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Question 1
Incorrect
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Which of the following is true regarding correlation coefficient?
Your Answer: It is represented by a small c
Correct Answer: It can assume any value between -1 and 1
Explanation:The degree of correlation is summarised by the correlation coefficient (r). This indicates how closely the points lie to a line drawn through the plotted data. In parametric data this is called Pearson’s correlation coefficient and can take any value between -1 to +1. A correlation of -1.0 indicates a perfect negative correlation, and a correlation of 1.0 indicates a perfect positive correlation.
For example
r = 1 – strong positive correlation (e.g. systolic blood pressure always increases with age)
r = 0 – no correlation (e.g. there is no correlation between systolic blood pressure and age)
r = – 1 – strong negative correlation (e.g. systolic blood pressure always decreases with age)
Whilst correlation coefficients give information about how one variable may increase or decrease as another variable increases they do not give information about how much the variable will change. They also do not provide information on cause and effect.
In contrast to the correlation coefficient, linear regression may be used to predict how much one variable changes when a second variable is changed.
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This question is part of the following fields:
- Statistical Methods
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Question 2
Incorrect
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Following an acute appendicectomy, a 6-year-old child is admitted to the recovery unit.
Your consultant has requested that you prescribe maintenance fluids for the next 12 hours. The child is 21 kg in weight.
What is the most suitable fluid volume to be prescribed?Your Answer: 610 ml
Correct Answer: 732 ml
Explanation:After a paediatric case, you’ll frequently have to calculate and prescribe maintenance fluids. The ‘4-2-1 rule’ should be used as a guideline:
1st 10 kg – 4 ml/kg/hr
2nd 10 kg – 2 ml/kg/hr
Subsequent kg – 1 ml/kg/hrHence
1st 10 kg = 4 × 10 = 40 ml
2nd 10 kg = 2 × 10 = 20 ml
Subsequent kg = 1 × 1 = 1 ml
Total = 61 ml/hr61 × 12 = 732 ml over 12 hrs.
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This question is part of the following fields:
- Physiology
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Question 3
Correct
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A 53-year old female with a diagnosis of anaplastic thyroid carcinoma is admitted in the surgery department for an elective total thyroidectomy with radical neck dissection. The operation is expected to last for 10 hours.
Which of the following is the most suitable humidifier to use in an anaesthetic circuit for this case?Your Answer: Heat and moisture exchanger (HME)
Explanation:Adequate humidification is vital to maintain homeostasis of the airway. Heat and moisture exchangers conserve some of the exhaled water, heat and return them to inspired gases. Many heat and moisture exchangers also perform bacterial/viral filtration and prevent inhalation of small particles. Heat and moisture exchangers are also called condenser humidifier, artificial nose, etc. Most of them are disposable devices with exchanging medium enclosed in a plastic housing. For adult and paediatric age group different dead space types are available. Heat and moisture exchangers are helpful during anaesthesia and ventilatory breathing system. To reduce the damage of the upper respiratory tract through cooling and dehydration inspiratory air can be heated and humidified, thus preventing the serious complications. Moreover, they are the most appropriate humidification devices used for routine anaesthesia.
Gases can be bubbled through water to increase humidity. Passing gas through water at room temperature causes the gas to cool due to latent heat of vaporisation. The water bath can be heated. This improves the efficiency of the device and also reduces the incidence of bacterial colonisation.
Nebulisers use a venturi system which employs the Bernoulli effect. A gas at high flow passes through a constriction causing the gas to accelerate, reducing its potential energy allowing other gases or liquids to be entrained. This can include medications or in the case of humidification, water vapour. The size of the water droplet produced by nebulisation determines where in the airway it is deposited. Standard nebulisers produced droplets of 4 microns in diameter and these are deposited in the upper airway and trachea. Efficacy can be improved by passing the droplets over an anvil which further reduces particle size. The most efficient form of nebuliser is the ultrasonic nebuliser. Here a transducer immersed in water and vibrated at a frequency of 3MHz produces1-2micron droplets. These particles easily reach the bronchioles and provide excellent humidification.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 4
Correct
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All of the following statements about cerebrospinal fluid are incorrect except:
Your Answer: Has a glucose concentration 2/3 that of the plasma glucose
Explanation:The pH of CSF is 7.31 which is lower than plasma.
