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Question 1
Incorrect
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Which of the following antibiotics inhibits protein synthesis in bacteria?
Your Answer: Penicillin
Correct Answer: Erythromycin
Explanation:Erythromycin binds to the 50s subunit of bacterial rRNA complex and inhibits protein synthesis.
Vancomycin binds to the acyl-D-ala-D-ala portion of the growing cell wall in a susceptible gram-positive bacterium. After binding, it prevents the cell wall from forming the cross-linking.
Trimethoprim binds to dihydrofolate reductase and inhibits the reduction of dihydrofolic acid to tetrahydrofolic acid. Tetrahydrofolic acid is an essential precursor in the thymidine synthesis pathway and interference with this pathway inhibits bacterial DNA synthesis.
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This question is part of the following fields:
- Pharmacology
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Question 2
Correct
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A 63-year old male who has heart failure has peripheral oedema and goes to the GP's office. The GP notes that he is fluid-overloaded. This causes his atrial myocytes to release atrial natriuretic peptide (ANP). ANP's main action is by which of these mechanisms?
Your Answer: Antagonist of angiotensin II
Explanation:Atrial natriuretic peptide (ANP) is secreted mainly from myocytes of right atrium and ventricle in response to increased blood volume.
It is secreted by both the right and left atria (right >> left).It is a 28 amino acid peptide hormone, which acts via cGMP
degraded by endopeptidases.It serves to promote the excretion of sodium, lowers blood pressure, and antagonise the actions of angiotensin II and aldosterone.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 3
Correct
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A 25-year old male with palpitations and dizziness presents to the emergency room. In the triage process, cardiac monitoring shows supraventricular tachycardia with a heart rate of 200 beats per minute. This high heart rate arises as a result of different specialised cells and nerve fibres in the heart which are responsible for conducting that action potential which is generated in the event of systole.
The fastest conduction velocity is carried out by which of the following?Your Answer: Purkinje fibres
Explanation:The correct answer is the Purkinje fibres, which conducts at a velocity of about 4m/sec.
The electrical conduction system of the heart starts with the SA node which generates spontaneous action potentials.
This is conducted across both atria by cell to cell conduction, and occurs at around 1 m/s. The only pathway for the action potential to enter the ventricles is through the AV node in a normal heart.
At this site, conduction is very slow at 0.05ms, which allows for the atria to completely contract and fill the ventricles with blood before the ventricles depolarise and contract.The action potentials are conducted through the Bundle of His from the AV node which then splits into the left and right bundle branches. This conduction is very fast, (,2m/s), and brings the action potential to the Purkinje fibres.
Purkinje fibres are specialised conducting cells which allow for a faster conduction speed of the action potential (,2-4m/s). This allows for a strong synchronized contraction from the ventricle and thus efficient generation of pressure in systole.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 4
Correct
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If a patient is to be placed under general anaesthesia using total intravenous technique with target-controlled infusions of propofol and remifentanil, what safety precaution is the most vital in this a scenario?
Your Answer: Cannula access site clearly visible and regularly checked
Explanation:According to the Safe Anaesthesia Liaison Group, the most important factor to consider the cannula access, and if the patient is properly receiving the total intravenous anaesthesia. The cannula access must be regularly checked for kinks, leaks and disconnections.
Below are the safety precautions and policies to be followed for total intravenous anaesthesia among children and adults:
When administering TIVA, a non-return valve must be used on any intravenous fluid line;
When using equipment, it is essential that clinical staff know its limitations and uses;
Sites of intravenous infusions should be visible so they may be monitored for disconnection, leaks or perivenous infusion into the subcutaneous tissues; and,
Organisations must give preference to clearly labelled intravenous connectors and valves. -
This question is part of the following fields:
- Pathophysiology
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Question 5
Incorrect
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The half-empty cylinder weighs 4.44 kg.
The tare weight of a nitrous oxide cylinder is 4 kg.
The molecular weight of nitrous oxide is 44gm.
Based on the data, how many litres of nitrous oxide remains in the cylinder for use?Your Answer: 112 litres
Correct Answer: 224 litres
Explanation:The Tare weight of a cylinder is the weight when it is empty. So,
Weight of cylinder – tare weight = weight of remaining N2O (g).
