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  • Question 1 - A randomized controlled trail has been conducted to compare two drugs used for...

    Correct

    • A randomized controlled trail has been conducted to compare two drugs used for the early management of acute severe asthma in the emergency department. After being allocated to the randomized groups, many patients have been excluded due to deleterious effect to the drugs.

      How the data would be analysed?

      Your Answer: Include the patients who drop out in the final data set

      Explanation:

      Randomized controlled trails will be analysed by the intention-to-treat (ITT) approach. It provides unbiased comparisons among the treatment groups. ITT analyses are done to avoid the effects of dropout, which may break the random assignment to the treatment groups in a study.

      ITT analysis is a comparison of the treatment groups that includes all patients as originally allocated after randomization.

      In order to include such participants in an analysis, outcome data could be imputed which involves making assumptions about the outcomes in the lost participants.

    • This question is part of the following fields:

      • Statistical Methods
      74
      Seconds
  • Question 2 - Regarding anaesthetic breathing circuits, which one would be the best for spontaneous breathing?...

    Correct

    • Regarding anaesthetic breathing circuits, which one would be the best for spontaneous breathing?

      Your Answer: Lack circuit

      Explanation:

      Among the breathing circuits, the Lack circuit is the most efficient for spontaneous breathing.

      An outer coaxial tube is present to deliver fresh air; exhaust air is routed to an inner tube, which is then delivered to a scavenging system. An expiratory valve is seen at the patient end, which is an advantage over other circuits. Moreover, the Lack circuit prevents rebreathing slightly greater than the alveolar minute ventilation at 4-5 litres per minute.

      The Bain circuit prevents rebreathing at 160-200ml/kg per minute, and is a co-axial version of the Mapleson D circuit.

      The Mapleson E circuit prevent rebreathing at a fresh gas flow (FGF) of approximately twice the patient’s normal minute volume. A modification of this, the Mapleson F, has a reservoir bag at the opposite end for the FGF. This circuit is appropriate for paediatric patients with a body weight less than 20 kg.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      20.4
      Seconds
  • Question 3 - Drug X, a new intravenous induction drug, is being administered as a bolus...

    Correct

    • 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
      359.4
      Seconds
  • Question 4 - A 58-year-old man, visits his general practitioner complaining of a lump in his...

    Incorrect

    • 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: Inferior and lateral to the pubic tubercle

      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.

    • This question is part of the following fields:

      • Anatomy
      20.3
      Seconds
  • Question 5 - A 53-year old female with a diagnosis of anaplastic thyroid carcinoma is admitted...

    Correct

    • A 53-year old female with a diagnosis of anaplastic thyroid carcinoma is admitted in the surgery department for an elective total thyroidectomy with radical neck dissection. The operation is expected to last for 10 hours.

      Which of the following is the most suitable humidifier to use in an anaesthetic circuit for this case?

      Your Answer: Heat and moisture exchanger (HME)

      Explanation:

      Adequate humidification is vital to maintain homeostasis of the airway. Heat and moisture exchangers conserve some of the exhaled water, heat and return them to inspired gases. Many heat and moisture exchangers also perform bacterial/viral filtration and prevent inhalation of small particles. Heat and moisture exchangers are also called condenser humidifier, artificial nose, etc. Most of them are disposable devices with exchanging medium enclosed in a plastic housing. For adult and paediatric age group different dead space types are available. Heat and moisture exchangers are helpful during anaesthesia and ventilatory breathing system. To reduce the damage of the upper respiratory tract through cooling and dehydration inspiratory air can be heated and humidified, thus preventing the serious complications. Moreover, they are the most appropriate humidification devices used for routine anaesthesia.

      Gases can be bubbled through water to increase humidity. Passing gas through water at room temperature causes the gas to cool due to latent heat of vaporisation. The water bath can be heated. This improves the efficiency of the device and also reduces the incidence of bacterial colonisation.

