risk Reduction (rounded to the next integer since number of patients can be integer only).
where ARR= (Risk factor associated with the new drug group) — (Risk factor associated with the currently available drug)
So,
ARR= (165/1050)-(132/1044)
ARR= (0.157-0.126)
ARR= 0.031
NNT= 1/0.031
NNT=32.3
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This question is part of the following fields:
- Statistical Methods
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Question 131
Incorrect
-
The leading cause of perioperative anaphylaxis per hundred thousand administrations is?
Your Answer:
Correct Answer: Teicoplanin
Explanation:The leading cause of perioperative anaphylaxis in the UK currently are antibiotics. They account for 46% of cases with identified causative agents. Co-amoxiclav and teicoplanin between them account for 89% of antibiotic-induced perioperative anaphylaxis
Neuromuscular blocking agents (NMBAs) are the second leading cause and account for 33% of case.
Chlorhexidine (0.78/100,000 administrations)
Co-amoxiclav (8.7/100,000 administrations)Suxamethonium (11.1/100,000 administrations)
Patent blue dye (14.6/100,000 administrations)
Teicoplanin (16.4/100,000 administrations)Anaphylaxis to chlorhexidine periop poses a significant risk in the healthcare setting because of its widespread use with some being fatal.
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This question is part of the following fields:
- Pharmacology
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Question 132
Incorrect
-
A study designed to examine the benefits of adding a new antiplatelet to aspirin after a myocardial infraction. The recorded results give us the percentage of patients that reported myocardial infraction within a three month period. The percentage was 4% and 3% for aspirin and the combination of drugs respectively.
How many further patients needed to be treated in order for one patient to avoid any more heart attacks during 3 months?Your Answer:
Correct Answer: 100
Explanation:Number needed to treat can be defined as the number of patients who need to be treated to prevent one additional bad outcome.
It can be found as:
NNT=1/Absolute Risk Reduction (rounded to the next integer since number of patients can be integer only).
where ARR= (Risk factor associated with the new drug group) — (Risk factor associated with the currently available drug)
So,
ARR= (0.04-0.03)
ARR= 0.01
NNT= 1/0.01
NNT=100
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This question is part of the following fields:
- Statistical Methods
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Question 133
Incorrect
-
The solutions that contains the most sodium is?
Your Answer:
Correct Answer: 3500 mL 0.9% N saline
Explanation:Sodium concentration for different fluids
3% N saline 513 mmol/L
5% N saline 856 mmol/L
0.9% N saline 154 mmol/L
Hartmann’s solution 131 mmol/L
0.45% N saline with 5% glucose 77 mmol/LThis means that:
500 mL 5% N saline contains 428 mmol of sodium
1000 mL 3% N saline contains 513 mmol of sodium
3500 mL 0.9% N saline contains 539 mmol of sodium
4000 mL Hartmann’s contains 524 mmol of sodium
6000 mL 0.45% N saline with 5% glucose contains 462 mmol of sodium. -
This question is part of the following fields:
- Physiology
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Question 134
Incorrect
-
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 135
Incorrect
-
The immediate physiological response to massive perioperative blood loss is:
Your Answer:
Correct 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 136
Incorrect
-
During the analysis phase, which of the provided options serves to control confounding factors?
Your Answer:
Correct Answer: Stratification
Explanation:During analytical stage a technique called stratification is used for controlling confounding variables. This technique involves sorting out the data into discernible groups.
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This question is part of the following fields:
- Statistical Methods
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Question 137
Incorrect
-
An inguinal hernia repair under general anaesthesia is scheduled for a fit 36-year-old man (75 kg). For perioperative and postoperative analgesia, you decide to perform an inguinal field block.
Which of the following local anaesthetic solutions is the most appropriate?Your Answer:
Correct Answer: 30 mL bupivacaine 0.5%
Explanation:Perioperative and postoperative analgesia can both be provided by an inguinal hernia field block. The Iliohypogastric and ilioinguinal nerves, as well as the skin, superficial fascia, and deeper structures, must be blocked for maximum effectiveness. The local anaesthetic should ideally have a long duration of action, be highly concentrated, and have a volume of at least 30 mL.
Plain bupivacaine has a maximum safe dose of 2 mg/kg body weight.
Because the patient weighs 75 kg, 150 mg bupivacaine can be safely administered. Both 30 mL 0.5 percent bupivacaine (150 mg) and 60 mL 0.25 percent bupivacaine (150 mg) are acceptable doses, but 30 mL 0.5 percent bupivacaine represents the optimal volume and strength, potentially providing a denser and longer block.
The maximum safe dose of plain lidocaine has been estimated to be between 3.5 and 5 mg/kg. The patient weighs 75 kg and can receive a maximum of 375 mg using the higher dosage regimen:
There are 200 mg of lidocaine in 10 mL of 2% lidocaine (and therefore 11 mL contains 220 mg)
200 mg of lidocaine is contained in 20 mL of 1% lidocaine.While alternatives are available, Although the doses of 11 mL lidocaine 2% and 20 mL lidocaine 1% are well within the dose limit, the volumes used are insufficient for effective field block for this surgery.
With 1 in 200,000 epinephrine, the maximum safe dose of lidocaine is 7 mg/kg. The patient can be given 525 mg in this case. Even with epinephrine, 60 mL of 1% lidocaine is 600 mg, which could be considered an overdose.
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This question is part of the following fields:
- Pharmacology
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Question 138
Incorrect
-
A 30-year-old man has been diagnosed with a direct inguinal hernia.
A direct inguinal hernia passes through Hesselbach's triangle. What forms the medial edge of this triangle?Your Answer:
Correct Answer: Rectus abdominis muscle
Explanation:The inguinal triangle of Hesselbach’s 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 139
Incorrect
-
What statement about endotoxins is true?
Your Answer:
Correct 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 140
Incorrect
-
How many unpaired branches leave the abdominal aorta to supply the abdominal viscera?
Your Answer:
Correct Answer: Three
Explanation:The abdominal arteries are divided into 3 branches;
– 3 main unpaired trunks (celiac trunk, superior mesenteric, inferior mesenteric arteries)
– 6 paired branches
– unpaired median sacral artery.We can group the abdominal aorta as follows;
-Ventral which includes: Coeliac trunk, superior mesenteric and inferior mesenteric arteries
-Lateral: Inferior phrenic, middle suprarenal, renal and gonadal arteries
-Dorsal: Lumbar and median sacral arteries
-Terminal : Right and left common iliac arteriesThe celiac trunk (L1) takes blood the foregut and its found posterior to the stomach. The unpaired superior mesenteric artery supplies blood to the mid-gut.
The paired renal arteries form the inferior suprarenal arteries. The renal arteries arise around L1/L2 and takes blood to either side of the kidneys.
The median sacral artery supplies blood to the lumbar vertebrae the L4 and L5.