4.44 kg – 4 kg = 0.44 kg
Here,
0.44 kg of nitrous oxide remains in the cylinderSince the molecular weight of nitrous oxide is 44 g and one mole of an ideal gas will occupy a volume of 22.4 litres at STP
Therefore amount left in the cylinder is several (gN2O/44) x 22.4 litres of N2O.(440/44) x 22.4 = 224 litres.
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This question is part of the following fields:
- Basic Physics
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Question 6
Correct
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With a 10-day history of severe vomiting, a 71-year-old man with a gastric outlet obstruction is admitted to the surgical ward.
The serum biochemical results listed below are available:
Sodium 128 mmol/L (137-144)
Potassium 2.6 mmol/L (3.5-4.9)
Chloride 50 mmol/L (95-107)
Urea 12 mmol/L (2.5-7.5)
Creatinine 180 µmol/L (60-110)
Which of the following do you think you are most likely to encounter?Your Answer: The standard base excess will be higher than actual base excess
Explanation:Hydrochloric acid is lost when you vomit for a long time (HCl). As a result, the following can be expected, in varying degrees of severity:
Hypokalaemia
Hypochloraemia
Increased bicarbonate to compensate for chloride loss and metabolic alkalosisThe alkalosis causes potassium to move from the intracellular to the extracellular compartment at first. Long-term vomiting and dehydration cause potassium to be excreted by the kidneys in order to conserve sodium. Dehydration can cause urea and creatinine levels to rise.
The actual base excess is always greater than the standard base excess.
The actual base excess (BE) is a measurement of a base’s contribution to a blood gas picture’s metabolic component. It’s the amount of base that needs to be added to a blood sample to bring the pH back to 7.4 after the respiratory component of a blood gas picture has been corrected (PaCO2 of 40 mmHg or 5.3 kPa). The BE has a normal range of +2 to 2. A large positive BE indicates a severe metabolic alkalosis, while a large negative BE indicates a severe metabolic acidosis. As a result, the actual BE in vitro is unaffected by CO2.
In vivo, however, standard BE is not independent of pCO2 because blood with haemoglobin acts as a better buffer than total ECF.
As a result, it is impossible to tell the difference between compensating for a respiratory disorder and compensating for the presence of a primary metabolic disorder.
The differences between in vitro and in vivo behaviour can be mostly eliminated if the BE is calculated for a haemoglobin concentration of 50 g/L (the ‘effective’ or virtual value of Hb if it was distributed throughout the extracellular space) rather than the actual haemoglobin. Because haemoglobin has a lower buffering capacity, the standard BE is higher than the actual BE. It reflects the BE better in the extracellular space rather than just the intravascular compartment.
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This question is part of the following fields:
- Pathophysiology
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Question 7
Correct
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Which of the following statements is true about monoamine oxidase (MOA) enzymes?
Your Answer: Type A and type B are found in the liver and brain
Explanation:Monoamine oxidase (MOA) enzymes are responsible for the catalyses of monoamine oxidative deamination. It assists the degradation of serotonin, norepinephrine (NE) and dopamine.
They are found in the mitochondria of most central and peripheral nerve tissues.
There are 2 different types:
Type A: Whose main function it to inactivate dopamine, tyramine, norepinephrine and 5-hydroxytryptamine. In addition to the nervous system, it is also found in the liver, brain gastrointestinal tract, pulmonary endothelium and placenta
Type B: Whose main function is to inactivate dopamine, tyramine, tryptamine and phenylethylamine. In addition to the nervous system, it is also found in the liver, brain (especially in the basal ganglia) and blood platelets. -
This question is part of the following fields:
- Pathophysiology
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Question 8
Incorrect
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A 30-year-old man has been stabbed in an area of the groin that contains the femoral triangle. He will undergo explorative surgery.
Which of the following makes the lateral wall of the femoral triangle?Your Answer: Adductor longus
Correct Answer: Sartorius
Explanation:The femoral triangle is a wedge-shaped area found within the superomedial aspect of the anterior thigh. It is a passageway for structures to leave and enter the anterior thigh.
Superior: Inguinal ligament
Medial: Adductor longus
Lateral: Sartorius
Floor: Iliopsoas, adductor longus and pectineusThe contents include: (medial to lateral)
Femoral vein
Femoral artery-pulse palpated at the mid inguinal point
Femoral nerve
Deep and superficial inguinal lymph nodes
Lateral cutaneous nerve
Great saphenous vein
Femoral branch of the genitofemoral nerve -
This question is part of the following fields:
- Anatomy
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Question 9
Incorrect
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A 58-year-old man, visits his general practitioner complaining of a lump in his groin. He explains he is otherwise well and reports no other symptoms. The lump is examined and is found to be soft, and can be reduced without causing the patient pain. The GP diagnoses an inguinal hernia. To determine the nature of the hernia, the GP reduced the lump and applies pressure on the deep inguinal ring.
