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  • Question 1 - Left ventricular afterload is mostly calculated from systemic vascular resistance.

    Which...

    Correct

    • Left ventricular afterload is mostly calculated from systemic vascular resistance.

      Which one of the following factors has most impact on systemic vascular resistance?

      Your Answer: Small arterioles

      Explanation:

      Systemic vascular resistance (SVR), also known as total peripheral resistance (TPR), is the amount of force exerted on circulating blood by the vasculature of the body. Three factors determine the force: the length of the blood vessels in the body, the diameter of the vessels, and the viscosity of the blood within them. The most important factor that determines the systemic vascular resistance (SVR) is the tone of the small arterioles.

      These are otherwise known as resistance arterioles. Their diameter ranges between 100 and 450 µm. Smaller resistance vessels, less than 100 µm in diameter (pre-capillary arterioles), play a less significant role in determining SVR. They are subject to autoregulation.

      Any change in the viscosity of blood and therefore flow (such as due to a change in haematocrit) might also have a small effect on the measured vascular resistance.

      Changes of blood temperature can also affect blood rheology and therefore flow through resistance vessels.

      Systemic vascular resistance (SVR) is measured in dynes·s·cm-5

      It can be calculated from the following equation:

      SVR = (mean arterial pressure − mean right atrial pressure) × 80 cardiac output

    • This question is part of the following fields:

      • Physiology
      20.3
      Seconds
  • Question 2 - A participant of a metabolism study is to be fed only granulated sugar...

    Incorrect

    • A participant of a metabolism study is to be fed only granulated sugar and water for 48 hours. What would be his expected respiratory quotient at the end of the study?

      Your Answer: 0.9

      Correct Answer: 1

      Explanation:

      The respiratory quotient is the ratio of CO2 produced to O2 consumed while food is being metabolized:

      RQ = CO2 eliminated/O2 consumed

      Most energy sources are food containing carbon, hydrogen and oxygen. Examples include fat, carbohydrates, protein, and ethanol. The normal range of respiratory coefficients for organisms in metabolic balance usually ranges from 1.0-0.7.

      Granulated sugar is a refined carbohydrate with no significant fat, protein or ethanol content.

      The RQ for carbohydrates is = 1.0

      The RQ for the rest of the compounds are:

      Fats RQ = 0.7
      The chemical composition of fats differs from that of carbohydrates in that fats contain considerably fewer oxygen atoms in proportion to atoms of carbon and hydrogen.

      Protein RQ = 0.8
      Due to the complexity of various ways in which different amino acids can be metabolized, no single RQ can be assigned to the oxidation of protein in the diet; however, 0.8 is a frequently utilized estimate.

    • This question is part of the following fields:

      • Physiology
      95.5
      Seconds
  • Question 3 - Concerning forced alkaline diuresis, which of the following statements is true? ...

    Incorrect

    • Concerning forced alkaline diuresis, which of the following statements is true?

      Your Answer: Increases the excretion of unionised drugs in the urine

      Correct Answer: Can be used in a barbiturate overdose

      Explanation:

      In situations of poisoning or drug overdose with acid dugs like salicylates and barbiturates, forced alkaline diuresis may be used.

      With regards to overdose with alkaline drugs, forced acid diuresis is used.

      By changing the pH of the urine, the ionised portion of the drug stays in the urine, and this prevents its diffusion back into the blood. Charged molecules do not readily cross biological membranes.

      The process involves the infusion of specific fluids at a rate of about 500ml per hour. This requires monitoring of the central venous pressure, urine output, plasma electrolytes, especially potassium, and blood gas analysis.

      The fluid regimen recommended is:
      500ml of 1.26% sodium bicarbonate (not 200ml of 8.4%)
      500ml of 5% dextrose and
      500ml of 0.9% sodium chloride.

    • This question is part of the following fields:

      • Physiology
      131.9
      Seconds
  • Question 4 - A mercury barometer can be used to determine absolute pressure. A mercury manometer...

    Incorrect

    • A mercury barometer can be used to determine absolute pressure. A mercury manometer can be used to check blood pressure. The SI units of length(mm) are used to measure pressure.

