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Question 1
Correct
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Which of the following statement regarding Adrenaline (Epinephrine) is not true?
Your Answer: Inhibits glycolysis in muscle
Explanation:Adrenaline acts on ?1, ?2,?1, and ?2 receptors and also on dopamine receptors (D1, D2) and have sympathomimetic effects.
Natural catecholamines are Adrenaline, Noradrenaline, and Dopamine
Adrenaline is a sympathomimetic amine with both alpha and beta-adrenergic stimulating properties.
Adrenaline is the drug of choice for anaphylactic shock
Adrenaline is also used in patients with cardiac arrest. The preferred route is i.v. followed by the intra-osseous and endotracheal route.Adrenaline is released by the adrenal glands, acts on ? 1 and 2, ? 1 and 2 receptors, and is responsible for fight or flight response.
It acts on ? 2 receptors in skeletal muscle vessels-causing vasodilation.
It acts on ? adrenergic receptors to inhibit insulin secretion by the pancreas. It also stimulates glycogenolysis in the liver and muscle, stimulates glycolysis in muscle.
It acts on ? adrenergic receptors to stimulate glucagon secretion in the pancreas
It also stimulates Adrenocorticotrophic Hormone (ACTH) and stimulates lipolysis by adipose tissue -
This question is part of the following fields:
- Pharmacology
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Question 2
Correct
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Drug X, a new intravenous induction drug, is being administered as a bolus at regular time intervals, and the following data were observed:
Time following injection (hours) vs Plasma concentration of drug X (mcg/mL)
2 / 400
6 / 100
10 / 25
14 / 6.25
Which of the following values estimate the plasma half-life (T½) of drug X?
Your Answer: 2 hours
Explanation:Half life (T½) is the time required to change the amount of drug in the body by one-half (or 50%) during elimination. The time course of a drug in the body will depend on both the volume of distribution and the clearance.
Extrapolating the values from the plasma concentration vs time:
Plasma concentration at 0 hours = 800 mcg/mL
Plasma concentration at 2 hours = 400 mcg/mL
Plasma concentration at 4 hours = 200 mcg/mL
Plasma concentration at 6 hours = 100 mcg/mL
Plasma concentration at 8 hours = 50 mcg/mL
Plasma concentration at 10 hours = 25 mcg/mL
Plasma concentration at 12 hours = 12.5 mcg/mL
Plasma concentration at 14 hours = 6.25 mcg/mL -
This question is part of the following fields:
- Statistical Methods
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Question 3
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)
9.2 kPa (10.5-13.1)
PaCO2 78 mmHg (35-45)
10.2 kPa (4.6-6.0)
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 4
Correct
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A 28-year-old girl complained of severe abdominal pain and hematemesis and was rushed into the emergency department. She has an increased heart rate of 120 beats per minute and blood pressure of 90/65. She has a history of taking Naproxen for her Achilles tendinopathy. On urgent endoscopy, she is diagnosed with a bleeding peptic ulcer.
The immediate treatment is to permanently stop the bleeding by performing embolization of the left gastric artery via an angiogram.
What level of the vertebra will be used as a radiological marker for the origin of the artery that supplies the left gastric artery during the angiogram?Your Answer: T12
Explanation:The left gastric artery is the smallest branch that originates from the coeliac trunk—the coeliac trunk branches of the abdominal aorta at the vertebral level of T12.
The left gastric artery runs along the superior portion of the lesser curvature of the stomach. A peptic ulcer that is serious enough to erode through the stomach mucosa into a branch of the left gastric artery can cause massive blood loss in the stomach, leading to hematemesis. The patient also takes Naproxen, a non-steroidal anti-inflammatory drug that is a common cause for peptic ulcers in otherwise healthy patients.
The left gastric artery is responsible for 85% of upper GI bleeds. In cases refractory to initial treatment, angiography is sometimes needed to embolise the vessel at its origin and stop bleeding. During an angiogram, the radiologist will enter the aorta via the femoral artery, ascend to the level of the 12th vertebrae and then enter the left gastric artery via the coeliac trunk.
The important landmarks of vessels arising from the abdominal aorta at different levels of vertebrae are:
T12 – Coeliac trunk
L1 – Left renal artery
L2 – Testicular or ovarian arteries
L3 – Inferior mesenteric artery
L4 – Bifurcation of the abdominal aorta
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This question is part of the following fields:
- Anatomy
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Question 5
Correct
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A study aimed at assessing the validity of a novel diagnostic test for heart failure is being performed. The curators are worried that not all the patients will get the prevalent gold standard test.
Which type of bias is that?Your Answer: Work-up bias
Explanation:Work up bias involves comparing the novel diagnostic test with the current standard test. A portion of the patients undergo the standard test while others undergo the new test as the standard test is costly. The result can be alteration in specify and sensitivity.
Selection bias is when randomisation is not achieved.
Attention bias refers to the person’s failure to consider various alternatives when he pre occupied by some other thoughts.
Instrument bias is related to the experience and extent of familiarization of the participating individuals with the test.
Co intervention bias is characterized by the groups receiving different co interventions.
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This question is part of the following fields:
- Statistical Methods
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Question 6
Correct
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At which of the following location is there no physiological oesophageal constriction?
Your Answer: Lower oesophageal sphincter
Explanation:The oesophagus is a muscular tube that connects the pharynx to the stomach. It begins at the lower border of the cricoid cartilage and C6 vertebra. It ends at T11.
The oesophagus has physiological constrictions at the following levels:
1. Cervical constriction: Pharyngo-oesophageal junction (15 cm from the incisor teeth) produced by the cricopharyngeal part of the inferior pharyngeal constrictor muscle
2. Thoracic constrictions:
i. where the oesophagus is first crossed by the arch of the aorta (22.5 cm from the incisor teeth)
ii. where the oesophagus is crossed by the left main bronchus (27.5 cm from the incisor teeth)
3. Diaphragmatic constriction: where the oesophagus passes through the oesophageal hiatus of the diaphragm (40 cm from the incisor teeth)Awareness of these constrictions is important for clinical purposes when it is required to pass instruments through the oesophagus into the stomach or when viewing radiographs of patients’ oesophagus.
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This question is part of the following fields:
- Anatomy
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Question 7
Correct
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Very large SI units are easily expressed using mathematical prefixes.
One terabyte is equal to which of the following numbers?Your Answer: 1,000,000,000,000 bytes
Explanation:To denote large measured units, the following SI mathematical prefixes are used:
1 deca = 10 bytes (101)
1 hecto (h) = 100 bytes
1 kilo (k)= 1,000 bytes
1 mega (M) = 1,000,000 bytes
1 giga (G) = 1,000,000,000 bytes
1 Tera (T) = 1,000,000,000,000 bytes
1 Peta (P) = 1,000,000,000,000,000 bytes -
This question is part of the following fields:
- Basic Physics
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Question 8
Correct
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Volunteers are being recruited for a new clinical trial of a novel drug treatment for Ulcerative colitis. The proposed study will enrol about 2000 people with ulcerative colitis. Testing will be performed to assess any reduction in disease severity with the new drug as compared to the current treatment available in the industry.
Which phase of clinical trial will this be?Your Answer: Phase 3
Explanation:This clinical trial consists over 1000 patients being evaluated for the response to a new treatment against a currently licensed treatment for ulcerative colitis. Therefore, it is comparing its efficacy to an established therapeutic or control in a larger population of volunteers. These are the characteristics of a phase III clinical trial.
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This question is part of the following fields:
- Statistical Methods
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Question 9
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 10
Incorrect
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The SI unit of measurement is kgm2s-2 in the System international d'unités (SI).
Which of the following derived units of measurement has this format?Your Answer:
Correct Answer: Energy
Explanation:The derived SI unit of force is Newton.
F = m·a (where a is acceleration)
F = 1 kg·m/s2The joule (J) is a converted unit of energy, work, or heat. When a force of one newton (N) is applied over a distance of one metre (Nm), the following amount of energy is expended:
J = 1 kg·m/s2·m =
J = 1 kg·m2/s2 or 1 kg·m2·s-2The unit of velocity is metres per second (m/s or ms-1).
The watt (W), or number of joules expended per second, is the SI unit of power:
J/s = kg·m2·s-2/s
J/s = kg·m2·s-3Pressure is measured in pascal (Pa) and is defined as force (N) per unit area (m2):
Pa = kg·m·s-2/m2
Pa = kg·m-1·s-2 -
This question is part of the following fields:
- Physiology
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Question 11
Incorrect
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A 50-year-old woman's blood pressure readings in the clinic are 170/109 mmHg, 162/100 mmHg and 175/107 mmHg and her routine haematology, biochemistry, and 12-lead ECG are normal.
She is assessed on the day of surgery prior to laparoscopic inguinal hernia repair and is found to be normally fit and well. Documentation of previous blood pressure measurements from her general practitioner in the primary healthcare setting are not available.
What is your next course of action?Your Answer:
Correct Answer: Proceed with scheduled surgery without treatment
Explanation:The AAGBI and the British Hypertension Society has published guidelines for the measurement of adult blood pressure and management of hypertension before elective surgery.
The objective is to ensure that patients admitted for elective surgery have a known systolic blood pressure below 160 mmHg and diastolic blood pressures below 100 mmHg. The primary health care teams, if possible, should ensure that this is the case and provide evidence to the pre-assessment clinic staff or on admission.
Avoiding cancellation on the day of surgery because of white coat hypertension is a secondary objective.
Patients with blood pressures below 180 mmHg systolic and 110 mmHg diastolic (measured in the preop assessment clinic), who present to pre-operative assessment clinics without documented evidence of primary care blood pressures should proceed to elective surgery.
In this question, the history/assessment does not appear to point to obvious end-organ damage so there is no indication for further investigation for secondary causes of hypertension or an echocardiogram at this point. Further review and treatment at this point is not required.
However, you should write to the patient’s GP and encourage serial blood pressure measurements in the primary health care setting.
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This question is part of the following fields:
- Pathophysiology
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Question 12
Incorrect
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Out of the following, which is NOT true regarding the external carotid?
Your Answer:
Correct Answer: It ends by bifurcating into the superficial temporal and ascending pharyngeal artery
Explanation:The external carotid artery has eight important branches:
Anterior surface:
1. Superior thyroid artery (first branch)
2. Lingual artery
3. Facial artery
Medial branch
4. Ascending pharyngeal artery
Posterior branches
5. Occipital artery
6. Posterior auricular artery
Terminal branches
7. Maxillary artery
8. Superficial temporal arteryThe external carotid has eight branches, 3 from its anterior surface ; thyroid, lingual and facial. The pharyngeal artery is a medial branch. The posterior auricular and occipital are posterior branches.
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This question is part of the following fields:
- Anatomy
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Question 13
Incorrect
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Which of the following statements about the central venous pressure (CVP) waveform is true?
Your Answer:
Correct Answer: Third degree heart block causes canon A waves
Explanation:The central venous pressure (CVP) waveform depicts changes of pressure within the right atrium. Different parts of the waveform are:
A wave: which represents atrial contraction. It is synonymous with the P wave seen during an ECG. It is often eliminated in the presence of atrial fibrillation, and increased tricuspid stenosis, pulmonary stenosis and pulmonary hypertension.
C wave: which represents right ventricle contraction at the point where the tricuspid valve bulges into the right atrium. It is synonymous with the QRS complex seen on ECG.
X descent: which represents relaxation of the atrial diastole and a decrease in atrial pressure, due to the downward movement of the right ventricle as it contracts. It is synonymous with the point before the T wave on ECG.
V wave: which represents an increase in atrial pressure just before the opening of the tricuspid valve. It is synonymous with the point after the T wave on ECG. It is increased in the background of a tricuspid regurgitation.
Y descent: which represents the emptying of the atrium as the tricuspid valve opens to allow for blood flow into the ventricle in early diastole.
Canon waves: which refer to large waves present on the trace that do not correspond to the A, V or C waves. They usually occur in a background of complete heart blocks or junctional arrythmias.
