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Question 1
Incorrect
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The following statements are about the conjugation of bilirubin. Which is true?
Your Answer: Occurs in the Kupfer cells of the liver
Correct Answer: Is catalysed by a glucuronyl transferase
Explanation:Bilirubin is formed by metabolizing heme, mostly from haemoglobin in red blood cells.
Bilirubin is conjugated to glucuronic acid in the hepatocytes by the glucuronyl transferase enzyme in order to enable it to become soluble and allow for its secretion across the canalicular membrane and into bile.
The conjugation process is increased by rifampicin and decreased by valproate.
Gilbert’s syndrome is caused by a decrease in glucuronyl transferase in the hepatic system, decreasing the transport of bilirubin into the hepatocyte, causing unconjugated bilirubinaemia.
Crigler-Najjer syndrome is caused by mutations in the genes responsible for hepatic glucuronyl transferase, decreasing the activity of the enzyme, meaning bilirubin cannot be conjugated, causing unconjugated bilirubinaemia.
Dubin-Johnson syndrome does not cause an impairment in the conjugation of bilirubin, but it blocks the transport of bilirubin out of the hepatocyte resulting in conjugated bilirubinaemia.
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This question is part of the following fields:
- Pathophysiology
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Question 2
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: α2γ2
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 3
Correct
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The spinal cord tracts that transmits the sensations of pain, crude temperature, and light touch is?
Your Answer: Spinothalamic
Explanation:Dorsal column (ascending tract) – Proprioception, vibration, discriminative
Spinocerebellar (ascending tract) – Subconscious muscle position and tone
Corticospinal (descending tract) – Voluntary muscle
Rubrospinal (descending tract) – Flexor muscle tone
Vestibulospinal (descending tract) – Reflexes and muscle tone
Reticulospinal(descending tract) – Voluntary movements, head position.-
Autonomic – Descending tract.
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This question is part of the following fields:
- Anatomy
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Question 4
Incorrect
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Which of these statements regarding the basilar artery and its branches is not true?
Your Answer: The superior cerebellar artery may be decompressed to treat trigeminal neuralgia
Correct Answer: The posterior inferior cerebellar artery is the largest of the cerebellar arteries arising from the basilar artery
Explanation:The posterior inferior cerebellar artery is the largest branch arising from the distal portion of the vertebral artery which forms the basilar artery. It is one of the arteries responsible for providing blood supply to the brain’s cerebellum.
The labyrinthine artery (auditory artery) is a long and slender artery which arises from the basilar artery and runs alongside the facial and vestibulocochlear nerves into the internal auditory meatus.
The posterior cerebellar artery is one of two cerebral arteries supplying the occipital lobe with oxygenated blood. It is usually bigger than the superior cerebellar artery. It is separated from the vessel near its origin by the oculomotor nerve.
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This question is part of the following fields:
- Anatomy
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Question 5
Correct
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A 10-year-old girl complains of right iliac fossa pain, and a provisional diagnosis of appendicitis is made.
Which of the following embryological structures gives rise to the appendix?Your Answer: Midgut
Explanation:The midgut gives rise to the appendix.
At week 6, the caecal diverticulum appears and is the precursor for the cecum and vermiform appendix. The cecum and appendix undergo rotation and descend into the right lower abdomen. The appendix can take up various positions:
1. Retrocecal appendix: behind the cecum
2. Retrocolic appendix: behind the ascending colon
3. Pelvic appendix: appendix descends into the pelvisThe appendix grows in length so that at birth, it is long and worm-shaped, or vermiform. After birth, the caecal wall grows unequally, and the appendix comes to lie on its medial side.
The midgut develops into the distal duodenum, jejunum, ileum, cecum, appendix, ascending colon, and proximal 2/3 of the transverse colon.
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This question is part of the following fields:
- Anatomy
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Question 6
Correct
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A 58-year-old man is being operated on for a radical gastrectomy for carcinoma of the stomach.
Which structure needs to be divided to gain access to the coeliac axis?Your Answer: Lesser omentum
Explanation:The lesser omentum will need to be divided. This forms one of the nodal stations that will need to be taken during a radical gastrectomy.
