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  • Question 1 - Which of the following anaesthetic agents is most suitable for inhalational induction in...

    Correct

    • Which of the following anaesthetic agents is most suitable for inhalational induction in an 8-year-old child for inhalational induction of anaesthesia before routine surgery?

      Your Answer: Sevoflurane at 4%

      Explanation:

      The ideal agent for this case should have low blood: gas coefficient, pleasant smell, and high oil: gas coefficient (potent with a low Minimum alveolar coefficient (MAC)). Among the given options, Sevoflurane is perfect with 0.692 blood: gas partition coefficient and is low pungency, and is sweet.

      Other drugs with their blood: gas partition coefficient and their smell are given as:
      Blood/gas partition coefficient MAC Smell
      Enflurane 1.8 1.68 Pungent, ethereal
      Desflurane 0.42 7 Pungent, ethereal
      Halothane 2.54 0.71 Sweet
      Isoflurane 1.4 1.15 Pungent, ethereal

    • This question is part of the following fields:

      • Pharmacology
      11.3
      Seconds
  • Question 2 - A post-operative patient was given paracetamol and pethidine for post-operative analgesia. A few...

    Correct

    • A post-operative patient was given paracetamol and pethidine for post-operative analgesia. A few hours later, the patient developed fever of 38°C, hypertension, and agitation.

      According to the patient's medical history, he is maintained on Levodopa and Selegiline for Parkinson's disease.

      Which of the following is the most probable cause of his manifestation?

      Your Answer: Pethidine

      Explanation:

      Selegiline is a monoamine oxidase inhibitor. Inhibition of monoamine oxidase leads to increased levels of norepinephrine and serotonin in the central nervous system.

      Pethidine, also known as meperidine, is a strong agonist at the mu and kappa receptors. It inhibits pain neurotransmission and blocks muscarinic-specific actions.

      Administering opioid analgesic is relatively contraindicated to individuals taking monoamine oxidase inhibitors. This is because of the high incidence of serotonin syndrome, which is characterized by fever, agitation, tremor, clonus, hyperreflexia and diaphoresis. Onset of symptoms is within hours, and the treatment is mainly through sedation, paralysis, intubation and ventilation.

      The clinical findings are more consistent with Serotonin syndrome rather than exacerbation of Parkinson’s. Parkinson’s Disease (PD) exacerbations are defined as patient-reported or caregiver-reported episodes of subacute worsening of PD motor function in 1 or more domains (bradykinesia, tremor, rigidity, or PD-related postural instability/gait disturbance) that caused a decline in functional status, developed over a period of < 2 months, did not fluctuate with medication timing, and are not caused by intentional adjustments of PD medications by the treating neurologist. Malignant hyperthermia usually occurs within minutes of administration of a volatile anaesthetic, such as halothane, or succinylcholine. There is massive release of calcium from the sarcoplasmic reticulum, leading to fever, acidosis, rhabdomyolysis, trismus, clonus, and hypertension. In sepsis, it more common for patients to present with hypotension rather than hypertension.

    • This question is part of the following fields:

      • Pharmacology
      69.7
      Seconds
  • Question 3 - Which of the following best explains the association between smoking and lower oxygen...

    Correct

    • Which of the following best explains the association between smoking and lower oxygen delivery to tissues?

      Your Answer: Left shift of the oxygen dissociation curve

      Explanation:

      Smoking is a major risk factor associated with perioperative respiratory and cardiovascular complications. Evidence also suggests that cigarette smoking causes imbalance in the prostaglandins and promotes vasoconstriction and excessive platelet aggregation. Two of the constituents of cigarette smoke, nicotine and carbon monoxide, have adverse cardiovascular effects. Carbon monoxide increases the incidence of arrhythmias and has a negative ionotropic effect both in animals and humans.

      Smoking causes an increase in carboxyhaemoglobin levels, resulting in a leftward shift in which appears to represent a risk factor for some of these cardiovascular complications.

      There are two mechanisms responsible for the leftward shift of oxyhaemoglobin dissociation curve when carbon monoxide is present in the blood. Carbon monoxide has a direct effect on oxyhaemoglobin, causing a leftward shift of the oxygen dissociation curve, and carbon monoxide also reduces the formation of 2,3-DPG by inhibiting glycolysis in the erythrocyte. Nicotine, on the other hand, has a stimulatory effect on the autonomic nervous system. The effects of nicotine on the cardiovascular system last less than 30 min.

    • This question is part of the following fields:

      • Physiology
      18.6
      Seconds
  • Question 4 - 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: Facial artery

      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
      32.8
      Seconds
  • Question 5 - Over the course of 10 minutes, one litre of 0.9% normal saline is...

    Correct

    • Over the course of 10 minutes, one litre of 0.9% normal saline is intravenously infused into a normally fit and well 58-year-old male. A catheter is used to measure urine output before and after the infusion. The patient is 70 kg in weight.

      The following data on urine output is obtained:

      50ml/hour Before the infusion
      200 ml/hour 1 hour following infusion
      90 ml/hour 2 hours after the infusion
      60 ml/hr 3 hours after the infusion

      Which of the following physiological responses is most likely to account for the sudden increase in urine output after a fluid bolus?

