00
Correct
00
Incorrect
00 : 00 : 0 00
Session Time
00 : 00
Average Question Time ( Mins)
  • Question 1 - Which of the following statements is true regarding drug dose and response? ...

    Incorrect

    • Which of the following statements is true regarding drug dose and response?

      Your Answer: Antagonists must have a higher receptor affinity than agonists

      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.

    • This question is part of the following fields:

      • Pharmacology
      44.2
      Seconds
  • Question 2 - A 42-year-old man presented with a bitemporal hemianopia with enlarged hands and feet....

    Incorrect

    • A 42-year-old man presented with a bitemporal hemianopia with enlarged hands and feet. On examination, he was found to be hypertensive.

      Which of the following correctly explains the cause of his visual field defect?

      Your Answer: Pituitary microadenoma secreting growth hormone (GH)

      Correct Answer: Pituitary macroadenoma secreting growth hormone (GH)

      Explanation:

      Pituitary macroadenoma is a benign tumour with growth larger than 10mm (those under 10mm are called microadenoma)

      Compression of optic chiasm by pituitary adenoma is responsible for causing visual field defects like bitemporal hemianopia, optic neuropathy.

    • This question is part of the following fields:

      • Pathophysiology
      7.4
      Seconds
  • Question 3 - A 28-year-old man is admitted to the critical care unit. He has been...

    Incorrect

    • A 28-year-old man is admitted to the critical care unit. He has been diagnosed with adult respiratory distress syndrome and is being ventilated. His haemodynamic condition is improved using a pulmonary artery flotation.

      His readings are listed below:

      Haemoglobin concentration: 10 g/dL
      Mixed venous oxygen saturation: 70%
      Mixed venous oxygen tensions (PvO2): 50 mmHg

      Estimate his mixed venous oxygen content (mL/100mL).

      Your Answer: 10.5

      Correct Answer: 9.5

      Explanation:

      Mixed venous oxygen content (CvO2) is the oxygen concentration in 100mL of mixed venous blood taken from the pulmonary artery. It is usually 12-17 mL/dL (70-75%). It is represented mathematically as:

      CvO2 = (1.34 x Hgb x SvO2 x 0.01) + (0.003 x PvO2)

      Where,

      1.34 = Huffner’s constant
      Hgb = Haemoglobin level (g/dL)
      SvO2 = % oxyhaemoglobin saturation of mixed venous blood
      PvO2 = 0.0225 = mL of O2 dissolved per 100mL plasma per kPa, or 0.003 mL per mmHg

      Therefore,

      CvO2 = (1.34 x 10 x 70 x 0.01) + (0.003 x 50)

      CvO2 = 9.38 + 0.15 = 9.53 mL/100mL

    • This question is part of the following fields:

      • Clinical Measurement
      672.2
      Seconds
  • Question 4 - The Control of Substances Hazardous to Health (COSHH) regulations recommend air supply rates...

    Correct

    • The Control of Substances Hazardous to Health (COSHH) regulations recommend air supply rates to specific environments. Which of the following statements is true?

      Your Answer: Preparation rooms receive a volume of 0.1 m3 of air per second

      Explanation:

      Control of Substances Hazardous to Health (COSHH) was established by government under the Health and Safety at Work act in 1989. Their employers work on identification and management of those substances that are dangerous to health. The implications for anaesthetists include gas scavenging, equipment contamination and environmental safety. Adequate ventilation is required in areas where anaesthetic gases are present. The minimum air supply that is legally required in each specific area is: Operating theatres: 0.65 m3/second. Anaesthetic rooms: 0.15 m3/s. Preparation rooms: 0.1 m3/s. Recovery rooms need 15 air changes per hour

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      16.3
      Seconds
  • Question 5 - Which of the following statements is true with regards to acetylcholine? ...

    Correct

    • Which of the following statements is true with regards to acetylcholine?

      Your Answer: Excess cholinesterase inhibitor medication causes cholinergic crisis

      Explanation:

      Myasthenic and cholinergic crises are two crises which are similar in their clinical presentation.