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This question is part of the following fields:
- Clinical Measurement
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Question 14
Incorrect
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Which of the following is the smallest value of pressure?
Your Answer:
Correct Answer: 14.69 psi
Explanation:The SI unit of pressure is the pascal (Pa) and it is equal to one newton (N) per square meter (m2) or N/m2.
1 atmosphere (atm) is the equivalent of:
101325 Pa760 mmHg
1.01325 bar
1033.23 cmH2O.
14.69 pounds per square inch (psi)
1013.25 millibar (mbar) or hectopascals (hPa), and14.69 psi is equal to one atmosphere. The other values are equal to two atmospheres of pressure.
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This question is part of the following fields:
- Basic Physics
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Question 15
Incorrect
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A 32-year-old man has multiple stab wounds to his abdomen and is rushed into the emergency. Resuscitative measures are performed, but the patient remains hypotensive.
Emergency laparotomy is performed, and it reveals a vessel is bleeding profusely at a certain level of lumbar vertebrae. The vessel is the testicular artery and is ligated.
At which lumbar vertebrae is the testicular artery identified?Your Answer:
Correct Answer: L2
Explanation:The important landmarks of vessels arising from the abdominal aorta at different levels of vertebrae are:
T12 – Coeliac trunk
L1 – Left renal artery
L2 – Testicular or ovarian arteries
L3 – Inferior mesenteric artery
L4 – Bifurcation of the abdominal aorta
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This question is part of the following fields:
- Anatomy
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Question 16
Incorrect
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What is NOT a feature of Propofol infusion syndrome?
Your Answer:
Correct 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 17
Incorrect
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An acidic drug with a pKA of 4.3 is injected intravenously into a patient.
At a normal physiological pH, the approximate ratio of ionised to unionised forms of this drug in the plasma is?Your Answer:
Correct Answer: 1000:01:00
Explanation:The pH at which the drug exists in 50 percent ionised and 50 percent unionised forms is known as the pKa.
To calculate the proportion of ionised to unionised form of an ACID, use the Henderson-Hasselbalch equation.
pH = pKa + log ([A-]/[HA])
or
pH = pKa + log [(salt)/(acid)]
pH = pKa + log ([ionised]/[unionised]).Hence, if the pKa − pH = 0, then 50% of drug is ionised and 50% is unionised.
In this example:
7.4 = 4.3 + log ([ionised]/[unionised])
7.4 − 4.3 = log ([ionised]/[unionised])
log 3.1 = log ([ionised]/[unionised])Simply put, the antilog is the inverse log calculation. In other words, if you know the logarithm of a number, you can use the antilog to find the value of the number. The antilogarithm’s definition is as follows:
y = antilog x = 10x
Antilog to the base 10 of 0 = 1, 1 = 10, 2 =100, 3 = 1000, and 4 = 10,000.
If you want to find the antilogarithm of 3.1, for a number between 3 and 4, the antilogarithm will return a value between 1000 and 10,000. The ratio is 1:1 if pKa = pH, that is, pH pKa = log 0. (50 percent ionised and unionised).
According to the above value, there is only one unionised molecule for every approximately 1000 (1259) ionised molecules of this drug in plasma, implying that this drug is largely ionised in plasma (99.99 percent ).
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This question is part of the following fields:
- Pharmacology
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Question 18
Incorrect
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A 72-year-old long-term rheumatoid arthritis patient is having shoulder replacement surgery.
He has chronic obstructive pulmonary disease with a limited exercise tolerance. He agrees to the procedure being performed with an interscalene brachial plexus block.
Which of the following neurological complications puts this patient at the greatest risk?Your Answer:
Correct Answer: Phrenic nerve block
Explanation:An ipsilateral phrenic nerve block will result from a successful interscalene block (ISB).
The phrenic nerve is the diaphragm’s sole motor supply, and ipsilateral hemidiaphragmatic paresis affects up to 100% of patients who receive ISBs. Phrenic nerve palsy is usually well tolerated and goes unnoticed by healthy people. However, forced vital capacity decreases by approximately 25%, which can produce ventilatory compromise in patients with limited pulmonary reserve, requiring assisted ventilation.
Vocal cord palsy occurs when the recurrent laryngeal nerve is inadvertently blocked, causing hoarseness and possibly acute respiratory insufficiency. Unless bilateral laryngeal nerve palsy occurs, which can cause severe laryngeal obstruction, this complication is usually of little consequence.
ISB can also cause cranial nerve X and XII palsy (Tapia’s syndrome). One-sided cord paralysis, aphonia, and the patient’s tongue deviating toward the block’s side are all symptoms.
When a local anaesthetic spreads to the stellate ganglion and its cervical sympathetic nerves, Horner’s syndrome can develop. Ptosis of the eyelid, miosis, and anhidrosis of the face are all symptoms. Horner’s syndrome, on the other hand, may not indicate that the brachial plexus is sufficiently blocked.
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This question is part of the following fields:
- Pathophysiology
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Question 19
Incorrect
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A patient under brachial plexus regional block complains of pain under the cuff after the torniquet is inflated.
Which nerve was most probably 'missed' by the local anaesthetic?Your Answer:
Correct Answer: Intercostobrachial nerve
Explanation:The area described in the question is supplied by the intercostobrachial nerve, which provides sensory innervation to the portions of the axilla, tail of the breast, lateral chest wall and medial side of the arm.
It is a common for it to be ‘missed’ during administration of local anaesthesia because of its very superficial anatomic course. It may be anesthetized by giving an analgesia from the upper border of the biceps at the anterior axillary fold, to the margin of the triceps by the axillary floor.
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This question is part of the following fields:
- Pathophysiology
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Question 20
Incorrect
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Which nerve is responsible for the direct innervation of the sinoatrial node?
Your Answer:
Correct Answer: None of the above
Explanation:The sinoatrial node receives innervation from multiple nerves arising from the complex cardiac plexus.
The cardiac plexus sends tiny branches into cardiac vessels, alongside the right and left coronary arteries.
The vagal efferent fibres originate from the vagal and accessory nerves in the brainstem, and then travel to the cardiac plexus within the heart. The resulting vagal discharge controls heart rate.
No singular nerve directly innervates the sinoatrial node.
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This question is part of the following fields:
- Anatomy
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Question 21
Incorrect
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A strict diet is mandatory for which of the following drugs for mood disorders?
Your Answer:
Correct Answer: Tranylcypromine
Explanation:Tranylcypromine is a monoamine oxidase inhibitor that binds irreversibly to target enzyme.
Monoamine oxidase inhibitors are responsible for blocking the monoamine oxidase enzyme. The monoamine oxidase enzyme breaks down different types of neurotransmitters from the brain: norepinephrine, serotonin, dopamine, and tyramine. MAOIs inhibit the breakdown of these neurotransmitters thus, increasing their levels and allowing them to continue to influence the cells that have been affected by depression.
There are two types of monoamine oxidase, A and B. The MAO A is mostly distributed in the placenta, gut, and liver, but MAO B is present in the brain, liver, and platelets. Serotonin and noradrenaline are substrates of MAO A, but phenylethylamine, methylhistamine, and tryptamine are substrates of MAO B. Dopamine and tyramine are metabolized by both MAO A and B. Selegiline and rasagiline are irreversible and selective inhibitors of MAO type B, but safinamide is a reversible and selective MAO B inhibitor.
MAOIs prevent the breakdown of tyramine found in the body and certain foods, drinks, and other medications. Patients that take MAOIs and consume tyramine-containing foods or drinks will exhibit high serum tyramine level. A high level of tyramine can cause a sudden increase in blood pressure, called the tyramine pressor response. Even though it is rare, a high tyramine level can trigger a cerebral haemorrhage, which can even result in death.
Eating foods with high tyramine can trigger a reaction that can have serious consequences. Patients should know that tyramine can increase with the aging of food; they should be encouraged to have fresh foods instead of leftovers or food prepared hours earlier. Examples of high levels of tyramine in food are types of fish and types of meat, including sausage, turkey, liver, and salami. Also, certain fruits can contain tyramine, like overripe fruits, avocados, bananas, raisins, or figs. Further examples are cheeses, alcohol, and fava beans; all of these should be avoided even after two weeks of stopping MAOIs. Anyone taking MAOIs is at risk for an adverse hypertensive reaction, with accompanying morbidity. Patients taking reversible MAOIs have fewer dietary restrictions.
Amitriptyline is a tricyclic antidepressant, and citalopram and escitalopram are selective serotonin reuptake inhibitors.
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This question is part of the following fields:
- Pharmacology
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Question 22
Incorrect
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A healthy 27-year old male who weighs 70kg has appendicitis. He is currently in the operating room and is being positioned to have a rapid sequence induction.
Prior to preoxygenation, the compartment likely to have the best oxygen reserve is:Your Answer:
Correct Answer: Red blood cells
Explanation:The following table shows the compartments and their relative oxygen reserve:
Compartment Factors Room air (mL) 100% O2 (mL)
Lung FAO2, FRC 630 2850
Plasma PaO2, DF, PV 7 45
Red blood cells Hb, TGV, SaO2 788 805
Myoglobin 200 200
Interstitial space 25 160Oxygen reserves in the body, with room air and after oxygenation.
FAO2-alveolar fraction of oxygen rises to 95% after administration of 100% oxygen (CO2 = 5%)
FRC- Functional residual capacity – (the most important store of oxygen in the body) – 2,500-3,000 mL in medium sized adults
PaO2-partial pressure of oxygen dissolved in arterial blood (80 mmHg breathing room air and 500 mmHg breathing 100% oxygen)
DF -dissolved form (0.3%)
PV-plasma volume (3L)
TG-total globular volume (5L)
Hb-haemoglobin concentration
SaO2-arterial oxygen concentration (98% breathing air and 100% when preoxygenated) -
This question is part of the following fields:
- Physiology
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Question 23
Incorrect
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Your manager asks you to inform patients that are suffering from a chronic pain about a trial that is going to be conducted in order to determine the efficacy of a novel analgesic. What phase is the trial currently in?
Your Answer:
Correct Answer: Phase 2
Explanation:Phase 0 trials assist the scientists in studying the behaviour of drugs in humans by micro dosing patients. They are used to speed up the developmental process. They have no measurable therapeutic effect and efficiency.
Phase 1 is associated with assessing whether a drug is safe to use or not. The process is extensive and can take up to several months. It also involves healthy participants (less than 100) that are paid to take part in the study. The side effects upon increasing dosage are also addressed by the study. The effects the drug has on humans including how its absorbed, metabolized and excreted are studied. Approximately 70% of the drugs pass this phase.
Phase 2 trials involve patients that are suffering from the disease under study and are associated with determining the efficiency and the optimum dosage of the drug.
Phase 3 also assesses the efficacy but at a higher scale with larger population sample.
Phase 4 trials are involved with the long term effects and side effects of the drug.
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This question is part of the following fields:
- Statistical Methods
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Question 24
Incorrect
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What is factually correct regarding correlation and regression?
Your Answer:
Correct Answer: Regression allows one variable to be predicted from another variable
Explanation:Linear regression, using a technique called curve fitting, allows us to make predictions regarding a certain variable.
Correlation coefficient gives us an idea whether or not the two parameters provide have any relation of some sort or not i.e. does change in one prompt any change in other?
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This question is part of the following fields:
- Statistical Methods
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Question 25
Incorrect
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What is the order of the anatomical components of the tracheobronchial tree from proximal to distal?
Your Answer:
Correct Answer: Bronchioles, terminal bronchioles, respiratory bronchioles, alveolar ducts, alveolar sacs
Explanation:The tracheobronchial tree is subdivided into the conducting and the respiratory zones.
The zones from proximal to distal are:
Trachea
Bronchi
Bronchioles
Terminal bronchioles
Respiratory bronchioles
Alveolar ducts
Alveolar sacsfrom the trachea to terminal bronchioles are the conducting zone while the respiratory zone is from the respiratory bronchioles to the alveola sacs
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This question is part of the following fields:
- Anatomy
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Question 26
Incorrect
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Following a near drowning accident, a 5-year-old child is admitted to the emergency department and advanced paediatric life support is started.
