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Question 1
Incorrect
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Which of the following statements is true regarding dopamine?
Your Answer: ? effects predominate at higher doses
Correct Answer: It can increase or decrease cAMP levels
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 cAMP). This increases g.f.r. In addition, DA exerts a natriuretic effect by D1 receptors on proximal tubular cells.
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 barrierāno CNS effects.
Dopamine is used in patients with cardiogenic or septic shock and severe CHF wherein it increases BP and urine outflow.
It is administered by i.v. infusion (0.2ā1 mg/min) which is regulated by monitoring BP and rate of urine formation
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This question is part of the following fields:
- Pharmacology
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Question 2
Incorrect
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A 54-year-old man weighing 70kg, underwent mesh repair for inguinal hernia under general anaesthesia. He was given intravenous co-amoxiclav (Augmentin) following which the patient developed widespread urticarial ras, became hypotensive (61/30 mmHg), and showed clinical signs of bronchospasm. Anaphylaxis is suspected in this patient.
Which one of the following is considered as best initial pharmacological treatment for this condition?Your Answer: Intramuscular adrenaline 0.5 mg
Correct Answer: Intravenous adrenaline 50 mcg
Explanation:The drug of choice for the treatment of anaphylaxis is adrenaline. It has an intravenous route of administration. Since the patient already has intravenous access, the intramuscular route is not appropriate.
Second-line pharmacological intervention includes the use of chlorpheniramine 10mg intravenous, Hydrocortisone 200mg.
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This question is part of the following fields:
- Pharmacology
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Question 3
Correct
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A study was concerned with finding out the normal reference range of IgE levels in adults was conducted. Presuming that the curve follows a normal distribution, what is the percentage of individuals having IgE levels greater than 2 standard deviations from mean?
Your Answer: 2.30%
Explanation:Since the data is normally distributed, 95.4% of the values lie with in 2 standard deviations from mean. The rest of the 4.6% are distributed symmetrically outside of that range which means 2.3% of the values lie above 2 standard deviations of the mean.
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This question is part of the following fields:
- Statistical Methods
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Question 4
Correct
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Which of the following descriptions best describes enflurane and isoflurane?
Your Answer: Have the same molecular formula but different structural formulae
Explanation:Structural isomers have a similar molecular formula, but they have a different structural formula as their atoms are arranged in a different manner. Such small changes lead to the differential pharmacological activity. Enflurane and isoflurane are two prime examples of structural isomers.
Stereoisomers are those substances that have a similar molecular and structural formula, but the arrangement spatially of atoms are different and have optical activity.
Enantiomers are a pair of stereoisomers, which are non-superimposable mirror images of each other. They also have chiral centres of molecular symmetry. Ketamine is considered as an example of racemic mixture (contain 50% R and 50% S enantiomers)
Geometric isomers contain a carbon-carbon double bond (i.e. C=C) or a rigid carbon-carbon single bond in a heterocyclic ring. Cis-atracurium is one example.
Dynamic isomers or Tautomers are a pait of unstable structural isomers, which are present in equilibrium. One isomer can easily change after the change in pH. Midazolam and thiopentone are their examples.
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This question is part of the following fields:
- Pharmacology
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Question 5
Correct
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The prostate and the rectum are separated by which anatomical plane?
Your Answer: Denonvilliers fascia
Explanation:The prostate is separated from the rectum by the Denonvilliers fascia (rectoprostatic fascia).
Waldeyers fascia functions to separate the rectum and the sacrum.
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This question is part of the following fields:
- Anatomy
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Question 6
Correct
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Which of the following does Lidocaine 1% solution equate to?
Your Answer: 1000 mg per 100 ml
Explanation:Lidocaine 1% is formulated as 1000 mg/100 mL.
% solution is based onĀ (grams of medicine) / 100 ml
% solution ~ (1000 mg) / 100 ml
% solution ~ 10 mg/ml
Examples:
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- LidocaineĀ 4% = 40 mg/mlĀ of Lidocaine
- LidocaineĀ 2% = 20 mg/mlĀ of Lidocaine
- LidocaineĀ 1% = 10 mg/mlĀ of Lidocaine
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This question is part of the following fields:
- Pharmacology
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Question 7
Correct
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A 35-year-old male presents to GP presenting an area of erythema which was around a recent cut on his right forearm. He was prescribed a short course of antibiotics and after 5 days again presented with progressive fatigue, headaches, and fevers.
