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Question 1
Correct
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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: 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 2
Correct
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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 variable
Explanation:A mercury barometer can be used to determine absolute pressure. A glass tube with one closed end serves as the barometer. The open end is inserted into a mercury-filled open vessel. The mercury in the container is pushed into the tube by atmospheric pressure exerted on its surface. Absolute pressure is the distance between the tube’s meniscus and the mercury surface.
Pressure is defined as force in newtons per unit area (F) (A).
Mass of mercury = area (A) × density (�) × length (L)
Pressure = ((A × � × L) × 9.8 m/s2)/A
Pressure = � × L x 9.8
Pressure is proportional to LThe numerator and denominator of the above equation, area (A), cancel out. The constants are density and the gravitational acceleration value.
The length is proportional to the applied pressure.
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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 is true in the Kreb's cycle?
Your Answer: Krebs' cycle can function under anaerobic conditions
Correct Answer: Alpha-ketoglutarate is a five carbon molecule
Explanation:Krebs’ cycle (tricarboxylic acid cycle or citric acid cycle) is a sequence of reactions to release stored energy through oxidation of acetyl coenzyme A (acetyl-CoA). Some of the products are carbon dioxide and hydrogen atoms.
The sequence of reactions, known collectively as oxidative phosphorylation, only occurs in the mitochondria (not cytoplasm).
The Krebs cycle can only take place when oxygen is present, though it does not require oxygen directly, because it relies on the by-products from the electron transport chain, which requires oxygen. It is therefore considered an aerobic process. It is the common pathway for the oxidation of carbohydrate, fat and some amino acids, required for the formation of adenosine triphosphate (ATP).
Pyruvate enters the mitochondria and is converted into acetyl-CoA. Acetyl-CoA is then condensed with oxaloacetate, to form citrate which is a six carbon molecule. Citrate is subsequently converted into isocitrate, alpha-ketoglutarate, succinyl-CoA, succinate, fumarate, malate and finally oxaloacetate.
The only five 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 4
Incorrect
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Of the following, which option best describes the muscle type that has the fastest twitch response to stimulation?
Your Answer: Type I skeletal muscle
Correct Answer: Type IIb skeletal muscle
Explanation:Human skeletal muscle is composed of a heterogeneous collection of muscle fibre types which differ histologically, biochemically and physiologically.
It can be biochemically classified into 2 groups. This is based on muscle fibre myosin ATPase histochemistry. These are:
Type 1 (slow twitch): Muscle fibres depend upon aerobic glycolytic metabolism and aerobic oxidative metabolism. They are rich in mitochondria, have a good blood supply, rich in myoglobin and are resistant to fatigue.
Type II (fast twitch): Muscle fibres are sub-divided into:
Type IIa – relies on aerobic/oxidative metabolism
Type IIb – relies on anaerobic/glycolytic metabolism.Fast twitch muscle fibres produce short bursts of power but are more easily fatigued.
Cardiac and smooth muscle twitches are relatively slow compared with skeletal muscle.
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This question is part of the following fields:
- Physiology
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Question 5
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: Replace the soda lime
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 6
Incorrect
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Which of the following statements is about the measurement of glomerular filtration rate (GFR) is correct?
Your Answer: Use of sodium as an indicator would lead to an erroneously high GFR
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 7
Incorrect
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Which of the following statement is true regarding the paediatric airway?
Your Answer: By the age of 1 year the vocal cords lie opposite C5/6
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 8
Correct
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All of the statements describing the blood brain barrier are false EXCEPT:
Your 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 9
Incorrect
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Left ventricular afterload is mostly calculated from systemic vascular resistance. Which one of the following factors has most impact on systemic vascular resistance?
Your Answer: Pre-capillary arterioles
Correct Answer: Small arterioles
Explanation:Systemic vascular resistance (SVR), also known as total peripheral resistance (TPR), is the amount of force exerted on circulating blood by the vasculature of the body. Three factors determine the force: the length of the blood vessels in the body, the diameter of the vessels, and the viscosity of the blood within them. The most important factor that determines the systemic vascular resistance (SVR) is the tone of the small arterioles.
These are otherwise known as resistance arterioles. Their diameter ranges between 100 and 450 µm. Smaller resistance vessels, less than 100 µm in diameter (pre-capillary arterioles), play a less significant role in determining SVR. They are subject to autoregulation.
Any change in the viscosity of blood and therefore flow (such as due to a change in haematocrit) might also have a small effect on the measured vascular resistance.
Changes of blood temperature can also affect blood rheology and therefore flow through resistance vessels.
Systemic vascular resistance (SVR) is measured in dynes·s·cm-5
It can be calculated from the following equation:
SVR = (mean arterial pressure ˆ’ mean right atrial pressure) × 80 cardiac output
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This question is part of the following fields:
- Physiology
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Question 10
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: Thyrotoxicosis
Correct Answer: Neonate
Explanation:The oxygen consumption rate (VO2) at rest is 3.5 ml/kg/minute (one metabolic equivalent or 1 MET).
3.86 ml/kg/minute 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 11
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: High altitude
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 12
Correct
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If a large volume of 0.9% N. saline is administered during resuscitation, it is most likely to cause?
