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Question 1
Incorrect
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Your manager asks you to inform patients that are suffering from a chronic pain about a trial that is going to be conducted in order to determine the efficacy of a novel analgesic. What phase is the trial currently in?
Your Answer: Phase 0
Correct Answer: Phase 2
Explanation:Phase 0 trials assist the scientists in studying the behaviour of drugs in humans by micro dosing patients. They are used to speed up the developmental process. They have no measurable therapeutic effect and efficiency.
Phase 1 is associated with assessing whether a drug is safe to use or not. The process is extensive and can take up to several months. It also involves healthy participants (less than 100) that are paid to take part in the study. The side effects upon increasing dosage are also addressed by the study. The effects the drug has on humans including how its absorbed, metabolized and excreted are studied. Approximately 70% of the drugs pass this phase.
Phase 2 trials involve patients that are suffering from the disease under study and are associated with determining the efficiency and the optimum dosage of the drug.
Phase 3 also assesses the efficacy but at a higher scale with larger population sample.
Phase 4 trials are involved with the long term effects and side effects of the drug.
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This question is part of the following fields:
- Statistical Methods
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Question 2
Correct
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A 77-year-old man is admitted to hospital for colorectal surgery. He is scheduled to undergo a preoperative assessment, which includes cardiopulmonary exercise test (CPX).
During the CPX, his maximum oxygen consumption (VO2 max) is determined to be 2,100 mL/minute. His weight is measured to be 100 kg.
Calculate the metabolic equivalent (MET) that is the best estimate for his VO2 max.Your Answer: 6 METs
Explanation:Metabolic equivalent (MET) measures the energy expenditure of an individual.
It is calculated mathematically by:
MET = (VO2 max/weight)/3.5 = 21/3.5 = 6 METs
Where 1 MET = 3.5 mL O2/kg/minute is utilized by the body.
Note:
1 MET Eating
Dressing
Use toilet
Walking slowly on level ground at 2-3 mph
2 METs Playing a musical instrument
Walking indoors around house
Light housework
4 METs Climbing a flight of stairs
Walking up hill
Running a short distance
Heavy housework, scrubbing floors, moving heavy furniture
Walking on level ground at 4 mph
Recreational activity, e.g. golf, bowling, dancing, tennis
6 METs Leisurely swimming
Leisurely cycling along the flat (8-10 mph)
8 METs Cycling along the flat (10-14 mph)
Basketball game
10 METs Moderate to hard swimming
Competitive football
Fast cycling (14-16 mph) -
This question is part of the following fields:
- Clinical Measurement
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Question 3
Incorrect
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All the following statements are false regarding local anaesthetic except
Your Answer: The onset of action is unrelated to pKa
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 4
Correct
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About the mechanism of action of bendroflumethiazide, Which of the following is correct?
Your Answer: Sodium-chloride symporter inhibitor
Explanation:Sodium-chloride symporter inhibitor.
The thiazide sensitive sodium chloride symporter is inhibited by thiazides at the proximal portion of the distal convoluted tubule leading to increased sodium and water excretion. Increased delivery of sodium to the distal portion of the distal convoluted tubule promotes potassium loss. This is why thiazides are associated with hyponatraemia and hypokalaemia.
Carbonic anhydrase inhibitors are used mainly in the treatment of glaucoma. They act on the proximal convoluted tubule to promote bicarbonate, sodium and potassium loss.
Sodium potassium chloride symporter is inhibited by Loop diuretics.
Epithelial sodium channels are inhibited by Amiloride.
Drugs which lead to nephrogenic diabetes insipidus such as lithium and demeclocycline, are Inhibitors of vasopressin. -
This question is part of the following fields:
- Pharmacology
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Question 5
Incorrect
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Which of the following statements is true regarding vecuronium?
Your Answer: Is mainly excreted in the urine
Correct Answer: Has a similar structure to rocuronium
Explanation:Vecuronium is used as a part of general anaesthesia to provide skeletal muscle relaxation during surgery or mechanical ventilation. It is a monoquaternary aminosteroid (not quaternary) non- depolarising neuromuscular blocking drug.
It has a structure similar to both rocuronium and pancuronium. The only difference is the substitution of specific groups on the steroid structure.
Vecuronium is not associated with the release of norepinephrine from sympathetic nerve endings. However, Pancuronium has norepinephrine releasing the property.
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This question is part of the following fields:
- Pharmacology
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Question 6
Incorrect
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Regarding the classification of breathing systems, which of the following is true?