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This question is part of the following fields:
- Anatomy
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Question 141
Incorrect
-
A 76-year-old man, presents to his general practitioner with a lump in his left groin. Upon examination, his doctor is able to diagnose a direct inguinal hernia. There are many structures present in the inguinal canal.
Where is the ilioinguinal nerve located in relation to the spermatic cord?Your Answer:
Correct Answer: Anterior to the spermatic cord
Explanation:The answer is anterior to the spermatic cord.
The inguinal canal in men contains the ilioinguinal nerve, the genitofemoral nerve and the spermatic cord.
The ilioinguinal nerve arises of the L1 nerve root with the Iliohypogastric nerve, before entering the inguinal canal from the side, through the muscles of the abdomen, travelling superficial to the spermatic cord.
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This question is part of the following fields:
- Anatomy
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Question 142
Incorrect
-
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:
Correct 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 143
Incorrect
-
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:
Correct 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 144
Incorrect
-
Which compound is secreted only from the adrenal medulla?
Your Answer:
Correct 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 145
Incorrect
-
A 24-year-old female, presents to the emergency department via ambulance. She has just been involved in a car accident. She is examined and undergoes various diagnostic investigations. Her X-ray report states that a fracture was noted on the surgical neck of her humerus.
What structure is most likely to the damaged as a result of a surgical neck fracture of the humerus?Your Answer:
Correct Answer: Axillary nerve
Explanation:Fractures to the surgical neck of the humerus are common place as it is the weakest point of the proximal humerus bone.
The structures most likely to be damaged are the axillary nerve and the posterior circumflex humeral artery as they surround the surgical neck.
The radial nerve runs along the radial groove, so injury to it would likely occur with a mid-shaft fracture of the humerus.
The brachial artery is most likely to be injured as a result of a supracondylar fracture of the humerus which increases the risk of volkmaan’s ischemic contractures.
Injury to the musculocutaneous nerve is least likely to happen and it very uncommon.
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This question is part of the following fields:
- Anatomy
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Question 146
Incorrect
-
According to the statements given which one is most correct concerning the spinal cord?
Your Answer:
Correct Answer: There are 31 pairs of spinal roots
Explanation:The cylindrical structure(spine), starts from the foramen magnum in medulla oblongata at the skull’s base. Its length varies in gender with men having 45cm and women having up to 43 cm. The spine contains 31 pairs of spinal nerves, named according to the spinal region:
– 8 cervical nerve pairs ( C1-C8)
– 12 thoracic nerve pairs ( T1-T12)
– 5 lumbar nerve pairs (L1-L5)
– 5 sacral nerves (S1-S5) and
– 1 coccygeal nerve pair.
These spinal nerves are classified as the peripheral nervous system though they branch from the spinal cord and central nervous system. They interact directly with the spinal cord to modulate the motor and sensory information from the peripheral’s region. As the nerves emerges form the spinal cords they are known as rootlets. They join to form nerve roots, and depending on their position, we have anterior nerve roots and posterior.The spinal cord is supplied by two posterior and one anterior spinal arteries which anastomoses caudally, at the conus medullaris.
The anterior spinal artery supplies the anterior two-thirds of the spinal cord and medulla. Disruption of the anterior spinal artery supply is characterised by ischaemia or infarction of motor tracts (corticospinal) and loss of pain and temperature sensation below the level of the lesion.
The posterior spinal arteries supply the posterior columns (posterior third of the spinal cord). Infarction of the posterior columns results in the loss of proprioceptive, vibration and two-point discrimination.
The spinal cord normally ends at the level of L1 or L2 in an adult and L3 in a newborn.
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This question is part of the following fields:
- Anatomy
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Question 147
Incorrect
-
A double blind placebo control clinical trial is done. Which of these is correct about it?
Your Answer:
Correct Answer: The clinician assessing the effects of the treatment does not know which treatment the patient has been given
Explanation:A ‘double blind crossover study’ happens when every patient receive both treatments.
It is incorrect to say that only half of the patients do not know which treatment they receive because in a double blind placebo control clinical trial ALL of the patients are blind to their treatment choice .
If some of the patients are not treated, they would be aware that they were not being treated and it could not be considered a blind trial.
In a double blind placebo control clinical trial both the clinician and the patient are blind to the treatment choice. The clinician assessing the effects of the treatment, therefore, does not know which treatment the patient has been given.
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This question is part of the following fields:
- Statistical Methods
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Question 148
Incorrect
-
An 80-year-old man will be operated on for an arterial bypass procedure to treat claudication and foot ulceration. The anterior tibial artery will be the target for distal arterial anastomosis.
Which structure is NOT closely related to the anterior tibial artery?Your Answer:
Correct Answer: Tibialis posterior
Explanation:The anterior tibial artery originates from the distal border of the popliteus. In the posterior compartment, it passes between the heads of the tibialis posterior and the oval aperture of the interosseous membrane to reach the anterior compartment.
On entry into the anterior compartment, it runs medially along the deep peroneal nerve.
The upper third of the artery courses between the tibialis anterior and extensor digitorum longus muscles, while the middle third runs between the tibialis anterior and extensor hallucis longus muscles.At the ankle, the anterior tibial artery is located approximately midway between the malleoli. It continues on the dorsum of the foot, lateral to extensor hallucis longus, as the dorsalis pedis artery.
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This question is part of the following fields:
- Anatomy
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Question 149
Incorrect
-
The following statements are about chronic obstructive pulmonary disease (COPD). Which is true?
Your Answer:
Correct Answer: Inhaled corticosteroid usage slows the decline in health status
Explanation:Chronic obstructive pulmonary disease (COPD) is an obstructive, inflammatory lung condition. It encompasses symptoms of emphysema, chronic bronchitis and asthma.
Inhaling high dose steroids are prescribed to treat COPD. They are effective at reducing symptoms and improving lung function. They also work to reduce the number of hospitalisations by decreasing the number of acute exacerbation events. Despite providing effective symptom relief, it cannot slow down the decline of FEV1 as COPD is an irreversible condition.
COPD reduces the FEV1 measurements, as well as the FEV1/FVC ratio.
Breathlessness is a major COPD symptom and can occur at any point in the disease progression, including at an FEV1 >50%.
FEV1 is used in COPD staging, and it is classed as follows:
>80%: Mild or stage I
50 – 79%: Moderate or stage II
30 – 49%: Severe or stage III
<30%: Very severe or stage IV
Patients with mild COPD are usually able to manage their condition on their own, however once the disease progresses to moderate, more GP visits are required, with those in the severe category requiring frequent hospitalisation.Asthma is correlated to an increase in transfer factor. COPD (emphysema) is correlated to a decreased transfer factor.
COPD predisposes to eventual pulmonary hypertension as a result of an increase in pulmonary vascular resistance.
