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  • Question 1 - The structure most likely to be damaged during cannulation of the subclavian vein...

    Incorrect

    • The structure most likely to be damaged during cannulation of the subclavian vein is?

      Your Answer: Pleura

      Correct Answer: Subclavian artery

      Explanation:

      The subclavian artery lies behind and partly above the subclavian vein. 3-4% of the time, it can be inadvertently cannulated during cannulation of the subclavian vein

      Because of its anatomical position, putting pressure on the subclavian artery is impossible so arresting bleeding with pressure when it is punctured is not viable.

      One of the consequences of subclavian vein cannulation (1%) is pleural puncture leading to a pneumothorax. This is because the apical pleura is inferior and caudal to the subclavian vein.

    • This question is part of the following fields:

      • Anatomy
      5.9
      Seconds
  • Question 2 - Which of the following is correct regarding correlation? ...

    Correct

    • Which of the following is correct regarding correlation?

      Your Answer: Complete absence of correlation is expressed by a value of 0

      Explanation:

      In statistical terms, correlation is used to denote association between two quantitative variables.

      The degree of association is measured by a correlation coefficient, denoted by r. The correlation coefficient is measured on a scale that varies from + 1 through 0 to €“ 1. Complete correlation between two variables is expressed by either + 1 or -1. When one variable increases as the other increases the correlation is positive; when one decreases as the other increases it is negative. Complete absence of correlation is represented by 0.

      The two methods are not synonymous as correlation measures the degree of relationship between two variables whereas regression analysis is about how one variable affects another or what changes it has on the other variable. Both are also shown by a different graphical representation.

    • This question is part of the following fields:

      • Statistical Methods
      21.4
      Seconds
  • Question 3 - Regarding renal autoregulation, which of the following best describes its process? ...

    Correct

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

      Your 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
      42.1
      Seconds
  • Question 4 - A young male is undergoing inguinal hernia repair. During the procedure, the surgeons...

    Incorrect

    • 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: Rectus abdominis muscle

      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.

    • This question is part of the following fields:

      • Anatomy
      15.9
      Seconds
  • Question 5 - In a normal healthy adult breathing 100 percent oxygen, which of the following...

    Incorrect

    • In a normal healthy adult breathing 100 percent oxygen, which of the following is the most likely cause of an alveolar-arterial (A-a) oxygen difference of 30 kPa?

      Your Answer: High altitude

      Correct Answer: Atelectasis

      Explanation:

      The ‘ideal’ alveolar PO2 minus arterial PO2 is the alveolar-arterial (A-a) oxygen difference.

      The ‘ideal’ alveolar PO2 is derived from the alveolar air equation and is the PO2 that the lung would have if there was no ventilation-perfusion (V/Q) inequality and it was exchanging gas at the same respiratory exchange ratio as real lung.

      The amount of oxygen in the blood is measured directly in the arteries.

      The A-a oxygen difference (or gradient) is a useful measure of shunt and V/Q mismatch, and it is less than 2 kPa in normal adults breathing air (15 mmHg). Because the shunt component is not corrected, the A-a difference increases when breathing 100 percent oxygen, and it can be up to 15 kPa (115 mmHg).

      An abnormally low or abnormally high V/Q ratio within the lung can cause an increased A-a difference, though the former is more common. Atelectasis, which results in a low V/Q ratio, is the most likely cause of an A-a difference in a healthy adult breathing 100 percent oxygen.

      Hypoventilation may cause an increase in alveolar (and thus arterial) CO2, lowering alveolar PO2 according to the alveolar air equation.

      The alveolar PO2 is also reduced at high altitude.

      Healthy people are unlikely to have a right-to-left shunt or an oxygen transport diffusion defect.

    • This question is part of the following fields:

      • Physiology
      41.6
      Seconds
  • Question 6 - A 43-year old woman, presented to the emergency department. She has suffered trauma...