What is the child's approximate weight, according to the preferred formulae of the Resuscitation Council (UK), the European Resuscitation Council, and the Royal College of Anaesthetists?Your Answer:
Correct Answer: 20-25kg
Explanation:For estimating a child’s weight, the Resuscitation Council (UK) and European Resuscitation Council teach the following formula:
Weight = (age + 4) × 2
The weight of the child will be around 20 kg.
This formula is used in the Primary FRCA exam by the Royal College of Anaesthetists.
In ‘developed’ countries, the traditional ‘APLS formula’ for estimating weight in children based on age (wt in kg = [age+4] x 2) is acknowledged as underestimating weight by 33.4 percent on average, with the degree of underestimation increasing with increasing age.
However, more recently, the APLS formula ‘Weight=3(age)+7’ has been found to provide a mean underestimate of only 6.9%. This formula is applicable to children aged 1 to 13 years.
The estimated weight based on age using this formula is 25 kg.
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This question is part of the following fields:
- Physiology
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Question 27
Incorrect
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Which of the following combinations of signs seen in a patient would most likely confirm ingestion of substances with anticholinesterase effects?
Your Answer:
Correct Answer: Bradycardia and miosis
Explanation:An acetylcholinesterase inhibitor or anticholinesterase is a chemical that inhibits the cholinesterase enzyme from breaking down acetylcholine (ACh) therefore increasing the level and duration of action of the neurotransmitter acetylcholine(ACh).
ACh stimulates postganglionic receptors to produce the following effects:
Salivation
Lacrimation
Defecation
Micturition
Sweating
Miosis
Bradycardia, and
Bronchospasm.Since these effects are produced by muscarine, they are referred to as muscarinic effects, and the postganglionic receptors are called muscarine receptors.
SLUD (Salivation, Lacrimation, Urination, Defecation – and emesis) is usually encountered only in cases of drug overdose or exposure to nerve gases. It is a syndrome of pathological effects indicating massive discharge of the parasympathetic nervous system.
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This question is part of the following fields:
- Pathophysiology
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Question 28
Incorrect
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Campylobacter is which type of bacteria?
Your Answer:
Correct Answer: sdgsdf
Explanation:Campylobacter is the commonest bacterial cause of infectious intestinal disease in the UK. The majority of cases are caused by the Gram-negative bacillus Campylobacter jejuni which is spread by the faecal-oral route. The incubation period is 1-6 days.
Features include a prodrome phase with headaches and malaise, then diarrhoea occurs which is often bloody.
There is often abdominal pain which may mimic appendicitis.It is usually self-limiting but treatment is warranted if the infection is severe or the infection occurs in an immunocompromised patient.
Severe infection comprises of high fever, bloody diarrhoea, or more than eight stools per day or symptoms last for more than one week.
This management would include antibiotics and the first-line antibiotic is clarithromycin.
Ciprofloxacin is an alternative but there are strains with decreased sensitivity to ciprofloxacin which can be frequently isolated.Complications include:
1.Guillain-Barre syndrome may follow Campylobacter
2. Jejuniinfections
3. Reactive arthritis
4. Septicaemia, endocarditis, arthritis -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 29
Incorrect
-
Which of the following is true in the Kreb's cycle?
Your Answer:
Correct Answer: Alpha-ketoglutarate is a five carbon molecule
Explanation:Krebs’ cycle (tricarboxylic acid cycle or citric acid cycle) is a sequence of reactions to release stored energy through oxidation of acetyl coenzyme A (acetyl-CoA). Some of the products are carbon dioxide and hydrogen atoms.
The sequence of reactions, known collectively as oxidative phosphorylation, only occurs in the mitochondria (not cytoplasm).
The Krebs cycle can only take place when oxygen is present, though it does not require oxygen directly, because it relies on the by-products from the electron transport chain, which requires oxygen. It is therefore considered an aerobic process. It is the common pathway for the oxidation of carbohydrate, fat and some amino acids, required for the formation of adenosine triphosphate (ATP).
Pyruvate enters the mitochondria and is converted into acetyl-CoA. Acetyl-CoA is then condensed with oxaloacetate, to form citrate which is a six carbon molecule. Citrate is subsequently converted into isocitrate, alpha-ketoglutarate, succinyl-CoA, succinate, fumarate, malate and finally oxaloacetate.
The only five carbon molecule in the cycle is Alpha-ketoglutarate.
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This question is part of the following fields:
- Physiology
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Question 30
Incorrect
-
Of the following, which is NOT a branch of the external carotid artery?
Your Answer:
Correct Answer: Mandibular artery
Explanation:The external carotid artery has eight important branches:
1. Superior thyroid artery
2. Ascending pharyngeal artery
3. Lingual artery
4. Facial artery
5. Occipital artery
6. Posterior auricular artery
7. Maxillary artery (terminal branch)
8. Superficial temporal artery (terminal branch)There is no mandibular artery but the first part of the maxillary artery is called the mandibular part as it is posterior to the lateral pterygoid muscle.
The maxillary artery is divided into three portions by its relation to the lateral pterygoid muscle:
first (mandibular) part: posterior to the lateral pterygoid muscle
second (pterygoid or muscular) part: within the lateral pterygoid muscle
third (pterygopalatine) part: anterior to the lateral pterygoid muscle -
This question is part of the following fields:
- Anatomy
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Question 31
Incorrect
-
A 45-year-old woman complains of pain in her upper abdomen to her physician. The pain comes intermittently in waves and gets worse after eating food. There are no associated complaints of fever or bowel problems.
The pain intensity is 6/10, and paracetamol relieves it a little. There is suspicion that part of the biliary tree is blocked.
Which area of the duodenum does this blocked tube open into?Your Answer:
Correct Answer: 2nd part of the duodenum
Explanation:The patient is likely suffering from biliary colic since her pain is intermittent and comes and goes in waves. Biliary colic pain gets worse after eating, especially fatty food as bile helps digest fats. Gallstones are the most common cause of biliary colic and are usually located in the cystic duct or common bile duct. But since this patient has no signs of jaundice or steatorrhea, the duct most likely blocked is the cystic duct.
The cystic duct drains the gallbladder and combines with the common hepatic duct to form the common bile duct. The common bile duct then merges with the pancreatic duct and opens into the second part of the duodenum (major duodenal papilla).
The duodenojejunal flexure is attached to the diaphragm by the ligament of Treitz and is not associated with any common pathology.
The fourth part of the duodenum passes very close to the abdominal aorta and can be compressed by an abdominal aortic aneurysm.
The third part of the duodenum can be affected by superior mesenteric artery syndrome, where the duodenum is compressed between the SMA and the aorta, often in cases of reduced body fat.
The first part of the duodenum is the most common location for peptic ulcers affecting this organ. -
This question is part of the following fields:
- Anatomy
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Question 32
Incorrect
-
Regarding aldosterone, one of the following is true.
Your Answer:
Correct Answer: Secretion is increased following haematemesis
Explanation:Aldosterone is produced in the zona glomerulosa of the adrenal cortex and acts to increase sodium reabsorption via intracellular mineralocorticoid receptors in the distal tubules and collecting ducts of the nephron.
Its release is stimulated by hypovolaemia, blood loss ,and low plasma sodium and is inhibited by hypertension and increased sodium. It is regulated by the renin-angiotensin system.
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This question is part of the following fields:
- Pathophysiology
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Question 33
Incorrect
-
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 34
Incorrect
-
What structure is most posterior at the porta hepatis?
Your Answer:
Correct Answer: Portal vein
Explanation:The structures in the porta hepatis from anterior to posterior are:
The ducts: Most anterior are the left and right hepatic ducts.
The arteries: Next are the left and right hepatic arteries
The veins: Next is the portal vein
The epiploic foramen of Winslow lies most posterior at the porta hepatis.
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This question is part of the following fields:
- Anatomy
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Question 35
Incorrect
-
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:
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 36
Incorrect
-
The muscle that lies behind the first part of the axillary nerve is?
Your Answer:
Correct Answer: Subscapularis
Explanation:The axillary nerve lies behind the axillary artery initially, and in front of the subscapularis. It passes downward to the lower border of the subscapularis muscle.
In company with the posterior humeral circumflex artery and vein, it winds backward through a quadrilateral space bounded above by the subscapularis (anterior) and teres minor (posterior), below by the teres major, medially by the long head of the triceps brachii, and laterally by the humerus (surgical neck).
It then divides into an anterior and a posterior part. The anterior division supplies the deltoid (anterior and middle heads) while the posterior division supplies the teres minor and posterior part of deltoid
The posterior division terminates as the superior lateral cutaneous nerve of the arm -
This question is part of the following fields:
- Anatomy
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Question 37
Incorrect
-
A 70-year old male has diverticular disease and is undergoing a sigmoid colectomy. His risk of developing a post operative would infection can be minimized by which of the following interventions?
Your Answer:
Correct Answer: Administration of single dose of broad spectrum antibiotics prior to the procedure
Explanation:Staphylococcus aureus infection is the most likely cause.
Surgical site infections (SSI) occur when there is a breach in tissue surfaces and allow normal commensals and other pathogens to initiate infection. They are a major cause of morbidity and mortality.
SSI comprise up to 20% of healthcare associated infections and approximately 5% of patients undergoing surgery will develop an SSI as a result.
The organisms are usually derived from the patient’s own body.Measures that may increase the risk of SSI include:
-Shaving the wound using a single use electrical razor with a disposable head
-Using a non iodine impregnated surgical drape if one is needed
-Tissue hypoxia
-Delayed prophylactic antibiotics administration in tourniquet surgery, patients with a prosthesis or valve, in clean-contaminated surgery of in contaminated surgery.Measures that may decrease the risk of SSI include:
1. Intraoperatively
– Prepare the skin with alcoholic chlorhexidine (Lowest incidence of SSI)
-Cover surgical site with dressingIn contrast to previous individual RCT’s, a recent meta analysis has confirmed that administration of supplementary oxygen does not reduce the risk of wound infection and wound edge protectors do not appear to confer benefit.
2. Post operatively
Tissue viability advice for management of surgical wounds healing by secondary intentionUse of diathermy for skin incisions
In the NICE guidelines the use of diathermy for skin incisions is not advocated. Several randomised controlled trials have been undertaken and demonstrated no increase in risk of SSI when diathermy is used. -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 38
Incorrect
-
You've been summoned to the recovery room to examine a 28-year-old man who has had an inguinal hernia repaired.
His vital signs are normal, but you notice that he has developed abnormal upper-limb movements due to muscle contractions that cause repetitive twisting movements.
What do you think is the most likely source for this patient's condition?Your Answer:
Correct Answer: Prochlorperazine
Explanation:Dystonia is characterised by repetitive twisting movements or abnormal postures. They are classified as either primary or secondary.
Primary dystonia is a genetic disorder that is inherited in an autosomal dominant pattern.
Secondary dystonia can be caused by focal brain lesions, Parkinson’s disease, or certain medications.The following drugs cause the most common drug-induced dystonic reactions:
Antipsychotics, antiemetics (especially prochlorperazine and metoclopramide), and antidepressants.Following the administration of the neuroleptic prochlorperazine, 16 percent of patients experience restlessness (akathisia) and 4% experience dystonia.
Several published reports have linked the anaesthetics thiopentone, fentanyl, and propofol to opisthotonos and other abnormal neurologic sequelae. Dystonias following a general anaesthetic are uncommon. Tramadol has been linked to serotonin syndrome, while remifentanil has been linked to muscle rigidity.
The following are some of the risk factors:
Positive family history
Male
Children
An episode of acute dystonia occurred previously.
Dopamine receptor (D2) antagonists at high doses and recent cocaine useDystonia is treated in a variety of ways, including:
Benztropine (as a first-line therapy):
1-2 mg intravenous injection for adults
Child: 0.02 mg/kg to 1 mg maximumBenzodiazepines are a type of benzodiazepine (second line treatment).
