00
Correct
00
Incorrect
00 : 00 : 00
Session Time
00 : 00
Average Question Time ( Mins)
  • Question 1 - A survey aimed at finding out mean glucose level in individuals that took...

    Incorrect

    • A survey aimed at finding out mean glucose level in individuals that took antipsychotics medicines was conducted. The results were as follows:

      Mean Value: 7mmol/L

      Standard Deviation: 6mmol/L

      Sample Size: 9

      Standard Error: 2mmol/L

      For a confidence interval of 95%, which of the option presents the correct range up to the nearest value?

      Your Answer: 5-9 mmol/L

      Correct Answer: 3-11 mmol/L

      Explanation:

      Key Point: While finding out confidence intervals, standard errors are used. Standard error and Standard deviation are two distinct entities and should not be confused.

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

      Multiply the standard error by 3.

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

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

      The range turns out to be 1-13 mmol/L.

      For a confidence interval of 68%, multiply the standard error with 1 and repeat the process. The range found for this interval is 3-11 mmol/L.

      For a 95% confidence interval. Standard Error is multiplied by 1.96 which gives us the limit ranging from 3.08 to 10.92 mmol/L which could be approximated to 3-11 mmol/L.

    • This question is part of the following fields:

      • Statistical Methods
      44.8
      Seconds
  • Question 2 - A normal woman at term, not in labour, has her arterial blood gas...

    Incorrect

    • A normal woman at term, not in labour, has her arterial blood gas analysed.

      Which set of results is most likely her own?

      Option - pH - PaCO2 - HCO3 - PaO2
      A - 7.35 - 28 mmHg (3.73 kPa) - 27 mmol/L - 104 mmHg (13.8kPa)
      B - 7.43 - 32 mmHg (4.27 kPa) - 21 mmol/L - 104 mmHg (13.8kPa)
      C - 7.44 - 36 mmHg (4.8 kPa) - 27 mmol/L - 104 mmHg (13.8kPa)
      D - 7.45 - 40 mmHg (5.33 kPa) - 21 mmol/L - 104 mmHg (13.8kPa)
      E - 7.46 - 44 mmHg (5.87kPa) - 21 mmol/L - 104 mmHg (13.8kPa)

      Your Answer: D

      Correct Answer: B

      Explanation:

      Due to an increased tidal volume with little change or slight increase in respiratory rate, Minute ventilation at term is increased by about 50%. Hypothalamic function are thought to influence by Progesterone, oestradiol and prostaglandins. This causes a mild compensated respiratory alkalosis.

      Maternal PaCO2 is usually decreased to about 32 mmHg (4.27 kPa) as a result of this increased alveolar ventilation at term . A compensatory decrease in serum bicarbonate from 27 to 21 mmol/L by renal excretion lessens the impact of maternal alkalosis.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      126.8
      Seconds
  • Question 3 - The average diastolic blood pressure of a control group was found out to...

    Incorrect

    • The average diastolic blood pressure of a control group was found out to be 80 with a standard deviation of 5 in a study aimed at exploring the efficiency of a novel anti-hypertensive drug. The trial was randomised.

      Making an assumption that the data is normally distributed, find out the number of patients that had diastolic blood pressure over 90.

      Your Answer:

      Correct Answer: 3%

      Explanation:

      Since the data is normally distributed, 95% of the values lie with in the interval 70 to 90. This can be calculated as follows:

      Interval= Mean ± ( 2 times standard deviation)
      = 80 ± 2(5)
      = 80 ± 10
      = 70 & 90

      The rest of the 5% are distributed symmetrically beyond 90 and below 70 which means 2.5% of the values lie above 90.

    • This question is part of the following fields:

      • Statistical Methods
      0
      Seconds
  • Question 4 - Standard error of the mean can be defined as: ...

    Incorrect

    • Standard error of the mean can be defined as:

      Your Answer:

      Correct Answer: Standard deviation / square root (number of patients)

      Explanation:

      The standard error of the mean (SEM) is a measure of the spread expected for the mean of the observations – i.e. how ‘accurate’ the calculated sample mean is from the true population mean. The relationship between the standard error of the mean and the standard deviation is such that, for a given sample size, the standard error of the mean equals the standard deviation divided by the square root of the sample size.

      SEM = SD / square root (n)

      where SD = standard deviation and n = sample size

    • This question is part of the following fields:

      • Statistical Methods
      0
      Seconds
  • Question 5 - Which statement is true when describing carbonic anhydrase? ...