    Incorrect

    • A 43-year old woman, presented to the emergency department. She has suffered trauma to her right orbital floor. On examination, it is noted that her right eye is deviated upwards when compared to her left. She also has a deliberate tilt in her head to the left in an attempt to compensate for loss of intorsion. This clinical sign is caused by damage to which of the following cranial nerves?

      Your Answer: Oculomotor nerve

      Correct Answer: Trochlear nerve

      Explanation:

      The trochlear nerve (CN IV) is the fourth and smallest cranial nerve. It’s role is to provide somatic motor innervation of the superior oblique muscle which is responsible for oculomotion.

      Injury to the trochlear nerve will result in vertical diplopia, which worsens when looking downwards or inwards. This diplopia presents as an upward deviation of the eye with a head tilt away from the site of the lesion.

      The abducens nerve (CN VI) provides somatic motor innervation for the lateral rectus muscle which functions to abduct the eye. Injury to this nerve will cause diplopia and an inability to abduct the eye, causing the patient to have to rotate their head to look sideways.

      The facial nerve (CN VII) provides sensory, motor and parasympathetic innervations. It’s motor aspect controls the muscles of facial expression. Damage will cause paralysis of facial expression.

      The oculomotor nerve (CN III) provides motor and parasympathetic innervations. Its motor component controls most of the other extraocular muscles. Damage to it will result in ptosis, dilatation of the pupil and a down and out eye position.

      The ophthalmic division of the trigeminal nerve (CN VI) is responsible for sensory innervation of skin, mucous membranes and sinuses of the upper face and scalp.

    • This question is part of the following fields:

      • Pathophysiology
      19.6
      Seconds
  • Question 7 - All of the following statements about that parasympathetic nervous system (PNS) are true...

    Correct

    • All of the following statements about that parasympathetic nervous system (PNS) are true except:

      Your Answer: The PNS has nicotinic receptors throughout the system

      Explanation:

      With regards to the autonomic nervous system (ANS)

      1. It is not under voluntary control
      2. It uses reflex pathways and different to the somatic nervous system.
      3. The hypothalamus is the central point of integration of the ANS. However, the gut can coordinate some secretions and information from the baroreceptors which are processed in the medulla.

      With regards to the central nervous system (CNS)
      1. There are myelinated preganglionic fibres which lead to the
      ganglion where the nerve cell bodies of the non-myelinated post ganglionic nerves are organised.
      2. From the ganglion, the post ganglionic nerves then lead on to the innervated organ.

      Most organs are under control of both systems although one system normally predominates.

      The nerves of the sympathetic nervous system (SNS) originate from the lateral horns of the spinal cord, pass into the anterior primary rami and then pass via the white rami communicates into the ganglia from T1-L2.

      There are short pre-ganglionic and long post ganglionic fibres.
      Pre-ganglionic synapses use acetylcholine (ACh) as a neurotransmitter on nicotinic receptors.
      Post ganglionic synapses uses adrenoceptors with norepinephrine / epinephrine as the neurotransmitter.
      However, in sweat glands, piloerector muscles and few blood vessels, ACh is still used as a neurotransmitter with nicotinic receptors.

      The ganglia form the sympathetic trunk – this is a collection of nerves that begin at the base of the skull and travel 2-3 cm lateral to the vertebrae, extending to the coccyx.

      There are cervical, thoracic, lumbar and sacral ganglia and visceral sympathetic innervation is by cardiac, coeliac and hypogastric plexi.

      Juxta glomerular apparatus, piloerector muscles and adipose tissue are all organs under sole sympathetic control.

      The PNS has a craniosacral outflow. It causes reduced arousal and cardiovascular stimulation and increases visceral activity.

      The cranial outflow consists of
      1. The oculomotor nerve (CN III) to the eye via the ciliary ganglion,
      2. Facial nerve (CN VII) to the submandibular, sublingual and lacrimal glands via the pterygopalatine and submandibular ganglions
      3. Glossopharyngeal (CN IX) to lungs, larynx and tracheobronchial tree via otic ganglion
      4. The vagus nerve (CN X), the largest contributor and carries ¾ of fibres covering innervation of the heart, lungs, larynx, tracheobronchial tree parotid gland and proximal gut to the splenic flexure, liver and pancreas

      The sacral outflow (S2 to S4) innervates the bladder, distal gut and genitalia.