      Why is pressure expressed in millimetres of mercury (mmHg)?

      Your Answer:

      Correct Answer: Pressure is directly proportional to length of the mercury column and is variable

      Explanation:

      A mercury barometer can be used to determine absolute pressure. A glass tube with one closed end serves as the barometer. The open end is inserted into a mercury-filled open vessel. The mercury in the container is pushed into the tube by atmospheric pressure exerted on its surface. Absolute pressure is the distance between the tube’s meniscus and the mercury surface.

      Pressure is defined as force in newtons per unit area (F) (A). 

      Mass of mercury = area (A) × density (ρ) × length (L)
      Pressure = ((A × ρ × L) × 9.8 m/s2)/A
      Pressure = ρ × L x 9.8
      Pressure is proportional to L

      The numerator and denominator of the above equation, area (A), cancel out. The constants are density and the gravitational acceleration value.

      The length is proportional to the applied pressure.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 5 - The Fick principle can be used to determine the blood flow to any...

    Incorrect

    • The Fick principle can be used to determine the blood flow to any organ of the body.

      At rest, which one of these organs has the highest blood flow (ml/min/100g)?

      Your Answer:

      Correct Answer: Thyroid gland

      Explanation:

      After the carotid body, the thyroid gland is the second most richly vascular organ in the body.

      The global blood flow to the thyroid gland can be measured using:
      1. Colour ultrasound sonography
      2. Quantitative perfusion maps using MRI of the thyroid gland using an arterial spin labelling (ASL) method.

      This table shows the blood flow to various organs of the body at rest:
      Organ Blood Flow(ml/minute/100g)
      Hepatoportal 58
      Kidney 420
      Brain 54
      Skin 13
      Skeletal muscle 2.7
      Heart 87
      Carotid body 2000
      Thyroid gland 560

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 6 - The biochemical assessment of malnutrition can be measured by the amount of plasma...

    Incorrect

    • The biochemical assessment of malnutrition can be measured by the amount of plasma proteins.

      In acute starvation, which of these plasma proteins is the most sensitive indicator?

      Your Answer:

      Correct Answer: Retinol binding globulin

      Explanation:

      The half life of Retinol binding protein (RBP) is 10-12 hours and therefore reflects more acute changes in protein metabolism than any of these proteins. Therefore it is not commonly used as a parameter for nutritional assessment.

      The half life of Transthyretin (thyroxine binding pre-albumin) is only one to two days and so levels are less sensitive and this protein is not an albumin precursor. 15 mg/dL represents early malnutrition and a need for nutritional support.

      Albumin levels have been frequently as a marker of nutrition but this is not a very sensitive marker. It’s half life more than 30 days and significant change takes some time to be noticed. Also, synthesis of albumin is decreased with the onset of the stress response after burns. Unrelated to nutritional status, the synthesis of acute phase proteins increases and that of albumin decreases.

      A more accurate indicator of protein stores is transferrin. It’s response to acute changes in protein status is much faster. The half life of serum transferrin is shorter (8-10 days) and there are smaller body stores than albumin. A low serum transferrin level is below 200 mg/dL and below 100 mg/dL is considered severe. Serum transferrin levels can also affect serum transferrin level.

      Fibronectin is used a nutritional marker but levels decrease after seven days of starvation. It is a glycoprotein which plays a role in enhancing the phagocytosis of foreign particles.

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 7 - Which statement is true when describing carbonic anhydrase? ...

    Incorrect

    • Which statement is true when describing carbonic anhydrase?

      Your Answer:

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

      Explanation:

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

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

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

      Spironolactone is a potassium diuretic and is an aldosterone antagonist.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 8 - The following statement is true with regards to the Nernst equation: ...

    Incorrect

    • The following statement is true with regards to the Nernst equation:

      Your Answer:

      Correct Answer: It is used to calculate the potential difference across a membrane when the individual ions are in equilibrium

      Explanation:

      The Nernst equation is used to calculate the membrane potential at which the ions are in equilibrium across the cell membrane.

      The normal resting membrane potential is -70 mV (not + 70 mV).