    Incorrect

    • Which statement is true when describing carbonic anhydrase?

      Your Answer:

      Correct Answer: Isoenzyme IV is found in the brush border of the proximal convoluted tubule

      Explanation:

      Carbonic anhydrase is an enzyme which contains zinc and can be found in:
      1. Erythrocytes
      2. Pulmonary endothelium
      3. The intestine
      4. Pancreas
      5. Cardiac muscle and skeletal muscle.

      To date, there have been seven isoenzymes identified. Of note, isoenzyme IV is found in the brush border of the proximal convoluted tubule and isoenzyme II is found within the luminal cells.

      Acetazolamides a carbonic anhydrase inhibitor and is used as prophylaxis against mountain sickness and in glaucoma management.

      Spironolactone is a potassium diuretic and is an aldosterone antagonist.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 6 - An 80 year old woman is due for cataract surgery.

    There are no...

    Incorrect

    • An 80 year old woman is due for cataract surgery.

      There are no contraindications to regional anaesthesia so a peribulbar block was performed. 8mls of 2% lidocaine was injected using an infratemporal approach. However, there is still movement of the globe after 5 mins.

      The least likely extraocular muscle to develop akinesia is:

      Your Answer:

      Correct Answer: Superior oblique

      Explanation:

      The fibrotendinous ring formed by the congregation of the rectus muscles at the apex of the orbit does not include superior oblique. This muscle is completely outside the ring and so it is the most difficult muscle to anaesthetise completely. A good grasp of the anatomy of the area being anaesthetised is important with all regional anaesthetic techniques so that potential problems and complications with a block can be anticipated.

      The borders of this pyramid whose apex points upwards and outwards of the bony orbit are as follows:
      Floor – Zygoma and Maxilla
      Roof – frontal bone
      Medial wall – maxilla, ethmoid, sphenoid and lacrimal bones.
      Lateral wall – greater wing of the sphenoid and the zygoma.

      The four recti muscles (superior, medial, lateral and inferior) originate from a tendinous ring (the annulus of Zinn) and extend anteriorly to insert beyond the equator of the globe. Bands of connective tissue are present between the rectus muscles forming a conical structure and hinder the passage of local anaesthetic.

      The superior oblique muscle is situated outside this ring and is the most difficult muscle to anaesthetise completely, particularly with a single inferotemporal peribulbar injection. An additional medial injection may help to prevent this.

      The cranial nerve supply to the extraocular muscles are:
      3rd (inferior oblique, inferior recti, medial and superior)
      4th (superior oblique), and
      6th (lateral rectus).

      The long and short ciliary nerves provide the sensory supply to the globe and these are branches of the nasociliary nerve, (which is itself a branch of the ophthalmic division of the trigeminal nerve).

      To achieve anaesthesia for the eye, these nerves which enter the fibrotendinous ring need to be fully blocked to anaesthetise the eye for surgery.

    • This question is part of the following fields:

      • Anatomy
      0
      Seconds
  • Question 7 - A 60 year old non insulin dependent diabetic on metformin undergoes hip arthroscopy...

    Incorrect

    • A 60 year old non insulin dependent diabetic on metformin undergoes hip arthroscopy under general anaesthesia.

      Her preoperative blood glucose is 6.5mmol/L. Anaesthesia is induced with 200 mg propofol and 100 mcg fentanyl and maintained with sevoflurane and air/oxygen mixture. she is given 8 mg dexamethasone, 40 mg parecoxib, 1 g paracetamol and 500 mL Hartmann's solution Intraoperatively.

      The procedure took thirty minutes and her blood glucose in recovery is 14 mmol/L.

      What is the most likely cause for her rise in blood sugar?

      Your Answer:

      Correct Answer: Stress response

      Explanation:

      A significant early feature of the metabolic response to trauma and surgery is hyperglycaemia. It is due to an increased glucose production and decreased glucose utilisation bought on by neuroendocrine stimulation. Catecholamines, Growth hormone, ACTH and cortisol, and Glucagon are all increased.

      There is also a decreased insulin sensitivity peripherally and an inhibition of insulin production from the beta cells of the pancreas. These changes lead to hyperglycaemia.

      The stress response to endoscopic surgery will only be prevented with use of high dose opioids or central neuraxial block at anaesthesia.
      To reduce the risk of inducing hyperchloremic acidosis, Ringer’s lactate/acetate or Hartmann’s solution is preferred to 0.9% sodium chloride as routine maintenance fluids.

