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  • Question 1 - A 27-year-old woman is admitted to the emergency room with an ectopic pregnancy...

    Incorrect

    • A 27-year-old woman is admitted to the emergency room with an ectopic pregnancy that has ruptured.

      The following is a description of the clinical examination:

      Anxious
      Capillary refill time of 3 seconds
      Cool peripheries
      Pulse 120 beats per minute
      Blood pressure 120/95 mmHg
      Respiratory rate 22 breaths per minute.

      Which of the following is the most likely explanation for these clinical findings?

      Your Answer: Reduction in blood volume of 30-40%

      Correct Answer: Reduction in blood volume of 15-30%

      Explanation:

      The following is the Advanced Trauma Life Support (ATLS) classification of haemorrhagic shock:

      Class I haemorrhage:
      It has blood loss up to 15%. There is very less tachycardia, and no changes in blood pressure, RR or pulse pressure. Usually, fluid replacement is not required.

      Class II haemorrhage:
      It has 15-30% blood loss, equivalent to 750 – 1500 ml. There is tachycardia, tachypnoea and a decrease in pulse pressure. Patient may be frightened, hostile and anxious. It can be stabilised by crystalloid and blood transfusion.

      Class III haemorrhage:
      There is 30-40% blood loss. It portrays inadequate perfusion, marked tachycardia, tachypnoea, altered mental state and fall in systolic pressure. It requires blood transfusion.

      Class IV haemorrhage:
      There is > 40% blood volume loss. It is a preterminal event, and the patient will die in minutes. It portrays tachycardia, significant depression in systolic pressure and pulse pressure, altered mental state, and cold clammy skin. There is need for rapid transfusion and surgical intervention.

    • This question is part of the following fields:

      • Physiology
      5.1
      Seconds
  • Question 2 - What is the percentage of values that lie within 3 standard deviations of...

    Incorrect

    • What is the percentage of values that lie within 3 standard deviations of the mean?

      Your Answer: 98.30%

      Correct Answer: 99.70%

      Explanation:

      99.7% of the values within 3 standard deviations of the mean.

      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.

      For a confidence interval of 68%, multiply the standard error with 1 and repeat the process. For a 95% confidence interval, Standard Error is multiplied by 1.96 to get the interval.

    • This question is part of the following fields:

      • Statistical Methods
      15.9
      Seconds
  • Question 3 - Which of the following explains the mode of action of Magnesium sulphate in...

    Incorrect

    • Which of the following explains the mode of action of Magnesium sulphate in preventing eclampsia in susceptible patients?

      Your Answer: Reduction of cerebral oedema formation opposing movement of solutes across capillaries

      Correct Answer: Dilatation of cerebral circulation due to calcium channel antagonism reducing cerebral vascular spasm

      Explanation:

      Magnesium is a unique calcium antagonist as it can act on most types of calcium channels in vascular smooth muscle and as such would be expected to decrease intracellular calcium. One major effect of decreased intracellular calcium would be inactivation of calmodulin-dependent myosin light chain kinase activity and decreased contraction, causing arterial relaxation that may subsequently lower peripheral and cerebral vascular resistance, relieve vasospasm, and decrease arterial blood pressure.

      The vasodilatory effect of MgSO4 has been investigated in a wide variety of vessels. For example, both in vivo and in vitro animal studies have shown that it is a vasodilator of large conduit arteries such as the aorta, as well as smaller resistance vessels including mesenteric, skeletal muscle, uterine, and cerebral arteries.

      The theory of cerebrovascular vasospasm as the aetiology of eclampsia seemed to be reinforced by transcranial Doppler (TCD) studies which suggested that MgSO4 treatment caused dilation in the cerebral circulation as well as in animal studies that used large cerebral arteries.

    • This question is part of the following fields:

      • Pathophysiology
      26.1
      Seconds
  • Question 4 - Following an uneventful laparoscopic right hemicolectomy, a previously fit and well 75-year-old male...

    Correct

    • Following an uneventful laparoscopic right hemicolectomy, a previously fit and well 75-year-old male is admitted to the critical care unit.

      You've been summoned to examine the patient because he's become oliguric.

      Which of the following is most likely to indicate that acute kidney injury is caused by a prerenal cause?

      Your Answer: Serum urea: creatinine ratio 200

      Explanation:

      Prerenal failure has a serum urea: creatinine ratio of >100, while acute kidney injury has a ratio of 40.
      In prerenal failure, ADH levels are typically high, resulting in water, urea, and sodium resorption. The fractional sodium excretion is less than 1%, but it is greater than 2% in acute tubular necrosis.
      Prerenal azotaemia has higher serum urea nitrogen/serum creatinine ratios (>20), whereas acute tubular necrosis has lower ratios (10-15). The normal range is between 12 and 20.
      Urinary sodium is less than 20 in prerenal failure and greater than 40 in acute tubular necrosis.
      Prerenal failure has a urine osmolality of >500, while acute tubular necrosis has an osmolality of 350.
      Prerenal failure has a urine/serum creatinine ratio of >40, while acute tubular necrosis has a urine/serum creatinine ratio of 20.

      The concentrations of serum urea or creatinine change in inverse proportion to glomerular filtration. Changes in serum creatinine concentrations are more reliable than changes in serum urea concentrations in predicting GFR. Creatinine is produced at a constant rate from creatine, and blood concentrations are almost entirely determined by GFR.

      A number of factors influence urea formation, including liver function, protein intake, and protein catabolism rate. Urea excretion is also influenced by hydration status, the amount of water reabsorption, and GFR.

      A high serum creatinine level, as well as a urine output of less than 10 mL/hour and the production of concentrated looking urine, do not necessarily indicate a specific cause of oliguria.

    • This question is part of the following fields:

      • Pathophysiology
      42.4
      Seconds
  • Question 5 - An 85-year old female is being investigated and treated for pancytopenia of unknown...

    Incorrect

    • An 85-year old female is being investigated and treated for pancytopenia of unknown origin. Her most recent blood test is shown below which shows that he has a low platelet count.

      Hb-102 g/l
      WBC - 2.9* 109/l
      Platelets - 7 * 109/l

      Which of the following normally stimulates platelet production?


      Your Answer: Erythropoietin

      Correct Answer: Thrombopoietin

      Explanation:

      Interleukin-4 is a cytokine which acts to regulate the responses of B and T cells.

      Erythropoietin is responsible for the signal that initiated red blood cell production.

      Granulocyte-colony stimulating factor stimulates the bone marrow to produce granulocytes.

      Interleukin-5 is a cytokine that stimulates the proliferation and activation of eosinophils.

      Thrombopoietin is the primary signal responsible for megakaryocyte and thus platelet production.
      Platelets are also called thrombocytes. They, like red blood cells, are also derived from myeloid stem cells. The process involves a megakaryocyte developing from a common myeloid progenitor cell. A megakaryocyte is a large cell with a multilobulated nucleus, this grows to become massive where it will then break up to form platelets.

      Immune cells are generated from haematopoietic stem cells in bone marrow. They generate two main types of progenitors, myeloid and lymphoid progenitor cells, from which all immune cells are derived.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      38.8
      Seconds
  • Question 6 - A 58-year-old man, visits his general practitioner complaining of a lump in his...

    Correct

    • A 58-year-old man, visits his general practitioner complaining of a lump in his groin. He explains he is otherwise well and reports no other symptoms. The lump is examined and is found to be soft, and can be reduced without causing the patient pain. The GP diagnoses an inguinal hernia. To determine the nature of the hernia, the GP reduced the lump and applies pressure on the deep inguinal ring.

      The deep inguinal ring has what anatomical landmark?

      Your Answer: Superior to the midpoint of the inguinal ligament

      Explanation:

      The deep inguinal ring lies approximately 1.5-2cm above the midpoint of the inguinal ligament, the halfway point between the anterior superior iliac spine and the pubic tubercle, next to the epigastric vessels.

      It is an important point in determining the nature of an inguinal hernia (direct or indirect). The patient is asked to cough after the hernia is reduced, with pressure applied to the deep inguinal ring. The hernia reappearing indicates it is direct, moving through the posterior wall of the inguinal canal.

      Inferior and lateral to the pubic tubercle is the normal anatomical position of the neck of a femoral hernia.

      Superior and medial to the pubic tubercle is the site of the superficial inguinal ring, and the normal anatomical position of the neck of an inguinal hernia.

      The mid-inguinal point is located halways between the pubic symphysis and the anterior superior iliac spine. It is the surface marking for taking the femoral pulse.

    • This question is part of the following fields:

      • Anatomy
      32.6
      Seconds
  • Question 7 - Which of the following statement is true or false regarding to the respiratory...

    Incorrect

    • Which of the following statement is true or false regarding to the respiratory tract?

      Your Answer: The lowest level of the pleural lower margin is at the tenth rib in the mid-clavicular line

      Correct Answer: The sympathetic innervation of the bronchi is derived from T2 - T4

      Explanation:

      The diaphragm has three opening through which different structures pass from the thoracic cavity to the abdominal cavity:

      Inferior vena cava passes at the level of T8.

      Oesophagus, oesophageal vessels and vagi at T10.

      Aorta, thoracic duct and azygous vein through T12.

      Sympathetic trunk and pulmonary branches of vagus nerve form a posterior pulmonary plexus at the root of the lung. Fibres continue posteriorly from superficial cardiac plexus to form Anterior pulmonary plexus. It contains vagi nerves and superficial cardiac plexus. These fibres then follow the blood vessel and bronchi into the lungs.

      The lower border of the pleura is at the level of:

      8th rib in the midclavicular line

      10th rib in the lower level of midaxillary line

      T12 at its termination.

      Both lungs have oblique fissure while right lung has transverse fissure too.

      The trachea expands from the lower edge of the cricoid cartilage (at the level of the 6th cervical vertebra) to the carina.

    • This question is part of the following fields:

      • Physiology
      25.8
      Seconds
  • Question 8 - Regarding the Valsalva manoeuvre, which of the following describes the cardiovascular changes in...

    Incorrect

    • Regarding the Valsalva manoeuvre, which of the following describes the cardiovascular changes in phase III in a normal patient?

      Your Answer: Raised intrathoracic pressure, increase in blood pressure, and increase 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.

    • This question is part of the following fields:

      • Pathophysiology
      15.7
      Seconds
  • Question 9 - If a large volume of 0.9% N. saline is administered during resuscitation, it...

    Correct

    • If a large volume of 0.9% N. saline is administered during resuscitation, it is most likely to cause?

      Your Answer: Hyperchloremic metabolic acidosis

      Explanation:

      Crystalloids recommended for fluid resuscitation include 0.9% N saline and Hartmann’s solution(a physiological solution). 0.9% N. saline is not a physiological solution for the following reasons:

      Compared with the normal range of 98-102 mmol/L, its chloride concentration is high (154 mmol/L)
      It lacks calcium, magnesium, glucose and potassium
      It does not have bicarbonate or bicarbonate precursor buffer necessary to maintain plasma pH within normal limits

      There is a difference in the activity (concentration) of strong ions at a physiological pH. This imbalance can explain abnormalities of acid base balance. A normal strong ion difference (SID) is in the order of 40.

      SID = ([Na+] + [K+] + [Ca2+] + [Mg2+]) – ([Cl-] + [lactate] + [SO42-])

      This imbalance is made up with the weaker anions to maintain electrical neutrality.
      Administration of a large volume of 0.9% normal saline during resuscitation results in excessive chloride administration and this impairs renal bicarbonate reabsorption. The SID of 0.9% normal saline is 0 (Na+ = 154mmol/L and Cl- = 154mmol/L = 154 – 154 = 0). A large volume of NS will decrease the plasma SID causing an acidosis.

      Other causes of a hyperchloremic acidosis are:

      Diabetic ketoacidosis
      Total Parenteral Nutrition
      Overdose of ammonium chloride and hydrochloric acid
      Gastrointestinal losses of bicarbonate like in diarrhoea and pancreatic fistula
      Proximal renal tubular acidosis with failure of bicarbonate reabsorption

    • This question is part of the following fields:

      • Physiology
      44.4
      Seconds
  • Question 10 - A 26-year-old doctor has recently been diagnosed with lung cancer. He would like...

    Correct

    • A 26-year-old doctor has recently been diagnosed with lung cancer. He would like to find out his survival time for the condition.

      Which statistical method is used to predict survival rate?

      Your Answer: Kaplan-Meier estimator

      Explanation:

      The Weibull distribution are used to describe various types of observed failures of the components. it is used in reliability and survival analysis.

      Regression Analysis is used to measure the relationship between among two or more variable. It determines the effect of independent variables on the dependent variables.

      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.

      A time series is a collection of observations of well-defined data obtained at regular interval of time.

      Kaplan-Meier estimator is used to estimate the survival function from lifetime data. It can be derived from maximum likelihood estimation of hazard function. It is most likely used to measure the fraction of patient’s life for a certain amount of time after treatment.

    • This question is part of the following fields:

      • Statistical Methods
      14.3
      Seconds
  • Question 11 - A double blind placebo control clinical trial is done. Which of these is...

    Incorrect

    • A double blind placebo control clinical trial is done. Which of these is correct about it?

      Your Answer: All patients receive a placebo

      Correct Answer: The clinician assessing the effects of the treatment does not know which treatment the patient has been given

      Explanation:

      A ‘double blind crossover study’ happens when every patient receive both treatments.

      It is incorrect to say that only half of the patients do not know which treatment they receive because in a double blind placebo control clinical trial ALL of the patients are blind to their treatment choice .

      If some of the patients are not treated, they would be aware that they were not being treated and it could not be considered a blind trial.

      In a double blind placebo control clinical trial both the clinician and the patient are blind to the treatment choice. The clinician assessing the effects of the treatment, therefore, does not know which treatment the patient has been given.