Midazolam:
1-2 mg intravenously, or 5-10 mg IV/PO diazepam
Antihistamines with anticholinergic activity (H1receptor antagonists):
Promethazine 25-50 mg IV/IM, or diphenhydramine 50 mg IV/IM (1 mg/kg in children) are used when benztropine is not available.
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This question is part of the following fields:
- Pharmacology
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Question 39
Incorrect
-
A 60-year old male has anaemia and is being investigated. The most common combination of globin chains in a normal adult is:
Your Answer:
Correct Answer: α2β2
Explanation:There are 4 different types of globin chains which surround 4 heme molecules in haemoglobin (Hb) – α (alpha), β (beta), γ (gamma), and δ (delta)
α chains are essential.
δ2β2 and β2γ2 are not found in a healthy adult.
97% of the Hb in a healthy adult is made of α2β2 (2 α chains and 2 β chains).
α2δ2 accounts for around 1.5-3% of the adult Hb.
α2γ2 accounts for less than 1%.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.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 40
Incorrect
-
Which of the following statements is correct regarding hypomagnesaemia?
Your Answer:
Correct Answer: Causes tetany
Explanation:The ECG changes seen in hypomagnesaemia include:
Prolonged PR interval
Prolonged QT interval
Flattening of T waves
ST segment depression
Prominent U wavesThese changes are almost the same as those of hypokalaemia.
There is an increased risk of digoxin toxicity and a risk of atrial and ventricular ectopic and ventricular arrhythmias.
There is impaired synthesis and release of parathyroid hormone (PTH) in chronic hypomagnesaemia leading to impaired target organ response to PTH. This produces secondary hypocalcaemia.
The use of potassium ‘wasting’ diuretics (e.g. loop diuretics like furosemide) may lead to Hypomagnesaemia.
A tall T wave is seen in hypermagnesemia.
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This question is part of the following fields:
- Pathophysiology
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Question 41
Incorrect
-
A 60-year-old man, presents to the emergency department with crushing pain in the central chest area, which radiates to his left arm and jaw. He also reports feelings of nausea with no other symptoms. Elevation of the ST-segment is noted in multiple chest leads upon ECG, leading to a diagnosis of ST-elevation MI.
What vessel gives rise to the coronary vessels?Your Answer:
Correct Answer: Ascending aorta
Explanation:The above mentioned patient presentation is one of an acute coronary syndrome.
The elevations noted in the ST-segments of multiple heart leads on ECG is diagnostic of an ST-elevation myocardial infarction.
The pulmonary artery branches to give rise to the right and left pulmonary arteries, which supply deoxygenated blood to the right and left lungs from the right ventricle.
The pulmonary veins do not form any bifurcations, and therefore do not give rise to any vessels. They travel to the left atrium from the lungs, carrying oxygenated blood.
The descending aorta continues from the aortic arch, and bifurcates to give off many branches, including the right and left common iliac arteries.
The coronary sinus is formed from the combination of four coronary veins, receiving blood supply from the great, middle, small and posterior cardiac veins, and transporting this venous blood into the right atrium.
The right and left aortic sinus give rise to the right and left coronary arteries, respectively. They branch of the ascending aorta, in the area just superior to the aortic valve.
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This question is part of the following fields:
- Anatomy
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Question 42
Incorrect
-
A study designed to examine the benefits of adding a new antiplatelet to aspirin after a myocardial infraction. The recorded results give us the percentage of patients that reported myocardial infraction within a three month period. The percentage was 4% and 3% for aspirin and the combination of drugs respectively.
How many further patients needed to be treated in order for one patient to avoid any more heart attacks during 3 months?Your Answer:
Correct Answer: 100
Explanation:Number needed to treat can be defined as the number of patients who need to be treated to prevent one additional bad outcome.
It can be found as:
NNT=1/Absolute Risk Reduction (rounded to the next integer since number of patients can be integer only).
where ARR= (Risk factor associated with the new drug group) — (Risk factor associated with the currently available drug)
So,
ARR= (0.04-0.03)
ARR= 0.01
NNT= 1/0.01
NNT=100
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This question is part of the following fields:
- Statistical Methods
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Question 43
Incorrect
-
An 80-year old female was taken to the emergency room for chest pain. She has a medical history of coronary artery disease and previous episodes of atrial fibrillation. She was immediately attached to the cardiac monitor, which showed tachycardia at 148 beats per minute. The 12-lead ECG revealed atrial fibrillation.
Digoxin was given as an anti-arrhythmic at 500 micrograms, which is higher than the maintenance dose routinely given. Why is this so?Your Answer:
Correct Answer: It has a high volume of distribution
Explanation:When the loading dose of Digoxin is given, the primary thing to consider is the volume of distribution. The volume of distribution is the proportionality factor that relates the total amount of drug in the body to the concentration. LD is computed as:
LD = Volume of distribution X (desired plasma concentration/bioavailability)
Digoxin is an anti-arrhythmic drug with a large volume of distribution and high bioavailability, and only a small percentage of Digoxin is bound to plasma proteins (,20%).
In the case, since the arrhythmia is not life-threatening, there is no need for the medication to work rapidly.
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This question is part of the following fields:
- Pharmacology
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Question 44
Incorrect
-
The Fick principle can be used to determine the blood flow to any organ of the body.
At rest, which one of these organs has the highest blood flow (ml/min/100g)?Your Answer:
Correct Answer: Thyroid gland
Explanation:After the carotid body, the thyroid gland is the second most richly vascular organ in the body.
The global blood flow to the thyroid gland can be measured using:
1. Colour ultrasound sonography
2. Quantitative perfusion maps using MRI of the thyroid gland using an arterial spin labelling (ASL) method.This table shows the blood flow to various organs of the body at rest:
Organ Blood Flow(ml/minute/100g)
Hepatoportal 58
Kidney 420
Brain 54
Skin 13
Skeletal muscle 2.7
Heart 87
Carotid body 2000
Thyroid gland 560 -
This question is part of the following fields:
- Physiology
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Question 45
Incorrect
-
A sevoflurane vaporiser with a 2 percent setting and a 200 kPa ambient pressure is used.
At this pressure, which of the following options best represents vaporiser output?Your Answer:
Correct Answer: The output is 1% because the saturated pressure of sevoflurane is unaffected by ambient pressure
Explanation:Ambient pressure has no effect on a volatile agent’s saturated vapour pressure (SVP). At a temperature of 20°C, the SVP of sevoflurane is approximately 21 kPa, or 21% of atmospheric pressure (100 kPa).
The SVP of sevoflurane remains the same when the ambient pressure is doubled to 200 kPa, but the output of the vaporiser is halved, now 21 percent of 200 kPa, equalling 10.5 percent. The vaporiser’s output has increased to 1%, but the partial pressure output has remained unchanged. The splitting ratio will not change because it is determined by temperature changes.
Calculations can be made as follows:
Vaporizer output % (ambient pressure) = % volatile (calibrated) x 100 kPa calibrated pressure/ambient pressure
2% = 2% (dialled) × 100/100
2% of 100 = 2 kPaAltitude, pressure 50 kPa
4% = 2% (dialled) × 100/50
4% of 50 = 2 kPaHigh pressure at 200 kPa
1% = 2% (dialled) × 100/200
1% of 200 = 2 kPaSevoflurane has a boiling point of 58°C and, unlike desflurane (which has a boiling point of 22.8°C), does not need to be heated and pressurised with a Tec 6 vaporiser.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 46
Incorrect
-
Given the following hormones, which of these will stimulate glycogenesis and gluconeogenesis?
Your Answer:
Correct Answer: Corticosteroids
Explanation:Insulin is the primary anabolic hormone that dominates regulation of metabolism during digestive phase. It promotes glucose uptake in skeletal myocytes and adipocytes, and other insulin-target cells. It promotes glycogenesis and inhibits gluconeogenesis.
Glucagon is the primary counterregulatory hormone that increases blood glucose levels, primarily through its effects on liver glucose output.
Similar to glucagon, growth hormone, catecholamines and corticosteroids are also counterregulatory factors released in response to decreased glucose concentrations. Growth hormone promotes glycogenolysis and inhibits gluconeogenesis; catecholamines stimulate glycogenolysis and gluconeogenesis; while corticosteroids stimulate glycogenesis and gluconeogenesis.
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This question is part of the following fields:
- Pathophysiology
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Question 47
Incorrect
-
Comparing pressure-volume curves in patients during an asthma attack with that of healthy subjects.
The increased resistive work of breathing in the patients with asthma is best indicated by?Your Answer:
Correct Answer: Larger hysteresis loop
Explanation:A major source of caloric expenditure and oxygen consumption in the body is work of breathing (WOB) and 70% of this is to overcome elastic forces. The remaining 30% is for flow-resistive work
In a normal patient breathing normally, the total area of hysteresis pressure volume curve represents the flow-resistive WOB.
The area of the expiratory resistive work increases during an asthma attack making the compliance curve larger in area. The larger the area the greater the work required to breathe.
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This question is part of the following fields:
- Physiology
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Question 48
Incorrect
-
A 2-year old male is admitted to the surgery ward for repair of an inguinal hernia. He weighs 10 kg. To provide post-operative analgesia, levobupivacaine was administered into the epidural space.
Given the information above, what is the most appropriate dose for the hernia repair?Your Answer:
Correct Answer: 0.25% 7.5 ml
Explanation:Caudal analgesia using bupivacaine is a widely employed technique for achieving both intraoperative and early postoperative pain relief. 0.5 ml/kg of 0.25% plain bupivacaine is favoured by many practitioners who employ this fixed scheme for procedures involving sacral dermatomes (circumcision, hypospadias repair) as well as lower thoracic dermatomes (orchidopexy). However, there are other dosing regimens for caudal blocks with variable analgesic success rates: These include 0.75 ml/kg, 1.0 ml/kg and 1.25 ml/kg.
A study indicated that plain bupivacaine 0.25% at a dose of 0.75 ml/kg compared to a dose of 0.5 ml/kg when administered for herniotomies provided improved quality of caudal analgesia with a low side effects profile. There were consistently more patients with favourable objective pain scale (OPS) scores at all timelines, increased the time to the analgesic request with similar postoperative consumption of paracetamol in the group of patients who received 0.75 ml/kg of 0.25% bupivacaine.
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This question is part of the following fields:
- Pharmacology
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Question 49
Incorrect
-
A graph is created to show the exponential relationship between bacterial growth (y-axis) and time (x-axis).
Which of the following statements is most true about this kind of exponential relationship?Your Answer:
Correct Answer: y = ex
Explanation:The relationship between bacterial growth and time is a tear-away exponential. The mathematical relationship between y and x in this case is:
y = ex
Where: the power is x, and the base is e.
Euler’s number (e) is a mathematical constant that is the base for all logarithms occurring naturally. Its value is 2.718.
The statement X increasing with an increase in Y is proportional to Y refers to the change in y in terms of x when considering any exponential relationship.
This is not a build-up exponential, and that is mathematically stated as y = 1-e-kt.
The negative x axis being a horizontal asymptote and the y intercept being 0, 1 are examples of tearaway exponentials , but do not describe an exponential process.
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This question is part of the following fields:
- Statistical Methods
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Question 50
Incorrect
-
Drug toxicity when using bupivacaine is most likely to occur when this local anaesthetic technique is performed.
Your Answer:
Correct Answer: Intercostal nerve block
Explanation:An intercostal nerve block is used for therapeutic and diagnostic purposes. Intercostal nerve blocks manage acute and chronic pain in the chest area. Common indications are chest wall surgery and shingles or postherpetic neuralgia.
An intercostal nerve block is also an effective option for the management of pain associated with chest trauma and rib fractures. These blocks have been shown to improve oxygenation and respiratory mechanics, and offer pain relief that is comparable to that of epidural analgesia.
