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  • Question 1 - Out of the following, which is NOT a part of the contents of...

    Correct

    • Out of the following, which is NOT a part of the contents of the porta hepatis?

      Your Answer: Cystic duct

      Explanation:

      The porta hepatis is a fissure in the inferior surface of the liver. All the neurovascular structures that enter and leave the porta hepatis are:
      1. hepatic portal vein
      2. hepatic artery
      3. hepatic ducts
      4. hepatic nerve plexus (It contains the sympathetic branch to the liver and gallbladder and the parasympathetic, hepatic branch of the vagus nerve.)

      These structures supply and drain the liver. Only the hepatic vein is not part of the porta hepatis.
      The porta hepatis is also surrounded by lymph nodes, that may enlarge to produce obstructive jaundice.
      These structures divide immediately after or within the porta hepatis to supply the functional left and right lobes of the liver.

      The cystic duct lies outside the porta hepatis and is an important landmark in laparoscopic cholecystectomy.

    • This question is part of the following fields:

      • Anatomy
      7.9
      Seconds
  • Question 2 - A new study is being carried out on the measurement of a new...

    Correct

    • A new study is being carried out on the measurement of a new cardiovascular disease biomarker, and its applications in preoperative screening. The data for this study is expected to be normally distributed.

      Which of the following statements is true about normal distributions?

      Your Answer: The mean, median and mode are the same value

      Explanation:

      The correct answer is the mean, median and mode of normally distributed data are the same value. This is as a result of the bell shaped curve which is equal on both sides.

      The bell-shape indicates that values around the mean are more frequent in occurrence than the values farther away.

      In a normal distribution:
      1) +/- one standard deviation of the mean accounts for 68% of the data.
      2) +/- two standard deviations of the mean accounts for 95% of the data.
      3) +/- three standard deviations of the mean accounts for 99.7% of the data.

    • This question is part of the following fields:

      • Statistical Methods
      18
      Seconds
  • Question 3 - A patient visits the radiology department for a magnetic resonance imaging (MRI) scan...

    Correct

    • A patient visits the radiology department for a magnetic resonance imaging (MRI) scan (MRI). The presence of metal implants must be ruled out prior to the scan.

      In a strong magnetic field, which of the following metals is the safest?

      Your Answer: Chromium

      Explanation:

      Ferromagnetism is the property of a substance that is magnetically attracted and can be magnetised indefinitely. A material is said to be paramagnetic if it is attracted to a magnetic field. A substance is said to be diamagnetic if it is repelled by a magnetic field.

      Cobalt, iron, gadolinium, neodymium, and nickel are ferromagnetic.

      Gadolinium is a ferromagnetic rare earth metal that is ferromagnetic below 20 degrees Celsius (its Curie temperature). MRI scans are enhanced with gadolinium-based contrast media.

      When ferromagnetic materials are exposed to a magnetic field, they can cause a variety of issues like magnetic field interactions, heating, and image artefacts.

      Titanium, lead, chromium, copper, aluminium, silver, gold, and tin are non ferromagnetic.

    • This question is part of the following fields:

      • Clinical Measurement
      42.5
      Seconds
  • Question 4 - A post-operative patient was brought to the recovery room after completion of dilation...

    Incorrect

    • A post-operative patient was brought to the recovery room after completion of dilation and curettage. Her medical history revealed that she was maintained on levodopa for Parkinson's disease. The nurses administered ondansetron 4 mg and dexamethasone 8 mg prior to transfer from the operating room to the recovery room. However, an additional antiemetic agent is warranted.

      Which of the following agents should be prescribed to the patient?

      Your Answer: Prochlorperazine 12 mg IM

      Correct Answer: Cyclizine 50 mg IV

      Explanation:

      The Beers criteria, a US set of criteria for good prescribing in the older patient, preclude the use of metoclopramide in Parkinson’s disease. The Adverse Reactions Register of the UK Committee on Safety of Medicines (CSM) for the years 1967 to 1982 contained 479 reports of extrapyramidal reactions in which metoclopramide was the suspected drug; 455 were for dystonic-dyskinetic reactions, 20 for parkinsonism and four for tardive dyskinesia. Effects can occur within days of initiation of treatment and may take months to wear off.

