00
Correct
00
Incorrect
00 : 00 : 00
Session Time
00 : 00
Average Question Time ( Secs)
  • Question 1 - The fluids with the highest osmolarity is? ...

    Incorrect

    • The fluids with the highest osmolarity is?

      Your Answer: 0.9% N. Saline

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

      Explanation:

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

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

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

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

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

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

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

    • This question is part of the following fields:

      • Physiology
      16.1
      Seconds
  • Question 2 - A patient was brought to the emergency room after passing black tarry stools....

    Incorrect

    • A patient was brought to the emergency room after passing black tarry stools. The initial diagnosis was upper gastrointestinal bleeding. The patient was placed on temporary nil per os (NPO) for the next 24 hours, his weight was 110 kg, and the required volume of intravenous fluid for the him was 3 litres. His electrolytes and other biochemistry studies were normal.

      If you were to choose the intravenous fluid regimen that would closely mimic his basic electrolyte and caloric requirements, which one would be the best answer?

      Your Answer: 3000 mL Hartmann's

      Correct Answer: 3000 mL 0.45% N. saline with 5% dextrose, each bag with 40 mmol of potassium

      Explanation:

      The patient in the case has a fluid volume requirement of 30 mL/kg/day. His basic electrolyte requirement per day is:

      Sodium at 2 mmol/kg/day x 110 = 220 mmol/day
      Potassium at 1 mmol/kg/day x 110 = 110 mmol/day

      His energy requirement per day is:

      35 kcal/kg/day x 110 kg = 3850 kcal/day

      One gram of glucose in fluid can provide approximately 4 kilocalories.

      The following are the electrolyte components of the different intravenous fluids:

      Fluid Na (mmol/L) K (mmol/L)
      0.9% Normal saline (NSS) 154 0
      0.45% NSS + 5% dextrose 77 0
      0.18% NSS + 4% dextrose 30 0
      Hartmann’s 131 5
      5% dextrose 0 0

      1000 mL of 5% dextrose has 50 g of glucose

      Option B is inadequate for his sodium and caloric requirements (30 mmol of Na+ and 560 kcal). It is adequate for his K+ requirement (120 mmol of K+).

      Option C is in excess of his Na+ requirement (462 mmol of Na+). Moreover, it does not provide any K+ replacement.

      Option D is inadequate for his caloric requirement (600 kcal) and K+ requirement (60 mmol of K+). Moreover it does not provide any Na+ replacement.

      Option E is in excess of his Na+ requirement (393 mmol of Na+), and is inadequate for his potassium requirement (15 mmol of K+)

      Option A has adequate amounts for his Na+ (231 mmol of Na+) and K+ (120 mmol of K+) requirements. It is inadequate for his caloric requirement (600 kcal).

    • This question is part of the following fields:

      • Physiology
      29.3
      Seconds
  • Question 3 - The biochemical assessment of malnutrition can be measured by the amount of plasma...

    Incorrect

    • The biochemical assessment of malnutrition can be measured by the amount of plasma proteins.

      In acute starvation, which of these plasma proteins is the most sensitive indicator?

      Your Answer: Transferrin

      Correct Answer: Retinol binding globulin

      Explanation:

      The half life of Retinol binding protein (RBP) is 10-12 hours and therefore reflects more acute changes in protein metabolism than any of these proteins. Therefore it is not commonly used as a parameter for nutritional assessment.

      The half life of Transthyretin (thyroxine binding pre-albumin) is only one to two days and so levels are less sensitive and this protein is not an albumin precursor. 15 mg/dL represents early malnutrition and a need for nutritional support.

      Albumin levels have been frequently as a marker of nutrition but this is not a very sensitive marker. It’s half life more than 30 days and significant change takes some time to be noticed. Also, synthesis of albumin is decreased with the onset of the stress response after burns. Unrelated to nutritional status, the synthesis of acute phase proteins increases and that of albumin decreases.

      A more accurate indicator of protein stores is transferrin. It’s response to acute changes in protein status is much faster. The half life of serum transferrin is shorter (8-10 days) and there are smaller body stores than albumin. A low serum transferrin level is below 200 mg/dL and below 100 mg/dL is considered severe. Serum transferrin levels can also affect serum transferrin level.

