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Question 1
Correct
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A 76-year-old man, presents to his general practitioner with a lump in his left groin. Upon examination, his doctor is able to diagnose a direct inguinal hernia. There are many structures present in the inguinal canal.
Where is the ilioinguinal nerve located in relation to the spermatic cord?Your Answer: Anterior to the spermatic cord
Explanation:The answer is anterior to the spermatic cord.
The inguinal canal in men contains the ilioinguinal nerve, the genitofemoral nerve and the spermatic cord.
The ilioinguinal nerve arises of the L1 nerve root with the Iliohypogastric nerve, before entering the inguinal canal from the side, through the muscles of the abdomen, travelling superficial to the spermatic cord.
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This question is part of the following fields:
- Anatomy
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Question 2
Incorrect
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A patient was brought to the emergency room after passing black tarry stools. The initial diagnosis was upper gastrointestinal bleeding. The patient was placed on temporary nil per os (NPO) for the next 24 hours, his weight was 110 kg, and the required volume of intravenous fluid for the him was 3 litres. His electrolytes and other biochemistry studies were normal.
If you were to choose the intravenous fluid regimen that would closely mimic his basic electrolyte and caloric requirements, which one would be the best answer?Your Answer: 3000 mL 0.9% N. saline
Correct Answer: 3000 mL 0.45% N. saline with 5% dextrose, each bag with 40 mmol of potassium
Explanation:The patient in the case has a fluid volume requirement of 30 mL/kg/day. His basic electrolyte requirement per day is:
Sodium at 2 mmol/kg/day x 110 = 220 mmol/day
Potassium at 1 mmol/kg/day x 110 = 110 mmol/dayHis energy requirement per day is:
35 kcal/kg/day x 110 kg = 3850 kcal/day
One gram of glucose in fluid can provide approximately 4 kilocalories.
The following are the electrolyte components of the different intravenous fluids:
Fluid Na (mmol/L) K (mmol/L)
0.9% Normal saline (NSS) 154 0
0.45% NSS + 5% dextrose 77 0
0.18% NSS + 4% dextrose 30 0
Hartmann’s 131 5
5% dextrose 0 01000 mL of 5% dextrose has 50 g of glucose
Option B is inadequate for his sodium and caloric requirements (30 mmol of Na+ and 560 kcal). It is adequate for his K+ requirement (120 mmol of K+).
Option C is in excess of his Na+ requirement (462 mmol of Na+). Moreover, it does not provide any K+ replacement.
Option D is inadequate for his caloric requirement (600 kcal) and K+ requirement (60 mmol of K+). Moreover it does not provide any Na+ replacement.
Option E is in excess of his Na+ requirement (393 mmol of Na+), and is inadequate for his potassium requirement (15 mmol of K+)
Option A has adequate amounts for his Na+ (231 mmol of Na+) and K+ (120 mmol of K+) requirements. It is inadequate for his caloric requirement (600 kcal).
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This question is part of the following fields:
- Physiology
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Question 3
Incorrect
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Which of the following anaesthetic agent is most potent with the lowest Minimal Alveolar Concentration (MAC)?
Your Answer: Sevoflurane
Correct Answer: Isoflurane
Explanation:The clinical potency of the anaesthetic agent is measured using minimal alveolar concentration(MAC).
MAC and oil: gas partition coefficient is inversely related. Anaesthetic agent Oil/gas partition coefficient and Minimal alveolar concentration (MAC) is given respectively as
Desflurane 18 6
Isoflurane 90 1.2
Nitrous oxide 1.4 104
Sevoflurane 53.4 2
Xenon 1.9 71With these data, we can conclude Isoflurane is the most potent with the highest oil/gas partition coefficient of 90 and the lowest MAC of 1.2
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This question is part of the following fields:
- Pharmacology
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Question 4
Correct
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Obeying Boyle's law and Charles's law is a characteristic feature of an ideal gas.
The gas which is most ideal out of the following options is?Your Answer: Helium
Explanation:The ideal gas equation makes the following assumptions:
The gas particles have a small volume in comparison to the volume occupied by the gas.
Between the gas particles, there are no forces of interaction.
Individual gas particle collisions, as well as gas particle collisions with container walls, are elastic, meaning momentum is conserved.
PV = nRT
Where:P = pressure
V = volume
n = moles of gas
T = temperature
R = universal gas constantHelium is a monoatomic gas with a small helium atom. The attractive forces between helium atoms are small because the helium atom is spherical and has no dipole moment. Because helium atoms are spherical, collisions between them approach the ideal state of elasticity.
Most real gases behave qualitatively like ideal gases at standard temperatures and pressures. When intermolecular forces and molecular size become important, the ideal gas model tends to fail at lower temperatures or higher pressures. It also fails to work with the majority of heavy gases.
Helium, argon, neon, and xenon are noble or inert gases that behave the most like an ideal gas. Xenon is a noble gas with a much larger atomic size than helium.
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This question is part of the following fields:
- Pharmacology
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Question 5
Incorrect
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The Medical Admissions unit receives a 71-year-old woman. She has type 2 diabetes, which she manages with diet, but she has been feeling ill for the past 48 hours.
