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  • Question 1 - A 39-year old man came to the Out-Patient department for symptoms of gastroesophageal...

    Incorrect

    • A 39-year old man came to the Out-Patient department for symptoms of gastroesophageal reflux disease. Medical history revealed he is on anti-epileptic medication Phenytoin. His plasma phenytoin levels are maintained between 10-12 mcg/mL (Therapeutic range: 10-20 mcg/mL). He is given a H2 antagonist receptor agent (Cimetidine) for his GERD symptoms.

      Upon follow-up, his plasma phenytoin levels increased to 38 mcg/mL.

      Regarding metabolism and elimination, which of the following best explains the pharmacokinetics of phenytoin at higher plasma levels?

      Your Answer: Elimination rate is proportional to the plasma concentration

      Correct Answer: Plasma concentration plotted against time is linear

      Explanation:

      Drug elimination is the termination of drug action, and may involve metabolism into inactive state and excretion out of the body. Duration of drug action is determined by the dose administered and the rate of elimination following the last dose.

      There are two types of elimination: first-order and zero-order elimination.

      In first-order elimination, the rate of elimination is proportionate to the concentration; the concentration decreases exponentially over time. It observes the characteristic half-life elimination, where the concentration decreases by 50% for every half-life.

      In zero-order elimination, the rate of elimination is constant regardless of concentration; the concentration decreases linearly over time. A constant amount of the drug being excreted over time, and it occurs when drugs have saturated their elimination mechanisms.

      Since phenytoin is observed in elevated levels, the elimination mechanisms for it has been saturated and, thus, will have to undergo zero-order elimination.

    • This question is part of the following fields:

      • Pharmacology
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  • Question 2 - A patient's ECG is abnormal, with an abnormal broad complex QRS complexes. This...

    Incorrect

    • A patient's ECG is abnormal, with an abnormal broad complex QRS complexes. This means either a ventricular origin problem or aberrant conduction. The normal resting membrane potential of the heart's ventricular contractile fibres is which of the following?

      Your Answer:

      Correct Answer: -90mV

      Explanation:

      The cardiac muscle’s contractile fibres have a much more stable resting potential than its conductive fibres. In the ventricular fibres it is -90mV and in the atrial fibres it is -80mV.

      The cardiac action potential has several phases which have different mechanisms of action as seen below:

      Phase 0: Rapid depolarisation – caused by a rapid sodium influx.
      These channels automatically deactivate after a few ms. (QRS complex)

      Phase 1: caused by early repolarisation and an efflux of potassium.

      Phase 2: Plateau – caused by a slow influx of calcium.

      Phase 3 – Final repolarisation – caused by an efflux of potassium.

      Phase 4 – Restoration of ionic concentrations – The resting potential is restored by Na+/K+ATPase.
      There is slow entry of Na+into the cell which decreases the potential difference until the threshold potential is reached. This then triggers a new action potential

      Of note, cardiac muscle remains contracted 10-15 times longer than skeletal muscle.

      Different sites have different conduction velocities:
      1. Atrial conduction – Spreads along ordinary atrial myocardial fibres at 1 m/sec

      2. AV node conduction – 0.05 m/sec

      3. Ventricular conduction – Purkinje fibres are of large diameter and achieve velocities of 2-4 m/sec, the fastest conduction in the heart. This allows a rapid and coordinated contraction of the ventricles

    • This question is part of the following fields:

      • Physiology And Biochemistry
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  • Question 3 - A pre-operative evaluation for a trans-sphenoidal pituitary adenectomy is being performed on a...

    Incorrect

    • A pre-operative evaluation for a trans-sphenoidal pituitary adenectomy is being performed on a 57-year-old woman. Her vision is causing her problems.

      A macroadenoma compressing the optic chiasm is visible on MRI.

      What is the most likely visual field defect to be discovered during an examination?

