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Question 1
Incorrect
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A patient on admission is given an infusion of 1000 mL of 10% glucose and 500 mL of 20% lipid over a 24 hour period.
Which of these best approximates to the energy input over this time period?Your Answer: 900 kcal
Correct Answer: 1300 kcal
Explanation:1% solution contains 1 g of substance per 100 mL.
A solution of 10% glucose is 10 g/100mL. Therefore 1000 mL of this glucose solution will contain 100 g.
1 g of glucose yields about 4 kcal of energy. One litre of 10% glucose will therefore release approximately 4x100g = 400 kcal of energy.
A solution of 20% fat is 20 g/100mL. Therefore 1000 mL of this fat solution will have 200 g and 500 mL will contain 100 g.
1 g of fat yields approximately 9 kcal. 500 mL of 20% fat therefore has the potential to yield 900 kcal of energy.
The total energy input over this 24 hour period is approximately 400kcal + 900kcal = 1300 kcal.
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This question is part of the following fields:
- Physiology
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Question 2
Incorrect
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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).
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This question is part of the following fields:
- Physiology
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Question 3
Incorrect
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Which of the following best explains the association between smoking and lower oxygen delivery to tissues?
Your Answer:
Correct Answer: Left shift of the oxygen dissociation curve
Explanation:Smoking is a major risk factor associated with perioperative respiratory and cardiovascular complications. Evidence also suggests that cigarette smoking causes imbalance in the prostaglandins and promotes vasoconstriction and excessive platelet aggregation. Two of the constituents of cigarette smoke, nicotine and carbon monoxide, have adverse cardiovascular effects. Carbon monoxide increases the incidence of arrhythmias and has a negative ionotropic effect both in animals and humans.
Smoking causes an increase in carboxyhaemoglobin levels, resulting in a leftward shift in which appears to represent a risk factor for some of these cardiovascular complications.
There are two mechanisms responsible for the leftward shift of oxyhaemoglobin dissociation curve when carbon monoxide is present in the blood. Carbon monoxide has a direct effect on oxyhaemoglobin, causing a leftward shift of the oxygen dissociation curve, and carbon monoxide also reduces the formation of 2,3-DPG by inhibiting glycolysis in the erythrocyte. Nicotine, on the other hand, has a stimulatory effect on the autonomic nervous system. The effects of nicotine on the cardiovascular system last less than 30 min.
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This question is part of the following fields:
- Physiology
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Question 4
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:The amount of energy required to change the temperature of 1 kg of a substance by 1°C is its specific heat capacity.
Energy required to increase the temperature of any object is given by the equation:
E = m × c × θ
Where:
E = energy required to heat an object
m = mass of object
c = specific heat capacity of the substance
θ = temperature changeThe specific heat capacity of water is 4181 J/(kg°C)
Therefore the energy required to raise 1kg of water by 10°C is:
1 kg × 4181 J/(kg°C) × 10°C = 41810 J
A 1 kW or 1000 W heater would be expected to supply 1000 J/s.
The approximate time it would take the 1 kW heater to raise the temperature of one Litre of water by 10°C is:
41810/1000 = 41.8 seconds.
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This question is part of the following fields:
- Physiology
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Question 5
Incorrect
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Regarding amide local anaesthetics, which one factor has the most significant effect on its duration of action?
Your Answer:
Correct Answer: Protein binding
Explanation:When drugs are bound to proteins, drugs cannot cross membranes and exert their effect. Only the free (unbound) drug can be absorbed, distributed, metabolized, excreted and exert pharmacologic effect. Thus, when amide local anaesthetics are bound to α1-glycoproteins, their duration of action are reduced.
The potency of local anaesthetics are affected by lipid solubility. Solubility influences the concentration of the drug in the extracellular fluid surrounding blood vessels. The brain, which is high in lipid content, will dissolve high concentration of lipid soluble drugs. When drugs are non-ionized and non-polarized, they are more lipid-soluble and undergo more extensive distribution. Hence allowing these drugs to penetrate the membrane of the target cells and exert their effect.
Tissue pKa and pH will determine the degree of ionization.
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This question is part of the following fields:
- Physiology
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Question 6
Incorrect
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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).
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This question is part of the following fields:
- Physiology
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Question 7
Incorrect
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Which statement best describes the bispectral index (BIS)?