      Your Answer: Increased glomerular filtration rate

      Explanation:

      The following are some basic assumptions:

      Extracellular fluid (ECF) makes up one-third of total body water (TBW), while intracellular fluid makes up the other two-thirds (ICF).
      One-quarter of ECF is plasma, and three-quarters is interstitial fluid (ISF).
      The volume receptors have a 7-10% blood volume change threshold. The osmoreceptors are sensitive to changes in osmolality of 1-2 percent.
      Prior to the transfusion, the plasma osmolality is normal (between 287 and 290 mOsm/kg).
      [Na+] in 0.9 percent N. saline is 154 mmol/L, which is similar to that of extracellular fluid. When given intravenously, this limits its distribution within the extracellular space, resulting in a plasma compartment:ISF volume ratio of 1:3.
      In this time frame, one litre of 0.9 percent N. saline will increase plasma volume by about 250 mL, which could be the threshold for activation of the volume receptors in the atria, resulting in the release of atrial natriuretic peptide (ANP).

      Because 0.9 percent N. saline is isosmotic, after a 1 L infusion, plasma osmolality will not change. No changes in antidiuretic hormone secretion will be detected by the hypothalamic osmoreceptors.

      Because normal saline is protein-free, the oncotic pressure in the blood is slightly reduced after the saline infusion. As a result, fluid movement into the ISF is favoured (Starling’s hypothesis), and the lower oncotic pressure causes an immediate increase in the glomerular filtration rate (GFR) and a reduction in water reabsorption in the proximal tubule.

      The flow of urine increases. There is no hormonal intermediary in this effect, so it is strictly local. Urine flow immediately increases. The fluid returns to the intravascular compartment, and urine flow continues until all of the transfused fluid has been excreted.

      Blood pressure changes associated with a 1 L fluid infusion are unlikely to affect high-pressure baroreceptors in the carotid sinus.

      The juxta-glomerular cells of the afferent arteriole are adjacent to the specialised cells (macula densa) of distal tubules. The sodium and chloride ions in the tubular fluid are detected by the macula densa. Renin release is inhibited when the tubular fluid contains too much sodium chloride. Hormonal changes take longer to manifest than physical changes that control glomerulotubular balance.
      Hypertonic saline, not 0.9 percent N saline, is an osmotic diuretic.

    • This question is part of the following fields:

      • Pathophysiology
      58.1
      Seconds
  • Question 6 - Where should you insert a needle to obtain a femoral artery sample to...

    Incorrect

    • Where should you insert a needle to obtain a femoral artery sample to be used for an arterial blood gas?

      Your Answer: 2cm inferomedially to the pubic tubercle

      Correct Answer: Mid inguinal point

      Explanation:

      The needle should be inserted just below the skin at the mid inguinal point which is the surface indicator for the femoral artery.

    • This question is part of the following fields:

      • Anatomy
      28.6
      Seconds
  • Question 7 - A randomized controlled trail has been conducted to compare two drugs used for...

    Incorrect

    • 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: For each patient who drops out, remove a patient from the other randomised group

      Correct 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
      54.5
      Seconds
  • Question 8 - A new study is being carried out on the measurement of a new...

    Incorrect

    • A new study is being carried out on the measurement of a new cardiovascular disease biomarker, and its applications in preoperative screening. The data for this study is expected to be normally distributed.

      Which of the following statements is true about normal distributions?

      Your Answer: The 95% confidence interval tells us how confident we are in the test

      Correct Answer: The mean, median and mode are the same value

      Explanation:

      The correct answer is the mean, median and mode of normally distributed data are the same value. This is as a result of the bell shaped curve which is equal on both sides.

      The bell-shape indicates that values around the mean are more frequent in occurrence than the values farther away.

      In a normal distribution:
      1) +/- one standard deviation of the mean accounts for 68% of the data.
      2) +/- two standard deviations of the mean accounts for 95% of the data.
      3) +/- three standard deviations of the mean accounts for 99.7% of the data.

    • This question is part of the following fields:

      • Statistical Methods
      53.3
      Seconds
  • Question 9 - A 58-year-old man is being operated on for a radical gastrectomy for carcinoma...

    Incorrect

    • A 58-year-old man is being operated on for a radical gastrectomy for carcinoma of the stomach.

      Which structure needs to be divided to gain access to the coeliac axis?

      Your Answer: Median arcuate ligament

      Correct Answer: Lesser omentum

      Explanation:

      The lesser omentum will need to be divided. This forms one of the nodal stations that will need to be taken during a radical gastrectomy.

      The celiac axis is the first branch of the abdominal aorta and supplies the entire foregut (mouth to the major duodenal papilla). It arises at the level of vertebra T12. It has three major branches:
      1. Left gastric
      2. Common hepatic
      3. Splenic arteries

    • This question is part of the following fields:

      • Anatomy
      21
      Seconds
  • Question 10 - A patient on admission is given an infusion of 1000 mL of 10%...

    Incorrect

    • A patient on admission is given an infusion of 1000 mL of 10% glucose and 500 mL of 20% lipid over a 24 hour period.

      Which of these best approximates to the energy input over this time period?

      Your Answer: 1100 kcal

      Correct Answer: 1300 kcal

      Explanation:

      1% solution contains 1 g of substance per 100 mL.

      A solution of 10% glucose is 10 g/100mL. Therefore 1000 mL of this glucose solution will contain 100 g.

      1 g of glucose yields about 4 kcal of energy. One litre of 10% glucose will therefore release approximately 4x100g = 400 kcal of energy.

      A solution of 20% fat is 20 g/100mL. Therefore 1000 mL of this fat solution will have 200 g and 500 mL will contain 100 g.

      1 g of fat yields approximately 9 kcal. 500 mL of 20% fat therefore has the potential to yield 900 kcal of energy.

      The total energy input over this 24 hour period is approximately 400kcal + 900kcal = 1300 kcal.

    • This question is part of the following fields:

      • Physiology
      19.4
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Pharmacology (2/2) 100%
Physiology (1/2) 50%
Anatomy (0/3) 0%
Pathophysiology (1/1) 100%
Statistical Methods (0/2) 0%
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