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This question is part of the following fields:
- Pathophysiology
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Question 150
Incorrect
-
A 40-year-old woman was diagnosed with hyperparathyroidism. She is undergoing a parathyroidectomy, and during the surgery, the inferior parathyroid gland is found to be enlarged. There is a vessel adjacent to this gland on its lateral side.
What is this vessel most likely to be?Your Answer:
Correct Answer: Common carotid artery
Explanation:There are four parathyroid glands that lie on the medial half of the posterior surface of each lobe of the thyroid gland, inside its sheath. There are two superior and two inferior parathyroid glands.
The common carotid artery is a lateral relation of the inferior parathyroid.
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This question is part of the following fields:
- Anatomy
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Question 151
Incorrect
-
Of the stated laws, which is correct?
Your Answer:
Correct Answer: Boyle's law states that at constant temperature the volume of a given mass of gas varies inversely with absolute pressure.
Explanation:Boyle’s law is correctly stated as it states that the volume of a gas of known mass is inversely proportional with absolute pressure, at a constant temperature.
Beer’s law states that radiation absorption by a solution of known thickness and concentration is identical to that of a solution of double thickness and half concentration.
Bougner’s (or Lambert’s) law states that every layer with the same thickness will absorb the same amount of radiation as it passes through.
Graham’s law states that the diffusion rate of a gas is inversely related to the square root of its molecular weight.
Raoult’s law states that the reduction of a solvent’s vapour pressure is directly proportional to the solute’s molar concentration.
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This question is part of the following fields:
- Statistical Methods
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Question 152
Incorrect
-
The following statement is true with regards to the Nernst equation:
Your Answer:
Correct Answer: It is used to calculate the potential difference across a membrane when the individual ions are in equilibrium
Explanation:The Nernst equation is used to calculate the membrane potential at which the ions are in equilibrium across the cell membrane.
The normal resting membrane potential is -70 mV (not + 70 mV).
The equation is:
E = RT/FZ ln {[X]o
/[X]i}Where:
E is the equilibrium potential
R is the universal gas constant
T is the absolute temperature
F is the Faraday constant
Z is the valency of the ion
[X]o is the extracellular concentration of ion X
[X]i is the intracellular concentration of ion X. -
This question is part of the following fields:
- Physiology
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Question 153
Incorrect
-
Regarding the Valsalva manoeuvre, which of the following describes the cardiovascular changes in phase III in a normal patient?
Your Answer:
Correct Answer: Normal intrathoracic pressure, decrease in blood pressure, and increase in heart rate
Explanation:When a person forcefully expires against a closed glottis, changes occur in intrathoracic pressure that dramatically affect venous return, cardiac output, arterial pressure, and heart rate. This forced expiratory effort is called a Valsalva maneuver.
Initially during a Valsalva, intrathoracic (intrapleural) pressure becomes very positive due to compression of the thoracic organs by the contracting rib cage. This increased external pressure on the heart and thoracic blood vessels compresses the vessels and cardiac chambers by decreasing the transmural pressure across their walls. Venous compression, and the accompanying large increase in right atrial pressure, impedes venous return into the thorax. This reduced venous return, and along with compression of the cardiac chambers, reduces cardiac filling and preload despite a large increase in intrachamber pressures. Reduced filling and preload leads to a fall in cardiac output by the Frank-Starling mechanism. At the same time, compression of the thoracic aorta transiently increases aortic pressure (phase I); however, aortic pressure begins to fall (phase II) after a few seconds because cardiac output falls. Changes in heart rate are reciprocal to the changes in aortic pressure due to the operation of the baroreceptor reflex. During phase I, heart rate decreases because aortic pressure is elevated; during phase II, heart rate increases as the aortic pressure falls.
When the person starts to breathe normally again, the intrathoracic pressure declines to normal levels, the aortic pressure briefly decreases as the external compression on the aorta is removed, and heart rate briefly increases reflexively (phase III). This is followed by an increase in aortic pressure (and reflex decrease in heart rate) as the cardiac output suddenly increases in response to a rapid increase in cardiac filling (phase IV). Aortic pressure also rises above normal because of a baroreceptor, sympathetic-mediated increase in systemic vascular resistance that occurred during the Valsava.
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This question is part of the following fields:
- Pathophysiology
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Question 154
Incorrect
-
You are shown the summary of a publication recommending use of their company's antiemetic to prevent postoperative nausea and vomiting by a pharmaceutical representative. You read the article, which is in a peer reviewed journal, and consider whether to change your current practice of using cyclizine intraoperatively.
Which type of publication will provide the best evidence on which to base changes to your practice?Your Answer:
Correct Answer: A prospective randomised double blind controlled trial against cyclizine in multiple centres
Explanation:A prospective randomised double blind controlled trial against cyclizine in multiple centres is the most likely to change your practice.
Case controlled studies are efficient in identifying an association between a drug treatment and outcome and are usually conducted retrospectively. They are generally less valued than prospective randomised trials. They cannot generate incidence data, are subject to bias, have difficult selection of controls and can be made more difficult if note keeping is not reliable.
The gold standard in intervention-based studies is randomised controlled double blind trials. Its features are:
Treating all intervention groups identically
Reduction of bias by random allocation to intervention groups
Patients and researchers unaware of which treatment was given until at completion of study
Patients analysed within the group to which they were allocated, and
Analysis focused on estimating the size of the difference in predefined outcomes between intervention groups.New healthcare interventions should be evaluated through properly designed randomised controlled trials (though there are some potential ethical disadvantages)
Conducting trials in multiple centres is an accepted way of evaluating a new drug as it may be the only way of recruiting sufficient number of patients within a reasonable time frame to satisfy the objectives of the trial. Type II statistical errors will occur if a small numbers of patients is used in study group.
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This question is part of the following fields:
- Statistical Methods
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Question 155
Incorrect
-
The prostate and the rectum are separated by which anatomical plane?
Your Answer:
Correct Answer: Denonvilliers fascia
Explanation:The prostate is separated from the rectum by the Denonvilliers fascia (rectoprostatic fascia).
Waldeyers fascia functions to separate the rectum and the sacrum.
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This question is part of the following fields:
- Anatomy
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Question 156
Incorrect
-
Which of the following statements is not correct regarding Noradrenaline?
Your Answer:
Correct Answer: Predominantly work through effects on ?-adrenergic receptors
Explanation:Noradrenaline also called norepinephrine belongs to the catecholamine family that functions in the brain and body as both a hormone and neurotransmitter.
They have sympathomimetic effects acting via adrenoceptors (?1, ?2,?1, ?2, ?3) or dopamine receptors (D1, D2).
May cause reflex bradycardia, reduce cardiac output and increase myocardial oxygen consumption
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This question is part of the following fields:
- Pharmacology
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Question 157
Incorrect
-
Of the following, which of these oxygen carrying molecules causes the greatest shift of the oxygen-dissociation curve to the left?
