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Question 1
Incorrect
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An ECG of a 30 year old woman revealed low voltage QRS complexes. This patient is most probably suffering from?
Your Answer: Ventricular tachycardia
Correct Answer: Pericardial effusion
Explanation:The QRS complex is associated with current that results in the contraction of both the ventricles. As ventricles have more muscle mass than the atria, they result in a greater deflection on the ECG. The normal duration of a QRS complex is 10s. A wide and deep Q wave depicts myocardial infarction. Abnormalities in the QRS complex maybe indicative of a bundle block, ventricular tachycardia or hypertrophy of the ventricles. Low voltage QRS complexes are characteristic of pericarditis or a pericardial effusion.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 2
Incorrect
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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: Increased capillary hydrostatic pressure
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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 3
Incorrect
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Which of the following factors will not affect the wound healing process in a young women who suffered serious burns to her chest and hands?
Your Answer: Steroids
Correct Answer: Vitamin A deficiency
Explanation:Healing can be sped-up or slowed down due to various reasons: 1. blood supply, 2. infection, 3. denervation, 4. collection of blood/hematoma, 5. mechanical stress, 6. foreign body, 7. techniques used during surgery and 8. dressing of the wound. Other systemic factors include 1. nutrition e.g. deficiency of zinc, vitamin C, protein deficiency, 2. metabolic status, 3. circulatory status and 4. hormonal influence
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 4
Incorrect
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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: Chronic renal failure
Correct 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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 5
Incorrect
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A 30 year old man suffered severe blood loss, approx. 20-30% of his blood volume. What changes are most likely seen in the pulmonary vascular resistance (PVR) and pulmonary artery pressure (PAP) respectively following this decrease in cardiac output?
Your Answer: No change No change
Correct Answer: Increase Decrease
Explanation:Hypovolemia will result in the activation of the sympathetic adrenal discharge resulting is a decrease pulmonary artery pressure and an elevated pulmonary vascular resistance.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 6
Correct
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During cardiac catheterisation, if the blood sample from the catheter shows an oxygen saturation of 70%, and the pressure ranging from 12 to 24 mm Hg, it implies that the catheter tip is located in the:
Your Answer: Pulmonary artery
Explanation:Normal values for various parameters are as follows:
Systolic arterial blood pressure (SBP): 90–140 mmHg.
Diastolic arterial blood pressure: 60–90 mmHg.
Mean arterial blood pressure (MAP): SBP + (2 × DBP)/3 (normal range: 70-105 mmHg).
Right atrial pressure (RAP): 2–6 mmHg.
Systolic right ventricular pressure (RVSP): 15–25 mmHg.
Diastolic right ventricular pressure (RVDP): 0–8 mmHg.
Pulmonary artery pressure (PAP): Systolic (PASP) is 15-25 mmHg and Diastolic (PADP) is 8–15 mmHg.
Pulmonary artery wedge pressure (PAWP): 6–12 mmHg.
Left atrial pressure (LAP): 6–12 mmHg.
Thus, the given value indicates that the position of catheter tip is likely to be in the pulmonary artery.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 7
Incorrect
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A 14-year old girl presented with a 2cm, mobile, cystic mass in the midline of her neck. Fine needle aspiration of the mass revealed clear fluid. This is most likely a case of:
Your Answer: Parathyroid cyst
Correct Answer: Thyroglossal duct cyst
Explanation:Thyroglossal cyst is the most common congenital thyroid anomaly which is clinically significant and affects women more than men. It is a vestigial remnant of developing thyroid. Although the thyroglossal cyst can develop anywhere along the thyroglossal duct, the most common site is in the midline between the isthmus of thyroid and hyoid bone, or just above the hyoid. Thyroglossal cysts are also associated with ectopic thyroid tissue. Clinically, the cyst moves upward with protrusion of the tongue. Rarely, the persistent duct can become malignant (thyroglossal duct carcinoma) where the cancerous cells arise in the ectopic thyroid tissue that are deposited along the duct. Exposure to radiation is a predisposing factor.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 8
Incorrect
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Which is a feature of the action of insulin?
