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Question 1
Correct
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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: 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 2
Correct
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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: 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 3
Correct
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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: 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 4
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: 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 (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 5
Incorrect
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In a normal healthy adult breathing 100 percent oxygen, which of the following is the most likely cause of an alveolar-arterial (A-a) oxygen difference of 30 kPa?
Your Answer: A right to left shunt
Correct Answer: Atelectasis
Explanation:The ‘ideal’ alveolar PO2 minus arterial PO2 is the alveolar-arterial (A-a) oxygen difference.
The ‘ideal’ alveolar PO2 is derived from the alveolar air equation and is the PO2 that the lung would have if there was no ventilation-perfusion (V/Q) inequality and it was exchanging gas at the same respiratory exchange ratio as real lung.
The amount of oxygen in the blood is measured directly in the arteries.
The A-a oxygen difference (or gradient) is a useful measure of shunt and V/Q mismatch, and it is less than 2 kPa in normal adults breathing air (15 mmHg). Because the shunt component is not corrected, the A-a difference increases when breathing 100 percent oxygen, and it can be up to 15 kPa (115 mmHg).
An abnormally low or abnormally high V/Q ratio within the lung can cause an increased A-a difference, though the former is more common. Atelectasis, which results in a low V/Q ratio, is the most likely cause of an A-a difference in a healthy adult breathing 100 percent oxygen.
Hypoventilation may cause an increase in alveolar (and thus arterial) CO2, lowering alveolar PO2 according to the alveolar air equation.
The alveolar PO2 is also reduced at high altitude.
Healthy people are unlikely to have a right-to-left shunt or an oxygen transport diffusion defect.
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This question is part of the following fields:
- Physiology
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Question 6
Correct
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A participant of a metabolism study is to be fed only granulated sugar and water for 48 hours. What would be his expected respiratory quotient at the end of the study?
Your Answer: 1
Explanation:The respiratory quotient is the ratio of CO2 produced to O2 consumed while food is being metabolized:
RQ = CO2 eliminated/O2 consumed
Most energy sources are food containing carbon, hydrogen and oxygen. Examples include fat, carbohydrates, protein, and ethanol. The normal range of respiratory coefficients for organisms in metabolic balance usually ranges from 1.0-0.7.
Granulated sugar is a refined carbohydrate with no significant fat, protein or ethanol content.
The RQ for carbohydrates is = 1.0
The RQ for the rest of the compounds are:
Fats RQ = 0.7
The chemical composition of fats differs from that of carbohydrates in that fats contain considerably fewer oxygen atoms in proportion to atoms of carbon and hydrogen.Protein RQ = 0.8
Due to the complexity of various ways in which different amino acids can be metabolized, no single RQ can be assigned to the oxidation of protein in the diet; however, 0.8 is a frequently utilized estimate. -
This question is part of the following fields:
- Physiology
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Question 7
Correct
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Metabolization of many drugs used in anaesthesia involves the cytochrome P450 (CYP) isoenzymes.
The CYP enzyme most likely to be subject to genetic variability and thus cause adverse drug reactions is which of these?Your Answer: CYP2D6
Explanation:Approximately 25% of phase-1 drug reactions is made responsible by CYP2D6.
As much as a 1,000-fold difference in the ability to metabolise drugs by CYP2D6 can happen between phenotypes, and this may result in adverse drug reactions (ADRs).
The metabolism of antiemetics, beta-blockers, codeine, tramadol, oxycodone, hydrocodone, tamoxifen, antidepressants, neuroleptics, and antiarrhythmics is also as a result of CYP2D6.
Patients who take drugs that are metabolised by CYP2D6 but have poor CYP2D6 metabolism are more likely to have ADRs. People with ultra-rapid CYP2D6 metabolism may have a decreased drug effect due to low plasma concentrations of these drugs.
All the other CYP enzymes are subject to genetic polymorphism. Variants are less likely to lead to adverse drug reactions.
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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 global cerebral blood flow (CBF) of 35 ml/100 g/min (Normal CBF = 54 ml/100 g/min) can lead to which of the following?
Your Answer:
Correct Answer: Poor prognostic EEG
Explanation:CBF is defined as the blood volume that flows per unit mass per unit time in brain tissue and is typically expressed in units of ml blood/100 g tissue/minute. The normal average CBF in adults human is about 50 ml/100 g/min, with lower values in the white matter (,20 ml/100 g/min) and greater values in the gray matter (,80 ml/100 g/min).
Low CBF levels between 30-40 ml/100 g/min may begin to show poor prognostic EEG. EEG findings consistently associated with a poor outcome are isoelectric EEG, low voltage EEG, and burst suppression (specifically burst suppression with identical bursts), as well as the absence of EEG reactivity.