    • This question is part of the following fields:

      • Statistical Methods
      29.2
      Seconds
  • Question 12 - The required sample size in a trial of a new therapeutic agent varies...

    Incorrect

    • The required sample size in a trial of a new therapeutic agent varies with?

      Your Answer: Type of statistical test to be employed

      Correct Answer: Level of statistical significance required

      Explanation:

      The level of statistical significance required influences the sample size used. This is because sample size is used in the calculation of SD/SE.

      Sample size does not affect

      The level of acceptance
      The alternative hypothesis with a general level set at p<0.05
      The test to be used.

      Experience of the investigator and the type of patient recruited should have no bearing on the required sample size.

    • This question is part of the following fields:

      • Statistical Methods
      25.4
      Seconds
  • Question 13 - Which of the following descriptions best describes enflurane and isoflurane? ...

    Incorrect

    • Which of the following descriptions best describes enflurane and isoflurane?

      Your Answer: Are a pair of stereoisomers that are non-superimposable mirror images of each other

      Correct 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.

    • This question is part of the following fields:

      • Pharmacology
      24.5
      Seconds
  • Question 14 - You draw a patient's blood sample from the median cubital vein in the...

    Incorrect

    • You draw a patient's blood sample from the median cubital vein in the antecubital fossa.

      Which of the following veins also connects to the cephalic vein other than the median cubital vein?

      Your Answer: Median vein

      Correct Answer: Basilic vein

      Explanation:

      The upper limb venous drainage is divided into superficial and deep. The superficial veins are accessible to draw blood for investigations. The cephalic, basilic, and median cubital veins are superficial veins.

      The median cubital vein connects the cephalic vein and basilic vein. It is located anteriorly in the antecubital fossa and is preferred for venepuncture due to its palpability and ease of access.

      The basilic vein and cephalic vein are the primary veins that drain the upper limb. They begin as the dorsal venous arch. The basilic vein originates from the ulnar side, while the cephalic vein originates from the radial side of the dorsal arch of the upper limb.

    • This question is part of the following fields:

      • Anatomy
      31.1
      Seconds
  • Question 15 - Under general anaesthesia, a 48-year-old patient is scheduled for some dental extractions. He...

    Incorrect

    • Under general anaesthesia, a 48-year-old patient is scheduled for some dental extractions. He tells you that he has a heart murmur and that he has always received antibiotic prophylaxis at the dentist. There are no allergies that he is aware of.

      Which antibiotic prophylaxis strategy is most appropriate for this patient?

      Your Answer: Give an oral dose of an appropriate antibiotic preoperatively

      Correct Answer: Prophylactic antibiotics are unnecessary for this patient

      Explanation:

      The National Institute for Health and Care Excellence (NICE) has published guidelines on infective endocarditis prophylaxis (IE). The goal was to create clear guidelines for antibiotic prophylaxis in patients undergoing dental procedures as well as certain non-dental interventional procedures. A number of studies have found an inconsistent link between recent interventional procedures and the development of infective endocarditis in both dental and non-dental procedures.

      Antibiotic prophylaxis against infective endocarditis is not advised or required in the following situations:

      Dental patients undergoing procedures
      Patients undergoing procedures involving the upper and lower gastrointestinal tracts, the genitourinary tract (including urological, gynaecological, and obstetric procedures, as well as childbirth), and the upper and lower respiratory tract (including ear, nose and throat procedures and bronchoscopy).

      Antibiotic resistance can be exacerbated by the indiscriminate use of prophylactic antibiotics, but this is not the primary reason for avoiding their use in these situations.

      To reduce the risk of endocarditis, any patient who is at risk of developing IE should be investigated and treated as soon as possible. Patients with the following conditions are at risk of developing IE:
      acquired valvular heart disease with regurgitation or stenosis
      previous valve replacement
      structural congenital heart disease
      past history of IE, or
      hypertrophic cardiomyopathy (HOCM)

      It would also be appropriate for high-risk dental procedures and those with severe gingival disease.

      Although this patient may not have structural heart disease, ABs should be administered on a case-by-case basis.

    • This question is part of the following fields:

      • Pharmacology
      38.5
      Seconds
  • Question 16 - Which of the following is a correct match for reflex and their root...

    Incorrect

    • Which of the following is a correct match for reflex and their root value?

      Your Answer: Biceps reflex: C7C8

      Correct Answer: Knee reflex: L3/L4

      Explanation:

      Reflexes are a routine part of clinical examination. Hyperreflexia (abnormally brisk reflexes) is the sign of upper motor neuron damage whereas diminished or absent jerks are most commonly due to lower motor neuron lesions. Reflexes may be Monosynaptic (deep tendon reflexes) or polysynaptic (superficial reflexes)

      Here are deep tendon reflexes with their nerve root
      Biceps = C5, C6
      Supinator (Brachioradialis) = C5, C6
      Triceps = C6, C7
      Knee reflex = L3,L4
      Ankle reflex = S1

      Polysynaptic superficial reflexes with their nerve root are listed below
      Planter response = S1-2
      Abdominal reflexes = T8-12
      Cremasteric reflex = L1-2

    • This question is part of the following fields:

      • Anatomy
      28.2
      Seconds
  • Question 17 - A 50-year-old female, known case of diabetes, has come in for a check-up...

    Incorrect

    • A 50-year-old female, known case of diabetes, has come in for a check-up at the diabetic foot clinic. The pulses of her feet are examined. The posterior tibial pulse and dorsalis pedis pulses are palpated.
      Which of the following artery continues as the dorsalis pedis artery?

      Your Answer: Peroneal artery

      Correct Answer: Anterior tibial artery

      Explanation:

      At the ankle joint, midway between the malleoli, the anterior tibial artery changes names, becoming the dorsalis pedis artery (dorsal artery of the foot).

      The dorsalis pedis artery is palpated against the underlying tarsals, immediately lateral to the tendon of extensor hallucis longus, from the midpoint between the malleoli to the proximal end of the first intermetatarsal space.

      The popliteal artery forms the anterior tibial artery.
      The tibioperoneal trunk is a branch of the popliteal artery.
      The peroneal artery (also known as the fibular artery) supplies the lateral compartment of the leg.
      The external iliac artery is formed from the common iliac artery at the level of the pelvis.

    • This question is part of the following fields:

      • Anatomy
      17.7
      Seconds
  • Question 18 - A 65-year-old man got operated on for carotid endarterectomy for his carotid artery...

    Incorrect

    • A 65-year-old man got operated on for carotid endarterectomy for his carotid artery disease. He is recovering well post-surgery. However, on follow-up in the ward, he has hoarseness of his voice.

      Which of the following explains the hoarseness?

      Your Answer: Damage to the hypoglossal nerve

      Correct Answer: Damage to the vagus

      Explanation:

      During carotid endarterectomy, injury to the vagus nerve or its branches can cause hoarseness. Injury to the vagus nerve can result in adductor vocal cord paralysis. It can also cause other symptoms like dysphagia or even vocal cord immobility.

      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. Because of the internal carotid artery relations, there is a risk of cranial nerve injury during the procedure involving one or more of the following nerves: CN IX, CN X (or its branch, the superior laryngeal nerve), CN XI, or CN XII.

      However, only damage to the vagus would account for speech difficulties.

    • This question is part of the following fields:

      • Anatomy
      12.8
      Seconds
  • Question 19 - Which of the following nerves is responsible for carrying taste sensation from the...

    Incorrect

    • Which of the following nerves is responsible for carrying taste sensation from the given part of the tongue?

      Your Answer: Posterior third of tongue - facial nerve

      Correct Answer: Anterior two thirds of tongue - facial nerve

      Explanation:

      Taste sensation from the anterior two-thirds of the tongue is carried by chorda tympani, a branch of the facial nerve.

      The general somatic sensation of the anterior two-third of the tongue is supplied by the lingual nerve, a branch of the mandibular nerve.

      Both general somatic sensation and taste from the posterior third of the tongue are carried by the glossopharyngeal nerve.

      All the muscles of the tongue except palatoglossus are supplied by the hypoglossal nerve whereas palatoglossus is supplied by the vagus nerve. (This is because palatoglossus is the only tongue muscle derived from the fourth branchial arch)

    • This question is part of the following fields:

      • Pathophysiology
      24.9
      Seconds
  • Question 20 - What can you see within the tunica media of a blood vessel on...

    Incorrect

    • What can you see within the tunica media of a blood vessel on examination?

      Your Answer: Vasa vasorum

      Correct Answer: Smooth muscle

      Explanation:

      The blood vessel well is divided into 3 parts, namely:

      The tunica intima, which is the deepest layer. It contains endothelial cells separated by gap junctions

      The tunica media, primarily consisting of the involuntary smooth muscle fibres, laid out in spiral layers with elastic fibres and connective tissue.

      The tunica adventitia, which is the most superficial layer. It consists of the vasa vasorum, fibroblast and collagen.

    • This question is part of the following fields:

      • Anatomy
      73.6
      Seconds
  • Question 21 - A 28-year male patient presents to the GP with a 2-day history of...

    Incorrect

    • A 28-year male patient presents to the GP with a 2-day history of abdominal pain and bloody diarrhoea. He reports that he was completely fine until one week ago when headache and general tiredness appeared. After further questioning, he revealed eating at a dodgy takeaway 3 days before the start of his symptoms.

      Which of the following diagnosis is most likely?

      Your Answer: Cholera

      Correct Answer: Campylobacter

      Explanation:

      Giardiasis is known to have a longer incubation time and doesn’t cause bloody diarrhoea.

      Cholera usually doesn’t cause bloody diarrhoea.

      Generally, most of the E.coli strains do not cause bloody diarrhoea.

      Diverticulitis can be a cause of bloody stool but the history here points out to an infectious cause.

      Campylobacter infection is the most probable cause as it is characterized by a prodrome, abdominal pain and bloody diarrhoea

    • This question is part of the following fields:

      • Physiology And Biochemistry
      48.9
      Seconds
  • Question 22 - A 31-year old Caucasian female came into the emergency department due to difficulty...

    Incorrect

    • A 31-year old Caucasian female came into the emergency department due to difficulty of breathing. History revealed exposure to room odorizes that are rich in alkyl nitrites. Upon physical examination, patient is tachypnoeic at 32 breaths per minute, desaturated at 88% while on a non-rebreather mask at 15 litres per minute oxygen. She was also noted to be cyanotic, however with clear breath sounds.

      Considering the history, what is the most probable cause of her difficulty of breathing?

      Your Answer: High arterial carboxyhaemoglobin concentration

      Correct Answer: Increased affinity of bound oxygen to haemoglobin

      Explanation:

      Amyl nitrate is part of the treatment of cyanide poisoning. The short acting nitrate causes oxidation of Fe2+ in haemoglobin to Fe3+ in methaemoglobin. Methaemoglobin combines with cyanide (cyanmethemoglobin), which reacts with sodium thiosulfate to convert nontoxic thiocyanate and methaemoglobin.

      Methaemoglobin is formed when the iron in haemoglobin is converted from the reduced state (Fe2+) to the oxidized state (Fe3+). The oxidized form of haemoglobin (Fe3+) does not bind oxygen as readily as Fe2+, but has high affinity for cyanide. It also results to high affinity of bound oxygen to haemoglobin, thus leading to tissue hypoxia. Arterial oxygen tension is normal despite observations of cyanosis and dyspnoea. Methemoglobinemia can be treated with methylene blue and vitamin C.

      Carboxyhaemoglobin can be due to carbon monoxide poisoning. In such cases, patients experience headache and dizziness, but do not develop cyanosis.

      2,3-diphosphoglycerate causes a shift in the oxygen dissociation curve to the right, decreasing haemoglobin’s affinity to oxygen to facilitate unloading of oxygen to the tissues.

    • This question is part of the following fields:

      • Pathophysiology
      30.3
      Seconds
  • Question 23 - A randomized study aimed at finding out the efficacy of a novel anticoagulant,...

    Incorrect

    • A randomized study aimed at finding out the efficacy of a novel anticoagulant, in preventing stroke in patients suffering from atrial fibrillation, relative to those already available in the market was performed. A 59 year old woman volunteered for it and was randomised to the treatment arm. A year later, following findings were reported:

      165 out of 1050 patients who were prescribed the already prevalent medicine had a stroke while the number of patients who had a single stroke after using the new drug was 132 out of 1044.

      In order to avoid one stroke case, what is the number of patients that need to be treated?

      Your Answer: 30

      Correct Answer: 32

      Explanation:

      Number needed to treat can be defined as the number of patients who need to be treated to prevent one additional bad outcome.

      It can be found as:

      NNT=1/Absolute Risk Reduction (rounded to the next integer since number of patients can be integer only).

      where ARR= (Risk factor associated with the new drug group) — (Risk factor associated with the currently available drug)

      So,

      ARR= (165/1050)-(132/1044)

      ARR= (0.157-0.126)

      ARR= 0.031

      NNT= 1/0.031

      NNT=32.3

    • This question is part of the following fields:

      • Statistical Methods
      18.7
      Seconds
  • Question 24 - Drug toxicity when using bupivacaine is most likely to occur when this local...

    Incorrect

    • Drug toxicity when using bupivacaine is most likely to occur when this local anaesthetic technique is performed.

      Your Answer: Sacral extradural (caudal) block

      Correct Answer: Intercostal nerve block

      Explanation:

      An intercostal nerve block is used for therapeutic and diagnostic purposes. Intercostal nerve blocks manage acute and chronic pain in the chest area. Common indications are chest wall surgery and shingles or postherpetic neuralgia.

      An intercostal nerve block is also an effective option for the management of pain associated with chest trauma and rib fractures. These blocks have been shown to improve oxygenation and respiratory mechanics, and offer pain relief that is comparable to that of epidural analgesia.