      Myasthenic crisis can be caused by:
      -lack of acetylcholine,
      -poor compliance with medication,
      -infection

      Cholinergic crisis can be caused by excess cholinesterase inhibitor medication (mimicking organophosphate poisoning) causing excess acetylcholine.

      Differentiation between the 2 crises is made by giving incremental doses of the short acting cholinesterase inhibitor, Edrophonium.
      This increase acetylcholine levels and will make a myasthenic crisis better and a cholinergic crisis worse.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      578.6
      Seconds
  • Question 6 - During a squint surgery, a 5-year-old child developed severe bradycardia as a result...

    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: Oculomotor nerve

      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.

    • This question is part of the following fields:

      • Pathophysiology
      28.4
      Seconds
  • Question 7 - A 65-year-old man got operated on for carotid endarterectomy for his carotid artery...

    Incorrect

    • A 65-year-old man got operated on for carotid endarterectomy for his carotid artery disease. He is recovering well post-surgery. However, on follow-up in the ward, he has hoarseness of his voice.

      Which of the following explains the hoarseness?

      Your Answer: Damage to the glossopharyngeal nerve

      Correct Answer: Damage to the vagus

      Explanation:

      During carotid endarterectomy, injury to the vagus nerve or its branches can cause hoarseness. Injury to the vagus nerve can result in adductor vocal cord paralysis. It can also cause other symptoms like dysphagia or even vocal cord immobility.

      Carotid endarterectomy is the procedure to relieve an obstruction in the carotid artery by opening the artery at its origin and stripping off the atherosclerotic plaque with the intima. Because of the internal carotid artery relations, there is a risk of cranial nerve injury during the procedure involving one or more of the following nerves: CN IX, CN X (or its branch, the superior laryngeal nerve), CN XI, or CN XII.

      However, only damage to the vagus would account for speech difficulties.

    • This question is part of the following fields:

      • Anatomy
      15.9
      Seconds
  • Question 8 - A 240 volt alternating current (AC) socket from a wall is used to...

    Incorrect

    • A 240 volt alternating current (AC) socket from a wall is used to charge a direct current (DC) cardiac defibrillator.

      Name the electrical component that converts AC to DC.

      Your Answer:

      Correct Answer: Rectifier

      Explanation:

      There are two types of defibrillators
      AC defibrillator
      DC defibrillator

      AC defibrillator,
      consists of a step-up transformer with primary and secondary winding and two switches. Since secondary coil consists of more turns of wire than the primary coil, it induces larger voltage. A voltage value ranging between 250V to 750V is applied for AC external defibrillator. And used to enable the charging of a capacitor.

      DC defibrillator,
      consists of auto transformer T1 that acts as primary of the high voltage transformer T2. Is an iron core that transfers energy between 2 circuits by electromagnetic induction. Transformers are used to isolate circuits, change impedance and alter voltage output. transformers do not convert AC to DC.

      Diode rectifier composed of 4 diodes made of semiconductor material allows current to flow only in one direction. Alternating current (AC) passing through these diodes produces direct current (DC). Capacitor stores the charge in the form of an electrostatic field.

      Capacitor is used to convert the rectified AC voltage to produce DC voltage but capacitors do not directly convert AC to DC.

      Inductor induces a counter electromotive force(emf) that reduces the capacitor discharge value.

      In step-down transformer primary coils has more turns of wire than secondary coil, so induced voltage is smaller in the secondary coil.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      0
      Seconds
  • Question 9 - Risk stratification is done prior to a major cardiac surgery using cardiopulmonary exercise...

    Incorrect

    • Risk stratification is done prior to a major cardiac surgery using cardiopulmonary exercise testing. Given the following options, which one is most likely to have the highest risk for post-operative cardiac morbidity?

      Your Answer:

      Correct Answer: Anaerobic threshold (AT) of less than 11 mL/kg/minute

      Explanation:

      The ventilatory anaerobic threshold (VAT), formerly referred to as the anaerobic threshold, is an index used to estimate exercise capacity. During the initial (aerobic) phase of CPET, which lasts until 50–60% of Vo2max is reached, expired ventilation (VE) increases linearly with Vo2 and reflects aerobically produced CO2 in the muscles. Blood lactate levels do not change substantially during this phase, since muscle lactic acid production is minimal.

