-
Question 1
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: 321-333
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
-
-
Question 2
Incorrect
-
Intracellular effectors are activated by receptors on the cell surface. These receptors receive signals that are relayed by second messenger systems. In the human body, which second messenger is most abundant?
Your Answer: Cyclic AMP (cAMP)
Correct Answer: Calcium ions
Explanation:Second messengers relay signals to target molecules in the cytoplasm or nucleus when an agonist interacts with a receptor on the cell surface. They also amplify the strength of the signal. The most ubiquitous and abundant second messenger is calcium and it regulates multiple cellular functions in the body.
These include:
Muscle contraction (skeletal, smooth and cardiac)
Exocytosis (neurotransmitter release at synapses and insulin secretion)
Apoptosis
Cell adhesion to the extracellular matrix
Lymphocyte activation
Biochemical changes mediated by protein kinase C.cAMP is either inhibited or stimulated by G proteins.
The receptors in the body that stimulate G proteins and increase cAMP include:
Beta (β1, β2, and β3)
Dopamine (D1 and D5)
Histamine (H2)
Glucagon
Vasopressin (V2).The second messenger for the action of nitric oxide (NO) and atrial natriuretic peptide (ANP) is cGMP.
The second messengers for angiotensin and thyroid stimulating hormone are inositol triphosphate (IP3) and diacylglycerol (DAG).
-
This question is part of the following fields:
- Physiology
-
-
Question 3
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: Transferrin
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
-
-
Question 4
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: Brain
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
-
-
Question 5
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: Pressure is directly proportional to length of the mercury column and is the only constant
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 (�) × length (L)
Pressure = ((A × � × L) × 9.8 m/s2)/A
Pressure = � × L x 9.8
Pressure is proportional to LThe 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
-
-
Question 6
Incorrect
-
The following is true about the extracellular fluid (ECF) in a normal adult woman weighing 60 kg.
Your Answer: Forms a greater proportion of the total body weight in the obese than in the lean woman
Correct Answer: Has a total volume of about 12 litres
Explanation:Total body water (TBW) is about 50% to 70% in adults depending on how much fat is present. ECF is relatively contracted in an obese person.
The simple rule is 60-40-20. (60% of weight = total body water, 40% of body weight is ICF and 20% is ECF)
For this woman, the total body water is 36 litres (0.6 × 60). ECF is 12 litres (1/3 of TBW) and 24 litres (2/3 of TBW) is intracellular fluid .
Sodium concentration is approximately 135-145 mmol/L in the ECF.
The ECF is made up of both intravascular and extravascular fluid and plasma proteins is found in both.
-
This question is part of the following fields:
- Physiology
-
-
Question 7
Correct
-
Which of the following, at a given PaO2, increases the oxygen content of arterial blood?
Your Answer: A reduced erythrocyte 2,3-diphosphoglycerate level
Explanation:The oxygen content of arterial blood can be calculated by the following equation:
(10 x haemoglobin x SaO2 x 1.34) + (PaO2 x 0.0225).
This is the sum of the oxygen bound to haemoglobin and the oxygen dissolved in the plasma.Oxygen content x cardiac output = The amount of oxygen delivered to the tissues in unit time which is known as the oxygen flux.
Any factor that increases the metabolic demand will encourage oxygen offloading from the haemoglobin in the tissues and this causes the oxygen dissociation curve (ODC) to shift to the right. This subsequently reduced the oxygen content of arterial blood.
Conditions like fever, metabolic or respiratory acidosis lowers the oxygen content and shifts the ODC to the right.
A low level of 2,3 diphosphoglycerate (2,3-DPG) is usually related to an increased oxygen content as there is less offloading, and so the ODC is shifted to the left.So for a given PaO2, a high blood oxygen content is related to any factors that can shift the ODC to the left and not to the right.
A low haematocrit usually means that there is a decreased haemoglobin concentration, and therefore is associated with decreased oxygen binding to haemoglobin.