      This technique, however, is limited by the relatively large doses of local anaesthetic required, and relatively high intravascular uptake from the intercostal space, increasing risk of local anaesthetic toxicity.

    • This question is part of the following fields:

      • Pharmacology
      25.5
      Seconds
  • Question 25 - A patient on admission is given an infusion of 1000 mL of 10%...

    Incorrect

    • A patient on admission is given an infusion of 1000 mL of 10% glucose and 500 mL of 20% lipid over a 24 hour period.

      Which of these best approximates to the energy input over this time period?

      Your Answer: 1600 kcal

      Correct Answer: 1300 kcal

      Explanation:

      1% solution contains 1 g of substance per 100 mL.

      A solution of 10% glucose is 10 g/100mL. Therefore 1000 mL of this glucose solution will contain 100 g.

      1 g of glucose yields about 4 kcal of energy. One litre of 10% glucose will therefore release approximately 4x100g = 400 kcal of energy.

      A solution of 20% fat is 20 g/100mL. Therefore 1000 mL of this fat solution will have 200 g and 500 mL will contain 100 g.

      1 g of fat yields approximately 9 kcal. 500 mL of 20% fat therefore has the potential to yield 900 kcal of energy.

      The total energy input over this 24 hour period is approximately 400kcal + 900kcal = 1300 kcal.

    • This question is part of the following fields:

      • Physiology
      25.7
      Seconds
  • Question 26 - A 19-year-old woman presents to the emergency department. She complains of symptoms indicative...

    Incorrect

    • A 19-year-old woman presents to the emergency department. She complains of symptoms indicative of an acute exacerbation of known 'brittle' asthma. On history, she reveals her asthma is normally controlled using inhalers and she has never had an acute exacerbation requiring hospitalisation.

      On her admission into the ICU, further examination and diagnostic investigations are conducted. Her readings are:

      Physical state: Alert, anxious and non-cyanotic.
      Respiratory rate: 30 breaths/min
      Pulse: 120 beats/min
      Blood pressure: 150/90 mmHg
      SPO2: 95% on air
      Auscultation: Quiet breath sounds at both lung bases

      What is the next most important step of investigation?

      Your Answer: Electrocardiograph

      Correct Answer: Peak expiratory flow rate

      Explanation:

      Peak expiratory flow rate (PEFR) is the maximum speed of air flow generated during a single forced exhaled breath. It is most useful when expressed as a percentage of the best value obtained from the patient.

      Forced expiratory volume over 1 second (FEV1) is a lung parameter measured using spirometry. It is the amount of air forced out of the lung in one exhaled breath. It is a more accurate measure of lung obstructions as it doesn’t rely on effort like PEFR

      PEFR and FEV1 are usually similar, but become more different in asthmatic patients as airflow becomes increasingly obstructed.

      Acute severe asthma is most often diagnosed on history taking and examinations:

      Respiratory rate: >25 breaths/min
      Heart rate: >110 beats/min
      PEFR: 33 – 50% predicted (<200L/min)
      Patient state: Unable to complete a sentence in a single breath.

      A chest x-ray is not routinely required, and is only indicated in specific circumstances, which are:

      If a pneumomediastinum or pneumothorax is suspected
      Possible life threatening asthma
      Possible consolidation
      Unresponsive asthma
      If ventilation is required.

      An echocardiograph (ECG) is not necessary in this case

      Routine haematological and biochemical investigations are not urgent in this case as any abnormalities they detect will be secondary to the patient’s presentation.

      An arterial blood gas (ABG) will only be indicated if SPO2 was <92% or if patient presented with life threatening symptoms.

    • This question is part of the following fields:

      • Clinical Measurement
      16.4
      Seconds
  • Question 27 - An elective left colectomy is being performed on a 60-year old male for...

    Incorrect

    • An elective left colectomy is being performed on a 60-year old male for left-sided colon cancer. The upper and lower parts of the descending colon are supplied by the left colic artery.

      Which of the following arteries gives rise to the left colic artery?

      Your Answer: Inferior rectal artery

      Correct Answer: Inferior mesenteric artery

      Explanation:

      The inferior mesenteric artery originates 3-4 cm above the bifurcation of the abdominal aorta. The left colic artery branches off the inferior mesenteric artery, arising close to its origin from the abdominal aorta. Other branches of IMA include the three sigmoid arteries that supply the sigmoid colon.

      The left colic artery branches off from IMA to supply the distal 1/3 of the transverse colon and the descending colon. It moves upwards posterior to the left colic mesentery and then travels anteriorly to the psoas major muscle, left ureter, and left internal spermatic vessels, before dividing into ascending and descending branches.

    • This question is part of the following fields:

      • Anatomy
      6.9
      Seconds
  • Question 28 - The Fick principle can be used to determine the blood flow to any...

    Incorrect

    • The Fick principle can be used to determine the blood flow to any organ of the body.

      At rest, which one of these organs has the highest blood flow (ml/min/100g)?

      Your Answer: Brain

      Correct Answer: Thyroid gland

      Explanation:

      After the carotid body, the thyroid gland is the second most richly vascular organ in the body.

      The global blood flow to the thyroid gland can be measured using:
      1. Colour ultrasound sonography
      2. Quantitative perfusion maps using MRI of the thyroid gland using an arterial spin labelling (ASL) method.

      This table shows the blood flow to various organs of the body at rest:
      Organ Blood Flow(ml/minute/100g)
      Hepatoportal 58
      Kidney 420
      Brain 54
      Skin 13
      Skeletal muscle 2.7
      Heart 87
      Carotid body 2000
      Thyroid gland 560

    • This question is part of the following fields:

      • Physiology
      11.7
      Seconds
  • Question 29 - An 80-year-old man will be operated on for an arterial bypass procedure to...

    Incorrect

    • An 80-year-old man will be operated on for an arterial bypass procedure to treat claudication and foot ulceration. The anterior tibial artery will be the target for distal arterial anastomosis.

      Which structure is NOT closely related to the anterior tibial artery?

      Your Answer: Interosseous membrane

      Correct Answer: Tibialis posterior

      Explanation:

      The anterior tibial artery originates from the distal border of the popliteus. In the posterior compartment, it passes between the heads of the tibialis posterior and the oval aperture of the interosseous membrane to reach the anterior compartment.

      On entry into the anterior compartment, it runs medially along the deep peroneal nerve.
      The upper third of the artery courses between the tibialis anterior and extensor digitorum longus muscles, while the middle third runs between the tibialis anterior and extensor hallucis longus muscles.

      At the ankle, the anterior tibial artery is located approximately midway between the malleoli. It continues on the dorsum of the foot, lateral to extensor hallucis longus, as the dorsalis pedis artery.

    • This question is part of the following fields:

      • Anatomy
      26.6
      Seconds
  • Question 30 - A sevoflurane vaporiser with a 2 percent setting and a 200 kPa ambient...

    Incorrect

    • 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 kPa

      Altitude, pressure 50 kPa
      4% = 2% (dialled) × 100/50
      4% of 50 = 2 kPa

      High pressure at 200 kPa
      1% = 2% (dialled) × 100/200
      1% of 200 = 2 kPa

      Sevoflurane 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.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      40.8
      Seconds
  • Question 31 - A 48-year-old woman has presented to the emergency with abdominal pain and distension...

    Incorrect

    • A 48-year-old woman has presented to the emergency with abdominal pain and distension complaints. She is a known case of diabetes mellitus type 2 and has a BMI of 28 kg/m². On investigations, the liver function tests (LFTs) show raised alanine transaminase (ALT).

      Liver ultrasound is performed next to visualize the blood flow into and out of the liver.

      Which blood vessel supplies approximately one-third of the blood supply to the liver?

      Your Answer: Inferior mesenteric artery

      Correct Answer: Hepatic artery proper

      Explanation:

      The liver receives blood supply from two sources.
      1. Hepatic artery proper
      It arises from the celiac trunk via the common hepatic artery and brings oxygenated blood to the liver.
      It contributes to approximately 30% of the blood supply of the liver.
      2. Hepatic portal vein – supplies the liver with partially deoxygenated blood, carrying nutrients absorbed from the small intestine. It gets tributaries from the inferior mesenteric vein, splenic vein, and superior mesenteric vein

      The inferior mesenteric artery supplies the hindgut.
      The superior mesenteric artery supplies the pancreas and intestine up to the proximal two-thirds of the transverse colon.
      The inferior phrenic artery supplies the inferior surface of the diaphragm and oesophagus.

    • This question is part of the following fields:

      • Anatomy
      28.2
      Seconds
  • Question 32 - A 74-year-old with a VVI pacemaker is undergoing a hip replacement.

    Which of the...

    Incorrect

    • A 74-year-old with a VVI pacemaker is undergoing a hip replacement.

      Which of the following is most likely to predispose him to an electrical hazard?

      Your Answer: A CVP line

      Correct Answer: Use of cutting unipolar diathermy

      Explanation:

      A single chamber pacemaker was implanted in the patient. In VVI mode, a pacemaker paces and senses the ventricle while being inhibited by a perceived ventricular event. The most likely electrical hazard from diathermy is electromagnetic interference (EMI).

      EMI has the potential to cause the following: Inhibition of pacing
      Asynchronous pacing
      Reset to backup mode
      Myocardial burns, and
      Trigger VF.

      Diathermy entails the implementation of high-frequency electrical currents to produce heat and either make incisions or induce coagulation. Monopolar cautery involves disposable cautery pencils and electrosurgical diathermy units. In typical monopolar cautery, an electrical plate is placed on the patient’s skin and acts as an electrode, while the current passes between the instrument and the plate. Monopolar diathermy can therefore interfere with implanted metal devices and pacemaker function.

      Bipolar diathermy, where the current passes between the forceps tips and not through the patient and is less likely to generate EMI.

      Whilst the presence of a CVP line may in theory predispose the patient to microshock, the use of prerequisite CF electrical equipment makes this very unlikely. The presence of a CVP line and pacemaker does not therefore unduly increase the risk of an electrical hazard.

      Isolating transformers are used to protect secondary circuits and individuals from electrical shocks. There is no step-up or step-down voltage (i.e. there is a ratio of 1 to 1 between the primary and secondary windings).

      A ground (or earth) wire is normally connected to the metal case of an operating table to protect patients from accidental electrocution. In the event that a fault allows a live wire to make contact with the metal table (broken cable, loose connection etc.) it becomes live. The earth will provide an immediate path for current to safely flow through and so the table remains safe to touch. Being a low resistance path, the earth lets a large current flow through it when the fault occurs ensuring that the fuse or RCD will quickly blow. Without an operating table earth, the patient is not at more risk of an electrical hazard because of the pacemaker.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      79.6
      Seconds
  • Question 33 - Which statement regarding the cardiac action potential is correct? ...

    Incorrect

    • Which statement regarding the cardiac action potential is correct?

      Your Answer: The absolute refractory period is in Phase 3

      Correct Answer:

      Explanation:

      Cardiac conduction

      Phase 0 – Rapid depolarization. Opening of fast sodium channels with large influx of sodium

      Phase 1 – Rapid partial depolarization. Opening of potassium channels and efflux of potassium ions. Sodium channels close and influx of sodium ions stop

      Phase 2 – Plateau phase with large influx of calcium ions. Offsets action of potassium channels. The absolute refractory period

      Phase 3 – Repolarization due to potassium efflux after calcium channels close. Relative refractory period

      Phase 4 – Repolarization continues as sodium/potassium pump restores the ionic gradient by pumping out 3 sodium ions in exchange for 2 potassium ions coming into the cell. Relative refractory period

    • This question is part of the following fields:

      • Physiology And Biochemistry
      27.4
      Seconds
  • Question 34 - The following are results of some pulmonary function tests:

    Measurement - Predicted result -...

    Correct

    • The following are results of some pulmonary function tests:

      Measurement - Predicted result - Test result
      Forced vital capacity (FVC) (btps) - 3.21 - 1.94
      Forced expiratory volume in 1 second (FEV1) (btps) - 2.77 - 1.82
      FEV1/FVC ratio % (btps) - 81.9 - 93.5
      Peak expiratory flow (PEF) (L/second) - 6.55 - 3.62
      Maximum voluntary ventilation (MVV) (L/minute) - 103 - 87.1

      Which statement applies to the results?

      Your Answer: The patient has a moderate restrictive pulmonary defect

      Explanation:

      Severity of a reduction in restrictive defect (%FVC) or obstructive defect (%FEV1/FVC) predicted are classified as follows:

      Mild 70-80%
      Moderate 60-69%
      Moderately severe 50-59%
      Severe 35-49%
      Very severe <35% This patient has a %FVC predicted of 60.4% and this corresponds to a moderate restrictive deficit. %FEV1/FVC ratio is 93.5%. FEV1/FVC ratio 80% < predicted and VC < 80% = mixed picture. FEV1/FVC ratio 80% < predicted and VC > 80% = obstructive picture.

      FEV1/FVC ratio 80% > predicted and VC > 80% = normal picture.

      FEV1/FVC ratio 80% > predicted and VC < 80% predicted= restrictive picture. The integrity of the alveolar-capillary barrier is measured by carbon monoxide transfer factor (TLCO) and carbon monoxide transfer coefficient (KCO). These values are seen to be reduced in emphysema, interstitial lung diseases and in pulmonary vascular pathology. However, the KCO (as % predicted) is high in extrapulmonary restriction (pleural, chest wall and respiratory neuromuscular disease), and in loss of lung units provided the structure of the lung remaining is normal. The KCO distinguishes extrapulmonary (high KCO) causes of ‘restriction’ from intrapulmonary causes (low KCO).