Her pulse rate is 110 beats per minute, her blood pressure is 90/50 mmHg, and she is clinically dehydrated. Her respiratory rate is 20 breaths per minute, and chest auscultation reveals no focal signs.
The following are the lab results:
Glucose 27.4 mmol/L (3.5-5.5)
Ketones 2.5 mmol/L (<0.1)
Urinary glucose is zero (dipstick) with ketones
A random blood glucose of 15.3 mmol/L was measured during a visit to the diabetic clinic one month prior to admission, according to her notes, and a urinary dipstick registered a high glucose and ketones++.
The discrepancy between plasma and urinary glucose measurements is best explained by which of the following physiological mechanisms?Your Answer: The renal plasma glucose threshold is abnormally low
Correct Answer: The glomerular filtration rate is abnormally low
Explanation:The glucose molecule enters the Bowman’s capsule freely and becomes part of the filtrate.
All glucose is reabsorbed in the proximal convoluted tubule when blood glucose concentrations are below a certain threshold (approximately 11 mmol/L) (PCT). Active transportation makes this possible. In the proximal tubular cells, sodium/glucose cotransporters (SGLT1 and SGLT2) are the proteins responsible.
Glucose does not normally appear in the urine below the renal threshold.
The renal glucose threshold is not set in stone and is affected by a variety of factors, including GFR, TmG, and the quantity of splay.
The different absorptive and filtering capacities of individual nephrons cause splay, which is the rounding of a glucose reabsorption curve.
The SGLT proteins have a high affinity for glucose, but not an infinite affinity. As a result, some glucose may escape reabsorption before the TmG. A decrease in renal threshold may be caused by an increase in splay.
Because the filtered glucose load is reduced and the PCT can reabsorb all of the filtered glucose despite hyperglycaemia, a low GFR causes an increase in TmG. In contrast, lowering the TmG lowers the threshold because the tubules’ ability to reabsorb glucose is reduced.
A reduction in GFR caused by severe dehydration and reduced perfusion pressure is the most obvious cause of the discrepancy between plasma and urinary glucose in this scenario.
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This question is part of the following fields:
- Pathophysiology
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Question 6
Incorrect
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At 37 weeks' gestation, a 29-year-old parturient is admitted to the labour ward. Her antenatal period was asymptomatic for her.
The haematological values listed below are available:
Hb concentration of 100 g/L (115-165)
200x109/L platelets (150-400)
MCV 81 fL (80-96)
Which of the following is the most likely reason for the problem?Your Answer:
Correct Answer: Iron deficiency
Explanation:This patient’s limited haematological profile includes mild normocytic anaemia and a normal platelet count.
Iron deficiency is the most common cause of anaemia during pregnancy. It affects 75 to 95 percent of patients. A haemoglobin level of less than 110 g/L in the first trimester and less than 105 g/L in the second and third trimesters is considered anaemia. There will usually be a low mean cell volume (MCV), mean cell haemoglobin (MCH), and mean cell haemoglobin concentration in addition to a low haemoglobin (MCHC). The MCV may be normal in mild cases of iron deficiency or coexisting vitamin B12 and folate deficiency.
To determine whether you have an iron deficiency, you’ll need to take more tests. Low serum ferritin (15 g/L) and less reliable indices like serum iron and total iron binding capacity are among them.
A number of factors contribute to iron deficiency in pregnancy, including:
Insufficient dietary iron to meet the mother’s and foetus’ nutritional needs
Multiple pregnancies
Blood loss, as well as
Absorption of iron from the gut is reduced.The volume of plasma increases by about 50% during pregnancy, but the mass of red blood cells (RBCs) increases by only 30%. Dilutional anaemia is the result of this situation. From the first trimester to delivery, the RBC mass increases linearly, while the plasma volume plateaus, stabilises, or falls slightly near term. As a result, between 28 and 34 weeks of pregnancy, haemoglobin concentrations are at their lowest. The effects of haemodilution will be negated in this patient because she is 37 weeks pregnant.
Vitamin B12 and folate deficiency are less common causes of anaemia in pregnancy. The diagnosis could be ruled out if the MVC is normal.
During pregnancy, the platelet count drops, especially in the third trimester. Gestational thrombocytopenia is the medical term for this condition. It’s due to a combination of factors, including haemodilution and increased platelet activation and clearance. Pre-eclampsia and HELLP syndrome are common causes of thrombocytopenia. Pre-eclampsia isn’t the only cause of anaemia during pregnancy.
A typical blood picture of a haemoglobinopathy like sickle cell disease shows quantitative and qualitative defects, with the former leading to a severe anaemia exacerbated by haemodilution and other factors that contribute to iron deficiency. Microcytic cells are the most common type.
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This question is part of the following fields:
- Pathophysiology
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Question 7
Incorrect
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Over the course of 10 minutes, one litre of 0.9% normal saline is intravenously infused into a normally fit and well 58-year-old male. A catheter is used to measure urine output before and after the infusion. The patient is 70 kg in weight.
The following data on urine output is obtained:
50ml/hour Before the infusion
200 ml/hour 1 hour following infusion
90 ml/hour 2 hours after the infusion
60 ml/hr 3 hours after the infusion
Which of the following physiological responses is most likely to account for the sudden increase in urine output after a fluid bolus?Your Answer:
Correct Answer: Increased glomerular filtration rate
Explanation:The following are some basic assumptions:
Extracellular fluid (ECF) makes up one-third of total body water (TBW), while intracellular fluid makes up the other two-thirds (ICF).
