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  • Question 1 - A 55-year-old woman presents for transsphenoidal surgery following a diagnosis of pituitary macroadenoma.

    Which...

    Correct

    • A 55-year-old woman presents for transsphenoidal surgery following a diagnosis of pituitary macroadenoma.

      Which of the following is the most common visual field defect caused by such lesions?

      Your Answer: Bitemporal hemianopia

      Explanation:

      Pituitary tumours that compress the optic chiasma primarily affect the neurones that decussate at this location. Bitemporal hemianopia is caused by neurones that emerge from the nasal half of the retina and transmit the temporal half of the visual field.

      The axons of ganglion cells in the retina form the optic nerve.

      It exits the orbit through the optic foramen and projects to the thalamic lateral geniculate body. The optic chiasma forms above the sella turcica as the nasal fibres decussate along the way. The optic radiation travels from the lateral geniculate body to the occipital cortex.

      Lesions at various points along this pathway cause the following visual field defects:

      Scotoma implies partial retinal or optic nerve damage.
      Monocular vision loss occurs when the optic nerve is completely damaged.
      Pathology at the optic chiasma causes bitemporal hemianopia.
      Cortical blindness with occipital cortex pathology and homonymous hemianopia with lesions compromising the optic radiation.

    • This question is part of the following fields:

      • Pathophysiology
      3143.4
      Seconds
  • Question 2 - A 50-year-old male is planned for elective parotidectomy for pleomorphic adenoma. The surgeon...

    Correct

    • A 50-year-old male is planned for elective parotidectomy for pleomorphic adenoma. The surgeon intends to use a nerve integrity monitor thus avoiding neuromuscular blockade. Which of the following nerves is liable to injury in parotidectomy?

      Your Answer: Facial nerve

      Explanation:

      Parotidectomy is basically an anatomical dissection. Identification of the facial nerve trunk is essential during parotid gland surgery because facial nerve injury is the most daunting potential complication of parotid gland surgery owing to the close relation between the gland and the extratemporal course of the facial nerve. After exiting the stylomastoid foramen, the facial nerve enters the substance of the parotid gland and then gives off five terminal branches:
      From superior to inferior, these are the:
      – Temporal branch supplying the extrinsic ear muscles, occipitofrontalis and orbicularis oculi
      – Zygomatic branch supplying orbicularis oculi
      – Buccal branch supplying buccinator and the lip muscles
      – Mandibular branch supplying the muscles of the lower lip and chin
      – Cervical branch supplying platysma.

      There are two approaches to identify the facial nerve trunk during parotidectomy—conventional antegrade dissection of the facial nerve, and retrograde dissection. Numerous soft tissue and bony landmarks have been proposed to assist the surgeon in the early identification of this nerve. Most commonly used anatomical landmarks to identify facial nerve trunk are stylomastoid foramen, tympanomastoid suture (TMS), posterior belly of digastric (PBD), tragal pointer (TP), mastoid process and peripheral branches of the facial nerve.

    • This question is part of the following fields:

      • Anatomy
      85.5
      Seconds
  • Question 3 - A bolus of alfentanil has a faster onset of action than an equal...

    Correct

    • 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 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
      188.6
      Seconds
  • Question 4 - Concerning the intercostal nerves, which one of the following is true? ...

    Incorrect

    • Concerning the intercostal nerves, which one of the following is true?

      Your Answer: Anteriorly they run in the costal groove on the upper margin of the rib

      Correct Answer: Each is connected to a ganglion of the sympathetic trunk

      Explanation:

      The intercostal nerves arise from the ventral rami of the first 11 thoracic spinal nerves. they course along the costal groove on the lower margin of the rib.

      The twelfth intercoastal nerve is called the subcostal nerve. This is because it is below the 12th rib.

      Each intercostal nerve is connected to a ganglion of the sympathetic trunk from which it carries preganglionic and postganglionic fibres that innervate blood vessels, sweat glands, and muscles.

      The lateral and medial pectoral nerves innervates pectoralis major muscle.