-
This question is part of the following fields:
- Physiology
-
-
Question 8
Correct
-
Given the following values: Expired tidal volume = 800 ml, Plateau pressure = 50 cmH2O, PEEP = 10 cmH2O. Compute for the static pulmonary compliance.
Your 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
-
-
Question 9
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/dayHis 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 01000 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
-
-
Question 10
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
-
-
Question 11
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
-
-
Question 12
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
-
-
Question 13
Incorrect
-
Of the following, which option best describes the muscle type that has the fastest twitch response to stimulation?
Your Answer:
Correct Answer: Type IIb skeletal muscle
Explanation:Human skeletal muscle is composed of a heterogeneous collection of muscle fibre types which differ histologically, biochemically and physiologically.
It can be biochemically classified into 2 groups. This is based on muscle fibre myosin ATPase histochemistry. These are:
Type 1 (slow twitch): Muscle fibres depend upon aerobic glycolytic metabolism and aerobic oxidative metabolism. They are rich in mitochondria, have a good blood supply, rich in myoglobin and are resistant to fatigue.
Type II (fast twitch): Muscle fibres are sub-divided into:
Type IIa – relies on aerobic/oxidative metabolism
Type IIb – relies on anaerobic/glycolytic metabolism.Fast twitch muscle fibres produce short bursts of power but are more easily fatigued.
Cardiac and smooth muscle twitches are relatively slow compared with skeletal muscle.
-
This question is part of the following fields:
- Physiology
-
-
Question 14
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 160Oxygen 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
-
-
Question 15
Incorrect
-
Which of the following statement is true regarding the paediatric airway?
Your Answer:
Correct Answer: The larynx is more anterior than in an adult
Explanation:In the neonatal stage, the tongue is usually large and comes to the normal size at the age of 1 year. The vocal cords lie inverse C4 and as it reaches the grown-up position inverse C5/6 by the age of 4 (not 1 year).
Due to the immature cricoid cartilage, the larynx lies more anterior in newborn children. That’s why the cricoid ring is the narrowest part of the paediatric respiratory tract, while in the adults the tightest portion of the respiratory route is vocal cords. The epiglottis is generally expansive and slants at a point of 45 degrees to the laryngeal opening.
The carina is the ridge of the cartilage in the trachea at the level of T2 in newborn (T4 in adults), that separates the openings of right and left main bronchi.
Neonates have a comparatively low number of alveoli and then this number gradually increases to a most extreme by the age of 8 (not 3 years).
Neonates are obligatory nose breathers and any hindrance can cause respiratory issues (e.g., choanal atresia).
-
This question is part of the following fields:
- Physiology
-
-
Question 16
Incorrect
-
Regarding the plateau phase of the cardiac potential, which electrolyte is the main determinant?
Your Answer:
Correct Answer: Ca2+
Explanation:The cardiac action potential has several phases which have different mechanisms of action as seen below:
Phase 0: Rapid depolarisation – caused by a rapid sodium influx.
These channels automatically deactivate after a few msPhase 1: caused by early repolarisation and an efflux of potassium.
Phase 2: Plateau – caused by a slow influx of calcium.
Phase 3 – Final repolarisation – caused by an efflux of potassium.
Phase 4 – Restoration of ionic concentrations – The resting potential is restored by Na+/K+ATPase.
There is slow entry of Na+into the cell which decreases the potential difference until the threshold potential is reached. This then triggers a new action potentialOf note, cardiac muscle remains contracted 10-15 times longer than skeletal muscle.
Different sites have different conduction velocities:
1. Atrial conduction – Spreads along ordinary atrial myocardial fibres at 1 m/sec2. AV node conduction – 0.05 m/sec
3. Ventricular conduction – Purkinje fibres are of large diameter and achieve velocities of 2-4 m/sec, the fastest conduction in the heart. This allows a rapid and coordinated contraction of the ventricles
-
This question is part of the following fields:
- Physiology
-
-
Question 17
Incorrect
-
A 61-year-old woman with myasthenia gravis is admitted to the ER with type II respiratory failure. There is a suspicion of myasthenic crisis. She is in a semiconscious state. Her blood pressure is 160/90 mmHg, pulse is 110 beats per minute, temperature is 37°C, and oxygen saturation is 84 percent. With a PaCO2 of 75 mmHg (10 kPa) breathing air, blood gas analysis confirms she is hypoventilating. Which of the following values is the most accurate representation of her alveolar oxygen tension (PAO2)?
