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  • Question 1 - The immediate physiological response to massive perioperative blood loss is: ...

    Correct

    • The immediate physiological response to massive perioperative blood loss is:

      Your Answer: Stimulation of baroreceptors in carotid sinus and aortic arch

      Explanation:

      With regards to compensatory response to blood loss, the following sequence of events take place:

      1. Decrease in venous return, right atrial pressure and cardiac output
      2. Baroreceptor reflexes (carotid sinus and aortic arch) are immediately activated
      3. There is decreased afferent input to the cardiovascular centre in medulla. This inhibits parasympathetic reflexes and increases sympathetic response
      4. This results in an increased cardiac output and increased SVR by direct sympathetic stimulation. There is increased circulating catecholamines and local tissue mediators (adenosine, potassium, NO2)
      5. Fluid moves into the intravascular space as a result of decreased capillary hydrostatic pressure absorbing interstitial fluid.

      A slower response is mounted by the hypothalamus-pituitary-adrenal axis.
      6. Reduced renal blood flow is sensed by the intra renal baroreceptors and this stimulates release of renin by the juxta-glomerular apparatus.
      7. There is cleavage of circulating Angiotensinogen to Angiotensin I, which is converted to Angiotensin II in the lungs (by Angiotensin Converting Enzyme ACE)

      Angiotensin II is a powerful vasoconstrictor that sets off other endocrine pathways.
      8. The adrenal cortex releases Aldosterone
      9. There is antidiuretic hormone release from posterior pituitary (also in response to hypovolaemia being sensed by atrial stretch receptors)
      10. This leads to sodium and water retention in the distal convoluted renal tubule to conserve fluid
      Fluid conservation is also aided by an increased amount of cortisol which is secreted in response to the increase in circulating catecholamines and sympathetic stimulation.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      20.7
      Seconds
  • Question 2 - A human's resting oxygen consumption (VO2) is typically 3.5 ml/kg/minute (one metabolic equivalent...

    Incorrect

    • A human's resting oxygen consumption (VO2) is typically 3.5 ml/kg/minute (one metabolic equivalent or 1 MET). Which of the following options is linked to the highest VO2 when a person is at rest?

      Your Answer: Dobutamine infusion in a fit adult

      Correct Answer: Neonate

      Explanation:

      The oxygen consumption rate (VO2) at rest is 3.5 ml/kg/minute (one metabolic equivalent or 1 MET).
      3.86 ml/kg/minute thyrotoxicosis

      Young children consume a lot of oxygen: around 7 ml/kg/min when they are born. The metabolic cost of breathing is higher in children than in adults, and it can account for up to 15% of total oxygen consumption. Similarly, an infant’s metabolic rate is nearly twice that of an adult, resulting in a larger alveolar minute volume and a lower FRC.

      At term, oxygen consumption at rest can increase by as much as 40% (5 ml/kg/minute) and can rise to 60% during labour.

      When compared to normal basal metabolism, sepsis syndrome increases VO2 and resting metabolic rate by 30% (4.55 ml/kg/minute). In septicaemic shock, VO2 decreases.

      Dobutamine hydrochloride was infused into 12 healthy male volunteers at a rate of 2 micrograms per minute per kilogramme, gradually increasing to 4 and 6 micrograms per minute per kilogramme. Dobutamine was infused for 20 minutes for each dose. VO2 increased by 10% to 15%. (3.85-4.0 ml/kg/min).

    • This question is part of the following fields:

      • Physiology
      23.5
      Seconds
  • Question 3 - Out of the following, which artery is NOT a branch of the hepatic...

    Incorrect

    • Out of the following, which artery is NOT a branch of the hepatic artery?

      Your Answer: Gastroduodenal artery

      Correct Answer: Pancreatic artery

      Explanation:

      The common hepatic artery arises from the celiac artery and has the following branches:
      1. hepatic artery proper that branches into –
      a. cystic artery to supply the gallbladder
      b. left and right hepatic arteries to supply the liver
      2. gastroduodenal artery that branches into
      a. right gastroepiploic artery
      b. superior pancreaticoduodenal artery
      3. right gastric artery

      The pancreatic artery is a branch of the splenic artery.

    • This question is part of the following fields:

      • Anatomy
      18.7
      Seconds
  • Question 4 - Arterial pressure waveforms give an indication of the operation of the heart and...

    Correct

    • Arterial pressure waveforms give an indication of the operation of the heart and the patient's clinical state. Which of the following listed characteristics of arterial waveforms is most indicative of myocardial contractility?

      Your Answer: Slope of the upstroke of the curve.