Your Answer: Promotes calcium release from bone
Correct Answer: Promotes protein synthesis
Explanation:Insulin is produced by the beta-cells of the islets of Langerhans in the pancreas. Its actions include:
– promoting uptake of glucose into cells
– glycogen synthesis (glycogenesis)
– protein synthesis
– stimulation of lipogenesis (fat formation).
– driving potassium into cells – used to treat hyperkaelamia.
Parathyroid hormone and activated vitamin D are the principal hormones involved in calcium/phosphate metabolism, rather than insulin.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 9
Incorrect
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Calculate the cardiac stroke volume of a patient whose oxygen consumption (measured by analysis of mixed expired gas) is 300 ml/min, arterial O2 content is 20 ml/100 ml blood, pulmonary arterial O2 content is 15 ml/100 ml blood and heart rate is 60/min.
Your Answer: 50 ml
Correct Answer: 100 ml
Explanation:By Fick’s principle, VO2 = Q × (CA (O2) − CV (O2)) where VO2 = O2 consumption, Q = cardiac output and CA(O2) and CV(O2) are arterial and mixed venous O2 content respectively. Thus, in the given problem, 300 ml O2/min = Q × (20−15) ml O2/100 ml. Thus, Q = 6000 ml blood/min. Then, we can calculate stroke volume by dividing the cardiac output with heart rate. Thus, stroke volume = 6000 ml/min divided by 60/min stroke volume = 100 ml.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 10
Incorrect
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The blood-brain barrier is a membrane that separates the circulating blood from the brain extracellular fluid in the central nervous system (CNS). Which of the following statements regarding the blood– brain barrier is CORRECT?
Your Answer: All regions of the brain have a blood-brain barrier
Correct Answer: It breaks down in areas of brain that are infected
Explanation:The blood–brain barrier is a membrane that controls the passage of substances from the blood into the central nervous system. It is a physical barrier between the local blood vessels and most parts of the central nervous system and stops many substances from travelling across it. During meningitis, the blood–brain barrier may be disrupted. This disruption may increase the penetration of various substances (including either toxins or antibiotics) into the brain. A few regions in the brain, including the circumventricular organs, do not have a blood–brain barrier.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 11
Incorrect
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A 25 year-old female medical student presents with fever, lack of appetite, rashes, sore throat and lymphadenopathy. Peripheral smear shows atypical lymphocytes. Which is the most likely organism responsible for this patient's condition?
Your Answer: Cytomegalovirus
Correct Answer: Epstein–Barr virus
Explanation:Epstein-Barr virus is in the herpes family of viruses and most people will become infected with EBV sometime during their lives. EBV commonly causes infectious mononucleosis, or mono, a contagious viral illness that initially attacks the lymph nodes in the neck and throat. When these tissues become less effective in fighting infection, sore throats, swelling of the nodes and fever may result.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 12
Incorrect
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Chronic obstructive pulmonary disease (COPD) is likely to result in:
Your Answer: Metabolic alkalosis
Correct Answer: Respiratory acidosis
Explanation:COPD leads to respiratory acidosis (chronic). This occurs due to hypoventilation which involves multiple causes, such as poor responsiveness to hypoxia and hypercapnia, increased ventilation/perfusion mismatch leading to increased dead space ventilation and decreased diaphragm function.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 13
Correct
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What is the normal duration of the ST segment?
Your 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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 14
Incorrect
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A young female in the 15th week of pregnancy presented to the emergency department with the passage of grape-like masses per vagina. Dilatation and curettage was done and microscopy revealed large avascular villi with trophoblastic proliferation. Which one of the following investigations is best recommended for her follow-up?