      Your Answer:

      Correct Answer: Bitemporal hemianopia

      Explanation:

      The pituitary gland plays a crucial role in the neuro-endocrine axis. It is located at the base of the skull in the sella turcica of the sphenoid bone. It is connected superiorly to the hypothalamus, third ventricle, and visual pathways, and laterally to the cavernous sinuses, internal carotid arteries, and cranial nerves III, IV, V, and VI.

      Pituitary tumours make up about 10-15% of all intracranial tumours. The majority of adenomas are benign. Over-secretion of pituitary hormones (most commonly prolactin, growth hormone, or ACTH), under-secretion of hormones, or localised or generalised pressure effects can all cause symptoms.

      Compression of the optic chiasm can result in visual field defects, the most common of which is bitemporal hemianopia. This is caused by compression of the nasal retinal fibres, which carry visual impulses from temporal vision across the optic chiasm to the contralateral sides before continuing to the optic tracts.

      The interruption of the visual pathways distal to the optic chiasm causes a homonymous visual field defect. The loss of the right or left halves of each eye’s visual field is referred to as homonymous hemianopia. It’s usually caused by a middle or posterior cerebral artery territory stroke that affects the occipital lobe’s optic radiation or visual cortex.

      Binasal hemianopia is a condition in which vision is lost in the inner half of both eyes (nasal or medial). It’s caused by compression of the temporal visual pathways, which don’t cross at the optic chiasm and instead continue to the ipsilateral optic tracts. Binasal hemianopia is a rare complication caused by the internal carotid artery impinging on the temporal (lateral) visual fibres.

      A monocular visual loss (that is, loss of vision in only one eye) can be caused by a variety of factors, but if caused by nerve damage, the damage would be proximal to the optic chiasm on the ipsilateral side.

      A central scotoma is another name for central visual field loss. Every normal mammalian eye has a scotoma, also known as a blind spot, in its field of vision. The optic disc is a region of the retina that lacks photoreceptor cells and is where the retinal ganglion cell axons that make up the optic nerve exit the retina. When both eyes are open, visual signals that are absent in one eye’s blind spot are provided for the other eye by the opposite visual cortex, even if the other eye is closed.

      Scotomata can be caused by a variety of factors, including demyelinating disease such as multiple sclerosis, damage to nerve fibre layer in the retina, methyl alcohol, ethambutol, quinine, nutritional deficiencies, and vascular blockages either in the retina or in the optic nerve.

      Bilateral scotoma can occur when a pituitary tumour compresses the optic chiasm, causing a bitemporal paracentral scotoma, which then spreads out to the periphery, causing bitemporal hemianopsia. A central scotoma in a pregnant woman could be a sign of severe pre-eclampsia.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 4 - A 66-year-old man, present to the emergency department with dyspepsia. On history taking,...

    Incorrect

    • A 66-year-old man, present to the emergency department with dyspepsia. On history taking, he admits to being a heavy smoker, and on testing is noted to be positive for a helicobacter pylori infection. A few evenings later, he suffers from haematemesis and collapses.

      What vessel is most likely to be involved?

      Your Answer:

      Correct Answer: Gastroduodenal artery

      Explanation:

      The most likely of the differential diagnosis in this case is a duodenal ulcer located on the posterior abdominal wall.

      These can cause an erosion of the abdominal wall, eventually affecting the gastroduodenal artery and resulting in major bleeding and haematemesis.

      Gastroduodenal artery supplies the pylorus, proximal part of the duodenum, and indirectly to the pancreatic head (via the anterior and posterior superior pancreaticoduodenal arteries)

    • This question is part of the following fields:

      • Anatomy
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  • Question 5 - With a cervical dilation of 7 cm, a 33-year-old term primigravida is in...

    Incorrect

    • With a cervical dilation of 7 cm, a 33-year-old term primigravida is in labour. She is otherwise in good health. She's been in labour for 14 hours and counting.

      The cardiotocograph shows late foetal pulse decelerations, and a pH of 7.24 was found in the recent foetal scalp blood sample.

      Which of the following is true about this patient's care and management?

      Your Answer:

      Correct Answer: Monitor for downward trend in fetal scalp blood pH as caesarean section is not indicated at the present time

      Explanation:

      Once the decision to deliver a baby by caesarean section has been made, it should be carried out with a level of urgency commensurate with the baby’s risk and the mother’s safety.

