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Question 1
Incorrect
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Which of the following statements about the central venous pressure (CVP) waveform is true?
Your Answer: Tricuspid regurgitation causes large A waves
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 2
Incorrect
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An arterial pressure transducer is supposedly in direct correlation to change, thus it is dependent on zero gradient drift and zero offset. Which of the following values will best compensate for the gradient drift?
Your Answer:
Correct Answer: 0 mmHg and 200 mmHg
Explanation:Since an arterial pressure transducer, and every other measuring apparatus, is prone to errors due to offset and gradient drifts, regular calibration is required to maintain accuracy of the instrument. The two-point calibration pressure values of 0 mmHg and 200 mmHg are within the physiologic range and can best compensate for the gradient drift.
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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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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 4
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 5
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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Question 6
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 7
Incorrect
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A 32-year-old male is admitted to the critical care unit. He has suffered a heroin overdose and requires intubation and ventilatory support.
What would be his predicted total static compliance (lung and chest wall) measurements.Your Answer:
Correct Answer: 100 ml/cmH2O
Explanation:Static lung compliance refers to the change in volume within the lung per given change in unit pressure. It is usually measured when air flow is absent, such as during pauses in inhalation and exhalation.
It is a combination of:
Chest wall compliance: normal value is 200 mL/cmH2O
Lung tissue compliance: normal value is 200 mL/ cmH2OIt is represented mathematically as:
1/Crs = 1/Cl + 1/Ccw
Where,
Crs = total compliance of the respiratory system
Cl = compliance of the lung
Ccw = compliance of the chest wallTherefore in this case:
1/Crs = 1/200 + 1/200
1/Crs = 0.005 + 0.005 = 0.01
1/Ct = 0.01
Rearranging equation gives:
Ct = 1/0.01 = 100 mL/cmH2O.
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This question is part of the following fields:
- Clinical Measurement
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Question 8
Incorrect
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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:
Correct 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 9
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 10
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 11
Incorrect
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A 50-year old man for septoplasty has a BMI of 32kg/m2 and neck circumference of 44 cm. He is troubled by intermittent nasal obstruction which causes his partner sleeps in a separate bedroom because of snoring. He currently on ramipril for hypertension.
Which of the following best approximates to his STOP-BANG score from the information above?Your Answer:
Correct Answer: 5
Explanation:The STOP-BANG questionnaire is used to screen patients for obstructive sleep apnoea (OSA).
The scoring system assigns one point for each feature.
S: Snoring (louder than talking or loud enough to be heard through closed doors)
T: Feeling tired, fatigued, or sleepy during daytime
O: Observed apnoeas during sleep
P: Hypertension
B: BMI more than 35 kg/m2
A: Age 50-years of age or greater
N: Neck circumference (male 17 inches / 43cm or greater and female 16 inches / 41 or greater)
G: Gender: MaleOur patient has a score of 5 ( O, P, A, N, G)
The score helps clinicians stratify patients for unrecognized OSA and target appropriate clinical management. It can also help triage patients for further investigation. A STOP-BANG score of 5-8 will identify patients with high probability of moderate to severe OSA in the surgical population.
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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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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 13
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 14
Incorrect
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A 27-year-old woman arrives at the emergency room after intentionally ingesting 2 g of amitriptyline.
A Glasgow coma score of 6 was discovered, as well as a pulse rate of 140 beats per minute and a blood pressure of 80/50 mmHg.
Which of the following ECG changes is most likely to indicate the onset of life-threatening arrhythmias?Your Answer:
Correct Answer: Prolongation of the QRS complex
Explanation:Arrhythmias and/or hypotension are the most common causes of death from tricyclic antidepressant (TCA) overdose.
The quinidine-like actions of tricyclic antidepressants on cardiac tissues are primarily responsible for their toxicity. Conduction through the His-Purkinje system and the myocardium slows as phase 0 depolarisation of the action potential slows. QRS prolongation and atrioventricular block are caused by slowed impulse conduction, which also contributes to ventricular arrhythmias and hypotension.
Arrhythmias can also be caused by abnormal repolarization, impaired automaticity, cholinergic blockade, and inhibition of neuronal catecholamine uptake, among other things.
