-
Question 1
Correct
-
Coagulation in the body (in vivo) is a process in which several proteins known as coagulation factors are activated in a cascade effect to stop bleeding. Which of the following initiates this cascade effect?
Your Answer: Tissue factor
Explanation:Tissue factor (TF), also known as ‘factor III’ or ‘thromboplastin’, is an anti-coagulation protein that initiates the extrinsic coagulation. TF acts as a transmembrane receptor for Factor VII/VIIa . It is expressed by endothelial cells but also certain tissues, such as the heart and brain.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 2
Correct
-
In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your 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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 3
Incorrect
-
A 62-year-old male smoker, presented with shortness of breath, chronic cough and haemoptysis over the last three months. He has developed a fat pad in the base of his neck, rounded face, acne and osteoporosis. Which of the following is the most likely pulmonary disease that is causing these symptoms and findings?
Your Answer: Squamous cell carcinoma
Correct Answer: Small-cell anaplastic carcinoma
Explanation:Small cell lung cancer is a highly aggressive form of lung cancer. It is thought to originate from neuroendocrine cells in the bronchus called Feyrter cells and is often associated to ectopic production of hormones like ADH and ACTH that result in paraneoplastic syndromes and Cushing’s syndrome.
-
This question is part of the following fields:
- Pathology
- Respiratory
-
-
Question 4
Correct
-
Which of the following muscles aid in inspiration?
Your 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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 5
Correct
-
The neurotransmitters adrenaline, noradrenaline and dopamine are derived from which amino acid?
Your Answer: Tyrosine
Explanation:Tyrosine is the precursor to adrenaline, noradrenaline and dopamine. Tyrosine hydroxylase converts tyrosine to DOPA, which is in turn converted to dopamine, then to noradrenaline and finally adrenaline.
-
This question is part of the following fields:
- Endocrine
- Physiology
-
-
Question 6
Correct
-
A patient with chronic renal disease, missed a day of his dialysis schedule., His serum potassium was 7.6 mmol/L when his electrolytes were checked. What is the ECG finding expected in this patient?
Your Answer: Tented T waves
Explanation:ECG characteristics of hyperkalaemia may show the following changes: P-waves are widened and of low amplitude due to slowing of conduction, widened QRS complex, QRS-T fusion, loss of ST segment and tall tented T waves.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 7
Correct
-
The renal cortex and medulla, if seen under the microscope, is lacking one of the following:
Your Answer: Squamous epithelium
Explanation:Capillaries, Henle’s loop, collecting ducts, Bertin columns and type IV collagen in glomerular basement membrane are all structures present in the renal cortex or medulla. The squamous epithelium is the only one that is lacking in both the renal cortex and medulla, because normally it is not found above the outer urethra.
-
This question is part of the following fields:
- Pathology
- Renal
-
-
Question 8
Correct
-
Endometrial hyperplasia is most likely to be associated with which of the following conditions?
Your Answer: Fibrothecoma
Explanation:A benign tumour arising from the ovarian stroma, fibrothecoma are bilateral in 10% cases. The thecoma component of the tumour can produce oestrogen leading to endometrial hyperplasia. The thecoma is rich in lipid content and is responsible for the yellowish appearance of the tumour. Meig’s syndrome is the presence of fibrothecoma with a right-sided hydrothorax.
-
This question is part of the following fields:
- Pathology
- Women's Health
-
-
Question 9
Correct
-
A 45-year-old man presents to the emergency department with an irregular pulse and shortness of breath. Electrocardiography findings show no P waves, normal QRS complexes and an irregularly irregular rhythm. The patient most probably has:
Your Answer: Atrial fibrillation
Explanation:Atrial fibrillation is one of the most common arrhythmias, characterised by an irregular and rapid heart rate. Due to the decreased cardiac output, atrial fibrillation increases the risk of heart failure. It can also lead to thrombus formation which may lead to thromboembolic events. Clinical findings include palpitations, shortness of breath, fatigue, chest pain and confusion. The diagnosis is made by electrocardiographic findings which include absent P wave, fibrillatory (f) waves between QRS complexes and irregularly irregular R-R intervals.
-
This question is part of the following fields:
- Cardiovascular
- Pathology
-
-
Question 10
Correct
-
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 11
Correct
-
A 14 year-old girl is found to have haemophilia B. What pathological problem does she have?
Your Answer: Deficiency of factor IX
Explanation:Haemophilia B (also known as Christmas disease) is due to a deficiency in factor IX. Haemophilia A is due to a deficiency in factor VIII.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 12
Correct
-
Leakage from a silicone breast implant can lead to:
Your Answer: Pain and contracture
Explanation:Breast implants are mainly: saline-filled and silicone gel-filled. Complications include haematoma, fluid collections, infection at the surgical site, pain, wrinkling, asymmetric appearance, wound dehiscence and thinning of the breast tissue.
-
This question is part of the following fields:
- Pathology
- Women's Health
-
-
Question 13
Correct
-
The chest X-ray of an 10-year-old boy, that presented with low-grade fever and cough, revealed hilar enlargement and parenchymal consolidation in the middle lobes. These X-ray findings are more typical for which of the following diagnoses?
Your Answer: Pulmonary tuberculosis
Explanation:Primary pulmonary tuberculosis is seen in patients exposed to Mycobacterium tuberculosis for the firs time. The main radiographic findings in primary pulmonary tuberculosis include homogeneous parenchymal consolidation typically in the lower and middle lobes, lymphadenopathy, miliary opacities and pleural effusion.
-
This question is part of the following fields:
- Pathology
- Respiratory
-
-
Question 14
Correct
-
A 34-year-old woman with severe burns, presented to casualty with a blood pressure of 75/40 mmHg and pulse of 172/minute. Obviously the patient is in shock. Which type of shock is it more likely to be?
Your Answer: Hypovolaemic shock
Explanation:Shock is a life-threatening condition that occurs when the organs and tissues of the body are not receiving a sufficient flow of blood. Lack of blood flow, oxygen and nutrients results in the inability to function properly and damage to many organs. Shock requires immediate treatment because, if left untreated the impaired tissue perfusion and cellular hypoxia can cause irreversible tissue injury, collapse, coma or even death. There are various types of physiological shock, including: cardiogenic (due to heart damage), hypovolaemic (due to low total volume of blood or plasma), neurogenic (due to nervous system damage), septic (due to infections) and anaphylactic shock (due to allergic reactions). Hypovolaemic shock can be caused by blood loss due to trauma, internal bleeding or other fluid loss due to severe burns, prolonged diarrhoea, vomiting and sweating.
-
This question is part of the following fields:
- Cardiovascular
- Pathology
-
-
Question 15
Correct
-
Decreased velocity of impulse conduction through the atrioventricular node (AV node) in the heart will lead to:
Your Answer: Increased PR interval
Explanation:AV node damage may lead to an increase in the PR interval to as high as 0.25 – 0.40 s (normal = 0.12 – 0.20 s). In the case of severe impairment, there might be a complete failure of passage of impulses leading to complete block. In this case, the atria and ventricles will beat independently of each other.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 16
Correct
-
Diuretics that act on the ascending limb of the loop of Henle produce:
Your 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 Henl̩ Рlow permeability to ions and urea, while being highly permeable to water
2. Thin ascending limb of loop of Henl̩ Рnot permeable to water, but it is permeable to ions
3. Medullary thick ascending limb of loop of Henlé – sodium (Na+), potassium (K+) and chloride (Cl–) ions are reabsorbed by active transport. K+ is passively transported along its concentration gradient through a K+ channel in the basolateral aspect of the cells, back into the lumen of the ascending limb.
4. The cortical thick ascending limb – the site of action where loop diuretics such as furosemide block the K+/Na+/2Cl− co-transporters = reduced active transport.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 17
Correct
-
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: An increase in the molecular weight of the substance
Explanation:Unless given IV, a drug must cross several semipermeable cell membranes before it reaches the systemic circulation. Drugs may cross cell membranes by diffusion, amongst other mechanisms. The rate of diffusion of a substance is proportional to the difference in the concentration of the diffusing substance between the two sides of the membrane, the temperature of the solution, the permeability of the membrane and, in the case of ions, the electrical potential difference between the two sides of the membrane.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 18
Correct
-
In the glomerulus of the kidney, the mesangium is a structure associated with the capillaries. It has extraglomerular mesangial cells that:
Your Answer: Form the juxtaglomerular apparatus in combination with the macula densa and juxtaglomerular cells
Explanation:The mesangium is an inner layer of the glomerulus, within the basement membrane surrounding the glomerular capillaries. The mesangial cells are phagocytic and secrete the amorphous basement membrane-like material known as the mesangial matrix. They are typically separated from the lumen of the capillaries by endothelial cells. The other type of cells in the mesangium are the extraglomerular mesangial cells which form the juxtaglomerular apparatus in combination with two other types of cells: the macula densa of the distal convoluted tubule and juxtaglomerular cells of the afferent arteriole. This apparatus controls blood pressure through the renin–angiotensin–aldosterone system.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 19
Correct
-
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: 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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 20
Incorrect
-
If a tumour is found in both lobes of the prostate, without nodal involvement or metastases, a histological grade of G2 and elevated PSA, what is the overall prostatic cancer stage?
Your Answer: Stage III
Correct Answer: Stage II
Explanation:The AJCC uses the TNM, Gleason score and PSA levels to determine the overall stage of prostatic cancer. This staging is as follows:
Stage I: T1, N0, M0, Gleason score 6 or less, PSA less than 10; or T2a, N0, M0, Gleason score 6 or less, PSA less than 10
Stage IIa: T1, N0, M0, Gleason score of 7, PSA less than 20; or T1, N0, M0, Gleason score of 6 or less, PSA at least 10 but less than 20; or T2a or T2b, N0, M0, Gleason score of 7 or less, PSA less than 20
Stage IIb: T2c, N0, M0, any Gleason score, any PSA; or T1 or T2, N0, M0, any Gleason score PSA of 20 or more; or T1 or T2, N0, M0, Gleason score of 8 or higher, any PSA
Stage III: T3, N0, M0, any Gleason score, any PSA Stage IV: T4, N0, M0,any Gleason score, any PSA; or any T, N1, M0,any Gleason score, any PSA; or Any T, any N, M1, any Gleason score, any PSA.
The patient in this case has a T2 N0 M0 G2 tumour, meaning it belongs in stage II
-
This question is part of the following fields:
- Pathology
- Urology
-
-
Question 21
Correct
-
A patient with testicular seminoma has the following tumour markers: LDH 1.3 times the reference levels, β-hCG 4500 mIU/ml and AFP 875 ng/ml. What's the serum tumour marker stage in this case?
Your Answer: S1
Explanation:According to AJCC guidelines, the serum tumour marker staging is the following:
S0: marker studies within normal limits
S1: lactate dehydrogenase (LDH) less than 1.5 times the reference range, beta-human chorionic gonadotrophin (β-hCG) <5000 mIU/ml, and alpha-fetoprotein (AFP) <1000 ng/ml S2: LDH 1.5–10 times the reference range, β-hCG 5000–50,000 mIU/ml or AFP 1000–10,000 ng/ml S3: LDH greater than 10 times the reference range, β-hCG >50,000 mIU/ml or AFP >10,000 ng/ml.
According to this, the patient’s tumour belongs to the S1 stage.
-
This question is part of the following fields:
- Pathology
- Urology
-
-
Question 22
Correct
-
Macrolides are a group of antibiotics commonly used to treat respiratory tract and soft-tissue infections. Which of the following antibiotics is a macrolide?
Your Answer: Erythromycin
Explanation:Erythromycin is a macrolide antibiotic used in the treatment of several bacterial infections, including respiratory tract infections, skin infections, chlamydia infections, pelvic inflammatory disease, and syphilis. It may also be used during pregnancy to prevent Group B streptococcal infection in the new-born.