The deep inguinal ring has what anatomical landmark?Your Answer: Superior and medial to anterior superior iliac spine
Correct Answer: Superior to the midpoint of the inguinal ligament
Explanation:The deep inguinal ring lies approximately 1.5-2cm above the midpoint of the inguinal ligament, the halfway point between the anterior superior iliac spine and the pubic tubercle, next to the epigastric vessels.
It is an important point in determining the nature of an inguinal hernia (direct or indirect). The patient is asked to cough after the hernia is reduced, with pressure applied to the deep inguinal ring. The hernia reappearing indicates it is direct, moving through the posterior wall of the inguinal canal.
Inferior and lateral to the pubic tubercle is the normal anatomical position of the neck of a femoral hernia.
Superior and medial to the pubic tubercle is the site of the superficial inguinal ring, and the normal anatomical position of the neck of an inguinal hernia.
The mid-inguinal point is located halways between the pubic symphysis and the anterior superior iliac spine. It is the surface marking for taking the femoral pulse.
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This question is part of the following fields:
- Anatomy
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Question 10
Incorrect
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A 68-year-old man with nausea and vomiting is admitted to the hospital.
For temporal arteritis, he takes 40 mg prednisolone orally in divided doses. His prescription chart will need to be adjusted to reflect his inability to take oral medications.
What is the equivalent dose of intravenous hydrocortisone to 40 mg oral prednisolone?Your Answer: 200 mg
Correct Answer: 160 mg
Explanation:Prednisolone 5 mg is the same as 20 mg hydrocortisone.
Prednisolone 40 mg is the same as 8 x 20 mg or 160 mg of prednisolone.
Mineralocorticoid effects and variations in action duration are not taken into account in these comparisons.
5 mg of prednisolone is the same as Dexamethasone 750 mcg, Hydrocortisone 20 mg, Methylprednisolone 4 mg, and Cortisone acetate 25 mg.
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This question is part of the following fields:
- Pharmacology
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Question 11
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: Blue body, blue shoulder, full cylinder; 4400 KPa, single gas, requires a single stage pressure regulator
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 12
Correct
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A 30-year old male has Von Willebrand's disease and attends the hospital to get an infusion of desmopressin acetate. The way this works is by stimulating the release of von Willebrand factor from cells, which in turn increases factor VIII and platelet plug formation in clotting.
In patients that have no clotting abnormalities, the substance that keeps the blood soluble and prevents platelet activation normally is which of these?Your Answer: Prostacyclin
Explanation:Even though aprotinin reduces fibrinolysis and therefore bleeding, there is an associated increased risk of death. It was withdrawn in 2007.
Protein C is dependent upon vitamin K and this may paradoxically increase the risk of thrombosis during the early phases of warfarin treatment.The coagulation cascade include two pathways which lead to fibrin formation:
1. Intrinsic pathway – these components are already present in the blood
Minor role in clotting
Subendothelial damage e.g. collagen
Formation of the primary complex on collagen by high-molecular-weight kininogen (HMWK), prekallikrein, and Factor 12
Prekallikrein is converted to kallikrein and Factor 12 becomes activated
Factor 12 activates Factor 11
Factor 11 activates Factor 9, which with its co-factor Factor 8a form the tenase complex which activates Factor 102. Extrinsic pathway – needs tissue factor that is released by damaged tissue)
In tissue damage:
Factor 7 binds to Tissue factor – this complex activates Factor 9
Activated Factor 9 works with Factor 8 to activate Factor 103. Common pathway
Activated Factor 10 causes the conversion of prothrombin to thrombin and this hydrolyses fibrinogen peptide bonds to form fibrin. It also activates factor 8 to form links between fibrin molecules.4. Fibrinolysis
Plasminogen is converted to plasmin to facilitate clot resorption -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 13
Incorrect
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What is the name of the space between the vocal cords?
Your Answer: Piriform recess
Correct Answer: Rima glottidis
Explanation:The rima glottidis is a narrow, triangle-shaped opening between the true vocal cords.
The vocal folds (true vocal cords) control sound production. The apex of each fold projects medially into the laryngeal cavity.