      Nebulisers use a venturi system which employs the Bernoulli effect. A gas at high flow passes through a constriction causing the gas to accelerate, reducing its potential energy allowing other gases or liquids to be entrained. This can include medications or in the case of humidification, water vapour. The size of the water droplet produced by nebulisation determines where in the airway it is deposited. Standard nebulisers produced droplets of 4 microns in diameter and these are deposited in the upper airway and trachea. Efficacy can be improved by passing the droplets over an anvil which further reduces particle size. The most efficient form of nebuliser is the ultrasonic nebuliser. Here a transducer immersed in water and vibrated at a frequency of 3MHz produces1-2micron droplets. These particles easily reach the bronchioles and provide excellent humidification.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      241.1
      Seconds
  • Question 6 - A 28-year-old girl complained of severe abdominal pain and hematemesis and was rushed...

    Incorrect

    • 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: L1

      Correct 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

    • This question is part of the following fields:

      • Anatomy
      18.4
      Seconds
  • Question 7 - Sugammadex binds to certain drugs that affect neuromuscular function during anaesthesia in a...

    Correct

    • Sugammadex binds to certain drugs that affect neuromuscular function during anaesthesia in a stereospecific, non-covalent, and irreversible manner.

      It has the greatest impact on the activity of which of the following drugs?

      Your Answer: Vecuronium

      Explanation:

      Sugammadex is a modified cyclodextrin that works as an aminosteroid neuromuscular blocking (nmb) reversal agent. By encapsulating each molecule in the plasma, it rapidly reverses rocuronium and, to a lesser extent, vecuronium-induced neuromuscular blockade. Consequently, a  concentration gradient favours the movement of these nmb agents away from the neuromuscular junction.  Pancuronium-induced neuromuscular blockade at low levels has also been reversed.

      By inhibiting voltage-dependent calcium channels at the neuromuscular junction, antibiotics in the aminoglycoside group potentiate neuromuscular blocking agents. This can be reversed by giving calcium but not neostigmine or sugammadex.

      Sugammadex will not reverse the effects of mivacurium, which belongs to the benzylisoquinolinium class of drugs.

      A phase II or desensitisation block occurs when the motor end-plate becomes less sensitive to acetylcholine as a result of an overdose or repeated administration of suxamethonium. The use of neostigmine has been shown to be effective in reversing this weakness.

    • This question is part of the following fields:

      • Pharmacology
      67.4
      Seconds
  • Question 8 - During a critical liver resection surgery, a 65-year-old man suffers uncontrolled hepatic bleeding....

    Correct

    • During a critical liver resection surgery, a 65-year-old man suffers uncontrolled hepatic bleeding. The 'Pringle manoeuvre is performed to stop the bleeding where the hepatic artery, portal vein, and common bile duct are clamped. These structures form the anterior boundary of the epiploic foramen.

      Which of the following vessels also contributes to the boundary of this region?

      Your Answer: Inferior vena cava

      Explanation:

      The epiploic foramen (foramen of Winslow or aditus to the lesser sac) is found behind the free right border of the lesser omentum. A short, 3 cm slit serves as the entrance to the lesser sac from the greater sac.

      The epiploic foramen has the following boundaries:
      Anteriorly: hepatoduodenal ligament, the bile duct (anteriorly on the right), the hepatic artery (anteriorly on the left), and the portal vein (posteriorly) together with nerves and lymphatics
      Superiorly: the peritoneum of the posterior layer of the hepatoduodenal ligament runs over the caudate process of the liver
      Posteriorly: inferior vena cava
      Floor: upper border of the first part of the duodenum
      The anterior and posterior walls of the foramen are normally
      apposed, which partly explains why patients can develop large fluid
      collections isolated to the greater or lesser sac

      Rapid control of the hepatic artery and portal vein can be obtained by compression of the free edge of the lesser omentum (a ‘Pringle’ manoeuvre), which is a potentially useful technique in liver trauma and surgery.

    • This question is part of the following fields:

      • Anatomy
      172.8
      Seconds
  • Question 9 - A project is being planned to assess the effects of a new anticoagulant...

    Correct

    • A project is being planned to assess the effects of a new anticoagulant on the coagulation cascade. The intrinsic pathway is being studied and the best measurement to be recorded is which of the following?

      Your Answer: aPTT

      Explanation:

      The intrinsic pathway is best assessed by the aPTT time.

      D-dimer is a fibrin degradation product which is raised in the presence of blood clots.

      A 50:50 mixing study is used to assess if a prolonged PT or aPTT is due to factor deficiency or a factor inhibitor.