Compared to plasma, it has a lower concentration of potassium, calcium, and protein and a higher concentration of sodium, chloride, bicarbonate and magnesium.
CSF usually has no cells present but if white cells are present, there should be no more than 4/ml.
The pressure of CSF should be less than 20 cm of water.
The concentration of glucose is approximately two-thirds of that of plasma, and it has a concentration of approximately 3.3-4 mmol/L.
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This question is part of the following fields:
- Physiology
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Question 5
Incorrect
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Concerning the intercostal nerves, which one of the following is true?
Your Answer: Anteriorly they run in the costal groove on the upper margin of the rib
Correct Answer: Each is connected to a ganglion of the sympathetic trunk
Explanation:The intercostal nerves arise from the ventral rami of the first 11 thoracic spinal nerves. they course along the costal groove on the lower margin of the rib.
The twelfth intercoastal nerve is called the subcostal nerve. This is because it is below the 12th rib.
Each intercostal nerve is connected to a ganglion of the sympathetic trunk from which it carries preganglionic and postganglionic fibres that innervate blood vessels, sweat glands, and muscles.
The lateral and medial pectoral nerves innervates pectoralis major muscle.
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This question is part of the following fields:
- Anatomy
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Question 6
Correct
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An adult and a 7-year-old child are anatomically and physiologically very different.
Which of the following physiological characteristics of a 5-year-old most closely resembles those of a healthy adult?Your Answer: Dead space ratio
Explanation:Whatever the age, the dead space ratio is 0.3. It’s the dead space (Vd) to tidal volume ratio (Vt).
The glottis is the narrowest point of the upper airway in an adult, while the cricoid ring is the narrowest point in a child.
A child’s airway resistance is much higher than an adult’s. The resistance to airflow increases as the diameter of a paediatric airway shrinks. The radius (r) to the power of 4 is inversely proportional to airway resistance (r4). As a result, paediatric patients are more susceptible to changes in airflow caused by a small reduction in airway diameter, such as caused by oedema.
The compliance of a newborn’s lungs is very low (5 mL/cmH2O), but it gradually improves as lung size and elasticity grow. Lung compliance in an adult is 200 mL/cmH2O.
In children, minute ventilation (mL/kg/minute) is much higher.
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This question is part of the following fields:
- Pathophysiology
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Question 7
Correct
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Regarding a paramagnetic oxygen analyser, the following statements are TRUE:
Your Answer: Utilises null deflection
Explanation:The electrons in the outer shell of an oxygen molecule are unpaired, thus it has paramagnetic properties and is attracted into a magnetic field.
It utilizes null deflection -True
Null deflection is a crucial principle in paramagnetic analysers (reflected beam of light on two photocells) which gives very accurate results (typically 0.1%).It can be used to measure the concentration of diamagnetic gases – False
Since most other gases are weakly diamagnetic they are repelled by a magnetic field (nitric oxide is also paramagnetic).Can measure gases dissolved in the blood – False
For accurate analysis the sample gas must be dried before passing into the analysis cell, for example, by passage through silica gel. Therefore, they are unsuitable to measure gases dissolved in blood.Does not require calibration – False
As with most measurement instruments paramagnetic analysers must be calibrated before use.E) The readings are unaffected by water vapour – False
Water vapour affects the readings hence for accurate analysis the sample gas must be dried before passing into the analysis cell, for example, by passage through silica gel. That is why they are unsuitable to measure dissolved blood gases. -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 8
Incorrect
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Which of the following options will likely play a major role in falling coronary blood flow?
Your Answer: Activation of sympathetic cardiac nerve fibres
Correct Answer: Intracoronary artery infusion of endothelin-1
Explanation:Endothelin-1 is considered as a powerful coronary vasoconstrictor, produced by the endothelium. It acts to counter the effects of Nitric oxide (NO).
