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  • Question 1 - One of the non-pharmacologic management of COPD is smoking cessation. Given a case...

    Correct

    • 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: 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
      138
      Seconds
  • Question 2 - 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: Fibronectin

      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
      13.8
      Seconds
  • Question 3 - 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: Thyrotoxicosis

      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
      27.3
      Seconds
  • Question 4 - Given the following values:

    Expired tidal volume = 800 ml
    Plateau pressure = 50 cmH2O
    PEEP...

    Incorrect

    • Given the following values:

      Expired tidal volume = 800 ml
      Plateau pressure = 50 cmH2O
      PEEP = 10 cmH2O

      Compute for the static pulmonary compliance.

      Your Answer:

      Correct Answer: 20 ml/cmH2O

      Explanation:

      Compliance of the respiratory system describes the expandability of the lungs and chest wall. There are two types of compliance: dynamic and static.

      Dynamic compliance describes the compliance measured during breathing, which involves a combination of lung compliance and airway resistance. Defined as the change in lung volume per unit change in pressure in the presence of flow.

      Static compliance describes pulmonary compliance when there is no airflow, like an inspiratory pause. Defined as the change in lung volume per unit change in pressure in the absence of flow.

      For example, if a person was to fill the lung with pressure and then not move it, the pressure would eventually decrease; this is the static compliance measurement. Dynamic compliance is measured by dividing the tidal volume, the average volume of air in one breath cycle, by the difference between the pressure of the lungs at full inspiration and full expiration. Static compliance is always a higher value than dynamic

      Static compliance can be computed using the formula:

      Cstat = Tidal volume/Plateau pressure – PEEP

      Substituting the values given,

      Cstat = 800/50-10
      Cstat = 20 ml/cmH2O

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 5 - Which one of the following factor affects the minimal alveolar concentration (MAC)? ...

    Incorrect

    • Which one of the following factor affects the minimal alveolar concentration (MAC)?

      Your Answer:

      Correct Answer: Hypoxaemia

      Explanation:

      The minimal alveolar concentration (MAC) is the concentration of an inhalation anaesthetic agent in the lung alveoli required to stop a response to the surgical stimulus in 50% of the patient.

      Following factors don’t affect the MAC of the inhaled anaesthetic agents:

      Gender, acidosis, alkalosis, hypothyroidism, hyperthyroidism, body weight, serum potassium level, and the duration of the anaesthesia.

      MAC increase in children, elevated temperature, high metabolic rate, sympathetic increase and chronic alcoholism.

      MAC decrease in low temperature, low oxygen level, old age, hypotension (<40 mmHg), depressant drugs e.g. opioids and low level of catecholamines; alpha methyl dopa. Carbon dioxide O2 at the pressure > 120mmHg is being used in anesthetic-Hinkman as an additive effect to decrease MAC, however, increase concentration of CO2 activates the sympathetic system resulting the MAC increases.

    • This question is part of the following fields:

      • Physiology
      0
      Seconds
  • Question 6 - 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
      Seconds
  • Question 7 - A 30-year old female athlete was brought to the Emergency Room for complaints...

    Incorrect

    • 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:

      Correct 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
      0
      Seconds
  • Question 8 - 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 9 - 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:

      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
      0
      Seconds
  • Question 10 - 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
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