      The thrombin time is a test used to assess fibrin formation from fibrinogen in plasma. Factors that prolong the thrombin time include heparin, fibrin degradation products, and fibrinogen deficiency.

      Intrinsic pathway – Best assessed by APTT. Factors 8,9,11,12 are involved. Prolonged aPTT can be seen in haemophilia and use of heparin.

      Extrinsic pathway – Best assessed by Increased PT. Factor 7 involved.

      Common pathway – Best assessed by APTT & PT. Factors 2,5,10 involved.

      Vitamin K dependent factors are factors 2,7,9,10

    • This question is part of the following fields:

      • Physiology And Biochemistry
      78.7
      Seconds
  • Question 10 - Which of the following statements best describes adenosine receptors? ...

    Incorrect

    • Which of the following statements best describes adenosine receptors?

      Your Answer: The A1 and A2 receptors are present centrally and peripherally

      Correct Answer:

      Explanation:

      Adenosine receptors are expressed on the surface of most cells.
      Four subtypes are known to exist which are A1, A2A, A2B and A3.

      Of these, the A1 and A2 receptors are present peripherally and centrally. There are agonists at the A1 receptors which are antinociceptive, which reduce the sensitivity to a painful stimuli for the individual. There are also agonists at the A2 receptors which are algogenic and activation of these results in pain.

      The role of adenosine and other A1 receptor agonists is currently under investigation for use in acute and chronic pain states.

    • This question is part of the following fields:

      • Physiology
      404.6
      Seconds
  • Question 11 - Regarding renal autoregulation, which of the following best describes its process? ...

    Incorrect

    • Regarding renal autoregulation, which of the following best describes its process?

      Your Answer: Renal vascular resistance increases when the mean arterial blood pressure falls to between 80 and 100 mmHg

      Correct Answer: Reduces the effect of changes in arterial blood pressure on renal Na+ excretion

      Explanation:

      Two mechanisms are responsible for autoregulation of RBF and GFR: one mechanism that responds to changes in arterial pressure and another that responds to changes in [NaCl] in tubular fluid. Both regulate the tone of the afferent arteriole. The pressure-sensitive mechanism, the so-called myogenic mechanism, is related to an intrinsic property of vascular smooth muscle: the tendency to contract when stretched. Accordingly, when arterial pressure rises and the renal afferent arteriole is stretched, the smooth muscle contracts in response. Because the increase in resistance of the arteriole offsets the increase in pressure, RBF, and therefore GFR, remains constant.

      The second mechanism responsible for autoregulation of GFR and RBF is the [NaCl]-dependent mechanism known as tubuloglomerular feedback. This mechanism involves a feedback loop in which a change in GFR leads to alteration in the concentration of NaCl in tubular fluid, which is sensed by the macula densa of the juxtaglomerular apparatus and converted into signals that affect afferent arteriolar resistance and thus the GFR (Fig. 33.19). For example, when the GFR increases and causes [NaCl] in tubular fluid in the loop of Henle to rise, more NaCl enters the macula densa cells in this segment (Fig. 33.20). This leads to an increase in formation and release of adenosine triphosphate (ATP) and adenosine (a metabolite of ATP) by macula densa cells, which causes vasoconstriction of the afferent arteriole and normalization of GFR. In contrast, when GFR and [NaCl] in tubule fluid decrease, less NaCl enters the macula densa cells, and both ATP and adenosine production and release decline. The fall in [ATP] and [adenosine] results in afferent arteriolar vasodilation, which returns GFR to normal. NO, a vasodilator produced by the macula densa, attenuates tubuloglomerular feedback, whereas angiotensin II enhances tubuloglomerular feedback. Thus the macula densa may release both vasoconstrictors (e.g., ATP and adenosine) and a vasodilator (e.g., NO) that oppose each other’s action at the level of the afferent arteriole. Production plus release of either vasoconstrictors or vasodilators ensures exquisite control over tubuloglomerular feedback.

      Renal autoregulation, thus, reduces the effect of changes in arterial blood pressure on renal sodium excretion.

    • This question is part of the following fields:

      • Pathophysiology
      116
      Seconds
  • Question 12 - The passage of glucose into the brain is facilitated by which transport method?...

