-
Question 1
Correct
-
What is the basic chemical reaction that takes place in the breakdown of complex foodstuffs?
Your Answer: Hydrolysis
Explanation:Breakdown of complex food into simpler compounds is achieved by hydrolysis, with the help of different enzymes specific for different compounds.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 2
Incorrect
-
A 78-year-old diabetic man undergoes renal function tests. Which of the following substances will be the most accurate for measuring glomerular filtration rate (GFR)?
Your Answer:
Correct Answer: Inulin
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal glomerular capillaries into the Bowman’s capsule per unit time. Clinically, this is often measured to determine renal function. Inulin was originally used as it is not reabsorbed by the kidney after glomerular filtration, therefore its rate of excretion is directly proportional to the rate of filtration of water and solutes across the glomerular filter. However, in clinical practice, creatinine clearance is used to measure GFR. Creatinine is an endogenous molecule, synthesised in the body, that is freely filtered by the glomerulus (but also secreted by the renal tubules in very small amounts). Creatinine clearance exceeds GFR due to creatinine secretion, and is therefore a close approximation of the GFR.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 3
Incorrect
-
C5a (a complement component) is a potent?
Your Answer:
Correct Answer: Anaphylotoxin
Explanation:C5a is a strong chemoattractant as well as an anaphylotoxin and is involved in the recruitment of inflammatory cells such as neutrophils, eosinophils, monocytes, and T lymphocytes. It is also involved in activation of phagocytic cells, release of granule-based enzymes and generation of oxidants. All of which contribute to innate immune functions.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 4
Incorrect
-
A suspected recreational drug user was brought to the Emergency department in an unconscious state, and was found to be hypoventilating. Which of the following set of arterial blood gas analysis report is most consistent with hypoventilation as the primary cause? pH, pa(CO2) (mmHg), pa(O2) (mmHg).
Your Answer:
Correct Answer: 7.28, 55, 81
Explanation:Hypoventilation (or respiratory depression) causes an increase in carbon dioxide (hypercapnia) and respiratory acidosis. It can result due to drugs such as alcohol, benzodiazepines, barbiturates, opiates, mechanical conditions or holding ones breath. Strong opioids such as heroin and fentanyl are commonly implicated and can lead to respiratory arrest. In recreational drug overdose, acute respiratory acidosis occurs with an increase in p(CO2) over 45 mm Hg and acidaemia (pH < 7.35)
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 5
Incorrect
-
Which of the following is a true statement regarding the pupil?
Your Answer:
Correct Answer: Phentolamine causes pupil constriction
Explanation:A balance between the sympathetic tone to the radial fibres of the iris and parasympathetic tone to the pupillary sphincter muscle determines the pupil size. Phentolamine (α-adrenergic receptor blocker) causes pupillary constriction. Dilatation of the pupil occurs with increased sympathetic activity, decreased parasympathetic activity during darkness or block of muscarinic receptors by atropine.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 6
Incorrect
-
A 30-year-old woman feels thirsty. This thirst is probably due to:
Your Answer:
Correct Answer: Increased level of angiotensin II
Explanation:Thirst is the basic need or instinct to drink. It arises from a lack of fluids and/or an increase in the concentration of certain osmolites such as salt. If the water volume of the body falls below a certain threshold or the osmolite concentration becomes too high, the brain signals thirst. Excessive thirst, known as polydipsia, along with excessive urination, known as polyuria, may be an indication of diabetes. Angiotensin II is a hormone that is a powerful dipsogen (i.e. it stimulates thirst) that acts via the subfornical organ. It increases secretion of ADH in the posterior pituitary and secretion of ACTH in the anterior pituitary.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 7
Incorrect
-
A 65-year-old man complains of headaches, weakness, cramps, and confusion; blood tests reveal he has severe hyponatremia. The most likely cause is:
Your Answer:
Correct Answer: Severe diarrhoea or vomiting
Explanation:Hyponatraemia occurs when the sodium level in the plasma falls below 135 mmol/l. Hyponatraemia is an abnormality that can occur in isolation or, more commonly as a complication of other medical illnesses. Severe hyponatraemia may cause osmotic shift of water from the plasma into the brain cells. Typical symptoms include nausea, vomiting, headache and malaise. As the hyponatraemia worsens, confusion, diminished reflexes, convulsions, stupor or coma may occur. The cause of hyponatremia is typically classified by a person’s fluid status into low volume, normal volume, and high volume. Low volume hyponatremia can occur from diarrhoea, vomiting, diuretics, and sweating.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 8
Incorrect
-
Purkinje fibres in the heart conduct action potentials at the rate of:
Your Answer:
Correct Answer: 1.5–4.0 m/s
Explanation:Purkinje fibres control the heart rate along with the sinoatrial node (SA node) and the atrioventricular node (AV node). The QRS complex is associated with the impulse passing through the Purkinje fibres. These fibres conduct action potential about six times faster than the velocity in normal cardiac muscle.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 9
Incorrect
-
Causes of metabolic acidosis with a normal anion gap include:
Your Answer:
Correct Answer: Diarrhoea
Explanation:Excess acid intake and excess bicarbonate loss as in diarrhoea, are causes of metabolic acidosis with a normal anion gap. The other conditions all result in an increased anion gap.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 10
Incorrect
-
What is the normal duration of PR interval on an electrocardiogram of a healthy individual?
Your Answer:
Correct Answer: 0.12–0.20 s
Explanation:PR interval extends from the beginning of the P-wave until the beginning of the QRS complex. The normal duration of the PR interval is 0.12-0.20 s. It can be prolonged in first degree heart block, and reduced in Wolff-Parkinson-White syndrome.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 11
Incorrect
-
Which of the following physiological changes will you see in a young man who has been trekking in the Himalayas for 3 years?
Your Answer:
Correct Answer: Increased renal excretion of HCO3 –
Explanation:The atmospheric pressure is lower at high altitudes as compared with sea level. This leads to a decrease in the partial pressure of oxygen. Once 2100 m (7000 feet) of altitude is reached, there is a drop in saturation of oxyhaemoglobin. The oxygen saturation of haemoglobin determines the oxygen content in the blood. The body physiological tries to adapt to high altitude by acclimatization. Immediate effects include hyperventilation, fluid loss (due to a decreased thirst drive), increase in heart rate and slightly lowered stroke volume. Long term effects include lower lactate production, compensatory alkali loss in urine, decrease in plasma volume, increased erythropoietin release and red cell mass, increased haematocrit, higher concentration of capillaries in striated muscle tissue, increase in myoglobin, increase in mitochondria, increase in aerobic enzyme concentration such as 2,3-DPG and pulmonary vasoconstriction.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 12
Incorrect
-
The most likely cause of a low p(O2) in arterial blood is:
Your Answer:
Correct Answer: Hypoxic hypoxia
Explanation:Hypoxia is when the whole body or a region is deprived of adequate oxygen supply. Different types of hypoxia include the following:
– Hypoxic hypoxia, which occurs due to poor oxygen supply, as a result of low partial pressure of oxygen in arterial blood. This could be due to low partial pressure of atmospheric oxygen (e.g., at high altitude), sleep apnoea, poor ventilation because of chronic obstructive pulmonary disease or respiratory arrest, or shunts. The other types of hypoxia have a normal partial pressure of oxygen.
– Anaemic hypoxia occurs due to low total oxygen content of the blood, with a normal arterial oxygen pressure.
– Hyperaemic hypoxia occurs due to poor delivery of oxygen to target tissues, such as in carbon monoxide poisoning or methemoglobinemia.
– Histotoxic hypoxia results due to inability of the cells to use the delivered oxygen due to disabled oxidative phosphorylation enzymes.
– Ischaemic (or stagnant) hypoxia occurs due to local flow restriction of well-oxygenated blood, seen in cases like cerebral ischaemia, ischaemic heart disease and intrauterine hypoxia.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 13
Incorrect
-
A sudden loud sound is more likely to result in cochlear damage than a slowly developing loud sound. This is because:
Your Answer:
Correct Answer: There is a latent period before the attenuation reflex can occur
Explanation:On transmission of a loud sound into the central nervous system, an attenuation reflex occurs after a latent period of 40-80 ms. This reflex contracts the two muscles that pull malleus and stapes closer, developing a high degree of rigidity in the entire ossicular chain. This reduces the ossicular conduction of low frequency sounds to the cochlea by 30-40 decibels. In this way, the cochlea is protected from damage due to loud sounds (these are low frequency sounds) when they develop slowly.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 14
Incorrect
-
The majority of gallstones are mainly composed of:
Your Answer:
Correct Answer: Cholesterol
Explanation:Bile salts are formed out of cholesterol in the liver cells. Occasionally, precipitation of cholesterol occurs resulting into cholesterol stones developing in the gall bladder.
