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Question 1
Correct
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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
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Question 2
Incorrect
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Cells use adenosine-5-triphosphate (ATP) as a coenzyme and is a source of energy.
Glucose metabolism produces the most ATP from which of the following biochemical processes?
Your Answer:
Correct Answer: Electron transport phosphorylation in the mitochondria
Explanation:Glycolysis occurs in the cytoplasm of the cell. It converts 1 glucose molecule (6-carbon) to pyruvate (two 3-carbon molecules) and produces 4 ATP molecules and 2NADH but uses 2 ATP in the process with an overall net energy production of 2 ATP.
Pyruvate is then oxidised to acetyl coenzyme A (generating 2 NADH per pyruvate molecule). This takes place in the mitochondria and then enters the Krebs cycle (citric acid cycle). It produces 2 ATP, 8 NADH and 2 FADH2 per glucose molecule.
Electron transport phosphorylation takes place in the mitochondria. The aim of this process is to break down NADH and FADH2 and also to pump H+ into the outer compartment of the mitochondria. It produces 32 ATP with an overall net production of 36ATP.
In anaerobic respiration which occurs in the cytoplasm, pyruvate is reduced to NAD producing 2 ATP.
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This question is part of the following fields:
- Physiology
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Question 3
Incorrect
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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.
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This question is part of the following fields:
- Physiology
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Question 4
Incorrect
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A transport ventilator connected to a size CD oxygen cylinder has a setting of air/oxygen entrainment ratio of 1:1 and a minute volume set at 10 litres/minute.
Which value best approximates to the FiO2?Your Answer:
Correct Answer: 0.6
Explanation:A nominal volume of 2 litres is contained in a CD cylinder. It has a pressure of 230 bar when full and contains litres 460 L of useable oxygen at STP.
For every 1000 mL 100% oxygen there will be an entrainment of 1000 mL or air (20% oxygen) in an air/oxygen mix.
The average concentration is, therefore, 120/2=60% or 0.6.
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This question is part of the following fields:
- Physiology
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Question 5
Incorrect
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The statement that best describes lactic acidosis is:
Your Answer:
Correct Answer: It can be precipitated by intravenous fructose
Explanation:An elevated arterial blood lactate level and an increase anion gap ([Na + K] – [Cl + HCO3]) of >20mmol gives rise to lactic acidosis. It can also be a result of overproduction and/or reduced metabolism of lactic acid.
The liver and kidney are the main sites of lactate metabolism, not skeletal muscle.
The two types of lactic acidosis that are known are:
Type A – due to tissue hypoxia, inadequate tissue perfusion and anaerobic glycolysis. These may be seen in cardiac arrest, shock, hypoxaemia and anaemia. The management of type A lactic acidosis involves reversing the underlying cause of the tissue hypoxia.
Type B – occurs in the absence of tissue hypoxia. Some of the causes of this include hepatic failure, renal failure, diabetes mellitus, pancreatitis and infection. Some drugs can also cause this lie aspirin, ethanol, methanol, biguanides and intravenous fructose.
The mainstay of treatment involves:
1. Optimising tissue oxygen delivery
2. Correcting the cause
3. Intravenous sodium bicarbonateIn resistant cases, peritoneal dialysis can be performed.
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This question is part of the following fields:
- Physiology
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Question 6
Incorrect
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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 7
Incorrect
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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.
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This question is part of the following fields:
- Physiology
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Question 8
Incorrect
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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 − 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 − 6.3 kPa)
PiO2 = 19.8Substituting:
PAO2 = 19.8 − 10/0.8
PAO2 = 19.8 − 12.5
PAO2 = 7.3k Pa -
This question is part of the following fields:
- Physiology
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Question 9
Incorrect
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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
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Question 10
Incorrect
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An orthopaedic surgery is scheduled for a 68-year-old man. He is normally in good shape. His routine biochemistry results are checked and found to be within normal limits.
Which of the following pairs has the greatest impact on his plasma osmolarity?Your Answer:
Correct Answer: Sodium and potassium cations
Explanation:The number of osmoles (Osm) of solute per litre (L) of solution (Osmol/L) is the unit of measurement for solute concentration. The calculated serum osmolality assumes that the primary solutes in the serum are sodium salts (chloride and bicarbonate), glucose, and urea nitrogen.
2 (Na + K) + Glucose + Urea (all in mmol/L) = calculated osmolarity
313 mOsm/L = 2 (144 + 6) + 9.5 + 3.5
Sodium and potassium ions clearly contribute the most to plasma osmolarity. Glucose and urea, on the other hand, are less so.
The osmolarity of normal serum is 285-295 mOsm/L. Temperature and pressure affect osmolality, and this calculated variable is less than osmolality for a given solution.
The number of osmoles (Osm) of solute per kilogramme (Osm/kg) is a measure of osmolality, which is also a measure of solute concentration. Temperature and pressure have no effect on the value. An osmometer is used to measure it in the lab. Osmometers rely on a solution’s colligative properties, such as a decrease in freezing point or a rise in vapour pressure.