      There are four types of caesarean section urgency:

      Category 1: A threat to the life of the mother or the foetus. 30 minutes to make a delivery decision
      Category 2 : Maternal or foetal compromise that is not immediately life threatening. In most cases, the decision to deliver is made within 75 minutes.
      Category 3 – Early delivery is required, but there is no risk to the mother or the foetus.
      Category 4: Elective delivery at a time that is convenient for both the mother and the maternity staff.

      There may be evidence of foetal compromise in the example above (late foetal pulse decelerations and a borderline pH).

      Blood samples from the foetus:
      normal: 7.25 or above
      borderline: 7.21 to 7.24
      abnormal: 7.20 or below

      When a foetal deceleration occurs, the mother should be given oxygen, kept in a left lateral position, and given a tocolytic if the foetal deceleration is hyper stimulating. Maintaining adequate hydration will reduce the likelihood of a caesarean section.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 6 - For a rapid sequence induction of anaesthesia, you are pre-oxygenating a patient using...

    Incorrect

    • For a rapid sequence induction of anaesthesia, you are pre-oxygenating a patient using 100% oxygen and a fresh gas flow equal to the patient's minute ventilation.

      Which would be the most suitable choice of anaesthetic breathing system in this situation?

      Your Answer:

      Correct Answer: Mapleson A system

      Explanation:

      The Mapleson A (Magill) and coaxial version of the Mapleson A system (Lack circuit) are more efficient for spontaneous breathing than any of the other Mapleson circuits. The fresh gas flow (FGF) required to prevent rebreathing is slightly greater than the alveolar minute ventilation (4-5 litres/minute). This is delivered to the patient through the outer coaxial tube and exhaust gases are moved to the scavenging system through the inner tube. In the Lack circuit, the expiratory valve is located close to the common gas outlet away from the patient end. This is the main advantage of the Lack circuit over the Mapleson A circuit.

      The Mapleson E circuit is a modification of the Ayres T piece and the FGF required to prevent rebreathing is 1.5-2 times the patient’s minute volume.

      The Bain circuit is the coaxial version of the Mapleson D circuit.

      The FGF for spontaneous respiration to avoid rebreathing is 160-200 ml/kg/minute.

      The FGF for controlled ventilation to avoid rebreathing is 70-100 ml/kg/min.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
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  • Question 7 - General anaesthesia is administered to a patient in a hospital in Lhasa which...

    Incorrect

    • General anaesthesia is administered to a patient in a hospital in Lhasa which is one of the highest cities in the world (at 11,975 feet). An Anaesthetic rotameter is normally calibrated at 20 C and 1 bar pressure and is known to be underread at altitude. The temperature of the theatre was 10 C.

      Which one of the following physical properties is responsible for the rotameter inaccuracy in these conditions?

      Your Answer:

      Correct Answer: Density of the gas

      Explanation:

      Since the gas is less dense at higher altitudes, the density of a gas influences flows when passing through the orifice. Due to this reason, for a given flow rate, the bobbin will not be forced as far up the rotameter tube.

      At higher altitudes, the volume of a fixed mass of gas increases, and therefore the molecules of gas are widely spaced resulting in a decrease in density with an increase in altitude.

      Viscosity is simply termed as friction of gas. The viscosity of a gas is important only at low flow rates when the flow characteristic of the gas is laminar.

      Charle’s law stated that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature (T) provided the pressure remains constant.

      Boyle’s law for a fixed amount of gas at constant temperature, the pressure (P) and volume (V) are inversely proportional.

    • This question is part of the following fields:

      • Basic Physics
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  • Question 8 - Regarding renal autoregulation, which of the following best describes its process? ...

    Incorrect

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

      Your Answer:

      Correct 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
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  • Question 9 - 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:

      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
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  • Question 10 - An 82-year old male has shortness of breath which is made worse when...