Your Answer:
Correct Answer: It decreases during normal sleep
Explanation:The bispectral index (BIS) is one of several systems used in anaesthesiology as of 2003 to measure the effects of specific anaesthetic drugs on the brain and to track changes in the patient’s level of sedation or hypnosis. It is a complex mathematical algorithm that allows a computer inside an anaesthesia monitor to analyse data from a patient’s electroencephalogram (EEG) during surgery. It is a dimensionless number (0-100) that is a summative measurement of time domain, frequency domain and high order spectral parameters derived from electroencephalogram (EEG) signals.
Sleep and anaesthesia have similar behavioural characteristics but are physiologically different but BIS monitors can be used to measure sleep depth. With increasing sleep depth during slow-wave sleep, BIS levels decrease. This correlates with changes in regional cerebral blood flow when measured using positron emission tomography (PET).
BIS shows a dose-response relationship with the intravenous and volatile anaesthetic agents. Opioids produce a clinical change in the depth of sedation or analgesia but fail to produce significant changes in the BIS. Ketamine increases CMRO2 and EEG activity.
BIS is unable to predict movement in response to a surgical stimulus. Some of these are spinal reflexes and not perceived by the cerebral cortex.
BIS is used during cardiopulmonary bypass to measure depth of anaesthesia and an index of cerebral perfusion. However, it cannot predict subtle or significant cerebral damage.
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This question is part of the following fields:
- Physiology
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Question 8
Incorrect
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The resistance to flow in a blood vessel is affected by the following except?
Your Answer:
Correct Answer: Thickness of the vessel wall
Explanation:The Poiseuille’s formula states that Q = ΔPÏ€r4/8Lη
where
ΔP = the pressure gradient along the vessel
Q = flow rate
r = radius of the vessel
η = coefficient of viscosity (haematocrit) of the blood
L = length of the blood vessel.Resistance is pressure difference/flow (ΔP/Q) (analogous to Ohm’s law)
Rearranging Poiseuille’s equation ΔP/Q =8Lη/Ï€r4
Although the stiffness of the vessel wall affects blood flow, the thickness of the vessel wall does not.
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This question is part of the following fields:
- Physiology
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Question 9
Incorrect
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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
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Question 10
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 11
Incorrect
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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.
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This question is part of the following fields:
- Physiology
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Question 12
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 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
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Question 13
Incorrect
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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.
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This question is part of the following fields:
- Physiology
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Question 14
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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Question 15
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 16
Incorrect
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All of the statements describing the blood brain barrier are false EXCEPT:
Your Answer:
Correct Answer: Inflammation alters its permeability
Explanation:The blood brain barrier (BBB) consists of the ultrafiltration barrier in the choroid plexus and the barrier around cerebral capillaries. The barrier is made by endothelial cells which line the interior of all blood vessels. In the capillaries that form the blood€“brain barrier, endothelial cells are wedged extremely close to each other, forming so-called tight junctions.
Outside of the BBB lies the hypothalamus, third and fourth ventricles and the chemoreceptor trigger zone (CTZ).
Water, oxygen and carbon dioxide cross the BBB freely but glucose is controlled. The ability of chemicals to cross the barrier is proportional to their lipid solubility, not their water solubility. It’s ability to cross is inversely proportional to their molecular size and charge.
In neonates, the BBB is less effective than in adults. This is why there is increased passage of opioids and bile salts (kernicterus) into the neonatal brain.
In meningitis, the effectiveness and permeability of the BBB is affected, and as a result, this effect helps the passage of antibiotics which would otherwise not normally be able to cross.
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This question is part of the following fields:
- Physiology
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Question 17
Incorrect
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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.
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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 25-year old man needs an emergency appendicectomy and has gone to the operating room. During general anaesthesia, ventilation is achieved using a circle system with a fresh gas flow (FGF) of 1L/min, with and air/oxygen and sevoflurane combination. The capnograph trace is normal.
Changes to the end tidal and baseline CO2 measurements at 10 and 20 mins respectively are seen on the capnograph below: (10 minutes and 20 minutes).
End-tidal CO2 4.9 kPa, 8.4 kPa.
Baseline end-tidal CO2 0.2 kPa, 2.4 kPa.
The other vitals were as follows:
Pulse 100-105 beats per minute,
Systolic blood pressure 120-133 mmHg,
O2 saturation 99%.