      During the latter half of exercise, anaerobic metabolism occurs because oxygen supply cannot keep up with the increasing metabolic requirements of exercising muscles. At this time, there is a significant increase in lactic acid production in the muscles and in the blood lactate concentration. The Vo2 at the onset of blood lactate accumulation is called the lactate threshold or the VAT. The VAT is also defined as the point at which minute ventilation increases disproportionately relative to Vo2, a response that is generally seen at 60–70% of Vo2max.

      The VAT is a useful measure as work below this level encompasses most daily living activities. The ability to achieve the VAT can help distinguish cardiac and non‐cardiac (pulmonary or musculoskeletal) causes of exercise limitation, since patients who fatigue before reaching VAT are likely to have a non‐cardiac problem.

      When VAT is detected, patients with PVo2 of ⩽10 ml/kg/min have a high event rate.

    • This question is part of the following fields:

      • Pathophysiology
      0
      Seconds
  • Question 10 - When describing the surface anatomy of the sacrum, which of the following anatomical...

    Incorrect

    • When describing the surface anatomy of the sacrum, which of the following anatomical landmarks refers to the base of an equilateral triangle is formed by the sacral hiatus?

      Your Answer:

      Correct Answer: A line connecting the posterior superior iliac spines

      Explanation:

      The apex of an equilateral triangle completed by the posterior superior iliac spines is where the sacral hiatus or sacrococcygeal membrane can normally located. The failure of posterior fusion of the laminae of the fourth and fifth sacral vertebrae allows the sacral canal to be accessible via the membrane.

      In adults, the spine of L4 usually lies on a line drawn between the highest points of the iliac crests (Tuffier’s line). A line connecting each anterior iliac spine, approximates to the L3/4 interspace in the sitting position. Both of these options are incorrect.

      A line connecting the greater trochanters is also incorrect.

      A line connecting the posterior superior iliac spines is correct, but in adults the presence of a sacral fat pad can still make identification of this landmark less straightforward.

      The processes of S5 are remnants only and form the sacral cornua, which are also used to help identify the sacral hiatus.

    • This question is part of the following fields:

      • Anatomy
      0
      Seconds
  • Question 11 - During positive pressure ventilation using positive end-expiratory pressure (PEEP), there is usually an...

    Incorrect

    • During positive pressure ventilation using positive end-expiratory pressure (PEEP), there is usually an associated reduction in cardiac output

      Which of the following is responsible?

      Your Answer:

      Correct Answer: Reduced venous return to the heart

      Explanation:

      The option that is most responsible is the progressive decrease in venous return of blood to the right atrium. The heart rate does not usually change with PEEP so the fall in cardiac output is due to a reduction in left ventricular (LV) stroke volume (SV).

      Note that the interventricular septum does shift toward the left and there is an increased pulmonary vascular resistance (PVR) from overdistention of alveolar air sacs that contribute to the reduction in cardiac output. Any increase in PVR will be associated with reduced pulmonary vascular capacitance.

    • This question is part of the following fields:

      • Pathophysiology
      0
      Seconds
  • Question 12 - A patient with a known history of asymptomatic ventriculoseptal defect (VSD) is to...

    Incorrect

    • A patient with a known history of asymptomatic ventriculoseptal defect (VSD) is to undergo an orthopaedic surgery under general anaesthesia. The rest of the patient's medical history, such as allergies and previous operations, are unremarkable.

      What is the best antibiotic prophylaxis prior to surgery?

      Your Answer:

      Correct Answer: No antibiotic prophylaxis required as the defect is repaired and no evidence of benefit from routine prophylaxis

      Explanation:

      According to the 2015 National Institute for Health and Care Excellence (NICE) Guidelines, antibiotic prophylaxis against infective endocarditis (IE) is not recommended routinely for people with any cardiac defect (corrected or uncorrected) due to lack of sufficient evidence regarding its benefits. Instead, antibiotic prophylaxis is recommended for those who are at risk of developing IE, such as those with acquired valvular heart disease with stenosis or regurgitation; hypertrophic cardiomyopathy; valve replacement; and previous IE.

