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Question 1
Correct
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The mechanism of action of streptokinase involves:
Your 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.
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This question is part of the following fields:
- General
- Physiology
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Question 2
Correct
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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: 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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 3
Correct
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A syndrome responsible for failure to absorb vitamin B12 from the GIT is called?
Your Answer: Pernicious anaemia
Explanation:Pernicious anaemia is a type of autoimmune disease in which antibodies form against the parietal cells or intrinsic factor. Intrinsic factor is required for the absorption of vitamin B12. Blood testing typically shows a macrocytic, normochromic anaemia and low levels of serum vitamin B12. A Schilling test can then be used to distinguish between pernicious anaemia, vitamin B12 malabsorption and vitamin B12 deficiency. Symptoms include shortness of breath, pallor and diarrhoea etc.
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This question is part of the following fields:
- General
- Physiology
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Question 4
Correct
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A 50 year old man on warfarin therapy following insertion of a pacemaker presented with epistaxis. Which of the following is true regarding blood coagulation?
Your Answer: Patients with haemophilia A usually have a normal bleeding time
Explanation:A prolonged bleeding time is seen in platelet disorders like thrombocytopenia. Patients with haemophilia A or B have a prolonged PTT but not a prolonged bleeding time.
Ca2+ is necessary for coagulation.
von Willebrand factor is an important part of the factor VIII complex and promotes platelet adhesion and aggregation.
DIC results in depleted coagulation factors and accumulation of fibrin.
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This question is part of the following fields:
- General
- Physiology
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Question 5
Correct
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QT interval in the electrocardiogram of a healthy individual is normally:
Your Answer: 0.40 s
Explanation:QT interval extends from beginning of the QRS complex till the end of he T-wave and normally lasts for 0.40 s. It is important in the diagnosis of long-QT and short-QT syndrome. The QT interval varies on the basis of heart rate and may need to be corrected.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 6
Correct
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Thalamic syndrome will most likely result in:
Your Answer: Hyperaesthesia
Explanation:Signs and symptoms of thalamic syndrome include contralateral hemi anaesthesia, burning or aching sensation in one half of a body (hyperaesthesia), often accompanied by mood swings.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 7
Incorrect
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Which of the following is a likely consequence of severe diarrhoea?
Your Answer: A metabolic alkalosis
Correct Answer: A decrease in the sodium content of the body
Explanation:Diarrhoea can occur due to any of the numerous aetiologies, which include infectious, drug-induced, food related, surgical, inflammatory, transit-related or malabsorption. Four mechanisms have been implicated in diarrhoea: increased osmotic load, increased secretion, inflammation and decreased absorption time. Diarrhoea can result in fluid loss with consequent dehydration, electrolyte loss (Na+, K+, Mg2+, Cl–) and even vascular collapse. Loss of bicarbonate ions can lead to a metabolic acidosis.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 8
Correct
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Which of the following compensatory parameters is responsible for causing an increase in the blood pressure in a 30 year old patient with a BP of 40 mmHg?
Your Answer: Baroreceptor reflex
Explanation:The baroreflex or baroreceptor reflex is one of the body’s homeostatic mechanisms for regulating blood pressure. It provides a negative feedback response in which an elevated blood pressure will causes blood pressure to decrease; similarly, decreased blood pressure depresses the baroreflex, causing blood pressure to rise. The system relies on specialised neurones (baroreceptors) in the aortic arch, carotid sinuses and elsewhere to monitor changes in blood pressure and relay them to the brainstem. Subsequent changes in blood pressure are mediated by the autonomic nervous system. Baroreceptors include those in the auricles of the heart and vena cava, but the most sensitive baroreceptors are in the carotid sinuses and aortic arch. The carotid sinus baroreceptors are innervated by the glossopharyngeal nerve (CN IX); the aortic arch baroreceptors are innervated by the vagus nerve (CN X).
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 9
Correct
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Depression of the normal coagulation system and excessive bleeding after surgery can occur in which of the following medical conditions?
Your Answer: Liver disease
Explanation:As most of the coagulation factors are synthesized in the liver, liver diseases like hepatitis or cirrhosis will depress the coagulation system. Vitamin K deficiency can also decrease the production of vitamin K dependent coagulation factors VII, XI, X and prothrombin.
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This question is part of the following fields:
- General
- Physiology
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Question 10
Correct
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The most likely cause of a low p(O2) in arterial blood is:
Your 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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 11
Incorrect
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During a normal respiratory exhalation, what is the recoil alveolar pressure?
Your Answer: Unchanged
Correct Answer: +10 cmH2O
Explanation:To determine compliance of the respiratory system, changes in transmural pressures (in and out) immediately across the lung or chest cage (or both) are measured simultaneously with changes in lung or thoracic cavity volume. Changes in lung or thoracic cage volume are determined using a spirometer with transmural pressures measured by pressure transducers. For the lung alone, transmural pressure is calculated as the difference between alveolar (pA; inside) and intrapleural (ppl; outside) pressure. To calculate chest cage compliance, transmural pressure is ppl (inside) minus atmospheric pressure (pB; outside). For the combined lung–chest cage, transmural pressure or transpulmonary pressure is computed as pA – pB. pA pressure is determined by having the subject deeply inhale a measured volume of air from a spirometer. Under physiological conditions the transpulmonary or recoil pressure is always positive; intrapleural pressure is always negative and relatively large, while alveolar pressure moves from slightly negative to slightly positive as a person breathes.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 12
Incorrect
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Which of the following can occur even in the absence of brainstem co-ordination?
Your Answer: Defecation
Correct Answer: Gastric emptying
Explanation:Although gastric emptying is under both neural and hormonal control, it does not require brainstem co-ordination. Increased motility of the orad stomach (decreased distensibility) or of the distal stomach (increased peristalsis), decreased pyloric tone, decreased duodenal motility or a combination of these, all increase the rate of gastric emptying. The major control mechanism for gastric emptying is through duodenal gastric feedback. The duodenum has receptors for the presence of acid, carbohydrate, fat and protein digestion products, osmolarity different from that of plasma, and distension. Activating these receptors decreases the rate of gastric emptying. Neural mechanisms involve both enteric and vagal pathways and a vagotomy impairs the gastric emptying regulation. CCK (cholecystokinin) slows gastric emptying at physiological levels of the hormone. Gastrin, secretin and glucose-1-phosphate also slow gastric emptying, but require higher doses.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 13
Correct
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A 55-year old gentleman presented to the doctor with worsening dysphagia for both solids and liquids over 6 months. This was associated with regurgitation of undigested food and occasional chest pain. Barium swallow revealed distal oesophageal dilatation with lack of peristalsis in the distal two-third oesophagus. The likely diagnosis is:
Your Answer: Achalasia
Explanation:Achalasia is an oesophageal motility disorder where inappropriate contractions in the oesophagus lead to reduced peristalsis and failure of the lower oesophageal sphincter to relax properly in response to swallowing. Classical triad of symptoms include dysphagia to fluids followed by solids, chest pain and regurgitation of undigested food. Other symptoms include belching, hiccups, weight loss and cough. Diagnosis is by:
– X-ray with a barium swallow or oesophagography : narrowing at the gastroesophageal junction (‘bird/parrot beak’ or ‘rat tail’ appearance) and various degrees of mega-oesophagus (oesophageal dilatation) as the oesophagus is gradually stretched by retained food. Effectiveness of treatment can be measured with a 5-minute timed barium swallow.
– Manometry – probe measures the pressure waves in different parts of oesophagus and stomach while swallowing.
– Endoscopy
– CT scan to exclude other causes like malignancy
– Pathological examination showing defect in the nerves which control oesophageal motility (myenteric plexus).
In Chagas disease, there is destruction of ganglion cells by Trypanosoma cruzi.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 14
Correct
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Question 15
Correct
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During pregnancy the uterus enlarges however after delivery it regresses to its original size. Which of the following organelles is responsible for this regression?
Your Answer: Lysosomes
Explanation:Lysosomes are formed by budding of the Golgi apparatus and contain enzymes which digest macromolecules. They are found in both plants and animals and are active in autophagic cell death, digestion after phagocytosis and for the cells own recycling process. They fuse with the molecules and release their content resulting in digestion.
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This question is part of the following fields:
- General
- Physiology
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Question 16
Correct
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A patient came into the emergency in a state of shock. His blood group is not known, but on testing it clotted when mixed with Type A antibodies. Which blood should be transfused?
Your Answer: B +ve
Explanation:There are two stages to determine the blood group, known as ABO typing. The first stage is called forward typing. In this method, RBCs are mixed with two separate solutions of type A or type B antibodies to see if they agglutinate. If this blood clumps, this indicates the presence of antigens within the blood sample. For example, a sample of type B blood will clump when tested with type A antibodies as it contains type B antigens. Group B – has only the B antigen on red cells (and A antibody in the plasma)
Group B – has only the B antigen on red cells (and A antibody in the plasma)
Group AB – has both A and B antigens on red cells (but neither A nor B antibody in the plasma)
Group O – has neither A nor B antigens on red cells (but both A and B antibody are in the plasma). Many people also have a Rh factor on the red blood cell’s surface. This is also an antigen and those who have it are called Rh+. Those who have not are called Rh–. A person with Rh– blood does not have Rh antibodies naturally in the blood plasma (as one can have A or B antibodies, for instance) but they can develop Rh antibodies in the blood plasma if they receive blood from a person with Rh+ blood, whose Rh antigens can trigger the production of Rh antibodies. A person with Rh+ blood can receive blood from a person with Rh– blood without any problems. The patient’s blood group is B positive as he has antigen B, antibody A and Rh antigens.
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This question is part of the following fields:
- General
- Physiology
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Question 17
Incorrect
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Which of the following is true about a patient who has undergone total colectomy and ileostomy?
Your Answer: This patient is at increased risk of anaemia due to malabsorption of vitamin B12
Correct Answer: Following total colectomy and ileostomy, the volume and water content of ileal discharge decreases over time
Explanation:After a patient has undergone total colectomy and ileostomy, the volume of ileal discharge, along with its water content gradually decreases over time. Post surgery, most patients can live a normal life. Iron and vitamin B12 absorption do not take place in the colon and hence are not affected significantly by a colectomy.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 18
Incorrect
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Which of the following is true regarding factor XI?
Your Answer: Is a member of the tissue factor (extrinsic) coagulation pathway
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.
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This question is part of the following fields:
- General
- Physiology
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Question 19
Correct
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The primary somatosensory cortex is located in the:
Your Answer: Postcentral gyrus
Explanation:The primary somatic sensory cortex is located in the postcentral gyrus and is the largest cortical receiving area for information from somatosensory receptors. Through corticocortical fibres, it then sends the information to other areas of the neocortex and further analysis takes place in the posterior parietal association cortex.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 20
Correct
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One sensitive indicator of heavy alcohol dependence is:
Your 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.
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This question is part of the following fields:
- Hepatobiliary
- Physiology
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Question 21
Correct
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Which of the following substances is most likely to cause pulmonary vasodilatation?
