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  • Question 1 - An 84-year-old woman has a fall. She fractures the neck of her femur...

    Correct

    • An 84-year-old woman has a fall. She fractures the neck of her femur and requires emergency surgery.

      On history and examination, she appears to also have a possible heart failure for which an echocardiogram is scheduled.

      Her measurements are:
      End-diastolic volume: 40mL (70-240)
      End-systolic volume: 30mL (16-140)

      Calculate her approximate ejection fraction.

      Your Answer: 25%

      Explanation:

      An echocardiogram provides real-time visualisation of cardiac structures. The ejection fraction (EF) is normally measured using this system.

      The ejection fraction (EF) can be deduced mathematically if the patient’s end-diastolic volume (EDV), end-systolic volume (ESV) and stroke volume (SV) are known, as:

      SV = EDV – ESV, and

      EF = SV/EDV x 100

      The normal range for EF is >55-70%.

      For this patient,

      SV= 40 – 30 = 10 mL, therefore

      EF = 10/40 x 100 = 25%

    • This question is part of the following fields:

      • Clinical Measurement
      26
      Seconds
  • Question 2 - Which of these statements is true about spirometry? ...

    Incorrect

    • Which of these statements is true about spirometry?

      Your Answer: Closing capacity is likely to equal the functional residual capacity in a 65-year-old patient in the supine position

      Correct Answer: A capacity is the sum of two or more volumes

      Explanation:

      Functional residual capacity (FRC) is 1.7 to 3.5L/kg

      A capacity is the sum of two or more volumes. The total lung capacity (TLC) is total sum of the volume of gas present in all lung compartments upon maximum inspiration. It is represented mathematically as:

      Total lung capacity (TLC) = Vital capacity (VC) + Residual volume (RV)

      The residual volume (RV) is the volume of gas still present within the lung post maximum exhalation. It cannot be measured by spirometry, but can be using a body plethysmograph and also with the helium dilution technique.

      Closing capacity (CC) is the volume of gas within the lungs at which small airways close upon expiration. It increases with age and is especially important when it surpasses the FRC as it causes changes in ventilation/perfusion mismatch and hypoxia.
      In the supine position, a patient with a normal body mass index and no history of lung pathology, the CC equals the FRC at approximately 44, and at approximately 66 at standing position.

    • This question is part of the following fields:

      • Clinical Measurement
      24.9
      Seconds
  • Question 3 - Of the following statements, which is true about the measurements of cardiac output...

    Incorrect

    • Of the following statements, which is true about the measurements of cardiac output using thermodilution?

      Your Answer: A thermistor is located at the tip of the pulmonary artery catheter

      Correct Answer: Cardiac output should be measured during the end-expiratory pause

      Explanation:

      Thermodilution is the most common dilution method used to measure cardiac output (CO) in a hospital setting.

      During the procedure, a Swan-Ganz catheter, which is a specialized catheter with a thermistor-tip, is inserted into the pulmonary artery via the peripheral vein. 5-10mL of a cold saline solution with a known temperature and volume is injected into the right atrium via a proximal catheter port. The solution is cooled as it mixes with the blood during its travel to the pulmonary artery. The temperature of the blood is the measured by the catheter and is profiled using a computer.

      The computer also uses the profile to measure cardiac output from the right ventricle, over several measurements until an average is selected.

      Cardiac output changes at each point of respiration, therefore to get an accurate measurement, the same point during respiration must be used at each procedure, this is usually the end of expiration, that is the end-expiratory pause.

    • This question is part of the following fields:

      • Clinical Measurement
      37.4
      Seconds
  • Question 4 - 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
      12.1
      Seconds
  • Question 5 - All of the following options describes a thermistor for the measurement of temperature...