Neuropeptide-Y, angiotensin1, cocaine, vasopressin, and nicotine are some other coronary vasoconstrictors.Chronotrophy and inotrophy occur after the activation of sympathetic nerve fibres, which in turn results in increasing the myocardial oxygen consumption, leading to increased coronary blood flow via local metabolic processes.
An alpha-receptor mediated coronary vasoconstrictor effect is also initiated that usually competes with vasodilation, resulting in decreased coronary vascular resistance. Some of the other dilators include hydrogen ions, CO2, potassium, and lactic acid. The action of endothelial NO synthase (eNOS) on L-arginine results in the formation of NO. This messenger also plays a vital role in the regulation of coronary blood flow via vasodilation, inhibition of platelet aggression, and decreasing vascular resistance.
Adenosine is considered as purine nucleoside that forms after the breakdown of adenosine triphosphate (ATP). Adenosine binds to adenosine type 2A (A2A) receptors in coronary vascular smooth muscles. These are coupled to the Gs protein. This mechanism leads to hyperpolarisation of muscle cells, resulting in relaxation and increased coronary blood flow.GTN is an veno and arteriolar dilator, which behaves as pro-drug with NO.
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This question is part of the following fields:
- Pathophysiology
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Question 9
Incorrect
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Which of the following organism is highly resistant to penicillin?
Your Answer: Beta-haemolytic Streptococci
Correct Answer: Escherichia coli
Explanation:Penicillinase is a narrow spectrum ?-lactamase that opens the ?-lactam ring and inactivates Penicillin G and some closely related congeners. The majority of Staphylococci and some strains of gonococci, B. subtilis, E. coli, and a few other bacteria produce penicillinase.
N. meningitidis is sensitive to penicillin and less than 20% resistance is found in pseudomonas.
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This question is part of the following fields:
- Pharmacology
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Question 10
Correct
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Which of the following statements is true about monoamine oxidase (MOA) enzymes?
Your Answer: Type A and type B are found in the liver and brain
Explanation:Monoamine oxidase (MOA) enzymes are responsible for the catalyses of monoamine oxidative deamination. It assists the degradation of serotonin, norepinephrine (NE) and dopamine.
They are found in the mitochondria of most central and peripheral nerve tissues.
There are 2 different types:
Type A: Whose main function it to inactivate dopamine, tyramine, norepinephrine and 5-hydroxytryptamine. In addition to the nervous system, it is also found in the liver, brain gastrointestinal tract, pulmonary endothelium and placenta
Type B: Whose main function is to inactivate dopamine, tyramine, tryptamine and phenylethylamine. In addition to the nervous system, it is also found in the liver, brain (especially in the basal ganglia) and blood platelets. -
This question is part of the following fields:
- Pathophysiology
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Question 11
Incorrect
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Very small SI units are easily expressed using mathematical prefixes.
One femtolitre is equal to which of the following volumes?Your Answer: 0.000, 001 L
Correct Answer: 0.000, 000, 000, 000, 001 L
Explanation:Small measurement units are denoted by the following SI mathematical prefixes:
1 deci = 0.1
1 milli = 0.001
1 micro = 0.000001
1 nano = 0.000000001
1 pico = 0.000000000001
1 femto = 0.000000000000001 (used to measure red blood cell volume)
1 atto = 0.000000000000000001 -
This question is part of the following fields:
- Basic Physics
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Question 12
Correct
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Which type of epithelium lines the luminal surface of the oesophagus?
Your Answer: Non keratinised stratified squamous epithelium
Explanation:Normally, the oesophagus is lined by non-keratinized stratified squamous epithelium. This epithelium can undergo metaplasia and convert to the columnar epithelium (stomach’s lining) in long-standing GERD that leads to Barret’s oesophagus.
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This question is part of the following fields:
- Anatomy
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Question 13
Correct
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A 10-year-old boy is undergoing investigations for coeliac disease. Tissue biopsies were taken from both the small and large intestinal linings.
