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  • Question 1 - In an experimental study, a healthy subject was given one litre of 5%...

    Incorrect

    • In an experimental study, a healthy subject was given one litre of 5% dextrose within a 15-minute period. Which of the following mechanisms is expected to affect the urine output?

      Your Answer: Osmotic diuresis

      Correct Answer: Inhibition of arginine vasopressin (AVP) secretion

      Explanation:

      Changes in the osmolality of body fluids (changes as minor as 1% are sufficient) play the most important role in regulating AVP secretion. The receptors that monitor changes in osmolality of body fluids (termed osmoreceptors) are distinct from the cells that synthesize and secrete AVP, and are located in the organum vasculosum of the lamina terminalis (OVLT) of the hypothalamus. The osmoreceptors sense changes in body osmolality by either shrinking or swelling. When the effective osmolality of the plasma increases, the osmoreceptors send signals to the AVP synthesizing/secreting cells located in the supraoptic and paraventricular nuclei of the hypothalamus, and AVP synthesis and secretion are stimulated. Conversely, when the effective osmolality of the plasma is reduced, secretion is inhibited. Because AVP is rapidly degraded in the plasma, circulating levels can be reduced to zero within minutes after secretion is inhibited.

      In this scenario, the osmolality of the plasma will decrease to an estimate of 2.5%, hence inhibition of AVP.

      Stimulation of atrial stretch receptors is incorrect because the increase in plasma volume is still below the threshold for its activation.

      Osmotic diuresis is incorrect because 5% dextrose is isotonic, hence osmotic diuresis is not probable.

      Renin is inhibited when an excess of NaCl in the tubular fluid is sensed by the macula densa.

    • This question is part of the following fields:

      • Physiology
      37.8
      Seconds
  • Question 2 - What can an outbreak of flu that has spread globally be termed as?...

    Correct

    • What can an outbreak of flu that has spread globally be termed as?

      Your Answer: Pandemic

      Explanation:

      An epidemic is declared when the increase in a give disease is above a certain level in a specific interval of time.

      An endemic is the general, usual level of a disease in a population at a particular time.

      A pandemic is an epidemic that is spread across many countries and continents.

    • This question is part of the following fields:

      • Statistical Methods
      16.7
      Seconds
  • Question 3 - A young male is operated on for an open inguinal hernia repair. During...

    Incorrect

    • A young male is operated on for an open inguinal hernia repair. During the procedure, the cord is mobilized, and the deep inguinal ring is located.

      What structure forms the lateral wall of the deep inguinal ring?

      Your Answer: External oblique aponeurosis

      Correct Answer: Transversalis fascia

      Explanation:

      The deep inguinal ring is the entrance of the inguinal canal. It is an opening in the transversalis fascia around 1 cm above the inguinal ligament. Therefore, the superolateral wall is made by the transervalis fascia.

      The inferior epigastric vessels run medially to the deep inguinal ring forming its inferomedial border.

      The inguinal canal extends obliquely from the deep inguinal ring to the superficial inguinal ring.
      An indirect inguinal hernia arises through the deep inguinal ring lateral to the inferior epigastric vessels.

    • This question is part of the following fields:

      • Anatomy
      36.3
      Seconds
  • Question 4 - Which structure does NOT lie in the posterior triangle of the neck? ...

    Incorrect

    • Which structure does NOT lie in the posterior triangle of the neck?

      Your Answer: Occipital lymph nodes

      Correct Answer: Internal jugular vein

      Explanation:

      The sternocleidomastoid muscle divides the neck into anterior and posterior triangles on both sides of the neck.

      The posterior triangle has the following boundaries:
      anteriorly – sternocleidomastoid muscle
      posteriorly – trapezius
      roof – investing layer of deep cervical fascia
      floor – prevertebral fascia overlying splenius capitis, levator scapulae, and the scalene muscles

      The contents of the posterior triangle are:
      1. fat
      2. lymph nodes (level V)
      3. accessory nerve
      4. cutaneous branches of the cervical plexus (A major branch of this plexus is the phrenic nerve, which arises from the anterior divisions of spinal nerves C3-C5)
      5. inferior belly of omohyoid
      6. branches of the thyrocervical trunk (transverse cervical and suprascapular arteries)
      7. third part of the subclavian artery
      8. external jugular vein

    • This question is part of the following fields:

      • Anatomy
      24.2
      Seconds
  • Question 5 - Drug A has a 1 L/kg volume of distribution and a 0.1 elimination...

    Incorrect

    • Drug A has a 1 L/kg volume of distribution and a 0.1 elimination rate constant (k).

      Drug B has a 2 L/kg volume of distribution and a 0.2 elimination rate constant (k).

      Which of the following statements best describes the pharmacokinetics of drug A in a single compartment?

      Your Answer: Drug A has the same clearance as drug B

      Correct Answer: Drug A has a lower clearance than drug B

      Explanation:

      The fall in plasma concentration of a drug with time decreases exponentially in a single compartment pharmacokinetic model (wash-out curve).