    Correct

    • The passage of glucose into the brain is facilitated by which transport method?

      Your Answer: Facilitated diffusion

      Explanation:

      Glucose transport is a highly regulated process accomplished mostly by facilitated diffusion using carrier proteins to cross cell membranes.

      There are many transporters, but the most important are known as glucose transporters (GLUTs).

      Stresses in various form of acute and chronic forms affect the activity of glucose transporters.
      They are responsive to many types of metabolic stress, including hypoxia, injury, hypoglycaemia, numerous metabolic inhibitors, stress hormones, and other influences such as growth factors.

      Numerous signalling pathways appear to be involved in transporter regulation.

      New evidence suggests that stresses regulating GLUTs are not only acute biological stresses. In addition, chronic low-grade inflammation, and their associated chronic diseases also lead to altered glucose transport. These include obesity, type 2 diabetes, cardiovascular disease, and the growth and spread of many tumours that are affected by altered glucose transporters. Some of these glucose transport effects are compensatory, while others are pathogenic.

      Ultimately, deliberate manipulation of GLUTs could be used as treatment for some of these chronic diseases.

    • This question is part of the following fields:

      • Physiology
      10.4
      Seconds
  • Question 13 - Which of these statements regarding the basilar artery and its branches is not...

    Incorrect

    • Which of these statements regarding the basilar artery and its branches is not true?

      Your Answer: The superior cerebellar artery may be decompressed to treat trigeminal neuralgia

      Correct Answer: The posterior inferior cerebellar artery is the largest of the cerebellar arteries arising from the basilar artery

      Explanation:

      The posterior inferior cerebellar artery is the largest branch arising from the distal portion of the vertebral artery which forms the basilar artery. It is one of the arteries responsible for providing blood supply to the brain’s cerebellum.

      The labyrinthine artery (auditory artery) is a long and slender artery which arises from the basilar artery and runs alongside the facial and vestibulocochlear nerves into the internal auditory meatus.

      The posterior cerebellar artery is one of two cerebral arteries supplying the occipital lobe with oxygenated blood. It is usually bigger than the superior cerebellar artery. It is separated from the vessel near its origin by the oculomotor nerve.

    • This question is part of the following fields:

      • Anatomy
      11
      Seconds
  • Question 14 - A peripheral nerve stimulator is used to stimulate the ulnar nerve at the...

    Correct

    • A peripheral nerve stimulator is used to stimulate the ulnar nerve at the wrist to indicate the degree of neuromuscular blockade.

      Which single muscle or group of muscles of the hand supplied by the ulnar nerve is best for monitoring the twitch function during neuromuscular blockade?

      Your Answer: Adductor pollicis

      Explanation:

      In anaesthesia, adductor pollicis neuromuscular monitoring with ulnar nerve stimulation is commonly used. It is the gold standard for measuring the degree of block and comparing neuromuscular blocking drugs and their effects on other muscles.

      Electrodes are usually placed over the ulnar nerve at the wrist to monitor the adductor pollicis.

      Neuromuscular blocking drugs have different sensitivity levels in different muscle groups.

      To achieve the same level of blockade, the diaphragm requires 1.4 to 2 times the amount of neuromuscular blocking agent as the adductor pollicis muscle. The small muscles of the larynx and the ocular muscles are two other respiratory muscles that are less resistant than the diaphragm (especially corrugator supercilii).

      The abdominal muscles, Orbicularis oculi, peripheral muscles of the limbs, Geniohyoid, Masseter, and Upper airway muscles are the most sensitive to neuromuscular blocking agents.

      The C8-T1 nerve roots, which are part of the medial cord of the brachial plexus, form the ulnar nerve. It enters the hand via the ulnar canal, superficial to the flexor retinaculum, after following the ulnar artery at the wrist.

      The nerve then splits into two branches: superficial and deep. The palmaris brevis is supplied by the superficial branch, which also provides palmar digital nerves to one and a half fingers. The dorsal surface of the medial/ulnar 1.5 fingers, as well as the corresponding skin over the hand, are also supplied by it (as well as the palmar surface).