      The equation is:
      E = RT/FZ ln {[X]o
      /[X]i}

      Where:
      E is the equilibrium potential
      R is the universal gas constant
      T is the absolute temperature
      F is the Faraday constant
      Z is the valency of the ion
      [X]o is the extracellular concentration of ion X
      [X]i is the intracellular concentration of ion X.

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 9 - In a normal healthy adult breathing 100 percent oxygen, which of the following...

    Incorrect

    • In a normal healthy adult breathing 100 percent oxygen, which of the following is the most likely cause of an alveolar-arterial (A-a) oxygen difference of 30 kPa?

      Your Answer:

      Correct Answer: Atelectasis

      Explanation:

      The ‘ideal’ alveolar PO2 minus arterial PO2 is the alveolar-arterial (A-a) oxygen difference.

      The ‘ideal’ alveolar PO2 is derived from the alveolar air equation and is the PO2 that the lung would have if there was no ventilation-perfusion (V/Q) inequality and it was exchanging gas at the same respiratory exchange ratio as real lung.

      The amount of oxygen in the blood is measured directly in the arteries.

      The A-a oxygen difference (or gradient) is a useful measure of shunt and V/Q mismatch, and it is less than 2 kPa in normal adults breathing air (15 mmHg). Because the shunt component is not corrected, the A-a difference increases when breathing 100 percent oxygen, and it can be up to 15 kPa (115 mmHg).

      An abnormally low or abnormally high V/Q ratio within the lung can cause an increased A-a difference, though the former is more common. Atelectasis, which results in a low V/Q ratio, is the most likely cause of an A-a difference in a healthy adult breathing 100 percent oxygen.

      Hypoventilation may cause an increase in alveolar (and thus arterial) CO2, lowering alveolar PO2 according to the alveolar air equation.

      The alveolar PO2 is also reduced at high altitude.

      Healthy people are unlikely to have a right-to-left shunt or an oxygen transport diffusion defect.

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 10 - Of the following, which of these oxygen carrying molecules causes the greatest shift...

    Incorrect

    • Of the following, which of these oxygen carrying molecules causes the greatest shift of the oxygen-dissociation curve to the left?

      Your Answer:

      Correct Answer: Myoglobin (Mb)

      Explanation:

      Myoglobin is a haemoglobin-like, iron-containing pigment that is found in muscle fibres. It has a high affinity for oxygen and it consists of a single alpha polypeptide chain. It binds only one oxygen molecule, unlike haemoglobin, which binds 4 oxygen molecules.

      The myoglobin ODC is a rectangular hyperbola. There is a very low P50 0.37 kPa (2.75 mmHg). This means that it needs a lower P50 to facilitate oxygen offloading from haemoglobin. It is low enough to be able to offload oxygen onto myoglobin where it is stored. Myoglobin releases its oxygen at the very low PO2 values found inside the mitochondria.

      P50 is defined as the affinity of haemoglobin for oxygen: It is the PO2 at which the haemoglobin becomes 50% saturated with oxygen. Normally, the P50 of adult haemoglobin is 3.47 kPa(26 mmHg).

      Foetal haemoglobin has 2 ? and 2 ?chains. The ODC is left shifted – this means that P50 lies between 2.34-2.67 kPa [18-20 mmHg]) compared with the adult curve and it has a higher affinity for oxygen. Foetal haemoglobin has no ? chains so this means that there is less binding of 2.3 diphosphoglycerate (2,3 DPG).

      Carbon monoxide binds to haemoglobin with an affinity more than 200-fold higher than that of oxygen. This therefore decreases the amount of haemoglobin that is available for oxygen transport. Carbon monoxide binding also increases the affinity of haemoglobin for oxygen, which shifts the oxygen-haemoglobin dissociation curve to the left and thus impedes oxygen unloading in the tissues.

      In sickle cell disease, (HbSS) has a P50 of 4.53 kPa(34 mmHg).

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 11 - The following is normally higher in concentration extracellularly than intracellularly ...

    Incorrect

    • The following is normally higher in concentration extracellularly than intracellularly

      Your Answer:

      Correct Answer: Sodium

      Explanation:

      The ions found in higher concentrations intracellularly than outside the cells are:

      ATP
      AMP
      Potassium
      Phosphate, and
      Magnesium Adenosine diphosphate (ADP)

      Sodium is a primarily extracellular ion.

