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Question 1
Correct
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A 33-year-old woman known to be hypothyroid and taking 150 mcg l-thyroxine daily is reviewed in the preoperative assessment clinic prior to a laparoscopic cholecystectomy.
She has required three increases in her thyroid replacement therapy in the last six months.
Her thyroid function tests are as follows:
TSH 11 (normal range 0.4-4mU/L)
T3 20 (normal range 9-25mU/L)
T4 6.2 (normal range 3.5-7.8mU/L)
What will explain this biochemical picture?Your Answer: Poor compliance with medication
Explanation:In patients with an intact hypothalamic-pituitary axis, serial TSH measurements are used to determine the adequacy of treatment with thyroid hormones . changes in TSH levels becoming apparent after approximately eight weeks of therapy with thyroid hormone replacement. Change in T3/T4 levels are seen before changes in TSH .
In patients taking thyroid replacement therapy, the most frequent reason for persistent elevation of serum TSH is poor compliance. Patients who do not regularly take their L-thyroxine try and catch up just before a visit to a clinician for blood test.
Tissue-level unresponsiveness to thyroid hormone is caused by mutation in the gene controlling a receptor for T3 and is rare.
Reduced responsiveness of target tissues to thyroid hormone aka resistance to thyroid hormones (rTH) occurs when there is a mutation in the thyroid hormone receptor ? gene. It is a rare autosomal dominant inherited syndrome of reduced end-organ responsiveness to thyroid hormone and has two types:
Generalised resistance (GrTH)
Pituitary resistance (PrTH)Patients with rTH have normal or slightly elevated serum thyroid stimulating hormone (TSH) level, elevated serum free thyroxine (FT4) and free triiodothyronine (FT3) concentrations.
Drugs that increase metabolism of thyroxine include:
Warfarin
Rifampin
Phenytoin
Phenobarbital
St John’s Wort
CarbamazepineThese drugs lower circulating thyroid hormones and would be associated with a raised TSH but low T3/T4.
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This question is part of the following fields:
- Pathophysiology
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Question 2
Correct
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An aged patient that has been suffering from diabetes criticised the health minister for his comments on incidence and prevalence. The minister had said that they both are two separate entities. It can be therefore inferred that the patient thinks that prevalence and incidence are the same thing.
Is he right?Your Answer: No. In chronic disease prevalence is greater than incidence.
Explanation:Only on rare occasions has it been found that the prevalence and incidence were same. Incidence can be greater than prevalence in acute cases only. In case of chronic diseases prevalence is far greater than incidence. One needs to have a deeper understanding of both the concepts to understand the health literature.
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This question is part of the following fields:
- Statistical Methods
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Question 3
Incorrect
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A previously fit 26-year-old is undergoing surgery to repair an inguinal hernia. He is breathing on his own, and a supraglottic airway is being maintained via a circle system with air/oxygen and sevoflurane.
With a fresh gas flow of 14 L/min, the end-tidal CO2 reading is 8.1 kPa. CO2 pressure is 1.9 kPa. The percentages of oxygen inhaled and exhaled are 38 and 33 percent, respectively.
What do you think is the most likely source for these readings?Your Answer: Leak in the expiratory limb
Correct Answer: Incompetent expiratory valve
Explanation:The patient is rebreathing carbon dioxide that has been exhaled.
Exhaustion of the soda lime and failure of the expiratory valve are the two most likely causes. A leak in the inspiratory limb is a less likely cause. Increased inhaled and exhaled carbon dioxide levels may appear with a normal-looking capnogram if the expiratory valve is ineffective.
The patient will exhale into both the inspiratory and expiratory limbs if the inspiratory valve is inoperable. A slanted downstroke inspiratory phase (as the patient inhales carbon dioxide-containing gas from the inspiratory limb) and increased end-tidal carbon dioxide can be seen on the capnogram.
Even if the soda lime were exhausted, a high fresh gas flow would be enough to prevent rebreathing. The difference in oxygen concentrations in inspired and expired breaths would be less pronounced.
Hypercapnia is caused by respiratory obstruction and malignant hyperthermia, but not by rebreathing.
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This question is part of the following fields:
- Pathophysiology
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Question 4
Correct
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Which of the following statements is correct regarding hypomagnesaemia?
Your Answer: Causes tetany
Explanation:The ECG changes seen in hypomagnesaemia include:
Prolonged PR interval
Prolonged QT interval
Flattening of T waves
ST segment depression
Prominent U wavesThese changes are almost the same as those of hypokalaemia.
There is an increased risk of digoxin toxicity and a risk of atrial and ventricular ectopic and ventricular arrhythmias.
There is impaired synthesis and release of parathyroid hormone (PTH) in chronic hypomagnesaemia leading to impaired target organ response to PTH. This produces secondary hypocalcaemia.
The use of potassium ‘wasting’ diuretics (e.g. loop diuretics like furosemide) may lead to Hypomagnesaemia.
A tall T wave is seen in hypermagnesemia.
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This question is part of the following fields:
- Pathophysiology
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Question 5
Correct
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What can an outbreak of flu that has spread globally be termed as?
Your Answer: Pandemic
Explanation:An epidemic is declared when the increase in a give disease is above a certain level in a specific interval of time.
An endemic is the general, usual level of a disease in a population at a particular time.
A pandemic is an epidemic that is spread across many countries and continents.
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This question is part of the following fields:
- Statistical Methods
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Question 6
Correct
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The equipment used for patient monitoring in theatre and intensive care settings have electrical safety requirements for the protection of hospital staff and patients.
Of the different classes of electrical equipment listed, which is least likely to cause a patient to suffer a microshock?Your Answer: II (CF)
Explanation:Microshock refers to ventricular fibrillation caused by miniscule amounts of currents or voltages (100-150 microamperes) passing through the myocardial tissue from external cables arising from electrical components within the cardiac muscle, for example, pacemaker electrodes or saline filled venous catheters.
The risk of shock changes with the construction of electrical equipment in question. The main classes of electrical equipment include: I: Appliances have a protective earth connected to an outer casing which prevents live elements from coming in contact with conductive elements. A fault in this equipment class will result in live elements coming in contact with the outer casing and allowing electrical flow into the protective earth. This triggers the protective fuse to disconnect the electric supply to the appliance.
II: These appliances have reinforced insulation. In the event of a fault which causes the first layer of insulation to fail, the second layer is able to prevent contact of live elements with outer casing.
III: These appliances have no insulation to provide safety, and rely solely on the use of separated extra low voltage source (SELV) which limits voltage to 25V AC or 60V DC allowing for a person to come in contact with it without risk of a shock under normal dry conditions. Under wet conditions, voltage supply should be lowered to reduce risk of shock. These devices have no risk of macroshocks, but some risk of microshocks.
Class I and II electrical appliances are further divided into subtypes developed to limit current leakage in the event of a singular fault:
B (body): Upper limit of current leakage is 500 µA. This current can cause skin tingling and microshocks, but is not sufficient to cause injury.
BF (body floating): These appliances have an isolating capacitor or transformer which separate the secondary circuit from the protective earth. The upper limit of current leakage is the same as type B.
CF (cardiac floating): Upper limit of leakage current during a singular fault is 50 microamps. It is least likely to result in a microshock -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 7
Incorrect
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With a 10-day history of severe vomiting, a 71-year-old man with a gastric outlet obstruction is admitted to the surgical ward.
The serum biochemical results listed below are available:
Sodium 128 mmol/L (137-144)
Potassium 2.6 mmol/L (3.5-4.9)
Chloride 50 mmol/L (95-107)
Urea 12 mmol/L (2.5-7.5)
Creatinine 180 µmol/L (60-110)
Which of the following do you think you are most likely to encounter?Your Answer: Bicarbonate 45-50 mmol/L
Correct Answer: The standard base excess will be higher than actual base excess
Explanation:Hydrochloric acid is lost when you vomit for a long time (HCl). As a result, the following can be expected, in varying degrees of severity:
Hypokalaemia
Hypochloraemia
Increased bicarbonate to compensate for chloride loss and metabolic alkalosisThe alkalosis causes potassium to move from the intracellular to the extracellular compartment at first. Long-term vomiting and dehydration cause potassium to be excreted by the kidneys in order to conserve sodium. Dehydration can cause urea and creatinine levels to rise.
The actual base excess is always greater than the standard base excess.
The actual base excess (BE) is a measurement of a base’s contribution to a blood gas picture’s metabolic component. It’s the amount of base that needs to be added to a blood sample to bring the pH back to 7.4 after the respiratory component of a blood gas picture has been corrected (PaCO2 of 40 mmHg or 5.3 kPa). The BE has a normal range of +2 to 2. A large positive BE indicates a severe metabolic alkalosis, while a large negative BE indicates a severe metabolic acidosis. As a result, the actual BE in vitro is unaffected by CO2.
In vivo, however, standard BE is not independent of pCO2 because blood with haemoglobin acts as a better buffer than total ECF.
As a result, it is impossible to tell the difference between compensating for a respiratory disorder and compensating for the presence of a primary metabolic disorder.
The differences between in vitro and in vivo behaviour can be mostly eliminated if the BE is calculated for a haemoglobin concentration of 50 g/L (the ‘effective’ or virtual value of Hb if it was distributed throughout the extracellular space) rather than the actual haemoglobin. Because haemoglobin has a lower buffering capacity, the standard BE is higher than the actual BE. It reflects the BE better in the extracellular space rather than just the intravascular compartment.
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This question is part of the following fields:
- Pathophysiology
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Question 8
Correct
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The following are results of some pulmonary function tests:
Measurement - Predicted result - Test result
Forced vital capacity (FVC) (btps) - 3.21 - 1.94
Forced expiratory volume in 1 second (FEV1) (btps) - 2.77 - 1.82
FEV1/FVC ratio % (btps) - 81.9 - 93.5
Peak expiratory flow (PEF) (L/second) - 6.55 - 3.62
Maximum voluntary ventilation (MVV) (L/minute) - 103 - 87.1
Which statement applies to the results?Your Answer: The patient has a moderate restrictive pulmonary defect
Explanation:Severity of a reduction in restrictive defect (%FVC) or obstructive defect (%FEV1/FVC) predicted are classified as follows:
Mild 70-80%
Moderate 60-69%
Moderately severe 50-59%
Severe 35-49%
Very severe <35% This patient has a %FVC predicted of 60.4% and this corresponds to a moderate restrictive deficit. %FEV1/FVC ratio is 93.5%. FEV1/FVC ratio 80% < predicted and VC < 80% = mixed picture. FEV1/FVC ratio 80% < predicted and VC > 80% = obstructive picture.FEV1/FVC ratio 80% > predicted and VC > 80% = normal picture.
FEV1/FVC ratio 80% > predicted and VC < 80% predicted= restrictive picture. The integrity of the alveolar-capillary barrier is measured by carbon monoxide transfer factor (TLCO) and carbon monoxide transfer coefficient (KCO). These values are seen to be reduced in emphysema, interstitial lung diseases and in pulmonary vascular pathology. However, the KCO (as % predicted) is high in extrapulmonary restriction (pleural, chest wall and respiratory neuromuscular disease), and in loss of lung units provided the structure of the lung remaining is normal. The KCO distinguishes extrapulmonary (high KCO) causes of ‘restriction’ from intrapulmonary causes (low KCO).
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This question is part of the following fields:
- Clinical Measurement
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Question 9
Correct
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The SI unit of measurement is kgm2s-2 in the System international d'unités (SI).
Which of the following derived units of measurement has this format?Your Answer: Energy
Explanation:The derived SI unit of force is Newton.
F = m·a (where a is acceleration)
F = 1 kg·m/s2The joule (J) is a converted unit of energy, work, or heat. When a force of one newton (N) is applied over a distance of one metre (Nm), the following amount of energy is expended:
J = 1 kg·m/s2·m =
J = 1 kg·m2/s2 or 1 kg·m2·s-2The unit of velocity is metres per second (m/s or ms-1).
The watt (W), or number of joules expended per second, is the SI unit of power:
J/s = kg·m2·s-2/s
J/s = kg·m2·s-3Pressure is measured in pascal (Pa) and is defined as force (N) per unit area (m2):
Pa = kg·m·s-2/m2
Pa = kg·m-1·s-2 -
This question is part of the following fields:
- Physiology
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Question 10
Correct
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Buffers are solutions that resist a change in pH when protons are produced or consumed. They consist of weak acids and their conjugate bases. Buffers are also present in our bodies, and they are known as physiologic buffers.
Which of these is the most effective buffer in the blood?Your Answer: Bicarbonate
Explanation:The first line of defence against acid-base disorder is buffering. The blood mainly utilizes bicarbonate ion (HCO3-) for its buffering capacity (total of 53%, plasma and red blood cells combined).
Strong acids, when acted upon by a buffer, release H+, which then combines to HCO3- and forms carbonic acid (H2CO3). When acted upon by the enzyme carbonic anhydrase, H2CO3 dissociates into H2O and CO.
The rest are the percentage of utilization for the following buffers:
Haemoglobin (by RBCs) – 35%
Plasma proteins (by plasma) – 7%
Organic phosphates (by RBCs) – 3%
Inorganic phosphates (by plasma) – 2% -
This question is part of the following fields:
- Pharmacology
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Question 11
Incorrect
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In reference to confounding variables, which among the given is not true?
Your Answer: Confounding factors are factors associated with both the exposure the outcome
Correct Answer: In the analytic stage of a study confounding can be controlled for by randomisation
Explanation:Randomisation can be used to provide control over the confounding variables during the design stage of a study however during analytical stage a technique called stratification is used for controlling confounding variables. Since the question asks for the information that is factually incorrect.
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This question is part of the following fields:
- Statistical Methods
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Question 12
Correct
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Compared to the parasympathetic nervous system (PNS), the sympathetic nervous system (SNS) has:
Your Answer: Nicotinic receptors in pre and post ganglionic synapses
Explanation:With regards to the autonomic nervous system (ANS)
1. It is not under voluntary control
2. It uses reflex pathways and different to the somatic nervous system.
3. The hypothalamus is the central point of integration of the ANS. However, the gut can coordinate some secretions and information from the baroreceptors which are processed in the medulla.With regards to the central nervous system (CNS)
1. There are myelinated preganglionic fibres which lead to the
ganglion where the nerve cell bodies of the non-myelinated post ganglionic nerves are organised.
2. From the ganglion, the post ganglionic nerves then lead on to the innervated organ.Most organs are under control of both systems although one system normally predominates.
The nerves of the sympathetic nervous system (SNS) originate from the lateral horns of the spinal cord, pass into the anterior primary rami and then pass via the white rami communicates into the ganglia from T1-L2.
There are short pre-ganglionic and long post ganglionic fibres.
Pre-ganglionic synapses use acetylcholine (ACh) as a neurotransmitter on nicotinic receptors.
Post ganglionic synapses uses adrenoceptors with norepinephrine / epinephrine as the neurotransmitter.
However, in sweat glands, piloerector muscles and few blood vessels, ACh is still used as a neurotransmitter with nicotinic receptors.The ganglia form the sympathetic trunk – this is a collection of nerves that begin at the base of the skull and travel 2-3 cm lateral to the vertebrae, extending to the coccyx.
There are cervical, thoracic, lumbar and sacral ganglia and visceral sympathetic innervation is by cardiac, coeliac and hypogastric plexi.
Juxta glomerular apparatus, piloerector muscles and adipose tissue are all organs under sole sympathetic control.
The PNS has a craniosacral outflow. It causes reduced arousal and cardiovascular stimulation and increases visceral activity.