      Fibronectin is used a nutritional marker but levels decrease after seven days of starvation. It is a glycoprotein which plays a role in enhancing the phagocytosis of foreign particles.

    • This question is part of the following fields:

      • Physiology
      16.6
      Seconds
  • Question 4 - Following a near drowning accident, a 5-year-old child is admitted to the emergency...

    Correct

    • Following a near drowning accident, a 5-year-old child is admitted to the emergency department and advanced paediatric life support is started.

      What is the child's approximate weight, according to the preferred formulae of the Resuscitation Council (UK), the European Resuscitation Council, and the Royal College of Anaesthetists?

      Your Answer: 20-25kg

      Explanation:

      For estimating a child’s weight, the Resuscitation Council (UK) and European Resuscitation Council teach the following formula:

      Weight = (age + 4) × 2

      The weight of the child will be around 20 kg.

      This formula is used in the Primary FRCA exam by the Royal College of Anaesthetists.

      In ‘developed’ countries, the traditional ‘APLS formula’ for estimating weight in children based on age (wt in kg = [age+4] x 2) is acknowledged as underestimating weight by 33.4 percent on average, with the degree of underestimation increasing with increasing age.

      However, more recently, the APLS formula ‘Weight=3(age)+7’ has been found to provide a mean underestimate of only 6.9%. This formula is applicable to children aged 1 to 13 years.

      The estimated weight based on age using this formula is 25 kg.

    • This question is part of the following fields:

      • Physiology
      42.1
      Seconds
  • Question 5 - Which plasma protein will bind the thyroid hormone triiodothyronine (T3) more readily? ...

    Correct

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

      Your Answer: Thyroxine binding globulin

      Explanation:

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

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

    • This question is part of the following fields:

      • Physiology
      4.1
      Seconds
  • Question 6 - A global cerebral blood flow (CBF) of 35 ml/100 g/min (Normal CBF =...

    Incorrect

    • A global cerebral blood flow (CBF) of 35 ml/100 g/min (Normal CBF = 54 ml/100 g/min) can lead to which of the following?

      Your Answer: Liberation of free radicals

      Correct Answer: Poor prognostic EEG

      Explanation:

      CBF is defined as the blood volume that flows per unit mass per unit time in brain tissue and is typically expressed in units of ml blood/100 g tissue/minute. The normal average CBF in adults human is about 50 ml/100 g/min, with lower values in the white matter (,20 ml/100 g/min) and greater values in the gray matter (,80 ml/100 g/min).

      Low CBF levels between 30-40 ml/100 g/min may begin to show poor prognostic EEG. EEG findings consistently associated with a poor outcome are isoelectric EEG, low voltage EEG, and burst suppression (specifically burst suppression with identical bursts), as well as the absence of EEG reactivity.

    • This question is part of the following fields:

      • Physiology
      39.2
      Seconds
  • Question 7 - If a large volume of 0.9% N. saline is administered during resuscitation, it...

    Correct

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

      Your Answer: Hyperchloremic metabolic acidosis

      Explanation:

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

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

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

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

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

      Other causes of a hyperchloremic acidosis are:

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

    • This question is part of the following fields:

      • Physiology
      33.2
      Seconds
  • Question 8 - A 57-year old lady is admitted to the Emergency Department with signs of...

    Correct

    • A 57-year old lady is admitted to the Emergency Department with signs of a subarachnoid haemorrhage.

      On admission, her GCS was 7. She has been intubated, sedated and is being ventilated and is waiting for a CT scan. Her Blood pressure is 140/70mmHg.

      The arterial blood gas analysis shows the following:

      pH 7.2 (7.35 - 7.45)
      PaO2 70 mmHg (80-100)
      9.2 kPa (10.5-13.1)
      PaCO2 78 mmHg (35-45)
      10.2 kPa (4.6-6.0)
      BE -3 mEq/L (-3 +/-3)
      Standard bic 27 mmol/L (21-27)
      SaO2 94%

      The most likely cause of an increase in the patient's global cerebral blood flow (CBF) is which of the following?


      Your Answer: Hypercapnia

      Explanation:

      PaCO2 is one of the most important factors that regulate cerebral vascular tone. CO2 induces cerebral vasodilatation and as a result, it increases CBF. Between 20 mmHg (2.7 kPa) and 80 mmHg (10.7 kPa), there is a linear increase of PaCO2.

