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Question 1
Incorrect
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Metabolization of many drugs used in anaesthesia involves the cytochrome P450 (CYP) isoenzymes. The CYP enzyme most likely to be subject to genetic variability and thus cause adverse drug reactions is which of these?
Your Answer: CYP2C19
Correct Answer: CYP2D6
Explanation:Approximately 25% of phase-1 drug reactions is made responsible by CYP2D6.
As much as a 1,000-fold difference in the ability to metabolise drugs by CYP2D6 can happen between phenotypes, and this may result in adverse drug reactions (ADRs).
The metabolism of antiemetics, beta-blockers, codeine, tramadol, oxycodone, hydrocodone, tamoxifen, antidepressants, neuroleptics, and antiarrhythmics is also as a result of CYP2D6.
Patients who take drugs that are metabolised by CYP2D6 but have poor CYP2D6 metabolism are more likely to have ADRs. People with ultra-rapid CYP2D6 metabolism may have a decreased drug effect due to low plasma concentrations of these drugs.
All the other CYP enzymes are subject to genetic polymorphism. Variants are less likely to lead to adverse drug reactions.
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This question is part of the following fields:
- Physiology
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Question 2
Incorrect
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The action potential in a muscle fibre is initiated by which of these ions?
Your Answer: Calcium ions
Correct Answer: Sodium ions
Explanation:The cardiac action potential has several phases which have different mechanisms of action as seen below:
Phase 0: Rapid depolarisation – caused by a rapid sodium influx.
These channels automatically deactivate after a few msPhase 1: caused by early repolarisation and an efflux of potassium.
Phase 2: Plateau – caused by a slow influx of calcium.
Phase 3 – Final repolarisation – caused by an efflux of potassium.
Phase 4 – Restoration of ionic concentrations – The resting potential is restored by Na+/K+ATPase.
There is slow entry of Na+into the cell which decreases the potential difference until the threshold potential is reached. This then triggers a new action potentialOf note, cardiac muscle remains contracted 10-15 times longer than skeletal muscle.
Different sites have different conduction velocities:
1. Atrial conduction – Spreads along ordinary atrial myocardial fibres at 1 m/sec2. AV node conduction – 0.05 m/sec
3. Ventricular conduction – Purkinje fibres are of large diameter and achieve velocities of 2-4 m/sec, the fastest conduction in the heart. This allows a rapid and coordinated contraction of the ventricles
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This question is part of the following fields:
- Physiology
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Question 3
Correct
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Which plasma protein will bind the thyroid hormone triiodothyronine (T3) more readily?
Your Answer: Thyroxine binding globulin
Explanation:Secreted T4 and T3 circulate in the bloodstream almost entirely bound to proteins. Normally only about 0.03% of total plasma T4 and 0.3% of total plasma T3 exist in the free state. Free T3 is biologically active and mediates the effects of thyroid hormone on peripheral tissues in addition to exerting negative feedback on the pituitary and hypothalamus. The major binding protein is thyroxine-binding globulin (TBG), which is synthesized in the liver and binds one molecule of T4 or T3. About 70% of circulating T4 and T3 is bound to TBGl 10% to 15% is bound to another specific thyroid-binding protein called transthyretin (TTR). Albumin binds 15% to 20%, and 3% to lipoproteins. Ordinarily only alterations in TBG concentration significantly affect total plasma T4 and T3 levels.
Two important biological functions have been ascribed to TBG. First, it maintains a large circulating reservoir of T4 that buffers any acute changes in thyroid gland function. Second, binding of plasma T4 and T3 to proteins prevents loss of these relatively small hormone molecules in urine and thereby helps conserve iodide. TTR transports T4 in CSF and provides thyroid hormones to the CNS.
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This question is part of the following fields:
- Physiology
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Question 4
Incorrect
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A 45-year old gentleman is in the operating room to have a knee arthroscopy under general anaesthesia. Induction is done using fentanyl 1mcg/kg and propofol 2mg/kg. A supraglottic airway is inserted and the mixture used to maintain anaesthesia is and air oxygen mixture and 2.5% sevoflurane. Using a Bain circuit, the patient breathes spontaneously and the fresh gas flow is 9L/min. Over the next 30 minutes, the end-tidal CO2 increase from 4.5kPa to 8.4kPa, and the baseline reading on the capnograph is 0kPa. The most appropriate action that should follow is:
Your Answer: Increase the fresh gas flow in the circuit
Correct Answer: Observe the patient for further change
Explanation:Such a high rise of end-tidal CO2 (EtCO2) in a patient who is spontaneously breathing is often encountered.
Close observation should occur for further rises in EtCO2 and other signs of malignant hyperthermia. If this were to rise even more, it might be wise to ensure that ventilatory support is available.