      The PNS has long preganglionic and short post ganglionic fibres.
      Preganglionic synapses, like in the SNS, use ACh as the neuro transmitter with nicotinic receptors.
      Post ganglionic synapses also use ACh as the neurotransmitter but have muscarinic receptors.

      Different types of these muscarinic receptors are present in different organs:
      There are:
      M1 = pupillary constriction, gastric acid secretion stimulation
      M2 = inhibition of cardiac stimulation
      M3 = visceral vasodilation, coronary artery constriction, increased secretions in salivary, lacrimal glands and pancreas
      M4 = brain and adrenal medulla
      M5 = brain

      The lacrimal glands are solely under parasympathetic control.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      59.5
      Seconds
  • Question 8 - A 35-year-old male presents to GP presenting an area of erythema which was...

    Incorrect

    • A 35-year-old male presents to GP presenting an area of erythema which was around a recent cut on his right forearm. He was prescribed a short course of antibiotics and after 5 days again presented with progressive fatigue, headaches, and fevers. On clinical examination: Oxygen saturation: 98% on room air, Respiratory rate: 22 per minute, Heart rate: 100 beats per minute, Blood pressure: 105/76 mmHg, Temperature: 38.2 degree Celsius. On physical examination, a dramatic increase in the area of erythema was noted. Blood culture was done in the patient and indicated the presence of bacterium containing beta-lactamase. Which of the following antibiotics was likely prescribed to the patient?

      Your Answer: Co-amoxiclav

      Correct Answer: Amoxicillin

      Explanation:

      Ciprofloxacin belongs to the quinolone group of antibiotics, and doxycycline and minocycline are tetracyclines. So, they are not affected by beta-lactamase.
      However, amoxicillin is a beta-lactam antibiotic and beta-lactamase cleaves the beta-lactam ring present in amoxicillin. This results in the breakdown of the antibiotic and thus the area of erythema dramatically increased.
      Co-amoxiclav contains amoxicillin and clavulanic acid which protects amoxicillin from beta-lactamase.

    • This question is part of the following fields:

      • Pharmacology
      25.8
      Seconds
  • Question 9 - 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
      16.9
      Seconds
  • Question 10 - Concerning the intercostal nerves, which one of the following is true? ...

    Incorrect

    • Concerning the intercostal nerves, which one of the following is true?

      Your Answer: The eleventh intercostal nerve is also known as the subcostal nerve

      Correct Answer: Each is connected to a ganglion of the sympathetic trunk

      Explanation:

      The intercostal nerves arise from the ventral rami of the first 11 thoracic spinal nerves. they course along the costal groove on the lower margin of the rib.

      The twelfth intercoastal nerve is called the subcostal nerve. This is because it is below the 12th rib.

      Each intercostal nerve is connected to a ganglion of the sympathetic trunk from which it carries preganglionic and postganglionic fibres that innervate blood vessels, sweat glands, and muscles.

      The lateral and medial pectoral nerves innervates pectoralis major muscle.

    • This question is part of the following fields:

      • Anatomy
      25.3
      Seconds
  • Question 11 - A 72-year old farmer is hospitalized with acute respiratory failure and autonomic dysfunction....

    Correct

    • A 72-year old farmer is hospitalized with acute respiratory failure and autonomic dysfunction. Suspected organophosphate poisoning. Which one is the best mechanism for acute toxicity caused by organophosphates?

      Your Answer: Inhibition of acetylcholinesterase

      Explanation:

      The toxicity of organophosphorus (OP) nerve agents is manifested through irreversible inhibition of acetylcholinesterase (AChE) at the cholinergic synapses, which stops nerve signal transmission, resulting in a cholinergic crisis and eventually death of the poisoned person. Oxime compounds used in nerve agent antidote regimen reactivate nerve agent-inhibited AChE and halt the development of this cholinergic crisis.