Your Answer: The Jackson Rees modification is a Mapleson D with an open ended bag
Correct Answer: The Conway classification describes a functional classification based on whether a CO2 absorber is required
Explanation:Breathing system is an assembly of components which connects patient’s airway to anaesthesia machine through which controlled composition of gas mixture is dispensed. It delivers gas to the patient, removes expired gas and controls the temperature and humidity of the inspired mixture. It allows spontaneous, controlled, or assisted respiration. It may also provide ports for gas sampling, airway pressure, flow and volume monitoring.
Breathing systems have been classified by Conway and Mapleson.
Conway suggested a functional classification:
– Circuits requiring a CO2 absorber
– Circuits not requiring a CO2 absorberWilliam Mapleson designated varying arrangements of breathing system components (masks, breathing tubes, fresh gas flow inlets, adjustable pressure-limiting valves, and reservoir bags) as Mapleson A-E circuits.
Mapleson A: Arranged as FGF inlet, reservoir bag, APL valve, mask.
In this circuit, because the reservoir bag is between the FGF inlet valve and the APL valve, expired gas from the patient may re-enter the system and fill the reservoir bag during controlled ventilation. This is the most efficient system for spontaneous breathing as the FGF must only be equal to a patient’s minute ventilation to prevent rebreathing.Mapleson B: Arranged as reservoir bag, FGF inlet, APL valve, mask.
In this circuit, the FGF inlet is closer to the APL valve, which helps prevent the rebreathing concern in the Mapleson A circuit as above during controlled ventilation.Mapleson C: Arranged as reservoir bag, FGF inlet, APL valve, mask.
In this circuit, the arrangement is the same as the Mapleson B circuit. However, this circuit is shorter as it does not contain elongated corrugated tubing. This circuit also has the FGF inlet close to the APL valve to aid in preventing rebreathing.Mapleson D: Arranged as reservoir bag, APL valve, FGF inlet, and mask.
In this circuit, the arrangement interchanges the FGF inlet and APL valve of the Mapleson A circuit. This system prevents rebreathing by directing FGF towards the APL valve rather than towards the patient during exhalation.Mapleson E: Arranged as corrugated tubing, FGF inlet, and mask.
In this circuit, there is no reservoir bag and no APL valve. Given the inability to alter the pressure of the circuit, this is ideal for spontaneously ventilating neonates or paediatric patients where low-pressure ventilation is desired. The system prevents rebreathing, similar to the Mapleson D circuit.Jackson Rees later modified the Mapleson E by adding an open ended bag, which has since become known as the Mapleson F.
Mapleson F: Arranged as APL valve directly connected to reservoir bag, corrugated tubing, FGF inlet, and mask.
The system prevents rebreathing similarly to Mapleson D by directing FGF towards the APL valve. -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 7
Incorrect
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The child-Pugh scoring system can be used, if risk classifying a patient with chronic liver disorder earlier to anaesthesia.
Which one is the best combination of clinical signs and examinations used within the Child-Pugh scoring system?Your Answer: Ascites, grade of encephalopathy, AST/ALT, bilirubin and INR
Correct Answer: Ascites, grade of encephalopathy, albumin, bilirubin and INR
Explanation:In the Child-Pugh classification system, the following 5 components are determined or calculated in order:
Ascites
Grade of encephalopathy
Serum bilirubin (?mol/L)
Serum Albumin (g/L)
Prothrombin time or INR
Raised liver enzymes are not the component of the classification system.
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This question is part of the following fields:
- Basic Physics
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Question 8
Correct
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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: 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 9
Correct
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Which compound is secreted only from the adrenal medulla?
Your Answer: Adrenaline
Explanation:The adrenal medulla comprises chromaffin cells (pheochromocytes), which are functionally equivalent to postganglionic sympathetic neurons. They synthesize, store and release the catecholamines noradrenaline (norepinephrine) and adrenaline (epinephrine) into the venous sinusoids.
The majority of the chromaffin cells synthesize adrenaline. -
This question is part of the following fields:
- Anatomy
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Question 10
Correct
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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: 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 11
Correct
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Venepuncture is being performed on the basilic vein in the cubital fossa. At which of the following points does the basilic vein pass deep under the muscle?
Your Answer: Midway up the humerus
Explanation:The basilic vein is one of the primary veins that drain the upper limb, like the cephalic vein. It begins as the dorsal venous arch. The basilic vein originates from the ulnar side of the dorsal arch of the upper limb passes along the posteromedial aspect of the forearm, moving towards the anterior surface of the elbow.
The basilic vein pierces the deep fascia at the elbow and joins the venae commitantes of the brachial vein to form the axillary vein.
The basilic vein passes deep under the muscles as it moves midway up the humerus. At the lower border of the teres major muscle, the anterior and posterior circumflex humeral veins feed into it.