Your Answer:
Correct Answer: 7.3
Explanation:The following is the alveolar gas equation:
PAO2 = PiO2 ˆ’ PaCO2/R
Where:
PAO2 is the partial pressure of oxygen in the alveoli.
PiO2 is the partial pressure of oxygen inhaled.
PaCO2 stands for partial pressure of carbon dioxide in the arteries.
The amount of carbon dioxide produced (200 mL/minute) divided by the amount of oxygen consumed (250 mL/minute) equals R = respiratory quotient. With a normal diet, the value is 0.8.By subtracting the partial pressure exerted by water vapour at body temperature, the PiO2 can be calculated:
PiO2 = 0.21 × (100 kPa ˆ’ 6.3 kPa)
PiO2 = 19.8Substituting:
PAO2 = 19.8 ˆ’ 10/0.8
PAO2 = 19.8 ˆ’ 12.5
PAO2 = 7.3k Pa -
This question is part of the following fields:
- Physiology
-
-
Question 18
Incorrect
-
Which of the statements below best describe the total cerebral flow (CBF) in an adult?
Your Answer:
Correct Answer: Accounts for 15% of the cardiac output
Explanation:While the brain only weighs 3% of the body weight, 15% of the cardiac output goes towards the brain.
Between mean arterial pressures (MAP) of 60-130 mmHg, autoregulation of cerebral blood flow (CBF) occurs. Exceeding this, the CBF is maintained at a constant level. This is controlled mainly by the PaCO2 level, and the autonomic nervous system has minimal role.
Beyond these limits, the CBF is directly proportional to the MAP, not the systolic blood pressure.
-
This question is part of the following fields:
- Physiology
-
-
Question 19
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
-
-
Question 20
Incorrect
-
A 30-year old female was anaesthetically induced for an elective open cholecystectomy. Upon mask ventilation, patient's oxygen saturation level dropped to 90% despite maximal head extension, jaw thrust and two handed mask seal. Intubation was performed twice but failed. Use of bougie also failed to localize the trachea. Oxygen levels continued to drop, but was maintained between 80 and 88% with mask ventilation. Which of the following options is the best action to take for this patient?
Your Answer:
Correct Answer: Insert a supraglottic airway
Explanation:A preplanned preinduction strategy includes the consideration of various interventions designed to facilitate intubation should a difficult airway occur. Non-invasive interventions intended to manage a difficult airway include, but are not limited to: (1) awake intubation, (2) video-assisted laryngoscopy, (3) intubating stylets or tube-changers, (4) SGA for ventilation (e.g., LMA, laryngeal tube), (5) SGA for intubation (e.g., ILMA), (6) rigid laryngoscopic blades of varying design and size, (7) fibreoptic-guided intubation, and (8) lighted stylets or light wands.
Most supraglottic airway devices (SADs) are designed for use during routine anaesthesia, but there are other roles such as airway rescue after failed tracheal intubation, use as a conduit to facilitate tracheal intubation and use by primary responders at cardiac arrest or other out-of-hospital emergencies. Supraglottic airway devices are intrinsically more invasive than use of a facemask for anaesthesia, but less invasive than tracheal intubation. Supraglottic airway devices can usefully be classified as first and second generation SADs and also according to whether they are specifically designed to facilitate tracheal intubation. First generation devices are simply €˜airway tubes’, whereas second generation devices incorporate specific design features to improve safety by protecting against regurgitation and aspiration.
-
This question is part of the following fields:
- Physiology
-
-
Question 21
Incorrect
-
The SI unit of energy is the joule. Energy can be kinetic, potential, electrical or chemical energy. Which of these correlates with the most energy?