Your Answer:
Correct Answer: Myoglobin (Mb)
Explanation:Myoglobin is a haemoglobin-like, iron-containing pigment that is found in muscle fibres. It has a high affinity for oxygen and it consists of a single alpha polypeptide chain. It binds only one oxygen molecule, unlike haemoglobin, which binds 4 oxygen molecules.
The myoglobin ODC is a rectangular hyperbola. There is a very low P50 0.37 kPa (2.75 mmHg). This means that it needs a lower P50 to facilitate oxygen offloading from haemoglobin. It is low enough to be able to offload oxygen onto myoglobin where it is stored. Myoglobin releases its oxygen at the very low PO2 values found inside the mitochondria.
P50 is defined as the affinity of haemoglobin for oxygen: It is the PO2 at which the haemoglobin becomes 50% saturated with oxygen. Normally, the P50 of adult haemoglobin is 3.47 kPa(26 mmHg).
Foetal haemoglobin has 2 ? and 2 ?chains. The ODC is left shifted – this means that P50 lies between 2.34-2.67 kPa [18-20 mmHg]) compared with the adult curve and it has a higher affinity for oxygen. Foetal haemoglobin has no ? chains so this means that there is less binding of 2.3 diphosphoglycerate (2,3 DPG).
Carbon monoxide binds to haemoglobin with an affinity more than 200-fold higher than that of oxygen. This therefore decreases the amount of haemoglobin that is available for oxygen transport. Carbon monoxide binding also increases the affinity of haemoglobin for oxygen, which shifts the oxygen-haemoglobin dissociation curve to the left and thus impedes oxygen unloading in the tissues.
In sickle cell disease, (HbSS) has a P50 of 4.53 kPa(34 mmHg).
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This question is part of the following fields:
- Physiology
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Question 158
Incorrect
-
Arrythmias can develop from abnormal conduction, which may be as a result of impaired blood flow in the coronary arteries which causes hypoxia. Phase 0 depolarisation can be slowed, and this leads to slower conduction speeds.
Rapid depolarisation in the cardiac action potential is caused by which movement of ions?Your Answer:
Correct Answer: Sodium influx
Explanation:Cardiac conduction
Phase 0 – Rapid depolarization. Opening of fast sodium channels with large influx of sodium
Phase 1 – Rapid partial depolarization. Opening of potassium channels and efflux of potassium ions. Sodium channels close and influx of sodium ions stop
Phase 2 – Plateau phase with large influx of calcium ions. Offsets action of potassium channels. The absolute refractory period
Phase 3 – Repolarization due to potassium efflux after calcium channels close. Relative refractory period
Phase 4 – Repolarization continues as sodium/potassium pump restores the ionic gradient by pumping out 3 sodium ions in exchange for 2 potassium ions coming into the cell. Relative refractory period
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 159
Incorrect
-
Which of the following statements is true with regards to the Krebs' cycle (also known as the tricarboxylic acid cycle or citric acid cycle)?
Your Answer:
Correct Answer: Alpha-ketoglutarate is a five carbon molecule
Explanation:Krebs’ cycle (tricarboxylic acid cycle or citric acid cycle) is a sequence of reactions in which acetyl coenzyme A (acetyl-CoA) is metabolised and this results in carbon dioxide and hydrogen atoms production.
This series of reactions occur in the mitochondria of eukaryotic cells, not the cytoplasm. The cycle requires oxygen and so, cannot function under anaerobic conditions.
It is the common pathway for carbohydrate, fat and some amino acids oxidation and is required for high energy phosphate bond formation in adenosine triphosphate (ATP).
When pyruvate enters the mitochondria, it is converted into acetyl-CoA. This represents the formation of a 2 carbon molecule from a 3 carbon molecule. There is loss of one CO2 but formation of one NADH molecule. Acetyl-CoA is condensed with oxaloacetate, the anion of a 4 carbon acid, to form citrate which is a 6 carbon molecule.
Citrate is then converted into isocitrate, alpha-ketoglutarate, succinyl-CoA, succinate, fumarate, malate and finally oxaloacetate.
The only 5 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 160
Incorrect
-
A 19-year-old woman presents to the emergency department. She complains of symptoms indicative of an acute exacerbation of known 'brittle' asthma. On history, she reveals her asthma is normally controlled using inhalers and she has never had an acute exacerbation requiring hospitalisation.
On her admission into the ICU, further examination and diagnostic investigations are conducted. Her readings are:
Physical state: Alert, anxious and non-cyanotic.
Respiratory rate: 30 breaths/min
Pulse: 120 beats/min
Blood pressure: 150/90 mmHg
SPO2: 95% on air
Auscultation: Quiet breath sounds at both lung bases
What is the next most important step of investigation?Your Answer:
Correct Answer: Peak expiratory flow rate
Explanation:Peak expiratory flow rate (PEFR) is the maximum speed of air flow generated during a single forced exhaled breath. It is most useful when expressed as a percentage of the best value obtained from the patient.
Forced expiratory volume over 1 second (FEV1) is a lung parameter measured using spirometry. It is the amount of air forced out of the lung in one exhaled breath. It is a more accurate measure of lung obstructions as it doesn’t rely on effort like PEFR
PEFR and FEV1 are usually similar, but become more different in asthmatic patients as airflow becomes increasingly obstructed.
Acute severe asthma is most often diagnosed on history taking and examinations:
Respiratory rate: >25 breaths/min
Heart rate: >110 beats/min
PEFR: 33 – 50% predicted (<200L/min)
Patient state: Unable to complete a sentence in a single breath.A chest x-ray is not routinely required, and is only indicated in specific circumstances, which are:
If a pneumomediastinum or pneumothorax is suspected
Possible life threatening asthma
Possible consolidation
Unresponsive asthma
If ventilation is required.An echocardiograph (ECG) is not necessary in this case
Routine haematological and biochemical investigations are not urgent in this case as any abnormalities they detect will be secondary to the patient’s presentation.
An arterial blood gas (ABG) will only be indicated if SPO2 was <92% or if patient presented with life threatening symptoms.
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This question is part of the following fields:
- Clinical Measurement
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Question 161
Incorrect
-
Regarding chest tube insertion, which of the following measurements is utilized when selecting a chest tube drain?
Your Answer:
Correct Answer: External circumference (mm)
Explanation:Selection of a chest drain will depend on the external circumference.
A cannula, whether intravenous or intra-arterial, are classified according to standard wire gauge, which refers to the number of wires that can fit into the same hole. If a cannula is labelled 22G, then 22 wires will fit into the standard size hole.
A more popular measurement than SWG nowadays is cross sectional area.
When the concern for selecting equipment is the rate of flow, then it is important to consider the diameter and the radius of a parallel sided tube. These, however, are not routinely considered when comparing sizes of a cannula.
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This question is part of the following fields:
- Pathophysiology
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Question 162
Incorrect
-
A delayed hypersensitivity reaction is type ____?