      Other antiemetics are available, such as cyclizine (Valoid), domperidone and ondansetron, which would be more appropriate to use in those with Parkinson’s disease.

      Cyclizine is a piperazine derivative with histamine H1 receptor antagonist and anticholinergic activity. It is used for the treatment of nausea, vomiting, (particularly opioid-induced vomiting), vertigo, motion sickness, and labyrinthine disorders.

      Prochlorperazine is an antipsychotic known to cause tardive dyskinesia, tremor and parkinsonian symptoms and is therefore likely to exacerbate Parkinson’s disease. Prochlorperazine is not favoured for older patients because of the increased risk of stroke and transient ischaemic attack (TIA).

      Droperidol and phenothiazine are also potent antagonists on D2 receptors and must also be avoided.

    • This question is part of the following fields:

      • Pharmacology
      35.1
      Seconds
  • Question 5 - A 30-year old female athlete was brought to the Emergency Room for complaints...

    Incorrect

    • A 30-year old female athlete was brought to the Emergency Room for complaints of light-headedness and nausea. Clinical chemistry studies were done and the results were the following:

      Na: 144 mmol/L (Reference: 137-144 mmol/L)
      K: 6 mmol/L (Reference: 3.5-4.9 mmol/L)
      Cl: 115 mmol/L (Reference: 95-107 mmol/L)
      HCO3: 24 mmol/L (Reference: 20-28 mmol/L)
      BUN: 9.5 mmol/L (Reference: 2.5-7.5 mmol/L)
      Crea: 301 µmol/l (Reference: 60 - 110 µmol/L)
      Glucose: 3.5 mmol/L (Reference: 3.0-6.0 mmol/L)

      Taking into consideration the values above, in which of the following ranges will his osmolarity fall into?

      Your Answer: 321-333

      Correct Answer: 300-313

      Explanation:

      Osmolarity refers to the osmotic pressure generated by the dissolved solute molecules in 1 L of solvent. Measurements of osmolarity are temperature dependent because the volume of the solvent varies with temperature. The higher the osmolarity of a solution, the more it attracts water from an opposite compartment.

      Osmolarity can be computed using the following formulas:

      Osmolarity = Concentration x number of dissociable particles; OR
      Plasma osmolarity (Posm) = 2([Na+]) + (glucose in mmol/L) + (BUN in mmol/L)

      Posm = 2 (144) + 3.5 + 9.5 = 301 mOsm/L

      Suppose there is electrical neutrality, the formula will double the cation activity to account for the anions.

      Plasma osmolarity (Posm) = 2([Na+] + [K+]) + (glucose in mmol/L) + (BUN in mmol/L)

      Posm = 2 (144 + 6) + 3.5 + 9.5 = 313 mOsm/L

    • This question is part of the following fields:

      • Physiology
      33.2
      Seconds
  • Question 6 - A 23-year-old man, has just undergone surgery under general anaesthesia. He has experienced...

    Incorrect

    • A 23-year-old man, has just undergone surgery under general anaesthesia. He has experienced a severe reaction to the anaesthetic agent resulting in malignant hyperthermia (MH) for which he has been referred for treatment.

      What investigation can be conducted to determine a patient's susceptibility to malignant hyperthermia?

      Your Answer: In vitro muscle contraction test using ryanodine

      Correct Answer: In vitro muscle contraction test using caffeine

      Explanation:

      Malignant hyperthermia (MH) is a autosomal dominant inherited medical condition which predisposes affected individuals to a clinical syndrome of hypermetabolism which involves abnormal ryanodine receptors in skeletal muscle causing a deregulation of calcium in muscle.

      It is a life threatening condition requiring immediate medical intervention. It often lies dormant until triggered in susceptible individuals mostly by volatile inhaled anaesthetic agents and succinylcholine which is a muscle relaxant.

      The signs and symptoms of MH are related to this hypermetabolism, which includes an increase in carbon dioxide production, metabolic and respiratory acidosis, accelerated oxygen consumption, heat production, activation of the sympathetic nervous system, hyperkalaemia, disseminated intravascular coagulation (DIC), and multiple organ dysfunction and failure.