    Correct

    • All of the following options describes a thermistor for the measurement of temperature except:

      Your Answer: Resistance of the bead increases exponentially as the temperature increases

      Explanation:

      There are different types of temperature measurement. These include:

      Thermistor – this is a type of semiconductor, meaning they have greater resistance than conducting materials, but lower resistance than insulating materials. There are small beads of semiconductor material (e.g. metal oxide) which are incorporated into a Wheatstone bridge circuit. As the temperature increases, the resistance of the bead decreases exponentially

      Thermocouple – Two different metals make up a thermocouple. Generally, in the form of two wires twisted, welded, or crimped together. Temperature is sensed by measuring the voltage. A potential difference is created that is proportional to the temperature at the junction (Seebeck effect)

      Platinum resistance thermometers (PTR) – uses platinum for determining the temperature. The principle used is that the resistance of platinum changes with the change of temperature. The thermometer measures the temperature over the range of 200°C to1200°C. Resistance in metals show a linear increase with temperature

      Tympanic thermometers – uses infrared radiation which is emitted by all living beings. It analyses the intensity and wavelength and then transduces the heat energy into a measurable electrical output

      Gauge/dial thermometers – Uses coils of different metals with different co-efficient of expansion. These either tighten or relax with changes in temperature, moving a lever on a calibrated dial.

    • This question is part of the following fields:

      • Clinical Measurement
      31.6
      Seconds
  • Question 6 - Systemic vascular resistance (multiplied by 80) to produce the units of dynes.s.cm-5 is...

    Incorrect

    • Systemic vascular resistance (multiplied by 80) to produce the units of dynes.s.cm-5 is represented by?

      Your Answer: Mean arterial pressure (MAP) - central venous pressure (CVP)/stroke volume (SV)

      Correct Answer: Mean arterial pressure (MAP) - central venous pressure (CVP)/cardiac output (CO)

      Explanation:

      Systemic vascular resistance (SVR) is a derived value based on:

      SVR = (MAP-CVP)/CO x 80

      = (60 -10)/5 x 80 = 800 dynes.s.cm-5

      A correction factor of 80 is needed in converting mmHg to dynes.s.cm-5
      Normal values is between 700 -1600 dynes.s.cm-5

      Pulmonary resistance (PVR) = (MPAP-PCWP)/CO x 80

      = (10 – 5)/5 x 80 = 80 dynes.s.cm-5

      To account for body size, cardiac index (CI) can be used instead of CO. CI = CO/body surface area (m2) or mL/minute/m2.
      N/B: either MAP and CVP, or MPAP and PCWP are used in calculation to get dynes.s.cm-5

    • This question is part of the following fields:

      • Clinical Measurement
      10.8
      Seconds
  • Question 7 - 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: 10 ml/cmH2O

      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
      14.8
      Seconds
  • Question 8 - A 50-year-old man is admitted in hospital. Over four hours, he produces 240...

    Incorrect

    • A 50-year-old man is admitted in hospital. Over four hours, he produces 240 mL of urine and has a plasma creatinine concentration is 10 mcg/mL. The normal concentration of creatinine in urine is 1.25 mg/mL.

      Calculate his approximate creatinine clearance.

      Your Answer: 24 ml/minute

      Correct Answer: 125 ml/minute

      Explanation:

      Creatinine clearance is a test used to approximate the glomerular filtration rate (GFR) as an assessment of kidney function.

      Creatinine is formed during the breakdown of dietary sources of meat and skeletal muscle. It is secreted at a consistent concentration and pace into the body’s circulation, and is easily filtered across the glomerulus without being reabsorbed or metabolized by the kidney.

      It is represented mathematically as:
      Creatinine clearance (CL) = U x V/P
      where,
      U: Urinary creatinine concentration (mg/mL)
      V: Volume of urine (mL/min)
      P: Plasma creatinine concentration (mg/mL)

      Therefore, in this case:
      CL: 1.25 x 1 = 125mL/min
      0.1

    • This question is part of the following fields:

      • Clinical Measurement
      49.1
      Seconds
  • Question 9 - The following statements are about capnography. Which of them is true? ...

    Incorrect

    • The following statements are about capnography. Which of them is true?

      Your Answer: Contains a quartz chamber

      Correct Answer: Collision broadening is due to presence of other polyatomic molecules

      Explanation:

      Capnography is the non-invasive measurement and pictorial display of inhaled and exhaled carbon dioxide (CO2) partial pressure.

      It is depicted graphically as the concentration of CO2 over time.

      It is used in disease diagnosis, determining disease severity, assessing response to treatment and is the best method to for indicating when an endotracheal tube is placed in the trachea after intubation.