The cranial outflow consists of
1. The oculomotor nerve (CN III) to the eye via the ciliary ganglion,
2. Facial nerve (CN VII) to the submandibular, sublingual and lacrimal glands via the pterygopalatine and submandibular ganglions
3. Glossopharyngeal (CN IX) to lungs, larynx and tracheobronchial tree via otic ganglion
4. The vagus nerve (CN X), the largest contributor and carries ¾ of fibres covering innervation of the heart, lungs, larynx, tracheobronchial tree parotid gland and proximal gut to the splenic flexure, liver and pancreasThe sacral outflow (S2 to S4) innervates the bladder, distal gut and genitalia.
The PNS has long preganglionic and short post ganglionic fibres.
Preganglionic synapses, like in the SNS, use ACh as the neuro transmitter with nicotinic receptors.
Post ganglionic synapses also use ACh as the neurotransmitter but have muscarinic receptors.Different types of these muscarinic receptors are present in different organs:
There are:
M1 = pupillary constriction, gastric acid secretion stimulation
M2 = inhibition of cardiac stimulation
M3 = visceral vasodilation, coronary artery constriction, increased secretions in salivary, lacrimal glands and pancreas
M4 = brain and adrenal medulla
M5 = brainThe lacrimal glands are solely under parasympathetic control.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 13
Incorrect
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Which of the following statements is an accurate fact about the vertebral column?
Your Answer: The vertebral artery passes through the foramen transversarium of C1-C7
Correct Answer: Herniation of intervertebral disc between the fifth and sixth cervical vertebrae will compress the sixth cervical nerve root
Explanation:The vertebral (spinal) column is the skeletal central axis made up of approximately 33 bones called the vertebrae.
Cervical disc herniations occur when some or all of the nucleus pulposus extends through the annulus fibrosus. The most commonly affected discs are the C5-C6 and C6-C7 discs. Each vertebrae has a corresponding nerve root which arises at a level above it. This means that a hernation of the C5-C6 disc will cause a compression of the C6 nerve root.
The foramen transversarium is a part of the transverse process of each cervical vertebrae, however, the vertebral artery only runs through the C1-C6 foramen transversarium.
The costal facets are the point of joint formation between a rib and a vertebrae. As such, they are only present on the transverse processes of T1-T10.
The lumbar vertebrae do not form a joint with the ribs, nor do they possess a foramina in their transverse process.
Intervertebral discs are thickest in the cervical and lumbar regions of the spinal column. However, there are no discs between C1 and C2.
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This question is part of the following fields:
- Pathophysiology
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Question 14
Correct
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A 50-year-old man has complained of persistent hoarseness and dry cough. He has a history of smoking 20 cigarettes per day. The examination reveals no significant clinical signs of cranial nerve damage.
Referred to an ENT specialist, the patient is explained how coughing is usually a defence mechanism of the body which is activated more than usual by the chemical irritants in cigarette smoke. However, the ENT doctor suspects a nerve involvement in the cough reflex as the patient also presents with hoarseness with the dry cough.
Which nerves is the ENT doctor suspecting to have been affected in this patient?Your Answer: CN IX and X
Explanation:Cough is an important defensive reflex that helps clear secretions and particulates from the airways. A complex reflex arc generates each cough.
The cough reflex begins with irritation of the cough receptors present in the epithelium of the trachea, main carina, branching points of large airways, and more distal smaller airways. These receptors are responsive to both mechanical and chemical stimuli.
Afferent pathway:
Impulses from stimulated receptors are transmitted via sensory nerve fibres of the vagus nerve (mainly) and glossopharyngeal nerve and travel to the medulla diffusely. CN 5 is also thought to contribute to the afferent limb. However, the vagus is the main nerve.Central pathway:
The cough centre is located in the upper brain stem and ponsEfferent pathway:
Impulses from the centre travel via the vagus, phrenic nerve, and spinal motor nerves to the diaphragm, abdominal wall, and muscles. -
This question is part of the following fields:
- Anatomy
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Question 15
Incorrect
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You have always been curious about the effects of statins. While going through a study, something ticks you off and makes you think that they are way more common then the data suggests and are mostly under reported. In search of some concrete evidence, you decide to conduct a study of your own. While doing research, you come across a recent study that highlights the long term effects of statins.
Which of the following types of study could that have been?Your Answer: Case-control study
Correct Answer: Clinical trial, Phase 4
Explanation:In general practice, majority of phase 3 trials and some of the trials conducted in phase 2 are randomized. Because phase 4 trials require a huge sample size, they are not randomized as much. The primal reason behind conducting phase 3 trials is to test the efficiency and safety in a significant sample population. At this point it is assumed that the drug is effective up to a certain extent.
During a case-control study, subjects that exhibit outcomes of interest are compared with those who don’t show the expected outcome. The extent of exposure to a particular risk factor is then matched between cases and controls. If the exposure among cases surpasses controls, it becomes a risk factor for the outcome that is being studied.
Pilot studies are conducted on a lower and much smaller level, to assess if a randomized controlled trial of the crucial components of a study will be plausible.
Phase 4 trials are the ones that are conducted after its established that the drug is effective and is approved by the regulating authority for use. These trials are concerned with the side effects and potential risks associated with the long term usage of the drug.
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This question is part of the following fields:
- Statistical Methods
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Question 16
Incorrect
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Which of the following best explains the statement Epinephrine is formulated as 1 in 1000 solution
Your Answer: 1 mg per 1000 ml solution
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 17
Incorrect
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The thyroid gland:
Your Answer: Is closely related to the internal carotid artery
Correct Answer: Internalises iodine through active transport
Explanation:The thyroid gland is a gland shaped like a butterfly which lies at the base of the anterior neck. It controls metabolism using hormone secretion.
Iodine is extremely important for the synthesis of hormones within the thyroid. It is internalised into the thyroid follicular cells via the sodium/iodide symporter (NIS).
The parathyroid glands are found posterior to the thyroid gland, with the recurrent laryngeal nerves running posteromedially.
The expected weight of a normal thyroid gland is about 30 grams.
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This question is part of the following fields:
- Pathophysiology
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Question 18
Incorrect
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One of the non-pharmacologic management of COPD is smoking cessation. Given a case of a 60-year old patient with history of smoking for 30 years and a FEV1 of 70%, what would be the most probable five-year course of his FEV1 if he ceases to smoke?
Your Answer: The FEV1 will return to normal within 5 years
Correct Answer: The FEV1 will decrease at the same rate as a non-smoker
Explanation:For this patient, his forced expiratory volume in 1 second (FEV1) will decrease at the same rate as a non-smoker.
There is a notable, but slow, decline in FEV1 when an individual reaches the age of 26. An average reduction of 30 mls every year in non-smokers, while a more significant reduction of 50-70 mls is observed in approximately 20% of smokers.
Considering the age of the patient, individuals who begin smoking cessation by the age of 60 are far less likely to achieve normal FEV1 levels, even in the next five years. It is expected that their FEV1 will be approximately 14% less than their peers of the same age.
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This question is part of the following fields:
- Physiology
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Question 19
Incorrect
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A radiologist is conducting an arch aortogram. She begins by entering the brachiocephalic artery using the angiography catheter. As she continues to advance the catheter, what vessels will the catheter enter?
Your Answer: Left subclavian artery
Correct Answer: Right subclavian artery
Explanation:As there is no brachiocephalic artery on the left side, the artery is entered by the catheter on the right side.
The brachiocephalic artery branches into the common carotid and the right subclavian artery, so the catheter is most likely to enter the right subclavian artery, or also possibly the right carotid.
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This question is part of the following fields:
- Anatomy
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Question 20
Incorrect
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The spinal cord in a neonate terminates at the lower border of:
Your Answer: L4
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 21
Incorrect
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A trail has analysed that a new screening test may increase the survival time of ovarian cancer patients. But analyst say that the apparent increase in the patients survival time is just because of earlier detection instead of actual improvement.
What kind of bias is in this experiment?Your Answer: Recall bias
Correct Answer: Lead time bias
Explanation:Observation bias occurs when the behaviour of an individual changes that results from their awareness of being observed.
Recall bias introduced when participants in a study are systematically more or less likely to recall and relate information on exposure depending on their outcome status.
Attrition bias is a systematic error caused by unequal loss of participants from a randomized controlled trial (RCT). In clinical trials, participants might dropout due to unsatisfactory treatment or efficacy, intolerable adverse events, or even death.
Selection bias introduced when the individuals are not chosen randomly to take a part in the study. It usually occurs when the research decides who is going to be studied, they are not the representative of the population.
Lead-time bias occurs when a disease is detected by a screening test at an earlier time point rather than it would have been diagnosed by its clinical appearance. In this bias, earlier detection improves the survival time in the intervention group.
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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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Which statement is true with regards to the cardiac action potential?
Your Answer: Opening of fast sodium channels with large influx of sodium initiates rapid repolarisation
Correct Answer: Repolarization due to potassium efflux after calcium channels close causes the relative refractory period to start
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 23
Incorrect
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A 20-year-old man has been diagnosed with mitral regurgitation. He will be treated with mitral valve repair.
What is true regarding the mitral valve?Your Answer: It has two anterior cusps
Correct Answer: Its closure is marked by the first heart sound
Explanation:The mitral valve is the valve between the left atrium and left ventricle. It opens when the heart is in diastole (relaxation) which allows blood to flow from the left atrium to the left ventricle. In systole (contraction), the mitral valve closes to prevent the backflow of blood from the left ventricle to the left atrium.
The mitral valve is located posterior to the sternum at the level of the 4th costal cartilage. It is best auscultated over the cardiac apex, where its closure marks the first heart sound.
The mitral valve anatomy is composed of five main structures:
1. Left atrial wall – the myocardium of the left atrial wall extends over the posterior leaflet of the mitral valve. (left atrial enlargement is one of the causes for mitral regurgitation)
2. Mitral annulus – a fibrous ring that connects with the anterior and posterior leaflets. It functions as a sphincter that contracts and reduces the surface area of the valve during systole (Annular dilatation can also lead to mitral regurgitation)
3. Mitral valve leaflets (cusps) – The mitral valve is the only valve in the heart with two cusps or leaflets. One anterior and one posterior.
i. The anterior leaflet is located posterior to the aortic root and is also anchored to the aortic root.
ii. The posterior leaflet is located posterior to the two commissural areas.
4. Chordae tendinae – The chordae tendinae connects both the cusps to the papillary muscles.
5. Papillary muscles – These muscles and their cords support the mitral valve, allowing the cusps to resist the pressure developed during contractions (pumping) of the left ventricleThe anterior and posterior cusps are attached to the chordae tendinae which itself is attached to the left ventricle via papillary muscle.
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This question is part of the following fields:
- Anatomy
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Question 24
Incorrect
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Which of the following is true about the bispectral index (BIS)?
Your Answer: Is reduced by intraoperative opioids
Correct Answer: Sevoflurane lowers BIS more than ketamine
Explanation:The bispectral index (BIS) monitors works to determine the level of consciousness of a patient by processing electroencephalographic (EEG) signals to obtain a value between 0 and 100, where 0 reflects no brain activity, and 100 reflects a patient is completely awake.
The general meaning of BIS values are:
>95: Patient is in an awake state.
65-85: Patient is in a sedated state.
40-65: Patient is in a state that is optimal for general surgery.
<40: Patient is in a deep hypnotic state It is important in measuring the depths of anaesthesia to prevent haemodynamic changes or patient awareness during surgery. The nature of anaesthetic agent used is a determinant factor in resultant BIS values. Intravenous agents, such as propofol, thiopental and midazolam, result in a deeper hypnotic state, whilst inhalation agents have a lesser hypnotic effect at the same BIS values. Certain agents result in inaccurate BIS values such as ketamine and nitrous oxide (NO). These two agents appear to increase the BIS value, whilst putting the patient in a deeper hypnotic state, and should therefore not be used with BIS monitoring. Hypothermia also affects the BIS value as it causes a 1.12 per °C decrease in body temperature. -
This question is part of the following fields:
- Clinical Measurement
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Question 25
Incorrect
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A 30-year-old man has been stabbed in an area of the groin that contains the femoral triangle. He will undergo explorative surgery.
Which of the following makes the lateral wall of the femoral triangle?Your Answer: Adductor magnus
Correct Answer: Sartorius
Explanation:The femoral triangle is a wedge-shaped area found within the superomedial aspect of the anterior thigh. It is a passageway for structures to leave and enter the anterior thigh.
Superior: Inguinal ligament
Medial: Adductor longus
Lateral: Sartorius
Floor: Iliopsoas, adductor longus and pectineusThe contents include: (medial to lateral)
Femoral vein
Femoral artery-pulse palpated at the mid inguinal point
Femoral nerve
Deep and superficial inguinal lymph nodes
Lateral cutaneous nerve
Great saphenous vein
Femoral branch of the genitofemoral nerve -
This question is part of the following fields:
- Anatomy
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Question 26
Incorrect
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You've been summoned to the paediatric ward after a 4-year-old child was discovered 'collapsed' in bed.
The child had been admitted the day before with febrile convulsions and was scheduled to be discharged. It is safe to approach the child.
What should your first life-saving action be?Your Answer: Open the airway
Correct Answer: Apply a gentle stimulus and ask the child if they are alright
Explanation:Paediatric life support differs from adult life support in that hypoxia is the primary cause of deterioration.
After checking for danger, the child should be given a gentle stimulus (such as holding the head and shaking the arm) and asked, Are you alright? according to current advanced paediatric life support (APLS) guidelines. Safety, Stimulate, Shout is a phrase that is frequently remembered. Any airway assessment should be preceded by these actions.
Although the algorithm includes five rescue breaths, they are performed after the airway assessment.
It is not recommended to ask parents to leave unless they are obstructing the resuscitation. A team member should be with them at all times to explain what is going on and answer any questions they may have.
CPR should not begin until the child has been properly assessed and rescue breaths have been administered.
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This question is part of the following fields:
- Pathophysiology
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Question 27
Incorrect
-
With regards to this state of matter which has a volume but no definite shape, particles are not tightly packed together. These are incompressible although there is free movement within the volume.
This statement best describes which one of the following states of matter?Your Answer: Solid
Correct Answer: Liquid
Explanation:The solid state of matter has a definite volume and shape and particles are packed closely together and are incompressible. Within this tight lattice, there is enough thermal energy to produce vibration of particles.
Liquids however have a volume but no definite shape. These particles are less tightly packed together. Although there is free movement within the volume, they are incompressible.
Gases, however, have no finite shape or volume and particles are free to move rapidly in a state of random motion. They are compressible and are completely shaped by the space in which they are held. Vapours exist as a gas phase in equilibrium with identical liquid or solid matter below its boiling point.
The most prevalent state of matter in the universe is plasma which is formed by heating atoms to very high temperatures to form ions.
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This question is part of the following fields:
- Basic Physics
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Question 28
Correct
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Given the following values:
Expired tidal volume = 800 ml
Plateau pressure = 50 cmH2O
PEEP = 10 cmH2O
Compute for the static pulmonary compliance.Your 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 29
Incorrect
-
From the following electromagnetic waves, which one has the shortest wavelength?
Your Answer: Visible light
Correct Answer: X rays
Explanation:Electromagnetic waves are categorized according to their frequency or equivalently according to their wavelength. Visible light makes up a small part of the full electromagnetic spectrum.
Electromagnetic waves with shorter wavelengths and higher frequencies include ultraviolet light, X-rays, and gamma rays. Electromagnetic waves with longer wavelengths and lower frequencies include infrared light, microwaves, and radio and televisions waves.