Your Answer:
Correct Answer: Energy released when 1 kg fat is metabolised to CO2 and water (the energy content of fat is 37 kJ/g)
Explanation:The derived unit of energy, work or amount of heat is joule (J). It is defined as the amount of energy expended if a force of one newton (N) is applied through a distance of one metre (N·m)
J = 1 kg·m/s2·m = 1 kg·m2/s2 or 1 kg·m2·s-2
Kinetic energy (KE) = ½ MV2
An object with a mass of 1500 kg moving at 30 m/s correlates to 675 kJ:
KE = ½ (1500) × (30)2 = 750 × 900 = 675 kJ
Total energy released when 1 kg fat is metabolised to CO2 and water is 37 MJ. 1 g fat produces 37 kJ/g, therefore 1 kg fat produces 37,000 × 1000 = 37 MJ.
Raising the temperature of 1 kg water from 0°C to 100°C correlates to 420 kJ. The amount of energy needed to change the temperature of 1 kg of the substance by 1°C is the specific heat capacity. We have 1 kg water therefore:
4,200 J × 100 = 420,000 J = 420 kJ
In order to calculate the energy involved in raising a 100 kg mass to a height of 1 km against gravity, we need to calculate the potential energy (PE) of the mass:
PE = mass × height attained × acceleration due to gravity
PE = 100 kg × 1000 m × 10 m/s2 = 1 MJThe heat generated when a direct current of 10 amps flows through a heating element for 10 seconds when the potential difference across the element is 1000 volts can be calculated by applying Joule’s law of heating:
Work done (WD) = V (potential difference) × I (current) × t (time)
WD = 10 × 10 × 1000 = 100 kJ -
This question is part of the following fields:
- Physiology
-
-
Question 22
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
-
-
Question 23
Incorrect
-
The fluids with the highest osmolarity is?
Your Answer:
Correct Answer: 0.45% N. Saline with 5% glucose
Explanation:The concentration of solute particles per litre (mosm/L) = the osmolarity of a solution. Changes in water content, ambient temperature, and pressure affects osmolarity. The osmolarity of any solution can be calculated by adding the concentration of key solutes in it.
Individual manufacturers of crystalloids and colloids may have different absolute values but they are similar to these.
0.45% N. Saline with 5% glucose:
Tonicity – hypertonic
Osmolarity – 405 mosm/L
Kilocalories (kCal) – 1070.9% N. Saline:
Tonicity – isotonic
Osmolarity – 308 mosm/L
Kilocalories (kCal) – 05% Dextrose:
Tonicity – isotonic
Osmolarity – 253 mosm/L
Kilocalories (kCal) – 170Gelofusine (154 mmol/L Na, 120 mmol/L Cl):
Tonicity – isotonic
Osmolarity – 274 mosm/L
Kilocalories (kCal) – 0Hartmann’s solution:
Tonicity – isotonic
Osmolarity – 273 mosm/L
Kilocalories (kCal) – 9 -
This question is part of the following fields:
- Physiology
-
-
Question 24
Incorrect
-
Which of the following statements best describes adenosine receptors?
Your Answer:
Correct Answer: The A1 and A2 receptors are present centrally and peripherally
Explanation:Adenosine receptors are expressed on the surface of most cells.
Four subtypes are known to exist which are A1, A2A, A2B and A3.Of these, the A1 and A2 receptors are present peripherally and centrally. There are agonists at the A1 receptors which are antinociceptive, which reduce the sensitivity to a painful stimuli for the individual. There are also agonists at the A2 receptors which are algogenic and activation of these results in pain.
The role of adenosine and other A1 receptor agonists is currently under investigation for use in acute and chronic pain states.
