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Question 1
Incorrect
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Regarding oxygen consumption, which of these organs has the highest consumption at rest?
Your Answer: Hepatoportal system
Correct Answer: Kidney
Explanation:Oxygen delivery is related to blood flow as most of the oxygen binds to haemoglobin in red blood cells, although a small amount is dissolved in the plasma. Blood flow per 100 g of tissue is greatest in the kidneys.
The following is the oxygen consumption rate of different organs in ml/minute/100g
Hepatoportal = 2.2
Kidney = 6.8
Brain = 3.7
Skin = 0.38
Skeletal muscle = 0.18
Heart = 11 -
This question is part of the following fields:
- Pathophysiology
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Question 2
Incorrect
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What is the number of valves between the superior vena cava and the right atrium?
Your Answer: One
Correct Answer: None
Explanation:The inflow of blood from the superior vena cava is directed towards the right atrioventricular orifice. It returns deoxygenated blood from all structures superior to the diaphragm, except the lungs and heart.
There are no valves in the superior vena cava which is why it is relatively easy to insert a CVP line from the internal jugular vein into the right atrium. The brachiocephalic vein is similar as it also has no valves.
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This question is part of the following fields:
- Anatomy
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Question 3
Correct
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Among the different classes of anti-arrhythmics, which one is the first line treatment for narrow complex AV nodal re-entry tachycardia?
Your Answer: Adenosine
Explanation:Adenosine is the first line for AV nodal re-entry tachycardia. An initial dose of 6 mg is given, and a consequent second dose or third dose of 12 mg is administered if the initial dose fails to terminate the arrhythmia.
Aside from Adenosine, a vagal manoeuvre (e.g. carotid massage) is done to help terminate the supraventricular arrhythmia.
Amiodarone is not a first-line drug for supraventricular tachycardias. Digoxin and Propranolol can be considered if the arrhythmia is of a narrow complex irregular type. Verapamil is an alternative to Adenosine if the latter is contraindicated.
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This question is part of the following fields:
- Pharmacology
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Question 4
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 5
Incorrect
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A survey aimed at finding out mean glucose level in individuals that took antipsychotics medicines was conducted. The results were as follows:
Mean Value: 7mmol/L
Standard Deviation: 6mmol/L
Sample Size: 9
Standard Error: 2mmol/L
For a confidence interval of 95%, which of the option presents the correct range up to the nearest value?Your Answer: 2-12 mmol/L
Correct Answer: 3-11 mmol/L
Explanation:Key Point: While finding out confidence intervals, standard errors are used. Standard error and Standard deviation are two distinct entities and should not be confused.
For 99.7% confidence interval, you can find the range as follows:
Multiply the standard error by 3.
Subtract the answer from mean value to get the lower limit.
Add the answer obtained in step 1 from the mean value to get the upper limit.
The range turns out to be 1-13 mmol/L.
For a confidence interval of 68%, multiply the standard error with 1 and repeat the process. The range found for this interval is 3-11 mmol/L.
For a 95% confidence interval. Standard Error is multiplied by 1.96 which gives us the limit ranging from 3.08 to 10.92 mmol/L which could be approximated to 3-11 mmol/L.
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This question is part of the following fields:
- Statistical Methods
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Question 6
Incorrect
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One of the causes of increased pulse pressure is when the aorta becomes less compliant because of age-related changes. Another cause of increased pulse pressure is which of the following?
Your Answer: Aortic stenosis
Correct Answer: Increased stroke volume
Explanation:Impaired ventricular relaxation reduces diastolic filling and therefore preload.
Decreased blood volume decreases preload due to reduced venous return.
Heart failure is characterized by reduced ejection fraction and therefore stroke volume.
Cardiac output = stroke volume x heart rate
Left ventricular ejection fraction = (stroke volume / end diastolic LV volume ) x 100%
Stroke volume = end diastolic LV volume – end systolic LV volume
Pulse pressure (is increased by stroke volume) = Systolic Pressure – Diastolic Pressure
Systemic vascular resistance = mean arterial pressure / cardiac output
Factors that increase pulse pressure include:
-a less compliant aorta (this tends to occur with advancing age)
-increased stroke volume
Aortic stenosis would decrease stroke volume as end systolic volume would increase.