Each vocal fold includes these vocal ligaments:
Vocalis muscle (most medial part of thyroarytenoid muscle)
The glottis is composed of the vocal folds, processes and rima glottidis.
The rima glottidis is the narrowest potential site within the larynx, as the vocal cords may be completely opposed, forming a complete barrier.
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This question is part of the following fields:
- Anatomy
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Question 14
Correct
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Which of the following statement is true regarding the paediatric airway?
Your Answer: The larynx is more anterior than in an adult
Explanation:In the neonatal stage, the tongue is usually large and comes to the normal size at the age of 1 year. The vocal cords lie inverse C4 and as it reaches the grown-up position inverse C5/6 by the age of 4 (not 1 year).
Due to the immature cricoid cartilage, the larynx lies more anterior in newborn children. That’s why the cricoid ring is the narrowest part of the paediatric respiratory tract, while in the adults the tightest portion of the respiratory route is vocal cords. The epiglottis is generally expansive and slants at a point of 45 degrees to the laryngeal opening.
The carina is the ridge of the cartilage in the trachea at the level of T2 in newborn (T4 in adults), that separates the openings of right and left main bronchi.
Neonates have a comparatively low number of alveoli and then this number gradually increases to a most extreme by the age of 8 (not 3 years).
Neonates are obligatory nose breathers and any hindrance can cause respiratory issues (e.g., choanal atresia).
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This question is part of the following fields:
- Physiology
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Question 15
Incorrect
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The typical fluid compartments in a normal 70kg male are:
Your Answer: interstitial>intravascular
Correct Answer: intracellular>extracellular
Explanation:Body fluid compartments in a 70kg male:
Total volume=42L (60% body weight)
Intracellular fluid compartment (ICF) =28L
Extracellular fluid compartment (ECF) = 14LECF comprises:
Intravascular fluid (plasma) = 3L
Extravascular fluid = 11LExtravascular fluids comprises:
Interstitial fluid = 10.5L
Transcellular fluid = 0.5L -
This question is part of the following fields:
- Physiology
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Question 16
Incorrect
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Which among the following is summed up by F statistic?
Your Answer: Chi squared test
Correct Answer: ANOVA
Explanation:ANOVA is based upon within group variance (i.e. the variance of the mean of a sample) and between group variance (i.e. the variance between means of different samples). The test works by finding out the ratio of the two variances mentioned above. (Commonly known as F statistic).
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This question is part of the following fields:
- Statistical Methods
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Question 17
Incorrect
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Which of the following is the maximum volume of 0.5% bupivacaine that should be administered to a 10kg child?
Your Answer: 3 ml
Correct Answer: 5 ml
Explanation:Bupivacaine is used to decrease sensation in a specific area. It is injected around a nerve that supplies the area, or into the spinal canal’s epidural space.
The maximum volume of 0.5% bupivacaine that should be administered to a 10kg child is 5 ml
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This question is part of the following fields:
- Pharmacology
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Question 18
Correct
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A 78-year-old man with a previous history of ischaemic heart disease is admitted to hospital. He is scheduled for a cardiopulmonary exercise test (CPX) before he undergoes an elective abdominal aneurysm repair.
What measurement obtained during a CPX test alone provides the best indication for postoperative mortality?Your Answer: Anaerobic threshold
Explanation:Cardiopulmonary exercise testing (CPX, CPEX, CPET) is a non-invasive testing method used to determine the performance of the heart, lungs and skeletal muscle. It measures the exercise tolerance of the patient.
The parameters measured include:
ECG and ST-segment analysis and blood pressure
Oxygen consumption (VO2)
Carbon dioxide production (VCO2)
Gas flows and volumes
Respiratory exchange ratio (RER)
Respiratory rate
Anaerobic threshold (AT)The anaerobic threshold (AT) is an estimate of exercise ability. Any measurement below 11 ml/kg/min is usually related with an increase in mortality, especially when there is a background of myocardial ischaemia occurring during the test.
Peak VO2 <20 mL/kg with a low AT have a correlation with postoperative complications and a 30 day mortality. The CPX test is used for risk-testing patients prior to surgery to determine the appropriate postoperative care facilities. The V slope measured in CPX testing represents VO2 versus VCO2 relationship. During AT, the ramp of V slope increases, but does not provide a picture of postoperative mortality.