      The ulnar nerve’s deep branch runs between the abductor and flexor digiti minimi, which it supplies. It also innervates the opponens, and with the deep palmar arch, it curves around the hook of the hamate and laterally across the palm. All of the interossei, the medial two lumbricals, the adductor pollicis, and, in most cases, the flexor pollicis brevis are supplied there.

    • This question is part of the following fields:

      • Anatomy
      216.4
      Seconds
  • Question 15 - Regarding laminar gas flow, which of the following options has the most influence...

    Incorrect

    • Regarding laminar gas flow, which of the following options has the most influence on laminar flow?

      Your Answer:

      Correct Answer: Diameter of tube

      Explanation:

      Laminar flow can be defined as the motion of a fluid where every particle in the fluid follows the same path of its previous particles. The following are properties of laminar flow of gas or fluids:

      1. Smooth unobstructed flow of gas through a tube of relatively uniform diameter
      2. Few directional changes
      3. Slow, steady flow through straight smooth, rigid, large calibre, cylindrical tube
      4. Outer layer flow slower than the centre due to friction, results in discrete cylindrical layers, or streamlines
      5. Double flow by doubling pressure as long as the flow pattern remains laminar

      Poiseuille’s Law relates the factors that determine laminar flow. It indicates the degree of resistance to fluid flow through a tube. The resistance to fluid flow through a tube is directly related to the length, flow and viscosity; and inversely related to the radius of the tube to the fourth power. This means that, when the radius is doubled, there is increase in flow by a factor of 16.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      0
      Seconds
  • Question 16 - A 63-year old male who has heart failure has peripheral oedema and goes...

    Incorrect

    • 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:

      Correct 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.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      0
      Seconds
  • Question 17 - Of the following, which is NOT a branch of the abdominal aorta? ...

    Incorrect

    • Of the following, which is NOT a branch of the abdominal aorta?

      Your Answer:

      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 L4

      The superior phrenic artery branches from the thoracic aorta.

    • This question is part of the following fields:

      • Anatomy
      0
      Seconds
  • Question 18 - The lung volume that is commonly measured indirectly is? ...

    Incorrect

    • 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.

    • This question is part of the following fields:

      • Pathophysiology
      0
      Seconds
  • Question 19 - A 56-year old man, presents to emergency department following a cardiac arrest. On...

    Incorrect

    • A 56-year old man, presents to emergency department following a cardiac arrest. On history and examination, he is found to be suffering from both metabolic and respiratory acidosis as a result of his cardiac arrest.

      What is the best way to reduce the risk of acidaemia during cardiac arrest

      Your Answer:

      Correct Answer: Chest compressions

      Explanation:

      Chest compressions are an essential part of cardiopulmonary resuscitation (CPR) which helps restore spontaneous circulation (ROSC).

      Sodium bicarbonate is only prescribed in patients with cardiac arrests as a result of an overdose of tricyclic antidepressants or hyperkalaemia. Its use causes the body to produce more CO2 which causes:

      Exacerbation of intracellular acidosis
      Negative inotropy to ischaemic myocardium
      Increased osmotic load of sodium into failing brain and body
      Shift of oxygen dissociation curve to the left.

      THAM is often used to treat metabolic acidosis as a result of cardiac bypass surgery and also cardiac arrest, when other standard methods have failed.

      Carbicarb (Na2CO3 0.33 molar NaHCO3 0.33 molar) has only mild effects on acidosis. It also causes an increase in arterial CO2 pressure and lactate concentration.

    • This question is part of the following fields:

      • Pathophysiology
      0
      Seconds
  • Question 20 - A radical neck dissection is being performed. The ENT surgeon wishes to expose...

    Incorrect

    • A radical neck dissection is being performed. The ENT surgeon wishes to expose the external carotid artery fully. He inserts a self-retaining retractor close to the origin of the external carotid artery.

      What structure lies posterolaterally to the external carotid at this point?

      Your Answer:

      Correct Answer: Internal carotid artery

      Explanation:

      External carotid artery originates at the upper border of the thyroid cartilage. It ascends and lies anterior to the internal carotid arteries and posterior to the posterior belly of the digastric muscle and stylohyoid muscle.

      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 artery

    • This question is part of the following fields:

      • Anatomy
      0
      Seconds
  • Question 21 - Which among the following is not true regarding disease rates? ...

    Incorrect

    • Which among the following is not true regarding disease rates?