      Explanation:

      Arterial pressure waveforms is an invasive form of monitoring cardiac parameters. It provides a lot of information on the performance of the heart from different sections, including:

      Cardiac measurements:

      Heart rate
      Systolic pressure
      Diastolic pressure
      Mean arterial pressure
      Pulse pressure
      Change in pulse amplitude corresponding to respiratory changes
      Slope of anacrotic limb associated with aortic stenosis

      From the shape of the arterial waveform displayed:

      Slope of anacrotic limb represents aortic valve and LVOT flow
      Indications of aortic stenosis (AS): Slurred wave, collapsing wave
      Rapid systolic decline in LVOTO
      Bisferiens wave in HOCM
      Low dicrotic notch in states with poor peripheral resistance
      Position and quality of dicrotic notch as a reflection of the damping coefficient

      For this question, the upstroke slope of the pressure wave is indicative of myocardial contractility and is mathematically represented as:

      dP/dt, which represents a change of pressure with regards to time.

    • This question is part of the following fields:

      • Clinical Measurement
      50.1
      Seconds
  • Question 5 - Which of the following statement is not true regarding the effects of Dopamine...

    Incorrect

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

      Your Answer: At higher rates of infusion, there is increased systemic vascular resistance and venous return

      Correct Answer: Decreasing gastric transit time

      Explanation:

      Moderately high doses of dopamine produce a positive inotropic (direct β1 and D1 action + that due to Noradrenaline release), but the little chronotropic effect on the heart.

      Vasoconstriction (α1 action) occurs only when large doses are infused.

      At doses normally employed, it raises cardiac output and systolic BP with little effect on diastolic BP. It has practically no effect on nonvascular α and β receptors; does not penetrate the blood-brain barrier€”no CNS effects.

    • This question is part of the following fields:

      • Pharmacology
      24.4
      Seconds
  • Question 6 - A previously fit 26-year-old is undergoing surgery to repair an inguinal hernia. He...

    Correct

    • A previously fit 26-year-old is undergoing surgery to repair an inguinal hernia. He is breathing on his own, and a supraglottic airway is being maintained via a circle system with air/oxygen and sevoflurane. With a fresh gas flow of 14 L/min, the end-tidal CO2 reading is 8.1 kPa. CO2 pressure is 1.9 kPa. The percentages of oxygen inhaled and exhaled are 38 and 33 percent, respectively. What do you think is the most likely source for these readings?

      Your Answer: Incompetent expiratory valve

      Explanation:

      The patient is rebreathing carbon dioxide that has been exhaled.

      Exhaustion of the soda lime and failure of the expiratory valve are the two most likely causes. A leak in the inspiratory limb is a less likely cause. Increased inhaled and exhaled carbon dioxide levels may appear with a normal-looking capnogram if the expiratory valve is ineffective.

      The patient will exhale into both the inspiratory and expiratory limbs if the inspiratory valve is inoperable. A slanted downstroke inspiratory phase (as the patient inhales carbon dioxide-containing gas from the inspiratory limb) and increased end-tidal carbon dioxide can be seen on the capnogram.

      Even if the soda lime were exhausted, a high fresh gas flow would be enough to prevent rebreathing. The difference in oxygen concentrations in inspired and expired breaths would be less pronounced.

      Hypercapnia is caused by respiratory obstruction and malignant hyperthermia, but not by rebreathing.

    • This question is part of the following fields:

      • Pathophysiology
      55.5
      Seconds
  • Question 7 - A 30-year old male has Von Willebrand's disease and attends the hospital to...

    Correct

    • A 30-year old male has Von Willebrand's disease and attends the hospital to get an infusion of desmopressin acetate. The way this works is by stimulating the release of von Willebrand factor from cells, which in turn increases factor VIII and platelet plug formation in clotting. In patients that have no clotting abnormalities, the substance that keeps the blood soluble and prevents platelet activation normally is which of these?

      Your Answer: Prostacyclin

      Explanation:

      Even though aprotinin reduces fibrinolysis and therefore bleeding, there is an associated increased risk of death. It was withdrawn in 2007.
      Protein C is dependent upon vitamin K and this may paradoxically increase the risk of thrombosis during the early phases of warfarin treatment.

      The coagulation cascade include two pathways which lead to fibrin formation:
      1. Intrinsic pathway – these components are already present in the blood
      Minor role in clotting
      Subendothelial damage e.g. collagen
      Formation of the primary complex on collagen by high-molecular-weight kininogen (HMWK), prekallikrein, and Factor 12
      Prekallikrein is converted to kallikrein and Factor 12 becomes activated
      Factor 12 activates Factor 11
      Factor 11 activates Factor 9, which with its co-factor Factor 8a form the tenase complex which activates Factor 10

      2. Extrinsic pathway – needs tissue factor that is released by damaged tissue)
      In tissue damage:
      Factor 7 binds to Tissue factor – this complex activates Factor 9
      Activated Factor 9 works with Factor 8 to activate Factor 10

      3. Common pathway
      Activated Factor 10 causes the conversion of prothrombin to thrombin and this hydrolyses fibrinogen peptide bonds to form fibrin. It also activates factor 8 to form links between fibrin molecules.