The celiac axis is the first branch of the abdominal aorta and supplies the entire foregut (mouth to the major duodenal papilla). It arises at the level of vertebra T12. It has three major branches:
1. Left gastric
2. Common hepatic
3. Splenic arteries -
This question is part of the following fields:
- Anatomy
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Question 7
Incorrect
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When there is a relation between two variables due to the existence of a confounding factor, this association is termed as:
Your Answer: Direct
Correct Answer: Indirect
Explanation:When the association among any two classes of object is defined by the presence of a third entity it is termed as indirect association. For an instance, the age of the employee may affect the rate of pay, which would have implications on job satisfaction. So, in this example, an indirect relationship between age and job satisfaction exists due to a third party i.e. rate of pay.
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This question is part of the following fields:
- Statistical Methods
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Question 8
Incorrect
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The whole water content of the body is calculated by multiplying body mass with 0.6. This water is diffused into distinct compartments.
Which fluid compartment can be measured indirectly?Your Answer: Plasma volume
Correct Answer: Intracellular volume
Explanation:The total body water content of a 70kg man is (70 × 0.6) = 42 litres. For a woman, the calculation is (70 × 0.55) = 38.5 litres.
For a man, it is subdivided into:
Extracellular fluid (ECF) = 14L (1/3)
Intracellular fluid (ICF) = 28L (2/3).The ECF volume is further divided into:
Interstitial fluid = 10.5 litres
Plasma = 3 litres
Transcellular fluid (CSF/synovial fluid) = 0.5 litres.Directly measured fluid compartments:
Heavy water (deuterium) can be used to measure total body water content, which is freely distributed.
Albumin labelled with a radioactive isotope or using a dye called Evans blue can be used to measure Plasma volume . They do not diffuse into red blood cells.
Radiolabelled (Cr-51) red blood cells can be used to measure total erythrocyte volume.
Inulin as the tracer can be used to measure ECF volume as it circulate freely in the interstitial and plasma volumes.Indirectly measured fluid compartments:
Total blood volume can be calculated with the level of haematocrit and the volume of total circulating red blood cells.
ICF volume can be calculated by subtracting ECF volume from total blood volume. -
This question is part of the following fields:
- Basic Physics
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Question 9
Correct
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A 63-year-old woman, is admitted into hospital. She has undergone a thoracoscopic sympathectomy.
To enable ease of access during surgery, her right arm has been abducted.
On examination, immediately after surgery, she is noted to have lost the ability to abduct her right arm, with the presence a weak lateral rotation in the same arm. She has also lost sensation in the outer aspect of the lower deltoid area of the skin.
Her symptoms are as a result of injury to a nerve during surgery. What nerve is it?Your Answer: Axillary nerve
Explanation:The axillary nerve arises from spinal roots C5-C6. It has both sensory and motor functions:
Sensory: Provides innervation to the skin over the lower deltoid area
Motor: Provides innervation to the teres minor (responsible for stabilisation of glenohumeral joint and external rotation of shoulder joint) and deltoid muscles (responsible for abduction of arms glenohumeral joint).
Injury to the axillary nerve will result in the patient being unable to abduct the arm beyond 15 degrees and a loss of sensory feeling over lower deltoid area.
These symptoms could also be a result of over-abduction of the arm (>90°) which would cause the head of the humerus to become dislocated.
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This question is part of the following fields:
- Pathophysiology
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Question 10
Incorrect
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A 78-year-old man with a previous history of ischaemic heart disease is admitted to hospital. He is scheduled for a cardiopulmonary exercise test (CPX) before he undergoes an elective abdominal aneurysm repair.
What measurement obtained during a CPX test alone provides the best indication for postoperative mortality?Your Answer: VO2 peak
Correct Answer: Anaerobic threshold
Explanation:Cardiopulmonary exercise testing (CPX, CPEX, CPET) is a non-invasive testing method used to determine the performance of the heart, lungs and skeletal muscle. It measures the exercise tolerance of the patient.
The parameters measured include:
ECG and ST-segment analysis and blood pressure
Oxygen consumption (VO2)
Carbon dioxide production (VCO2)
Gas flows and volumes
Respiratory exchange ratio (RER)
Respiratory rate
Anaerobic threshold (AT)The anaerobic threshold (AT) is an estimate of exercise ability. Any measurement below 11 ml/kg/min is usually related with an increase in mortality, especially when there is a background of myocardial ischaemia occurring during the test.
Peak VO2 <20 mL/kg with a low AT have a correlation with postoperative complications and a 30 day mortality. The CPX test is used for risk-testing patients prior to surgery to determine the appropriate postoperative care facilities. The V slope measured in CPX testing represents VO2 versus VCO2 relationship. During AT, the ramp of V slope increases, but does not provide a picture of postoperative mortality.