This is because of an increase in afterload, an increase in resistance that the heart must pump against due to a hard stenotic valve. -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 7
Incorrect
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The lung volume that is commonly measured indirectly is?
Your Answer: Forced expiratory volume in 1 second
Correct Answer: Functional residual capacity
Explanation:The functional residual capacity (FRC) is the volume in the lungs at the end of passive expiration. It is determined by opposing forces of the expanding chest wall and the elastic recoil of the lung. A normal FRC = 1.7 to 3.5 L. It a marker for lung function, and, during this time, the alveolar pressure is equal to the atmospheric pressure.
FRC cannot be measured by spirometry because it contains the residual volume.
Tidal volume, inspiratory reserve volume, forced expiratory volume in 1 second, and vital capacity can be measured directly.
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This question is part of the following fields:
- Pathophysiology
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Question 8
Incorrect
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Of the following, which is NOT a branch of the abdominal aorta?
Your Answer: Renal artery
Correct Answer: Superior phrenic artery
Explanation:The abdominal aorta begins at the level of the body of T12 near the midline, as a continuation of the thoracic aorta. It descends and bifurcates at the level of L4 into the common iliac arteries.
The branches of the abdominal aorta (with their vertebra level) are:
1. Inferior phrenic arteries: T12 (upper border)
2. Coeliac artery: T12
3. Superior mesenteric artery: L1
4. Middle suprarenal arteries: L1
5. Renal arteries: Between L1 and L2
6. Gonadal arteries: L2 (in males, it is the testicular artery, and in females, the ovarian artery)
7. Inferior mesenteric artery: L3
8. Median sacral artery: L4
9. Lumbar arteries: Between L1 and L4The superior phrenic artery branches from the thoracic aorta.
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This question is part of the following fields:
- Anatomy
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Question 9
Incorrect
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With regards to the repolarisation phase of the myocardial action potential, which of the following is responsible?
Your Answer: Slow efflux of calcium
Correct Answer: Efflux of potassium
Explanation:Cardiac conduction
Phase 0 – Rapid depolarization. Opening of fast sodium channels with large influx of sodium
Phase 1 – Rapid partial depolarization. Opening of potassium channels and efflux of potassium ions. Sodium channels close and influx of sodium ions stop
Phase 2 – Plateau phase with large influx of calcium ions. Offsets action of potassium channels. The absolute refractory period
Phase 3 – Repolarization due to potassium efflux after calcium channels close. Relative refractory period
Phase 4 – Repolarization continues as sodium/potassium pump restores the ionic gradient by pumping out 3 sodium ions in exchange for 2 potassium ions coming into the cell. Relative refractory period
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 10
Correct
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All of the following statements about calcium channel antagonists are incorrect except:
Your Answer: May cause potentiation of muscle relaxants
Explanation:Calcium channel blocker (CCB) blocks L-type of voltage-gated calcium channels present in blood vessels and the heart. By inhibiting the calcium channels, these agents decrease the frequency of opening of calcium channels activity of the heart, decrease heart rate, AV conduction, and contractility.
Three groups of CCBs include
1) Phenylalkylamines: Verapamil, Norverapamil
2) Benzothiazepines : Diltiazem
3) Dihydropyridine : Nifedipine, Nicardipine, Nimodipine, Nislodipine, Nitrendipine, Isradipine, Lacidipine, Felodipine and Amlodipine.Even though verapamil as good absorption from GIT, its oral bioavailability is low due to high first-pass metabolism.
Nimodipine is a Cerebro-selective CCB, used to reverse the compensatory vasoconstriction after sub-arachnoid haemorrhage and is more lipid soluble analogue of nifedipine
Calcium channel antagonist can potentiate the effect of non-depolarising muscle relaxants.
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This question is part of the following fields:
- Pharmacology
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Question 11
Correct
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The average diastolic blood pressure of a control group was found out to be 80 with a standard deviation of 5 in a study aimed at exploring the efficiency of a novel anti-hypertensive drug. The trial was randomised.