Your 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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 22
Correct
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The Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity. According to this equation, the buffering capacity of the system is at maximum when the number of free anions compared with undissociated acid is:
Your Answer: Equal
Explanation:In 1908, Lawrence Joseph Henderson wrote an equation describing the use of carbonic acid as a buffer solution. Later, Karl Albert Hasselbalch re-expressed that formula in logarithmic terms, resulting in the Henderson–Hasselbalch equation. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid–base reactions. Two equivalent forms of the equation are: pH = pKa + log10 [A–]/[HA] or pH = pKa + log10 [base]/[acid]. Here, pKa is − log10(Ka) where Ka is the acid dissociation constant, that is: pKa = –log10(Ka) = –log10 ([H3 O+][A–]/[HA]) for the reaction: HA + H2 O ≈ A– + H3 O+ In these equations, A– denotes the ionic form of the relevant acid. Bracketed quantities such as [base] and [acid] denote the molar concentration of the quantity enclosed. Maximum buffering capacity is found when pH = pKa or when the number of free anions to undissociated acid is equal and buffer range is considered to be at a pH = pKa ± 1.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 23
Correct
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Electrophoresis is used to detect antibodies (immunoglobulins) in a blood sample from the umbilical artery of a new born. Which antibodies have the highest percentage in a new-born?
Your Answer: IgG
Explanation:IgG is a monomeric immunoglobulin. It is formed by two heavy chains and two light chains and has two binding sites. Its is the most abundant antibody that is equally distributed in the blood and the tissues. It is the only antibody that can pass through the placenta and thus the only antibody present in the baby after it is born. There are four subclasses: IgG1 (66%), IgG2 (23%), IgG3 (7%) and IgG4 (4%). IgG1, IgG3 and IgG4 cross the placenta easily
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This question is part of the following fields:
- General
- Physiology
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Question 24
Correct
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Action potentials are used extensively by the nervous system to communicate between neurones and muscles or glands. What happens during the activation of a nerve cell membrane?
Your Answer: Sodium ions flow inward
Explanation:During the generation of an action potential, the membrane gets depolarized which cause the voltage gated sodium channels to open and sodium diffuses inside the neuron, resulting in the membrane potential moving towards a positive value. This positive potential will then open the voltage gated potassium channels and cause more K+ to move out decreasing the membrane potential and restoring the membrane potential to its resting value.
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This question is part of the following fields:
- General
- Physiology
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Question 25
Correct
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Which of these substances is secreted by pericytes in the juxtaglomerular cells?
Your 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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 26
Correct
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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: 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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 27
Correct
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Selective destruction of which of the following cells will affect antibody synthesis?
Your Answer: Plasma cells
Explanation:Plasma cell are memory cells. After the antigen Is engulfed by the B cells it is presented to the CD4+ helper cells via the MCH II receptor and this leads to their activation which in turn stimulates the B cells to form antibodies against that specific antigen. Some B cells differentiate into plasma cells also called memory cells that get activated after subsequent infection.
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This question is part of the following fields:
- General
- Physiology
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Question 28
Correct
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Which of the following will increase blood pressure and cause hypokalaemia?
Your Answer: Angiotensin II
Explanation:Angiotensin is a peptide that is released in response to a decrease in blood volume and blood pressure. It has multiple functions but mainly acts to cause vasoconstriction, increase BP and release aldosterone from the adrenal cortex. It is a powerful vasoconstrictor and release of aldosterone causes increased retention of sodium and excretion of potassium.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 29
Incorrect
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As per the Poiseuille-Hagen formula, doubling the diameter of a vessel will change the resistance of the vessel from 16 peripheral resistance units (PRU) to:
Your Answer: 32 PRU
Correct Answer: 1 PRU
Explanation:Poiseuille-Hagen formula for flow in along narrow tube states that F = (PA– PB) × (Π/8) × (1/η) × (r4/l) where F = flow, PA– PB = pressure difference between the two ends of the tube, η = viscosity, r = radius of tube and L = length of tube. Also, flow is given by pressure difference divided by resistance. Hence, R = 8ηL ÷ Πr4. Hence, the resistance of the vessel changes in inverse proportion to the fourth power of the diameter. So, if the diameter of the vessel is increased to twice the original, it will lead to decrease in resistance to one-sixteenth its initial value.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 30
Correct
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A 40 year old man suffered severe trauma following an MVA. His BP is 70/33 mmhg, heart rate of 140 beats/mins and very feeble pulse. He was transfused 3 units of blood resulting in his BP returning to 100/70 and his heart rate to 90 beats/min. What decreased following transfusion?
Your Answer: Total peripheral resistance
Explanation:The patient is in hypovolemic shock, he is transfused with blood to replace the volume lost. It is important not only to replace fluids but stop active bleeding in resuscitation. Fluid replacement will result in a decreased sympathetic discharge and adequate ventricular filling thus reducing total peripheral resistance and increasing cardiac output and cardiac filling pressures.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 31
Correct
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A patient with a long standing lower motor neuron lesion will have:
Your 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.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 32
Correct
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A 60 year old patient with a history of carcinoma of the head of the pancreas, and obstructive jaundice presents with a spontaneous nose bleed and easy bruising. What is the most likely reason for this?
Your Answer: Vitamin-K-dependent clotting factors deficiency
Explanation:Vitamin K is a fat soluble vitamin requiring fat metabolism to function properly to allow for its absorption. People with obstructive jaundice develop vitamin k deficiency as fat digestion is impaired. Vit K causes carboxylation of glutamate residue and hence regulates blood coagulation including: prothrombin (factor II), factors VII, IX, X, protein C, protein S and protein Z.
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This question is part of the following fields:
- General
- Physiology
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Question 33
Correct
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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: 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.
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This question is part of the following fields:
- General
- Physiology
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Question 34
Correct
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Normally, the O2 transfer in the lungs from alveolar to capillary is perfusion-limited. In which of the following situations does it become a diffusion-limited process?
Your Answer: Pulmonary oedema
Explanation:Normally, the transfer of oxygen from air spaces to blood takes place across the alveolar-capillary membrane by simple diffusion and depends entirely on the amount of blood flow (perfusion-limited process). Diseases that affect this diffusion will transform the normal process to a diffusion limited process. Thus, the diseases which cause a thickened barrier (such as pulmonary oedema due to increased extravascular lung water or asbestosis) will limit the diffusion of oxygen. Chronic obstructive lung diseases will have little effect on diffusion. Inhaling hyperbaric gas mixtures might overcome the diffusion limitation in patients with mild asbestosis or interstitial oedema, by increasing the driving force. Strenuous (not mild) exercise might also favour diffusion limitation and decrease passage time. Increasing the rate of ventilation will not have this affect but will only maintain a high oxygen gradient from air to blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 35
Correct
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What is the role of factor VII in coagulation?
Your Answer: Initiates the process of coagulation in conjunction with tissue factor
Explanation:The main role of factor VII is to initiate the process of coagulation along with tissue factor (TF). TF is found in the blood vessels and is not normally exposed to the bloodstream. When a vessel is injured tissue factor is exposed to blood and circulating factor VII. Factor VII is converted to VIIa by TF.
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This question is part of the following fields:
- General
- Physiology
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Question 36
Correct
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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: 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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 37
Incorrect
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A 50-year old gentleman who suffered a stroke was brought to the emergency department by his relatives. The patient however denied the presence of paralysis of his left upper and lower limbs. What is the most likely site of the lesion in this patient?
Your Answer:
Correct Answer: Right posterior parietal cortex
Explanation:A large injury to the non-dominant parietal cortex can make the patient neglect or refuse to acknowledge the presence of paralysis on the contralateral side. This can also involve the perception of the external world. Smaller injuries in this area which involve the precentral gyrus (primary motor cortex) or postcentral gyrus (primary sensory cortex) cause contralateral spastic paralysis or contralateral loss of tactile sensation respectively. A lesion in posterior inferior gyrus of the dominant frontal lobe results in motor aphasia. Involvement of the posterior superior gyrus of the dominant frontal lobe produces sensory aphasia.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 38
Incorrect
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Hormones of the anterior pituitary include which of the following?
Your Answer:
Correct Answer: Prolactin
Explanation:The anterior pituitary gland (adenohypophysis or pars distalis) synthesizes and secretes:
1. FSH (follicle-stimulating hormone)
2. LH (luteinizing hormone)
3. Growth hormone
4. Prolactin
5. ACTH (adrenocorticotropic hormone)
6. TSH (thyroid-stimulating hormone).
The posterior pituitary gland (neurohypophysis) stores and secretes 2 hormones produced by the hypothalamus:
1. ADH (antidiuretic hormone or vasopressin)
2. Oxytocin
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 39
Incorrect
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Which of the following can lead to haemolytic anaemia?
Your Answer:
Correct Answer: Presence of haemoglobin S
Explanation:Haemoglobin S is an abnormal type of haemoglobin seen in sickle cell anaemia. This allows for the haemoglobin to crystalize within the RBC upon exposure to low partial pressures of oxygen. This results in rupture of the RBCs as they pass through microcirculation, especially in the spleen. This can cause blockage of the vessel down stream and ischaemic death of tissues, accompanied by severe pain.
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This question is part of the following fields:
- General
- Physiology
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Question 40
Incorrect
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A 23-year-old woman decides to donate a kidney through a kidney chain. Which of the following indices would be expected to be decreased in the donor after full recovery from the operation?
Your Answer:
Correct Answer: Creatinine clearance
Explanation:Since medication to prevent rejection is so effective, donors do not need to be similar to their recipient. Most donated kidneys come from deceased donors; however, the utilisation of living donors is on the rise. Most problems encountered with live donation are associated with the donor. Firstly, there are the potentially harmful investigative procedures carried out in the assessment phase, the most hazardous being renal angiography, where there is cannulation of the artery and injection of a radio-opaque dye to determine the blood supply to the kidney. Secondly, there are the short-term risks of nephrectomy surgery. According to the literature, there is a mortality rate of between 1 in 1600 and 1 in 3000, but this is no more than is associated with any anaesthetic. In the initial postoperative period creatinine clearance may be decreased but this recovers fully over a few weeks to months. Long-term complications include prolonged wound pain.
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This question is part of the following fields:
- Physiology
- Renal
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Question 41
Incorrect
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Which of the following conditions is characterized by generalised oedema due to effusion of fluid into the extracellular space?
Your Answer:
Correct Answer: Anasarca
Explanation:Anasarca (or ‘generalised oedema’) is a condition characterised by widespread swelling of the skin due to effusion of fluid into the extracellular space. It is usually caused by liver failure (cirrhosis of the liver), renal failure/disease, right-sided heart failure, as well as severe malnutrition/protein deficiency.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 42
Incorrect
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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.
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This question is part of the following fields:
- General
- Physiology
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Question 43
Incorrect
-
Which organ is responsible for the secretion of enzymes that aid in digestion of complex starches?
Your Answer:
Correct Answer: Pancreas
Explanation:α-amylase is secreted by the pancreas, which is responsible for hydrolysis of starch, glycogen and other carbohydrates into simpler compounds.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 45
Incorrect
-
Multiple cells were labelled using a fluorescent dye that doesn’t cross the cell membrane. One cell in the middle was bleached with a light that destroys the dye, but the cell soon recovers its stain. The presence of which structures best explains this?
Your Answer:
Correct Answer: Gap junctions
Explanation:Gap junctions are attachments between cells that permit intercellular communication e.g. they permit current flow and electrical coupling between myocardial cells. They allow direct electrical transmission among cells and also permit certain substance to pass through as well. They are either homotypic, formed by two identical hemichannels or heterotypic, formed by different hemichannels.
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This question is part of the following fields:
- General
- Physiology
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Question 46
Incorrect
-
How are amino acids transported across the luminal surface of the small intestinal epithelium?