A lot would depend on whether surgery was almost completed. At this stage of anaesthesia, it would be inappropriate to administer opioid antagonists or respiratory stimulants.
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This question is part of the following fields:
- Physiology
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Question 5
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 apical units are better ventilated
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 6
Incorrect
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Which of the following statement is true or false regarding to the respiratory tract?
Your Answer: The trachea starts at the lower end of the thyroid cartilage at the level of the sixth cervical vertebra
Correct Answer: The sympathetic innervation of the bronchi is derived from T2 - T4
Explanation:The diaphragm has three opening through which different structures pass from the thoracic cavity to the abdominal cavity:
Inferior vena cava passes at the level of T8.
Oesophagus, oesophageal vessels and vagi at T10.
Aorta, thoracic duct and azygous vein through T12.
Sympathetic trunk and pulmonary branches of vagus nerve form a posterior pulmonary plexus at the root of the lung. Fibres continue posteriorly from superficial cardiac plexus to form Anterior pulmonary plexus. It contains vagi nerves and superficial cardiac plexus. These fibres then follow the blood vessel and bronchi into the lungs.
The lower border of the pleura is at the level of:
8th rib in the midclavicular line
10th rib in the lower level of midaxillary line
T12 at its termination.
Both lungs have oblique fissure while right lung has transverse fissure too.
The trachea expands from the lower edge of the cricoid cartilage (at the level of the 6th cervical vertebra) to the carina.
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This question is part of the following fields:
- Physiology
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Question 7
Incorrect
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A 72-year old farmer is hospitalized with acute respiratory failure and autonomic dysfunction. Suspected organophosphate poisoning. Which one is the best mechanism for acute toxicity caused by organophosphates?
Your Answer: Inhibition of butrylcholinesterase
Correct Answer: Inhibition of acetylcholinesterase
Explanation:The toxicity of organophosphorus (OP) nerve agents is manifested through irreversible inhibition of acetylcholinesterase (AChE) at the cholinergic synapses, which stops nerve signal transmission, resulting in a cholinergic crisis and eventually death of the poisoned person. Oxime compounds used in nerve agent antidote regimen reactivate nerve agent-inhibited AChE and halt the development of this cholinergic crisis.
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This question is part of the following fields:
- Physiology
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Question 8
Correct
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A 57-year old lady is admitted to the Emergency Department with signs of a subarachnoid haemorrhage. On admission, her GCS was 7. She has been intubated, sedated and is being ventilated and is waiting for a CT scan. Her Blood pressure is 140/70mmHg. The arterial blood gas analysis shows the following: pH 7.2 (7.35 - 7.45), PaO2 70 mmHg (80-100), PaCO2 78 mmHg (35-45), BE -3 mEq/L (-3 +/-3), Standard bic 27 mmol/L (21-27), SaO2 94%. The most likely cause of an increase in the patient's global cerebral blood flow (CBF) is which of the following?
Your Answer: Hypercapnia
Explanation:PaCO2 is one of the most important factors that regulate cerebral vascular tone. CO2 induces cerebral vasodilatation and as a result, it increases CBF. Between 20 mmHg (2.7 kPa) and 80 mmHg (10.7 kPa), there is a linear increase of PaCO2.
Sometimes, there are areas where auto regulation has failed locally but not globally. Similarly, local vs. systemic acidosis will have similar effects. When the PaO2 falls below 50 mmHg (6.5 kPa), the CBF progressively increases.
An increase in the cerebral metabolic rate for oxygen (CMRO2) and therefore CBF can be caused by hyperthermia.
A late feature of cerebral injury is hyperthermia secondary to hypothalamic injury. Therefore this is not the most likely cause of an increased CBF in this scenario. -
This question is part of the following fields:
- Physiology
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Question 9
Incorrect
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The fluids with the highest osmolarity is?
Your Answer: 0.9% N. Saline
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 10
Incorrect
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An intravenous infusion is started with a 500 mL bag of 0.18 percent N. saline and 4% dextrose. Which of the following best describes its make-up?
Your Answer:
Correct Answer: Osmolarity 284 mOsmol/L, sodium 15 mequivalents and glucose 20 g
Explanation:30 mmol Na+ and 30 mmol Cl- are found in 1 litre of 0.18 percent N. saline with 4% dextrose. Percent (percent) refers to the number of grammes of a compound per 100 mL, so a litre of 4 percent dextrose solution contains 40 grammes.
As a result, a 500 mL bag of 1/5th N. saline and 4% dextrose contains approximately 15 mequivalents of sodium and 20 g of glucose. It is hypotonic due to its osmolarity of 284.
Because of the risk of hyponatraemia, it is no longer considered the crystalloid of choice for fluid maintenance in children.
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This question is part of the following fields:
- Physiology
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