      A straight line is produced when the logarithm (ln) of a drug’s plasma concentration is plotted against time because a constant proportion of the drug is removed from the plasma per unit time. The line’s gradient or slope can be expressed mathematically as k. (the rate constant). The gradient is related to the half life (T1/2) because it can be used to predict a drug’s plasma concentration at any time.

      According to the following formula, clearance (CL), volume of distribution (Vd), and elimination rate constant (k) are mathematically related.

      CL = Vd x k

      For drug A, CL = 1 x 0.1 = 0.1units per minute

      For drug B, Cl = 2 x 0.2 = 0.4 units per minute

      Hence, it is proved that Drug A has a lower clearance than drug B.

    • This question is part of the following fields:

      • Pharmacology
      72.8
      Seconds
  • Question 6 - 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: 800 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
      70.6
      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: Abandon the trial if more than 5% of patients drop out

      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
      45.1
      Seconds
  • Question 8 - All of the following statements are true regarding calcium except: ...

    Incorrect

    • All of the following statements are true regarding calcium except:

      Your Answer: Trousseau's sign is when inflation of tourniquet induces carpopedal spasm

      Correct Answer: Serum calcium accounts for 10% of total body calcium stores

      Explanation:

      Calcium is a very important ion and is involved in:
      -cell homeostasis
      -coagulation
      -muscle contraction
      -neuronal impulse transmission/membrane stabilization
      -bone formation and skeletal strength
      -secretion processes

      99% is found in bone and 1% in the plasma. Of the 1% that is in the plasma
      -45% is free ionized calcium
      -45% is bound to proteins, mainly Albumin
      -10% is present as an anion complex

      Reduced levels of IONIZED calcium give rise to features of hypocalcaemia , resulting in increased excitability of membranes. This results when the total calcium concentration goes below 2 mmol/L.

      Features of mild to moderate hypocalcaemia are:
      -paraesthesia (peri-oral, fingers)
      -tetany
      -spasm
      -muscle cramps
      -ECG changes (prolonged QT)
      -Trousseau’s sign (inflation of tourniquet induces carpopedal spasm)
      -Chvostek’s sign (tapping the facial nerve – cranial nerve VII – causes facial muscle twitch/spasm)

      Features of severe hypocalcaemia are:
      -cardiogenic shock and congestive cardiac failure due to reduced myocardial contractility
      respiratory distress due to bronchospasm, agitation, confusion, seizures

      Features of hypercalcaemia (remember ‘bones, stones, groans and psychic moans’):
      -Abdominal pain
      -Vomiting
      -Constipation
      -Polyuria
      -Polydipsia
      -Depression
      -Lethargy
      -Anorexia
      -Weight loss
      -Hypertension
      -Confusion
      -Pyrexia
      -Calcification in the cornea
      -Renal stones
      -Renal failure
      -Decreased Q-T interval
      -Cardiac shock/collapse

    • This question is part of the following fields:

      • Physiology And Biochemistry
      40.2
      Seconds
  • Question 9 - An acidic drug with a pKA of 4.3 is injected intravenously into a...

    Correct

    • An acidic drug with a pKA of 4.3 is injected intravenously into a patient.

      At a normal physiological pH, the approximate ratio of ionised to unionised forms of this drug in the plasma is?

      Your Answer: 1000:01:00

      Explanation:

      The pH at which the drug exists in 50 percent ionised and 50 percent unionised forms is known as the pKa.

      To calculate the proportion of ionised to unionised form of an ACID, use the Henderson-Hasselbalch equation.

      pH = pKa + log ([A-]/[HA])

      or

      pH = pKa + log [(salt)/(acid)]
      pH = pKa + log ([ionised]/[unionised]).

      Hence, if the pKa − pH = 0, then 50% of drug is ionised and 50% is unionised.

      In this example:

      7.4 = 4.3 + log ([ionised]/[unionised])
      7.4 − 4.3 = log ([ionised]/[unionised])
      log 3.1 = log ([ionised]/[unionised])

      Simply put, the antilog is the inverse log calculation. In other words, if you know the logarithm of a number, you can use the antilog to find the value of the number. The antilogarithm’s definition is as follows:

      y = antilog x = 10x

      Antilog to the base 10 of 0 = 1, 1 = 10, 2 =100, 3 = 1000, and 4 = 10,000.

      If you want to find the antilogarithm of 3.1, for a number between 3 and 4, the antilogarithm will return a value between 1000 and 10,000. The ratio is 1:1 if pKa = pH, that is, pH pKa = log 0. (50 percent ionised and unionised).

      According to the above value, there is only one unionised molecule for every approximately 1000 (1259) ionised molecules of this drug in plasma, implying that this drug is largely ionised in plasma (99.99 percent ).

    • This question is part of the following fields:

      • Pharmacology
      33.8
      Seconds
  • Question 10 - Buffers are solutions that resist a change in pH when protons are produced...

