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  • Question 1 - Which of the following statements about the central venous pressure (CVP) waveform is...

    Correct

    • Which of the following statements about the central venous pressure (CVP) waveform is true?

      Your Answer: Third degree heart block causes canon A waves

      Explanation:

      The central venous pressure (CVP) waveform depicts changes of pressure within the right atrium. Different parts of the waveform are:

      A wave: which represents atrial contraction. It is synonymous with the P wave seen during an ECG. It is often eliminated in the presence of atrial fibrillation, and increased tricuspid stenosis, pulmonary stenosis and pulmonary hypertension.

      C wave: which represents right ventricle contraction at the point where the tricuspid valve bulges into the right atrium. It is synonymous with the QRS complex seen on ECG.

      X descent: which represents relaxation of the atrial diastole and a decrease in atrial pressure, due to the downward movement of the right ventricle as it contracts. It is synonymous with the point before the T wave on ECG.

      V wave: which represents an increase in atrial pressure just before the opening of the tricuspid valve. It is synonymous with the point after the T wave on ECG. It is increased in the background of a tricuspid regurgitation.

      Y descent: which represents the emptying of the atrium as the tricuspid valve opens to allow for blood flow into the ventricle in early diastole.

      Canon waves: which refer to large waves present on the trace that do not correspond to the A, V or C waves. They usually occur in a background of complete heart blocks or junctional arrythmias.

    • This question is part of the following fields:

      • Clinical Measurement
      17.7
      Seconds
  • Question 2 - Which of the following, at a given PaO2, increases the oxygen content of...

    Correct

    • Which of the following, at a given PaO2, increases the oxygen content of arterial blood?

      Your Answer: A reduced erythrocyte 2,3-diphosphoglycerate level

      Explanation:

      The oxygen content of arterial blood can be calculated by the following equation:
      (10 x haemoglobin x SaO2 x 1.34) + (PaO2 x 0.0225).
      This is the sum of the oxygen bound to haemoglobin and the oxygen dissolved in the plasma.

      Oxygen content x cardiac output = The amount of oxygen delivered to the tissues in unit time which is known as the oxygen flux.

      Any factor that increases the metabolic demand will encourage oxygen offloading from the haemoglobin in the tissues and this causes the oxygen dissociation curve (ODC) to shift to the right. This subsequently reduced the oxygen content of arterial blood.

      Conditions like fever, metabolic or respiratory acidosis lowers the oxygen content and shifts the ODC to the right.
      A low level of 2,3 diphosphoglycerate (2,3-DPG) is usually related to an increased oxygen content as there is less offloading, and so the ODC is shifted to the left.

      So for a given PaO2, a high blood oxygen content is related to any factors that can shift the ODC to the left and not to the right.

      A low haematocrit usually means that there is a decreased haemoglobin concentration, and therefore is associated with decreased oxygen binding to haemoglobin.

    • This question is part of the following fields:

      • Physiology
      18.5
      Seconds
  • Question 3 - A patient is being prepped for major bowel surgery. Alice, a final-year medical...

    Incorrect

    • A patient is being prepped for major bowel surgery. Alice, a final-year medical student, observes the surgery but is asked to scrub in and assist the anaesthetist during intubation. The anaesthetist inserts the laryngoscope and asks Alex to locate the larynx. What anatomical landmark corresponds to the position of the larynx?

      Your Answer: C5

      Correct Answer: C3-C6

      Explanation:

      The larynx is an air passage, sphincter, and organ of phonation that extends from the tongue to the trachea. It lies in the anterior part of the neck at the vertebral levels C3 to C6.

      Important anatomical landmarks:
      C1-C2 – Atlas and axis, respectively

      C3-C6 – Larynx

      C5 – Thyroid cartilage

      T5-T7 – Pulmonary hilum

      T12-L1 – Duodenum.

    • This question is part of the following fields:

      • Anatomy
      20.5
      Seconds
  • Question 4 - In a study lasting over a period of two years, in which the...

    Correct

    • In a study lasting over a period of two years, in which the mean age of 800 patients was 82 years, the efficacy of hip protectors in reducing femoral neck fractures was discussed. Both experimental and control group had 400 members. Instances of fractures reported over the two year time duration were 10 for the control group (that were prescribed hip protector) and 20 for the control group. What is the value of Absolute Risk Reduction?

      Your Answer: 0.025

      Explanation:

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

      So,

      ARR= (10/400)-(20/400)

      ARR= 0.025-0.05

      ARR= 0.025 (Numerical Value)

    • This question is part of the following fields:

      • Statistical Methods
      161.2
      Seconds
  • Question 5 - Which of the following statements is true with regards to acetylcholine? ...

    Correct

    • Which of the following statements is true with regards to acetylcholine?

      Your Answer: Excess cholinesterase inhibitor medication causes cholinergic crisis

      Explanation:

      Myasthenic and cholinergic crises are two crises which are similar in their clinical presentation.

      Myasthenic crisis can be caused by:
      -lack of acetylcholine,
      -poor compliance with medication,
      -infection

      Cholinergic crisis can be caused by excess cholinesterase inhibitor medication (mimicking organophosphate poisoning) causing excess acetylcholine.

      Differentiation between the 2 crises is made by giving incremental doses of the short acting cholinesterase inhibitor, Edrophonium.
      This increase acetylcholine levels and will make a myasthenic crisis better and a cholinergic crisis worse.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      44.6
      Seconds
  • Question 6 - An acidic drug with a pKA of 4.3 is injected intravenously into a...

    Incorrect

    • An acidic drug with a pKA of 4.3 is injected intravenously into a patient. At a normal physiological pH, the approximate ratio of ionised to unionised forms of this drug in the plasma is?

