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  • Question 1 - A 60-year-old man had previously been diagnosed with Type 2 diabetes. He had...

    Correct

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

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

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

      Your Answer: Chromaffin cells of the adrenal medulla

      Explanation:

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

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

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

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

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

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

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

    • This question is part of the following fields:

      • Anatomy
      207.9
      Seconds
  • Question 2 - 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
      283.1
      Seconds
  • Question 3 - 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
      20.8
      Seconds
  • Question 4 - A 55-year-old businesswoman presents to the emergency department complaining of shortness of breath...

    Incorrect

    • A 55-year-old businesswoman presents to the emergency department complaining of shortness of breath and pleuritic chest pain. Her work requires her to travel internationally frequently. The consultant makes a diagnosis and treats her.

      Now, the consultant recommends placing a filter that will prevent future incidents. A needle is placed into the femoral vein and passed up into the abdomen to insert the filter.

      What is true regarding the organ where the filter is placed for this patient's condition?

      Your Answer: It is attached to the liver via an omentum

      Correct Answer: It is located posteriorly to the peritoneum

      Explanation:

      The patient likely suffers from pulmonary embolism due to her history of frequent international travels. A filter is placed in the inferior vena cava to decrease the risk of future episodes of pulmonary embolism. The IVC filter is a small, wiry device that can catch blood clots and stop them from going into the heart and lungs. Your IVC is a major vessel that brings deoxygenated blood from the lower body to the heart, from where it is pumped into the lungs.

      The filter is placed via a thin catheter inserted into the femoral vein in the groin. The catheter is gently moved up into your IVC, and a filter is introduced.

      The IVC is a retroperitoneal organ.

    • This question is part of the following fields:

      • Anatomy
      414.7
      Seconds
  • Question 5 - The production of carbon dioxide and water occurs during cellular respiration, which involves...

    Incorrect

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

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

      Your Answer: Low carbohydrate, low fat and high protein

      Correct Answer: Low carbohydrate, high fat and low protein

      Explanation:

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

      CO2 produced / O2 consumed = RQ

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

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

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

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

    • This question is part of the following fields:

      • Pathophysiology
      133.3
      Seconds
  • Question 6 - A 46-year-old woman is listed for clipping of a cerebral aneurysm, following a...

    Correct

    • A 46-year-old woman is listed for clipping of a cerebral aneurysm, following a diagnosis of surgical third nerve palsy.

      Which of the following clinical findings correlate with surgical third nerve palsy?

      Your Answer: Ptosis, inferolateral rotation of globe and mydriasis

      Explanation:

      Ptosis and mydriasis are visible in surgical third nerve palsy, and the eye looks ‘down and out.’ The loss of innervation to all of the major structures supplied by the oculomotor nerve is reflected in these characteristics.

      Ptosis is caused by the paralysis of the levator palpebrae superioris in oculomotor nerve palsy. Due to the unopposed actions of the superior oblique and lateral rectus muscles, the eye rotates down and out.

      Mydriasis is caused by surgical (compressive) causes of third nerve palsy, which disrupt the parasympathetic pupillomotor fibres on the nerve’s periphery.

      Medical (ischaemic) causes of a third nerve palsy, on the other hand, leave the superficial parasympathetic fibres relatively unaffected and the pupil unaffected.

      Horner’s syndrome is characterised by ptosis, anhidrosis, and miosis, which are caused by a loss of sympathetic innervation to the tarsal muscle of the upper lid, facial skin, and dilator pupillae, respectively.

    • This question is part of the following fields:

      • Pathophysiology
      250.8
      Seconds
  • Question 7 - Regarding the anatomical relations of the first rib, one of the following is...

    Incorrect

    • Regarding the anatomical relations of the first rib, one of the following is right

      Your Answer: The upper trunk of the brachial plexus grooves the upper surface

      Correct Answer: The subclavius muscle attaches to the upper surface

      Explanation:

      The first rib is an atypical rib. It is short, wide, and flattened and lies in an oblique plane.

      It has a small scalene tubercle on its medial border which marks the point of attachment of scalenus anterior. The lower surface lies on the pleura and is smooth.

      The tubercle on the upper surface separates an anterior groove for the subclavian vein and a posterior groove for the subclavian artery and lower trunk of the brachial plexus.

      Scalenus medius is attached to a roughened area posterior to the groove for the subclavian artery.