The next most important immediate step is which of the following?Your Answer:
Correct Answer: Increase the FGF
Explanation:This scenario describes rebreathing management.
Changes is exhaustion of the soda lime and a progressive rise in circuit deadspace is the most likely explanation for the capnograph.
It is important that the soda lime canister is inspected for a change in colour of the granules. Initially fresh gas flow should be increased and then if necessary, replace the soda lime granules. Other strategies include changing to another circuit or bypassing the soda lime canister after the fresh gas flow is increased.
Any other causes of increased equipment deadspace should be excluded.
Intraoperative hypercarbia can be caused by:
1. Hypoventilation – Breathing spontaneously; drugs which include anaesthetic agents, opioids, residual neuromuscular blockade, pre-existing respiratory or neuromuscular disease and cerebrovascular accident.
2. Controlled ventilation- circuit leaks, disconnection, miscalculation of patient’s minute volume.
3. Rebreathing – Soda lime exhaustion with circle, inadequate fresh gas flow into Mapleson circuits, increased breathing system deadspace.
4. Endogenous source – Tourniquet release, hypermetabolic states (MH or thyroid storm) and release of vascular clamps.
5. Exogenous source – Absorption of CO2 absorption from the pneumoperitoneum. -
This question is part of the following fields:
- Physiology
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Question 19
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 20
Incorrect
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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).
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This question is part of the following fields:
- Physiology
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Question 21
Incorrect
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The single most important prerequisite for accuracy in measuring basal metabolic rate (BMR) using indirect calorimetry is performing the test:
Your Answer:
Correct Answer: In a neutral thermal environment
Explanation:The basal metabolic rate (BMR) is the amount of energy required to maintain basic bodily functions in the resting state. The unit is Watt (Joule/second) or calories per unit time.
Indirect calorimetry measures O2 consumption and CO2 production where gases are collected in a canopy which is the gold standard, Douglas bag, face-mask dilution technique or interfaced with a ventilator.
The BMR can be calculated using the Weir formula:
Metabolic rate (kcal per day) = 1.44 (3.94 VO2 + 1.11 VCO2)
The BMR should be measured while lying down and at rest with the following conditions met:
It should follow a 12 -hour fast
No stimulants ingested within a 12-hour period
It should be done in a neutral thermal environment (between 20°C-25°C) -
This question is part of the following fields:
- Physiology
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Question 22
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 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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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 25
Incorrect
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A 27-year-old woman takes part in a study looking into the effects of different dietary substrates on metabolism. She receives a 24-hour ethyl alcohol infusion.
A constant volume, closed system respirometer is used to measure CO2 production and consumption. The production of carbon dioxide is found to be 200 mL/minute.
Which of the following values most closely resembles her anticipated O2 consumption at the conclusion of the trial?Your Answer:
Correct Answer: 300 mL/minute
Explanation:The respiratory quotient (RQ) is the ratio of CO2 produced by the body to O2 consumed in a given amount of time.
CO2 produced / O2 consumed = RQ
CO2 is produced at a rate of 200 mL per minute, while O2 is consumed at a rate of 250 mL per minute. An RQ of around 0.8 is typical for a mixed diet.
The RQ will change depending on the energy substrates consumed in the diet. Granulated sugar is a refined carbohydrate that contains 99.999 percent carbohydrate and no lipids, proteins, minerals, or vitamins.
Glucose and other hexose sugars (glucose and other hexose sugars):
RQ=1Fats:
RQ = 0.7Proteins:
Approximately 0.9 RQEthyl alcohol is a type of alcohol.
200/300 = 0.67 RQ
For complete oxidation, lipids and alcohol require more oxygen than carbohydrates.
When carbohydrate is converted to fat, the RQ can rise above 1.0. Fat deposition and weight gain are likely to occur in these circumstances.
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This question is part of the following fields:
- Physiology
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Question 26
Incorrect
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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/LThis 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 27
Incorrect
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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.
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This question is part of the following fields:
- Physiology
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Question 28
Incorrect
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Which of the following, at a given PaO2, increases the oxygen content of arterial blood?
Your Answer:
Correct 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.
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This question is part of the following fields:
- Physiology
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Question 29
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 30
Incorrect
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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
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This question is part of the following fields:
- Physiology
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