Your Answer: Pelvic ultrasound
Correct Answer: Serum β-hCG
Explanation:Trophoblast is the layer of cells surrounding the blastocyst and that later develops into the chorion and amnion. Gestational trophoblastic disease is a tumour arising from this trophoblast. It can occur during or after either an intrauterine or ectopic pregnancy. If it occurs in a pregnant woman, it usually leads to spontaneous abortion, eclampsia or fetal death. It can be either malignant or benign.
In suspected cases, investigations include measurement of serum beta subunit of human chorionic gonadotrophin (β-hCG) and pelvic ultrasound. Confirmatory test is a biopsy. Post-removal, the disease is classified clinically to assess further treatment. To assess the presence of metastases, further work-up includes computed tomography of the brain, chest, abdomen and pelvis. Chemotherapy is usually needed for persistent disease. If at least three consecutive, weekly serum β-hCG measurements are normal, treatment is considered successful. Follow-up is also done by measuring β-hCG.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 15
Incorrect
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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).
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This question is part of the following fields:
- Physiology
- Renal
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Question 16
Incorrect
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Which tumour marker is associated with medullary carcinoma of thyroid?
Your Answer: Alpha-fetoprotein
Correct Answer: Calcitonin
Explanation:Medullary carcinoma of thyroid accounts for 3% of thyroid cancers. It arises from the parafollicular cells (C cells) of the thyroid gland that produce calcitonin. It is often familial and caused by mutation of ret proto-oncogene, but can occasionally be sporadic. The familial cases can also occur as part of MEN syndromes IIA and IIB. The high calcitonin leads to down-regulation of the receptors, which does not affect the calcium levels significantly. Medullary carcinoma of thyroid shows characteristic amyloid deposits that stain positively with Congo red. The initial presentation consists of an asymptomatic thyroid nodule. Many cases are diagnosed due to routine screening of relatives of patients with MEN IIA and IIB. Medullary carcinoma can also cause ectopic production of other hormones/peptides such as adrenocorticotrophic hormone, vasoactive intestinal polypeptide, kallikreins and serotonin.
Metastasis from medullary carcinoma spread via the lymphatics to cervical and mediastinal nodes, and can also affect the liver, lungs and bone. Diagnosis is by raised serum calcitonin levels. A provocative test with calcium (15 mg/kg intravenously over 4 hours) also aids in diagnosis by leading to excessive secretion of calcitonin. X-ray might also show dense, conglomerate calcification.
CA-125 is frequently elevated in ovarian carcinomas. CA 15-3 is often associated with breast carcinomas. Alpha-fetoprotein is seen raised in hepatomas and gonadal tumours. Elevated HCG is associated with normal pregnancies, gonadal tumours, and choriocarcinomas. Thyroglobulin is used for surveillance in papillary carcinoma of thyroid. CA 19-9 is used in the management of pancreatic cancer.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 17
Incorrect
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A 35-year-old woman in her 37th week of pregnancy complains of urinary incontinence. She is most likely to have:
Your Answer: Urge incontinence
Correct Answer: Stress incontinence
Explanation:Urinary incontinence is the involuntary excretion of urine from one’s body. It is often temporary and it almost always results from an underlying medical condition. Several types include:
– Stress incontinence is the voiding of urine following increased abdominal pressure e.g. laughing, coughing, pregnancy etc. It is the most common form of incontinence in women, most commonly due to pelvic floor muscle weakness, physical changes from pregnancy, childbirth and menopause. In men it is a common problem following a prostatectomy. Most lab results such as urine analysis, cystometry and postvoid residual volume are normal.
– Urge incontinence is involuntary loss of urine occurring for no apparent reason while suddenly feeling the need or urge to urinate. The most common cause of urge incontinence are involuntary and inappropriate detrusor muscle contractions.
– Functional incontinence – occurs when a person does not recognise the need to go to the toilet, recognise where the toilet is or get to the toilet in time. The urine loss may be large. Causes of functional incontinence include confusion, dementia, poor eyesight, poor mobility, poor dexterity or unwillingness. t
– Overflow incontinence – sometimes people find that they cannot stop their bladders from constantly dribbling or continuing to dribble for some time after they have passed urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 18
Correct
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Which organ is most vulnerable to haemorrhagic shock?