      The wavelength of IR light usually absorbed by nitrous oxide is between 4.4-4.6?m (very close to that of CO2). Its absorption of wavelengths at 3.9 ?m is much weaker. It causes a measurable deficit of 0.1% for every 10% of nitrous oxide. The maximal wavelength of infrared (IR) light absorbed by carbon monoxide is 4.7 ?m. The volatile agents have strong absorption bands at 3.3 ?m and throughout the ranges 8-12 ?m.

      IR light is not absorbed by oxygen (O2), but O2 and CO2 molecules are constantly colliding which interrupts the absorption of IR light by CO2. This increases the band of absorption, that is the Collison or pressure broadening). An oxygen percentage of 95 will result in a 0.5 percentage fall in CO2 measure.

      IR light is also absorbed by water vapour which will result in an overlap of the absorption band, collision broadening and a dilution of partial pressure. This is why water trap and water permeable tubing is recommended for use as it reduces measurement inaccuracies.

      The use of multi-gas analysers of modern gases also help reduce the effects of collision broadening.

      Beer’s law is also applied in this system as an increase in the concentrations of CO2 causes a decrease in the amount of IR able to pass through the gas. This IR light is what generated the signal that is analysed for display.

      The capnograph can indicate oesophageal intubation, but cannot determine if it is endotracheal or endobronchial. For this, auscultation is used.

    • This question is part of the following fields:

      • Clinical Measurement
      25.8
      Seconds
  • Question 10 - Which of the following statements about intra-arterial blood pressure monitoring is true? ...

    Correct

    • Which of the following statements about intra-arterial blood pressure monitoring is true?

      Your Answer: Fluid-filled tubing conducts the intravascular pressure wave from the catheter tip to the transducer

      Explanation:

      Intra-arterial blood pressure monitoring is a common place procedure in the ICU. It is used to provide accurate beat-to-beat information using a pressure wave displayed on a monitor.

      It involves catheter insertion in a peripheral artery (most commonly the radial, brachial and dorsalis pedis arteries). Each subsequent contraction of cardiac muscles results in pressure wave which induces a mechanical motion of flow in the catheter. This mechanical motion is then passed on to a transducer through a rigid fluid-filled tubing. The transducer is the able to process this mechanical motion into electrical signals which are displayed as arterial waves and pressure represented numerically on the monitor.

      The transducer should be placed at the same level as the heart on the phlebostatic axis, and at the level of the atria (the 4th intercostal space, in the mid-axillary line).

      Air bubbles and catheter tubing with longer lengths result in wave dampening (rounding of the resulting pressure waves). This dampening causes a decrease in systolic pressure, and an increase in diastolic pressure.

    • This question is part of the following fields:

      • Clinical Measurement
      20.8
      Seconds
  • Question 11 - A 77-year-old man is admitted to hospital for colorectal surgery. He is scheduled...

    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: 10 METs

      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
      19.3
      Seconds
  • Question 12 - Which statement is the most accurate when describing electrical equipment and shock? ...

    Incorrect

    • Which statement is the most accurate when describing electrical equipment and shock?

      Your Answer: Class I has double insulation for all equipment

      Correct Answer: Type CF is considered to safe for direct connection with the heart

      Explanation:

      There are different classes of electrical equipment that can be classified in the table below:

      Class 1 – provides basic protection only. It must be connected to earth and insulated from the mains supply

      Class II – provides double insulation for all equipment. It does not require an earth.

      Class III – uses safety extra low voltage (SELV) which does not exceed 24 V AC. There is no risk of gross electrocution but risk of microshock exists.

      Type B – All of above with low leakage currents (0.5mA for Class IB, 0.1 mA for Class IIB)

      Type BF – Same as with other equipment but has ‘floating circuit’ which means that the equipment applied to patient is isolated from all its other parts.

      Type CF – Class I or II equipment with ‘floating circuits’ that is considered to be safe for direct connection with the heart. There are extremely low leakage currents (0.05mA for Class I CF and 0.01mA for Class II CF)

    • This question is part of the following fields:

      • Clinical Measurement
      10
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Clinical Measurement (3/12) 25%
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