    Incorrect

    • An 82-year old male has shortness of breath which is made worse when he lies down but investigations have revealed a normal ejection fraction. Why might this be?

      Your Answer:

      Correct Answer: He has diastolic dysfunction

      Explanation:

      Decreased stroke volume causes decreased ejection fraction which results in diastolic dysfunction.
      Ejection fraction is not a useful measure in someone with diastolic dysfunction because stroke volume may be reduced whilst end-diastolic volume may be reduced.
      Diastolic dysfunction may arise with reduced heart compliance.

      Ejection fraction measures of the proportion of blood leaving the ventricles with each beat and is calculated as follows:
      Stroke volume / end-diastolic volume.

      A healthy ejection fraction is usually taken as 60% (based on a stroke volume of 70ml and end-diastolic volume of 120ml).

      Respiratory inspiration causes a decreased pressure in the thoracic cavity, which in turn causes more blood to flow into the atrium.

      Sitting up decreases venous because of the action of gravity on blood in the venous system.
      Hypotension also decreases venous return.
      A less compliant aorta, like in aortic stenosis increases end systolic left ventricular volume which decreases stroke volume.

      Systemic vascular resistance = mean arterial pressure / cardiac output.
      Increased vascular resistance impedes the flow of blood back to the heart.

      Increased venous return increases end diastolic LV volume as there is more blood returning to the ventricles.

    • This question is part of the following fields:

      • Physiology And Biochemistry
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  • Question 11 - Which statement is the most accurate when describing electrical equipment and shock? ...

    Incorrect

    • Which statement is the most accurate when describing electrical equipment and shock?

      Your Answer:

      Correct Answer: Type CF is considered to safe for direct connection with the heart

      Explanation:

      There are different classes of electrical equipment that can be classified in the table below:

      Class 1 – provides basic protection only. It must be connected to earth and insulated from the mains supply

      Class II – provides double insulation for all equipment. It does not require an earth.

      Class III – uses safety extra low voltage (SELV) which does not exceed 24 V AC. There is no risk of gross electrocution but risk of microshock exists.

      Type B – All of above with low leakage currents (0.5mA for Class IB, 0.1 mA for Class IIB)

      Type BF – Same as with other equipment but has ‘floating circuit’ which means that the equipment applied to patient is isolated from all its other parts.

      Type CF – Class I or II equipment with ‘floating circuits’ that is considered to be safe for direct connection with the heart. There are extremely low leakage currents (0.05mA for Class I CF and 0.01mA for Class II CF)

    • This question is part of the following fields:

      • Clinical Measurement
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  • Question 12 - Concerning the trachea, which of these is true? ...

    Incorrect

    • Concerning the trachea, which of these is true?

      Your Answer:

      Correct Answer: In an adult is approximately 15 cm long

      Explanation:

      In an adult, the trachea is approximately 15 cm long. It extends at the level of the 6th cervical vertebra, from the lower border of the cricoid cartilage.

      The trachea terminates between T4 and T6 at the carina or bronchial bifurcation. This variation is because of changes during respiration.

      The trachea has 16-20 C-shaped cartilaginous rings that maintain its patency.

      The trachea is first of the 23 generations of air passages in the tracheobronchial tree (not 25), from the trachea to the alveoli..

      The inferior thyroid arteries which are branches of the thyrocervical trunk, arise from the first part of the subclavian artery and supplies the trachea.

    • This question is part of the following fields:

      • Anatomy
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  • Question 13 - Which of the given statements is true about standard error of the mean?...

    Incorrect

    • Which of the given statements is true about standard error of the mean?

      Your Answer:

      Correct Answer: Gets smaller as the sample size increases

      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

      Therefore, the SEM gets smaller as the sample size (n) increases.

      If we want to depict how widely scattered some measurements are, we use the standard deviation. For indicating the uncertainty around the estimate of the mean, we use the standard error of the mean. The standard error is most useful as a means of calculating a confidence interval. For a large sample, a 95% confidence interval is obtained as the values 1.96×SE either side of the mean.