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This question is part of the following fields:
- Anatomy
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Question 12
Incorrect
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A 10-year-old girl complains of right iliac fossa pain, and a provisional diagnosis of appendicitis is made.
Which of the following embryological structures gives rise to the appendix?Your Answer: Hindgut
Correct Answer: Midgut
Explanation:The midgut gives rise to the appendix.
At week 6, the caecal diverticulum appears and is the precursor for the cecum and vermiform appendix. The cecum and appendix undergo rotation and descend into the right lower abdomen. The appendix can take up various positions:
1. Retrocecal appendix: behind the cecum
2. Retrocolic appendix: behind the ascending colon
3. Pelvic appendix: appendix descends into the pelvisThe appendix grows in length so that at birth, it is long and worm-shaped, or vermiform. After birth, the caecal wall grows unequally, and the appendix comes to lie on its medial side.
The midgut develops into the distal duodenum, jejunum, ileum, cecum, appendix, ascending colon, and proximal 2/3 of the transverse colon.
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This question is part of the following fields:
- Anatomy
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Question 13
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: A line connecting the greater trochanters
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 14
Incorrect
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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: The amount of splay in the glucose reabsorption curve is abnormally increased
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 15
Incorrect
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An 80-year old female was taken to the emergency room for chest pain. She has a medical history of coronary artery disease and previous episodes of atrial fibrillation. She was immediately attached to the cardiac monitor, which showed tachycardia at 148 beats per minute. The 12-lead ECG revealed atrial fibrillation.
Digoxin was given as an anti-arrhythmic at 500 micrograms, which is higher than the maintenance dose routinely given. Why is this so?Your Answer: It has a high clearance
Correct Answer: It has a high volume of distribution
Explanation:When the loading dose of Digoxin is given, the primary thing to consider is the volume of distribution. The volume of distribution is the proportionality factor that relates the total amount of drug in the body to the concentration. LD is computed as:
LD = Volume of distribution X (desired plasma concentration/bioavailability)
Digoxin is an anti-arrhythmic drug with a large volume of distribution and high bioavailability, and only a small percentage of Digoxin is bound to plasma proteins (,20%).
In the case, since the arrhythmia is not life-threatening, there is no need for the medication to work rapidly.
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This question is part of the following fields:
- Pharmacology
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Question 16
Incorrect
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A 61-year-old woman with myasthenia gravis is admitted to the ER with type II respiratory failure. There is a suspicion of myasthenic crisis.
She is in a semiconscious state. Her blood pressure is 160/90 mmHg, pulse is 110 beats per minute, temperature is 37°C, and oxygen saturation is 84 percent.
With a PaCO2 of 75 mmHg (10 kPa) breathing air, blood gas analysis confirms she is hypoventilating.
Which of the following values is the most accurate representation of her alveolar oxygen tension (PAO2)?Your Answer: 8.3
Correct Answer: 7.3
Explanation:The following is the alveolar gas equation:
PAO2 = PiO2 − PaCO2/R
Where:
PAO2 is the partial pressure of oxygen in the alveoli.
PiO2 is the partial pressure of oxygen inhaled.
PaCO2 stands for partial pressure of carbon dioxide in the arteries.
The amount of carbon dioxide produced (200 mL/minute) divided by the amount of oxygen consumed (250 mL/minute) equals R = respiratory quotient. With a normal diet, the value is 0.8.By subtracting the partial pressure exerted by water vapour at body temperature, the PiO2 can be calculated:
PiO2 = 0.21 × (100 kPa − 6.3 kPa)
PiO2 = 19.8Substituting:
PAO2 = 19.8 − 10/0.8
PAO2 = 19.8 − 12.5
PAO2 = 7.3k Pa -
This question is part of the following fields:
- Physiology
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Question 17
Incorrect
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In a diagnosis of a compensated respiratory acidosis, which of the following arterial blood gas results is likely to be seen?
Your Answer: pH = 7.34
PaCO2 = 7.2 kPa
HCO3 = 29Correct Answer:
Explanation:During normal tissue metabolism, there is production of CO2 (acid) which is then expired by the lungs. If metabolism switches from aerobic to anaerobic due to a lack of oxygen, the tissues are unable to completely oxidise sugars to CO2. As a consequence, the sugars can only be partially oxidised to lactic acid. Since lactic acid cannot be expired by the lungs, it remains in the circulation leading to metabolic acidosis.
Also, normal tissue metabolism leads to the production of some amount of acid from the breakdown of proteins. These acids are excreted from the body by kidney filtration. Renal failure will therefore results in acidosis after several days.
