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Question 1
Correct
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A common renal adverse effect of non-steroidal anti-inflammatory drugs is?
Your Answer: Haemodynamic renal insufficiency
Explanation:Prostaglandins do not play a major role in regulating RBF in healthy resting individuals. However, during pathophysiological conditions such as haemorrhage and reduced extracellular fluid volume (ECVF), prostaglandins (PGI2, PGE1, and PGE2) are produced locally within the kidneys and serve to increase RBF without changing GFR. Prostaglandins increase RBF by dampening the vasoconstrictor effects of both sympathetic activation and angiotensin II. These effects are important because they prevent severe and potentially harmful vasoconstriction and renal ischemia. Synthesis of prostaglandins is stimulated by ECVF depletion and stress (e.g. surgery, anaesthesia), angiotensin II, and sympathetic nerves.
Non-steroidal anti-inflammatory drugs (NSAIDs), such as ibuprofen and naproxen, potently inhibit prostaglandin synthesis. Thus administration of these drugs during renal ischemia and hemorrhagic shock is contraindicated because, by blocking the production of prostaglandins, they decrease RBF and increase renal ischemia. Prostaglandins also play an increasingly important role in maintaining RBF and GFR as individuals age. Accordingly, NSAIDs can significantly reduce RBF and GFR in the elderly.
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This question is part of the following fields:
- Physiology
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Question 2
Correct
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Which of the following, at a given PaO2, increases the oxygen content of arterial blood?
Your Answer: A reduced erythrocyte 2,3-diphosphoglycerate level
Explanation:The oxygen content of arterial blood can be calculated by the following equation:
(10 x haemoglobin x SaO2 x 1.34) + (PaO2 x 0.0225).
This is the sum of the oxygen bound to haemoglobin and the oxygen dissolved in the plasma.Oxygen content x cardiac output = The amount of oxygen delivered to the tissues in unit time which is known as the oxygen flux.
Any factor that increases the metabolic demand will encourage oxygen offloading from the haemoglobin in the tissues and this causes the oxygen dissociation curve (ODC) to shift to the right. This subsequently reduced the oxygen content of arterial blood.
Conditions like fever, metabolic or respiratory acidosis lowers the oxygen content and shifts the ODC to the right.
A low level of 2,3 diphosphoglycerate (2,3-DPG) is usually related to an increased oxygen content as there is less offloading, and so the ODC is shifted to the left.So for a given PaO2, a high blood oxygen content is related to any factors that can shift the ODC to the left and not to the right.
A low haematocrit usually means that there is a decreased haemoglobin concentration, and therefore is associated with decreased oxygen binding to haemoglobin.
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This question is part of the following fields:
- Physiology
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Question 3
Correct
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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 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 4
Correct
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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: 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 − 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 5
Incorrect
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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: 0.7
Correct 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 6
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: Glucose and urea molecules
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 7
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 8
Correct
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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 decrease at the same rate as a non-smoker
Explanation:For this patient, his forced expiratory volume in 1 second (FEV1) will decrease at the same rate as a non-smoker.
There is a notable, but slow, decline in FEV1 when an individual reaches the age of 26. An average reduction of 30 mls every year in non-smokers, while a more significant reduction of 50-70 mls is observed in approximately 20% of smokers.
Considering the age of the patient, individuals who begin smoking cessation by the age of 60 are far less likely to achieve normal FEV1 levels, even in the next five years. It is expected that their FEV1 will be approximately 14% less than their peers of the same age.
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This question is part of the following fields:
- Physiology
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Question 9
Incorrect
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A 30-year old female was anaesthetically induced for an elective open cholecystectomy. Upon mask ventilation, patient's oxygen saturation level dropped to 90% despite maximal head extension, jaw thrust and two handed mask seal. Intubation was performed twice but failed. Use of bougie also failed to localize the trachea. Oxygen levels continued to drop, but was maintained between 80 and 88% with mask ventilation.