One-quarter of ECF is plasma, and three-quarters is interstitial fluid (ISF).
The volume receptors have a 7-10% blood volume change threshold. The osmoreceptors are sensitive to changes in osmolality of 1-2 percent.
Prior to the transfusion, the plasma osmolality is normal (between 287 and 290 mOsm/kg).
[Na+] in 0.9 percent N. saline is 154 mmol/L, which is similar to that of extracellular fluid. When given intravenously, this limits its distribution within the extracellular space, resulting in a plasma compartment:ISF volume ratio of 1:3.
In this time frame, one litre of 0.9 percent N. saline will increase plasma volume by about 250 mL, which could be the threshold for activation of the volume receptors in the atria, resulting in the release of atrial natriuretic peptide (ANP).Because 0.9 percent N. saline is isosmotic, after a 1 L infusion, plasma osmolality will not change. No changes in antidiuretic hormone secretion will be detected by the hypothalamic osmoreceptors.
Because normal saline is protein-free, the oncotic pressure in the blood is slightly reduced after the saline infusion. As a result, fluid movement into the ISF is favoured (Starling’s hypothesis), and the lower oncotic pressure causes an immediate increase in the glomerular filtration rate (GFR) and a reduction in water reabsorption in the proximal tubule.
The flow of urine increases. There is no hormonal intermediary in this effect, so it is strictly local. Urine flow immediately increases. The fluid returns to the intravascular compartment, and urine flow continues until all of the transfused fluid has been excreted.
Blood pressure changes associated with a 1 L fluid infusion are unlikely to affect high-pressure baroreceptors in the carotid sinus.
The juxta-glomerular cells of the afferent arteriole are adjacent to the specialised cells (macula densa) of distal tubules. The sodium and chloride ions in the tubular fluid are detected by the macula densa. Renin release is inhibited when the tubular fluid contains too much sodium chloride. Hormonal changes take longer to manifest than physical changes that control glomerulotubular balance.
Hypertonic saline, not 0.9 percent N saline, is an osmotic diuretic. -
This question is part of the following fields:
- Pathophysiology
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Question 8
Incorrect
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An 81-year old man was admitted to the Pulmonology Ward due to chronic obstructive airway disease (COPD) exacerbation. Upon physical examination, he was hypertensive at 140/90 mmHg, tachycardic at 114 beats per minute, and tachypnoeic at 33 breaths per minute.
Arterial blood gas analysis was obtained and showed the following results:
pH: 7.25 (Reference range: 7.35-7.45)
PaO2: 73 mmHg (9.7 kPa) (Reference range: 11.3-12.6 kPa)
PaCO2: 56 mmHg (7.5 kPa) (Reference range: 4.7-6.0 kPa)
SaO2: 90%
Standard bicarbonate: 29 mmol/L (Reference range: 20-28)
BE: +4 mmol/L (± 2)
Which of the following options has the most significant impact on his respiratory rate?Your Answer:
Correct Answer: CSF pH
Explanation:The arterial blood gas analysis indicates presence of acute respiratory acidosis.
Central chemoreceptors are located in the ventral medulla and respond directly to presence of hydrogen ions in the CSF. When stimulated, it causes an increase in respiratory rate.
It is believed that hydrogen ions may be the only important direct stimulus for these neurons, however, CO2 is believed to stimulate these neurons secondarily by changing the hydrogen ion concentration.
Changes in O2 concentration have virtually no direct effect on the respiratory centre itself to alter respiratory drive. Although, O2 changes do have an indirect effect by acting through the peripheral chemoreceptors.
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This question is part of the following fields:
- Pathophysiology
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Question 9
Incorrect
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An otherwise fit 7-year-old boy for an elective tonsillectomy is seen holding a bottle of sugared orange squash at anaesthetic assessment . He appears to have consumed 120 mL of the bottle's contents.
What is the minimal safe fasting time prior to proceeding with a general anaesthetic in this patient?Your Answer:
Correct Answer: 1 hour
Explanation: -
This question is part of the following fields:
- Pathophysiology
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Question 10
Incorrect
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Which of the following best describes the phenomenon of higher partial pressures of oxygen in the alveoli (PAO2) in the lung apices when in the upright position?
Your Answer:
Correct Answer: The ventilation/perfusion (V/Q) ratio is greater than in the basal units
Explanation:The ventilation/perfusion ratio varies in different areas of the lung. In an upright individual, although both ventilation and perfusion increase from the apex to the base of the lung, the increase in ventilation is less than the increase in blood flow. As a result, the normal V̇ /Q̇ ratio at the apex of the lung is much greater than 1 (ventilation exceeds perfusion), whereas the V̇ /Q̇ ratio at the base of the lung is much less than 1 (perfusion exceeds ventilation).
There is more volume in the alveoli found in the apices than in the bases of the lungs. This is due to the weight of the lung stretching the apical alveoli to the maximum size. Also, the weight of the lungs pull themselves away from the chest wall, creating a negative intrapleural pressure. These factors, however, do not directly affect the PAO2.