    • This question is part of the following fields:

      • Anatomy
      347.6
      Seconds
  • Question 5 - 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
      65
      Seconds
  • Question 6 - What is the name of the space between the vocal cords? ...

    Incorrect

    • What is the name of the space between the vocal cords?

      Your Answer: Rima vestibuli

      Correct Answer: Rima glottidis

      Explanation:

      The rima glottidis is a narrow, triangle-shaped opening between the true vocal cords.

      The vocal folds (true vocal cords) control sound production. The apex of each fold projects medially into the laryngeal cavity.

      Each vocal fold includes these vocal ligaments:

      Vocalis muscle (most medial part of thyroarytenoid muscle)

      The glottis is composed of the vocal folds, processes and rima glottidis.

      The rima glottidis is the narrowest potential site within the larynx, as the vocal cords may be completely opposed, forming a complete barrier.

    • This question is part of the following fields:

      • Anatomy
      22.7
      Seconds
  • Question 7 - All of the following statements about cerebrospinal fluid are incorrect except: ...

    Incorrect

    • All of the following statements about cerebrospinal fluid are incorrect except:

      Your Answer: Has approximately 100 red cells/ml

      Correct Answer: Has a glucose concentration 2/3 that of the plasma glucose

      Explanation:

      The pH of CSF is 7.31 which is lower than plasma.

      Compared to plasma, it has a lower concentration of potassium, calcium, and protein and a higher concentration of sodium, chloride, bicarbonate and magnesium.

      CSF usually has no cells present but if white cells are present, there should be no more than 4/ml.

      The pressure of CSF should be less than 20 cm of water.

      The concentration of glucose is approximately two-thirds of that of plasma, and it has a concentration of approximately 3.3-4 mmol/L.

    • This question is part of the following fields:

      • Physiology
      158.4
      Seconds
  • Question 8 - Which of the following closely estimates the interstitial oncotic pressure acting on a...

    Incorrect

    • Which of the following closely estimates the interstitial oncotic pressure acting on a pulmonary capillary?

      Your Answer: 0 mmHg

      Correct Answer: 17 mmHg

      Explanation:

      The starling forces operate to maintain a homeostatic flow across the pulmonary capillary bed.

      The outward driving force comprises of the capillary hydrostatic pressure (13 mmHg), negative interstitial fluid pressure (zero to slightly negative), and interstitial colloid osmotic pressure (17 mmHg). The inward driving force is controlled by the plasma colloid osmotic pressure (25 mmHg).

    • This question is part of the following fields:

      • Basic Physics
      419
      Seconds
  • Question 9 - What feature is found in the flowmeters of modern anaesthetic machines ? ...

    Incorrect

    • What feature is found in the flowmeters of modern anaesthetic machines ?

      Your Answer: The reading of the flowmeter is taken from the midpoint of the bobbin

      Correct Answer: The use of stannic oxide increases the accuracy of flowmeters

      Explanation:

      Flowmeters measure the rate at which a specific gas, that the flowmeter has been calibrated for, passes through. This calibration is done at room temperature and standard atmospheric pressure with an accuracy of +/- 2%.

      Reading the flowmeter is done from the top of a bobbin (the midpoint of a ball). Oxygen is the last gas to be added downstream to the mixture delivered to the back bar as a safety feature. This prevents delivery of a hypoxic mixture.

      Inaccurate flow measurements occur when the bobbin sticks to the inside wall of the flowmeter. Stannic oxide has been used as a successful antistatic substance thus, reducing the aforementioned risk.

      Carbon dioxide being easily delivered is found on some older machines, but those attached flowmeters are limited by a maximum flow of 500 ml /min. Thus avoiding the delivery of a hypercarbic mixture.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      9995.9
      Seconds
  • Question 10 - Which of the following is incorrect with regards to atrial natriuretic peptide? ...

    Incorrect

    • Which of the following is incorrect with regards to atrial natriuretic peptide?