      Though it has been suggested that administration of Hartmann’s solution to patients with type 2 diabetes leads to hyperglycaemia, one Litre of Hartmann’s solution would yield a maximum of 14.5 mmol of glucose. A rapid infusion of this volume would increase the plasma glucose by no more than 1 mmol/L..

      Dexamethasone, a glucocorticoid, produces hyperglycaemia by stimulating gluconeogenesis . Glucocorticoids are agonists of intracellular glucocorticoid receptors. Their effects are mainly mediated via altered protein synthesis via gene transcription and so the onset of action is slow. The onset of action of dexamethasone is about one to four hours and therefore would NOT contribute to the hyperglycaemia in this patient in the time given.

      0.9% Normal saline with or without adrenaline is the usual irrigation fluid. With this type of surgery, systemic absorption is unlikely to occur.

      Fentanyl is not likely the primary cause of hyperglycaemia in this patient. In high doses (50 mcg/Kg) it has been shown to reduce the hyperglycaemic responses to surgery.

    • This question is part of the following fields:

      • Pathophysiology
      0
      Seconds
  • Question 8 - Which of the following correctly explains the mechanism of sevoflurane preconditioning? ...

    Incorrect

    • Which of the following correctly explains the mechanism of sevoflurane preconditioning?

      Your Answer:

      Correct Answer: Opening of mitochondrial KATP channels

      Explanation:

      Sevoflurane is highly fluorinated methyl isopropyl ether widely used as an inhalational anaesthetic. It is suggested that sevoflurane preconditioning occurs via the opening of mitochondrial Potassium ATP dependent channel similar to that of Ischemic Preconditioning protection.

    • This question is part of the following fields:

      • Pharmacology
      0
      Seconds
  • Question 9 - You've been summoned to the paediatric ward after a 4-year-old child was discovered...

    Incorrect

    • You've been summoned to the paediatric ward after a 4-year-old child was discovered 'collapsed' in bed.

      The child had been admitted the day before with febrile convulsions and was scheduled to be discharged. It is safe to approach the child.

      What should your first life-saving action be?

      Your Answer:

      Correct Answer: Apply a gentle stimulus and ask the child if they are alright

      Explanation:

      Paediatric life support differs from adult life support in that hypoxia is the primary cause of deterioration.

      After checking for danger, the child should be given a gentle stimulus (such as holding the head and shaking the arm) and asked, Are you alright? according to current advanced paediatric life support (APLS) guidelines. Safety, Stimulate, Shout is a phrase that is frequently remembered. Any airway assessment should be preceded by these actions.

      Although the algorithm includes five rescue breaths, they are performed after the airway assessment.

      It is not recommended to ask parents to leave unless they are obstructing the resuscitation. A team member should be with them at all times to explain what is going on and answer any questions they may have.

      CPR should not begin until the child has been properly assessed and rescue breaths have been administered.

    • This question is part of the following fields:

      • Pathophysiology
      0
      Seconds
  • Question 10 - Obeying Boyle's law and Charles's law is a characteristic feature of an ideal...

    Incorrect

    • Obeying Boyle's law and Charles's law is a characteristic feature of an ideal gas.

      The gas which is most ideal out of the following options is?

      Your Answer:

      Correct Answer: Helium

      Explanation:

      The ideal gas equation makes the following assumptions:

      The gas particles have a small volume in comparison to the volume occupied by the gas.
      Between the gas particles, there are no forces of interaction.
      Individual gas particle collisions, as well as gas particle collisions with container walls, are elastic, meaning momentum is conserved.
      PV = nRT
      Where:

      P = pressure
      V = volume
      n = moles of gas
      T = temperature
      R = universal gas constant

      Helium is a monoatomic gas with a small helium atom. The attractive forces between helium atoms are small because the helium atom is spherical and has no dipole moment. Because helium atoms are spherical, collisions between them approach the ideal state of elasticity.

      Most real gases behave qualitatively like ideal gases at standard temperatures and pressures. When intermolecular forces and molecular size become important, the ideal gas model tends to fail at lower temperatures or higher pressures. It also fails to work with the majority of heavy gases.

      Helium, argon, neon, and xenon are noble or inert gases that behave the most like an ideal gas. Xenon is a noble gas with a much larger atomic size than helium.

    • This question is part of the following fields:

      • Pharmacology
      0
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Statistical Methods (0/1) 0%
Physiology And Biochemistry (0/1) 0%
Passmed