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  • Question 1 - A delayed hypersensitivity reaction is type ____? ...

    Correct

    • A delayed hypersensitivity reaction is type ____?

      Your Answer: IV

      Explanation:

      Type I – immediate hypersensitivity reaction

      Examples are: Atopy, urticaria, Anaphylaxis, Asthma( IgE mediated).

      Type II – Antibody mediated cytotoxic reaction

      Examples are: Autoimmune haemolytic anaemia, Thrombocytopenia( IgM or IgG mediated).

      Type III – Immune complex mediated reaction

      Examples are: Serum sickness,SLE – IgG., Farmers lungs, rheumatoid arthritis

      Type IV – Delayed hypersensitivity reaction

      Examples are: Contact dermatitis, drug allergies.

      Type V – Autoimmune

      Graves’
      Myasthenia – IgM or IgG.

    • This question is part of the following fields:

      • Pathophysiology
      4.3
      Seconds
  • Question 2 - In North America, there have been reports of paediatric patients dying after undergoing...

    Incorrect

    • In North America, there have been reports of paediatric patients dying after undergoing adenotonsillectomy for obstructive sleep apnoea.

      Respiratory depression/obstruction is thought to be the cause of death. The codeine dose was 0.5-1 mg/kg, given every 4-6 hours.

      In this group of patients, which of the following is the most likely cause of respiratory depression and obstruction?

      Your Answer: Overdose of codeine

      Correct Answer: Exaggerated metabolism of codeine

      Explanation:

      Codeine is easily absorbed from the gastrointestinal tract and converted to morphine and norcodeine in the liver via O- and N-demethylation. Morphine and norcodeine are excreted almost entirely by the kidney, primarily as conjugates with glucuronic acid.

      By glucuronidation, phase II metabolism enzyme UDP-glucuronosyl transferase-2B7 converts morphine to morphine-3-glucuronide (M3G) and morphine-6-glucuronide (M6G) (UGT2B7).

      Approximately 60% of morphine is converted to M3G, with the remaining 6-10% converted to M6G. M3G is inactive, but M6G is said to be 4 to 650 times more potent on the MOP receptor than morphine.

      When codeine is consumed, cytochrome P450 2D6 in the liver converts it to morphine (CYP2D6).

      Some people have DNA variations that increase the activity of this enzyme, causing codeine to be converted to morphine more quickly and completely than in others. After taking codeine, these ultra-rapid metabolisers are more likely to have higher than normal levels of morphine in their blood.

      Respiratory depression/obstruction can be caused by high levels of morphine and M6G, especially in people who have a history of obstructive sleep apnoea. The estimated number of ultra-rapid metabolisers ranges from 1 to 7 per 100 people, but some ethnic groups may have as many as 28 per 100 people.

    • This question is part of the following fields:

      • Pathophysiology
      81.1
      Seconds
  • Question 3 - The following statements are about burns patients. Which one is true? ...

    Incorrect

    • The following statements are about burns patients. Which one is true?

      Your Answer:

      Correct Answer: High protein diets may improve survival

      Explanation:

      Patients who have sustained thermal injuries are at high risk of becoming hypercatabolic with larger cardiac outputs and oxygen consumptions.

      The hypermetabolic states increase with an increase in the burn severity and surface area of the skin affected. A patient with thermal injuries affecting 60% of the total surface area of the body will have twice the normal metabolic rate.

      The optimal temperature for nursing patients with burn injuries is 30°C to conserve the energy usage. The areas affected by the burn injuries should be covered to reduce loss of fluid via evaporation. Resetting hypothalamic thermoregulation will cause a 1-2°C increase in core temperature.

      Burn injuries will have an immediate effect on the intestine, destroying the barrier function and allowing for the movement of bacteria and endotoxins within hours.

      Enteral nutrition allows for the delivery of nutrients directly to the stomach or intestine. It has correlation with a dampened hypermetabolic response to a thermal and injury, especially when initiated early as it helps to protect the integrity of the mucosal lining and prevents the movement of bacteria into circulation.

      Diet changes have been linked to reduced mortality due to burn injuries. Diets high in protein especially (calorie: nitrogen ratio of 100: 1), have the highest correlation with improved survival rates.

      Parenteral feeds may be required alongside enteral nutrition, even with the increased risks of infection.

    • This question is part of the following fields:

      • Pathophysiology
      0
      Seconds
  • Question 4 - A 30-year-old woman admitted following a tonsillectomy has developed stridor with a respiratory...

    Incorrect

    • A 30-year-old woman admitted following a tonsillectomy has developed stridor with a respiratory rate of 22 breaths per minute and obstructive movements of the chest and abdomen that is in a see-saw pattern .

      Her SpO2 is 92% on 60% oxygen with pulse rate 120 beats per minute while her blood pressure is 180/90mmHg. She is repeatedly trying to remove the oxygen mask and appears anxious.

      Her pharynx is suctioned and CPAP applied with 100% oxygen via a Mapleson C circuit.

      Which of these is the most appropriate next step in her management?