Which of the following is found in the small intestine lining but not in that of the large intestine in a normal biopsy?Your Answer: Villi
Explanation:The small and large intestinal walls are composed of the following common layers:
1. Mucosa
2. Submucosa
3. Muscularis Externa
4. AdventitiaIntestinal villi are highly vascular projections of the mucosal surface that cover the entire small intestinal mucosa. They increase the lumen’s surface area, which aids in absorption and digestion, the primary functions of the small intestine. Villi are large and most abundant in the duodenum and jejunum.
In both the small and large intestines, the muscularis mucosae are found within the mucosa. The myenteric nerve plexus is found innervating the muscularis externa. The mucosa is lined with columnar epithelial cells, and goblet cells may be present to secrete mucins.
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This question is part of the following fields:
- Anatomy
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Question 14
Correct
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A 30-year-old woman with a BMI of 24 kg/m2 consumes four glasses of wine on an empty stomach. Her serum alcohol are levels measured over the following five hours. The serum alcohol level of 30-year-old man with the same BMI and alcohol consumption is also measured over the same duration.
The peak concentration of alcohol is found to be greater in the woman than in the man.
Which of these offers best explanation for this observation?Your Answer: Lower volume of distribution
Explanation:The blood alcohol concentration depends on:
-The rate of alcohol absorption from the gastrointestinal tract
-The volume of distribution of alcohol in the body, and
-The rate of elimination of alcohol from the body.Total body water is approximately 50% in a female as compared to 60% in a typical male. This means that the volume of distribution of alcohol is lower in female compared with men. This is the principal reason for higher peak in alcohol levels.
About 4% of ingested alcohol is metabolised by the liver accounting for first pass metabolism and 0.4% is metabolised by gastric alcohol dehydrogenase (ADH). The absorbed alcohol is NOT distributed to fat cells but it is distributed throughout the water compartments (plasma, interstitial and intracellular) of the body. Women have very little gastric ADH, which further influences this exaggerated rise.
85-98% of the alcohol is oxidised by the liver to acetaldehyde and then to acetate. The metabolic pathway initially observes first order kinetics and then saturation or zero order kinetics leading to peaks in alcohol levels.
Clearance of ethanol per unit lean body mass is lower in male. The calculated alcohol elimination rate and liver volume per kilogram of lean body mass were 33% and 38% higher in women than in men, respectively.
Available evidence in the literature about the relationship of alcohol metabolism to the phases of the menstrual cycle is conflicting.
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This question is part of the following fields:
- Pathophysiology
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Question 15
Correct
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Which of the following statements is true about an acute pulmonary embolism?
Your Answer: Thrombolysis administered through a peripheral vein is as effective as through a pulmonary artery catheter
Explanation:Acute pulmonary embolism occurs when a blood clot becomes embedded in a pulmonary artery and restricts lung blood flow.
Thrombolysis is recommended in patients with extremely compromised circulation rather than reduced oxygen in the blood. It is effective when administered via a peripheral vein or a pulmonary artery catheter.
Anticoagulant therapy (heparin use) decreases the risk of further embolic evens and decreases constriction of pulmonary vessels.
An ECG may be normal in patients with an acute pulmonary embolism.
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This question is part of the following fields:
- Pathophysiology
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Question 16
Incorrect
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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:
Correct 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 17
Incorrect
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Which of the following statements is true regarding the relation to the liver?
Your Answer:
Correct Answer: The caudate lobe is superior to the porta hepatis
Explanation:Ligamentum venosum is an anterior relation of the liver: The ligamentum venosum, the fibrous remnant of the ductus venosus of the fetal circulation, lies posterior to the liver. It lies in the fossa for ductus venosus that separates the caudate lobe and the left lobe of the liver.
The portal triad contains three important tubes: 1. Proper hepatic artery 2. Hepatic portal vein 3. Bile ductules It also contains lymphatic vessels and a branch of the vagus nerve.
The bare area of the liver is a large triangular area that is devoid of any peritoneal covering. The bare area is attached directly to the diaphragm by loose connective tissue. This nonperitoneal area is created by a wide separation between the coronary ligaments.
