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Question 1
Correct
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Which one of the following patients presenting for elective surgery has an American Society of Anaesthesiologists (ASA) preoperative physical status grading of III?
Your Answer: A 50-year old man with a BMI of 41 with a reduced exercise tolerance
Explanation:The ASA physical status classification system is a system for assessing the fitness of patients before surgery. It was last updated in October 2014.
ASA I A normal healthy patient
ASA II A patient with mild systemic disease
ASA III A patient with severe systemic disease
ASA IV A patient with severe systemic disease that is a constant threat to life
ASA V A moribund patient who is not expected to survive without the operation
ASA VI A declared brain-dead patient whose organs are being removed for donor purposesA 20-year old woman who is 39-weeks pregnant with no other medical conditions – ASA II
A 35-year-old man with a BMI of 29 with a good exercise tolerance who smokes-ASA II
A 50-year old man with a BMI of 41 with a reduced exercise tolerance -ASA III
A 65-year old woman with a BMI of 34 with treated hypertension with no functional limitations-ASA II
A 73-year old man who has had a TIA ten-weeks ago but has a good exercise tolerance and is a non-smoker-ASA IV
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This question is part of the following fields:
- Clinical Measurement
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Question 2
Incorrect
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A 77-year-old woman is scheduled for day case cataract surgery under local anaesthesia. She has no cardiac or respiratory problems. Lisinopril is being used to treat her hypertension, which is under control.
Which of the following preoperative investigations are the most appropriate for this patient?Your Answer: FBC, urea and electrolytes, ECG and chest x ray
Correct Answer: No investigations
Explanation:Because the patient has mild systemic disease, he is ASA 2 and the procedure will be performed under local anaesthesia.
The following factors should be considered when requesting preoperative investigations:
Indications derived from a preliminary clinical examination
Whether or not a general anaesthetic will be used, the possibility of asymptomatic abnormalities, and the scope of the surgery.No special investigations are needed if the patient has no history of significant systemic disease and no abnormal findings on examination during the nurse-led assessment.
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This question is part of the following fields:
- Clinical Measurement
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Question 3
Incorrect
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The statement that best describes temperature management is:
Your Answer:
Correct Answer: Gauge thermometers use coils of different metals with different co-efficients of expansion which either tighten or relax with changes in temperature
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.
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This question is part of the following fields:
- Clinical Measurement
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Question 4
Incorrect
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Arterial pressure waveforms give an indication of the operation of the heart and the patient's clinical state.
Which of the following listed characteristics of arterial waveforms is most indicative of myocardial contractility?Your Answer:
Correct Answer: Slope of the upstroke of the curve.
Explanation:Arterial pressure waveforms is an invasive form of monitoring cardiac parameters. It provides a lot of information on the performance of the heart from different sections, including:
Cardiac measurements:
Heart rate
Systolic pressure
Diastolic pressure
Mean arterial pressure
Pulse pressure
Change in pulse amplitude corresponding to respiratory changes
Slope of anacrotic limb associated with aortic stenosisFrom the shape of the arterial waveform displayed:
Slope of anacrotic limb represents aortic valve and LVOT flow
Indications of aortic stenosis (AS): Slurred wave, collapsing wave
Rapid systolic decline in LVOTO
Bisferiens wave in HOCM
Low dicrotic notch in states with poor peripheral resistance
Position and quality of dicrotic notch as a reflection of the damping coefficientFor this question, the upstroke slope of the pressure wave is indicative of myocardial contractility and is mathematically represented as:
dP/dt, which represents a change of pressure with regards to time.
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This question is part of the following fields:
- Clinical Measurement
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Question 5
Incorrect
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A 49-year-old woman is admitted to hospital. She is scheduled for surgery and is undergoing preoperative assessment.
As part of the preoperative assessment, her functional capacity is estimated. At 50kg in weight, she is able to cycle along a flat surface at a speed of 10-14 miles/hour (8 metabolic equivalents or METs).
Provide the best estimated value of oxygen consumption (VO2) for eight METs.Your Answer:
Correct Answer: 1400 mL/minute
Explanation:Oxygen consumption (VO2) refers to the optimal amount of oxygen used by the body during exercise.