      Early signs of MH to look out for in patients includes an uptick in end-tidal carbon dioxide (even with increasing minute ventilation), tachycardia, muscle rigidity, tachypnoea, and hyperkalaemia. Later signs include fever, myoglobinuria, and multiple organ failure.

      In vitro muscle contracture test (IVCT) is the standard for determining individual susceptibility to MH. It is conducted by measuring the force of muscle contraction after exposing the patient’s muscle sample to halothane and caffeine., the sample is normally taken from the vastus medialis or lateralis under regional anaesthesia.

    • This question is part of the following fields:

      • Clinical Measurement
      16.6
      Seconds
  • Question 7 - The whole water content of the body is calculated by multiplying body mass...

    Incorrect

    • The whole water content of the body is calculated by multiplying body mass with 0.6. This water is diffused into distinct compartments.

      Which fluid compartment can be measured indirectly?

      Your Answer: Total body water

      Correct Answer: Intracellular volume

      Explanation:

      The total body water content of a 70kg man is (70 × 0.6) = 42 litres. For a woman, the calculation is (70 × 0.55) = 38.5 litres.

      For a man, it is subdivided into:

      Extracellular fluid (ECF) = 14L (1/3)
      Intracellular fluid (ICF) = 28L (2/3).

      The ECF volume is further divided into:

      Interstitial fluid = 10.5 litres
      Plasma = 3 litres
      Transcellular fluid (CSF/synovial fluid) = 0.5 litres.

      Directly measured fluid compartments:

      Heavy water (deuterium) can be used to measure total body water content, which is freely distributed.
      Albumin labelled with a radioactive isotope or using a dye called Evans blue can be used to measure Plasma volume . They do not diffuse into red blood cells.
      Radiolabelled (Cr-51) red blood cells can be used to measure total erythrocyte volume.
      Inulin as the tracer can be used to measure ECF volume as it circulate freely in the interstitial and plasma volumes.

      Indirectly measured fluid compartments:

      Total blood volume can be calculated with the level of haematocrit and the volume of total circulating red blood cells.
      ICF volume can be calculated by subtracting ECF volume from total blood volume.

    • This question is part of the following fields:

      • Basic Physics
      20.5
      Seconds
  • Question 8 - During a squint surgery, a 5-year-old child developed severe bradycardia as a result...

    Correct

    • 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: 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.

    • This question is part of the following fields:

      • Pathophysiology
      36.6
      Seconds
  • Question 9 - In reference to confounding variables, which among the given is not true? ...

    Incorrect

    • In reference to confounding variables, which among the given is not true?

      Your Answer: Stratification is a technique used to control for confounding

      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.

    • This question is part of the following fields:

      • Statistical Methods
      33.7
      Seconds
  • Question 10 - Which of the following may indicate an inadequate reversal of non-depolarising neuromuscular blockade?...

    Correct

    • Which of the following may indicate an inadequate reversal of non-depolarising neuromuscular blockade?

      Your Answer: Post tetanic count of 5

      Explanation:

      A post-tetanic count of 5 denotes a deep neuromuscular blockade.

      Post tetanic count (PTC) is a well-established method of evaluating neuromuscular recovery during intense neuromuscular blockade. It cam ne used when there is no response to single twitch, tetanic, or train-of-four (TOF) stimulation to assess the intensity of neuromuscular blockade and to estimate the duration after which the first twitch in the TOF (T1) is likely to reappear.

      During a nondepolarizing block, the high frequency of tetanic stimulation will induce a transient increase in the amount of acetylcholine released from the presynaptic nerve ending, such that the intensity of subsequent muscle contractions will be increased (potentiated) briefly (period of post-tetanic potentiation, which may last 2 to 5 min. The neuromuscular response to stimulation during post tetanic potentiation can be used to gauge the depth of block when TOF stimulation otherwise evokes no responses. The number of post tetanic responses is inversely proportional to the depth of block: fewer post tetanic contractions denote a deeper block. When the post tetanic count (PTC) is 6 to 8, recovery to TOF count = 1 is likely imminent from an intermediate-duration blocking agent; when the PTC is 0, the depth of block is profound, and no additional NMBA should be administered.

    • This question is part of the following fields:

      • Pathophysiology
      26.1
      Seconds
  • Question 11 - A 33-year old man was referred to you because of difficulty moving his...

