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Question 1
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With regards to this state of matter which has a volume but no definite shape, particles are not tightly packed together. These are incompressible although there is free movement within the volume. This statement best describes which one of the following states of matter?
Your Answer: Liquid
Explanation:The solid state of matter has a definite volume and shape and particles are packed closely together and are incompressible. Within this tight lattice, there is enough thermal energy to produce vibration of particles.
Liquids however have a volume but no definite shape. These particles are less tightly packed together. Although there is free movement within the volume, they are incompressible.
Gases, however, have no finite shape or volume and particles are free to move rapidly in a state of random motion. They are compressible and are completely shaped by the space in which they are held. Vapours exist as a gas phase in equilibrium with identical liquid or solid matter below its boiling point.
The most prevalent state of matter in the universe is plasma which is formed by heating atoms to very high temperatures to form ions.
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This question is part of the following fields:
- Basic Physics
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Question 2
Incorrect
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Which of the statements below best describe the total cerebral flow (CBF) in an adult?
Your Answer: Is mainly regulated by the autonomic nervous system
Correct Answer: Accounts for 15% of the cardiac output
Explanation:While the brain only weighs 3% of the body weight, 15% of the cardiac output goes towards the brain.
Between mean arterial pressures (MAP) of 60-130 mmHg, autoregulation of cerebral blood flow (CBF) occurs. Exceeding this, the CBF is maintained at a constant level. This is controlled mainly by the PaCO2 level, and the autonomic nervous system has minimal role.
Beyond these limits, the CBF is directly proportional to the MAP, not the systolic blood pressure.
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This question is part of the following fields:
- Physiology
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Question 3
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 4
Correct
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Which of the following is correct about the action of bile salts?
Your Answer: Emulsification of lipids
Explanation:The emulsification and absorption of fats requires Bile salts.
Absorption of fats is associated with the activation of lipases in the intestine.
Bile salts are involved in fat soluble vitamin absorption and are reabsorbed in the terminal ileum (B12 is NOT fat soluble).
Although Vitamin B12 is also absorbed in the terminal ileum, it is a water soluble vitamin (as are B1, nicotinic acid, folic acid and vitamin C) .
The gastric parietal cells secretes Intrinsic factor that is essential for the absorption of B12.
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This question is part of the following fields:
- Pathophysiology
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Question 5
Correct
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A 70-year-old man presents with central crushing chest pain that radiates to the jaw in the emergency department. He has associated symptoms of nausea and diaphoresis. A 12 lead ECG is performed. ST-elevation is observed in leads V2-V4. The diagnosis of anteroseptal ST-elevation myocardial infarction is made. Which coronary vessel is responsible for this condition and runs in the interventricular septum on the anterior surface of the heart to reach the apex?
Your Answer: Left anterior descending artery
Explanation:The heart receives blood supply from coronary arteries. The right and left coronary arteries branch off the aorta and supply oxygenated blood to all heart muscle parts.
The left main coronary artery branches into:
1. Circumflex artery – supplies the left atrium, side, and back of the left ventricle. The left marginal artery arises from the left circumflex artery. It travels along the obtuse margin of the heart.
2. Left Anterior Descending (LAD) artery – supplies the front and bottom of the left ventricle and front of the interventricular septumThe left anterior descending coronary artery is the largest coronary artery. It courses anterior to the interventricular septum in the anterior interventricular groove, extending from the base of the heart to its apex. Around the apex, the LAD anastomosis with the terminal branches of the posterior descending artery (branch of the right coronary artery).
Atherosclerosis or thrombotic occlusion of LAD causes myocardial infarction in large areas of the anterior, septal, and apical portions of the heart muscle. It can lead to a serious deterioration in heart performance.Occlusion of the LAD causes anteroseptal myocardial infarction, which is evident on the ECG with changes in leads V1-V4. Occlusion of the left circumflex artery causes lateral, posterior, or anterolateral MI. However, as it does not run towards the apex in the interventricular septum of the heart, it is not the correct answer for this question.
The right coronary artery branches into:
1. Right marginal artery
2. Posterior descending arteryThe right coronary artery supplies the right atrium, right ventricle, interatrial septum, and the inferior posterior third of the interventricular septum. Occlusion of the right coronary artery causes inferior MI, which is indicated on ECG with changes in leads II, III, and aVF.
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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 49-year-old female has presented to her physician with complaints of a lump in her groin area. The lump is painless and is more prominent in coughing. On examination, the lump's location is inferior and lateral to the pubic tubercle. This points towards the diagnosis of femoral hernia, where part of her intestines has entered the femoral canal, causing a bulge in the femoral triangle. The femoral triangle is an anatomical region in the upper thigh. Name the structures found in the femoral triangle, laterally to medially.