On clinical examination:
Oxygen saturation: 98% on room air
Respiratory rate: 22 per minute
Heart rate: 100 beats per minute
Blood pressure: 105/76 mmHg
Temperature: 38.2 degree Celsius
On physical examination, a dramatic increase in the area of erythema was noted.
Blood culture was done in the patient and indicated the presence of bacterium containing beta-lactamase. Which of the following antibiotics was likely prescribed to the patient?Your Answer: Amoxicillin
Explanation:Ciprofloxacin belongs to the quinolone group of antibiotics, and doxycycline and minocycline are tetracyclines. So, they are not affected by beta-lactamase.
However, amoxicillin is a beta-lactam antibiotic and beta-lactamase cleaves the beta-lactam ring present in amoxicillin. This results in the breakdown of the antibiotic and thus the area of erythema dramatically increased.
Co-amoxiclav contains amoxicillin and clavulanic acid which protects amoxicillin from beta-lactamase. -
This question is part of the following fields:
- Pharmacology
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Question 8
Correct
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At sea level, Sevoflurane is administered via a plenum vaporiser. 100 mL of the fresh gas flow is bypassed into the vaporising chamber. Temperature within the vaporising chamber is maintained at 20°C.
The following fresh gas flows approximates best for the delivery of 1% sevoflurane.Your Answer: 2.7 L/minute
Explanation:The equation for calculating vaporiser output is:
Vaporiser output (VO) mL = Carrier gas flow (mL/minute) Ć SVP of agent (kPa)
Ambient pressure (kPa) ā SVP of agent (kPa)The saturated vapour pressure of sevoflurane at 1 atm (100 kPa) and 20°C is 21 kPa.
VO = (100 mL Ć 21 kPa)/(100 kPa ā 21kPa) for sevoflurane,
VO = 26.6 mL26.6 mL of 100% sevoflurane and 100 mL bypass carrier gas is being added to the fresh gas flow per minute.
2660 mL of 1% sevoflurane and 100 mL bypass carrier gas is approximately 2.7 L/minute.
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This question is part of the following fields:
- Pharmacology
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Question 9
Correct
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Diagnosis of the neuroleptic malignant syndrome is best supported by which of the following statement?
Your Answer: Increased Creatine Kinase
Explanation:The neuroleptic malignant syndrome is a rare complication in response to neuroleptic or antipsychotic medication.
The main features are:
– Elevated creatinine kinase
– Hyperthermia and tachycardia
– Altered mental state
– Increased white cell count
– Insidious onset over 1-3 days
– Extrapyramidal dysfunction (muscle rigidity, tremor, dystonia)
– Autonomic dysfunction (Labile blood pressure, sweating, salivation, urinary incontinence)Management is supportive ICU care, anticholinergic drugs, increasing dopaminergic activity with Amantadine, L-dopa, and dantrolene, and non- depolarising neuromuscular blockade drugs
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This question is part of the following fields:
- Pharmacology
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Question 10
Correct
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Question 11
Correct
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Of the following statements, which is true about the measurements of cardiac output using thermodilution?
Your Answer: Cardiac output should be measured during the end-expiratory pause
Explanation:Thermodilution is the most common dilution method used to measure cardiac output (CO) in a hospital setting.
During the procedure, a Swan-Ganz catheter, which is a specialized catheter with a thermistor-tip, is inserted into the pulmonary artery via the peripheral vein. 5-10mL of a cold saline solution with a known temperature and volume is injected into the right atrium via a proximal catheter port. The solution is cooled as it mixes with the blood during its travel to the pulmonary artery. The temperature of the blood is the measured by the catheter and is profiled using a computer.
The computer also uses the profile to measure cardiac output from the right ventricle, over several measurements until an average is selected.
Cardiac output changes at each point of respiration, therefore to get an accurate measurement, the same point during respiration must be used at each procedure, this is usually the end of expiration, that is the end-expiratory pause.
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This question is part of the following fields:
- Clinical Measurement
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Question 12
Correct
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Which of the following is a characteristic of a type 1B antiarrhythmic agent such as Lidocaine?
Your Answer: Shortens refractory period
Explanation:The action of class 1 anti-arrhythmic is sodium channel blockade. Subclasses of this action reflect effects on the action potential duration (APD) and the kinetics of sodium channel blockade.
Drugs with class 1A prolong the APD and refractory period, and dissociate from the channel with intermediate kinetics.
Drugs with class 1B action shorten the APD in some tissues of the heart, shorten the refractory period, and dissociate from the channel with rapid kinetics.