      The upper surface gives attachment anteriorly to the subclavius muscle and costoclavicular ligament.

    • This question is part of the following fields:

      • Anatomy
      448.5
      Seconds
  • Question 8 - Regarding adrenocorticotropic hormone (ACTH) one of these is true. ...

    Incorrect

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

      Your Answer: Production is governed by the pituitary

      Correct Answer: Is increased in the maternal plasma in pregnancy

      Explanation:

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

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

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

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

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

    • This question is part of the following fields:

      • Pathophysiology
      352.2
      Seconds
  • Question 9 - A patient visits the radiology department for a magnetic resonance imaging (MRI) scan...

    Correct

    • A patient visits the radiology department for a magnetic resonance imaging (MRI) scan (MRI). The presence of metal implants must be ruled out prior to the scan.

      In a strong magnetic field, which of the following metals is the safest?

      Your Answer: Chromium

      Explanation:

      Ferromagnetism is the property of a substance that is magnetically attracted and can be magnetised indefinitely. A material is said to be paramagnetic if it is attracted to a magnetic field. A substance is said to be diamagnetic if it is repelled by a magnetic field.

      Cobalt, iron, gadolinium, neodymium, and nickel are ferromagnetic.

      Gadolinium is a ferromagnetic rare earth metal that is ferromagnetic below 20 degrees Celsius (its Curie temperature). MRI scans are enhanced with gadolinium-based contrast media.

      When ferromagnetic materials are exposed to a magnetic field, they can cause a variety of issues like magnetic field interactions, heating, and image artefacts.

      Titanium, lead, chromium, copper, aluminium, silver, gold, and tin are non ferromagnetic.

    • This question is part of the following fields:

      • Clinical Measurement
      3.8
      Seconds
  • Question 10 - About low molecular weight (LMW) heparin, Which of these is true? ...

    Correct

    • About low molecular weight (LMW) heparin, Which of these is true?

      Your Answer: Is excreted in the urine

      Explanation:

      Low molecular weight heparin (LMWH) creates a complex by binding to antithrombin. This complex binds with and inactivates factor Xa.

      There is less risk of bleeding with LMWH because it binds less to platelets, endothelium and von Willebrand factor.

      LMW binds Xa more readily. The shorter chains are less likely to bind both antithrombin and thrombin.

      There is need for monitoring in renal impairment because LMHW is excreted in the urine (and partly by hepatic metabolism)

      LMWH have been shown to be as efficacious as unfractionated heparin. It is also safer and have improved inpatient stay and reduced hospital cost.

    • This question is part of the following fields:

      • Pharmacology
      933.7
      Seconds
  • Question 11 - All of the following statements are true about blood clotting except: ...

    Correct

    • All of the following statements are true about blood clotting except:

      Your Answer: Administration of aprotinin during liver transplantation surgery prolongs survival

      Explanation:

      Even though aprotinin reduces fibrinolysis and therefore bleeding, there is an associated increased risk of death. It was withdrawn in 2007.
      Protein C is dependent upon vitamin K and this may paradoxically increase the risk of thrombosis during the early phases of warfarin treatment.

      The coagulation cascade include two pathways which lead to fibrin formation:
      1. Intrinsic pathway – these components are already present in the blood
      Minor role in clotting
      Subendothelial damage e.g. collagen
      Formation of the primary complex on collagen by high-molecular-weight kininogen (HMWK), prekallikrein, and Factor 12
      Prekallikrein is converted to kallikrein and Factor 12 becomes activated
      Factor 12 activates Factor 11
      Factor 11 activates Factor 9, which with its co-factor Factor 8a form the tenase complex which activates Factor 10

      2. Extrinsic pathway – needs tissue factor that is released by damaged tissue)
      In tissue damage:
      Factor 7 binds to Tissue factor – this complex activates Factor 9
      Activated Factor 9 works with Factor 8 to activate Factor 10

      3. Common pathway
      Activated Factor 10 causes the conversion of prothrombin to thrombin and this hydrolyses fibrinogen peptide bonds to form fibrin. It also activates factor 8 to form links between fibrin molecules.

      4. Fibrinolysis
      Plasminogen is converted to plasmin to facilitate clot resorption

    • This question is part of the following fields:

      • Physiology And Biochemistry
      157.4
      Seconds
  • Question 12 - Concerning the anterior pituitary gland, one of following is true. ...