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 12 - Which of the following statements is true with regards to the Krebs' cycle...

    Incorrect

    • Which of the following statements is true with regards to the Krebs' cycle (also known as the tricarboxylic acid cycle or citric acid cycle)?

      Your Answer:

      Correct Answer: Alpha-ketoglutarate is a five carbon molecule

      Explanation:

      Krebs’ cycle (tricarboxylic acid cycle or citric acid cycle) is a sequence of reactions in which acetyl coenzyme A (acetyl-CoA) is metabolised and this results in carbon dioxide and hydrogen atoms production.

      This series of reactions occur in the mitochondria of eukaryotic cells, not the cytoplasm. The cycle requires oxygen and so, cannot function under anaerobic conditions.

      It is the common pathway for carbohydrate, fat and some amino acids oxidation and is required for high energy phosphate bond formation in adenosine triphosphate (ATP).

      When pyruvate enters the mitochondria, it is converted into acetyl-CoA. This represents the formation of a 2 carbon molecule from a 3 carbon molecule. There is loss of one CO2 but formation of one NADH molecule. Acetyl-CoA is condensed with oxaloacetate, the anion of a 4 carbon acid, to form citrate which is a 6 carbon molecule.

      Citrate is then converted into isocitrate, alpha-ketoglutarate, succinyl-CoA, succinate, fumarate, malate and finally oxaloacetate.

      The only 5 carbon molecule in the cycle is alpha-ketoglutarate.

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 13 - All of the following statements about cerebrospinal fluid are incorrect except: ...

    Incorrect

    • All of the following statements about cerebrospinal fluid are incorrect except:

      Your Answer:

      Correct Answer: Has a glucose concentration 2/3 that of the plasma glucose

      Explanation:

      The pH of CSF is 7.31 which is lower than plasma.

      Compared to plasma, it has a lower concentration of potassium, calcium, and protein and a higher concentration of sodium, chloride, bicarbonate and magnesium.

      CSF usually has no cells present but if white cells are present, there should be no more than 4/ml.

      The pressure of CSF should be less than 20 cm of water.

      The concentration of glucose is approximately two-thirds of that of plasma, and it has a concentration of approximately 3.3-4 mmol/L.

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 14 - The renal glomerulus is able to filter 180 litres of blood per day,...

    Incorrect

    • The renal glomerulus is able to filter 180 litres of blood per day, as determined by the starling forces present in the glomerulus. Ninety-nine percent of which is reabsorbed thereafter.

      Water is reabsorbed in the highest proportion in which segment of the nephron?

      Your Answer:

      Correct Answer: Proximal convoluted tubule

      Explanation:

      Sixty-seven percent of filtered water is reabsorbed in the proximal tubule. The driving force for water reabsorption is a transtubular osmotic gradient established by reabsorption of solutes (e.g., NaCl, Na+-glucose).

      Henle’s loop reabsorbs approximately 25% of filtered NaCl and 15% of filtered water. The thin ascending limb reabsorbs NaCl by a passive mechanism, and is impermeable to water. Reabsorption of water, but not NaCl, in the descending thin limb increases the concentration of NaCl in the tubule fluid entering the ascending thin limb. As the NaCl-rich fluid moves toward the cortex, NaCl diffuses out of the tubule lumen across the ascending thin limb and into the medullary interstitial fluid, down a concentration gradient as directed from the tubule fluid to the interstitium. This mechanism is known as the counter current multiplier.

      The distal tubule and collecting duct reabsorb approximately 8% of filtered NaCl, secrete variable amounts of K+ and H+, and reabsorb a variable amount of water (approximately 8%-17%).

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 15 - A patient was brought to the emergency room after passing black tarry stools....

    Incorrect

    • A patient was brought to the emergency room after passing black tarry stools. The initial diagnosis was upper gastrointestinal bleeding. The patient was placed on temporary nil per os (NPO) for the next 24 hours, his weight was 110 kg, and the required volume of intravenous fluid for the him was 3 litres. His electrolytes and other biochemistry studies were normal.