The porta hepatis is a fissure in the inferior surface of the liver. All the neurovascular structures (except the hepatic veins) and hepatic ducts enter or leave the liver via the porta hepatis. It contains the sympathetic branch to the liver and gallbladder and the parasympathetic, hepatic branch of the vagus nerve. The caudate lobe (segment I) lies in the lesser sac on the inferior surface of the liver between the IVC on the right, the ligamentum venosum on the left, and the porta hepatis in front
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This question is part of the following fields:
- Anatomy
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Question 18
Incorrect
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Volunteers are being recruited for a new clinical trial of a novel drug treatment for Ulcerative colitis. The proposed study will enrol about 2000 people with ulcerative colitis. Testing will be performed to assess any reduction in disease severity with the new drug as compared to the current treatment available in the industry.
Which phase of clinical trial will this be?Your Answer:
Correct Answer: Phase 3
Explanation:This clinical trial consists over 1000 patients being evaluated for the response to a new treatment against a currently licensed treatment for ulcerative colitis. Therefore, it is comparing its efficacy to an established therapeutic or control in a larger population of volunteers. These are the characteristics of a phase III clinical trial.
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This question is part of the following fields:
- Statistical Methods
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Question 19
Incorrect
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Regarding amide local anaesthetics, which one factor has the most significant effect on its duration of action?
Your Answer:
Correct Answer: Protein binding
Explanation:When drugs are bound to proteins, drugs cannot cross membranes and exert their effect. Only the free (unbound) drug can be absorbed, distributed, metabolized, excreted and exert pharmacologic effect. Thus, when amide local anaesthetics are bound to ?1-glycoproteins, their duration of action are reduced.
The potency of local anaesthetics are affected by lipid solubility. Solubility influences the concentration of the drug in the extracellular fluid surrounding blood vessels. The brain, which is high in lipid content, will dissolve high concentration of lipid soluble drugs. When drugs are non-ionized and non-polarized, they are more lipid-soluble and undergo more extensive distribution. Hence allowing these drugs to penetrate the membrane of the target cells and exert their effect.
Tissue pKa and pH will determine the degree of ionization.
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This question is part of the following fields:
- Physiology
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Question 20
Incorrect
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A 31-year old Caucasian female came into the emergency department due to difficulty of breathing. History revealed exposure to room odorizes that are rich in alkyl nitrites. Upon physical examination, patient is tachypnoeic at 32 breaths per minute, desaturated at 88% while on a non-rebreather mask at 15 litres per minute oxygen. She was also noted to be cyanotic, however with clear breath sounds.
Considering the history, what is the most probable cause of her difficulty of breathing?Your Answer:
Correct Answer: Increased affinity of bound oxygen to haemoglobin
Explanation:Amyl nitrate is part of the treatment of cyanide poisoning. The short acting nitrate causes oxidation of Fe2+ in haemoglobin to Fe3+ in methaemoglobin. Methaemoglobin combines with cyanide (cyanmethemoglobin), which reacts with sodium thiosulfate to convert nontoxic thiocyanate and methaemoglobin.
Methaemoglobin is formed when the iron in haemoglobin is converted from the reduced state (Fe2+) to the oxidized state (Fe3+). The oxidized form of haemoglobin (Fe3+) does not bind oxygen as readily as Fe2+, but has high affinity for cyanide. It also results to high affinity of bound oxygen to haemoglobin, thus leading to tissue hypoxia. Arterial oxygen tension is normal despite observations of cyanosis and dyspnoea. Methemoglobinemia can be treated with methylene blue and vitamin C.
Carboxyhaemoglobin can be due to carbon monoxide poisoning. In such cases, patients experience headache and dizziness, but do not develop cyanosis.
2,3-diphosphoglycerate causes a shift in the oxygen dissociation curve to the right, decreasing haemoglobin’s affinity to oxygen to facilitate unloading of oxygen to the tissues.
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This question is part of the following fields:
- Pathophysiology
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Question 21
Incorrect
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Which peripheral nerve of the foot is often utilized to evaluate for neuromuscular blockade?
