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Question 1
Incorrect
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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: Mitochondria
Correct 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 2
Correct
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When one is silently counting, what part of his brain will show increased regional cerebral blood flow (rCBF)?
Your 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 3
Correct
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A 56-year-old man undergoes tests to determine his renal function. His results over a period of 24 hours were:
Urine flow rate: 2. 0 ml/min
Urine inulin: 1.0 mg/ml
Plasma inulin: 0.01 mg/ml
Urine urea: 260 mmol/l
Plasma urea: 7 mmol/l
What is the glomerular filtration rate?Your Answer: 200 ml/min
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal (kidney) glomerular capillaries into the Bowman’s capsule per unit time. GFR is equal to the inulin clearance because inulin is freely filtered into Bowman’s capsule but is not reabsorbed or secreted. 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. Thus, glomerular filtration rate = (1.0 × 2. 0)/0.01 = 200 ml/min.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 4
Incorrect
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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: Cardiac muscle cells
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 5
Correct
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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: 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 6
Incorrect
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After having donated a unit of blood. The blood bank will prefer to use which of the following anticoagulants to store the blood?
Your Answer: Heparin
Correct Answer: Citrate
Explanation:Calcium is necessary for coagulation to occur. Citrate being a chelator and combining with calcium ions to form un-ionised compound will prevent coagulation. Following transfusion the citrate is removed by the liver with in a few minutes. Oxalate also works on the same principle but it is toxic to the body.
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This question is part of the following fields:
- General
- Physiology
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Question 7
Correct
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A 40 year old man suffered severe trauma following a MVA. His BP is 72/30 mmhg, heart rate of 142 beats/mins and very feeble pulse. He was transfused 3 units of blood and his BP returned to 100/70 and his heart rate slowed to 90 beats/min. What decreased after transfusion?
Your Answer: Total peripheral resistance
Explanation:The patient is in hypovolemic shock, he is transfused with blood, this fluid resuscitation will result in a decreased sympathetic discharge and adequate ventricular filling which will result in the decreases TPR with an increased CO and cardiac filling pressures
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 8
Incorrect
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A lesion involving the lateral geniculate nucleus of the thalamus is likely to affect:
Your Answer: Touch
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 9
Incorrect
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Calculate the cardiac output of a patient with the following measurements: oxygen uptake 200 ml/min, oxygen concentration in the peripheral vein 7 vol%, oxygen concentration in the pulmonary artery 10 vol% and oxygen concentration in the aorta 15 vol%.
Your Answer: 1500 ml/min
Correct Answer: 4000 ml/min
Explanation:The Fick’s principle states that the uptake of a substance by an organ equals the arteriovenous difference of the substance multiplied by the blood flowing through the organ. We can thus calculate the pulmonary blood flow with pulmonary arterial (i.e., mixed venous) oxygen content, aortic oxygen content and oxygen uptake. The pulmonary blood flow, systemic blood flow and cardiac output can be considered the same assuming there are no intracardiac shunts. Thus, we can calculate the cardiac output. Cardiac output = oxygen uptake/(aortic − mixed venous oxygen content) = 200 ml/min/(15 ml O2/100 ml − 10 ml O2/100 ml) = 200 ml/min/(5 ml O2/100 ml) = 200 ml/min/0.05 = 4000 ml/min.
It is crucial to remember to use pulmonary arterial oxygen content and not peripheral vein oxygen content, when calculating the cardiac output by Fick’s method.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 10
Incorrect
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A patient presents with loss of fine touch and sense of proprioception in the lower part of the body (below T6). He is likely to have a lesion involving:
Your Answer: Descending corticospinal tract
Correct Answer: Gracile nucleus
Explanation:The gracile nucleus is located in the medulla oblongata and is one of the dorsal column nuclei involved in the sensation of fine touch and proprioception. It contains second-order neurons of the dorsal column–medial lemniscus system, that receive inputs from sensory neurones of the dorsal root ganglia and send axons that synapse in the thalamus.
The gracile nucleus and fasciculus carry epicritic, kinaesthetic and conscious proprioceptive information from the lower part of the body (below the level of T6 in the spinal cord). Similar information from the upper part of body (above T6, except for face and ear) is carried by the cuneate nucleus and fasciculus. The information from face and ear is carried by the primary sensory trigeminal nucleus.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 11
Incorrect
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Decreased velocity of impulse conduction through the atrioventricular node (AV node) in the heart will lead to:
Your Answer: Atrial fibrillation
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 12
Incorrect
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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: A +ve
Correct 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 13
Incorrect
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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: Ipsilateral spastic paralysis, contralateral loss of vibration and proprioception (position sense) and contralateral loss of pain and temperature sensation beginning one or two segments below the lesion
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
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Question 14
Correct
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With respect to far accommodation, which of the following is a TRUE statement?