    Incorrect

    • Concerning the anterior pituitary gland, one of following is true.

      Your Answer:

      Correct Answer: Produces glycoproteins

      Explanation:

      The posterior pituitary and the hypothalamus are connected by the pituitary stalk. It contains in the pituitary sella and has the optic chiasm and hypothalamus as superior relations.

      The anterior pituitary produces thyroid-stimulating hormone (TSH), luteinising hormone (LH) and follicle-stimulating hormone (FSH) . These hormones are Glycoproteins and share a common alpha subunit with unique beta subunits.

      The secretion of pituitary hormones are pulsatile. Examples are LH, adrenocorticotropic hormone (ACTH) and growth hormone (GH).

    • This question is part of the following fields:

      • Pathophysiology
      0
      Seconds
  • Question 13 - Which statement is true about the autonomic nervous system? ...

    Incorrect

    • Which statement is true about the autonomic nervous system?

      Your Answer:

      Correct Answer: Preganglionic synapse utilise Acetylcholine as the neurotransmitter in both parasympathetic and sympathetic systems

      Explanation:

      With regards to the autonomic nervous system (ANS)

      1. It is not under voluntary control
      2. It uses reflex pathways and different to the somatic nervous system.
      3. The hypothalamus is the central point of integration of the ANS. However, the gut can coordinate some secretions and information from the baroreceptors which are processed in the medulla.

      With regards to the central nervous system (CNS)
      1. There are myelinated preganglionic fibres which lead to the
      ganglion where the nerve cell bodies of the non-myelinated post ganglionic nerves are organised.
      2. From the ganglion, the post ganglionic nerves then lead on to the innervated organ.

      Most organs are under control of both systems although one system normally predominates.

      The nerves of the sympathetic nervous system (SNS) originate from the lateral horns of the spinal cord, pass into the anterior primary rami and then pass via the white rami communicates into the ganglia from T1-L2.

      There are short pre-ganglionic and long post ganglionic fibres.
      Pre-ganglionic synapses use acetylcholine (ACh) as a neurotransmitter on nicotinic receptors.
      Post ganglionic synapses uses adrenoceptors with norepinephrine / epinephrine as the neurotransmitter.
      However, in sweat glands, piloerector muscles and few blood vessels, ACh is still used as a neurotransmitter with nicotinic receptors.

      The ganglia form the sympathetic trunk – this is a collection of nerves that begin at the base of the skull and travel 2-3 cm lateral to the vertebrae, extending to the coccyx.

      There are cervical, thoracic, lumbar and sacral ganglia and visceral sympathetic innervation is by cardiac, coeliac and hypogastric plexi.

      Juxta glomerular apparatus, piloerector muscles and adipose tissue are all organs under sole sympathetic control.

      The PNS has a craniosacral outflow. It causes reduced arousal and cardiovascular stimulation and increases visceral activity.

      The cranial outflow consists of
      1. The oculomotor nerve (CN III) to the eye via the ciliary ganglion,
      2. Facial nerve (CN VII) to the submandibular, sublingual and lacrimal glands via the pterygopalatine and submandibular ganglions
      3. Glossopharyngeal (CN IX) to lungs, larynx and tracheobronchial tree via otic ganglion
      4. The vagus nerve (CN X), the largest contributor and carries ¾ of fibres covering innervation of the heart, lungs, larynx, tracheobronchial tree parotid gland and proximal gut to the splenic flexure, liver and pancreas

      The sacral outflow (S2 to S4) innervates the bladder, distal gut and genitalia.

      The PNS has long preganglionic and short post ganglionic fibres.
      Preganglionic synapses, like in the SNS, use ACh as the neuro transmitter with nicotinic receptors.
      Post ganglionic synapses also use ACh as the neurotransmitter but have muscarinic receptors.

      Different types of these muscarinic receptors are present in different organs:
      There are:
      M1 = pupillary constriction, gastric acid secretion stimulation
      M2 = inhibition of cardiac stimulation
      M3 = visceral vasodilation, coronary artery constriction, increased secretions in salivary, lacrimal glands and pancreas
      M4 = brain and adrenal medulla
      M5 = brain

      The lacrimal glands are solely under parasympathetic control.

    • This question is part of the following fields:

      • Physiology And Biochemistry
      0
      Seconds
  • Question 14 - Which of the following best describes the phenomenon of higher partial pressures of...