Your Answer: Femoral nerve, femoral artery, femoral vein, empty space, lymphatics
Explanation:The femoral triangle is a wedge-shaped area found within the superomedial aspect of the anterior thigh. It is a passageway for structures to leave and enter the anterior thigh.
Superior: Inguinal ligament
Medial: Adductor longus
Lateral: Sartorius
Floor: Iliopsoas, adductor longus and pectineusThe contents include: (medial to lateral)
Femoral vein
Femoral artery-pulse palpated at the mid inguinal point
Femoral nerve
Deep and superficial inguinal lymph nodes
Lateral cutaneous nerve
Great saphenous vein
Femoral branch of the genitofemoral nerve. -
This question is part of the following fields:
- Anatomy
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Question 7
Correct
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Which one of the following factor affects the minimal alveolar concentration (MAC)?
Your Answer: Hypoxaemia
Explanation:The minimal alveolar concentration (MAC) is the concentration of an inhalation anaesthetic agent in the lung alveoli required to stop a response to the surgical stimulus in 50% of the patient.
Following factors don’t affect the MAC of the inhaled anaesthetic agents:
Gender, acidosis, alkalosis, hypothyroidism, hyperthyroidism, body weight, serum potassium level, and the duration of the anaesthesia.
MAC increase in children, elevated temperature, high metabolic rate, sympathetic increase and chronic alcoholism.
MAC decrease in low temperature, low oxygen level, old age, hypotension ( 120mmHg is being used in anesthetic-Hinkman as an additive effect to decrease MAC, however, increase concentration of CO2 activates the sympathetic system resulting the MAC increases.
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This question is part of the following fields:
- Physiology
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Question 8
Incorrect
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You've been summoned to help resuscitate a 6-year-old child who has suffered a cardiac arrest. The ECG monitor shows electrical activity that isn't pulsed. Which of the following statements is the most appropriate during resuscitation?
Your Answer: The rate of chest compressions should be 120-140 per minute
Correct Answer: The dose of intravenous adrenaline is 180 mcg
Explanation:To begin, one must determine the child’s approximate weight. There are a variety of formulas to choose from. It is acceptable to use the advanced paediatric life support formula:
(Age + 4) 2 = Weight
A 5-year-old child will weigh around 18 kilogrammes.
10 mcg/kg (0.1 ml/kg of 1 in 10 000 adrenaline) = 180 mcg is the appropriate dose of intravenous or intraosseous adrenaline.
The correct energy level to deliver is 4 J/kg, which equals 72 joules.
The pad size that is appropriate for this patient is 8-12 cm. For an infant, a 4.5 cm pad is appropriate.
To allow adequate separation in infants and small children, the pads should be placed anteriorly and posteriorly on the chest.
When using a bag and mask to ventilate, take two breaths for every 15 chest compressions. If chest compressions are being applied intubated and without interruption, a ventilation rate of 10-20 breaths per minute should be given.
Chest compressions should be done at a rate of 100-120 per minute, the same as an adult.
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This question is part of the following fields:
- Pharmacology
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Question 9
Correct
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Among the different classes of anti-arrhythmics, which one is the first line treatment for narrow complex AV nodal re-entry tachycardia?
Your Answer: Adenosine
Explanation:Adenosine is the first line for AV nodal re-entry tachycardia. An initial dose of 6 mg is given, and a consequent second dose or third dose of 12 mg is administered if the initial dose fails to terminate the arrhythmia.
Aside from Adenosine, a vagal manoeuvre (e.g. carotid massage) is done to help terminate the supraventricular arrhythmia.
Amiodarone is not a first-line drug for supraventricular tachycardias. Digoxin and Propranolol can be considered if the arrhythmia is of a narrow complex irregular type. Verapamil is an alternative to Adenosine if the latter is contraindicated.
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This question is part of the following fields:
- Pharmacology
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Question 10
Correct
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A 30-year-old man has been stabbed in an area of the groin that contains the femoral triangle. He will undergo explorative surgery. Which of the following makes the lateral wall of the femoral triangle?
Your Answer: Sartorius
Explanation:The femoral triangle is a wedge-shaped area found within the superomedial aspect of the anterior thigh. It is a passageway for structures to leave and enter the anterior thigh.
Superior: Inguinal ligament
Medial: Adductor longus
Lateral: Sartorius
Floor: Iliopsoas, adductor longus and pectineusThe contents include: (medial to lateral)
Femoral vein
Femoral artery-pulse palpated at the mid inguinal point
Femoral nerve
Deep and superficial inguinal lymph nodes
Lateral cutaneous nerve
Great saphenous vein
Femoral branch of the genitofemoral nerve. -
This question is part of the following fields:
- Anatomy
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Question 11
Correct
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An experiment is designed to investigate that how three diets having different sugar content affect the body weight to a different level. Which one of the following test will determine a statistically significant difference among the diets?