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This question is part of the following fields:
- Clinical Measurement
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Question 11
Incorrect
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A 70-year-old male presented to an outpatient clinic with a complaint of a lump in his groin. Physical examination reveals the lumps increase in size while coughing and reduces in size after lying down flat. Based on his age and examination, a diagnosis of direct inguinal hernia was made.
Which structures does the bowel pass through in order to be classed as direct inguinal hernia?Your Answer: Femoral ring
Correct Answer: Hesselbach's triangle
Explanation:A hernia is a protrusion of the abdominal viscera through a defect in the abdominal wall. Inguinal hernias are of two types; Indirect inguinal hernia and Direct inguinal hernia.
– Indirect inguinal hernia is common at young age commonly due to a patent processes vaginalis and bowel passes through the deep inguinal ring lateral to the inferior epigastric artery.
– Direct hernia forms as a result of the weakening of the posterior wall of the inguinal canal more specifically within a region called ‘Hasselbach triangle. It is defined medially by the rectus abdominis muscle, laterally by the epigastric vessels, and inferiorly by the inguinal ligament.Direct and indirect hernias can be differentiated based on their relation to the inferior epigastric artery. Direct inguinal hernia lies medial to it while indirect inguinal hernia lies lateral to the inferior epigastric artery.
The femoral ring is the site of the femoral hernia.
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This question is part of the following fields:
- Anatomy
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Question 12
Incorrect
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Which of the following intravenous induction agents is best for the patient with acute intermittent porphyria requiring rapid sequence induction for emergency surgery?
Your Answer: Diazepam
Correct Answer: Propofol
Explanation:Propofol is considered a safe drug to use in porphyria because even if causes mild elevation of porphyrins inpatient, it does not cause any symptoms.
Since barbiturates are inducers of ALA synthetase, they are contraindicated in porphyria patients. So, thiopentone most not be used.
Etomidate is a potent inhibitor of adrenal 11 beta-hydroxylase and 17 alpha-hydroxylase reducing cortisol and aldosterone synthesis in the adrenal cortex and has been associated with exacerbations of porphyria in animal studies and it is advisable not to use it in this condition.
Ketamine should be reserved for the hemodynamically unstable patient, however, it is a safe drug.
Diazepam is safe in porphyria but is not usually used for a rapid sequence induction.
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This question is part of the following fields:
- Pharmacology
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Question 13
Incorrect
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Which among the following is not true regarding disease rates?
Your Answer: Attributable risk is equal to the disease rate in exposed people minus that in unexposed people
Correct Answer: The odds ratio is synonymous with the risk ratio
Explanation:The relative risk (also known as risk ratio [RR]) is the ratio of risk of an event in one group (e.g., exposed group) versus the risk of the event in the other group (e.g., nonexposed group).
The odds ratio (OR) is the ratio of odds of an event in one group versus the odds of the event in the other group.
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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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Your manager asks you to inform patients that are suffering from a chronic pain about a trial that is going to be conducted in order to determine the efficacy of a novel analgesic. What phase is the trial currently in?
Your Answer: Phase 1
Correct Answer: Phase 2
Explanation:Phase 0 trials assist the scientists in studying the behaviour of drugs in humans by micro dosing patients. They are used to speed up the developmental process. They have no measurable therapeutic effect and efficiency.
Phase 1 is associated with assessing whether a drug is safe to use or not. The process is extensive and can take up to several months. It also involves healthy participants (less than 100) that are paid to take part in the study. The side effects upon increasing dosage are also addressed by the study. The effects the drug has on humans including how its absorbed, metabolized and excreted are studied. Approximately 70% of the drugs pass this phase.
Phase 2 trials involve patients that are suffering from the disease under study and are associated with determining the efficiency and the optimum dosage of the drug.
Phase 3 also assesses the efficacy but at a higher scale with larger population sample.
Phase 4 trials are involved with the long term effects and side effects of the drug.
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This question is part of the following fields:
- Statistical Methods
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Question 15
Correct
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Which of the following statements is true about data analysed from a new drug trial?
Your Answer: The data could be evaluated using the chi square test
Explanation:The data is ideal for chi square test evaluation as it will help determine if observed outcomes are in line with expected outcomes, and also if results are significant or due to chance.
The student’s t test is not ideal as it requires comparison of means from different populations, rather than proportions.