      4. Fibrinolysis
      Plasminogen is converted to plasmin to facilitate clot resorption

    • This question is part of the following fields:

      • Physiology And Biochemistry
      163.1
      Seconds
  • Question 8 - A 50-year old female came to the Obstetric and Gynaecology department for an...

    Incorrect

    • A 50-year old female came to the Obstetric and Gynaecology department for an elective hysterectomy under general anaesthesia. Upon physical examination, she was noted to be obese (BMI = 40). Regarding the optimal dose of thiopentone, which of the following parameters must be considered for the computation?

      Your Answer: Ideal body weight

      Correct Answer: Lean body weight

      Explanation:

      Using a lean body weight metric encompasses a more scientific approach to weight-based dosing. Lean body weight reflects the weight of all €˜non-fat’ body components, including muscle and vascular organs such as the liver and kidneys. As lean body weight contributes to approximately 99% of a drug’s clearance, it is useful for guiding dosing in obesity.

      This metric has undergone a number of transformations. The most commonly cited formula derived by Cheymol is not optimal for dosing across body compositions and can even produce a negative result. A new formula has been developed that appears stable across different body sizes, in particular the obese to morbidly obese.

      A practical downfall of the calculation of lean body weight (and other body size descriptors) is the numerical complexity, which may not be palatable to a busy clinician. Often limited time is available for prescribing and an immediate calculation is required. Lean body weight calculators are available online, for example in the Therapeutic Guidelines.

      Using total body weight assumes that the pharmacokinetics of the drug are linearly scalable from normal-weight patients to those who are obese. This is inaccurate. For example, we cannot assume that a 150 kg patient eliminates a drug twice as fast as a 75 kg patient and therefore double the dose. Clinicians are alert to toxicities with higher doses, for example nephro- and neurotoxicity with some antibiotics and chemotherapeutics, and bleeding with anticoagulants. Arbitrary dose reductions or €˜caps’ are used to avoid these toxicities, but if too low can result in sub-therapeutic exposure and treatment failure.

      Body surface area is traditionally used to dose chemotherapeutics. It is a function of weight and height and has been shown to correlate with cardiac output, blood volume and renal function. However, it is controversial in patients at extremes of size because it does not account for varying body compositions. As a consequence, some older drugs such as cyclophosphamide, paclitaxel and doxorubicin were €˜capped’ (commonly at 2 m^2) potentially resulting in sub-therapeutic treatment. Recent guidelines suggest that unless there is a justifiable reason to reduce the dose (e.g. renal disease), total body weight should be used in the calculation of body surface area, until further research is done. Little research into dosing based on body surface area has been conducted for other medicines.

      Ideal body weight was developed for insurance purposes not for drug dosing. It is a function of height and gender only and, like body surface area, does not take into account body composition. Using ideal body weight, all patients of the same height and sex would receive the same dose, which is inadequate and generally results in under-dosing. For example a male who has a total body weight of 150 kg and a height of 170 cm will have the same ideal body weight as a male who is 80 kg and 170 cm tall. Both could potentially receive a mg/kg dose based on 65 kg (ideal body weight).

    • This question is part of the following fields:

      • Pharmacology
      17.5
      Seconds
  • Question 9 - An older woman has been brought into the emergency department with symptoms of...

    Correct

    • An older woman has been brought into the emergency department with symptoms of a stroke. A CT angiogram is performed for diagnosis, which displays narrowing in the artery that supplies the right common carotid. Which of the following artery is the cause of stroke in this patient?

      Your Answer: Brachiocephalic artery

      Explanation:

      The arch of aorta gives rise to three main branches:
      1. Brachiocephalic artery
      2. Left common carotid artery
      3. Left subclavian artery

      The brachiocephalic artery then gives rise to the right subclavian artery and the right common carotid artery.

      The right common carotid artery arises from the brachiocephalic trunk posterior to the sternoclavicular joint.

      The coeliac trunk is a branch of the abdominal aorta.
      The ascending aorta supplies the coronary arteries.

    • This question is part of the following fields:

      • Anatomy
      30.6
      Seconds
  • Question 10 - A 50-year-old man is admitted in hospital. Over four hours, he produces 240...

    Correct

    • A 50-year-old man is admitted in hospital. Over four hours, he produces 240 mL of urine and has a plasma creatinine concentration is 10 mcg/mL. The normal concentration of creatinine in urine is 1.25 mg/mL. Calculate his approximate creatinine clearance.

      Your Answer: 125 ml/minute

      Explanation:

      Creatinine clearance is a test used to approximate the glomerular filtration rate (GFR) as an assessment of kidney function.

      Creatinine is formed during the breakdown of dietary sources of meat and skeletal muscle. It is secreted at a consistent concentration and pace into the body’s circulation, and is easily filtered across the glomerulus without being reabsorbed or metabolized by the kidney.