    • This question is part of the following fields:

      • Pharmacology
      0
      Seconds
  • Question 13 - A 78-year-old man with a previous history of ischaemic heart disease is admitted...

    Incorrect

    • A 78-year-old man with a previous history of ischaemic heart disease is admitted to hospital. He is scheduled for a cardiopulmonary exercise test (CPX) before he undergoes an elective abdominal aneurysm repair.

      What measurement obtained during a CPX test alone provides the best indication for postoperative mortality?

      Your Answer:

      Correct Answer: Anaerobic threshold

      Explanation:

      Cardiopulmonary exercise testing (CPX, CPEX, CPET) is a non-invasive testing method used to determine the performance of the heart, lungs and skeletal muscle. It measures the exercise tolerance of the patient.

      The parameters measured include:

      ECG and ST-segment analysis and blood pressure
      Oxygen consumption (VO2)
      Carbon dioxide production (VCO2)
      Gas flows and volumes
      Respiratory exchange ratio (RER)
      Respiratory rate
      Anaerobic threshold (AT)

      The anaerobic threshold (AT) is an estimate of exercise ability. Any measurement below 11 ml/kg/min is usually related with an increase in mortality, especially when there is a background of myocardial ischaemia occurring during the test.

      Peak VO2 <20 mL/kg with a low AT have a correlation with postoperative complications and a 30 day mortality. The CPX test is used for risk-testing patients prior to surgery to determine the appropriate postoperative care facilities. The V slope measured in CPX testing represents VO2 versus VCO2 relationship. During AT, the ramp of V slope increases, but does not provide a picture of postoperative mortality.

    • This question is part of the following fields:

      • Clinical Measurement
      0
      Seconds
  • Question 14 - Which of the following is a feature of a central venous pressure waveform?...

    Incorrect

    • Which of the following is a feature of a central venous pressure waveform?

      Your Answer:

      Correct Answer: An a wave due to atrial contraction

      Explanation:

      The central venous pressure (CVP) waveform depicts changes of pressure within the right atrium. Different parts of the waveform are:

      A wave: which represents atrial contraction. It is synonymous with the P wave seen during an ECG. It is often eliminated in the presence of atrial fibrillation, and increased tricuspid stenosis, pulmonary stenosis and pulmonary hypertension.

      C wave: which represents right ventricle contraction at the point where the tricuspid valve bulges into the right atrium. It is synonymous with the QRS complex seen on ECG.

      X descent: which represents relaxation of the atrial diastole and a decrease in atrial pressure, due to the downward movement of the right ventricle as it contracts. It is synonymous with the point before the T wave on ECG.

      V wave: which represents an increase in atrial pressure just before the opening of the tricuspid valve. It is synonymous with the point after the T wave on ECG. It is increased in the background of a tricuspid regurgitation.

      Y descent: which represents the emptying of the atrium as the tricuspid valve opens to allow for blood flow into the ventricle in early diastole.

    • This question is part of the following fields:

      • Pathophysiology
      0
      Seconds
  • Question 15 - Which of the following is true about the bispectral index (BIS)? ...

    Incorrect

    • Which of the following is true about the bispectral index (BIS)?

      Your Answer:

      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
      0
      Seconds
  • Question 16 - In which of the following situations will a regional fall in cerebral blood...

    Incorrect

    • In which of the following situations will a regional fall in cerebral blood flow occur, suppose there is no changes in the mean arterial pressure (MAP)?

      Your Answer:

      Correct Answer: Hyperoxia

      Explanation:

      The response of cerebral blood flow (CBF) to hyperoxia (PaO2 >15 kPa, 113 mmHg), the cerebral oxygen vasoreactivity is less well defined. A study originally described, using a nitrous oxide washout technique, a reduction in CBF of 13% and a moderate increase in cerebrovascular resistance in subjects inhaling 85-100% oxygen. Subsequent human studies, using a variety of differing methods, have also shown CBF reductions with hyperoxia, although the reported extent of this change is variable. Another study assessed how supra-atmospheric pressures influenced CBF, as estimated by changes in middle cerebral artery flow velocity (MCAFV) in healthy individuals. Atmospheric pressure alone had no effect on MCAFV if PaO2 was kept constant. Increases in PaO2 did lead to a significant reduction in MCAFV; however, there were no further reductions in MCAFV when oxygen was increased from 100% at 1 atmosphere of pressure to 100% oxygen at 2 atmospheres of pressure. This suggests that the ability of cerebral vasculature to constrict in response to increasing partial pressure of oxygen is limited.