Drugs with class 1C action have minimal effects on the APD and the refractory period, and dissociate from the channel with slow kinetics.
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This question is part of the following fields:
- Pharmacology
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Question 13
Correct
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A young male is undergoing inguinal hernia repair. During the procedure, the surgeons approach the inguinal canal and expose the superficial inguinal ring.
Which structure forms the lateral edge of the superficial inguinal ring?Your Answer: External oblique aponeurosis
Explanation:The superficial inguinal ring is an opening in the aponeurosis of the external oblique muscle, just above and lateral to the pubic crest.
The superficial ring resembles a triangle more than a ring with the base lying on the pubic crest and its apex pointing towards the anterior superior iliac spine. The sides of the triangle are crura of the opening in the external oblique aponeurosis. The lateral crura of the triangle is attached to the pubic tubercle. The medial crura of the triangle is attached to the pubic crest.
The external oblique aponeurosis forms the anterior wall of the inguinal canal and also the lateral edge of the superficial inguinal ring. The rectus abdominis lies posteromedially, and the transversalis posterior to this.
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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 40-year old female comes to the GP's office with unexplained weight gain, cold intolerance and fatigue. Her thyroid function tests are performed as there is a suspicion of hypothyroidism. A negative feedback mechanism is incorporated in the control of thyroid hormone release. All of choices below are also controlled by a negative feedback loop except:
Your Answer: Clotting cascade
Explanation:The correct answer is the clotting cascade, which occurs via a positive feedback mechanism. As clotting factors are attracted to a site, their presence attracts further clotting factors. This continues until a functioning clot is formed.
This patient has presented with symptoms of hypothyroidism and symptoms include weight gain, lethargy, cold intolerance, dry skin, coarse hair and constipation. It can be treated by replacing the missing thyroid hormone with levothyroxine which is a synthetic version of thyroxine (T4).
Serum carbon dioxide (CO2) is controlled via a negative feedback mechanism as well. Chemoreceptors can detect when the serum CO2 is high, and send an impulse to the respiratory centre of the brain to increase the respiratory rate. As a result, more CO2 is exhaled which lowers the serum concentration.
Cortisol is also released according to a negative feedback mechanism. Cortisol acts on both the hypothalamus and the anterior pituitary. Its action serve to decrease the formation of corticotrophin releasing hormone (CRH) and adrenocorticotropic hormone (ACTH), respectively. CRH acts on the anterior pituitary to release ACTH. This then acts on the adrenal gland to cause the release of cortisol. Thus, inhibition of CRH and ACTH formation results in high levels of cortisol which inhibit its further release.
Blood pressure (BP) is controlled via a negative feedback mechanism. Low BP results in renin-angiotensin-aldosterone system (RAAS) activation. This leads to vasoconstriction and retention of salt and water which increased BP.
Blood sugar is controlled via a negative feedback mechanism. A rise in blood sugar causes insulin to be released. Insulin acts to transport glucose into the cell which lowers blood sugar. -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 15
Correct
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A strict diet is mandatory for which of the following drugs for mood disorders?
Your Answer: Tranylcypromine
Explanation:Tranylcypromine is a monoamine oxidase inhibitor that binds irreversibly to target enzyme.
Monoamine oxidase inhibitors are responsible for blocking the monoamine oxidase enzyme. The monoamine oxidase enzyme breaks down different types of neurotransmitters from the brain: norepinephrine, serotonin, dopamine, and tyramine. MAOIs inhibit the breakdown of these neurotransmitters thus, increasing their levels and allowing them to continue to influence the cells that have been affected by depression.
There are two types of monoamine oxidase, A and B. The MAO A is mostly distributed in the placenta, gut, and liver, but MAO B is present in the brain, liver, and platelets. Serotonin and noradrenaline are substrates of MAO A, but phenylethylamine, methylhistamine, and tryptamine are substrates of MAO B. Dopamine and tyramine are metabolized by both MAO A and B. Selegiline and rasagiline are irreversible and selective inhibitors of MAO type B, but safinamide is a reversible and selective MAO B inhibitor.
MAOIs prevent the breakdown of tyramine found in the body and certain foods, drinks, and other medications. Patients that take MAOIs and consume tyramine-containing foods or drinks will exhibit high serum tyramine level. A high level of tyramine can cause a sudden increase in blood pressure, called the tyramine pressor response. Even though it is rare, a high tyramine level can trigger a cerebral haemorrhage, which can even result in death.