Your Answer: Hyperchloremic metabolic acidosis
Explanation:Crystalloids recommended for fluid resuscitation include 0.9% N saline and Hartmann’s solution(a physiological solution). 0.9% N. saline is not a physiological solution for the following reasons:
Compared with the normal range of 98-102 mmol/L, its chloride concentration is high (154 mmol/L)
It lacks calcium, magnesium, glucose and potassium
It does not have bicarbonate or bicarbonate precursor buffer necessary to maintain plasma pH within normal limitsThere is a difference in the activity (concentration) of strong ions at a physiological pH. This imbalance can explain abnormalities of acid base balance. A normal strong ion difference (SID) is in the order of 40.
SID = ([Na+] + [K+] + [Ca2+] + [Mg2+]) – ([Cl-] + [lactate] + [SO42-])
This imbalance is made up with the weaker anions to maintain electrical neutrality.
Administration of a large volume of 0.9% normal saline during resuscitation results in excessive chloride administration and this impairs renal bicarbonate reabsorption. The SID of 0.9% normal saline is 0 (Na+ = 154mmol/L and Cl- = 154mmol/L = 154 – 154 = 0). A large volume of NS will decrease the plasma SID causing an acidosis.Other causes of a hyperchloremic acidosis are:
Diabetic ketoacidosis
Total Parenteral Nutrition
Overdose of ammonium chloride and hydrochloric acid
Gastrointestinal losses of bicarbonate like in diarrhoea and pancreatic fistula
Proximal renal tubular acidosis with failure of bicarbonate reabsorption -
This question is part of the following fields:
- Physiology
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Question 13
Correct
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Regarding amide local anaesthetics, which one factor has the most significant effect on its duration of action?
Your 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 14
Incorrect
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Concerning forced alkaline diuresis, which of the following statements is true?
Your Answer: Is used in poisoning or overdoses with alkaline drugs
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
Correct
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The typical fluid compartments in a normal 70kg male are:
Your Answer: intracellular>extracellular
Explanation:Body fluid compartments in a 70kg male:
Total volume=42L (60% body weight)
Intracellular fluid compartment (ICF) =28L
Extracellular fluid compartment (ECF) = 14LECF comprises:
Intravascular fluid (plasma) = 3L
Extravascular fluid = 11LExtravascular fluids comprises:
Interstitial fluid = 10.5L
Transcellular fluid = 0.5L -
This question is part of the following fields:
- Physiology
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Question 16
Incorrect
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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: CYP3A4
Correct 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 17
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: Irreversible brain damage
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 18
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: 8.3
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 19
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 20
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: 1 mg IV adrenaline and 300 mg IV amiodarone should be administered
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 21
Correct
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A 27-year old man loses 1000ml blood after being stabbed on his thigh. The most impactful physiological response occurring at the start to combat the decline in the intravascular blood volume of this man is?
Your Answer: Venoconstriction
Explanation:In contrast to the arterial system, which contains 15% of the circulating blood volume, the body’s veins contain 70% of it.
In severe haemorrhage, when sympathetic stimulation causes venoconstriction, venous tone is important in maintaining the return of blood to the heart.
Because the liver receives about 30% of the resting cardiac output, it is a very vascular organ. The hepatic vascular system is dynamic, which means it can store and release blood in large amounts – it acts as a reservoir within the general circulation.
In a normal situation, the liver contains 10-15% of total blood volume, with the sinusoids accounting for roughly 60% of that. The liver dynamically adjusts its blood volume when blood is lost and can eject enough blood to compensate for a moderate amount of haemorrhage.
In the portal venous and hepatic arterial systems, sympathetic nerves constrict the presinusoidal resistance vessels. More importantly, sympathetic stimulation lowers the portal system’s capacitance, allowing blood to flow more efficiently to the heart.
Net transcapillary absorption of interstitial fluid from skeletal muscle into the intravascular space compensates for blood loss effectively during haemorrhage. The decrease in capillary hydrostatic pressure (Pc), caused by reflex adrenergic readjustment of the ratio of pre- to postcapillary resistance, is primarily responsible for fluid absorption. Within a few hours of blood loss, these fluid shifts become significant, further diluting haemoglobin and plasma proteins.
Albumin synthesis begins to increase after 48 hours.
The juxtamedullary complex releases renin in response to a drop in mean arterial pressure, which causes an increase in aldosterone level and, eventually, sodium and water resorption. Increased antidiuretic hormone (ADH) levels also contribute to water retention.
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This question is part of the following fields:
- Physiology
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Question 22
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: The basal lung units are better perfused than apical units
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 23
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: Albumin
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 24
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: Immediate transfusion of four units of O -ve blood
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%):
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 25
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: Increased end-diastolic volume
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 26
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: 288-294
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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Question 27
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: 15-19 kg
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 28
Incorrect
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Regarding bilirubin, which one of the following statement is true?
Your Answer: Conjugated bilirubin is lipid soluble
Correct Answer: Conjugated bilirubin is stored in the gall bladder
Explanation:Bilirubin is the tetrapyrrole and a catabolic product of heme. 70-90% of bilirubin is end product of haemoglobin degradation in the liver.
Bilirubin circulates in the blood in 2 forms; unconjugated and conjugated bilirubin.
Unconjugated bilirubin is insoluble in water. It travels through the bloodstream to the liver, where it changes from insoluble into a soluble form (i.e.; unconjugated into conjugated form).
This conjugated bilirubin travels from the liver into the small intestine and the gut bacteria convert bilirubin into urobilinogen and then into urobilin (not urobilin to urobilinogen). A very small amount passes into the kidneys and is excreted in urine.
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This question is part of the following fields:
- Physiology
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Question 29
Correct
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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: 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 30
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: The FEV1 will return to normal within 5 years
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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