    Correct

    • Buffers are solutions that resist a change in pH when protons are produced or consumed. They consist of weak acids and their conjugate bases. Buffers are also present in our bodies, and they are known as physiologic buffers.

      Which of these is the most effective buffer in the blood?

      Your Answer: Bicarbonate

      Explanation:

      The first line of defence against acid-base disorder is buffering. The blood mainly utilizes bicarbonate ion (HCO3-) for its buffering capacity (total of 53%, plasma and red blood cells combined).

      Strong acids, when acted upon by a buffer, release H+, which then combines to HCO3- and forms carbonic acid (H2CO3). When acted upon by the enzyme carbonic anhydrase, H2CO3 dissociates into H2O and CO.

      The rest are the percentage of utilization for the following buffers:
      Haemoglobin (by RBCs) – 35%
      Plasma proteins (by plasma) – 7%
      Organic phosphates (by RBCs) – 3%
      Inorganic phosphates (by plasma) – 2%

    • This question is part of the following fields:

      • Pharmacology
      29.7
      Seconds
  • Question 11 - Using a negative feedback loop, Haem production is controlled by which of these...

    Incorrect

    • Using a negative feedback loop, Haem production is controlled by which of these enzymes?

      Your Answer: Uroporphyrinogen decarboxylase

      Correct Answer: ALA synthetase

      Explanation:

      Heme a exists in cytochrome a and heme c in cytochrome c; they are both involved in the process of oxidative phosphorylation. 5′-Aminolevulinic acid synthase (ALA-S) is the regulated enzyme for heme synthesis in the liver and erythroid cells.

      There are two forms of ALA Synthase, ALAS1, and ALAS2.

    • This question is part of the following fields:

      • Physiology
      45.2
      Seconds
  • Question 12 - Which of the following would most likely explain a failed post-operative analgesia via...

    Correct

    • Which of the following would most likely explain a failed post-operative analgesia via local anaesthesia of a neck abscess?

      Your Answer: pKA

      Explanation:

      For the local anaesthetic base to be stable in solution, it is formulated as a hydrochloride salt. As such, the molecules exist in a quaternary, water-soluble state at the time of injection. However, this form will not penetrate the neuron. The time for onset of local anaesthesia is therefore predicated on the proportion of molecules that convert to the tertiary, lipid-soluble structure when exposed to physiologic pH (7.4).

      The ionization constant (pKa) for the anaesthetic predicts the proportion of molecules that exists in each of these states. By definition, the pKa of a molecule represents the pH at which 50% of the molecules exist in the lipid-soluble tertiary form and 50% in the quaternary, water-soluble form. The pKa of all local anaesthetics is >7.4 (physiologic pH), and therefore a greater proportion the molecules exists in the quaternary, water-soluble form when injected into tissue having normal pH of 7.4.

      Furthermore, the acidic environment associated with inflamed tissues favours the quaternary, water-soluble configuration even further. Presumably, this accounts for difficulty when attempting to anesthetize inflamed or infected tissues; fewer molecules exist as tertiary lipid-soluble forms that can penetrate nerves.

    • This question is part of the following fields:

      • Physiology
      35.8
      Seconds
  • Question 13 - A double blind placebo control clinical trial is done. Which of these is...

    Incorrect

    • A double blind placebo control clinical trial is done. Which of these is correct about it?

      Your Answer: Some of the patients know they are not treated

      Correct Answer: The clinician assessing the effects of the treatment does not know which treatment the patient has been given

      Explanation:

      A ‘double blind crossover study’ happens when every patient receive both treatments.

      It is incorrect to say that only half of the patients do not know which treatment they receive because in a double blind placebo control clinical trial ALL of the patients are blind to their treatment choice .

      If some of the patients are not treated, they would be aware that they were not being treated and it could not be considered a blind trial.

      In a double blind placebo control clinical trial both the clinician and the patient are blind to the treatment choice. The clinician assessing the effects of the treatment, therefore, does not know which treatment the patient has been given.

    • This question is part of the following fields:

      • Statistical Methods
      44
      Seconds
  • Question 14 - A 45-year-old woman complains of pain in her upper abdomen to her physician....

    Incorrect

    • A 45-year-old woman complains of pain in her upper abdomen to her physician. The pain comes intermittently in waves and gets worse after eating food. There are no associated complaints of fever or bowel problems.

      The pain intensity is 6/10, and paracetamol relieves it a little. There is suspicion that part of the biliary tree is blocked.

      Which area of the duodenum does this blocked tube open into?

      Your Answer: Duodenojejunal flexure

      Correct Answer: 2nd part of the duodenum

      Explanation:

      The patient is likely suffering from biliary colic since her pain is intermittent and comes and goes in waves. Biliary colic pain gets worse after eating, especially fatty food as bile helps digest fats. Gallstones are the most common cause of biliary colic and are usually located in the cystic duct or common bile duct. But since this patient has no signs of jaundice or steatorrhea, the duct most likely blocked is the cystic duct.