Eating foods with high tyramine can trigger a reaction that can have serious consequences. Patients should know that tyramine can increase with the aging of food; they should be encouraged to have fresh foods instead of leftovers or food prepared hours earlier. Examples of high levels of tyramine in food are types of fish and types of meat, including sausage, turkey, liver, and salami. Also, certain fruits can contain tyramine, like overripe fruits, avocados, bananas, raisins, or figs. Further examples are cheeses, alcohol, and fava beans; all of these should be avoided even after two weeks of stopping MAOIs. Anyone taking MAOIs is at risk for an adverse hypertensive reaction, with accompanying morbidity. Patients taking reversible MAOIs have fewer dietary restrictions.
Amitriptyline is a tricyclic antidepressant, and citalopram and escitalopram are selective serotonin reuptake inhibitors.
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This question is part of the following fields:
- Pharmacology
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Question 16
Incorrect
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All of the following statements about intravenous induction agents are false except:
Your Answer: Barbiturates include thiopental and methohexitone.
Correct Answer:
Explanation:Thiopental is a new British Approved Name for thiopentone and is thio-barbiturate.
Methohexitone is an oxy- barbiturate. Both thiopental and methohexitone are intravenous induction agents.Ketamine cannot cause loss of consciousness in less than 30 seconds. At least 30 seconds is needed to cause loss of consciousness following intravenous administration.
Etomidate is an imidazole but it is not used in the Intensive Care unit for sedation because it has an antidepressant effect on the steroid axis.
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This question is part of the following fields:
- Pharmacology
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Question 17
Correct
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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: 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 18
Correct
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A new study is being carried out on the measurement of a new cardiovascular disease biomarker, and its applications in preoperative screening. The data for this study is expected to be normally distributed.
Which of the following statements is true about normal distributions?Your Answer: The mean, median and mode are the same value
Explanation:The correct answer is the mean, median and mode of normally distributed data are the same value. This is as a result of the bell shaped curve which is equal on both sides.
The bell-shape indicates that values around the mean are more frequent in occurrence than the values farther away.
In a normal distribution:
1) +/- one standard deviation of the mean accounts for 68% of the data.
2) +/- two standard deviations of the mean accounts for 95% of the data.
3) +/- three standard deviations of the mean accounts for 99.7% of the data. -
This question is part of the following fields:
- Statistical Methods
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Question 19
Correct
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What statement about endotoxins is true?
Your Answer: Can often survive autoclaving
Explanation:Endotoxins are the lipopolysaccharides found in the outer cell wall of Gram-negative bacteria. They are responsible for providing the structure and stability of the cell wall.
They cannot be destroyed by normal sterilisation as they are heat stable molecules. They require the use of certain sterilant such as superoxide, peroxide and hypochlorite to be neutralised.
They stimulate strong immune responses, but can only be destroyed partially by specific antibodies. Repeat infections occur as memory T cells cannot be formed.
It can cause septicaemia and associated symptoms such as fever, shock, hypotension and nausea.
It activates the alternative complement pathway and the coagulation pathway using secreted cytokines.
It is not involved in botulism as clostridium botulinum, the responsible organism, secretes a neurotoxic exotoxin.
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This question is part of the following fields:
- Pathophysiology
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Question 20
Correct
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A 72-year-old man complains of severe, central abdominal pain that radiates to the back. He has a past medical history of an abdominal aortic aneurysm.
A focused abdominal ultrasonography test (FAST) is performed, revealing diffuse dilatation of the abdominal aorta. The most prominent dilatation is at the bifurcation site of abdominal aorta into the iliac arteries.
What vertebra level corresponds to the site of the most prominent dilatation as evident on the FAST scan?Your Answer: L4
Explanation:The important landmarks of vessels arising from the abdominal aorta at different levels of vertebrae are:
T12 – Coeliac trunk
L1 – Left renal artery
L2 – Testicular or ovarian arteries
L3 – Inferior mesenteric artery
L4 – Bifurcation of the abdominal aorta
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This question is part of the following fields:
- Anatomy
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Question 21
Correct
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An acidic drug with a pKA of 4.3 is injected intravenously into a patient.
At a normal physiological pH, the approximate ratio of ionised to unionised forms of this drug in the plasma is?Your Answer: 1000:01:00
Explanation:The pH at which the drug exists in 50 percent ionised and 50 percent unionised forms is known as the pKa.