-
This question is part of the following fields:
- Pathology
- Pharmacology
-
-
Question 23
Correct
-
A 25-year old man presented to the clinic with swelling of the penis. His uncircumcised penis was erythematous and oedematous. The foreskin could not be retracted over the glans. Which of the following agents is the likely cause of his condition?
Your Answer: Staphylococcus aureus
Explanation:Inflammation of the glans penis is known as balanitis. Associated involvement of the foreskin is then known as balanoposthitis. More likely to occur in men who have a tight foreskin that is difficult to pull back, or poor hygiene.
-
This question is part of the following fields:
- Pathology
- Urology
-
-
Question 24
Incorrect
-
A medical student is told a substance is freely filtered but is not metabolised, secreted, or stored in the kidney. It has a plasma concentration of 1000 mg/l and its urine excretion rate is 25 mg/min, and the inulin clearance is 100 ml/min. What is the rate of tubular reabsorption of the substance?
Your Answer: 65 mg/min
Correct Answer: 75 mg/min
Explanation:Reabsorption or tubular reabsorption is the process by which the nephron removes water and solutes from the tubular fluid (pre-urine) and returns them to the circulating blood. To calculate the reabsorption rate of substance Z we use the following equation: excretion = (filtration + secretion) – reabsorption. As this substance is freely filtered, its filtration rate is equal to that of inulin. So 25 = (100 + 0) – reabsorption. Reabsorption = 100 – 25 therefore reabsorption = 75 mg/min.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 25
Correct
-
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: 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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 26
Correct
-
A 45-year-old man with short bowel syndrome requires parenteral nutrition. The solution of choice for parenteral nutrition is:
Your Answer: Crystalline amino acids
Explanation:Total parenteral nutrition (TPN), is the practice of feeding a person intravenously, circumventing the gut. It is normally used in the following situations: surgery, when feeding by mouth is not possible, when a person’s digestive system cannot absorb nutrients due to chronic disease or if a person’s nutrient requirement cannot be met by enteral feeding and supplementation. A sterile bag of nutrient solution, between 500 ml and 4L, is provided. The pump infuses a small amount (0.1–10 ml/h) continuously to keep the vein open. The nutrient solution consists of water, glucose, salts, amino acids, vitamins and sometimes emulsified fats. Ideally each patient is assessed individually before commencing on parenteral nutrition, and a team consisting of doctors, nurses, clinical pharmacists and dietitians evaluate the patient’s individual data and decide what formula to use and at what rate.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 27
Correct
-
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: 30 mmHg
Explanation:According to Bohr’s equation, VD/VT = (pA(CO2) − pE(CO2))/pA(CO2), where pE(CO2) is mixed expired CO2 and pA(CO2) is alveolar CO2pressure. Normally, the pa(CO2) is virtually identical to pA(CO2). Thus, VD/VT = (pa(CO2)) − pE(CO2)/pa(CO2). By Fowler’s method, VD/VT= 0.25. In the given problem, (pa(CO2) − pE(CO2)/pa(CO2) = (40 − pE(CO2)/40 = 0.25. Thus, pE(CO2) = 30 mmHg. If there is a great perfusion/ventilation inequality, pE(CO2) could be significantly lower than 30 mm Hg, and the patient’s physiological dead space would exceed the anatomical dead space.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 28
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: Distal convoluted tubule
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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 29
Correct
-
Etoposide is a chemotherapeutic agent used in the treatment of different types of cancer. Which of the following is the correct indication for this drug?
Your Answer: Lung cancer
Explanation:Etoposide phosphate is an inhibitor of the enzyme topoisomerase II. It is used as a form of chemotherapy for malignancies such as lung cancer, testicular cancer, lymphoma, non-lymphocytic leukaemia and glioblastoma multiforme. Side effects are very common and can include low blood cell counts, vomiting, loss of appetite, diarrhoea, hair loss, and fever.
-
This question is part of the following fields:
- Pathology
- Pharmacology
-
-
Question 30
Correct
-
After being admitted to the hospital, a 60-year-old man is administered sodium nitroprusside. Which class of drugs does nitroprusside belong to?
Your Answer: Vasodilators
Explanation:Sodium nitroprusside is a potent peripheral vasodilator that affects both arterioles and venules. It is often administered intravenously to patients who are experiencing a hypertensive emergency. It reduces both total peripheral resistance as well as venous return, so decreasing both preload and afterload. For this reason it can be used in severe cardiogenic heart failure where this combination of effects can act to increase cardiac output. It is administered by intravenous infusion. Onset is typically immediate and effects last for up to ten minutes. The duration of treatment should not exceed 72 hours.
-
This question is part of the following fields:
- Pathology
- Pharmacology
-
-
Question 31
Correct
-
Which of the following chemotherapy regimens is most likely to be used in colorectal cancer?
Your Answer: FOLFOX
Explanation:Chemotherapy regimens are often identified by acronyms, identifying the agents used in the drug combination. However, the letters used are not consistent across regimens. FOLFOX is a chemotherapy regimen used for the treatment of colorectal cancer, made up of the following drugs: • FOL: fluorouracil (5-fluorouracil or 5-FU) • F: folinic acid (leucovorin) • OX: oxaliplatin (Eloxatin®).
-
This question is part of the following fields:
- Pathology
- Pharmacology
-
-
Question 32
Correct
-
A 34-year-old man is receiving chemotherapy for non-Hodgkin's lymphoma. Which of the following chemotherapy regimens would be used in this case?
Your Answer: CHOP
Explanation:CHOP is the acronym for a chemotherapy regimen used in the treatment of non-Hodgkin’s lymphoma, comprising cyclophosphamide, hydroxyrubicin (adriamycin), vincristine and prednisone. This regimen can also be combined with the monoclonal antibody rituximab if the lymphoma is of B cell origin; this combination is called R-CHOP.
-
This question is part of the following fields:
- Pathology
- Pharmacology
-
-
Question 33
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:
- Physiology
- Renal
-
-
Question 34
Correct
-
A 18-year old girl presents to her doctor with an excessively enlarged left breast as compared to the right breast since puberty. The most likely cause for this is:
Your Answer: Virginal breast hypertrophy
Explanation:Virginal breast hypertrophy’ is the term assigned to excessive growth of breasts during puberty and is a common phenomenon. It is also known as ‘juvenile macromastia’ or ‘ juvenile gigantomastia’. The breast hypertrophy often starts with menarche and occasionally occurs in growth spurts. These spurts can cause physical discomfort, red and itchy skin or pain in the breasts. The breasts can also grow continuously over several years and lead to overdevelopment of a normal breast. Nipples also undergo enlargement along with the breasts.
-
This question is part of the following fields:
- Pathology
- Women's Health
-
-
Question 35
Incorrect
-
A 45 year-old male, with behavioural changes developed euvolemic hyponatraemia. Which of the following conditions most likely predisposed the patient to develop euvolemic hyponatraemia?
Your Answer: Protracted vomiting
Correct Answer: Psychosis
Explanation:In euvolemic hyponatraemia, there is volume expansion in the body, there is no oedema, but hyponatremia occurs. Causes include: state of severe pain or nausea, psychosis, brain trauma, SIADH, hypothyroidism and glucocorticoid deficiency.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 36
Correct
-
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: 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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 37
Correct
-
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.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 38
Correct
-
Which of the following is NOT a Gram-negative rod?
Your Answer: Clostridium tetani
Explanation:Gram-positive rods include Clostridia, Listeria and diphtheroids.
Gram-negative rods include Escherichia coli, Klebsiella, Yersinia, Haemohilus, Pseudomonas, Shigella, Legionella, proteus and Salmonella
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 39
Correct
-
Which of the following is the cause of flattened (notched) T waves on electrocardiogram (ECG)?
Your Answer: Hypokalaemia
Explanation:The T-wave is formed due to ventricular repolarisation. Normally, it is seen as a positive wave. It can be normally inverted (negative) in V1 (occasionally in V2-3 in African-Americans/Afro-Caribbeans). Hyperacute T-waves are the earliest ECG change of acute myocardial infarction. ECG findings of hyperkalaemia include high, tent-shaped T-waves, a small P-wave and a wide QRS complex. Hypokalaemia results in flattened (notched) T-waves, U-waves, ST-segment depression and prolonged QT interval.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 40
Incorrect
-
Which of the following conditions of the breast is most often associated with use of oral contraceptives?
Your Answer: Galactocele
Correct Answer: Cyst formation
Explanation:Breast cysts are common and are smooth, firm, mobile lumps that can sometimes be tender. Cysts can also be bilateral and are known to recur in 10% of cases. They are more common in women in the third and fourth decades and most often disappear after menopause. Cysts are filled with fluid from the breast which occurs due to the normal menstrual cycle of a woman. The end of the menstrual cycle causes breast cells to swell. After the menses, the cells shrink and the released fluid disappears. However, in some cases this fluid forms a cyst. Incidence of cysts was also linked to use of oral contraceptives.
-
This question is part of the following fields:
- Pathology
- Women's Health
-
-
Question 41
Incorrect
-
QT interval in the electrocardiogram of a healthy individual is normally:
Your Answer: 0.20 s
Correct 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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 42
Correct
-
If a 68-year-old man is diagnosed with a testicular seminoma that reaches the tunica albuginea and involves the tunica vaginalis, with retroperitoneal lymph nodes greater than 5cm, LDH 1.4 times the reference levels, β-hCG 4250 mIU/ml and AFP 780 ng/ml, what's the clinical stage in this case?
Your Answer: Stage IIC
Explanation:According to the AJCC, the clinical staging for testicular seminoma is:
Stage IA: T1 N0 M0 S0
Stage IB: T2/3/4 N0 M0 S0
Stage IC: any T N0 M0 S1/2/3
Stage IIA: any T N1 M0 S0/1
Stage IIB: any T N2 M0 S0/1
Stage IIC: any T N3 M0 S0/1
Stage IIIA: any T any N M1a S0/1
Stage IIIB: any T any N M0/1a S2
Stage IIIC: any T any N M1a/1b S3.
The patient in this case has IIC stage -
This question is part of the following fields:
- Pathology
- Urology
-
-
Question 43
Incorrect
-
A 54-year-old woman is re-admitted to the hospital with shortness of breath and sharp chest pain 2 weeks after surgical cholecystectomy. The most probable cause of these clinical findings is:
Your Answer: Pleural effusion
Correct Answer: Pulmonary embolus
Explanation:Pulmonary embolism is caused by the sudden blockage of a major lung blood vessel, usually by a blood clot. Symptoms include sudden sharp chest pain, cough, dyspnoea, palpitations, tachycardia or loss of consciousness. Risk factors for developing pulmonary embolism include long periods of inactivity, recent surgery, trauma, pregnancy, oral contraceptives, oestrogen replacement, malignancies and venous stasis.
-
This question is part of the following fields:
- Pathology
- Respiratory
-
-
Question 44
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: Decreased haematocrit
Correct Answer: Increased renal excretion of HCO3 –
Explanation:The atmospheric pressure is lower at high altitudes as compared with sea level. This leads to a decrease in the partial pressure of oxygen. Once 2100 m (7000 feet) of altitude is reached, there is a drop in saturation of oxyhaemoglobin. The oxygen saturation of haemoglobin determines the oxygen content in the blood. The body physiological tries to adapt to high altitude by acclimatization. Immediate effects include hyperventilation, fluid loss (due to a decreased thirst drive), increase in heart rate and slightly lowered stroke volume. Long term effects include lower lactate production, compensatory alkali loss in urine, decrease in plasma volume, increased erythropoietin release and red cell mass, increased haematocrit, higher concentration of capillaries in striated muscle tissue, increase in myoglobin, increase in mitochondria, increase in aerobic enzyme concentration such as 2,3-DPG and pulmonary vasoconstriction.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 45
Incorrect
-
A 78-year-old woman was brought to the emergency department with decreased consciousness, weakness and dehydration. Which serum electrolyte would most likely be low in this patient?