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This question is part of the following fields:
- Pathophysiology
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Question 11
Incorrect
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In the adrenal gland:
Your Answer:
Correct Answer: Catecholamine release is mediated by cholinergic nicotinic transmission
Explanation:The adrenal (suprarenal) gland is composed of two main parts: the adrenal cortex, which is the largest and outer part of the gland, and the adrenal medulla. The adrenal cortex consists of three zones: 1. Zona glomerulosa (outermost layer) is responsible for the production of mineralocorticoids, mainly aldosterone, which regulates blood pressure and electrolyte balance. 2. Zona fasciculata (middle layer) is responsible for the production of glucocorticoids, predominantly cortisol, which increases blood sugar levels via gluconeogenesis, suppresses the immune system, and aids in metabolism. It also produces 11-deoxycorticosterone and corticosterone in addition to cortisol. 3. Zona reticularis (innermost layer) is responsible for the production of gonadocorticoids, mainly dehydroepiandrosterone (DHEA), which serves as the starting material for many other important hormones produced by the adrenal gland, such as oestrogen, progesterone, testosterone, and cortisol. It is also responsible for administering these hormones to the reproductive regions of the body.
The adrenal medulla majorly secretes epinephrine (adrenaline), and norepinephrine in small quantity. Both hormones have similar functions and initiate the flight or fight response.
Catecholamine is mediated by cholinergic nicotinic transmission through changes in sympathetic nervous system (T5 – T11), being increased during stress and hypoglycaemia.
Blood supply to the adrenal gland is by these three arteries: superior suprarenal arteries, middle suprarenal artery and inferior suprarenal artery. Venous drainage is via the suprarenal vein to the left renal vein or directly to the inferior vena cava on the right side. There is no portal (venous) system between cortex and medulla.
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This question is part of the following fields:
- Anatomy
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Question 12
Incorrect
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A 60-year old male has anaemia and is being investigated. The most common combination of globin chains in a normal adult is:
Your Answer:
Correct Answer: α2β2
Explanation:There are 4 different types of globin chains which surround 4 heme molecules in haemoglobin (Hb) – α (alpha), β (beta), γ (gamma), and δ (delta)
α chains are essential.
δ2β2 and β2γ2 are not found in a healthy adult.
97% of the Hb in a healthy adult is made of α2β2 (2 α chains and 2 β chains).
α2δ2 accounts for around 1.5-3% of the adult Hb.
α2γ2 accounts for less than 1%.With respect to oxygen transport in cells, almost all oxygen is transported within erythrocytes. There is limited solubility and only 1% is carried as solution. Thus, the amount of oxygen transported depends upon haemoglobin concentration and its degree of saturation.
Haemoglobin is a globular protein composed of 4 subunits. Haem is made up of a protoporphyrin ring surrounding an iron atom in its ferrous state. The iron can form two additional bonds – one is with oxygen and the other with a polypeptide chain. There are two alpha and two beta subunits to this polypeptide chain in an adult and together these form globin. Globin cannot bind oxygen but can bind to CO2 and hydrogen ions. The beta chains are able to bind to 2,3 diphosphoglycerate. The oxygenation of haemoglobin is a reversible reaction. The molecular shape of haemoglobin is such that binding of one oxygen molecule facilitates the binding of subsequent molecules.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 13
Incorrect
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In reference to confounding variables, which among the given is not true?
Your Answer:
Correct Answer: In the analytic stage of a study confounding can be controlled for by randomisation
Explanation:Randomisation can be used to provide control over the confounding variables during the design stage of a study however during analytical stage a technique called stratification is used for controlling confounding variables. Since the question asks for the information that is factually incorrect.
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This question is part of the following fields:
- Statistical Methods
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Question 14
Incorrect
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A chain smoker is interested in knowing how many years of his life would be lessened by smoking. You tell him explicitly that precise determination is impossible but you can tell him the proportion of people who died due to smoking. Which of the following epidemiological term is apt in this regard?
Your Answer:
Correct Answer: Attributable risk - the rate in the exposed group minus the rate in the unexposed group
Explanation:Attributable proportion is the proportion of disease that is caused due to exposure. It refers to the proportion of disease that would be eradicated from a particular population if the disease rate was diminished to match that of the unexposed group.
Risk ratio (relative risk) compares the probability of an event in an exposed (experimental) group to that of an event in the unexposed (control) group. Thus two are not the same.
The attributable risk is the rate of a disease in an exposed group to that of a group that has not been exposed to it i.e. how many deaths did the exposure cause.
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This question is part of the following fields:
- Statistical Methods
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Question 15
Incorrect
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Modified rapid sequence induction with 1mg/kg rocuronium was done is a 42-year-old male for emergency appendicectomy. The patient weight was 70 kg and a failed intubation indicated immediate reversal of neuromuscular blockade.
Which of the following is the correct dose of sugammadex for this patient?Your Answer:
Correct Answer: 1120 mg
Explanation:Sugammadex is used for immediate reversal of rocuronium-induced neuromuscular blockade.
It is used at a dose of 16 mg/kg.Since the patient in the question is 70 kg, the required dose of sugammadex can be calculated as:
16×70 = 1120 mg.Sugammadex selectively binds rocuronium or vecuronium, thereby reversing their neuromuscular blocking action. Due to its 1:1 binding of rocuronium or vecuronium, it can reverse any depth of neuromuscular block.