      It is represented mathematically as:
      Creatinine clearance (CL) = U x V/P
      where,
      U: Urinary creatinine concentration (mg/mL)
      V: Volume of urine (mL/min)
      P: Plasma creatinine concentration (mg/mL)

      Therefore, in this case:
      CL: 1.25 x 1 = 125mL/min.

    • This question is part of the following fields:

      • Clinical Measurement
      197.1
      Seconds
  • Question 11 - 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
      28.8
      Seconds
  • Question 12 - A 30-year old female athlete was brought to the Emergency Room for complaints...

    Correct

    • A 30-year old female athlete was brought to the Emergency Room for complaints of light-headedness and nausea. Clinical chemistry studies were done and the results were the following: Na: 144 mmol/L (Reference: 137-144 mmol/L), K: 6 mmol/L (Reference: 3.5-4.9 mmol/L), Cl: 115 mmol/L (Reference: 95-107 mmol/L), HCO3: 24 mmol/L (Reference: 20-28 mmol/L), BUN: 9.5 mmol/L (Reference: 2.5-7.5 mmol/L), Crea: 301 µmol/l (Reference: 60 - 110 µmol/L), Glucose: 3.5 mmol/L (Reference: 3.0-6.0 mmol/L). Taking into consideration the values above, in which of the following ranges will his osmolarity fall into?

      Your Answer: 300-313

      Explanation:

      Osmolarity refers to the osmotic pressure generated by the dissolved solute molecules in 1 L of solvent. Measurements of osmolarity are temperature dependent because the volume of the solvent varies with temperature. The higher the osmolarity of a solution, the more it attracts water from an opposite compartment.

      Osmolarity can be computed using the following formulas:

      Osmolarity = Concentration x number of dissociable particles; OR
      Plasma osmolarity (Posm) = 2([Na+]) + (glucose in mmol/L) + (BUN in mmol/L)

      Posm = 2 (144) + 3.5 + 9.5 = 301 mOsm/L

      Suppose there is electrical neutrality, the formula will double the cation activity to account for the anions.

      Plasma osmolarity (Posm) = 2([Na+] + [K+]) + (glucose in mmol/L) + (BUN in mmol/L)

      Posm = 2 (144 + 6) + 3.5 + 9.5 = 313 mOsm/L

    • This question is part of the following fields:

      • Physiology
      81
      Seconds
  • Question 13 - You're summoned to the emergency room, where a 39-year-old man has been admitted...

    Incorrect

    • You're summoned to the emergency room, where a 39-year-old man has been admitted following a cardiac arrest. He was rescued from a river, but little else is known about him. CPR is being performed on the patient, who has been intubated. He's received three DC shocks and is still in VF. A rectal temperature of 29.5°C is taken with a low-reading thermometer. Which of the following statements about his resuscitation is correct?

      Your Answer: 1 mg IV adrenaline and 300 mg IV amiodarone should be administered

      Correct Answer: No further DC shocks and no drugs should be given until his core temperature is greater than 30°C

      Explanation:

      The guidelines for the management of cardiac arrest in hypothermic patients published by the UK Resuscitation Council differ slightly from the standard algorithm.

      In a patient with a core temperature of less than 30°C, do the following:

      If you’re on the shockable side of the algorithm (VF/VT), you should give three DC shocks.
      Further shocks are not recommended until the patient has been rewarmed to a temperature of more than 30°C because the rhythm is refractory and unlikely to change.
      There should be no drugs given because they will be ineffective.

      In a patient with a core temperature of 30°C to 35°C, do the following:

      DC shocks are used as usual.
      Because they are metabolised much more slowly, the time between drug doses should be doubled.

      Active rewarming and protection against hyperthermia should be given to the patient.

      Option e is false because there is insufficient information to determine whether resuscitation should be stopped.

    • This question is part of the following fields:

      • Physiology
      777.5
      Seconds
  • Question 14 - Seven days after undergoing an inguinal hernia repair, a 24-year old male presents...

    Incorrect

    • Seven days after undergoing an inguinal hernia repair, a 24-year old male presents with a wound that is erythematous, tender and has a purulent discharge. The most likely cause of this is which of the following?

      Your Answer: Infection with Streptococcus pyogenes

      Correct Answer: Infection with Staphylococcus aureus

      Explanation:

      Staphylococcus aureus infection is the most likely cause.

      Surgical site infections (SSI) occur when there is a breach in tissue surfaces and allow normal commensals and other pathogens to initiate infection. They are a major cause of morbidity and mortality.

      SSI comprise up to 20% of healthcare associated infections and approximately 5% of patients undergoing surgery will develop an SSI as a result.
      The organisms are usually derived from the patient’s own body.

      Measures that may increase the risk of SSI include:
      -Shaving the wound using a single use electrical razor with a disposable head
      -Using a non iodine impregnated surgical drape if one is needed
      -Tissue hypoxia
      -Delayed prophylactic antibiotics administration in tourniquet surgery, patients with a prosthesis or valve, in clean-contaminated surgery of in contaminated surgery.