    Incorrect

    • Which of the following best describes the phenomenon of higher partial pressures of oxygen in the alveoli (PAO2) in the lung apices when in the upright position?

      Your Answer:

      Correct Answer: The ventilation/perfusion (V/Q) ratio is greater than in the basal units

      Explanation:

      The ventilation/perfusion ratio varies in different areas of the lung. In an upright individual, although both ventilation and perfusion increase from the apex to the base of the lung, the increase in ventilation is less than the increase in blood flow. As a result, the normal V̇ /Q̇ ratio at the apex of the lung is much greater than 1 (ventilation exceeds perfusion), whereas the V̇ /Q̇ ratio at the base of the lung is much less than 1 (perfusion exceeds ventilation).

      There is more volume in the alveoli found in the apices than in the bases of the lungs. This is due to the weight of the lung stretching the apical alveoli to the maximum size. Also, the weight of the lungs pull themselves away from the chest wall, creating a negative intrapleural pressure. These factors, however, do not directly affect the PAO2.

    • This question is part of the following fields:

      • Pathophysiology
      0
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  • Question 15 - After consuming 12 g of paracetamol, a 37-year-old man is admitted to the...

    Incorrect

    • After consuming 12 g of paracetamol, a 37-year-old man is admitted to the medical admissions unit. He has hepatocellular necrosis in both clinical and biochemical aspects.

      The most significant reason for paracetamol causing toxicity is?

      Your Answer:

      Correct Answer: Glutathione is rapidly exhausted

      Explanation:

      Phase I and phase II metabolism are used by the liver to break down paracetamol.

      1st Phase:

      Prostaglandin synthetase and cytochrome P450 (CYP1A2, CYP2E2, CYP3A4 and CYP2D6) to N-acetyl-p-benzoquinoneimine (NAPQI) and N-acetylbenzo-semiquinoneimine. NAPQI is a toxic metabolite that binds to the sulfhydryl groups of cellular proteins in hepatocytes, making it toxic. This can result in centrilobular necrosis.

      Glutathione and glutathione transferases prevent NAPQI from binding to hepatocytes at low paracetamol doses by preferentially binding to these toxic metabolites. The cysteine and mercapturic acid conjugates are then excreted in the urine. Depletion of glutathione occurs at higher doses of paracetamol, resulting in high levels of NAPQI and the risk of hepatocellular damage. Hepatotoxicity would not be an issue if the body’s glutathione stores were sufficient.

      N-acetylcysteine is a precursor for glutathione synthesis and is the drug of choice for the treatment of paracetamol overdose.

      Phase II:

      Conjugation with glucuronic acid to paracetamol glucuronide is the most common method of metabolism and excretion, accounting for 60% of renally excreted metabolites. Paracetamol sulphate (35%), unchanged paracetamol (5%), and mercapturic acid are among the other renally excreted metabolites (3 percent ). The capacity of conjugation pathways is limited. The capacity of the sulphate conjugation pathway is lower than that of the glucuronidation pathway.

      Because of the low pH in the stomach, paracetamol absorption is minimal (pKa value is 9.5). Paracetamol is absorbed quickly and completely in the alkaline environment of the small intestine. Oral bioavailability is extremely high, approaching 100%.

      As a result, measuring paracetamol levels in plasma after an injury is important. Peak plasma concentrations are reached after 30-60 minutes, with a volume of distribution of 0.95 L/kg. It binds to plasma proteins at a rate of 10% to 25%.

    • This question is part of the following fields:

      • Pharmacology
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  • Question 16 - In endurance athletes, which of the following physiological adaptations to exercise is the...

    Incorrect

    • In endurance athletes, which of the following physiological adaptations to exercise is the best predictor of performance?

      Your Answer:

      Correct Answer: Velocity of blood lactate accumulation

      Explanation:

      Multiple regression analysis revealed that velocity of lactate accumulation (VOBLA) accounted for 92 percent of the variation in marathon running velocity (VM), and VOBLA plus training volume prior to the marathon accounted for 96 percent of the variation. Percent ST muscle fibre distribution (r = 0.55-0.69) and capillary density (r = 052-0.63) were found to be positively correlated with all performance variables. As a result, marathon running performance was linked to VOBLA and the ability to run at a pace close to it during the race. The percent ST, capillary density, and training volume were all related to these properties.