      Your Answer: Antagonises actions of angiotensin II and aldosterone

      Correct Answer: Secreted mainly by the left atrium

      Explanation:

      Atrial natriuretic peptide (ANP) is secreted mainly from myocytes of right atrium and ventricle in response to increased blood volume.
      It is secreted by both the right and left atria (right >> left).

      It is a 28 amino acid peptide hormone, which acts via cGMP
      degraded by endopeptidases.

      It serves to promote the excretion of sodium, lowers blood pressure, and antagonise the actions of angiotensin II and aldosterone.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      123.1
      Seconds
  • Question 11 - 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: Right marginal 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
      77.3
      Seconds
  • Question 12 - A 50-year old man for septoplasty has a BMI of 32kg/m2 and neck...

    Incorrect

    • A 50-year old man for septoplasty has a BMI of 32kg/m2 and neck circumference of 44 cm. He is troubled by intermittent nasal obstruction which causes his partner sleeps in a separate bedroom because of snoring. He currently on ramipril for hypertension.

      Which of the following best approximates to his STOP-BANG score from the information above?

      Your Answer: 7

      Correct Answer: 5

      Explanation:

      The STOP-BANG questionnaire is used to screen patients for obstructive sleep apnoea (OSA).

      The scoring system assigns one point for each feature.

      S: Snoring (louder than talking or loud enough to be heard through closed doors)
      T: Feeling tired, fatigued, or sleepy during daytime
      O: Observed apnoeas during sleep
      P: Hypertension
      B: BMI more than 35 kg/m2
      A: Age 50-years of age or greater
      N: Neck circumference (male 17 inches / 43cm or greater and female 16 inches / 41 or greater)
      G: Gender: Male

      Our patient has a score of 5 ( O, P, A, N, G)

      The score helps clinicians stratify patients for unrecognized OSA and target appropriate clinical management. It can also help triage patients for further investigation. A STOP-BANG score of 5-8 will identify patients with high probability of moderate to severe OSA in the surgical population.

    • This question is part of the following fields:

      • Clinical Measurement
      361.9
      Seconds
  • Question 13 - A 70-year-old man presents with bilateral buttock claudication that spreads down the thigh...

    Incorrect

    • 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: L2

      Correct Answer: L4

      Explanation:

      The bifurcation of the abdominal aorta into common iliac arteries occurs at the level of L4. The bifurcation is a common site for atherosclerotic plaques as it is an area of high turbulence.

      Leriche Syndrome is an aortoiliac occlusive disease and affects the distal abdominal aorta, iliac arteries, and femoropopliteal vessels. It has a triad of symptoms:
      1. Claudication (cramping lower extremities pain that is reproducible by exercise)
      2. Impotence (reduced penile arterial flow)
      3. Absent/weak femoral pulses (hallmark)

      T12 – aorta enters the diaphragm with the thoracic duct and azygous veins

      L2 – testicular or ovarian arteries branch off the aorta

      L3 – inferior mesenteric artery

    • This question is part of the following fields:

      • Anatomy
      796.2
      Seconds
  • Question 14 - A weakly acidic drug with a pKa of 8.4 is injected intravenously into...

    Incorrect

    • A weakly acidic drug with a pKa of 8.4 is injected intravenously into a patient.

      At a normal physiological pH, the percentage of this drug unionised in the plasma is?

      Your Answer: 99

      Correct Answer: 90

      Explanation:

      Primary FRCA is concerned with two issues. The first is a working knowledge of the Henderson-Hasselbalch equation, and the second is a working knowledge of logarithms and antilogarithms.

      The pH at which the drug exists in 50 percent ionised and 50 percent unionised forms is known as the pKa.

      To calculate the proportion of ionised to unionised form of a drug, use the Henderson-Hasselbalch equation.

      pH = pKa + log ([A-]/[HA])

      or

      pH = pKa + log [(salt)/(acid)]
      pH = pKa + log ([ionised]/[unionised])

      Hence, if the pKa − pH = 0, then 50% of drug is ionised and 50% is unionised.