Pearson’s coefficient of linear regression is not ideal as it requires the plotting of a linear regression.
The numbers should be analysed before determining if there are any statistical conclusions that can be drawn from the population.
Statistical analysis is always required to determine the performance of any treatment during a clinical drug trial. Conclusions cannot be drawn simply by looking at the data.
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This question is part of the following fields:
- Statistical Methods
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Question 16
Correct
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Of the following, which of these oxygen carrying molecules causes the greatest shift of the oxygen-dissociation curve to the left?
Your Answer: Myoglobin (Mb)
Explanation:Myoglobin is a haemoglobin-like, iron-containing pigment that is found in muscle fibres. It has a high affinity for oxygen and it consists of a single alpha polypeptide chain. It binds only one oxygen molecule, unlike haemoglobin, which binds 4 oxygen molecules.
The myoglobin ODC is a rectangular hyperbola. There is a very low P50 0.37 kPa (2.75 mmHg). This means that it needs a lower P50 to facilitate oxygen offloading from haemoglobin. It is low enough to be able to offload oxygen onto myoglobin where it is stored. Myoglobin releases its oxygen at the very low PO2 values found inside the mitochondria.
P50 is defined as the affinity of haemoglobin for oxygen: It is the PO2 at which the haemoglobin becomes 50% saturated with oxygen. Normally, the P50 of adult haemoglobin is 3.47 kPa(26 mmHg).
Foetal haemoglobin has 2 ? and 2 ?chains. The ODC is left shifted – this means that P50 lies between 2.34-2.67 kPa [18-20 mmHg]) compared with the adult curve and it has a higher affinity for oxygen. Foetal haemoglobin has no ? chains so this means that there is less binding of 2.3 diphosphoglycerate (2,3 DPG).
Carbon monoxide binds to haemoglobin with an affinity more than 200-fold higher than that of oxygen. This therefore decreases the amount of haemoglobin that is available for oxygen transport. Carbon monoxide binding also increases the affinity of haemoglobin for oxygen, which shifts the oxygen-haemoglobin dissociation curve to the left and thus impedes oxygen unloading in the tissues.
In sickle cell disease, (HbSS) has a P50 of 4.53 kPa(34 mmHg).
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This question is part of the following fields:
- Physiology
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Question 17
Correct
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Blood pressure monitoring is a requirement for in-patient care. Different factors can result in an inaccurate blood pressure reading, including the damping of an arterial waveform.
How does a damped arterial waveform affect blood pressure measurements?Your Answer: The systolic pressure is lower and the diastolic pressure higher with the same mean
Explanation:Damping is the reduction of energy in a system achieved by reducing the amplitude of oscillations. It is necessary to some degree to prevent wave overshoots.
Critical damping usually causes the system to be slow, so optimal damping is normally applied to provide a balance between speed and distortion.
Damping can cause errors if excessive (overdamping) or inadequate (Underdamping). The amount of damping in a system can be determined using the damping coefficient (D), where:
Undamped: 0
Critically damped: 1
Optimally damped: 0.64An overdamped system will cause an artificial decrease in the systolic blood pressure, and an artificial increase in the diastolic blood pressure.
An underdamped system will cause an artificial increase in systolic blood pressure and an artificial decrease in diastolic blood pressure.
Damping can be caused by a number of factors affecting different parts of the system, including:
The tubing/cannula: The presence of air bubbles, increased blood viscosity or formation of blood clots.
The diaphragm/tubing: Increased malleability/compliance
The tubing: Presence of kinks, narrowing or too many ports of injection.The answer here is a damped system will have a low systolic pressure, a high diastolic pressure with a normal mean pressure.
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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 70-year-old man presents with bilateral buttock claudication that spreads down the thigh and erectile dysfunction in a vascular clinic.
The left femoral pulse is not palpable on examination, and the right is weakly palpable. Leriche syndrome is diagnosed as the blood flow at the abdominal aortic bifurcation is blocked due to atherosclerosis. He is prepared for aortoiliac bypass surgery.
Which vertebral level will you find the affected artery that requires bypassing?Your Answer: L1
Correct Answer: L4
Explanation:The bifurcation of the abdominal aorta into common iliac arteries occurs at the level of L4. The bifurcation is a common site for atherosclerotic plaques as it is an area of high turbulence.