      Your Answer: 10,000:1

      Correct Answer: 1000:01:00

      Explanation:

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

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

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

      or

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

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

      In this example:

      7.4 = 4.3 + log ([ionised]/[unionised])
      7.4 ˆ’ 4.3 = log ([ionised]/[unionised])
      log 3.1 = log ([ionised]/[unionised])

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

      y = antilog x = 10x

      Antilog to the base 10 of 0 = 1, 1 = 10, 2 =100, 3 = 1000, and 4 = 10,000.

      If you want to find the antilogarithm of 3.1, for a number between 3 and 4, the antilogarithm will return a value between 1000 and 10,000. The ratio is 1:1 if pKa = pH, that is, pH pKa = log 0. (50 percent ionised and unionised).

      According to the above value, there is only one unionised molecule for every approximately 1000 (1259) ionised molecules of this drug in plasma, implying that this drug is largely ionised in plasma (99.99 percent ).

    • This question is part of the following fields:

      • Pharmacology
      173
      Seconds
  • Question 7 - Standard error of the mean can be defined as: ...

    Correct

    • Standard error of the mean can be defined as:

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

      Explanation:

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

      SEM = SD / square root (n)

      where SD = standard deviation and n = sample size

    • This question is part of the following fields:

      • Statistical Methods
      11.1
      Seconds
  • Question 8 - A 5-year old male has ingested a peanut and has developed urticaria, vomiting...

    Correct

    • A 5-year old male has ingested a peanut and has developed urticaria, vomiting and hypotension. The pathogenesis of this condition is derived from predominant cells of which cell line?

      Your Answer: Common myeloid progenitor

      Explanation:

      A is correct. Common myeloid progenitor cells are involved in the anaphylaxis reaction.
      B is incorrect. The common lymphoid lineage gives rise to T-cells, B-cell and NK cells.
      C is incorrect as megakaryocytes give rise to platelets.
      D is incorrect – Neural crest cells give rise to various cells throughout the body, including melanocytes, enterochromaffin cells and Schwann cells. However, they do not give rise to mast cells.
      E is incorrect. Reticulocytes give rise to erythrocytes.

      This is a classic case of anaphylaxis. In this situation, IgE previously raised against antigens (in this case peanut antigen) bind to mast cells, and this causes them to degranulate.
      There is release of vasoactive substances like histamine into the blood, and this is responsible for the symptoms seen. Therefore, the main type of cells involved in the pathogenesis of the disease is mast cells.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      13.6
      Seconds
  • Question 9 - A 40-year old farmer came into the emergency room with a chief complaint...

    Incorrect

    • A 40-year old farmer came into the emergency room with a chief complaint of 4 episodes of non-bloody diarrhoea. This was associated with frequent urination, vomiting and salivation. History also revealed frequent use of insecticides. Upon physical examination, there was miosis and bradycardia. Given the different types of bonds, which is the most likely bond formed between insecticide poisoning and receptors?

      Your Answer: Ionic

      Correct Answer: Covalent

      Explanation:

      Organophosphate poisoning occurs most often due to accidental exposure to toxic amounts of pesticides. Signs and symptoms include diarrhoea, urination, miosis, bradycardia, emesis, lacrimation, lethargy and salivation.

      Organophosphates are classified as indirect acting cholinomimetics, and their mode of action involves: (1) the inhibition of acetylcholinesterase (AChE) by forming a stable covalent bond on the active site serine; and, (2) amplification of endogenously release acetylcholine (ACh), hence the clinical manifestation.

      There are 4 types of bonds or interactions: ionic, covalent, hydrogen bonds, and van der Waals interactions. Ionic and covalent bonds are strong interactions that require a larger energy input to break apart. When an element donates an electron from its outer shell, a positive ion is formed. The element accepting the electron is now negatively charged. Because positive and negative charges attract, these ions stay together and form an ionic bond. Covalent bonds form when an electron is shared between two elements and are the strongest and most common form of chemical bond in living organisms. Covalent bonds form between the elements that make up the biological molecules in our cells. Unlike ionic bonds, covalent bonds do not dissociate in water.

      When polar covalent bonds containing a hydrogen atom form, the hydrogen atom in that bond has a slightly positive charge. This is because the shared electron is pulled more strongly toward the other element and away from the hydrogen nucleus. Because the hydrogen atom is slightly positive, it will be attracted to neighbouring negative partial charges. When this happens, a weak interaction occurs between the slightly positive charge of the hydrogen atom of one molecule and the slightly negative charge of the other molecule. This interaction is called a hydrogen bond.

    • This question is part of the following fields:

      • Pathophysiology
      34.2
      Seconds
  • Question 10 - A meta analysis takes into consideration five studies each of which is aimed...

    Incorrect

    • A meta analysis takes into consideration five studies each of which is aimed at finding out the relation between a novel drug and upper gastro intestinal bleeding. The relative risk of getting an upper gastrointestinal bleed matched to a control population is recorded by each study. Which among the following studies provides the most compelling evidence that the bleeding is not caused by the new drug?

      Your Answer: Dennis et al

      Correct Answer: Atkinson et al

      Explanation:

      The research study conducted by Atkinson et al makes the most compelling case for the drug as it indicates the relative risk levels to be lower than 1. Also the study employs large square and a narrow confidence interval that are an indicative of a well performed study.

    • This question is part of the following fields:

      • Statistical Methods
      18.4
      Seconds
  • Question 11 - The tip of a pulmonary artery flotation catheter becomes wedged when threaded through...

    Correct

    • The tip of a pulmonary artery flotation catheter becomes wedged when threaded through the chambers of the heart and the pulmonary artery. Which of the following options best describes the sequence of pressures measured at the catheter's tip during its passage through a normal patient's pulmonary artery?