      Your Answer:

      Correct Answer: Administer intravenous propofol 0.5 mg/kg

      Explanation:

      Continuous closure of the vocal cords resulting in partial or complete airway obstruction is called Laryngospasm. It is a reflex that helps protect against pulmonary aspiration.

      Predisposing factors include: Hyperactive airway disease, Insufficient depth of anaesthesia, Inexperience of the anaesthetist, Airway irritation, Smoking, Shared airway surgery and Paediatric patients

      Its primary treatment includes checking for blood or stomach aspirate in the pharynx, removing any triggering stimulation, relieving any possible supra-glottic component to airway obstruction and application of CPAP with 100% oxygen.

      In this patient, all the above has been done and the next treatment of choice is the administration of a rapidly acting intravenous anaesthetic agent such as propofol (0.5 mg/kg) in increments as it has been reported to relieve laryngospasm in approximately 75% of cases. Administering suxamethonium to an awake patient would be inappropriate at this stage.

      Magnesium and lidocaine are used for prevention rather than acute treatment of laryngospasm. Superior laryngeal nerve blocks have been reported to successfully treat recurrent laryngospasm but it is not the next logical step in index patient.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 5 - What is the most sensitive method of detecting an intra-operative air embolism? ...

    Incorrect

    • What is the most sensitive method of detecting an intra-operative air embolism?

      Your Answer:

      Correct Answer: Transoesophageal echocardiogram

      Explanation:

      An intra-operative air embolism occurs when air becomes trapped in the blood vessels during surgery.

      A transoesophageal echocardiography (OE) uses invasive echocardiography to monitor the integrity and performance of the heart. It is the gold standard as it provides real-time imaging of the heart to enable early diagnosis and treatment.

      Precordial doppler ultrasonography can also be used to detect into-operative air emboli. It is non-invasive and more practical, but is less sensitive.

      A change in end-tidal CO2 could be indicative of and increase in physiological dead-space, but could also be indicative of any processes that reduces the excretion or increases the production of CO2, making it non-specific.

      A transoesophageal stethoscope can be used to listen for the classic mill-wheel murmur produced by a large air embolus.

    • This question is part of the following fields:

      • Pathophysiology
      0
      Seconds
  • Question 6 - The production of carbon dioxide and water occurs during cellular respiration, which involves...

    Incorrect

    • The production of carbon dioxide and water occurs during cellular respiration, which involves an energy substrate and oxygen. For a patient, the respiratory quotient is calculated as 0.7.

      Which of the following energy substrate combinations is the most likely in this patient's diet?

      Your Answer:

      Correct Answer: Low carbohydrate, high fat and low protein

      Explanation:

      The respiratory quotient (RQ) is the proportion of CO2 produced by the body to O2 consumed per unit of time.

      CO2 produced / O2 consumed = RQ

      CO2 is produced at a rate of 200 mL per minute, while O2 is consumed at a rate of 250 mL per minute. An RQ of around 0.8 is typical for a mixed diet.

      The RQ will change depending on the energy substrates consumed in the diet.

      Granulated sugar is a refined carbohydrate that contains 99.999 percent carbohydrate and no lipids, proteins, minerals, or vitamins.

      Glucose and other hexose sugars – RQ = 1
      Fats – RQ = 0.7
      Proteins – RQ is 0.9
      Ethyl alcohol – RQ = 0.67

    • This question is part of the following fields:

      • Pathophysiology
      0
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  • Question 7 - Which of the following is an expected change in pulmonary function seen during...

    Incorrect

    • Which of the following is an expected change in pulmonary function seen during a moderate asthma attack?

      Your Answer:

      Correct Answer: Decreased forced expiratory volume in 1 sec (FEV1)

      Explanation:

      Asthma is a lung condition that causes reversible narrowing and swelling of airway passages. It is classified by the frequency and severity of symptoms.

      The following are symptoms of moderate asthma:

      Symptoms include cough, wheezing, chest tightness, or difficulty breathing which occurs daily
      Decreased activity levels due to flare-ups
      Night-time symptoms 5 or more times a month
      Lung function test FEV1 is 60-80% of predicted normal values
      Peak flow has more than 30% variability

      With moderate asthma attacks, the arterial pCO2 levels may decrease, but as severity increases, so does the pCO2, reaching normal levels, and then exceeding them in severe asthma attacks.

      Airway obstruction increases the functional residual capacity.

      Concentration of serum bicarbonate would not increase in moderate asthma, but it could possibly increase in life-threatening asthma via the same mechanism as what increases arterial PCO2.

      FEV1 is a good measure of airway obstruction. and is reduced in acute asthma attacks.

      In the case of a pneumothorax, a decrease in arterial PO2 is higher.

    • This question is part of the following fields:

      • Pathophysiology
      0
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  • Question 8 - The cardiac muscle will primarily utilize which metabolic substrate to produce energy when...

    Incorrect

    • The cardiac muscle will primarily utilize which metabolic substrate to produce energy when at rest?