Your Answer:
Correct Answer: Posterior tibial nerve
Explanation:The posterior tibial nerve lies on the posterior surface of the tibialis posterior and, lower down the leg, on the posterior surface of the tibia. The nerve accompanies the posterior tibial artery and lies at first on its medial side, then crosses posterior to it, and finally lies on its lateral side. The nerve, with the artery, passes behind the medial malleolus, between the tendons of the flexor digitorum longus and the flexor hallucis longus.
It gives off muscular branches to the soleus, flexor digitorum longus, flexor hallucis longus, and tibialis posterior. A medial calcaneal branches off to supply the skin over the medial surface of the heel, and an articular nerve to supply the ankle joint. Finally, it terminates to become the medial and lateral plantar nerves.
The saphenous nerve is a branch of the femoral nerve that gives off branches that supply the skin on the posteromedial surface of the leg.
The sural nerve is a branch of the tibial nerve that supplies the skin on the lower part of the posterolateral surface of the leg.
The superficial peroneal nerve is one of the terminal branches of the common peroneal nerve. It arises in the substance of the peroneus longus muscle on the lateral side of the neck of the fibular. It ascends between the peroneus longus and brevis muscles, and in the lower part of the leg it becomes cutaneous. Muscular branches of the superficial peroneal nerve supply the peroneus longus and brevis muscles, while medial and lateral cutaneous branches are distributed to the skin on the lower part of the leg and dorsum of the foot. In addition, the cutaneous branches supply the dorsal surfaces of the skin of all the toes, except the adjacent sides of the first and second toes and the lateral side of the little toe.
The superficial peroneal, sural and saphenous nerves cannot be used to assess neuromuscular blocks since they are sensory nerves.
The deep peroneal nerve enters the dorsum of the foot by passing deep to the extensor retinacula on the lateral side of the dorsalis pedis artery. It divides into terminal, medial, and lateral branches. The medial branch supplies the skin of the adjacent sides of the big and second toes. The lateral branch supplies the extensor digitorum brevis muscle. Both terminal branches give articular branches to the joints of the foot. This nerve is too deep to use for neuromuscular blockade assessment
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This question is part of the following fields:
- Anatomy
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Question 22
Incorrect
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The diaphragm is a muscle that is relatively resistant to non-depolarizing neuromuscular blockade's effects.
When these muscle relaxants are used, which of the following peripheral nerve stimulator twitch patterns is best for monitoring the return of diaphragmatic function?Your Answer:
Correct Answer: Post-tetanic count stimulation
Explanation:Certain skeletal muscles are more resistant to the effects of neuromuscular blocking agents, both non-depolarizing and depolarizing. The diaphragm is the most resistant. The muscles of the larynx and the corrugator supercilii are less resistant. The abdominal, orbicularis oris, and limb peripheral muscles are the most sensitive muscles.
Twitch stimulation patterns:
Supramaximal single stimulus:
The frequency ranges from 1 Hz to 0.1 Hz (one every second to one every 10 seconds)
The response is proportional to the frequency of the event.
It has limited clinical utility because it only tells you whether or not a patient is paralysed (no information on degree of paralysis).Over the course of 0.5 seconds (2 Hz), four supramaximal stimulate were applied:
It is possible to see ‘fade’ and use it as a basis for evaluation.
This stimulation pattern is used to determine the degree of blockade (1-2 twitches is appropriate for abdominal surgery)
If the train of four (TOF) count is 1-2, reversal agents can be used in conjunction with medium-acting neuromuscular blocking agents.Ratio of TOF:
This is the ratio of the 4th twitch amplitude to the 1st twitch amplitude.
The ratio decreases with non-depolarising block and is inversely proportional to the degree of block, allowing objective measurement of residual neuromuscular blockade.
To achieve adequate reversal, the ratio (as measured by accelerography) must be between 0.7 and 0.9.Count of twitches after a tetanic experience(PTC):
50 Hz for 5 seconds, then a 3 second pause, followed by a single 1 Hz twitch stimulus.
When the TOF count is zero, this stimulation pattern is used to assess deep blockade (that is, in neurosurgery, microsurgery or ophthalmic surgery when even small movements of a patient will disturb the surgical field)
It gives an estimate of how long it will take for the response to return to single twitches, allowing assessment of blocks that are too deep for any other technique.