      The cystic duct drains the gallbladder and combines with the common hepatic duct to form the common bile duct. The common bile duct then merges with the pancreatic duct and opens into the second part of the duodenum (major duodenal papilla).

      The duodenojejunal flexure is attached to the diaphragm by the ligament of Treitz and is not associated with any common pathology.
      The fourth part of the duodenum passes very close to the abdominal aorta and can be compressed by an abdominal aortic aneurysm.
      The third part of the duodenum can be affected by superior mesenteric artery syndrome, where the duodenum is compressed between the SMA and the aorta, often in cases of reduced body fat.
      The first part of the duodenum is the most common location for peptic ulcers affecting this organ.

    • This question is part of the following fields:

      • Anatomy
      60.2
      Seconds
  • Question 15 - Which of the following statements is TRUE regarding an epidural set? ...

    Incorrect

    • Which of the following statements is TRUE regarding an epidural set?

      Your Answer: The distal end of the catheter has an open rounded tip and two or more side ports

      Correct Answer: 19G Tuohy needles have 0.5 cm markings

      Explanation:

      A paediatric 19G Tuohy catheter is available that is 5cm in length and has 0.5cm markings

      18G Tuohy catheters are generally 9 to 10cm to hub

      Distal end of catheter is angled (15 to 30 degrees) and closed to avoid puncturing the dura

      Epidural mesh are usually 0.2 microns and are used to filter bacteria and viruses to ensure sterility of procedure

      Transparent catheters are 90cm long with diameters depending on gauge size. It has 1cm graduations from 5 to 20cm to ensure they have been inserted amply and removed completely. Distal end is smooth which can be open or closed (with lateral openings)

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      31.3
      Seconds
  • Question 16 - A 28-year-old woman who is 36 weeks pregnant presented with significant proteinuria and...

    Correct

    • A 28-year-old woman who is 36 weeks pregnant presented with significant proteinuria and severe headache. On examination, the blood pressure recorded was consistently raised at 190/110. Subsequently, she was admitted to the labour ward as a case of pre-eclampsia.

      A loading dose of IV magnesium sulphate 4g is given, followed by a 1g/hour infusion.

      The most suitable parameter to access magnesium toxicity is?

      Your Answer: Tendon reflexes

      Explanation:

      Magnesium is a very important cation due to its various physiological roles in the body. This includes:
      – playing the role of a cofactor in many enzymatic reactions
      – influencing hormone receptor binding
      – affecting calcium channels
      – impact on cardiac, vascular and neural cells

      Magnesium sulphate is used as first line in the treatment of eclampsia. Moreover, it has some preventive role in patients with severe pre-eclampsia. All the clinical effects of magnesium are in line with its plasma concentration.

      The first sign of magnesium toxicity in obstetric patients is the loss of patellar reflex, which is regularly monitored during treatment. The other options are all late signs of magnesium toxicity.

      Whenever there is a doubt, serum magnesium levels should always be monitored.

      The table below correlates the effects of increased levels of magnesium on the body:

      Plasma Concentration
      (mmol/L) Effect
      0.7-1.2 Normal
      4-8 Decreased deep tendon reflexes, nausea, headache, weakness, malaise, lethargy and facial flushing
      5-10 ECG changes (prolonged PR, prolonged QT, and widened QRS)
      10 Muscle weakness, loss of deep tendon reflexes, hypotension
      15 SA/AV nodal block, respiratory paralysis and depression
      20 Cardiac arrest

    • This question is part of the following fields:

      • Pathophysiology
      43
      Seconds
  • Question 17 - Which of the following may indicate an inadequate reversal of non-depolarising neuromuscular blockade?...

    Correct

    • Which of the following may indicate an inadequate reversal of non-depolarising neuromuscular blockade?

      Your Answer: Post tetanic count of 5

      Explanation:

      A post-tetanic count of 5 denotes a deep neuromuscular blockade.

      Post tetanic count (PTC) is a well-established method of evaluating neuromuscular recovery during intense neuromuscular blockade. It cam ne used when there is no response to single twitch, tetanic, or train-of-four (TOF) stimulation to assess the intensity of neuromuscular blockade and to estimate the duration after which the first twitch in the TOF (T1) is likely to reappear.

      During a nondepolarizing block, the high frequency of tetanic stimulation will induce a transient increase in the amount of acetylcholine released from the presynaptic nerve ending, such that the intensity of subsequent muscle contractions will be increased (potentiated) briefly (period of post-tetanic potentiation, which may last 2 to 5 min. The neuromuscular response to stimulation during post tetanic potentiation can be used to gauge the depth of block when TOF stimulation otherwise evokes no responses. The number of post tetanic responses is inversely proportional to the depth of block: fewer post tetanic contractions denote a deeper block. When the post tetanic count (PTC) is 6 to 8, recovery to TOF count = 1 is likely imminent from an intermediate-duration blocking agent; when the PTC is 0, the depth of block is profound, and no additional NMBA should be administered.