Your Answer:
Correct Answer: IV
Explanation:Type I – immediate hypersensitivity reaction
Examples are: Atopy, urticaria, Anaphylaxis, Asthma( IgE mediated).
Type II – Antibody mediated cytotoxic reaction
Examples are: Autoimmune haemolytic anaemia, Thrombocytopenia( IgM or IgG mediated).
Type III – Immune complex mediated reaction
Examples are: Serum sickness,SLE – IgG., Farmers lungs, rheumatoid arthritis
Type IV – Delayed hypersensitivity reaction
Examples are: Contact dermatitis, drug allergies.
Type V – Autoimmune
Graves’
Myasthenia – IgM or IgG. -
This question is part of the following fields:
- Pathophysiology
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Question 163
Incorrect
-
Dinamap is an automated blood pressure monitoring device. Which of these statements best fit its properties?
Your Answer:
Correct Answer: The cuff should be positioned at the same level as the heart
Explanation:Dinamap continuously measures the systolic, diastolic and mean arterial pressure along with pulse rate, thereby providing a continuous monitoring of the blood pressure using the osscillitonometric principle of measurement.
The device loses accuracy towards the extremes of BP and is more accurate with systolic compared with diastolic pressure. In arrhythmias such as AF, the devices are also inaccurate due to the major fluctuations associated with the individual pulse pressure variations.
The manual BP device is still the gold standard for BP measurement and monitoring.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 164
Incorrect
-
A 54-year-old man weighing 70kg, underwent mesh repair for inguinal hernia under general anaesthesia. He was given intravenous co-amoxiclav (Augmentin) following which the patient developed widespread urticarial ras, became hypotensive (61/30 mmHg), and showed clinical signs of bronchospasm. Anaphylaxis is suspected in this patient.
Which one of the following is considered as best initial pharmacological treatment for this condition?Your Answer:
Correct Answer: Intravenous adrenaline 50 mcg
Explanation:The drug of choice for the treatment of anaphylaxis is adrenaline. It has an intravenous route of administration. Since the patient already has intravenous access, the intramuscular route is not appropriate.
Second-line pharmacological intervention includes the use of chlorpheniramine 10mg intravenous, Hydrocortisone 200mg.
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This question is part of the following fields:
- Pharmacology
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Question 165
Incorrect
-
The arterial-venous oxygen (a-vO2) difference is the difference between the oxygen content of arterial blood (CaO2) and mixed-venous blood (CvO2).
At rest, which organ has the greatest a-vO2 difference?Your Answer:
Correct Answer: Heart
Explanation:At rest, the heart has the greatest a-vO2 difference, a high capillary to myocyte ratio, short diffusion distances, and a high mitochondrial density. The flow of blood through the coronary arteries is also tightly controlled. At rest, 70-80 percent of the oxygen available to the cardiac muscle is extracted, increasing to 90 percent during exercise.
The a-vO2 difference indicates the body’s or an individual organ’s ability to extract oxygen from the blood.
CaO2 is influenced by a number of factors, including Hb concentration, PaO2 and pulmonary diffusion capacity.
CvO2 is influenced by a number of factors, including capillary density, regional blood flow, heart, resting skeletal muscle, kidney, intestine and skin.
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This question is part of the following fields:
- Pathophysiology
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Question 166
Incorrect
-
Which among the following is summed up by F statistic?
Your Answer:
Correct Answer: ANOVA
Explanation:ANOVA is based upon within group variance (i.e. the variance of the mean of a sample) and between group variance (i.e. the variance between means of different samples). The test works by finding out the ratio of the two variances mentioned above. (Commonly known as F statistic).
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This question is part of the following fields:
- Statistical Methods
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Question 167
Incorrect
-
A 42-year-old man presented with a bitemporal hemianopia with enlarged hands and feet. On examination, he was found to be hypertensive.
Which of the following correctly explains the cause of his visual field defect?Your Answer:
Correct Answer: Pituitary macroadenoma secreting growth hormone (GH)
Explanation:Pituitary macroadenoma is a benign tumour with growth larger than 10mm (those under 10mm are called microadenoma)
Compression of optic chiasm by pituitary adenoma is responsible for causing visual field defects like bitemporal hemianopia, optic neuropathy.
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This question is part of the following fields:
- Pathophysiology
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Question 168
Incorrect
-
Of the following statements, which is true about the measurements of cardiac output using thermodilution?
Your Answer:
Correct Answer: Cardiac output should be measured during the end-expiratory pause
Explanation:Thermodilution is the most common dilution method used to measure cardiac output (CO) in a hospital setting.
During the procedure, a Swan-Ganz catheter, which is a specialized catheter with a thermistor-tip, is inserted into the pulmonary artery via the peripheral vein. 5-10mL of a cold saline solution with a known temperature and volume is injected into the right atrium via a proximal catheter port. The solution is cooled as it mixes with the blood during its travel to the pulmonary artery. The temperature of the blood is the measured by the catheter and is profiled using a computer.
The computer also uses the profile to measure cardiac output from the right ventricle, over several measurements until an average is selected.
Cardiac output changes at each point of respiration, therefore to get an accurate measurement, the same point during respiration must be used at each procedure, this is usually the end of expiration, that is the end-expiratory pause.
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This question is part of the following fields:
- Clinical Measurement
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Question 169
Incorrect
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Of the following, which is NOT a branch of the subclavian artery?
Your Answer:
Correct Answer: Superior thyroid artery
Explanation:The left subclavian artery originates from the aortic arch, while the right subclavian artery originates from the brachiocephalic artery.
The subclavian artery gives off branches on both sides of the body:
1. Vertebral artery
2. Internal thoracic artery
3. Thyrocervical trunk
4. Costocervical trunk
5. Dorsal scapular arteryThe superior thyroid artery is the first branch of the external carotid artery. The other branches of the external carotid artery are:
1. Superior thyroid artery
2. Ascending pharyngeal artery
3. Lingual artery
4. Facial artery
5. Occipital artery
6. Posterior auricular artery
7. Maxillary artery
8. Superficial temporal artery -
This question is part of the following fields:
- Anatomy
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Question 170
Incorrect
-
With regards to oxygen delivery in the body, which of these statements is true?
Your Answer:
Correct 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 171
Incorrect
-
A radical neck dissection is being performed. The ENT surgeon wishes to expose the external carotid artery fully. He inserts a self-retaining retractor close to the origin of the external carotid artery.
What structure lies posterolaterally to the external carotid at this point?Your Answer:
Correct Answer: Internal carotid artery
Explanation:External carotid artery originates at the upper border of the thyroid cartilage. It ascends and lies anterior to the internal carotid arteries and posterior to the posterior belly of the digastric muscle and stylohyoid muscle.