This technique, however, is limited by the relatively large doses of local anaesthetic required, and relatively high intravascular uptake from the intercostal space, increasing risk of local anaesthetic toxicity.
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This question is part of the following fields:
- Pharmacology
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Question 51
Incorrect
-
When compared to unipolar diathermy, which of the following is more specific to bipolar diathermy?
Your Answer:
Correct Answer: Has a power output of up to 140 joules per second
Explanation:Electrocautery, also known as diathermy, is a technique for coagulation, tissue cutting, and fulguration that uses a high-frequency current to generate heat (cell destruction from dehydration).
The two electrodes in bipolar diathermy are the tips of forceps, and current passes between the tips rather than through the patient. Bipolar diathermy’s power output (40-140 W) is lower than unipolar diathermy’s typical output (400 W). There is no earthing in the bipolar circuit.
A cutting electrode and a indifferent electrode in the form of a metal plate are used in unipolar diathermy. The high-frequency current completes a circuit by passing through the patient from the active electrode to the metal plate. When used correctly, the current density at the indifferent electrode is low, and the patient is unlikely to be burned. Between the patient plate and the earth is placed an isolating capacitor. This has a low impedance to a high frequency current, such as diathermy current, and is used in modern diathermy machines. The capacitor has a high impedance to current at 50 Hz, which protects the patient from electrical shock.
High frequency currents (500 KHz – 1 MHz) are used in both unipolar and bipolar diathermy, which can cause tissue damage and interfere with pacemaker function (less so with bipolar diathermy).
The effect of diathermy is determined by the current density and waveform employed. The current is a pulsed square wave pattern in coagulation mode and a continuous square wave pattern in cutting mode.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 52
Incorrect
-
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 53
Incorrect
-
Which of the following is true regarding correlation coefficient?
Your Answer:
Correct Answer: It can assume any value between -1 and 1
Explanation:The degree of correlation is summarised by the correlation coefficient (r). This indicates how closely the points lie to a line drawn through the plotted data. In parametric data this is called Pearson’s correlation coefficient and can take any value between -1 to +1. A correlation of -1.0 indicates a perfect negative correlation, and a correlation of 1.0 indicates a perfect positive correlation.
For example
r = 1 – strong positive correlation (e.g. systolic blood pressure always increases with age)
r = 0 – no correlation (e.g. there is no correlation between systolic blood pressure and age)
r = – 1 – strong negative correlation (e.g. systolic blood pressure always decreases with age)
Whilst correlation coefficients give information about how one variable may increase or decrease as another variable increases they do not give information about how much the variable will change. They also do not provide information on cause and effect.
In contrast to the correlation coefficient, linear regression may be used to predict how much one variable changes when a second variable is changed.
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This question is part of the following fields:
- Statistical Methods
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Question 54
Incorrect
-
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 55
Incorrect
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With regards to arterial oxygen content, which of the following contributes most from a quantitative perspective?
Your Answer:
Correct Answer: Haemoglobin concentration
Explanation:The amount of oxygen carried by 100 ml of blood is called the arterial oxygen content (CaO2)and is normally 17-24 ml/dL and can be determined by this equation:
CaO2 = oxygen bound to haemoglobin + oxygen dissolved in plasma
CaO2 = (1.34 × Hgb × SaO2 × 0.01) + (0.003 × PaO2)
where:
1.34 = Huffner’s constant (D) – Huffner’s constant does not change and its magnitude relatively small.
Hgb is the haemoglobin level in g/dL and SaO2 is the percent oxyhaemoglobin saturation of arterial blood
PaO2 is (0.0225 = ml of O2 dissolved per 100 ml plasma per kPa, or 0.003 ml per mmHg).Quantitatively, the amount of oxygen dissolved in plasma is 0.3 mL/dL.
Henry’s law states that at constant temperature, the amount of gas dissolved at equilibrium in a given quantity of a liquid is proportional to the pressure of the gas in contact with the liquid.
Given a haemoglobin concentration of 15 g/dL and a SaO2 of 100% and a PaO2 of 13.3 kPa, the amount of oxygen bound to haemoglobin is 20.4 mL/100mL.
Cardiac output is an important determinant of oxygen delivery but does not influence the oxygen content of blood.
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This question is part of the following fields:
- Basic Physics
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Question 56
Incorrect
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All of the following are causes of hypalbuminaemia except:
Your Answer:
Correct Answer: Starvation
Explanation:Major surgery induces the systemic inflammatory response and this causes endothelial leakage and a low albumin level.
Albumin is a single polypeptide which is made but not stored in the liver. Therefore, levels are a reflection of synthetic activity. It is negatively charged and very soluble.
Only 40% of albumin is intravascular, and the rest in the in interstitial compartment.
If there was normal liver function during starvation, albumin will be maintained and proteolysis will occur elsewhere.
It is not catabolised during starvation.
Starvation and malnutrition may, however, present as part of other disease processes that are associated with hypalbuminaemia.Causes of low albumin are
1. Decreased production (hepatic dysfunction)
2. Increased loss (renal dysfunction)
3. Redistribution (endothelial leak/damage)
4. Increased catabolism (very rare) -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 57
Incorrect
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All of the following statements are false regarding insulin except:
Your Answer:
Correct Answer: Can be detected in the lymph
Explanation:Insulin is secreted from the ? cells of the pancreas. It consists of 51 amino acids arranged in two chains. It interacts with cell surface receptors (not the nuclear receptors and thus mechanism of action is not similar to steroids).
Since insulin can pass from plasma to interstitium and lymphatics, it can be measured in lymph but the concentrations here can be up to 30% less than that of plasma.It decreases blood glucose by stimulating the entry of glucose in muscle and fat (by increasing the synthesis of Glucose transporters)
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This question is part of the following fields:
- Pharmacology
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Question 58
Incorrect
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Concerning the trachea, which of these is true?
Your Answer:
Correct Answer: In an adult is approximately 15 cm long
Explanation:In an adult, the trachea is approximately 15 cm long. It extends at the level of the 6th cervical vertebra, from the lower border of the cricoid cartilage.
The trachea terminates between T4 and T6 at the carina or bronchial bifurcation. This variation is because of changes during respiration.
The trachea has 16-20 C-shaped cartilaginous rings that maintain its patency.
The trachea is first of the 23 generations of air passages in the tracheobronchial tree (not 25), from the trachea to the alveoli..
The inferior thyroid arteries which are branches of the thyrocervical trunk, arise from the first part of the subclavian artery and supplies the trachea.
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This question is part of the following fields:
- Anatomy
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Question 59
Incorrect
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Which of the following drug can be the first-line drug for both broad and narrow complex tachyarrhythmia?
Your Answer:
Correct Answer: Amiodarone
Explanation:Amiodarone is the longest-acting anti-arrhythmic drug. It possesses the action of all classes of antiarrhythmic drugs (Sodium channel blockade, Beta blockade, Potassium channel blockade, and Calcium channel blockade). Due to this property, it has the widest anti-arrhythmic spectrum and thus can be used in both broad and narrow complex tachyarrhythmia.
Adenosine is shortest acting anti-arrhythmic drug.
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This question is part of the following fields:
- Pharmacology
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Question 60
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:
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 61
Incorrect
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All the following statements are false regarding local anaesthetic except
Your Answer:
Correct Answer: Potency is directly related to lipid solubility
Explanation:The potency of local anaesthetics is directly proportional to lipid solubility because they need to penetrate the lipid-soluble membrane to enter the cell.
Protein binding has a direct relationship with the duration of action because the higher the ability of the drug to bind with membrane protein, the higher is the duration of action.
Higher the pKa of a drug, slower the onset of action. Because a drug with higher pKa will be more ionized than the one with lower pKa at a given pH. Local anaesthetics are weak bases, and unionized form diffuses more rapidly across the nerve membrane than the protonated form. As a result drugs with higher pKa will be more ionized will diffuse less across the nerve membrane.
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This question is part of the following fields:
- Pharmacology
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Question 62
Incorrect
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Which of the following statements is true regarding vecuronium?
Your Answer:
Correct Answer: Has a similar structure to rocuronium
Explanation:Vecuronium is used as a part of general anaesthesia to provide skeletal muscle relaxation during surgery or mechanical ventilation. It is a monoquaternary aminosteroid (not quaternary) non- depolarising neuromuscular blocking drug.
It has a structure similar to both rocuronium and pancuronium. The only difference is the substitution of specific groups on the steroid structure.
Vecuronium is not associated with the release of norepinephrine from sympathetic nerve endings. However, Pancuronium has norepinephrine releasing the property.
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This question is part of the following fields:
- Pharmacology
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Question 63
Incorrect
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The right coronary artery supplies blood to all the following, except which?
Your Answer:
Correct Answer: The circumflex artery
Explanation:The right coronary artery supplies the right ventricle, the right atrium, the sinoatrial (SA) node and the atrioventricular (AV) node.
The circumflex artery originates from the left coronary artery and is supplied by it.
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This question is part of the following fields:
- Anatomy
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Question 64
Incorrect
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An 80 year old woman is due for cataract surgery.
There are no contraindications to regional anaesthesia so a peribulbar block was performed. 8mls of 2% lidocaine was injected using an infratemporal approach. However, there is still movement of the globe after 5 mins.
The least likely extraocular muscle to develop akinesia is:Your Answer:
Correct Answer: Superior oblique
Explanation:The fibrotendinous ring formed by the congregation of the rectus muscles at the apex of the orbit does not include superior oblique. This muscle is completely outside the ring and so it is the most difficult muscle to anaesthetise completely. A good grasp of the anatomy of the area being anaesthetised is important with all regional anaesthetic techniques so that potential problems and complications with a block can be anticipated.
The borders of this pyramid whose apex points upwards and outwards of the bony orbit are as follows:
Floor – Zygoma and Maxilla
Roof – frontal bone
Medial wall – maxilla, ethmoid, sphenoid and lacrimal bones.
Lateral wall – greater wing of the sphenoid and the zygoma.The four recti muscles (superior, medial, lateral and inferior) originate from a tendinous ring (the annulus of Zinn) and extend anteriorly to insert beyond the equator of the globe. Bands of connective tissue are present between the rectus muscles forming a conical structure and hinder the passage of local anaesthetic.
The superior oblique muscle is situated outside this ring and is the most difficult muscle to anaesthetise completely, particularly with a single inferotemporal peribulbar injection. An additional medial injection may help to prevent this.
The cranial nerve supply to the extraocular muscles are:
3rd (inferior oblique, inferior recti, medial and superior)
4th (superior oblique), and
6th (lateral rectus).The long and short ciliary nerves provide the sensory supply to the globe and these are branches of the nasociliary nerve, (which is itself a branch of the ophthalmic division of the trigeminal nerve).
To achieve anaesthesia for the eye, these nerves which enter the fibrotendinous ring need to be fully blocked to anaesthetise the eye for surgery.
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This question is part of the following fields:
- Anatomy
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Question 65
Incorrect
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Standard error of the mean can be defined as:
Your Answer:
Correct Answer: Standard deviation / square root (number of patients)
Explanation:The standard error of the mean (SEM) is a measure of the spread expected for the mean of the observations – i.e. how ‘accurate’ the calculated sample mean is from the true population mean. The relationship between the standard error of the mean and the standard deviation is such that, for a given sample size, the standard error of the mean equals the standard deviation divided by the square root of the sample size.
SEM = SD / square root (n)
where SD = standard deviation and n = sample size
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This question is part of the following fields:
- Statistical Methods
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Question 66
Incorrect
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A 55-year-old man with a ventricular rate of 210 beats per minute is admitted to the emergency department with atrial fibrillation. The patient develops ventricular fibrillation shortly after receiving pharmacotherapy to treat his arrhythmia, from which he is successfully resuscitated.
He has a PR interval of 40 Ms, a prominent delta wave in lead I, and a QRS duration of 120 Ms, according to an ECG from a previous admission.