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This question is part of the following fields:
- Physiology
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Question 9
Incorrect
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Which of the following statements is true with regards to the Krebs' cycle (also known as the tricarboxylic acid cycle or citric acid cycle)?
Your Answer:
Correct Answer: Alpha-ketoglutarate is a five carbon molecule
Explanation:Krebs’ cycle (tricarboxylic acid cycle or citric acid cycle) is a sequence of reactions in which acetyl coenzyme A (acetyl-CoA) is metabolised and this results in carbon dioxide and hydrogen atoms production.
This series of reactions occur in the mitochondria of eukaryotic cells, not the cytoplasm. The cycle requires oxygen and so, cannot function under anaerobic conditions.
It is the common pathway for carbohydrate, fat and some amino acids oxidation and is required for high energy phosphate bond formation in adenosine triphosphate (ATP).
When pyruvate enters the mitochondria, it is converted into acetyl-CoA. This represents the formation of a 2 carbon molecule from a 3 carbon molecule. There is loss of one CO2 but formation of one NADH molecule. Acetyl-CoA is condensed with oxaloacetate, the anion of a 4 carbon acid, to form citrate which is a 6 carbon molecule.
Citrate is then converted into isocitrate, alpha-ketoglutarate, succinyl-CoA, succinate, fumarate, malate and finally oxaloacetate.
The only 5 carbon molecule in the cycle is alpha-ketoglutarate.
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This question is part of the following fields:
- Physiology
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Question 10
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 11
Incorrect
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A 25-year old lady is in the operating room and has had general anaesthesia for a knee arthroscopy.
Induction was done with fentanyl 1mcg/kg and propofol 2mg/kg. A supra-glottic airway was inserted and using and air oxygen mixture with 2.5% sevoflurane, her anaesthesia was maintained. The patient is allowed to spontaneously breathe using a Bain circuit, and the fresh gas flow is 9L/min. Over the next 30 minutes, the end-tidal Co2 rises from 4.5kPa to 8.4kPa, and the baseline reading on the capnograph is 0kPa.
The most appropriate initial action is which of the following?Your Answer:
Correct Answer: Hypoventilation
Explanation:The commonest and most likely cause of a gradual rise in end-tidal CO2 (EtCO2) occurring during anaesthesia in a spontaneously breathing patient is hypoventilation. This occurs from the respiratory depressant effects of the opioid and sevoflurane.
Malignant hyperthermia should be sought if the EtCO2 shows further progressive rise.
Causes of rebreathing and a rise in the baseline of the capnograph can be caused by exhausted soda lime and inadequate fresh gas flow into the Bain circuit.
A sudden rise in EtCO2 can be caused deflation of the tourniquet.
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This question is part of the following fields:
- Physiology
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Question 12
Incorrect
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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:
Correct Answer: 20 ml/cmH2O
Explanation:Compliance of the respiratory system describes the expandability of the lungs and chest wall. There are two types of compliance: dynamic and static.
Dynamic compliance describes the compliance measured during breathing, which involves a combination of lung compliance and airway resistance. Defined as the change in lung volume per unit change in pressure in the presence of flow.
Static compliance describes pulmonary compliance when there is no airflow, like an inspiratory pause. Defined as the change in lung volume per unit change in pressure in the absence of flow.
For example, if a person was to fill the lung with pressure and then not move it, the pressure would eventually decrease; this is the static compliance measurement. Dynamic compliance is measured by dividing the tidal volume, the average volume of air in one breath cycle, by the difference between the pressure of the lungs at full inspiration and full expiration. Static compliance is always a higher value than dynamic
Static compliance can be computed using the formula:
Cstat = Tidal volume/Plateau pressure – PEEP
Substituting the values given,
Cstat = 800/50-10
Cstat = 20 ml/cmH2O -
This question is part of the following fields:
- Physiology
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Question 13
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 14
Incorrect
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A human's resting oxygen consumption (VO2) is typically 3.5 ml/kg/minute (one metabolic equivalent or 1 MET).
Which of the following options is linked to the highest VO2 when a person is at rest?Your Answer:
Correct Answer: Neonate
Explanation:The oxygen consumption rate (VO2) at rest is 3.5 ml/kg/minute (one metabolic equivalent or 1 MET).
3.86 ml/kg/minute thyrotoxicosisYoung children consume a lot of oxygen: around 7 ml/kg/min when they are born. The metabolic cost of breathing is higher in children than in adults, and it can account for up to 15% of total oxygen consumption. Similarly, an infant’s metabolic rate is nearly twice that of an adult, resulting in a larger alveolar minute volume and a lower FRC.
At term, oxygen consumption at rest can increase by as much as 40% (5 ml/kg/minute) and can rise to 60% during labour.
When compared to normal basal metabolism, sepsis syndrome increases VO2 and resting metabolic rate by 30% (4.55 ml/kg/minute). In septicaemic shock, VO2 decreases.