Which of the following options is the best action to take for this patient?Your Answer: Perform a cannula cricothyroidotomy
Correct Answer: Insert a supraglottic airway
Explanation:A preplanned preinduction strategy includes the consideration of various interventions designed to facilitate intubation should a difficult airway occur. Non-invasive interventions intended to manage a difficult airway include, but are not limited to: (1) awake intubation, (2) video-assisted laryngoscopy, (3) intubating stylets or tube-changers, (4) SGA for ventilation (e.g., LMA, laryngeal tube), (5) SGA for intubation (e.g., ILMA), (6) rigid laryngoscopic blades of varying design and size, (7) fibreoptic-guided intubation, and (8) lighted stylets or light wands.
Most supraglottic airway devices (SADs) are designed for use during routine anaesthesia, but there are other roles such as airway rescue after failed tracheal intubation, use as a conduit to facilitate tracheal intubation and use by primary responders at cardiac arrest or other out-of-hospital emergencies. Supraglottic airway devices are intrinsically more invasive than use of a facemask for anaesthesia, but less invasive than tracheal intubation. Supraglottic airway devices can usefully be classified as first and second generation SADs and also according to whether they are specifically designed to facilitate tracheal intubation. First generation devices are simply ‘airway tubes’, whereas second generation devices incorporate specific design features to improve safety by protecting against regurgitation and aspiration.
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This question is part of the following fields:
- Physiology
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Question 10
Correct
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Which one of the following factor affects the minimal alveolar concentration (MAC)?
Your 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 11
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: 3000 mL 5% dextrose, each bag with 20 mmol of potassium
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 12
Correct
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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: 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 13
Incorrect
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In the fetal circulation, the cerebral and coronary circulations are preferentially supplied by oxygen-rich blood over other organs. This is possible because of which phenomenon?
Your Answer: The ductus venosus allows 60% of well oxygenated blood from the placenta to bypass the liver into the inferior vena cava
Correct Answer: Well oxygenated blood from the inferior vena cava is preferentially streamed across the patent foramen ovale
Explanation:During fetal development, blood oxygenated by the placenta flows to the foetus through the umbilical vein, bypasses the fetal liver through the ductus venosus, and returns to the fetal heart through the inferior vena cava.
Blood returning from the inferior vena cava then enters the right atrium and is preferentially shunted to the left atrium through the patent foramen ovale. Blood in the left atrium is then pumped from the left ventricle to the aorta. The oxygenated blood ejected through the ascending aorta is preferentially directed to the fetal coronary and cerebral circulations.
Deoxygenated blood returns from the superior vena cava to the right atrium and ventricle to be pumped into the pulmonary artery. Fetal pulmonary vascular resistance (PVR), however, is higher than fetal systemic vascular resistance (SVR); this forces deoxygenated blood to mostly bypass the fetal lungs. This poorly oxygenated blood enters the aorta through the patent ductus arteriosus and mixes with the well-oxygenated blood in the descending aorta. The mixed blood in the descending aorta then returns to the placenta for oxygenation through the two umbilical arteries.
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This question is part of the following fields:
- Physiology
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Question 14
Incorrect
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Regarding anti diuretic hormone (ADH), one of the following statements is correct:
Your Answer: Increases the reabsorption of water in the proximal tubule
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 15
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: Reduction in blood volume of 0-15%
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 16
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: DC shocks should be given on alternate cycles
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 17
Incorrect
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The SI unit of energy is the joule. Energy can be kinetic, potential, electrical or chemical energy.
Which of these correlates with the most energy?Your Answer: An object with a mass of 1500 kg moving at 30 m/s
Correct Answer: Energy released when 1 kg fat is metabolised to CO2 and water (the energy content of fat is 37 kJ/g)
Explanation:The derived unit of energy, work or amount of heat is joule (J). It is defined as the amount of energy expended if a force of one newton (N) is applied through a distance of one metre (N·m)
J = 1 kg·m/s2·m = 1 kg·m2/s2 or 1 kg·m2·s-2
Kinetic energy (KE) = ½ MV2
An object with a mass of 1500 kg moving at 30 m/s correlates to 675 kJ:
KE = ½ (1500) × (30)2 = 750 × 900 = 675 kJ
Total energy released when 1 kg fat is metabolised to CO2 and water is 37 MJ. 1 g fat produces 37 kJ/g, therefore 1 kg fat produces 37,000 × 1000 = 37 MJ.