Which of the following drugs is most likely to be involved in this patient's development of ventricular fibrillation?Your Answer:
Correct Answer: Digoxin
Explanation:The Wolff-Parkinson-White syndrome (WPWS) is linked to an additional electrical conduction pathway between the atria and ventricles. This accessory pathway (bundle of Kent), unlike the atrioventricular (AV) node, is incapable of slowing down a rapid rate of atrial depolarization. In other words, a short circuit bypasses the AV node. Patients with a rapid ventricular response or narrow complex AV re-entry tachycardia are more likely to develop atrial fibrillation or flutter.
Digoxin can promote impulse transmission through this accessory pathway if a patient with WPWS develops atrial fibrillation because it works by blocking the AV node. This can cause ventricular fibrillation and an extremely rapid ventricular rate. As a result, it’s not advised.
Adenosine, beta-blockers, and calcium channel blockers, among other drugs that interfere with AV nodal conduction, are also generally contraindicated.
The class III antiarrhythmic drugs amiodarone and ibutilide (K+ channel block) and procainamide (Na+ channel block) are the drugs of choice.
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This question is part of the following fields:
- Pharmacology
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Question 67
Incorrect
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The lung volume that is commonly measured indirectly is?
Your Answer:
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 68
Incorrect
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Over the course of 10 minutes, a normally fit and well 22-year-old male receives a 1 litre intravenous bolus of 20% albumin.
Which of the following primary physiological responses in this patient has the highest chance to influence a change in urine output?
Your Answer:
Correct Answer: Stimulation of atrial natriuretic peptide (ANP) secretion
Explanation:The renal effects of atrial natriuretic peptide (ANP) secretion are as follows:
Increased glomerular filtration rate by dilating the afferent glomerular arteriole. Moreover, it constricts the efferent glomerular arteriole, and relaxes the mesangial cells.
Reduces sodium reabsorption in the collecting ducts and distal convoluted tubule.
The renin-angiotensin system (RAS) is inhibited.
Blood flow in the vasa recta is increased.Because plasma osmolality is unlikely to change, hypothalamic osmoreceptors are unaffected.
The plasma protein has a molecular weight of 66 kDa, is not normally filtered into the proximal convoluted tubule, and has no osmotic diuretic effect.
The following are some basic assumptions:
Extracellular fluid (ECF) makes up one-third of total body water (TBW), while intracellular fluid makes up the other two-thirds (ICF)
One-quarter plasma and three-quarters interstitial fluid make up ECF (ISF)
The volume receptors in the atria have a 7-10% blood volume change threshold.
The osmoreceptors are sensitive to changes in osmolality of 1-2 percent.
The normal plasma osmolality before the transfusion is 287-290 mOsm/kg.
The plasma protein solution is a colloid that is only delivered to the intravascular compartment. The tonicity remains unchanged.
The blood volume increases by 20%, from 5,000 mls to 6,000 mls. This is higher than the volume receptor threshold of 7 to 10%. -
This question is part of the following fields:
- Pathophysiology
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Question 69
Incorrect
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Suppose the afterload and myocardial contractility remain unchanged, which of the following factors in the pressure-volume loop indicates an increase in the preload of the left ventricle?
Your Answer:
Correct Answer: Increased end-diastolic volume
Explanation:If the afterload and myocardiac contractility remains unchanged, an increase in the preload can be attributed to an increase in end-diastolic volume.
Preload can be defined as the initial stretching of the cardiac myocytes prior to contraction. Preload, therefore, is related to muscle sarcomere length. Because sarcomere length cannot be determined in the intact heart, other indices of preload are used such as ventricular end-diastolic volume or pressure. When venous return to the heart is increased, the end-diastolic pressure and volume of the ventricles are increased, which stretches the sarcomeres, thereby increasing their preload.
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This question is part of the following fields:
- Basic Physics
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Question 70
Incorrect
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A 62-year-old woman, presents to emergency department with an ischaemic left colon.
Multiple arteries arise from the aorta at the level of the L3 vertebrae, which is most likely to be involved in this pathology?Your Answer:
Correct Answer: Inferior mesenteric artery
Explanation:The inferior mesenteric artery arises from the abdominal aorta at the level of the L3 vertebrae and supplies blood to the final third of the transverse colon, the descending colon, the sigmoid colon and the uppermost part of the rectum.
It is the artery most likely to affect the left colon.
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This question is part of the following fields:
- Anatomy
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Question 71
Incorrect
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Substitution at different positions of the barbituric ring give rise to different pharmacologic properties.
Substitution with and at which specific site of the ring affects lipid solubility the most?Your Answer:
Correct Answer: Sulphur atom at position 2
Explanation:Barbiturates are derived from barbituric acid, which itself is nondepressant, but appropriate side-chain substitutions result in CNS depressant activity that varies in potency and duration with carbon chain length, branching, and saturation.
Oxybarbiturates retain an oxygen atom on number 2-carbon atom of the barbituric acid ring.
Thiobarbiturates replace this oxygen atom with a sulphur atom, which confers greater lipid solubility. Generally speaking, a substitution such as sulphuration that increases lipid solubility is associated with greater hypnotic potency and more rapid onset, but shorter duration of action.
Addition of a methyl group to the nitrogen atom of the barbituric acid ring, as with oxybarbiturate methohexital, also results in a compound with a short duration of action.
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This question is part of the following fields:
- Pharmacology
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Question 72
Incorrect
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One litre of water at 0°C and a pressure of 1 bar is in a water-bath. A 1 kW element is used in heating it.
Given that the specific heat capacity of water is 4181 J/(kg°C) or J/(kg K), how long will it take to raise the temperature of the water by 10°C?Your Answer:
Correct Answer: 42 seconds
Explanation: -
This question is part of the following fields:
- Physiology
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Question 73
Incorrect
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All the following statements are false regarding carbamazepine except
Your Answer:
Correct Answer: Has neurotoxic side effects
Explanation:Phenytoin, Carbamazepine, and Valproate act by inhibiting the sodium channels when these are open. These drugs also prolong the inactivated stage of these channels (Sodium channels are refractory to stimulation till these reach the closed/ resting phase from inactivated phase)
Carbamazepine is the drug of choice for partial seizures and trigeminal neuralgia
It can have neurotoxic side effects. Major neurotoxic effects include dizziness, headache, ataxia, vertigo, and diplopia
After single oral doses of carbamazepine, the absorption is fairly complete and the elimination half-life is about 35 hours (range 18 to 65 hours). During multiple dosing, the half-life is decreased to 10-20 hours, probably due to autoinduction of the oxidative metabolism of the drug.
It is metabolized in liver into active metabolite, carbamazepine-10,11-epoxide.
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This question is part of the following fields:
- Pharmacology
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Question 74
Incorrect
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An emergency appendicectomy is being performed on a 20 year old man. For maintenance of anaesthesia, he is being ventilated using a circle system with a fresh gas flow (FGF) of 1 L/min (air/oxygen and sevoflurane). The trace on the capnograph shows a normal shape.
The table below demonstrates the changes in the end-tidal and baseline carbon dioxide measurements of the capnograph at 10 and 20 minutes of anaesthesia maintenance.
End-tidal CO2: 4.9 kPa vs 8.4kPa (10 minutes vs 20 minutes)
Baseline end-tidal CO2: 0.2 kPa vs 2.4kPa
Pulse 100-107 beats per minute, systolic blood pressure 125-133 mmHg and oxygen saturation 98-99%.
Which of the following is the single most important immediate course of action?Your Answer:
Correct Answer: Increase the FGF
Explanation:End-tidal carbon dioxide (ETCO2) monitoring has been an important factor in reducing anaesthesia-related mortality and morbidity. Hypercarbia, or hypercapnia, occurs when levels of CO2 in the blood become abnormally high (Paco2 >45 mm Hg). Hypercarbia is confirmed by arterial blood gas analysis. When using capnography to approximate Paco2, remember that the normal arterial–end-tidal carbon dioxide gradient is roughly 5 mm Hg. Hypercarbia, therefore, occurs when PETco2 is greater than 40 mm Hg.
The most likely explanation for the changes in capnograph is either exhaustion of the soda lime and a progressive rise in circuit dead space.
Inspect the soda lime canister for a change in colour of the granules. To overcome soda lime exhaustion, the first step is to increase the fresh gas flow (FGF) (Option A). Then, if need arises, replace the soda lime granules. Other strategies that can work are changing to another circuit or bypassing the soda lime canister, but remember that both these strategies are employed only after increasing FGF first. Exclude other causes of equipment deadspace too.
There are also other causes for hypercarbia to develop intraoperatively:
1. Hypoventilation is the most common cause of hypercapnia. A. Inadequate ventilation can occur with spontaneous breathing due to drugs like anaesthetic agents, opioids, residual NMDs, chronic respiratory or neuromuscular disease, cerebrovascular accident.
B. In controlled ventilation, hypercapnia due to circuit leaks, disconnection or miscalculation of patient’s minute volume.
2. Rebreathing – Soda lime exhaustion with circle, inadequate fresh gas flow into Mapleson circuits and increased breathing system deadspace.
3. Endogenous source – Tourniquet release, hypermetabolic states (MH or thyroid storm) and release of vascular clamps.
4. Exogenous source – Absorption of CO2 from pneumoperitoneum. -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 75
Incorrect
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Which of the following is true regarding the mechanism of action of daptomycin?
Your Answer:
Correct Answer: Interferes with the outer membrane of gram positive bacteria resulting in cell death
Explanation:Daptomycin alters the curvature of the membrane, which creates holes that leak ions. This causes rapid depolarization, resulting in loss of membrane potential. Thus it interferes with the outer membrane of gram-positive bacteria resulting in cell death.
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This question is part of the following fields:
- Pharmacology
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Question 76
Incorrect
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Which of the following vertebral levels is the site where the aorta perforates the diaphragm?
Your Answer:
Correct Answer: T12
Explanation:The diaphragm divides the thoracic cavity from the abdominal cavity. Structures penetrate the diaphragm at different vertebral levels through openings in the diaphragm to communicate between the two cavities. The diaphragm has openings at three vertebral levels:
T8: vena cava, terminal branches of the right phrenic nerve
T10: oesophagus, vagal trunks, left anterior phrenic vessels, oesophageal branches of the left gastric vessels
T12: descending aorta, thoracic duct, azygous and hemi-azygous vein -
This question is part of the following fields:
- Anatomy
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Question 77
Incorrect
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An 18-year old female was brought into the emergency room because of active seizures. The informant reported that it has been more than 5 minutes since the patient started seizing. The attending physician gave an initial diagnosis of status epilepticus.
According to the paramedics who brought in the patient, 10 mg of diazepam was given rectally. Upon physical examination, she was normotensive at 120/80 mmHg; tachycardic at 138 beats per minute; tachypnoeic at 24 breaths per minute; and well-saturated at 99% on high flow oxygen. Her random blood glucose level was normal at 7.0 mmol/L.
Given this situation and an initial diagnosis of status epilepticus, what would be the best initial anti-epileptic drug to administer to the patient?Your Answer:
Correct Answer: Lorazepam
Explanation:Lorazepam is an intermediate-acting benzodiazepine that binds to the GABA-A receptor subunit to increase the frequency of chloride channel opening and facilitate membrane hyperpolarization. It is the preferred treatment for status epilepticus, although Diazepam can also be used as an alternative.
Lorazepam has a longer duration of action than Diazepam, and binds with greater affinity to the GABA-A receptor subunit.
Phenobarbital is a barbiturate that acts on the GABA-A receptor site to increase the duration of chloride channel opening. Barbiturates, particularly phenobarbital, is considered the drug of choice for seizures in infants.
Phenytoin is a sodium-channel blocker that is given for generalized tonic-clonic seizures, partial seizures, and status epilepticus. Phenytoin is preferred in prolonged therapy for status epilepticus because it is less sedating.
Propofol or thiopentone is preferred when airway protection is required.
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This question is part of the following fields:
- Pharmacology
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Question 78
Incorrect
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Which of the following statements is true regarding dopamine?