It is calculated mathematically by:
VO2 = 3.5 x 50 x 8 = 1400 mL/kg/minute
where,
1 MET = 3.5 mL O2/kg/minute is utilized by the body.
Note:
1 MET Eating
Dressing
Use toilet
Walking slowly on level ground at 2-3 mph
2 METs Playing a musical instrument
Walking indoors around house
Light housework
4 METs Climbing a flight of stairs
Walking up hill
Running a short distance
Heavy housework, scrubbing floors, moving heavy furniture
Walking on level ground at 4 mph
Recreational activity, e.g. golf, bowling, dancing, tennis
6 METs Leisurely swimming
Leisurely cycling along the flat (8-10 mph)
8 METs Cycling along the flat (10-14 mph)
Basketball game
10 METs Moderate to hard swimming
Competitive football
Fast cycling (14-16 mph) -
This question is part of the following fields:
- Clinical Measurement
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Question 6
Incorrect
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Which of the following statements is false when considering the electrical safety of equipment?
Your Answer:
Correct Answer: The low leakage currents limit is 0.01mA for Class I CF
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)
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This question is part of the following fields:
- Clinical Measurement
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Question 7
Incorrect
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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:
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.
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This question is part of the following fields:
- Clinical Measurement
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Question 8
Incorrect
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Of the following statements, which is true about the measurements of cardiac output using thermodilution?
Your Answer:
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.
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This question is part of the following fields:
- Clinical Measurement
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Question 9
Incorrect
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With regards to devices for temperature management, all of these are used EXCEPT:
Your Answer:
Correct Answer: Thermistors use the resistance of a semiconductor bead which 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.
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This question is part of the following fields:
- Clinical Measurement
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Question 10
Incorrect
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The statement that best describes the classification of theatre equipment in terms of electrical safety is:
Your Answer:
Correct Answer: A floating circuit is equipment applied to patient that is isolated from all its other parts
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)
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This question is part of the following fields:
- Clinical Measurement
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Question 11
Incorrect
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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:
Correct 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%
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This question is part of the following fields:
- Clinical Measurement
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Question 12
Incorrect
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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:
Correct 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.
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This question is part of the following fields:
- Clinical Measurement
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Question 13
Incorrect
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A 64-year-old man is admitted to the critical care unit. He has a recent medical history of faecal peritonitis for which a laparotomy was performed. His vitals have been monitored using an invasive pulmonary artery flotation catheter.
His vital readings are:
Temperature: 38.1°C
Blood pressure: 79/51 mmHg (mean 58 mmHg)
Pulmonary artery pressure: 19/6 mmHg (mean 10 mmHg)
Pulmonary capillary occlusion pressure: 5 mmHg
Central venous pressure: 12 mmHg
Cardiac output: 5 L/min
Mixed venous oxygen saturation: 82%
Calculate his approximate pulmonary vascular resistance.
Note: A correction factor of 80 is require to convert mmHg to dynes·s·cm-5Your Answer:
Correct Answer: 80 dynes·s·cm-5
Explanation:Pulmonary vascular resistance (PVR) refers to the resistance to blood flow to the left atrium from the pulmonary artery.
It is derived mathematically by:PVR = MPAP – PCWP
CO
where,
MPAP: Mean pulmonary artery pressure
PCWP: Pulmonary capillary occlusion pressure
CO: Cardiac outputFor this patient:
PVR = 10 – 5 = 1mmHg
5Remember, multiply by correction factor 80 to change units:
PVR = 1mmHg x 80 = 80 dynes·s·cm-5
Normal values range between 20-130 dynes·s·cm-5
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This question is part of the following fields:
- Clinical Measurement
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Question 14
Incorrect
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Which of the following statements about intra-arterial blood pressure monitoring is true?
Your Answer:
Correct 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.
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This question is part of the following fields:
- Clinical Measurement
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Question 15
Incorrect
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Which of these statements is true about spirometry?
Your Answer:
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
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Question 16
Incorrect
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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.