Your Answer:
Correct Answer: Co-transport with sodium ions
Explanation:Once complex peptides are broken down into amino acids by the peptidases present in the brush border of small intestine, they are ready for absorption by at least four sodium-dependent amino acid co-transporters – one each for acidic, basic, neutral and amino acids, present on the luminal plasma membrane. These transporters first bind sodium and can then bind the amino acids. Thus, amino acid absorption is totally dependent on the electrochemical gradient of sodium across the epithelium. The basolateral membrane in contrast, possesses additional transporters to carry amino acids from the cell into the blood, but these are sodium-independent.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 47
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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 48
Incorrect
-
Calculate the pulmonary vascular resistance in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer:
Correct Answer: 2.0 resistance units (mmHg/l per min)
Explanation:Pulmonary vascular resistance (PVR) = (Mean pulmonary artery pressure – Pulmonary capillary wedge pressure) divided by Cardiac output. To get cardiac output, Fick’s principle needs to be applied which states that VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CA = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24–0.16, CO = 500/0.8, CO = 6.25 l/min. To calculate mean pulmonary artery pressure, we use the formula: Mean pulmonary artery pressure = Diastolic pressure + 1/3(Systolic pressure – Diastolic pressure). Thus, Mean pulmonary artery pressure = 15 + 1/3(25 – 15) = 15 + 3. 33 = 18.33. Substituting these values in the first formula, PVR = 18.3–5/6.25 = 13.5/5.25 = 2.0 resistance units (mmHg/l per min) approximately.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 49
Incorrect
-
When does the heart rate decrease?
Your Answer:
Correct Answer: Pressure on the eyeball
Explanation:Various vagotonic manoeuvres (e.g. Valsalva manoeuvre, carotid sinus massage, pressure on eyeballs, ice-water facial immersion, swallowing of ice-cold water) result in increased parasympathetic tone through the vagus nerve which results in a decrease in heart rate. These manoeuvres may be clinically useful in terminating supraventricular arrhythmias.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 50
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.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 51
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.
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This question is part of the following fields:
- Physiology
- Respiratory
-
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Question 52
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.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 53
Incorrect
-
A 31-year-old woman is diagnosed with adrenal hyperplasia, and laboratory samples are taken to measure serum aldosterone and another substance. Which is most likely to be the other test that was prescribed to this patient?
Your Answer:
Correct Answer: Plasma renin
Explanation:The evaluation of a patient in whom hyperaldosteronism is first to determine that hyperaldosteronism is present (serum aldosterone) and, if it is present, to differentiate primary from secondary causes of hyperaldosteronism. The aldosterone-to-renin ratio (ARR) is the most sensitive means of differentiating primary from secondary causes of hyperaldosteronism as it is abnormally increased in primary hyperaldosteronism, and decreased or normal but with high renin levels in secondary hyperaldosteronism.
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This question is part of the following fields:
- Physiology
- Renal
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Question 54
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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 55
Incorrect
-
What is the most likely cause of prolonged bleeding time in a 40 year old women admitted for a laparoscopic cholecystectomy?
Your Answer:
Correct Answer: Thrombocytopaenia
Explanation:Bleeding time is related to platelet function, thus a decrease in platelet function, as seen in thrombocytopenia, DIC and von Willebrand disease in which platelet aggregation is defective, leads to an increase in bleeding time. It is not affected by a decrease or deficiency of any other clotting factors. Aspirin and other COX inhibitors prolong bleeding time along with warfarin and heparin.
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This question is part of the following fields:
- General
- Physiology
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Question 56
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.
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This question is part of the following fields:
- General
- Physiology
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Question 57
Incorrect
-
A 76-year-old man with a urinary tract obstruction due to prostatic hyperplasia develops acute renal failure. Which of the following physiological abnormalities of acute renal failure will be most life threatening for this patient?
Your Answer:
Correct Answer: Acidosis
Explanation:Acute renal failure (ARF) is a rapid loss of renal function due to damage to the kidneys, resulting in retention of nitrogenous (urea and creatinine) and non-nitrogenous waste products that are normally excreted by the kidney. This accumulation may be accompanied by metabolic disturbances, such as metabolic acidosis and hyperkalaemia, changes in body fluid balance and effects on many other organ systems. Metabolic acidosis and hyperkalaemia are the two most serious biochemical manifestations of acute renal failure and may require medical treatment with sodium bicarbonate administration and antihyperkalaemic measures. If not appropriately treated these can be life-threatening. ARF is diagnosed on the basis of characteristic laboratory findings, such as elevated blood urea nitrogen and creatinine, or inability of the kidneys to produce sufficient amounts of urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 58
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.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 59
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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 60
Incorrect
-
There are several mechanisms involved in the transport of sodium ions from blood to interstitial fluid of the muscle cells. Which of the following mechanisms best describes this phenomenon?
Your Answer:
Correct Answer: Diffusion through channels between endothelial cells
Explanation:Capillaries are the smallest of the body’s blood vessels, measuring 5–10 μm and they help to enable the exchange of water, oxygen, carbon dioxide, and many other nutrients and waste substances between the blood and the tissues surrounding them. The walls of capillaries are composed of only a single layer of cells, the endothelium. Ion channels are pore-forming proteins that help to establish and control the small voltage gradient that exists across the plasma membrane of all living cells by allowing the flow of ions down their electrochemical gradient. An ion channel is an integral membrane protein or more typically an assembly of several proteins. The archetypal channel pore is just one or two atoms wide at its narrowest point. It conducts a specific ion such as sodium or potassium and conveys them through the membrane in single file.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 61
Incorrect
-
A teenage Somalian boy presents with a complaint of an enlarged lower jaw. His blood film shows blast cells and macrophages. Which virus is responsible for this?
Your Answer:
Correct Answer: Epstein–Barr virus
Explanation:Burkitt’s lymphoma is a type of non-Hodgkin’s lymphoma. Histologically it is characterised by a starry sky appearance due to numerous neoplastic macrophages which are required to clear the rapidly dividing tumour cells/blast cells. Burkitt’s lymphoma commonly affects the jaw bone, forming a huge tumour mass. It is associated with translocation of c-myc gene and has three types: 1) endemic/African type, 2)sporadic and 3)immunodeficiency-associated. The first type is strongly associated with EBV.
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This question is part of the following fields:
- General
- Physiology
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Question 62
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 63
Incorrect
-
In the kidney, the macula densa is an area of closely packed specialized cells lining the wall of the:
Your Answer:
Correct Answer: Distal convoluted tubule
Explanation:In the kidney, the macula densa is an area of closely packed specialised cells lining the region of the distal convoluted tubule (DCT) lying next to the glomerular vascular pole. The cells of the macula densa are sensitive to the ionic content and water volume of the fluid in the DCT, producing signals that promote renin secretion by other cells of the juxtaglomerular apparatus.
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This question is part of the following fields:
- Physiology
- Renal
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Question 64
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.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 65
Incorrect
-
Which of the following enzymes is secreted by the small intestinal mucosa?
Your Answer:
Correct Answer: Lactase
Explanation:Lactase, an enzyme belonging to β-galactosidase family of enzymes, brings about the hydrolysis of the disaccharide lactose into galactose and glucose. In humans, it is present along the brush border membrane of the cells lining the small intestinal villi. Deficiency of lactase causes lactose intolerance.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 66
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.
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This question is part of the following fields:
- General
- Physiology
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Question 67
Incorrect
-
A 50 year old man was admitted to the surgical ICU following a hemicolectomy for carcinoma of the caecum. A full blood count revealed: haematocrit = 30%, erythrocytes = 4 × 106/μ, haemoglobin level = 8 g/dl. To determine the likely cause of his anaemia, red blood cell indices were calculated. Which RBC indices are correct?
Your Answer:
Correct Answer: MCHC = haemoglobin concentration/haematocrit
Explanation:Mean corpuscular haemoglobin concentration (MCHC) is calculated simply by dividing the haemoglobin concentration (8 g/dl) by the haematocrit (0.3). The normal range is 31–36 g/dl. This patient has a hypochromic anaemia (MCHC = 8/0.3 = 26.7 g/dl). Dividing the haemoglobin concentration × 10 by erythrocyte number yields mean corpuscular haemoglobin (MCH). Normal range is 25.4–34.6 pg/cell and this patient has a significantly reduced cellular haemoglobin content (MCH = 8 × 10/4 = 20 pg/cell). Mean corpuscular volume (MCV) is calculated by dividing haematocrit × 1000 by erythrocyte number (4 × 106/μl). Normal range is 80–100 fl and this patient has a microcytic anaemia (MCV = 0.3 × 1000/4 = 75 fl). Microcytic, hypochromic anaemia is characteristic for iron-deficiency.
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This question is part of the following fields:
- General
- Physiology
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Question 68
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 69
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.
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This question is part of the following fields:
- General
- Physiology
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Question 70
Incorrect
-
Atractyloside is an inhibitor of electron transport chain. It is expected to have little or no effect on the functioning of which of the following cell types?
Your Answer:
Correct Answer: Red blood cells
Explanation:Electron transport chain is a series of electron carriers that are embedded in the mitochondrial membrane. It is the place where ATP is made. Inhibiting the electron transport chain will stop production of ATP. Red blood cells are the only cell in the given option which do not contain ATP.
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This question is part of the following fields:
- General
- Physiology
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Question 71
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.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 72
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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 73
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 74
Incorrect
-
Ventricular filling follows a delay caused by?
Your Answer:
Correct Answer: AV node
Explanation:The AV node is a conducting tissue found between the atria and the ventricles of the heart. It conducts electrical signal from the atria to the ventricles and acts a delaying mechanism preventing the atria and the ventricles from contracting at the same time. This decremental conduction prevents premature ventricular contraction in cases such as atrial fibrillation. A delay in the AV node is the reason for the PR segment seen on the ECG. In certain types of supraventricular tachycardia, a person could have two AV nodes; this will cause a loop in electrical current and uncontrollably rapid heart beat. When this electricity catches up with itself, it will dissipate and return to a normal heart rate.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 75
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 76
Incorrect
-
Myoglobin is released as a result of rhabdomyolysis from damaged skeletal muscles. What function do they perform in the muscle?
Your Answer:
Correct Answer: Acts like haemoglobin and binds with O2
Explanation:Myoglobin is a pigmented globular protein made up of 153 amino acids with a prosthetic group containing haem around which the apoprotein folds. It is the primary oxygen carrying protein of the muscles. The binding of oxygen to myoglobin is unaffected by the oxygen pressure as it has an instant tendency to bind given its hyperbolic oxygen curve. It releases oxygen at very low pO2 levels.
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This question is part of the following fields:
- General
- Physiology
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Question 77
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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 78
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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 79
Incorrect
-
What causes a reduction in pulmonary functional residual capacity?
Your Answer:
Correct Answer: Pulmonary fibrosis
Explanation:Pulmonary functional residual capacity (FRC) is = volume of air present in the lungs at the end of passive expiration.
Obstructive diseases (e.g. emphysema, chronic bronchitis, asthma) = an increase in FRC due to an increase in lung compliance and air trapping.
Restrictive diseases (e.g. pulmonary fibrosis) result in stiffer, less compliant lungs and a reduction in FRC.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 80
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 81
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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 82
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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 83
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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 84
Incorrect
-
Which of the following muscles aid in inspiration?
Your Answer:
Correct Answer: Diaphragm and external intercostals
Explanation:The diaphragm and external intercostals are muscles of inspiration as they increase the volume of thoracic cavity and reduce the intrathoracic pressure. Muscles of expiration include abdominal muscles and internal intercostals.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 85
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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 86
Incorrect
-
Calculate the resistance of the artery if the pressure at one end is 60 mmHg, pressure at the other end is 20 mm Hg and the flow rate in the artery is 200 ml/min.