      Your Answer:

      Correct Answer: Fatty acids

      Explanation:

      Approximately 70% of the heart’s ATP requirement is met by cardiac mitochondria through beta-oxidation of fatty acids at rest. The remaining 30% is supplied by glucose.

      Amino acids and ketones, in the presence of ketoacidosis, may supply at most 10% of the ATP requirement. And, when in high levels, lactate may also contribute to the ATP requirement of the heart, particularly during moments of high muscular activity.

    • This question is part of the following fields:

      • Pathophysiology
      0
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  • Question 9 - Which of the following is correct regarding nitric oxide? ...

    Incorrect

    • Which of the following is correct regarding nitric oxide?

      Your Answer:

      Correct Answer: Is produced by both inducible and constitutive forms of nitric oxide synthetase

      Explanation:

      Nitric oxide is generated from L-arginine by nitric oxide synthase. It is produced in response to haemodynamic stress by the vascular endothelium, and it produces both smooth muscle relaxation and reduced vascular resistance.

      Nitric oxide may be inactivated through interaction with other oxygen free radicals, (e.g. oxidised low-density lipoprotein (LDL)).

      Nitric oxide causes the production of the second messenger, cyclic guanosine monophosphate (cGMP).

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 10 - Which of the following best describes the phenomenon of higher partial pressures of...

    Incorrect

    • Which of the following best describes the phenomenon of higher partial pressures of oxygen in the alveoli (PAO2) in the lung apices when in the upright position?

      Your Answer:

      Correct Answer: The ventilation/perfusion (V/Q) ratio is greater than in the basal units

      Explanation:

      The ventilation/perfusion ratio varies in different areas of the lung. In an upright individual, although both ventilation and perfusion increase from the apex to the base of the lung, the increase in ventilation is less than the increase in blood flow. As a result, the normal V̇ /Q̇ ratio at the apex of the lung is much greater than 1 (ventilation exceeds perfusion), whereas the V̇ /Q̇ ratio at the base of the lung is much less than 1 (perfusion exceeds ventilation).

      There is more volume in the alveoli found in the apices than in the bases of the lungs. This is due to the weight of the lung stretching the apical alveoli to the maximum size. Also, the weight of the lungs pull themselves away from the chest wall, creating a negative intrapleural pressure. These factors, however, do not directly affect the PAO2.

    • This question is part of the following fields:

      • Pathophysiology
      0
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  • Question 11 - The following statements are about the cervical plexus. Which one is true? ...

    Incorrect

    • The following statements are about the cervical plexus. Which one is true?

      Your Answer:

      Correct Answer: Recurrent laryngeal nerve block is a complication of a cervical plexus block

      Explanation:

      The cervical plexus is a complex network of nerves within the head and neck region, providing nerve innervation to regions within the head, neck and trunk.

      It is comprised of nerves arising from the anterior primary rami of the C1-C4 nerve roots.

      The cervical plexus gives off superficial and deep branches. The superficial branches penetrate through the deep fascia at the centre point of the posterior border of the sternocleidomastoid. It provides sensory innervation from the lower border of the mandible to the 2nd rib. The deep branches provide motor innervation to the neck and diaphragmatic muscles.

      Cervical plexus block is surgically relevant as it is used to provide regional anaesthesia for procedures in the neck region. The anaesthesia should be injected into the centre point of the posterior border of the sternocleidomastoid. Complications arise when anaesthesia is instead injected into the wrong point, including into the vertebral artery, subarachnoid and epidural spaces, blockade of phrenic and recurrent laryngeal nerves, and the cervical sympathetic plexus.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 12 - The Medical Admissions unit receives a 71-year-old woman. She has type 2 diabetes,...

    Incorrect

    • The Medical Admissions unit receives a 71-year-old woman. She has type 2 diabetes, which she manages with diet, but she has been feeling ill for the past 48 hours.

      Her pulse rate is 110 beats per minute, her blood pressure is 90/50 mmHg, and she is clinically dehydrated. Her respiratory rate is 20 breaths per minute, and chest auscultation reveals no focal signs.

      The following are the lab results:

      Glucose 27.4 mmol/L (3.5-5.5)
      Ketones 2.5 mmol/L (<0.1)
      Urinary glucose is zero (dipstick) with ketones

      A random blood glucose of 15.3 mmol/L was measured during a visit to the diabetic clinic one month prior to admission, according to her notes, and a urinary dipstick registered a high glucose and ketones++.

      The discrepancy between plasma and urinary glucose measurements is best explained by which of the following physiological mechanisms?

      Your Answer:

      Correct Answer: The glomerular filtration rate is abnormally low

      Explanation:

      The glucose molecule enters the Bowman’s capsule freely and becomes part of the filtrate.

      All glucose is reabsorbed in the proximal convoluted tubule when blood glucose concentrations are below a certain threshold (approximately 11 mmol/L) (PCT). Active transportation makes this possible. In the proximal tubular cells, sodium/glucose cotransporters (SGLT1 and SGLT2) are the proteins responsible.