      If you were to choose the intravenous fluid regimen that would closely mimic his basic electrolyte and caloric requirements, which one would be the best answer?

      Your Answer:

      Correct Answer: 3000 mL 0.45% N. saline with 5% dextrose, each bag with 40 mmol of potassium

      Explanation:

      The patient in the case has a fluid volume requirement of 30 mL/kg/day. His basic electrolyte requirement per day is:

      Sodium at 2 mmol/kg/day x 110 = 220 mmol/day
      Potassium at 1 mmol/kg/day x 110 = 110 mmol/day

      His energy requirement per day is:

      35 kcal/kg/day x 110 kg = 3850 kcal/day

      One gram of glucose in fluid can provide approximately 4 kilocalories.

      The following are the electrolyte components of the different intravenous fluids:

      Fluid Na (mmol/L) K (mmol/L)
      0.9% Normal saline (NSS) 154 0
      0.45% NSS + 5% dextrose 77 0
      0.18% NSS + 4% dextrose 30 0
      Hartmann’s 131 5
      5% dextrose 0 0

      1000 mL of 5% dextrose has 50 g of glucose

      Option B is inadequate for his sodium and caloric requirements (30 mmol of Na+ and 560 kcal). It is adequate for his K+ requirement (120 mmol of K+).

      Option C is in excess of his Na+ requirement (462 mmol of Na+). Moreover, it does not provide any K+ replacement.

      Option D is inadequate for his caloric requirement (600 kcal) and K+ requirement (60 mmol of K+). Moreover it does not provide any Na+ replacement.

      Option E is in excess of his Na+ requirement (393 mmol of Na+), and is inadequate for his potassium requirement (15 mmol of K+)

      Option A has adequate amounts for his Na+ (231 mmol of Na+) and K+ (120 mmol of K+) requirements. It is inadequate for his caloric requirement (600 kcal).

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 16 - One of the non-pharmacologic management of COPD is smoking cessation. Given a case...

    Incorrect

    • One of the non-pharmacologic management of COPD is smoking cessation. Given a case of a 60-year old patient with history of smoking for 30 years and a FEV1 of 70%, what would be the most probable five-year course of his FEV1 if he ceases to smoke?

      Your Answer:

      Correct Answer: The FEV1 will decrease at the same rate as a non-smoker

      Explanation:

      For this patient, his forced expiratory volume in 1 second (FEV1) will decrease at the same rate as a non-smoker.

      There is a notable, but slow, decline in FEV1 when an individual reaches the age of 26. An average reduction of 30 mls every year in non-smokers, while a more significant reduction of 50-70 mls is observed in approximately 20% of smokers.

      Considering the age of the patient, individuals who begin smoking cessation by the age of 60 are far less likely to achieve normal FEV1 levels, even in the next five years. It is expected that their FEV1 will be approximately 14% less than their peers of the same age.

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 17 - The most abundant intracellular ion is? ...

    Incorrect

    • The most abundant intracellular ion is?

      Your Answer:

      Correct Answer: Phosphate

      Explanation:

      Phosphate is the principal anion of the intracellular fluid, most of which is bound to either lipids or proteins. They dissociate or associate with different compounds, depending on the enzymatic reaction, thus forming a constantly shifting pool.

      Calcium and magnesium are also present intracellularly, however in lesser amounts than phosphate.

      Sodium is the most abundant extracellular cation, and Chloride and is the most abundant extracellular anion.

    • This question is part of the following fields:

      • Physiology
      0
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  • Question 18 - A 27-year-old woman is admitted to the emergency room with an ectopic pregnancy...

    Incorrect

    • A 27-year-old woman is admitted to the emergency room with an ectopic pregnancy that has ruptured.

      The following is a description of the clinical examination:

      Anxious
      Capillary refill time of 3 seconds
      Cool peripheries
      Pulse 120 beats per minute
      Blood pressure 120/95 mmHg
      Respiratory rate 22 breaths per minute.

      Which of the following is the most likely explanation for these clinical findings?

      Your Answer:

      Correct Answer: Reduction in blood volume of 15-30%

      Explanation:

      The following is the Advanced Trauma Life Support (ATLS) classification of haemorrhagic shock:

      Class I haemorrhage:
      It has blood loss up to 15%. There is very less tachycardia, and no changes in blood pressure, RR or pulse pressure. Usually, fluid replacement is not required.