    • This question is part of the following fields:

      • Physiology
      13.7
      Seconds
  • Question 12 - A bolus of alfentanil has a faster onset of action than an equal...

    Incorrect

    • A bolus of alfentanil has a faster onset of action than an equal dose of fentanyl. Which of the following statements most accurately describes the difference?

      Your Answer: The lipid solubility of fentanyl is more than alfentanil

      Correct Answer: The pKa of alfentanil is less than that of fentanyl

      Explanation:

      Unionised molecules are more likely than ionised molecules to cross membranes (such as the blood-brain barrier).

      Because alfentanil and fentanyl are weak bases, the Henderson-Hasselbalch equation says that the ratio of ionised to unionised molecules is determined by the parent compound’s pKa in relation to physiological pH.

      Alfentanil has a pKa of 6.5, while fentanyl has a pKa of 8.4.
      At a pH of 7.4, 89 percent of alfentanil is unionised, whereas 9% of fentanyl is.

      As a result, alfentanil has a faster onset than fentanyl.

      Fentanyl is 83% plasma protein bound
      Alfentanil is 90% plasma protein bound.

      Alfentanil’s pharmacokinetics are affected by its higher plasma protein binding. Because alfentanil has a low hepatic extraction ratio (0.4), clearance is determined by the degree of protein binding rather than the time it takes to take effect.

    • This question is part of the following fields:

      • Pharmacology
      28.2
      Seconds
  • Question 13 - Regarding the anatomical relations of the scalenus muscles, which of these is true?...

    Incorrect

    • Regarding the anatomical relations of the scalenus muscles, which of these is true?

      Your Answer: The phrenic nerve passes behind scalenus anterior

      Correct Answer: The trunks of the brachial plexus emerge from the lateral border of scalenus anterior

      Explanation:

      The ascending cervical artery lies media the phrenic nerve on scalenus anterior and can easily be mistaken for the phrenic nerve at operation.

      The phrenic nerve passes across scalenus anterior and medius inferiorly.

      The subclavian artery is separated from the vein by the scalenus anterior.

      The brachiocephalic vein is formed at the medial border of scalenus anterior by the subclavian vein and the internal jugular vein.

      Emerging from the lateral border of scalenus anterior are the trunks of the brachial plexus .

    • This question is part of the following fields:

      • Anatomy
      2647.5
      Seconds
  • Question 14 - A 76-year-old man, presents to his general practitioner with a lump in his...

    Incorrect

    • A 76-year-old man, presents to his general practitioner with a lump in his left groin. Upon examination, his doctor is able to diagnose a direct inguinal hernia. There are many structures present in the inguinal canal. Where is the ilioinguinal nerve located in relation to the spermatic cord?

      Your Answer: Medial to the spermatic cord

      Correct Answer: Anterior to the spermatic cord

      Explanation:

      The answer is anterior to the spermatic cord.

      The inguinal canal in men contains the ilioinguinal nerve, the genitofemoral nerve and the spermatic cord.

      The ilioinguinal nerve arises of the L1 nerve root with the Iliohypogastric nerve, before entering the inguinal canal from the side, through the muscles of the abdomen, travelling superficial to the spermatic cord.

    • This question is part of the following fields:

      • Anatomy
      36.7
      Seconds
  • Question 15 - A 20-year old lady has been having excessive bruising and bleeding of her...

    Incorrect

    • A 20-year old lady has been having excessive bruising and bleeding of her gums. She is under investigation for the extrinsic pathway of coagulation. Which is the best investigation to order?

      Your Answer: aPTT time

      Correct Answer: Prothrombin time (PT)

      Explanation:

      The extrinsic pathway is best assessed by the PT 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
      11.3
      Seconds
  • Question 16 - Which of the following statements is true about an acute pulmonary embolism? ...

    Incorrect

    • Which of the following statements is true about an acute pulmonary embolism?