A palpable post-tetanic count (PTC) of 2 indicates no twitch response for about 20-30 minutes, and a PTC of 5 indicates no twitch response for about 10-15 minutes.This is without a doubt the best way to keep track of paralysis in patients who need to avoid diaphragmatic movement. It’s best to use drug infusions and aim for a PTC of 2. After a tetanic stimulus, acetylcholine is mobilised, causing post-tetanic potentiation.
Stimulation in Two Bursts:
750 milliseconds between two short bursts of 50 Hz
This stimulation pattern is used to assess small amounts of residual blockade manually (tactile). -
This question is part of the following fields:
- Clinical Measurement
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Question 23
Incorrect
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Which of the following statements is correct regarding hypomagnesaemia?
Your Answer:
Correct Answer: Causes tetany
Explanation:The ECG changes seen in hypomagnesaemia include:
Prolonged PR interval
Prolonged QT interval
Flattening of T waves
ST segment depression
Prominent U wavesThese changes are almost the same as those of hypokalaemia.
There is an increased risk of digoxin toxicity and a risk of atrial and ventricular ectopic and ventricular arrhythmias.
There is impaired synthesis and release of parathyroid hormone (PTH) in chronic hypomagnesaemia leading to impaired target organ response to PTH. This produces secondary hypocalcaemia.
The use of potassium ‘wasting’ diuretics (e.g. loop diuretics like furosemide) may lead to Hypomagnesaemia.
A tall T wave is seen in hypermagnesemia.
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This question is part of the following fields:
- Pathophysiology
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Question 24
Incorrect
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Which one of the following patients presenting for elective surgery has an American Society of Anaesthesiologists (ASA) preoperative physical status grading of III?
Your Answer:
Correct Answer: A 50-year old man with a BMI of 41 with a reduced exercise tolerance
Explanation:The ASA physical status classification system is a system for assessing the fitness of patients before surgery. It was last updated in October 2014.
ASA I A normal healthy patient
ASA II A patient with mild systemic disease
ASA III A patient with severe systemic disease
ASA IV A patient with severe systemic disease that is a constant threat to life
ASA V A moribund patient who is not expected to survive without the operation
ASA VI A declared brain-dead patient whose organs are being removed for donor purposesA 20-year old woman who is 39-weeks pregnant with no other medical conditions – ASA II
A 35-year-old man with a BMI of 29 with a good exercise tolerance who smokes-ASA II
A 50-year old man with a BMI of 41 with a reduced exercise tolerance -ASA III
A 65-year old woman with a BMI of 34 with treated hypertension with no functional limitations-ASA II
A 73-year old man who has had a TIA ten-weeks ago but has a good exercise tolerance and is a non-smoker-ASA IV
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This question is part of the following fields:
- Clinical Measurement
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Question 25
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:
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 26
Incorrect
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All of the following statements about calcium channel antagonists are incorrect except:
Your Answer:
Correct Answer: May cause potentiation of muscle relaxants
Explanation:Calcium channel blocker (CCB) blocks L-type of voltage-gated calcium channels present in blood vessels and the heart. By inhibiting the calcium channels, these agents decrease the frequency of opening of calcium channels activity of the heart, decrease heart rate, AV conduction, and contractility.
Three groups of CCBs include
1) Phenylalkylamines: Verapamil, Norverapamil
2) Benzothiazepines : Diltiazem
3) Dihydropyridine : Nifedipine, Nicardipine, Nimodipine, Nislodipine, Nitrendipine, Isradipine, Lacidipine, Felodipine and Amlodipine.Even though verapamil as good absorption from GIT, its oral bioavailability is low due to high first-pass metabolism.
Nimodipine is a Cerebro-selective CCB, used to reverse the compensatory vasoconstriction after sub-arachnoid haemorrhage and is more lipid soluble analogue of nifedipine
Calcium channel antagonist can potentiate the effect of non-depolarising muscle relaxants.
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This question is part of the following fields:
- Pharmacology
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Question 27
Incorrect
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A trail has analysed that a new screening test may increase the survival time of ovarian cancer patients. But analyst say that the apparent increase in the patients survival time is just because of earlier detection instead of actual improvement.