    • This question is part of the following fields:

      • Pathophysiology
      34.1
      Seconds
  • Question 18 - Which of the statements below best describe the total cerebral flow (CBF) in...

    Incorrect

    • Which of the statements below best describe the total cerebral flow (CBF) in an adult?

      Your Answer: Below a mean arterial pressure of 60 mmHg the cerebral blood flow is proportional to the systolic blood pressure

      Correct Answer: Accounts for 15% of the cardiac output

      Explanation:

      While the brain only weighs 3% of the body weight, 15% of the cardiac output goes towards the brain.

      Between mean arterial pressures (MAP) of 60-130 mmHg, autoregulation of cerebral blood flow (CBF) occurs. Exceeding this, the CBF is maintained at a constant level. This is controlled mainly by the PaCO2 level, and the autonomic nervous system has minimal role.

      Beyond these limits, the CBF is directly proportional to the MAP, not the systolic blood pressure.

    • This question is part of the following fields:

      • Physiology
      39.3
      Seconds
  • Question 19 - A 28-year-old girl, Sam, has suffered a road traffic accident. She is brought...

    Correct

    • A 28-year-old girl, Sam, has suffered a road traffic accident. She is brought to the emergency department and undergoes investigations and treatment.

      On X-ray of the humerus, she has a mid-shaft fracture.

      What structure is at the highest risk of damage with a mid-shaft humeral fracture?

      Your Answer: Radial nerve

      Explanation:

      Mid-shaft fractures of the humerus usually occur after a direct blow to the upper arm, which can occur after a fall or RTAs.
      The most important clinical significance of a mid-shaft humeral fracture is an injury to the radial nerve. The radial nerve originates from the brachial plexus and has roots of C5-T1. It crosses the spiral groove on the posterior side of the shaft of the humerus.
      On examination, the patient may have a wrist drop, loss or weakness of finger extension, and decreased or absent sensation to the posterior forearm, digits 1 to 3, and the radial half of the fourth digit.

      The humeral shaft has two compartments:
      1. Anterior:
      Brachial artery and vein
      Biceps brachii, brachialis, coracobrachialis
      Musculocutaneous, median, and ulnar nerves
      2. Posterior:
      Radial nerve
      Triceps

      Other significant nerve injuries are:
      1. Axillary nerve – surgical neck fracture of the humerus
      2. Brachial Artery – supracondylar fracture of the humerus
      3. Axillary artery – surgical neck fracture of the humerus, but is relatively uncommon

    • This question is part of the following fields:

      • Anatomy
      36.4
      Seconds
  • Question 20 - A new drug treatment has been developed for Crohn's disease. The pharmaceutical company...

    Incorrect

    • A new drug treatment has been developed for Crohn's disease. The pharmaceutical company behind this, is planning to conduct a trial and is looking for hiring around 200 individuals that are suffering from Crohn's disease. The aim would be to determine if there is any decline in the disease activity in response to the drug and compare it with a placebo.

      What phase is the trial in?

      Your Answer: Phase 0

      Correct Answer: Phase 2

      Explanation:

      The study is being conducted on a smaller level with only 200 participants and is determining the effectiveness of the drug in comparison to a placebo. These characteristics are in accordance with the second phase of trial.

    • This question is part of the following fields:

      • Statistical Methods
      31.1
      Seconds
  • Question 21 - An 85-year old female is being investigated and treated for pancytopenia of unknown...

    Incorrect

    • An 85-year old female is being investigated and treated for pancytopenia of unknown origin. Her most recent blood test is shown below which shows that he has a low platelet count.

      Hb-102 g/l
      WBC - 2.9* 109/l
      Platelets - 7 * 109/l

      Which of the following normally stimulates platelet production?


      Your Answer: Granulocyte-colony stimulating factor

      Correct Answer: Thrombopoietin

      Explanation:

      Interleukin-4 is a cytokine which acts to regulate the responses of B and T cells.

      Erythropoietin is responsible for the signal that initiated red blood cell production.

      Granulocyte-colony stimulating factor stimulates the bone marrow to produce granulocytes.

      Interleukin-5 is a cytokine that stimulates the proliferation and activation of eosinophils.

      Thrombopoietin is the primary signal responsible for megakaryocyte and thus platelet production.
      Platelets are also called thrombocytes. They, like red blood cells, are also derived from myeloid stem cells. The process involves a megakaryocyte developing from a common myeloid progenitor cell. A megakaryocyte is a large cell with a multilobulated nucleus, this grows to become massive where it will then break up to form platelets.

      Immune cells are generated from haematopoietic stem cells in bone marrow. They generate two main types of progenitors, myeloid and lymphoid progenitor cells, from which all immune cells are derived.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      32.7
      Seconds
  • Question 22 - A 47-year-old woman presented with chief complaints of dysuria, urinary frequency, and malaise....