To calculate the proportion of ionised to unionised form of an ACID, use the Henderson-Hasselbalch equation.
pH = pKa + log ([A-]/[HA])
or
pH = pKa + log [(salt)/(acid)]
pH = pKa + log ([ionised]/[unionised]).Hence, if the pKa ā pH = 0, then 50% of drug is ionised and 50% is unionised.
In this example:
7.4 = 4.3 + log ([ionised]/[unionised])
7.4 ā 4.3 = log ([ionised]/[unionised])
log 3.1 = log ([ionised]/[unionised])Simply put, the antilog is the inverse log calculation. In other words, if you know the logarithm of a number, you can use the antilog to find the value of the number. The antilogarithm’s definition is as follows:
y = antilog x = 10x
Antilog to the base 10 of 0 = 1, 1 = 10, 2 =100, 3 = 1000, and 4 = 10,000.
If you want to find the antilogarithm of 3.1, for a number between 3 and 4, the antilogarithm will return a value between 1000 and 10,000. The ratio is 1:1 if pKa = pH, that is, pH pKa = log 0. (50 percent ionised and unionised).
According to the above value, there is only one unionised molecule for every approximately 1000 (1259) ionised molecules of this drug in plasma, implying that this drug is largely ionised in plasma (99.99 percent ).
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This question is part of the following fields:
- Pharmacology
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Question 22
Correct
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An 80-year-old presents to the emergency department with symptoms raising suspicion of mesenteric ischemia. To diagnose the condition, an angiogram is performed. The radiologist needs to cannulate the coeliac axis from the aorta for the angiogram.
What vertebral level does the coeliac axis originate from the aorta?
Your Answer: T12
Explanation:Mesenteric ischemia is ischemia of the blood vessels of the intestines. It can be life-threatening especially if the small intestine is involved.
A critical factor for survival of acute mesenteric ischemia is early diagnosis and intervention. Angiography uses X-ray and contrast dye to image arteries and identify the severity of ischemia or obstruction.
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 arteriesThere are some important landmarks of vessels at different levels of vertebrae that need to be memorized.
T12 – Coeliac trunk
L1 – Left renal artery
L2 – Testicular or ovarian arteries
L3 – Inferior mesenteric artery
L4 – Bifurcation of the abdominal aorta
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This question is part of the following fields:
- Anatomy
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Question 23
Correct
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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: 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 24
Incorrect
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In a study lasting over a period of two years, in which the mean age of 800 patients was 82 years, the efficacy of hip protectors in reducing femoral neck fractures was discussed.
Both experimental and control group had 400 members. Instances of fractures reported over the two year time duration were 10 for the control group (that were prescribed hip protector) and 20 for the control group.
What is the value of Absolute Risk Reduction?Your Answer: 0.5
Correct Answer: 0.025
Explanation:ARR= (Risk factor associated with the new drug group) — (Risk factor associated with the currently available drug)
So,
ARR= (10/400)-(20/400)
ARR= 0.025-0.05
ARR= 0.025 (Numerical Value)
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This question is part of the following fields:
- Statistical Methods
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Question 25
Correct
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During positive pressure ventilation using positive end-expiratory pressure (PEEP), there is usually an associated reduction in cardiac output
Which of the following is responsible?
Your Answer: Reduced venous return to the heart
Explanation:The option that is most responsible is the progressive decrease in venous return of blood to the right atrium. The heart rate does not usually change with PEEP so the fall in cardiac output is due to a reduction in left ventricular (LV) stroke volume (SV).
Note that the interventricular septum does shift toward the left and there is an increased pulmonary vascular resistance (PVR) from overdistention of alveolar air sacs that contribute to the reduction in cardiac output. Any increase in PVR will be associated with reduced pulmonary vascular capacitance.
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This question is part of the following fields:
- Pathophysiology
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Question 26
Correct
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Among the following options which one compares variance within the group and variance between groups?
Your Answer: ANOVA
Explanation:ANOVA is based upon within group variance (i.e. the variance of the mean of a sample) and between group variance (i.e. the variance between means of different samples). The test works by finding out the ratio of the two variances mentioned above. (Commonly known as F statistic).
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This question is part of the following fields:
- Statistical Methods
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Question 27
Correct
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A double-blinded randomised controlled trial is proposed to assess the effectiveness of a new blood pressure medication.
Which type of bias can be avoided by ensuring the patient and doctor are blinded?Your Answer: Expectation bias
Explanation:Observers may subconsciously measure or report data in a way that favours the expected study outcome. Therefore, by blinding the study we can eliminate expectation bias.