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This question is part of the following fields:
- Pharmacology
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Question 16
Incorrect
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Following a near drowning accident, a 5-year-old child is admitted to the emergency department and advanced paediatric life support is started.
What is the child's approximate weight, according to the preferred formulae of the Resuscitation Council (UK), the European Resuscitation Council, and the Royal College of Anaesthetists?Your Answer:
Correct Answer: 20-25kg
Explanation:For estimating a child’s weight, the Resuscitation Council (UK) and European Resuscitation Council teach the following formula:
Weight = (age + 4) × 2
The weight of the child will be around 20 kg.
This formula is used in the Primary FRCA exam by the Royal College of Anaesthetists.
In ‘developed’ countries, the traditional ‘APLS formula’ for estimating weight in children based on age (wt in kg = [age+4] x 2) is acknowledged as underestimating weight by 33.4 percent on average, with the degree of underestimation increasing with increasing age.
However, more recently, the APLS formula ‘Weight=3(age)+7’ has been found to provide a mean underestimate of only 6.9%. This formula is applicable to children aged 1 to 13 years.
The estimated weight based on age using this formula is 25 kg.
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This question is part of the following fields:
- Physiology
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Question 17
Incorrect
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A 19-year-old woman presents to the emergency department. She complains of symptoms indicative of an acute exacerbation of known 'brittle' asthma. On history, she reveals her asthma is normally controlled using inhalers and she has never had an acute exacerbation requiring hospitalisation.
On her admission into the ICU, further examination and diagnostic investigations are conducted. Her readings are:
Physical state: Alert, anxious and non-cyanotic.
Respiratory rate: 30 breaths/min
Pulse: 120 beats/min
Blood pressure: 150/90 mmHg
SPO2: 95% on air
Auscultation: Quiet breath sounds at both lung bases
What is the next most important step of investigation?Your Answer:
Correct Answer: Peak expiratory flow rate
Explanation:Peak expiratory flow rate (PEFR) is the maximum speed of air flow generated during a single forced exhaled breath. It is most useful when expressed as a percentage of the best value obtained from the patient.
Forced expiratory volume over 1 second (FEV1) is a lung parameter measured using spirometry. It is the amount of air forced out of the lung in one exhaled breath. It is a more accurate measure of lung obstructions as it doesn’t rely on effort like PEFR
PEFR and FEV1 are usually similar, but become more different in asthmatic patients as airflow becomes increasingly obstructed.
Acute severe asthma is most often diagnosed on history taking and examinations:
Respiratory rate: >25 breaths/min
Heart rate: >110 beats/min
PEFR: 33 – 50% predicted (<200L/min)
Patient state: Unable to complete a sentence in a single breath.A chest x-ray is not routinely required, and is only indicated in specific circumstances, which are:
If a pneumomediastinum or pneumothorax is suspected
Possible life threatening asthma
Possible consolidation
Unresponsive asthma
If ventilation is required.An echocardiograph (ECG) is not necessary in this case
Routine haematological and biochemical investigations are not urgent in this case as any abnormalities they detect will be secondary to the patient’s presentation.
An arterial blood gas (ABG) will only be indicated if SPO2 was <92% or if patient presented with life threatening symptoms.
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This question is part of the following fields:
- Clinical Measurement
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Question 18
Incorrect
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A 45-year-old woman gives a two-week history of fatigue, muscle cramps, and paraesthesia of her fingers and toes. She has low serum calcium and low serum parathyroid hormone levels on investigations.
She appears slightly confused, likely due to hypocalcaemia, and cannot give a full account of her past medical history, but can recall that she recently was admitted to the hospital.
What is the most likely cause of her hypoparathyroidism?Your Answer:
Correct Answer: Thyroidectomy
Explanation:The most common cause of hypoparathyroidism is injury or removing the parathyroid glands. They can be injured accidentally during surgery to remove the thyroid as they are located posterior to the thyroid gland.
A result of both low parathyroid hormone and low calcium is likely to mean that the parathyroid glands are not responding to hypocalcaemia. The hypocalcaemia can cause confusion, and the stay in the hospital is likely to refer to her surgery.
While a parathyroid adenoma is fairly common and can cause hypoparathyroidism, it much more likely causes hyperparathyroidism.
Chronic kidney disease is likely to cause hypocalcaemia, which would increase parathyroid hormone production in an attempt to increase calcium levels, causing hyperparathyroidism. Vitamin D is activated by the kidneys and then binds to calcium to be absorbed in the terminal ileum so that a deficiency would cause hyperparathyroidism.
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This question is part of the following fields:
- Anatomy
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Question 19
Incorrect
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In North America, there have been reports of paediatric patients dying after undergoing adenotonsillectomy for obstructive sleep apnoea.
Respiratory depression/obstruction is thought to be the cause of death. The codeine dose was 0.5-1 mg/kg, given every 4-6 hours.
In this group of patients, which of the following is the most likely cause of respiratory depression and obstruction?Your Answer:
Correct Answer: Exaggerated metabolism of codeine
Explanation:Codeine is easily absorbed from the gastrointestinal tract and converted to morphine and norcodeine in the liver via O- and N-demethylation. Morphine and norcodeine are excreted almost entirely by the kidney, primarily as conjugates with glucuronic acid.