      Glucose does not normally appear in the urine below the renal threshold.

      The renal glucose threshold is not set in stone and is affected by a variety of factors, including GFR, TmG, and the quantity of splay.

      The different absorptive and filtering capacities of individual nephrons cause splay, which is the rounding of a glucose reabsorption curve.

      The SGLT proteins have a high affinity for glucose, but not an infinite affinity. As a result, some glucose may escape reabsorption before the TmG. A decrease in renal threshold may be caused by an increase in splay.

      Because the filtered glucose load is reduced and the PCT can reabsorb all of the filtered glucose despite hyperglycaemia, a low GFR causes an increase in TmG. In contrast, lowering the TmG lowers the threshold because the tubules’ ability to reabsorb glucose is reduced.

      A reduction in GFR caused by severe dehydration and reduced perfusion pressure is the most obvious cause of the discrepancy between plasma and urinary glucose in this scenario.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 13 - A 43-year-old patient was brought to the emergency department with a traumatic amputation...

    Incorrect

    • A 43-year-old patient was brought to the emergency department with a traumatic amputation of his leg at mid-thigh level. Resuscitation with 1 L gelofusine was done and four units of packed red blood cells were given before theatre. Thirty minutes following blood transfusion, the patient became flushed, breathless, hypotensive, develops haemoglobinuria, and had a fever of 38oC.

      Which one of the following correctly explains the patient signs and symptoms?

      Your Answer:

      Correct Answer: Activation of classic complement pathway

      Explanation:

      This may be the classical case of blood transfusion reaction due to ABO incompatibility.

      Here red cells are destroyed in the bloodstream with the release of haemoglobin in circulation (causing haemoglobinuria). Here, IgM or IgG anti-A or anti-B antibody can cause rapid activation of complement cascade usually the classical pathway. This is called intravascular haemolysis.

      There may be extravascular haemolysis by cells of the mononuclear phagocyte system situated in the liver and spleen. Extravascular red cell destruction can increase breakdown products of haemoglobin, such as bilirubin and urobilinogen.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 14 - With a 10-day history of severe vomiting, a 71-year-old man with a gastric...

    Incorrect

    • With a 10-day history of severe vomiting, a 71-year-old man with a gastric outlet obstruction is admitted to the surgical ward.

      The serum biochemical results listed below are available:

      Sodium 128 mmol/L (137-144)
      Potassium 2.6 mmol/L (3.5-4.9)
      Chloride 50 mmol/L (95-107)
      Urea 12 mmol/L (2.5-7.5)
      Creatinine 180 µmol/L (60-110)

      Which of the following do you think you are most likely to encounter?

      Your Answer:

      Correct Answer: The standard base excess will be higher than actual base excess

      Explanation:

      Hydrochloric acid is lost when you vomit for a long time (HCl). As a result, the following can be expected, in varying degrees of severity:

      Hypokalaemia
      Hypochloraemia
      Increased bicarbonate to compensate for chloride loss and metabolic alkalosis

      The alkalosis causes potassium to move from the intracellular to the extracellular compartment at first. Long-term vomiting and dehydration cause potassium to be excreted by the kidneys in order to conserve sodium. Dehydration can cause urea and creatinine levels to rise.

      The actual base excess is always greater than the standard base excess.

      The actual base excess (BE) is a measurement of a base’s contribution to a blood gas picture’s metabolic component. It’s the amount of base that needs to be added to a blood sample to bring the pH back to 7.4 after the respiratory component of a blood gas picture has been corrected (PaCO2 of 40 mmHg or 5.3 kPa). The BE has a normal range of +2 to 2. A large positive BE indicates a severe metabolic alkalosis, while a large negative BE indicates a severe metabolic acidosis. As a result, the actual BE in vitro is unaffected by CO2.

      In vivo, however, standard BE is not independent of pCO2 because blood with haemoglobin acts as a better buffer than total ECF.

      As a result, it is impossible to tell the difference between compensating for a respiratory disorder and compensating for the presence of a primary metabolic disorder.

      The differences between in vitro and in vivo behaviour can be mostly eliminated if the BE is calculated for a haemoglobin concentration of 50 g/L (the ‘effective’ or virtual value of Hb if it was distributed throughout the extracellular space) rather than the actual haemoglobin. Because haemoglobin has a lower buffering capacity, the standard BE is higher than the actual BE. It reflects the BE better in the extracellular space rather than just the intravascular compartment.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 15 - One of the following sets of arterial blood gases best demonstrates compensated respiratory...

    Incorrect

    • One of the following sets of arterial blood gases best demonstrates compensated respiratory acidosis?

      Your Answer:

      Correct Answer: pH=7.36, PaCO2=8.5kPa, PaO2=7.5kPa, HCO3- = 43mmol/L

      Explanation:

      pH=7.36, PaCO2=8.5kPa, PaO2=7.5kPa, HCO3- = 43mmol/L is the correct answer.