      Increases in arterial blood CO2 tension (PaCO2) elicit marked cerebral vasodilation.

      CBF increases with general anaesthesia, ketamine anaesthesia, and hypoviscosity.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 17 - In an experimental study, a healthy subject was given one litre of 5%...

    Incorrect

    • In an experimental study, a healthy subject was given one litre of 5% dextrose within a 15-minute period. Which of the following mechanisms is expected to affect the urine output?

      Your Answer:

      Correct Answer: Inhibition of arginine vasopressin (AVP) secretion

      Explanation:

      Changes in the osmolality of body fluids (changes as minor as 1% are sufficient) play the most important role in regulating AVP secretion. The receptors that monitor changes in osmolality of body fluids (termed osmoreceptors) are distinct from the cells that synthesize and secrete AVP, and are located in the organum vasculosum of the lamina terminalis (OVLT) of the hypothalamus. The osmoreceptors sense changes in body osmolality by either shrinking or swelling. When the effective osmolality of the plasma increases, the osmoreceptors send signals to the AVP synthesizing/secreting cells located in the supraoptic and paraventricular nuclei of the hypothalamus, and AVP synthesis and secretion are stimulated. Conversely, when the effective osmolality of the plasma is reduced, secretion is inhibited. Because AVP is rapidly degraded in the plasma, circulating levels can be reduced to zero within minutes after secretion is inhibited.

      In this scenario, the osmolality of the plasma will decrease to an estimate of 2.5%, hence inhibition of AVP.

      Stimulation of atrial stretch receptors is incorrect because the increase in plasma volume is still below the threshold for its activation.

      Osmotic diuresis is incorrect because 5% dextrose is isotonic, hence osmotic diuresis is not probable.

      Renin is inhibited when an excess of NaCl in the tubular fluid is sensed by the macula densa.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 18 - Which of the following is a characteristic of a type 1B antiarrhythmic agent...

    Incorrect

    • Which of the following is a characteristic of a type 1B antiarrhythmic agent such as Lidocaine?

      Your Answer:

      Correct Answer: Shortens refractory period

      Explanation:

      The action of class 1 anti-arrhythmic is sodium channel blockade. Subclasses of this action reflect effects on the action potential duration (APD) and the kinetics of sodium channel blockade.

      Drugs with class 1A prolong the APD and refractory period, and dissociate from the channel with intermediate kinetics.

      Drugs with class 1B action shorten the APD in some tissues of the heart, shorten the refractory period, and dissociate from the channel with rapid kinetics.

      Drugs with class 1C action have minimal effects on the APD and the refractory period, and dissociate from the channel with slow kinetics.

    • This question is part of the following fields:

      • Pharmacology
      0
      Seconds
  • Question 19 - Which of the following statements is about the measurement of glomerular filtration rate...

    Incorrect

    • Which of the following statements is about the measurement of glomerular filtration rate (GFR) is correct?

      Your Answer:

      Correct Answer: The result matches clearance of the indicator if it is renally inert

      Explanation:

      The measurements of GFR are done using renally inert indicators like inulin, where passive rate of filtration at the glomerulus = rate of excretion. Normal value is about 180 litres per day.

      GFR is altered by renal blood flow but blood flow does not need to be measured.

      The reabsorption of Na leads to a low excretion rate and low urine concentration and therefore its use as an indicator would lead to an erroneously LOW GFR.

      If there is tubular secretion of any solute, the clearance value will be higher than that of inulin. This will be either due to tubular reabsorption or the solute not being freely filtered at the glomerulus.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 20 - Which of the following lung parameters can be measured directly using spirometry? ...

    Incorrect

    • Which of the following lung parameters can be measured directly using spirometry?