      Class II haemorrhage:
      It has 15-30% blood loss, equivalent to 750 – 1500 ml. There is tachycardia, tachypnoea and a decrease in pulse pressure. Patient may be frightened, hostile and anxious. It can be stabilised by crystalloid and blood transfusion.

      Class III haemorrhage:
      There is 30-40% blood loss. It portrays inadequate perfusion, marked tachycardia, tachypnoea, altered mental state and fall in systolic pressure. It requires blood transfusion.

      Class IV haemorrhage:
      There is > 40% blood volume loss. It is a preterminal event, and the patient will die in minutes. It portrays tachycardia, significant depression in systolic pressure and pulse pressure, altered mental state, and cold clammy skin. There is need for rapid transfusion and surgical intervention.

    • This question is part of the following fields:

      • Physiology
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  • Question 19 - Useful diagnostic information can be obtained from measuring the osmolality of biological fluids....

    Incorrect

    • Useful diagnostic information can be obtained from measuring the osmolality of biological fluids.

      Of the following physical principles, which is the most accurate and reliable method of measuring osmolality?

      Your Answer:

      Correct Answer: Depression of freezing point

      Explanation:

      Colligative properties are properties of solutions that depend on the number of dissolved particles in solution. They do not depend on the identities of the solutes.

      All of the above have colligative properties with the exception of depression of melting point.

      The osmolality from the concentration of a substance in a solution is measured by an osmometer. The freezing point of a solution can determines concentration of a solution and this can be measured by using a freezing point osmometer. This is applicable as depression of freezing point is directly correlated to concentration.

      Vapour pressure osmometers, which measure vapour pressure, may miss certain volatiles such as CO2, ammonia and alcohol that are in the solution

      The use of a freezing point osmometer provides the most accurate and reliable results for the majority of applications.

      Colligative properties does not include melting point depression . Mixtures of substances in which the liquid phase components are insoluble, display a melting point depression and a melting range or interval instead of a fixed melting point.

      The magnitude of the melting point depression depends on the mixture composition.

      The melting point depression is used to determine the purity and identity of compounds. EMLA (eutectic mixture of local anaesthetics) cream is a mixture of lidocaine and prilocaine and is used as a topical local anaesthetic. The melting point of the combined drugs is lower than that individually and is below room temperature (18°C).

    • This question is part of the following fields:

      • Physiology
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  • Question 20 - The solutions that contains the most sodium is? ...

    Incorrect

    • The solutions that contains the most sodium is?

      Your Answer:

      Correct Answer: 3500 mL 0.9% N saline

      Explanation:

      Sodium concentration for different fluids
      3% N saline 513 mmol/L
      5% N saline 856 mmol/L
      0.9% N saline 154 mmol/L
      Hartmann’s solution 131 mmol/L
      0.45% N saline with 5% glucose 77 mmol/L

      This means that:

      500 mL 5% N saline contains 428 mmol of sodium
      1000 mL 3% N saline contains 513 mmol of sodium
      3500 mL 0.9% N saline contains 539 mmol of sodium
      4000 mL Hartmann’s contains 524 mmol of sodium
      6000 mL 0.45% N saline with 5% glucose contains 462 mmol of sodium.

    • This question is part of the following fields:

      • Physiology
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  • Question 21 - Which of the following statements is true about fluid balance? ...

    Incorrect

    • Which of the following statements is true about fluid balance?

      Your Answer:

      Correct Answer: After intravenous administration of crystalloids, the distribution of these fluids throughout the body depends on its osmotic activity

      Explanation:

      When there is capillary leakage as seen in dependent oedema or ascites, oncotic pressure becomes a problem.

      The intracellular sodium concentration is very sensitive to the extracellular sodium concentrations. When there is an imbalance, osmosis occurs resulting in shifts in water between the two compartments.

      The microvascular endothelium relies upon osmosis and other processes as it is not freely permeable to water.

    • This question is part of the following fields:

      • Physiology
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  • Question 22 - Metabolization of many drugs used in anaesthesia involves the cytochrome P450 (CYP) isoenzymes.