Your Answer: HCO3 −
Correct Answer: Na+
Explanation:Hyponatremia is a sodium level below 135 mEq/L. Signs and symptoms may include: nausea with vomiting, fatigue, headache or confusion, cramps or spasm, irritability and restlessness and severe cases may lead to seizures and comma.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 46
Incorrect
-
A 38-year old woman presents to the clinic with a 2 cm eczema-like lesion on the areolar region of her left breast, for 5 months. Biopsy of the lesion showed large cells at the dermal-epidermal junction with positive staining for mucin. What is the likely diagnosis?
Your Answer: Cystosarcoma phyllodes
Correct Answer: Paget’s disease of the breast
Explanation:Paget’s disease of the breast or nipple resembles eczema in appearance with an underlying carcinoma typically. The disease is usually unilateral and presents with inflammation, oozing and crusting along with a non-healing ulcer. Treatment is often delayed due to the innocuous appearance but can be fatal. It results due to spread of neoplastic cells from the ducts of the mammary gland to the epithelium.
-
This question is part of the following fields:
- Pathology
- Women's Health
-
-
Question 47
Correct
-
How much blood can the pulmonary vessels of a 45-year-old healthy man accommodate when he is at rest?
Your 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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 48
Incorrect
-
A 34-year-old Asian male presents with cervical lymphadenopathy. The patient is suspected to have tuberculous lymphadenopathy. Excision biopsy of one of the nodes showed granulomatous inflammation. Which histopathologic feature is most likely consistent with the diagnosis of tuberculosis?
Your Answer: Liquefactive necrosis
Correct Answer: Caseation necrosis
Explanation:The granulomas of tuberculosis tend to contain necrosis (caseating tubercles), but non-necrotizing granulomas may also be present. Multinucleated giant cells with nuclei arranged like a horseshoe (Langhans giant cells) and foreign body giant cells are often present, but are not specific for tuberculosis. A definitive diagnosis of tuberculosis requires identification of the causative organism by microbiological cultures.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 49
Incorrect
-
Renin is secreted by pericytes in the vicinity of the afferent arterioles of the kidney from the juxtaglomerular cells. Plasma renin levels are decreased in patients with:
Your Answer: Renal artery stenosis
Correct Answer: Primary aldosteronism
Explanation:Primary aldosteronism, also known as primary hyperaldosteronism or Conn’s syndrome, is excess production of the hormone aldosterone by the adrenal glands resulting in low renin levels. Most patients with primary aldosteronism (Conn’s syndrome) have an adrenal adenoma. The increased plasma aldosterone concentration leads to increased renal Na+ reabsorption, which results in plasma volume expansion. The increase in plasma volume suppresses renin release from the juxtaglomerular apparatus and these patients usually have low plasma renin levels. Salt restriction and upright posture decrease renal perfusion pressure and therefore increases renin release from the juxtaglomerular apparatus. Secondary aldosteronism is due to elevated renin levels and may be caused by heart failure or renal artery stenosis.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 50
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: Creatinine
Correct Answer: Glucose
Explanation:Under normal conditions the renal clearance of glucose is zero, since glucose is completely reabsorbed in the renal tubules and not excreted. Glycosuria – the excretion of glucose into the urine- is nearly always caused by elevated blood glucose levels, most commonly due to untreated diabetes mellitus.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 51
Incorrect
-
A 65-year-old female is taking different medications for various medical conditions. Which medication would most likely predispose the patient to develop hyperkalaemia?
Your Answer: B Carbenicillin
Correct Answer: Spironolactone
Explanation:The most important potential side effect of spironolactone is hyperkalaemia (high potassium levels), which, in severe cases, can be life-threatening. Hyperkalaemia in these patients can present as a non anion-gap metabolic acidosis.Â
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 52
Incorrect
-
Laboratory findings in a patient with dark urine and yellowish skin revealed a prolonged prothrombin time. Which of the following is the most likely cause of this finding?
Your Answer: Disseminated intravascular coagulopathyÂ
Correct Answer: Liver damage
Explanation:Various conditions may prolong the prothrombin time (PT), including: warfarin use, vitamin K deficiency, liver disease, disseminated intravascular coagulopathy, hypofibrinogenemia, heparin infusion, massive blood transfusion and hypothermia. Liver disease causes prolonging of PT due to diminished synthesis of clotting factors. Dark urine colour and jaundice are indicators of the presence of a liver disease in this patient.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 53
Incorrect
-
A 25-year-old woman complains of generalised swelling and particularly puffiness around the eyes which is worst in the morning. Laboratory studies showed:
Blood urea nitrogen (BUN) = 30 mg/dl
Creatinine = 2. 8 mg/dl
Albumin = 2. 0 mg/dl
Alanine transaminase (ALT) = 25 U/l
Bilirubin = 1 mg/dl
Urine analysis shows 3+ albumin and no cells.
Which of the following is the most likely diagnosis?Your Answer: Interstitial renal disease
Correct Answer: Nephrotic syndrome
Explanation:Nephrotic syndrome is a disorder in which the glomeruli have been damaged, characterized by:
– Proteinuria (>3.5 g per 1.73 m2 body surface area per day, or > 40 mg per square meter body surface area per hour in children)
– Hypoalbuminemia (< 2,5 g/dl) – Hyperlipidaemia, and oedema (generalized anasarca).
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 54
Incorrect
-
A 17-year-old boy, who had developed shortness of breath and a loss of appetite over the last month, was referred to a haematologist because he presented with easy bruising and petechiae. His prothrombin time, platelet count, partial thromboplastin and bleeding time were all normal. Which of the following would explain the presence of the petechiae and easy bruising tendency?
Your Answer: Uraemia
Correct Answer: Scurvy
Explanation:Scurvy is a condition caused by a dietary deficiency of vitamin C, also known as ascorbic acid. Humans are unable to synthesize vitamin C, therefore the quantity of it that the body needs has to come from the diet. The presence of an adequate quantity of vitamin C is required for normal collagen synthesis. In scurvy bleeding tendency is due to capillary fragility and not coagulation defects, therefore blood tests are normal.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 55
Incorrect
-
Which condition presents with a positive urine dipstick test for blood, but no blood cells on urine microscopy?
Your Answer: Renal papillary necrosis
Correct Answer: Myoglobinuria
Explanation:Myoglobinuria, or presence of myoglobulin in the urine is seen due to rhabdomyolysis (muscle destruction). Common causes of rhabdomyolysis include trauma, electrical injuries, burns, venom and drugs. Damaged muscle leads to release of myoglobin in the blood. Ideally, the released myoglobin gets filtered and excreted by the kidneys. However, excess myoglobin can occlude the renal filtration system leading to acute tubular necrosis and acute renal dysfunction.
-
This question is part of the following fields:
- Pathology
- Renal
-
-
Question 56
Correct
-
When does the heart rate decrease?
Your 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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 57
Incorrect
-
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: 1.65 l/min
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 58
Incorrect
-
Which of the following coagulation factors is responsible for the formation of a complex with tissue factor to activate factors IX and X?
Your Answer: Factor XI
Correct Answer: Factor VII
Explanation:Factor VII, also known as proconvertin or stable factor, is a vitamin K–dependent protein that plays a central role in haemostasis and coagulation. Tissue factor is a protein that is normally not exposed on the surface of intact blood vessels. Damage to the vascular lumen leads to tissue factor exposure. The exposed tissue factor binds to factor VII. This facilitates the activation of factor VII to factor VIIa.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 59
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: Metabolic acidosis
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:
- Physiology
- Respiratory
-
-
Question 60
Incorrect
-
Nephrotic syndrome is a condition that causes proteinuria, hypoalbuminemia and oedema. Which of the following is the cause of the oedema in these patients?
Your Answer: Arterial occlusion
Correct Answer: Decreased oncotic pressure
Explanation:The glomeruli of the kidneys are the parts that normally filter the blood. They consist of capillaries that are fenestrated and allow fluid, salts and other small solutes to flow through, but normally not proteins. In nephrotic syndrome, the glomeruli become damaged allowing small proteins, such as albumin to pass through the kidneys into urine. Oedema usually occurs due to salt and water retention by the diseased kidneys as well as due to the reduced colloid oncotic pressure (because of reduced albumin in the plasma). Lower serum oncotic pressure causes fluid to accumulate in the interstitial tissues.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 61
Incorrect
-
A patient is suspected to have Blastomyces dermatidis infection. The patient contracted the disease most likely through which port of entry?
Your Answer: Anus
Correct Answer: Respiratory tract
Explanation:Blastomycosis disease is a fungal infection acquired through inhalation of the spores. It caused by the organism Blastomyces dermatitidis and manifests as a primary lung infection in about 70% of cases. The onset is relatively slow and symptoms are suggestive of pneumonia.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 62
Incorrect
-
A 60-year-old woman complains of left sided headaches which have been recurring for several years. She recently suffered from a focal seizure for the first time a few days ago. A CT scan shows a mass in the left hemisphere of the brain. The most likely diagnosis is:
Your Answer: Medulloblastoma
Correct Answer: Meningioma
Explanation:Meningiomas are a common benign intracranial tumour, and their incidence is higher in women between the ages of 40-60 years old. Many of these tumours are asymptomatic and are diagnosed incidentally, although some of them may have malignant presentations (less than 2% of cases). These benign tumours can develop wherever there is dura, over the convexities near the venous sinuses, along the base of the skull, in the posterior fossa and, within the ventricles.
-
This question is part of the following fields:
- Neurology
- Pathology
-
-
Question 63
Incorrect
-
In a young, sexually active male, what is the most common cause of non-gonococcal urethritis?
Your Answer: Treponema
Correct Answer: Chlamydia
Explanation:Non-gonococcal urethritis is most commonly caused by Chlamydia trachomatis (50% cases). Less common organisms include Mycoplasma genitalium, Urea urealyticum and Trichomonas vaginalis. Chlamydia is also the commonest cause of non-gonococcal cervicitis in women and proctitis in both sexes.
-
This question is part of the following fields:
- Pathology
- Renal
-
-
Question 64
Correct
-
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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 65
Correct
-
A patient with this condition has extracellular fluid volume expansion:
Your Answer: Nephrotic syndrome
Explanation:Nephrotic syndrome is a syndrome comprising of signs of nephrosis, including proteinuria, hypoalbuminemia, and oedema. It is a component of glomerulonephritis, in which different degrees of proteinuria occur. Essentially, loss of protein through the kidneys leads to low protein levels in the blood , which causes water to be drawn into soft tissues (oedema). Severe hypoalbuminemia can also cause a variety of secondary problems, such as water in the abdominal cavity (ascites), around the heart or lung (pericardial effusion, pleural effusion), high cholesterol, loss of molecules regulating coagulation (hence increased risk of thrombosis). The most common sign is excess fluid in the body due to the serum hypoalbuminemia. Lower serum oncotic pressure causes fluid to accumulate in the interstitial tissues. Sodium and water retention aggravates the oedema.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 66
Incorrect
-
A 24-year-old patient with recurrent episodes of deep vein thrombosis presents again to the clinic. Deficiency of which of the following blood proteins is the most probable cause of this episode?
Your Answer: Von Willebrand factorÂ
Correct Answer: Antithrombin III
Explanation:Antithrombin III (ATIII) is a blood protein that acts by inhibiting blood coagulation by neutralizing the enzymatic activity of thrombin.
Antithrombin III deficiency is an autosomal dominant disorder that leads to an increased risk of venous and arterial thrombosis. Clinical manifestations typically appear in young adulthood.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 67
Incorrect
-
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: Exactly half
Correct 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.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 68
Incorrect
-
A 78-year-old man who has been bedridden for a month is prescribed griseofulvin. What class of drugs does griseofulvin belong to?