      Sometimes, there are areas where auto regulation has failed locally but not globally. Similarly, local vs. systemic acidosis will have similar effects. When the PaO2 falls below 50 mmHg (6.5 kPa), the CBF progressively increases.

      An increase in the cerebral metabolic rate for oxygen (CMRO2) and therefore CBF can be caused by hyperthermia.
      A late feature of cerebral injury is hyperthermia secondary to hypothalamic injury. Therefore this is not the most likely cause of an increased CBF in this scenario.

    • This question is part of the following fields:

      • Physiology
      24.2
      Seconds
  • Question 9 - Regarding bilirubin, which one of the following statement is true? ...

    Incorrect

    • Regarding bilirubin, which one of the following statement is true?

      Your Answer: Has a steroidal structure

      Correct Answer: Conjugated bilirubin is stored in the gall bladder

      Explanation:

      Bilirubin is the tetrapyrrole and a catabolic product of heme. 70-90% of bilirubin is end product of haemoglobin degradation in the liver.

      Bilirubin circulates in the blood in 2 forms; unconjugated and conjugated bilirubin.

      Unconjugated bilirubin is insoluble in water. It travels through the bloodstream to the liver, where it changes from insoluble into a soluble form (i.e.; unconjugated into conjugated form).

      This conjugated bilirubin travels from the liver into the small intestine and the gut bacteria convert bilirubin into urobilinogen and then into urobilin (not urobilin to urobilinogen). A very small amount passes into the kidneys and is excreted in urine.

    • This question is part of the following fields:

      • Physiology
      17.1
      Seconds
  • Question 10 - Using a negative feedback loop, Haem production is controlled by which of these...

    Incorrect

    • Using a negative feedback loop, Haem production is controlled by which of these enzymes?

      Your Answer: Protoporphyrinogen oxidase

      Correct Answer: ALA synthetase

      Explanation:

      Heme a exists in cytochrome a and heme c in cytochrome c; they are both involved in the process of oxidative phosphorylation. 5′-Aminolevulinic acid synthase (ALA-S) is the regulated enzyme for heme synthesis in the liver and erythroid cells.

      There are two forms of ALA Synthase, ALAS1, and ALAS2.

    • This question is part of the following fields:

      • Physiology
      10.6
      Seconds
  • Question 11 - Which of the following is true in the Kreb's cycle? ...

    Correct

    • Which of the following is true in the Kreb's cycle?

      Your Answer: Alpha-ketoglutarate is a five carbon molecule

      Explanation:

      Krebs’ cycle (tricarboxylic acid cycle or citric acid cycle) is a sequence of reactions to release stored energy through oxidation of acetyl coenzyme A (acetyl-CoA). Some of the products are carbon dioxide and hydrogen atoms.

      The sequence of reactions, known collectively as oxidative phosphorylation, only occurs in the mitochondria (not cytoplasm).

      The Krebs cycle can only take place when oxygen is present, though it does not require oxygen directly, because it relies on the by-products from the electron transport chain, which requires oxygen. It is therefore considered an aerobic process. It is the common pathway for the oxidation of carbohydrate, fat and some amino acids, required for the formation of adenosine triphosphate (ATP).

      Pyruvate enters the mitochondria and is converted into acetyl-CoA. Acetyl-CoA is then condensed with oxaloacetate, to form citrate which is a six carbon molecule. Citrate is subsequently converted into isocitrate, alpha-ketoglutarate, succinyl-CoA, succinate, fumarate, malate and finally oxaloacetate.

      The only five carbon molecule in the cycle is Alpha-ketoglutarate.

    • This question is part of the following fields:

      • Physiology
      3.8
      Seconds
  • Question 12 - Which of the following statements best describes adenosine receptors? ...

    Incorrect

    • Which of the following statements best describes adenosine receptors?

      Your Answer: Agonists at the A1 receptors are algogenic

      Correct Answer:

      Explanation:

      Adenosine receptors are expressed on the surface of most cells.
      Four subtypes are known to exist which are A1, A2A, A2B and A3.

      Of these, the A1 and A2 receptors are present peripherally and centrally. There are agonists at the A1 receptors which are antinociceptive, which reduce the sensitivity to a painful stimuli for the individual. There are also agonists at the A2 receptors which are algogenic and activation of these results in pain.