      Your Answer: 0-12 mmHg, 2-25 mmHg, 12-25 mmHg and 8-12 mmHg

      Explanation:

      The tricuspid valve allows the tip of a pulmonary artery catheter to pass through the right atrium and into the right ventricle.

      The balloon will be inflated before crossing the pulmonary valve and entering the pulmonary artery, where it will eventually wedge or occlude the artery, providing an indirect measure of left atrial pressure.

      0-12 mmHg in the right atrium
      2-25 mmHg in the right ventricle
      12-25 mmHg in the pulmonary artery
      8-12 mmHg is the occlusion pressure

    • This question is part of the following fields:

      • Physiology And Biochemistry
      19.3
      Seconds
  • Question 12 - Drug toxicity when using bupivacaine is most likely to occur when this local...

    Correct

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

      Your Answer: Intercostal nerve block

      Explanation:

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

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

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

    • This question is part of the following fields:

      • Pharmacology
      3.2
      Seconds
  • Question 13 - Which of the following is a correct match for reflex and their root...

    Correct

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

      Your Answer: Knee reflex: L3/L4

      Explanation:

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

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

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

    • This question is part of the following fields:

      • Anatomy
      12.2
      Seconds
  • Question 14 - A 68-year-old man presents worried about his risk of motor neurone disease. No...

    Correct

    • 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: 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
      41.3
      Seconds
  • Question 15 - A 72-year-old man complains of severe, central abdominal pain that radiates to the...

    Correct

    • 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: 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
      6.2
      Seconds
  • Question 16 - In a normal healthy adult breathing 100 percent oxygen, which of the following...

    Correct

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

      Your Answer: Atelectasis

      Explanation:

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

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

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

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

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

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

      The alveolar PO2 is also reduced at high altitude.

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

    • This question is part of the following fields:

      • Physiology
      7.3
      Seconds
  • Question 17 - While administering a general anaesthetic to a 65-year-old man booked for a hip...

    Correct

    • While administering a general anaesthetic to a 65-year-old man booked for a hip hemiarthroplasty, with a weight 70 kg, and an ASA 1 score, you give 1 g of paracetamol IV but notice that he had received the same dose on the ward one hour prior. What is the most appropriate subsequent management of this patient?

      Your Answer: Do nothing and give the next doses of paracetamol at standard 6 hour intervals

      Explanation:

      After ingestion of more than 150 mg/kg paracetamol within 24 hours, hepatotoxicity can occur but can also develop rarely after ingestion of doses as low as 75 mg/kg within 24 hours. Hepatocellular damage will not occur in this patient and therefore no need to engage management pathway for paracetamol overdose. If his weight was <33 kg or he already had a history of impaired liver function, then the management would bde different.

      Subsequent post-operative doses will be a standard dose of 1 g 6 hourly.

      This is a drug administration error and should be reported as an incident even though the patient will not be harmed.

    • This question is part of the following fields:

      • Pharmacology
      27.7
      Seconds
  • Question 18 - What is the number of valves between the superior vena cava and the...

    Incorrect

    • What is the number of valves between the superior vena cava and the right atrium?

      Your Answer: One

      Correct Answer: None

      Explanation:

      The inflow of blood from the superior vena cava is directed towards the right atrioventricular orifice. It returns deoxygenated blood from all structures superior to the diaphragm, except the lungs and heart.

      There are no valves in the superior vena cava which is why it is relatively easy to insert a CVP line from the internal jugular vein into the right atrium. The brachiocephalic vein is similar as it also has no valves.

    • This question is part of the following fields:

      • Anatomy
      3.4
      Seconds
  • Question 19 - A 32-year-old male is admitted to the critical care unit. He has suffered...

    Incorrect

    • A 32-year-old male is admitted to the critical care unit. He has suffered a heroin overdose and requires intubation and ventilatory support. What would be his predicted total static compliance (lung and chest wall) measurements.

      Your Answer: 20 ml/cmH2O

      Correct Answer: 100 ml/cmH2O

      Explanation:

      Static lung compliance refers to the change in volume within the lung per given change in unit pressure. It is usually measured when air flow is absent, such as during pauses in inhalation and exhalation.

      It is a combination of:

      Chest wall compliance: normal value is 200 mL/cmH2O
      Lung tissue compliance: normal value is 200 mL/ cmH2O

      It is represented mathematically as:

      1/Crs = 1/Cl + 1/Ccw

      Where,

      Crs = total compliance of the respiratory system
      Cl = compliance of the lung
      Ccw = compliance of the chest wall

      Therefore in this case:

      1/Crs = 1/200 + 1/200

      1/Crs = 0.005 + 0.005 = 0.01

      1/Ct = 0.01

      Rearranging equation gives:

      Ct = 1/0.01 = 100 mL/cmH2O.

    • This question is part of the following fields:

      • Clinical Measurement
      11.8
      Seconds
  • Question 20 - A new volatile anaesthetic agent has been approved for use in clinical testing....

    Correct

    • A new volatile anaesthetic agent has been approved for use in clinical testing. It's a non-irritating, sweet-smelling substance. It has a molecular weight of 170, a 0.6 blood:gas partition coefficient, and a 180 oil:gas partition coefficient. An oxidative pathway converts 2% of the substance to trifluoroacetic acid. Which of the following statements best describes this agent's pharmacological profile?

      Your Answer: It has a lower molecular weight than isoflurane

      Explanation:

      Because enflurane is much less soluble in blood and has a blood: gas partition coefficient of 1.8, both wash-in and wash-out should be faster.

      Sevoflurane’s sweet-smelling, non-irritant nature, combined with a low blood: gas partition coefficient, would result in similar offset and onset characteristics.