An increased acidosis stimulates the brain’s respiratory centres to increase the respiratory rate. This lowers the CO2 in the blood, leading to a decrease in its acidity. Renal excretion removes the excess acid, resulting in a normal pH, and a reduced PaCO2 and HCO3.
pH PaCO2 (kPa) HCO3
Compensated respiratory acidosis 7.34 7.2 29
Acute respiratory acidosis 7.25 7.3 22
Compensated metabolic acidosis 7.34 3.6 14
Metabolic acidosis 7.21 5.3 15
Metabolic alkalosis 7.51 5.1 30 -
This question is part of the following fields:
- Pathophysiology
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Question 18
Correct
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Question 19
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: Serum carbon dioxide
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 20
Incorrect
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An 80-year-old presents to the emergency department with symptoms raising suspicion of mesenteric ischemia. To diagnose the condition, an angiogram is performed. The radiologist needs to cannulate the coeliac axis from the aorta for the angiogram.
What vertebral level does the coeliac axis originate from the aorta?
Your Answer: L2
Correct Answer: T12
Explanation:Mesenteric ischemia is ischemia of the blood vessels of the intestines. It can be life-threatening especially if the small intestine is involved.
A critical factor for survival of acute mesenteric ischemia is early diagnosis and intervention. Angiography uses X-ray and contrast dye to image arteries and identify the severity of ischemia or obstruction.
The celiac axis is the first branch of the abdominal aorta and supplies the entire foregut (mouth to the major duodenal papilla). It arises at the level of vertebra T12. It has three major branches:
1. Left gastric
2. Common hepatic
3. Splenic arteriesThere are some important landmarks of vessels at different levels of vertebrae that need to be memorized.
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 21
Incorrect
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A 26-year-old doctor has recently been diagnosed with lung cancer. He would like to find out his survival time for the condition.
Which statistical method is used to predict survival rate?Your Answer: Student t-test
Correct Answer: Kaplan-Meier estimator
Explanation:The Weibull distribution are used to describe various types of observed failures of the components. it is used in reliability and survival analysis.
Regression Analysis is used to measure the relationship between among two or more variable. It determines the effect of independent variables on the dependent variables.
Student t-test is one of the most commonly used method to test the hypothesis. It determines the significant difference between the means of two different groups.
A time series is a collection of observations of well-defined data obtained at regular interval of time.
Kaplan-Meier estimator is used to estimate the survival function from lifetime data. It can be derived from maximum likelihood estimation of hazard function. It is most likely used to measure the fraction of patient’s life for a certain amount of time after treatment.
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This question is part of the following fields:
- Statistical Methods
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Question 22
Incorrect
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A 64-year old male has shortness of breath on exertion and presented to the cardiology clinic. He has a transthoracic echo performed to help in assessing the function of his heart.
How can this echo aid in calculating cardiac output?Your Answer: (stroke volume / end diastolic LV volume ) * 100%
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 23
Incorrect
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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: 1.25%
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 24
Correct
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Which of the following is true about the patellar reflex?
Your Answer: Is abolished immediately after transection of the spinal cord at T6
Explanation:The patellar (knee jerk) reflex is a monosynaptic stretch reflex arising from L2-L4 nerve roots. It occurs after a tap on the patellar tendon which causes the spindles of the quadriceps muscles to stretch.
The afferent nerve pathway occurred through A gamma fibres.
Wesphal’s sign refers to a reduction, or absence of the patellar reflex. It is often indicated of a neurological disease affecting the PNS.
A transection of the spinal cord results in a degree of shock which causes all reflexes to be reduced or completely absent, and required a period of approximately 6 weeks to recover.
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This question is part of the following fields:
- Pathophysiology
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Question 25
Incorrect
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A 35-year old male is found to be bradycardic in the emergency room. His cardiac muscle will most likely stay in a prolonged phase 4 state of the cardiac action potential. During phase 4 of the cardiac action potential, which of these occurs?
Your Answer: Slow calcium influx
Correct Answer: Na+/K+ ATPase acts
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 26
Correct
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A graph was plotted after administration of fentanyl infusion to a patient. The following are the x- and y-axis of the graph:
X-axis: Dose of fentanyl
Y-axis: Mu receptor occupancy, measured using positron emission tomography
Given the data above, what would be the best representation of the graph if the data on the x-axis are converted to logarithms?Your Answer: Rectangular hyperbola to sigmoid curve
Explanation:The dose-response curve plots the graph of the dose (drug concentration) versus the response. As doses increase, the response increment diminishes; finally, doses may be reached at which no further increase in response can be achieved. This relation between drug concentration and effect is traditionally described by a hyperbolic curve. When the x-axis is plotted in log scale, the graph yields a sigmoid curve.