Different electromagnetic waves according to their wavelength from shorter to longer are X-rays, ultraviolet radiations, visible light, infrared radiation, radio waves. X-ray among electromagnetic waves has the shortest wavelength and higher frequency with wavelengths ranging from 10*-8 to 10* -12 and corresponding frequencies.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 30
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: Sat up at 45 degrees breathing air
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 31
Incorrect
-
Which of following statements is true regarding the comparison of fentanyl and alfentanil?
Your Answer: Alfentanil has a higher clearance (ml/kg/minute)
Correct Answer: Fentanyl is more potent than alfentanil
Explanation:Fentanyl is a pethidine congener, 80–100 times more potent than morphine, both in analgesia and respiratory depression. Fentanyl is ten times more potent than alfentanil.
Alfentanil has a more rapid onset than fentanyl even if fentanyl is more lipid-soluble because both are basic compounds and alfentanil has lower pKa, so a greater proportion of alfentanil is unionized and is more available to cross membranes.
Elimination of alfentanil is higher than fentanyl due to its lower volume of distribution.
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This question is part of the following fields:
- Pharmacology
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Question 32
Incorrect
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In order to determine if there is any correlation among systolic blood pressure and the age of a person.
Which among the provided options is false regarding the calculation of correlation coefficient, r ?Your Answer: r may lie anywhere between -1 and 1
Correct Answer: May be used to predict systolic blood pressure for a given age
Explanation:Correlation doesn’t justify causality. 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? It has nothing to do with predictions. For that purpose linear regression is used.
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This question is part of the following fields:
- Statistical Methods
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Question 33
Incorrect
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Which measure of central tendency is most useful for a continuous, non-skewed data?
Your Answer: Median
Correct Answer: Mean
Explanation:Mean, also known as the average, is the most common measure of central tendency. It is the sum of all observed values divided by the number of observation. It is not useful for skewed data, which has an abnormal distribution. It is useful, instead, for numerical data that have symmetric distribution. It reflects the contributions of each data in the group, and are sensitive to outliers.
The median is the value that falls in the middle position when the observations are ranked in order from the smallest to the largest. If the number of observations is odd, the median is the middle number. If it is even, the median is the average of the two middle numbers. Unlike the mean, the median is useful on skewed data, and can be used for ordinal or numerical data if skewed.
The mode is the value that occurs with the greatest frequency in a set of observations, and is utilized for bimodal distribution.
The variance and the standard deviation are not measures of central tendency, but of dispersion.
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This question is part of the following fields:
- Statistical Methods
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Question 34
Incorrect
-
A 55-year-old woman presents for transsphenoidal surgery following a diagnosis of pituitary macroadenoma.
Which of the following is the most common visual field defect caused by such lesions?Your Answer: Homonymous hemianopia
Correct Answer: Bitemporal hemianopia
Explanation:Pituitary tumours that compress the optic chiasma primarily affect the neurones that decussate at this location. Bitemporal hemianopia is caused by neurones that emerge from the nasal half of the retina and transmit the temporal half of the visual field.
The axons of ganglion cells in the retina form the optic nerve.
It exits the orbit through the optic foramen and projects to the thalamic lateral geniculate body. The optic chiasma forms above the sella turcica as the nasal fibres decussate along the way. The optic radiation travels from the lateral geniculate body to the occipital cortex.
Lesions at various points along this pathway cause the following visual field defects:
Scotoma implies partial retinal or optic nerve damage.
Monocular vision loss occurs when the optic nerve is completely damaged.
Pathology at the optic chiasma causes bitemporal hemianopia.
Cortical blindness with occipital cortex pathology and homonymous hemianopia with lesions compromising the optic radiation. -
This question is part of the following fields:
- Pathophysiology
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Question 35
Incorrect
-
Which one of the following statement is true regarding United Kingdom gas cylinders?
Your Answer: The filling ratio for nitrous oxide is 75%
Correct Answer: Tensile tests are performed on sections of one cylinder in every hundred
Explanation:Medical gas cylinders are made up of molybdenum steel but not cast iron. They are checked and assessed at a regular interval.
At least one cylinder in each hundred are tested for tensile, pressure, smash, twist and straightening.
Nitrous Oxide cylinders contain a mixture of liquid and vapour at a pressure of approx. 4500 kPa or 45 Bar. Carbon dioxide cylinder contain gas at the pressure of 5000kPa.
The filling ratio is the ratio of mass of liquified gas in the cylinder to the mass of water required to fill the cylinder at the temperature of 15ºC. In the united kingdom, filling ratio of liquid nitrous oxide is 0.75. The cylinders are usually attached to the anaesthetic machine. As nitrous oxide is an N-methyl-d-aspartate receptor antagonist that may reduce the incidence of chronic post-surgical pain.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 36
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: Slow depolarisation due to influx of sodium
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 37
Correct
-
The following statement is true with regards to the Nernst equation:
Your 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 38
Incorrect
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Which of the following statement is true regarding hypoxic pulmonary vasoconstriction (HPV)?
Your Answer: Is increased by volatile anaesthetic agents
Correct Answer: 20 parts per million (ppm) of nitric oxide will reduce hypoxic pulmonary vasoconstriction
Explanation:Hypoxic Pulmonary vasoconstriction (HPV) reflects the constriction of small pulmonary arteries in response to hypoxic alveoli (.i.e.; PO2 below 80-100mmHg or 11-13kPa).
These blood vessels become independent of the nerve stimulus, when blood with a high PO2 flows through the lung which contains a low alveolar PO2.
Thus a low PO2 within the alveoli has been shown to impact on hypoxic pulmonary vasoconstriction (HPV) more than a low PO2 within the blood.
HPV results in the blood flow being directed away from poorly ventilated areas of the lung and helps to reduce the ventilation/perfusion mismatch (not increase).
In animals, volatile anaesthetic agents can diminish HPV, while in adults, the evidence proves less persuading, in spite of the fact that it certainly doesn’t strengthen the effects.
HPV response will be suppressed by 20 parts per million (ppm) of nitric oxide.
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This question is part of the following fields:
- Physiology
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Question 39
Incorrect
-
These proprietary preparations of local anaesthetic are available in your hospital:
Solution A contains 10 mL 0.5% bupivacaine (plain), and
Solution B contains 10 mL 0.5% bupivacaine with adrenaline 1 in 200,000.
What is the pharmacokinetic difference between the two solutions?Your Answer: The protein binding of solution A is higher than solution B
Correct Answer: The onset of action of solution A is quicker than solution B
Explanation:The reasons for adding adrenaline to a local anaesthetic solution are:
1. To Increase the duration of block
2. To reduce absorption of the local anaesthetic into the circulation
3. To Increase the upper safe limit of local anaesthetic (2.5 mg/kg instead of 2 mg/kg, in this case).The addition of adrenaline to bupivacaine does not affect its potency, lipid solubility, protein binding, or pKa(8.1 with or without adrenaline).
The pH of bupivacaine is between 5-7. Premixed with adrenaline, it is 3.3-5.5.
The onset of a local anaesthetic and its ability to penetrate membranes depends upon degree of ionisation. Compared with the ionised fraction, unionised local anaesthetic readily penetrates tissue membranes to site of action. The onset of action of solution B is slower. this is because the relationship between pKa(8.1) and pH(3.3-5.5) of the solution results in a greater proportion of ionised local anaesthetic molecules compared with solution A. -
This question is part of the following fields:
- Pharmacology
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Question 40
Incorrect
-
All of the following are true when describing the autonomic nervous system except:
Your Answer: The parasympathetic nervous system has long preganglionic and short post ganglionic fibres
Correct Answer: Juxta glomerular apparatus, piloerector muscles and adipose tissue are all organs under sole parasympathetic control
Explanation:With regards to the autonomic nervous system (ANS)
1. It is not under voluntary control
2. It uses reflex pathways and different to the somatic nervous system.
3. The hypothalamus is the central point of integration of the ANS. However, the gut can coordinate some secretions and information from the baroreceptors which are processed in the medulla.With regards to the central nervous system (CNS)
1. There are myelinated preganglionic fibres which lead to the
ganglion where the nerve cell bodies of the non-myelinated post ganglionic nerves are organised.
2. From the ganglion, the post ganglionic nerves then lead on to the innervated organ.Most organs are under control of both systems although one system normally predominates.
The nerves of the sympathetic nervous system (SNS) originate from the lateral horns of the spinal cord, pass into the anterior primary rami and then pass via the white rami communicates into the ganglia from T1-L2.
There are short pre-ganglionic and long post ganglionic fibres.
Pre-ganglionic synapses use acetylcholine (ACh) as a neurotransmitter on nicotinic receptors.
Post ganglionic synapses uses adrenoceptors with norepinephrine / epinephrine as the neurotransmitter.
However, in sweat glands, piloerector muscles and few blood vessels, ACh is still used as a neurotransmitter with nicotinic receptors.The ganglia form the sympathetic trunk – this is a collection of nerves that begin at the base of the skull and travel 2-3 cm lateral to the vertebrae, extending to the coccyx.
There are cervical, thoracic, lumbar and sacral ganglia and visceral sympathetic innervation is by cardiac, coeliac and hypogastric plexi.
Juxta glomerular apparatus, piloerector muscles and adipose tissue are all organs under sole sympathetic control.
The PNS has a craniosacral outflow. It causes reduced arousal and cardiovascular stimulation and increases visceral activity.
The cranial outflow consists of
1. The oculomotor nerve (CN III) to the eye via the ciliary ganglion,
2. Facial nerve (CN VII) to the submandibular, sublingual and lacrimal glands via the pterygopalatine and submandibular ganglions
3. Glossopharyngeal (CN IX) to lungs, larynx and tracheobronchial tree via otic ganglion
4. The vagus nerve (CN X), the largest contributor and carries ¾ of fibres covering innervation of the heart, lungs, larynx, tracheobronchial tree parotid gland and proximal gut to the splenic flexure, liver and pancreasThe sacral outflow (S2 to S4) innervates the bladder, distal gut and genitalia.
The PNS has long preganglionic and short post ganglionic fibres.
Preganglionic synapses, like in the SNS, use ACh as the neuro transmitter with nicotinic receptors.
Post ganglionic synapses also use ACh as the neurotransmitter but have muscarinic receptors.Different types of these muscarinic receptors are present in different organs:
There are:
M1 = pupillary constriction, gastric acid secretion stimulation
M2 = inhibition of cardiac stimulation
M3 = visceral vasodilation, coronary artery constriction, increased secretions in salivary, lacrimal glands and pancreas
M4 = brain and adrenal medulla
M5 = brainThe lacrimal glands are solely under parasympathetic control.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 41
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: E3V2M5=10
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 42
Correct
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During 2015 it was reported in the New England Journal of Medicine that the usage of empagliflozin(a sodium-glucose-co-transporter 2 inhibitor) caused a decrease in the cardiovascular deaths, non fatal heart attacks and strokes in patients suffering from type 2 diabetes. The results were published per 1000 patient years. With the above mentioned drug, the event rate turned out to be 37.3/1000 patient years whereas the placebo had an event rate of 43.9/1000 patient years.
How many further patients need to be treated with empagliflozin to avoid any further incidence of cardiovascular death or non fatal myocardial infraction and non fatal stroke?Your Answer: 150
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= (43.9-37.3)
ARR= 6.6
NNT= 1000/6.6
NNT=151.5
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This question is part of the following fields:
- Statistical Methods
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Question 43
Incorrect
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All of the following are responses to massive haemorrhage except which of the following?
Your Answer: Reduced renal blood flow stimulates release of renin by the juxta-glomerular apparatus
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 44
Incorrect
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Comparing pressure-volume curves in patients during an asthma attack with that of healthy subjects.
The increased resistive work of breathing in the patients with asthma is best indicated by?Your Answer:
Correct Answer: Larger hysteresis loop
Explanation:A major source of caloric expenditure and oxygen consumption in the body is work of breathing (WOB) and 70% of this is to overcome elastic forces. The remaining 30% is for flow-resistive work
In a normal patient breathing normally, the total area of hysteresis pressure volume curve represents the flow-resistive WOB.
The area of the expiratory resistive work increases during an asthma attack making the compliance curve larger in area. The larger the area the greater the work required to breathe.
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This question is part of the following fields:
- Physiology
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Question 45
Incorrect
-
A 68-year-old man with nausea and vomiting is admitted to the hospital.
For temporal arteritis, he takes 40 mg prednisolone orally in divided doses. His prescription chart will need to be adjusted to reflect his inability to take oral medications.
What is the equivalent dose of intravenous hydrocortisone to 40 mg oral prednisolone?Your Answer:
Correct Answer: 160 mg
Explanation:Prednisolone 5 mg is the same as 20 mg hydrocortisone.
Prednisolone 40 mg is the same as 8 x 20 mg or 160 mg of prednisolone.
Mineralocorticoid effects and variations in action duration are not taken into account in these comparisons.
5 mg of prednisolone is the same as Dexamethasone 750 mcg, Hydrocortisone 20 mg, Methylprednisolone 4 mg, and Cortisone acetate 25 mg.
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This question is part of the following fields:
- Pharmacology
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Question 46
Incorrect
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Which of the following is incorrect with regards to atrial natriuretic peptide?
Your Answer:
Correct Answer: Secreted mainly by the left atrium
Explanation:Atrial natriuretic peptide (ANP) is secreted mainly from myocytes of right atrium and ventricle in response to increased blood volume.
It is secreted by both the right and left atria (right >> left).It is a 28 amino acid peptide hormone, which acts via cGMP
degraded by endopeptidases.It serves to promote the excretion of sodium, lowers blood pressure, and antagonise the actions of angiotensin II and aldosterone.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 47
Incorrect
-
One of two divisions of the autonomic nervous system is the sympathetic nervous system. It is both anatomically and physiologically different from the parasympathetic nervous system.
Which best describes the anatomical layout of the sympathetic nervous system?Your Answer:
Correct Answer: Short myelinated preganglionic neurones from T1-L5 in lateral horns of grey matter of spinal cord, synapse in sympathetic ganglia (neurotransmitter - acetyl choline), long unmyelinated postganglionic neurones, synapse with effector organ (neurotransmitter - adrenaline or noradrenaline)
Explanation:The autonomic nervous system is divided into the sympathetic and parasympathetic nervous system. They are anatomically and physiologically different.
The sympathetic nervous system arises from the thoracolumbar outflow (T1-L5 ) at the lateral horns of grey matter of the spinal cord. Their preganglionic neurones are usually short myelinated and synapse in ganglia lateral to the vertebral column and have acetyl choline (Ach) as the neurotransmitter. Their postganglionic neurones are longer and unmyelinated and synapse with effector organ where the neurotransmitter is either adrenaline or noradrenaline.
The outflow of the parasympathetic nervous system is craniosacral. The cranial part originates from the midbrain and medulla (cranial nerves III, VII, IX and X) and the sacral outflow is from S2, S3 and S4. Their preganglionic neurones are usually long myelinated and synapse in ganglia close to the target organ and has Ach as its neurotransmitter. The unmyelinated postganglionic neurones is shorter and they synapse with effector organ. The neurotransmitter here is also Ach.
Both sympathetic and parasympathetic preganglionic neurons are cholinergic. Only the postganglionic parasympathetic neurons are cholinergic.
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This question is part of the following fields:
- Anatomy
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Question 48
Incorrect
-
A 27-year-old woman arrives at the emergency room after intentionally ingesting 2 g of amitriptyline.