      Isoflurane and enflurane have a molecular weight of 184.

      The oil: gas partition coefficient on a volatile agent is a measure of lipid solubility, potency, and thus MAC. Halothane has an oil: gas partition coefficient of 220 and a MAC of 0.74. One would expect the MAC to be higher with an oil gas partition coefficient of 180 (less lipid soluble).

      The conversion of halothane (20%) to trifluoroacetic acid via oxidative metabolism has been linked to the development of hepatitis.

      P450 2E1 converts sevoflurane to hexafluoroisopropanol, which results in the release of inorganic fluoride ions. It’s the only fluorinated volatile anaesthetic that doesn’t break down into trifluoracetic acid.

      Desflurane is likely to cause airway irritation, which can lead to coughing, apnoea, and laryngospasm, despite its low blood:gas partition coefficient (0.42).

    • This question is part of the following fields:

      • Pharmacology
      85.4
      Seconds
  • Question 21 - Which of the statements below best describe the total cerebral flow (CBF) in...

    Correct

    • Which of the statements below best describe the total cerebral flow (CBF) in an adult?

      Your Answer: Accounts for 15% of the cardiac output

      Explanation:

      While the brain only weighs 3% of the body weight, 15% of the cardiac output goes towards the brain.

      Between mean arterial pressures (MAP) of 60-130 mmHg, autoregulation of cerebral blood flow (CBF) occurs. Exceeding this, the CBF is maintained at a constant level. This is controlled mainly by the PaCO2 level, and the autonomic nervous system has minimal role.

      Beyond these limits, the CBF is directly proportional to the MAP, not the systolic blood pressure.

    • This question is part of the following fields:

      • Physiology
      20.1
      Seconds
  • Question 22 - The thyroid gland: ...

    Correct

    • The thyroid gland:

      Your Answer: Internalises iodine through active transport

      Explanation:

      The thyroid gland is a gland shaped like a butterfly which lies at the base of the anterior neck. It controls metabolism using hormone secretion.

      Iodine is extremely important for the synthesis of hormones within the thyroid. It is internalised into the thyroid follicular cells via the sodium/iodide symporter (NIS).

      The parathyroid glands are found posterior to the thyroid gland, with the recurrent laryngeal nerves running posteromedially.

      The expected weight of a normal thyroid gland is about 30 grams.

    • This question is part of the following fields:

      • Pathophysiology
      10.7
      Seconds
  • Question 23 - Heights of 100 individuals(adults) who were administered steroids at any stage during childhood...

    Incorrect

    • Heights of 100 individuals(adults) who were administered steroids at any stage during childhood was studied. The mean height was found to be 169cm with the data having a standard deviation of 16cm. What will be the standard error associated with the mean?

      Your Answer: 0.16

      Correct Answer: 1.6

      Explanation:

      Standard error can be calculated by the following formula:
      Standard Error= (Standard Deviation)/ˆš(Sample Size)
      = (16) / ˆš(100)
      = 16 / 10
      = 1.6

    • This question is part of the following fields:

      • Statistical Methods
      37.4
      Seconds
  • Question 24 - In order to determine if there is any correlation among systolic blood pressure...

    Incorrect

    • In order to determine if there is any correlation among systolic blood pressure and the age of a person. Which among the provided options is false regarding the calculation of correlation coefficient, r ?

      Your Answer: r may lie anywhere between -1 and 1

      Correct Answer: May be used to predict systolic blood pressure for a given age

      Explanation:

      Correlation doesn’t justify causality. Correlation coefficient gives us an idea whether or not the two parameters provide have any relation of some sort or not i.e. does change in one prompt any change in other? It has nothing to do with predictions. For that purpose linear regression is used.

    • This question is part of the following fields:

      • Statistical Methods
      41
      Seconds
  • Question 25 - Which of the following can be measured directly using spirometry? ...

    Correct

    • Which of the following can be measured directly using spirometry?

      Your Answer: Vital capacity

      Explanation:

      Spirometry measures the total volume of air that can be forced out in one maximum breath, that is the total lung capacity (TLC), to maximal expiration, that is the residual volume (RV).

      It is conducted using a spirometer which is capable of measuring lung volumes using techniques of dilution.

      During spirometry, the following measurements can be determined:
      Forced vital capacity (FVC)/vital capacity (VC): The maximum volume of air exhaled in one single forced breathe.
      Forced expiratory volume in one second (FEV1)
      FEV1/FVC ratio
      Peak expiratory flow (PEF): the maximum amount of air flow exhaled in one blow.
      Forced expiratory flow (mid expiratory flow): the flow at 25%, 50% and 75% of FVC
      Inspiratory vital capacity (IVC): The maximum volume of air inhaled after a full total expiration.

      Anatomical dead space is measured using a single breath nitrogen washout called the Fowler’s method.

      Residual volume and total lung capacity are both measured using the body plethysmograph or helium dilution

      The functional residual capacity is usually measured using a nitrogen washout or the helium dilution technique.

    • This question is part of the following fields:

      • Clinical Measurement
      10.2
      Seconds
  • Question 26 - A doctor has recorded the number of times the patient did not attend...

    Correct

    • A doctor has recorded the number of times the patient did not attend (DNA) the clinic for a study over a 10 month period. Number of DNAs in 10 Months: 1st Month: 0, 2nd Month: 3, 3rd Month: 1, 4th Month: 45, 5th Month: 2, 6th Month: 0, 7th Month: 1, 8th Month: 4, 9th Month: 4, 10th Month: 2. Which among the following is the most apt way of summarizing mean value?

      Your Answer: Median

      Explanation:

      Variance and standard deviation indicate the dispersion of the plot from mean value and thus are not really helpful in summarizing the mean.