      Since the pH is the lower limit of normal, it is compensated despite a raised PaCO2. Retention of bicarbonate ions by the renal system suggests this process is chronic.

      pH=7.24, PaCO2=3.5kPa, PaO2=8.5kPa, HCO3- =18mmol/L represents an acute uncompensated metabolic acidosis

      The remaining stems are degrees of uncompensated respiratory acidosis and therefore incorrect.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 16 - Given the following hormones, which of these will stimulate glycogenesis and gluconeogenesis? ...

    Incorrect

    • Given the following hormones, which of these will stimulate glycogenesis and gluconeogenesis?

      Your Answer:

      Correct Answer: Corticosteroids

      Explanation:

      Insulin is the primary anabolic hormone that dominates regulation of metabolism during digestive phase. It promotes glucose uptake in skeletal myocytes and adipocytes, and other insulin-target cells. It promotes glycogenesis and inhibits gluconeogenesis.

      Glucagon is the primary counterregulatory hormone that increases blood glucose levels, primarily through its effects on liver glucose output.

      Similar to glucagon, growth hormone, catecholamines and corticosteroids are also counterregulatory factors released in response to decreased glucose concentrations. Growth hormone promotes glycogenolysis and inhibits gluconeogenesis; catecholamines stimulate glycogenolysis and gluconeogenesis; while corticosteroids stimulate glycogenesis and gluconeogenesis.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 17 - The following statements are about changes that occur at birth. Which is accurate?...

    Incorrect

    • The following statements are about changes that occur at birth. Which is accurate?

      Your Answer:

      Correct Answer: The systemic vascular resistance rises

      Explanation:

      The umbilical vein closes once the umbilical cord is clamped following birth. This causes a rise in systemic vascular resistance, closing the ductus venosus.

      Upon birth, the pulmonary vascular resistance is decreased as the lungs are aerated.

      At birth, there is a rise in oxygen tension which causes the corresponding constriction of the ductus arteriosus. This prevents a left to right shunt as it stops aortic blood and blood from the pulmonary artery from mixing. The ventricles do no have an opening connecting them.

      The foramen ovale closes soon after birth. It is the septum opening between the left and right atrium.

      An adult’s cardiac output is expected to be 5 L/min

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 18 - A 52-year old man was placed under general anaesthesia for an emergent open...

    Incorrect

    • A 52-year old man was placed under general anaesthesia for an emergent open cholecystectomy. As part of the induction, suxamethonium was administered at 1.5mg/kg. Post-operatively, there was failure to restore muscle twitch responses over a course of five hours.

      Clinical chemistry studies were obtained and showed the following results:
      Butrylcholinesterase (BChE) activity: 49 U/L (Reference range: 3300-10,300 U/L)
      Dibucaine number: <4% (Reference range: 83-88%)

      The attending physician gave an initial diagnosis of Suxamethonium Apnoea.

      What is the most probable phenotype of BChE of the patient?

      Your Answer:

      Correct Answer: S (silent)

      Explanation:

      Silent (S) is the most probable phenotype of the patient. In S phenotype, patients have significantly reduced levels of BChE, the lowest among the four phenotypes. Because of this, individuals with S phenotype are subjected to long periods of apnoea. In addition, their dibucaine number is very low.

      Other BChE phenotypes are the following:

      Usual (U)
      Atypical (A)
      Fluoride-resistant (F)

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 19 - The following statements are about chronic obstructive pulmonary disease (COPD). Which is true?...

    Incorrect

    • The following statements are about chronic obstructive pulmonary disease (COPD). Which is true?

      Your Answer:

      Correct Answer: Inhaled corticosteroid usage slows the decline in health status

      Explanation:

      Chronic obstructive pulmonary disease (COPD) is an obstructive, inflammatory lung condition. It encompasses symptoms of emphysema, chronic bronchitis and asthma.

      Inhaling high dose steroids are prescribed to treat COPD. They are effective at reducing symptoms and improving lung function. They also work to reduce the number of hospitalisations by decreasing the number of acute exacerbation events. Despite providing effective symptom relief, it cannot slow down the decline of FEV1 as COPD is an irreversible condition.

      COPD reduces the FEV1 measurements, as well as the FEV1/FVC ratio.

      Breathlessness is a major COPD symptom and can occur at any point in the disease progression, including at an FEV1 >50%.

      FEV1 is used in COPD staging, and it is classed as follows:
      >80%: Mild or stage I
      50 – 79%: Moderate or stage II
      30 – 49%: Severe or stage III
      <30%: Very severe or stage IV
      Patients with mild COPD are usually able to manage their condition on their own, however once the disease progresses to moderate, more GP visits are required, with those in the severe category requiring frequent hospitalisation.

      Asthma is correlated to an increase in transfer factor. COPD (emphysema) is correlated to a decreased transfer factor.

      COPD predisposes to eventual pulmonary hypertension as a result of an increase in pulmonary vascular resistance.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 20 - In a diagnosis of a compensated respiratory acidosis, which of the following arterial...

    Incorrect

    • In a diagnosis of a compensated respiratory acidosis, which of the following arterial blood gas results is likely to be seen?