Raising the temperature of 1 kg water from 0°C to 100°C correlates to 420 kJ. The amount of energy needed to change the temperature of 1 kg of the substance by 1°C is the specific heat capacity. We have 1 kg water therefore:
4,200 J × 100 = 420,000 J = 420 kJ
In order to calculate the energy involved in raising a 100 kg mass to a height of 1 km against gravity, we need to calculate the potential energy (PE) of the mass:
PE = mass × height attained × acceleration due to gravity
PE = 100 kg × 1000 m × 10 m/s2 = 1 MJThe heat generated when a direct current of 10 amps flows through a heating element for 10 seconds when the potential difference across the element is 1000 volts can be calculated by applying Joule’s law of heating:
Work done (WD) = V (potential difference) × I (current) × t (time)
WD = 10 × 10 × 1000 = 100 kJ -
This question is part of the following fields:
- Physiology
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Question 18
Incorrect
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The statement that best describes lactic acidosis is:
Your Answer:
Correct Answer: It can be precipitated by intravenous fructose
Explanation:An elevated arterial blood lactate level and an increase anion gap ([Na + K] – [Cl + HCO3]) of >20mmol gives rise to lactic acidosis. It can also be a result of overproduction and/or reduced metabolism of lactic acid.
The liver and kidney are the main sites of lactate metabolism, not skeletal muscle.
The two types of lactic acidosis that are known are:
Type A – due to tissue hypoxia, inadequate tissue perfusion and anaerobic glycolysis. These may be seen in cardiac arrest, shock, hypoxaemia and anaemia. The management of type A lactic acidosis involves reversing the underlying cause of the tissue hypoxia.
Type B – occurs in the absence of tissue hypoxia. Some of the causes of this include hepatic failure, renal failure, diabetes mellitus, pancreatitis and infection. Some drugs can also cause this lie aspirin, ethanol, methanol, biguanides and intravenous fructose.
The mainstay of treatment involves:
1. Optimising tissue oxygen delivery
2. Correcting the cause
3. Intravenous sodium bicarbonateIn resistant cases, peritoneal dialysis can be performed.
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This question is part of the following fields:
- Physiology
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Question 19
Incorrect
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An intravenous infusion is started with a 500 mL bag of 0.18 percent N. saline and 4% dextrose.
Which of the following best describes its make-up?Your Answer:
Correct Answer: Osmolarity 284 mOsmol/L, sodium 15 mequivalents and glucose 20 g
Explanation:30 mmol Na+ and 30 mmol Cl- are found in 1 litre of 0.18 percent N. saline with 4% dextrose. Percent (percent) refers to the number of grammes of a compound per 100 mL, so a litre of 4 percent dextrose solution contains 40 grammes.
As a result, a 500 mL bag of 1/5th N. saline and 4% dextrose contains approximately 15 mequivalents of sodium and 20 g of glucose. It is hypotonic due to its osmolarity of 284.
Because of the risk of hyponatraemia, it is no longer considered the crystalloid of choice for fluid maintenance in children.
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This question is part of the following fields:
- Physiology
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Question 20
Incorrect
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Which of the following 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 21
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 22
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 23
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 24
Incorrect
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Which of the following statement is true or false regarding to the respiratory tract?
Your Answer:
Correct Answer: The sympathetic innervation of the bronchi is derived from T2 - T4
Explanation:The diaphragm has three opening through which different structures pass from the thoracic cavity to the abdominal cavity:
Inferior vena cava passes at the level of T8.
Oesophagus, oesophageal vessels and vagi at T10.
Aorta, thoracic duct and azygous vein through T12.