Recall bias is a systematic error that occurs when the study participants omit details or do not remember previous events or experiences accurately.
Verification can occur during investigations when there is a difference in testing strategy between groups of individuals, which might lead to biasness due to differing ways of verifying the disease of interest.
Nonresponse bias is the bias that occurs when the people who respond to a survey differ significantly from the people who do not respond to the survey.
A distortion that modifies an association between an exposure and an outcome because a factor is independently associated with the exposure and the outcome. Randomization is the best way to reduce the risk of confounding.
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This question is part of the following fields:
- Statistical Methods
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Question 28
Correct
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After a bariatric surgery, average weight loss observed in patients is 18 kg. The standard deviation was found to be 3 kg. What is the percentage of patients that lie between 9 and 27 kg?
Note: Assume that the curve is normally distributed.Your Answer: 99.70%
Explanation:9 & 27 can be obtained by subtracting and adding 9 from the mean. 9 is three times the standard deviation and we know that 99.7% values lie within 3 standard deviations from the mean. We can find the interval for 99.7% to verify in the following way:
For 99.7% confidence interval, you can find the range as follows:
1. Multiply the standard error by 3.
2. Subtract the answer from mean value to get the lower limit.
3. Add the answer obtained in step 1 from the mean value to get the upper limit.
4. The range turns out to be 9-27 kg.
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This question is part of the following fields:
- Statistical Methods
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Question 29
Correct
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The action potential in a muscle fibre is initiated by which of these ions?
Your Answer: Sodium ions
Explanation:The cardiac action potential has several phases which have different mechanisms of action as seen below:
Phase 0: Rapid depolarisation – caused by a rapid sodium influx.
These channels automatically deactivate after a few msPhase 1: caused by early repolarisation and an efflux of potassium.
Phase 2: Plateau – caused by a slow influx of calcium.
Phase 3 – Final repolarisation – caused by an efflux of potassium.
Phase 4 – Restoration of ionic concentrations – The resting potential is restored by Na+/K+ATPase.
There is slow entry of Na+into the cell which decreases the potential difference until the threshold potential is reached. This then triggers a new action potentialOf note, cardiac muscle remains contracted 10-15 times longer than skeletal muscle.
Different sites have different conduction velocities:
1. Atrial conduction – Spreads along ordinary atrial myocardial fibres at 1 m/sec2. AV node conduction – 0.05 m/sec
3. Ventricular conduction – Purkinje fibres are of large diameter and achieve velocities of 2-4 m/sec, the fastest conduction in the heart. This allows a rapid and coordinated contraction of the ventricles
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This question is part of the following fields:
- Physiology
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Question 30
Incorrect
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A 28-year-old girl complained of severe abdominal pain and hematemesis and was rushed into the emergency department. She has an increased heart rate of 120 beats per minute and blood pressure of 90/65. She has a history of taking Naproxen for her Achilles tendinopathy. On urgent endoscopy, she is diagnosed with a bleeding peptic ulcer.
The immediate treatment is to permanently stop the bleeding by performing embolization of the left gastric artery via an angiogram.
What level of the vertebra will be used as a radiological marker for the origin of the artery that supplies the left gastric artery during the angiogram?Your Answer:
Correct Answer: T12
Explanation:The left gastric artery is the smallest branch that originates from the coeliac trunkāthe coeliac trunk branches of the abdominal aorta at the vertebral level of T12.
The left gastric artery runs along the superior portion of the lesser curvature of the stomach. A peptic ulcer that is serious enough to erode through the stomach mucosa into a branch of the left gastric artery can cause massive blood loss in the stomach, leading to hematemesis. The patient also takes Naproxen, a non-steroidal anti-inflammatory drug that is a common cause for peptic ulcers in otherwise healthy patients.
The left gastric artery is responsible for 85% of upper GI bleeds. In cases refractory to initial treatment, angiography is sometimes needed to embolise the vessel at its origin and stop bleeding. During an angiogram, the radiologist will enter the aorta via the femoral artery, ascend to the level of the 12th vertebrae and then enter the left gastric artery via the coeliac trunk.
The important landmarks of vessels arising from the abdominal aorta at different levels of vertebrae are:
T12 – Coeliac trunk
L1 – Left renal artery
L2 – Testicular or ovarian arteries
L3 – Inferior mesenteric artery
L4 – Bifurcation of the abdominal aorta
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This question is part of the following fields:
- Anatomy
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