Making an assumption that the data is normally distributed, find out the number of patients that had diastolic blood pressure over 90.Your Answer: 3%
Explanation:Since the data is normally distributed, 95% of the values lie with in the interval 70 to 90. This can be calculated as follows:
Interval= Mean ± ( 2 times standard deviation)
= 80 ± 2(5)
= 80 ± 10
= 70 & 90The rest of the 5% are distributed symmetrically beyond 90 and below 70 which means 2.5% of the values lie above 90.
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This question is part of the following fields:
- Statistical Methods
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Question 12
Correct
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Anaesthetic gas concentrations can be measured using a refractometer. The main principal which allows it to be used for this purpose is which of the following?
Your Answer: Refraction
Explanation:Refractometers measure the degree to which the light changes direction, called the angle of refraction. A refractometer takes the refraction angles and correlates them to refractive index (nD) values that have been established. Using these values, you can determine the concentrations of solutions.
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This question is part of the following fields:
- Basic Physics
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Question 13
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: 45%
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 14
Incorrect
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The external urethral sphincter arises from which nerve root?
Your Answer: S1, S2, S3
Correct Answer: S2, S3, S4
Explanation:The external urethral sphincter functions to provide voluntary control of urine flow from the bladder to the urethra.
It receives its innervation from the branches of the pudendal nerve which originate from S2, S3 and S4.
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This question is part of the following fields:
- Anatomy
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Question 15
Correct
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The parameter that is indirectly measured from a blood gas analysis is?
Your Answer: Standard bicarbonate
Explanation:Automated blood gas analysers are commonly used to analyse blood gas samples, and they measure specific components of the arterial blood gas sample, whether directly or indirectly.
The following are the components of arterial blood gas:
pH = measured (directly determined) acid-base balance of the blood
PaO2 = measured partial pressure of oxygen in arterial blood
PaCO2 = measured partial pressure of carbon dioxide in arterial blood
HCO3 = calculated (indirectly determined) concentration of bicarbonate in arterial blood
Base excess/deficit = calculated relative excess or deficit of base in arterial blood
SaO2 = calculated arterial oxygen saturation unless a co-oximetry is obtained, in which case it is measured
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This question is part of the following fields:
- Pathophysiology
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Question 16
Incorrect
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Which of the following statements is true regarding the relation to the liver?
Your Answer: The liver is completely covered by peritoneum
Correct Answer: The caudate lobe is superior to the porta hepatis
Explanation:Ligamentum venosum is an anterior relation of the liver: The ligamentum venosum, the fibrous remnant of the ductus venosus of the fetal circulation, lies posterior to the liver. It lies in the fossa for ductus venosus that separates the caudate lobe and the left lobe of the liver.
The portal triad contains three important tubes: 1. Proper hepatic artery 2. Hepatic portal vein 3. Bile ductules It also contains lymphatic vessels and a branch of the vagus nerve.
The bare area of the liver is a large triangular area that is devoid of any peritoneal covering. The bare area is attached directly to the diaphragm by loose connective tissue. This nonperitoneal area is created by a wide separation between the coronary ligaments.
The porta hepatis is a fissure in the inferior surface of the liver. All the neurovascular structures (except the hepatic veins) and hepatic ducts enter or leave the liver via the porta hepatis. It contains the sympathetic branch to the liver and gallbladder and the parasympathetic, hepatic branch of the vagus nerve. The caudate lobe (segment I) lies in the lesser sac on the inferior surface of the liver between the IVC on the right, the ligamentum venosum on the left, and the porta hepatis in front
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This question is part of the following fields:
- Anatomy
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Question 17
Incorrect
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A 58-year-old man is being operated on for a radical gastrectomy for carcinoma of the stomach.
Which structure needs to be divided to gain access to the coeliac axis?Your Answer: Falciform ligament
Correct Answer: Lesser omentum
Explanation:The lesser omentum will need to be divided. This forms one of the nodal stations that will need to be taken during a radical gastrectomy.
The celiac axis is the first branch of the abdominal aorta and supplies the entire foregut (mouth to the major duodenal papilla). It arises at the level of vertebra T12. It has three major branches:
1. Left gastric
2. Common hepatic
3. Splenic arteries -
This question is part of the following fields:
- Anatomy
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Question 18
Correct
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What is NOT a feature of Propofol infusion syndrome?