By glucuronidation, phase II metabolism enzyme UDP-glucuronosyl transferase-2B7 converts morphine to morphine-3-glucuronide (M3G) and morphine-6-glucuronide (M6G) (UGT2B7).
Approximately 60% of morphine is converted to M3G, with the remaining 6-10% converted to M6G. M3G is inactive, but M6G is said to be 4 to 650 times more potent on the MOP receptor than morphine.
When codeine is consumed, cytochrome P450 2D6 in the liver converts it to morphine (CYP2D6).
Some people have DNA variations that increase the activity of this enzyme, causing codeine to be converted to morphine more quickly and completely than in others. After taking codeine, these ultra-rapid metabolisers are more likely to have higher than normal levels of morphine in their blood.
Respiratory depression/obstruction can be caused by high levels of morphine and M6G, especially in people who have a history of obstructive sleep apnoea. The estimated number of ultra-rapid metabolisers ranges from 1 to 7 per 100 people, but some ethnic groups may have as many as 28 per 100 people.
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This question is part of the following fields:
- Pathophysiology
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Question 20
Incorrect
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An emergency appendicectomy is being performed on a 20 year old man. For maintenance of anaesthesia, he is being ventilated using a circle system with a fresh gas flow (FGF) of 1 L/min (air/oxygen and sevoflurane). The trace on the capnograph shows a normal shape.
The table below demonstrates the changes in the end-tidal and baseline carbon dioxide measurements of the capnograph at 10 and 20 minutes of anaesthesia maintenance.
End-tidal CO2: 4.9 kPa vs 8.4kPa (10 minutes vs 20 minutes)
Baseline end-tidal CO2: 0.2 kPa vs 2.4kPa
Pulse 100-107 beats per minute, systolic blood pressure 125-133 mmHg and oxygen saturation 98-99%.
Which of the following is the single most important immediate course of action?Your Answer:
Correct Answer: Increase the FGF
Explanation:End-tidal carbon dioxide (ETCO2) monitoring has been an important factor in reducing anaesthesia-related mortality and morbidity. Hypercarbia, or hypercapnia, occurs when levels of CO2 in the blood become abnormally high (Paco2 >45 mm Hg). Hypercarbia is confirmed by arterial blood gas analysis. When using capnography to approximate Paco2, remember that the normal arterial–end-tidal carbon dioxide gradient is roughly 5 mm Hg. Hypercarbia, therefore, occurs when PETco2 is greater than 40 mm Hg.
The most likely explanation for the changes in capnograph is either exhaustion of the soda lime and a progressive rise in circuit dead space.
Inspect the soda lime canister for a change in colour of the granules. To overcome soda lime exhaustion, the first step is to increase the fresh gas flow (FGF) (Option A). Then, if need arises, replace the soda lime granules. Other strategies that can work are changing to another circuit or bypassing the soda lime canister, but remember that both these strategies are employed only after increasing FGF first. Exclude other causes of equipment deadspace too.
There are also other causes for hypercarbia to develop intraoperatively:
1. Hypoventilation is the most common cause of hypercapnia. A. Inadequate ventilation can occur with spontaneous breathing due to drugs like anaesthetic agents, opioids, residual NMDs, chronic respiratory or neuromuscular disease, cerebrovascular accident.
B. In controlled ventilation, hypercapnia due to circuit leaks, disconnection or miscalculation of patient’s minute volume.
2. Rebreathing – Soda lime exhaustion with circle, inadequate fresh gas flow into Mapleson circuits and increased breathing system deadspace.
3. Endogenous source – Tourniquet release, hypermetabolic states (MH or thyroid storm) and release of vascular clamps.
4. Exogenous source – Absorption of CO2 from pneumoperitoneum. -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 21
Incorrect
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A post-operative patient was given paracetamol and pethidine for post-operative analgesia. A few hours later, the patient developed fever of 38°C, hypertension, and agitation.
According to the patient's medical history, he is maintained on Levodopa and Selegiline for Parkinson's disease.
Which of the following is the most probable cause of his manifestation?Your Answer:
Correct Answer: Pethidine
Explanation:Selegiline is a monoamine oxidase inhibitor. Inhibition of monoamine oxidase leads to increased levels of norepinephrine and serotonin in the central nervous system.
Pethidine, also known as meperidine, is a strong agonist at the mu and kappa receptors. It inhibits pain neurotransmission and blocks muscarinic-specific actions.
Administering opioid analgesic is relatively contraindicated to individuals taking monoamine oxidase inhibitors. This is because of the high incidence of serotonin syndrome, which is characterized by fever, agitation, tremor, clonus, hyperreflexia and diaphoresis. Onset of symptoms is within hours, and the treatment is mainly through sedation, paralysis, intubation and ventilation.