The osmolar gap (OG) is calculated as follows:
OG = osmolaRity calculated from measured serum osmolaLity
Excess alcohols, lipids, and proteins in the blood can all contribute to the difference.
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This question is part of the following fields:
- Physiology
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Question 11
Incorrect
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Comparing pressure-volume curves in patients during an asthma attack with that of healthy subjects.
The increased resistive work of breathing in the patients with asthma is best indicated by?Your Answer:
Correct Answer: Larger hysteresis loop
Explanation:A major source of caloric expenditure and oxygen consumption in the body is work of breathing (WOB) and 70% of this is to overcome elastic forces. The remaining 30% is for flow-resistive work
In a normal patient breathing normally, the total area of hysteresis pressure volume curve represents the flow-resistive WOB.
The area of the expiratory resistive work increases during an asthma attack making the compliance curve larger in area. The larger the area the greater the work required to breathe.
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This question is part of the following fields:
- Physiology
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Question 12
Incorrect
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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
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This question is part of the following fields:
- Physiology
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Question 13
Incorrect
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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:
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 (?1, ?2, and ?3)
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).
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This question is part of the following fields:
- Physiology
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Question 14
Incorrect
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In which of the following situations will a regional fall in cerebral blood flow occur, suppose there is no changes in the mean arterial pressure (MAP)?
Your Answer:
Correct Answer: Hyperoxia
Explanation:The response of cerebral blood flow (CBF) to hyperoxia (PaO2 >15 kPa, 113 mmHg), the cerebral oxygen vasoreactivity is less well defined. A study originally described, using a nitrous oxide washout technique, a reduction in CBF of 13% and a moderate increase in cerebrovascular resistance in subjects inhaling 85-100% oxygen. Subsequent human studies, using a variety of differing methods, have also shown CBF reductions with hyperoxia, although the reported extent of this change is variable. Another study assessed how supra-atmospheric pressures influenced CBF, as estimated by changes in middle cerebral artery flow velocity (MCAFV) in healthy individuals. Atmospheric pressure alone had no effect on MCAFV if PaO2 was kept constant. Increases in PaO2 did lead to a significant reduction in MCAFV; however, there were no further reductions in MCAFV when oxygen was increased from 100% at 1 atmosphere of pressure to 100% oxygen at 2 atmospheres of pressure. This suggests that the ability of cerebral vasculature to constrict in response to increasing partial pressure of oxygen is limited.
Increases in arterial blood CO2 tension (PaCO2) elicit marked cerebral vasodilation.
CBF increases with general anaesthesia, ketamine anaesthesia, and hypoviscosity.
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This question is part of the following fields:
- Physiology
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Question 15
Incorrect
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The following is true about the extracellular fluid (ECF) in a normal adult woman weighing 60 kg.
Your Answer:
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.
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This question is part of the following fields:
- Physiology
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Question 16
Incorrect
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In the erect position, the partial pressure of oxygen in the alveoli (PAO2) is higher in the apical lung units than in the basal lung units.
What is the most significant reason for this?Your Answer:
Correct Answer: The V/Q ratio of apical units is greater than that of basal units
Explanation:In any alveolar unit, the V/Q ratio affects alveolar oxygen (PAO2) and carbon dioxide tension (PACO2).
The partial pressure of alveolar carbon dioxide (PACO2) is plotted against the partial pressure of alveolar oxygen in a Ventilation-Perfusion (V/Q) ratio graph (PAO2). Given a set of model assumptions, the curve represents all of the possible values for PACO2 and PAO2 that an individual alveolus could have.
In the case of an infinity V/Q ratio (ventilation but no perfusion or dead space), the PACO2 of the alveolus will equal zero, while the PAO2 will approach that of external air (150mmmHg). At the apex of the lung, the V/Q ratio is 3.3, compared to 0.67 at the base.
PACO2 and PAO2 approach the partial pressures for these gases in the venous blood when the V/Q ratio is zero (no ventilation but perfusion). At the base of the lung, the V/Q ratio is 0.67, whereas at the apex, it is 3.3.
PAO2 at the apex is typically 132mmHg, and PACO2 is typically 28mmHg.
The average PAO2 at the base is 89 mmHg, while the average PACO2 is 42 mmHg.
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This question is part of the following fields:
- Physiology
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Question 17
Incorrect
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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.
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This question is part of the following fields:
- Physiology
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Question 18
Incorrect
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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
Incorrect
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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%).
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This question is part of the following fields:
- Physiology
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Question 20
Incorrect
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Which of the following statement is true or false regarding to the respiratory tract?
Your Answer:
Correct Answer: The sympathetic innervation of the bronchi is derived from T2 - T4
Explanation:The diaphragm has three opening through which different structures pass from the thoracic cavity to the abdominal cavity:
Inferior vena cava passes at the level of T8.
Oesophagus, oesophageal vessels and vagi at T10.
Aorta, thoracic duct and azygous vein through T12.
Sympathetic trunk and pulmonary branches of vagus nerve form a posterior pulmonary plexus at the root of the lung. Fibres continue posteriorly from superficial cardiac plexus to form Anterior pulmonary plexus. It contains vagi nerves and superficial cardiac plexus. These fibres then follow the blood vessel and bronchi into the lungs.