Your Answer: Antimycobacterial
Correct Answer: Antifungal
Explanation:Griseofulvin is an antifungal drug. It is administered orally, and it is used to treat ringworm infections of the skin and nails. It binds to keratin in keratin precursor cells and makes them resistant to fungal infections. Griseofulvin works by interfering with fungal mitosis.
-
This question is part of the following fields:
- Pathology
- Pharmacology
-
-
Question 69
Incorrect
-
Which of the following variables are needed to calculate inspiratory reserve volume of a patient?
Your Answer: Tidal volume 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:
- Physiology
- Respiratory
-
-
Question 70
Incorrect
-
A 65-year old gentleman presents to the clinic with chronic back pain and weight loss. His blood count shows a white blood cell count of 10 × 109/l, with a differential count of 66 polymorphonuclear leukocytes, 7 bands, 3 metamyelocytes, 3 myelocytes, 14 lymphocytes, 7 monocytes, and 5 nucleated red blood cells. The haemoglobin is 13 g/dl with a haematocrit of 38.1%, a mean corpuscular volume of 82 fl, and a platelet count of 126 × 109/l. What is the likely diagnosis?
Your Answer: Chronic lymphocytic leukaemia
Correct Answer: Metastatic carcinoma
Explanation:The peripheral blood findings suggest a leucoerythroblastic picture, the common causes of which in a 65-year old gentleman includes prostatic or lung malignancy.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 71
Incorrect
-
Driving pressure is considered to be a strong predictor of mortality in patients with ARDS. What is the normal mean intravascular driving pressure for the respiratory circulation?
Your Answer: 250 mmHg
Correct Answer: 10 mmHg
Explanation:Driving pressure is the difference between inflow and outflow pressure. For the pulmonary circulation, this is the difference between pulmonary arterial (pa) and left atrial pressure (pLA). Normally, mean driving pressure is about 10 mmHg, computed by subtracting pLA (5 mmHg) from pA (15 mmHg). This is in contrast to a mean driving pressure of nearly 100 mmHg in the systemic circulation.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 72
Incorrect
-
Loperamide is a drug used to treat diarrhoea. What is the mechanism of action of loperamide?
Your Answer: Competitively inhibits H2 receptors
Correct Answer: Opiate agonist
Explanation:Loperamide is an opioid-receptor agonist and acts on the mu opioid receptors in the myenteric plexus of large intestine. It works by decreasing the motility of the circular and longitudinal smooth muscles of the intestinal wall. It is often used for this purpose in gastroenteritis, inflammatory bowel disease, and short bowel syndrome.
-
This question is part of the following fields:
- Pathology
- Pharmacology
-
-
Question 73
Incorrect
-
Cyclophosphamide is used as a chemotherapy and immunosuppressant agent and is indicated in various diseases. One of the most severe complications of its use is cancer of the:
Your Answer: Colon
Correct Answer: Urinary bladder
Explanation:Cyclophosphamide is used to treat various types of cancer and autoimmune disorders. The main use of cyclophosphamide is in combination with other chemotherapy agents in the treatment of lymphomas, some forms of leukaemia and some solid tumours. Side-effects include nausea and vomiting, bone marrow suppression, stomach ache, diarrhoea, darkening of the skin/nails, alopecia, lethargy, and haemorrhagic cystitis. Cyclophosphamide is itself carcinogenic, potentially causing transitional cell carcinoma of the bladder as a long-term complication.
-
This question is part of the following fields:
- Pathology
- Pharmacology
-
-
Question 74
Correct
-
A 29-year-old woman presents to the doctor complaining of cough, shortness of breath, fever and weight loss. Chest X-ray revealed bilateral hilar and mediastinal lymph node enlargement and bilateral pulmonary opacities. Non-caseating granulomas were found on histological examination. The most likely diagnosis is:
Your Answer: Sarcoidosis
Explanation:Sarcoidosis is an inflammatory disease of unknown aetiology that affects multiple organs but predominantly the lungs and intrathoracic lymph nodes. Systemic and pulmonary symptoms may both be present. Pulmonary involvement is confirmed by a chest X-ray and other imaging studies. The main histological finding is the presence of non-caseating granulomas.
-
This question is part of the following fields:
- Pathology
- Respiratory
-
-
Question 75
Correct
-
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: Increased residual volume
Explanation:Chronic obstructive pulmonary disease (COPD) is a type of obstructive lung disease characterized by long-term poor airflow. The main symptoms include shortness of breath and cough with sputum production. The best diagnostic test for evaluating patients with suspected chronic obstructive pulmonary disease (COPD) is lung function measured with spirometry. Key spirometrical measures may be obtained with a portable office spirometer and should include forced vital capacity (FVC) and the normal forced expiratory volume in the first second of expiration (FEV1). The ratio of FEV1 to forced vital capacity (FEV1/FVC) normally exceeds 0.75. Patients with COPD typically present with obstructive airflow. Complete pulmonary function testing may show increased total lung capacity, functional residual capacity and residual volume. A substantial loss of lung surface area available for effective oxygen exchange causes diminished carbon monoxide diffusion in the lung (DLco) in patients with emphysema. Tobacco smoking is the most common cause of COPD, with factors such as air pollution and genetics playing a smaller role.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 76
Incorrect
-
A 45-year old gentleman presented to the emergency department at 5.00 AM with pain in his left flank. The pain began suddenly and presented in waves throughout the night. Urine examination was normal except for presence of blood and few white blood cells. The pH and specific gravity of the urine were also found to be within normal range. What is the likely diagnosis?
Your Answer: Membranous glomerulonephritis
Correct Answer: Ureteric calculus
Explanation:A calculus in the ureter, if less than 5mm in diameter is likely to pass spontaneously. However, a larger calculus irritates the ureter and may become lodged, leading to hydroureter and/or hydronephrosis. Likely sites where the calculus might get lodged, include pelviureteric junction, distal ureter at the level of iliac vessels and the vesicoureteric junction. An obstruction can result in reduced glomerular filtration. There can be deterioration in renal function due to hydronephrosis and a raised glomerular pressure, leading to poor renal blood flow. Permanent renal dysfunction usually takes about 4 weeks to occur. Secondary infection can also occur in chronic obstruction.
-
This question is part of the following fields:
- Pathology
- Renal
-
-
Question 77
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: 12 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:
- Cardiovascular
- Physiology
-
-
Question 78
Incorrect
-
A patient is diagnosed with Brucellosis. What is the mode of transmission of this disease?
Your Answer: Ticks
Correct Answer: Unpasteurised milk
Explanation:Brucellosis is a highly contagious zoonosis caused by ingestion of unpasteurized milk or undercooked meat from infected animals, or close contact with their secretions.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 79
Incorrect
-
A 55-year-old male chronic smoker is diagnosed with non-small-cell cancer. His right lung underwent complete atelectasis and he has a 7cm tumour involving the chest wall. What is the stage of the lung cancer of this patient?
Your Answer: T4
Correct Answer: T3
Explanation:Non-small-cell lung cancer is staged through TNM classification. The stage of this patient is T3 because based on the TNM classification the tumour is staged T3 if > 7 cm or one that directly invades any of the following: Chest wall (including superior sulcus tumours), diaphragm, phrenic nerve, mediastinal pleura, or parietal pericardium; or the tumour is in the main bronchus < 2 cm distal to the carina but without involvement of the carina, Or it is associated with atelectasis/obstructive pneumonitis of the entire lung or separate tumour nodule(s) in the same lobe.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 80
Incorrect
-
The gynaecologist suspects that her patient has a cervical cancer. What particular test should be done on this patient to screen for cervical cancer?
Your Answer: Blood culture
Correct Answer: Pap smear
Explanation:Worldwide, approximately 500,000 new cases of cervical cancer and 274,000 deaths are attributable to cervical cancer yearly. This makes cervical cancer the second most common cause of death from cancer in women. The mainstay of cervical cancer screening has been the Papanicolaou test (Pap smear).
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 81
Incorrect
-
Painful erections along with deviation of the penis to one side when erect are seen in which of the following conditions?
Your Answer: Priapism
Correct Answer: Peyronie’s disease
Explanation:Peyronie’s disease leads to development of fibrous plaques in the penile soft tissue and occurs in 1% of men, most commonly affecting white males above 40 years age. It is a connective tissue disorder named after a French surgeon, François de la Peyronie who first described it. Symptoms include pain, hard lesions on the penis, abnormal curvature of erect penis, narrowing/shortening, painful sexual intercourse and in later stages, erectile dysfunction. 30% cases report fibrosis in other elastic tissues such as Dupuytren’s contractures of the hand. There is likely a genetic predisposition as increased incidence is noted among the male relatives of an affected individual.
-
This question is part of the following fields:
- Pathology
- Urology
-
-
Question 82
Incorrect
-
An alcoholic patient was found to have hypomagnesaemia on blood tests. Which of the following clinical features will have prompted the doctor to check the serum magnesium level in this patient?
Your Answer: Vomiting
Correct Answer: Seizures
Explanation:Hypomagnesaemia is a condition characterised by a low level of magnesium in the blood. The normal range for serum magnesium level is 0.75-1.05 mmol/l. In hypomagnesaemia serum levels of magnesium are less than 0.75 mmol/l. The cardiovascular and nervous systems are the most commonly affected. Neuromuscular manifestations include symptoms like tremor, tetany, weakness, apathy, delirium, a positive Chvostek and Trousseau sign, nystagmus and seizures. Cardiovascular manifestations include electrocardiographic abnormalities and arrhythmias e.g. ventricular fibrillation.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 83
Incorrect
-
Staphylococcus aureus can be identified in the laboratory based on the clotting of plasma. Which microbial product is responsible for this activity?
Your Answer: Plasmin
Correct Answer: Coagulase
Explanation:Staphylococcus aureus is the most pathogenic species and is implicated in a variety of infections. S. aureus can be identified due to its production of coagulase. The staphylococcal enzyme coagulase will cause inoculated citrated rabbit plasma to gel or coagulate. The coagulase converts soluble fibrinogen in the plasma into insoluble fibrin.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 84
Incorrect
-
The proximal tubule is the portion of the ductal system of the nephron of the kidney which leads from Bowman's capsule to the loop of Henle. Which of the following is most likely to be seen in a sample of fluid leaving the proximal tubule?
Your Answer: It will be hypertonic compared with plasma
Correct Answer: It will have no amino acids
Explanation:The proximal tubule is the portion of the duct system of the nephron leading from Bowman’s capsule to the loop of Henlé. The most distinctive characteristic of the proximal tubule is its brush border (or ‘striated border’). The luminal surface of the epithelial cells of this segment of the nephron is covered with densely packed microvilli forming a border which greatly increases the luminal surface area of the cells, presumably facilitating their reabsorptive function. Glucose, amino acids, inorganic phosphate, and some other solutes are100% reabsorbed via secondary active transport through co-transporters driven by the sodium gradient out of the nephron.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 85
Incorrect
-
The physician suggested lifestyle modification for his patient because his present condition could increase his risk for the development of adenocarcinoma of the oesophagus. What is the most common predisposing factor for the development of adenocarcinoma of the oesophagus?
Your Answer: Tylosis palmaris
Correct Answer: Gastro-oesophageal reflux disease
Explanation:Barret’s oesophagus is attributed primarily to gastro-oesophageal reflux disease. The chronic acidic environment damages the squamous epithelial lining of the oesophagus, and subsequently undifferentiated pluripotent stem cells develop into columnar epithelium, this is then known as Barret’s oesophagitis.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 86
Incorrect
-
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: Arterial CO2
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 87
Correct
-
Question 88
Incorrect
-
Hormones of the anterior pituitary include which of the following?
Your Answer: Thyroxin
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
-
This question is part of the following fields:
- Endocrine
- Physiology
-
-
Question 89
Incorrect
-
Elevated mean corpuscular volume with hypersegmented neutrophils and low reticulocyte index is seen in on the blood count of a middle-aged lady about to undergo elective surgery. On enquiry, she mentions feeling tired for a few months. Which of the following investigations should be carried out in her to reach a diagnosis?