Your Answer: ANOVA
Explanation:Chi-square test is used to determine the statistically significant different between categorical variables. It also determines the difference between expected frequencies and the observed frequencies.
Mann Whitney U test is used to determine the statistically significant different between two independent groups.
Wilcoxon’s test is the test of dependency. it determines the statistically significant difference between two dependent groups.
Student t-test is one of the most commonly used method to test the hypothesis. It determines the significant difference between the means of two different groups.
ANOVA (analysis of variance) is similar to student’s t-test.
ANOVA is a statistical method used to determines the statistically significant difference between the mean of more than two group. In this experiment as we are dealing with three different group, ANOVA is most suitable test to determine the difference between each groups.
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This question is part of the following fields:
- Statistical Methods
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Question 12
Correct
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Which of the following antiplatelet drugs would be best for rapid offset action?
Your Answer: Epoprostenol
Explanation:Epoprostenol has a half-life of only 42 seconds and has rapid offset. It is used for the treatment of pulmonary hypertension.
Aspirin inhibits the COX enzyme irreversibly. It inhibits thromboxane synthesis but does not inhibit the enzyme thromboxane synthetase.
Ticlopidine, clopidogrel and prasugrel act as irreversible antagonists of P2 Y12 receptor of Adenosine Diphosphate (ADP). These drugs interfere with the activation of platelets by ADP and fibrinogen. Both aspirin and clopidogrel act irreversibly so they are not correct.
Paclitaxel is a long-acting antiproliferative agent used for the prevention of restenosis (recurrent narrowing) of coronary and peripheral stents and is not the correct answer.
Tirofiban has the next shortest duration of action after epoprostenol. If epoprostenol is not given in the question, it would be the best answer.
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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 77-year-old man, is scheduled for an angiogram to investigate gastro-intestinal bleeding. The radiologist performing the angiogram inserts the catheter into the coeliac axis. What level of the vertebrae does the coeliac axis normally arise from the aorta?
Your Answer: T12
Explanation:The coeliac axis refers to one of the splanchnic arteries located within the abdomen.
It arises from the aorta almost horizontally at the level of the T12 vertebrae.
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This question is part of the following fields:
- Anatomy
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Question 14
Incorrect
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All of the following are true when describing the autonomic nervous system except:
Your Answer: The parasympathetic nervous system has long preganglionic and short post ganglionic fibres
Correct Answer: Juxta glomerular apparatus, piloerector muscles and adipose tissue are all organs under sole parasympathetic control
Explanation:With regards to the autonomic nervous system (ANS)
1. It is not under voluntary control
2. It uses reflex pathways and different to the somatic nervous system.
3. The hypothalamus is the central point of integration of the ANS. However, the gut can coordinate some secretions and information from the baroreceptors which are processed in the medulla.With regards to the central nervous system (CNS)
1. There are myelinated preganglionic fibres which lead to the
ganglion where the nerve cell bodies of the non-myelinated post ganglionic nerves are organised.
2. From the ganglion, the post ganglionic nerves then lead on to the innervated organ.Most organs are under control of both systems although one system normally predominates.
The nerves of the sympathetic nervous system (SNS) originate from the lateral horns of the spinal cord, pass into the anterior primary rami and then pass via the white rami communicates into the ganglia from T1-L2.
There are short pre-ganglionic and long post ganglionic fibres.
Pre-ganglionic synapses use acetylcholine (ACh) as a neurotransmitter on nicotinic receptors.
Post ganglionic synapses uses adrenoceptors with norepinephrine / epinephrine as the neurotransmitter.
However, in sweat glands, piloerector muscles and few blood vessels, ACh is still used as a neurotransmitter with nicotinic receptors.The ganglia form the sympathetic trunk – this is a collection of nerves that begin at the base of the skull and travel 2-3 cm lateral to the vertebrae, extending to the coccyx.
There are cervical, thoracic, lumbar and sacral ganglia and visceral sympathetic innervation is by cardiac, coeliac and hypogastric plexi.
Juxta glomerular apparatus, piloerector muscles and adipose tissue are all organs under sole sympathetic control.
The PNS has a craniosacral outflow. It causes reduced arousal and cardiovascular stimulation and increases visceral activity.
The cranial outflow consists of
1. The oculomotor nerve (CN III) to the eye via the ciliary ganglion,
2. Facial nerve (CN VII) to the submandibular, sublingual and lacrimal glands via the pterygopalatine and submandibular ganglions
3. Glossopharyngeal (CN IX) to lungs, larynx and tracheobronchial tree via otic ganglion
4. The vagus nerve (CN X), the largest contributor and carries ¾ of fibres covering innervation of the heart, lungs, larynx, tracheobronchial tree parotid gland and proximal gut to the splenic flexure, liver and pancreasThe sacral outflow (S2 to S4) innervates the bladder, distal gut and genitalia.