      The role of adenosine and other A1 receptor agonists is currently under investigation for use in acute and chronic pain states.

    • This question is part of the following fields:

      • Physiology
      15.6
      Seconds
  • Question 13 - Regarding anti diuretic hormone (ADH), one of the following statements is correct: ...

    Incorrect

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

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

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

      Explanation:

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

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

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

    • This question is part of the following fields:

      • Physiology
      18
      Seconds
  • Question 14 - In which of the following situations will a regional fall in cerebral blood...

    Incorrect

    • In which of the following situations will a regional fall in cerebral blood flow occur, suppose there is no changes in the mean arterial pressure (MAP)?

      Your Answer: Ketamine anaesthesia

      Correct Answer: Hyperoxia

      Explanation:

      The response of cerebral blood flow (CBF) to hyperoxia (PaO2 >15 kPa, 113 mmHg), the cerebral oxygen vasoreactivity is less well defined. A study originally described, using a nitrous oxide washout technique, a reduction in CBF of 13% and a moderate increase in cerebrovascular resistance in subjects inhaling 85-100% oxygen. Subsequent human studies, using a variety of differing methods, have also shown CBF reductions with hyperoxia, although the reported extent of this change is variable. Another study assessed how supra-atmospheric pressures influenced CBF, as estimated by changes in middle cerebral artery flow velocity (MCAFV) in healthy individuals. Atmospheric pressure alone had no effect on MCAFV if PaO2 was kept constant. Increases in PaO2 did lead to a significant reduction in MCAFV; however, there were no further reductions in MCAFV when oxygen was increased from 100% at 1 atmosphere of pressure to 100% oxygen at 2 atmospheres of pressure. This suggests that the ability of cerebral vasculature to constrict in response to increasing partial pressure of oxygen is limited.

      Increases in arterial blood CO2 tension (PaCO2) elicit marked cerebral vasodilation.

      CBF increases with general anaesthesia, ketamine anaesthesia, and hypoviscosity.

    • This question is part of the following fields:

      • Physiology
      10.8
      Seconds
  • Question 15 - The SI unit of measurement is kgm2s-2 in the System international d'unités (SI).

    Which...

    Incorrect

    • The SI unit of measurement is kgm2s-2 in the System international d'unités (SI).

      Which of the following derived units of measurement has this format?

      Your Answer: Force

      Correct Answer: Energy

      Explanation:

      The derived SI unit of force is Newton.
      F = m·a (where a is acceleration)
      F = 1 kg·m/s2

      The joule (J) is a converted unit of energy, work, or heat. When a force of one newton (N) is applied over a distance of one metre (Nm), the following amount of energy is expended:

      J = 1 kg·m/s2·m =
      J = 1 kg·m2/s2 or 1 kg·m2·s-2

      The unit of velocity is metres per second (m/s or ms-1).

      The watt (W), or number of joules expended per second, is the SI unit of power:

      J/s = kg·m2·s-2/s
      J/s = kg·m2·s-3

      Pressure is measured in pascal (Pa) and is defined as force (N) per unit area (m2):
      Pa = kg·m·s-2/m2
      Pa = kg·m-1·s-2

    • This question is part of the following fields:

      • Physiology
      14.2
      Seconds
  • Question 16 - Cells use adenosine-5-triphosphate (ATP) as a coenzyme and is a source of energy.

    Glucose...

    Correct

    • Cells use adenosine-5-triphosphate (ATP) as a coenzyme and is a source of energy.

      Glucose metabolism produces the most ATP from which of the following biochemical processes?



      Your Answer: Electron transport phosphorylation in the mitochondria

      Explanation:

      Glycolysis occurs in the cytoplasm of the cell. It converts 1 glucose molecule (6-carbon) to pyruvate (two 3-carbon molecules) and produces 4 ATP molecules and 2NADH but uses 2 ATP in the process with an overall net energy production of 2 ATP.

      Pyruvate is then oxidised to acetyl coenzyme A (generating 2 NADH per pyruvate molecule). This takes place in the mitochondria and then enters the Krebs cycle (citric acid cycle). It produces 2 ATP, 8 NADH and 2 FADH2 per glucose molecule.