Your Answer: HIV antibody test
Correct Answer: Serum vitamin B12 and folate
Explanation:Elevated levels of MCV indicates megaloblastic anaemia, which are associated with hypersegmented neutrophils. Likely causes include vitamin B12 or folate deficiency. Megaloblastic anaemia results from defective synthesis of DNA. As RNA production continues, the cells enlarge with a large nucleus. The cytoplasmic maturity becomes greater than nuclear maturity. Megaloblasts are produced initially in the marrow, before blood. Dyspoiesis makes erythropoiesis ineffective, causing direct hyperbilirubinemia and hyperuricemia. As all cell lines are affected, reticulocytopenia, thrombocytopenia and leukopenia develop. Large, oval blood cells (macro-ovalocytes) are released in the circulation, along with presence of hypersegmented neutrophils.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 90
Incorrect
-
A 47-year old-woman diagnosed with pancreatitis presented to the emergency department complaining of a worsening shortness of breath, fever, agitation and cough. Oxygen saturation was 67% in room air. Her respiratory status continued to deteriorate therefore she was intubated. She was admitted to the intensive care unit for management. Chest X-ray demonstrated bilateral perihilar opacities. The patient failed conventional treatment and died several days later. At autopsy, the lung shows growth of type 2 pneumocytes and thickened alveolar walls. What is the most probable diagnosis?
Your Answer: Chronic bronchitis
Correct Answer: Adult respiratory distress syndrome
Explanation:Acute (or adult) respiratory distress syndrome (ARDS) is a life-threatening lung condition characterised by a non-cardiogenic pulmonary oedema that leads to acute respiratory failure. The most common risk factors for ARDS include trauma with direct lung injury, sepsis, pneumonia, pancreatitis, burns, drug overdose, massive blood transfusion and shock. Acute onset of dyspnoea with hypoxemia, anxiety and agitation is typical. Chest X ray most commonly demonstrates bilateral pulmonary infiltrates. Histological changes include the exudative, proliferative and fibrotic phase. ARDS is mainly a clinical diagnosis.
-
This question is part of the following fields:
- Pathology
- Respiratory
-
-
Question 91
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: 51Cr-EDTA
Correct Answer: Inulin
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal glomerular capillaries into the Bowman’s capsule per unit time. Clinically, this is often measured to determine renal function. Inulin was originally used as it is not reabsorbed by the kidney after glomerular filtration, therefore its rate of excretion is directly proportional to the rate of filtration of water and solutes across the glomerular filter. However, in clinical practice, creatinine clearance is used to measure GFR. Creatinine is an endogenous molecule, synthesised in the body, that is freely filtered by the glomerulus (but also secreted by the renal tubules in very small amounts). Creatinine clearance exceeds GFR due to creatinine secretion, and is therefore a close approximation of the GFR.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 92
Incorrect
-
After finding elevated PSA levels, a 69-year-old man undergoes a needle biopsy and is diagnosed with prostatic cancer. What is the stage of this primary tumour?
Your Answer: T1a
Correct Answer: T1c
Explanation:The AJCC uses a TNM system to stage prostatic cancer, with categories for the primary tumour, regional lymph nodes and distant metastases:
TX: cannot evaluate the primary tumour T0: no evidence of tumour
T1: tumour present, but not detectable clinically or with imaging T1a: tumour was incidentally found in less than 5% of prostate tissue resected (for other reasons)
T1b: tumour was incidentally found in more than 5% of prostate tissue resected
T1c: tumour was found in a needle biopsy performed due to an elevated serum prostate-specific antigen
T2: the tumour can be felt (palpated) on examination, but has not spread outside the prostate
T2a: the tumour is in half or less than half of one of the prostate gland’s two lobes
T2b: the tumour is in more than half of one lobe, but not both
T2c: the tumour is in both lobes
T3: the tumour has spread through the prostatic capsule (if it is only part-way through, it is still T2)
T3a: the tumour has spread through the capsule on one or both sides
T3b: the tumour has invaded one or both seminal vesicles
T4: the tumour has invaded other nearby structures.
In this case, the tumour has a T1c stage.
-
This question is part of the following fields:
- Pathology
- Urology
-
-
Question 93
Correct
-
A 45-year-old pregnant woman develops high blood pressure at 20 weeks. She complains of headaches and swollen feet, and a test reveals proteinuria (350 mg/day). Which of the following is the most likely diagnosis?
Your Answer: Pre-eclampsia
Explanation:Pre-eclampsia (PE) is a disorder of pregnancy characterized by the onset of high blood pressure (two separate readings taken at least 6 h apart of 140/90 or more) and often a significant amount of protein in the urine (>300 mg of protein in a 24-h urine sample). While blood pressure elevation is the most visible sign of the disease, it involves generalised damage to the maternal endothelium of the kidneys and liver, with the release of vasopressive factors only secondary to the original damage. Pre-eclampsia may develop at varying times within pregnancy and its progress differs among patients; most cases present pre-term. It has no known cure apart from ending the pregnancy (induction of labour or abortion). It may also present up to 6 weeks post partum. Risk factors for pre-eclampsia include obesity, prior hypertension, older age, and diabetes mellitus.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 94
Correct
-
Extracellular body fluid as compared with intracellular body fluid:
Your Answer: Is relatively rich in glucose
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. Ensuring the right amount of body water is part of fluid balance, an aspect of homeostasis. The extracellular fluid (ECF) includes all fluids outside the cells. This fluid can be divided into three fluid departments: interstitial (in the tissue spaces) fluid, blood plasma and lymph, and specialised compartments called transcellular fluid. The extracellular fluid surrounds all the cells in the body and is in equilibrium with the intracellular fluid. So, its composition must remain fairly constant even though substances are passing into and out of the cells. The interstitial fluid, though called a fluid, is in a reality a gel-like composition made up of: water, proteoglycan molecules and collagen. The extracellular fluid constitutes 40% of total body water, with intracellular fluid making up the remaining 60%. It is relatively rich in glucose.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 95
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: Low haematocrit
Correct Answer: Severe diarrhoea or vomiting
Explanation:Hyponatraemia occurs when the sodium level in the plasma falls below 135 mmol/l. Hyponatraemia is an abnormality that can occur in isolation or, more commonly as a complication of other medical illnesses. Severe hyponatraemia may cause osmotic shift of water from the plasma into the brain cells. Typical symptoms include nausea, vomiting, headache and malaise. As the hyponatraemia worsens, confusion, diminished reflexes, convulsions, stupor or coma may occur. The cause of hyponatremia is typically classified by a person’s fluid status into low volume, normal volume, and high volume. Low volume hyponatremia can occur from diarrhoea, vomiting, diuretics, and sweating.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 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: 2000 ml/min
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 97
Incorrect
-
A 34-year-old woman has been suffering from headaches, fever, vomiting, and confusion for the last 5 days. A CT scan reveals an oedematous mass with ring enhancement in the left temporal region. It is biopsied, revealing glial cells, necrosis, neutrophils and lymphocytes. What is the most likely diagnosis?
Your Answer: Vascular malformation
Correct Answer: Cerebral abscess
Explanation:A cerebral abscess can result from direct extension of cranial infections, penetrating head trauma, haematogenous spread, or for unknown causes. An abscess forms when an area of cerebral inflammation becomes necrotic and encapsulated by glial cells and fibroblasts. Oedema around the abscess can increase the intracranial pressure. Symptoms result from increased intracranial pressure and mass effects. It is most frequent in the third decade of life, and when it occurs in children, it is usually associated with congenital heart disease.
-
This question is part of the following fields:
- Neurology
- Pathology
-
-
Question 98
Incorrect
-
A 27 year-old male patient was admitted to the hospital due to recurrent fever for the past 2 weeks. The patient claimed that he is an intravenous drug user. Following work up, the patient was diagnosed with infective endocarditis. Which is the most likely organism responsible for this diagnosis?
Your Answer: Escherichia coli
Correct Answer: Staphylococcus aureus
Explanation:Acute bacterial endocarditis is a fulminant illness lasting over days to weeks (<2weeks). It is most likely due to Staphylococcus aureus especially in intravenous drug abusers.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 99
Incorrect
-
Which organ is most vulnerable to haemorrhagic shock?
Your Answer: Brain
Correct Answer: Kidneys
Explanation:At rest, the brain receives 15% cardiac output, muscles 15%, gastrointestinal tract 30% and kidneys receive 20%. However, if normalised by weight, the largest specific blood flow is received by the kidneys at rest (400 ml/min x 100g), making them highly vulnerable in the case of a haemorrhagic shock.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 100
Incorrect
-
A 76 year old man who presented with lower back pain is diagnosed with prostatic carcinoma that has metastasized to his lumber spine. Which of the following markers is characteristically elevated?
Your Answer: CEA
Correct Answer: PSA
Explanation:Spread of prostatic carcinoma is common to the lumbar spine and pelvis. This results in osteoblastic metastases that will present as lower back pain with increased alkaline phosphatase, prostatic acid phosphates and PSA. PSA is more specific and a PSA > 10 ng/ml for any age is worrisome.
-
This question is part of the following fields:
- Neoplasia; Urology
- Pathology
-
-
Question 101
Incorrect
-
A Jewish man was diagnosed with haemophilia C. Which of the following factors is deficient in this form of haemophilia?
Your Answer: Factor VIII
Correct Answer: Factor XI
Explanation:Haemophilia C, also known as plasma thromboplastin antecedent (PTA) deficiency or Rosenthal syndrome, is a condition caused by the deficiency of the coagulation factor XI. The condition is rare and it is usually found in Ashkenazi Jews.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 102
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: Urinary sodium
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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 103
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: Bowel perforation
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 104
Incorrect
-
Intracellular shifting of hydrogen ions can generate a metabolic alkalosis. In which of the following conditions is metabolic alkalosis caused by this mechanism ?
Your Answer: Vomiting
Correct Answer: Hypokalaemia
Explanation:Metabolic alkalosis is characterized by a primary increase in the concentration of serum bicarbonate ions. This may occur as a consequence of a loss of hydrogen ions or a gain in bicarbonate. Hydrogen ions may be lost through the kidneys or the GI tract, as for example during vomiting, nasogastric suction or use of diuretics. Intracellular shifting of hydrogen ions develops mainly during hypokalaemia to maintain neutrality. Gain in bicarbonate ions may develop during administration of sodium bicarbonate in high amounts or in amounts that exceed the capacity of excretion of the kidneys, as seen in renal failure. Fluid losses may be another cause of metabolic alkalosis, causing the reduction of extracellular fluid volume.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 105
Incorrect
-
A 7-year-old boys undergoes a testicular biopsy after a tumour is found in his right testis. Elements similar to hair and teeth are found in it. What kind of tumour is this?
Your Answer: Leydig cell tumour
Correct Answer: Teratoma
Explanation:A teratoma is a tumour containing tissue elements that are similar to normal derivatives of more than one germ layer. They usually contain skin, hair, teeth and bone tissue and are more common in children, behaving as a benign tumour. After puberty, they are regarded as malignant and can metastasise.
-
This question is part of the following fields:
- Pathology
- Urology
-
-
Question 106
Correct
-
The most likely cause of prominent U waves on the electrocardiogram (ECG) of a patient is:
Your Answer: Hypokalaemia
Explanation:The U-wave, not always visible in ECGs, is thought to represent repolarisation of papillary muscles or Purkinje fibres. When seen, it is very small and occurs after the T-wave. Inverted U-waves indicate myocardial ischaemia or left ventricular volume overload. Prominent U-waves are most commonly seen in hypokalaemia. Other causes include hypercalcaemia, thyrotoxicosis, digitalis exposure, adrenaline and class 1A and 3 anti-arrhythmic agents. It can also be seen in congenital long-QT syndrome and in intracranial haemorrhage.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 107
Incorrect
-
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: Increased FEV1/FVC
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 108
Incorrect
-
A biopsy taken from the respiratory passage of a 37 year old male, chronic smoker will mostly likely show which cellular adaptation?