Your Answer: The ciliary muscles are relaxed
Explanation:Myopia or near-sightedness is a disease due to elongated eyeballs or too strong a lens. For far accommodation (focus of a distant object onto the retina), the lens needs to decrease its refractive power, or in other words, increase its focal length. This is done by relaxation of ciliary muscles which tightens the zonular fibres and flattening of the lens. Relaxation of the zonular fibres, rounding of the lens, shortening of the focal length and constriction of the pupil occurs during near accommodation.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 15
Correct
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Which of the following variables are needed to calculate inspiratory reserve volume of a patient?
Your Answer: Tidal volume, vital capacity and expiratory reserve volume
Explanation:Vital capacity = inspiratory reserve volume + tidal volume + expiratory reserve volume. Thus, inspiratory reserve volume can be calculated if tidal volume, vital capacity and expiratory reserve volume are known.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 16
Incorrect
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After surgery, a patient developed a stitch granuloma . Which leukocyte in the peripheral blood will become an activated macrophage in this granuloma?
Your Answer: Basophil
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 17
Incorrect
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Diuretics that act on the ascending limb of the loop of Henle produce:
Your Answer: Reduced passive transport of chloride
Correct Answer: Reduced active transport of sodium
Explanation:The loop of Henlé leads from the proximal convoluted tubule to the distal convoluted tubule. Its primary function uses a counter current multiplier mechanism in the medulla to reabsorb water and ions from the urine. It can be divided into four parts:
1. Descending limb of loop of Henlé – low permeability to ions and urea, while being highly permeable to water
2. Thin ascending limb of loop of Henlé – not permeable to water, but it is permeable to ions
3. Medullary thick ascending limb of loop of Henlé – sodium (Na+), potassium (K+) and chloride (Cl–) ions are reabsorbed by active transport. K+ is passively transported along its concentration gradient through a K+ channel in the basolateral aspect of the cells, back into the lumen of the ascending limb.
4. The cortical thick ascending limb – the site of action where loop diuretics such as furosemide block the K+/Na+/2Cl− co-transporters = reduced active transport.
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This question is part of the following fields:
- Physiology
- Renal
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Question 18
Incorrect
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Glucose is the most important source of energy for cellular respiration. The transport of glucose in the renal tubular cells occurs via:
Your Answer: Facilitated diffusion
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 19
Incorrect
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Myoglobin is released as a result of rhabdomyolysis from damaged skeletal muscles. What function do they perform in the muscle?
Your Answer: Produces ATP
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 20
Incorrect
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The blood-brain barrier is a membrane that separates the circulating blood from the brain extracellular fluid in the central nervous system (CNS). Which of the following statements regarding the blood– brain barrier is CORRECT?
Your Answer: It permits carbon dioxide to pass via facilitated diffusion
Correct Answer: It breaks down in areas of brain that are infected
Explanation:The blood–brain barrier is a membrane that controls the passage of substances from the blood into the central nervous system. It is a physical barrier between the local blood vessels and most parts of the central nervous system and stops many substances from travelling across it. During meningitis, the blood–brain barrier may be disrupted. This disruption may increase the penetration of various substances (including either toxins or antibiotics) into the brain. A few regions in the brain, including the circumventricular organs, do not have a blood–brain barrier.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 21
Correct
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Post-total gastrectomy, there will be a decreased production of which of the following enzymes?
Your Answer: Pepsin
Explanation:Pepsin is a protease that is released from the gastric chief cells and acts to degrade proteins into peptides. Released as pepsinogen, it is activated by hydrochloric acid and into pepsin itself. Gastrin and the vagus nerve trigger the release of pepsinogen and HCl when a meal is ingested. Pepsin functions optimally in an acidic environment, especially at a pH of 2.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 22
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 23
Incorrect
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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: 60 mmHg
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 24
Correct
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During strenuous exercise, what else occurs besides tachycardia?
Your 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 25
Correct
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A patient is diagnosed with Conn’s syndrome. Aldosterone is secreted from where?
Your Answer: Zona glomerulosa of the adrenal cortex
Explanation:The adrenal gland comprises an outer cortex and an inner medulla, which represent two developmentally and functionally independent endocrine glands.
The adrenal medulla secretes adrenaline (70%) and noradrenaline (30%)
The adrenal cortex consists of three layers (remembered by the mnemonic GFR):
G = zona glomerulosa – secretes aldosterone
F = zona fasciculata – secretes cortisol and sex steroids
R = zona reticularis – secretes cortisol and sex steroids.
Aldosterone facilitates the reabsorption of sodium and water and the excretion of potassium and hydrogen ions from the distal convoluted tubule and collecting ducts. Conn’s syndrome is characterized by increased aldosterone secretion from the adrenal glands.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 26
Correct
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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: 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 27
Incorrect
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Hormones of the anterior pituitary include which of the following?
Your Answer: Antidiuretic hormone (ADH)
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 28
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 29
Correct
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In the kidney, the macula densa is an area of closely packed specialized cells lining the wall of the:
Your 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 30
Incorrect
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Which of the following conditions causes an elevation of the pH in the tissues with elevated arterial CO2 content?
Your Answer: Diabetic ketoacidosis
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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SESSION STATS - PERFORMANCE PER SPECIALTY