      Your Answer:

      Correct Answer:

      Explanation:

      During normal tissue metabolism, there is production of CO2 (acid) which is then expired by the lungs. If metabolism switches from aerobic to anaerobic due to a lack of oxygen, the tissues are unable to completely oxidise sugars to CO2. As a consequence, the sugars can only be partially oxidised to lactic acid. Since lactic acid cannot be expired by the lungs, it remains in the circulation leading to metabolic acidosis.

      Also, normal tissue metabolism leads to the production of some amount of acid from the breakdown of proteins. These acids are excreted from the body by kidney filtration. Renal failure will therefore results in acidosis after several days.

      An increased acidosis stimulates the brain’s respiratory centres to increase the respiratory rate. This lowers the CO2 in the blood, leading to a decrease in its acidity. Renal excretion removes the excess acid, resulting in a normal pH, and a reduced PaCO2 and HCO3.

      pH PaCO2 (kPa) HCO3
      Compensated respiratory acidosis 7.34 7.2 29
      Acute respiratory acidosis 7.25 7.3 22
      Compensated metabolic acidosis 7.34 3.6 14
      Metabolic acidosis 7.21 5.3 15
      Metabolic alkalosis 7.51 5.1 30

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 21 - Regarding chest tube insertion, which of the following measurements is utilized when selecting...

    Incorrect

    • Regarding chest tube insertion, which of the following measurements is utilized when selecting a chest tube drain?

      Your Answer:

      Correct Answer: External circumference (mm)

      Explanation:

      Selection of a chest drain will depend on the external circumference.

      A cannula, whether intravenous or intra-arterial, are classified according to standard wire gauge, which refers to the number of wires that can fit into the same hole. If a cannula is labelled 22G, then 22 wires will fit into the standard size hole.

      A more popular measurement than SWG nowadays is cross sectional area.

      When the concern for selecting equipment is the rate of flow, then it is important to consider the diameter and the radius of a parallel sided tube. These, however, are not routinely considered when comparing sizes of a cannula.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 22 - Iron is one of the most important micronutrients in the body.

    Out of...

    Incorrect

    • Iron is one of the most important micronutrients in the body.

      Out of the following, which one has the most abundant storage of iron in the body?

      Your Answer:

      Correct Answer: Haemoglobin

      Explanation:

      Iron is a necessary micronutrient for proper erythropoietic function, oxidative metabolism, and cellular immune responses. Although dietary iron absorption (1-2 mg/d) is tightly controlled, it is only just balanced by losses.

      The adult body contains 35-45 mg/kg iron (about 4-5 g)

      Iron can be found in a variety of forms, including haemoglobin, ferritin, haemosiderin, myoglobin, haem enzymes, and transferrin bound proteins.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 23 - A 59-year-old smoker booked for an emergency laparotomy is in the anaesthetic room...

    Incorrect

    • A 59-year-old smoker booked for an emergency laparotomy is in the anaesthetic room prior to intubation. He is breathing room air and an arterial blood gas is obtained on insertion of an arterial cannula and sent for analysis.

      The following results are available:

      Haemoglobin 75 g/L
      PaO2 10.7 kPa
      PaCO2 5.2 kPa

      After intravenous induction, intubation is difficult and he rapidly begins to de-saturate.

      Which of the following is most effective in prolonging the oxygen de-saturation time?

      Your Answer:

      Correct Answer: Pre-oxygenation with 100% O2 for three minutes

      Explanation:

      Breathing 100% oxygen for three minutes will provide the best reservoir of oxygen during apnoea by oxygenating the functional residual capacity (FRC).

      Sitting at 45 degrees might increase the FRC and improve oxygen reserve but not compared with 100% oxygenation.

      The following table compares the oxygen reserves in the body following pre-oxygenation with room air and 100% oxygen:

      Compartment Factors Room air (mL) 100% O2 (mL)
      Lung FAO2, FRC 630 2850
      Plasma PaO2, DF, PV 7 45
      Red blood cells Hb, TGV, SaO2 788 805
      Myoglobin – 200 200
      Interstitial space – 25 160

      FAO2 = alveolar fraction of oxygen.
      FRC = Functional residual capacity.
      PaO2 = partial pressure of oxygen dissolved in arterial blood
      DF = dissolved form.
      PV = plasma volume.
      TG = total globular volume .
      Hb = haemoglobin concentration.
      SaO2 = arterial oxygen saturation

      Stopping smoking one month prior to surgery will not be more effective than pre-oxygenation with 100% oxygen though it may reduce postoperative pulmonary complications. Note that both long term and short term abstinence reduces pulse rate and blood pressure thus reducing oxygen consumption and also reduce carboxyhaemoglobin levels.

      Blood transfusion will not make a big difference in oxygen reserve, particularly if a blood transfusion is administered within 12-24-hours before surgery.

      Heliox (79% helium and 21% oxygen) despite its lower viscosity is unlikely to be more effective than 100% oxygen .

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 24 - Concerning the anterior pituitary gland, one of following is true. ...