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This question is part of the following fields:
- Clinical Measurement
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Question 19
Correct
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A 70-year-old man presents with central crushing chest pain that radiates to the jaw in the emergency department. He has associated symptoms of nausea and diaphoresis.
A 12 lead ECG is performed. ST-elevation is observed in leads V2-V4. The diagnosis of anteroseptal ST-elevation myocardial infarction is made.
Which coronary vessel is responsible for this condition and runs in the interventricular septum on the anterior surface of the heart to reach the apex?Your Answer: Left anterior descending artery
Explanation:The heart receives blood supply from coronary arteries. The right and left coronary arteries branch off the aorta and supply oxygenated blood to all heart muscle parts.
The left main coronary artery branches into:
1. Circumflex artery – supplies the left atrium, side, and back of the left ventricle. The left marginal artery arises from the left circumflex artery. It travels along the obtuse margin of the heart.
2. Left Anterior Descending (LAD) artery – supplies the front and bottom of the left ventricle and front of the interventricular septumThe left anterior descending coronary artery is the largest coronary artery. It courses anterior to the interventricular septum in the anterior interventricular groove, extending from the base of the heart to its apex. Around the apex, the LAD anastomosis with the terminal branches of the posterior descending artery (branch of the right coronary artery).
Atherosclerosis or thrombotic occlusion of LAD causes myocardial infarction in large areas of the anterior, septal, and apical portions of the heart muscle. It can lead to a serious deterioration in heart performance.Occlusion of the LAD causes anteroseptal myocardial infarction, which is evident on the ECG with changes in leads V1-V4. Occlusion of the left circumflex artery causes lateral, posterior, or anterolateral MI. However, as it does not run towards the apex in the interventricular septum of the heart, it is not the correct answer for this question.
The right coronary artery branches into:
1. Right marginal artery
2. Posterior descending arteryThe right coronary artery supplies the right atrium, right ventricle, interatrial septum, and the inferior posterior third of the interventricular septum. Occlusion of the right coronary artery causes inferior MI, which is indicated on ECG with changes in leads II, III, and aVF.
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This question is part of the following fields:
- Anatomy
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Question 20
Correct
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Out of the following, which therapy for acute severe asthma or life-threatening asthma has been NOT been approved in recent guidelines?
Your Answer: Heliox
Explanation:Recommendations from the British Thoracic Society for acute severe asthma or life-threatening asthma are:
1. Give controlled supplementary oxygen to all hypoxemic patients with acute severe asthma titrated to maintain a SpO₂ level of 94 98%.
2. Use high-dose inhaled ?₂ agonists as first-line agents in patients with acute asthma and administer them as early as possible. Reserve
intravenous ?₂ agonists for those patients in whom inhaled therapy cannot be used reliably.
3. Give steroids in adequate doses to all patients with an acute asthma attack.
4. Add nebulized ipratropium bromide (0.5 mg 4–6 hourly) to ?₂ agonist treatment for acute severe or life-threatening asthma or those with a poor initial response to ?₂ agonist therapy.
5. Consider aminophylline for children with severe or life-threatening asthma unresponsive to maximal doses of bronchodilators and steroids.A review (including 12 case reports, three RCTs, and five other observational studies) of ketamine use in adults and children in status asthmaticus reported that ketamine is a potential bronchodilator. Still, prospective trials are needed before conclusions about effectiveness can be drawn.
Heliox has no place in the current guidelines issued by the British Thoracic Society.
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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 74-year-old with a VVI pacemaker is undergoing a hip replacement.
Which of the following is most likely to predispose him to an electrical hazard?Your Answer: Use of bipolar diathermy
Correct Answer: Use of cutting unipolar diathermy
Explanation:A single chamber pacemaker was implanted in the patient. In VVI mode, a pacemaker paces and senses the ventricle while being inhibited by a perceived ventricular event. The most likely electrical hazard from diathermy is electromagnetic interference (EMI).
EMI has the potential to cause the following: Inhibition of pacing
Asynchronous pacing
Reset to backup mode
Myocardial burns, and
Trigger VF.Diathermy entails the implementation of high-frequency electrical currents to produce heat and either make incisions or induce coagulation. Monopolar cautery involves disposable cautery pencils and electrosurgical diathermy units. In typical monopolar cautery, an electrical plate is placed on the patient’s skin and acts as an electrode, while the current passes between the instrument and the plate. Monopolar diathermy can therefore interfere with implanted metal devices and pacemaker function.