      Your Answer: Embolectomy is more effective than thrombolysis in improving survival

      Correct Answer: Thrombolysis administered through a peripheral vein is as effective as through a pulmonary artery catheter

      Explanation:

      Acute pulmonary embolism occurs when a blood clot becomes embedded in a pulmonary artery and restricts lung blood flow.

      Thrombolysis is recommended in patients with extremely compromised circulation rather than reduced oxygen in the blood. It is effective when administered via a peripheral vein or a pulmonary artery catheter.

      Anticoagulant therapy (heparin use) decreases the risk of further embolic evens and decreases constriction of pulmonary vessels.

      An ECG may be normal in patients with an acute pulmonary embolism.

    • This question is part of the following fields:

      • Pathophysiology
      48.8
      Seconds
  • Question 17 - Two different anti-viral treatments are being evaluated for COVID-19 in a clinical study....

    Incorrect

    • Two different anti-viral treatments are being evaluated for COVID-19 in a clinical study. Which of the following statistical method should be opted to compare survival time with?

      Your Answer: Absolute risk reduction

      Correct Answer: Hazard ratio

      Explanation:

      The hazard ratio (HR) is simply a comparison of two hazards in a study. It provides an estimate of the ratio of the hazard rates between the experimental group and a control group over the entire study duration. It is typically used when analysing survival over time, hence is the most suitable statistical method in this case.

      An odds ratio is a statistic that quantifies the strength of the association between two events, A and B. It is the €œmeasure of association€� for a case-control study.

      The Pearson product-moment correlation coefficient (Pearson’s correlation, for short) is a measure of the strength and direction of association that exists between two variables. An example would be if scientists wanted to evaluate the relationship between quality of certain population of rice and their genetic make-up.

      Relative risk is the ratio of the risks for an event for the exposure group to the risks for the non-exposure group. Thus relative risk provides an increase or decrease in the likelihood of an event based on some exposure. Relative risk measures the association between the exposure and the outcome.

      Absolute risk reduction is the number of percentage points your own risk goes down if you do a preventive act such as stop drinking alcohol. It depends on what your risk factors are to begin with.

    • This question is part of the following fields:

      • Statistical Methods
      16.2
      Seconds
  • Question 18 - The cardiac tissue type that that has the highest conduction velocity is: ...

    Correct

    • The cardiac tissue type that that has the highest conduction velocity is:

      Your Answer: Purkinje fibres

      Explanation:

      Potassium maintains the resting potential of cardiac myocytes, with depolarization triggered by a rapid influx of sodium ions, and repolarization due to efflux of potassium. A slow influx of calcium is responsible for the longer duration of a cardiac action potential compared with skeletal muscle.

      The cardiac action potential has several phases which have different mechanisms of action as seen below:

      Phase 0: Rapid depolarisation – caused by a rapid sodium influx.
      These channels automatically deactivate after a few ms.

      Phase 1: caused by early repolarisation and an efflux of potassium.

      Phase 2: Plateau – caused by a slow influx of calcium.

      Phase 3 – Final repolarisation – caused by an efflux of potassium.

      Phase 4 – Restoration of ionic concentrations – The resting potential is restored by Na+/K+ATPase.
      There is slow entry of Na+into the cell which decreases the potential difference until the threshold potential is reached. This then triggers a new action potential

      Of note, cardiac muscle remains contracted 10-15 times longer than skeletal muscle.

      Different sites have different conduction velocities:
      1. Atrial conduction – Spreads along ordinary atrial myocardial fibres at 1 m/sec

      2. AV node conduction – 0.05 m/sec

      3. Ventricular conduction – Purkinje fibres are of large diameter and achieve velocities of 2-4 m/sec, the fastest conduction in the heart. This allows a rapid and coordinated contraction of the ventricles

    • This question is part of the following fields:

      • Physiology And Biochemistry
      7.8
      Seconds
  • Question 19 - 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: Haemoglobin (HbF)

      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 α and 2 γchains. 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 β 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
      42.8
      Seconds
  • Question 20 - A new proton pump inhibitor (PPI) is being evaluated in elderly patients who...