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This question is part of the following fields:
- Clinical Measurement
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Question 17
Incorrect
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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:
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
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This question is part of the following fields:
- Clinical Measurement
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Question 18
Incorrect
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A patient has a myocardial infarction with anterior ST elevation. There are the following observations:
Cardiac output 2.0 L/min
Blood pressure 80/60 mmHg
CVP 20 mmHg
SpO2 91% on 4 L/min oxygen
What is the most logical physiological explanation for these findings?Your Answer:
Correct Answer: Biventricular failure
Explanation:The occlusion of the left anterior descending (LAD) coronary artery causes anterior ST elevation myocardial infarction (STEMI). It has the worst prognosis of all the infarct locations due to its larger infarct size. It has a higher rate of total mortality (27 percent versus 11 percent), heart failure (41 percent versus 15 percent), and a lower ejection fraction on admission than an inferior myocardial infarction (38 percent versus 55 percent ).
The LAD artery supplies the majority of the interventricular septum, as well as the anterior, lateral, and apical walls of the left ventricle, as well as the majority of the right and left bundle branches and the bicuspid valve’s anterior papillary muscle (left ventricle).
The left or right ventricle’s end-diastolic volume (EDV) is the volume of blood in each chamber at the end of diastole before systole. Preload is synonymous with the EDV.
120 mL is a typical left ventricular EDV (range 65-240 mL). The EDV of the right ventricle in a typical range is (100-160 mL).
With an ejection fraction (EF) of less than 45 percent, the patient is most likely suffering from systolic dysfunction. Increases in right and left ventricular end-diastolic pressures and volumes are likely with a reduced EF because the ventricles are not adequately emptied. The left atrium and the pulmonary vasculature are affected by the increased pressures on the left side of the heart.
By causing an imbalance of the Starling forces acting across the capillaries, increased hydrostatic pressure in the pulmonary circulation favours the development of pulmonary oedema. With cardiogenic pulmonary oedema, capillary permeability is likely to remain unchanged.
Biventricular failure will result as a result of the pressure changes being transmitted to the right side of the circulation. The patient’s systemic vascular resistance is likely to be elevated as well, but it is not the most likely cause of his symptoms. The patient is suffering from cardiogenic shock as a result of biventricular failure. The patient has low cardiac output and is hypotensive. Right ventricular filling pressures are elevated, indicating right ventricular dysfunction.
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This question is part of the following fields:
- Clinical Measurement
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Question 19
Incorrect
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Which of the following options will cause an artificial increase in pulse oximeter (SpO2) readings?
Your Answer:
Correct Answer: Heavy smoker
Explanation:A pulse oximeter is a piece of medical equipment used as a non-invasive method of measuring the oxygen saturation of blood.
It works by measuring the ratio of absorption of red and infrared light in a section of blood flow, as red light is largely absorbed by deoxygenated blood, and infrared light is largely absorbed by oxygenated blood.
Pulse oximetry relies on photoplethysmography (PPG) waveforms. The oximeter has 2 sides, with different functions. One side houses light-emitting diodes which are responsible for transmitting 2 light wavelengths, 660nm for red light and 940nm for near infrared light. The other side is a photodetector. The light emitted travels through the body and the amount that is not absorbed is measured by the photodetector.
Smokers often have increased levels of carboxy haemoglobin (COHb). This leads to artificial increases in pulse oximeter readings as it is unable to differentiate between COHb and oxyhaemoglobin (O2HB) as they both absorb red light at 660nm. Every 1% increase of circulating carboxyhaemoglobin, results in a correlative 1% increase in oximeter readings.
Prilocaine toxicity will cause an artificial decrease in oximeter readings. This is because prilocaine metabolites cause methemoglobinemia (MetHB), which are dysfunctional haemoglobins unable to properly transport oxygen. In this case, a laboratory multiwavelength co-oximeter is recommended for a more accurate reading.
Anaemia will not affect oximeter readings as long as haemoglobins in the blood are normal.
Sickle cell disease does not affect oximeter readings despite its ability to cause hypoxia and shift the oxygen dissociation curve to the right.
Brown-red fingernail polish will cause an underestimation of pulse oximeter readings.
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This question is part of the following fields:
- Clinical Measurement
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Question 20
Incorrect
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Which of the following is a correctly stated fundamental (base) SI unit?
Your Answer:
Correct Answer: A metre is the unit of length
Explanation:The international system of units, or system international d’unites (SI) is a collection of measurements derived from expanding the metric system.