      Measures that may decrease the risk of SSI include:
      1. Intraoperatively
      – Prepare the skin with alcoholic chlorhexidine (Lowest incidence of SSI)
      -Cover surgical site with dressing

      In contrast to previous individual RCT’s, a recent meta analysis has confirmed that administration of supplementary oxygen does not reduce the risk of wound infection and wound edge protectors do not appear to confer benefit.

      2. Post operatively
      Tissue viability advice for management of surgical wounds healing by secondary intention

      Use of diathermy for skin incisions
      In the NICE guidelines the use of diathermy for skin incisions is not advocated. Several randomised controlled trials have been undertaken and demonstrated no increase in risk of SSI when diathermy is used.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      33.1
      Seconds
  • Question 15 - What can you see within the tunica media of a blood vessel on...

    Correct

    • What can you see within the tunica media of a blood vessel on examination?

      Your Answer: Smooth muscle

      Explanation:

      The blood vessel well is divided into 3 parts, namely:

      The tunica intima, which is the deepest layer. It contains endothelial cells separated by gap junctions

      The tunica media, primarily consisting of the involuntary smooth muscle fibres, laid out in spiral layers with elastic fibres and connective tissue.

      The tunica adventitia, which is the most superficial layer. It consists of the vasa vasorum, fibroblast and collagen.

    • This question is part of the following fields:

      • Anatomy
      15.6
      Seconds
  • Question 16 - A 20-year old male was involved in an accident and has presented to...

    Correct

    • A 20-year old male was involved in an accident and has presented to the Emergency Department with a pelvic crush injury. The clinical exam according to ATLS protocol revealed the following: Airway-patent, Breathing - respiratory rate 25 breaths per minute. Breath sounds are vesicular and there are no added sounds. Circulation - Capillary refill time - 4 seconds. Peripheries are cool. Pulse 125 beats/min. BP - 125/95 mmHg. Disability - GSC 15, anxious and in pain. Secondary survey reveals no other injuries. The patient is administered high flow oxygen and IV access is established. The most appropriate IV fluid regimen in this case will be which of the following?

      Your Answer: Judicious infusion of Hartmann's solution to maintain a systolic blood pressure greater than 90mmHg

      Explanation:

      These clinical signs suggest that 15-30% of circulating blood volume has been lost.

      Pelvic fractures are associated with significant haemorrhage (>2000 ml) that can be concealed. This may require aggressive fluid resuscitation which is initially with crystalloids and then blood. What is also important is including stabilisation of the fracture(s) and pain relief.

      The Advanced Trauma Life Support (ATLS) classification of haemorrhagic shock is as follows:

      Class I haemorrhage (blood loss up to 15%):
      40% blood volume loss):
      Preterminal event patient will die in minutes
      Marked tachycardia, significant depression in systolic pressure and very narrow pulse pressure (or unobtainable diastolic pressure)
      Mental state is markedly depressed
      Skin cold and pale.
      Needs rapid transfusion and immediate surgical intervention.

      A blood loss of >50% results in loss of consciousness, pulse and blood pressure.

      Fluid resuscitation following trauma is a controversial area.

      This clinical scenario points to a 15-30% blood loss. However, further crystalloid and blood replacement may be required after assessing the clinical situation. There is increasing evidence to suggest that transfusion of large volumes of crystalloid in the hospital setting are likely to be deleterious to the patient and hypotensive resuscitation and judicious blood and blood product resuscitation is a more appropriate option. A ratio of 1 unit of plasma to 1 unit of red blood cells is used to replace fluid volume in adults.

      This patient does not require immediate transfusion of O negative blood and there is time for a formal crossmatch. The argument about colloids versus crystalloids has existed for decades. However, while they have a role in fluid resuscitation, they are not first line.

      There is a risk of anaphylaxis, Hypernatraemia, and acute renal injury with colloidal solutions.

    • This question is part of the following fields:

      • Physiology
      95.6
      Seconds
  • Question 17 - With respect to the peripheral nerve stimulators, which one is used to perform...

    Correct

    • With respect to the peripheral nerve stimulators, which one is used to perform nerve blocking?

      Your Answer: Using an insulated needle improves the success rate for the block

      Explanation:

      The nerve stimulators deliver a stimulus lasting for 1-2 milliseconds (not second) to perform nerve blockage.

      There are just 2 leads (not 3); one for the skin and other for the needle.

      Prior to the administration of the local anaesthesia, a current of 0.25 – 0.5 mA (not 1-2mA) at the frequency of 1-2 Hz is preferred.

      If the needle tip is close to the nerve, muscular contraction could be possible at the lowest possible current.

      Insulated needles have improved the block success rate, as the current is only conducting through needle tip.

      Stimulus to the femoral nerve which is placed in the mid lingual line causes withdrawer of the quadriceps and knee extension, that’s the dancing patella ( not plantar flexion).