    Incorrect

    • A 33-year old man was referred to you because of difficulty moving his limbs.

      History revealed that he was placed under anaesthesia for a major surgery 12 hours prior to the referral. Other symptoms were noted such as anxiousness, agitation, and fever of 38°C. Upon physical examination, he was tachycardic at 119 beats per minute. Moreover, his medical history showed that he was on Fluoxetine for clinical depression.

      The nurses reported that, because of his frequent complaints of axillary pain, he was given tramadol with paracetamol.

      Which of the following is responsible for his clinical features?

      Your Answer: Neuroleptic malignant syndrome

      Correct Answer: Tramadol

      Explanation:

      Tramadol is weak agonist at the mu receptor. It inhibits the neuronal reuptake of serotonin and norepinephrine, and inhibits pain neurotransmission. It is given for moderate pain, chronic pain syndromes, and neuropathic pain.

      Fluoxetine is a selective serotonin reuptake inhibitor (SSRI). It inhibits the neuronal reuptake of serotonin by inhibiting the serotonin transporter (SERT). It is the drug of choice for major depressive disorder, and is given for other psychiatric disorders such as anxiety, obsessive-compulsive, post-traumatic stress, and phobias.

      When tramadol is given with SSRIs, serotonin syndrome may occur. Serotonin syndrome is characterized by fever, agitation, tremors, clonus, hyperreflexia and diaphoresis. The onset of symptoms may occur within a few hours, and the first-line treatment is sedation, paralysis, intubation and ventilation.

    • This question is part of the following fields:

      • Pharmacology
      32.8
      Seconds
  • Question 12 - A 28-year male patient presents to the GP with a 2-day history of...

    Incorrect

    • A 28-year male patient presents to the GP with a 2-day history of abdominal pain and bloody diarrhoea. He reports that he was completely fine until one week ago when headache and general tiredness appeared. After further questioning, he revealed eating at a dodgy takeaway 3 days before the start of his symptoms.

      Which of the following diagnosis is most likely?

      Your Answer: E. coli

      Correct Answer: Campylobacter

      Explanation:

      Giardiasis is known to have a longer incubation time and doesn’t cause bloody diarrhoea.

      Cholera usually doesn’t cause bloody diarrhoea.

      Generally, most of the E.coli strains do not cause bloody diarrhoea.

      Diverticulitis can be a cause of bloody stool but the history here points out to an infectious cause.

      Campylobacter infection is the most probable cause as it is characterized by a prodrome, abdominal pain and bloody diarrhoea

    • This question is part of the following fields:

      • Physiology And Biochemistry
      50.6
      Seconds
  • Question 13 - Because this benzodiazepine has a half-life of 2-4 hours, it is preferred for...

    Correct

    • Because this benzodiazepine has a half-life of 2-4 hours, it is preferred for clinical use.

      This benzodiazepine has which of the following properties that no other benzodiazepine has?

      Your Answer: It is water soluble at a pH of 3.5 and lipid soluble at a pH of 7.4

      Explanation:

      Midazolam is the benzodiazepine in question. It’s the only benzodiazepine that undergoes tautomeric transformation (dynamic isomerism). The molecule is ionised and water soluble at pH 3.5, but when injected into the body at pH 7.4, it becomes unionised and lipid soluble, allowing it to easily pass through the blood brain barrier.

      The half-life of midazolam is only 2-4 hours.

      It is a GABAA receptor agonist because it is a benzodiazepine. GABAA receptors are found in abundance throughout the central nervous system, particularly in the cerebral cortex, hippocampus, thalamus, basal ganglia, and limbic system. GABAA receptors are ligand-gated ion channels, with the inhibitory neurotransmitter gamma-aminobutyric acid as the endogenous agonist. It is a pentameric protein (2, 2 and one subunit) that spans the cell membrane, and when the agonist interacts with the alpha subunit, a conformational change occurs, allowing chloride ions to enter the cell, resulting in neuronal hyperpolarization.

      For status epilepticus, midazolam is not the drug of choice. Lorazepam is the benzodiazepine of choice for status epilepticus.

    • This question is part of the following fields:

      • Pharmacology
      18.7
      Seconds
  • Question 14 - A 72-year old man is experiencing a cardiac risk evaluation for the management...