Bipolar diathermy, where the current passes between the forceps tips and not through the patient and is less likely to generate EMI.
Whilst the presence of a CVP line may in theory predispose the patient to microshock, the use of prerequisite CF electrical equipment makes this very unlikely. The presence of a CVP line and pacemaker does not therefore unduly increase the risk of an electrical hazard.
Isolating transformers are used to protect secondary circuits and individuals from electrical shocks. There is no step-up or step-down voltage (i.e. there is a ratio of 1 to 1 between the primary and secondary windings).
A ground (or earth) wire is normally connected to the metal case of an operating table to protect patients from accidental electrocution. In the event that a fault allows a live wire to make contact with the metal table (broken cable, loose connection etc.) it becomes live. The earth will provide an immediate path for current to safely flow through and so the table remains safe to touch. Being a low resistance path, the earth lets a large current flow through it when the fault occurs ensuring that the fuse or RCD will quickly blow. Without an operating table earth, the patient is not at more risk of an electrical hazard because of the pacemaker.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 22
Correct
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An arterial pressure transducer is supposedly in direct correlation to change, thus it is dependent on zero gradient drift and zero offset. Which of the following values will best compensate for the gradient drift?
Your Answer: 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 23
Correct
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Concerning the pathway of endothelial nitric oxide (eNO), one of the following best describes it.
Your Answer: Stimulation of guanylyl cyclase, increases cGMP concentration leading to vasodilation
Explanation:Nitric oxide (NO), an endothelial-derived relaxant factor (EDRF), is a powerful vasodilator. Its cell-signalling molecule is calcium-dependant and generated endogenous by nitric oxide synthetases from the precursor L-arginine, oxygen and NADPH. Three main isoforms have been isolated and they are inducible (iNO), neuronal (nNO) and endothelial (eNO).
Endothelial NO stimulates intracellular guanylyl cyclase which generates cyclic GMP (cGMP) from its action on guanylyl tri-phosphate (GTP). The cGMP goes on to activate protein kinase G (PKG). PKG phosphorylates cell membrane proteins that regulate intracellular calcium concentrations and level of calcium sensitisation.
Smooth muscle vasodilatation results from:
1. Light chain phosphatase activation.
2. Inhibition of calcium entry into the cell (reducing Ca2+ concentrations) and
3. Hyperpolarisation of cells by activation of H+ channels. -
This question is part of the following fields:
- Pathophysiology
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Question 24
Incorrect
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What separates the tunica media from the tunica adventitia in a blood vessel?
Your Answer: Internal elastic lamina
Correct Answer: External elastic lamina
Explanation:Blood vessels (except capillaries and venules) have three distinctive layers (innermost to outermost):
1. Tunica intima
2. Tunica media
3. Tunica adventitiaThe tunica media contains smooth muscle cells and is separated from the intima layer by the internal elastic lamina and the adventitia by the external elastic lamina.
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This question is part of the following fields:
- Anatomy
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Question 25
Correct
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Which drug, if given to a pregnant woman, can lead to deleterious fetal effects due to its ability to cross the placenta?
Your Answer: Atropine
Explanation:It is well known that atropine will cross the placenta and that maternal administration results in an increase in fetal heart rate.
Atropine is highly selective for muscarinic receptors. Its potency at nicotinic receptors is much lower, and actions at non-muscarinic receptors are generally undetectable clinically. Atropine does not distinguish among the M1, M2, and M3 subgroups of muscarinic receptors. In contrast, other antimuscarinic drugs are moderately selective for one or another of these subgroups. Most synthetic antimuscarinic drugs are considerably less selective than atropine in interactions with nonmuscarinic receptors.
A study on glycopyrrolate, a quaternary ammonium salt, was found to have a fetal: maternal serum concentration ratio of 0.4 indicating partial transfer.
Heparin, suxamethonium, and vecuronium do not cross the placenta.
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This question is part of the following fields:
- Pharmacology
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Question 26
Incorrect
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Of the following, which is NOT a branch of the abdominal aorta?
Your Answer: Superior mesenteric artery
Correct Answer: Superior phrenic artery
Explanation:The abdominal aorta begins at the level of the body of T12 near the midline, as a continuation of the thoracic aorta. It descends and bifurcates at the level of L4 into the common iliac arteries.