      A 95% confidence interval:

      lower limit = mean – (1.96 * SEM)

      upper limit = mean + (1.96 * SEM)

      Results such as mean value are often presented along with a confidence interval. For example, in a study the mean height in a sample taken from a population is 183cm. You know that the standard error (SE) (the standard deviation of the mean) is 2cm. This gives a 95% confidence interval of 179-187cm (+/- 2 SE).

      Hence, it would be wrong to say that confidence levels do not apply to standard error of the mean.

    • This question is part of the following fields:

      • Statistical Methods
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  • Question 14 - Which among the following is summed up by F statistic? ...

    Incorrect

    • Which among the following is summed up by F statistic?

      Your Answer:

      Correct Answer: ANOVA

      Explanation:

      ANOVA is based upon within group variance (i.e. the variance of the mean of a sample) and between group variance (i.e. the variance between means of different samples). The test works by finding out the ratio of the two variances mentioned above. (Commonly known as F statistic).

    • This question is part of the following fields:

      • Statistical Methods
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  • Question 15 - 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:

      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
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  • Question 16 - Which measure of central tendency is most useful for a continuous, non-skewed data?...

    Incorrect

    • Which measure of central tendency is most useful for a continuous, non-skewed data?

      Your Answer:

      Correct Answer: Mean

      Explanation:

      Mean, also known as the average, is the most common measure of central tendency. It is the sum of all observed values divided by the number of observation. It is not useful for skewed data, which has an abnormal distribution. It is useful, instead, for numerical data that have symmetric distribution. It reflects the contributions of each data in the group, and are sensitive to outliers.

      The median is the value that falls in the middle position when the observations are ranked in order from the smallest to the largest. If the number of observations is odd, the median is the middle number. If it is even, the median is the average of the two middle numbers. Unlike the mean, the median is useful on skewed data, and can be used for ordinal or numerical data if skewed.

      The mode is the value that occurs with the greatest frequency in a set of observations, and is utilized for bimodal distribution.

      The variance and the standard deviation are not measures of central tendency, but of dispersion.

    • This question is part of the following fields:

      • Statistical Methods
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  • Question 17 - Over the course of 10 minutes, one litre of 0.9% normal saline is...

    Incorrect

    • Over the course of 10 minutes, one litre of 0.9% normal saline is intravenously infused into a normally fit and well 58-year-old male. A catheter is used to measure urine output before and after the infusion. The patient is 70 kg in weight.

      The following data on urine output is obtained:

      50ml/hour Before the infusion
      200 ml/hour 1 hour following infusion
      90 ml/hour 2 hours after the infusion
      60 ml/hr 3 hours after the infusion

      Which of the following physiological responses is most likely to account for the sudden increase in urine output after a fluid bolus?

      Your Answer:

      Correct Answer: Increased glomerular filtration rate

      Explanation:

      The following are some basic assumptions:

      Extracellular fluid (ECF) makes up one-third of total body water (TBW), while intracellular fluid makes up the other two-thirds (ICF).
      One-quarter of ECF is plasma, and three-quarters is interstitial fluid (ISF).
      The volume receptors have a 7-10% blood volume change threshold. The osmoreceptors are sensitive to changes in osmolality of 1-2 percent.
      Prior to the transfusion, the plasma osmolality is normal (between 287 and 290 mOsm/kg).
      [Na+] in 0.9 percent N. saline is 154 mmol/L, which is similar to that of extracellular fluid. When given intravenously, this limits its distribution within the extracellular space, resulting in a plasma compartment:ISF volume ratio of 1:3.
      In this time frame, one litre of 0.9 percent N. saline will increase plasma volume by about 250 mL, which could be the threshold for activation of the volume receptors in the atria, resulting in the release of atrial natriuretic peptide (ANP).

      Because 0.9 percent N. saline is isosmotic, after a 1 L infusion, plasma osmolality will not change. No changes in antidiuretic hormone secretion will be detected by the hypothalamic osmoreceptors.

