-
Question 1
Incorrect
-
Which of the following variables are needed to calculate inspiratory reserve volume of a patient?
Your Answer: Vital capacity and expiratory reserve volume
Correct 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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 2
Incorrect
-
A lesion involving the lateral geniculate nucleus of the thalamus is likely to affect:
Your Answer: Hearing
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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 3
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: Compensated respiratory acidosis
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).
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 4
Incorrect
-
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: von Willebrand factor suppresses blood coagulation
Correct 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. -
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 5
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: Anomalous left coronary artery
Correct Answer: Tetralogy of Fallot
Explanation:The most common congenital cyanotic heart disease and the most common cause of blue baby syndrome, Tetralogy of Fallot shows four cardiac malformations occurring together. These are ventricular septal defect (VSD), pulmonary stenosis (right ventricular outflow obstruction), overriding aorta (degree of which is variable), and right ventricular hypertrophy. The primary determinant of severity of disease is the degree of pulmonary stenosis. Tetralogy of Fallot is seen in 3-6 per 10,000 births and is responsible for 5-7% congenital heart defects, with slightly higher incidence in males. It has also been associated with chromosome 22 deletions and DiGeorge syndrome. It gives rise to right-to-left shunt leading to poor oxygenation of blood. Primary symptom is low oxygen saturation in the blood with or without cyanosis at birth of within first year of life. Affected children ay develop acute severe cyanosis or ‘tet spells’ (sudden, marked increase in cyanosis, with syncope, and may result in hypoxic brain injury and death). Other symptoms include heart murmur, failure to gain weight, poor development, clubbing, dyspnoea on exertion and polycythaemia. Chest X-ray reveals characteristic coeur-en-sabot (boot-shaped) appearance of the heart. Treatment consists of immediate care for cyanotic spells and Blalock–Taussig shunt (BT shunt) followed by corrective surgery.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 6
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: Muscle glycogen
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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 7
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: 0 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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 8
Correct
-
When at rest, which of the following will be higher in a marathon runner compared to an untrained individual?
Your Answer: Cardiac stroke volume
Explanation:Cardiac muscle hypertrophy is seen in trained athletes as compared to the normal population. This hypertrophy results in higher stroke volume at rest and increased cardiac reserve (maximum cardiac output during exercise). However, the cardiac output at rest is almost the same in both trained and untrained people. This is because in trained athletes, the heart rate is slower, even up to 40-50 beats/min. There is minimal affect of athletic training on oxygen consumption and respiratory rate.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 9
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: Gut
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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 10
Correct
-
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, 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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 11
Incorrect
-
Production of pain is most likely associated with:
Your Answer: Acetylcholine
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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 12
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: Respiratory alkalosis
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).
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 13
Incorrect
-
Which of the following organelles have the capacity to regenerate and spontaneously replicate?
Your Answer: Vacuole
Correct Answer: Mitochondrion
Explanation:A mitochondria is a membrane bound organelle found in eukaryotic cells. They are called the powerhouse of the cell and are the place where ATP is formed from energy generated through metabolism. They are capable of replication as well as repair and regeneration.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 14
Incorrect
-
What is the mostly likely cause of prolonged activated partial thromboplastin time (aPPT) ?
Your Answer: Liver disease
Correct Answer: Heparin therapy
Explanation:The partial thromboplastin time (PTT) or activated partial thromboplastin time (aPTT) is an indicator for measuring the efficacy of both the intrinsic and common coagulation pathway. Prolonged aPTT may indicate: use of heparin, antiphospholipid antibody and coagulation factor deficiency (e.g., haemophilia). Deficiencies of factors VIII, IX, XI and XII and rarely von Willebrand factor (if causing a low factor VIII level) may lead to a prolonged aPTT correcting on mixing studies.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 15
Incorrect
-
what is the cause of a prolonged PT(prothrombin time)?
Your Answer: Antiphospholipid antibody
Correct Answer: Liver disease
Explanation:PT measure the intrinsic pathway of coagulation. It determines the measure of the warfarin dose regime, liver disease and vit K deficiency status along with the clotting tendency of blood. PT measured factors are II,V,VII,X and fibrinogen. It is used along with aPTT which measure the intrinsic pathway.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 16
Incorrect
-
The gradual depolarization in-between action potentials in pacemaker tissue is a result of?