Efficacy (Emax) and potency (EC50) can be derived from this curve. Emax is the maximal effect achievable, with increasing concentration of a drug. EC50 is the concentration of the drug, wherein half of the maximal effect is achieved.
When the graph is plotted using a log [response/1-response] against log dose, the sigmoid curve becomes a straight line (Hill plot). A graph that transforms from a straight line to exponential curve is mathematically incorrect. A graph that transforms from either a wash-in or wash-out exponential curve to a straight line comes from an initial set of data plotted against time, to a logarithmic transformation of the initial data set against time.
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This question is part of the following fields:
- Statistical Methods
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Question 27
Incorrect
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A double blind placebo control clinical trial is done. Which of these is correct about it?
Your Answer: All patients receive a placebo
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 28
Incorrect
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Among the following, which statement is true regarding electrical safety in an operation theatre?
Your Answer: Wet skin reduces the effect of electrocution
Correct Answer: The higher the frequency of the current the less risk to the patient
Explanation:The operating theatre is an unusual place with several applications of electrical equipment to the human body. This can lead to potential dangers associated with it that need to be prevented. Electrical safety in the operation theatre is the understanding of how these potential dangers can occur and how they can be prevented.
Electricity can cause morbidity or mortality by one of the following ways:
(i) electrocution
(ii) burns
(iii) ignition of a flammable material, causing a fire or explosion.Electrocution is dependant on factors like duration of contact with electric current, the current pathway and the frequency and size of current.
Option A: The higher the frequency, the less effects of electrocution on the body.
Option B & D: Equipment can be classified in classes and types.
The class designation describes the method used for protection against electrocution. Class I is basic protection, class II is double insulation and class III is safety extra low voltage.
The type designation describes the degree of protection based on the maximum permissible leakage currents under normal and fault conditions.
Type B:
can be class I, II or III but the maximum leakage current must not exceed 100 µA. It is therefore not suitable for direct connection to the heart.
Type BF
Similar to type B, but uses an isolated (or floating) circuit.
Type CF
Only type CF protect against microshock as they allow leakage currents of 0.05 mA per electrode for class I and 0.01 mA for class II. Microshock is a small leakage current that can cause harm because of direct connection to the heart via transvenous lines or wires, bypassing the impedance of the skin, leading to ventricular fibrillation. Microshock current of 100 ?A is sufficient to cause VF.Option C: A 75mA electrocution can cause ventricular fibrillation. Use the following as a general guide to understand the effect of current size on the body.
1 mA – tingling pain
5 mA – pain
15 mA – tonic muscular contraction
50 mA – respiratory muscle paralysis
75 mA – ventricular fibrillation.Option E: Wet skin reduces the resistance to current flow and therefore increases the effects of electrocution.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 29
Incorrect
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An 80-year old lady has a background history of a previous myocardial infarction which has left permanent damage to her heart's conduction system. The part of the conduction system with the highest velocities is damaged, and this has resulted in desynchronisation of the ventricles. The part of the heart that conducts the fastest is which of the following?
Your Answer: Atrioventricular node
Correct Answer: Purkinje fibres
Explanation:The electrical conduction system of the heart starts with the SA node which generates spontaneous action potentials.
This is conducted across both atria by cell to cell conduction, and occurs at around 1 m/s. The only pathway for the action potential to enter the ventricles is through the AV node in a normal heart.
At this site, conduction is very slow at 0.05ms, which allows for the atria to completely contract and fill the ventricles with blood before the ventricles depolarise and contract.The action potentials are conducted through the Bundle of His from the AV node which then splits into the left and right bundle branches. This conduction is very fast, (,2m/s), and brings the action potential to the Purkinje fibres.
Purkinje fibres are specialised conducting cells which allow for a faster conduction speed of the action potential (,2-4m/s). This allows for a strong synchronized contraction from the ventricle and thus efficient generation of pressure in systole.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 30
Incorrect
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Concerning the anterior pituitary gland, one of following is true.
Your Answer:
Correct Answer: Produces glycoproteins
Explanation:The posterior pituitary and the hypothalamus are connected by the pituitary stalk. It contains in the pituitary sella and has the optic chiasm and hypothalamus as superior relations.
The anterior pituitary produces thyroid-stimulating hormone (TSH), luteinising hormone (LH) and follicle-stimulating hormone (FSH) . These hormones are Glycoproteins and share a common alpha subunit with unique beta subunits.
The secretion of pituitary hormones are pulsatile. Examples are LH, adrenocorticotropic hormone (ACTH) and growth hormone (GH).
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This question is part of the following fields:
- Pathophysiology
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