These cholesterol gallstones are the most common type and account for 80% of all gallstones. Another type, accounting for 20% gallstones is pigment stones which are composed of bilirubin and calcium salts. Occasionally, stones of mixed origin are also seen.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 15
Incorrect
-
If the blood flow is constant, oxygen extraction by tissues will show the greatest decrease due to which of the following interventions?
Your Answer:
Correct Answer: Tissue cooling
Explanation:With a constant blood flow to a given tissue bed, there will be an increase in oxygen extraction by the tissue with the following; an increase in tissue metabolism and oxygen requirements: warming (or fever), exercise, catecholamines and thyroxine. With cooling, the demand for oxygen decreases, leading to decreased oxygen extraction.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 16
Incorrect
-
The anatomical dead space in a patient with low oxygen saturation, is 125 ml, with a tidal volume of 500 ml and pa(CO2) of 40 mm Hg. The dead space was determined by Fowler's method. If we assume that the patient's lungs are healthy, what will his mixed expired CO2 tension [pE(CO2)] be?
Your Answer:
Correct Answer: 30 mmHg
Explanation:According to Bohr’s equation, VD/VT = (pA(CO2) − pE(CO2))/pA(CO2), where pE(CO2) is mixed expired CO2 and pA(CO2) is alveolar CO2pressure. Normally, the pa(CO2) is virtually identical to pA(CO2). Thus, VD/VT = (pa(CO2)) − pE(CO2)/pa(CO2). By Fowler’s method, VD/VT= 0.25. In the given problem, (pa(CO2) − pE(CO2)/pa(CO2) = (40 − pE(CO2)/40 = 0.25. Thus, pE(CO2) = 30 mmHg. If there is a great perfusion/ventilation inequality, pE(CO2) could be significantly lower than 30 mm Hg, and the patient’s physiological dead space would exceed the anatomical dead space.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 17
Incorrect
-
Which of the following is true regarding factor XI?
Your Answer:
Correct Answer: Deficiency causes haemophilia C
Explanation:Factor XI is also known as plasma thromboplastin and is one of the enzymes of the coagulation cascade. It is produced in the liver and is a serine protease. It is activated by factor XIIa, thrombin and by itself. Deficiency of factor XI causes the rare type of haemophilia C. Low levels of factor XI also occur in other disease states, including Noonan syndrome. High levels of factor XI have been seen in thrombosis.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 18
Incorrect
-
A 7-year-old girl is given cephalexin to treat an infection and develops hives, with localised facial oedema. Which of the following conditions will cause localised oedema?
Your Answer:
Correct Answer: Angio-oedema
Explanation:Angio-oedema, is the rapid swelling of the skin, mucosa and submucosal tissues. The underlying mechanism typically involves histamine or bradykinin. The version related to histamine is to due an allergic reaction to agents such as insect bites, food, or medications. The version related to bradykinin may occur due to an inherited C1 esterase inhibitor deficiency, medications e.g. angiotensin converting enzyme inhibitors, or a lymphoproliferative disorder.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 19
Incorrect
-
Which of the following proteins prevents red blood cells (RBCs) from bursting when they pass through capillaries?
Your Answer:
Correct Answer: Spectrin
Explanation:Spectrin is a structural protein found in the cytoskeleton that lines the intercellular side of the membrane of cells which include RBCs. They maintain the integrity and structure of the cell. It is arranged into a hexagonal arrangement formed from tetramers of spectrin and associated with short actin filaments that form junctions allowing the RBC to distort its shape.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 20
Incorrect
-
A 50 year old woman presented with excessive bleeding after an inguinal hernia repair. Labs are suggestive of a primary haemostasis defect. Deficiency of which of the following is most likely to cause it?
Your Answer:
Correct Answer: Platelets
Explanation:Primary haemostatic control means the first line of defence against immediate bleeding. This is carried out by the platelets. They immediately form a haemostatic plug at the site of injury. Coagulation starts within 20s after an injury to the blood vessel which damage the endothelial cells. Secondary haemostasis follows which includes activation of the coagulation factors to form fibrin strands which mesh together forming the platelet plug. Platelets interact with platelet collagen receptor, glycoprotein Ia/IIa and to collagen fibres in the vascular endothelium. This adhesion is mediated by von Willebrand factor (vWF), which forms links between the platelet glycoprotein Ib/IX/V and collagen fibrils. The platelets are then activated and release the contents of their granules into the plasma, in turn activating other platelets and white blood cells.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 21
Incorrect
-
A medical student is told a substance is freely filtered but is not metabolised, secreted, or stored in the kidney. It has a plasma concentration of 1000 mg/l and its urine excretion rate is 25 mg/min, and the inulin clearance is 100 ml/min. What is the rate of tubular reabsorption of the substance?
Your Answer:
Correct Answer: 75 mg/min
Explanation:Reabsorption or tubular reabsorption is the process by which the nephron removes water and solutes from the tubular fluid (pre-urine) and returns them to the circulating blood. To calculate the reabsorption rate of substance Z we use the following equation: excretion = (filtration + secretion) – reabsorption. As this substance is freely filtered, its filtration rate is equal to that of inulin. So 25 = (100 + 0) – reabsorption. Reabsorption = 100 – 25 therefore reabsorption = 75 mg/min.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 22
Incorrect
-
Which is a feature of the action of insulin?
Your Answer:
Correct Answer: Promotes protein synthesis
Explanation:Insulin is produced by the beta-cells of the islets of Langerhans in the pancreas. Its actions include:
– promoting uptake of glucose into cells
– glycogen synthesis (glycogenesis)
– protein synthesis
– stimulation of lipogenesis (fat formation).
– driving potassium into cells – used to treat hyperkaelamia.
Parathyroid hormone and activated vitamin D are the principal hormones involved in calcium/phosphate metabolism, rather than insulin.
-
This question is part of the following fields:
- Endocrine
- Physiology
-
-
Question 23
Incorrect
-
The mechanism of action of streptokinase involves:
Your Answer:
Correct Answer: Direct conversion of plasminogen to plasmin
Explanation:Streptokinase is an enzyme that is produced by group A beta haemolytic streptococcus and is an effective and cost efficient method for the dissolution of a clot used in cases of MI and pulmonary embolism. It works by directly converting plasminogen to plasmin which breaks down the blood components in the clot and fibrin, dissolving the clot. Streptokinase is a bacterial product and thus the body will develop immunity against it.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 24
Incorrect
-
Glucose is the most important source of energy for cellular respiration. The transport of glucose in the renal tubular cells occurs via:
Your Answer:
Correct Answer: Secondary active transport with sodium
Explanation:In 1960, Robert K. Crane presented for the first time his discovery of the sodium-glucose cotransport as the mechanism for glucose absorption. Glucose transport through biological membranes requires specific transport proteins. Transport of glucose through the apical membrane of renal tubular as well as intestinal epithelial cells depends on the presence of secondary active Na+–glucose symporters, SGLT-1 and SGLT-2, which concentrate glucose inside the cells, using the energy provided by co-transport of Na+ ions down their electrochemical gradient.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 25
Incorrect
-
Dysarthria, nystagmus and a tremor worsening with directed movement are likely to be seen in:
Your Answer:
Correct Answer: Cerebellar disease
Explanation:The given symptoms are seen in diseases affecting the cerebellum. A cerebellar tremor is a slow tremor that occurs at the end of a purposeful movement. It is seen in cerebellar disease, such as multiple sclerosis or some inherited degenerative disorders and chronic alcoholism. Classically, tremors are produced in the same side of the body as a one-sided lesion. Cerebellar disease can also result in a wing-beating’ type of tremor called rubral or Holmes’ tremor – a combination of rest, action and postural tremors. Other signs of cerebellar disease include dysarthria (speech problems), nystagmus (rapid, involuntary rolling of the eyes), gait problems and postural tremor of the trunk and neck.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 26
Incorrect
-
Which of these substances is secreted by pericytes in the juxtaglomerular cells?
Your Answer:
Correct Answer: Renin
Explanation:The juxtaglomerular cells synthesise, store and secrete the enzyme renin in the kidney. They are specialised smooth muscle cells in the wall of the afferent arteriole that delivers blood to the glomerulus and thus play a critical role in the renin– angiotensin system and so in renal autoregulation.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 27
Incorrect
-
Which of the following brings about a reduction in gastric blood flow?
Your Answer:
Correct Answer: Vasopressin
Explanation:Gastric blood flow is increased by vagal stimulation, gastrin, histamine and acetylcholine as they stimulate gastric section and the production of vasodilator metabolites. Acetylcholine and histamine also have a direct action on the gastric arterioles. Similarly, gastric blood flow is reduced by inhibitors of secretion – catecholamines, secretin and vasopressin.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 28
Incorrect
-
Some substances, such as Chromium-51 and Technetium-99, are freely filtered but not secreted or absorbed by the kidney. In these cases, their clearance rate is equal to:
Your Answer:
Correct Answer: Glomerular filtration rate
Explanation:If a substance passes through the glomerular membrane with perfect ease, the glomerular filtrate contains virtually the same concentration of the substance as does the plasma and if the substance is neither secreted nor reabsorbed by the tubules, all of the filtered substance continues on into the urine. Glomerular filtration rate (GFR) describes the flow rate of filtered fluid through the kidney.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 29
Incorrect
-
Which of the following organs is most likely to have dendritic cells?