The branches of the abdominal aorta (with their vertebra level) are:
1. Inferior phrenic arteries: T12 (upper border)
2. Coeliac artery: T12
3. Superior mesenteric artery: L1
4. Middle suprarenal arteries: L1
5. Renal arteries: Between L1 and L2
6. Gonadal arteries: L2 (in males, it is the testicular artery, and in females, the ovarian artery)
7. Inferior mesenteric artery: L3
8. Median sacral artery: L4
9. Lumbar arteries: Between L1 and L4The superior phrenic artery branches from the thoracic aorta.
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This question is part of the following fields:
- Anatomy
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Question 27
Incorrect
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An study on post-operative nausea and vomiting (PONV) among paediatric patients who underwent tonsillectomy showed a decrease in incidence from 10% to 5% following a new management protocol.
Which of the following best estimates the numbers needed to treat (NNT) for one additional patient to benefit from the new management of PONV?Your Answer: 10
Correct Answer: 20
Explanation:The Number Needed to Treat (NNT) is the number of patients you need to treat to prevent one additional bad outcome. For example, if a drug has an NNT of 5, it means you have to treat 5 people with the drug to prevent one additional bad outcome.
To calculate the NNT, you need to know the Absolute Risk Reduction (ARR); the NNT is the inverse of the ARR:
NNT = 1/ARR
Where ARR = CER (Control Event Rate) – EER (Experimental Event Rate).
NNTs are always rounded up to the nearest whole number.
In this case, the NNT can be computed as follows:
ARR = 10% – 5% = 0.05
NNT = 1/0.05 = 20
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This question is part of the following fields:
- Statistical Methods
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Question 28
Incorrect
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Infrared radiation is absorbed by asymmetric, polyatomic polar molecules like carbon dioxide (CO2).
When measuring CO2 in gas samples, mainstream capnography uses this physical principle.
Which of the following gases is most likely to interfere with carbon dioxide's absorption spectrum?Your Answer: Isoflurane
Correct Answer: Nitrous oxide
Explanation:Carbon dioxide absorbs the most infrared (IR) light between the wavelengths of 4.2-4.4m (4.26m is ideal).
Nitrous oxide absorbs infrared light at wavelengths of 4.4-4.6m (very similar to CO2) and less so at 3.9m.
At a frequency of 4.7m, carbon monoxide absorbs the most IR light.
At 3.3 m and throughout the ranges 8-12 m, the volatile agents have strong absorption bands.
Although oxygen does not absorb infrared light, it collides with CO2 molecules, interfering with absorption. The absorption band is widened as a result of this (so called collision or pressure broadening). A drop of 0.5 percent in measured CO2 can be caused by 95% oxygen.
Nitrous oxide causes a greater inaccuracy of 0.1 percent per ten percent of nitrous oxide.
Water vapour absorbs infrared light as well, resulting in absorption band overlap, collision broadening, and partial pressure dilution. Water traps and water permeable tubing are used to reduce inaccuracies.
Collision broadening is compensated for in modern gas multi-gas analysers.
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This question is part of the following fields:
- Pharmacology
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Question 29
Incorrect
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An 80-year-old female presents to the emergency department with symptoms consistent with mesenteric ischemia. She is quickly shifted to the operation theatre for an emergency laparotomy.
On exploration, the segment of the colon from the splenic flexure down to the rectum is ischemic.
The artery blocked in this scenario arises at what vertebral level from the aorta?Your Answer: T12
Correct Answer: L3
Explanation:The hindgut is from the distal third of the transverse colon down to the upper one-third of the anal canal. The inferior mesenteric artery supplies the hindgut.
The inferior mesenteric artery arises from the aorta behind the inferior border of the third part of the duodenum 3–4 cm above the aortic bifurcation, at the third lumbar vertebra level.
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 30
Incorrect
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A 30-year-old man has been diagnosed with a direct inguinal hernia.
A direct inguinal hernia passes through Hesselbach's triangle. What forms the medial edge of this triangle?Your Answer:
Correct Answer: Rectus abdominis muscle
Explanation:The inguinal triangle of Hesselbach’s is an important clinical landmark on the posterior wall of the inguinal canal. It has the following relations:
Inferiorly – medial third of the inguinal ligament
Medially – lower lateral border of the rectus abdominis
Laterally – inferior epigastric vesselsDirect inguinal hernia is when the bowel bulges directly through the abdominal wall. These hernias usually protrude through Hesselbach’s triangle
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This question is part of the following fields:
- Anatomy
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