    Correct

    • A 47-year-old woman presented with chief complaints of dysuria, urinary frequency, and malaise. On urine culture and sensitivity testing, E.coli was detected with resistance to ampicillin.
      What is the mechanism of resistance to ampicillin?

      Your Answer: Beta-lactamase production

      Explanation:

      Ampicillin belongs to the family of penicillin. Resistance to this group of drugs is due to ?-lactamase production which opens the ?-lactam ring and inactivates Penicillin G and some closely related congeners. The majority of Staphylococci and some strains of gonococci, B. subtilis, E. coli, and a few other bacteria produce penicillinase.

      Resistance to cephalosporins is due to changes in penicillin-binding proteins.

      Resistance to macrolides are due to post-transcriptional methylation of 23s bacterial ribosomal RNA

      Resistance to fluoroquinolones is due to mutations in DNA gyrase.

    • This question is part of the following fields:

      • Pharmacology
      32.6
      Seconds
  • Question 23 - A 60-year old male has anaemia and is being investigated. The most common...

    Incorrect

    • A 60-year old male has anaemia and is being investigated. The most common combination of globin chains in a normal adult is:

      Your Answer: δ2β2

      Correct Answer: α2β2

      Explanation:

      There are 4 different types of globin chains which surround 4 heme molecules in haemoglobin (Hb) – α (alpha), β (beta), γ (gamma), and δ (delta)
      α chains are essential.
      δ2β2 and β2γ2 are not found in a healthy adult.
      97% of the Hb in a healthy adult is made of α2β2 (2 α chains and 2 β chains).
      α2δ2 accounts for around 1.5-3% of the adult Hb.
      α2γ2 accounts for less than 1%.

      With respect to oxygen transport in cells, almost all oxygen is transported within erythrocytes. There is limited solubility and only 1% is carried as solution. Thus, the amount of oxygen transported depends upon haemoglobin concentration and its degree of saturation.

      Haemoglobin is a globular protein composed of 4 subunits. Haem is made up of a protoporphyrin ring surrounding an iron atom in its ferrous state. The iron can form two additional bonds – one is with oxygen and the other with a polypeptide chain. There are two alpha and two beta subunits to this polypeptide chain in an adult and together these form globin. Globin cannot bind oxygen but can bind to CO2 and hydrogen ions. The beta chains are able to bind to 2,3 diphosphoglycerate. The oxygenation of haemoglobin is a reversible reaction. The molecular shape of haemoglobin is such that binding of one oxygen molecule facilitates the binding of subsequent molecules.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      26
      Seconds
  • Question 24 - The diaphragm is a muscle that is relatively resistant to non-depolarizing neuromuscular blockade's...

    Correct

    • The diaphragm is a muscle that is relatively resistant to non-depolarizing neuromuscular blockade's effects.

      When these muscle relaxants are used, which of the following peripheral nerve stimulator twitch patterns is best for monitoring the return of diaphragmatic function?

      Your Answer: Post-tetanic count stimulation

      Explanation:

      Certain skeletal muscles are more resistant to the effects of neuromuscular blocking agents, both non-depolarizing and depolarizing. The diaphragm is the most resistant. The muscles of the larynx and the corrugator supercilii are less resistant. The abdominal, orbicularis oris, and limb peripheral muscles are the most sensitive muscles.

      Twitch stimulation patterns:

      Supramaximal single stimulus:

      The frequency ranges from 1 Hz to 0.1 Hz (one every second to one every 10 seconds)
      The response is proportional to the frequency of the event.
      It has limited clinical utility because it only tells you whether or not a patient is paralysed (no information on degree of paralysis).

      Over the course of 0.5 seconds (2 Hz), four supramaximal stimulate were applied:

      It is possible to see ‘fade’ and use it as a basis for evaluation.
      This stimulation pattern is used to determine the degree of blockade (1-2 twitches is appropriate for abdominal surgery)
      If the train of four (TOF) count is 1-2, reversal agents can be used in conjunction with medium-acting neuromuscular blocking agents.

      Ratio of TOF:

      This is the ratio of the 4th twitch amplitude to the 1st twitch amplitude.
      The ratio decreases with non-depolarising block and is inversely proportional to the degree of block, allowing objective measurement of residual neuromuscular blockade.
      To achieve adequate reversal, the ratio (as measured by accelerography) must be between 0.7 and 0.9.

      Count of twitches after a tetanic experience(PTC):

      50 Hz for 5 seconds, then a 3 second pause, followed by a single 1 Hz twitch stimulus.
      When the TOF count is zero, this stimulation pattern is used to assess deep blockade (that is, in neurosurgery, microsurgery or ophthalmic surgery when even small movements of a patient will disturb the surgical field)
      It gives an estimate of how long it will take for the response to return to single twitches, allowing assessment of blocks that are too deep for any other technique.
      A palpable post-tetanic count (PTC) of 2 indicates no twitch response for about 20-30 minutes, and a PTC of 5 indicates no twitch response for about 10-15 minutes.