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      32.7
      Seconds
  • Question 18 - The typical fluid compartments in a normal 70kg male are: ...

    Incorrect

    • The typical fluid compartments in a normal 70kg male are:

      Your Answer: intracellular>interstitial

      Correct Answer: intracellular>extracellular

      Explanation:

      Body fluid compartments in a 70kg male:
      Total volume=42L (60% body weight)
      Intracellular fluid compartment (ICF) =28L
      Extracellular fluid compartment (ECF) = 14L

      ECF comprises:
      Intravascular fluid (plasma) = 3L
      Extravascular fluid = 11L

      Extravascular fluids comprises:
      Interstitial fluid = 10.5L
      Transcellular fluid = 0.5L

    • This question is part of the following fields:

      • Physiology
      39.3
      Seconds
  • Question 19 - 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: A

      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
      110.2
      Seconds
  • Question 20 - A 74-year-old with a VVI pacemaker is undergoing a hip replacement. Which of...

    Correct

    • A 74-year-old with a VVI pacemaker is undergoing a hip replacement. Which of the following is most likely to predispose him to an electrical hazard?

      Your Answer: Use of cutting unipolar diathermy

      Explanation:

      A single chamber pacemaker was implanted in the patient. In VVI mode, a pacemaker paces and senses the ventricle while being inhibited by a perceived ventricular event. The most likely electrical hazard from diathermy is electromagnetic interference (EMI).

      EMI has the potential to cause the following: Inhibition of pacing
      Asynchronous pacing
      Reset to backup mode
      Myocardial burns, and
      Trigger VF.

      Diathermy entails the implementation of high-frequency electrical currents to produce heat and either make incisions or induce coagulation. Monopolar cautery involves disposable cautery pencils and electrosurgical diathermy units. In typical monopolar cautery, an electrical plate is placed on the patient’s skin and acts as an electrode, while the current passes between the instrument and the plate. Monopolar diathermy can therefore interfere with implanted metal devices and pacemaker function.

      Bipolar diathermy, where the current passes between the forceps tips and not through the patient and is less likely to generate EMI.

      Whilst the presence of a CVP line may in theory predispose the patient to microshock, the use of prerequisite CF electrical equipment makes this very unlikely. The presence of a CVP line and pacemaker does not therefore unduly increase the risk of an electrical hazard.

      Isolating transformers are used to protect secondary circuits and individuals from electrical shocks. There is no step-up or step-down voltage (i.e. there is a ratio of 1 to 1 between the primary and secondary windings).

      A ground (or earth) wire is normally connected to the metal case of an operating table to protect patients from accidental electrocution. In the event that a fault allows a live wire to make contact with the metal table (broken cable, loose connection etc.) it becomes live. The earth will provide an immediate path for current to safely flow through and so the table remains safe to touch. Being a low resistance path, the earth lets a large current flow through it when the fault occurs ensuring that the fuse or RCD will quickly blow. Without an operating table earth, the patient is not at more risk of an electrical hazard because of the pacemaker.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      37.6
      Seconds
  • Question 21 - Which of the following causes the right-sided shift of the oxygen haemoglobin dissociation...

    Incorrect

    • Which of the following causes the right-sided shift of the oxygen haemoglobin dissociation curve?

      Your Answer: Hypothermia

      Correct Answer: Chronic iron deficiency anaemia

      Explanation:

      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.

      The oxygen dissociation curve (ODC) describes the relationship between the percentage of saturated haemoglobin and partial pressure of oxygen in the blood.
      Of note, it is not affected by haemoglobin concentration.

      Chronic anaemia causes 2, 3 DPG levels to increase, hence shifting the curve to the right

      Haldane effect – Causes the ODC to shift to the left. For a given oxygen tension there is increased saturation of Hb with oxygen i.e. Decreased oxygen delivery to tissues.
      This can be caused by:
      -HbF, methaemoglobin, carboxyhaemoglobin
      -low [H+] (alkali)
      -low pCO2
      -ow 2,3-DPG
      -ow temperature

      Bohr effect – causes the ODC to shifts to the right = for given oxygen tension there is reduced saturation of Hb with oxygen i.e. Enhanced oxygen delivery to tissues. This can be caused by:
      – raised [H+] (acidic)
      – raised pCO2
      -raised 2,3-DPG
      -raised temperature

    • This question is part of the following fields:

      • Physiology And Biochemistry
      17.2
      Seconds
  • Question 22 - A 26-year-old doctor has recently been diagnosed with lung cancer. He would like...

    Incorrect

    • A 26-year-old doctor has recently been diagnosed with lung cancer. He would like to find out his survival time for the condition. Which statistical method is used to predict survival rate?

      Your Answer: Time series

      Correct Answer: Kaplan-Meier estimator

      Explanation:

      The Weibull distribution are used to describe various types of observed failures of the components. it is used in reliability and survival analysis.