    Correct

    • A 72-year old man is experiencing a cardiac risk evaluation for the management of obstructive umbilical hernia. Echocardiogram demonstrates an aortic valve area=0.59cm with a pressure of 70mmHg. Five years ago, he had mild myocardial infarction complicated with pulmonary oedema. Now he encounters angina with little exertion.

      Which of the following factor is the foremost profoundly weighted using Deysky's cardiac risk scoring system in this case?

      Your Answer: Aortic stenosis

      Explanation:

      Detsky’s Modified cardiac risk classification system in patients undergoing non-cardiac surgery:

      Age more than 70: 05 points

      History of myocardial infarction:

      Less than 6 months: 10 points
      More than 6 months: 5 points

      Angina Pectoris:

      Angina with minimal exertion: 10 points

      Angina at any level of exertion: 20 points

      Pulmonary Oedema:

      Within 7 days: 10 points
      At any time: 5 points

      Suspected aortic valve stenosis with valve area <0.6cm2: 20 points Arrhythmia: Any rhythm other than sinus or sinus with premature atrial complexes (PACs): 5 points More than 5 premature ventricular contractions: 5 points
      Emergency Surgery: 10 points
      Deficient general medical condition: 5 points

      Risk classification:

      Grade I: 0-15 points = low risk
      Grade II: 15-30 points = moderate risk
      Grade III: >30 points = high risk

    • This question is part of the following fields:

      • Pathophysiology
      11.3
      Seconds
  • Question 15 - Which statement is correct concerning breathing systems? ...

    Incorrect

    • Which statement is correct concerning breathing systems?

      Your Answer: In a coaxial Mapleson A system (or Lack) the inner tube has a diameter of 20 mm and the outer tube has a diameter of 30 mm

      Correct Answer: The reservoir bag can limit the pressure in the breathing system to about 40 cm of water

      Explanation:

      Mapleson classified breathing systems into A, B, C, D and E. Jackson-Rees subsequently modified the Mapleson E by adding a double-ended bag to the end of the reservoir tubing, creating the Mapleson F. A Mapleson E or T-piece does not have a reservoir bag.

      A Mapleson A system is a very efficient system for use during spontaneous ventilation. However, it is not suitable for use with patients less than 25 kg, due to the increased dead space at the distal / patient end. This system can be modified into a Lack system or coaxial Mapleson A, where the fresh gas flows through an outer tube (30 mm) and exhaled gases flow through the inner tube (14 mm).

      The adjustable pressure limiting valve (APL) or expiratory valve allows exhaled gas and excess fresh gas to leave the breathing system. It is a one-way, adjustable spring-loaded valve, and gases escape when the pressure in the system exceeds the valve opening pressure. During spontaneous ventilation a pressure of less than 1 cm of water (0.1 kPa) is needed when the valve is in the open position (not 2 cm of H2O).

      The reservoir bag is highly compliant and when over inflated, the rubber bag can limit the pressure in the system to about 40 cm of H2O.

      This is due to the law of Laplace, which states that the pressure will fall as the radius of the bag increases:

      Pressure = 2 x tension/radius.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      27.3
      Seconds
  • Question 16 - Concerning the anterior pituitary gland, one of following is true. ...

    Incorrect

    • Concerning the anterior pituitary gland, one of following is true.

      Your Answer: Is connected to the thalamus by neural tissue

      Correct Answer: Produces glycoproteins

      Explanation:

      The posterior pituitary and the hypothalamus are connected by the pituitary stalk. It contains in the pituitary sella and has the optic chiasm and hypothalamus as superior relations.

      The anterior pituitary produces thyroid-stimulating hormone (TSH), luteinising hormone (LH) and follicle-stimulating hormone (FSH) . These hormones are Glycoproteins and share a common alpha subunit with unique beta subunits.

      The secretion of pituitary hormones are pulsatile. Examples are LH, adrenocorticotropic hormone (ACTH) and growth hormone (GH).

    • This question is part of the following fields:

      • Pathophysiology
      26.3
      Seconds
  • Question 17 - Pressure volume loop represents the compliance of left ventricle.

    Considering there...

    Incorrect

    • Pressure volume loop represents the compliance of left ventricle.