The external carotid artery has eight important branches:
Anterior surface:
1. Superior thyroid artery (first branch)
2. Lingual artery
3. Facial artery
Medial branch
4. Ascending pharyngeal artery
Posterior branches
5. Occipital artery
6. Posterior auricular artery
Terminal branches
7. Maxillary artery
8. Superficial temporal artery -
This question is part of the following fields:
- Anatomy
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Question 172
Incorrect
-
A 67-year-old man, presents with an embolus in the lower limbs. He has previous medical history of atrial fibrillation. After examination and diagnostic investigations, he is scheduled for a transpopliteal embolectomy, which will require the surgeons to explore the central region of the popliteal fossa.
What structures will the surgeons come across after incising the deep fascia?Your Answer:
Correct Answer: Tibial nerve
Explanation:The tibial nerve lies on top of the vessels contained within the inferior aspect of the popliteal fossa.
In the superior aspect of the fossa, the tibial nerve runs lateral to the vessels, before then travelling superficial to the vessels, and then finally changing course to lie medial to the vessels.
The popliteal artery is the most deep structure present in the popliteal fossa
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This question is part of the following fields:
- Anatomy
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Question 173
Incorrect
-
With respect to the peripheral nerve stimulators, which one is used to perform nerve blocking?
Your Answer:
Correct Answer:
Explanation:The nerve stimulators deliver a stimulus lasting for 1-2 milliseconds (not second) to perform nerve blockage.
There are just 2 leads (not 3); one for the skin and other for the needle.
Prior to the administration of the local anaesthesia, a current of 0.25 – 0.5 mA (not 1-2mA) at the frequency of 1-2 Hz is preferred.
If the needle tip is close to the nerve, muscular contraction could be possible at the lowest possible current.
Insulated needles have improved the block success rate, as the current is only conducting through needle tip.
Stimulus to the femoral nerve which is placed in the mid lingual line causes withdrawer of the quadriceps and knee extension, that’s the dancing patella ( not plantar flexion).
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 174
Incorrect
-
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:
Correct 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 175
Incorrect
-
Regarding a drug whose elimination exhibits first-order kinetics, which of the following statements is correct?
Your Answer:
Correct Answer: The rate of elimination is proportional to plasma concentration
Explanation:The elimination of phenytoin follows first order kinetics. Plasma concentrations determine the rate of elimination. The relationship between drug X plasma concentration and time is described by an exponential process in the following equation used to describe the rate of elimination:
C = C0. e-kt
C=drug concentration, C0= drug concentration at time zero (extrapolated), k = rate constant and t=time
As enzyme systems become saturated when phenytoin concentrations are above the usual range, clearance of the medication becomes zero-order. The medication is metabolised at a constant pace, regardless of its plasma levels. Aspirin and ethyl alcohol are two more significant examples of medications that operate in this way.
A plot of drug concentration with time is a washout exponential curve.
A graph of concentration with time is a straight line i.e. Zero-order kinetics
The amount eliminated per unit time is constant defines the point at which zero order kinetics commences.
Elimination involves a rate-limiting reaction operating at its maximal velocity is incorrect.
The half life of the drug is proportional to the drug concentration in the plasma corresponds to a definition of first-order kinetics.
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This question is part of the following fields:
- Pharmacology
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Question 176
Incorrect
-
All of the following statements are false regarding propranolol except:
Your Answer:
Correct Answer: Has a plasma half life of 3-6 hours.
Explanation:Propranolol is a nonselective beta-blocker with a half-life of 3 to 6 hours.
Since it is lipid-soluble it crosses the blood-brain barrier and causes Central Nervous System side effects like sedation, nightmares, and depression.
They are contraindicated in asthma, Congestive heart failure, and diabetes.
It has a large volume of distribution with no intrinsic sympathomimetic action.
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This question is part of the following fields:
- Pharmacology
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Question 177
Incorrect
-
Following a physical assault, a 28-year-old man is admitted to the emergency room. A golf club has struck him in the head.
There is a large haematoma on the scalp, as well as a bleeding wound. In response to painful stimuli, he opens his eyes and makes deliberate movements. Because of inappropriate responses, a history is impossible to construct, but words can be discerned.
Which of the options below best describes his current Glasgow Coma Scale (GCS)?Your Answer:
Correct Answer: E2V3M5=10
Explanation:The Glasgow Coma Scale (GCS) has been used in outcome models as a measure of physiological derangement and as a tool for assessing head trauma.
Eye opening (E):
4 Spontaneously
3 Responds to voice
2 Responds to painful stimulus
1 No response.Best verbal response (V):
5 Orientated, converses normally
4 Confused, disoriented conversation, but able to answer basic questions
3 Inappropriate responses, words discernible
2 Incomprehensible speech
1 Makes no sounds.Best motor response (M):
6 Obeys commands for movement
5 Purposeful movement to painful stimulus
4 Withdraws from pain
3 Abnormal (spastic) flexor response to painful stimuli, decorticate posture
2 Extensor response to painful stimuli, decerebrate posture
1 No response.In this case, GCS = 2+3+5 = 10.
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This question is part of the following fields:
- Pathophysiology
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Question 178
Incorrect
-
Which of the following is correct about the action of bile salts?
Your Answer:
Correct Answer: Emulsification of lipids
Explanation:The emulsification and absorption of fats requires Bile salts.
Absorption of fats is associated with the activation of lipases in the intestine.
Bile salts are involved in fat soluble vitamin absorption and are reabsorbed in the terminal ileum (B12 is NOT fat soluble).
Although Vitamin B12 is also absorbed in the terminal ileum, it is a water soluble vitamin (as are B1, nicotinic acid, folic acid and vitamin C) .
The gastric parietal cells secretes Intrinsic factor that is essential for the absorption of B12.
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This question is part of the following fields:
- Pathophysiology
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Question 179
Incorrect
-
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 180
Incorrect
-
The cardiac tissue type that that has the highest conduction velocity is:
Your Answer:
Correct Answer: Purkinje fibres
Explanation:Potassium maintains the resting potential of cardiac myocytes, with depolarization triggered by a rapid influx of sodium ions, and repolarization due to efflux of potassium. A slow influx of calcium is responsible for the longer duration of a cardiac action potential compared with skeletal muscle.
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 ms.Phase 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 And Biochemistry
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Question 181
Incorrect
-
Which of the following statements is true about fluid balance?
Your Answer:
Correct Answer: After intravenous administration of crystalloids, the distribution of these fluids throughout the body depends on its osmotic activity
Explanation:When there is capillary leakage as seen in dependent oedema or ascites, oncotic pressure becomes a problem.
The intracellular sodium concentration is very sensitive to the extracellular sodium concentrations. When there is an imbalance, osmosis occurs resulting in shifts in water between the two compartments.
The microvascular endothelium relies upon osmosis and other processes as it is not freely permeable to water.
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This question is part of the following fields:
- Physiology
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Question 182
Incorrect
-
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:
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 183
Incorrect
-
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:
Correct Answer: 42 seconds
Explanation: -
This question is part of the following fields:
- Physiology
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Question 184
Incorrect
-
The muscle that lies behind the first part of the axillary nerve is?