Your Answer: Changes in permeability of cells to the principal extracellular anions, Cl− and HCO3 −
Correct Answer: A combination of gradual inactivation outward IK along with the presence of an inward ‘funny’ current (If) due to opening of channels permeable to both Na+ and K+ ions
Explanation:One of the characteristic features of the pacemaker cell is the generation of a gradual diastolic depolarization also called the pacemaker potential. In phase 0, the upstroke of the action potential caused by an increase in the Ca2+ conductance, an influx of calcium occurs and a positive membrane potential is generated. The next is phase 3 which is repolarization caused by increased K+ conductance as a result of outwards K+ current. Phase 4 is a slow depolarization which accounts for the pacemaker activity, caused by increased conductance of Na+, inwards Na+ current called IF. it is turned on by repolarization.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 17
Incorrect
-
A 35-year-old ultra marathon runner becomes severely dehydrated and collapses. This patient most likely has:
Your Answer: Low water permeability of collecting duct tubular cells
Correct Answer: Decreased baroreceptor firing rate
Explanation:Baroreceptors are sensors located in the blood vessels of all vertebrate animals. They sense the blood pressure and relay the information to the brain, so that a proper blood pressure can be maintained. Acute dehydration results in decreased plasma volume and increased plasma osmolarity, since more water than salt is lost in sweat. The decrease in plasma volume leads to an inhibition of the baroreceptors and a lower firing rate. The increase in plasma osmolarity leads to increased ADH secretion and high plasma ADH levels, which increases water permeability of collecting duct cells. Therefore more water is reabsorbed by the kidneys and renal water excretion is low.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 18
Correct
-
What is the pH of freshly formed saliva at ultimate stimulation?
Your Answer: 8
Explanation:Saliva has four major components: mucus (lubricant), α-amylase (enzyme that initiates digestion of starch), lingual lipase (enzyme that begins fat digestion), and a slightly alkaline electrolyte solution for moistening food. As the secretion rate of saliva increases, its osmolality increases. Moreover, the pH changes from slightly acidic (at rest) to basic (pH 8) at ultimate stimulation. This occurs due to increase of HCO3-. Amylase and mucus also increase in concentration after stimulation.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 19
Correct
-
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: 7.28, 55, 81
Explanation:Hypoventilation (or respiratory depression) causes an increase in carbon dioxide (hypercapnia) and respiratory acidosis. It can result due to drugs such as alcohol, benzodiazepines, barbiturates, opiates, mechanical conditions or holding ones breath. Strong opioids such as heroin and fentanyl are commonly implicated and can lead to respiratory arrest. In recreational drug overdose, acute respiratory acidosis occurs with an increase in p(CO2) over 45 mm Hg and acidaemia (pH < 7.35)
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 20
Incorrect
-
In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your Answer: Decrease in temperature
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
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. -
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 21
Incorrect
-
Which of the following is a likely consequence of severe diarrhoea?
Your Answer: An increase in the chloride content of the body
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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 22
Correct
-
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 in the intervening period between power failure and hand ventilation? Arterial p(CO2), pH
Your Answer: Increase, Decrease
Explanation:Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. Respiratory acidosis can be acute or chronic. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg). The given problem represents acute respiratory acidosis and thus, will show a increase in arterial p(CO2) and decrease in pH.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 23
Incorrect
-
A 30-year-old woman feels thirsty. This thirst is probably due to:
Your Answer: Increased ICF volume
Correct Answer: Increased level of angiotensin II
Explanation:Thirst is the basic need or instinct to drink. It arises from a lack of fluids and/or an increase in the concentration of certain osmolites such as salt. If the water volume of the body falls below a certain threshold or the osmolite concentration becomes too high, the brain signals thirst. Excessive thirst, known as polydipsia, along with excessive urination, known as polyuria, may be an indication of diabetes. Angiotensin II is a hormone that is a powerful dipsogen (i.e. it stimulates thirst) that acts via the subfornical organ. It increases secretion of ADH in the posterior pituitary and secretion of ACTH in the anterior pituitary.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 24
Incorrect
-
Post-total gastrectomy, there will be a decreased production of which of the following enzymes?
Your Answer: Chymotrypsin
Correct 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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 25
Correct
-
Which of the following can lead to haemolytic anaemia?
Your 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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 26
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: Work of breathing
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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 27
Incorrect
-
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: 4 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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 28
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: Fructokinase
Correct Answer: Glucose-6-phosphatase
Explanation:The most common glycogen storage disorder is von Gierke’s disease or glycogen storage disease type I. It results from a deficiency of enzyme glucose-6-phosphatase which affects the ability of liver to produce free glucose from glycogen and gluconeogenesis; leading to severe hypoglycaemia. There is also increased glycogen storage in the liver and kidneys causing enlargement and various problems in their functioning. The disease also causes lactic acidosis and hyperlipidaemia. The main treatment includes frequent or continuous feedings of corn-starch or other carbohydrates.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 29
Correct
-
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: 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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 30
Incorrect
-
Decreased velocity of impulse conduction through the atrioventricular node (AV node) in the heart will lead to:
Your Answer: Increased heart rate
Correct Answer: Increased PR interval
Explanation:AV node damage may lead to an increase in the PR interval to as high as 0.25 – 0.40 s (normal = 0.12 – 0.20 s). In the case of severe impairment, there might be a complete failure of passage of impulses leading to complete block. In this case, the atria and ventricles will beat independently of each other.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
SESSION STATS - PERFORMANCE PER SPECIALTY