Your Answer:
Correct Answer: Skin
Explanation:Dendritic cells are part of the immune system and they function mainly as antigen presenting cells. They are present in small quantities in tissues which are in contact in the external environment. Mainly in the skin and to a lesser extent in the lining of the nose, lungs, stomach and intestines. In the skin they are known as Langerhans cells.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 30
Incorrect
-
what is the cause of a prolonged PT(prothrombin time)?
Your Answer:
Correct Answer: Liver disease
Explanation:PT measure the intrinsic pathway of coagulation. It determines the measure of the warfarin dose regime, liver disease and vit K deficiency status along with the clotting tendency of blood. PT measured factors are II,V,VII,X and fibrinogen. It is used along with aPTT which measure the intrinsic pathway.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 31
Incorrect
-
When the pitch of a sound increases, what is the physiological response seen in the listener?
Your Answer:
Correct Answer: The location of maximal basilar membrane displacement moves toward the base of the cochlea
Explanation:An increase in the frequency of sound waves results in a change in the position of maximal displacement of the basilar membrane in the cochlea. Low pitch sound produces maximal displacement towards the cochlear apex and greatest activation of hair cells there. With an increasing pitch, the site of greatest displacement moves towards the cochlear base. However, increased amplitude of displacement, increase in the number of activated hair cells, increased frequency of discharge of units in the auditory nerve and increase in the range of frequencies to which such units respond, are all seen in increases in the intensity or a sound stimulus.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 32
Incorrect
-
After a prolonged coronary artery bypass surgery, a 60-year old gentleman was transfused 3 units of fresh-frozen plasma and 2 units of packed red cells. Two days later, the nurse noticed that he was tachypnoeic and chest X-ray showed signs consistent with adult respiratory distress syndrome. Which of the following variables will be low in this patient?
Your Answer:
Correct Answer: Compliance of the lung
Explanation:Acute or adult respiratory distress syndrome (ARDS) is a reaction to several forms of lung injuries and is commonly associated with sepsis and SIRS (systemic inflammatory response syndrome), severe traumatic injury, severe head injury, narcotics overdose, drowning, pulmonary contusion, and multiple blood transfusions. There is an increase in risk due to pre-existing liver disease or coagulation abnormalities. It results due to indirect toxic effects of neutrophil-derived inflammatory mediators in the lungs. ARDS is defined by the 1994 American–European Consensus Committee as the acute onset of bilateral infiltrates on chest X-ray, a partial pressure of arterial oxygen (pa(O2)) to fraction of inspired oxygen Fi(O2) ratio of less than 200 mmHg and a pulmonary artery occlusion pressure of less than 18 or the absence of clinical evidence of left arterial hypertension. ARDS is basically pulmonary oedema in the absence of volume overload or poor left ventricular function. This is different from acute lung injury, which shows a pa(O2)/Fi(O2) ratio of less than 300 mmHg. Pathogenesis of ARDS starts from damage to alveolar epithelium and vascular endothelium, causing increased permeability. Damage to surfactant-producing type II cells disrupts the production and function of pulmonary surfactant, causing micro atelectasis and poor gas exchange. There is a decrease in lung compliance and increase in work of breathing. Eventually, there is resorption of alveolar oedema, regeneration of epithelial cells, proliferation and differentiation of type II alveolar cells and alveolar remodelling. Some show resolution and some progress to fibrosing alveolitis, which involves the deposition of collagen in alveolar, vascular and interstitial spaces.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 33
Incorrect
-
A recognised side-effect of prefrontal leukotomy is:
Your Answer:
Correct Answer: Confusion
Explanation:Used previously as a treatment for psychiatric disorders, prefrontal leucotomy severs the connection between the prefrontal cortical association area and the thalamus. This leads to functional isolation of the prefrontal and orbitofrontal association cortex. Thus, along with the desired reduction in anger and frustration, undesirable side effects included changes in mood and affect, as well as confusion.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 34
Incorrect
-
Work of breathing (WOB) is the energy expended to inhale and exhale a breathing gas. Normally, maximal amount of work of breathing is required to overcome:
Your Answer:
Correct Answer: Elastic lung compliance
Explanation:The forces of elastance (compliance), frictional resistance and inertia have been identified as the forces that oppose lung inflation and deflation. The normal relaxed state of the lung and chest is partially empty. Further exhalation requires muscular work. Inhalation is an active process requiring work. About 60–66% of the total work performed by the respiratory muscles is used to overcome the elastic or compliance characteristics of the lung–chest cage, 30–35% is used to overcome frictional resistance and only 2–5% of the work is used for inertia.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 35
Incorrect
-
Which of the following substances is most likely to cause pulmonary vasodilatation?
Your Answer:
Correct Answer: Nitric oxide
Explanation:In the body, nitric oxide is synthesised from arginine and oxygen by various nitric oxide synthase (NOS) enzymes and by sequential reduction of inorganic nitrate. The endothelium of blood vessels uses nitric oxide to signal the surrounding smooth muscle to relax, so dilating the artery and increasing blood flow. Nitric oxide/oxygen blends are used in critical care to promote capillary and pulmonary dilation to treat primary pulmonary hypertension in neonatal patients post-meconium aspiration and related to birth defects.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 36
Incorrect
-
A 45-year-old man with short bowel syndrome requires parenteral nutrition. The solution of choice for parenteral nutrition is:
Your Answer:
Correct Answer: Crystalline amino acids
Explanation:Total parenteral nutrition (TPN), is the practice of feeding a person intravenously, circumventing the gut. It is normally used in the following situations: surgery, when feeding by mouth is not possible, when a person’s digestive system cannot absorb nutrients due to chronic disease or if a person’s nutrient requirement cannot be met by enteral feeding and supplementation. A sterile bag of nutrient solution, between 500 ml and 4L, is provided. The pump infuses a small amount (0.1–10 ml/h) continuously to keep the vein open. The nutrient solution consists of water, glucose, salts, amino acids, vitamins and sometimes emulsified fats. Ideally each patient is assessed individually before commencing on parenteral nutrition, and a team consisting of doctors, nurses, clinical pharmacists and dietitians evaluate the patient’s individual data and decide what formula to use and at what rate.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 37
Incorrect
-
What is the pH of freshly formed saliva at ultimate stimulation?
Your Answer:
Correct Answer: 8
Explanation:Saliva has four major components: mucus (lubricant), α-amylase (enzyme that initiates digestion of starch), lingual lipase (enzyme that begins fat digestion), and a slightly alkaline electrolyte solution for moistening food. As the secretion rate of saliva increases, its osmolality increases. Moreover, the pH changes from slightly acidic (at rest) to basic (pH 8) at ultimate stimulation. This occurs due to increase of HCO3-. Amylase and mucus also increase in concentration after stimulation.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 38
Incorrect
-
A 77-year-old woman's renal function is tested. The following results were obtained during a 24-h period:
Urine flow rate: 2. 0 ml/min
Urine inulin: 0.5 mg/ml
Plasma inulin: 0.02 mg/ml
Urine urea: 220 mmol/l
Plasma urea: 5 mmol/l.
What is the urea clearance?Your Answer:
Correct Answer: 88 ml/min
Explanation:Urea is reabsorbed in the inner medullary collecting ducts of the nephrons. The clearance (C) of any substance can be calculated as follows: C = (U × V)/P, where U and P are the urine and plasma concentrations of the substance, respectively and V is the urine flow rate. So, glomerular filtration rate = (0.220 × 2. 0)/0.005 = 88 ml/min.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 39
Incorrect
-
A 70-year-old male who has smoked since his teens complains of progressive shortness of breath and a persistent cough. He is diagnosed with COPD. Which of the following abnormalities is most likely to be present in his pulmonary function tests?