Your Answer:
Correct Answer: It can increase or decrease cAMP levels
Explanation:Dopamine (DA) is a dopaminergic (D1 and D2) as well as adrenergic ? and?1 (but not ?2 ) agonist.
The D1 receptors in renal and mesenteric blood vessels are the most sensitive: i.v. infusion of a low dose of DA dilates these vessels (by raising intracellular cAMP). This increases g.f.r. In addition, DA exerts a natriuretic effect by D1 receptors on proximal tubular cells.
Moderately high doses produce a positive inotropic (direct?1 and D1 action + that due to NA release), but the little chronotropic effect on the heart.
Vasoconstriction (?1 action) occurs only when large doses are infused.
At doses normally employed, it raises cardiac output and systolic BP with little effect on diastolic BP. It has practically no effect on nonvascular ? and ? receptors; does not penetrate the blood-brain barrier—no CNS effects.
Dopamine is used in patients with cardiogenic or septic shock and severe CHF wherein it increases BP and urine outflow.
It is administered by i.v. infusion (0.2–1 mg/min) which is regulated by monitoring BP and rate of urine formation
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This question is part of the following fields:
- Pharmacology
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Question 79
Incorrect
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A 68-year-old man presents worried about his risk of motor neurone disease. No symptoms have developed, but his father suffered from motor neurone disease. Recently, his cousin has also been diagnosed with amyotrophic lateral sclerosis. He searched the internet for screening tests for motor neurone disease and found a blood test called ‘neuron’, and requests to have it done. You search this blood test and find a prospective study going on evaluating the potential benefits of this blood test. On average, this test diagnosed patients with the disease 8 months earlier than the patients who are diagnosed on the basis of their clinical symptoms. The patients diagnosed using this neuron test also survived, on average, 48 months from the diagnosis, whereas the patients diagnosed clinically survived an average of 39 months from the diagnosis. Considering the clear benefits, you decide to have it done on the patient.
Which of the following options best relate to the above scenario?Your Answer:
Correct Answer: Lead-time bias
Explanation:Hypochondriasis is an illness anxiety disorder, and describes excessively worriedness about the presence of a disease. While the woman is concerned about her possibility of developing motor neurone disease, she understands that no symptoms have yet appeared. Hypochondriasis involves patients who refuse to accept that they don’t have the disease, even if the results come back negative.
Late Look Bias occurs when the data is gathered or analysed at an inappropriate time e.g. when many of the subjects suffering from a fatal disease have died. This type of biasness might occur in some retrospective studies of motor neurone disease, but is not applicable to this prospective study.
In procedure bias, the researcher decides assignment of a treatment versus control and assigns particular patients to one group or the other non-randomly. This is unlikely to have occurred in this case, although it is not mentioned specifically. Of all the options, lead time-bias is a better answer.
The Hawthorne Effect refers to groups modifying their behaviour simply because they are aware of being observed. Any differences in the behaviour have not been mentioned in the question, and it is highly unlikely that a change in patient’s behaviour would have affected their length of survival in this case.
The correct option is lead-time bias. Even if the new blood test diagnoses the disease earlier, it doesn’t affect the outcome, as the survival time was still on average 43 months from the onset of symptoms in both groups. With the help of blood test, the disease was only detected 8 months earlier.
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This question is part of the following fields:
- Statistical Methods
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Question 80
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 81
Incorrect
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The phenomenon that the patients behaved in a different manner when they know that they are being observed is termed as?
Your Answer:
Correct Answer: Hawthorne effect
Explanation:Hawthorne effect explains the change in any behavioural aspect owing to the awareness that the person is being observed.
Simpson’s Paradox explains the association developed when the data from several groups is combined to form a single larger group.The remaining terms are made up.
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This question is part of the following fields:
- Statistical Methods
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Question 82
Incorrect
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A laceration to the upper lateral margin of the popliteal fossa will pose the greatest risk of injury for which nerve?
Your Answer:
Correct Answer: Common peroneal nerve
Explanation:The common peroneal (fibular) nerve descends obliquely along the lateral side of the popliteal fossa to the fibular head, medial to biceps femoris.
The sural nerve exits at the fossa’s lower inferolateral aspect and is more at risk in short saphenous vein surgery.
The tibial nerve lies more medially and is even less likely to be injured in this location.
The boundaries of the popliteal fossa are:
Superolateral – the biceps femoris tendon
Superomedial – semimembranosus reinforced by semitendinosus
Inferomedial and inferolateral – medial and lateral heads of gastrocnemiusThe contents of the Popliteal fossa are:
1. The popliteal artery
2. The popliteal vein
3. The Tibial nerve and common Fibular nerve
4. Posterior femoral cutaneous nerve: descends and pierces the roof
5. Small saphenous vein
6. popliteal lymph nodes
7. fat -
This question is part of the following fields:
- Anatomy
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Question 83
Incorrect
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Pressure volume loop represents the compliance of left ventricle.
Considering there is no change in preload and myocardial contractility, which physiological change may result an increase in left ventricular afterload?Your Answer:
Correct Answer: Increased end-systolic volume
Explanation:If there is no change in preload and myocardial contractility, there will be decrease in end-diastolic volume and stroke volume. So there must be increase in end-systolic volume.
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This question is part of the following fields:
- Physiology
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Question 84
Incorrect
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All of the following statements are false regarding tetracyclines except:
Your Answer:
Correct Answer:
Explanation:Tetracyclines inhibit protein synthesis through reversible binding to bacterial 30s ribosomal subunits (not 50s) which prevent binding of new incoming amino acids (aminoacyl-tRNA) and thus interfere with peptide growth.
They penetrate macrophages and are thus a drug of choice for treating infections due to intracellular organisms.
Tetracycline does not inhibit transpeptidation. Meanwhile, it is chloramphenicol which is responsible for inhibiting transpeptidation.
Tetracycline can get deposited in growing bone and teeth due to its calcium-binding effect and thus causes dental discoloration and dental hypoplasia. Due to this reason, they should be avoided in pregnant or lactating mothers.
Simultaneous administration of aluminium hydroxide can impede the absorption of tetracyclines.
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This question is part of the following fields:
- Pharmacology
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Question 85
Incorrect
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A post-operative patient was given paracetamol and pethidine for post-operative analgesia. A few hours later, the patient developed fever of 38°C, hypertension, and agitation.
According to the patient's medical history, he is maintained on Levodopa and Selegiline for Parkinson's disease.
Which of the following is the most probable cause of his manifestation?Your Answer:
Correct Answer: Pethidine
Explanation:Selegiline is a monoamine oxidase inhibitor. Inhibition of monoamine oxidase leads to increased levels of norepinephrine and serotonin in the central nervous system.
Pethidine, also known as meperidine, is a strong agonist at the mu and kappa receptors. It inhibits pain neurotransmission and blocks muscarinic-specific actions.
Administering opioid analgesic is relatively contraindicated to individuals taking monoamine oxidase inhibitors. This is because of the high incidence of serotonin syndrome, which is characterized by fever, agitation, tremor, clonus, hyperreflexia and diaphoresis. Onset of symptoms is within hours, and the treatment is mainly through sedation, paralysis, intubation and ventilation.
The clinical findings are more consistent with Serotonin syndrome rather than exacerbation of Parkinson’s. Parkinson’s Disease (PD) exacerbations are defined as patient-reported or caregiver-reported episodes of subacute worsening of PD motor function in 1 or more domains (bradykinesia, tremor, rigidity, or PD-related postural instability/gait disturbance) that caused a decline in functional status, developed over a period of < 2 months, did not fluctuate with medication timing, and are not caused by intentional adjustments of PD medications by the treating neurologist. Malignant hyperthermia usually occurs within minutes of administration of a volatile anaesthetic, such as halothane, or succinylcholine. There is massive release of calcium from the sarcoplasmic reticulum, leading to fever, acidosis, rhabdomyolysis, trismus, clonus, and hypertension. In sepsis, it more common for patients to present with hypotension rather than hypertension.
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This question is part of the following fields:
- Pharmacology
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Question 86
Incorrect
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Which of the following statement is true regarding the mechanism of action of rifampicin?
Your Answer:
Correct Answer: Inhibit RNA synthesis
Explanation:Rifampicin is a derivative of a rifamycin (other derivatives are rifabutin and rifapentine). It is bactericidal against both dividing and non-dividing mycobacterium and acts by inhibiting DNA-dependent RNA polymerase. Thus this drug inhibits RNA synthesis.
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This question is part of the following fields:
- Pharmacology
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Question 87
Incorrect
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Prior to an urgent appendicectomy, a 49-year-old man requires a rapid sequence induction.
His BMI is equal to 50.
Which of the following formulas is the most appropriate for calculating a suxamethonium dose in order to achieve optimal intubating conditions?Your Answer:
Correct Answer: 1-1.5 × actual body weight (mg)
Explanation:The usual method of calculating the dose of a drug to be given to patients of normal weight is to use total body weight (TBW). This is because the lean body weight (LBW) and ideal body weight (IBW) dosing scalars are similar in these patients.
Because the LBW and fat mass do not increase in proportion in patients with morbid obesity, this is not the case. Drugs that are lipid soluble, such as propofol or thiopentone, can cause a relative overdose. Lean body mass is a better scalar in these situations.
Suxamethonium has a small volume of distribution, so the dose is best calculated using the TBW to ensure optimal and deep intubating conditions. The higher dose was justified because these patients’ plasma cholinesterase activity was elevated.
Other scalars include:
The dose of highly lipid soluble drugs like benzodiazepines, thiopentone, and propofol can be calculated using lean body weight (LBW). The formula LBW = IBW + 20% can be used on occasion.
Fentanyl, rocuronium, atracurium, vecuronium, morphine, paracetamol, bupivacaine, and lidocaine are all administered with LBW.
Formulas can be used to calculate the ideal body weight (IBW). There are a number of drawbacks, including the fact that patients of the same height receive the same dose, and the formulae do not account for changes in body composition associated with obesity. Because IBW is typically lower than LBW, administering a drug based on IBW may result in underdosing. The body mass index (BMI) isn’t used to calculate drug dosage directly.
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This question is part of the following fields:
- Pharmacology
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Question 88
Incorrect
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You are approached by a drug rep who tells you about a new drug. The dosage and side effects of the drug are being determined in a trial. The representative asks you to refer participants for the trial.
What type of participants should you refer? In which phase of trials is the drug currently in?Your Answer:
Correct Answer: Healthy participants, Phase 1
Explanation:Phase 2 trials involve patients that are suffering from the disease under study and are associated with determining the efficiency and the optimum dosage of the drug.
Phase 0 trials assist the scientists in studying the behaviour of drugs in humans by micro dosing patients. They are used to speed up the developmental process. They have no measurable therapeutic effect and efficiency.
Phase 1 is associated with assessing whether a drug is safe to use or not. The process is extensive and can take up to several months. It also involves healthy participants (less than 100) that are paid to take part in the study. The side effects upon increasing dosage are also addressed by the study. The effects the drug has on humans including how its absorbed, metabolized and excreted are studied. Approximately 70% of the drugs pass this phase.
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This question is part of the following fields:
- Statistical Methods
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Question 89
Incorrect
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The following is normally higher in concentration extracellularly than intracellularly
Your Answer:
Correct Answer: Sodium
Explanation:The ions found in higher concentrations intracellularly than outside the cells are:
ATP
AMP
Potassium
Phosphate, and
Magnesium Adenosine diphosphate (ADP)Sodium is a primarily extracellular ion.
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This question is part of the following fields:
- Physiology
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Question 90
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:
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 91
Incorrect
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Which one of the following factor affects the minimal alveolar concentration (MAC)?
Your Answer:
Correct Answer: Hypoxaemia
Explanation:The minimal alveolar concentration (MAC) is the concentration of an inhalation anaesthetic agent in the lung alveoli required to stop a response to the surgical stimulus in 50% of the patient.
Following factors don’t affect the MAC of the inhaled anaesthetic agents:
Gender, acidosis, alkalosis, hypothyroidism, hyperthyroidism, body weight, serum potassium level, and the duration of the anaesthesia.