Dobutamine hydrochloride was infused into 12 healthy male volunteers at a rate of 2 micrograms per minute per kilogramme, gradually increasing to 4 and 6 micrograms per minute per kilogramme. Dobutamine was infused for 20 minutes for each dose. VO2 increased by 10% to 15%. (3.85-4.0 ml/kg/min).
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This question is part of the following fields:
- Physiology
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Question 15
Incorrect
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Following a near drowning accident, a 5-year-old child is admitted to the emergency department and advanced paediatric life support is started.
What is the child's approximate weight, according to the preferred formulae of the Resuscitation Council (UK), the European Resuscitation Council, and the Royal College of Anaesthetists?Your Answer:
Correct Answer: 20-25kg
Explanation:For estimating a child’s weight, the Resuscitation Council (UK) and European Resuscitation Council teach the following formula:
Weight = (age + 4) × 2
The weight of the child will be around 20 kg.
This formula is used in the Primary FRCA exam by the Royal College of Anaesthetists.
In ‘developed’ countries, the traditional ‘APLS formula’ for estimating weight in children based on age (wt in kg = [age+4] x 2) is acknowledged as underestimating weight by 33.4 percent on average, with the degree of underestimation increasing with increasing age.
However, more recently, the APLS formula ‘Weight=3(age)+7’ has been found to provide a mean underestimate of only 6.9%. This formula is applicable to children aged 1 to 13 years.
The estimated weight based on age using this formula is 25 kg.
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This question is part of the following fields:
- Physiology
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Question 16
Incorrect
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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.
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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 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 18
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:
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 19
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 20
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 21
Incorrect
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Regarding anti diuretic hormone (ADH), one of the following statements is correct:
Your Answer:
Correct Answer: Increases the total amount of electrolyte free water in the body
Explanation:The major action of ADH is to increase reabsorption of osmotically unencumbered water from the glomerular filtrate and decreases the volume of urine passed. The osmolarity of urine is increased to a maximum of four times that of plasma (approx. 1200 mOsm/kg) by Increasing water reabsorption.
Chronic water loading, Lithium, potassium deficiency, cortisol and calcium excess, all blunt the action of ADH. This leads to nephrogenic diabetes insipidus.
ADH’s primary site of action is the distal tubule and collecting duct.
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This question is part of the following fields:
- Physiology
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Question 22
Incorrect
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A morbidly obese (BMI=48) patient has the following co-morbidities: type II diabetes mellitus and hypertension. It is recommended for the patient to undergo bariatric surgery.
If the patient is laid flat for induction of anaesthesia, what physiologic changes of the respiratory system is the most important to consider?Your Answer:
Correct Answer: Functional residual capacity will decrease
Explanation:A decrease in the functional residual capacity (FRC) is the most important physiologic change to consider for such patients.
FRC is the sum of the expiratory reserve volume and the residual volume. It is the resting volume of the lung, and is an important marker for lung function. During this time, the alveolar pressure is equal to the atmospheric pressure. When morbidly obese individuals lie supine, the FRC decreases by as much as 40% because the abdominal contents push the diaphragm into the thoracic cavity.
Chest wall compliance is expected to reduce because of fat deposition surrounding adjacent structures.
Inspiratory reserve volume (IRV) is expected to increase, and peak expiratory flow is expected to decrease, however the decrease in FRC is more important to consider because of the risk of hypoxia secondary to premature airway closure and ventilation-perfusion mismatch.
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This question is part of the following fields:
- Physiology
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Question 23
Incorrect
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Which statement is true when describing carbonic anhydrase?
Your Answer:
Correct Answer: Isoenzyme IV is found in the brush border of the proximal convoluted tubule
Explanation:Carbonic anhydrase is an enzyme which contains zinc and can be found in:
1. Erythrocytes
2. Pulmonary endothelium
3. The intestine
4. Pancreas
5. Cardiac muscle and skeletal muscle.To date, there have been seven isoenzymes identified. Of note, isoenzyme IV is found in the brush border of the proximal convoluted tubule and isoenzyme II is found within the luminal cells.
Acetazolamides a carbonic anhydrase inhibitor and is used as prophylaxis against mountain sickness and in glaucoma management.
Spironolactone is a potassium diuretic and is an aldosterone antagonist.
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This question is part of the following fields:
- Physiology
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Question 24
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 25
Incorrect
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Using a negative feedback loop, Haem production is controlled by which of these enzymes?
Your Answer:
Correct Answer: ALA synthetase
Explanation:Heme a exists in cytochrome a and heme c in cytochrome c; they are both involved in the process of oxidative phosphorylation. 5′-Aminolevulinic acid synthase (ALA-S) is the regulated enzyme for heme synthesis in the liver and erythroid cells.