    • This question is part of the following fields:

      • Clinical Measurement
      9.4
      Seconds
  • Question 35 - During 2015 it was reported in the New England Journal of Medicine that...

    Correct

    • During 2015 it was reported in the New England Journal of Medicine that the usage of empagliflozin(a sodium-glucose-co-transporter 2 inhibitor) caused a decrease in the cardiovascular deaths, non fatal heart attacks and strokes in patients suffering from type 2 diabetes. The results were published per 1000 patient years. With the above mentioned drug, the event rate turned out to be 37.3/1000 patient years whereas the placebo had an event rate of 43.9/1000 patient years.

      How many further patients need to be treated with empagliflozin to avoid any further incidence of cardiovascular death or non fatal myocardial infraction and non fatal stroke?

      Your Answer: 150

      Explanation:

      Number needed to treat can be defined as the number of patients who need to be treated to prevent one additional bad outcome.

      It can be found as:

      NNT=1/Absolute Risk Reduction (rounded to the next integer since number of patients can be integer only).

      where ARR= (Risk factor associated with the new drug group) — (Risk factor associated with the currently available drug)

      So,

      ARR= (43.9-37.3)

      ARR= 6.6

      NNT= 1000/6.6

      NNT=151.5

    • This question is part of the following fields:

      • Statistical Methods
      27.6
      Seconds
  • Question 36 - Regarding adrenocorticotropic hormone (ACTH) one of these is true. ...

    Incorrect

    • Regarding adrenocorticotropic hormone (ACTH) one of these is true.

      Your Answer: Production is governed by the pituitary

      Correct Answer: Is increased in the maternal plasma in pregnancy

      Explanation:

      ACTH production is stimulated through the secretion of corticotropin-releasing hormone (CRH) from the hypothalamic nuclei.

      ACTH secretion has a circadian rhythm. A high level of cortisol in the body stops its production. ACTH is secreted maximally in the morning and concentrations are lowest at midnight.

      ACTH can be expressed in the placenta, the pituitary and other tissues.

      Conditions where ACTH concentrations rise include: stress, disease and pregnancy.

      Glucocorticoids (not mineralocorticoids – aldosterone) switch off ACTH production through a negative feedback loop .

    • This question is part of the following fields:

      • Pathophysiology
      19.4
      Seconds
  • Question 37 - Arterial pressure waveforms give an indication of the operation of the heart and...

    Incorrect

    • Arterial pressure waveforms give an indication of the operation of the heart and the patient's clinical state.

      Which of the following listed characteristics of arterial waveforms is most indicative of myocardial contractility?

      Your Answer: Calculation of mean arterial pressure.

      Correct Answer: Slope of the upstroke of the curve.

      Explanation:

      Arterial pressure waveforms is an invasive form of monitoring cardiac parameters. It provides a lot of information on the performance of the heart from different sections, including:

      Cardiac measurements:

      Heart rate
      Systolic pressure
      Diastolic pressure
      Mean arterial pressure
      Pulse pressure
      Change in pulse amplitude corresponding to respiratory changes
      Slope of anacrotic limb associated with aortic stenosis

      From the shape of the arterial waveform displayed:

      Slope of anacrotic limb represents aortic valve and LVOT flow
      Indications of aortic stenosis (AS): Slurred wave, collapsing wave
      Rapid systolic decline in LVOTO
      Bisferiens wave in HOCM
      Low dicrotic notch in states with poor peripheral resistance
      Position and quality of dicrotic notch as a reflection of the damping coefficient

      For this question, the upstroke slope of the pressure wave is indicative of myocardial contractility and is mathematically represented as:

      dP/dt, which represents a change of pressure with regards to time.

    • This question is part of the following fields:

      • Clinical Measurement
      25.6
      Seconds
  • Question 38 - Standard error of the mean can be defined as: ...

    Incorrect

    • Standard error of the mean can be defined as:

      Your Answer: Number of patients / square root (mean)

      Correct Answer: Standard deviation / square root (number of patients)

      Explanation:

      The standard error of the mean (SEM) is a measure of the spread expected for the mean of the observations – i.e. how ‘accurate’ the calculated sample mean is from the true population mean. The relationship between the standard error of the mean and the standard deviation is such that, for a given sample size, the standard error of the mean equals the standard deviation divided by the square root of the sample size.

      SEM = SD / square root (n)

      where SD = standard deviation and n = sample size

    • This question is part of the following fields:

      • Statistical Methods
      7.9
      Seconds
  • Question 39 - What statement about endotoxins is true? ...

    Correct

    • 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.

    • This question is part of the following fields:

      • Pathophysiology
      7.5
      Seconds
  • Question 40 - Which plasma protein will bind the thyroid hormone triiodothyronine (T3) more readily? ...

    Incorrect

    • Which plasma protein will bind the thyroid hormone triiodothyronine (T3) more readily?

      Your Answer: Serum lipoproteins

      Correct Answer: Thyroxine binding globulin

      Explanation:

      Secreted T4 and T3 circulate in the bloodstream almost entirely bound to proteins. Normally only about 0.03% of total plasma T4 and 0.3% of total plasma T3 exist in the free state. Free T3 is biologically active and mediates the effects of thyroid hormone on peripheral tissues in addition to exerting negative feedback on the pituitary and hypothalamus. The major binding protein is thyroxine-binding globulin (TBG), which is synthesized in the liver and binds one molecule of T4 or T3. About 70% of circulating T4 and T3 is bound to TBGl 10% to 15% is bound to another specific thyroid-binding protein called transthyretin (TTR). Albumin binds 15% to 20%, and 3% to lipoproteins. Ordinarily only alterations in TBG concentration significantly affect total plasma T4 and T3 levels.

      Two important biological functions have been ascribed to TBG. First, it maintains a large circulating reservoir of T4 that buffers any acute changes in thyroid gland function. Second, binding of plasma T4 and T3 to proteins prevents loss of these relatively small hormone molecules in urine and thereby helps conserve iodide. TTR transports T4 in CSF and provides thyroid hormones to the CNS.

    • This question is part of the following fields:

      • Physiology
      7.2
      Seconds
  • Question 41 - Which of the following statements is true regarding drug dose and response? ...

    Correct

    • Which of the following statements is true regarding drug dose and response?

      Your Answer: Intrinsic activity determines maximal response

      Explanation:

      There are two types of drug dose-response relationships, namely, the graded dose-response and the quantal dose-response relationships.

      Drug response curves are plotted as percentage response again LOG drug concentration. This graph is sigmoid in shape.

      Agonists are drugs with high affinity and high intrinsic activity. Meanwhile, the antagonist is a drug with high affinity but no intrinsic activity. Intrinsic activity determines the maximal response. The maximal response can be achieved even by activation of a small proportion of receptor sites.

    • This question is part of the following fields:

      • Pharmacology
      14.2
      Seconds
  • Question 42 - Activation of which of the following GABA A receptor subunit leads to anxiolytic...

    Incorrect

    • Activation of which of the following GABA A receptor subunit leads to anxiolytic effects of Benzodiazepines?

      Your Answer: Epsilon

      Correct Answer: Alpha

      Explanation:

    • This question is part of the following fields:

      • Pharmacology
      14.8
      Seconds
  • Question 43 - The fluids with the highest osmolarity is? ...

    Incorrect

    • The fluids with the highest osmolarity is?

      Your Answer: 0.9% N. Saline

      Correct Answer: 0.45% N. Saline with 5% glucose

      Explanation:

      The concentration of solute particles per litre (mosm/L) = the osmolarity of a solution. Changes in water content, ambient temperature, and pressure affects osmolarity. The osmolarity of any solution can be calculated by adding the concentration of key solutes in it.

      Individual manufacturers of crystalloids and colloids may have different absolute values but they are similar to these.

      0.45% N. Saline with 5% glucose:
      Tonicity – hypertonic
      Osmolarity – 405 mosm/L
      Kilocalories (kCal) – 107

      0.9% N. Saline:
      Tonicity – isotonic
      Osmolarity – 308 mosm/L
      Kilocalories (kCal) – 0

      5% Dextrose:
      Tonicity – isotonic
      Osmolarity – 253 mosm/L
      Kilocalories (kCal) – 170

      Gelofusine (154 mmol/L Na, 120 mmol/L Cl):
      Tonicity – isotonic
      Osmolarity – 274 mosm/L
      Kilocalories (kCal) – 0

      Hartmann’s solution:
      Tonicity – isotonic
      Osmolarity – 273 mosm/L
      Kilocalories (kCal) – 9

    • This question is part of the following fields:

      • Physiology
      61.4
      Seconds
  • Question 44 - A 68-year-old man has suffered a myocardial infarction. He has a heart rate...

    Incorrect

    • A 68-year-old man has suffered a myocardial infarction. He has a heart rate of 40 beats per minute currently.
      Your senior attending explains that the slow heart rate is due to the damage to the conduction pathways between the sinoatrial and atrioventricular nodes. His ventricles are being paced by the AV node alone.

      What artery supplies the AV node in the majority of patients?

      Your Answer: Left circumflex artery

      Correct Answer: Right coronary artery

      Explanation:

      The AV node has an intrinsic firing rate of 40-60 beats per minute which is clinically significant in cases of damage to the conducting pathways as patients continue to have a ventricular rate of 40-60. Patients who have an AV node supplied by the right coronary are said to be right dominant. The remaining 10% are left dominant and supplied by the left circumflex.

      The right coronary artery supplies the right atrium, right ventricle, interatrial septum, and the inferior posterior third of the interventricular septum. It also supplies the atrioventricular node + sinoatrial node in most patients. The posterior descending artery supplies the posterior third of the interventricular septum.

      The heart receives blood supply from coronary arteries. The right and left coronary arteries branch off the aorta and supply oxygenated blood to all heart muscle parts.

      The left main coronary artery branches into:
      1. Circumflex artery – supplies the left atrium, side, and back of the left ventricle. The left marginal artery arises from the left circumflex artery. It travels along the obtuse margin of the heart.
      The left marginal artery, a branch of the circumflex artery, supplies the left ventricle.
      2. Left Anterior Descending (LAD) artery – supplies the front and bottom of the left ventricle and front of the interventricular septum

      The right coronary artery branches into:
      1. Right marginal artery
      2. Posterior descending artery

    • This question is part of the following fields:

      • Anatomy
      10.2
      Seconds
  • Question 45 - A bolus of alfentanil has a faster onset of action than an equal...

    Incorrect

    • A bolus of alfentanil has a faster onset of action than an equal dose of fentanyl.

      Which of the following statements most accurately describes the difference?

      Your Answer: The plasma clearance of fentanyl is more than that of alfentanil

      Correct Answer: The pKa of alfentanil is less than that of fentanyl

      Explanation:

      Unionised molecules are more likely than ionised molecules to cross membranes (such as the blood-brain barrier).

      Because alfentanil and fentanyl are weak bases, the Henderson-Hasselbalch equation says that the ratio of ionised to unionised molecules is determined by the parent compound’s pKa in relation to physiological pH.

      Alfentanil has a pKa of 6.5, while fentanyl has a pKa of 8.4.
      At a pH of 7.4, 89 percent of alfentanil is unionised, whereas 9% of fentanyl is.

      As a result, alfentanil has a faster onset than fentanyl.

      Fentanyl is 83% plasma protein bound
      Alfentanil is 90% plasma protein bound.

      Alfentanil’s pharmacokinetics are affected by its higher plasma protein binding. Because alfentanil has a low hepatic extraction ratio (0.4), clearance is determined by the degree of protein binding rather than the time it takes to take effect.

    • This question is part of the following fields:

      • Pharmacology
      38.1
      Seconds
  • Question 46 - Which of the following options will best reflect the adequacy of preoxygenation prior...

    Incorrect

    • Which of the following options will best reflect the adequacy of preoxygenation prior to rapid sequence induction of a patient?

      Your Answer: Expired partial pressure of carbon dioxide (EtCO2)

      Correct Answer: Expired fraction of oxygen (FEO2)

      Explanation:

      The most important determinant of preoxygenation adequacy is expired fraction of oxygen. Denitrogenating of the functional residual capacity is the purpose of preoxygenation. This is dependent on three vital factors: (1) respiratory rate; (2) inspired volume, and; (3) inspired oxygen concentration (FiO2).

      Arterial oxygen saturation does not efficiently determine adequacy of preoxygenation because of its inability to measure tissue reserves. Arterial partial pressure of oxygen is also unsuitable for determining preoxygenation adequacy. Moreover, the absence of central cyanosis is a very crude sign of low tissue oxygenation.

    • This question is part of the following fields:

      • Pathophysiology
      24.1
      Seconds
  • Question 47 - A 60-year-old man had previously been diagnosed with Type 2 diabetes. He had...

    Incorrect

    • A 60-year-old man had previously been diagnosed with Type 2 diabetes. He had recently started gliclazide, a sulphonyl urea, as his diabetes was not controlled by metformin alone.

      Now, he presents to his physician with complaints of anxiety, sweating, and palpitations since the morning. On physical examination, he is pale and clammy and has mydriasis and increased bowel sounds.

      Which biological site primarily synthesizes the hormone responsible for this patient's condition?

      Your Answer: Post-ganglionic neurones of the sympathetic nervous system

      Correct Answer: Chromaffin cells of the adrenal medulla

      Explanation:

      This patient has been shifted to a sulfonylurea drug whose most common side effect is hypoglycaemia. Similar symptoms can arise in a patient on insulin too. The signs and symptoms are consistent with a hypoglycaemic attack and include tachycardia, altered consciousness, and behaviour. This needs to be treated as an emergency with rapid correction of the blood glucose level using glucose or IV 20% dextrose.