MAC increase in children, elevated temperature, high metabolic rate, sympathetic increase and chronic alcoholism.
MAC decrease in low temperature, low oxygen level, old age, hypotension (<40 mmHg), depressant drugs e.g. opioids and low level of catecholamines; alpha methyl dopa. Carbon dioxide O2 at the pressure > 120mmHg is being used in anesthetic-Hinkman as an additive effect to decrease MAC, however, increase concentration of CO2 activates the sympathetic system resulting the MAC increases.
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This question is part of the following fields:
- Physiology
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Question 92
Incorrect
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A young male is undergoing inguinal hernia repair. During the procedure, the surgeons approach the inguinal canal and expose the superficial inguinal ring.
Which structure forms the lateral edge of the superficial inguinal ring?Your Answer:
Correct Answer: External oblique aponeurosis
Explanation:The superficial inguinal ring is an opening in the aponeurosis of the external oblique muscle, just above and lateral to the pubic crest.
The superficial ring resembles a triangle more than a ring with the base lying on the pubic crest and its apex pointing towards the anterior superior iliac spine. The sides of the triangle are crura of the opening in the external oblique aponeurosis. The lateral crura of the triangle is attached to the pubic tubercle. The medial crura of the triangle is attached to the pubic crest.
The external oblique aponeurosis forms the anterior wall of the inguinal canal and also the lateral edge of the superficial inguinal ring. The rectus abdominis lies posteromedially, and the transversalis posterior to this.
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This question is part of the following fields:
- Anatomy
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Question 93
Incorrect
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All of the following statements are false regarding propranolol except:
Your Answer:
Correct Answer: Has a plasma half life of 3-6 hours.
Explanation:Propranolol is a nonselective beta-blocker with a half-life of 3 to 6 hours.
Since it is lipid-soluble it crosses the blood-brain barrier and causes Central Nervous System side effects like sedation, nightmares, and depression.
They are contraindicated in asthma, Congestive heart failure, and diabetes.
It has a large volume of distribution with no intrinsic sympathomimetic action.
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This question is part of the following fields:
- Pharmacology
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Question 94
Incorrect
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An older woman has been brought into the emergency department with symptoms of a stroke. A CT angiogram is performed for diagnosis, which displays narrowing in the artery that supplies the right common carotid. Which of the following artery is the cause of stroke in this patient?
Your Answer:
Correct Answer: Brachiocephalic artery
Explanation:The arch of aorta gives rise to three main branches:
1. Brachiocephalic artery
2. Left common carotid artery
3. Left subclavian arteryThe brachiocephalic artery then gives rise to the right subclavian artery and the right common carotid artery.
The right common carotid artery arises from the brachiocephalic trunk posterior to the sternoclavicular joint.
The coeliac trunk is a branch of the abdominal aorta.
The ascending aorta supplies the coronary arteries. -
This question is part of the following fields:
- Anatomy
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Question 95
Incorrect
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A 70-year-old man will have a PICC line inserted as he requires long-term parenteral nutrition. To gain venous access, the line is inserted into the basilic vein at the elbow region.
As the catheter tip advances into the basilic vein, which venous structure will it first encounter?
Your Answer:
Correct Answer: Axillary vein
Explanation:A peripherally inserted central catheter (PICC) line is a long, thin tube inserted into the vein of a patient’s arm to gain access to the large central veins near the heart. PICC line is indicated for parenteral nutrition or to deliver medications. They can be used for medium-term venous access, defined as anywhere between several weeks to 6 months.
The veins of choice for PICC are:
1. Basilic
2. Brachial
3. Cephalic
4. Medial cubital veinThe vein of choice is the right basilic vein as it has a large circumference and is located superficially. It has the most straight route to the final destination of PICC (SVC or Right atrium). It courses through the axillary vein, then the subclavian, and finally settles into the SVC. It also has the least number of valves and a shallow angle of insertion when compared to the other veins.
The basilic vein drains the medial end of the dorsal arch of the upper limb, passes along the medial aspect of the forearm, and pierces the deep fascia at the elbow. The basilic vein joins the venae comitantes of the brachial artery to form the axillary vein at the elbow.
The posterior circumflex humeral vein is encountered before the axillary vein. However, a PICC line is unlikely to enter this structure because of its entry angle into the basilic vein. -
This question is part of the following fields:
- Anatomy
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Question 96
Incorrect
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When an inotrope is given to the body, it has the following effects on the cardiovascular system:
The automaticity of the sino-atrial node increases
Lusitropy is accelerated
Dromotropy is increased
Chronotropy is increased
Inotropy increases
There is increased excitability of the conducting system
The most probably mechanism of action of this compound is?Your Answer:
Correct Answer: Increase in intracellular calcium influenced by a conformational change of a Gs protein
Explanation:A beta-1 adrenoreceptor agonist is most likely the ligand that causes increased automaticity, increased chronotropy, increased excitability, and increased inotropy on the sino-atrial node. However, alpha-1 adrenoreceptor effects may cause an increase in systemic vascular resistance. Noradrenaline, adrenaline, dopamine, and ephedrine are examples of drugs with mixed alpha and beta effects.
Adrenaline, noradrenaline, dopamine, dopexamine, dobutamine, ephedrine, and isoprenaline are examples of drugs that have some beta-1 activity. The beta-1 receptor is a G protein-coupled metabotropic receptor. When the beta-1 agonist binds to the cell surface membrane, it causes a conformational change in the Gs unit, which triggers a cAMP-dependent pathway and a calcium influx into the cell.
Catecholamines also help to relax the heart muscle (positive lusitropy). Dromotropy is the ability to increase the atrioventricular (AV) node’s conduction velocity.
Inodilators cause an increase in intracellular calcium as a result of phosphodiesterase III (PDIII) inhibition. Milrinone, enoximone, and amrinone are some examples. Positive inotropy is caused by increased calcium entry into the myocytes. Lusitropy is also increased by phosphodiesterase inhibitors. Increased cAMP inhibits myosin light chain kinase, resulting in reduced phosphorylation of vascular smooth muscle myosin, lowering systemic and pulmonary vascular resistance.
The mechanism of action of alpha-1 adrenoreceptor agonists is an increase in intracellular calcium caused by an increase in inositol triphosphate (IP3). IP3 is a second messenger that causes an increase in systemic vascular resistance by stimulating the influx of Ca2+ into smooth muscle cells. Reflex bradycardia can occur as a result of the subsequent increase in blood pressure. Phenylephrine and metaraminol are examples of pure alpha-1 agonists.
Levosimendin is a novel inotrope that makes myocytes more sensitive to intracellular Ca2+. It causes a positive inotropy without changing heart rate or oxygen consumption significantly.
The Na-K-ATPase membrane pump in the myocardium is inhibited by digoxin. This inhibition promotes sodium-calcium exchange, resulting in an increase in intracellular Ca2+ and increased contraction force. The parasympathetic effects of digoxin on the AV node result in bradycardia. Systemic vascular resistance will not be affected by it.
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This question is part of the following fields:
- Pathophysiology
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Question 97
Incorrect
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International colour coding is used on medical gas cylinders. Other characteristics also play a role in determining the gas's identity within a cylinder.
Which of the following options best describes a cylinder containing analgesics for obstetrics?Your Answer:
Correct Answer: Blue body, blue/white shoulder, full cylinder; 13700 KPa, gas mixture, requires a dual stage pressure regulator
Explanation:The body of the Entonox cylinder is usually blue (occasionally white), with blue and white shoulders. Entonox contains a 50:50 mixture of oxygen and nitrous oxide, with a full cylinder pressure of 13700 KPa (137 bar). The cylinder is equipped with a two-stage pressure regulator for safe operation.
The cylinder body and shoulder of nitrous oxide are (French) blue.
In today’s anaesthetic workstations, carbon dioxide cylinders are no longer used.
The body of an oxygen cylinder is black, with a white shoulder.
The white Heliox (21 percent oxygen and 79 percent helium) cylinder has a brown and white shoulder. The administration of this gas mixture, which is less dense than air, is used to reduce turbulence (stridor) of inspiratory flow in patients with upper airway obstruction.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 98
Incorrect
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A log-dose response curve is plotted after drug A is given. The shape of this curve is sigmoid, with a maximum response of 100%.
The log-dose response curve of drug A shifts to the right with a maximum response of 100 percent when drug B is administered.
What does this mean in terms of drug B?Your Answer:
Correct Answer: Drug B has affinity for the receptor but has no intrinsic efficacy
Explanation:Drug A is a pure agonist for the receptor, with high intrinsic efficacy and affinity, according to the log-dose response curve.
Drug B, on the other hand, works as a competitive antagonist. It binds to the receptor but has no inherent efficacy. Drug A’s efficacy will not change, but its potency will be reduced.
A partial agonist is a drug with partial intrinsic efficacy and affinity for the receptor. Giving a partial agonist after a pure agonist will not increase receptor occupancy or decrease receptor activity, and thus will not affect drug A’s efficacy. The inverse agonist flumazenil can reverse all benzodiazepines.
An inverse agonist is a drug that binds to the receptor but has the opposite pharmacological effect.
A non-competitive antagonist is a drug that has affinity for a receptor but has different pharmacological effects and reduces the efficacy of an agonist for that receptor.
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This question is part of the following fields:
- Pharmacology
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Question 99
Incorrect
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A 68-year-old man is to be operated.
His past history is significant for a stroke, and some residual neurological deficit. The cranial nerves are examined clinically. He is unable to rotate his head to the left side when resistance is applied. Moreover, there is tongue wasting on the right side. There are no unusual sensory signs and symptoms.
The most likely reason for these clinical findings is?Your Answer:
Correct Answer: Damage to hypoglossal (XII) and spinal accessory (XI) nerves
Explanation:The upper five cervical segments of the spinal cord give rise to the XI cranial nerve. They connect with a few smaller branches before exiting the skull through the jugular foramen. The sternomastoid and trapezius muscles get their motor supply from the accessory root. Except for the palatoglossus, the hypoglossal nerve supplies motor supply to all tongue muscles.
The inability to shrug the shoulder on the affected side and rotate the head to the side against resistance is caused by damage to the spinal accessory nerve. This is due to the trapezius and sternomastoid muscles’ weakness.
The hypoglossal nerve is damaged, resulting in tongue wasting and inability to move from side to side.
The stylopharyngeus receives motor supply from the glossopharyngeal nerve. It also carries taste sensory fibres from the back third of the tongue, as well as the carotid sinus, carotid body, pharynx, and middle ear.
Motor supply to the larynx, pharynx, and palate; parasympathetic innervation to the heart, lung, and gut; and sensory fibres from the epiglottis and valleculae are all provided by the vagus nerve.
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This question is part of the following fields:
- Pathophysiology
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Question 100
Incorrect
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A 40-year-old obese woman has complaints of heartburn and regurgitation that is worse on lying flat. The doctor suspects gastroesophageal reflux due to a hiatus hernia. Lifestyle modifications to lose weight and antacids are prescribed to her.
At which level of the diaphragm will you find an opening for this problem?Your Answer:
Correct Answer: T10
Explanation:Hiatus is an opening in the diaphragm. A hiatal hernia is a protrusion of the upper part of the stomach through an opening in the diaphragm, the oesophageal hiatus, into the thorax. The oesophageal hiatus occurs at the level of T10 in the right crus of the diaphragm.
Other important openings in the diaphragm:
T8: vena cava, terminal branches of the right phrenic nerve
T10: oesophagus, vagal trunks, left anterior phrenic vessels, oesophageal branches of the left gastric vessels
T12: descending aorta, thoracic duct, azygous and hemi-azygous veinAn opening in the diaphragm is called a hiatus. The oesophageal hiatus is at vertebral level T10. A hiatus hernia is where the stomach bulges through the oesophageal hiatus hence the name – hiatus hernia.
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This question is part of the following fields:
- Anatomy
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