Your Answer: Mucous hyperplasia
Correct Answer: Stratified squamous metaplasia
Explanation:Metaplasia is a change in the cell type caused in part due to an extrinsic stress on the organ. It involves a change in the surface epithelium from one cell type to the another, most commonly squamous to columnar. This is a reversible process, and removal of the stress should theoretically reverse the surface epithelium back to normal morphology. Respiratory tract metaplasia is a classic example, in which the normal pseudostratified columnar epithelium is replaced by stratified squamous epithelium to better cope with the stress. Under continuous stress metaplasia can progress to dysplasia which is a disordered growth of cells eventually leading to the development of carcinoma.
-
This question is part of the following fields:
- Cell Injury & Wound Healing; Respiratory
- Pathology
-
-
Question 109
Correct
-
Under normal conditions, what is the major source of energy of cardiac muscles?
Your Answer: Fatty acids
Explanation:Under basal conditions, most of the energy needed by cardiac muscle for metabolism is derived from fats (60%), 35% by carbohydrates, and 5% by ketones and amino acids. However, after intake of large amounts of glucose, lactate and pyruvate are mainly used. During prolonged starvation, fat acts as the primary source. 50% of the used lipids are sourced from circulating fatty acids.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 110
Incorrect
-
A patient is diagnosed with lung cancer. His physician told him that his lung cancer type is aggressive. It can grow rapidly and may undergo early metastasis, however it is very sensitive to chemotherapy and radiotherapy. Which lung cancer type is most likely present
Your Answer: Adenocarcinoma
Correct Answer: Small-cell carcinoma
Explanation:Small-cell lung carcinoma (SCLC) is a type of highly malignant cancer that most commonly arises within the lung. SCLC usually metastasizes widely very early on in the natural history of the tumour, and in nearly all cases responds dramatically to chemotherapy and/or radiotherapy. Surgery has no role in the treatment of this disease.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 111
Incorrect
-
What is the percentage of bone calcium that is freely exchangeable with the extracellular fluid that is available for buffering changes in the calcium ion balance?
Your Answer: 20%
Correct Answer: 1%
Explanation:Around 1% of calcium in the body is available for buffering changes in calcium ion balance. These are mainly derived from the bone that are freely exchangeable with extracellular fluid.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 112
Incorrect
-
Fine-needle aspiration is a type of biopsy procedure. When performing a fine-needle aspiration of the lungs, which is the most common complication of the procedure?
Your Answer: Chronic pain after the procedure
Correct Answer: Pneumothorax
Explanation:Pneumothorax is the most common complication of a fine-needle aspiration procedure. Various factors, such as lesion size, have been associated with increased risk of pneumothorax .
-
This question is part of the following fields:
- Pathology
- Respiratory
-
-
Question 113
Correct
-
A 15-year-old girl was diagnosed with bacterial meningitis. Gram staining of the spinal fluid shows numerous polymorphonuclear neutrophils and Gram-positive cocci. Which is the empiric drug of choice to be given to the patient until the antibiotic sensitivity report is available?
Your Answer: Ceftriaxone
Explanation:Acute meningitis is a medical emergency. All suspects should receive their first dose of antibiotics immediately and be transferred to hospital as soon as possible. If facilities for blood culture and/or lumbar puncture (LP) are immediately available, they should be performed before administration
of the first dose of antibiotics (see contraindications to LP below). Neither procedure should lead to a significant delay in antibiotic administration.
Administer ceftriaxone 80-100 mg/kg (maximum 2 g, 12 hourly) intravenously. The intramuscular or intraosseous route can be used if there is no vascular access. Penicillin allergy is not a contraindication to ceftriaxone in acute meningitis (C-1). Omit ceftriaxone only if there has been documented ceftriaxone anaphylaxis. Give chloramphenicol 25 mg/kg (maximum 500 mg) intravenously instead, if available. Administer adequate analgesia and transfer the patient immediately to hospital, detailing all administered
medication in the referral letter. -
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 114
Incorrect
-
A 22-year old man presented with a mass in his left scrotum which was more prominent when standing and felt like a 'bag of worms'. Examination revealed a non-tender mass along the spermatic cord. Also, the right testis was larger than the left testis. What is the likely diagnosis?
Your Answer: Haematocele
Correct Answer: Varicocele
Explanation:Varicocele refers to dilatation and increased tortuosity of the pampiniform plexus – which is a network of veins found in spermatic cord that drain the testicle. Defective valves or extrinsic compression can result in outflow obstruction and cause dilatation near the testis. Normal diameter of the small vessels ranges from 0.5 – 1.5mm. A varicocele is a dilatation more than 2mm.
The plexus travels from the posterior aspect of testis into the inguinal canal with other structures forming the spermatic cord. They then form the testicular veins out of which the right testicular vein drains into the inferior vena cava and the left into the left renal vein.
It affects 15-20% men, and 40% of infertile males. Usually diagnosed in 15-25 years of age, they are rarely seen after 40 years of age. Because of the vertical path taken by the left testicular vein to drain into left renal vein, 98% idiopathic varicoceles occur on the left side. It is bilateral in 70% cases. Right-sided varicoceles are rare.
Symptoms include pain or heaviness in the testis, infertility, testicular atrophy, a palpable mass, which is non-tender and along the spermatic cord (resembling a ‘bag of worms’). The testis on the affected side might be smaller.
Diagnosis can be made by ultrasound. Provocative measures such as Valsalva manoeuvre or making the patient stand up to increase the dilatation by increasing the intra-abdominal venous pressure.
-
This question is part of the following fields:
- Pathology
- Urology
-
-
Question 115
Incorrect
-
A 24 year old mother is breastfeeding her first child. Which of the following cellular adaptations occurred in her breast tissue to allow her to do this?
Your Answer: Lobular atrophy
Correct Answer: Lobular hyperplasia
Explanation:Under the influence of oestrogen in pregnancy, there is an increase in the number of lobules which will facilitate lactation.
Steatocytes occur due to loss of weight and nutritional deficit.
Metaplasia is a normal physiological process which is due to a change in normal epithelium with another type.
Lobular atrophy will result in a decreased capacity to provide milk.
-
This question is part of the following fields:
- Cell Injury & Wound Healing; Female Health
- Pathology
-
-
Question 116
Incorrect
-
A 70-year-old female who underwent coronary artery bypass graft developed post-operative acute renal failure. Her urinary catheter was left in place to monitor urine output. 6 days later she developed fever and chills. She also complained of suprapubic and left flank pain. She is found to have developed acute ascending pyelonephritis. Which of the following organism was most likely isolated during urine culture?
Your Answer: Staphylococcus aureus
Correct Answer: Escherichia coli with pili
Explanation:The most common cause of urinary tract infection is Escherichia coli. Pilated strains of E. coli ascend the urethra to infect the kidney and the bladder. Catheters have been associated with an increased risk of UTIs.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 117
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: Increase, Increase
Correct Answer: Increase, Decrease
Explanation:Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. Respiratory acidosis can be acute or chronic. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg). The given problem represents acute respiratory acidosis and thus, will show a increase in arterial p(CO2) and decrease in pH.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 118
Incorrect
-
Purkinje fibres in the heart conduct action potentials at the rate of:
Your Answer: 12.5 - 15.0 m/s
Correct Answer: 1.5–4.0 m/s
Explanation:Purkinje fibres control the heart rate along with the sinoatrial node (SA node) and the atrioventricular node (AV node). The QRS complex is associated with the impulse passing through the Purkinje fibres. These fibres conduct action potential about six times faster than the velocity in normal cardiac muscle.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 119
Incorrect
-
Chronic obstructive pulmonary disease (COPD) is likely to result in:
Your Answer: Metabolic acidosis
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 120
Correct
-
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 121
Incorrect
-
Calculate the total peripheral resistance for a patient with a blood pressure of 130/70 mm HG and cardiac output of 5 litres / min?
Your Answer: 16 mmHg × min/l
Correct Answer: 18 mmHg × min/l
Explanation:Total peripheral resistance = Mean arterial pressure/Cardiac output. And the mean arterial pressure = Diastolic pressure + 1/3 (Systolic pressure – Diastolic pressure), i.e., 70 + 1/3 (130-70) = 90 mmHg. Therefore, total peripheral resistance = 90 mmHg/5 l per min = 18 mmHg × min/l.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 122
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: Is due to a large atrial septal defect
Correct Answer: Is normal and due to shunted blood
Explanation:Blood that bypasses the ventilated parts of lung and enters the arterial circulation directly is known as shunted blood. It happens in normal people due to mixing of arterial blood with bronchial and some myocardial venous blood (which drains into the left heart). Diffusion limitation and reaction velocity with haemoglobin are immeasurably small. CO2 unloading will not affect the difference between alveolar and arterial p(O2). A large VSD will result in much lower arterial O2 as compared to alveolar O2.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 123
Incorrect
-
A 40-year-old woman is suspected to have an ovarian cancer. Which tumour marker should be requested to confirm the diagnosis?
Your Answer: AFP
Correct Answer: CA-125
Explanation:CA-125 is a protein that is used as a tumour marker. This substance is found in high concentration in patients with ovarian cancer. It is the only tumour marker recommended for clinical use in the diagnosis and management of ovarian cancer.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 124
Incorrect
-
 A 45-year-old man presented to the doctor complaining of a flank pain and episodes of haematuria. Abdominal ultrasound revealed a left renal mass and the patient underwent a nephrectomy. Histopathological pattern was triphasic with blastemal, epithelial, and stromal components. The pathologist suggested the tumour resulted from the lack of a tumour suppressor gene on chromosome 11. Which of the following tumours is the pathologist most likely suggesting?
Your Answer: Angiomyolipoma
Correct Answer: Wilms’ tumour
Explanation:Wilms’ tumour is one of the most common malignant tumours of childhood but it can also rarely be found in adults. In biopsy, classical histopathological findings include the triphasic pattern composed by blastemal, epithelial, and stromal elements. First symptoms in children include an abdominal palpable mass, while in adults pain and haematuria are the most common complaints. Deletions of tumours’ suppressor genes on chromosome 11 are usually associated with Wilms’ tumour.
-
This question is part of the following fields:
- Pathology
- Renal
-
-
Question 125
Incorrect
-
Vincristine is a chemotherapy agent used to treat a number of types of cancer. Which of the following is a recognised major side-effect of vincristine?
Your Answer: Respiratory suppression
Correct Answer: Peripheral neuropathy
Explanation:Vincristine is an alkaloid chemotherapeutic agent. It is used to treat a number of types of cancer including acute lymphocytic leukaemia, acute myeloid leukaemia, Hodgkin’s disease, neuroblastoma, and small cell lung cancer among others. The main side-effects of vincristine are peripheral neuropathy and constipation.
-
This question is part of the following fields:
- Pathology
- Pharmacology
-
-
Question 126
Incorrect
-
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: 250 ml/min
Correct 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.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 127
Incorrect
-
Which of these illnesses is most likely to precede Guillain-Barré syndrome?
Your Answer: Epilepsy
Correct Answer: Viral pneumonia
Explanation:Guillain–Barré syndrome (GBS) is characterized by a rapid-onset muscle weakness caused by the immune system damaging the peripheral nervous system. In about two-thirds of patients, the syndrome begins 5 days to 3 weeks after an infectious disease, surgery or vaccination. Infection is the trigger in over 50% of patients; common pathogens include Campylobacter jejuni, enteric viruses, herpesviruses (including cytomegalovirus and those causing infectious mononucleosis) and Mycoplasma species. The underlying mechanism involves an autoimmune disorder in which the body’s immune system mistakenly attacks the peripheral nerves and damages their myelin insulation, although the cause for this is still unknown.