    Correct

    • A new proton pump inhibitor (PPI) is being evaluated in elderly patients who are taking aspiring. Study designed has 120 patients receiving the PPI, while a control group of 240 individuals is given the standard PPI. Over a span of 6 years, 24 of the group receiving the new PPI had an upper GI bleed compared to 60 individuals who received the standard PPI. How would you calculate the absolute risk reduction?

      Your Answer: 5%

      Explanation:

      Absolute risk reduction = (Control event rate) – (Experimental event rate)

      Experimental event rate = 24 / 120 = 0.2

      Control event rate = 60 / 240 = 0.25

      Absolute risk reduction = 0.25 – 0.2 = 0.05 = 5% reduction

    • This question is part of the following fields:

      • Statistical Methods
      42.7
      Seconds
  • Question 21 - A 40-year old female comes to the GP's office with unexplained weight gain,...

    Correct

    • A 40-year old female comes to the GP's office with unexplained weight gain, cold intolerance and fatigue. Her thyroid function tests are performed as there is a suspicion of hypothyroidism. A negative feedback mechanism is incorporated in the control of thyroid hormone release. All of choices below are also controlled by a negative feedback loop except:

      Your Answer: Clotting cascade

      Explanation:

      The correct answer is the clotting cascade, which occurs via a positive feedback mechanism. As clotting factors are attracted to a site, their presence attracts further clotting factors. This continues until a functioning clot is formed.

      This patient has presented with symptoms of hypothyroidism and symptoms include weight gain, lethargy, cold intolerance, dry skin, coarse hair and constipation. It can be treated by replacing the missing thyroid hormone with levothyroxine which is a synthetic version of thyroxine (T4).

      Serum carbon dioxide (CO2) is controlled via a negative feedback mechanism as well. Chemoreceptors can detect when the serum CO2 is high, and send an impulse to the respiratory centre of the brain to increase the respiratory rate. As a result, more CO2 is exhaled which lowers the serum concentration.

      Cortisol is also released according to a negative feedback mechanism. Cortisol acts on both the hypothalamus and the anterior pituitary. Its action serve to decrease the formation of corticotrophin releasing hormone (CRH) and adrenocorticotropic hormone (ACTH), respectively. CRH acts on the anterior pituitary to release ACTH. This then acts on the adrenal gland to cause the release of cortisol. Thus, inhibition of CRH and ACTH formation results in high levels of cortisol which inhibit its further release.

      Blood pressure (BP) is controlled via a negative feedback mechanism. Low BP results in renin-angiotensin-aldosterone system (RAAS) activation. This leads to vasoconstriction and retention of salt and water which increased BP.
      Blood sugar is controlled via a negative feedback mechanism. A rise in blood sugar causes insulin to be released. Insulin acts to transport glucose into the cell which lowers blood sugar.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      17.8
      Seconds
  • Question 22 - Which of the following is incorrect with regards to atrial natriuretic peptide? ...

    Correct

    • Which of the following is incorrect with regards to atrial natriuretic peptide?

      Your Answer: Secreted mainly by the left atrium

      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
      15.3
      Seconds
  • Question 23 - A 28-year-old girl complained of severe abdominal pain and hematemesis and was rushed...

    Correct

    • 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.

    • This question is part of the following fields:

      • Anatomy
      30.2
      Seconds
  • Question 24 - The following statements are about capnography. Which of them is true? ...

    Incorrect

    • The following statements are about capnography. Which of them is true?

      Your Answer: Utilises Lambert's law

      Correct Answer: Collision broadening is due to presence of other polyatomic molecules

      Explanation:

      Capnography is the non-invasive measurement and pictorial display of inhaled and exhaled carbon dioxide (CO2) partial pressure.

      It is depicted graphically as the concentration of CO2 over time.

      It is used in disease diagnosis, determining disease severity, assessing response to treatment and is the best method to for indicating when an endotracheal tube is placed in the trachea after intubation.