      Regression Analysis is used to measure the relationship between among two or more variable. It determines the effect of independent variables on the dependent variables.

      Student t-test is one of the most commonly used method to test the hypothesis. It determines the significant difference between the means of two different groups.

      A time series is a collection of observations of well-defined data obtained at regular interval of time.

      Kaplan-Meier estimator is used to estimate the survival function from lifetime data. It can be derived from maximum likelihood estimation of hazard function. It is most likely used to measure the fraction of patient’s life for a certain amount of time after treatment.

    • This question is part of the following fields:

      • Statistical Methods
      19.1
      Seconds
  • Question 23 - Which of the following is the best marker of mast cell degranulation in...

    Correct

    • Which of the following is the best marker of mast cell degranulation in a patient with suspected anaphylaxis who became hypotensive and developed widespread urticarial rash after administration of Augmentin?

      Your Answer: Tryptase

      Explanation:

      Mast cell tryptase is a reliable marker of mast cell degranulation. Tryptase is a protease enzyme that acts via widespread protease-activated receptors (PARs).

    • This question is part of the following fields:

      • Pharmacology
      12.9
      Seconds
  • Question 24 - A 45-year-old woman complains of pain in her upper abdomen to her physician....

    Correct

    • 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: 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
      31.9
      Seconds
  • Question 25 - Clearance techniques are used to assess renal glomerular function. Which of the following...

    Correct

    • Clearance techniques are used to assess renal glomerular function. Which of the following is the most accurate marker for glomerular filtration rate measurement?

      Your Answer: Inulin

      Explanation:

      The perfect glomerular filtration marker is:

      The human body is not harmed by it.
      Chemical or physical methods are used to accurately measure
      Extracellular fluid (ECF) compartment is freely and evenly diffusible.
      Inability to access the intracellular fluid (ICF) compartment
      Filtration in the kidney is the only way to remove it from the blood.

      The ideal marker should not be reabsorbed into the bloodstream by the renal tubules or other urinary system components.

      Creatinine is an endogenous substance that is filtered freely by the glomerulus and secreted by the proximal tubule. As a result, creatinine clearance consistently underestimates GFR. In healthy people, this overestimation ranges from 10% to 40%, but it is higher and more unpredictable in patients with chronic kidney disease.

      The gold standard method of inulin clearance necessitates an intravenous infusion and several hours of timed urine collection, making it costly and time-consuming. Inulin is hard to come by and is difficult to mix and keep as a solution.

      Exogenous filtration markers include the following:

      Although plasma clearance of 51chromium EDTA is a widely used method in Europe, tubular reabsorption can occur.
      Because 125I-iothalamate can be excreted by renal tubules in the urine, it cannot be used in patients who have an iodine assay.

      Radioactive substances must be stored, administered, and disposed of according to these methods.

      The glomerulus filters para-aminohippuric acid (PAH) freely, and any that remains in the peritubular capillaries is secreted into the proximal convoluted tubules. This marker is used to determine the amount of blood flowing through the kidneys.

    • This question is part of the following fields:

      • Pathophysiology
      18.6
      Seconds
  • Question 26 - A 32-year-old male is admitted to the critical care unit. He has suffered...

    Correct

    • A 32-year-old male is admitted to the critical care unit. He has suffered a heroin overdose and requires intubation and ventilatory support. What would be his predicted total static compliance (lung and chest wall) measurements.

      Your Answer: 100 ml/cmH2O

      Explanation:

      Static lung compliance refers to the change in volume within the lung per given change in unit pressure. It is usually measured when air flow is absent, such as during pauses in inhalation and exhalation.

      It is a combination of:

      Chest wall compliance: normal value is 200 mL/cmH2O
      Lung tissue compliance: normal value is 200 mL/ cmH2O

      It is represented mathematically as:

      1/Crs = 1/Cl + 1/Ccw

      Where,

      Crs = total compliance of the respiratory system
      Cl = compliance of the lung
      Ccw = compliance of the chest wall

      Therefore in this case:

      1/Crs = 1/200 + 1/200

      1/Crs = 0.005 + 0.005 = 0.01

      1/Ct = 0.01

      Rearranging equation gives:

      Ct = 1/0.01 = 100 mL/cmH2O.

    • This question is part of the following fields:

      • Clinical Measurement
      26.5
      Seconds
  • Question 27 - The statement that best describes the classification of theatre equipment in terms of...

    Incorrect

    • The statement that best describes the classification of theatre equipment in terms of electrical safety is:

      Your Answer: Class I equipment do not need to be connected to earth

      Correct Answer: A floating circuit is equipment applied to patient that is isolated from all its other parts

      Explanation:

      There are different classes of electrical equipment that can be classified in the table below:

      Class 1 – provides basic protection only. It must be connected to earth and insulated from the mains supply

      Class II – provides double insulation for all equipment. It does not require an earth.

      Class III – uses safety extra low voltage (SELV) which does not exceed 24 V AC. There is no risk of gross electrocution but risk of microshock exists.