      Your Answer:

      Correct Answer: The odds ratio is synonymous with the risk ratio

      Explanation:

      The relative risk (also known as risk ratio [RR]) is the ratio of risk of an event in one group (e.g., exposed group) versus the risk of the event in the other group (e.g., nonexposed group).

      The odds ratio (OR) is the ratio of odds of an event in one group versus the odds of the event in the other group.

    • This question is part of the following fields:

      • Statistical Methods
      0
      Seconds
  • Question 22 - Of the following, which of these oxygen carrying molecules causes the greatest shift...

    Incorrect

    • Of the following, which of these oxygen carrying molecules causes the greatest shift of the oxygen-dissociation curve to the left?

      Your Answer:

      Correct Answer: Myoglobin (Mb)

      Explanation:

      Myoglobin is a haemoglobin-like, iron-containing pigment that is found in muscle fibres. It has a high affinity for oxygen and it consists of a single alpha polypeptide chain. It binds only one oxygen molecule, unlike haemoglobin, which binds 4 oxygen molecules.

      The myoglobin ODC is a rectangular hyperbola. There is a very low P50 0.37 kPa (2.75 mmHg). This means that it needs a lower P50 to facilitate oxygen offloading from haemoglobin. It is low enough to be able to offload oxygen onto myoglobin where it is stored. Myoglobin releases its oxygen at the very low PO2 values found inside the mitochondria.

      P50 is defined as the affinity of haemoglobin for oxygen: It is the PO2 at which the haemoglobin becomes 50% saturated with oxygen. Normally, the P50 of adult haemoglobin is 3.47 kPa(26 mmHg).

      Foetal haemoglobin has 2 ? and 2 ?chains. The ODC is left shifted – this means that P50 lies between 2.34-2.67 kPa [18-20 mmHg]) compared with the adult curve and it has a higher affinity for oxygen. Foetal haemoglobin has no ? chains so this means that there is less binding of 2.3 diphosphoglycerate (2,3 DPG).

      Carbon monoxide binds to haemoglobin with an affinity more than 200-fold higher than that of oxygen. This therefore decreases the amount of haemoglobin that is available for oxygen transport. Carbon monoxide binding also increases the affinity of haemoglobin for oxygen, which shifts the oxygen-haemoglobin dissociation curve to the left and thus impedes oxygen unloading in the tissues.

      In sickle cell disease, (HbSS) has a P50 of 4.53 kPa(34 mmHg).

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 23 - When describing the surface anatomy of the sacrum, which of the following anatomical...

    Incorrect

    • When describing the surface anatomy of the sacrum, which of the following anatomical landmarks refers to the base of an equilateral triangle is formed by the sacral hiatus?

      Your Answer:

      Correct Answer: A line connecting the posterior superior iliac spines

      Explanation:

      The apex of an equilateral triangle completed by the posterior superior iliac spines is where the sacral hiatus or sacrococcygeal membrane can normally located. The failure of posterior fusion of the laminae of the fourth and fifth sacral vertebrae allows the sacral canal to be accessible via the membrane.

      In adults, the spine of L4 usually lies on a line drawn between the highest points of the iliac crests (Tuffier’s line). A line connecting each anterior iliac spine, approximates to the L3/4 interspace in the sitting position. Both of these options are incorrect.

      A line connecting the greater trochanters is also incorrect.

      A line connecting the posterior superior iliac spines is correct, but in adults the presence of a sacral fat pad can still make identification of this landmark less straightforward.

      The processes of S5 are remnants only and form the sacral cornua, which are also used to help identify the sacral hiatus.

    • This question is part of the following fields:

      • Anatomy
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  • Question 24 - Which of the following statement is not true regarding the effects of Dopamine...

    Incorrect

    • Which of the following statement is not true regarding the effects of Dopamine infusions?

      Your Answer:

      Correct Answer: Decreasing gastric transit time

      Explanation:

      Moderately high doses of dopamine produce a positive inotropic (direct?1 and D1 action + that due to Noradrenaline 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.

    • This question is part of the following fields:

      • Pharmacology
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  • Question 25 - Which of the following statements is true regarding the Wrights Respirometer? ...

    Incorrect

    • Which of the following statements is true regarding the Wrights Respirometer?