What kind of bias is in this experiment?Your Answer:
Correct Answer: Lead time bias
Explanation:Observation bias occurs when the behaviour of an individual changes that results from their awareness of being observed.
Recall bias introduced when participants in a study are systematically more or less likely to recall and relate information on exposure depending on their outcome status.
Attrition bias is a systematic error caused by unequal loss of participants from a randomized controlled trial (RCT). In clinical trials, participants might dropout due to unsatisfactory treatment or efficacy, intolerable adverse events, or even death.
Selection bias introduced when the individuals are not chosen randomly to take a part in the study. It usually occurs when the research decides who is going to be studied, they are not the representative of the population.
Lead-time bias occurs when a disease is detected by a screening test at an earlier time point rather than it would have been diagnosed by its clinical appearance. In this bias, earlier detection improves the survival time in the intervention group.
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This question is part of the following fields:
- Statistical Methods
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Question 28
Incorrect
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The passage of glucose into the brain is facilitated by which transport method?
Your Answer:
Correct Answer: Facilitated diffusion
Explanation:Glucose transport is a highly regulated process accomplished mostly by facilitated diffusion using carrier proteins to cross cell membranes.
There are many transporters, but the most important are known as glucose transporters (GLUTs).
Stresses in various form of acute and chronic forms affect the activity of glucose transporters.
They are responsive to many types of metabolic stress, including hypoxia, injury, hypoglycaemia, numerous metabolic inhibitors, stress hormones, and other influences such as growth factors.Numerous signalling pathways appear to be involved in transporter regulation.
New evidence suggests that stresses regulating GLUTs are not only acute biological stresses. In addition, chronic low-grade inflammation, and their associated chronic diseases also lead to altered glucose transport. These include obesity, type 2 diabetes, cardiovascular disease, and the growth and spread of many tumours that are affected by altered glucose transporters. Some of these glucose transport effects are compensatory, while others are pathogenic.
Ultimately, deliberate manipulation of GLUTs could be used as treatment for some of these chronic diseases.
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This question is part of the following fields:
- Physiology
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Question 29
Incorrect
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Regarding the blood supply of the heart, are the following statements true?
Your Answer:
Correct Answer: The left coronary artery originates from the left posterior aortic sinus
Explanation:The left coronary artery arises from the left posterior aortic sinus and divides into the circumflex arteries and the left anterior descending (LAD) artery.
The right coronary artery arises from the anterior aortic sinus and supplies:
– the right ventricle
– part of the interventricular septum
– the atrioventricular (A-V) node and
– in 85% of cases the inferior part of the left ventricle.
The right coronary artery provides a posterior interventricular branch and a marginal branch that anastomoses with the LAD at the apex.The oblique vein together with the small, middle and great cardiac veins drain into the coronary sinus, which drains into the right atrium.
The anterior cardiac vein drains directly into the right atrium.
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This question is part of the following fields:
- Anatomy
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Question 30
Incorrect
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The production of carbon dioxide and water occurs during cellular respiration, which involves an energy substrate and oxygen. For a patient, the respiratory quotient is calculated as 0.7.
Which of the following energy substrate combinations is the most likely in this patient's diet?Your Answer:
Correct Answer: Low carbohydrate, high fat and low protein
Explanation:The respiratory quotient (RQ) is the proportion of CO2 produced by the body to O2 consumed per unit of time.
CO2 produced / O2 consumed = RQ
CO2 is produced at a rate of 200 mL per minute, while O2 is consumed at a rate of 250 mL per minute. An RQ of around 0.8 is typical for a mixed diet.
The RQ will change depending on the energy substrates consumed in the diet.
Granulated sugar is a refined carbohydrate that contains 99.999 percent carbohydrate and no lipids, proteins, minerals, or vitamins.
Glucose and other hexose sugars – RQ = 1
Fats – RQ = 0.7
Proteins – RQ is 0.9
Ethyl alcohol – RQ = 0.67 -
This question is part of the following fields:
- Pathophysiology
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