A Glasgow coma score of 6 was discovered, as well as a pulse rate of 140 beats per minute and a blood pressure of 80/50 mmHg.
Which of the following ECG changes is most likely to indicate the onset of life-threatening arrhythmias?Your Answer:
Correct Answer: Prolongation of the QRS complex
Explanation:Arrhythmias and/or hypotension are the most common causes of death from tricyclic antidepressant (TCA) overdose.
The quinidine-like actions of tricyclic antidepressants on cardiac tissues are primarily responsible for their toxicity. Conduction through the His-Purkinje system and the myocardium slows as phase 0 depolarisation of the action potential slows. QRS prolongation and atrioventricular block are caused by slowed impulse conduction, which also contributes to ventricular arrhythmias and hypotension.
Arrhythmias can also be caused by abnormal repolarization, impaired automaticity, cholinergic blockade, and inhibition of neuronal catecholamine uptake, among other things.
Acidaemia, hypotension, and hyperthermia can all exacerbate toxicity.
The anticholinergic effects of tricyclic antidepressants, as well as the blockade of neuronal catecholamine reuptake, cause sinus tachycardia. Sinus tachycardia is usually well tolerated and does not require treatment. It can be difficult to tell the difference between sinus tachycardia and ventricular tachycardia with QRS prolongation.
A QRS duration of more than 100 milliseconds indicates a higher risk of arrhythmia and should be treated with systemic sodium bicarbonate.
The tricyclic is dissociated from myocardial sodium channels by serum alkalinization, and the extracellular sodium load improves sodium channel function.
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This question is part of the following fields:
- Clinical Measurement
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Question 49
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 50
Incorrect
-
Which drug, if given to a pregnant woman, can lead to deleterious fetal effects due to its ability to cross the placenta?
Your Answer:
Correct Answer: Atropine
Explanation:It is well known that atropine will cross the placenta and that maternal administration results in an increase in fetal heart rate.
Atropine is highly selective for muscarinic receptors. Its potency at nicotinic receptors is much lower, and actions at non-muscarinic receptors are generally undetectable clinically. Atropine does not distinguish among the M1, M2, and M3 subgroups of muscarinic receptors. In contrast, other antimuscarinic drugs are moderately selective for one or another of these subgroups. Most synthetic antimuscarinic drugs are considerably less selective than atropine in interactions with nonmuscarinic receptors.
A study on glycopyrrolate, a quaternary ammonium salt, was found to have a fetal: maternal serum concentration ratio of 0.4 indicating partial transfer.
Heparin, suxamethonium, and vecuronium do not cross the placenta.
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This question is part of the following fields:
- Pharmacology
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Question 51
Incorrect
-
A 79-year-old female complains of painful legs, especially in her thigh region. The pain starts after walking and settles with rest. She occasionally has to take paracetamol to relieve the pain. She is a known case of hyperlipidaemia, type 2 diabetes mellitus, hypertension, and depression.
Her physician makes a provisional diagnosis of claudication of the femoral artery, which is a continuation of the external iliac artery.
Which of the following anatomical landmarks does the external iliac artery cross to become the femoral artery?Your Answer:
Correct Answer: Inguinal ligament
Explanation:The external iliac artery is the larger of the two branches of the common iliac artery. It forms the main blood supply to the lower limbs. The common iliac bifurcates into the internal and external iliac artery anterior to the sacroiliac joint.
The external iliac artery courses on the medial border of the psoas major muscles and exits the pelvic girdle posterior to the inguinal ligament. Here, midway between the anterior superior iliac spine and the pubic symphysis, the external iliac artery becomes the femoral artery and descends along the anteromedial part of the thigh in the femoral triangle.
The pectineus forms the posterior border of the femoral canal.
The femoral vein forms the lateral border of the femoral canal.
The medial border of the adductor longus muscle forms the medial wall of the femoral triangle.
The medial border of the sartorius muscle forms the lateral wall of the femoral triangle. -
This question is part of the following fields:
- Anatomy
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Question 52
Incorrect
-
A 73-year-old man, presents with abdominal pain, constipation and blood on defecation. He is diagnosed with a distal sigmoid colon carcinoma.
Which artery is most likely to provide its blood supply?Your Answer:
Correct Answer: Inferior mesenteric artery
Explanation:The inferior mesenteric artery supplies blood to the hindgut, which includes the sigmoid colon.
Note that during high anterior resection of distal sigmoid colon tumours, the inferior mesenteric artery is ligated, interrupting blood supply.
The branches of the internal iliac artery, particularly the middle rectal branch, are essential in retaining vascularity of the rectal stump.
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This question is part of the following fields:
- Anatomy
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Question 53
Incorrect
-
Gag reflex was assessed as a part of brain stem death in a 22-year-old man with severe traumatic brain injury.
Which of the following nerves forms the afferent limb of this reflex?Your Answer:
Correct Answer: Glossopharyngeal nerve
Explanation:The gag reflex is a protective mechanism that prevents any foreign material to enter the aerodigestive tract.
This reflex has afferent (sensory) and effect (motor) components.
– Glossopharyngeal nerve form the afferent limb
– Vagus nerve form the efferent limb -
This question is part of the following fields:
- Pathophysiology
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Question 54
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 55
Incorrect
-
An 82-year old male has shortness of breath which is made worse when he lies down but investigations have revealed a normal ejection fraction. Why might this be?
Your Answer:
Correct Answer: He has diastolic dysfunction
Explanation:Decreased stroke volume causes decreased ejection fraction which results in diastolic dysfunction.
Ejection fraction is not a useful measure in someone with diastolic dysfunction because stroke volume may be reduced whilst end-diastolic volume may be reduced.
Diastolic dysfunction may arise with reduced heart compliance.Ejection fraction measures of the proportion of blood leaving the ventricles with each beat and is calculated as follows:
Stroke volume / end-diastolic volume.A healthy ejection fraction is usually taken as 60% (based on a stroke volume of 70ml and end-diastolic volume of 120ml).
Respiratory inspiration causes a decreased pressure in the thoracic cavity, which in turn causes more blood to flow into the atrium.
Sitting up decreases venous because of the action of gravity on blood in the venous system.
Hypotension also decreases venous return.
A less compliant aorta, like in aortic stenosis increases end systolic left ventricular volume which decreases stroke volume.Systemic vascular resistance = mean arterial pressure / cardiac output.
Increased vascular resistance impedes the flow of blood back to the heart.Increased venous return increases end diastolic LV volume as there is more blood returning to the ventricles.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 56
Incorrect
-
Which muscle separates the subclavian artery and the subclavian vein?
Your Answer:
Correct Answer: Scalenus anterior
Explanation:The subclavian artery and vein have a similar path throughout their course, with the subclavian vein running anterior to the subclavian artery. The artery and vein are separated by the insertion of the scalenus anterior muscle.
There are three scalene muscles, found on each side of the neck:
1. Anterior scalene
2. Middle scalene
3. Posterior scaleneThe scalenus anterior muscle is the anterior most of the three scalene muscles. It originates from the transverse processes of vertebrae C3-C6 and is inserted in the first rib.
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This question is part of the following fields:
- Anatomy
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Question 57
Incorrect
-
Which of these anaesthetics has the best chance of preventing HPV (hypoxic pulmonary vasoconstriction)?
Your Answer:
Correct Answer: Desflurane 2 MAC
Explanation:Resistance pulmonary arteries constrict in response to alveolar and airway hypoxia, diverting blood to better-oxygenated alveoli.
In atelectasis, pneumonia, asthma, and adult respiratory distress syndrome, hypoxic pulmonary vasoconstriction optimises O2 uptake. Hypoxic pulmonary vasoconstriction helps maintain systemic oxygenation during single-lung anaesthesia.
A redox-based O2 sensor within pulmonary artery smooth muscle cells is involved in hypoxic pulmonary vasoconstriction. The production of reactive oxygen species by smooth muscle cells in the pulmonary artery varies in proportion to PaO2. Hypoxic removal of these redox second messengers inhibits voltage-gated potassium channels, depolarizing smooth muscle cells in the pulmonary artery.
L-type calcium channels are activated by depolarization, which raises cytosolic calcium and causes hypoxic pulmonary vasoconstriction. Some anaesthetics suppress this response, increasing the risk of further deterioration in ventilation perfusion mismatch.
Agents that inhibit HPV are ether, halothane, and desflurane (>1.6 MAC).
Agents with no effect on HPV include thiopentone, fentanyl, desflurane (1MAC), isoflurane (<1.5MAC), sevoflurane(1MAC), and propofol. -
This question is part of the following fields:
- Pharmacology
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Question 58
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 59
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 60
Incorrect
-
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:
Correct 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 61
Incorrect
-
A study aimed at assessing the validity of a novel diagnostic test for heart failure is being performed. The curators are worried that not all the patients will get the prevalent gold standard test.
Which type of bias is that?Your Answer:
Correct Answer: Work-up bias
Explanation:Work up bias involves comparing the novel diagnostic test with the current standard test. A portion of the patients undergo the standard test while others undergo the new test as the standard test is costly. The result can be alteration in specify and sensitivity.
Selection bias is when randomisation is not achieved.
Attention bias refers to the person’s failure to consider various alternatives when he pre occupied by some other thoughts.
Instrument bias is related to the experience and extent of familiarization of the participating individuals with the test.
Co intervention bias is characterized by the groups receiving different co interventions.
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This question is part of the following fields:
- Statistical Methods
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Question 62
Incorrect
-
Which of the following statements is true regarding drug dose and response?
Your Answer:
Correct Answer: Intrinsic activity determines maximal response
Explanation:There are two types of drug dose-response relationships, namely, the graded dose-response and the quantal dose-response relationships.
Drug response curves are plotted as percentage response again LOG drug concentration. This graph is sigmoid in shape.
Agonists are drugs with high affinity and high intrinsic activity. Meanwhile, the antagonist is a drug with high affinity but no intrinsic activity. Intrinsic activity determines the maximal response. The maximal response can be achieved even by activation of a small proportion of receptor sites.
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This question is part of the following fields:
- Pharmacology
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Question 63
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 64
Incorrect
-
Regarding the following induction agents, which one is cleared at the fastest rate from the plasma?
Your Answer:
Correct Answer: Propofol
Explanation:Propofol is cleared at the fastest rate at the rate of 60ml/kg/min.
Clearance rate of other drugs are as follows:
– Thiopental: 3.5 ml/kg/min
– Methohexitone: 11 ml/kg/min
– Ketamine: 17 ml/kg/min
– Etomidate: 10-20 ml/kg/min -
This question is part of the following fields:
- Pharmacology
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Question 65
Incorrect
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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 66
Incorrect
-
A common renal adverse effect of non-steroidal anti-inflammatory drugs is?
Your Answer:
Correct Answer: Haemodynamic renal insufficiency
Explanation:Prostaglandins do not play a major role in regulating RBF in healthy resting individuals. However, during pathophysiological conditions such as haemorrhage and reduced extracellular fluid volume (ECVF), prostaglandins (PGI2, PGE1, and PGE2) are produced locally within the kidneys and serve to increase RBF without changing GFR. Prostaglandins increase RBF by dampening the vasoconstrictor effects of both sympathetic activation and angiotensin II. These effects are important because they prevent severe and potentially harmful vasoconstriction and renal ischemia. Synthesis of prostaglandins is stimulated by ECVF depletion and stress (e.g. surgery, anaesthesia), angiotensin II, and sympathetic nerves.
Non-steroidal anti-inflammatory drugs (NSAIDs), such as ibuprofen and naproxen, potently inhibit prostaglandin synthesis. Thus administration of these drugs during renal ischemia and hemorrhagic shock is contraindicated because, by blocking the production of prostaglandins, they decrease RBF and increase renal ischemia. Prostaglandins also play an increasingly important role in maintaining RBF and GFR as individuals age. Accordingly, NSAIDs can significantly reduce RBF and GFR in the elderly.
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This question is part of the following fields:
- Physiology
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Question 67
Incorrect
-
A 46-year old man was taken to the emergency room due to slow, laboured breathing. A relative reported that he's maintained on codeine 60 mg, taken orally every 6 hours for severe pain from oesophageal cancer. His creatinine was elevated, and glomerular filtration rate was severely decreased at 27 ml/minute.
Given the scenario above, which of the metabolites of codeine is the culprit for his clinical findings?Your Answer:
Correct Answer: Morphine-6-glucuronide
Explanation:Accumulation of morphine-6-glucuronide is a risk factor for opioid toxicity during morphine treatment. Morphine is metabolized in the liver to morphine-6-glucuronide and morphine-3-glucuronide, both of which are excreted by the kidneys. In the setting of renal failure, these metabolites can accumulate, resulting in a lowering of the seizure threshold. However, it does not occur in all patients with renal insufficiency, which is the most common reason for accumulation of morphine-6-glucuronide; this suggests that other risk factors can contribute to morphine-6-glucuronide toxicity.
The active metabolites of codeine are morphine and the morphine metabolite morphine-6-glucuronide. The enzyme systems responsible for this metabolism are: CYP2D for codeine and UGT2B7 for morphine, codeine-6-gluronide, and morphine-6-glucuronide. Both of these systems are subject to genetic variation. Some patients are ultrarapid metabolizers of codeine and produce higher levels of morphine and active metabolites in a very short period of time after administration. These increased levels will produce increased side effects, especially drowsiness and central nervous system depression.
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This question is part of the following fields:
- Pharmacology
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Question 68
Incorrect
-
A new volatile anaesthetic agent has been approved for use in clinical testing.
It's a non-irritating, sweet-smelling substance. It has a molecular weight of 170, a 0.6 blood:gas partition coefficient, and a 180 oil:gas partition coefficient. An oxidative pathway converts 2% of the substance to trifluoroacetic acid.
Which of the following statements best describes this agent's pharmacological profile?Your Answer:
Correct Answer: It has a lower molecular weight than isoflurane
Explanation:Because enflurane is much less soluble in blood and has a blood: gas partition coefficient of 1.8, both wash-in and wash-out should be faster.
Sevoflurane’s sweet-smelling, non-irritant nature, combined with a low blood: gas partition coefficient, would result in similar offset and onset characteristics.
Isoflurane and enflurane have a molecular weight of 184.
The oil: gas partition coefficient on a volatile agent is a measure of lipid solubility, potency, and thus MAC. Halothane has an oil: gas partition coefficient of 220 and a MAC of 0.74. One would expect the MAC to be higher with an oil gas partition coefficient of 180 (less lipid soluble).
The conversion of halothane (20%) to trifluoroacetic acid via oxidative metabolism has been linked to the development of hepatitis.
P450 2E1 converts sevoflurane to hexafluoroisopropanol, which results in the release of inorganic fluoride ions. It’s the only fluorinated volatile anaesthetic that doesn’t break down into trifluoracetic acid.
Desflurane is likely to cause airway irritation, which can lead to coughing, apnoea, and laryngospasm, despite its low blood:gas partition coefficient (0.42).
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This question is part of the following fields:
- Pharmacology
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Question 69
Incorrect
-
In a normal healthy adult breathing 100 percent oxygen, which of the following is the most likely cause of an alveolar-arterial (A-a) oxygen difference of 30 kPa?
Your Answer:
Correct Answer: Atelectasis
Explanation:The ‘ideal’ alveolar PO2 minus arterial PO2 is the alveolar-arterial (A-a) oxygen difference.
The ‘ideal’ alveolar PO2 is derived from the alveolar air equation and is the PO2 that the lung would have if there was no ventilation-perfusion (V/Q) inequality and it was exchanging gas at the same respiratory exchange ratio as real lung.