Your Answer: Hypotriglyceridaemia
Explanation:Propofol infusion syndrome is a rare but extremely dangerous complication of propofol administration
Common organ systems affected by PRIS include the following:
1. cardiovascular
widening of QRS complex, Brugada syndrome-like patterns (particularly type 1), ventricular tachyarrhythmias, cardiogenic shock, and asystole2. hepatic
Liver enzymes elevation, hepatomegaly, and steatosis3. skeletal muscular
myopathy and overt rhabdomyolysis4. renal
Hyperkalaemia, acute kidney injury5. metabolic
High anion gap metabolic acidosis (due to elevation in lactic acid) -
This question is part of the following fields:
- Anatomy
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Question 19
Incorrect
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A 33-year-old woman known to be hypothyroid and taking 150 mcg l-thyroxine daily is reviewed in the preoperative assessment clinic prior to a laparoscopic cholecystectomy.
She has required three increases in her thyroid replacement therapy in the last six months.
Her thyroid function tests are as follows:
TSH 11 (normal range 0.4-4mU/L)
T3 20 (normal range 9-25mU/L)
T4 6.2 (normal range 3.5-7.8mU/L)
What will explain this biochemical picture?Your Answer: Tissue level unresponsiveness to thyroid hormone
Correct Answer: Poor compliance with medication
Explanation:In patients with an intact hypothalamic-pituitary axis, serial TSH measurements are used to determine the adequacy of treatment with thyroid hormones . changes in TSH levels becoming apparent after approximately eight weeks of therapy with thyroid hormone replacement. Change in T3/T4 levels are seen before changes in TSH .
In patients taking thyroid replacement therapy, the most frequent reason for persistent elevation of serum TSH is poor compliance. Patients who do not regularly take their L-thyroxine try and catch up just before a visit to a clinician for blood test.
Tissue-level unresponsiveness to thyroid hormone is caused by mutation in the gene controlling a receptor for T3 and is rare.
Reduced responsiveness of target tissues to thyroid hormone aka resistance to thyroid hormones (rTH) occurs when there is a mutation in the thyroid hormone receptor ? gene. It is a rare autosomal dominant inherited syndrome of reduced end-organ responsiveness to thyroid hormone and has two types:
Generalised resistance (GrTH)
Pituitary resistance (PrTH)Patients with rTH have normal or slightly elevated serum thyroid stimulating hormone (TSH) level, elevated serum free thyroxine (FT4) and free triiodothyronine (FT3) concentrations.
Drugs that increase metabolism of thyroxine include:
Warfarin
Rifampin
Phenytoin
Phenobarbital
St John’s Wort
CarbamazepineThese drugs lower circulating thyroid hormones and would be associated with a raised TSH but low T3/T4.
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This question is part of the following fields:
- Pathophysiology
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Question 20
Incorrect
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A 63-year old man has palpitations and goes to the emergency room. An ECG shows tall tented T waves, which corresponds to phase 3 of the cardiac action potential.
The shape of the T wave is as a result of which of the following?Your Answer: Repolarisation due to efflux of calcium
Correct Answer: Repolarisation due to efflux of potassium
Explanation:Cardiac conduction
Phase 0 – Rapid depolarization. Opening of fast sodium channels with large influx of sodium
Phase 1 – Rapid partial depolarization. Opening of potassium channels and efflux of potassium ions. Sodium channels close and influx of sodium ions stop
Phase 2 – Plateau phase with large influx of calcium ions. Offsets action of potassium channels. The absolute refractory period
Phase 3 – Repolarization due to potassium efflux after calcium channels close. Relative refractory period
Phase 4 – Repolarization continues as sodium/potassium pump restores the ionic gradient by pumping out 3 sodium ions in exchange for 2 potassium ions coming into the cell. Relative refractory period
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 21
Incorrect
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Intracellular effectors are activated by receptors on the cell surface. These receptors receive signals that are relayed by second messenger systems.