      In this example:
      7.4 = 8.4 + log ([ionised]/[unionised])
      7.4 − 8.4 = log ([ionised]/[unionised])
      log −1 = log ([ionised]/[unionised])

      Simply put, the antilog is the inverse log calculation. In other words, if you know the logarithm of a number, you can use the antilog to find the value of the number. The antilogarithm’s definition is as follows:

      y = antilog x = 10x

      Antilog to the base 10 of 0 = 1, −1 = 0.1, −2 = 0.01, −3 = 0.001 and, −4 = 0.0001.

      [A-]/[HA] = 0.1

      Assuming that we can apply the approximation [A-] << [HA} then this means the acid is 0.1 x 100% = 10% ionised so the percentage of (non-ionized) acid will be 100% – 10% = 90%

    • This question is part of the following fields:

      • Pharmacology
      68.2
      Seconds
  • Question 15 - A 72-year-old man complains of severe, central abdominal pain that radiates to the...

    Incorrect

    • A 72-year-old man complains of severe, central abdominal pain that radiates to the back. He has a past medical history of an abdominal aortic aneurysm.

      A focused abdominal ultrasonography test (FAST) is performed, revealing diffuse dilatation of the abdominal aorta. The most prominent dilatation is at the bifurcation site of abdominal aorta into the iliac arteries.

      What vertebra level corresponds to the site of the most prominent dilatation as evident on the FAST scan?

      Your Answer: L2

      Correct Answer: L4

      Explanation:

      The important landmarks of vessels arising from the abdominal aorta at different levels of vertebrae are:

      T12 – Coeliac trunk

      L1 – Left renal artery

      L2 – Testicular or ovarian arteries

      L3 – Inferior mesenteric artery

      L4 – Bifurcation of the abdominal aorta

    • This question is part of the following fields:

      • Anatomy
      401
      Seconds
  • Question 16 - 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
      284.9
      Seconds
  • Question 17 - Patient’s having disease (Test Positive: 60, Test Negative:40)


    Patient’s not having...

    Correct

    • Patient’s having disease (Test Positive: 60, Test Negative:40)


      Patient’s not having the disease (Test Positive:20, Test Negative: 80)

      This is a result of a new tumour marker blood test, that was performed on 200 women for breast cancer screening. The director of the screening programme ask you to evaluate the observations and inform them the specificity of this new test.

      Which one of the following figure you will relay to the programme director?

      Your Answer: 80%

      Explanation:

      The positive predictive value is the ratio of patients truly diagnosed as positive to all those who had positive test results. In this case, this is 60/(60+20)=75%.

      The negative predictive value is the ratio of patients truly diagnosed as negative to all those who had negative test results. In this case, this is 80/(80+40)=67%.

      The sensitivity is the ratio of patients with the disease who test positive i.e. true positive patients to the total number of people with the disease. In this case, this is 60/(60+40)=60%.

      The specificity is the ratio of people who don’t have the disease who test negative i.e. true negatives to the total number of people without the disease. In this case, this is 80/(20+80)=80%.

      70% is not the result of any screening measurements

    • This question is part of the following fields:

      • Statistical Methods
      267.1
      Seconds
  • Question 18 - A 58-year-old man, visits his general practitioner complaining of a lump in his...

    Incorrect

    • 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: Inferior and lateral to the pubic tubercle

      Correct 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
      259.6
      Seconds
  • Question 19 - A 68-year-old man presents worried about his risk of motor neurone disease. No...

    Incorrect

    • A 68-year-old man presents worried about his risk of motor neurone disease. No symptoms have developed, but his father suffered from motor neurone disease. Recently, his cousin has also been diagnosed with amyotrophic lateral sclerosis. He searched the internet for screening tests for motor neurone disease and found a blood test called ‘neuron’, and requests to have it done. You search this blood test and find a prospective study going on evaluating the potential benefits of this blood test. On average, this test diagnosed patients with the disease 8 months earlier than the patients who are diagnosed on the basis of their clinical symptoms. The patients diagnosed using this neuron test also survived, on average, 48 months from the diagnosis, whereas the patients diagnosed clinically survived an average of 39 months from the diagnosis. Considering the clear benefits, you decide to have it done on the patient.