Acidaemia, hypotension, and hyperthermia can all exacerbate toxicity.
The anticholinergic effects of tricyclic antidepressants, as well as the blockade of neuronal catecholamine reuptake, cause sinus tachycardia. Sinus tachycardia is usually well tolerated and does not require treatment. It can be difficult to tell the difference between sinus tachycardia and ventricular tachycardia with QRS prolongation.
A QRS duration of more than 100 milliseconds indicates a higher risk of arrhythmia and should be treated with systemic sodium bicarbonate.
The tricyclic is dissociated from myocardial sodium channels by serum alkalinization, and the extracellular sodium load improves sodium channel function.
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This question is part of the following fields:
- Clinical Measurement
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Question 15
Incorrect
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The following haemodynamic data is available from a patient with pulmonary artery catheter inserted:
Pulse rate - 100 beats per minute
Blood pressure - 120/70mmHg
Mean central venous pressure (MCVP) - 10mmHg
Right ventricular pressure (RVP) - 30/4 mmHg
Mean pulmonary artery wedge pressure (MPAWP) - 12mmHg
Which value best approximates the patient's coronary perfusion pressure?Your Answer:
Correct Answer: 58mmHg
Explanation:Coronary perfusion pressure(CPP), the difference between aortic diastolic pressure (Pdiastolic) and the left ventricular end-diastolic pressure (LVEDP), is mainly determined by the formula:
CPP = Pdiastolic -LVEDP
where
Pdiastolic is the lowest pressure in the aorta before left ventricular ejection and
LVEDP is measured directly during a cardiac catheterisation or indirectly using a pulmonary artery catheter. The pulmonary artery occlusion or wedge pressure approximates best with LVEDP.Using this patient’s haemodynamic data:
CPP = Pdiastolic – MPAWP
COO = 70 – 12 = 58mmHg -
This question is part of the following fields:
- Clinical Measurement
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Question 16
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 17
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 18
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 19
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 20
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 21
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 22
Incorrect
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Which of the following statements about closing capacity is true?
Your Answer:
Correct Answer: It is less than the functional residual capacity in a 30-year-old
Explanation:Closing capacity refers to volume of gas within the lungs at which the conducting small airways begin to close, that is, the point during expiration when small airways close.
It is calculated mathematically as:
Closing capacity = Closing volume (CV) + Residual volume (RV)
Functional residual capacity (FRC) is the volume of gas still present within the lungs post expiration.
Closing capacity is lower than the functional residual capacity in younger adults, but begins to rise to eventually equal, and then exceed it with increasing age (at about middle age), increasing intrabdominal pressure, decreasing blood flow in the pulmonary system and parenchymal disease within the pulmonary system.
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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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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 24
Incorrect
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A 20-year-old female presents to the emergency department. She complains of increased shortness of breath and wheezing over the last 48 hours. On examination, she is found to have tachycardia, tachypnoea, and oxygen saturation at 91% on air. She admits to a previous medical history of asthma, diagnosed 4 years ago. She requires further investigations for diagnosis.
Which of the following is true about the assessment of a patient with symptomatic asthma?Your Answer:
Correct Answer: Oxygen saturations of 91% on air would be an indication for performing arterial blood gases
Explanation:A patient presenting with symptomatic asthma should be assessed for severity to determine appropriate management options. Indications of acute severe asthma are:
Peak expiratory flow rate (PEFR): 33-50% best/predicted
Respiratory rate: ≥25/min
Heart rate: ≥110/min
Inability to finish a complete sentence in a single breath.Oxygen saturation should be measured. Any measurement of an oxygen saturation of 92% or less, either on air or on oxygen, indicates severe, life threatening asthma, and requires an arterial blood gas (ABG) to detect normo- or hypercarbia.
A chest x-ray would not be routine as it will not provide any relevant information. It is only required in specific cases, including:
Diagnosis of a subcutaneous emphysema
Indications of a unilateral pneumothorax
Indications of a lobar collapse of consolidation
Treatment-resistance life-threatening asthma
If mechanical ventilation is indicatedA peak expiratory flow rate (PEFR) can provide relevant information to help distinguish between acute, moderate, severe and life threatening asthma. However, it is not necessary as other parameters exist that can also help make the same distinction.