      Considering there is no change in preload and myocardial contractility, which physiological change may result an increase in left ventricular afterload?

      Your Answer: Reduced cardiac work

      Correct Answer: Increased end-systolic volume

      Explanation:

      If there is no change in preload and myocardial contractility, there will be decrease in end-diastolic volume and stroke volume. So there must be increase in end-systolic volume.

    • This question is part of the following fields:

      • Physiology
      31.2
      Seconds
  • Question 18 - The phenomenon that the patients behaved in a different manner when they know...

    Correct

    • The phenomenon that the patients behaved in a different manner when they know that they are being observed is termed as?

      Your Answer: Hawthorne effect

      Explanation:

      Hawthorne effect explains the change in any behavioural aspect owing to the awareness that the person is being observed.
      Simpson’s Paradox explains the association developed when the data from several groups is combined to form a single larger group.

      The remaining terms are made up.

    • This question is part of the following fields:

      • Statistical Methods
      16.2
      Seconds
  • Question 19 - A 19-year-old woman presents to the emergency department. She complains of symptoms indicative...

    Incorrect

    • A 19-year-old woman presents to the emergency department. She complains of symptoms indicative of an acute exacerbation of known 'brittle' asthma. On history, she reveals her asthma is normally controlled using inhalers and she has never had an acute exacerbation requiring hospitalisation.

      On her admission into the ICU, further examination and diagnostic investigations are conducted. Her readings are:

      Physical state: Alert, anxious and non-cyanotic.
      Respiratory rate: 30 breaths/min
      Pulse: 120 beats/min
      Blood pressure: 150/90 mmHg
      SPO2: 95% on air
      Auscultation: Quiet breath sounds at both lung bases

      What is the next most important step of investigation?

      Your Answer: Erect chest x ray

      Correct Answer: Peak expiratory flow rate

      Explanation:

      Peak expiratory flow rate (PEFR) is the maximum speed of air flow generated during a single forced exhaled breath. It is most useful when expressed as a percentage of the best value obtained from the patient.

      Forced expiratory volume over 1 second (FEV1) is a lung parameter measured using spirometry. It is the amount of air forced out of the lung in one exhaled breath. It is a more accurate measure of lung obstructions as it doesn’t rely on effort like PEFR

      PEFR and FEV1 are usually similar, but become more different in asthmatic patients as airflow becomes increasingly obstructed.

      Acute severe asthma is most often diagnosed on history taking and examinations:

      Respiratory rate: >25 breaths/min
      Heart rate: >110 beats/min
      PEFR: 33 – 50% predicted (<200L/min)
      Patient state: Unable to complete a sentence in a single breath.

      A chest x-ray is not routinely required, and is only indicated in specific circumstances, which are:

      If a pneumomediastinum or pneumothorax is suspected
      Possible life threatening asthma
      Possible consolidation
      Unresponsive asthma
      If ventilation is required.

      An echocardiograph (ECG) is not necessary in this case

      Routine haematological and biochemical investigations are not urgent in this case as any abnormalities they detect will be secondary to the patient’s presentation.

      An arterial blood gas (ABG) will only be indicated if SPO2 was <92% or if patient presented with life threatening symptoms.

    • This question is part of the following fields:

      • Clinical Measurement
      28.3
      Seconds
  • Question 20 - Very small SI units are easily expressed using mathematical prefixes.

    One femtolitre is equal...

    Correct

    • Very small SI units are easily expressed using mathematical prefixes.

      One femtolitre is equal to which of the following volumes?

      Your Answer: 0.000, 000, 000, 000, 001 L

      Explanation:

      Small measurement units are denoted by the following SI mathematical prefixes:

      1 deci = 0.1
      1 milli = 0.001
      1 micro = 0.000001
      1 nano = 0.000000001
      1 pico = 0.000000000001
      1 femto = 0.000000000000001 (used to measure red blood cell volume)
      1 atto = 0.000000000000000001

    • This question is part of the following fields:

      • Basic Physics
      8.4
      Seconds
  • Question 21 - The resistance to flow in a blood vessel is affected by the following...

    Correct

    • The resistance to flow in a blood vessel is affected by the following except?