      Which of the following options best relate to the above scenario?

      Your Answer: Late-look bias

      Correct Answer: Lead-time bias

      Explanation:

      Hypochondriasis is an illness anxiety disorder, and describes excessively worriedness about the presence of a disease. While the woman is concerned about her possibility of developing motor neurone disease, she understands that no symptoms have yet appeared. Hypochondriasis involves patients who refuse to accept that they don’t have the disease, even if the results come back negative.

      Late Look Bias occurs when the data is gathered or analysed at an inappropriate time e.g. when many of the subjects suffering from a fatal disease have died. This type of biasness might occur in some retrospective studies of motor neurone disease, but is not applicable to this prospective study.

      In procedure bias, the researcher decides assignment of a treatment versus control and assigns particular patients to one group or the other non-randomly. This is unlikely to have occurred in this case, although it is not mentioned specifically. Of all the options, lead time-bias is a better answer.

      The Hawthorne Effect refers to groups modifying their behaviour simply because they are aware of being observed. Any differences in the behaviour have not been mentioned in the question, and it is highly unlikely that a change in patient’s behaviour would have affected their length of survival in this case.

      The correct option is lead-time bias. Even if the new blood test diagnoses the disease earlier, it doesn’t affect the outcome, as the survival time was still on average 43 months from the onset of symptoms in both groups. With the help of blood test, the disease was only detected 8 months earlier.

    • This question is part of the following fields:

      • Statistical Methods
      116.8
      Seconds
  • Question 20 - During a stabbing incident, a 30-year-old injured his inferior vena cava. What number...

    Incorrect

    • During a stabbing incident, a 30-year-old injured his inferior vena cava. What number of functional valves can be usually found in this vessel?

      Your Answer: 3

      Correct Answer: 0

      Explanation:

      The inferior vena cava is formed by the union of the right and left common iliac veins. The inferior vena cava has no functional valves like the one-way valves commonly found in many veins. The forward flow to the heart is driven by the differential pressure created by normal respiration.

      The absence of functional valves has an important clinical role when cannulating during cardiopulmonary bypass.

      There is a valve that is non-functioning called the eustachian valve that lies at the junction of the IVC and the right atrium. This valve has a role to help direct the flow of oxygen-rich blood through the right atrium to the left atrium via the foramen ovale during fetal life. It has no specific function in adult life.

    • This question is part of the following fields:

      • Anatomy
      29.2
      Seconds
  • Question 21 - In medical testing, there are true negative, true positive, false positive and false...

    Incorrect

    • In medical testing, there are true negative, true positive, false positive and false negative results for some test.

      How are the sensitivity of these predictive tests calculated?

      Your Answer: True negatives / (true negatives + false positives)

      Correct Answer: True positives / (true positives + false negatives)

      Explanation:

      The following terms are used in medical testing:

      True negative – The test is negative and the patient does not have the disease.
      True positive – The test is positive and the patient has the disease.
      False positive – The test is positive but the patient does not have the disease.
      False negative – The test is negative but the patient has the disease.

      The sensitivity of a predictive test = true positives / (true positives + false negatives).

      The specificity of a test = true negatives / (false positives + true negatives).

      The negative predictive value of a test = true negatives / (false negatives + true negatives).

    • This question is part of the following fields:

      • Statistical Methods
      19.1
      Seconds
  • Question 22 - A 71-year-old man, presents with central crushing pain in his chest to the...

    Incorrect

    • A 71-year-old man, presents with central crushing pain in his chest to the emergency department. On examination, he complains of nausea and is notably sweating. On ECG, elevation in the ST-segment is noted in multiple chest leads, as well as sinus bradycardia. A myocardial infarction can cause a sinus bradycardia.