    The...

    Incorrect

    • Metabolization of many drugs used in anaesthesia involves the cytochrome P450 (CYP) isoenzymes.

      The CYP enzyme most likely to be subject to genetic variability and thus cause adverse drug reactions is which of these?

      Your Answer:

      Correct Answer: CYP2D6

      Explanation:

      Approximately 25% of phase-1 drug reactions is made responsible by CYP2D6.

      As much as a 1,000-fold difference in the ability to metabolise drugs by CYP2D6 can happen between phenotypes, and this may result in adverse drug reactions (ADRs).

      The metabolism of antiemetics, beta-blockers, codeine, tramadol, oxycodone, hydrocodone, tamoxifen, antidepressants, neuroleptics, and antiarrhythmics is also as a result of CYP2D6.

      Patients who take drugs that are metabolised by CYP2D6 but have poor CYP2D6 metabolism are more likely to have ADRs. People with ultra-rapid CYP2D6 metabolism may have a decreased drug effect due to low plasma concentrations of these drugs.

      All the other CYP enzymes are subject to genetic polymorphism. Variants are less likely to lead to adverse drug reactions.

    • This question is part of the following fields:

      • Physiology
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  • Question 23 - Which of the following would most likely explain a failed post-operative analgesia via...

    Incorrect

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

      Your Answer:

      Correct Answer: pKA

      Explanation:

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

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

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

    • This question is part of the following fields:

      • Physiology
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  • Question 24 - A common renal adverse effect of non-steroidal anti-inflammatory drugs is? ...

    Incorrect

    • A common renal adverse effect of non-steroidal anti-inflammatory drugs is?

      Your Answer:

      Correct Answer: Haemodynamic renal insufficiency

      Explanation:

      Prostaglandins do not play a major role in regulating RBF in healthy resting individuals. However, during pathophysiological conditions such as haemorrhage and reduced extracellular fluid volume (ECVF), prostaglandins (PGI2, PGE1, and PGE2) are produced locally within the kidneys and serve to increase RBF without changing GFR. Prostaglandins increase RBF by dampening the vasoconstrictor effects of both sympathetic activation and angiotensin II. These effects are important because they prevent severe and potentially harmful vasoconstriction and renal ischemia. Synthesis of prostaglandins is stimulated by ECVF depletion and stress (e.g. surgery, anaesthesia), angiotensin II, and sympathetic nerves.

      Non-steroidal anti-inflammatory drugs (NSAIDs), such as ibuprofen and naproxen, potently inhibit prostaglandin synthesis. Thus administration of these drugs during renal ischemia and hemorrhagic shock is contraindicated because, by blocking the production of prostaglandins, they decrease RBF and increase renal ischemia. Prostaglandins also play an increasingly important role in maintaining RBF and GFR as individuals age. Accordingly, NSAIDs can significantly reduce RBF and GFR in the elderly.

    • This question is part of the following fields:

      • Physiology
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  • Question 25 - Which of the following statements is about the measurement of glomerular filtration rate...

    Incorrect

    • Which of the following statements is about the measurement of glomerular filtration rate (GFR) is correct?

      Your Answer:

      Correct Answer: The result matches clearance of the indicator if it is renally inert

      Explanation:

      The measurements of GFR are done using renally inert indicators like inulin, where passive rate of filtration at the glomerulus = rate of excretion. Normal value is about 180 litres per day.

      GFR is altered by renal blood flow but blood flow does not need to be measured.

      The reabsorption of Na leads to a low excretion rate and low urine concentration and therefore its use as an indicator would lead to an erroneously LOW GFR.

      If there is tubular secretion of any solute, the clearance value will be higher than that of inulin. This will be either due to tubular reabsorption or the solute not being freely filtered at the glomerulus.

    • This question is part of the following fields:

      • Physiology
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  • Question 26 - An intravenous infusion is started with a 500 mL bag of 0.18 percent...

    Incorrect

    • An intravenous infusion is started with a 500 mL bag of 0.18 percent N. saline and 4% dextrose.

      Which of the following best describes its make-up?