Your Answer:
Correct Answer: 0.2
Explanation:Flow in any vessel = Effective perfusion pressure divided by resistance, where effective perfusion pressure is the mean intraluminal pressure at the arterial end minus the mean pressure at the venous end. Thus, in the given problem, resistance = (60 − 20)/200 = 0.2 mmHg/ml per min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 87
Incorrect
-
What is a major source of fuel being oxidised by the skeletal muscles of a man who has undergone starvation for 7 days?
Your Answer:
Correct Answer: Serum fatty acids
Explanation:Starvation is the most extreme form of malnutrition. Prolonged starvation can lead to permanent organ damage and can be fatal. Starved individuals eventually lose significant fat and muscle mass as the body uses these for energy.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 88
Incorrect
-
Which of the following causes the maximum increase in the secretion of antidiuretic hormone (ADH)?
Your Answer:
Correct Answer: Increased plasma osmolarity
Explanation:The most potent stimulus for ADH release is increased plasma osmolarity. Decreased plasma volume is a less potent stimulus in comparison. However, decrease blood volume and arterial pressure due to severe haemorrhage does lead to ADH secretion. Hypothalamic releasing factors do not control the release of posterior pituitary hormones ADH and oxytocin.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 89
Incorrect
-
Which of the following will show decreased hearing when tested by air conduction but normal hearing when tested by bone conduction?
Your Answer:
Correct Answer: Fibrosis causing fixation of the ossicles
Explanation:As the cochlea is embedded into bone, the vibrations from the bone are transmitted directly to the fluid in the cochlea. Hence, any damage to the ossicles or tympanic membrane will not show an abnormal result on bone conduction test.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 90
Incorrect
-
The primary area involved in the pathology of Parkinson's disease is:
Your Answer:
Correct Answer: Substantia nigra
Explanation:Parkinson’s disease is a degenerative, movement disorder of the central nervous system, and is typically characterized by muscle rigidity, tremor and bradykinesia (in extreme cases, akinesia). Secondary symptoms include high-level cognitive dysfunction and subtle language problems.
Parkinson’s disease is also called ‘primary Parkinsonism’ or ‘idiopathic Parkinson’s disease and is the most common cause of Parkinsonism, a group of similar symptoms. The disorder is caused due to loss of pigmented dopaminergic cells in the pars compacta region of the substantia nigra.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 91
Incorrect
-
A child defecates a few minutes after being fed by the mother. This is most likely due to:
Your Answer:
Correct Answer: Gastrocolic reflex
Explanation:The gastrocolic reflex is a physiological reflex that involves increase in colonic motility in response to stretch in the stomach and by-products of digestion in the small intestine. It is shown to be uneven in its distribution throughout the colon, with the sigmoid colon affected more than the right side of the colon in terms of a phasic response. Various neuropeptides have been proposed as mediators of this reflex, such as serotonin, neurotensin, cholecystokinin and gastrin.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 92
Incorrect
-
Which of the following associations is correctly matched with the body's defence mechanism in fighting infection?
Your Answer:
Correct Answer: Specific cellular mechanism → cytotoxic T cells
Explanation:The immune system has certain levels of defence against pathogens. First line includes simple barriers such as skin, mucosa and stomach acid that prevent the pathogen from entering into the body. If this barrier is breached then the innate immune system is activated which includes leukocytes (macrophages, neutrophils, mast cells, eosinophils, basophils, natural killer cells). If the pathogens invade the second layer of defence then the third layer, adaptive immunity is activated, which includes B and T lymphocytes. B cells provide a humoral response whereas cytotoxic T cells have specific cellular mechanisms. They maintain a memory of past infections and are activated faster following a recurrence.
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This question is part of the following fields:
- General
- Physiology
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Question 93
Incorrect
-
Calculate the cardiac stroke volume of a patient whose oxygen consumption (measured by analysis of mixed expired gas) is 300 ml/min, arterial O2 content is 20 ml/100 ml blood, pulmonary arterial O2 content is 15 ml/100 ml blood and heart rate is 60/min.
Your Answer:
Correct Answer: 100 ml
Explanation:By Fick’s principle, VO2 = Q × (CA (O2) − CV (O2)) where VO2 = O2 consumption, Q = cardiac output and CA(O2) and CV(O2) are arterial and mixed venous O2 content respectively. Thus, in the given problem, 300 ml O2/min = Q × (20−15) ml O2/100 ml. Thus, Q = 6000 ml blood/min. Then, we can calculate stroke volume by dividing the cardiac output with heart rate. Thus, stroke volume = 6000 ml/min divided by 60/min stroke volume = 100 ml.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 94
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.
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This question is part of the following fields:
- General
- Physiology
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Question 95
Incorrect
-
In a hypertensive patient with secondary hyperaldosteronism, aldosterone is released mainly in response to:
Your Answer:
Correct Answer: Angiotensin II
Explanation:Secondary hyperaldosteronism in hypertension is either due to primary renin overproduction by the kidneys or renin overproduction secondary to decreased renal blood flow. The main stimulus for aldosterone release are adrenocorticotrophic hormone (ACTH), angiotensin II and high plasma K+ levels. Low plasma Na+ might also stimulate the adrenal cortex. Fluid overload will reduce aldosterone secretion. Atrial natriuretic peptide is secreted under conditions of expanded extracellular volume and will not lead to aldosterone secretion.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 96
Incorrect
-
Whilst snorkelling, a 30-year old gentleman has the respiratory rate of 10/min, tidal volume of 550 ml and an effective anatomical dead space of 250 ml. What is his alveolar ventilation?
Your Answer:
Correct Answer: 3000 ml/min
Explanation:Alveolar ventilation is the amount of air reaching the alveoli per minute. Alveolar ventilation = respiratory rate × (tidal volume – anatomical dead space volume). Thus, alveolar ventilation = 10 × (550 − 250) = 3000 ml/min.
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This question is part of the following fields:
- Physiology
- Respiratory
-
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Question 97
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).
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This question is part of the following fields:
- Gastroenterology
- Physiology
-
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Question 98
Incorrect
-
Which of the following is involved in vitamin B12 absorption?
Your Answer:
Correct Answer: Intrinsic factor
Explanation:Absorption of vitamin B12 is by an active transport process and occurs in the ileum. Most cobalamins are bound to proteins and are released in the stomach due to low pH and pepsin. The cobalamins then bind to R proteins, i.e. haptocorrin (HC) secreted from salivary glands and gastric juice. Another cobalamin binding protein is Intrinsic factor (IF) secreted from the gastric parietal cells. The cobalamin-HC complex is digested by pancreatic proteases in the intestinal lumen, and the free cobalamin then binds to IF. The complex then reaches a transmembrane receptor in the ileum and undergoes endocytosis. Cobalamin is then released intracellularly and binds to transcobalamin II (TC II). The newly formed complex then exits the ileal cell and enters the blood circulation.
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This question is part of the following fields:
- Gastroenterology
- Physiology
-
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Question 99
Incorrect
-
Which portion of the renal tubule absorbs amino acids and glucose?
Your Answer:
Correct Answer: Proximal convoluted tubule
Explanation:In relation to the morphology of the kidney as a whole, the convoluted segments of the proximal tubules are confined entirely to the renal cortex. Glucose, amino acids, inorganic phosphate and some other solutes are reabsorbed via secondary active transport in the proximal renal tubule through co-transport channels driven by the sodium gradient.
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This question is part of the following fields:
- Physiology
- Renal
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Question 100
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)
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This question is part of the following fields:
- Physiology
- Respiratory
-
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Question 101
Incorrect
-
What is the normal amount of oxygen that is carried in the blood?
Your Answer:
Correct Answer: 20 ml oxygen/100 ml blood
Explanation:Normally, 100 ml of blood contains 15g haemoglobin and a single gram of haemoglobin can bind to 1.34 ml oxygen when 100% saturated. Thus, 15 × 1.34 = 20 ml O2/100 ml blood. The haemoglobin in venous blood that is leaving the tissues is about 75% saturated with oxygen, and hence it carries about 15 ml O2/100 ml venous blood. This implies that for each 10 ml of blood, 5 ml oxygen is transported to the tissues. With a p(O2) > 100 mm Hg, only 3 ml of oxygen is dissolved in every one litre of plasma. By increasing the pA(O2) by breathing 100% oxygen, one can add an extra amount of oxygen in the plasma, but the amount of oxygen carried by haemoglobin will not increase significantly as it is already > 95% saturated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 102
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.
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This question is part of the following fields:
- General
- Physiology
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Question 103
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 104
Incorrect
-
The bronchial circulation is a part of the circulatory system that supplies nutrients and oxygen to the pulmonary parenchyma. What percentage of cardiac output is received by bronchial circulation?
Your Answer:
Correct Answer: 2%
Explanation:The bronchial circulation is part of the systemic circulation and receives about 2% of the cardiac output from the left heart. Bronchial arteries arise from branches of the aorta, intercostal, subclavian or internal mammary arteries. The bronchial arteries supply the tracheobronchial tree with both nutrients and O2. It is complementary to the pulmonary circulation that brings deoxygenated blood to the lungs and carries oxygenated blood away from them in order to oxygenate the rest of the body.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 105
Incorrect
-
A 34-year-old woman is diagnosed with cerebral oedema after suffering a severe head trauma. Which of the following conditions is not likely to be associated with the extracellular oedema?
Your Answer:
Correct Answer: Increased plasma colloid osmotic pressure
Explanation:Cerebral oedema is extracellular fluid accumulation in the brain. Increased capillary permeability, increased capillary pressure, increased interstitial fluid colloid osmotic pressure and lymphatic blockage would increase fluid movement into the interstitial spaces. Increased plasma colloid osmotic pressure, however, would oppose fluid movement from the capillaries into the interstitial compartment.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
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Question 106
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.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 107
Incorrect
-
Which of the following cytokines produced by the T cell induce MHC-II proteins?
Your Answer:
Correct Answer: γ-Interferon
Explanation:Interferons elicit a non-specific antiviral activity by inducing specific RNA synthesis and expression of proteins in neighbouring cells. Common interferon inducers are viruses, double-stranded RNA and micro-organisms. INF-γ is produced mainly by CD4+, CD8+ T cells and less commonly by B cells and natural killer cells. INF-γ has antiviral and antiparasitic activity but its main biological activity appears to be immunomodulatory. Among its many functions are activation of macrophages and enhanced expression of MHC-II proteins or macrophages.
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This question is part of the following fields:
- General
- Physiology
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Question 108
Incorrect
-
Severe abdominal pain radiating to the back, along with increased serum amylase levels, is seen in which of the following conditions?
Your Answer:
Correct Answer: Pancreatitis
Explanation:The primary test for diagnosis and monitoring of pancreatitis is amylase. Increased plasma levels of amylase can be found in: salivary trauma (including anaesthetic intubation), mumps, pancreatitis and renal failure. However, a rise in the total amylase levels over 10 times the upper limit of normal (ULN) is suggestive of pancreatitis; 5–10 times the ULN may indicate ileus or duodenal disease or renal failure. Lower levels are commonly found in salivary gland disease.
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This question is part of the following fields:
- Gastroenterology
- Physiology
-
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Question 109
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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 110
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.
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This question is part of the following fields:
- General
- Physiology
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Question 111
Incorrect
-
During cardiac catheterisation, if the blood sample from the catheter shows an oxygen saturation of 70%, and the pressure ranging from 12 to 24 mm Hg, it implies that the catheter tip is located in the:
Your Answer:
Correct Answer: Pulmonary artery
Explanation:Normal values for various parameters are as follows:
Systolic arterial blood pressure (SBP): 90–140 mmHg.