      The sinoatrial (SA) node and the atrioventricular (AV) node receive arterial supply from which vessel?

      Your Answer: Left circumflex artery

      Correct Answer: Right coronary artery

      Explanation:

      The left marginal artery comes off the left circumflex artery, and runs alongside the heart.

      The left circumflex artery is one of the bifurcations of the left coronary artery, and eventually forms the left marginal artery.

      An occlusion in the left circumflex artery often results in a lateral MI.

      The right marginal artery originates from the right coronary artery.

      The left anterior descending artery (LAD) is another bifurcation of the left coronary artery. An occlusion in the LAD would often result in an anteroseptal MI as is diagnosed on ECG by noting changes in leads V1-V4.

      The right coronary artery originates from the right aortic sinus of the ascending aorta, and bifurcates to give rise to many branches, including the sinoatrial artery which supplies the sinoatrial (SA) node in 50-70% of cases, the artery of the atrioventricular (AV) node in 50-60% of cases, the right acute marginal artery which supplies the right ventricle. It also supplies the right atrium, interatrial septum and the posterior inferior third of the interventricular septum.

      Arrhythmias and inferior MI often occurs as a result of an occlusion in the right coronary artery, and can be diagnosed by ECG changes in leads II, III and aVF.

    • This question is part of the following fields:

      • Anatomy
      39.7
      Seconds
  • Question 23 - Systemic vascular resistance (multiplied by 80) to produce the units of dynes.s.cm-5 is...

    Incorrect

    • Systemic vascular resistance (multiplied by 80) to produce the units of dynes.s.cm-5 is represented by?

      Your Answer: Mean arterial pressure (MAP) - central venous pressure (CVP)/stroke volume (SV)

      Correct Answer: Mean arterial pressure (MAP) - central venous pressure (CVP)/cardiac output (CO)

      Explanation:

      Systemic vascular resistance (SVR) is a derived value based on:

      SVR = (MAP-CVP)/CO x 80

      = (60 -10)/5 x 80 = 800 dynes.s.cm-5

      A correction factor of 80 is needed in converting mmHg to dynes.s.cm-5
      Normal values is between 700 -1600 dynes.s.cm-5

      Pulmonary resistance (PVR) = (MPAP-PCWP)/CO x 80

      = (10 – 5)/5 x 80 = 80 dynes.s.cm-5

      To account for body size, cardiac index (CI) can be used instead of CO. CI = CO/body surface area (m2) or mL/minute/m2.
      N/B: either MAP and CVP, or MPAP and PCWP are used in calculation to get dynes.s.cm-5

    • This question is part of the following fields:

      • Clinical Measurement
      13.4
      Seconds
  • Question 24 - Which of the following is true regarding a laryngoscope? ...

    Incorrect

    • Which of the following is true regarding a laryngoscope?

      Your Answer: The McCoy laryngoscope is based on the standard Robertshaw's blade

      Correct Answer: The Wisconsin and Seward are examples of straight blade laryngoscopes

      Explanation:

      Direct laryngoscopy are performed using laryngoscopes and they can be classed according to the shape of the blade as curved or straight.

      Miller, Soper, Wisconsin and Seward are examples of straight blade laryngoscopes. Straight blades are commonly used for intubating neonates and infants but can be used in adults too.

      The tip of the miller blade is advanced over the epiglottis to the tracheal entrance then lifted in order to view the vocal cords.

      The RIGHT-SIDED Macintosh blade is used in adults while the left-sided blade may be used in conditions that make intubation with standard blade difficult e.g. facial deformities.

      The McCoy laryngoscope is based on the STANDARD MACINTOSH blade not Robertshaw’s. It has a lever operated hinged tip, which improves the view during laryngoscopy.

      Polio blade is mounted at an angle of 120-135 degrees to the handle. Originally designed for use during the polio epidemic ​in intubation patients within iron lung ventilators, it is now useful in patients with conditions like breast hypertrophy, barrel chest, and restricted neck mobility.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      172.5
      Seconds
  • Question 25 - Drug X, a new intravenous induction drug, is being administered as a bolus...