      Because normal saline is protein-free, the oncotic pressure in the blood is slightly reduced after the saline infusion. As a result, fluid movement into the ISF is favoured (Starling’s hypothesis), and the lower oncotic pressure causes an immediate increase in the glomerular filtration rate (GFR) and a reduction in water reabsorption in the proximal tubule.

      The flow of urine increases. There is no hormonal intermediary in this effect, so it is strictly local. Urine flow immediately increases. The fluid returns to the intravascular compartment, and urine flow continues until all of the transfused fluid has been excreted.

      Blood pressure changes associated with a 1 L fluid infusion are unlikely to affect high-pressure baroreceptors in the carotid sinus.

      The juxta-glomerular cells of the afferent arteriole are adjacent to the specialised cells (macula densa) of distal tubules. The sodium and chloride ions in the tubular fluid are detected by the macula densa. Renin release is inhibited when the tubular fluid contains too much sodium chloride. Hormonal changes take longer to manifest than physical changes that control glomerulotubular balance.
      Hypertonic saline, not 0.9 percent N saline, is an osmotic diuretic.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 18 - A 45-year old gentleman is in the operating room to have a knee...

    Incorrect

    • A 45-year old gentleman is in the operating room to have a knee arthroscopy under general anaesthesia.

      Induction is done using fentanyl 1mcg/kg and propofol 2mg/kg. A supraglottic airway is inserted and the mixture used to maintain anaesthesia is and air oxygen mixture and 2.5% sevoflurane. Using a Bain circuit, the patient breathes spontaneously and the fresh gas flow is 9L/min. Over the next 30 minutes, the end-tidal CO2 increase from 4.5kPa to 8.4kPa, and the baseline reading on the capnograph is 0kPa.

      The most appropriate action that should follow is:

      Your Answer:

      Correct Answer: Observe the patient for further change

      Explanation:

      Such a high rise of end-tidal CO2 (EtCO2) in a patient who is spontaneously breathing is often encountered.

      Close observation should occur for further rises in EtCO2 and other signs of malignant hyperthermia. If this were to rise even more, it might be wise to ensure that ventilatory support is available.

      A lot would depend on whether surgery was almost completed. At this stage of anaesthesia, it would be inappropriate to administer opioid antagonists or respiratory stimulants.

    • This question is part of the following fields:

      • Physiology
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  • Question 19 - Intracellular effectors are activated by receptors on the cell surface. These receptors receive...

    Incorrect

    • Intracellular effectors are activated by receptors on the cell surface. These receptors receive signals that are relayed by second messenger systems.

      In the human body, which second messenger is most abundant?

      Your Answer:

      Correct Answer: Calcium ions

      Explanation:

      Second messengers relay signals to target molecules in the cytoplasm or nucleus when an agonist interacts with a receptor on the cell surface. They also amplify the strength of the signal. The most ubiquitous and abundant second messenger is calcium and it regulates multiple cellular functions in the body.

      These include:
      Muscle contraction (skeletal, smooth and cardiac)
      Exocytosis (neurotransmitter release at synapses and insulin secretion)
      Apoptosis
      Cell adhesion to the extracellular matrix
      Lymphocyte activation
      Biochemical changes mediated by protein kinase C.

      cAMP is either inhibited or stimulated by G proteins.

      The receptors in the body that stimulate G proteins and increase cAMP include:

      Beta (?1, ?2, and ?3)
      Dopamine (D1 and D5)
      Histamine (H2)
      Glucagon
      Vasopressin (V2).

      The second messenger for the action of nitric oxide (NO) and atrial natriuretic peptide (ANP) is cGMP.

      The second messengers for angiotensin and thyroid stimulating hormone are inositol triphosphate (IP3) and diacylglycerol (DAG).

    • This question is part of the following fields:

      • Physiology
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  • Question 20 - A 56-year old man, presents to emergency department following a cardiac arrest. On...

    Incorrect

    • A 56-year old man, presents to emergency department following a cardiac arrest. On history and examination, he is found to be suffering from both metabolic and respiratory acidosis as a result of his cardiac arrest.