There are seven base units, which are:
Metre (m): a unit of length
Second (s): a unit of time
Kilogram (kg): a unit of mass
Ampere (A): a unit of electrical current
Kelvin (K): a unit of thermodynamic temperature
Candela (cd): a unit of luminous intensity
Mole (mol): a unit of substance. -
This question is part of the following fields:
- Clinical Measurement
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Question 21
Incorrect
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Regarding sine wave damping, which one would approach equilibrium the fastest at zero amplitude, without overshoot?
Your Answer:
Correct Answer: Critical damping
Explanation:A damped sine wave is a smooth, periodic oscillation with an amplitude that approaches zero as time goes to infinity. In other words, the wave gets flatter as the x-values become larger.
Critical damping is defined as the threshold between overdamping and underdamping. In the case of critical damping, the oscillator returns to the equilibrium position as quickly as possible, without oscillating, and passes it once at most.
In overdamping, the system moves slowly towards the equilibrium. An underdamped system moves quickly to equilibrium, but will oscillate about the equilibrium point as it does so.
Optimal damping has a damping coefficient of around 0.64-0.7. It maximizes frequency response, minimizes overshoot of oscillations, and minimizes phase and amplitude distortion.
In an undamped system, the amplitude of the waves that are being generated remain unchanged and constant over time.
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This question is part of the following fields:
- Clinical Measurement
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Question 22
Incorrect
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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:
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 mmHgTherefore,
CvO2 = (1.34 x 10 x 70 x 0.01) + (0.003 x 50)
CvO2 = 9.38 + 0.15 = 9.53 mL/100mL
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This question is part of the following fields:
- Clinical Measurement
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Question 23
Incorrect
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A pulmonary function tests results are:
Measurement Predicted result Test result
Forced vital capacity (FVC) (btps) 3.85 2.36
Forced expiratory volume in 1 second (FEV1) (btps) 3.34 0.97
FEV1/FVC ratio % (btps) 85.1 39.9
Peak expiratory flow (PEF) (L/second) 7.33 2.11
Maximum voluntary ventilation (MVV) (L/minute) 116 44.4
What does this indicate?Your Answer:
Correct Answer: Moderate restrictive and severe obstructive picture
Explanation:Severity of a reduction in restrictive defect (%FVC) or obstructive defect (%FEV1/FVC) predicted are classified as follows:
Mild 70-80%
Moderate 60-69%
Moderately severe 50-59%
Severe 35-49%
Very severe <35% This patient has a mixed deficit with a severe obstructive deficit as %FEV1/FVC predicted is 46.9% and a moderate restrictive deficit as %FVC of predicted is 61.3 FEV1/FVC ratio 80% < predicted and VC < 80% = mixed picture. FEV1/FVC ratio 80% < predicted and VC > 80% = obstructive picture.FEV1/FVC ratio 80% > predicted and VC > 80% = normal picture.
FEV1/FVC ratio 80% > predicted and VC < 80% predicted= restrictive picture.
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This question is part of the following fields:
- Clinical Measurement
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Question 24
Incorrect
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A 76-year-old female, presents to her GP. She complains of fatigue and increased shortness of breath. On examination, she is noted to have pallor, an increased respiratory and heart rate. Her GP requests further diagnostic investigations, including a full blood count (FBC) which finds decreased MCV and MCHC.
What is the most likely cause of her symptoms?Your Answer:
Correct Answer: Iron deficiency
Explanation:The patient’s diagnosis is microcytic hypochromic anaemia which is often as a result of iron deficiency and thalassaemia.
Macrocytic anaemia is often caused by folate and B12 deficiencies and alcohol abuse.
Normocytic normochromic anaemia is often caused by acute blood loss, haemolytic anaemia, anaemia of chronic disease and leucoerythroblastic anaemias.
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This question is part of the following fields:
- Clinical Measurement
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Question 25
Incorrect
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Which of the following statements about the central venous pressure (CVP) waveform is true?
Your Answer:
Correct Answer: Third degree heart block causes canon A waves
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.
Canon waves: which refer to large waves present on the trace that do not correspond to the A, V or C waves. They usually occur in a background of complete heart blocks or junctional arrythmias.