      Your Answer:

      Correct Answer: Osmolarity 284 mOsmol/L, sodium 15 mequivalents and glucose 20 g

      Explanation:

      30 mmol Na+ and 30 mmol Cl- are found in 1 litre of 0.18 percent N. saline with 4% dextrose. Percent (percent) refers to the number of grammes of a compound per 100 mL, so a litre of 4 percent dextrose solution contains 40 grammes.

      As a result, a 500 mL bag of 1/5th N. saline and 4% dextrose contains approximately 15 mequivalents of sodium and 20 g of glucose. It is hypotonic due to its osmolarity of 284.

      Because of the risk of hyponatraemia, it is no longer considered the crystalloid of choice for fluid maintenance in children.

    • This question is part of the following fields:

      • Physiology
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  • Question 27 - Which of the following statements is true with regards to 2,3-diphosphoglycerate (2,3-DPG)? ...

    Incorrect

    • Which of the following statements is true with regards to 2,3-diphosphoglycerate (2,3-DPG)?

      Your Answer:

      Correct Answer: Production is increased in heart failure

      Explanation:

      During glycolysis, 2,3-diphosphoglycerate (2,3-DPG) is
      created in erythrocytes by the Rapoport-Luebering shunt.

      The production of 2,3-DPG increases for several conditions
      in the presence of decreased peripheral tissue O2 availability.
      Some of these conditions include hypoxaemia, chronic lung
      disease anaemia, and congestive heart failure. Thus,
      2,3-DPG production is likely an important adaptive mechanism.

      High levels of 2,3-DPG cause a shift of the curve to the right.
      Low levels of 2,3-DPG cause a shift of the curve to the left,
      as seen in states such as septic shock and hypophosphatemia.

    • This question is part of the following fields:

      • Physiology
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  • Question 28 - A healthy 27-year old male who weighs 70kg has appendicitis. He is currently...

    Incorrect

    • A healthy 27-year old male who weighs 70kg has appendicitis. He is currently in the operating room and is being positioned to have a rapid sequence induction.

      Prior to preoxygenation, the compartment likely to have the best oxygen reserve is:

      Your Answer:

      Correct Answer: Red blood cells

      Explanation:

      The following table shows the compartments and their relative oxygen reserve:
      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

      Oxygen reserves in the body, with room air and after oxygenation.

      FAO2-alveolar fraction of oxygen rises to 95% after administration of 100% oxygen (CO2 = 5%)
      FRC- Functional residual capacity – (the most important store of oxygen in the body) – 2,500-3,000 mL in medium sized adults
      PaO2-partial pressure of oxygen dissolved in arterial blood (80 mmHg breathing room air and 500 mmHg breathing 100% oxygen)
      DF -dissolved form (0.3%)
      PV-plasma volume (3L)
      TG-total globular volume (5L)
      Hb-haemoglobin concentration
      SaO2-arterial oxygen concentration (98% breathing air and 100% when preoxygenated)

    • This question is part of the following fields:

      • Physiology
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  • Question 29 - In an experimental study, a healthy subject was given one litre of 5%...

    Incorrect

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

      Your Answer:

      Correct Answer: Inhibition of arginine vasopressin (AVP) secretion

      Explanation:

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

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

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

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

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

    • This question is part of the following fields:

      • Physiology
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  • Question 30 - A 72-year old farmer is hospitalized with acute respiratory failure and autonomic dysfunction....

    Incorrect

    • A 72-year old farmer is hospitalized with acute respiratory failure and autonomic dysfunction. Suspected organophosphate poisoning.

      Which one is the best mechanism for acute toxicity caused by organophosphates?

      Your Answer:

      Correct Answer: Inhibition of acetylcholinesterase

      Explanation:

      The toxicity of organophosphorus (OP) nerve agents is manifested through irreversible inhibition of acetylcholinesterase (AChE) at the cholinergic synapses, which stops nerve signal transmission, resulting in a cholinergic crisis and eventually death of the poisoned person. Oxime compounds used in nerve agent antidote regimen reactivate nerve agent-inhibited AChE and halt the development of this cholinergic crisis.

    • This question is part of the following fields:

      • Physiology
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Physiology (1/3) 33%
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