Diastolic arterial blood pressure: 60–90 mmHg.
Mean arterial blood pressure (MAP): SBP + (2 × DBP)/3 (normal range: 70-105 mmHg).
Right atrial pressure (RAP): 2–6 mmHg.
Systolic right ventricular pressure (RVSP): 15–25 mmHg.
Diastolic right ventricular pressure (RVDP): 0–8 mmHg.
Pulmonary artery pressure (PAP): Systolic (PASP) is 15-25 mmHg and Diastolic (PADP) is 8–15 mmHg.
Pulmonary artery wedge pressure (PAWP): 6–12 mmHg.
Left atrial pressure (LAP): 6–12 mmHg.
Thus, the given value indicates that the position of catheter tip is likely to be in the pulmonary artery.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 112
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.
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This question is part of the following fields:
- General
- Physiology
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Question 113
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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
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Question 114
Incorrect
-
Lung compliance is increased by:
Your Answer:
Correct Answer: Emphysema
Explanation:Lung compliance is increased by emphysema, acute asthma and increasing age and decreased by alveolar oedema, pulmonary hypertension, atelectasis and pulmonary fibrosis.
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This question is part of the following fields:
- Physiology
- Respiratory
-
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Question 115
Incorrect
-
Which of the following is true about myasthenia gravis?
Your Answer:
Correct Answer: Response of skeletal muscle to nerve stimulation is weakened
Explanation:An autoimmune disorder, myasthenia gravis leads to progressive muscle weakness. It occurs due to formation of antibodies against the nicotinic acetylcholine (ACh) receptor of the motor endplate, which leads to impaired neuromuscular transmission. Thus, nerve stimulation will lead to a weakened muscle response, but direct electrical stimulation will bring about a normal response. Diagnostic test includes improvement of muscle weakness by small doses of acetylcholinesterase inhibitors (physostigmine or edrophonium). However, a large dose of physostigmine worsens the weakness due to desensitisation of the endplate to persistent Ach. One of the investigative tools includes radiolabelled snake venom α-bungarotoxin. It is an in vitro study performed on muscle biopsy specimens and used to quantify the number of ACh receptors at the motor endplate.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 116
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.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 117
Incorrect
-
Which of the following is responsible for the maximum increase in total peripheral resistance on sympathetic stimulation?
Your Answer:
Correct Answer: Arterioles
Explanation:Arterioles are also known as the resistance vessels as they are responsible for approximately half the resistance of the entire systemic circulation. They are richly innervated by the autonomic nervous system and hence, will bring about the maximum increase in peripheral resistance on sympathetic stimulation.
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This question is part of the following fields:
- Cardiovascular
- Physiology
-
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Question 118
Incorrect
-
Carbon dioxide is principally transported in the blood in which form?
Your Answer:
Correct Answer: Bicarbonate
Explanation:Carbon dioxide is transported in the blood in various forms:
– Bicarbonate (80–90%)
– Carbamino compounds (5–10%)
– Physically dissolved in solution (5%).
Carbon dioxide is carried on the haemoglobin molecule as carbamino-haemoglobin; carboxyhaemoglobin is the combination of haemoglobin with carbon monoxide.
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This question is part of the following fields:
- Physiology
- Respiratory; Cardiovascular
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Question 119
Incorrect
-
Chest X-ray of a 45-year old gentleman with a week history of pleurisy showed a small pneumothorax with moderate-sized pleural effusion. Arterial blood gas analysis showed p(CO2) = 23 mmHg, p(O2) = 234.5 mmHg, standard bicarbonate = 16 mmol/l. What are we most likely dealing with?
Your Answer:
Correct Answer: Compensated respiratory alkalosis
Explanation:Normal pH with low p(CO2) and low standard bicarbonate could indicate either compensated respiratory alkalosis or a compensated metabolic acidosis. However, the history of hyperventilation for 5 days (pleurisy) favours compensated respiratory alkalosis. Compensated metabolic acidosis would have been likely in a diabetic patient with fever, vomiting and high glucose (diabetic ketoacidosis).
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This question is part of the following fields:
- Physiology
- Respiratory
-
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Question 120
Incorrect
-
In which situation is a stretch reflex such as knee jerk likely to be exaggerated?
Your Answer:
Correct Answer: In upper motor neuron lesion
Explanation:A stretch reflex is a monosynaptic reflex that causes muscle contraction in response to stretching within that muscle. The sensory apparatus in a muscle that are sensitive to stretch are the muscle spindles. The patellar (knee jerk) reflex is an example. In upper motor neuron lesions, the stretch reflexes tend to be brisk due to loss of inhibitory signals on gamma neurons through the lateral reticulospinal tract.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 121
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.
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This question is part of the following fields:
- Physiology
- Renal
-
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Question 122
Incorrect
-
Basal Metabolic Rate (BMR) will most likely be reduced by which of the following?
Your Answer:
Correct Answer: Decrease in body temperature
Explanation:The basal metabolic rate (BMR) is defined as the rate of calorie consumption after an overnight fast, in the absence of any muscular activity, with the patient in a restful state. Various factors affect the BMR including weight, body surface area and age. The BMR is 30 kcal/m2 per hour at birth; at age 2, the rate is 57 kcal/m2 per hour; and at age 20, 41 kcal/m2 per hour. After this, the BMR decreases by 10% between 20-60 years of age. Women are known to have a 10% lower BMR than men (due to higher fat content). A one-degree change in body temperature leads to a 10% change in BMR in the same direction. However, shivering and increasing ambient temperature brings about a rise in BMR, and so does stress, physical activity, caffeine, theophylline and hyperthyroidism. Also, thermogenesis induced by diet results in increased metabolic rate and hence, BMR should be ideally measured after overnight fasting.
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This question is part of the following fields:
- Gastroenterology
- Physiology
-
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Question 123
Incorrect
-
During strenuous exercise, what else occurs besides tachycardia?
Your Answer:
Correct Answer: Increased stroke volume
Explanation:During strenuous exercise there is an increase in:
– Heart rate, stroke volume and therefore cardiac output. (CO = HR x SV)
– Respiratory rate (hyperventilation) which will lead to a reduction in Paco2.
– Oxygen demand of skeletal muscle, therefore leading to a reduction in mixed venous blood oxygen concentration.
Renal blood flow is autoregulated, so renal blood flow is preserved and will tend to remain the same. Mean arterial blood pressure is a function of cardiac output and total peripheral resistance and will increase with exercise, mainly as a result of the increase in cardiac output that occurs.
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This question is part of the following fields:
- Cardiovascular
- Physiology
-
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Question 124
Incorrect
-
A 54-year-old woman with amyotrophic lateral sclerosis is diagnosed with respiratory acidosis. The patient’s renal excretion of potassium would be expected to:
Your Answer:
Correct Answer: Fall, since tubular secretion of potassium is inversely coupled to acid secretion
Explanation:Respiratory acidosis is a medical emergency in which decreased ventilation (hypoventilation) increases the concentration of carbon dioxide in the blood and decreases the blood’s pH (a condition generally called acidosis). Secretion of acid and potassium by the renal tubule are inversely related. So, increased excretion of H+ during renal compensation for respiratory acidosis will result in decreased secretion (or increased retention) of potassium ions, with the result that the body’s potassium store rises. An increase in K+ excretion would be associated with renal compensation for respiratory alkalosis. The filtered load of K+depends only on K+ plasma concentration and glomerular filtration rate, not on plasma pH.
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This question is part of the following fields:
- Physiology
- Renal
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Question 125
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.
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This question is part of the following fields:
- Neurology
- Physiology
-
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Question 126
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.
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This question is part of the following fields:
- Physiology
- Respiratory
-
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Question 127
Incorrect
-
Pain in the right upper quadrant of the abdomen on ingestion of a fatty meal is seen in a condition which involves which of the following substances?
Your Answer:
Correct Answer: Cholecystokinin
Explanation:The clinical scenario described here favours the presence of gallstones. During food ingestion, vagal discharges stimulate gallbladder contraction. Moreover, presence of fat and amino acids in the intestinal lumen stimulates the release of cholecystokinin (CCK) in the duodenum. This causes sustained gallbladder contraction and relaxation of the sphincter of Oddi. If gallstones are present, there will be inflammation in the gallbladder and CCK will aggravate it due to contractions.
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This question is part of the following fields:
- Gastroenterology
- Physiology
-
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Question 128
Incorrect
-
A 23 year old woman is Rh -ve and she delivered a baby with a Rh+ blood group. What measure can be performed to prevent Rh incompatibility in the next pregnancy?
Your Answer:
Correct Answer: Immunoglobulin D
Explanation:Rh disease is also known as erythroblastosis fetalis and is a disease of the new-born. In mild states it can cause anaemia with reticulocytosis and in severe forms causes severe anaemia, morbus hemolytcus new-born and hydrops fetalis. RBCs of the Rh+ baby can cross the placenta and enter into the maternal blood. As she is Rh- her body will form antibodies against the D antigen which will pass through the placenta in subsequent pregnancies.
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This question is part of the following fields:
- General
- Physiology
-
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Question 129
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
-
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Question 130
Incorrect
-
A TRUE statement regarding abolition of the cephalic phase of pancreatic secretion is that it:
Your Answer:
Correct Answer: Will result after vagotomy
Explanation:Recognition and integration of the sight, smell and taste of food triggers the cephalic phase of pancreatic secretion, causing an increase in pancreatic HCO3- and enzyme secretion. The degree of enzyme secretion in this phase is about 50% of the maximal response seen with exogenous CCK and secretin. The vagus nerve regulates the secretion through the cholinergic fibres innervating the acinar cells of the pancreas, and through peptidergic nerve fibres, which innervate ductal cells.
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This question is part of the following fields:
- Gastroenterology
- Physiology
-
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Question 131
Incorrect
-
Which of the following substances will enhance the activity of antithrombin III?
Your Answer:
Correct Answer: Heparin
Explanation:Antithrombin III is a glycoprotein that inactivates multiple enzymes involved in the coagulation system. It inactivates factor X, factor IX, factor II, factor VII, factor XI and factor XII. Its activity is greatly increased by the action of heparin.
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This question is part of the following fields:
- General
- Physiology
-
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Question 132
Incorrect
-
What is the basic chemical reaction that takes place in the breakdown of complex foodstuffs?
Your Answer:
Correct Answer: Hydrolysis
Explanation:Breakdown of complex food into simpler compounds is achieved by hydrolysis, with the help of different enzymes specific for different compounds.
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This question is part of the following fields:
- Gastroenterology
- Physiology
-
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Question 133
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.
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This question is part of the following fields:
- General
- Physiology
-
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Question 134
Incorrect
-
A medical student is asked to calculate the net pressure difference in a capillary wall, considering: Interstitial fluid hydrostatic pressure = –3 mmHg, Plasma colloid osmotic pressure = 28 mmHg, Capillary hydrostatic pressure = 17 mmHg, Interstitial fluid colloid osmotic pressure = 8 mmHg, and Filtration coefficient = 1. Which is the correct answer?