The clinical findings are more consistent with Serotonin syndrome rather than exacerbation of Parkinson’s. Parkinson’s Disease (PD) exacerbations are defined as patient-reported or caregiver-reported episodes of subacute worsening of PD motor function in 1 or more domains (bradykinesia, tremor, rigidity, or PD-related postural instability/gait disturbance) that caused a decline in functional status, developed over a period of < 2 months, did not fluctuate with medication timing, and are not caused by intentional adjustments of PD medications by the treating neurologist. Malignant hyperthermia usually occurs within minutes of administration of a volatile anaesthetic, such as halothane, or succinylcholine. There is massive release of calcium from the sarcoplasmic reticulum, leading to fever, acidosis, rhabdomyolysis, trismus, clonus, and hypertension. In sepsis, it more common for patients to present with hypotension rather than hypertension.
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This question is part of the following fields:
- Pharmacology
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Question 22
Incorrect
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The required sample size in a trial of a new therapeutic agent varies with?
Your Answer:
Correct Answer: Level of statistical significance required
Explanation:The level of statistical significance required influences the sample size used. This is because sample size is used in the calculation of SD/SE.
Sample size does not affect
The level of acceptance
The alternative hypothesis with a general level set at p<0.05
The test to be used.Experience of the investigator and the type of patient recruited should have no bearing on the required sample size.
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This question is part of the following fields:
- Statistical Methods
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Question 23
Incorrect
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A new drug treatment has been developed for Crohn's disease. The pharmaceutical company behind this, is planning to conduct a trial and is looking for hiring around 200 individuals that are suffering from Crohn's disease. The aim would be to determine if there is any decline in the disease activity in response to the drug and compare it with a placebo.
What phase is the trial in?Your Answer:
Correct Answer: Phase 2
Explanation:The study is being conducted on a smaller level with only 200 participants and is determining the effectiveness of the drug in comparison to a placebo. These characteristics are in accordance with the second phase of trial.
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This question is part of the following fields:
- Statistical Methods
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Question 24
Incorrect
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Which of the following vertebral levels is the site where the aorta perforates the diaphragm?
Your Answer:
Correct Answer: T12
Explanation:The diaphragm divides the thoracic cavity from the abdominal cavity. Structures penetrate the diaphragm at different vertebral levels through openings in the diaphragm to communicate between the two cavities. The diaphragm has openings at three vertebral levels:
T8: vena cava, terminal branches of the right phrenic nerve
T10: oesophagus, vagal trunks, left anterior phrenic vessels, oesophageal branches of the left gastric vessels
T12: descending aorta, thoracic duct, azygous and hemi-azygous vein -
This question is part of the following fields:
- Anatomy
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Question 25
Incorrect
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A pre-operative evaluation for a trans-sphenoidal pituitary adenectomy is being performed on a 57-year-old woman. Her vision is causing her problems.
A macroadenoma compressing the optic chiasm is visible on MRI.
What is the most likely visual field defect to be discovered during an examination?Your Answer:
Correct Answer: Bitemporal hemianopia
Explanation:The pituitary gland plays a crucial role in the neuro-endocrine axis. It is located at the base of the skull in the sella turcica of the sphenoid bone. It is connected superiorly to the hypothalamus, third ventricle, and visual pathways, and laterally to the cavernous sinuses, internal carotid arteries, and cranial nerves III, IV, V, and VI.
Pituitary tumours make up about 10-15% of all intracranial tumours. The majority of adenomas are benign. Over-secretion of pituitary hormones (most commonly prolactin, growth hormone, or ACTH), under-secretion of hormones, or localised or generalised pressure effects can all cause symptoms.
Compression of the optic chiasm can result in visual field defects, the most common of which is bitemporal hemianopia. This is caused by compression of the nasal retinal fibres, which carry visual impulses from temporal vision across the optic chiasm to the contralateral sides before continuing to the optic tracts.
The interruption of the visual pathways distal to the optic chiasm causes a homonymous visual field defect. The loss of the right or left halves of each eye’s visual field is referred to as homonymous hemianopia. It’s usually caused by a middle or posterior cerebral artery territory stroke that affects the occipital lobe’s optic radiation or visual cortex.
Binasal hemianopia is a condition in which vision is lost in the inner half of both eyes (nasal or medial). It’s caused by compression of the temporal visual pathways, which don’t cross at the optic chiasm and instead continue to the ipsilateral optic tracts. Binasal hemianopia is a rare complication caused by the internal carotid artery impinging on the temporal (lateral) visual fibres.
A monocular visual loss (that is, loss of vision in only one eye) can be caused by a variety of factors, but if caused by nerve damage, the damage would be proximal to the optic chiasm on the ipsilateral side.
A central scotoma is another name for central visual field loss. Every normal mammalian eye has a scotoma, also known as a blind spot, in its field of vision. The optic disc is a region of the retina that lacks photoreceptor cells and is where the retinal ganglion cell axons that make up the optic nerve exit the retina. When both eyes are open, visual signals that are absent in one eye’s blind spot are provided for the other eye by the opposite visual cortex, even if the other eye is closed.
Scotomata can be caused by a variety of factors, including demyelinating disease such as multiple sclerosis, damage to nerve fibre layer in the retina, methyl alcohol, ethambutol, quinine, nutritional deficiencies, and vascular blockages either in the retina or in the optic nerve.
Bilateral scotoma can occur when a pituitary tumour compresses the optic chiasm, causing a bitemporal paracentral scotoma, which then spreads out to the periphery, causing bitemporal hemianopsia. A central scotoma in a pregnant woman could be a sign of severe pre-eclampsia.