Leriche Syndrome is an aortoiliac occlusive disease and affects the distal abdominal aorta, iliac arteries, and femoropopliteal vessels. It has a triad of symptoms:
1. Claudication (cramping lower extremities pain that is reproducible by exercise)
2. Impotence (reduced penile arterial flow)
3. Absent/weak femoral pulses (hallmark)T12 – aorta enters the diaphragm with the thoracic duct and azygous veins
L2 – testicular or ovarian arteries branch off the aorta
L3 – inferior mesenteric artery
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This question is part of the following fields:
- Anatomy
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Question 19
Incorrect
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All of the following statements about dopamine are FALSE except:
Your Answer: When given as an infusion dopamine reduces the progression to renal failure.
Correct Answer:
Explanation:Dopamine (DA) is a dopaminergic (D1 and D2) as well as adrenergic ? and ?1 (but not ?2 )agonist.
The D1 receptors in renal and mesenteric blood vessels are the most sensitive: i.v. infusion of a low dose of DA dilates these vessels (by raising intracellular cyclic adenosine monophosphate).
Moderately high doses produce a positive inotropic (direct ?1 and D1 action + that due to NA release), but the little chronotropic effect on the heart.
Vasoconstriction (?1 action) occurs only when large doses are infused.
At doses normally employed, it raises cardiac output and systolic BP with little effect on diastolic BP. It has practically no effect on nonvascular ? and ? receptors; does not penetrate the blood-brain barrier – no Central nervous system effects.
Dopamine is less arrhythmogenic than adrenaline
Regarding dopamine part of the dose is converted to Noradrenaline in sympathetic nerve terminals.
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This question is part of the following fields:
- Pharmacology
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Question 20
Incorrect
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Which of the following is the best associated option with Kaplan-Meier survival plot?
Your Answer: An accurate prediction of decreasing survival with time after an event.
Correct Answer: An estimate of decreasing survival with time after an event.
Explanation:Also known as the “product limit estimate’’, the Kaplan-Meier survival plot is used to estimate the true survival function from the collected data.
Using this plot, probabilities of occurrence of an event at a certain point in time can be computed. The successive probabilities are multiplied by any earlier computed probabilities to get the final estimate. For a given population, the survival probability at any particular time on the plot = (number of subjects living at the start – number of subjects who died)/number of subjects living at the start.
The description of a scatter plot is a graphical representation using Cartesian coordinates to display values for more than two variables for data set. It is used for to assess the relationship between 2 different variables.
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This question is part of the following fields:
- Statistical Methods
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Question 21
Incorrect
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Which of the following is included in monosynaptic reflexes?
Your Answer: The withdrawal reflex.
Correct Answer: The patellar or knee jerk reflex.
Explanation:Monosynaptic reflexes is a type of reflex arc providing direct communication between motor and sensory innervation in a muscle. It occurs very quickly as it arises and ends in the same muscle. Examples include: biceps reflex, brachioradialis reflex, extensor digitorum reflex, triceps reflex, Achilles reflex and patellar reflex.
Polysynaptic reflexes facilitates contraction and inhibition in muscle by providing communication between multiple muscles.
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This question is part of the following fields:
- Pathophysiology
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Question 22
Incorrect
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A sevoflurane vaporiser with a 2 percent setting and a 200 kPa ambient pressure is used.
At this pressure, which of the following options best represents vaporiser output?Your Answer: The output is 1% because the splitting ratio alters
Correct Answer: The output is 1% because the saturated pressure of sevoflurane is unaffected by ambient pressure
Explanation:Ambient pressure has no effect on a volatile agent’s saturated vapour pressure (SVP). At a temperature of 20°C, the SVP of sevoflurane is approximately 21 kPa, or 21% of atmospheric pressure (100 kPa).
The SVP of sevoflurane remains the same when the ambient pressure is doubled to 200 kPa, but the output of the vaporiser is halved, now 21 percent of 200 kPa, equalling 10.5 percent. The vaporiser’s output has increased to 1%, but the partial pressure output has remained unchanged. The splitting ratio will not change because it is determined by temperature changes.
Calculations can be made as follows:
Vaporizer output % (ambient pressure) = % volatile (calibrated) x 100 kPa calibrated pressure/ambient pressure
2% = 2% (dialled) × 100/100
2% of 100 = 2 kPaAltitude, pressure 50 kPa
4% = 2% (dialled) × 100/50
4% of 50 = 2 kPaHigh pressure at 200 kPa
1% = 2% (dialled) × 100/200
1% of 200 = 2 kPaSevoflurane has a boiling point of 58°C and, unlike desflurane (which has a boiling point of 22.8°C), does not need to be heated and pressurised with a Tec 6 vaporiser.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 23
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: Phase 0
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 24
Correct
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The average diastolic blood pressure of a control group was found out to be 80 with a standard deviation of 5 in a study aimed at exploring the efficiency of a novel anti-hypertensive drug. The trial was randomised.