Your Answer:
Correct Answer: Increased residual volume
Explanation:Chronic obstructive pulmonary disease (COPD) is a type of obstructive lung disease characterized by long-term poor airflow. The main symptoms include shortness of breath and cough with sputum production. The best diagnostic test for evaluating patients with suspected chronic obstructive pulmonary disease (COPD) is lung function measured with spirometry. Key spirometrical measures may be obtained with a portable office spirometer and should include forced vital capacity (FVC) and the normal forced expiratory volume in the first second of expiration (FEV1). The ratio of FEV1 to forced vital capacity (FEV1/FVC) normally exceeds 0.75. Patients with COPD typically present with obstructive airflow. Complete pulmonary function testing may show increased total lung capacity, functional residual capacity and residual volume. A substantial loss of lung surface area available for effective oxygen exchange causes diminished carbon monoxide diffusion in the lung (DLco) in patients with emphysema. Tobacco smoking is the most common cause of COPD, with factors such as air pollution and genetics playing a smaller role.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 40
Incorrect
-
Decreased velocity of impulse conduction through the atrioventricular node (AV node) in the heart will lead to:
Your Answer:
Correct Answer: Increased PR interval
Explanation:AV node damage may lead to an increase in the PR interval to as high as 0.25 – 0.40 s (normal = 0.12 – 0.20 s). In the case of severe impairment, there might be a complete failure of passage of impulses leading to complete block. In this case, the atria and ventricles will beat independently of each other.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 41
Incorrect
-
During cardiac catheterisation in a 20-year old man, the following data is obtained: Pressure (mmHg), O2 saturation (%) Right atrium 7 (N = 5) 90 (N = 75), Right ventricle 35/7 (N = 25/5) 90 (N = 75), Pulmonary artery 35/8 (N = 25/15), 90 (N = 75), Left atrium 7 (N = 9) 95 (N = 95), Left ventricle 110/7 (N = 110/9) 95 (N = 95), Aorta 110/75 (N = 110/75) 95 (N = 95) where N = Normal value. What is the likely diagnosis?
Your Answer:
Correct Answer: Atrial septal defect
Explanation:A congenital heart disease, ASD or atrial septal defect leads to a communication between the right and left atria due to a defect in the interatrial septum. This leads to mixing of arterial and venous blood from the right and left side of the heart. The hemodynamic significance of this defect depends on the presence of shunting of blood. Normally, the left side of the heart has higher pressure than the right as the left side has to pump blood throughout the body. A large ASD (> 9 mm) will result in a clinically significant left-to-right shunt, causing volume overload of the right atrium and ventricle, eventually leading to heart failure. Cardiac catheterization would reveal very high oxygen saturation in the right atrium, right ventricle and pulmonary artery. Eventually, the left-to-right shunt will lead to pulmonary hypertension and increased afterload in the right ventricle, along with the increased preload due to the shunted blood. This will either cause right ventricular failure, or raise the pressure in the right side of the heart to equal or more than that in the left. Elevation of right atrial pressure to that of left atrial pressure would thus lead to diminishing or complete cessation of the shunt. If left uncorrected, there will be reversal of the shunt, known as Eisenmenger syndrome, resulting in clinical signs of cyanosis as the oxygen-poor blood form right side of the heart will mix with the blood in left side and reach the peripheral vascular system.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 42
Incorrect
-
A 56-year-old woman weighs 75 kg. In this patient, total body water, intracellular fluid and extracellular fluid are respectively:
Your Answer:
Correct Answer: 45 l, 30 l, 15 l
Explanation:The percentages of body water contained in various fluid compartments add up to total body water (TBW). This water makes up a significant fraction of the human body, both by weight and by volume. The total body water (TBW) content of humans is approximately 60% of body weight. Two-thirds is located in the intracellular and one-third in the extracellular compartment. So, in a 75-kg individual, TBW = 60 × 75/100 = 45 l. Intracellular content = 2/3 × 45 = 30 l and extracellular content = 1/3 × 45 = 15 l.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 43
Incorrect
-
The renal tubule is the portion of the nephron that contains the fluid that has been filtered by the glomerulus. Which of the following substances is actively secreted into the renal tubules?
Your Answer:
Correct Answer: Potassium
Explanation:The renal corpuscle filters out solutes from the blood, delivering water and small solutes to the renal tubule for modification. In normal circumstances more than 90% of the filtered load of K is reabsorbed by the proximal tubules and loops of Henlé and almost all K appearing in the urine has been secreted by the late distal tubules and collecting tubules. So the rate of excretion is usually independent of the rate of filtration, but is closely tied to the rate of secretion and control of K excretion, largely accomplished by control of the secretion rate. Around 65–70% of the filtered potassium is reabsorbed along with water in the proximal tubule and the concentration of potassium in the tubular fluid varies little from that of the plasma.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 44
Incorrect
-
A 76-year-old woman is diagnosed with diabetes mellitus after a urine test revealed she has glucosuria. Glucosuria may occur due to inadequate glucose reabsorption at:
Your Answer:
Correct Answer: Proximal convoluted tubule
Explanation:Glucose is reabsorbed almost 100% via sodium–glucose transport proteins (apical) and GLUT (basolateral) in the proximal convoluted tubule. Glycosuria or glucosuria is a condition of osmotic diuresis typical in those suffering from diabetes mellitus. Due to a lack of insulin, plasma glucose levels are above normal. This leads to saturation of receptors in the kidneys and glycosuria usually at plasma glucose levels above 11 mmol/l. Rarely, glycosuria is due to an intrinsic problem with glucose reabsorption within the kidneys (such as Fanconi syndrome), producing a condition termed renal glycosuria.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 45
Incorrect
-
A glycogen storage disorder is characterised by increased liver glycogen with a normal structure and no increase in serum glucose after oral intake of a protein-rich diet. Deficiency of which of the following enzymes is responsible for this disorder?
Your Answer:
Correct Answer: Glucose-6-phosphatase
Explanation:The most common glycogen storage disorder is von Gierke’s disease or glycogen storage disease type I. It results from a deficiency of enzyme glucose-6-phosphatase which affects the ability of liver to produce free glucose from glycogen and gluconeogenesis; leading to severe hypoglycaemia. There is also increased glycogen storage in the liver and kidneys causing enlargement and various problems in their functioning. The disease also causes lactic acidosis and hyperlipidaemia. The main treatment includes frequent or continuous feedings of corn-starch or other carbohydrates.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 46
Incorrect
-
Which of the following will be a likely sequelae of complete ileal resection?
Your Answer:
Correct Answer: Vitamin B12 deficiency
Explanation:The ileum is a part of the small intestine and has a pH of around 7-8 (neutral or slightly alkaline). Its main function is absorption of products of digestion. The ileal wall has multiple villi, which in turn have numerous microvilli. This increases the surface area available for absorption significantly. The cells lining the ileum contain multiple enzymes such as protease and carbohydrase, which aid in the final stages of digestion. Villi contain lacteals which absorb the products of fat digestion, fatty acids and glycerol. Thus, ileal resection will lead to their decreased absorption and increased fat content in the stool. The ileum is also responsible for absorption of vitamin B12.
Maximum water absorption occurs in the colon followed by the jejunum. Hence, ileal resection is less likely to lead to fluid volume deficiency. Also, most minerals (like calcium, iron etc.) are absorbed in the duodenum, and thus will not be affected by ileal resection.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 47
Incorrect
-
A 40 year old man suffered severe trauma following a MVA. His BP is 72/30 mmhg, heart rate of 142 beats/mins and very feeble pulse. He was transfused 3 units of blood and his BP returned to 100/70 and his heart rate slowed to 90 beats/min. What decreased after transfusion?
Your Answer:
Correct Answer: Total peripheral resistance
Explanation:The patient is in hypovolemic shock, he is transfused with blood, this fluid resuscitation will result in a decreased sympathetic discharge and adequate ventricular filling which will result in the decreases TPR with an increased CO and cardiac filling pressures
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 48
Incorrect
-
A 49-year-old woman with acute renal failure has a total plasma [Ca2+] = 2. 5 mmol/l and a glomerular filtration rate of 160 l/day. What is the estimated daily filtered load of calcium?
Your Answer:
Correct Answer: 240 mmol/day
Explanation:Calcium is the most abundant mineral in the human body. The average adult body contains in total approximately 1 kg of calcium of which 99% is in the skeleton in the form of calcium phosphate salts. The extracellular fluid (ECF) contains approximately 22 mmol, of which about 9 mmol is in the plasma. About 40% of total plasma Ca2+ is bound to proteins and not filtered at the glomerular basement membrane. Therefore, the estimated daily filtered load is 1.5 mmol/l × 160 l/day = 240 mmol/day. The exact amount of free versus total Ca2+ depends on the blood pH: free Ca2+ increases during acidosis and decreases during alkalosis.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 49
Incorrect
-
A 40-year old lady with a flail chest due to trauma was breathing with the help of a mechanical ventilator in the ICU, and was heavily sedated on muscle relaxants. Due to sudden power failure, a nurse began to hand-ventilate the patient with a Ambu bag. What change will occur in the following parameters: (Arterial p(CO2), pH) in the intervening period between power failure and hand ventilation?
Your Answer:
Correct Answer: Increase, Decrease
Explanation:Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. Respiratory acidosis can be acute or chronic. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg). The given problem represents acute respiratory acidosis and thus, will show a increase in arterial p(CO2) and decrease in pH.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 50
Incorrect
-
When at rest, which of the following will be higher in a marathon runner compared to an untrained individual?