The amount of oxygen in the blood is measured directly in the arteries.
The A-a oxygen difference (or gradient) is a useful measure of shunt and V/Q mismatch, and it is less than 2 kPa in normal adults breathing air (15 mmHg). Because the shunt component is not corrected, the A-a difference increases when breathing 100 percent oxygen, and it can be up to 15 kPa (115 mmHg).
An abnormally low or abnormally high V/Q ratio within the lung can cause an increased A-a difference, though the former is more common. Atelectasis, which results in a low V/Q ratio, is the most likely cause of an A-a difference in a healthy adult breathing 100 percent oxygen.
Hypoventilation may cause an increase in alveolar (and thus arterial) CO2, lowering alveolar PO2 according to the alveolar air equation.
The alveolar PO2 is also reduced at high altitude.
Healthy people are unlikely to have a right-to-left shunt or an oxygen transport diffusion defect.
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This question is part of the following fields:
- Physiology
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Question 70
Incorrect
-
A 32-year-old male is admitted to the critical care unit. He has suffered a heroin overdose and requires intubation and ventilatory support.
What would be his predicted total static compliance (lung and chest wall) measurements.Your Answer:
Correct Answer: 100 ml/cmH2O
Explanation:Static lung compliance refers to the change in volume within the lung per given change in unit pressure. It is usually measured when air flow is absent, such as during pauses in inhalation and exhalation.
It is a combination of:
Chest wall compliance: normal value is 200 mL/cmH2O
Lung tissue compliance: normal value is 200 mL/ cmH2OIt is represented mathematically as:
1/Crs = 1/Cl + 1/Ccw
Where,
Crs = total compliance of the respiratory system
Cl = compliance of the lung
Ccw = compliance of the chest wallTherefore in this case:
1/Crs = 1/200 + 1/200
1/Crs = 0.005 + 0.005 = 0.01
1/Ct = 0.01
Rearranging equation gives:
Ct = 1/0.01 = 100 mL/cmH2O.
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This question is part of the following fields:
- Clinical Measurement
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Question 71
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 72
Incorrect
-
The required sample size in a trial of a new therapeutic agent varies with?
Your Answer:
Correct Answer: Level of statistical significance required
Explanation:The level of statistical significance required influences the sample size used. This is because sample size is used in the calculation of SD/SE.
Sample size does not affect
The level of acceptance
The alternative hypothesis with a general level set at p<0.05
The test to be used.Experience of the investigator and the type of patient recruited should have no bearing on the required sample size.
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This question is part of the following fields:
- Statistical Methods
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Question 73
Incorrect
-
Heights of 100 individuals(adults) who were administered steroids at any stage during childhood was studied. The mean height was found to be 169cm with the data having a standard deviation of 16cm. What will be the standard error associated with the mean?
Your Answer:
Correct Answer: 1.6
Explanation:Standard error can be calculated by the following formula:
Standard Error= (Standard Deviation)/√(Sample Size)
= (16) / √(100)
= 16 / 10
= 1.6 -
This question is part of the following fields:
- Statistical Methods
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Question 74
Incorrect
-
A healthy 27-year old male who weighs 70kg has appendicitis. He is currently in the operating room and is being positioned to have a rapid sequence induction.
Prior to preoxygenation, the compartment likely to have the best oxygen reserve is:Your Answer:
Correct Answer: Red blood cells
Explanation:The following table shows the compartments and their relative oxygen reserve:
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 160Oxygen reserves in the body, with room air and after oxygenation.
FAO2-alveolar fraction of oxygen rises to 95% after administration of 100% oxygen (CO2 = 5%)
FRC- Functional residual capacity – (the most important store of oxygen in the body) – 2,500-3,000 mL in medium sized adults
PaO2-partial pressure of oxygen dissolved in arterial blood (80 mmHg breathing room air and 500 mmHg breathing 100% oxygen)
DF -dissolved form (0.3%)
PV-plasma volume (3L)
TG-total globular volume (5L)
Hb-haemoglobin concentration
SaO2-arterial oxygen concentration (98% breathing air and 100% when preoxygenated) -
This question is part of the following fields:
- Physiology
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Question 75
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 76
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 77
Incorrect
-
Bacteria and viruses that are smaller than 0.1 ?m in diameter can be filtered out using heat and moisture exchanger (HME) with a typical pore size 0.2 ?m.
Choose the most appropriate mechanisms of particle capture for most bacteria and viruses.Your Answer:
Correct Answer: Diffusion
Explanation:Warming, humidifying, and filtering inspired anaesthetic gases is done by heat and moisture exchangers (HME) and breathing system filters. They are made of glass fibres materials and are supported by a sturdy frame. Pleating increases the surface area to reduce resistance to air flow and boost efficiency.
Filters’ effectiveness is determined by the amount and size of particles they keep out of the patient’s airway. The efficiency of filters might be classified as 95, 99.95, or 99.97 percent. Pores with a diameter of 0.2 µm are common. The following are examples of typical particle sizes:
Red blood cell – 5 µm
Lymphocyte – 5-8 µm
Viruses – 0.02-0.3 µm
Bacteria – 0.5-1 µm
Depending on particle size, gas flow speed, and charge, particles are collected via a number of processes. Mechanical sieve, interception, diffusion, electrostatic filtration, and inertial impaction are some of the options:Sieve:
The diameter of the particle the filter is supposed to collect is smaller than the apertures of the filter’s fibres.Interception:
When a particle following a gas streamline approaches a fibre within one radius of itself, it becomes attached and captured.
Diffusion:A particle’s random (Brownian) zig-zag path or motion causes it to collide with a fibre.
By attracting and capturing a particle from within the gas flow, it generates a lower-concentration patch within the gas flow into which another particle diffuses, only to be captured. At low gas velocities and with smaller particles (0.1µm diameter), this is more common.Electrostatic:
These filters use large diameter fibre media and rely on electrostatic charges to improve fine particle removal effectiveness.
Impaction due to inertia:
When a particle is too large to respond fast to abrupt changes in streamline direction near a filter fibre, this happens. Because of its inertia, the particle will continue on its original course and collide with the filter fibre. When high gas velocities and dense fibre packing of the filter media are present, this sort of filtration mechanism is most prevalent.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 78
Incorrect
-
A 52-year-old patient is brought to ER with a chief complaint of chest pain for two hours. Chest pain was tightness in nature, located in the centre of the chest and radiate into the neck and left arm. The patient otherwise looks fit and well.
Just after admitting the patient, he suffered VF cardiac arrest and is immediately defibrillated with the return of spontaneous circulation (ROSC).
On clinical examination following was the finding:
BP: 82/45 mmHg
Heart rate: 120 beats/min
Oxygen saturation on air: 25%
Heart sounds: Normal
There is no sign of pulmonary oedema. The patient is anxious, cold, and clammy.
A 12 lead ECG was done which revealed a sinus rhythm of 120 with ST-segment depression and T wave inversion in leads II, III, and aVF. Which of the following is considered best for the initial treatment of the patient?Your Answer:
Correct Answer: Oral aspirin
Explanation:This is a classical case of unstable angina or NSTEMI (Non-ST-elevation myocardial infarction). As soon as the diagnosis of unstable angina or NSTEMI is made the initial treatment is Aspirin and antithrombin therapy.
Betablocker is known to reduce mortality from acute myocardial infarction by reducing oxygen demand. If there is no contraindication (heart block, bradycardia, hypotension, severe left ventricular dysfunction, and asthma), a beta-blocker should be given early. This patient has hypotension and therefore metoprolol is contraindicated.
If three doses of nitroglycerine tablets or Nitrolingual sprays and intravenous beta-blockers too cannot relieve the symptoms intravenous Glyceryl Trinitrate (GTN) should be considered provided that there is no hypotension. But in this case, the patient is hypotensive, and therefore, it is contraindicated.
If the symptoms are not relieved after three serial doses of nitroglycerine or if symptoms recur despite adequate anti-anginal treatment morphine sulphate is indicated.
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This question is part of the following fields:
- Pathophysiology
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Question 79
Incorrect
-
Which is correct about normal distribution?
Your Answer:
Correct Answer: Mean = mode = median
Explanation:The normal distribution is a symmetrical, bell-shaped distribution in which the mean, median and mode are all equal.
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This question is part of the following fields:
- Statistical Methods
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Question 80
Incorrect
-
A 4-year-old boy with status epilepticus was brought to ER and has already received two doses of intravenous lorazepam but is still continuing to have seizures.
Which of the following drug would be best for his treatment?Your Answer:
Correct Answer: Phenytoin 20 mg/kg IV
Explanation:When the convulsion lasts for five or more than five minutes, or if there are recurrent episodes of convulsions in a 5 minute period without returning to the baseline, it is termed as Status Epilepticus.
The first priority in the patient with seizures is maintaining the airway, breathing, and circulation.Guideline for the management of Status Epilepticus in children by Advanced Life Support Group is as follow:
Step 1 (Five minutes after the start of seizures):
If intravascular access is available start treatment with lorazepam 0.1 mg/kg IV
If no intravascular access then give buccal midazolam 0.5 mg/kg or rectal diazepam 0.5 mg/kg.Step 2 (Ten minutes after the start of seizure):
If the convulsions continue then a second dose of benzodiazepine should be given. Senior should be called on-site and phenytoin should be prepared.
No more than two doses or benzodiazepines should be given (including any doses given before arrival at the hospital)
If still no IV access then obtain intraosseous access (IO).Step 3 (Ten minutes after step 2)
Senior help along with anaesthetic/ICU help should be sought
Phenytoin 20 mg/kg IV over 20 minutes
If the seizure stops before the full dose of phenytoin is given then the infusion should be completed as this provides up to 24 hours of anticonvulsant effect
In children already receiving phenytoin as treatment for epilepsy then an alternative is phenobarbitone 20 mg/kg IV over five minutes
Once the phenytoin is started, senior staff may wish to give rectal paraldehyde 0.4 mg/kg although this is no longer included in the routine algorithm recommended by APLS.Step 4 (20 minutes after step 3)
If 20 minutes after starting phenytoin the child remains in status epilepticus then rapid sequence induction of anaesthesia with thiopentone and a short acting paralysing agent is needed and the child transferred to paediatric intensive care.
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This question is part of the following fields:
- Pathophysiology
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Question 81
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 82
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 83
Incorrect
-
A controlled retrospective study's level of evidence is?
Your Answer:
Correct Answer: Level 3
Explanation:Level 1 – High-quality randomised controlled trial with statistically significant difference or no statistically significant difference but narrow confidence intervals (prospective controlled)
Level 2 – Prospective comparative study (prospective uncontrolled)
Level 3 – Case-control study, retrospective comparative study (retrospective controlled)
Level 4 – Case series (retrospective uncontrolled)
Level 5 – Expert opinion.
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This question is part of the following fields:
- Statistical Methods
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Question 84
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 85
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 86
Incorrect
-
Out of the following, which is NOT true regarding the external carotid?
Your Answer:
Correct Answer: It ends by bifurcating into the superficial temporal and ascending pharyngeal artery
Explanation: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 arteryThe external carotid has eight branches, 3 from its anterior surface ; thyroid, lingual and facial. The pharyngeal artery is a medial branch. The posterior auricular and occipital are posterior branches.
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This question is part of the following fields:
- Anatomy
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Question 87
Incorrect
-
Which of the following statements is not true regarding Adrenaline?
Your Answer:
Correct Answer: Exerts its effect by decreasing intracellular calcium
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 88
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 89
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 90
Incorrect
-
A study of blood pressure measurements is being performed in patients with chronic kidney disease.
Considering that the results are normally distributed, what percentage of values lie within two standard deviations of the mean blood pressure reading?Your Answer:
Correct Answer: 95.40%
Explanation:Normal distribution, also called Gaussian distribution, the most common distribution function for independent, randomly generated variables, and describes the spread for many biological and clinical measurements.
Properties of the Normal distribution
symmetrical i.e. Mean = mode = median
68.3% of values lie within 1 SD of the mean
95.4% of values lie within 2 SD of the mean
99.7% of values lie within 3 SD of the mean
The empirical rule, or the 68-95-99.7 rule, tells you where most of the values lie in a normal distribution: Around 68% of values are within 1 standard deviation of the mean.
Around 95% of values are within 2 standard deviations of the mean. Around 99.7% of values are within 3 standard deviations of the mean.
the standard deviation (SD) is a measure of how much dispersion exists from the mean.SD = square root (variance)
The empirical rule, or the 68-95-99.7 rule states where most of the values lie in a normal distribution. Around 68% of values fall within 1 S.D of the mean, about 95% within 2 S.D of the mean, and about 99.7% of values within 3 S.D of the mean. Therefore, 95.4% is the most reasonable answer if results are normally distributed.
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This question is part of the following fields:
- Statistical Methods
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Question 91
Incorrect
-
Which of the following statements best describes adenosine receptors?
Your Answer:
Correct Answer:
Explanation:Adenosine receptors are expressed on the surface of most cells.
Four subtypes are known to exist which are A1, A2A, A2B and A3.Of these, the A1 and A2 receptors are present peripherally and centrally. There are agonists at the A1 receptors which are antinociceptive, which reduce the sensitivity to a painful stimuli for the individual. There are also agonists at the A2 receptors which are algogenic and activation of these results in pain.
The role of adenosine and other A1 receptor agonists is currently under investigation for use in acute and chronic pain states.
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This question is part of the following fields:
- Physiology
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Question 92
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 93
Incorrect
-
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 94
Incorrect
-
During a stabbing incident, a 30-year-old injured his inferior vena cava. What number of functional valves can be usually found in this vessel?
Your Answer:
Correct Answer: 0
Explanation:The inferior vena cava is formed by the union of the right and left common iliac veins. The inferior vena cava has no functional valves like the one-way valves commonly found in many veins. The forward flow to the heart is driven by the differential pressure created by normal respiration.
The absence of functional valves has an important clinical role when cannulating during cardiopulmonary bypass.
There is a valve that is non-functioning called the eustachian valve that lies at the junction of the IVC and the right atrium. This valve has a role to help direct the flow of oxygen-rich blood through the right atrium to the left atrium via the foramen ovale during fetal life. It has no specific function in adult life.
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This question is part of the following fields:
- Anatomy
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Question 95
Incorrect
-
Which of the following statements is true about monoamine oxidase (MOA) enzymes?
Your Answer:
Correct Answer: Type A and type B are found in the liver and brain
Explanation:Monoamine oxidase (MOA) enzymes are responsible for the catalyses of monoamine oxidative deamination. It assists the degradation of serotonin, norepinephrine (NE) and dopamine.
They are found in the mitochondria of most central and peripheral nerve tissues.
There are 2 different types:
Type A: Whose main function it to inactivate dopamine, tyramine, norepinephrine and 5-hydroxytryptamine. In addition to the nervous system, it is also found in the liver, brain gastrointestinal tract, pulmonary endothelium and placenta
Type B: Whose main function is to inactivate dopamine, tyramine, tryptamine and phenylethylamine. In addition to the nervous system, it is also found in the liver, brain (especially in the basal ganglia) and blood platelets. -
This question is part of the following fields:
- Pathophysiology
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Question 96
Incorrect
-
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:
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 97
Incorrect
-
Which of the following nerves is responsible for carrying taste sensation from the given part of the tongue?
Your Answer:
Correct Answer: Anterior two thirds of tongue - facial nerve
Explanation:Taste sensation from the anterior two-thirds of the tongue is carried by chorda tympani, a branch of the facial nerve.