-
This question is part of the following fields:
- Neurology
- Pathology
-
-
Question 128
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: 2.5 resistance units (mmHg/l per min)
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 129
Incorrect
-
A 30-year-old woman feels thirsty. This thirst is probably due to:
Your Answer: Hypoglycaemia
Correct Answer: Increased level of angiotensin II
Explanation:Thirst is the basic need or instinct to drink. It arises from a lack of fluids and/or an increase in the concentration of certain osmolites such as salt. If the water volume of the body falls below a certain threshold or the osmolite concentration becomes too high, the brain signals thirst. Excessive thirst, known as polydipsia, along with excessive urination, known as polyuria, may be an indication of diabetes. Angiotensin II is a hormone that is a powerful dipsogen (i.e. it stimulates thirst) that acts via the subfornical organ. It increases secretion of ADH in the posterior pituitary and secretion of ACTH in the anterior pituitary.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 130
Incorrect
-
A 45-year-old woman was brought to the emergency department due to fever and chills. She has a history of recurrent UTI and complains of dysuria and urinary frequency. Urinary white blood cell count is >200 cell/high power field. If urine culture is performed, what is the most likely organism that will grow?
Your Answer: Staphylococcus aureus
Correct Answer: Escherichia coli
Explanation:The pathogen that most likely causes recurrent urinary tract infection in young women are E. coli, Enterococcus and Staphylococcus saprophyticus.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 131
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: Alveolar–arterial p(O2) difference
Correct Answer: Compliance of the lung
Explanation:Acute or adult respiratory distress syndrome (ARDS) is a reaction to several forms of lung injuries and is commonly associated with sepsis and SIRS (systemic inflammatory response syndrome), severe traumatic injury, severe head injury, narcotics overdose, drowning, pulmonary contusion, and multiple blood transfusions. There is an increase in risk due to pre-existing liver disease or coagulation abnormalities. It results due to indirect toxic effects of neutrophil-derived inflammatory mediators in the lungs. ARDS is defined by the 1994 American–European Consensus Committee as the acute onset of bilateral infiltrates on chest X-ray, a partial pressure of arterial oxygen (pa(O2)) to fraction of inspired oxygen Fi(O2) ratio of less than 200 mmHg and a pulmonary artery occlusion pressure of less than 18 or the absence of clinical evidence of left arterial hypertension. ARDS is basically pulmonary oedema in the absence of volume overload or poor left ventricular function. This is different from acute lung injury, which shows a pa(O2)/Fi(O2) ratio of less than 300 mmHg. Pathogenesis of ARDS starts from damage to alveolar epithelium and vascular endothelium, causing increased permeability. Damage to surfactant-producing type II cells disrupts the production and function of pulmonary surfactant, causing micro atelectasis and poor gas exchange. There is a decrease in lung compliance and increase in work of breathing. Eventually, there is resorption of alveolar oedema, regeneration of epithelial cells, proliferation and differentiation of type II alveolar cells and alveolar remodelling. Some show resolution and some progress to fibrosing alveolitis, which involves the deposition of collagen in alveolar, vascular and interstitial spaces.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 132
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: 75 ml
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 133
Incorrect
-
The chest X-ray of a 72 year old patient reveals the presence of a round lesion containing an air-fluid level in the left lung. These findings are most probably suggestive of:
Your Answer: Bronchiectasis
Correct Answer: Lung abscess
Explanation:Lung abscesses are collections of pus within the lung that arise most commonly as a complication of aspiration pneumonia caused by oral anaerobes. Older patients are more at risk due to poor oral hygiene, gingivitis an inability to handle their oral secretions due to other diseases. Chest X-ray most commonly reveals the appearance of an irregularly shaped cavity with an air-fluid level.
-
This question is part of the following fields:
- Pathology
- Respiratory
-
-
Question 134
Incorrect
-
A 70-year-old man suffers an ischaemic stroke and develops a left homonymous hemianopia. Where is the likely infarct?
Your Answer: Left frontal lobe
Correct Answer: Right occipital lobe
Explanation:The man has a left homonymous hemianopia which means he is unable to view objects in the left visual field. This information is processed by the right primary visual cortex which lies in the right occipital lobe.
-
This question is part of the following fields:
- Neurology
- Pathology
-
-
Question 135
Incorrect
-
A 32-year-old man presented with a metabolic acidosis and increased anion gap. What is the most likely cause of the changes of the anion gap in this patient?
Your Answer: Lithium intoxication
Correct Answer: Lactic acidosis
Explanation:High anion gap in metabolic acidosis is caused generally by the elevation of the levels of acids like ketones, lactate, sulphates in the body, which consume the bicarbonate ions. Other causes of a high anion gap include overdosing on salicylates, uraemia, rhabdomyolysis, hypocalcaemia, hypomagnesaemia, or ingestion of toxins such as ethylene glycol, methanol, propyl alcohol, cyanide and iron.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 136
Incorrect
-
Medulloblastoma usually occurs in children between 5 to 9 years old. Where does medulloblastoma commonly originate from?
Your Answer: Pons
Correct Answer: Cerebellar vermis
Explanation:Medulloblastoma is the most common malignant brain tumour in children, accounting for 10-20% of primary CNS neoplasms. Most of the tumours originate in the cerebellar vermis.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 137
Incorrect
-
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 allows 5-hydroxytryptophan (5HT) to cross to a very limited degree
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.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 138
Incorrect
-
A 5-year-old child diagnosed with nephrotic syndrome developed generalised oedema. What is the mechanism for the development of oedema in patients with nephrotic syndrome?
Your Answer: Lymphatic obstruction
Correct Answer: Decreased colloid osmotic pressure
Explanation:The development of oedema in nephrotic syndrome has traditionally been viewed as an underfill mechanism. According to this view, urinary loss of protein results in hypoalbuminemia and decreased plasma oncotic pressure. As a result, plasma water translocates out of the intravascular space and results in a decrease in intravascular volume. In response to the underfilled circulation, effector mechanisms are then activated that signal the kidney to secondarily retain salt and water. While an underfill mechanism may be responsible for oedema formation in a minority of patients, recent clinical and experimental findings would suggest that oedema formation in most nephrotic patients is the result of primary salt retention. Direct measurements of blood and plasma volume or measurement of neurohumoral markers that indirectly reflect effective circulatory volume are mostly consistent with either euvolemia or a volume expanded state. The ability to maintain plasma volume in the setting of a decreased plasma oncotic pressure is achieved by alterations in transcapillary exchange mechanisms known to occur in the setting of hypoalbuminemia that limit excessive capillary fluid filtration.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 139
Correct
-
In a study, breast lumps were analysed to determine the characteristic of malignant neoplasm on biopsy. What microscopic findings are suggestive of malignancy?
Your Answer: Invasion
Explanation:Invasion is suggestive of malignancy and an even better option would have been metastasis. Pleomorphism is found in both benign and malignant neoplasms along with atypia and anaplasia. A height nuclear/cytoplasmic ratio is suggestive of malignancy but not the best indicator. Malignant tumours are aggressive and growth rapidly. Necrosis can be seen in benign tumours if they deplete their blood supply.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 140
Correct
-
A 45 year old man who complains of chronic post prandial, burning epigastric pain undergoes a gastrointestinal endoscopy. There is no apparent mass or haemorrhage and a biopsy is taken from the lower oesophageal mucosa just above the gastro-oesophageal junction. The results reveal the presence of columnar cells interspersed with goblet cells. Which change best explains the above mentioned histology?
Your Answer: Metaplasia
Explanation:Metaplasia is the transformation of one type of epithelium into another as a means to better cope with external stress on that epithelium. In this case metaplasia occurs due to the inflammation resulting from gastro-oesophageal reflux disease. Dysplasia is disordered cellular growth. Hyperplasia is an increase in cell number but not cell type i.e. transformation. Carcinoma is characterized by cellular atypia. Ischaemia would result in necrosis with ulceration. Carcinoma insitu involves dysplastic atypical cells with the basement membrane intact and atrophy would mean a decrease in number of cells.
-
This question is part of the following fields:
- Cell Injury & Wound Healing; Gastrointestinal
- Pathology
-
-
Question 141
Correct
-
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: Elastic lung compliance
Explanation:The forces of elastance (compliance), frictional resistance and inertia have been identified as the forces that oppose lung inflation and deflation. The normal relaxed state of the lung and chest is partially empty. Further exhalation requires muscular work. Inhalation is an active process requiring work. About 60–66% of the total work performed by the respiratory muscles is used to overcome the elastic or compliance characteristics of the lung–chest cage, 30–35% is used to overcome frictional resistance and only 2–5% of the work is used for inertia.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 142
Incorrect
-
A significantly elevated white cell count of 50 x 109/l with 5% blasts and raised leucocyte alkaline phosphatase is seen in which of the following conditions?
Your Answer: Chronic lymphocytic leukaemia
Correct Answer: Leukaemoid reaction
Explanation:Non-neoplastic proliferation of leucocytes causes an increase in leukocyte alkaline phosphatase (LAP). This is referred to as ‘leukemoid reaction’ because of the similarity to leukaemia with an increased white cell count (>50 × 109/l) with immature forms. Causes of leukemoid reaction includes haemorrhage, drugs (glucocorticoids, all-trans retinoic acid etc), infections such as tuberculosis and pertussis, and as a paraneoplastic phenomenon. Leukemoid reaction can also be seen in infancy as a feature of trisomy 21. This is usually a benign condition, but can be a response to a disease state. Differential diagnosis include chronic myelogenous leukaemia (CML).
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 143
Incorrect
-
During strenuous exercise, what else occurs besides tachycardia?
Your Answer: No change to blood pressure
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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 144
Incorrect
-
Increased resistance to flow of blood in cerebral vessels is most likely seen in:
Your Answer: Epileptic seizure
Correct Answer: Elevation in systemic arterial pressure from 100 to 130 mmHg
Explanation:Constant cerebral blood flow is maintained by autoregulation in the brain, which causes an increase in local vascular resistance to offset an increase in blood pressure. There will be an increase in cerebral blood flow (and decrease in resistance to cerebral blood flow) with a decrease in arterial oxygen or an increase in arterial CO2. Similarly, a decrease in viscosity will also increase the blood flow. Due to increased brain metabolism and activity during a seizure, there will also be an increase in the cerebral blood flow.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 145
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:
- Cardiovascular
- Physiology
-
-
Question 146
Incorrect
-
A 36-year-old woman suddenly suffers from a generalized seizure. She was previously healthy. An emergency CT scan reveals a mass in the posterior fossa, with distortion of the lateral ventricles. After removing the tumour, the biopsy reveals it contains glial fibrillary acidic protein (GEAP). What's the most likely diagnosis?
Your Answer: Meningioma
Correct Answer: Astrocytoma
Explanation:Astrocytomas are primary intracranial tumours derived from astrocyte cells of the brain. They can arise in the cerebral hemispheres, in the posterior fossa, in the optic nerve and, rarely, in the spinal cord. These tumours express glial fibrillary acidic protein (GFAP). In almost half of cases, the first symptom of an astrocytoma is the onset of a focal or generalised seizure. Between 60% and 75% of patients will have recurrent seizures during the course of their illness. Secondary clinical sequelae may be caused by elevated intracranial pressure (ICP) cause by the direct mass effect, increased blood volume, or increased cerebrospinal fluid (CSF) volume. CT will usually show distortion of the third and lateral ventricles, with displacement of the anterior and middle cerebral arteries. Histological diagnosis with tissue biopsy will normally reveal an infiltrative character suggestive of the slow growing nature of the tumour.
-
This question is part of the following fields:
- Neurology
- Pathology
-
-
Question 147
Correct
-
Anthrax is an infection caused by the bacterium Bacillus anthracis. Anthrax spores have been used as a biological warfare weapon. What is the drug of choice in treating anthrax infection?