Your Answer:
Correct Answer: Cardiac stroke volume
Explanation:Cardiac muscle hypertrophy is seen in trained athletes as compared to the normal population. This hypertrophy results in higher stroke volume at rest and increased cardiac reserve (maximum cardiac output during exercise). However, the cardiac output at rest is almost the same in both trained and untrained people. This is because in trained athletes, the heart rate is slower, even up to 40-50 beats/min. There is minimal affect of athletic training on oxygen consumption and respiratory rate.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 51
Incorrect
-
A neonate with failure to pass meconium is being evaluated. His abdomen is distended and X-ray films of the abdomen show markedly dilated small bowel and colon loops. The likely diagnosis is:
Your Answer:
Correct Answer: Aganglionosis in the rectum
Explanation:Hirschsprung’s disease (also known as aganglionic megacolon) leads to colon enlargement due to bowel obstruction by an aganglionic section of bowel that starts at the anus. A blockage is created by a lack of ganglion cells needed for peristalsis that move the stool. 1 in 5000 children suffer from this disease, with boys affected four times more commonly than girls. It develops in the fetus in early stages of pregnancy. Symptoms include not having a first bowel movement (meconium) within 48 hours of birth, repeated vomiting and a swollen abdomen. Two-third of cases are diagnosed within 3 months of birth. Some children may present with delayed toilet training and some might not show symptoms till early childhood. Diagnosis is by barium enema and rectal biopsy (showing lack of ganglion cells).
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 52
Incorrect
-
Which of the following is the cause of flattened (notched) T waves on electrocardiogram (ECG)?
Your Answer:
Correct Answer: Hypokalaemia
Explanation:The T-wave is formed due to ventricular repolarisation. Normally, it is seen as a positive wave. It can be normally inverted (negative) in V1 (occasionally in V2-3 in African-Americans/Afro-Caribbeans). Hyperacute T-waves are the earliest ECG change of acute myocardial infarction. ECG findings of hyperkalaemia include high, tent-shaped T-waves, a small P-wave and a wide QRS complex. Hypokalaemia results in flattened (notched) T-waves, U-waves, ST-segment depression and prolonged QT interval.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 53
Incorrect
-
What is the reason for a deranged thrombin clotting time?
Your Answer:
Correct Answer: Heparin therapy
Explanation:Thrombic clotting time is also known as thrombin time. It is clinically performed to determine the therapeutic levels of heparin. After plasma is isolated from the blood, bovine thrombin is added to it and the time it takes from the addition to clot is recorded. The reference interval is usually <21s. deranged results are indicative of heparin therapy, hypofibrinogenemia, hyperfibrinogenaemia or lupus anticoagulant.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 54
Incorrect
-
An 18 year-old with an iron deficient diet was prescribed an iron supplement by her GP. Lack of iron often results in:
Your Answer:
Correct Answer: Hypochromic anaemia
Explanation:Iron deficiency anaemia is the most common type of anaemia. It can occur due to deficiency of iron from decreased intake, increased loss or inadequate absorption. An MCV less than 80 will indicated iron deficiency anaemia. On the smear the RBCs will be microcytic hypochromic and will also show poikilocytosis. Iron profile tests are important to make a diagnosis. Clinically the patient will be pale and lethargic.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 55
Incorrect
-
Under normal conditions, what is the major source of energy of cardiac muscles?
Your Answer:
Correct Answer: Fatty acids
Explanation:Under basal conditions, most of the energy needed by cardiac muscle for metabolism is derived from fats (60%), 35% by carbohydrates, and 5% by ketones and amino acids. However, after intake of large amounts of glucose, lactate and pyruvate are mainly used. During prolonged starvation, fat acts as the primary source. 50% of the used lipids are sourced from circulating fatty acids.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 56
Incorrect
-
A brain tumour causing blockage of the hypophyseal portal system is likely to result in an increased secretion of which of the following hormones?
Your Answer:
Correct Answer: Prolactin
Explanation:The hypophyseal portal system links the hypothalamus and the anterior pituitary. With the help of this system, the anterior pituitary receives releasing and inhibitory hormones from the hypothalamus and regulates the action of other endocrine glands. One of the inhibitory hormones carried by this system is the prolactin-inhibitory hormone. In the absence of this hormone which might occur in case of a blockage of the system, prolactin secretion increases to about three times normal levels.
-
This question is part of the following fields:
- Endocrinology
- Physiology
-
-
Question 57
Incorrect
-
A lesion involving the lateral portion of the dorsal columns at the level of the nape of the neck will most likely affect:
Your Answer:
Correct Answer: Vibratory sensations from the ipsilateral arm
Explanation:At the level mentioned in the question, the lateral portion of dorsal columns comprises of the fasciculus cuneatus. Axons carrying the sensations of touch, vibration and proprioception from the ipsilateral arm enter the spinal cord and ascend in the fasciculus cuneatus, synapsing in the nucleus cuneatus of the caudal medulla. Secondary neurons from this nucleus give rise to internal arcuate fibres, which decussate and ascend to the thalamus as the medial lemniscus. Tertiary neurons from there project to the ipsilateral somatosensory cortex. Thus, any damage to the fasciculus cuneatus will result in a deficit in tactile, proprioceptive and vibratory sensations in the ipsilateral arm, and not the contralateral arm.
Fine motor control of the fingers is mainly carried by the ipsilateral lateral corticospinal tract in the lateral funiculus of the cord. Motor control of the contralateral foot is carried by the ipsilateral corticospinal tract in the lateral funiculus of the cord. Lack of sweating of the face could be produced by interruption of sympathetic innervation. Proprioception from the ipsilateral leg is carried by the fasciculus gracilis in the medial part of the dorsal columns.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 58
Incorrect
-
Which of the following is a likely cause of jaundice?
Your Answer:
Correct Answer: Hepatic disease if plasma albumin is low and serum aminotransferase elevations > 500 units
Explanation:Jaundice can occur due to any of the possible causes and treatment depends upon diagnosing the correct condition. Mild hyperbilirubinemia with normal levels of aminotransferase and alkaline phosphatase is often unconjugated (e.g., due to haemolysis or Gilbert’s syndrome rather than hepatobiliary disease). Moderate or severe hyperbilirubinemia along with increased urinary bilirubin (bilirubinuria), high alkaline phosphatase or aminotransferase levels suggest hepatobiliary disease. Hyperbilirubinemia produced by any hepatobiliary disease is largely conjugated. In this case, other blood tests include hepatitis serology for suspected hepatitis, prothrombin time (PT) or international normalised ratio (INR), albumin and globulin levels, and antimitochondrial antibody levels (suspected primary biliary cirrhosis). Low albumin and high globulin levels suggest chronic rather than acute liver disease. In cases where there is only a an elevation of alkaline phosphatase, γ-glutamyl transpeptidase (GGT) levels should be checked – the levels of which will be found high in hepatobiliary disease, but not in bone disorder which can also lead to elevated alkaline phosphatase levels. In diseases of hepatobiliary origin, aminotransferase elevations > 500 units suggest a hepatocellular cause, whereas disproportionate increases of alkaline phosphatase (e.g., alkaline phosphatase > 3 times normal and aminotransferase < 200 units) suggest cholestasis. Because hepatobiliary disease alone rarely causes bilirubin levels > 30 mg/dl, higher levels are suggestive of a combination of severe hepatobiliary disease and haemolysis or renal dysfunction. Imaging is best for diagnosing infiltrative and cholestatic causes of jaundice. Liver biopsy is rarely needed, but can be of use in intrahepatic cholestasis and in some types of hepatitis.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 59
Incorrect
-
Destruction of the ventromedial nucleus of the hypothalamus will result in:
Your Answer:
Correct Answer: Loss of satiety
Explanation:The ventromedial nucleus of the hypothalamus is divided into an anterior and a superior part. The anterior part controls the female sexual drive, whereas the superior part is responsible for satiety. Destruction of the superior part of the nucleus will result in overeating, as no signal tells the body that it is satisfied.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 60
Incorrect
-
Intravenous diazepam was administered to a man who was brought to the emergency department with status epilepticus. He was administered 15 l/min oxygen via a reservoir bag mask. Blood investigations showed sodium = 140 mmol/l, potassium = 4 mmol/l and chloride = 98 mmol/l. His arterial blood gas analysis revealed pH 7.08, p(CO2)= 61.5 mmHg, p(O2) = 111 mmHg and standard bicarbonate = 17 mmol/l. This patient had:
Your Answer:
Correct Answer: Mixed acidosis
Explanation:Acidosis with high p(CO2) and low standard bicarbonate indicates mixed acidosis. Lower p(O2) is due to breathing of 70% oxygen. The prolonged seizures lead to lactic acidosis and the intravenous diazepam is responsible for the respiratory acidosis. Treatment includes airway manoeuvres and oxygen, assisted ventilation if needed, and treatment with fluids.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 61
Incorrect
-
Which of the following substances brings about a dilatation of the gastrointestinal resistance vessels?