      Your Answer:

      Correct Answer: Vital capacity

      Explanation:

      Spirometry measures the total volume of air that can be forced out in one maximum breath, that is the total lung capacity (TLC), to maximal expiration, that is the residual volume (RV).

      It is conducted using a spirometer which is capable of measuring lung volumes using techniques of dilution.

      During spirometry, the following measurements can be determined:
      Forced vital capacity (FVC)/vital capacity (VC): The maximum volume of air exhaled in one single forced breathe.
      Forced expiratory volume in one second (FEV1)
      FEV1/FVC ratio
      Peak expiratory flow (PEF): the maximum amount of air flow exhaled in one blow.
      Forced expiratory flow (mid expiratory flow): the flow at 25%, 50% and 75% of FVC
      Inspiratory vital capacity (IVC): The maximum volume of air inhaled after a full total expiration.

      Anatomical dead space is measured using a single breath nitrogen washout called the Fowler’s method.

      Residual volume and total lung capacity are both measured using the body plethysmograph or helium dilution

      The functional residual capacity is usually measured using a nitrogen washout or the helium dilution technique.

    • This question is part of the following fields:

      • Clinical Measurement
      0
      Seconds
  • Question 21 - A project is being planned to assess the effects of a new anticoagulant...

    Incorrect

    • 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

    • This question is part of the following fields:

      • Physiology And Biochemistry
      0
      Seconds
  • Question 22 - Which vessel is the first to branch from the external carotid artery? ...

    Incorrect

    • Which vessel is the first to branch from the external carotid artery?

      Your Answer:

      Correct Answer: Superior thyroid artery

      Explanation:

      The superior thyroid artery is the first branch of the external carotid artery. The other branches of the external carotid artery are:
      1. Superior thyroid artery
      2. Ascending pharyngeal artery
      3. Lingual artery
      4. Facial artery
      5. Occipital artery
      6. Posterior auricular artery
      7. Maxillary artery
      8. Superficial temporal artery

      The inferior thyroid artery is derived from the thyrocervical trunk.

    • This question is part of the following fields:

      • Anatomy
      0
      Seconds
  • Question 23 - A 5-year old male has ingested a peanut and has developed urticaria, vomiting...

    Incorrect

    • A 5-year old male has ingested a peanut and has developed urticaria, vomiting and hypotension. The pathogenesis of this condition is derived from predominant cells of which cell line?

      Your Answer:

      Correct Answer: Common myeloid progenitor

      Explanation:

      A is correct. Common myeloid progenitor cells are involved in the anaphylaxis reaction.
      B is incorrect. The common lymphoid lineage gives rise to T-cells, B-cell and NK cells.
      C is incorrect as megakaryocytes give rise to platelets.
      D is incorrect – Neural crest cells give rise to various cells throughout the body, including melanocytes, enterochromaffin cells and Schwann cells. However, they do not give rise to mast cells.
      E is incorrect. Reticulocytes give rise to erythrocytes.

      This is a classic case of anaphylaxis. In this situation, IgE previously raised against antigens (in this case peanut antigen) bind to mast cells, and this causes them to degranulate.
      There is release of vasoactive substances like histamine into the blood, and this is responsible for the symptoms seen. Therefore, the main type of cells involved in the pathogenesis of the disease is mast cells.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      0
      Seconds
  • Question 24 - A measuring system's response to change is complex, yet it can be mathematically modelled.

    Which of the following terms best characterises a pressure transducer's responsiveness to blood pressure changes?

    ...

    Incorrect

    • A measuring system's response to change is complex, yet it can be mathematically modelled.

      Which of the following terms best characterises a pressure transducer's responsiveness to blood pressure changes?

      Your Answer:

      Correct Answer: Dynamic second-order response

      Explanation:

      The static-response defines how a measuring system behaves while it is in equilibrium (i.e. when the measured values are not changing). If the value being measured changes over time, the reaction of a measuring system will change as well which would be a dynamic response.
      The dynamic response of a measuring system can be subdivided into zero-order, first-order and second-order responses:

      Zero-order:
      Consider a thermometer that has been left in a room for a week. The thermometer will display the current ambient temperature when you enter the room.