      Your Answer: Thickness of the vessel wall

      Explanation:

    • This question is part of the following fields:

      • Physiology
      11.1
      Seconds
  • Question 22 - Which of the following statements is an accurate fact about the vertebral column?...

    Incorrect

    • Which of the following statements is an accurate fact about the vertebral column?

      Your Answer: T10-T11 vertebrae have no facets on the transverse processes

      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.

    • This question is part of the following fields:

      • Pathophysiology
      23.8
      Seconds
  • Question 23 - A 50-year-old female is having her central venous pressure (CVP) measured. A long...

    Incorrect

    • A 50-year-old female is having her central venous pressure (CVP) measured. A long femoral line was inserted that passes from the common iliac vein into the inferior vena cava.

      At which level of vertebra does this occur?

      Your Answer: L2

      Correct Answer: L5

      Explanation:

      The inferior vena cava is formed by the union of the right and left common iliac veins. This occurs at the L5 vertebral level. The IVC courses along the right anterolateral side of the vertebral column and ascends through the central tendon of the diaphragm at the T8 vertebral level.

    • This question is part of the following fields:

      • Anatomy
      28.1
      Seconds
  • Question 24 - Of the following, which of these oxygen carrying molecules causes the greatest shift...

    Incorrect

    • Of the following, which of these oxygen carrying molecules causes the greatest shift of the oxygen-dissociation curve to the left?

      Your Answer: Carboxyhaemoglobin (HbCO)

      Correct Answer: Myoglobin (Mb)

      Explanation:

      Myoglobin is a haemoglobin-like, iron-containing pigment that is found in muscle fibres. It has a high affinity for oxygen and it consists of a single alpha polypeptide chain. It binds only one oxygen molecule, unlike haemoglobin, which binds 4 oxygen molecules.

      The myoglobin ODC is a rectangular hyperbola. There is a very low P50 0.37 kPa (2.75 mmHg). This means that it needs a lower P50 to facilitate oxygen offloading from haemoglobin. It is low enough to be able to offload oxygen onto myoglobin where it is stored. Myoglobin releases its oxygen at the very low PO2 values found inside the mitochondria.

      P50 is defined as the affinity of haemoglobin for oxygen: It is the PO2 at which the haemoglobin becomes 50% saturated with oxygen. Normally, the P50 of adult haemoglobin is 3.47 kPa(26 mmHg).

      Foetal haemoglobin has 2 ? and 2 ?chains. The ODC is left shifted – this means that P50 lies between 2.34-2.67 kPa [18-20 mmHg]) compared with the adult curve and it has a higher affinity for oxygen. Foetal haemoglobin has no ? chains so this means that there is less binding of 2.3 diphosphoglycerate (2,3 DPG).

      Carbon monoxide binds to haemoglobin with an affinity more than 200-fold higher than that of oxygen. This therefore decreases the amount of haemoglobin that is available for oxygen transport. Carbon monoxide binding also increases the affinity of haemoglobin for oxygen, which shifts the oxygen-haemoglobin dissociation curve to the left and thus impedes oxygen unloading in the tissues.

      In sickle cell disease, (HbSS) has a P50 of 4.53 kPa(34 mmHg).

    • This question is part of the following fields:

      • Physiology
      16.5
      Seconds
  • Question 25 - Of the following, which is NOT a branch of the external carotid artery?...

    Incorrect

    • Of the following, which is NOT a branch of the external carotid artery?

      Your Answer: Superior thyroid artery

      Correct Answer: Mandibular artery

      Explanation:

      The external carotid artery has eight important branches:
      1. Superior thyroid artery
      2. Ascending pharyngeal artery
      3. Lingual artery
      4. Facial artery
      5. Occipital artery
      6. Posterior auricular artery
      7. Maxillary artery (terminal branch)
      8. Superficial temporal artery (terminal branch)

      There is no mandibular artery but the first part of the maxillary artery is called the mandibular part as it is posterior to the lateral pterygoid muscle.
      The maxillary artery is divided into three portions by its relation to the lateral pterygoid muscle:
      first (mandibular) part: posterior to the lateral pterygoid muscle
      second (pterygoid or muscular) part: within the lateral pterygoid muscle
      third (pterygopalatine) part: anterior to the lateral pterygoid muscle

    • This question is part of the following fields:

      • Anatomy
      7.7
      Seconds
  • Question 26 - Which of the following statement is true regarding the mechanism of action of...