      Range is the difference between maximum and minimum value that is 45 in this case.

      The mean in this case is 6.2 due to the presence of an outlier 45. In the presence of outlier mean can be misleading as it is quite sensitive to skewness in data.

      Mode is the most frequent value. In this case mode has 4 values: 0,1,2,4.

      In case of skewedness, median is the most apt representative of the mean as it is not affected by outliers. In this case since the data set has even values i.e. 10. Median is the average of the 5th & 6th entry after arranging the data in ascending order like that in case of the question (0,0,1,1,2,2,3,4,4,45). This turns out to be 2.

    • This question is part of the following fields:

      • Statistical Methods
      104.2
      Seconds
  • Question 27 - Rocuronium is substituted for succinylcholine during induction of anaesthesia for a caesarean section...

    Incorrect

    • Rocuronium is substituted for succinylcholine during induction of anaesthesia for a caesarean section delivery. Which of the following feature of rocuronium ensures the neonate shows no clinical signs of muscle relaxation?

      Your Answer: Highly protein bound

      Correct Answer: Highly ionised

      Explanation:

      Drugs cross the placenta by Simple, Ion channel and Facilitated diffusion; Exocytosis and Endocytosis, Osmosis, and Active transport (primary and secondary)

      The following factors influence rate of diffusion across the placenta:

      Protein binding
      Degree of ionisation
      Placental blood flow
      Maternal and foetal blood pH
      Materno-foetal concentration gradient.
      Thickness of placental membrane
      Molecular weight of drug <600 Daltons cross by diffusion
      Lipid solubility (lipid soluble molecules readily diffuse across the placenta)

      Rocuronium has a F/M ratios of 0.16, a 30% plasma protein binding, low lipid solubility, a low volume of distribution (0.25L/kg), and a high molecular weight (530Da).

    • This question is part of the following fields:

      • Pharmacology
      23.5
      Seconds
  • Question 28 - The following foetal anatomical features functionally closes earliest at birth? ...

    Incorrect

    • The following foetal anatomical features functionally closes earliest at birth?

      Your Answer: Ductus venosus

      Correct Answer: Foramen ovale

      Explanation:

      Foramen ovale, ductus arteriosus (DA) and ductus venosus (DV) are the three important cardiac shunts in-utero.

      At birth the umbilical vessels constrict in response to stretch as they are clamped. Blood flow through the ductus venosus (DV) decreases but the DV closes passively in 3-10 days.

      As the pulmonary circulation is established, there is a drastic fall in pulmonary vascular resistance and an increased pulmonary blood flow. This increases flow and pressure in the Left Atrium that exceeds that of the right atrium. The difference in pressure usually leads to the IMMEDIATE closure of the foramen ovale.

      The DA is functionally closed within the first 36-hours of birth in a healthy full-term newborn. Subsequent endothelial and fibroblast proliferation leads to permanent anatomical closure within 2 – 3 weeks.

      Oxygenated blood from the placenta passes via the umbilical vein to the liver. Blood also bypasses the liver via the ductus venosus into the inferior vena cava (IVC). The Crista dividens is a tissue flap situated at the junction of the IVC and the right atrium (RA). This flap directs the oxygen-rich blood, along the posterior aspect of the IVC, through the foramen ovale into the left atrium (LA).

      The Eustachian valve also known as the valve of The IVC is a remnant of the crista dividens.

    • This question is part of the following fields:

      • Anatomy
      21.1
      Seconds
  • Question 29 - A global cerebral blood flow (CBF) of 35 ml/100 g/min (Normal CBF =...

    Correct

    • 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: 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
      6.7
      Seconds
  • Question 30 - A 20-year-old boy is undergoing surgery for indirect inguinal hernia repair. The deep...

    Correct

    • A 20-year-old boy is undergoing surgery for indirect inguinal hernia repair. The deep inguinal ring is exposed and held with a retractor at its medial aspect during the procedure. What structure is most likely to lie under the retractor on the medial side?

      Your Answer: Inferior epigastric artery

      Explanation:

      The deep inguinal ring is the entrance of the inguinal canal. It is an opening in the transversalis fascia around 1 cm above the inguinal ligament. Therefore, the superolateral wall is made by the transervalis fascia.

      The inferior epigastric vessels run medially to the deep inguinal ring forming its inferomedial border.

      The inguinal canal extends obliquely from the deep inguinal ring to the superficial inguinal ring.
      An indirect inguinal hernia arises through the deep inguinal ring lateral to the inferior epigastric vessels.

    • This question is part of the following fields:

      • Anatomy
      21.8
      Seconds
  • Question 31 - Which statement is false in regards to the Circle of Willis? ...

    Incorrect

    • Which statement is false in regards to the Circle of Willis?

      Your Answer: Does not include the middle cerebral artery

      Correct Answer: Majority of blood passing through the vessels mix together

      Explanation:

      There is minimum mixing of blood passing through the vessels.

      The cerebral hemispheres are supplied by arteries that make up the Circle of Willis. The Circle of Willis is formed by the anastomosis of the two internal carotid arteries and two vertebral arteries. It lies in the subarachnoid space within the basal cisterns that surround the optic chiasma and infundibulum.

      Each half of the circle is formed by:
      1. Anterior communicating artery
      2. Anterior cerebral artery
      3. Internal carotid artery
      4. Posterior communicating artery
      5. Posterior cerebral arteries and the termination of the basilar artery

      The circle and its branches supply; the corpus striatum, internal capsule, diencephalon, and midbrain.

    • This question is part of the following fields:

      • Anatomy
      26.8
      Seconds
  • Question 32 - Which of the following statements is true regarding prazosin? ...