In the human body, which second messenger is most abundant?Your Answer: Inositol triphosphate
Correct Answer: Calcium ions
Explanation:Second messengers relay signals to target molecules in the cytoplasm or nucleus when an agonist interacts with a receptor on the cell surface. They also amplify the strength of the signal. The most ubiquitous and abundant second messenger is calcium and it regulates multiple cellular functions in the body.
These include:
Muscle contraction (skeletal, smooth and cardiac)
Exocytosis (neurotransmitter release at synapses and insulin secretion)
Apoptosis
Cell adhesion to the extracellular matrix
Lymphocyte activation
Biochemical changes mediated by protein kinase C.cAMP is either inhibited or stimulated by G proteins.
The receptors in the body that stimulate G proteins and increase cAMP include:
Beta (?1, ?2, and ?3)
Dopamine (D1 and D5)
Histamine (H2)
Glucagon
Vasopressin (V2).The second messenger for the action of nitric oxide (NO) and atrial natriuretic peptide (ANP) is cGMP.
The second messengers for angiotensin and thyroid stimulating hormone are inositol triphosphate (IP3) and diacylglycerol (DAG).
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This question is part of the following fields:
- Physiology
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Question 22
Incorrect
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Which of the following causes the right-sided shift of the oxygen haemoglobin dissociation curve?
Your Answer: Decreased 2,3-DPG in transfused red cells
Correct Answer: Chronic iron deficiency anaemia
Explanation:With respect to oxygen transport in cells, almost all oxygen is transported within erythrocytes. There is limited solubility and only 1% is carried as solution. Thus, the amount of oxygen transported depends upon haemoglobin concentration and its degree of saturation.
Haemoglobin is a globular protein composed of 4 subunits. Haem is made up of a protoporphyrin ring surrounding an iron atom in its ferrous state. The iron can form two additional bonds – one is with oxygen and the other with a polypeptide chain.
There are two alpha and two beta subunits to this polypeptide chain in an adult and together these form globin. Globin cannot bind oxygen but can bind to CO2 and hydrogen ions.
The beta chains are able to bind to 2,3 diphosphoglycerate. The oxygenation of haemoglobin is a reversible reaction. The molecular shape of haemoglobin is such that binding of one oxygen molecule facilitates the binding of subsequent molecules.The oxygen dissociation curve (ODC) describes the relationship between the percentage of saturated haemoglobin and partial pressure of oxygen in the blood.
Of note, it is not affected by haemoglobin concentration.Chronic anaemia causes 2, 3 DPG levels to increase, hence shifting the curve to the right
Haldane effect – Causes the ODC to shift to the left. For a given oxygen tension there is increased saturation of Hb with oxygen i.e. Decreased oxygen delivery to tissues.
This can be caused by:
-HbF, methaemoglobin, carboxyhaemoglobin
-low [H+] (alkali)
-low pCO2
-ow 2,3-DPG
-ow temperatureBohr effect – causes the ODC to shifts to the right = for given oxygen tension there is reduced saturation of Hb with oxygen i.e. Enhanced oxygen delivery to tissues. This can be caused by:
– raised [H+] (acidic)
– raised pCO2
-raised 2,3-DPG
-raised temperature -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 23
Incorrect
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A 50-year-old female is having her central venous pressure (CVP) measured. A long femoral line was inserted that passes from the common iliac vein into the inferior vena cava.
At which level of vertebra does this occur?Your Answer: S1
Correct Answer: L5
Explanation:The inferior vena cava is formed by the union of the right and left common iliac veins. This occurs at the L5 vertebral level. The IVC courses along the right anterolateral side of the vertebral column and ascends through the central tendon of the diaphragm at the T8 vertebral level.
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This question is part of the following fields:
- Anatomy
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Question 24
Incorrect
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The fluids with the highest osmolarity is?
Your Answer: Hartmann's solution
Correct Answer: 0.45% N. Saline with 5% glucose
Explanation:The concentration of solute particles per litre (mosm/L) = the osmolarity of a solution. Changes in water content, ambient temperature, and pressure affects osmolarity. The osmolarity of any solution can be calculated by adding the concentration of key solutes in it.
Individual manufacturers of crystalloids and colloids may have different absolute values but they are similar to these.