Your Answer:
Correct Answer: Vasoactive intestinal peptide
Explanation:Gastric vasoconstrictors include catecholamines, angiotensin II and vasopressin. Vasodilators include vasoactive intestinal peptide and the hormones; gastrin, cholecystokinin and glucagon.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 62
Incorrect
-
Which statement is correct regarding secretions from the adrenal glands?
Your Answer:
Correct Answer: Aldosterone is producd by the zona glomerulosa
Explanation:The secretions of the adrenal glands by zone are:
Zona glomerulosa – aldosterone
Zona fasciculata – cortisol and testosterone
Zona reticularis – oestradiol and progesterone
Adrenal medulia – adrenaline, noradrenaline and dopamine.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 63
Incorrect
-
A 26-year-old female patient had the following blood report: RBC count = 4. 0 × 106/μl, haematocrit = 27% and haemoglobin = 11 g/dl, mean corpuscular volume (MCV) = 90 fl, mean corpuscular haemoglobin concentration (MCHC) = 41 g/dl. Further examination of blood sample revealed increased osmotic fragility of the erythrocytes. Which of the following is the most likely cause of this patient’s findings?
Your Answer:
Correct Answer: Spherocytosis
Explanation:Spherocytes are small rounded RBCs. It is due to an inherited defect of the RBC cytoskeleton membrane tethering proteins. Membrane blebs form that are lost over time and cells become round instead of biconcave. As it is a normochromic anaemia, the MCV is normal. it is diagnosed by osmotic fragility test which reveals increased fragility in a hypotonic solution.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 64
Incorrect
-
An experiment was conducted in which the skeletal muscle protein (not smooth muscle) involved in contraction was selectively inhibited. Which protein was inhibited?
Your Answer:
Correct Answer: Troponin
Explanation:The mechanism of contraction of smooth muscles is different from that of skeletal muscles in which the contractile protein is troponin whilst in smooth muscle contraction is a protein called calmodulin. Calmodulin reacts with calcium ions and stimulates the formation of myosin crossbridges.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 65
Incorrect
-
Which organ is most vulnerable to haemorrhagic shock?
Your Answer:
Correct Answer: Kidneys
Explanation:At rest, the brain receives 15% cardiac output, muscles 15%, gastrointestinal tract 30% and kidneys receive 20%. However, if normalised by weight, the largest specific blood flow is received by the kidneys at rest (400 ml/min x 100g), making them highly vulnerable in the case of a haemorrhagic shock.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 66
Incorrect
-
Which type of contractions are responsible for the propulsion of chyme along the small intestine?
Your Answer:
Correct Answer: Segmentation
Explanation:Two major types of intestinal contractions are segmentation and peristalsis:
Segmentation occurs most frequently and primarily involves circular muscle. It is essentially a contraction of 2- or 3-cm long intestinal segments while the muscle on either side of it relaxes. Chyme in the segment is displaced in both directions. As the contracted segment relaxes, the previously relaxed segments on either side may contract. This efficiently mixes the chyme with the digestive secretions and exposes the mucosal absorptive surface to the luminal contents. It also serves a propulsive function and contributes to the movement of chyme.
Peristalsis is a propulsive wave of contraction that is initiated by intestinal distension. It is short lived and travels only a few centimetres before dying out. The combined effects of intestinal peristalsis and segmentation provide for both adequate mixing of the intestinal contents and slow, steady movement of chyme.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 67
Incorrect
-
Diffusion is the movement of molecules from a region of high concentration to a region of low concentration. Which of these changes will decrease the rate of diffusion of a substance?
Your Answer:
Correct Answer: An increase in the molecular weight of the substance
Explanation:Unless given IV, a drug must cross several semipermeable cell membranes before it reaches the systemic circulation. Drugs may cross cell membranes by diffusion, amongst other mechanisms. The rate of diffusion of a substance is proportional to the difference in the concentration of the diffusing substance between the two sides of the membrane, the temperature of the solution, the permeability of the membrane and, in the case of ions, the electrical potential difference between the two sides of the membrane.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 68
Incorrect
-
After surgery, a patient developed a stitch granuloma . Which leukocyte in the peripheral blood will become an activated macrophage in this granuloma?
Your Answer:
Correct Answer: Monocyte
Explanation:Monocytes are leukocytes that protect the body against infections and move to the site of infection within 8-12 hours to deal with it. They are produced in the bone marrow and shortly after being produced are released into the blood stream where they circulate until an infection is detected. When called upon they leave the circulation and transform into macrophages within the tissue fluid and thus gain the capability to phagocytose the offending substance. Monocyte count is part of a complete blood picture. Monocytosis is the state of excess monocytes in the peripheral blood and may be indicative of various disease states. Examples of processes that can increase a monocyte count include: • chronic inflammation • stress response • hyperadrenocorticism • immune-mediated disease • pyogranulomatous disease • necrosis • red cell regeneration.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 69
Incorrect
-
Which of the following organelles have the capacity to regenerate and spontaneously replicate?
Your Answer:
Correct Answer: Mitochondrion
Explanation:A mitochondria is a membrane bound organelle found in eukaryotic cells. They are called the powerhouse of the cell and are the place where ATP is formed from energy generated through metabolism. They are capable of replication as well as repair and regeneration.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 70
Incorrect
-
What is the normal duration of the ST segment?
Your Answer:
Correct Answer: 0.08 s
Explanation:The ST segment lies between the QRS complex and the T-wave. The normal duration of the ST segment is 0.08 s. ST-segment elevation or depression may indicate myocardial ischaemia or infarction.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 71
Incorrect
-
If a catheter is placed in the main pulmonary artery of a healthy 30-year-old woman, which of the following will be its mean pulmonary arterial pressure?
Your Answer:
Correct Answer: 15 mmHg
Explanation:The pulmonary artery pressure (PA pressure) is a measure of the blood pressure found in the main pulmonary artery. The hydrostatic pressure of the pulmonary circulation refers to the actual pressure inside pulmonary vessels relative to atmospheric pressure. Hydrostatic (blood pressure) in the pulmonary vascular bed is low compared with that of similar systemic vessels. The mean pulmonary arterial pressure is about 15 mmHg (ranging from about 13 to 19 mmHg) and is much lower than the average systemic arterial pressure of 90 mmHg.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 72
Incorrect
-
The neurotransmitters adrenaline, noradrenaline and dopamine are derived from which amino acid?
Your Answer:
Correct Answer: Tyrosine
Explanation:Tyrosine is the precursor to adrenaline, noradrenaline and dopamine. Tyrosine hydroxylase converts tyrosine to DOPA, which is in turn converted to dopamine, then to noradrenaline and finally adrenaline.
-
This question is part of the following fields:
- Endocrine
- Physiology
-
-
Question 73
Incorrect
-
Calculate the cardiac output of a patient with the following measurements: oxygen uptake 200 ml/min, oxygen concentration in the peripheral vein 7 vol%, oxygen concentration in the pulmonary artery 10 vol% and oxygen concentration in the aorta 15 vol%.
Your Answer:
Correct Answer: 4000 ml/min
Explanation:The Fick’s principle states that the uptake of a substance by an organ equals the arteriovenous difference of the substance multiplied by the blood flowing through the organ. We can thus calculate the pulmonary blood flow with pulmonary arterial (i.e., mixed venous) oxygen content, aortic oxygen content and oxygen uptake. The pulmonary blood flow, systemic blood flow and cardiac output can be considered the same assuming there are no intracardiac shunts. Thus, we can calculate the cardiac output. Cardiac output = oxygen uptake/(aortic − mixed venous oxygen content) = 200 ml/min/(15 ml O2/100 ml − 10 ml O2/100 ml) = 200 ml/min/(5 ml O2/100 ml) = 200 ml/min/0.05 = 4000 ml/min.
It is crucial to remember to use pulmonary arterial oxygen content and not peripheral vein oxygen content, when calculating the cardiac output by Fick’s method.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 74
Incorrect
-
A patient with a long standing lower motor neuron lesion will have:
Your Answer:
Correct Answer: Muscle wasting
Explanation:Lower motor neurons (LMNs) connect the brainstem and spinal cord to muscle fibres. Damage to lower motor neurons is indicated by abnormal electromyographic potentials, fasciculations, paralysis, weakening and wasting of skeletal muscles.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 75
Incorrect
-
Different substances have different renal clearance rates. Which of the following substances should have the lowest renal clearance rate in a healthy patient?
Your Answer:
Correct Answer: Glucose
Explanation:Under normal conditions the renal clearance of glucose is zero, since glucose is completely reabsorbed in the renal tubules and not excreted. Glycosuria – the excretion of glucose into the urine- is nearly always caused by elevated blood glucose levels, most commonly due to untreated diabetes mellitus.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 76
Incorrect
-
Which of the following is likely to induce secretion of glucagon?