      What is the best way to reduce the risk of acidaemia during cardiac arrest

      Your Answer:

      Correct Answer: Chest compressions

      Explanation:

      Chest compressions are an essential part of cardiopulmonary resuscitation (CPR) which helps restore spontaneous circulation (ROSC).

      Sodium bicarbonate is only prescribed in patients with cardiac arrests as a result of an overdose of tricyclic antidepressants or hyperkalaemia. Its use causes the body to produce more CO2 which causes:

      Exacerbation of intracellular acidosis
      Negative inotropy to ischaemic myocardium
      Increased osmotic load of sodium into failing brain and body
      Shift of oxygen dissociation curve to the left.

      THAM is often used to treat metabolic acidosis as a result of cardiac bypass surgery and also cardiac arrest, when other standard methods have failed.

      Carbicarb (Na2CO3 0.33 molar NaHCO3 0.33 molar) has only mild effects on acidosis. It also causes an increase in arterial CO2 pressure and lactate concentration.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 21 - A 20-year old lady has been having excessive bruising and bleeding of her...

    Incorrect

    • A 20-year old lady has been having excessive bruising and bleeding of her gums. She is under investigation for the extrinsic pathway of coagulation. Which is the best investigation to order?

      Your Answer:

      Correct Answer: Prothrombin time (PT)

      Explanation:

      The extrinsic pathway is best assessed by the PT time.

      D-dimer is a fibrin degradation product which is raised in the presence of blood clots.

      A 50:50 mixing study is used to assess if a prolonged PT or aPTT is due to factor deficiency or a factor inhibitor.

      The thrombin time is a test used to assess fibrin formation from fibrinogen in plasma. Factors that prolong the thrombin time include heparin, fibrin degradation products, and fibrinogen deficiency.

      Intrinsic pathway – Best assessed by APTT. Factors 8,9,11,12 are involved. Prolonged aPTT can be seen in haemophilia and use of heparin.

      Extrinsic pathway – Best assessed by Increased PT. Factor 7 involved.

      Common pathway – Best assessed by APTT & PT. Factors 2,5,10 involved.

      Vitamin K dependent factors are factors 2,7,9,10

    • This question is part of the following fields:

      • Physiology And Biochemistry
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  • Question 22 - Which of the following facts about IgE is true? ...

    Incorrect

    • Which of the following facts about IgE is true?

      Your Answer:

      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
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  • Question 23 - The ED95 of muscle relaxants is the dose required to reduce twitch height...

    Incorrect

    • 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:

      Correct 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
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  • Question 24 - 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:

      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
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  • Question 25 - A 32-year-old man has multiple stab wounds to his abdomen and is rushed...

    Incorrect

    • A 32-year-old man has multiple stab wounds to his abdomen and is rushed into the emergency. Resuscitative measures are performed, but the patient remains hypotensive.

      Emergency laparotomy is performed, and it reveals a vessel is bleeding profusely at a certain level of lumbar vertebrae. The vessel is the testicular artery and is ligated.

      At which lumbar vertebrae is the testicular artery identified?

      Your Answer:

      Correct Answer: L2

      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
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  • Question 26 - The thebesian veins contribute to the venous drainage of the heart. Into which...

    Incorrect

    • The thebesian veins contribute to the venous drainage of the heart. Into which of the following structures do they primarily drain?

      Your Answer:

      Correct Answer: Atrium

      Explanation:

      The heart has two venous drainage systems:
      1. Greater venous system – it parallels the coronary arterial circulation and provides 70% venous drainage to the heart
      2. Lesser venous system – includes the thebasian veins and provides up to 30% of the venous drainage to the heart

      Thebasian veins (also called venae cordis minimae) are the smallest coronary veins and run in the myocardial layer of the heart. They serve to drain the myocardium and are present in all four heart chambers. They are more abundant on the right side of the heart and, more specifically, are most abundant in the right atrium. Thebesian veins drain the subendocardial myocardium either directly, via connecting intramural arteries and veins, or indirectly, via subendocardial sinusoidal spaces.