The lower border of the pleura is at the level of:
8th rib in the midclavicular line
10th rib in the lower level of midaxillary line
T12 at its termination.
Both lungs have oblique fissure while right lung has transverse fissure too.
The trachea expands from the lower edge of the cricoid cartilage (at the level of the 6th cervical vertebra) to the carina.
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This question is part of the following fields:
- Physiology
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Question 21
Incorrect
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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.
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This question is part of the following fields:
- Physiology
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Question 22
Incorrect
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Pressure volume loop represents the compliance of left ventricle.
Considering there is no change in preload and myocardial contractility, which physiological change may result an increase in left ventricular afterload?Your Answer:
Correct Answer: Increased end-systolic volume
Explanation:If there is no change in preload and myocardial contractility, there will be decrease in end-diastolic volume and stroke volume. So there must be increase in end-systolic volume.
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This question is part of the following fields:
- Physiology
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Question 23
Incorrect
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During exercise, muscle blood flow can increase by 20 to 50 times.
Which mechanism is the most important for increased blood flow?Your Answer:
Correct Answer: Local autoregulation
Explanation:Skeletal muscle blood flow is in the range of 1-4 ml/min per 100 g when at rest. Blood flow can reach 50-100 ml/min per 100 g during exercise. With maximal vasodilation, blood flow can increase 20 to 50 times.
The adrenal medulla releases catecholamines and increases neural sympathetic activity during exercise. Normally, alpha-1 and alpha-2 would cause vasoconstriction in the muscle groups being used, but vasodilatory metabolites override these effects, resulting in a so-called functional sympathectomy. Local hypoxia and hypercarbia, nitric oxide, K+ ions, adenosine, and lactate are some of the stimuli that cause vasodilation.
However, the splanchnic and cutaneous circulations, which supply inactive muscles, vasoconstrict.
Sympathetic cholinergic innervation of skeletal muscle arteries is found in some species (such as cats and dogs, but not humans). Vasodilation is induced by stimulating smooth muscle beta-2 adrenoreceptors, but at rest, the alpha-adrenoreceptor effects of adrenaline and noradrenaline predominate. During exercise, the skeletal muscle pump promotes venous emptying, but it does not necessarily increase blood flow.
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This question is part of the following fields:
- Physiology
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Question 24
Incorrect
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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.
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This question is part of the following fields:
- Physiology
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Question 25
Incorrect
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One litre of water at 0°C and a pressure of 1 bar is in a water-bath. A 1 kW element is used in heating it.
Given that the specific heat capacity of water is 4181 J/(kg°C) or J/(kg K), how long will it take to raise the temperature of the water by 10°C?Your Answer:
Correct Answer: 42 seconds
Explanation: -
This question is part of the following fields:
- Physiology
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Question 26
Incorrect
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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.
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This question is part of the following fields:
- Physiology
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Question 27
Incorrect
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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 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.
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This question is part of the following fields:
- Physiology
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Question 28
Incorrect
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In the fetal circulation, the cerebral and coronary circulations are preferentially supplied by oxygen-rich blood over other organs. This is possible because of which phenomenon?
Your Answer:
Correct Answer: Well oxygenated blood from the inferior vena cava is preferentially streamed across the patent foramen ovale
Explanation:During fetal development, blood oxygenated by the placenta flows to the foetus through the umbilical vein, bypasses the fetal liver through the ductus venosus, and returns to the fetal heart through the inferior vena cava.
Blood returning from the inferior vena cava then enters the right atrium and is preferentially shunted to the left atrium through the patent foramen ovale. Blood in the left atrium is then pumped from the left ventricle to the aorta. The oxygenated blood ejected through the ascending aorta is preferentially directed to the fetal coronary and cerebral circulations.
Deoxygenated blood returns from the superior vena cava to the right atrium and ventricle to be pumped into the pulmonary artery. Fetal pulmonary vascular resistance (PVR), however, is higher than fetal systemic vascular resistance (SVR); this forces deoxygenated blood to mostly bypass the fetal lungs. This poorly oxygenated blood enters the aorta through the patent ductus arteriosus and mixes with the well-oxygenated blood in the descending aorta. The mixed blood in the descending aorta then returns to the placenta for oxygenation through the two umbilical arteries.
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This question is part of the following fields:
- Physiology
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Question 29
Incorrect
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Anaesthetic awareness is most probable in general anaesthesia for which surgical operation?
Your Answer:
Correct Answer: Emergency surgery for major trauma
Explanation:Awareness during general anaesthesia is a frightening experience, which may result in serious emotional injury and post-traumatic stress disorder.
The incidence of awareness during general anaesthesia with current anaesthetic agents and techniques has been reported as 0.2-0.4% in nonobstetric and noncardiac surgery, as 0.4% during caesarean section, and as 1.5% in cardiac surgery.
The incidence during major trauma surgery is higher. Incidence of recall has been reported to be as high as 11-43% in major trauma cases.
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This question is part of the following fields:
- Physiology
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Question 30
Incorrect
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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
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