    Incorrect

    • Concerning the anterior pituitary gland, one of following is true.

      Your Answer:

      Correct Answer: Produces glycoproteins

      Explanation:

      The posterior pituitary and the hypothalamus are connected by the pituitary stalk. It contains in the pituitary sella and has the optic chiasm and hypothalamus as superior relations.

      The anterior pituitary produces thyroid-stimulating hormone (TSH), luteinising hormone (LH) and follicle-stimulating hormone (FSH) . These hormones are Glycoproteins and share a common alpha subunit with unique beta subunits.

      The secretion of pituitary hormones are pulsatile. Examples are LH, adrenocorticotropic hormone (ACTH) and growth hormone (GH).

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 25 - Which of the following is correct about the action of bile salts? ...

    Incorrect

    • Which of the following is correct about the action of bile salts?

      Your Answer:

      Correct Answer: Emulsification of lipids

      Explanation:

      The emulsification and absorption of fats requires Bile salts.

      Absorption of fats is associated with the activation of lipases in the intestine.

      Bile salts are involved in fat soluble vitamin absorption and are reabsorbed in the terminal ileum (B12 is NOT fat soluble).

      Although Vitamin B12 is also absorbed in the terminal ileum, it is a water soluble vitamin (as are B1, nicotinic acid, folic acid and vitamin C) .

      The gastric parietal cells secretes Intrinsic factor that is essential for the absorption of B12.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 26 - Regarding the Valsalva manoeuvre, which of the following describes the cardiovascular changes in...

    Incorrect

    • Regarding the Valsalva manoeuvre, which of the following describes the cardiovascular changes in phase III in a normal patient?

      Your Answer:

      Correct Answer: Normal intrathoracic pressure, decrease in blood pressure, and increase in heart rate

      Explanation:

      When a person forcefully expires against a closed glottis, changes occur in intrathoracic pressure that dramatically affect venous return, cardiac output, arterial pressure, and heart rate. This forced expiratory effort is called a Valsalva maneuver.

      Initially during a Valsalva, intrathoracic (intrapleural) pressure becomes very positive due to compression of the thoracic organs by the contracting rib cage. This increased external pressure on the heart and thoracic blood vessels compresses the vessels and cardiac chambers by decreasing the transmural pressure across their walls. Venous compression, and the accompanying large increase in right atrial pressure, impedes venous return into the thorax. This reduced venous return, and along with compression of the cardiac chambers, reduces cardiac filling and preload despite a large increase in intrachamber pressures. Reduced filling and preload leads to a fall in cardiac output by the Frank-Starling mechanism. At the same time, compression of the thoracic aorta transiently increases aortic pressure (phase I); however, aortic pressure begins to fall (phase II) after a few seconds because cardiac output falls. Changes in heart rate are reciprocal to the changes in aortic pressure due to the operation of the baroreceptor reflex. During phase I, heart rate decreases because aortic pressure is elevated; during phase II, heart rate increases as the aortic pressure falls.

      When the person starts to breathe normally again, the intrathoracic pressure declines to normal levels, the aortic pressure briefly decreases as the external compression on the aorta is removed, and heart rate briefly increases reflexively (phase III). This is followed by an increase in aortic pressure (and reflex decrease in heart rate) as the cardiac output suddenly increases in response to a rapid increase in cardiac filling (phase IV). Aortic pressure also rises above normal because of a baroreceptor, sympathetic-mediated increase in systemic vascular resistance that occurred during the Valsava.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 27 - A 66-year-old man with a haemorrhagic stroke is admitted to the medical admissions...

    Incorrect

    • A 66-year-old man with a haemorrhagic stroke is admitted to the medical admissions unit.

      He has been taking warfarin for a long time because of atrial fibrillation. His INR at the time of admission was 9.1.

      Which of the following treatment options is the most effective in managing his condition?

      Your Answer:

      Correct Answer: Prothrombin complex concentrate

      Explanation:

      Haemorrhage, including intracranial bleeding, is a common and potentially fatal side effect of warfarin therapy, and reversing anticoagulation quickly and completely can save lives. When complete and immediate correction of the coagulation defect is required in orally anticoagulated patients with life-threatening haemorrhage, clotting factor concentrates are the only viable option.

      For rapid reversal of vitamin K anticoagulants, prothrombin complex concentrates (PCC) are recommended. They contain the vitamin K-dependent clotting factors II, VII, IX, and X and are derived from human plasma. They can be used as an adjunctive therapy in patients with major bleeding because they normalise vitamin K dependent clotting factors and restore haemostasis.

      The most common treatments are fresh frozen plasma (FFP) and vitamin K. The efficacy of this approach is questioned due to the variable content of vitamin K-dependent clotting factors in FFP and the effects of dilution. Significant intravascular volume challenge, as well as the possibility of rare complications like transfusion-associated lung injury or blood-borne infection, are all potential issues.

      To avoid anaphylactic reactions, vitamin K should be given as a slow intravenous infusion over 30 minutes. Regardless of the route of administration, the reversal of INRs with vitamin K can take up to 24 hours to reach its maximum effect.