Your Answer:
Correct Answer: Low serum concentration of glucose
Explanation:The most potent stimulus for secretion of glucagon is hypoglycaemia whereas hyperglycaemia is a stimulus for insulin release. Glucagon secretion also occurs in response to high levels of amino acids. Somatostatin inhibits glucagon secretion. Parasympathetic stimulation increases pancreatic acinar secretion, but not of α-cells.
-
This question is part of the following fields:
- Endocrinology
- Physiology
-
-
Question 77
Incorrect
-
A 38-year-old woman with end-stage renal disease, is undergoing haemodialysis. She has normocytic normochromic anaemia. What is the best treatment for her?
Your Answer:
Correct Answer: Erythropoietin
Explanation:E erythropoietin (EPO) is a hormone that is released by the kidney. It is responsible for the regulation of red blood cell production in the body. It can be made using recombinant technology and is used in the treatment of anaemia of chronic renal failure and in patients under going chemotherapy
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 78
Incorrect
-
Diuretics that act on the ascending limb of the loop of Henle produce:
Your Answer:
Correct Answer: Reduced active transport of sodium
Explanation:The loop of Henlé leads from the proximal convoluted tubule to the distal convoluted tubule. Its primary function uses a counter current multiplier mechanism in the medulla to reabsorb water and ions from the urine. It can be divided into four parts:
1. Descending limb of loop of Henlé – low permeability to ions and urea, while being highly permeable to water
2. Thin ascending limb of loop of Henlé – not permeable to water, but it is permeable to ions
3. Medullary thick ascending limb of loop of Henlé – sodium (Na+), potassium (K+) and chloride (Cl–) ions are reabsorbed by active transport. K+ is passively transported along its concentration gradient through a K+ channel in the basolateral aspect of the cells, back into the lumen of the ascending limb.
4. The cortical thick ascending limb – the site of action where loop diuretics such as furosemide block the K+/Na+/2Cl− co-transporters = reduced active transport.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 79
Incorrect
-
The transmembrane proteins responsible for resting membrane potential of vascular smooth muscle cells was blocked by a drug. Which of the following transmembrane proteins were blocked by this drug?
Your Answer:
Correct Answer: K+ channels
Explanation:The resting membrane potential is due to selective permeability of the membrane to potassium ions. The Na/K pump is responsible for the generation of a gradient across the membrane and it is due to the inherent ability of the K channels to allow diffusion back into the nerve at rest which charges the cells. In reality, the resting membrane potential is more positive because of small contributions by Na+ channels, Cl− channels and non-selective cation channels.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 80
Incorrect
-
A 40-year old gentleman, known with a history of peptic ulcer disease, was brought to the clinic in a dehydrated state with persistent vomiting. His blood investigations revealed:
- sodium = 142 mmol/l
- potassium = 2.6 mmol/l
- chloride = 85 mmol/l
- pH = 7.55
- p(CO2) = 50 mmHg
- p(O2) = 107 mmHg
- standard bicarbonate = 40 mmol/l
Your Answer:
Correct Answer: Metabolic alkalosis
Explanation:High pH with high standard bicarbonate indicates metabolic alkalosis. The pa(CO2) was appropriately low in compensation. This is hypokalaemic hypochloraemic metabolic acidosis due to prolonged vomiting. Treatment includes treating the cause and intravenous sodium chloride with potassium.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 81
Incorrect
-
A 55 year old lady underwent an uneventful appendicectomy. Two hours later, her arterial blood gas analysis on room revealed pH: 7.30, p(CO2): 53 mmHg and p(O2): 79 mmHg. What is the most likely cause of these findings?
Your Answer:
Correct Answer: Alveolar hypoventilation
Explanation:In the given problem, there is respiratory acidosis due to hypercapnia from a low respiratory rate and/or volume (hypoventilation). Causes of hypoventilation include conditions impairing the central nervous system (CNS) respiratory drive, impaired neuromuscular transmission and other causes of muscular weakness (drugs and sedatives), along with obstructive, restrictive and parenchymal pulmonary disorders. Hypoventilation leads to hypoxia and hypercapnia reduces the arterial pH. Severe acidosis leads to pulmonary arteriolar vasoconstriction, systemic vascular dilatation, reduced myocardial contractility, hyperkalaemia, hypotension and cardiac irritability resulting in arrhythmias. Raised carbon dioxide concentration also causes cerebral vasodilatation and raised intracranial pressure. Over time, buffering and renal compensation occurs. However, this might not be seen in acute scenarios where the rise in p(CO2) occurs rapidly.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 82
Incorrect
-
Which of the following will be a seen in a patient with a plasma thyroid-stimulating hormone (TSH) level of 14 mU/l (normal < 5 mU/l) and a low T3 resin uptake of 19% (normal 25–35%)?
Your Answer:
Correct Answer: Periorbital swelling and lethargy
Explanation:Low T3 resin uptake combined with raised TSH is indicative of hypothyroidism. Signs and symptoms include dull expression, facial puffiness, lethargy, periorbital swelling due to infiltration with mucopolysaccharides, bradycardia and cold intolerance. Anxiety, palpitations, tachycardia, raised body temperature, heat intolerance and weight loss are all seen in hyperthyroidism.
-
This question is part of the following fields:
- Endocrinology
- Physiology
-
-
Question 83
Incorrect
-
A 34-year old gentleman presented with acute pancreatitis to the emergency department. On enquiry, there was found to be a history of recurrent pancreatitis, eruptive xanthomas and raised plasma triglyceride levels associated with chylomicrons. Which of the following will be found deficient in this patient?
Your Answer:
Correct Answer: Lipoprotein lipase
Explanation:The clinical features mentioned here suggest the diagnosis of hypertriglyceridemia due to lipoprotein lipase (LPL) deficiency. LPL aids in hydrolysing the lipids in lipoproteins into free fatty acids and glycerol. Apo-CII acts as a co-factor. Deficiency of this enzyme leads to hypertriglyceridemia.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 84
Incorrect
-
A 15-day old male baby was brought to the emergency department with sweating and his lips turning blue while feeding. He was born full term. On examination, his temperature was 37.9°C, blood pressure 75/45 mmHg, pulse was 175/min, and respiratory rate was 42/min. A harsh systolic ejection murmur could be heard at the left upper sternal border. X-ray chest showed small, boot-shaped heart with decreased pulmonary vascular markings. He most likely has:
Your Answer:
Correct Answer: Tetralogy of Fallot
Explanation:The most common congenital cyanotic heart disease and the most common cause of blue baby syndrome, Tetralogy of Fallot shows four cardiac malformations occurring together. These are ventricular septal defect (VSD), pulmonary stenosis (right ventricular outflow obstruction), overriding aorta (degree of which is variable), and right ventricular hypertrophy. The primary determinant of severity of disease is the degree of pulmonary stenosis. Tetralogy of Fallot is seen in 3-6 per 10,000 births and is responsible for 5-7% congenital heart defects, with slightly higher incidence in males. It has also been associated with chromosome 22 deletions and DiGeorge syndrome. It gives rise to right-to-left shunt leading to poor oxygenation of blood. Primary symptom is low oxygen saturation in the blood with or without cyanosis at birth of within first year of life. Affected children ay develop acute severe cyanosis or ‘tet spells’ (sudden, marked increase in cyanosis, with syncope, and may result in hypoxic brain injury and death). Other symptoms include heart murmur, failure to gain weight, poor development, clubbing, dyspnoea on exertion and polycythaemia. Chest X-ray reveals characteristic coeur-en-sabot (boot-shaped) appearance of the heart. Treatment consists of immediate care for cyanotic spells and Blalock–Taussig shunt (BT shunt) followed by corrective surgery.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 85
Incorrect
-
A blood sample from a patient with polycythaemia vera will show which of the following abnormalities?