Sympathetic trunk and pulmonary branches of vagus nerve form a posterior pulmonary plexus at the root of the lung. Fibres continue posteriorly from superficial cardiac plexus to form Anterior pulmonary plexus. It contains vagi nerves and superficial cardiac plexus. These fibres then follow the blood vessel and bronchi into the lungs.
The lower border of the pleura is at the level of:
8th rib in the midclavicular line
10th rib in the lower level of midaxillary line
T12 at its termination.
Both lungs have oblique fissure while right lung has transverse fissure too.
The trachea expands from the lower edge of the cricoid cartilage (at the level of the 6th cervical vertebra) to the carina.
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This question is part of the following fields:
- Physiology
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Question 25
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 26
Incorrect
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The fluids with the highest osmolarity is?
Your Answer:
Correct Answer: 0.45% N. Saline with 5% glucose
Explanation:The concentration of solute particles per litre (mosm/L) = the osmolarity of a solution. Changes in water content, ambient temperature, and pressure affects osmolarity. The osmolarity of any solution can be calculated by adding the concentration of key solutes in it.
Individual manufacturers of crystalloids and colloids may have different absolute values but they are similar to these.
0.45% N. Saline with 5% glucose:
Tonicity – hypertonic
Osmolarity – 405 mosm/L
Kilocalories (kCal) – 1070.9% N. Saline:
Tonicity – isotonic
Osmolarity – 308 mosm/L
Kilocalories (kCal) – 05% Dextrose:
Tonicity – isotonic
Osmolarity – 253 mosm/L
Kilocalories (kCal) – 170Gelofusine (154 mmol/L Na, 120 mmol/L Cl):
Tonicity – isotonic
Osmolarity – 274 mosm/L
Kilocalories (kCal) – 0Hartmann’s solution:
Tonicity – isotonic
Osmolarity – 273 mosm/L
Kilocalories (kCal) – 9 -
This question is part of the following fields:
- Physiology
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Question 27
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 28
Incorrect
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Intracellular effectors are activated by receptors on the cell surface. These receptors receive signals that are relayed by second messenger systems.
In the human body, which second messenger is most abundant?Your Answer:
Correct Answer: Calcium ions
Explanation:Second messengers relay signals to target molecules in the cytoplasm or nucleus when an agonist interacts with a receptor on the cell surface. They also amplify the strength of the signal. The most ubiquitous and abundant second messenger is calcium and it regulates multiple cellular functions in the body.
These include:
Muscle contraction (skeletal, smooth and cardiac)
Exocytosis (neurotransmitter release at synapses and insulin secretion)
Apoptosis
Cell adhesion to the extracellular matrix
Lymphocyte activation
Biochemical changes mediated by protein kinase C.cAMP is either inhibited or stimulated by G proteins.
The receptors in the body that stimulate G proteins and increase cAMP include:
Beta (?1, ?2, and ?3)
Dopamine (D1 and D5)
Histamine (H2)
Glucagon
Vasopressin (V2).The second messenger for the action of nitric oxide (NO) and atrial natriuretic peptide (ANP) is cGMP.
The second messengers for angiotensin and thyroid stimulating hormone are inositol triphosphate (IP3) and diacylglycerol (DAG).
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This question is part of the following fields:
- Physiology
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Question 29
Incorrect
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Which of the following statements is about the measurement of glomerular filtration rate (GFR) is correct?
Your Answer:
Correct Answer: The result matches clearance of the indicator if it is renally inert
Explanation:The measurements of GFR are done using renally inert indicators like inulin, where passive rate of filtration at the glomerulus = rate of excretion. Normal value is about 180 litres per day.
GFR is altered by renal blood flow but blood flow does not need to be measured.
The reabsorption of Na leads to a low excretion rate and low urine concentration and therefore its use as an indicator would lead to an erroneously LOW GFR.
If there is tubular secretion of any solute, the clearance value will be higher than that of inulin. This will be either due to tubular reabsorption or the solute not being freely filtered at the glomerulus.
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This question is part of the following fields:
- Physiology
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Question 30
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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