      First-order:
      Consider the use of a mercury thermometer to check a patient’s temperature. It is comprised of a mercury column that expands as it warms up. The scale’s initial temperature is room temperature, but when it’s placed under the patient’s tongue, the temperature readings rise until they reach body temperature.

      Second-order
      Consider putting weights on a mechanical weighing scale. The weight as reported on the measuring dial, will wobble around the correct value at first until reaching equilibrium. An example of this is in clinical practice is the direct measurement of arterial pressure with a transducer. The value of the input fluctuates around a central point.

      Drift is the progressive deterioration of a measurement system’s precision. With time, the measurement deviates from the genuine, calibrated value. The graph between this measurement and the real value should, ideally, be linear (e.g. on the y-axis the measured end-tidal CO2 against true value of the end-tidal CO2). Drift is split into three types: zero-offset, gradient, and zonal drift.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      0
      Seconds
  • Question 25 - Which compound is secreted only from the adrenal medulla? ...

    Incorrect

    • Which compound is secreted only from the adrenal medulla?

      Your Answer:

      Correct Answer: Adrenaline

      Explanation:

      The adrenal medulla comprises chromaffin cells (pheochromocytes), which are functionally equivalent to postganglionic sympathetic neurons. They synthesize, store and release the catecholamines noradrenaline (norepinephrine) and adrenaline (epinephrine) into the venous sinusoids.
      The majority of the chromaffin cells synthesize adrenaline.

    • This question is part of the following fields:

      • Anatomy
      0
      Seconds
  • Question 26 - Which of the following is an expected change in pulmonary function seen during...

    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% variability

      With 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.

    • This question is part of the following fields:

      • Pathophysiology
      0
      Seconds
  • Question 27 - Which of the following statements is TRUE regarding an epidural set? ...

    Incorrect

    • Which of the following statements is TRUE regarding an epidural set?

      Your Answer:

      Correct Answer: 19G Tuohy needles have 0.5 cm markings

      Explanation:

      A paediatric 19G Tuohy catheter is available that is 5cm in length and has 0.5cm markings

      18G Tuohy catheters are generally 9 to 10cm to hub

      Distal end of catheter is angled (15 to 30 degrees) and closed to avoid puncturing the dura

      Epidural mesh are usually 0.2 microns and are used to filter bacteria and viruses to ensure sterility of procedure

      Transparent catheters are 90cm long with diameters depending on gauge size. It has 1cm graduations from 5 to 20cm to ensure they have been inserted amply and removed completely. Distal end is smooth which can be open or closed (with lateral openings)

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      0
      Seconds
  • Question 28 - A weakly acidic drug with a pKa of 8.4 is injected intravenously into...

    Incorrect

    • A weakly acidic drug with a pKa of 8.4 is injected intravenously into a patient.

      At a normal physiological pH, the percentage of this drug unionised in the plasma is?

      Your Answer:

      Correct Answer: 90

      Explanation:

      Primary FRCA is concerned with two issues. The first is a working knowledge of the Henderson-Hasselbalch equation, and the second is a working knowledge of logarithms and antilogarithms.

      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 a drug, 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 = 8.4 + log ([ionised]/[unionised])
      7.4 − 8.4 = log ([ionised]/[unionised])
      log −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 = 0.1, −2 = 0.01, −3 = 0.001 and, −4 = 0.0001.

      [A-]/[HA] = 0.1

      Assuming that we can apply the approximation [A-] << [HA} then this means the acid is 0.1 x 100% = 10% ionised so the percentage of (non-ionized) acid will be 100% – 10% = 90%

    • This question is part of the following fields:

      • Pharmacology
      0
      Seconds
  • Question 29 - Heights of 100 individuals(adults) who were administered steroids at any stage during childhood...

    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
      0
      Seconds
  • Question 30 - A 32-year-old male is admitted to the critical care unit. He has suffered...

    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/ cmH2O

      It 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 wall

      Therefore 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.

    • This question is part of the following fields:

      • Clinical Measurement
      0
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Pathophysiology (1/2) 50%
Clinical Measurement (1/1) 100%
Anaesthesia Related Apparatus (2/2) 100%
Physiology And Biochemistry (1/1) 100%
Anatomy (0/1) 0%
Passmed