      Another metabolic adaptation compared to normal people is the early selection of fat for oxidation by muscle, especially when glucose availability is limited during high-intensity exercise. This helps to delay the onset of muscle fatigue, but it does not prevent VOBLA.

      For a given level of exercise, training can also result in cardiovascular adaptation, such as increased heart size, increased contractility, and a slower heart rate. All of these factors contribute to an increase in maximal oxygen consumption (VO2 max), but genetic factors, despite intensive training, play a large role in an athlete’s performance.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 17 - What structure is most critical in providing support for the duodenojejunal flexure? ...

    Incorrect

    • What structure is most critical in providing support for the duodenojejunal flexure?

      Your Answer:

      Correct Answer: Ligament of Treitz

      Explanation:

      The duodenojejunal flexure is the point where the duodenum becomes the jejunum.

      The ligament of Treitz, which arises from the right crus of diaphragm, provides suspension for support.

      Between the ileum and the caecum is the ligament of Treves.

    • This question is part of the following fields:

      • Anatomy
      0
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  • Question 18 - During positive pressure ventilation using positive end-expiratory pressure (PEEP), there is usually an...

    Incorrect

    • During positive pressure ventilation using positive end-expiratory pressure (PEEP), there is usually an associated reduction in cardiac output

      Which of the following is responsible?

      Your Answer:

      Correct Answer: Reduced venous return to the heart

      Explanation:

      The option that is most responsible is the progressive decrease in venous return of blood to the right atrium. The heart rate does not usually change with PEEP so the fall in cardiac output is due to a reduction in left ventricular (LV) stroke volume (SV).

      Note that the interventricular septum does shift toward the left and there is an increased pulmonary vascular resistance (PVR) from overdistention of alveolar air sacs that contribute to the reduction in cardiac output. Any increase in PVR will be associated with reduced pulmonary vascular capacitance.

    • This question is part of the following fields:

      • Pathophysiology
      0
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  • Question 19 - Over the course of 10 minutes, one litre of 0.9% normal saline is...

    Incorrect

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

      The following data on urine output is obtained:

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

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

      Your Answer:

      Correct Answer: Increased glomerular filtration rate

      Explanation:

      The following are some basic assumptions:

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

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

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

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

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

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

    • This question is part of the following fields:

      • Pathophysiology
      0
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  • Question 20 - How many unpaired branches leave the abdominal aorta to supply the abdominal viscera?...

    Incorrect

    • How many unpaired branches leave the abdominal aorta to supply the abdominal viscera?

      Your Answer:

      Correct Answer: Three

      Explanation:

      The abdominal arteries are divided into 3 branches;
      – 3 main unpaired trunks (celiac trunk, superior mesenteric, inferior mesenteric arteries)
      – 6 paired branches
      – unpaired median sacral artery.

      We can group the abdominal aorta as follows;
      -Ventral which includes: Coeliac trunk, superior mesenteric and inferior mesenteric arteries
      -Lateral: Inferior phrenic, middle suprarenal, renal and gonadal arteries
      -Dorsal: Lumbar and median sacral arteries
      -Terminal : Right and left common iliac arteries

      The celiac trunk (L1) takes blood the foregut and its found posterior to the stomach. The unpaired superior mesenteric artery supplies blood to the mid-gut.

      The paired renal arteries form the inferior suprarenal arteries. The renal arteries arise around L1/L2 and takes blood to either side of the kidneys.

      The median sacral artery supplies blood to the lumbar vertebrae the L4 and L5.

    • This question is part of the following fields:

      • Anatomy
      0
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  • Question 21 - Which of the given statements is true about standard error of the mean?...

    Incorrect

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

      Your Answer:

      Correct Answer: Gets smaller as the sample size increases

      Explanation:

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

      SEM = SD / square root (n)

      where SD = standard deviation and n = sample size

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

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

      A 95% confidence interval:

      lower limit = mean – (1.96 * SEM)

      upper limit = mean + (1.96 * SEM)

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

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

    • This question is part of the following fields:

      • Statistical Methods
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  • Question 22 - The tip of a pulmonary artery flotation catheter becomes wedged when threaded through...

    Incorrect

    • 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:

      Correct 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
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  • Question 23 - A normal woman at term, not in labour, has her arterial blood gas...