Making an assumption that the data is normally distributed, find out the number of patients that had diastolic blood pressure over 90.Your Answer: 3%
Explanation:Since the data is normally distributed, 95% of the values lie with in the interval 70 to 90. This can be calculated as follows:
Interval= Mean ± ( 2 times standard deviation)
= 80 ± 2(5)
= 80 ± 10
= 70 & 90The rest of the 5% are distributed symmetrically beyond 90 and below 70 which means 2.5% of the values lie above 90.
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This question is part of the following fields:
- Statistical Methods
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Question 25
Incorrect
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Regarding the Valsalva manoeuvre, which of the following describes the cardiovascular changes in phase III in a normal patient?
Your Answer: Normal intrathoracic pressure, increase in blood pressure, and decrease in heart rate
Correct Answer: Normal intrathoracic pressure, decrease in blood pressure, and increase in heart rate
Explanation:When a person forcefully expires against a closed glottis, changes occur in intrathoracic pressure that dramatically affect venous return, cardiac output, arterial pressure, and heart rate. This forced expiratory effort is called a Valsalva maneuver.
Initially during a Valsalva, intrathoracic (intrapleural) pressure becomes very positive due to compression of the thoracic organs by the contracting rib cage. This increased external pressure on the heart and thoracic blood vessels compresses the vessels and cardiac chambers by decreasing the transmural pressure across their walls. Venous compression, and the accompanying large increase in right atrial pressure, impedes venous return into the thorax. This reduced venous return, and along with compression of the cardiac chambers, reduces cardiac filling and preload despite a large increase in intrachamber pressures. Reduced filling and preload leads to a fall in cardiac output by the Frank-Starling mechanism. At the same time, compression of the thoracic aorta transiently increases aortic pressure (phase I); however, aortic pressure begins to fall (phase II) after a few seconds because cardiac output falls. Changes in heart rate are reciprocal to the changes in aortic pressure due to the operation of the baroreceptor reflex. During phase I, heart rate decreases because aortic pressure is elevated; during phase II, heart rate increases as the aortic pressure falls.
When the person starts to breathe normally again, the intrathoracic pressure declines to normal levels, the aortic pressure briefly decreases as the external compression on the aorta is removed, and heart rate briefly increases reflexively (phase III). This is followed by an increase in aortic pressure (and reflex decrease in heart rate) as the cardiac output suddenly increases in response to a rapid increase in cardiac filling (phase IV). Aortic pressure also rises above normal because of a baroreceptor, sympathetic-mediated increase in systemic vascular resistance that occurred during the Valsava.
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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 72-year old man is experiencing a cardiac risk evaluation for the management of obstructive umbilical hernia. Echocardiogram demonstrates an aortic valve area=0.59cm with a pressure of 70mmHg. Five years ago, he had mild myocardial infarction complicated with pulmonary oedema. Now he encounters angina with little exertion.
Which of the following factor is the foremost profoundly weighted using Deysky's cardiac risk scoring system in this case?Your Answer: Angina with mild exertion
Correct Answer: Aortic stenosis
Explanation:Detsky’s Modified cardiac risk classification system in patients undergoing non-cardiac surgery:
Age more than 70: 05 points
History of myocardial infarction:
Less than 6 months: 10 points
More than 6 months: 5 pointsAngina Pectoris:
Angina with minimal exertion: 10 points
Angina at any level of exertion: 20 points
Pulmonary Oedema:
Within 7 days: 10 points
At any time: 5 pointsSuspected aortic valve stenosis with valve area <0.6cm2: 20 points Arrhythmia: Any rhythm other than sinus or sinus with premature atrial complexes (PACs): 5 points More than 5 premature ventricular contractions: 5 points
Emergency Surgery: 10 points
Deficient general medical condition: 5 pointsRisk classification:
Grade I: 0-15 points = low risk
Grade II: 15-30 points = moderate risk
Grade III: >30 points = high risk -
This question is part of the following fields:
- Pathophysiology
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Question 27
Incorrect
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A 66-year-old man with a haemorrhagic stroke is admitted to the medical admissions unit.