    Incorrect

    • Drug X, a new intravenous induction drug, is being administered as a bolus at regular time intervals, and the following data were observed:

      Time following injection (hours) vs Plasma concentration of drug X (mcg/mL)
      2 / 400
      6 / 100
      10 / 25
      14 / 6.25

      Which of the following values estimate the plasma half-life (T½) of drug X?

      Your Answer: 8 hours

      Correct Answer: 2 hours

      Explanation:

      Half life (T½) is the time required to change the amount of drug in the body by one-half (or 50%) during elimination. The time course of a drug in the body will depend on both the volume of distribution and the clearance.

      Extrapolating the values from the plasma concentration vs time:

      Plasma concentration at 0 hours = 800 mcg/mL
      Plasma concentration at 2 hours = 400 mcg/mL
      Plasma concentration at 4 hours = 200 mcg/mL
      Plasma concentration at 6 hours = 100 mcg/mL
      Plasma concentration at 8 hours = 50 mcg/mL
      Plasma concentration at 10 hours = 25 mcg/mL
      Plasma concentration at 12 hours = 12.5 mcg/mL
      Plasma concentration at 14 hours = 6.25 mcg/mL

    • This question is part of the following fields:

      • Statistical Methods
      362.1
      Seconds
  • Question 26 - The fluids with the highest osmolarity is? ...

    Incorrect

    • The fluids with the highest osmolarity is?

      Your Answer: Hartmann's solution

      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
      42.1
      Seconds
  • Question 27 - Which of the following facts about IgE is true? ...

    Incorrect

    • Which of the following facts about IgE is true?

      Your Answer: Is involved in type 2 hypersensitivity

      Correct Answer: Is increased in the serum of atopic individuals

      Explanation:

      Immunoglobulin E (IgE) are an antibody subtype produced by the immune system. They are the least abundant type and function in parasitic infections and allergy responses.

      The most predominant type of immunoglobulin is IgG. It is able to be transmitted across the placenta to provide immunity to the foetus.

      IgE is involved in the type I hypersensitivity reaction as it stimulates mast cells to release histamine. It has no role in type 2 hypersensitivity.

      Its concentration in the serum is normally the least abundant, however certain reactions cause a rise in its concentration, such as atopy, but not in acute asthma.

    • This question is part of the following fields:

      • Pathophysiology
      121.7
      Seconds
  • Question 28 - A 25-year-old soldier is shot in the abdomen. He has multiple injuries, including...

    Correct

    • A 25-year-old soldier is shot in the abdomen. He has multiple injuries, including a major disruption to the abdominal aorta. The bleeding is torrential and needs to be controlled by placing a vascular clamp immediately inferior to the diaphragm.

      During this manoeuvre, which vessel may be injured?

      Your Answer: Inferior phrenic arteries

      Explanation:

      The inferior phrenic nerves are at the highest risk of damage as they are the first branches of the abdominal aorta. The potential space at the level of the diaphragmatic hiatus is a potentially useful site for aortic occlusion. However, leaving the clamp applied for more than 10 -15 minutes usually leads to poor outcomes.

      The superior phrenic artery branches from the thoracic aorta.

      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.

      The branches of the abdominal aorta (with their vertebra level) are:
      1. Inferior phrenic arteries: T12 (upper border)
      2. Coeliac artery: T12
      3. Superior mesenteric artery: L1
      4. Middle suprarenal arteries: L1
      5. Renal arteries: Between L1 and L2
      6. Gonadal arteries: L2 (in males, it is the testicular artery, and in females, the ovarian artery)
      7. Inferior mesenteric artery: L3
      8. Median sacral artery: L4
      9. Lumbar arteries: Between L1 and L4

    • This question is part of the following fields:

      • Anatomy
      301.3
      Seconds
  • Question 29 - The tissue layer in a patient is infiltrated with local anaesthetic (marcaine 0.125%)...