The general somatic sensation of the anterior two-third of the tongue is supplied by the lingual nerve, a branch of the mandibular nerve.
Both general somatic sensation and taste from the posterior third of the tongue are carried by the glossopharyngeal nerve.
All the muscles of the tongue except palatoglossus are supplied by the hypoglossal nerve whereas palatoglossus is supplied by the vagus nerve. (This is because palatoglossus is the only tongue muscle derived from the fourth branchial arch)
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This question is part of the following fields:
- Pathophysiology
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Question 98
Incorrect
-
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 99
Incorrect
-
A new drug treatment has been developed for Crohn's disease. The pharmaceutical company behind this, is planning to conduct a trial and is looking for hiring around 200 individuals that are suffering from Crohn's disease. The aim would be to determine if there is any decline in the disease activity in response to the drug and compare it with a placebo.
What phase is the trial in?Your Answer:
Correct Answer: Phase 2
Explanation:The study is being conducted on a smaller level with only 200 participants and is determining the effectiveness of the drug in comparison to a placebo. These characteristics are in accordance with the second phase of trial.
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This question is part of the following fields:
- Statistical Methods
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Question 100
Incorrect
-
What structure lies deepest within the popliteal fossa?
Your Answer:
Correct Answer: Popliteal artery
Explanation:The popliteal fossa is the shallow, diamond-shaped depression located in the back of the knee joint.
The structures that lie within in from superficial to deep are:
The tibial and common fibular nerve: Most superficial. They arise from the sciatic nerve.
The popliteal vein
The popliteal artery: Lies deepest. It arises from the femoral arteryBoundaries of the popliteal fossa:
Laterally
Biceps femoris above, lateral head of gastrocnemius and plantaris belowMedially
Semimembranosus and semitendinosus above, medial head of gastrocnemius belowFloor
Popliteal surface of the femur, posterior ligament of knee joint and popliteus muscleRoof
Superficial and deep fascia -
This question is part of the following fields:
- Anatomy
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Question 101
Incorrect
-
A 65-year-old man has been diagnosed with transitional cell carcinoma of the left kidney. He will be operated on, and as part of the surgery, the left renal artery has to be located and dissected.
Which of the following vertebral levels gives rise to this artery?
Your Answer:
Correct Answer: L1
Explanation:The renal arteries branch from the abdominal aorta just below the origin of the superior mesenteric artery. The right renal artery is higher and longer than the left renal artery. The left renal artery passes behind the left renal vein, the body of the pancreas, and the splenic vein.
The important landmarks of vessels arising from the abdominal aorta at different levels of vertebrae are:
T10 – oesophageal opening in the diaphragm
T12 – Coeliac trunk, aortic hiatus in the diaphragm
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 102
Incorrect
-
Which of the following ionic changes is associated with the ventricular myocyte action potential's initial repolarization phase?
Your Answer:
Correct Answer: Ceased Na+ and increase K+ conductances
Explanation:The Purkinje system, as well as the action potentials of ventricular and atrial myocytes, have the same ionic changes. It lasts about 200 milliseconds and has a resting membrane potential, as well as fast depolarisation and plateau phases.
There are five stages to the process:
Increased Na+ and decreased K+ conductance in Phase 0 (rapid depolarisation).
1st phase (initial repolarisation) : Na+ conductance decreased, while K+ conductance increased.
Phase two (plateau phase) : Ca2+ conductance increased
Phase three (repolarisation phase) : Lower Ca2+ conductance and higher K+ conductance
4th Phase (resting membrane potential) : K+ conductance increased, Na+ conductance decreased, and Ca2+ conductance decreased. -
This question is part of the following fields:
- Pathophysiology
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Question 103
Incorrect
-
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 104
Incorrect
-
The right coronary artery supplies blood to all the following, except which?
Your Answer:
Correct Answer: The circumflex artery
Explanation:The right coronary artery supplies the right ventricle, the right atrium, the sinoatrial (SA) node and the atrioventricular (AV) node.
The circumflex artery originates from the left coronary artery and is supplied by it.
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This question is part of the following fields:
- Anatomy
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Question 105
Incorrect
-
A 25-year-old soldier is shot in the abdomen. He has multiple injuries, including a major disruption to the abdominal aorta. The bleeding is torrential and needs to be controlled by placing a vascular clamp immediately inferior to the diaphragm.
During this manoeuvre, which vessel may be injured?Your Answer:
Correct Answer: Inferior phrenic arteries
Explanation:The inferior phrenic nerves are at the highest risk of damage as they are the first branches of the abdominal aorta. The potential space at the level of the diaphragmatic hiatus is a potentially useful site for aortic occlusion. However, leaving the clamp applied for more than 10 -15 minutes usually leads to poor outcomes.
The superior phrenic artery branches from the thoracic aorta.
The abdominal aorta begins at the level of the body of T12 near the midline, as a continuation of the thoracic aorta. It descends and bifurcates at the level of L4 into the common iliac arteries.
The branches of the abdominal aorta (with their vertebra level) are:
1. Inferior phrenic arteries: T12 (upper border)
2. Coeliac artery: T12
3. Superior mesenteric artery: L1
4. Middle suprarenal arteries: L1
5. Renal arteries: Between L1 and L2
6. Gonadal arteries: L2 (in males, it is the testicular artery, and in females, the ovarian artery)
7. Inferior mesenteric artery: L3
8. Median sacral artery: L4
9. Lumbar arteries: Between L1 and L4 -
This question is part of the following fields:
- Anatomy
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Question 106
Incorrect
-
Cells use adenosine-5-triphosphate (ATP) as a coenzyme and is a source of energy.
Glucose metabolism produces the most ATP from which of the following biochemical processes?
Your Answer:
Correct Answer: Electron transport phosphorylation in the mitochondria
Explanation:Glycolysis occurs in the cytoplasm of the cell. It converts 1 glucose molecule (6-carbon) to pyruvate (two 3-carbon molecules) and produces 4 ATP molecules and 2NADH but uses 2 ATP in the process with an overall net energy production of 2 ATP.
Pyruvate is then oxidised to acetyl coenzyme A (generating 2 NADH per pyruvate molecule). This takes place in the mitochondria and then enters the Krebs cycle (citric acid cycle). It produces 2 ATP, 8 NADH and 2 FADH2 per glucose molecule.
Electron transport phosphorylation takes place in the mitochondria. The aim of this process is to break down NADH and FADH2 and also to pump H+ into the outer compartment of the mitochondria. It produces 32 ATP with an overall net production of 36ATP.
In anaerobic respiration which occurs in the cytoplasm, pyruvate is reduced to NAD producing 2 ATP.
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This question is part of the following fields:
- Physiology
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Question 107
Incorrect
-
Which of the following vertebral levels is the site where the aorta perforates the diaphragm?
Your Answer:
Correct Answer: T12
Explanation:The diaphragm divides the thoracic cavity from the abdominal cavity. Structures penetrate the diaphragm at different vertebral levels through openings in the diaphragm to communicate between the two cavities. The diaphragm has openings at three vertebral levels:
T8: vena cava, terminal branches of the right phrenic nerve
T10: oesophagus, vagal trunks, left anterior phrenic vessels, oesophageal branches of the left gastric vessels
T12: descending aorta, thoracic duct, azygous and hemi-azygous vein -
This question is part of the following fields:
- Anatomy
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Question 108
Incorrect
-
Which of the following statements is correct regarding opioid receptors?
Your Answer:
Correct Answer: Binding with an opioid agonist increases potassium conductance
Explanation:Opioid receptors are a large family of seven transmembrane domain receptors. They are of four types:
1) Delta opioid receptor
2) Mu opioid receptor
3) Kappa opioid receptor
4) Orphan receptor-like 1
They contain about 372-400 amino acids and thus their molecular weight is different.
Opioid receptor activation reduces the intracellular cAMP formation and opens K+ channels (mainly through µ and δ receptors) or suppresses voltage-gated N-type Ca2+ channels (mainly κ receptor). These actions result in neuronal hyperpolarization and reduced availability of intracellular Ca2+ which results in decreased neurotransmitter release by cerebral, spinal, and myenteric neurons (e.g. glutamate from primary nociceptive afferents).
However, other mechanisms and second messengers may also be involved, particularly in the long-term
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This question is part of the following fields:
- Pharmacology
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Question 109
Incorrect
-
Which of the following statements is true regarding alfentanil?
Your Answer:
Correct Answer: Is less lipid soluble than fentanyl
Explanation:Alfentanil is less lipid-soluble than fentanyl and thus is less permeable to the membrane making it less potent.
Alfentanil is a phenylpiperidine opioid analgesic with rapid onset and shorter duration of action.
Alfentanil has less volume of distribution due to its high plasma protein binding (92%)
It can cause respiratory depression and can cause sedation
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This question is part of the following fields:
- Pharmacology
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Question 110
Incorrect
-
General anaesthesia is administered to a patient in a hospital in Lhasa which is one of the highest cities in the world (at 11,975 feet). An Anaesthetic rotameter is normally calibrated at 20 C and 1 bar pressure and is known to be underread at altitude. The temperature of the theatre was 10 C.
Which one of the following physical properties is responsible for the rotameter inaccuracy in these conditions?Your Answer:
Correct Answer: Density of the gas
Explanation:Since the gas is less dense at higher altitudes, the density of a gas influences flows when passing through the orifice. Due to this reason, for a given flow rate, the bobbin will not be forced as far up the rotameter tube.
At higher altitudes, the volume of a fixed mass of gas increases, and therefore the molecules of gas are widely spaced resulting in a decrease in density with an increase in altitude.
Viscosity is simply termed as friction of gas. The viscosity of a gas is important only at low flow rates when the flow characteristic of the gas is laminar.
Charle’s law stated that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature (T) provided the pressure remains constant.
Boyle’s law for a fixed amount of gas at constant temperature, the pressure (P) and volume (V) are inversely proportional.
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This question is part of the following fields:
- Basic Physics
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Question 111
Incorrect
-
Which of the following 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 112
Incorrect
-
A 30-year old male has Von Willebrand's disease and attends the hospital to get an infusion of desmopressin acetate. The way this works is by stimulating the release of von Willebrand factor from cells, which in turn increases factor VIII and platelet plug formation in clotting.
In patients that have no clotting abnormalities, the substance that keeps the blood soluble and prevents platelet activation normally is which of these?Your Answer:
Correct Answer: Prostacyclin
Explanation:Even though aprotinin reduces fibrinolysis and therefore bleeding, there is an associated increased risk of death. It was withdrawn in 2007.
Protein C is dependent upon vitamin K and this may paradoxically increase the risk of thrombosis during the early phases of warfarin treatment.The coagulation cascade include two pathways which lead to fibrin formation:
1. Intrinsic pathway – these components are already present in the blood
Minor role in clotting
Subendothelial damage e.g. collagen
Formation of the primary complex on collagen by high-molecular-weight kininogen (HMWK), prekallikrein, and Factor 12
Prekallikrein is converted to kallikrein and Factor 12 becomes activated
Factor 12 activates Factor 11
Factor 11 activates Factor 9, which with its co-factor Factor 8a form the tenase complex which activates Factor 102. Extrinsic pathway – needs tissue factor that is released by damaged tissue)
In tissue damage:
Factor 7 binds to Tissue factor – this complex activates Factor 9
Activated Factor 9 works with Factor 8 to activate Factor 103. Common pathway
Activated Factor 10 causes the conversion of prothrombin to thrombin and this hydrolyses fibrinogen peptide bonds to form fibrin. It also activates factor 8 to form links between fibrin molecules.4. Fibrinolysis
Plasminogen is converted to plasmin to facilitate clot resorption -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 113
Incorrect
-
In the adrenal gland:
Your Answer:
Correct Answer: Catecholamine release is mediated by cholinergic nicotinic transmission
Explanation:The adrenal (suprarenal) gland is composed of two main parts: the adrenal cortex, which is the largest and outer part of the gland, and the adrenal medulla. The adrenal cortex consists of three zones: 1. Zona glomerulosa (outermost layer) is responsible for the production of mineralocorticoids, mainly aldosterone, which regulates blood pressure and electrolyte balance. 2. Zona fasciculata (middle layer) is responsible for the production of glucocorticoids, predominantly cortisol, which increases blood sugar levels via gluconeogenesis, suppresses the immune system, and aids in metabolism. It also produces 11-deoxycorticosterone and corticosterone in addition to cortisol. 3. Zona reticularis (innermost layer) is responsible for the production of gonadocorticoids, mainly dehydroepiandrosterone (DHEA), which serves as the starting material for many other important hormones produced by the adrenal gland, such as oestrogen, progesterone, testosterone, and cortisol. It is also responsible for administering these hormones to the reproductive regions of the body.
The adrenal medulla majorly secretes epinephrine (adrenaline), and norepinephrine in small quantity. Both hormones have similar functions and initiate the flight or fight response.
Catecholamine is mediated by cholinergic nicotinic transmission through changes in sympathetic nervous system (T5 – T11), being increased during stress and hypoglycaemia.
Blood supply to the adrenal gland is by these three arteries: superior suprarenal arteries, middle suprarenal artery and inferior suprarenal artery. Venous drainage is via the suprarenal vein to the left renal vein or directly to the inferior vena cava on the right side. There is no portal (venous) system between cortex and medulla.
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This question is part of the following fields:
- Anatomy
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Question 114
Incorrect
-
Regarding the use of soda lime as part of a modern circle system with a vaporiser outside the circuit (VOC), which of the following is its most deleterious consequence?
Your Answer:
Correct Answer: Carbon monoxide formation
Explanation:When using dry soda lime for VOCs, very high amounts of carbon monoxide may be produced, regardless of the inhalational anaesthetic agent used. The carbon monoxide produced is sufficient enough to cause cytotoxic and anaemic hypoxia. To prevent this, soda lime canisters are shaken well to even out the packing of granules. This can help to evenly distribute gas flow for proper CO2 absorption and ventilation.
Compound A is formed when dry soda lime, or soda lime in high temperature, reacts with the inhalational anaesthetic Sevoflurane. Animal studies have shown renal toxicity in rats, but renal adverse effects in humans are yet to be observed.
When monitors are not employed with VOCs, deleterious effects are not for certain. However, monitors not employed with vaporiser inside the circuit (VIC) can lead to significant adverse events.
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This question is part of the following fields:
- Pathophysiology
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Question 115
Incorrect
-
A 46-year-old woman is listed for clipping of a cerebral aneurysm, following a diagnosis of surgical third nerve palsy.
Which of the following clinical findings correlate with surgical third nerve palsy?Your Answer:
Correct Answer: Ptosis, inferolateral rotation of globe and mydriasis
Explanation:Ptosis and mydriasis are visible in surgical third nerve palsy, and the eye looks ‘down and out.’ The loss of innervation to all of the major structures supplied by the oculomotor nerve is reflected in these characteristics.
Ptosis is caused by the paralysis of the levator palpebrae superioris in oculomotor nerve palsy. Due to the unopposed actions of the superior oblique and lateral rectus muscles, the eye rotates down and out.
Mydriasis is caused by surgical (compressive) causes of third nerve palsy, which disrupt the parasympathetic pupillomotor fibres on the nerve’s periphery.
Medical (ischaemic) causes of a third nerve palsy, on the other hand, leave the superficial parasympathetic fibres relatively unaffected and the pupil unaffected.
Horner’s syndrome is characterised by ptosis, anhidrosis, and miosis, which are caused by a loss of sympathetic innervation to the tarsal muscle of the upper lid, facial skin, and dilator pupillae, respectively.