      In a hypoglycaemic attack, the body undergoes stress and releases hormones to increase blood glucose levels. These include:
      Glucagon
      Cortisol
      Adrenaline

      Adrenaline or epinephrine is the hormone responsible for this patient’s condition and is primarily produced in the medulla of the adrenal gland. It functions primarily to raise cardiac output and raise blood glucose levels in the blood.

      Alpha-cells of the islets of Langerhans produce the hormone glucagon, which has opposing effects to insulin.

      Follicular cells of the thyroid gland produce and secrete thyroid hormones. Thyroid hormones can cause similar symptoms, but it is unlikely with the patient’s medical history.

      Post-ganglionic neurons of the sympathetic nervous system use norepinephrine as a neurotransmitter. Adrenaline can be made in these cells, but it is not their primary production site.

      Zona fasciculata of the adrenal cortex is the main site for the production of cortisol.

    • This question is part of the following fields:

      • Anatomy
      10.9
      Seconds
  • Question 48 - The lung volume that is commonly measured indirectly is? ...

    Incorrect

    • The lung volume that is commonly measured indirectly is?

      Your Answer: Vital capacity

      Correct Answer: Functional residual capacity

      Explanation:

      The functional residual capacity (FRC) is the volume in the lungs at the end of passive expiration. It is determined by opposing forces of the expanding chest wall and the elastic recoil of the lung. A normal FRC = 1.7 to 3.5 L. It a marker for lung function, and, during this time, the alveolar pressure is equal to the atmospheric pressure.

      FRC cannot be measured by spirometry because it contains the residual volume.

      Tidal volume, inspiratory reserve volume, forced expiratory volume in 1 second, and vital capacity can be measured directly.

    • This question is part of the following fields:

      • Pathophysiology
      18.6
      Seconds
  • Question 49 - Radical prostatectomy is being performed on a 60-year-old man for carcinoma of the...

    Incorrect

    • Radical prostatectomy is being performed on a 60-year-old man for carcinoma of the prostate gland.

      What is the direct blood supply of the prostate?

      Your Answer: External iliac artery

      Correct Answer: Inferior vesical artery

      Explanation:

      The prostate gland is primarily supplied by the inferior vesical artery, which branches off from the anterior division of the internal iliac artery. The inferior vesical artery supplies the base of the bladder, the distal ureters, and the prostate. The branches to the prostate communicate with the corresponding vessels of the opposite side.

      The inferior vesical artery branches into two main arteries:
      1. Urethral artery – supplies the transition zone and is the main arterial supply for the adenomas in BPH
      2. Capsular artery – supplies the glandular tissue

      The venous drainage of the prostate is from the prostatic venous plexus, which drains into the paravertebral veins.

    • This question is part of the following fields:

      • Anatomy
      10.8
      Seconds
  • Question 50 - A pharmaceutical company has developed a new drug considered a breakthrough in treating...

    Incorrect

    • A pharmaceutical company has developed a new drug considered a breakthrough in treating ovarian cancer.

      The efficacy of this drug can be assessed by which phase of a clinical trial?

      Your Answer: Phase III

      Correct Answer: Phase IIa

      Explanation:

      Phase IIa studies are usually pilot studies designed to demonstrate clinical efficacy or biological activity (‘proof of concept’ studies) whereas phase IIb studies determine the optimal dose at which the drug shows biological activity with minimal side-effects (definite dose-finding studies).

      Phase III and Phase IV studies are performed on larger set of participants (usually hundreds to thousands) when safety and efficacy have been established.

    • This question is part of the following fields:

      • Statistical Methods
      21.1
      Seconds
  • Question 51 - With regards to this state of matter which has a volume but no...

    Incorrect

    • 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: Plasma

      Correct 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.

    • This question is part of the following fields:

      • Basic Physics
      12.5
      Seconds
  • Question 52 - Which of the following best explains the association between smoking and lower oxygen...

    Incorrect

    • Which of the following best explains the association between smoking and lower oxygen delivery to tissues?

      Your Answer: Narrowing of small airways

      Correct Answer: Left shift of the oxygen dissociation curve

      Explanation:

      Smoking is a major risk factor associated with perioperative respiratory and cardiovascular complications. Evidence also suggests that cigarette smoking causes imbalance in the prostaglandins and promotes vasoconstriction and excessive platelet aggregation. Two of the constituents of cigarette smoke, nicotine and carbon monoxide, have adverse cardiovascular effects. Carbon monoxide increases the incidence of arrhythmias and has a negative ionotropic effect both in animals and humans.

      Smoking causes an increase in carboxyhaemoglobin levels, resulting in a leftward shift in which appears to represent a risk factor for some of these cardiovascular complications.

      There are two mechanisms responsible for the leftward shift of oxyhaemoglobin dissociation curve when carbon monoxide is present in the blood. Carbon monoxide has a direct effect on oxyhaemoglobin, causing a leftward shift of the oxygen dissociation curve, and carbon monoxide also reduces the formation of 2,3-DPG by inhibiting glycolysis in the erythrocyte. Nicotine, on the other hand, has a stimulatory effect on the autonomic nervous system. The effects of nicotine on the cardiovascular system last less than 30 min.

    • This question is part of the following fields:

      • Physiology
      31.2
      Seconds
  • Question 53 - The physiological properties of a fast glycolytic (fast twitch) muscle fibre are characterised...

    Incorrect

    • The physiological properties of a fast glycolytic (fast twitch) muscle fibre are characterised by which of the following?

      Your Answer: Generally fatigue resistant

      Correct Answer: Synthesis of ATP is brought about by anaerobic respiration

      Explanation:

      Muscle fibre myosin ATPase histochemistry is used to divide the biochemical classification into two groups: type 1 and type II.

      Type I (slow twitch) muscle fibres rely on aerobic glycolytic and aerobic oxidative metabolism to function. They have a lot of mitochondria, a good blood supply, a lot of myoglobin, and they don’t get tired easily.

      Because they contain more motor units, Type II (fast twitch) muscle fibres are thicker. They are more easily fatigued, but produce powerful bursts. The capillary networks and mitochondria are less dense in these white muscle fibres than in type I fibres. They have a low myoglobin content as well.

      Muscle fibres of type II (fast twitch) are divided into three types:

      Type IIa – aerobic/oxidative metabolism is used.
      Type IIb – anaerobic/glycolytic metabolism is used by these fibres.

      When compared to skeletal muscle, cardiac and smooth muscle twitch at a slower rate.

    • This question is part of the following fields:

      • Pharmacology
      13.2
      Seconds
  • Question 54 - What is NOT a feature of Propofol infusion syndrome? ...

    Correct

    • What is NOT a feature of Propofol infusion syndrome?

      Your Answer: Hypotriglyceridaemia

      Explanation:

      Propofol infusion syndrome is a rare but extremely dangerous complication of propofol administration

      Common organ systems affected by PRIS include the following:
      1. cardiovascular
      widening of QRS complex, Brugada syndrome-like patterns (particularly type 1), ventricular tachyarrhythmias, cardiogenic shock, and asystole

      2. hepatic
      Liver enzymes elevation, hepatomegaly, and steatosis

      3. skeletal muscular
      myopathy and overt rhabdomyolysis

      4. renal
      Hyperkalaemia, acute kidney injury

      5. metabolic
      High anion gap metabolic acidosis (due to elevation in lactic acid)

    • This question is part of the following fields:

      • Anatomy
      14.2
      Seconds
  • Question 55 - A 30-year-old man has been stabbed in an area of the groin that...

    Incorrect

    • 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: Adductor longus

      Correct 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 pectineus

      The 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
      78.6
      Seconds
  • Question 56 - A 70-year-old man presents with bilateral buttock claudication that spreads down the thigh...

    Correct

    • 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: 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

    • This question is part of the following fields:

      • Anatomy
      41.4
      Seconds
  • Question 57 - Given the following values:

    Expired tidal volume = 800 ml
    Plateau pressure = 50 cmH2O
    PEEP...

    Incorrect

    • Given the following values:

      Expired tidal volume = 800 ml
      Plateau pressure = 50 cmH2O
      PEEP = 10 cmH2O

      Compute for the static pulmonary compliance.

      Your Answer: 2 ml/cmH2O

      Correct Answer: 20 ml/cmH2O

      Explanation:

      Compliance of the respiratory system describes the expandability of the lungs and chest wall. There are two types of compliance: dynamic and static.

      Dynamic compliance describes the compliance measured during breathing, which involves a combination of lung compliance and airway resistance. Defined as the change in lung volume per unit change in pressure in the presence of flow.

      Static compliance describes pulmonary compliance when there is no airflow, like an inspiratory pause. Defined as the change in lung volume per unit change in pressure in the absence of flow.

      For example, if a person was to fill the lung with pressure and then not move it, the pressure would eventually decrease; this is the static compliance measurement. Dynamic compliance is measured by dividing the tidal volume, the average volume of air in one breath cycle, by the difference between the pressure of the lungs at full inspiration and full expiration. Static compliance is always a higher value than dynamic

      Static compliance can be computed using the formula:

      Cstat = Tidal volume/Plateau pressure – PEEP

      Substituting the values given,

      Cstat = 800/50-10
      Cstat = 20 ml/cmH2O

    • This question is part of the following fields:

      • Physiology
      41.5
      Seconds
  • Question 58 - The ED95 of muscle relaxants is the dose required to reduce twitch height...

    Correct

    • The ED95 of muscle relaxants is the dose required to reduce twitch height by 95% in half of the target population. The dose of non-depolarizing muscle relaxants used for intubation is 2-3 times the ED95.

      For procedures that need a short duration of muscle relaxation and abrupt recovery, the short-acting drug Mivacurium is given at less than 2 times the ED95. What is the explanation for Mivacurium being an exception to this rule?

      Your Answer: Dose related histamine release occurs which frequently leads to tachycardia and hypotension

      Explanation:

      Mivacurium, when administered at doses greater than 0.2 mg/kg,increases the risk for hypotension, tachycardia, and erythema. This is due to the ability of mivacurium to release histamine with increasing dose. Contrary to this fact, anaphylaxis is rare for mivacurium because of the short duration of histamine release.

      The effective dose 50 (ED50) of mivacurium is between 0.08-0.15 mg/kg. It is administered slowly to prevent and decrease the risk of developing adverse effects.

      Mivacurium has a high potency thus a longer duration of action, however this is not the answer that we are looking for.

      Although drug metabolism takes longer for mivacurium than succinylcholine, it has no effect on the dose required for intubation.

    • This question is part of the following fields:

      • Pharmacology
      18.4
      Seconds
  • Question 59 - Which statement is correct about the Mapleson anaesthetic breathing circuits? ...

    Incorrect

    • Which statement is correct about the Mapleson anaesthetic breathing circuits?

      Your Answer: Mapleson B is efficient for both spontaneous and positive pressure ventilation

      Correct Answer: Mapleson A is most efficient for spontaneous ventilation

      Explanation:

      Mapleson breathing system (or circuit) analysed five different arrangements of components of the breathing system:
      Mapleson A – It is the most efficient for spontaneous respiration. The flow of fresh gas required is 70-85 ml/kg/min, i.e., approximately 5-6 lit./min fresh gas flow for an average adult.
      Mapleson B and C – inefficient for both SV and PPV; requires gas flow of two to three times minute volume (100 ml/kg/min). Not commonly used but category C may be used for emergency resuscitation.
      Mapleson D – efficient for PPV at gas flow equivalent to patient’s minute volume; the Bain’s circuit is a coaxial version of the Mapleson D
      Mapleson E and F – for paediatric use; requires gas flow at two to three times the patient’s minute volume. The Mapleson F consists of an open-ended reservoir bag (Jackson-Rees modification).

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      32.4
      Seconds
  • Question 60 - A study of blood pressure measurements is being performed in patients with chronic...

    Correct

    • A study of blood pressure measurements is being performed in patients with chronic kidney disease.

      Considering that the results are normally distributed, what percentage of values lie within two standard deviations of the mean blood pressure reading?

      Your Answer: 95.40%

      Explanation:

      Normal distribution, also called Gaussian distribution, the most common distribution function for independent, randomly generated variables, and describes the spread for many biological and clinical measurements.

      Properties of the Normal distribution

      symmetrical i.e. Mean = mode = median

      68.3% of values lie within 1 SD of the mean

      95.4% of values lie within 2 SD of the mean

      99.7% of values lie within 3 SD of the mean

      The empirical rule, or the 68-95-99.7 rule, tells you where most of the values lie in a normal distribution: Around 68% of values are within 1 standard deviation of the mean.

      Around 95% of values are within 2 standard deviations of the mean. Around 99.7% of values are within 3 standard deviations of the mean.
      the standard deviation (SD) is a measure of how much dispersion exists from the mean.

      SD = square root (variance)

      The empirical rule, or the 68-95-99.7 rule states where most of the values lie in a normal distribution. Around 68% of values fall within 1 S.D of the mean, about 95% within 2 S.D of the mean, and about 99.7% of values within 3 S.D of the mean. Therefore, 95.4% is the most reasonable answer if results are normally distributed.

    • This question is part of the following fields:

      • Statistical Methods
      29.3
      Seconds
  • Question 61 - A 77-year-old man, is scheduled for an angiogram to investigate gastro-intestinal bleeding. The...

    Incorrect

    • 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: L4

      Correct 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

    • This question is part of the following fields:

      • Anatomy
      25.1
      Seconds
  • Question 62 - Which among the following is not true regarding disease rates? ...

    Incorrect

    • Which among the following is not true regarding disease rates?

      Your Answer: The terms risk ratio and relative risk are synonymous

      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.

    • This question is part of the following fields:

      • Statistical Methods
      51.8
      Seconds
  • Question 63 - A 70-year-old man presents with central crushing chest pain that radiates to the...