0.45% N. Saline with 5% glucose:
Tonicity – hypertonic
Osmolarity – 405 mosm/L
Kilocalories (kCal) – 1070.9% N. Saline:
Tonicity – isotonic
Osmolarity – 308 mosm/L
Kilocalories (kCal) – 05% Dextrose:
Tonicity – isotonic
Osmolarity – 253 mosm/L
Kilocalories (kCal) – 170Gelofusine (154 mmol/L Na, 120 mmol/L Cl):
Tonicity – isotonic
Osmolarity – 274 mosm/L
Kilocalories (kCal) – 0Hartmann’s solution:
Tonicity – isotonic
Osmolarity – 273 mosm/L
Kilocalories (kCal) – 9 -
This question is part of the following fields:
- Physiology
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Question 25
Incorrect
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With a cervical dilation of 7 cm, a 33-year-old term primigravida is in labour. She is otherwise in good health. She's been in labour for 14 hours and counting.
The cardiotocograph shows late foetal pulse decelerations, and a pH of 7.24 was found in the recent foetal scalp blood sample.
Which of the following is true about this patient's care and management?Your Answer: The patient requires a category 2 caesarean section under spinal anaesthetic
Correct Answer: Monitor for downward trend in fetal scalp blood pH as caesarean section is not indicated at the present time
Explanation:Once the decision to deliver a baby by caesarean section has been made, it should be carried out with a level of urgency commensurate with the baby’s risk and the mother’s safety.
There are four types of caesarean section urgency:
Category 1: A threat to the life of the mother or the foetus. 30 minutes to make a delivery decision
Category 2 : Maternal or foetal compromise that is not immediately life threatening. In most cases, the decision to deliver is made within 75 minutes.
Category 3 – Early delivery is required, but there is no risk to the mother or the foetus.
Category 4: Elective delivery at a time that is convenient for both the mother and the maternity staff.There may be evidence of foetal compromise in the example above (late foetal pulse decelerations and a borderline pH).
Blood samples from the foetus:
normal: 7.25 or above
borderline: 7.21 to 7.24
abnormal: 7.20 or belowWhen a foetal deceleration occurs, the mother should be given oxygen, kept in a left lateral position, and given a tocolytic if the foetal deceleration is hyper stimulating. Maintaining adequate hydration will reduce the likelihood of a caesarean section.
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This question is part of the following fields:
- Pathophysiology
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Question 26
Incorrect
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Given the following values:
Expired tidal volume = 800 ml
Plateau pressure = 50 cmH2O
PEEP = 10 cmH2O
Compute for the static pulmonary compliance.Your Answer: 200 ml/cmH2O
Correct Answer: 20 ml/cmH2O
Explanation:Compliance of the respiratory system describes the expandability of the lungs and chest wall. There are two types of compliance: dynamic and static.
Dynamic compliance describes the compliance measured during breathing, which involves a combination of lung compliance and airway resistance. Defined as the change in lung volume per unit change in pressure in the presence of flow.
Static compliance describes pulmonary compliance when there is no airflow, like an inspiratory pause. Defined as the change in lung volume per unit change in pressure in the absence of flow.
For example, if a person was to fill the lung with pressure and then not move it, the pressure would eventually decrease; this is the static compliance measurement. Dynamic compliance is measured by dividing the tidal volume, the average volume of air in one breath cycle, by the difference between the pressure of the lungs at full inspiration and full expiration. Static compliance is always a higher value than dynamic
Static compliance can be computed using the formula:
Cstat = Tidal volume/Plateau pressure – PEEP
Substituting the values given,
Cstat = 800/50-10
Cstat = 20 ml/cmH2O -
This question is part of the following fields:
- Physiology
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Question 27
Incorrect
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In the erect position, the partial pressure of oxygen in the alveoli (PAO2) is higher in the apical lung units than in the basal lung units.
What is the most significant reason for this?Your Answer: The basal units are better ventilated than apical units
Correct Answer: The V/Q ratio of apical units is greater than that of basal units
Explanation:In any alveolar unit, the V/Q ratio affects alveolar oxygen (PAO2) and carbon dioxide tension (PACO2).