Your Answer:
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 185
Incorrect
-
Which of the following statements is true regarding the Wrights Respirometer?
Your Answer:
Correct Answer: Measures the minute volume to within an accuracy of +/- 10%
Explanation:A Wrights Respirometer measures the volume of air exhaled over the course of one minute of normal breathing
It is unidirectional and measures tidal volume and minute volume of gas flow in one direction. It is placed at the expiratory side (lower pressure than inspiratory side therefore lower chances of gas leaks)
Slits are arranged such that incoming gas will rotate the vane at a rate of 150 revolutions per litre of flowing gas
The Wright respirometer tends to over-read at high flow rates and under-read at low flows because of mechanical causes like friction and inertia and the accumulation of water vapour
The ideal flow for accurate readings is 2 L/min for the respirometer. The respirometer reads the tidal volume and minute volume with a ±5–10% accuracy within the range of 4–24 L/min.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 186
Incorrect
-
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:
Correct 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 187
Incorrect
-
All the following statements are false regarding local anaesthetic except
Your Answer:
Correct Answer: Potency is directly related to lipid solubility
Explanation:The potency of local anaesthetics is directly proportional to lipid solubility because they need to penetrate the lipid-soluble membrane to enter the cell.
Protein binding has a direct relationship with the duration of action because the higher the ability of the drug to bind with membrane protein, the higher is the duration of action.
Higher the pKa of a drug, slower the onset of action. Because a drug with higher pKa will be more ionized than the one with lower pKa at a given pH. Local anaesthetics are weak bases, and unionized form diffuses more rapidly across the nerve membrane than the protonated form. As a result drugs with higher pKa will be more ionized will diffuse less across the nerve membrane.
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This question is part of the following fields:
- Pharmacology
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Question 188
Incorrect
-
The Control of Substances Hazardous to Health (COSHH) regulations recommend air supply rates to specific environments. Which of the following statements is true?
Your Answer:
Correct Answer: Preparation rooms receive a volume of 0.1 m3 of air per second
Explanation:Control of Substances Hazardous to Health (COSHH) was established by government under the Health and Safety at Work act in 1989. Their employers work on identification and management of those substances that are dangerous to health. The implications for anaesthetists include gas scavenging, equipment contamination and environmental safety. Adequate ventilation is required in areas where anaesthetic gases are present. The minimum air supply that is legally required in each specific area is: Operating theatres: 0.65 m3/second. Anaesthetic rooms: 0.15 m3/s. Preparation rooms: 0.1 m3/s. Recovery rooms need 15 air changes per hour
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 189
Incorrect
-
Which of the following statements is correct about a characteristic that is normally distributed in a population?
Your Answer:
Correct Answer: There will be approximately equal numbers who have more or less of the characteristic than the mean
Explanation:68% of the population will be found in one standard deviation (SD) above plus one SD below the mean. Two SDs above plus two SDs below the mean will include 95% of the population.
The median can be greater or less than the mean as it is simply the mid point of the data after the data is arranged. Half the data are above and half below the median .
The mode is a true score, unlike the mean or the median. It is the most common score or the score obtained from the largest number of subjects in any given data.
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This question is part of the following fields:
- Statistical Methods
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Question 190
Incorrect
-
The incidence and prevalence of a diabetes in a locality are being measured by a health care professional. This is what he found:
Year People with Diabetes New Cases
2017 150,000 2000
2018 150,000 4000
What can be stated regarding the incidence and prevalence of the disease in that area?Your Answer:
Correct Answer: Incidence increasing, prevalence equal
Explanation:Incidence tells us about the number of new cases that have been reported while prevalence gives us the idea of existing cases.
In this particular instance, the parameter of the study i.e. the total number of cases has not changed thus the prevalence of the disease remains same. Although, more cases have been reported in the second instance as a result of which incidence has increased.
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This question is part of the following fields:
- Statistical Methods
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Question 191
Incorrect
-
Which of the following statements is true about the in-hospital management of ventricular fibrillation?
Your Answer:
Correct Answer: Amiodarone may be administered following a third DC shock
Explanation:Ventricular fibrillation (VT) is an arrhythmia caused by a distortion in the organized contraction of the ventricles leading to an inability to pump blood out into the body.
Amiodarone is an anti arrhythmic drug used for the treatment of ventricular and atrial fibrillations. It is the gold standard of treatment for refractory pulseless ventricular tachycardia (VT) and ventricular fibrillation (VF).
Guidelines for emergency treatment state that only the rescuer carrying out chest compressions on the patient may stand near the defibrillator as it charges.
Cardio-pulmonary resuscitation (CPR) during cardiac arrest is required for 2 minute cycles.
Hypovolaemia is as a cause of pulseless electrical activity (PEA) can be reversed using fluid resuscitation, whereas hypotension during cardiac arrest is either persistent or undetectable and is therefore irreversible.
Hyperkalaemia and hypocalcaemia are treated using calcium salts, but calcium chloride is often preferred over calcium gluconate.
During a pulseless VT or VF, a single precordial thump will be effective if administered within the first seconds of the occurrence of a shockable rhythm.
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This question is part of the following fields:
- Pathophysiology
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Question 192
Incorrect
-
What is factually correct regarding correlation and regression?
Your Answer:
Correct Answer: Regression allows one variable to be predicted from another variable
Explanation:Linear regression, using a technique called curve fitting, allows us to make predictions regarding a certain variable.
Correlation coefficient gives us an idea whether or not the two parameters provide have any relation of some sort or not i.e. does change in one prompt any change in other?
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This question is part of the following fields:
- Statistical Methods
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Question 193
Incorrect
-
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:
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 194
Incorrect
-
A 70-year-old man presents with central crushing chest pain that radiates to the jaw in the emergency department. He has associated symptoms of nausea and diaphoresis.
A 12 lead ECG is performed. ST-elevation is observed in leads V2-V4. The diagnosis of anteroseptal ST-elevation myocardial infarction is made.
Which coronary vessel is responsible for this condition and runs in the interventricular septum on the anterior surface of the heart to reach the apex?Your Answer:
Correct Answer: Left anterior descending artery
Explanation:The heart receives blood supply from coronary arteries. The right and left coronary arteries branch off the aorta and supply oxygenated blood to all heart muscle parts.
The left main coronary artery branches into:
1. Circumflex artery – supplies the left atrium, side, and back of the left ventricle. The left marginal artery arises from the left circumflex artery. It travels along the obtuse margin of the heart.
2. Left Anterior Descending (LAD) artery – supplies the front and bottom of the left ventricle and front of the interventricular septumThe left anterior descending coronary artery is the largest coronary artery. It courses anterior to the interventricular septum in the anterior interventricular groove, extending from the base of the heart to its apex. Around the apex, the LAD anastomosis with the terminal branches of the posterior descending artery (branch of the right coronary artery).