Your Answer:
Correct Answer: 0 mmHg
Explanation:The rate of filtration at any point along a capillary depends on a balance of forces sometimes called Starling’s forces after the physiologist who first described their operation in detail. The Starling principle of fluid exchange is key to understanding how plasma fluid (solvent) within the bloodstream (intravascular fluid) moves to the space outside the bloodstream (extravascular space). Fluid movement = k[(pc– pi)–(Πc– Πi)] where k = capillary filtration coefficient, pc = capillary hydrostatic pressure, pi= interstitial hydrostatic pressure, Πc = capillary colloid osmotic pressure, Πi = interstitial colloid osmotic pressure. Therefore: 1 × [capillary hydrostatic pressure (17) – interstitial fluid hydrostatic pressure (–3)] – [plasma colloid osmotic pressure (28) – interstitial fluid colloid osmotic pressure (8)] = 0 mmHg
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
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Question 135
Incorrect
-
When one is silently counting, what part of his brain will show increased regional cerebral blood flow (rCBF)?
Your Answer:
Correct Answer: Supplementary motor area
Explanation:Regional cerebral blood flow (rCBF) increases in the superior speech cortex (supplementary motor area) during periods of silent counting, whereas speaking aloud will do so in the motor cortex and medial temporal lobe, along with the superior speech cortex.
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This question is part of the following fields:
- Neurology
- Physiology
-
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Question 136
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.
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This question is part of the following fields:
- Physiology
- Respiratory
-
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Question 137
Incorrect
-
Which of the following is a potential cause of a positive D-dimer assay?
Your Answer:
Correct Answer: Deep venous thrombosis
Explanation:A D-dimer test is performed to detect and diagnose thrombotic conditions and thrombosis. A negative result would rule out thrombosis and a positive result although not diagnostic, is highly suspicious of thrombotic conditions like a deep vein thrombosis, pulmonary embolism as well as DIC.
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This question is part of the following fields:
- General
- Physiology
-
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Question 138
Incorrect
-
A 30-year-old man is brought to the emergency department suffering from extreme dehydration, and subsequent hypotension and tachycardia. Which part of the kidney will compensate for this loss?
Your Answer:
Correct Answer: Collecting ducts
Explanation:The collecting duct system of the kidney consists of a series of tubules and ducts that physically connect nephrons to a minor calyx or directly to the renal pelvis. The collecting duct system is the last component of the kidney to influence the body’s electrolyte and fluid balance. In humans, the system accounts for 4–5% of the kidney’s reabsorption of sodium and 5% of the kidney’s reabsorption of water. At times of extreme dehydration, over 24% of the filtered water may be reabsorbed in the collecting duct system.
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This question is part of the following fields:
- Physiology
- Renal
-
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Question 139
Incorrect
-
A blood sample of a 58 year old male patient, who underwent an abdominal aortic aneurysm repair, was sent to the laboratory. The laboratory technician said that the patient’s blood agglutinates with antisera anti-A and anti-D, while the patient’s serum agglutinates cells of blood group B. What is the blood group of this patient?
Your Answer:
Correct Answer: A positive
Explanation:Group A – has only the A antigen on red cells (and B antibody in the plasma)
Group B – has only the B antigen on red cells (and A antibody in the plasma)
Group AB – has both A and B antigens on red cells (but neither A nor B antibody in the plasma)
Group O – has neither A nor B antigens on red cells (but both A and B antibody are in the plasma). Many people also have a so-called Rh factor on the red blood cell’s surface. This is also an antigen and those who have it are called Rh+. A person with Rh– blood does not have Rh antibodies naturally in the blood plasma (as one can have A or B antibodies, for instance) but they can develop Rh antibodies in the blood plasma if they receive blood from a person with Rh+ blood, whose Rh antigens can trigger the production of Rh antibodies. A person with Rh+ blood can receive blood from a person with Rh– blood without any problems. In this scenario the person has blood group A+ as he has A antigen, anti B antibody and Rh antigen
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This question is part of the following fields:
- General
- Physiology
-
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Question 140
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.
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This question is part of the following fields:
- General
- Physiology
-
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Question 141
Incorrect
-
A lesion involving the lateral geniculate nucleus of the thalamus is likely to affect:
Your Answer:
Correct Answer: Vision
Explanation:The lateral geniculate nucleus (LGN) of the thalamus is the primary processor of visual information in the central nervous system. The LGN receives information directly from the retina and sends projections directly to the primary visual cortex. The LGN likely helps the visual system focus its attention on the most important information.
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This question is part of the following fields:
- Neurology
- Physiology
-
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Question 142
Incorrect
-
Which of the following clinical signs will be demonstrated in a case of Brown-Séquard syndrome due to hemisection of the spinal cord at mid-thoracic level?
Your Answer:
Correct Answer: Ipsilateral spastic paralysis, ipsilateral loss of vibration and proprioception (position sense) and contralateral loss of pain and temperature sensation beginning one or two segments below the lesion
Explanation:Brown–Séquard syndrome results due to lateral hemisection of the spinal cord and results in a loss of motricity (paralysis and ataxia) and sensation. The hemisection of the cord results in a lesion of each of the three main neural systems: the principal upper motor neurone pathway of the corticospinal tract, one or both dorsal columns and the spinothalamic tract. As a result of the injury to these three main brain pathways the patient will present with three lesions. The corticospinal lesion produces spastic paralysis on the same side of the body (the loss of moderation by the upper motor neurons). The lesion to fasciculus gracilis or fasciculus cuneatus results in ipsilateral loss of vibration and proprioception (position sense). The loss of the spinothalamic tract leads to pain and temperature sensation being lost from the contralateral side beginning one or two segments below the lesion. At the lesion site, all sensory modalities are lost on the same side, and an ipsilateral flaccid paralysis.
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This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 143
Incorrect
-
Where is factor VIII predominantly synthesised?
Your Answer:
Correct Answer: Vascular endothelium
Explanation:Factor VIII is an important part of the coagulation cascade. Deficiency causes haemophilia A. It is synthesised predominantly by the vascular endothelium and is not affected by liver disease. In the circulation it is bound to von Willebrand factor and it forms a stable complex with it. It is activated by thrombin or factor Xa and acts as a co factor to factor IXa to activate factor X which is a co factor to factor Va. Thrombin cleaves fibrinogen in fibrin and forms a meshwork to trap RBC and platelets to form a clot.
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This question is part of the following fields:
- General
- Physiology
-
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Question 144
Incorrect
-
Which is the site of action of the drug omeprazole?
Your Answer:
Correct Answer: H+/K+ ATPase
Explanation:H+/K+-ATPase or ‘proton pump’ located in the canalicular membrane plays a major role in acid secretion. The ATPase here is magnesium-dependent. Omeprazole is a proton pump inhibitor and blocks H+/K+- ATPase.
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This question is part of the following fields:
- Gastroenterology
- Physiology
-
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Question 145
Incorrect
-
In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your Answer:
Correct Answer: Exercise
Explanation:The oxygen-haemoglobin dissociation curve plots saturated haemoglobin against the oxygen tension and is usually a sigmoid plot. Each molecule of haemoglobin can bind to four molecules of oxygen reversibly. Factors that can influence the binding include: pH, concentration of 2,3-diphosphoglycerate (2,3-DPG), temperature, type of haemoglobin molecules, and presence of toxins, especially carbon monoxide. Shape of the curve is due to interaction of bound oxygen molecules with the incoming molecules. The binding of first molecule is difficult, with easier binding of the second and third molecule and increase in difficulty with the fourth molecule – partly as a result of crowding and partly as a natural tendency of oxygen to dissociate.
Left shift of curve indicates haemoglobin’s increased affinity for oxygen (seen at lungs). Right shift indicates decreased affinity and is seen with an increase in body temperature, hydrogen ions, 2,3-diphosphoglycerate (DPG), carbon dioxide concentration and exercise. Under normal resting conditions in a healthy individual, the normal position of the curve is at a pH of 7.4. A shift in the position of the curve with a change in pH is called the Bohr effect. Left shift occurs in acute alkalosis, decrease in p(CO2), decrease in temperature and decrease in 2,3-DPG. The fetal haemoglobin curve is to the left of the adult haemoglobin to allow for oxygen diffusion across the placenta. The curve for myoglobin is even further to the left. Carbon monoxide has a much higher affinity for haemoglobin than oxygen does. Thus, carbon monoxide poisoning leads to hypoxia.
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This question is part of the following fields:
- Physiology
- Respiratory
-
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Question 146
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.
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This question is part of the following fields:
- Neurology
- Physiology
-
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Question 147
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.
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This question is part of the following fields:
- General
- Physiology
-
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Question 148
Incorrect
-
Which of the following will be affected by a lesion in the posterior column-medial lemniscus system?
Your Answer:
Correct Answer: Fine touch
Explanation:The posterior column–medial lemniscus (PCML) pathway is a sensory pathway that transmits fine touch and conscious proprioceptive information from the body to the brain. As the posterior columns are also known as dorsal columns, the pathway is also called the dorsal column–medial lemniscus system or DCML.
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This question is part of the following fields:
- Neurology
- Physiology
-
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Question 149
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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
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Question 150
Incorrect
-
Whilst snorkelling, a 30-year old gentleman has the respiratory rate of 10/min, tidal volume of 550 ml and an effective anatomical dead space of 250 ml. Which of the following will bring about a maximum increase in his alveolar ventilation?
Your Answer:
Correct Answer: A 2x increase in tidal volume and a shorter snorkel
Explanation:Alveolar ventilation = respiratory rate × (tidal volume − anatomical dead space volume). Increase in respiratory rate simply causes movement of air in the anatomical dead space, with no contribution to the alveolar ventilation. By use of a shorter snorkel, the effective anatomical dead space will decrease and will cause a maximum rise in alveolar ventilation along with doubling of tidal volume.
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This question is part of the following fields:
- Physiology
- Respiratory
-
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Question 151
Incorrect
-
In what form are fats primarily transported in the body?
Your Answer:
Correct Answer: Free fatty acids
Explanation:Fat is mainly transported in the body as free fatty acids. Once out of the adipose cell, the free fatty acids get ionized and combine with albumin.
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This question is part of the following fields:
- Gastroenterology
- Physiology
-
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Question 152
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.
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This question is part of the following fields:
- Neurology
- Physiology
-
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Question 153
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 154
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.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 155
Incorrect
-
The most important difference between interstitial fluid and plasma is the:
Your Answer:
Correct Answer: Protein concentration
Explanation:Interstitial fluid (or tissue fluid or intercellular fluid) is a solution that surrounds the cells of multicellular animals. It is the main component of the extracellular fluid, which also includes plasma, lymph and transcellular fluid. Plasma, the major component in blood, communicates freely with interstitial fluid through pores and intercellular clefts in capillary endothelium. Interstitial fluid consists of a water solvent containing amino acids, sugars, fatty acids, coenzymes, hormones, neurotransmitters, salts, as well as waste products from the cells. Red blood cells, platelets and plasma proteins cannot pass through the walls of the capillaries. The resulting mixture that does pass through is essentially blood plasma without the plasma proteins. Tissue fluid also contains certain types of white blood cells. Once the extracellular fluid collects into small vessels it is considered to be lymph, and the vessels that carry it back to the blood are called the lymphatic vessels. The lymphatic system returns protein and excess interstitial fluid to the circulation.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
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Question 156
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
-
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Question 157
Incorrect
-
Which part of the nephron would have to be damaged to stop the reabsorption of the majority of salt and water?
Your Answer:
Correct Answer: Proximal tubule
Explanation:The proximal tubule is the portion of the duct system of the nephron of the kidney which leads from Bowman’s capsule to the loop of Henle. It is conventionally divided into the proximal convoluted tubule (PCT) and the proximal straight tubule (PST). The proximal tubule reabsorbs the majority (about two-thirds) of filtered salt and water. This is done in an essentially iso-osmotic manner. Both the luminal salt concentration and the luminal osmolality remain constant (and equal to plasma values) along the entire length of the proximal tubule. Water and salt are reabsorbed proportionally because the water is dependent on and coupled with the active reabsorption of Na+. The water permeability of the proximal tubule is high and therefore a significant transepithelial osmotic gradient is not possible.