There are two forms of ALA Synthase, ALAS1, and ALAS2.
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This question is part of the following fields:
- Physiology
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Question 26
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 27
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 28
Incorrect
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A 20-year old male was involved in an accident and has presented to the Emergency Department with a pelvic crush injury.
The clinical exam according to ATLS protocol revealed the following:
Airway-patent
Breathing - respiratory rate 25 breaths per minute. Breath sounds are vesicular and there are no added sounds.
Circulation - Capillary refill time - 4 seconds. Peripheries are cool. Pulse 125 beats/min. BP - 125/95 mmHg.
Disability - GSC 15, anxious and in pain.
Secondary survey reveals no other injuries. The patient is administered high flow oxygen and IV access is established.
The most appropriate IV fluid regimen in this case will be which of the following?Your Answer:
Correct Answer: Judicious infusion of Hartmann's solution to maintain a systolic blood pressure greater than 90mmHg
Explanation:These clinical signs suggest that 15-30% of circulating blood volume has been lost.
Pelvic fractures are associated with significant haemorrhage (>2000 ml) that can be concealed. This may require aggressive fluid resuscitation which is initially with crystalloids and then blood. What is also important is including stabilisation of the fracture(s) and pain relief.
The Advanced Trauma Life Support (ATLS) classification of haemorrhagic shock is as follows:
Class I haemorrhage (blood loss up to 15%):
<750 ml of blood loss
Minimal tachycardia
No changes in blood pressure, RR or pulse pressure
Patients do not normally not require fluid replacement as will be restored in 24 hours, but in trauma, this needs to be correct.Class II haemorrhage (15-30% blood volume loss):
Uncomplicated haemorrhage requiring crystalloid resuscitation
Represents about 750 – 1500 ml of blood loss
Tachycardia, tachypnoea and a decrease in pulse pressure (due to a rise in diastolic component due action of catecholamines).
There are minimal systolic pressure changes.
There may be associated anxiety, fright or hostilityClass III haemorrhage (30-40% blood volume loss):
Complicated haemorrhagic state – crystalloid and probably blood replacement are required
There are classical signs of inadequate perfusion, marked tachycardia, tachypnoea, significant changes in mental state and measurable fall in systolic pressure.
Almost always require blood transfusion, but decision based on patient initial response to fluid resuscitation.Class IV haemorrhage (> 40% blood volume loss):
Preterminal event patient will die in minutes
Marked tachycardia, significant depression in systolic pressure and very narrow pulse pressure (or unobtainable diastolic pressure)
Mental state is markedly depressed
Skin cold and pale.
Needs rapid transfusion and immediate surgical intervention.A blood loss of >50% results in loss of consciousness, pulse and blood pressure.
Fluid resuscitation following trauma is a controversial area.
This clinical scenario points to a 15-30% blood loss. However, further crystalloid and blood replacement may be required after assessing the clinical situation. There is increasing evidence to suggest that transfusion of large volumes of crystalloid in the hospital setting are likely to be deleterious to the patient and hypotensive resuscitation and judicious blood and blood product resuscitation is a more appropriate option. A ratio of 1 unit of plasma to 1 unit of red blood cells is used to replace fluid volume in adults.
This patient does not require immediate transfusion of O negative blood and there is time for a formal crossmatch. The argument about colloids versus crystalloids has existed for decades. However, while they have a role in fluid resuscitation, they are not first line.
There is a risk of anaphylaxis, Hypernatraemia, and acute renal injury with colloidal solutions.
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This question is part of the following fields:
- Physiology
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Question 29
Incorrect
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Which one of the following factor affects the minimal alveolar concentration (MAC)?
Your Answer:
Correct Answer: Hypoxaemia
Explanation:The minimal alveolar concentration (MAC) is the concentration of an inhalation anaesthetic agent in the lung alveoli required to stop a response to the surgical stimulus in 50% of the patient.
Following factors don’t affect the MAC of the inhaled anaesthetic agents:
Gender, acidosis, alkalosis, hypothyroidism, hyperthyroidism, body weight, serum potassium level, and the duration of the anaesthesia.
MAC increase in children, elevated temperature, high metabolic rate, sympathetic increase and chronic alcoholism.
MAC decrease in low temperature, low oxygen level, old age, hypotension (<40 mmHg), depressant drugs e.g. opioids and low level of catecholamines; alpha methyl dopa. Carbon dioxide O2 at the pressure > 120mmHg is being used in anesthetic-Hinkman as an additive effect to decrease MAC, however, increase concentration of CO2 activates the sympathetic system resulting the MAC increases.
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This question is part of the following fields:
- Physiology
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Question 30
Incorrect
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The action potential in a muscle fibre is initiated by which of these ions?
Your Answer:
Correct Answer: Sodium ions
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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