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This question is part of the following fields:
- Pathophysiology
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Question 116
Incorrect
-
Patient’s having disease (Test Positive: 60, Test Negative:40)
Patient’s not having the disease (Test Positive:20, Test Negative: 80)
This is a result of a new tumour marker blood test, that was performed on 200 women for breast cancer screening. The director of the screening programme ask you to evaluate the observations and inform them the specificity of this new test.
Which one of the following figure you will relay to the programme director?Your Answer:
Correct Answer: 80%
Explanation:The positive predictive value is the ratio of patients truly diagnosed as positive to all those who had positive test results. In this case, this is 60/(60+20)=75%.
The negative predictive value is the ratio of patients truly diagnosed as negative to all those who had negative test results. In this case, this is 80/(80+40)=67%.
The sensitivity is the ratio of patients with the disease who test positive i.e. true positive patients to the total number of people with the disease. In this case, this is 60/(60+40)=60%.
The specificity is the ratio of people who don’t have the disease who test negative i.e. true negatives to the total number of people without the disease. In this case, this is 80/(20+80)=80%.
70% is not the result of any screening measurements
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This question is part of the following fields:
- Statistical Methods
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Question 117
Incorrect
-
A 25 year-old female came to the out-patient department with complaints of vaginal discharge with a distinct fishy odour. She was later diagnosed with bacterial vaginosis and was prescribed to take metronidazole.
The mechanism of action of metronidazole is?Your Answer:
Correct Answer: Interferes with bacterial DNA synthesis
Explanation:Metronidazole is a nitroimidazole antiprotozoal drug that is selectively absorbed by anaerobic bacteria and sensitive protozoa. Once taken up be anaerobes, it is nonenzymatically reduced by reacting with reduced ferredoxin. This reduction results in products that accumulate in and are toxic to anaerobic cells. The metabolites of metronidazole are taken up into bacterial DNA, forming unstable molecules. This action occurs only when metronidazole is partially reduced, and, because this reduction usually happens only in anaerobic cells, it has relatively little effect on human cells or aerobic bacteria.
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This question is part of the following fields:
- Pharmacology
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Question 118
Incorrect
-
What is the percentage of values that lie within 3 standard deviations of the mean?
Your Answer:
Correct Answer: 99.70%
Explanation:99.7% of the values within 3 standard deviations of the mean.
For 99.7% confidence interval, you can find the range as follows:
1. Multiply the standard error by 3.
2. Subtract the answer from mean value to get the lower limit.
3. Add the answer obtained in step 1 from the mean value to get the upper limit.
For a confidence interval of 68%, multiply the standard error with 1 and repeat the process. For a 95% confidence interval, Standard Error is multiplied by 1.96 to get the interval.
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This question is part of the following fields:
- Statistical Methods
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Question 119
Incorrect
-
Regarding amide local anaesthetics, which one factor has the most significant effect on its duration of action?
Your Answer:
Correct Answer: Protein binding
Explanation:When drugs are bound to proteins, drugs cannot cross membranes and exert their effect. Only the free (unbound) drug can be absorbed, distributed, metabolized, excreted and exert pharmacologic effect. Thus, when amide local anaesthetics are bound to ?1-glycoproteins, their duration of action are reduced.
The potency of local anaesthetics are affected by lipid solubility. Solubility influences the concentration of the drug in the extracellular fluid surrounding blood vessels. The brain, which is high in lipid content, will dissolve high concentration of lipid soluble drugs. When drugs are non-ionized and non-polarized, they are more lipid-soluble and undergo more extensive distribution. Hence allowing these drugs to penetrate the membrane of the target cells and exert their effect.
Tissue pKa and pH will determine the degree of ionization.
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This question is part of the following fields:
- Physiology
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Question 120
Incorrect
-
International colour coding is used on medical gas cylinders. Other characteristics also play a role in determining the gas's identity within a cylinder.
Which of the following options best describes a cylinder containing analgesics for obstetrics?Your Answer:
Correct Answer: Blue body, blue/white shoulder, full cylinder; 13700 KPa, gas mixture, requires a dual stage pressure regulator
Explanation:The body of the Entonox cylinder is usually blue (occasionally white), with blue and white shoulders. Entonox contains a 50:50 mixture of oxygen and nitrous oxide, with a full cylinder pressure of 13700 KPa (137 bar). The cylinder is equipped with a two-stage pressure regulator for safe operation.
The cylinder body and shoulder of nitrous oxide are (French) blue.
In today’s anaesthetic workstations, carbon dioxide cylinders are no longer used.
The body of an oxygen cylinder is black, with a white shoulder.
The white Heliox (21 percent oxygen and 79 percent helium) cylinder has a brown and white shoulder. The administration of this gas mixture, which is less dense than air, is used to reduce turbulence (stridor) of inspiratory flow in patients with upper airway obstruction.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 121
Incorrect
-
A 31-year old Caucasian female came into the emergency department due to difficulty of breathing. History revealed exposure to room odorizes that are rich in alkyl nitrites. Upon physical examination, patient is tachypnoeic at 32 breaths per minute, desaturated at 88% while on a non-rebreather mask at 15 litres per minute oxygen. She was also noted to be cyanotic, however with clear breath sounds.
Considering the history, what is the most probable cause of her difficulty of breathing?Your Answer:
Correct Answer: Increased affinity of bound oxygen to haemoglobin
Explanation:Amyl nitrate is part of the treatment of cyanide poisoning. The short acting nitrate causes oxidation of Fe2+ in haemoglobin to Fe3+ in methaemoglobin. Methaemoglobin combines with cyanide (cyanmethemoglobin), which reacts with sodium thiosulfate to convert nontoxic thiocyanate and methaemoglobin.
Methaemoglobin is formed when the iron in haemoglobin is converted from the reduced state (Fe2+) to the oxidized state (Fe3+). The oxidized form of haemoglobin (Fe3+) does not bind oxygen as readily as Fe2+, but has high affinity for cyanide. It also results to high affinity of bound oxygen to haemoglobin, thus leading to tissue hypoxia. Arterial oxygen tension is normal despite observations of cyanosis and dyspnoea. Methemoglobinemia can be treated with methylene blue and vitamin C.
Carboxyhaemoglobin can be due to carbon monoxide poisoning. In such cases, patients experience headache and dizziness, but do not develop cyanosis.
2,3-diphosphoglycerate causes a shift in the oxygen dissociation curve to the right, decreasing haemoglobin’s affinity to oxygen to facilitate unloading of oxygen to the tissues.
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This question is part of the following fields:
- Pathophysiology
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Question 122
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 123
Incorrect
-
Regarding aldosterone, one of the following is true.
Your Answer:
Correct Answer: Secretion is increased following haematemesis
Explanation:Aldosterone is produced in the zona glomerulosa of the adrenal cortex and acts to increase sodium reabsorption via intracellular mineralocorticoid receptors in the distal tubules and collecting ducts of the nephron.
Its release is stimulated by hypovolaemia, blood loss ,and low plasma sodium and is inhibited by hypertension and increased sodium. It is regulated by the renin-angiotensin system.
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This question is part of the following fields:
- Pathophysiology
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Question 124
Incorrect
-
Which of the following is an expected change in pulmonary function seen during a moderate asthma attack?
Your Answer:
Correct Answer: Decreased forced expiratory volume in 1 sec (FEV1)
Explanation:Asthma is a lung condition that causes reversible narrowing and swelling of airway passages. It is classified by the frequency and severity of symptoms.
The following are symptoms of moderate asthma:
Symptoms include cough, wheezing, chest tightness, or difficulty breathing which occurs daily
Decreased activity levels due to flare-ups
Night-time symptoms 5 or more times a month
Lung function test FEV1 is 60-80% of predicted normal values
Peak flow has more than 30% variabilityWith moderate asthma attacks, the arterial pCO2 levels may decrease, but as severity increases, so does the pCO2, reaching normal levels, and then exceeding them in severe asthma attacks.
Airway obstruction increases the functional residual capacity.
Concentration of serum bicarbonate would not increase in moderate asthma, but it could possibly increase in life-threatening asthma via the same mechanism as what increases arterial PCO2.
FEV1 is a good measure of airway obstruction. and is reduced in acute asthma attacks.
In the case of a pneumothorax, a decrease in arterial PO2 is higher.
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This question is part of the following fields:
- Pathophysiology
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Question 125
Incorrect
-
In the United Kingdom, a new breast cancer screening test is being conducted compared to the conventional use of mammography. This test predicts that if the breast cancer is diagnosed at an earlier stage, it could improve the survival rate but the overall results remains constant.
This is an example of what kind of bias?Your Answer:
Correct Answer: Lead time bias
Explanation:Recall bias introduced when participants in a study are systematically more or less likely to recall and relate information on exposure depending on their outcome status.
In procedure bias, the researcher decides assignment of a treatment versus control and assigns particular patients to one group or the other non-randomly. This is unlikely to have occurred in this case, although it is not mentioned specifically.
Self Selection or volunteer bias occur when those subjects are selected to participate in the study who are not the representative of the entire target population. those subjects may be from high socio-economic status and practice those activities or lifestyle that improves their health.
Lead-time bias occurs when a disease is detected by a screening test at an earlier time point rather than it would have been diagnosed by its clinical appearance. In this bias, earlier detection improves the survival time in the intervention group.
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This question is part of the following fields:
- Statistical Methods
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Question 126
Incorrect
-
A 53-year-old-male is being operated on for a right hemicolectomy. In the procedure, the ileocolic artery is ligated. Which vessel does this artery originate from?
Your Answer:
Correct Answer: Superior mesenteric artery
Explanation:The ileocolic artery is the terminal branch of the superior mesenteric artery. It supplies:
1. terminal ileum
2. proximal right colon
3. cecum
4. appendix (via its branch of the appendicular artery)As veins accompany arteries in the mesentery and are lined by lymphatics, high ligation is the norm in cancer resections—the ileocolic artery branches off the SMA near the duodenum.
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This question is part of the following fields:
- Anatomy
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Question 127
Incorrect
-
A patient's ECG is abnormal, with an abnormal broad complex QRS complexes. This means either a ventricular origin problem or aberrant conduction. The normal resting membrane potential of the heart's ventricular contractile fibres is which of the following?
Your Answer:
Correct Answer: -90mV
Explanation:The cardiac muscle’s contractile fibres have a much more stable resting potential than its conductive fibres. In the ventricular fibres it is -90mV and in the atrial fibres it is -80mV.
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. (QRS complex)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 128
Incorrect
-
Regarding the plateau phase of the cardiac potential, which electrolyte is the main determinant?
Your Answer:
Correct Answer: Ca2+
Explanation:The cardiac action potential has several phases which have different mechanisms of action as seen below:
Phase 0: Rapid depolarisation – caused by a rapid sodium influx.
These channels automatically deactivate after a few msPhase 1: caused by early repolarisation and an efflux of potassium.
Phase 2: Plateau – caused by a slow influx of calcium.
Phase 3 – Final repolarisation – caused by an efflux of potassium.
Phase 4 – Restoration of ionic concentrations – The resting potential is restored by Na+/K+ATPase.
There is slow entry of Na+into the cell which decreases the potential difference until the threshold potential is reached. This then triggers a new action potentialOf note, cardiac muscle remains contracted 10-15 times longer than skeletal muscle.
Different sites have different conduction velocities:
1. Atrial conduction – Spreads along ordinary atrial myocardial fibres at 1 m/sec2. AV node conduction – 0.05 m/sec
3. Ventricular conduction – Purkinje fibres are of large diameter and achieve velocities of 2-4 m/sec, the fastest conduction in the heart. This allows a rapid and coordinated contraction of the ventricles
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This question is part of the following fields:
- Physiology
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Question 129
Incorrect
-
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 130
Incorrect
-
An 18-year old female was brought into the emergency room because of active seizures. The informant reported that it has been more than 5 minutes since the patient started seizing. The attending physician gave an initial diagnosis of status epilepticus.
According to the paramedics who brought in the patient, 10 mg of diazepam was given rectally. Upon physical examination, she was normotensive at 120/80 mmHg; tachycardic at 138 beats per minute; tachypnoeic at 24 breaths per minute; and well-saturated at 99% on high flow oxygen. Her random blood glucose level was normal at 7.0 mmol/L.
Given this situation and an initial diagnosis of status epilepticus, what would be the best initial anti-epileptic drug to administer to the patient?Your Answer:
Correct Answer: Lorazepam
Explanation:Lorazepam is an intermediate-acting benzodiazepine that binds to the GABA-A receptor subunit to increase the frequency of chloride channel opening and facilitate membrane hyperpolarization. It is the preferred treatment for status epilepticus, although Diazepam can also be used as an alternative.
Lorazepam has a longer duration of action than Diazepam, and binds with greater affinity to the GABA-A receptor subunit.
Phenobarbital is a barbiturate that acts on the GABA-A receptor site to increase the duration of chloride channel opening. Barbiturates, particularly phenobarbital, is considered the drug of choice for seizures in infants.
Phenytoin is a sodium-channel blocker that is given for generalized tonic-clonic seizures, partial seizures, and status epilepticus. Phenytoin is preferred in prolonged therapy for status epilepticus because it is less sedating.
Propofol or thiopentone is preferred when airway protection is required.
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This question is part of the following fields:
- Pharmacology
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Question 131
Incorrect
-
A 45-year-old man is being operated on for emergency laparotomy as he presented with bowel perforation. During the surgery, the marginal artery of Drummond is encountered and preserved.
Which of the following two arteries fuse to form the marginal artery of Drummond?Your Answer:
Correct Answer: Superior mesenteric artery and inferior mesenteric artery
Explanation:The arteries of the midgut (superior mesenteric artery) and hindgut (inferior mesenteric artery) give off terminal branches that form an anastomotic vessel called the marginal artery of Drummond. It runs in the inner margins of the colon and gives off short terminal branches to the bowel wall.
The marginal artery is formed by the main branches and arcades arising from the ileocolic, right colic, middle colic, and left colic arteries. It is most apparent in the ascending, transverse, and descending colons and poorly developed in the sigmoid colon.
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This question is part of the following fields:
- Anatomy
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Question 132
Incorrect
-
The spinal cord tracts that transmits the sensations of pain, crude temperature, and light touch is?
Your Answer:
Correct Answer: Spinothalamic
Explanation:Dorsal column (ascending tract) – Proprioception, vibration, discriminative
Spinocerebellar (ascending tract) – Subconscious muscle position and tone
Corticospinal (descending tract) – Voluntary muscle
Rubrospinal (descending tract) – Flexor muscle tone
Vestibulospinal (descending tract) – Reflexes and muscle tone
Reticulospinal(descending tract) – Voluntary movements, head position.-
Autonomic – Descending tract.
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This question is part of the following fields:
- Anatomy
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Question 133
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 134
Incorrect
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A laceration to the upper lateral margin of the popliteal fossa will pose the greatest risk of injury for which nerve?
Your Answer:
Correct Answer: Common peroneal nerve
Explanation:The common peroneal (fibular) nerve descends obliquely along the lateral side of the popliteal fossa to the fibular head, medial to biceps femoris.
The sural nerve exits at the fossa’s lower inferolateral aspect and is more at risk in short saphenous vein surgery.
The tibial nerve lies more medially and is even less likely to be injured in this location.