-
This question is part of the following fields:
- Physiology
-
-
Question 25
Incorrect
-
The typical fluid compartments in a normal 70kg male are:
Your Answer:
Correct Answer: intracellular>extracellular
Explanation:Body fluid compartments in a 70kg male:
Total volume=42L (60% body weight)
Intracellular fluid compartment (ICF) =28L
Extracellular fluid compartment (ECF) = 14LECF comprises:
Intravascular fluid (plasma) = 3L
Extravascular fluid = 11LExtravascular fluids comprises:
Interstitial fluid = 10.5L
Transcellular fluid = 0.5L -
This question is part of the following fields:
- Physiology
-
-
Question 26
Incorrect
-
Following an acute appendicectomy, a 6-year-old child is admitted to the recovery unit. Your consultant has requested that you prescribe maintenance fluids for the next 12 hours. The child is 21 kg in weight. What is the most suitable fluid volume to be prescribed?
Your Answer:
Correct Answer: 732 ml
Explanation:After a paediatric case, you’ll frequently have to calculate and prescribe maintenance fluids. The ‘4-2-1 rule’ should be used as a guideline:
1st 10 kg – 4 ml/kg/hr
2nd 10 kg – 2 ml/kg/hr
Subsequent kg – 1 ml/kg/hrHence
1st 10 kg = 4 × 10 = 40 ml
2nd 10 kg = 2 × 10 = 20 ml
Subsequent kg = 1 × 1 = 1 ml
Total = 61 ml/hr61 × 12 = 732 ml over 12 hrs.
-
This question is part of the following fields:
- Physiology
-
-
Question 27
Incorrect
-
You're summoned to the emergency room, where a 39-year-old man has been admitted following a cardiac arrest. He was rescued from a river, but little else is known about him. CPR is being performed on the patient, who has been intubated. He's received three DC shocks and is still in VF. A rectal temperature of 29.5°C is taken with a low-reading thermometer. Which of the following statements about his resuscitation is correct?
Your Answer:
Correct Answer: No further DC shocks and no drugs should be given until his core temperature is greater than 30°C
Explanation:The guidelines for the management of cardiac arrest in hypothermic patients published by the UK Resuscitation Council differ slightly from the standard algorithm.
In a patient with a core temperature of less than 30°C, do the following:
If you’re on the shockable side of the algorithm (VF/VT), you should give three DC shocks.
Further shocks are not recommended until the patient has been rewarmed to a temperature of more than 30°C because the rhythm is refractory and unlikely to change.
There should be no drugs given because they will be ineffective.In a patient with a core temperature of 30°C to 35°C, do the following:
DC shocks are used as usual.
Because they are metabolised much more slowly, the time between drug doses should be doubled.Active rewarming and protection against hyperthermia should be given to the patient.
Option e is false because there is insufficient information to determine whether resuscitation should be stopped.
-
This question is part of the following fields:
- Physiology
-
-
Question 28
Incorrect
-
Which of the following statement is true regarding hypoxic pulmonary vasoconstriction (HPV)?
Your Answer:
Correct Answer: 20 parts per million (ppm) of nitric oxide will reduce hypoxic pulmonary vasoconstriction
Explanation:Hypoxic Pulmonary vasoconstriction (HPV) reflects the constriction of small pulmonary arteries in response to hypoxic alveoli (.i.e.; PO2 below 80-100mmHg or 11-13kPa).
These blood vessels become independent of the nerve stimulus, when blood with a high PO2 flows through the lung which contains a low alveolar PO2.
Thus a low PO2 within the alveoli has been shown to impact on hypoxic pulmonary vasoconstriction (HPV) more than a low PO2 within the blood.
HPV results in the blood flow being directed away from poorly ventilated areas of the lung and helps to reduce the ventilation/perfusion mismatch (not increase).
In animals, volatile anaesthetic agents can diminish HPV, while in adults, the evidence proves less persuading, in spite of the fact that it certainly doesn’t strengthen the effects.
HPV response will be suppressed by 20 parts per million (ppm) of nitric oxide.
-
This question is part of the following fields:
- Physiology
-
-
Question 29
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
-
-
Question 30
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
-
00
Correct
00
Incorrect
00
:
00
:
00
Session Time
00
:
00
Average Question Time (
Mins)