      Type B – All of above with low leakage currents (0.5mA for Class IB, 0.1 mA for Class IIB)

      Type BF – Same as with other equipment but has ‘floating circuit’ which means that the equipment applied to patient is isolated from all its other parts.

      Type CF – Class I or II equipment with ‘floating circuits’ that is considered to be safe for direct connection with the heart. There are extremely low leakage currents (0.05mA for Class I CF and 0.01mA for Class II CF).

    • This question is part of the following fields:

      • Clinical Measurement
      14.1
      Seconds
  • Question 28 - Regarding basal metabolic rate (BMR), which of the following is the most efficient...

    Correct

    • Regarding basal metabolic rate (BMR), which of the following is the most efficient regulator of BMR?

      Your Answer: Hypothalamus

      Explanation:

      The hypothalamus is primarily responsible for the regulation of the basal metabolic rate. It releases thyrotropin releasing hormones (TRH) in response to low levels of triiodothyronine (T3) and thyroxine (T4). The TRH acts on the pituitary gland to release thyroid stimulating hormone, which will stimulate the thyroid gland to synthesize more T3 and T4.

      Basal metabolic rate refers to the energy expended by an individual in a resting, post-absorptive state. It represents the energy required to carry out normal body functions, such as respiration.

    • This question is part of the following fields:

      • Pathophysiology
      31.7
      Seconds
  • Question 29 - A 31-year old Caucasian female came into the emergency department due to difficulty...

    Correct

    • A 31-year old Caucasian female came into the emergency department due to difficulty of breathing. History revealed exposure to room odorizes that are rich in alkyl nitrites. Upon physical examination, patient is tachypnoeic at 32 breaths per minute, desaturated at 88% while on a non-rebreather mask at 15 litres per minute oxygen. She was also noted to be cyanotic, however with clear breath sounds. Considering the history, what is the most probable cause of her difficulty of breathing?

      Your Answer: Increased affinity of bound oxygen to haemoglobin

      Explanation:

      Amyl nitrate is part of the treatment of cyanide poisoning. The short acting nitrate causes oxidation of Fe2+ in haemoglobin to Fe3+ in methaemoglobin. Methaemoglobin combines with cyanide (cyanmethemoglobin), which reacts with sodium thiosulfate to convert nontoxic thiocyanate and methaemoglobin.

      Methaemoglobin is formed when the iron in haemoglobin is converted from the reduced state (Fe2+) to the oxidized state (Fe3+). The oxidized form of haemoglobin (Fe3+) does not bind oxygen as readily as Fe2+, but has high affinity for cyanide. It also results to high affinity of bound oxygen to haemoglobin, thus leading to tissue hypoxia. Arterial oxygen tension is normal despite observations of cyanosis and dyspnoea. Methemoglobinemia can be treated with methylene blue and vitamin C.

      Carboxyhaemoglobin can be due to carbon monoxide poisoning. In such cases, patients experience headache and dizziness, but do not develop cyanosis.

      2,3-diphosphoglycerate causes a shift in the oxygen dissociation curve to the right, decreasing haemoglobin’s affinity to oxygen to facilitate unloading of oxygen to the tissues.

    • This question is part of the following fields:

      • Pathophysiology
      108.4
      Seconds
  • Question 30 - At which of the following location is there no physiological oesophageal constriction? ...

    Incorrect

    • At which of the following location is there no physiological oesophageal constriction?

      Your Answer: Cricoid cartilage

      Correct Answer: Lower oesophageal sphincter

      Explanation:

      The oesophagus is a muscular tube that connects the pharynx to the stomach. It begins at the lower border of the cricoid cartilage and C6 vertebra. It ends at T11.

      The oesophagus has physiological constrictions at the following levels:
      1. Cervical constriction: Pharyngo-oesophageal junction (15 cm from the incisor teeth) produced by the cricopharyngeal part of the inferior pharyngeal constrictor muscle
      2. Thoracic constrictions:
      i. where the oesophagus is first crossed by the arch of the aorta (22.5 cm from the incisor teeth)
      ii. where the oesophagus is crossed by the left main bronchus (27.5 cm from the incisor teeth)
      3. Diaphragmatic constriction: where the oesophagus passes through the oesophageal hiatus of the diaphragm (40 cm from the incisor teeth)

      Awareness of these constrictions is important for clinical purposes when it is required to pass instruments through the oesophagus into the stomach or when viewing radiographs of patients’ oesophagus.

    • This question is part of the following fields:

      • Anatomy
      28.1
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Physiology And Biochemistry (3/6) 50%
Physiology (2/5) 40%
Anatomy (3/5) 60%
Clinical Measurement (3/4) 75%
Pharmacology (1/3) 33%
Pathophysiology (4/4) 100%
Anaesthesia Related Apparatus (2/2) 100%
Statistical Methods (0/1) 0%
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