    Incorrect

    • Which of the following statements is true regarding prazosin?

      Your Answer: Has central sympathomimetic activity.

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

      Explanation:

      Selective α1 -Blockers like prazosin, terazosin, doxazosin, and alfuzosin cause a decrease in blood pressure with lesser tachycardia than nonselective blockers (due to lack of α2 blocking action.

      The major adverse effect of these drugs is postural hypotension. It is seen with the first few doses or on-dose escalation (First dose effect).

      Its half-life is approximately three hours.

      It is excreted primarily through bile and faeces (not through kidneys)

    • This question is part of the following fields:

      • Pharmacology
      99.8
      Seconds
  • Question 33 - After a bariatric surgery, average weight loss observed in patients is 18 kg....

    Correct

    • After a bariatric surgery, average weight loss observed in patients is 18 kg. The standard deviation was found to be 3 kg. What is the percentage of patients that lie between 9 and 27 kg? Note: Assume that the curve is normally distributed.

      Your Answer: 99.70%

      Explanation:

      9 & 27 can be obtained by subtracting and adding 9 from the mean. 9 is three times the standard deviation and we know that 99.7% values lie within 3 standard deviations from the mean. We can find the interval for 99.7% to verify in the following way:

      For 99.7% confidence interval, you can find the range as follows:

      1. Multiply the standard error by 3.

      2. Subtract the answer from mean value to get the lower limit.

      3. Add the answer obtained in step 1 from the mean value to get the upper limit.

      4. The range turns out to be 9-27 kg.

    • This question is part of the following fields:

      • Statistical Methods
      66
      Seconds
  • Question 34 - Which among the following is not true regarding disease rates? ...

    Incorrect

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

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

      Correct Answer: The odds ratio is synonymous with the risk ratio

      Explanation:

      Phase 1 is associated with assessing whether a drug is safe to use or not. The process is extensive and can take up to several months. It also involves healthy participants (less than 100) that are paid to take part in the study.

      The side effects upon increasing dosage are also addressed by the study. The effects the drug has on humans including how its absorbed, metabolized and excreted are studied. Approximately 70% of the drugs pass this phase.

    • This question is part of the following fields:

      • Statistical Methods
      52.3
      Seconds
  • Question 35 - A 50-year-old man is admitted in hospital. Over four hours, he produces 240...

    Incorrect

    • A 50-year-old man is admitted in hospital. Over four hours, he produces 240 mL of urine and has a plasma creatinine concentration is 10 mcg/mL. The normal concentration of creatinine in urine is 1.25 mg/mL. Calculate his approximate creatinine clearance.

      Your Answer: 12.5 ml/minute

      Correct Answer: 125 ml/minute

      Explanation:

      Creatinine clearance is a test used to approximate the glomerular filtration rate (GFR) as an assessment of kidney function.

      Creatinine is formed during the breakdown of dietary sources of meat and skeletal muscle. It is secreted at a consistent concentration and pace into the body’s circulation, and is easily filtered across the glomerulus without being reabsorbed or metabolized by the kidney.

      It is represented mathematically as:
      Creatinine clearance (CL) = U x V/P
      where,
      U: Urinary creatinine concentration (mg/mL)
      V: Volume of urine (mL/min)
      P: Plasma creatinine concentration (mg/mL)

      Therefore, in this case:
      CL: 1.25 x 1 = 125mL/min.

    • This question is part of the following fields:

      • Clinical Measurement
      344.9
      Seconds
  • Question 36 - Which one is true with respect to the first rib? ...

    Correct

    • Which one is true with respect to the first rib?

      Your Answer: Scalenus anterior is inserted onto the scalene tubercle

      Explanation:

      Specific knowledge of the anatomical relationship is required to address this examination question.

      The first rib is small and thick and contains a single facet that articulates at the costovertebral joint. It consist of a head, neck and shaft but a discrete angle is deficit. Along the side the shaft is indented with a groove for the subclavian artery and the lower brachial plexus trunk. Front to the scalene tubercle is a space for the subclavian vein.

      The first rib has the scalenus front muscle joined to the scalene tubercle, isolating the subclavian vein (anteriorly) from the subclavian artery (posteriorly). This anatomical relationship is of major significance with respect to subclavian vein cannulation.

      The 1st rib has the following relationships:

      superior: lower trunk of the brachial plexus, subclavian vessels, clavicle.

      inferior: intercostal vessels and nerves

      posterior and inferior: pleura

      anterior: sympathetic trunk (over neck)

      superior intercostal artery, ventral T1 nerve root.

    • This question is part of the following fields:

      • Anatomy
      9.5
      Seconds
  • Question 37 - A 27-year-old woman arrives at the emergency room after intentionally ingesting 2 g...

    Correct

    • A 27-year-old woman arrives at the emergency room after intentionally ingesting 2 g of amitriptyline. A Glasgow coma score of 6 was discovered, as well as a pulse rate of 140 beats per minute and a blood pressure of 80/50 mmHg. Which of the following ECG changes is most likely to indicate the onset of life-threatening arrhythmias?

      Your Answer: Prolongation of the QRS complex

      Explanation:

      Arrhythmias and/or hypotension are the most common causes of death from tricyclic antidepressant (TCA) overdose.

      The quinidine-like actions of tricyclic antidepressants on cardiac tissues are primarily responsible for their toxicity. Conduction through the His-Purkinje system and the myocardium slows as phase 0 depolarisation of the action potential slows. QRS prolongation and atrioventricular block are caused by slowed impulse conduction, which also contributes to ventricular arrhythmias and hypotension.

      Arrhythmias can also be caused by abnormal repolarization, impaired automaticity, cholinergic blockade, and inhibition of neuronal catecholamine uptake, among other things.