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This question is part of the following fields:
- Clinical Measurement
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Question 26
Incorrect
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The following are results of some pulmonary function tests:
Measurement - Predicted result - Test result
Forced vital capacity (FVC) (btps) - 3.21 - 1.94
Forced expiratory volume in 1 second (FEV1) (btps) - 2.77 - 1.82
FEV1/FVC ratio % (btps) - 81.9 - 93.5
Peak expiratory flow (PEF) (L/second) - 6.55 - 3.62
Maximum voluntary ventilation (MVV) (L/minute) - 103 - 87.1
Which statement applies to the results?Your Answer:
Correct Answer: The patient has a moderate restrictive pulmonary defect
Explanation:Severity of a reduction in restrictive defect (%FVC) or obstructive defect (%FEV1/FVC) predicted are classified as follows:
Mild 70-80%
Moderate 60-69%
Moderately severe 50-59%
Severe 35-49%
Very severe <35% This patient has a %FVC predicted of 60.4% and this corresponds to a moderate restrictive deficit. %FEV1/FVC ratio is 93.5%. FEV1/FVC ratio 80% < predicted and VC < 80% = mixed picture. FEV1/FVC ratio 80% < predicted and VC > 80% = obstructive picture.FEV1/FVC ratio 80% > predicted and VC > 80% = normal picture.
FEV1/FVC ratio 80% > predicted and VC < 80% predicted= restrictive picture. The integrity of the alveolar-capillary barrier is measured by carbon monoxide transfer factor (TLCO) and carbon monoxide transfer coefficient (KCO). These values are seen to be reduced in emphysema, interstitial lung diseases and in pulmonary vascular pathology. However, the KCO (as % predicted) is high in extrapulmonary restriction (pleural, chest wall and respiratory neuromuscular disease), and in loss of lung units provided the structure of the lung remaining is normal. The KCO distinguishes extrapulmonary (high KCO) causes of ‘restriction’ from intrapulmonary causes (low KCO).
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This question is part of the following fields:
- Clinical Measurement
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Question 27
Incorrect
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Which measurements is the most accurate for predicting fluid responsiveness?
Your Answer:
Correct Answer: Change in stroke volume following passive straight leg raise
Explanation:The passive leg raising (PLR) manoeuvre is a method of altering left and right ventricular preload and it is done with real-time measurement of stroke volume. It is a simple, quick, relatively unbiased, and accurate bedside test to guide fluid management and avoid fluid overload.
Pulse pressure variation (PPV), Stroke volume variation (SVV), superior vena cava diameter variation (threshold 36%) and end-expiratory occlusion test are used for dynamic tests of fluid responsiveness.
PPV is derived peripherally from the arterial pressure waveform.
Stroke volume variation (SVV) can be derived peripherally through pulse contour analysis of the arterial waveform. PPV and SVV have a threshold of 12% but since they are not used in patients who have cardiac arrhythmias, are spontaneous breathing, and in ventilated patients with low lung compliance and tidal volumes, they are of limited value.
The tests of fluid responsiveness’ accuracy is determined by calculating the area under the receiver operating characteristic curve (UROC) obtained by plotting the sensitivity of the parameter in predicting fluid responsiveness vs. 1-specificity.
Under optimal conditions, the ability to determine the need for fluid is best with PPV>SVV>LVEDA>CVP.
Central venous pressure (CVP) is a static test of preload (not preload responsiveness) and a key determinant of cardiac function. The left ventricular end-diastolic area (LVEDA) a static test of fluid responsiveness, is derived using echocardiography
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This question is part of the following fields:
- Clinical Measurement
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Question 28
Incorrect
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If the speed of an ultrasound in soft tissue is 1540 meters per second, what is the estimated wavelength produced if the frequency of a generated ultrasound is 10 megahertz?
Your Answer:
Correct Answer: 0.15 millimetre
Explanation:Wavelength can be computed as follows:
Wavelength = velocity/frequency
In the given problem, the values stated are:
Frequency = 10 x 10^6
Velocity = 1540 meters per secondWavelength = 1540/(10×10^6)
Wavelength = 1540/10,000,000 meters
Wavelength = 0.15 millimetres -
This question is part of the following fields:
- Clinical Measurement
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Question 29
Incorrect
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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:
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.
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This question is part of the following fields:
- Clinical Measurement
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Question 30
Incorrect
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Which statement is the most accurate when describing electrical equipment and shock?
Your Answer:
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)
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This question is part of the following fields:
- Clinical Measurement
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