      Your Answer:

      Correct Answer: Measures the minute volume to within an accuracy of +/- 10%

      Explanation:

      A Wrights Respirometer measures the volume of air exhaled over the course of one minute of normal breathing

      It is unidirectional and measures tidal volume and minute volume of gas flow in one direction. It is placed at the expiratory side (lower pressure than inspiratory side therefore lower chances of gas leaks)

      Slits are arranged such that incoming gas will rotate the vane at a rate of 150 revolutions per litre of flowing gas

      The Wright respirometer tends to over-read at high flow rates and under-read at low flows because of mechanical causes like friction and inertia and the accumulation of water vapour

      The ideal flow for accurate readings is 2 L/min for the respirometer. The respirometer reads the tidal volume and minute volume with a ±5–10% accuracy within the range of 4–24 L/min.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
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  • Question 26 - A strain gauge operates on which of the following principles? ...

    Incorrect

    • A strain gauge operates on which of the following principles?

      Your Answer:

      Correct Answer: Varying resistance of a wire when stretched

      Explanation:

      The principle by which a strain gauge works is that when a wire is stretched, it becomes longer and thinner, and as a result, its resistance increases.

      A strain gauge, which is used in pressure transducers, acts as a resistor. When the pressure in a pressure transducer changes, the diaphragm moves, changing the tension in the resistance wire and thus changing the resistance.

      Changes in current flow through the resistor are amplified and displayed as a pressure change measure.

      A Wheatstone bridge, on the other hand, is frequently used to measure or monitor these changes in resistance.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
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  • Question 27 - A 25-year old male with palpitations and dizziness presents to the emergency room....

    Incorrect

    • 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:

      Correct 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.

    • This question is part of the following fields:

      • Physiology And Biochemistry
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  • Question 28 - After a bariatric surgery, average weight loss observed in patients is 18 kg....

    Incorrect

    • After a bariatric surgery, average weight loss observed in patients is 18 kg. The standard deviation was found to be 3 kg. What is the percentage of patients that lie between 9 and 27 kg?

      Note: Assume that the curve is normally distributed.

      Your Answer:

      Correct Answer: 99.70%

      Explanation:

      9 & 27 can be obtained by subtracting and adding 9 from the mean. 9 is three times the standard deviation and we know that 99.7% values lie within 3 standard deviations from the mean. We can find the interval for 99.7% to verify in the following way:

      For 99.7% confidence interval, you can find the range as follows:

      1. Multiply the standard error by 3.

      2. Subtract the answer from mean value to get the lower limit.

      3. Add the answer obtained in step 1 from the mean value to get the upper limit.

      4. The range turns out to be 9-27 kg.

    • This question is part of the following fields:

      • Statistical Methods
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  • Question 29 - One of the following sets of arterial blood gases best demonstrates compensated respiratory...

    Incorrect

    • One of the following sets of arterial blood gases best demonstrates compensated respiratory acidosis?

      Your Answer:

      Correct Answer: pH=7.36, PaCO2=8.5kPa, PaO2=7.5kPa, HCO3- = 43mmol/L

      Explanation:

      pH=7.36, PaCO2=8.5kPa, PaO2=7.5kPa, HCO3- = 43mmol/L is the correct answer.

      Since the pH is the lower limit of normal, it is compensated despite a raised PaCO2. Retention of bicarbonate ions by the renal system suggests this process is chronic.

      pH=7.24, PaCO2=3.5kPa, PaO2=8.5kPa, HCO3- =18mmol/L represents an acute uncompensated metabolic acidosis

      The remaining stems are degrees of uncompensated respiratory acidosis and therefore incorrect.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 30 - A 40-year-old obese woman has complaints of heartburn and regurgitation that is worse...

    Incorrect

    • 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 vein

      An 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.

    • This question is part of the following fields:

      • Anatomy
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SESSION STATS - PERFORMANCE PER SPECIALTY

Statistical Methods (2/2) 100%
Anaesthesia Related Apparatus (2/2) 100%
Anatomy (2/5) 40%
Pharmacology (1/1) 100%
Physiology And Biochemistry (1/1) 100%
Physiology (1/2) 50%
Pathophysiology (0/1) 0%
Passmed