      Acidaemia, hypotension, and hyperthermia can all exacerbate toxicity.

      The anticholinergic effects of tricyclic antidepressants, as well as the blockade of neuronal catecholamine reuptake, cause sinus tachycardia. Sinus tachycardia is usually well tolerated and does not require treatment. It can be difficult to tell the difference between sinus tachycardia and ventricular tachycardia with QRS prolongation.

      A QRS duration of more than 100 milliseconds indicates a higher risk of arrhythmia and should be treated with systemic sodium bicarbonate.

      The tricyclic is dissociated from myocardial sodium channels by serum alkalinization, and the extracellular sodium load improves sodium channel function.

    • This question is part of the following fields:

      • Clinical Measurement
      12.5
      Seconds
  • Question 38 - A young male is operated on for an open inguinal hernia repair. During...

    Incorrect

    • A young male is operated on for an open inguinal hernia repair. During the procedure, the cord is mobilized, and the deep inguinal ring is located. What structure forms the lateral wall of the deep inguinal ring?

      Your Answer: External oblique aponeurosis

      Correct Answer: Transversalis fascia

      Explanation:

      The deep inguinal ring is the entrance of the inguinal canal. It is an opening in the transversalis fascia around 1 cm above the inguinal ligament. Therefore, the superolateral wall is made by the transervalis fascia.

      The inferior epigastric vessels run medially to the deep inguinal ring forming its inferomedial border.

      The inguinal canal extends obliquely from the deep inguinal ring to the superficial inguinal ring.
      An indirect inguinal hernia arises through the deep inguinal ring lateral to the inferior epigastric vessels.

    • This question is part of the following fields:

      • Anatomy
      18.6
      Seconds
  • Question 39 - You draw a patient's blood sample from the median cubital vein in the...

    Correct

    • You draw a patient's blood sample from the median cubital vein in the antecubital fossa. Which of the following veins also connects to the cephalic vein other than the median cubital vein?

      Your Answer: Basilic vein

      Explanation:

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

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

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

    • This question is part of the following fields:

      • Anatomy
      46.5
      Seconds
  • Question 40 - Which of the following statements is true about monoamine oxidase (MOA) enzymes? ...

    Correct

    • Which of the following statements is true about monoamine oxidase (MOA) enzymes?

      Your Answer: Type A and type B are found in the liver and brain

      Explanation:

      Monoamine oxidase (MOA) enzymes are responsible for the catalyses of monoamine oxidative deamination. It assists the degradation of serotonin, norepinephrine (NE) and dopamine.

      They are found in the mitochondria of most central and peripheral nerve tissues.

      There are 2 different types:

      Type A: Whose main function it to inactivate dopamine, tyramine, norepinephrine and 5-hydroxytryptamine. In addition to the nervous system, it is also found in the liver, brain gastrointestinal tract, pulmonary endothelium and placenta
      Type B: Whose main function is to inactivate dopamine, tyramine, tryptamine and phenylethylamine. In addition to the nervous system, it is also found in the liver, brain (especially in the basal ganglia) and blood platelets.

    • This question is part of the following fields:

      • Pathophysiology
      31.4
      Seconds
  • Question 41 - A 27-year-old woman takes part in a study looking into the effects of...

    Correct

    • A 27-year-old woman takes part in a study looking into the effects of different dietary substrates on metabolism. She receives a 24-hour ethyl alcohol infusion. A constant volume, closed system respirometer is used to measure CO2 production and consumption. The production of carbon dioxide is found to be 200 mL/minute. Which of the following values most closely resembles her anticipated O2 consumption at the conclusion of the trial?

      Your Answer: 300 mL/minute

      Explanation:

      The respiratory quotient (RQ) is the ratio of CO2 produced by the body to O2 consumed in a given amount of time.

      CO2 produced / O2 consumed = RQ

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

      The RQ will change depending on the energy substrates consumed in the diet. Granulated sugar is a refined carbohydrate that contains 99.999 percent carbohydrate and no lipids, proteins, minerals, or vitamins.

      Glucose and other hexose sugars (glucose and other hexose sugars):
      RQ=1

      Fats:
      RQ = 0.7

      Proteins:
      Approximately 0.9 RQ

      Ethyl alcohol is a type of alcohol.

      200/300 = 0.67 RQ

      For complete oxidation, lipids and alcohol require more oxygen than carbohydrates.

      When carbohydrate is converted to fat, the RQ can rise above 1.0. Fat deposition and weight gain are likely to occur in these circumstances.

    • This question is part of the following fields:

      • Physiology
      136.7
      Seconds
  • Question 42 - Which of the following facts about IgE is true? ...

    Incorrect

    • Which of the following facts about IgE is true?

      Your Answer: Is increased acutely in an asthmatic attack

      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
      30.1
      Seconds
  • Question 43 - 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), PaCO2 78 mmHg (35-45), 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
      35.5
      Seconds
  • Question 44 - A 64-year-old man is admitted to the critical care unit. He has a...

    Incorrect

    • A 64-year-old man is admitted to the critical care unit. He has a recent medical history of faecal peritonitis for which a laparotomy was performed. His vitals have been monitored using an invasive pulmonary artery flotation catheter. His vital readings are: Temperature: 38.1°C. Blood pressure: 79/51 mmHg (mean 58 mmHg), Pulmonary artery pressure: 19/6 mmHg (mean 10 mmHg). Pulmonary capillary occlusion pressure: 5 mmHg, Central venous pressure: 12 mmHg, Cardiac output: 5 L/min, Mixed venous oxygen saturation: 82%. Calculate his approximate pulmonary vascular resistance. Note: A correction factor of 80 is require to convert mmHg to dynes·s·cm-5

      Your Answer: 40 dynes·s·cm-5

      Correct Answer: 80 dynes·s·cm-5

      Explanation:

      Pulmonary vascular resistance (PVR) refers to the resistance to blood flow to the left atrium from the pulmonary artery.
      It is derived mathematically by:

      PVR = MPAP – PCWP
      CO
      where,
      MPAP: Mean pulmonary artery pressure
      PCWP: Pulmonary capillary occlusion pressure
      CO: Cardiac output

      For this patient:
      PVR = 10 – 5 = 1mmHg
      5

      Remember, multiply by correction factor 80 to change units:

      PVR = 1mmHg x 80 = 80 dynes·s·cm-5

      Normal values range between 20-130 dynes·s·cm-5.