Atherosclerosis or thrombotic occlusion of LAD causes myocardial infarction in large areas of the anterior, septal, and apical portions of the heart muscle. It can lead to a serious deterioration in heart performance.Occlusion of the LAD causes anteroseptal myocardial infarction, which is evident on the ECG with changes in leads V1-V4. Occlusion of the left circumflex artery causes lateral, posterior, or anterolateral MI. However, as it does not run towards the apex in the interventricular septum of the heart, it is not the correct answer for this question.
The right coronary artery branches into:
1. Right marginal artery
2. Posterior descending arteryThe right coronary artery supplies the right atrium, right ventricle, interatrial septum, and the inferior posterior third of the interventricular septum. Occlusion of the right coronary artery causes inferior MI, which is indicated on ECG with 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 195
Incorrect
-
A 27-year-old woman is admitted to the emergency room with an ectopic pregnancy that has ruptured.
The following is a description of the clinical examination:
Anxious
Capillary refill time of 3 seconds
Cool peripheries
Pulse 120 beats per minute
Blood pressure 120/95 mmHg
Respiratory rate 22 breaths per minute.
Which of the following is the most likely explanation for these clinical findings?Your Answer:
Correct Answer: Reduction in blood volume of 15-30%
Explanation:The following is the Advanced Trauma Life Support (ATLS) classification of haemorrhagic shock:
Class I haemorrhage:
It has blood loss up to 15%. There is very less tachycardia, and no changes in blood pressure, RR or pulse pressure. Usually, fluid replacement is not required.Class II haemorrhage:
It has 15-30% blood loss, equivalent to 750 – 1500 ml. There is tachycardia, tachypnoea and a decrease in pulse pressure. Patient may be frightened, hostile and anxious. It can be stabilised by crystalloid and blood transfusion.Class III haemorrhage:
There is 30-40% blood loss. It portrays inadequate perfusion, marked tachycardia, tachypnoea, altered mental state and fall in systolic pressure. It requires blood transfusion.Class IV haemorrhage:
There is > 40% blood volume loss. It is a preterminal event, and the patient will die in minutes. It portrays tachycardia, significant depression in systolic pressure and pulse pressure, altered mental state, and cold clammy skin. There is need for rapid transfusion and surgical intervention. -
This question is part of the following fields:
- Physiology
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Question 196
Incorrect
-
A 25-year-old man, has been scheduled for a splenectomy. For this procedure, he requires a urethral catheter.
Where does resistance first occur during the insertion of a catheter?Your Answer:
Correct Answer: Membranous urethra
Explanation:The membranous urethra is the shortest part of the urethra and the least dilatable part of it.
This is as a result of it being surrounded by the external urethral sphincter which is made up of striated muscle and controls voluntary urine flow from the bladder to the urethra.
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This question is part of the following fields:
- Anatomy
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Question 197
Incorrect
-
A study was concerned with finding out the normal reference range of IgE levels in adults was conducted. Presuming that the curve follows a normal distribution, what is the percentage of individuals having IgE levels greater than 2 standard deviations from mean?
Your Answer:
Correct Answer: 2.30%
Explanation:Since the data is normally distributed, 95.4% of the values lie with in 2 standard deviations from mean. The rest of the 4.6% are distributed symmetrically outside of that range which means 2.3% of the values lie above 2 standard deviations of the mean.
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This question is part of the following fields:
- Statistical Methods
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Question 198
Incorrect
-
Which of the following is true when testing for statistical significance?
Your Answer:
Correct Answer: The probability associated with a type I error is the significance level
Explanation:The null hypothesis states that there is no significant difference between two groups.
The alternative hypothesis states that there is a significant difference between two groups.
A type I error (false positive) occurs when a null hypothesis is rejected when it should be accepted.
A type II error (false negative) occurs when the alternative hypothesis is rejected when it should be accepted.
The probability determines the rejection of a null hypothesis.
The level of significance is set at p <0.05.
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This question is part of the following fields:
- Statistical Methods
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Question 199
Incorrect
-
A pharmaceutical company has developed a new drug considered a breakthrough in treating ovarian cancer.
The efficacy of this drug can be assessed by which phase of a clinical trial?Your Answer:
Correct Answer: Phase IIa
Explanation:Phase IIa studies are usually pilot studies designed to demonstrate clinical efficacy or biological activity (‘proof of concept’ studies) whereas phase IIb studies determine the optimal dose at which the drug shows biological activity with minimal side-effects (definite dose-finding studies).
Phase III and Phase IV studies are performed on larger set of participants (usually hundreds to thousands) when safety and efficacy have been established.
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This question is part of the following fields:
- Statistical Methods
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Question 200
Incorrect
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Which peripheral nerve of the foot is often utilized to evaluate for neuromuscular blockade?
Your Answer:
Correct Answer: Posterior tibial nerve
Explanation:The posterior tibial nerve lies on the posterior surface of the tibialis posterior and, lower down the leg, on the posterior surface of the tibia. The nerve accompanies the posterior tibial artery and lies at first on its medial side, then crosses posterior to it, and finally lies on its lateral side. The nerve, with the artery, passes behind the medial malleolus, between the tendons of the flexor digitorum longus and the flexor hallucis longus.
It gives off muscular branches to the soleus, flexor digitorum longus, flexor hallucis longus, and tibialis posterior. A medial calcaneal branches off to supply the skin over the medial surface of the heel, and an articular nerve to supply the ankle joint. Finally, it terminates to become the medial and lateral plantar nerves.
The saphenous nerve is a branch of the femoral nerve that gives off branches that supply the skin on the posteromedial surface of the leg.
The sural nerve is a branch of the tibial nerve that supplies the skin on the lower part of the posterolateral surface of the leg.
The superficial peroneal nerve is one of the terminal branches of the common peroneal nerve. It arises in the substance of the peroneus longus muscle on the lateral side of the neck of the fibular. It ascends between the peroneus longus and brevis muscles, and in the lower part of the leg it becomes cutaneous. Muscular branches of the superficial peroneal nerve supply the peroneus longus and brevis muscles, while medial and lateral cutaneous branches are distributed to the skin on the lower part of the leg and dorsum of the foot. In addition, the cutaneous branches supply the dorsal surfaces of the skin of all the toes, except the adjacent sides of the first and second toes and the lateral side of the little toe.
The superficial peroneal, sural and saphenous nerves cannot be used to assess neuromuscular blocks since they are sensory nerves.
The deep peroneal nerve enters the dorsum of the foot by passing deep to the extensor retinacula on the lateral side of the dorsalis pedis artery. It divides into terminal, medial, and lateral branches. The medial branch supplies the skin of the adjacent sides of the big and second toes. The lateral branch supplies the extensor digitorum brevis muscle. Both terminal branches give articular branches to the joints of the foot. This nerve is too deep to use for neuromuscular blockade assessment
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This question is part of the following fields:
- Anatomy
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