    Incorrect

    • A normal woman at term, not in labour, has her arterial blood gas analysed.

      Which set of results is most likely her own?

      Option - pH - PaCO2 - HCO3 - PaO2
      A - 7.35 - 28 mmHg (3.73 kPa) - 27 mmol/L - 104 mmHg (13.8kPa)
      B - 7.43 - 32 mmHg (4.27 kPa) - 21 mmol/L - 104 mmHg (13.8kPa)
      C - 7.44 - 36 mmHg (4.8 kPa) - 27 mmol/L - 104 mmHg (13.8kPa)
      D - 7.45 - 40 mmHg (5.33 kPa) - 21 mmol/L - 104 mmHg (13.8kPa)
      E - 7.46 - 44 mmHg (5.87kPa) - 21 mmol/L - 104 mmHg (13.8kPa)

      Your Answer:

      Correct Answer: B

      Explanation:

      Due to an increased tidal volume with little change or slight increase in respiratory rate, Minute ventilation at term is increased by about 50%. Hypothalamic function are thought to influence by Progesterone, oestradiol and prostaglandins. This causes a mild compensated respiratory alkalosis.

      Maternal PaCO2 is usually decreased to about 32 mmHg (4.27 kPa) as a result of this increased alveolar ventilation at term . A compensatory decrease in serum bicarbonate from 27 to 21 mmol/L by renal excretion lessens the impact of maternal alkalosis.

    • This question is part of the following fields:

      • Physiology And Biochemistry
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  • Question 24 - The following statements are about burns patients. Which one is true? ...

    Incorrect

    • The following statements are about burns patients. Which one is true?

      Your Answer:

      Correct Answer: High protein diets may improve survival

      Explanation:

      Patients who have sustained thermal injuries are at high risk of becoming hypercatabolic with larger cardiac outputs and oxygen consumptions.

      The hypermetabolic states increase with an increase in the burn severity and surface area of the skin affected. A patient with thermal injuries affecting 60% of the total surface area of the body will have twice the normal metabolic rate.

      The optimal temperature for nursing patients with burn injuries is 30°C to conserve the energy usage. The areas affected by the burn injuries should be covered to reduce loss of fluid via evaporation. Resetting hypothalamic thermoregulation will cause a 1-2°C increase in core temperature.

      Burn injuries will have an immediate effect on the intestine, destroying the barrier function and allowing for the movement of bacteria and endotoxins within hours.

      Enteral nutrition allows for the delivery of nutrients directly to the stomach or intestine. It has correlation with a dampened hypermetabolic response to a thermal and injury, especially when initiated early as it helps to protect the integrity of the mucosal lining and prevents the movement of bacteria into circulation.

      Diet changes have been linked to reduced mortality due to burn injuries. Diets high in protein especially (calorie: nitrogen ratio of 100: 1), have the highest correlation with improved survival rates.

      Parenteral feeds may be required alongside enteral nutrition, even with the increased risks of infection.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 25 - Which of the following statements is true about the in-hospital management of ventricular...

    Incorrect

    • Which of the following statements is true about the in-hospital management of ventricular fibrillation?

      Your Answer:

      Correct Answer: Amiodarone may be administered following a third DC shock

      Explanation:

      Ventricular fibrillation (VT) is an arrhythmia caused by a distortion in the organized contraction of the ventricles leading to an inability to pump blood out into the body.

      Amiodarone is an anti arrhythmic drug used for the treatment of ventricular and atrial fibrillations. It is the gold standard of treatment for refractory pulseless ventricular tachycardia (VT) and ventricular fibrillation (VF).

      Guidelines for emergency treatment state that only the rescuer carrying out chest compressions on the patient may stand near the defibrillator as it charges.

      Cardio-pulmonary resuscitation (CPR) during cardiac arrest is required for 2 minute cycles.

      Hypovolaemia is as a cause of pulseless electrical activity (PEA) can be reversed using fluid resuscitation, whereas hypotension during cardiac arrest is either persistent or undetectable and is therefore irreversible.

      Hyperkalaemia and hypocalcaemia are treated using calcium salts, but calcium chloride is often preferred over calcium gluconate.

      During a pulseless VT or VF, a single precordial thump will be effective if administered within the first seconds of the occurrence of a shockable rhythm.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 26 - Under general anaesthesia, a 48-year-old patient is scheduled for some dental extractions. He...