      Electron transport phosphorylation takes place in the mitochondria. The aim of this process is to break down NADH and FADH2 and also to pump H+ into the outer compartment of the mitochondria. It produces 32 ATP with an overall net production of 36ATP.

      In anaerobic respiration which occurs in the cytoplasm, pyruvate is reduced to NAD producing 2 ATP.

    • This question is part of the following fields:

      • Physiology
      14.7
      Seconds
  • Question 17 - Left ventricular afterload is mostly calculated from systemic vascular resistance.

    Which...

    Incorrect

    • Left ventricular afterload is mostly calculated from systemic vascular resistance.

      Which one of the following factors has most impact on systemic vascular resistance?

      Your Answer: Capillaries

      Correct Answer: Small arterioles

      Explanation:

      Systemic vascular resistance (SVR), also known as total peripheral resistance (TPR), is the amount of force exerted on circulating blood by the vasculature of the body. Three factors determine the force: the length of the blood vessels in the body, the diameter of the vessels, and the viscosity of the blood within them. The most important factor that determines the systemic vascular resistance (SVR) is the tone of the small arterioles.

      These are otherwise known as resistance arterioles. Their diameter ranges between 100 and 450 µm. Smaller resistance vessels, less than 100 µm in diameter (pre-capillary arterioles), play a less significant role in determining SVR. They are subject to autoregulation.

      Any change in the viscosity of blood and therefore flow (such as due to a change in haematocrit) might also have a small effect on the measured vascular resistance.

      Changes of blood temperature can also affect blood rheology and therefore flow through resistance vessels.

      Systemic vascular resistance (SVR) is measured in dynes·s·cm-5

      It can be calculated from the following equation:

      SVR = (mean arterial pressure − mean right atrial pressure) × 80 cardiac output

    • This question is part of the following fields:

      • Physiology
      15.4
      Seconds
  • Question 18 - Which of the following statement is true regarding hypoxic pulmonary vasoconstriction (HPV)? ...

    Incorrect

    • Which of the following statement is true regarding hypoxic pulmonary vasoconstriction (HPV)?

      Your Answer: The predominant stimulus is a low PO2 in the pulmonary arterial blood

      Correct Answer: 20 parts per million (ppm) of nitric oxide will reduce hypoxic pulmonary vasoconstriction

      Explanation:

      Hypoxic Pulmonary vasoconstriction (HPV) reflects the constriction of small pulmonary arteries in response to hypoxic alveoli (.i.e.; PO2 below 80-100mmHg or 11-13kPa).

      These blood vessels become independent of the nerve stimulus, when blood with a high PO2 flows through the lung which contains a low alveolar PO2.

      Thus a low PO2 within the alveoli has been shown to impact on hypoxic pulmonary vasoconstriction (HPV) more than a low PO2 within the blood.

      HPV results in the blood flow being directed away from poorly ventilated areas of the lung and helps to reduce the ventilation/perfusion mismatch (not increase).

      In animals, volatile anaesthetic agents can diminish HPV, while in adults, the evidence proves less persuading, in spite of the fact that it certainly doesn’t strengthen the effects.

      HPV response will be suppressed by 20 parts per million (ppm) of nitric oxide.

    • This question is part of the following fields:

      • Physiology
      17.9
      Seconds
  • Question 19 - 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: Give doxapram

      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
      51.2
      Seconds
  • Question 20 - Which of the following statements is about the measurement of glomerular filtration rate...

    Correct

    • Which of the following statements is about the measurement of glomerular filtration rate (GFR) is correct?

      Your Answer: The result matches clearance of the indicator if it is renally inert

      Explanation:

      The measurements of GFR are done using renally inert indicators like inulin, where passive rate of filtration at the glomerulus = rate of excretion. Normal value is about 180 litres per day.

      GFR is altered by renal blood flow but blood flow does not need to be measured.

      The reabsorption of Na leads to a low excretion rate and low urine concentration and therefore its use as an indicator would lead to an erroneously LOW GFR.

      If there is tubular secretion of any solute, the clearance value will be higher than that of inulin. This will be either due to tubular reabsorption or the solute not being freely filtered at the glomerulus.

    • This question is part of the following fields:

      • Physiology
      20.6
      Seconds

SESSION STATS - PERFORMANCE PER SPECIALTY

Physiology (7/20) 35%
Passmed