The boundaries of the popliteal fossa are:
Superolateral – the biceps femoris tendon
Superomedial – semimembranosus reinforced by semitendinosus
Inferomedial and inferolateral – medial and lateral heads of gastrocnemiusThe contents of the Popliteal fossa are:
1. The popliteal artery
2. The popliteal vein
3. The Tibial nerve and common Fibular nerve
4. Posterior femoral cutaneous nerve: descends and pierces the roof
5. Small saphenous vein
6. popliteal lymph nodes
7. fat -
This question is part of the following fields:
- Anatomy
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Question 135
Incorrect
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Which of the following drugs would cause the most clinical concern if accidentally administered intravenously to a 4-year-old boy?
Your Answer:
Correct Answer: 20 mg codeine
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.
The following are the appropriate doses of the drugs listed above:
Gentamicin (once daily) – 5-7 mg/kg = 90-126 mg and subsequent dose modified according to plasma levels
Ondansetron – 0.1 mg/kg, but a maximum of 4 mg as a single dose = 1.8 mg
Codeine should be administered orally at a dose of 1 mg/kg rather than intravenously, as the latter can cause ‘dangerous’ hypotension due to histamine release.
15 mg/kg paracetamol = 270 mg orally or intravenously (a loading dose of 20 mg/kg, or 360 mg, is sometimes recommended, which is not far short of the doses listed above).
Cefuroxime – the initial intravenous dose is 20 mg/kg (360 mg) depending on the indication (again, similar to the dose given in the answer options above). -
This question is part of the following fields:
- Pharmacology
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Question 136
Incorrect
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Which of the following statement is true regarding the mechanism of action of doxycycline?
Your Answer:
Correct Answer: Inhibit 30S subunit of ribosomes
Explanation:Doxycycline belongs to the family of tetracyclines and inhibits protein synthesis through reversible binding to bacterial 30s ribosomal subunits, which prevent binding of new incoming amino acids (aminoacyl-tRNA) and thus interfere with peptide growth.
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This question is part of the following fields:
- Pharmacology
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Question 137
Incorrect
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While inspecting the caecum, what structure will be identified at the point at which all the taeniae coli converge?
Your Answer:
Correct Answer: Appendix base
Explanation:The taeniae coli are the three outer muscular bands of the cecum, ascending colon, transverse colon, and descending colon.
The taeniae coli converge at the base of the appendix in the cecum where they form a complete longitudinal layer. In the ascending and descending colon, the bands are located anteriorly, posteromedially, and posterolateral.
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This question is part of the following fields:
- Anatomy
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Question 138
Incorrect
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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 139
Incorrect
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A 35-year-old female, presents to the emergency department via ambulance. The paramedics have noted the patient's symptoms as unilateral left-sided weakness of the upper and lower limbs, homonymous hemianopia and dysphasia.
She has previous personal and family history of deep vein thromboses.
The report of her CT scan suggests a stroke involving the middle cerebral artery.
Post recovery, she undergoes further diagnostic investigations to determine the cause of a stroke at her young age. She is eventually diagnosed with a hypercoagulable state disease called Factor V Leiden thrombophilia.
An emboli in the middle cerebral artery results in dysfunction of which areas of the brain?Your Answer:
Correct Answer: Frontal, temporal and parietal lobes
Explanation:The middle cerebral artery is a part of the circle of Willis system of anastomosis within the brain, and the most often affected by brain pathology.
The primary function of the middle cerebral artery is providing oxygenated blood to related regions of the brain. It achieves this by giving off different branches to supply different brain regions, namely:
The cortical branches: which supplies the primary motor and somatosensory cortical areas of some parts of the face, trunk and upper limbs.
The small central branches: which supply the basal ganglia and internal capsule via the lenticulostriate vessels.
The superior division: which supplies the lateral inferior frontal lobe, including the Broca area which is responsible for production of speech, language comprehension, and writing.
The inferior division: which supplies the superior temporal gyrus, including Wernicke’s area which controls speech comprehension and language development.
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This question is part of the following fields:
- Anatomy
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Question 140
Incorrect
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A 30-year-old woman admitted following a tonsillectomy has developed stridor with a respiratory rate of 22 breaths per minute and obstructive movements of the chest and abdomen that is in a see-saw pattern .
Her SpO2 is 92% on 60% oxygen with pulse rate 120 beats per minute while her blood pressure is 180/90mmHg. She is repeatedly trying to remove the oxygen mask and appears anxious.
Her pharynx is suctioned and CPAP applied with 100% oxygen via a Mapleson C circuit.
Which of these is the most appropriate next step in her management?Your Answer:
Correct Answer: Administer intravenous propofol 0.5 mg/kg
Explanation:Continuous closure of the vocal cords resulting in partial or complete airway obstruction is called Laryngospasm. It is a reflex that helps protect against pulmonary aspiration.
Predisposing factors include: Hyperactive airway disease, Insufficient depth of anaesthesia, Inexperience of the anaesthetist, Airway irritation, Smoking, Shared airway surgery and Paediatric patients
Its primary treatment includes checking for blood or stomach aspirate in the pharynx, removing any triggering stimulation, relieving any possible supra-glottic component to airway obstruction and application of CPAP with 100% oxygen.
In this patient, all the above has been done and the next treatment of choice is the administration of a rapidly acting intravenous anaesthetic agent such as propofol (0.5 mg/kg) in increments as it has been reported to relieve laryngospasm in approximately 75% of cases. Administering suxamethonium to an awake patient would be inappropriate at this stage.
Magnesium and lidocaine are used for prevention rather than acute treatment of laryngospasm. Superior laryngeal nerve blocks have been reported to successfully treat recurrent laryngospasm but it is not the next logical step in index patient.
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This question is part of the following fields:
- Pathophysiology
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Question 141
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 142
Incorrect
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A paediatric patient was referred to the surgery department after an initial assessment of acute gastroenteritis was proven otherwise to be a case acute appendicitis. History revealed multiple episodes of non-bloody emesis. In the paediatric ward, the patient had already undergone fluid resuscitation and replacement, and electrolytes were already corrected. Other pertinent laboratory studies were the following:
Serum Na: 138 mmol/l
Blood glucose: 6.4 mmol/l
If the patient weighed 25 kg, which intravenous fluid maintenance regimen would be best for the child?
Your Answer:
Correct Answer: 65 ml/hr Hartmann's solution with 0% glucose
Explanation:Maintenance therapy aims to replace water and electrolytes lost under ordinary conditions. In the perioperative period, maintenance fluid administration may not sufficiently account for the increased fluid requirements caused by third-space losses into the interstitium and gut. Specific recommendations vary with the patient, the procedure, and the type and amount of fluid administered during the operation. The fluid for maintenance therapy replaces deficits arising primarily from insensible losses and urinary or gastrointestinal (GI) losses.
The maintenance fluid volume can be computed using the Holliday-Segar method.
Body weight Fluid volume
first 10 kg 4 ml/kg/hr
next 10-20 kg 2 ml/kg/hr
>20 kg 1 ml/kg/hrIn the past few years, there has been growing recognition of the increased risk of hyponatremia in hospitalized children in intensive care and postoperative settings who receive hypotonic maintenance fluids. Several studies, including a randomized controlled trial and a Cochrane analysis, found that the use of isotonic fluids is associated with fewer electrolyte derangements and concluded that isotonic maintenance fluids are preferable to hypotonic solutions in hospitalized children.
A European consensus statement suggests that an intraoperative fluid should have an osmolarity close to the physiologic range in children in order to avoid hyponatremia, an addition of 1-2.5% in order to avoid hypoglycaemia, lipolysis or hyperglycaemia and should also include metabolic anions as bicarbonate precursors to prevent hyperchloremic acidosis.
A rate of 40 ml/hr is suboptimal.
If 0.9% NaCl with 0% glucose is given at a rate of 65 ml/hr, despite of the correct infusion rate, large volumes can lead to hyperchloremic acidosis.
If 0.18% NaCl with 4% glucose is given at a rate of 65 ml/hr, infusion of this fluid regimen can lead to hyponatremia because of its hypotonicity.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 143
Incorrect
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Which of the following statement is true regarding the paediatric airway?
Your Answer:
Correct Answer: The larynx is more anterior than in an adult
Explanation:In the neonatal stage, the tongue is usually large and comes to the normal size at the age of 1 year. The vocal cords lie inverse C4 and as it reaches the grown-up position inverse C5/6 by the age of 4 (not 1 year).
Due to the immature cricoid cartilage, the larynx lies more anterior in newborn children. That’s why the cricoid ring is the narrowest part of the paediatric respiratory tract, while in the adults the tightest portion of the respiratory route is vocal cords. The epiglottis is generally expansive and slants at a point of 45 degrees to the laryngeal opening.
The carina is the ridge of the cartilage in the trachea at the level of T2 in newborn (T4 in adults), that separates the openings of right and left main bronchi.
Neonates have a comparatively low number of alveoli and then this number gradually increases to a most extreme by the age of 8 (not 3 years).
Neonates are obligatory nose breathers and any hindrance can cause respiratory issues (e.g., choanal atresia).
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This question is part of the following fields:
- Physiology
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Question 144
Incorrect
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A 70-year-old male is brought to the Emergency department with:
Pulse rate: 32 beats per minute
Blood pressure: 82/35 mmHg
12 lead ECG shows a sinus bradycardia of 35 beats per minute with no evidence of myocardial ischemia or infarction. There was no chest pain but the patient feels light-headed.
Which of the following would be the best initial treatment for this condition?Your Answer:
Correct Answer: Atropine
Explanation:Based on the presenting symptoms and clinical examination, it is a case of an adult sinus bradycardia with adverse signs. The first pharmacological treatment for this condition is atropine 500mcg intravenously and if necessary repeat every three to five minutes up to a maximum of 3 mg.
If the bradycardia does not subside even after the administration of atropine, cardiac pacing should be considered. If pacing cannot be achieved promptly, we should consider the use of second-line drugs like adrenaline, dobutamine, or isoprenaline.
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This question is part of the following fields:
- Pharmacology
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Question 145
Incorrect
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A project is being planned to assess the effects of a new anticoagulant on the coagulation cascade. The intrinsic pathway is being studied and the best measurement to be recorded is which of the following?
Your Answer:
Correct Answer: aPTT
Explanation:The intrinsic pathway is best assessed by the aPTT time.
D-dimer is a fibrin degradation product which is raised in the presence of blood clots.
A 50:50 mixing study is used to assess if a prolonged PT or aPTT is due to factor deficiency or a factor inhibitor.
The thrombin time is a test used to assess fibrin formation from fibrinogen in plasma. Factors that prolong the thrombin time include heparin, fibrin degradation products, and fibrinogen deficiency.
Intrinsic pathway – Best assessed by APTT. Factors 8,9,11,12 are involved. Prolonged aPTT can be seen in haemophilia and use of heparin.
Extrinsic pathway – Best assessed by Increased PT. Factor 7 involved.
Common pathway – Best assessed by APTT & PT. Factors 2,5,10 involved.
Vitamin K dependent factors are factors 2,7,9,10
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 146
Incorrect
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A 45-year old male who was involved in a road traffic accident has had to receive a large blood transfusion of whole blood which is two weeks old. Which of these best describes the oxygen carrying capacity of this blood?
Your Answer:
Correct Answer: It will have an increased affinity for oxygen
Explanation:With respect to oxygen transport in cells, almost all oxygen is transported within erythrocytes. There is limited solubility and only 1% is carried as solution. Thus, the amount of oxygen transported depends upon haemoglobin concentration and its degree of saturation.
Haemoglobin is a globular protein composed of 4 subunits. Haem is made up of a protoporphyrin ring surrounding an iron atom in its ferrous state. The iron can form two additional bonds – one is with oxygen and the other with a polypeptide chain.
There are two alpha and two beta subunits to this polypeptide chain in an adult and together these form globin. Globin cannot bind oxygen but can bind to CO2 and hydrogen ions.
The beta chains are able to bind to 2,3 diphosphoglycerate. The oxygenation of haemoglobin is a reversible reaction. The molecular shape of haemoglobin is such that binding of one oxygen molecule facilitates the binding of subsequent molecules.The oxygen dissociation curve (ODC) describes the relationship between the percentage of saturated haemoglobin and partial pressure of oxygen in the blood.
Of note, it is not affected by haemoglobin concentration.Chronic anaemia causes 2, 3 DPG levels to increase, hence shifting the curve to the right
Haldane effect – Causes the ODC to shift to the left. For a given oxygen tension there is increased saturation of Hb with oxygen i.e. Decreased oxygen delivery to tissues.
This can be caused by:
-HbF, methaemoglobin, carboxyhaemoglobin
-low [H+] (alkali)
-low pCO2
-ow 2,3-DPG
-ow temperatureBohr effect – causes the ODC to shifts to the right = for given oxygen tension there is reduced saturation of Hb with oxygen i.e. Enhanced oxygen delivery to tissues. This can be caused by:
– raised [H+] (acidic)
– raised pCO2
-raised 2,3-DPG
-raised temperature -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 147
Incorrect
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You performed pelvic ultrasound of Mrs Aciman as she had pelvic bloating and intermittent pain. The ultrasound shows a complex ovarian cyst and the radiologist who reported the results has advised urgent consultation with a gynaecologist. Upon breaking the news to Mrs Aciman you learn that she recently had a blood test done that was normal. You explain it to her that the test performed (Ca-125) is not always perfect and is only able to detect 80% of the cancer cases while the remaining 20% go undetected.
Which statistical term appropriately explains the 80% in this example.Your Answer:
Correct Answer: Sensitivity
Explanation:Tests are used to confirm the presence of a particular disease. However the results can be misleading at times since most of the tests have some limitations associated with them.
Sensitivity is the correct term that refers to the probability of a positive test. The others are explained below:False Positive rates refer to the proportion of the patients who don’t have the condition being detected as positive.
False Negative rates refer to the proportion of the patients who have the condition being detected as negative (like the 20% of the patients that went undetected by the Ca-125 test).
Specificity describes the ability of a test to detect and pick up people without the disease. Absolute risk ratio compares the rate of two separate outcomes.
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This question is part of the following fields:
- Statistical Methods
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Question 148
Incorrect
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The most sensitive indicator of mild obstructive airway disease is?
Your Answer:
Correct Answer: Forced expiratory flow (FEF25-75%)
Explanation:The volume expired in the first second of maximal expiration after a maximal inspiration is known as forced expiratory volume in one second (FEV1), and it indicates how quickly full lungs can be emptied. It is the most commonly measured parameter for bronchoconstriction assessment.
The maximum volume of air exhaled after a maximal inspiration is known as the ‘slow’ vital capacity (VC). VC is normally equal to FVC after a forced vital capacity (FVC) or slow vital capacity (VC) manoeuvre, unless there is an airflow obstruction, in which case VC is usually higher than FVC.
The FEV1/FVC (Tiffeneau index) is a clinically useful index of airflow restriction that can be used to distinguish between restrictive and obstructive respiratory disorders.
The average expired flow over the middle half (25-75 percent) of the FVC manoeuvre is the forced expiratory volume (FEF25-75). The airflow from the resistance bronchioles corresponds to this. It’s a more sensitive indicator of mild small airway narrowing than FEV1, but it’s difficult to tell if the VC (or FVC) is decreasing or increasing.
The maximum expiratory flow rate achieved is called the peak expiratory flow (PEF), which is usually 8-14 L/second.
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This question is part of the following fields:
- Pathophysiology
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Question 149
Incorrect
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You are given an intravenous induction agent. The following are its characteristics:
A racemic mixture of cyclohexanone rings with one chiral centre
Local anaesthetic properties.
Which of the following statements about its primary mechanism of action is most accurate?Your Answer:
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 150
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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