    Incorrect

    • 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 marginal artery

      Correct 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 septum

      The 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 artery

      The 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.

    • This question is part of the following fields:

      • Anatomy
      4.8
      Seconds
  • Question 64 - A 60-year old male has anaemia and is being investigated. The most common...

    Incorrect

    • 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: δ2β2

      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.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      40.2
      Seconds
  • Question 65 - An older woman has been brought into the emergency department with symptoms of...

    Incorrect

    • An older woman has been brought into the emergency department with symptoms of a stroke. A CT angiogram is performed for diagnosis, which displays narrowing in the artery that supplies the right common carotid. Which of the following artery is the cause of stroke in this patient?

      Your Answer: Coeliac trunk

      Correct Answer: Brachiocephalic artery

      Explanation:

      The arch of aorta gives rise to three main branches:
      1. Brachiocephalic artery
      2. Left common carotid artery
      3. Left subclavian artery

      The brachiocephalic artery then gives rise to the right subclavian artery and the right common carotid artery.

      The right common carotid artery arises from the brachiocephalic trunk posterior to the sternoclavicular joint.

      The coeliac trunk is a branch of the abdominal aorta.
      The ascending aorta supplies the coronary arteries.

    • This question is part of the following fields:

      • Anatomy
      42.9
      Seconds
  • Question 66 - Pressure volume loop represents the compliance of left ventricle.

    Considering there...

    Incorrect

    • Pressure volume loop represents the compliance of left ventricle.

      Considering there is no change in preload and myocardial contractility, which physiological change may result an increase in left ventricular afterload?

      Your Answer: Increased stroke volume

      Correct Answer: Increased end-systolic volume

      Explanation:

      If there is no change in preload and myocardial contractility, there will be decrease in end-diastolic volume and stroke volume. So there must be increase in end-systolic volume.

    • This question is part of the following fields:

      • Physiology
      29.9
      Seconds
  • Question 67 - Which of the following drugs would cause the most clinical concern if accidentally...

    Incorrect

    • Which of the following drugs would cause the most clinical concern if accidentally administered intravenously to a 4-year-old boy?

      Your Answer: 4 mg ondansetron

      Correct Answer: 20 mg codeine

      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.

      The following are the appropriate doses of the drugs listed above:

      Gentamicin (once daily) – 5-7 mg/kg = 90-126 mg and subsequent dose modified according to plasma levels
      Ondansetron – 0.1 mg/kg, but a maximum of 4 mg as a single dose = 1.8 mg
      Codeine should be administered orally at a dose of 1 mg/kg rather than intravenously, as the latter can cause ‘dangerous’ hypotension due to histamine release.
      15 mg/kg paracetamol = 270 mg orally or intravenously (a loading dose of 20 mg/kg, or 360 mg, is sometimes recommended, which is not far short of the doses listed above).
      Cefuroxime – the initial intravenous dose is 20 mg/kg (360 mg) depending on the indication (again, similar to the dose given in the answer options above).

    • This question is part of the following fields:

      • Pharmacology
      41
      Seconds
  • Question 68 - Which of the following is true regarding Noradrenaline (Norepinephrine)? ...

    Incorrect

    • Which of the following is true regarding Noradrenaline (Norepinephrine)?

      Your Answer: Increases systolic but decreases diastolic blood pressure

      Correct Answer: Sympathomimetic effects work mainly through ?1 but also ? receptors

      Explanation:

      Noradrenaline acts as a sympathomimetic effect via alpha as well as a beta receptor. However, they have weak ?2 action.

      Natural catecholamines are Adrenaline, Noradrenaline, and Dopamine

    • This question is part of the following fields:

      • Pharmacology
      29.4
      Seconds
  • Question 69 - Regarding renal autoregulation, which of the following best describes its process? ...

    Correct

    • Regarding renal autoregulation, which of the following best describes its process?

      Your Answer: Reduces the effect of changes in arterial blood pressure on renal Na+ excretion

      Explanation:

      Two mechanisms are responsible for autoregulation of RBF and GFR: one mechanism that responds to changes in arterial pressure and another that responds to changes in [NaCl] in tubular fluid. Both regulate the tone of the afferent arteriole. The pressure-sensitive mechanism, the so-called myogenic mechanism, is related to an intrinsic property of vascular smooth muscle: the tendency to contract when stretched. Accordingly, when arterial pressure rises and the renal afferent arteriole is stretched, the smooth muscle contracts in response. Because the increase in resistance of the arteriole offsets the increase in pressure, RBF, and therefore GFR, remains constant.

      The second mechanism responsible for autoregulation of GFR and RBF is the [NaCl]-dependent mechanism known as tubuloglomerular feedback. This mechanism involves a feedback loop in which a change in GFR leads to alteration in the concentration of NaCl in tubular fluid, which is sensed by the macula densa of the juxtaglomerular apparatus and converted into signals that affect afferent arteriolar resistance and thus the GFR (Fig. 33.19). For example, when the GFR increases and causes [NaCl] in tubular fluid in the loop of Henle to rise, more NaCl enters the macula densa cells in this segment (Fig. 33.20). This leads to an increase in formation and release of adenosine triphosphate (ATP) and adenosine (a metabolite of ATP) by macula densa cells, which causes vasoconstriction of the afferent arteriole and normalization of GFR. In contrast, when GFR and [NaCl] in tubule fluid decrease, less NaCl enters the macula densa cells, and both ATP and adenosine production and release decline. The fall in [ATP] and [adenosine] results in afferent arteriolar vasodilation, which returns GFR to normal. NO, a vasodilator produced by the macula densa, attenuates tubuloglomerular feedback, whereas angiotensin II enhances tubuloglomerular feedback. Thus the macula densa may release both vasoconstrictors (e.g., ATP and adenosine) and a vasodilator (e.g., NO) that oppose each other’s action at the level of the afferent arteriole. Production plus release of either vasoconstrictors or vasodilators ensures exquisite control over tubuloglomerular feedback.

      Renal autoregulation, thus, reduces the effect of changes in arterial blood pressure on renal sodium excretion.

    • This question is part of the following fields:

      • Pathophysiology
      17.5
      Seconds
  • Question 70 - Of the following, which of these oxygen carrying molecules causes the greatest shift...

    Incorrect

    • Of the following, which of these oxygen carrying molecules causes the greatest shift of the oxygen-dissociation curve to the left?

      Your Answer: Haemoglobin (HbF)

      Correct 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).

    • This question is part of the following fields:

      • Physiology
      11.3
      Seconds
  • Question 71 - Which of the following statements is the most correct about ketamine? ...

    Incorrect

    • Which of the following statements is the most correct about ketamine?

      Your Answer: It causes a state of dissociative analgesia

      Correct Answer: The S (+) isomer is more potent that the R (-) isomer

      Explanation:

      Ketamine, a phencyclidine derivative, is an antagonist at the NMDA receptor. It causes depression of the CNS that is dose dependent and induces a dissociative anaesthetic state with profound analgesia and amnesia.

      Ketamine has a chiral centre usually presented as a racemic mixture with two optical isomers, S (+) and R (-) forms. These isomers are in equal proportions. The S (+) isomer is about three times more potent than the R (-) form. The S (+) form is less likely to cause emergence delirium and hallucinations.

      Ketamine is extensively metabolised by hepatic microsomal cytochrome P450 enzymes producing norketamine as its main metabolite. Norketamine has a one third to one fifth as potency as its parent compound.
      It increases the CMRO2, cerebral blood flow and potentially increase intracranial pressure.

    • This question is part of the following fields:

      • Pharmacology
      18.3
      Seconds
  • Question 72 - The main site of storage of thyroid hormones in the thyroid gland is?...

    Incorrect

    • The main site of storage of thyroid hormones in the thyroid gland is?

      Your Answer: In free form

      Correct Answer: Thyroglobulin

      Explanation:

      The follicle is the functional unit of the thyroid gland. The follicular cells surround the follicle which is filled with colloid. Suspended within the colloid is the is a pro-hormone complex thyroglobulin.

      The synthesis and storage of thyroid hormones is done by follicular cells and the thyroglobulin within the colloid.

      Iodide ions (I−) are actively transported against a concentration gradient into the follicular cell under the influence of thyroid stimulating hormone (TSH). It then undergoes oxidation to active iodine catalysed by thyroid peroxidase (TPO). The synthesis of thyroglobulin is in the follicular cells and it contains up to 140 tyrosine residues. The tyrosine residues of thyroglobulin and active iodine are merged to form mono- and di-iodotyrosines (MIT and DIT). The iodinated thyroglobulin is then taken up into the colloid where it is stored and dimerised. Two DIT molecules are joined to produce thyroxine (T4) while one MIT and one DIT molecule are joined to produce tri-iodotyrosine (T3) by a process catalysed by TPO.

      Thyroglobulin droplets are taken up as vesicles into follicular cells by pinocytosis. This process is stimulated by TSH. When these vesicles fuse with lysosomes, hydrolysis of the thyroglobulin molecules and subsequent release of T4 and T3 into the circulation occurs.

    • This question is part of the following fields:

      • Pathophysiology
      45.4
      Seconds
  • Question 73 - An intravenous drug infusion is started at a rate of 20 ml/hour. The...

    Incorrect

    • An intravenous drug infusion is started at a rate of 20 ml/hour. The drug concentration in the syringe is 5 mg/mL. The drug's plasma clearance is 20 L/hour.

      Which of the following values, assuming that the infusion rate remains constant, best approximates the drug's plasma concentration at steady state?

      Your Answer: 50 mg/mL

      Correct Answer: 5 mcg/mL

      Explanation:

      When a drug is given via intravenous infusion, the plasma concentration rises exponentially as a wash-in curve until it reaches steady-state concentration (the point at which the infusion rate is balanced by the elimination rate or clearance). To reach this steady state, the drug will take 4-5 half-lives.

      Cpss (target plasma concentration at steady state) and clearance (CL) in ml/minute or litre/hour are the two factors that determine the infusion rate or dose (ID) in mg/hour of a drug.

      ID = Cpss × CL

      We know the infusion rate is 20 ml/hour in this case. The drug’s concentration is 5 mg/mL. The patient is receiving 100 mg of the drug per hour, with a 20 L/hour clearance rate.

      ID = Cpss × 20

      Therefore,

      Cpss = 100 mg/20000 ml

      Cpss = 0.005 mg/mL or 5 mcg/mL

    • This question is part of the following fields:

      • Pharmacology
      15.5
      Seconds
  • Question 74 - A 70-year-old man collapsed at home. He was brought into the emergency department...

    Incorrect

    • A 70-year-old man collapsed at home. He was brought into the emergency department in an ambulance. His wife tells you that he complained of sudden lower back pain just before he collapsed.

      He is pale and hypotensive. You suspect a ruptured abdominal aortic aneurysm.
      What vertebral level does this affected vessel terminate?

      Your Answer: T5

      Correct Answer: L4

      Explanation:

      The abdominal aorta begins at the level of the body of T12 near the midline, as a continuation of the thoracic aorta. It descends and bifurcates at the level of L4 into the common iliac arteries.

      An abdominal aortic aneurysm is a swelling in the abdominal aorta. It most commonly occurs in men over 65 years old of age. Smoking, diabetes, hypertension, and hypercholesterolemia are other risk factors contributing to the disease.

      The NHS screening program for abdominal aortic aneurysms involves an ultrasound test for men aged 65 or over if they have not undergone screening for a one-off screening test.

    • This question is part of the following fields:

      • Anatomy
      21.3
      Seconds
  • Question 75 - An adult and a 7-year-old child are anatomically and physiologically very different.

    Which of...

    Incorrect

    • An adult and a 7-year-old child are anatomically and physiologically very different.

      Which of the following physiological characteristics of a 5-year-old most closely resembles those of a healthy adult?

      Your Answer: Lung compliance mL/cmH2O

      Correct Answer: Dead space ratio

      Explanation:

      Whatever the age, the dead space ratio is 0.3. It’s the dead space (Vd) to tidal volume ratio (Vt).

      The glottis is the narrowest point of the upper airway in an adult, while the cricoid ring is the narrowest point in a child.

      A child’s airway resistance is much higher than an adult’s. The resistance to airflow increases as the diameter of a paediatric airway shrinks. The radius (r) to the power of 4 is inversely proportional to airway resistance (r4). As a result, paediatric patients are more susceptible to changes in airflow caused by a small reduction in airway diameter, such as caused by oedema.

      The compliance of a newborn’s lungs is very low (5 mL/cmH2O), but it gradually improves as lung size and elasticity grow. Lung compliance in an adult is 200 mL/cmH2O.

      In children, minute ventilation (mL/kg/minute) is much higher.

    • This question is part of the following fields:

      • Pathophysiology
      23.9
      Seconds
  • Question 76 - A 20-year-old female presents to the emergency department. She complains of increased shortness...

    Incorrect

    • A 20-year-old female presents to the emergency department. She complains of increased shortness of breath and wheezing over the last 48 hours. On examination, she is found to have tachycardia, tachypnoea, and oxygen saturation at 91% on air. She admits to a previous medical history of asthma, diagnosed 4 years ago. She requires further investigations for diagnosis.

      Which of the following is true about the assessment of a patient with symptomatic asthma?

      Your Answer: An electrocardiogram (ECG) is not indicated

      Correct Answer: Oxygen saturations of 91% on air would be an indication for performing arterial blood gases

      Explanation:

      A patient presenting with symptomatic asthma should be assessed for severity to determine appropriate management options. Indications of acute severe asthma are:

      Peak expiratory flow rate (PEFR): 33-50% best/predicted
      Respiratory rate: ≥25/min
      Heart rate: ≥110/min
      Inability to finish a complete sentence in a single breath.