      The wavelength of IR light usually absorbed by nitrous oxide is between 4.4-4.6μm (very close to that of CO2). Its absorption of wavelengths at 3.9 μm is much weaker. It causes a measurable deficit of 0.1% for every 10% of nitrous oxide. The maximal wavelength of infrared (IR) light absorbed by carbon monoxide is 4.7 μm. The volatile agents have strong absorption bands at 3.3 μm and throughout the ranges 8-12 μm.

      IR light is not absorbed by oxygen (O2), but O2 and CO2 molecules are constantly colliding which interrupts the absorption of IR light by CO2. This increases the band of absorption, that is the Collison or pressure broadening). An oxygen percentage of 95 will result in a 0.5 percentage fall in CO2 measure.

      IR light is also absorbed by water vapour which will result in an overlap of the absorption band, collision broadening and a dilution of partial pressure. This is why water trap and water permeable tubing is recommended for use as it reduces measurement inaccuracies.

      The use of multi-gas analysers of modern gases also help reduce the effects of collision broadening.

      Beer’s law is also applied in this system as an increase in the concentrations of CO2 causes a decrease in the amount of IR able to pass through the gas. This IR light is what generated the signal that is analysed for display.

      The capnograph can indicate oesophageal intubation, but cannot determine if it is endotracheal or endobronchial. For this, auscultation is used.

    • This question is part of the following fields:

      • Clinical Measurement
      26
      Seconds
  • Question 25 - Which compound is secreted only from the adrenal medulla? ...

    Correct

    • Which compound is secreted only from the adrenal medulla?

      Your Answer: Adrenaline

      Explanation:

      The adrenal medulla comprises chromaffin cells (pheochromocytes), which are functionally equivalent to postganglionic sympathetic neurons. They synthesize, store and release the catecholamines noradrenaline (norepinephrine) and adrenaline (epinephrine) into the venous sinusoids.
      The majority of the chromaffin cells synthesize adrenaline.

    • This question is part of the following fields:

      • Anatomy
      8.3
      Seconds
  • Question 26 - Regarding thermocouple, which of the following best describes its properties? ...

    Incorrect

    • Regarding thermocouple, which of the following best describes its properties?

      Your Answer: Resistance of the measuring junction varies exponentially with temperature

      Correct Answer: The electromotive force at the measuring junction is proportional to temperature

      Explanation:

      Thermocouples are based on the Seebeck effect, i.e. a small thermoelectric current is generated when two different metal wires are put into contact at both ends with their junctions having a different temperature. If one junction is open, a contact electromotive force is generated. The current, or the electromotive force, is to a first approximation proportional to the temperature difference ΔT between the two junctions. A better approximation is obtained with a MacLaurin expansion with the second power of ΔT.

      Two wires bonded together with different coefficients of expansion can be used as a switch for thermostatic control.

      The resistance at the measuring junction is irrelevant.

      The resistance of a thermistor varies exponentially with temperature, while the resistance of a measuring junction varies linearly with temperature. Incorporated into a Wheatstone bridge is this unknown resistance, which is used to indirectly measure temperature.

    • This question is part of the following fields:

      • Basic Physics
      18.7
      Seconds
  • Question 27 - An emergency appendicectomy is being performed on a 20 year old man. For...

    Correct

    • 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: 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
      35.6
      Seconds
  • Question 28 - What statement about endotoxins is true? ...

    Incorrect

    • What statement about endotoxins is true?

      Your Answer: Elicit an antibody response which may protect the host from future attack

      Correct Answer: Can often survive autoclaving

      Explanation:

      Endotoxins are the lipopolysaccharides found in the outer cell wall of Gram-negative bacteria. They are responsible for providing the structure and stability of the cell wall.

      They cannot be destroyed by normal sterilisation as they are heat stable molecules. They require the use of certain sterilant such as superoxide, peroxide and hypochlorite to be neutralised.

      They stimulate strong immune responses, but can only be destroyed partially by specific antibodies. Repeat infections occur as memory T cells cannot be formed.

      It can cause septicaemia and associated symptoms such as fever, shock, hypotension and nausea.