      Reversal of anticoagulation in patients with warfarin-associated intracranial haemorrhage may be considered with factor VIIa (recombinant), but its use is controversial. There are concerns about thromboembolic events following treatment, as well as questions about assessing efficacy in changes in the INR. If the drug is to be administered, patients should be screened for an increased risk of thrombosis before the drug is given.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 28 - A 30-year-old woman with a BMI of 24 kg/m2 consumes four glasses of...

    Incorrect

    • A 30-year-old woman with a BMI of 24 kg/m2 consumes four glasses of wine on an empty stomach. Her serum alcohol are levels measured over the following five hours. The serum alcohol level of 30-year-old man with the same BMI and alcohol consumption is also measured over the same duration.

      The peak concentration of alcohol is found to be greater in the woman than in the man.

      Which of these offers best explanation for this observation?

      Your Answer:

      Correct Answer: Lower volume of distribution

      Explanation:

      The blood alcohol concentration depends on:

      -The rate of alcohol absorption from the gastrointestinal tract
      -The volume of distribution of alcohol in the body, and
      -The rate of elimination of alcohol from the body.

      Total body water is approximately 50% in a female as compared to 60% in a typical male. This means that the volume of distribution of alcohol is lower in female compared with men. This is the principal reason for higher peak in alcohol levels.

      About 4% of ingested alcohol is metabolised by the liver accounting for first pass metabolism and 0.4% is metabolised by gastric alcohol dehydrogenase (ADH). The absorbed alcohol is NOT distributed to fat cells but it is distributed throughout the water compartments (plasma, interstitial and intracellular) of the body. Women have very little gastric ADH, which further influences this exaggerated rise.

      85-98% of the alcohol is oxidised by the liver to acetaldehyde and then to acetate. The metabolic pathway initially observes first order kinetics and then saturation or zero order kinetics leading to peaks in alcohol levels.

      Clearance of ethanol per unit lean body mass is lower in male. The calculated alcohol elimination rate and liver volume per kilogram of lean body mass were 33% and 38% higher in women than in men, respectively.

      Available evidence in the literature about the relationship of alcohol metabolism to the phases of the menstrual cycle is conflicting.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 29 - If a patient is to be placed under general anaesthesia using total intravenous...

    Incorrect

    • If a patient is to be placed under general anaesthesia using total intravenous technique with target-controlled infusions of propofol and remifentanil, what safety precaution is the most vital in this a scenario?

      Your Answer:

      Correct Answer: Cannula access site clearly visible and regularly checked

      Explanation:

      According to the Safe Anaesthesia Liaison Group, the most important factor to consider the cannula access, and if the patient is properly receiving the total intravenous anaesthesia. The cannula access must be regularly checked for kinks, leaks and disconnections.

      Below are the safety precautions and policies to be followed for total intravenous anaesthesia among children and adults:

      When administering TIVA, a non-return valve must be used on any intravenous fluid line;
      When using equipment, it is essential that clinical staff know its limitations and uses;
      Sites of intravenous infusions should be visible so they may be monitored for disconnection, leaks or perivenous infusion into the subcutaneous tissues; and,
      Organisations must give preference to clearly labelled intravenous connectors and valves.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 30 - Risk stratification is done prior to a major cardiac surgery using cardiopulmonary exercise...

    Incorrect

    • Risk stratification is done prior to a major cardiac surgery using cardiopulmonary exercise testing. Given the following options, which one is most likely to have the highest risk for post-operative cardiac morbidity?

      Your Answer:

      Correct Answer: Anaerobic threshold (AT) of less than 11 mL/kg/minute

      Explanation:

      The ventilatory anaerobic threshold (VAT), formerly referred to as the anaerobic threshold, is an index used to estimate exercise capacity. During the initial (aerobic) phase of CPET, which lasts until 50–60% of Vo2max is reached, expired ventilation (VE) increases linearly with Vo2 and reflects aerobically produced CO2 in the muscles. Blood lactate levels do not change substantially during this phase, since muscle lactic acid production is minimal.

      During the latter half of exercise, anaerobic metabolism occurs because oxygen supply cannot keep up with the increasing metabolic requirements of exercising muscles. At this time, there is a significant increase in lactic acid production in the muscles and in the blood lactate concentration. The Vo2 at the onset of blood lactate accumulation is called the lactate threshold or the VAT. The VAT is also defined as the point at which minute ventilation increases disproportionately relative to Vo2, a response that is generally seen at 60–70% of Vo2max.

      The VAT is a useful measure as work below this level encompasses most daily living activities. The ability to achieve the VAT can help distinguish cardiac and non‐cardiac (pulmonary or musculoskeletal) causes of exercise limitation, since patients who fatigue before reaching VAT are likely to have a non‐cardiac problem.

      When VAT is detected, patients with PVo2 of ⩽10 ml/kg/min have a high event rate.

    • This question is part of the following fields:

      • Pathophysiology
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