      Oxygen saturation should be measured. Any measurement of an oxygen saturation of 92% or less, either on air or on oxygen, indicates severe, life threatening asthma, and requires an arterial blood gas (ABG) to detect normo- or hypercarbia.

      A chest x-ray would not be routine as it will not provide any relevant information. It is only required in specific cases, including:
      Diagnosis of a subcutaneous emphysema
      Indications of a unilateral pneumothorax
      Indications of a lobar collapse of consolidation
      Treatment-resistance life-threatening asthma
      If mechanical ventilation is indicated

      A peak expiratory flow rate (PEFR) can provide relevant information to help distinguish between acute, moderate, severe and life threatening asthma. However, it is not necessary as other parameters exist that can also help make the same distinction.

      An ECG is indicated in this case as the patient has tachycardia and tachypnoea which are indicative of acute severe asthma. The ECG would indicate if arrhythmia is also present which would suggest life-threatening asthma.

    • This question is part of the following fields:

      • Clinical Measurement
      12.8
      Seconds
  • Question 77 - Which vessel is the first to branch from the external carotid artery? ...

    Incorrect

    • Which vessel is the first to branch from the external carotid artery?

      Your Answer: Lingual artery

      Correct Answer: Superior thyroid artery

      Explanation:

      The superior thyroid artery is the first branch of the external carotid artery. The other branches of the external carotid artery are:
      1. Superior thyroid artery
      2. Ascending pharyngeal artery
      3. Lingual artery
      4. Facial artery
      5. Occipital artery
      6. Posterior auricular artery
      7. Maxillary artery
      8. Superficial temporal artery

      The inferior thyroid artery is derived from the thyrocervical trunk.

    • This question is part of the following fields:

      • Anatomy
      32
      Seconds
  • Question 78 - Which of the following options will likely play a major role in falling...

    Incorrect

    • Which of the following options will likely play a major role in falling coronary blood flow?

      Your Answer: Peripheral venous infusion of glyceryl trinitrate (GTN)

      Correct Answer: Intracoronary artery infusion of endothelin-1

      Explanation:

      Endothelin-1 is considered as a powerful coronary vasoconstrictor, produced by the endothelium. It acts to counter the effects of Nitric oxide (NO).
      Neuropeptide-Y, angiotensin1, cocaine, vasopressin, and nicotine are some other coronary vasoconstrictors.

      Chronotrophy and inotrophy occur after the activation of sympathetic nerve fibres, which in turn results in increasing the myocardial oxygen consumption, leading to increased coronary blood flow via local metabolic processes.

      An alpha-receptor mediated coronary vasoconstrictor effect is also initiated that usually competes with vasodilation, resulting in decreased coronary vascular resistance. Some of the other dilators include hydrogen ions, CO2, potassium, and lactic acid. The action of endothelial NO synthase (eNOS) on L-arginine results in the formation of NO. This messenger also plays a vital role in the regulation of coronary blood flow via vasodilation, inhibition of platelet aggression, and decreasing vascular resistance.
      Adenosine is considered as purine nucleoside that forms after the breakdown of adenosine triphosphate (ATP). Adenosine binds to adenosine type 2A (A2A) receptors in coronary vascular smooth muscles. These are coupled to the Gs protein. This mechanism leads to hyperpolarisation of muscle cells, resulting in relaxation and increased coronary blood flow.

      GTN is an veno and arteriolar dilator, which behaves as pro-drug with NO.

    • This question is part of the following fields:

      • Pathophysiology
      16.5
      Seconds
  • Question 79 - Which type of epithelium lines the luminal surface of the oesophagus? ...

    Correct

    • Which type of epithelium lines the luminal surface of the oesophagus?

      Your Answer: Non keratinised stratified squamous epithelium

      Explanation:

      Normally, the oesophagus is lined by non-keratinized stratified squamous epithelium. This epithelium can undergo metaplasia and convert to the columnar epithelium (stomach’s lining) in long-standing GERD that leads to Barret’s oesophagus.

    • This question is part of the following fields:

      • Anatomy
      20.8
      Seconds
  • Question 80 - A study aimed at assessing the validity of a novel diagnostic test for...

    Incorrect

    • A study aimed at assessing the validity of a novel diagnostic test for heart failure is being performed. The curators are worried that not all the patients will get the prevalent gold standard test.

      Which type of bias is that?

      Your Answer: Instrument bias

      Correct Answer: Work-up bias

      Explanation:

      Work up bias involves comparing the novel diagnostic test with the current standard test. A portion of the patients undergo the standard test while others undergo the new test as the standard test is costly. The result can be alteration in specify and sensitivity.

      Selection bias is when randomisation is not achieved.

      Attention bias refers to the person’s failure to consider various alternatives when he pre occupied by some other thoughts.

      Instrument bias is related to the experience and extent of familiarization of the participating individuals with the test.

      Co intervention bias is characterized by the groups receiving different co interventions.

    • This question is part of the following fields:

      • Statistical Methods
      32.2
      Seconds
  • Question 81 - How data is collected for the Delphi survey technique? ...

    Incorrect

    • How data is collected for the Delphi survey technique?

      Your Answer: Observations

      Correct Answer: Questionnaires

      Explanation:

      The Delphi is a group facilitation technique that seeks to obtain consensus on the opinions of `experts’ through a series of structured questionnaires (commonly referred to as rounds). By using successive questionnaires, opinions are considered in a non-adversarial manner, with the current status of the groups’ collective opinion being repeatedly fed back. Studies employing the Delphi make use of individuals who have knowledge of the topic being investigated

    • This question is part of the following fields:

      • Statistical Methods
      60.6
      Seconds
  • Question 82 - A 55-year-old man with a ventricular rate of 210 beats per minute is...

    Incorrect

    • A 55-year-old man with a ventricular rate of 210 beats per minute is admitted to the emergency department with atrial fibrillation. The patient develops ventricular fibrillation shortly after receiving pharmacotherapy to treat his arrhythmia, from which he is successfully resuscitated.

      He has a PR interval of 40 Ms, a prominent delta wave in lead I, and a QRS duration of 120 Ms, according to an ECG from a previous admission.

      Which of the following drugs is most likely to be involved in this patient's development of ventricular fibrillation?

      Your Answer: Procainamide

      Correct Answer: Digoxin

      Explanation:

      The Wolff-Parkinson-White syndrome (WPWS) is linked to an additional electrical conduction pathway between the atria and ventricles. This accessory pathway (bundle of Kent), unlike the atrioventricular (AV) node, is incapable of slowing down a rapid rate of atrial depolarization. In other words, a short circuit bypasses the AV node. Patients with a rapid ventricular response or narrow complex AV re-entry tachycardia are more likely to develop atrial fibrillation or flutter.

      Digoxin can promote impulse transmission through this accessory pathway if a patient with WPWS develops atrial fibrillation because it works by blocking the AV node. This can cause ventricular fibrillation and an extremely rapid ventricular rate. As a result, it’s not advised.

      Adenosine, beta-blockers, and calcium channel blockers, among other drugs that interfere with AV nodal conduction, are also generally contraindicated.

      The class III antiarrhythmic drugs amiodarone and ibutilide (K+ channel block) and procainamide (Na+ channel block) are the drugs of choice.

    • This question is part of the following fields:

      • Pharmacology
      9.6
      Seconds
  • Question 83 - A 30-year old lady has a sub total thyroidectomy. On the 5th post-operative...

    Incorrect

    • A 30-year old lady has a sub total thyroidectomy. On the 5th post-operative day, the wound becomes erythematous and there is a purulent discharge. The most likely organism causing this is:

      Your Answer: Haemophilus influenzae

      Correct Answer: Staphylococcus aureus

      Explanation:

      Staphylococcus aureus infection is the most likely cause.

      Surgical site infections (SSI) occur when there is a breach in tissue surfaces and allow normal commensals and other pathogens to initiate infection. They are a major cause of morbidity and mortality.

      SSI comprise up to 20% of healthcare associated infections and approximately 5% of patients undergoing surgery will develop an SSI as a result.
      The organisms are usually derived from the patient’s own body.

      Measures that may increase the risk of SSI include:
      -Shaving the wound using a single use electrical razor with a disposable head
      -Using a non iodine impregnated surgical drape if one is needed
      -Tissue hypoxia
      -Delayed prophylactic antibiotics administration in tourniquet surgery, patients with a prosthesis or valve, in clean-contaminated surgery of in contaminated surgery.

      Measures that may decrease the risk of SSI include:
      1. Intraoperatively
      – Prepare the skin with alcoholic chlorhexidine (Lowest incidence of SSI)
      -Cover surgical site with dressing

      In contrast to previous individual RCT’s, a recent meta analysis has confirmed that administration of supplementary oxygen does not reduce the risk of wound infection and wound edge protectors do not appear to confer benefit.

      2. Post operatively
      Tissue viability advice for management of surgical wounds healing by secondary intention

      Use of diathermy for skin incisions
      In the NICE guidelines the use of diathermy for skin incisions is not advocated. Several randomised controlled trials have been undertaken and demonstrated no increase in risk of SSI when diathermy is used.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      17.6
      Seconds
  • Question 84 - Weight of all of your patients in the ICU is analysed, and shows...

    Incorrect

    • Weight of all of your patients in the ICU is analysed, and shows that your date set is skewed.

      Which of the following will correctly show the average weight of your patients?

      Your Answer: Mean

      Correct Answer: Median

      Explanation:

      The question mentions a quantitative, ratio scale data set. The use of mean would be ideal under normal circumstances, however, in this situation median is preferred as it is less sensitive to the skewness of data. The median is usually preferred to other measures of central tendency when your data set is skewed (i.e., forms a skewed distribution)

    • This question is part of the following fields:

      • Statistical Methods
      7.4
      Seconds
  • Question 85 - An emergency appendicectomy is being performed on a 20 year old man. For...

    Correct

    • An emergency appendicectomy is being performed on a 20 year old man. For maintenance of anaesthesia, he is being ventilated using a circle system with a fresh gas flow (FGF) of 1 L/min (air/oxygen and sevoflurane). The trace on the capnograph shows a normal shape.

      The table below demonstrates the changes in the end-tidal and baseline carbon dioxide measurements of the capnograph at 10 and 20 minutes of anaesthesia maintenance.  
      End-tidal CO2: 4.9 kPa vs 8.4kPa (10 minutes vs 20 minutes)
      Baseline end-tidal CO2: 0.2 kPa vs 2.4kPa

      Pulse 100-107 beats per minute, systolic blood pressure 125-133 mmHg and oxygen saturation 98-99%. 

      Which of the following is the single most important immediate course of action?

      Your Answer: Increase the FGF

      Explanation:

      End-tidal carbon dioxide (ETCO2) monitoring has been an important factor in reducing anaesthesia-related mortality and morbidity. Hypercarbia, or hypercapnia, occurs when levels of CO2 in the blood become abnormally high (Paco2 >45 mm Hg). Hypercarbia is confirmed by arterial blood gas analysis. When using capnography to approximate Paco2, remember that the normal arterial–end-tidal carbon dioxide gradient is roughly 5 mm Hg. Hypercarbia, therefore, occurs when PETco2 is greater than 40 mm Hg.

      The most likely explanation for the changes in capnograph is either exhaustion of the soda lime and a progressive rise in circuit dead space.

      Inspect the soda lime canister for a change in colour of the granules. To overcome soda lime exhaustion, the first step is to increase the fresh gas flow (FGF) (Option A). Then, if need arises, replace the soda lime granules. Other strategies that can work are changing to another circuit or bypassing the soda lime canister, but remember that both these strategies are employed only after increasing FGF first. Exclude other causes of equipment deadspace too.

      There are also other causes for hypercarbia to develop intraoperatively:
      1. Hypoventilation is the most common cause of hypercapnia. A. Inadequate ventilation can occur with spontaneous breathing due to drugs like anaesthetic agents, opioids, residual NMDs, chronic respiratory or neuromuscular disease, cerebrovascular accident.
      B. In controlled ventilation, hypercapnia due to circuit leaks, disconnection or miscalculation of patient’s minute volume.
      2. Rebreathing – Soda lime exhaustion with circle, inadequate fresh gas flow into Mapleson circuits and increased breathing system deadspace.
      3. Endogenous source – Tourniquet release, hypermetabolic states (MH or thyroid storm) and release of vascular clamps.
      4. Exogenous source – Absorption of CO2 from pneumoperitoneum.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      24.1
      Seconds
  • Question 86 - Considering research studies, which of the following is considered as a limitation of...

    Incorrect

    • Considering research studies, which of the following is considered as a limitation of the Delphi method?

      Your Answer: It is very expensive

      Correct Answer: Potential low response rates

      Explanation:

      The Delphi technique was developed in the 1950s and is a widely used and accepted method for achieving convergence of opinion concerning real-world knowledge solicited from experts within certain topic areas. Choosing the appropriate subjects is the most important step in the entire process because it directly relates to the quality of the results generated, despite this, there is no exact criterion currently listed in the literature concerning the selection of Delphi participants.

      Therefore, due to the potential scarcity of qualified participants and the relatively small number of subjects used in a Delphi study, the ability to achieve and maintain an ideal response rate can either ensure or jeopardize the validity of a Delphi study.

    • This question is part of the following fields:

      • Statistical Methods
      34.4
      Seconds
  • Question 87 - Sugammadex binds to certain drugs that affect neuromuscular function during anaesthesia in a...

    Correct

    • Sugammadex binds to certain drugs that affect neuromuscular function during anaesthesia in a stereospecific, non-covalent, and irreversible manner.

      It has the greatest impact on the activity of which of the following drugs?