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This question is part of the following fields:
- Physiology
- Renal
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Question 158
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.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 159
Incorrect
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Calculate the cardiac output in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer:
Correct Answer: 6.25 l/min
Explanation:As per Fick’s principle, VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CAO2 = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24 – 0.16, CO = 500/0.8, CO = 6.25 l/min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 160
Incorrect
-
Renal function is an indication of the state of the kidney, measured by glomerular filtration rate (GFR). In a healthy person, GFR would be greatly increased by:
Your Answer:
Correct Answer: Substantial increases in renal blood flow
Explanation:An increase in the rate of renal blood flow (RBF) greatly increases the glomerular filtration rate (GFR). The more plasma available (from increased RBF), the more filtrate is formed. Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal (kidney) glomerular capillaries into the Bowman’s capsule per unit time. Central to the physiologic maintenance of GFR is the differential basal tone of the afferent and efferent arterioles.
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This question is part of the following fields:
- Physiology
- Renal
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Question 161
Incorrect
-
Calculate the stroke volume in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer:
Correct Answer: 90 ml
Explanation:Fick’s principle states that, VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CAO2 = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24–0.16, CO = 500/0.8, CO = 6.25 l/min. Cardiac output is also given by product of stroke volume and heart rate. Thus, stroke volume = cardiac output / heart rate = 6.25/70 × 1000 stroke volume = 90 ml approximately.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 162
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.
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This question is part of the following fields:
- General
- Physiology
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Question 163
Incorrect
-
After a severe asthma attack, a 26-year-old woman is left in a markedly hypoxic state. In which of the following organs are the arterial beds most likely to be vasoconstricted due to the hypoxia?
Your Answer:
Correct Answer: Lungs
Explanation:Hypoxic pulmonary vasoconstriction is a local response to hypoxia resulting primarily from constriction of small muscular pulmonary arteries in response to reduced alveolar oxygen tension. This unique response of pulmonary arterioles results in a local adjustment of perfusion to ventilation. This means that if a bronchiole is obstructed, the lack of oxygen causes contraction of the pulmonary vascular smooth muscle in the corresponding area, shunting blood away from the hypoxic region to better-ventilated regions. The purpose of hypoxic pulmonary vasoconstriction is to distribute blood flow regionally to increase the overall efficiency of gas exchange between air and blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 164
Incorrect
-
Calculate the stroke volume in a patient admitted for coronary bypass surgery, with the following parameters pre-operatively:
Oxygen consumption = 300 ml/min
Arterial oxygen content = 20 ml/100 ml blood
Pulmonary arterial oxygen content = 15 ml/100 ml blood and Heart rate = 100 beats/min.Your Answer:
Correct Answer: 60 ml
Explanation:By Fick’s principle, cardiac output can be calculated as follows: VO2 = CO × (CAO2– CVO2) where VO2= oxygen consumption, CO = cardiac output, CAO2 = arterial oxygen content and CvO2 = mixed venous oxygen content. Thus, in the given problem, 300 ml/min = CO × (20 – 15) ml/100 ml CO = 300 × 100/5 ml/min CO = 6000 ml/min. Also, cardiac output = stroke volume × heart rate. Thus, 6000 ml/min = stroke volume × 100 beats/min. Hence, stroke volume = 6000/100 ml/min which is 60 ml/min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 165
Incorrect
-
upon stroking the plantar surface of a patient's foot, extension of toes was noted. This is likely to be accompanied with:
Your Answer:
Correct Answer: Spasticity
Explanation:An upper motor neuron lesion affects the neural pathway above the anterior horn cell or motor nuclei of the cranial nerves, whereas a lower motor neurone lesion affects nerve fibres travelling from the anterior horn of the spinal cord to the relevant muscles. An upper motor neurone lesions results in the following:
– Spasticity in the extensor muscles (lower limbs) or flexor muscles (upper limbs).
– ‘clasp-knife’ response where initial resistance to movement is followed by relaxation
– Weakness in the flexors (lower limbs) or extensors (upper limbs) with no muscle wasting
– Brisk tendon jerk reflexes
– Positive Babinski sign (on stimulation of the sole of the foot, the big toe is raised rather than curled downwards)
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This question is part of the following fields:
- Neurology
- Physiology
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Question 166
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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 167
Incorrect
-
Chronic obstructive pulmonary disease (COPD) is likely to result in:
Your Answer:
Correct Answer: Respiratory acidosis
Explanation:COPD leads to respiratory acidosis (chronic). This occurs due to hypoventilation which involves multiple causes, such as poor responsiveness to hypoxia and hypercapnia, increased ventilation/perfusion mismatch leading to increased dead space ventilation and decreased diaphragm function.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 168
Incorrect
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What is the normal glomerular filtration rate?
Your Answer:
Correct Answer: 125 mL/min
Explanation:The normal glomerular filtration rate (GFR) in humans is 125 mL/min. After the age of 40, GFR decreases progressively by about 0.4–1.2 mL/min per year.
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This question is part of the following fields:
- Physiology
- Renal
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Question 169
Incorrect
-
Arterial blood gas analysis of a man admitted with acute exacerbation of chronic obstructive pulmonary disease (COPD) showed the following: pH = 7.28, p(CO2) = 65.5 mmHg, p(O2)= 60 mmHg and standard bicarbonate = 30.5 mmol/l. This patient had:
Your Answer:
Correct Answer: Respiratory acidosis
Explanation:Acidosis with high p(CO2) and normal standard bicarbonate indicates respiratory acidosis, commonly seen in acute worsening of COPD patients. 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. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with a 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).
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 170
Incorrect
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A 30 year old female suffered from mismatched transfusion induced haemolysis. Which substance will be raised in the plasma of this patient?
Your Answer:
Correct Answer: Bilirubin
Explanation:Bilirubin is a yellow pigment that is formed due to the break down of RBCs. Haemolysis results in haemoglobin that is broken down into a haem portion and globin which is converted into amino acids and used again. Haem is converted into unconjugated bilirubin in the macrophages and shunted to the liver. In the liver it is conjugated with glucuronic acid making it water soluble and thus excreted in the urine. Its normal levels are from 0.2-1 mg/dl. Increased bilirubin causes jaundice and yellowish discoloration of the skin.
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This question is part of the following fields:
- General
- Physiology
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Question 171
Incorrect
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Which of the following is a true statement regarding secretion of gastric acid?
Your Answer:
Correct Answer: Acetylcholine increases gastric acid secretion
Explanation:Gastric acid secretion is increased by acetylcholine, histamine and gastrin, with the help of cAMP as a secondary messenger. They increase H+ and Cl- secretion by increasing the number of H+/K+ ATPase molecules and Cl- channels. In contrast, gastric acid secretion is decreased by somatostatin, epidermal growth factor and prostaglandins.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 172
Incorrect
-
A 35-year-old woman in her 37th week of pregnancy complains of urinary incontinence. She is most likely to have:
Your Answer:
Correct Answer: Stress incontinence
Explanation:Urinary incontinence is the involuntary excretion of urine from one’s body. It is often temporary and it almost always results from an underlying medical condition. Several types include:
– Stress incontinence is the voiding of urine following increased abdominal pressure e.g. laughing, coughing, pregnancy etc. It is the most common form of incontinence in women, most commonly due to pelvic floor muscle weakness, physical changes from pregnancy, childbirth and menopause. In men it is a common problem following a prostatectomy. Most lab results such as urine analysis, cystometry and postvoid residual volume are normal.
– Urge incontinence is involuntary loss of urine occurring for no apparent reason while suddenly feeling the need or urge to urinate. The most common cause of urge incontinence are involuntary and inappropriate detrusor muscle contractions.
– Functional incontinence – occurs when a person does not recognise the need to go to the toilet, recognise where the toilet is or get to the toilet in time. The urine loss may be large. Causes of functional incontinence include confusion, dementia, poor eyesight, poor mobility, poor dexterity or unwillingness. t
– Overflow incontinence – sometimes people find that they cannot stop their bladders from constantly dribbling or continuing to dribble for some time after they have passed urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 173
Incorrect
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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) = 80–100 fl, mean corpuscular haemoglobin concentration (MCHC) = 31–37 g/dl. Which of the following is correct regarding this patient’s erythrocytes:
Your Answer:
Correct Answer: Normal MCV
Explanation:MCV is the mean corpuscular volume and it is calculated from the haematocrit and the RBC count. It is normally 90 fl. Mean corpuscular haemoglobin concentration (MCHC) [g/dl] = haemoglobin [g/dl]/haematocrit = 11/0.27 = 41 g/dl and is higher than normal range (32 to 36 g/dL).
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This question is part of the following fields:
- General
- Physiology
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Question 174
Incorrect
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Production of pain is most likely associated with:
Your Answer:
Correct Answer: Substance P
Explanation:Substance P is a short-chain polypeptide that functions as a neurotransmitter and as a neuromodulator, and is thus, a neuropeptide. It has been linked with pain regulation, mood disorders, stress, reinforcement, neurogenesis, respiratory rhythm, neurotoxicity, nausea and emesis. It is also a potent vasodilator as it brings about release of nitric oxide from the endothelium. Its release can also cause hypotension.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 175
Incorrect
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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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 176
Incorrect
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Gastric acid secretion is stimulated by which of the following?
Your Answer:
Correct Answer: Gastrin
Explanation:Gastric acid secretion is stimulated by three factors:
– Acetylcholine, from parasympathetic neurones of the vagus nerve that innervate parietal cells directly
– Gastrin, produced by pyloric G-cells
– Histamine, produced by mast cells.
Gastric acid is inhibited by three factors:
– Somatostatin
– Secretin
– Cholecystokinin
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 177
Incorrect
-
After a total colectomy and ileotomy, a 50-year old diabetic man who was a known case of diabetic nephropathy had persistent metabolic acidosis. The patient appeared well perfused, with normal vital signs and normal fluid balance. Investigations revealed:
Sodium = 132 mmol/l
Potassium = 6.6 mmol/l
Creatinine = 185 μmol/l (2.16 mg/dl)
Chloride = 109 μmol/l
8am cortisol = 500 nmol/l (18 μg/dl)
pH = 7.29, p(CO2) = 27 mmHg
p(O2) = 107 mmHg
standard bicarbonate = 12 mmol/l.
What is the likely causes of his acidosis?Your Answer:
Correct Answer: Renal tubular acidosis
Explanation:Acidosis here is due to low bicarbonate. The low p(CO2) is seen in compensation. The anion gap is normal, ruling out intra-abdominal ischaemia (which leads to metabolic acidosis). If it was a gastrointestinal aetiology, low potassium would be seen. The history of diabetic nephropathy predisposes to renal tubular acidosis. Type 4 (hyporeninaemic hypoaldosteronism) is associated with high potassium and is found in diabetic and hypertensive renal disease.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 178
Incorrect
-
A 50-year-old woman goes to the doctor complaining of myalgia, muscle cramps, and weakness; she is diagnosed with severe hypokalaemia. Which of the following is the most common cause of hypokalaemia?