The PNS has long preganglionic and short post ganglionic fibres.
Preganglionic synapses, like in the SNS, use ACh as the neuro transmitter with nicotinic receptors.
Post ganglionic synapses also use ACh as the neurotransmitter but have muscarinic receptors.Different types of these muscarinic receptors are present in different organs:
There are:
M1 = pupillary constriction, gastric acid secretion stimulation
M2 = inhibition of cardiac stimulation
M3 = visceral vasodilation, coronary artery constriction, increased secretions in salivary, lacrimal glands and pancreas
M4 = brain and adrenal medulla
M5 = brainThe lacrimal glands are solely under parasympathetic control.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 15
Correct
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Which of the following statements is true regarding prazosin?
Your Answer: Is a selective alpha 1 adrenergic receptor antagonist.
Explanation:Selective α1 -Blockers like prazosin, terazosin, doxazosin, and alfuzosin cause a decrease in blood pressure with lesser tachycardia than nonselective blockers (due to lack of α2 blocking action.
The major adverse effect of these drugs is postural hypotension. It is seen with the first few doses or on-dose escalation (First dose effect).
Its half-life is approximately three hours.
It is excreted primarily through bile and faeces (not through kidneys)
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This question is part of the following fields:
- Pharmacology
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Question 16
Correct
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Which among the given choices can be used to describe a persistent and expected level of disease in a particular population?
Your Answer: Endemic
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.
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This question is part of the following fields:
- Statistical Methods
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Question 17
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 ˆ’ pH = 0, then 50% of drug is ionised and 50% is unionised.
In this example:
7.4 = 4.3 + log ([ionised]/[unionised])
7.4 ˆ’ 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 18
Correct
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Which of the following best describes the phenomenon of higher partial pressures of oxygen in the alveoli (PAO2) in the lung apices when in the upright position?
Your Answer: The ventilation/perfusion (V/Q) ratio is greater than in the basal units
Explanation:The ventilation/perfusion ratio varies in different areas of the lung. In an upright individual, although both ventilation and perfusion increase from the apex to the base of the lung, the increase in ventilation is less than the increase in blood flow. As a result, the normal V̇ /Q̇ ratio at the apex of the lung is much greater than 1 (ventilation exceeds perfusion), whereas the V̇ /Q̇ ratio at the base of the lung is much less than 1 (perfusion exceeds ventilation).
There is more volume in the alveoli found in the apices than in the bases of the lungs. This is due to the weight of the lung stretching the apical alveoli to the maximum size. Also, the weight of the lungs pull themselves away from the chest wall, creating a negative intrapleural pressure. These factors, however, do not directly affect the PAO2.
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This question is part of the following fields:
- Pathophysiology
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Question 19
Incorrect
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A 46-year-old woman is listed for clipping of a cerebral aneurysm, following a diagnosis of surgical third nerve palsy. Which of the following clinical findings correlate with surgical third nerve palsy?
Your Answer: Ptosis, superolateral rotation of globe and mydriasis
Correct Answer: Ptosis, inferolateral rotation of globe and mydriasis
Explanation:Ptosis and mydriasis are visible in surgical third nerve palsy, and the eye looks ‘down and out.’ The loss of innervation to all of the major structures supplied by the oculomotor nerve is reflected in these characteristics.
Ptosis is caused by the paralysis of the levator palpebrae superioris in oculomotor nerve palsy. Due to the unopposed actions of the superior oblique and lateral rectus muscles, the eye rotates down and out.
Mydriasis is caused by surgical (compressive) causes of third nerve palsy, which disrupt the parasympathetic pupillomotor fibres on the nerve’s periphery.
Medical (ischaemic) causes of a third nerve palsy, on the other hand, leave the superficial parasympathetic fibres relatively unaffected and the pupil unaffected.
Horner’s syndrome is characterised by ptosis, anhidrosis, and miosis, which are caused by a loss of sympathetic innervation to the tarsal muscle of the upper lid, facial skin, and dilator pupillae, respectively.
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This question is part of the following fields:
- Pathophysiology
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Question 20
Correct
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Heights of 100 individuals(adults) who were administered steroids at any stage during childhood was studied. The mean height was found to be 169cm with the data having a standard deviation of 16cm. What will be the standard error associated with the mean?
Your Answer: 1.6
Explanation:Standard error can be calculated by the following formula:
Standard Error= (Standard Deviation)/ˆš(Sample Size)
= (16) / ˆš(100)
= 16 / 10
= 1.6 -
This question is part of the following fields:
- Statistical Methods
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