      Your Answer: Vecuronium

      Explanation:

      Sugammadex is a modified cyclodextrin that works as an aminosteroid neuromuscular blocking (nmb) reversal agent. By encapsulating each molecule in the plasma, it rapidly reverses rocuronium and, to a lesser extent, vecuronium-induced neuromuscular blockade. Consequently, a  concentration gradient favours the movement of these nmb agents away from the neuromuscular junction.  Pancuronium-induced neuromuscular blockade at low levels has also been reversed.

      By inhibiting voltage-dependent calcium channels at the neuromuscular junction, antibiotics in the aminoglycoside group potentiate neuromuscular blocking agents. This can be reversed by giving calcium but not neostigmine or sugammadex.

      Sugammadex will not reverse the effects of mivacurium, which belongs to the benzylisoquinolinium class of drugs.

      A phase II or desensitisation block occurs when the motor end-plate becomes less sensitive to acetylcholine as a result of an overdose or repeated administration of suxamethonium. The use of neostigmine has been shown to be effective in reversing this weakness.

    • This question is part of the following fields:

      • Pharmacology
      68.2
      Seconds
  • Question 88 - A 28-year-old man is admitted to the critical care unit. He has been...

    Correct

    • A 28-year-old man is admitted to the critical care unit. He has been diagnosed with adult respiratory distress syndrome and is being ventilated. His haemodynamic condition is improved using a pulmonary artery flotation.

      His readings are listed below:

      Haemoglobin concentration: 10 g/dL
      Mixed venous oxygen saturation: 70%
      Mixed venous oxygen tensions (PvO2): 50 mmHg

      Estimate his mixed venous oxygen content (mL/100mL).

      Your Answer: 9.5

      Explanation:

      Mixed venous oxygen content (CvO2) is the oxygen concentration in 100mL of mixed venous blood taken from the pulmonary artery. It is usually 12-17 mL/dL (70-75%). It is represented mathematically as:

      CvO2 = (1.34 x Hgb x SvO2 x 0.01) + (0.003 x PvO2)

      Where,

      1.34 = Huffner’s constant
      Hgb = Haemoglobin level (g/dL)
      SvO2 = % oxyhaemoglobin saturation of mixed venous blood
      PvO2 = 0.0225 = mL of O2 dissolved per 100mL plasma per kPa, or 0.003 mL per mmHg

      Therefore,

      CvO2 = (1.34 x 10 x 70 x 0.01) + (0.003 x 50)

      CvO2 = 9.38 + 0.15 = 9.53 mL/100mL

    • This question is part of the following fields:

      • Clinical Measurement
      10
      Seconds
  • Question 89 - Which of the following vertebral levels is the site where the aorta perforates...

    Incorrect

    • Which of the following vertebral levels is the site where the aorta perforates the diaphragm?

      Your Answer: T10

      Correct Answer: T12

      Explanation:

      The diaphragm divides the thoracic cavity from the abdominal cavity. Structures penetrate the diaphragm at different vertebral levels through openings in the diaphragm to communicate between the two cavities. The diaphragm has openings at three vertebral levels:

      T8: vena cava, terminal branches of the right phrenic nerve
      T10: oesophagus, vagal trunks, left anterior phrenic vessels, oesophageal branches of the left gastric vessels
      T12: descending aorta, thoracic duct, azygous and hemi-azygous vein

    • This question is part of the following fields:

      • Anatomy
      61.1
      Seconds
  • Question 90 - Levels of serum potassium in around 1000 patients that were on ACE inhibitor...

    Incorrect

    • Levels of serum potassium in around 1000 patients that were on ACE inhibitor were measured. The mean value was calculated to be 4.6mmol/L and a standard deviation of 0.3mmol/L was recorded.

      Which among the given options is correct?

      Your Answer: 95.4% of values lie between 4.3 and 4.9 mmol/l

      Correct Answer: 68.3% of values lie between 4.3 and 4.9 mmol/l

      Explanation:

      Its known that 68.3% of the total values of a normally distributed variable are found within a range of 1 standard deviation from the mean which makes the range to be 4.3 to 4.9 mmol/L.

    • This question is part of the following fields:

      • Statistical Methods
      16.3
      Seconds
  • Question 91 - The clavipectoral fascia is penetrated by the cephalic vein to terminate in which...

    Incorrect

    • The clavipectoral fascia is penetrated by the cephalic vein to terminate in which of the listed veins?

      Your Answer: Internal jugular

      Correct Answer: Axillary

      Explanation:

      The cephalic vein is a superficial vein that runs through the forearm and the arm, before draining into the axillary vein where it terminates.

    • This question is part of the following fields:

      • Anatomy
      26
      Seconds
  • Question 92 - Regarding anti diuretic hormone (ADH), one of the following statements is correct: ...

    Incorrect

    • Regarding anti diuretic hormone (ADH), one of the following statements is correct:

      Your Answer: Increases the reabsorption of water in the proximal tubule

      Correct Answer: Increases the total amount of electrolyte free water in the body

      Explanation:

      The major action of ADH is to increase reabsorption of osmotically unencumbered water from the glomerular filtrate and decreases the volume of urine passed. The osmolarity of urine is increased to a maximum of four times that of plasma (approx. 1200 mOsm/kg) by Increasing water reabsorption.

      Chronic water loading, Lithium, potassium deficiency, cortisol and calcium excess, all blunt the action of ADH. This leads to nephrogenic diabetes insipidus.

      ADH’s primary site of action is the distal tubule and collecting duct.

    • This question is part of the following fields:

      • Physiology
      36
      Seconds
  • Question 93 - An study on post-operative nausea and vomiting (PONV) among paediatric patients who underwent...

    Correct

    • An study on post-operative nausea and vomiting (PONV) among paediatric patients who underwent tonsillectomy showed a decrease in incidence from 10% to 5% following a new management protocol.

      Which of the following best estimates the numbers needed to treat (NNT) for one additional patient to benefit from the new management of PONV?

      Your Answer: 20

      Explanation:

      The Number Needed to Treat (NNT) is the number of patients you need to treat to prevent one additional bad outcome. For example, if a drug has an NNT of 5, it means you have to treat 5 people with the drug to prevent one additional bad outcome.

      To calculate the NNT, you need to know the Absolute Risk Reduction (ARR); the NNT is the inverse of the ARR:

      NNT = 1/ARR

      Where ARR = CER (Control Event Rate) – EER (Experimental Event Rate).

      NNTs are always rounded up to the nearest whole number.

      In this case, the NNT can be computed as follows:

      ARR = 10% – 5% = 0.05

      NNT = 1/0.05 = 20

    • This question is part of the following fields:

      • Statistical Methods
      66.7
      Seconds
  • Question 94 - In a normal healthy adult breathing 100 percent oxygen, which of the following...

    Incorrect

    • In a normal healthy adult breathing 100 percent oxygen, which of the following is the most likely cause of an alveolar-arterial (A-a) oxygen difference of 30 kPa?

      Your Answer: Hypoventilation

      Correct Answer: Atelectasis

      Explanation:

      The ‘ideal’ alveolar PO2 minus arterial PO2 is the alveolar-arterial (A-a) oxygen difference.

      The ‘ideal’ alveolar PO2 is derived from the alveolar air equation and is the PO2 that the lung would have if there was no ventilation-perfusion (V/Q) inequality and it was exchanging gas at the same respiratory exchange ratio as real lung.

      The amount of oxygen in the blood is measured directly in the arteries.

      The A-a oxygen difference (or gradient) is a useful measure of shunt and V/Q mismatch, and it is less than 2 kPa in normal adults breathing air (15 mmHg). Because the shunt component is not corrected, the A-a difference increases when breathing 100 percent oxygen, and it can be up to 15 kPa (115 mmHg).

      An abnormally low or abnormally high V/Q ratio within the lung can cause an increased A-a difference, though the former is more common. Atelectasis, which results in a low V/Q ratio, is the most likely cause of an A-a difference in a healthy adult breathing 100 percent oxygen.

      Hypoventilation may cause an increase in alveolar (and thus arterial) CO2, lowering alveolar PO2 according to the alveolar air equation.

      The alveolar PO2 is also reduced at high altitude.

      Healthy people are unlikely to have a right-to-left shunt or an oxygen transport diffusion defect.

    • This question is part of the following fields:

      • Physiology
      35.1
      Seconds
  • Question 95 - The production of carbon dioxide and water occurs during cellular respiration, which involves...

    Incorrect

    • The production of carbon dioxide and water occurs during cellular respiration, which involves an energy substrate and oxygen. For a patient, the respiratory quotient is calculated as 0.7.

      Which of the following energy substrate combinations is the most likely in this patient's diet?

      Your Answer: Low carbohydrate, low fat and high protein

      Correct Answer: Low carbohydrate, high fat and low protein

      Explanation:

      The respiratory quotient (RQ) is the proportion of CO2 produced by the body to O2 consumed per unit of time.

      CO2 produced / O2 consumed = RQ

      CO2 is produced at a rate of 200 mL per minute, while O2 is consumed at a rate of 250 mL per minute. An RQ of around 0.8 is typical for a mixed diet.

      The RQ will change depending on the energy substrates consumed in the diet.

      Granulated sugar is a refined carbohydrate that contains 99.999 percent carbohydrate and no lipids, proteins, minerals, or vitamins.

      Glucose and other hexose sugars – RQ = 1
      Fats – RQ = 0.7
      Proteins – RQ is 0.9
      Ethyl alcohol – RQ = 0.67

    • This question is part of the following fields:

      • Pathophysiology
      28.3
      Seconds
  • Question 96 - During positive pressure ventilation using positive end-expiratory pressure (PEEP), there is usually an...

    Incorrect

    • 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: Increased pulmonary vascular resistance

      Correct 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.

    • This question is part of the following fields:

      • Pathophysiology
      18.6
      Seconds
  • Question 97 - A 25-year-old male has tonsillitis and is in considerable pain.

    Which nerve is responsible...

    Incorrect

    • A 25-year-old male has tonsillitis and is in considerable pain.

      Which nerve is responsible for the sensory innervation of the tonsillar fossa?

      Your Answer: Facial nerve

      Correct Answer: Glossopharyngeal nerve

      Explanation:

      A tonsillar sinus or fossa is a space that is bordered by the triangular fold of the palatoglossal and palatopharyngeal arches in the lateral wall of the oral cavity. The palatine tonsils are in these sinuses.

      The glossopharyngeal nerve is the main sensory nerve for the tonsillar fossa. The tonsillar branches of the glossopharyngeal nerve supply the palatine tonsils forming a plexus around it. Filaments from this plexus are distributed to the soft palate and fauces where they communicate with the palatine nerves. A lesser contribution is made by the lesser palatine nerve. Because of this otalgia may occur following tonsillectomy.

    • This question is part of the following fields:

      • Anatomy
      15.9
      Seconds
  • Question 98 - Which of the following statements is true regarding prazosin? ...

    Incorrect

    • Which of the following statements is true regarding prazosin?

      Your Answer: Causes elevation of renin concentrations.

      Correct Answer: Is a selective alpha 1 adrenergic receptor antagonist.

      Explanation:

      Selective ?1 -Blockers like prazosin, terazosin, doxazosin, and alfuzosin cause a decrease in blood pressure with lesser tachycardia than nonselective blockers (due to lack of ?2 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)

    • This question is part of the following fields:

      • Pharmacology
      24.9
      Seconds
  • Question 99 - With respect to the peripheral nerve stimulators, which one is used to perform...

    Incorrect

    • With respect to the peripheral nerve stimulators, which one is used to perform nerve blocking?

      Your Answer: Prior to injecting the local anaesthetic the ideal current is 1-2 mili amperes at a frequency of 1-2 Hz

      Correct Answer:

      Explanation:

      The nerve stimulators deliver a stimulus lasting for 1-2 milliseconds (not second) to perform nerve blockage.

      There are just 2 leads (not 3); one for the skin and other for the needle.

      Prior to the administration of the local anaesthesia, a current of 0.25 – 0.5 mA (not 1-2mA) at the frequency of 1-2 Hz is preferred.

      If the needle tip is close to the nerve, muscular contraction could be possible at the lowest possible current.

      Insulated needles have improved the block success rate, as the current is only conducting through needle tip.

      Stimulus to the femoral nerve which is placed in the mid lingual line causes withdrawer of the quadriceps and knee extension, that’s the dancing patella ( not plantar flexion).

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      66.6
      Seconds
  • Question 100 - Provided below is an abstract of a study conducted recently.

    A consensus...

    Incorrect

    • Provided below is an abstract of a study conducted recently.

      A consensus was developed among international experts. A total of 27 experts were invited. 91% of them decided to show up. A systematic review was performed. This comprised of open ended questions and the participants were encouraged to provide suggestions by e-mail. In the second phase google forms were used. Participants were asked to rate survey items on a scale of 5 points. Items that were rated critical by no less than 80% of the experts were included. Items that were rendered important by 65-79% of experts were inducted in the next survey for re rating. Items that were rated below 65% were rejected.

      Which of the following methods was used in the study from which the abstract has been taken?

      Your Answer:

      Correct Answer: The Delphi method

      Explanation:

      The process used in the study is Delphi method. This method kicks off with an open ended questionnaire and uses its responses as a survey instrument for the next round in which each of the participants is asked to rate the items that the investigators have summarized on the basis of the data collected in the first round.

      Any disagreement is further discussed in phases to come on the basis of information obtained from previous phases.

    • This question is part of the following fields:

      • Statistical Methods
      0
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Statistical Methods (13/17) 76%
Pathophysiology (13/15) 87%
Physiology And Biochemistry (4/5) 80%
Anatomy (18/24) 75%
Physiology (9/12) 75%
Pharmacology (13/15) 87%
Clinical Measurement (4/5) 80%
Anaesthesia Related Apparatus (4/5) 80%
Basic Physics (0/1) 0%
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