    Incorrect

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

      Which antibiotic prophylaxis strategy is most appropriate for this patient?

      Your Answer:

      Correct Answer: Prophylactic antibiotics are unnecessary for this patient

      Explanation:

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

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

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

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

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

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

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

    • This question is part of the following fields:

      • Pharmacology
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  • Question 27 - A patient admitted to the hospital is on oxygen via a venturi mask....

    Incorrect

    • A patient admitted to the hospital is on oxygen via a venturi mask.
      The air entrainment ratio is 1:9 i.e. 1 litre of 100% oxygen from the source entrains 9 litres of air from the atmosphere). The flow rate of 100% oxygen is 6L/minute.

      Based on the given data which of the following value approximates the oxygen concentration delivered to the patient?

      Your Answer:

      Correct Answer: 28%

      Explanation:

      The formula for calculating air: oxygen entrainment ratio is given as :
      100% − FiO2 = air/oxygen entrainment ratio
      Since FiO2 − 21% and the entrainment ratio is already known. Substituting the values in the equation: x = FiO2.

      100 − x = 9
      x − 21
      100 − x = 9(x − 21)
      100 − x = 9x − 189
      10x = 289
      x = 289/10
      x = 28.9%

    • This question is part of the following fields:

      • Basic Physics
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  • Question 28 - Concerning platelets one of the following is true ...

    Incorrect

    • Concerning platelets one of the following is true

      Your Answer:

      Correct Answer: Are formed in the bone marrow from megakaryocytes

      Explanation:

      Platelets are fragments of megakaryocytes and they are encapsulated by membrane.

      They have no nucleus but are metabolically active and are able to express membrane receptors and release stored substances when triggered. adenosine diphosphate and serotonin are 2 of its content.

      Because they have no nucleus, they are not able to produce new proteins. This is why aspirin and other drugs affect function for their entire lifespan after exposure. Its lifespan is approximately 9-10 days in normal individuals.

      Platelets does NOT PRODUCE prostacyclin but are able to produce nitric oxide, prostaglandins and thromboxane.

    • This question is part of the following fields:

      • Pathophysiology
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  • Question 29 - A randomized controlled trail has been conducted to compare two drugs used for...

    Incorrect

    • A randomized controlled trail has been conducted to compare two drugs used for the early management of acute severe asthma in the emergency department. After being allocated to the randomized groups, many patients have been excluded due to deleterious effect to the drugs.

      How the data would be analysed?

      Your Answer:

      Correct Answer: Include the patients who drop out in the final data set

      Explanation:

      Randomized controlled trails will be analysed by the intention-to-treat (ITT) approach. It provides unbiased comparisons among the treatment groups. ITT analyses are done to avoid the effects of dropout, which may break the random assignment to the treatment groups in a study.

      ITT analysis is a comparison of the treatment groups that includes all patients as originally allocated after randomization.

      In order to include such participants in an analysis, outcome data could be imputed which involves making assumptions about the outcomes in the lost participants.

    • This question is part of the following fields:

      • Statistical Methods
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  • Question 30 - A new clinical trial evaluates the effect of a new drug Z on...

    Incorrect

    • A new clinical trial evaluates the effect of a new drug Z on all-cause mortality. The rate of death in the group receiving this drug is 8%, compared with 16% in the control group.

      What is the number needed to treat with drug Z to prevent death?

      Your Answer:

      Correct Answer: 13

      Explanation:

      Number needed to treat is a measure of the impact of a treatment or intervention that is often used to communicate results to patients, clinicians, the public and policymakers. It states how many patients need to be treated for one additional patient to experience an adverse outcome (e.g. a death).

      It is calculated as the inverse of the absolute risk reduction and is rounded to the next highest whole number.

      The absolute risk reduction is 8% (16% – 8%). 100/8 = 12.5, so rounding up the next integer this gives at NNT of 13. i.e. you would need to give the new drug to 13 people to ensure that you prevented one death.

    • This question is part of the following fields:

      • Statistical Methods
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SESSION STATS - PERFORMANCE PER SPECIALTY

Anatomy (1/3) 33%
Pathophysiology (2/4) 50%
Statistical Methods (1/1) 100%
Clinical Measurement (1/1) 100%
Pharmacology (1/1) 100%
Physiology And Biochemistry (1/1) 100%
Passmed