He has been taking warfarin for a long time because of atrial fibrillation. His INR at the time of admission was 9.1.
Which of the following treatment options is the most effective in managing his condition?Your Answer: Vitamin K
Correct Answer: Prothrombin complex concentrate
Explanation:Haemorrhage, including intracranial bleeding, is a common and potentially fatal side effect of warfarin therapy, and reversing anticoagulation quickly and completely can save lives. When complete and immediate correction of the coagulation defect is required in orally anticoagulated patients with life-threatening haemorrhage, clotting factor concentrates are the only viable option.
For rapid reversal of vitamin K anticoagulants, prothrombin complex concentrates (PCC) are recommended. They contain the vitamin K-dependent clotting factors II, VII, IX, and X and are derived from human plasma. They can be used as an adjunctive therapy in patients with major bleeding because they normalise vitamin K dependent clotting factors and restore haemostasis.
The most common treatments are fresh frozen plasma (FFP) and vitamin K. The efficacy of this approach is questioned due to the variable content of vitamin K-dependent clotting factors in FFP and the effects of dilution. Significant intravascular volume challenge, as well as the possibility of rare complications like transfusion-associated lung injury or blood-borne infection, are all potential issues.
To avoid anaphylactic reactions, vitamin K should be given as a slow intravenous infusion over 30 minutes. Regardless of the route of administration, the reversal of INRs with vitamin K can take up to 24 hours to reach its maximum effect.
Reversal of anticoagulation in patients with warfarin-associated intracranial haemorrhage may be considered with factor VIIa (recombinant), but its use is controversial. There are concerns about thromboembolic events following treatment, as well as questions about assessing efficacy in changes in the INR. If the drug is to be administered, patients should be screened for an increased risk of thrombosis before the drug is given.
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This question is part of the following fields:
- Pathophysiology
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Question 28
Incorrect
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After establishing a cardiopulmonary bypass, the right atrium is opened to repair the tricuspid valve.
Out of the following, which is NOT a part of the right atrium?Your Answer: Crista terminalis
Correct Answer: Trabeculae carnae
Explanation:The right atrium receives blood supply from the SVC, IVC, and coronary sinus. It forms the right border of the heart.
The interior of the right atrium has 5 distinct features:
1. Sinus venarum – smooth, thin-walled posterior part of the right atrium where the SVC, IVC, and coronary sinus open
2. Musculi pectinati – an anterior rough, wall of pectinate muscles
3. Tricuspid valve orifice – the opening through which the right atrium empties blood into the right ventricle
4. Crista terminalis – separates the rough (musculi pectinati) from the smooth (sinus venarum) internally
5. Fossa ovalis – a thumbprint size depression in the interatrial septum which is a remnant of the oval foramen and its valve in the foetusThe trabeculae carneae are irregular muscular elevations that form the interior of the right ventricle.
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This question is part of the following fields:
- Anatomy
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Question 29
Incorrect
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Which of the following statements is true regarding antibiotics?
Your Answer: Penicillin is active against the bacterial cell membrane
Correct Answer: Staphylococcus aureus colonises the nasopharynx in >20% of the general population
Explanation:Staphylococcus aureus colonizes the nasopharynx in >20% of the general population.
Methicillin-resistant Staphylococcus aureus (MRSA) is resistant to flucloxacillin.
Trimethoprim binds to dihydrofolate reductase and inhibits the reduction of dihydrofolic acid (DHF) to tetrahydrofolic acid (THF). THF is an essential precursor in the thymidine synthesis pathway and interference with this pathway inhibits bacterial DNA synthesis.
All ?-lactam antibiotics like penicillin interfere with the synthesis of the bacterial cell walls. The ?-lactam antibiotics inhibit the transpeptidases so that cross-linking (which maintains the close-knit structure of the cell wall) does not take place
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This question is part of the following fields:
- Pharmacology
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Question 30
Incorrect
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While inspecting the caecum, what structure will be identified at the point at which all the taeniae coli converge?
Your Answer:
Correct Answer: Appendix base
Explanation:The taeniae coli are the three outer muscular bands of the cecum, ascending colon, transverse colon, and descending colon.
The taeniae coli converge at the base of the appendix in the cecum where they form a complete longitudinal layer. In the ascending and descending colon, the bands are located anteriorly, posteromedially, and posterolateral.
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This question is part of the following fields:
- Anatomy
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