      This is without a doubt the best way to keep track of paralysis in patients who need to avoid diaphragmatic movement. It’s best to use drug infusions and aim for a PTC of 2. After a tetanic stimulus, acetylcholine is mobilised, causing post-tetanic potentiation.

      Stimulation in Two Bursts:

      750 milliseconds between two short bursts of 50 Hz
      This stimulation pattern is used to assess small amounts of residual blockade manually (tactile).

    • This question is part of the following fields:

      • Clinical Measurement
      34.9
      Seconds
  • Question 25 - A 50-year-old woman is brought into the emergency department in an ambulance. She...

    Incorrect

    • A 50-year-old woman is brought into the emergency department in an ambulance. She was found collapsed on the street. She has visual and oculomotor deficits on examination, but her motor function is intact.

      A digital subtraction angiography is performed that shows occlusion of the basilar artery at the site where the vertebral arteries fuse to form the basilar artery.

      Which anatomical landmark corresponds to this site of occlusion?

      Your Answer: The base of the midbrain

      Correct Answer: The base of the pons

      Explanation:

      The basilar artery is a large vessel that is formed by the union of the vertebral arteries at the junction of the medulla and pons. It lies in the pontine cistern and follows a shallow groove on the ventral pontine surface, extending to the upper border of the pons.

      The basilar artery then bifurcates into the two posterior cerebral arteries that form part of the Circle of Willis.

    • This question is part of the following fields:

      • Anatomy
      44.9
      Seconds
  • Question 26 - Which of the following explains the mode of action of Magnesium sulphate in...

    Correct

    • Which of the following explains the mode of action of Magnesium sulphate in preventing eclampsia in susceptible patients?

      Your Answer: Dilatation of cerebral circulation due to calcium channel antagonism reducing cerebral vascular spasm

      Explanation:

      Magnesium is a unique calcium antagonist as it can act on most types of calcium channels in vascular smooth muscle and as such would be expected to decrease intracellular calcium. One major effect of decreased intracellular calcium would be inactivation of calmodulin-dependent myosin light chain kinase activity and decreased contraction, causing arterial relaxation that may subsequently lower peripheral and cerebral vascular resistance, relieve vasospasm, and decrease arterial blood pressure.

      The vasodilatory effect of MgSO4 has been investigated in a wide variety of vessels. For example, both in vivo and in vitro animal studies have shown that it is a vasodilator of large conduit arteries such as the aorta, as well as smaller resistance vessels including mesenteric, skeletal muscle, uterine, and cerebral arteries.

      The theory of cerebrovascular vasospasm as the aetiology of eclampsia seemed to be reinforced by transcranial Doppler (TCD) studies which suggested that MgSO4 treatment caused dilation in the cerebral circulation as well as in animal studies that used large cerebral arteries.

    • This question is part of the following fields:

      • Pathophysiology
      42.6
      Seconds
  • Question 27 - A 27-year old man loses 1000ml blood after being stabbed on his thigh....

    Correct

    • A 27-year old man loses 1000ml blood after being stabbed on his thigh.

      The most impactful physiological response occurring at the start to combat the decline in the intravascular blood volume of this man is?

      Your Answer: Venoconstriction

      Explanation:

      In contrast to the arterial system, which contains 15% of the circulating blood volume, the body’s veins contain 70% of it.

      In severe haemorrhage, when sympathetic stimulation causes venoconstriction, venous tone is important in maintaining the return of blood to the heart.

      Because the liver receives about 30% of the resting cardiac output, it is a very vascular organ. The hepatic vascular system is dynamic, which means it can store and release blood in large amounts – it acts as a reservoir within the general circulation.

      In a normal situation, the liver contains 10-15% of total blood volume, with the sinusoids accounting for roughly 60% of that. The liver dynamically adjusts its blood volume when blood is lost and can eject enough blood to compensate for a moderate amount of haemorrhage.

      In the portal venous and hepatic arterial systems, sympathetic nerves constrict the presinusoidal resistance vessels. More importantly, sympathetic stimulation lowers the portal system’s capacitance, allowing blood to flow more efficiently to the heart.

      Net transcapillary absorption of interstitial fluid from skeletal muscle into the intravascular space compensates for blood loss effectively during haemorrhage. The decrease in capillary hydrostatic pressure (Pc), caused by reflex adrenergic readjustment of the ratio of pre- to postcapillary resistance, is primarily responsible for fluid absorption. Within a few hours of blood loss, these fluid shifts become significant, further diluting haemoglobin and plasma proteins.

      Albumin synthesis begins to increase after 48 hours.

      The juxtamedullary complex releases renin in response to a drop in mean arterial pressure, which causes an increase in aldosterone level and, eventually, sodium and water resorption. Increased antidiuretic hormone (ADH) levels also contribute to water retention.

    • This question is part of the following fields:

      • Physiology
      36.1
      Seconds
  • Question 28 - A 4-year-old boy with status epilepticus was brought to ER and has already...