Your Answer:
Correct Answer: Prolonged vomiting
Explanation:Potassium is one of the body’s major ions. Nearly 98% of the body’s potassium is intracellular. The ratio of intracellular to extracellular potassium is important in determining the cellular membrane potential. Small changes in the extracellular potassium level can have profound effects on the function of the cardiovascular and neuromuscular systems. Hypokalaemia may result from conditions as varied as renal or gastrointestinal (GI) losses, inadequate diet, transcellular shift (movement of potassium from serum into cells) and medications. The important causes of hypokalaemia are:
Renal losses: renal tubular acidosis, hyperaldosteronism, magnesium depletion, leukaemia (mechanism uncertain).
GI losses: vomiting or nasogastric suctioning, diarrhoea, enemas or laxative use, ileal loop.
Medication effects: diuretics (most common cause), β-adrenergic agonists, steroids, theophylline, aminoglycosides.
Transcellular shift: insulin, alkalosis.
Severe hypokalaemia, with serum potassium concentrations of 2.5–3 meq/l, may cause muscle weakness, myalgia, tremor, muscle cramps and constipation.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 179
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.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 180
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.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 181
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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 182
Incorrect
-
Rapid Eye Movement (REM) sleep is typically associated with:
Your Answer:
Correct Answer: Penile erections
Explanation:Normal sleep comprises of alternate cycles between slow-wave sleep (non-REM sleep) and REM sleep. REM sleep is characterized by increased metabolic brain activity and EEG desynchronization. Somnambulism (sleepwalking), enuresis (bedwetting) and night terrors all occur during slow-wave sleep or during arousal from slow-wave sleep. In comparison, REM sleep is characterized by hypotonia of major muscle groups (excluding ocular muscles), dreams, nightmares and penile erection.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 183
Incorrect
-
Which of the following conditions causes an elevation of the pH in the tissues with elevated arterial CO2 content?
Your Answer:
Correct Answer: Metabolic alkalosis
Explanation:Metabolic alkalosis is a metabolic condition in which the pH of tissue is elevated beyond the normal range (7.35-7.45). This is the result of decreased hydrogen ion concentration, leading to increased bicarbonate, or alternatively a direct result of increased bicarbonate concentrations. Normally, arterial pa(CO2) increases by 0.5–0.7 mmHg for every 1 mEq/l increase in plasma bicarbonate concentration, a compensatory response that is very quick. If the change in pa(CO2) is not within this range, then a mixed acid–base disturbance occurs.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 184
Incorrect
-
A 45-year old gentleman presents with diarrhoea for two weeks. He has no history of fever and the diarrhoea stops on fasting. Which is the most likely type of diarrhoea that he is suffering from?
Your Answer:
Correct Answer: Osmotic
Explanation:The different types of diarrhoea are:
1. Secretory diarrhoea – Due to increased secretion or decreased absorption. There is minimal to no structural damage in this type. The most common cause is cholera toxin which stimulates secretion of anions (especially chloride), with sodium and water.
2. Osmotic diarrhoea – Due to increased osmotic load, there is water loss. This occurs in cases of maldigestion syndromes, such as coeliac or pancreatic disease.
3. Motility-related diarrhoea – Occurs in cases of abnormal gastrointestinal motility. Due to increased motility, there is poor absorption and this leads to diarrhoea. This is seen post-vagotomy or in diabetic neuropathy.
4. Inflammatory diarrhoea – Due to damage to the mucosa or brush border, there is a loss of protein-rich fluids and poor absorption. Features of all the above three types can be seen in this type. Aetiology includes bacterial, viral, parasitic infections or autoimmune problems including inflammatory bowel disease.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 185
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.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 186
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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 187
Incorrect
-
If a 55-year old gentleman who has suffered a stroke, develops a tremor in his fingers that worsens on reaching for an object, what part of his brain is likely to be involved?
Your Answer:
Correct Answer: Cerebellum
Explanation:The cerebellum plays an important role in the integration of sensory perception and motor output. Multiple neural pathways link the cerebellum with the motor cortex and the spinocerebellar tract. The cerebellum uses the constant feedback on body position to fine-tune the movements and integrates these pathways. The patient described here has a characteristic cerebellar tremor that is a slow, broad tremor of the extremities and occurs at the end of a purposeful movement.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 188
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 189
Incorrect
-
Mechanical distortion, and not K+ channels are responsible for distortion of which of the following structures?
Your Answer:
Correct Answer: Pacinian corpuscle
Explanation:Pacinian corpuscles are a type of mechanoreceptor, sensitive to deep pressure, touch and high-frequency vibration. The Pacinian corpuscles are ovoid and about 1 mm long. In the centre of the corpuscle is the inner bulb, which is a fluid-filled cavity with a single afferent unmyelinated nerve ending. Any deformation in the corpuscle causes the generation of action potentials by opening of pressure-sensitive sodium ion channels in the axon membrane. This allows influx of sodium ions, creating a receptor potential (independent of potassium channels).
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This question is part of the following fields:
- Neurology
- Physiology
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Question 190
Incorrect
-
Most of the coagulation factors are serine proteases. Which of the following is not one of them?
Your Answer:
Correct Answer: Factor XIII
Explanation:Serine protease coagulation factors include: thrombin, plasmin, Factors X, XI and XII. Factor VIII and factor V are glycoproteins and factor XIII is a transglutaminase.
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This question is part of the following fields:
- General
- Physiology
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Question 191
Incorrect
-
Which of the following is responsible for the activation of pepsinogen released in the stomach?
Your Answer:
Correct Answer: Acid pH and pepsin
Explanation:Pepsinogen is the inactive precursor of pepsin. Once secreted, it comes in contact with hydrochloric acid and pepsin, previously formed, and undergoes cleavage to form active pepsin.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 192
Incorrect
-
How much blood can the pulmonary vessels of a 45-year-old healthy man accommodate when he is at rest?
Your Answer:
Correct Answer: 500 ml
Explanation:Pulmonary circulation is the portion of the cardiovascular system which carries deoxygenated blood away from the heart, to the lungs, and returns oxygenated blood back to the heart. The vessels of the pulmonary circulation are very compliant (easily distensible) and so typically accommodate about 500 ml of blood in an adult man. This large lung blood volume can serve as a reservoir for the left ventricle, particularly during periods when left ventricular output momentarily exceeds venous return.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 193
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.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 194
Incorrect
-
For calculation of cardiac output by Fick's principle, which of the following vessels is the best source of venous blood to determine the arterial-to-venous oxygen tension difference?
Your Answer:
Correct Answer: Pulmonary artery
Explanation:Fick’s principle states that the total uptake (or release) of a substance by peripheral tissues is equal to the product of the blood flow to the peripheral tissues and the arterial– venous concentration difference (gradient) of the substance. It is used to measure the cardiac output, and the formula is Cardiac output = oxygen consumption divided by arteriovenous oxygen difference. Assuming there are no shunts across the pulmonary system, the pulmonary blood flow equals the systemic blood flow. The arterial and venous blood oxygen content is measured by sampling from the pulmonary artery (low oxygen content) and pulmonary vein (high oxygen content). Peripheral arterial blood is used as a surrogate for the pulmonary vein.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 195
Incorrect
-
Acute respiratory distress syndrome (ARDS) is a medical condition that occurs in critically ill patients, and can be triggered by events such as trauma and sepsis. Which of the following variables is most likely to be lower than normal in a patient with ARDS?
Your Answer:
Correct Answer: Lung compliance
Explanation:Acute (or Adult) respiratory distress syndrome (ARDS) is a medical condition occurring in critically ill patients characterized by widespread inflammation in the lungs. The development of acute respiratory distress syndrome (ARDS) starts with damage to the alveolar epithelium and vascular endothelium, resulting in increased permeability to plasma and inflammatory cells. These cells pass into the interstitium and alveolar space, resulting in pulmonary oedema. Damage to the surfactant-producing type II cells and the presence of protein-rich fluid in the alveolar space disrupt the production and function of pulmonary surfactant, leading to micro atelectasis and impaired gas exchange. The pathophysiological consequences of lung oedema in ARDS include a decrease in lung volumes, compliance and large intrapulmonary shunts. ARDS may be seen in the setting of pneumonia, sepsis, following trauma, multiple blood transfusions, severe burns, severe pancreatitis, near-drowning, drug reactions, or inhalation injuries.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 196
Incorrect
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A 56 year old gentleman, who is a chronic smoker presents to the clinic with dyspnoea. His Chest X-ray shows intercostal space widening with increased blackening bilaterally. What is the most likely finding on his pulmonary function test?
Your Answer:
Correct Answer: Increased total lung capacity
Explanation:In patients suspected of having chronic obstructive pulmonary disease, pulmonary function testing (PFT) is useful to confirm airway obstruction, to quantify its severity and reversibility, for following disease progression and monitoring response to treatment. These tests include:
FEV1 – volume of air forcefully expired during the first second after a full breath
Forced vital capacity (FVC) – total volume of air expired with maximal force and flow-volume loops. The hallmark of airway obstruction is reduction of FEV1, FVC and the ratio of FEV1/FVC, with a concave pattern in expiratory tracing on the flow-volume loop. FEV1 and forced vital capacity (FVC) are easily measured with office spirometry and are useful to assess the severity of disease. Other parameters include increased total lung capacity, functional residual capacity and residual volume, which can help distinguish chronic obstructive pulmonary disease (COPD) from restrictive pulmonary disease where these values are lower than normal. Along with these, other tests are decreased vital capacity; and decreased diffusing capacity for carbon monoxide (DLCO). DLCO is non-specific and can be low in other disorders that affect the pulmonary vascular bed, such as interstitial lung disease. DLCO is however useful to distinguish COPD from asthma, in which DLCO is normal or elevated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 197
Incorrect
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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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 198
Incorrect
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A 79-year-old has been bedridden for 2 months after suffering from a stroke. She suddenly developed shortness of breath and chest pain, and was diagnosed with a pulmonary embolism. Which of the following is most likely to increase in this case?
Your Answer:
Correct Answer: Ventilation/perfusion ratio
Explanation:Pulmonary embolism (PE) is a blockage of an artery in the lungs by an embolus that has travelled from elsewhere in the body through the bloodstream. The change in cardiopulmonary function is proportional to the extent of the obstruction, which varies with the size and number of emboli obstructing the pulmonary arteries. The resulting physiological changes may include pulmonary hypertension with right ventricular failure and shock, dyspnoea with tachypnoea and hyperventilation, arterial hypoxaemia and pulmonary infarction. Consequent alveolar hyperventilation is manifested by a lowered pa(CO2). After occlusion of the pulmonary artery, areas of the lung are ventilated but not perfused, resulting in wasted ventilation with an increased ventilation/perfusion ratio – the physiological hallmark of PE – contributing to a further hyperventilatory state. The risk of blood clots is increased by cancer, prolonged bed rest, smoking, stroke, certain genetic conditions, oestrogen-based medication, pregnancy, obesity, and post surgery.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 199
Incorrect
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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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 200
Incorrect
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A 43-year-old diabetic man complains of headaches, palpitations, anxiety, abdominal pain and weakness. He is administered sodium bicarbonate used to treat:
Your Answer:
Correct Answer: Metabolic acidosis
Explanation:Sodium bicarbonate is indicated in the management of metabolic acidosis, which may occur in severe renal disease, uncontrolled diabetes, circulatory insufficiency due to shock or severe dehydration, extracorporeal circulation of blood, cardiac arrest and severe primary lactic acidosis. Bicarbonate is given at 50-100 mmol at a time under scrupulous monitoring of the arterial blood gas readings. This intervention, however, has some serious complications including lactic acidosis, and in those cases, should be used with great care.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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SESSION STATS - PERFORMANCE PER SPECIALTY