      It activates the alternative complement pathway and the coagulation pathway using secreted cytokines.

      It is not involved in botulism as clostridium botulinum, the responsible organism, secretes a neurotoxic exotoxin.

    • This question is part of the following fields:

      • Pathophysiology
      15.8
      Seconds
  • Question 29 - A patient was brought to the emergency room after passing black tarry stools....

    Incorrect

    • A patient was brought to the emergency room after passing black tarry stools. The initial diagnosis was upper gastrointestinal bleeding. The patient was placed on temporary nil per os (NPO) for the next 24 hours, his weight was 110 kg, and the required volume of intravenous fluid for the him was 3 litres. His electrolytes and other biochemistry studies were normal. If you were to choose the intravenous fluid regimen that would closely mimic his basic electrolyte and caloric requirements, which one would be the best answer?

      Your Answer: 3000 mL Hartmann's

      Correct Answer: 3000 mL 0.45% N. saline with 5% dextrose, each bag with 40 mmol of potassium

      Explanation:

      The patient in the case has a fluid volume requirement of 30 mL/kg/day. His basic electrolyte requirement per day is:

      Sodium at 2 mmol/kg/day x 110 = 220 mmol/day
      Potassium at 1 mmol/kg/day x 110 = 110 mmol/day

      His energy requirement per day is:

      35 kcal/kg/day x 110 kg = 3850 kcal/day

      One gram of glucose in fluid can provide approximately 4 kilocalories.

      The following are the electrolyte components of the different intravenous fluids:

      Fluid Na (mmol/L) K (mmol/L)
      0.9% Normal saline (NSS) 154 0
      0.45% NSS + 5% dextrose 77 0
      0.18% NSS + 4% dextrose 30 0
      Hartmann’s 131 5
      5% dextrose 0 0

      1000 mL of 5% dextrose has 50 g of glucose

      Option B is inadequate for his sodium and caloric requirements (30 mmol of Na+ and 560 kcal). It is adequate for his K+ requirement (120 mmol of K+).

      Option C is in excess of his Na+ requirement (462 mmol of Na+). Moreover, it does not provide any K+ replacement.

      Option D is inadequate for his caloric requirement (600 kcal) and K+ requirement (60 mmol of K+). Moreover it does not provide any Na+ replacement.

      Option E is in excess of his Na+ requirement (393 mmol of Na+), and is inadequate for his potassium requirement (15 mmol of K+)

      Option A has adequate amounts for his Na+ (231 mmol of Na+) and K+ (120 mmol of K+) requirements. It is inadequate for his caloric requirement (600 kcal).

    • This question is part of the following fields:

      • Physiology
      117.8
      Seconds
  • Question 30 - Transthoracic echocardiogram (TTE) can be used to investigate the function of the heart...

    Correct

    • Transthoracic echocardiogram (TTE) can be used to investigate the function of the heart in patients with suspected heart failure. The aim is to measure the ejection fraction, but to do that, the stroke volume must first be measured. How is stroke volume calculated?

      Your Answer: End diastolic volume - end systolic volume

      Explanation:

      Cardiac output = stroke volume x heart rate

      Left ventricular ejection fraction = (stroke volume / end diastolic LV volume ) x 100%

      Stroke volume = end diastolic LV volume – end systolic LV volume

      Pulse pressure = Systolic Pressure – Diastolic Pressure

      Systemic vascular resistance = mean arterial pressure / cardiac output
      Factors that increase pulse pressure include:
      -a less compliant aorta (this tends to occur with advancing age)
      -increased stroke volume

    • This question is part of the following fields:

      • Physiology And Biochemistry
      15.9
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Anatomy (2/8) 25%
Statistical Methods (2/3) 67%
Pathophysiology (1/4) 25%
Physiology (1/4) 25%
Physiology And Biochemistry (5/6) 83%
Pharmacology (0/2) 0%
Clinical Measurement (0/1) 0%
Basic Physics (0/1) 0%
Anaesthesia Related Apparatus (1/1) 100%
Passmed