    Incorrect

    • The tissue layer in a patient is infiltrated with local anaesthetic (marcaine 0.125%) with 1 in 120,000 adrenaline as part of an enhanced recovery programme for primary hip replacement surgery. The total volume of solution is 120mL.

      What is the appropriate combination of constituents in the final solution?

      Your Answer: 10mL 0.5% bupivacaine, 1mL 1 in 1,000 adrenaline and 109mL 0.9% N. Saline

      Correct Answer: 30mL 0.5% bupivacaine, 1mL 1 in 1,000 adrenaline and 89mL 0.9% N. Saline

      Explanation:

      30mL 0.5% bupivacaine, 1mL 1 in 1,000 adrenaline and 89mL 0.9% N. Saline is the correct answer.
      Initial concentration of bupivacaine is 0.5% with a volume of 30mL

      The volume is doubled (60mL) by the addition of 0.9% N. saline (30mls) and the concentration of bupivacaine is halved to (0.25%).

      If the volume is doubled again (120mL) by the addition of further 0.9% N. saline (59mls) the final concentration of bupivacaine is halved again to 0.125%. Total N. saline = 89mls

      The 1 mL of 1 in 1000 adrenaline has also been diluted into the final volume of 120 mL making it a 1 in 120000 concentration.

    • This question is part of the following fields:

      • Pharmacology
      62.9
      Seconds
  • Question 30 - A 53-year old female with a diagnosis of anaplastic thyroid carcinoma is admitted...

    Incorrect

    • A 53-year old female with a diagnosis of anaplastic thyroid carcinoma is admitted in the surgery department for an elective total thyroidectomy with radical neck dissection. The operation is expected to last for 10 hours.

      Which of the following is the most suitable humidifier to use in an anaesthetic circuit for this case?

      Your Answer:

      Correct Answer: Heat and moisture exchanger (HME)

      Explanation:

      Adequate humidification is vital to maintain homeostasis of the airway. Heat and moisture exchangers conserve some of the exhaled water, heat and return them to inspired gases. Many heat and moisture exchangers also perform bacterial/viral filtration and prevent inhalation of small particles. Heat and moisture exchangers are also called condenser humidifier, artificial nose, etc. Most of them are disposable devices with exchanging medium enclosed in a plastic housing. For adult and paediatric age group different dead space types are available. Heat and moisture exchangers are helpful during anaesthesia and ventilatory breathing system. To reduce the damage of the upper respiratory tract through cooling and dehydration inspiratory air can be heated and humidified, thus preventing the serious complications. Moreover, they are the most appropriate humidification devices used for routine anaesthesia.

      Gases can be bubbled through water to increase humidity. Passing gas through water at room temperature causes the gas to cool due to latent heat of vaporisation. The water bath can be heated. This improves the efficiency of the device and also reduces the incidence of bacterial colonisation.

      Nebulisers use a venturi system which employs the Bernoulli effect. A gas at high flow passes through a constriction causing the gas to accelerate, reducing its potential energy allowing other gases or liquids to be entrained. This can include medications or in the case of humidification, water vapour. The size of the water droplet produced by nebulisation determines where in the airway it is deposited. Standard nebulisers produced droplets of 4 microns in diameter and these are deposited in the upper airway and trachea. Efficacy can be improved by passing the droplets over an anvil which further reduces particle size. The most efficient form of nebuliser is the ultrasonic nebuliser. Here a transducer immersed in water and vibrated at a frequency of 3MHz produces1-2micron droplets. These particles easily reach the bronchioles and provide excellent humidification.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
      0
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Pathophysiology (1/2) 50%
Anatomy (6/10) 60%
Pharmacology (2/3) 67%
Physiology (1/3) 33%
Basic Physics (0/1) 0%
Anaesthesia Related Apparatus (1/3) 33%
Physiology And Biochemistry (0/1) 0%
Clinical Measurement (0/2) 0%
Statistical Methods (2/4) 50%
Passmed