An ECG is indicated in this case as the patient has tachycardia and tachypnoea which are indicative of acute severe asthma. The ECG would indicate if arrhythmia is also present which would suggest life-threatening asthma.
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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 can be measured directly using spirometry?
Your Answer:
Correct Answer: Vital capacity
Explanation:Spirometry measures the total volume of air that can be forced out in one maximum breath, that is the total lung capacity (TLC), to maximal expiration, that is the residual volume (RV).
It is conducted using a spirometer which is capable of measuring lung volumes using techniques of dilution.
During spirometry, the following measurements can be determined:
Forced vital capacity (FVC)/vital capacity (VC): The maximum volume of air exhaled in one single forced breathe.
Forced expiratory volume in one second (FEV1)
FEV1/FVC ratio
Peak expiratory flow (PEF): the maximum amount of air flow exhaled in one blow.
Forced expiratory flow (mid expiratory flow): the flow at 25%, 50% and 75% of FVC
Inspiratory vital capacity (IVC): The maximum volume of air inhaled after a full total expiration.Anatomical dead space is measured using a single breath nitrogen washout called the Fowler’s method.
Residual volume and total lung capacity are both measured using the body plethysmograph or helium dilution
The functional residual capacity is usually measured using a nitrogen washout or the helium dilution technique.
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This question is part of the following fields:
- Clinical Measurement
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Question 26
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 27
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 28
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:
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 29
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 30
Incorrect
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The diaphragm is a muscle that is relatively resistant to non-depolarizing neuromuscular blockade's effects.
When these muscle relaxants are used, which of the following peripheral nerve stimulator twitch patterns is best for monitoring the return of diaphragmatic function?Your Answer:
Correct Answer: Post-tetanic count stimulation
Explanation:Certain skeletal muscles are more resistant to the effects of neuromuscular blocking agents, both non-depolarizing and depolarizing. The diaphragm is the most resistant. The muscles of the larynx and the corrugator supercilii are less resistant. The abdominal, orbicularis oris, and limb peripheral muscles are the most sensitive muscles.
Twitch stimulation patterns:
Supramaximal single stimulus:
The frequency ranges from 1 Hz to 0.1 Hz (one every second to one every 10 seconds)
The response is proportional to the frequency of the event.
It has limited clinical utility because it only tells you whether or not a patient is paralysed (no information on degree of paralysis).Over the course of 0.5 seconds (2 Hz), four supramaximal stimulate were applied:
It is possible to see ‘fade’ and use it as a basis for evaluation.
This stimulation pattern is used to determine the degree of blockade (1-2 twitches is appropriate for abdominal surgery)
If the train of four (TOF) count is 1-2, reversal agents can be used in conjunction with medium-acting neuromuscular blocking agents.Ratio of TOF:
This is the ratio of the 4th twitch amplitude to the 1st twitch amplitude.
The ratio decreases with non-depolarising block and is inversely proportional to the degree of block, allowing objective measurement of residual neuromuscular blockade.
To achieve adequate reversal, the ratio (as measured by accelerography) must be between 0.7 and 0.9.Count of twitches after a tetanic experience(PTC):
50 Hz for 5 seconds, then a 3 second pause, followed by a single 1 Hz twitch stimulus.
When the TOF count is zero, this stimulation pattern is used to assess deep blockade (that is, in neurosurgery, microsurgery or ophthalmic surgery when even small movements of a patient will disturb the surgical field)
It gives an estimate of how long it will take for the response to return to single twitches, allowing assessment of blocks that are too deep for any other technique.
A palpable post-tetanic count (PTC) of 2 indicates no twitch response for about 20-30 minutes, and a PTC of 5 indicates no twitch response for about 10-15 minutes.This is without a doubt the best way to keep track of paralysis in patients who need to avoid diaphragmatic movement. It’s best to use drug infusions and aim for a PTC of 2. After a tetanic stimulus, acetylcholine is mobilised, causing post-tetanic potentiation.
Stimulation in Two Bursts:
750 milliseconds between two short bursts of 50 Hz
This stimulation pattern is used to assess small amounts of residual blockade manually (tactile). -
This question is part of the following fields:
- Clinical Measurement
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