    • This question is part of the following fields:

      • Clinical Measurement
      261.2
      Seconds
  • Question 45 - The most abundant intracellular ion is? ...

    Incorrect

    • The most abundant intracellular ion is?

      Your Answer: Calcium

      Correct Answer: Phosphate

      Explanation:

      Phosphate is the principal anion of the intracellular fluid, most of which is bound to either lipids or proteins. They dissociate or associate with different compounds, depending on the enzymatic reaction, thus forming a constantly shifting pool.

      Calcium and magnesium are also present intracellularly, however in lesser amounts than phosphate.

      Sodium is the most abundant extracellular cation, and Chloride and is the most abundant extracellular anion.

    • This question is part of the following fields:

      • Physiology
      15.5
      Seconds
  • Question 46 - Which of the given statements is true about standard error of the mean?...

    Correct

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

      Your 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
      18.3
      Seconds
  • Question 47 - All of the following statements are false regarding propranolol except: ...

    Incorrect

    • All of the following statements are false regarding propranolol except:

      Your Answer: Is a cardio-selective beta adrenoceptor antagonist.

      Correct Answer: Has a plasma half life of 3-6 hours.

      Explanation:

      Propranolol is a nonselective beta-blocker with a half-life of 3 to 6 hours.

      Since it is lipid-soluble it crosses the blood-brain barrier and causes Central Nervous System side effects like sedation, nightmares, and depression.

      They are contraindicated in asthma, Congestive heart failure, and diabetes.

      It has a large volume of distribution with no intrinsic sympathomimetic action.

    • This question is part of the following fields:

      • Pharmacology
      15.2
      Seconds
  • Question 48 - For a rapid sequence induction of anaesthesia, you are pre-oxygenating a patient using...

    Correct

    • 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: 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
      7.9
      Seconds
  • Question 49 - The diaphragm is a muscle that is relatively resistant to non-depolarizing neuromuscular blockade's...

    Incorrect

    • The diaphragm is a muscle that is relatively resistant to non-depolarizing neuromuscular blockade's effects. When these muscle relaxants are used, which of the following peripheral nerve stimulator twitch patterns is best for monitoring the return of diaphragmatic function?

      Your Answer: Train-of-four stimulation

      Correct Answer: Post-tetanic count stimulation

      Explanation:

      Certain skeletal muscles are more resistant to the effects of neuromuscular blocking agents, both non-depolarizing and depolarizing. The diaphragm is the most resistant. The muscles of the larynx and the corrugator supercilii are less resistant. The abdominal, orbicularis oris, and limb peripheral muscles are the most sensitive muscles.

      Twitch stimulation patterns:

      Supramaximal single stimulus:

      The frequency ranges from 1 Hz to 0.1 Hz (one every second to one every 10 seconds)
      The response is proportional to the frequency of the event.
      It has limited clinical utility because it only tells you whether or not a patient is paralysed (no information on degree of paralysis).

      Over the course of 0.5 seconds (2 Hz), four supramaximal stimulate were applied:

      It is possible to see ‘fade’ and use it as a basis for evaluation.
      This stimulation pattern is used to determine the degree of blockade (1-2 twitches is appropriate for abdominal surgery)
      If the train of four (TOF) count is 1-2, reversal agents can be used in conjunction with medium-acting neuromuscular blocking agents.

      Ratio of TOF:

      This is the ratio of the 4th twitch amplitude to the 1st twitch amplitude.
      The ratio decreases with non-depolarising block and is inversely proportional to the degree of block, allowing objective measurement of residual neuromuscular blockade.
      To achieve adequate reversal, the ratio (as measured by accelerography) must be between 0.7 and 0.9.

      Count of twitches after a tetanic experience(PTC):

      50 Hz for 5 seconds, then a 3 second pause, followed by a single 1 Hz twitch stimulus.
      When the TOF count is zero, this stimulation pattern is used to assess deep blockade (that is, in neurosurgery, microsurgery or ophthalmic surgery when even small movements of a patient will disturb the surgical field)
      It gives an estimate of how long it will take for the response to return to single twitches, allowing assessment of blocks that are too deep for any other technique.
      A palpable post-tetanic count (PTC) of 2 indicates no twitch response for about 20-30 minutes, and a PTC of 5 indicates no twitch response for about 10-15 minutes.

      This is without a doubt the best way to keep track of paralysis in patients who need to avoid diaphragmatic movement. It’s best to use drug infusions and aim for a PTC of 2. After a tetanic stimulus, acetylcholine is mobilised, causing post-tetanic potentiation.

      Stimulation in Two Bursts:

      750 milliseconds between two short bursts of 50 Hz
      This stimulation pattern is used to assess small amounts of residual blockade manually (tactile).

    • This question is part of the following fields:

      • Clinical Measurement
      23.5
      Seconds
  • Question 50 - Which of the following statements is the most correct about ketamine? ...

    Correct

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

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

      Explanation:

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

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

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

    • This question is part of the following fields:

      • Pharmacology
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