    Incorrect

    • A 4-year-old boy with status epilepticus was brought to ER and has already received two doses of intravenous lorazepam but is still continuing to have seizures.

      Which of the following drug would be best for his treatment?

      Your Answer: Thiopentone 4 mg/kg IV

      Correct Answer: Phenytoin 20 mg/kg IV

      Explanation:

      When the convulsion lasts for five or more than five minutes, or if there are recurrent episodes of convulsions in a 5 minute period without returning to the baseline, it is termed as Status Epilepticus.
      The first priority in the patient with seizures is maintaining the airway, breathing, and circulation.

      Guideline for the management of Status Epilepticus in children by Advanced Life Support Group is as follow:

      Step 1 (Five minutes after the start of seizures):

      If intravascular access is available start treatment with lorazepam 0.1 mg/kg IV
      If no intravascular access then give buccal midazolam 0.5 mg/kg or rectal diazepam 0.5 mg/kg.

      Step 2 (Ten minutes after the start of seizure):

      If the convulsions continue then a second dose of benzodiazepine should be given. Senior should be called on-site and phenytoin should be prepared.
      No more than two doses or benzodiazepines should be given (including any doses given before arrival at the hospital)
      If still no IV access then obtain intraosseous access (IO).

      Step 3 (Ten minutes after step 2)

      Senior help along with anaesthetic/ICU help should be sought
      Phenytoin 20 mg/kg IV over 20 minutes
      If the seizure stops before the full dose of phenytoin is given then the infusion should be completed as this provides up to 24 hours of anticonvulsant effect
      In children already receiving phenytoin as treatment for epilepsy then an alternative is phenobarbitone 20 mg/kg IV over five minutes
      Once the phenytoin is started, senior staff may wish to give rectal paraldehyde 0.4 mg/kg although this is no longer included in the routine algorithm recommended by APLS.

      Step 4 (20 minutes after step 3)

      If 20 minutes after starting phenytoin the child remains in status epilepticus then rapid sequence induction of anaesthesia with thiopentone and a short acting paralysing agent is needed and the child transferred to paediatric intensive care.

    • This question is part of the following fields:

      • Pathophysiology
      20.6
      Seconds
  • Question 29 - All of the following statements are false regarding salmeterol except: ...

    Incorrect

    • All of the following statements are false regarding salmeterol except:

      Your Answer: Has similar potency to salbutamol at the beta-1 receptor

      Correct Answer: Is more potent than salbutamol at the beta-2 receptor

      Explanation:

      Salmeterol is a long-acting Beta 2 selective agonist. Therefore it is only used for prophylaxis whereas salbutamol is a short-acting Beta 2 agonist and is thus used for the treatment of acute attacks of asthma.

      Salmeterol is 15 times more potent than salbutamol at the Beta 2 receptor but 4 times less potent at the Beta 1 receptor.

      Tachyphylaxis to the unwanted side effects commonly occurs, but not to bronchodilation.

    • This question is part of the following fields:

      • Pharmacology
      25.9
      Seconds
  • Question 30 - You've been summoned to help resuscitate a 6-year-old child who has suffered a...

    Incorrect

    • You've been summoned to help resuscitate a 6-year-old child who has suffered a cardiac arrest. The ECG monitor shows electrical activity that isn't pulsed.

      Which of the following statements is the most appropriate during resuscitation?

      Your Answer: The rate of chest compressions should be 120-140 per minute

      Correct Answer: The dose of intravenous adrenaline is 180 mcg

      Explanation:

      To begin, one must determine the child’s approximate weight. There are a variety of formulas to choose from. It is acceptable to use the advanced paediatric life support formula:

      (Age + 4) 2 = Weight

      A 5-year-old child will weigh around 18 kilogrammes.

      10 mcg/kg (0.1 ml/kg of 1 in 10 000 adrenaline) = 180 mcg is the appropriate dose of intravenous or intraosseous adrenaline.

      The correct energy level to deliver is 4 J/kg, which equals 72 joules.

      The pad size that is appropriate for this patient is 8-12 cm. For an infant, a 4.5 cm pad is appropriate.

      To allow adequate separation in infants and small children, the pads should be placed anteriorly and posteriorly on the chest.

      When using a bag and mask to ventilate, take two breaths for every 15 chest compressions. If chest compressions are being applied intubated and without interruption, a ventilation rate of 10-20 breaths per minute should be given.

      Chest compressions should be done at a rate of 100-120 per minute, the same as an adult.

    • This question is part of the following fields:

      • Pharmacology
      43.5
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Physiology (2/6) 33%
Statistical Methods (1/4) 25%
Anatomy (1/5) 20%
Pharmacology (3/6) 50%
Physiology And Biochemistry (0/3) 0%
Anaesthesia Related Apparatus (0/1) 0%
Pathophysiology (3/4) 75%
Clinical Measurement (1/1) 100%
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