The partial pressure of alveolar carbon dioxide (PACO2) is plotted against the partial pressure of alveolar oxygen in a Ventilation-Perfusion (V/Q) ratio graph (PAO2). Given a set of model assumptions, the curve represents all of the possible values for PACO2 and PAO2 that an individual alveolus could have.
In the case of an infinity V/Q ratio (ventilation but no perfusion or dead space), the PACO2 of the alveolus will equal zero, while the PAO2 will approach that of external air (150mmmHg). At the apex of the lung, the V/Q ratio is 3.3, compared to 0.67 at the base.
PACO2 and PAO2 approach the partial pressures for these gases in the venous blood when the V/Q ratio is zero (no ventilation but perfusion). At the base of the lung, the V/Q ratio is 0.67, whereas at the apex, it is 3.3.
PAO2 at the apex is typically 132mmHg, and PACO2 is typically 28mmHg.
The average PAO2 at the base is 89 mmHg, while the average PACO2 is 42 mmHg.
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This question is part of the following fields:
- Physiology
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Question 28
Incorrect
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All of the following are responses to massive haemorrhage except which of the following?
Your Answer: Fluid moves into the intravascular space due to decreased capillary hydrostatic pressure
Correct Answer: Decreased cardiac output by increased direct parasympathetic stimulation
Explanation:With regards to compensatory response to blood loss, the following sequence of events take place:
1. Decrease in venous return, right atrial pressure and cardiac output
2. Baroreceptor reflexes (carotid sinus and aortic arch) are immediately activated
3. There is decreased afferent input to the cardiovascular centre in medulla. This inhibits parasympathetic reflexes and increases sympathetic response
4. This results in an increased cardiac output and increased SVR by direct sympathetic stimulation. There is increased circulating catecholamines and local tissue mediators (adenosine, potassium, NO2)
5. Fluid moves into the intravascular space as a result of decreased capillary hydrostatic pressure absorbing interstitial fluid.A slower response is mounted by the hypothalamus-pituitary-adrenal axis.
6. Reduced renal blood flow is sensed by the intra renal baroreceptors and this stimulates release of renin by the juxta-glomerular apparatus.
7. There is cleavage of circulating Angiotensinogen to Angiotensin I, which is converted to Angiotensin II in the lungs (by Angiotensin Converting Enzyme ACE)Angiotensin II is a powerful vasoconstrictor that sets off other endocrine pathways.
8. The adrenal cortex releases Aldosterone
9. There is antidiuretic hormone release from posterior pituitary (also in response to hypovolaemia being sensed by atrial stretch receptors)
10. This leads to sodium and water retention in the distal convoluted renal tubule to conserve fluid
Fluid conservation is also aided by an increased amount of cortisol which is secreted in response to the increase in circulating catecholamines and sympathetic stimulation. -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 29
Correct
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A new proton pump inhibitor (PPI) is being evaluated in elderly patients who are taking aspiring. Study designed has 120 patients receiving the PPI, while a control group of 240 individuals is given the standard PPI. Over a span of 6 years, 24 of the group receiving the new PPI had an upper GI bleed compared to 60 individuals who received the standard PPI.
How would you calculate the absolute risk reduction?Your Answer: 5%
Explanation:Absolute risk reduction = (Control event rate) – (Experimental event rate)
Experimental event rate = 24 / 120 = 0.2
Control event rate = 60 / 240 = 0.25
Absolute risk reduction = 0.25 – 0.2 = 0.05 = 5% reduction
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This question is part of the following fields:
- Statistical Methods
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Question 30
Incorrect
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Calculation of the left ventricular ejection fraction is determined by which of the following equations?
Your Answer:
Correct Answer: Stroke volume / end diastolic LV volume
Explanation:Cardiac output = stroke volume x heart rate
Left ventricular ejection fraction = (stroke volume / end diastolic LV volume ) x 100%
Stroke volume = end diastolic LV volume – end systolic LV volume
Pulse pressure = Systolic Pressure – Diastolic Pressure
Systemic vascular resistance = mean arterial pressure / cardiac output
Factors that increase pulse pressure include:
-a less compliant aorta (this tends to occur with advancing age)
-increased stroke volume -
This question is part of the following fields:
- Physiology And Biochemistry
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