Your Answer:
Correct Answer: High platelet count
Explanation:Polycythaemia is a condition that results in an increase in the total number of red blood cells (RBCs) in the blood. It can be due to a myeloproliferative syndrome, chronically low oxygen levels or rarely malignancy. In primary polycythaemia/ polycythaemia vera the increase is due to an abnormality in the bone marrow, resulting in increased RBCs, white blood cells (WBCs) and platelets. In secondary polycythaemia the increase occurs due to high levels of erythropoietin either artificially or naturally. The increase is about 6-8 million/cm3 of blood. A type of secondary polycythaemia is physiological polycythaemia where people living in high altitudes who are exposed to hypoxic conditions produce more erythropoietin as a compensatory mechanism for thin oxygen and low oxygen partial pressure.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 86
Incorrect
-
The most likely cause of prominent U waves on the electrocardiogram (ECG) of a patient is:
Your Answer:
Correct Answer: Hypokalaemia
Explanation:The U-wave, not always visible in ECGs, is thought to represent repolarisation of papillary muscles or Purkinje fibres. When seen, it is very small and occurs after the T-wave. Inverted U-waves indicate myocardial ischaemia or left ventricular volume overload. Prominent U-waves are most commonly seen in hypokalaemia. Other causes include hypercalcaemia, thyrotoxicosis, digitalis exposure, adrenaline and class 1A and 3 anti-arrhythmic agents. It can also be seen in congenital long-QT syndrome and in intracranial haemorrhage.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 87
Incorrect
-
In the glomerulus of the kidney, the mesangium is a structure associated with the capillaries. It has extraglomerular mesangial cells that:
Your Answer:
Correct Answer: Form the juxtaglomerular apparatus in combination with the macula densa and juxtaglomerular cells
Explanation:The mesangium is an inner layer of the glomerulus, within the basement membrane surrounding the glomerular capillaries. The mesangial cells are phagocytic and secrete the amorphous basement membrane-like material known as the mesangial matrix. They are typically separated from the lumen of the capillaries by endothelial cells. The other type of cells in the mesangium are the extraglomerular mesangial cells which form the juxtaglomerular apparatus in combination with two other types of cells: the macula densa of the distal convoluted tubule and juxtaglomerular cells of the afferent arteriole. This apparatus controls blood pressure through the renin–angiotensin–aldosterone system.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 88
Incorrect
-
Routine evaluation of a 38 year old gentleman showed a slightly lower arterial oxygen [pa(O2)] than the alveolar oxygen [pA(O2)]. This difference is:
Your Answer:
Correct Answer: Is normal and due to shunted blood
Explanation:Blood that bypasses the ventilated parts of lung and enters the arterial circulation directly is known as shunted blood. It happens in normal people due to mixing of arterial blood with bronchial and some myocardial venous blood (which drains into the left heart). Diffusion limitation and reaction velocity with haemoglobin are immeasurably small. CO2 unloading will not affect the difference between alveolar and arterial p(O2). A large VSD will result in much lower arterial O2 as compared to alveolar O2.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 89
Incorrect
-
An ECG of a 30 year old woman revealed low voltage QRS complexes. This patient is most probably suffering from?
Your Answer:
Correct Answer: Pericardial effusion
Explanation:The QRS complex is associated with current that results in the contraction of both the ventricles. As ventricles have more muscle mass than the atria, they result in a greater deflection on the ECG. The normal duration of a QRS complex is 10s. A wide and deep Q wave depicts myocardial infarction. Abnormalities in the QRS complex maybe indicative of a bundle block, ventricular tachycardia or hypertrophy of the ventricles. Low voltage QRS complexes are characteristic of pericarditis or a pericardial effusion.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 90
Incorrect
-
What will the destruction of endoplasmic reticulum stop?
Your Answer:
Correct Answer: Synthesis of proteins
Explanation:The rough endoplasmic reticulum is the factory for the manufacturing of proteins. It contains ribosomes attached to it and transports proteins that are destined for membranes and secretions. The rough ER is connected to the nuclear envelope and to the cisternae of the Golgi apparatus by vesicles that shuttle between the two compartments.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 91
Incorrect
-
A neurotransmitter of the nigrostriatal pathway is:
Your Answer:
Correct Answer: Dopamine
Explanation:Dopamine acts as a neurotransmitter in the brain, activating dopamine receptors. It is also a neurohormone released from the hypothalamus. It plays an important role in the reward system. It is believed that dopamine provides a teaching signal to parts of the brain responsible for acquiring new motor sequences (behaviours), by activation of dopamine neurons when an unexpected reward is presented. Loss of dopamine neurones in the nigrostriatal pathway causes Parkinson’s disease. In the frontal lobes, dopamine controls the flow of information from other areas of the brain, and thus, dopamine disorders in this region can cause a decline in neurocognitive functions, especially memory, attention and problem solving. Reduced dopamine concentrations in the prefrontal cortex are thought to contribute to attention-deficit disorder and some symptoms of schizophrenia. Dopamine is also the primary neuroendocrine regulator of the secretion of prolactin from the anterior pituitary gland. Dopamine is also commonly associated with the pleasure system of the brain. This plays a key role in understanding the mechanism of action of drugs (such as cocaine and the amphetamines), which seem to be directly or indirectly related to the increase of dopamine.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 92
Incorrect
-
What percentage of the cardiac output is delivered to the brain?
Your Answer:
Correct Answer: 15%
Explanation:Among all body organs, the brain is most susceptible to ischaemia. Comprising of only 2.5% of total body weight, the brain receives 15% of the cardiac output. Oxygen extraction is also higher with venous oxygen levels approximating 13 vol%, and arteriovenous oxygen difference of 7 vol%.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 93
Incorrect
-
One sensitive indicator of heavy alcohol dependence is:
Your Answer:
Correct Answer: Elevated serum gamma-glutamyl transpeptidase
Explanation:Elevated serum gamma-glutamyl transpeptidase (GGT) may be the only laboratory abnormality in patients who are dependent on alcohol. Heavy drinkers may also have an increased MCV.
-
This question is part of the following fields:
- Hepatobiliary
- Physiology
-
-
Question 94
Incorrect
-
Question 95
Incorrect
-
Nephrotic syndrome is a condition that causes proteinuria, hypoalbuminemia and oedema. Which of the following is the cause of the oedema in these patients?
Your Answer:
Correct Answer: Decreased oncotic pressure
Explanation:The glomeruli of the kidneys are the parts that normally filter the blood. They consist of capillaries that are fenestrated and allow fluid, salts and other small solutes to flow through, but normally not proteins. In nephrotic syndrome, the glomeruli become damaged allowing small proteins, such as albumin to pass through the kidneys into urine. Oedema usually occurs due to salt and water retention by the diseased kidneys as well as due to the reduced colloid oncotic pressure (because of reduced albumin in the plasma). Lower serum oncotic pressure causes fluid to accumulate in the interstitial tissues.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 96
Incorrect
-
In a cardiac cycle, what event does the closing of atrioventricular (AV) valves coincide with?
Your Answer:
Correct Answer: First heart sound
Explanation:In the cardiac cycle, the closing of the atrioventricular (AV) valves coincides with the onset of ventricular systole. This event marks the beginning of the isovolumetric contraction phase, where the ventricles begin to contract, but the volume of blood in the ventricles remains the same because both the AV valves and the semilunar valves (aortic and pulmonary valves) are closed. The closing of the AV valves produces the first heart sound, known as “S1” or “lub.”
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 97
Incorrect
-
Which statement is correct regarding coagulation?
Your Answer:
Correct Answer: Thrombin converts fibrinogen to fibrin
Explanation:Coagulation of blood is a complex process and an important part of haemostasis. There are two main pathways related to coagulation: the contact activation pathway/intrinsic pathway and tissue factor/extrinsic pathway. The extrinsic pathway is activated by external trauma that causes blood to escape from the vascular system. This pathway is quicker than the intrinsic pathway and involves factor VII. The intrinsic pathway is activated by trauma inside the vascular system, and initiated by platelets, exposed endothelium, chemicals, or collagen. This pathway is slower than the extrinsic pathway, but more important. It involves factors XII, XI, IX, VIII. Both pathways meet to finish the formation of a clot in what is known as the common pathway. The common pathway involves factors I, II, V, and X. They converge on the common pathway in which activation of prothrombin to thrombin leads to conversion of fibrinogen to fibrin and clot formation.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 98
Incorrect
-
Signals pass through neuromuscular junctions via the neurotransmitter acetylcholine. After release from the skeletal neuromuscular junction, acetylcholine:
Your Answer:
Correct Answer: Causes postsynaptic depolarisation
Explanation:Acetylcholine is released from the presynaptic membrane into the cleft where it binds to the ion gated channels on the post synaptic membrane, causing them to open. This results in sodium entering into the fibre and further depolarizing it, creating an action potential.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 99
Incorrect
-
A cerebellar tremor can be differentiated from a Parkinsonian tremor in that:
Your Answer:
Correct Answer: It only occurs during voluntary movements
Explanation:Cerebellar disease leads to intention tremors, which is absent at rest and appears at the onset of voluntary movements. In comparison, Parkinson’s tremor is present at rest. Frequency of tremor is a less reliable means to differentiate between the two as the oscillation amplitude of the tremor is not constant throughout a voluntary action.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 100
Incorrect
-
Driving pressure is considered to be a strong predictor of mortality in patients with ARDS. What is the normal mean intravascular driving pressure for the respiratory circulation?
Your Answer:
Correct Answer: 10 mmHg
Explanation:Driving pressure is the difference between inflow and outflow pressure. For the pulmonary circulation, this is the difference between pulmonary arterial (pa) and left atrial pressure (pLA). Normally, mean driving pressure is about 10 mmHg, computed by subtracting pLA (5 mmHg) from pA (15 mmHg). This is in contrast to a mean driving pressure of nearly 100 mmHg in the systemic circulation.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
SESSION STATS - PERFORMANCE PER SPECIALTY