    • This question is part of the following fields:

      • Anatomy
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  • Question 27 - Regarding the carbon dioxide monitoring, which of the following statements is correct? ...

    Incorrect

    • Regarding the carbon dioxide monitoring, which of the following statements is correct?

      Your Answer:

      Correct Answer: Carbon dioxide absorbs infrared radiation at 4.28 µm

      Explanation:

      Carbon dioxide (CO2), is a carbonic gas made up of two dissimilar atoms, namely one carbon atom and two oxygen atoms. Capnography is a technique used to measure carbon dioxide during a respiratory cycle, and it consists in calculating the concentration of the partial pressure of CO2, through the absorption of the infrared light, namely that CO2 absorbs infrared radiation at a wavelength of 4.28 µm.

      End-tidal CO2 (ETCO2), referring to the level of the carbon dioxide released at the end of an exhaled breath, is required to be continuously monitored, especially in ventilated patients, as it is a sensitive and a non invasive technique that provides immediate information about ventilation, circulation, and metabolism functions. ETCO2 is normally lower than the arterial partial pressure and varies between 0.6 and 0.7 kPa.

      There are two methods used to measure carbon dioxide. The sidestream capnometer method samples gases at a set flow rate (150-200 mL/min) from a sampling area through small diameter tubing, and the mainstream analyser method that uses a direct measurement of the patient exhaled CO2 by a relatively large and heavy sensors. Sidestram method allows the analysis of multiple gases and anaesthetic vapours comparing to the mainstream method that does not allow the measurement of other gases.

    • This question is part of the following fields:

      • Anaesthesia Related Apparatus
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  • Question 28 - What is the most sensitive method of detecting an intra-operative air embolism? ...

    Incorrect

    • What is the most sensitive method of detecting an intra-operative air embolism?

      Your Answer:

      Correct Answer: Transoesophageal echocardiogram

      Explanation:

      An intra-operative air embolism occurs when air becomes trapped in the blood vessels during surgery.

      A transoesophageal echocardiography (OE) uses invasive echocardiography to monitor the integrity and performance of the heart. It is the gold standard as it provides real-time imaging of the heart to enable early diagnosis and treatment.

      Precordial doppler ultrasonography can also be used to detect into-operative air emboli. It is non-invasive and more practical, but is less sensitive.

      A change in end-tidal CO2 could be indicative of and increase in physiological dead-space, but could also be indicative of any processes that reduces the excretion or increases the production of CO2, making it non-specific.

      A transoesophageal stethoscope can be used to listen for the classic mill-wheel murmur produced by a large air embolus.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 29 - The statement that best describes temperature management is: ...

    Incorrect

    • The statement that best describes temperature management is:

      Your Answer:

      Correct Answer: Gauge thermometers use coils of different metals with different co-efficients of expansion which either tighten or relax with changes in temperature

      Explanation:

      There are different types of temperature measurement. These include:

      Thermistor – this is a type of semiconductor, meaning they have greater resistance than conducting materials, but lower resistance than insulating materials. There are small beads of semiconductor material (e.g. metal oxide) which are incorporated into a Wheatstone bridge circuit. As the temperature increases, the resistance of the bead decreases exponentially

      Thermocouple – Two different metals make up a thermocouple. Generally, in the form of two wires twisted, welded, or crimped together. Temperature is sensed by measuring the voltage. A potential difference is created that is proportional to the temperature at the junction (Seebeck effect)

      Platinum resistance thermometers (PTR) – uses platinum for determining the temperature. The principle used is that the resistance of platinum changes with the change of temperature. The thermometer measures the temperature over the range of 200°C to1200°C. Resistance in metals show a linear increase with temperature

      Tympanic thermometers – uses infrared radiation which is emitted by all living beings. It analyses the intensity and wavelength and then transduces the heat energy into a measurable electrical output

      Gauge/dial thermometers – Uses coils of different metals with different co-efficient of expansion. These either tighten or relax with changes in temperature, moving a lever on a calibrated dial.

    • This question is part of the following fields:

      • Clinical Measurement
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  • Question 30 - 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:

      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
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