Your Answer: Ciprofloxacin
Explanation:Early antibiotic treatment of anthrax is essential. A delay may significantly lessen the chances for survival of the patient. Treatment for anthrax infection include large doses of intravenous and oral antibiotics, such as fluoroquinolones (ciprofloxacin), doxycycline, erythromycin, vancomycin, or penicillin.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 148
Correct
-
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).
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 149
Incorrect
-
Lung compliance is increased by:
Your Answer: Atelectasis
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 150
Incorrect
-
Which of the following conditions causes an elevation of the pH in the tissues with elevated arterial CO2 content?
Your Answer: Respiratory alkalosis
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.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 151
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: 600 mmol/day
Correct Answer: 240 mmol/day
Explanation:Calcium is the most abundant mineral in the human body. The average adult body contains in total approximately 1 kg of calcium of which 99% is in the skeleton in the form of calcium phosphate salts. The extracellular fluid (ECF) contains approximately 22 mmol, of which about 9 mmol is in the plasma. About 40% of total plasma Ca2+ is bound to proteins and not filtered at the glomerular basement membrane. Therefore, the estimated daily filtered load is 1.5 mmol/l × 160 l/day = 240 mmol/day. The exact amount of free versus total Ca2+ depends on the blood pH: free Ca2+ increases during acidosis and decreases during alkalosis.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 152
Incorrect
-
A 55-year old lady underwent a major surgery for repair of an aortic aneurysm. Her blood pressure was low throughout the intra-operative and the post-operative period, along with increasing serum creatinine and urea. Microscopic examination of her urine showed multiple granular and hyaline casts. What is the likely condition the patient is suffering from?
Your Answer: Renal vein thrombosis
Correct Answer: Acute tubular necrosis
Explanation:The most common predisposing factor leading to acute tubular necrosis is ischemia, typically seen in hospitalized patients with low blood pressure.
-
This question is part of the following fields:
- Pathology
- Renal
-
-
Question 153
Incorrect
-
Which of the following tumours is very radiosensitive?
Your Answer: Gastric carcinoma
Correct Answer: Seminoma
Explanation:Seminoma is the most radiosensitive tumour and responds well to radiation therapy after unilateral orchidectomy. The ipsilateral inguinal areas are routinely not treated however, depending on the stage, the mediastinum and the left supraclavicular regions may also be irradiated.
-
This question is part of the following fields:
- Neoplasia; Urology
- Pathology
-
-
Question 154
Incorrect
-
A 30 year old male has a painless and transilluminant swelling at the upper pole of his left testi. There is a negative cough test. Which of the following is the likely diagnosis?
Your Answer:
Correct Answer: Spermatocoele
Explanation:Spermatocele, also known as a spermatic cyst is a cystic mass usually occurring at the upper pole of the testis. Differential diagnosis included hydrocele as both are cystic, painless and transilluminant. Ultrasound is a useful modality. If symptomatic or large, surgical excision can be done.
-
This question is part of the following fields:
- Pathology
- Urology
-
-
Question 155
Incorrect
-
A 55 year old lady underwent an uneventful appendicectomy. Two hours later, her arterial blood gas analysis on room revealed pH: 7.30, p(CO2): 53 mmHg and p(O2): 79 mmHg. What is the most likely cause of these findings?
Your Answer:
Correct Answer: Alveolar hypoventilation
Explanation:In the given problem, there is respiratory acidosis due to hypercapnia from a low respiratory rate and/or volume (hypoventilation). Causes of hypoventilation include conditions impairing the central nervous system (CNS) respiratory drive, impaired neuromuscular transmission and other causes of muscular weakness (drugs and sedatives), along with obstructive, restrictive and parenchymal pulmonary disorders. Hypoventilation leads to hypoxia and hypercapnia reduces the arterial pH. Severe acidosis leads to pulmonary arteriolar vasoconstriction, systemic vascular dilatation, reduced myocardial contractility, hyperkalaemia, hypotension and cardiac irritability resulting in arrhythmias. Raised carbon dioxide concentration also causes cerebral vasodilatation and raised intracranial pressure. Over time, buffering and renal compensation occurs. However, this might not be seen in acute scenarios where the rise in p(CO2) occurs rapidly.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 156
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
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 157
Incorrect
-
The extent of cancer growth can be described through staging. What is taken into consideration when staging a cancer?
Your Answer:
Correct Answer: Local invasion
Explanation:Cancer stage is based on four characteristics: the size of cancer, whether the cancer is invasive or non-invasive, whether the cancer has spread to the lymph nodes, and whether the cancer has spread to other parts of the body, in this case beyond the breast. Staging is important as it is often a good predictor of outcomes and treatment is adjusted accordingly.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 158
Incorrect
-
A 68-year-old woman complains of headaches, dizziness, and memory loss. About a month ago, she fell from a staircase but only suffered mild head trauma. What is the most likely diagnosis in this case?
Your Answer:
Correct Answer: Chronic subdural haematoma
Explanation:A quarter to a half of patients with chronic subdural haematoma have no identifiable history of head trauma. If a patient does have a history of head trauma, it usually is mild. The average time between head trauma and chronic subdural haematoma diagnosis is 4–5 weeks. Symptoms include decreased level of consciousness, balance problems, cognitive dysfunction and memory loss, motor deficit (e.g. hemiparesis), headache or aphasia. Some patients present acutely. They usually result from tears in bridging veins which cross the subdural space, and may cause an increase in intracranial pressure (ICP).
-
This question is part of the following fields:
- Neurology
- Pathology
-
-
Question 159
Incorrect
-
A 39-year-old man, after radiological evaluation and thoracentesis, was found to have chylothorax. What is the most probable cause of this diagnose?
Your Answer:
Correct Answer: Mediastinal malignant lymphoma
Explanation:Chylothorax is a potentially lethal condition characterized by fluid (chyle) accumulation in the pleural cavity, resulting from disruption of lymphatic drainage in the thoracic duct. Chyle is a fluid rich in triglycerides and chylomicrons and can originate from the thorax, the abdomen or both. Malignant tumours, especially lymphoma, are the most common causes of nontraumatic chylothorax. Bronchogenic carcinoma and trauma are the most common causes after lymphomas. Other rare causes of chylothorax are; granulomatous diseases, tuberculosis, congenital malformations, nephrotic syndrome, hypothyroidism, cirrhosis, decompensated heart failure and idiopathic chylothorax.
-
This question is part of the following fields:
- Pathology
- Respiratory
-
-
Question 160
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 161
Incorrect
-
A 30 year old female presented in the emergency with an irregular pulse. Her ECG showed absent P-waves with irregular RR interval. What is the most likely diagnosis?
Your Answer:
Correct Answer: Atrial fibrillation
Explanation:Atrial fibrillation is one of the most common cardiac arrhythmias. It is often asymptomatic but may present with symptoms of palpitations, fainting, chest pain and heart failure. Characteristic findings are: absence of P-waves, unorganised electrical activity in their place, irregularity of RR interval due to irregular conduction of impulses to the ventricles and if paroxysmal AF is suspected, episodes may be documented with the use of Holter monitoring
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 162
Incorrect
-
Tumours derived from all three germ-cell layers in new-borns usually occur in which of the following sites?
Your Answer:
Correct Answer: Sacrococcygeal area
Explanation:A teratoma is a tumour with tissue or organ components resembling normal derivatives of more than one germ layer. It is derived from all three cell layers. The most common location of teratoma in new-born infants is in the sacrococcygeal area.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 163
Incorrect
-
A 4-year-old child was brought to a paediatrician for consult due to a palpable mass in his abdomen. The child has poor appetite and regularly complains of abdominal pain. The child was worked up and diagnosed with a tumour. What is the most likely diagnosis ?
Your Answer:
Correct Answer: Nephroblastoma
Explanation:Nephroblastoma is also known as Wilms’ tumour. It is a cancer of the kidneys that typically occurs in children. The median age of diagnose is approximately 3.5 years. With the current treatment, approximately 80-90% of children with Wilms’ tumour survive.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 164
Incorrect
-
Which of these infectious agents tends to affect people under 20 and over 40 years old, can cause acute encephalitis with cerebral oedema and petechial haemorrhages, along with haemorrhagic lesions of the temporal lobe. A lumbar puncture will reveal clear cerebrospinal fluid with an elevated lymphocyte count?
Your Answer:
Correct Answer: Herpes simplex virus
Explanation:Haemorrhagic lesions of the temporal lobe are typical of Herpes simplex encephalitis (HSE). It tends to affect patients aged under 20 or over 40 years, and is often fatal if left untreated. In acute encephalitis, cerebral oedema and petechial haemorrhages occur and direct viral invasion of the brain usually damages neurones. The majority of cases of herpes encephalitis are caused by herpes simplex virus-1 (HSV-1), and about 10% of cases of herpes encephalitis are due to HSV-2, which is typically spread through sexual contact.
-
This question is part of the following fields:
- Neurology
- Pathology
-
-
Question 165
Incorrect
-
A 26-year-old female sought consultation due to excessive vaginal discharge. Vaginal smear showed numerous bacilli under the microscope. The organism was non-pathogenic. What is the most likely organism:
Your Answer:
Correct Answer: Lactobacillus species
Explanation:Lactobacillus is a Gram-positive facultative bacteria. It is commonly present in the vagina and the gastrointestinal tract. Colonization of Lactobacillus is usually benign and it makes up a small portion of the gastrointestinal flora.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 166
Incorrect
-
The pleural cavity is the space between the two pulmonary pleurae which cover the lungs. What is the normal amount of pleural fluid?
Your Answer:
Correct Answer: 10 ml
Explanation:Pleural fluid is a serous fluid produced by the serous membrane covering normal pleurae. Most fluid is produced by the parietal circulation (intercostal arteries) via bulk flow and reabsorbed by the lymphatic system. The total volume of fluid present in the intrapleural space is estimated to be only 2–10 ml. A small amount of protein is present in intrapleural fluid. Normally, the rate of reabsorption increases as a physiological response to accumulating fluid.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 167
Incorrect
-
The periphery of a haematoma is infiltrated by fibroblasts, collagen and new vasculature. This process is best described as?
Your Answer:
Correct Answer: Organisation of the haematoma
Explanation:Formation of granulation tissue at the periphery of the hematoma is a normal process leading to resolution. This granulation tissue is composed of new capillaries, fibroblasts and collagen. Lysis of a blood clot can occur, but the actual process of this response is known as organization, wherein the scar tissue will become part of the vessels. This is followed by recanalization and embolization which can lead to eventual complications. Proliferation of a clot will occur due to an imbalance in the clotting and lysing systems. Thrombosis has nothing to do with the process described above.
-
This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
-
-
Question 168
Incorrect
-
If a catheter is placed in the main pulmonary artery of a healthy 30-year-old woman, which of the following will be its mean pulmonary arterial pressure?
Your Answer:
Correct Answer: 15 mmHg
Explanation:The pulmonary artery pressure (PA pressure) is a measure of the blood pressure found in the main pulmonary artery. The hydrostatic pressure of the pulmonary circulation refers to the actual pressure inside pulmonary vessels relative to atmospheric pressure. Hydrostatic (blood pressure) in the pulmonary vascular bed is low compared with that of similar systemic vessels. The mean pulmonary arterial pressure is about 15 mmHg (ranging from about 13 to 19 mmHg) and is much lower than the average systemic arterial pressure of 90 mmHg.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 169
Incorrect
-
Which of the following will increase blood pressure and cause hypokalaemia?
Your Answer:
Correct 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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 170
Incorrect
-
What is the normal duration of the ST segment?
Your Answer:
Correct Answer: 0.08 s
Explanation:The ST segment lies between the QRS complex and the T-wave. The normal duration of the ST segment is 0.08 s. ST-segment elevation or depression may indicate myocardial ischaemia or infarction.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
SESSION STATS - PERFORMANCE PER SPECIALTY