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This question is part of the following fields:
- Pathophysiology
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Question 26
Incorrect
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A 70-year-old male is brought to the Emergency department with:
Pulse rate: 32 beats per minute
Blood pressure: 82/35 mmHg
12 lead ECG shows a sinus bradycardia of 35 beats per minute with no evidence of myocardial ischemia or infarction. There was no chest pain but the patient feels light-headed.
Which of the following would be the best initial treatment for this condition?Your Answer:
Correct Answer: Atropine
Explanation:Based on the presenting symptoms and clinical examination, it is a case of an adult sinus bradycardia with adverse signs. The first pharmacological treatment for this condition is atropine 500mcg intravenously and if necessary repeat every three to five minutes up to a maximum of 3 mg.
If the bradycardia does not subside even after the administration of atropine, cardiac pacing should be considered. If pacing cannot be achieved promptly, we should consider the use of second-line drugs like adrenaline, dobutamine, or isoprenaline.
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This question is part of the following fields:
- Pharmacology
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Question 27
Incorrect
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An air ambulance brought a young boy involved in a fight to the emergency department.
On examination, his blood pressure cannot be recorded. He suffers a stab wound to his chest that has penetrated the left atrium and the artery that supplies it.
Which artery is most likely damaged in this scenario?Your Answer:
Correct Answer: Left coronary artery
Explanation:The left atrium is supplied by the left coronary artery and its major branch the left circumflex.
The heart receives blood supply from coronary arteries. The right and left coronary arteries branch off the aorta and supply oxygenated blood to all heart muscle parts.
The left main coronary artery branches into:
1. Circumflex artery – supplies the left atrium, side, and back of the left ventricle. The left marginal artery arises from the left circumflex artery. It travels along the obtuse margin of the heart.
The left marginal artery, a branch of the circumflex artery, supplies the left ventricle.
2. Left Anterior Descending (LAD) artery – supplies the front and bottom of the left ventricle and front of the interventricular septumThe right coronary artery branches into:
1. Right marginal artery
2. Posterior descending arteryThe right coronary artery supplies the right atrium, right ventricle, interatrial septum, and the inferior posterior third of the interventricular septum. It also supplies the atrioventricular node + sinoatrial node in most patients. The posterior descending artery supplies the posterior third of the interventricular septum.
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This question is part of the following fields:
- Anatomy
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Question 28
Incorrect
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An 80-year-old female suffered a TIA 2 weeks ago. She has been admitted to the vascular ward as she will be undergoing carotid endarterectomy tomorrow morning. To explain the procedure and its complications, the surgeon gives her information about the procedure, telling her the artery will be tied during the operation.
She inquires about the areas supplied by the different arteries. You explain that the internal carotid artery supplies the brain while the external carotid artery ascends the neck and bifurcates into two arteries. One of these arteries is the superficial temporal artery. Which of the following is the second branch?Your Answer:
Correct Answer: Maxillary artery
Explanation:Carotid endarterectomy is the procedure to relieve an obstruction in the carotid artery by opening the artery at its origin and stripping off the atherosclerotic plaque with the intima. This procedure is performed to prevent further episodes, especially in patients who have suffered ischemic strokes or transient ischemic attacks.
The external carotid artery terminates by dividing into the superficial temporal and maxillary branches. The maxillary artery is the larger of the two terminal branches and arises posterior to the neck of the mandible.
The other arteries mentioned in the answer options branch off from the following:
Temporal arteries from the maxillary artery
Middle meningeal artery from the maxillary artery
Lingual artery from the anterior aspect of the external carotid artery
Facial artery from the anterior aspect of the external carotid artery -
This question is part of the following fields:
- Anatomy
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Question 29
Incorrect
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A 30 year old male was the victim of an electrocution injury and has been treated. The option that best describes the current levels for this injury is:
Your Answer:
Correct Answer: Tonic muscle contraction - 15 mA
Explanation:There are different effects of electrocution and these can be shown in the table below.
Current Effect
1 mA Tingling
5 mA Pain
15 mA Tonic muscle contraction
50 mA Respiratory arrest
100 mA Ventricular fibrillation and cardiac arrest -
This question is part of the following fields:
- Clinical Measurement
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Question 30
Incorrect
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A mercury barometer can be used to determine absolute pressure. A mercury manometer can be used to check blood pressure. The SI units of length(mm) are used to measure pressure.
Why is pressure expressed in millimetres of mercury (mmHg)?Your Answer:
Correct Answer: Pressure is directly proportional to length of the mercury column and is variable
Explanation:A mercury barometer can be used to determine absolute pressure. A glass tube with one closed end serves as the barometer. The open end is inserted into a mercury-filled open vessel. The mercury in the container is pushed into the tube by atmospheric pressure exerted on its surface. Absolute pressure is the distance between the tube’s meniscus and the mercury surface.
Pressure is defined as force in newtons per unit area (F) (A).
Mass of mercury = area (A) × density (ρ) × length (L)
Pressure = ((A × ρ × L) × 9.8 m/s2)/A
Pressure = ρ × L x 9.8
Pressure is proportional to LThe numerator and denominator of the above equation, area (A), cancel out. The constants are density and the gravitational acceleration value.
The length is proportional to the applied pressure.
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This question is part of the following fields:
- Physiology
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