    Incorrect

    • Which of the following statement is true regarding the mechanism of action of macrolides?

      Your Answer: Inhibits RNA synthesis

      Correct Answer: Inhibits protein synthesis

      Explanation:

      The mechanism of action of macrolides is inhibition of bacterial protein synthesis by preventing peptidyltransferase from adding to the growing peptide which is attached to tRNA to the next amino acid.

    • This question is part of the following fields:

      • Pharmacology
      5.3
      Seconds
  • Question 27 - Which of the following statements is true with regards to the Krebs' cycle...

    Correct

    • Which of the following statements is true with regards to the Krebs' cycle (also known as the tricarboxylic acid cycle or citric acid cycle)?

      Your Answer: Alpha-ketoglutarate is a five carbon molecule

      Explanation:

      Krebs’ cycle (tricarboxylic acid cycle or citric acid cycle) is a sequence of reactions in which acetyl coenzyme A (acetyl-CoA) is metabolised and this results in carbon dioxide and hydrogen atoms production.

      This series of reactions occur in the mitochondria of eukaryotic cells, not the cytoplasm. The cycle requires oxygen and so, cannot function under anaerobic conditions.

      It is the common pathway for carbohydrate, fat and some amino acids oxidation and is required for high energy phosphate bond formation in adenosine triphosphate (ATP).

      When pyruvate enters the mitochondria, it is converted into acetyl-CoA. This represents the formation of a 2 carbon molecule from a 3 carbon molecule. There is loss of one CO2 but formation of one NADH molecule. Acetyl-CoA is condensed with oxaloacetate, the anion of a 4 carbon acid, to form citrate which is a 6 carbon molecule.

      Citrate is then converted into isocitrate, alpha-ketoglutarate, succinyl-CoA, succinate, fumarate, malate and finally oxaloacetate.

      The only 5 carbon molecule in the cycle is alpha-ketoglutarate.

    • This question is part of the following fields:

      • Physiology
      21.6
      Seconds
  • Question 28 - Regarding the emergency oxygen flush, which is true? ...

    Correct

    • Regarding the emergency oxygen flush, which is true?

      Your Answer: May lead to awareness if used inappropriately

      Explanation:

      When the emergency oxygen flush is pressed, 100% oxygen is supplied from the common gas outlet. This gas bypasses BOTH flowmeters and vaporisers. The flow of oxygen is usually 45 l/min at a PRESSURE OF 400 kPa.

      There is an increased risk of pulmonary barotrauma when the emergency flush is pressed, especially when anaesthetising paediatric patients.

      The inappropriate use of the flush causes dilution of anaesthetic gases and this increases the possibility of anaesthetic awareness .

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      10
      Seconds
  • Question 29 - Following are some examples of induction agents. Which one has the longest elimination...

    Incorrect

    • Following are some examples of induction agents. Which one has the longest elimination half-life?

      Your Answer: Methohexitone

      Correct Answer: Thiopental

      Explanation:

      Thiopental has the longest elimination half-life of 6-15 hours.

      Elimination half-life of other drugs are given as:
      – Propofol: 5-12 h
      – Methohexitone: 3-5 h
      – Ketamine: 2 h
      – Etomidate: 1-4 h

    • This question is part of the following fields:

      • Pharmacology
      10.4
      Seconds
  • Question 30 - A 20-year old lady has been having excessive bruising and bleeding of her...

    Correct

    • A 20-year old lady has been having excessive bruising and bleeding of her gums. She is under investigation for the extrinsic pathway of coagulation. Which is the best investigation to order?

      Your Answer: Prothrombin time (PT)

      Explanation:

      The extrinsic pathway is best assessed by the PT 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

    • This question is part of the following fields:

      • Physiology And Biochemistry
      18.5
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Anatomy (1/3) 33%
Statistical Methods (2/3) 67%
Clinical Measurement (1/3) 33%
Pharmacology (1/5) 20%
Physiology (2/5) 40%
Basic Physics (1/2) 50%
Pathophysiology (3/5) 60%
Physiology And Biochemistry (1/2) 50%
Anaesthesia Related Apparatus (1/2) 50%
Passmed