Your Answer: Kidneys
Explanation:At rest, the brain receives 15% cardiac output, muscles 15%, gastrointestinal tract 30% and kidneys receive 20%. However, if normalised by weight, the largest specific blood flow is received by the kidneys at rest (400 ml/min x 100g), making them highly vulnerable in the case of a haemorrhagic shock.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 19
Incorrect
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The Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity. According to this equation, the buffering capacity of the system is at maximum when the number of free anions compared with undissociated acid is:
Your Answer: One-third less
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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 20
Incorrect
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Which of the following is NOT a nutritional factor involved in wound healing:
Your Answer: Vitamin A
Correct Answer: Vitamin B3
Explanation:Vitamin B6 is required for collagen cross-links.
Vitamin A is required for epithelial cell proliferation.
Zinc is required for RNA and DNA synthesis.
Copper is required for cross-linking of collagen.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Physiology
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Question 21
Incorrect
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Extracellular body fluid as compared with intracellular body fluid:
Your Answer: Has greater volume
Correct 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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 22
Correct
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If the blood flow is constant, oxygen extraction by tissues will show the greatest decrease due to which of the following interventions?
Your Answer: Tissue cooling
Explanation:With a constant blood flow to a given tissue bed, there will be an increase in oxygen extraction by the tissue with the following; an increase in tissue metabolism and oxygen requirements: warming (or fever), exercise, catecholamines and thyroxine. With cooling, the demand for oxygen decreases, leading to decreased oxygen extraction.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 23
Correct
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What is the primary function of the Kupffer cells found in the liver?
Your Answer: Recycling of old red blood cells
Explanation:Kupffer cells found in the liver are part of the monocyte-reticular system. They are specialised macrophages and primarily function to recycle old and damaged RBCs. The RBCs are phagocytosed and the haemoglobin is broken down into haem and globin. The haem is further broken down into iron that is recycled and bilirubin that is conjugated with glucuronic acid and excreted in the bile.
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This question is part of the following fields:
- Inflammation & Immunology; Hepatobiliary
- Pathology
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Question 24
Incorrect
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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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 25
Incorrect
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A 50-year-old gentleman was recently diagnosed with hypertension, with no other abnormalities on physical examination. Further investigations revealed the following :
Na+ 144 mmol/l
K+ 3.0 mmol/l
Cl- 107 mmol/l
Bicarbonate 25 mmol/l.
Blood glucose 5.8 mmol/l.
What is the likely diagnosis?Your Answer: Congenital adrenal hyperplasia
Correct Answer: Conn syndrome
Explanation:Overproduction of aldosterone (a mineralocorticoid) by the adrenal glands is known as Conn’s syndrome. It can be either due to an aldosterone-secreting adrenal adenoma (50-60% cases) or adrenal gland hyperplasia (40-50% cases). Excess aldosterone leads to sodium and water retention, along with potassium excretion. This leads to arterial (non-essential) hypertension. Conn’s syndrome is the commonest cause of primary hyperaldosteronism. Other symptoms include muscle cramps, headache (due to hypokalaemia) and metabolic alkalosis, which occurs due to increased secretion of H+ ions by the kidney. The raised pH of the blood traps calcium leading to symptoms of hypocalcaemia, which can be mimicked by liquorice ingestion and Liddle syndrome. To diagnose Conn’s syndrome, the ratio of renin and aldosterone is measured. Due to suppression of renin secretion, there is low renin to aldosterone ratio (<0.05). However, anti-hypertensives may affect the test results and should be withdrawn for 6 weeks. Computed tomography can also be done to detect the presence of adrenal adenoma. Cushing’s syndrome does not cause hypokalaemia with normal serum glucose levels. Nelson’s syndrome refers to increased ACTH secretion due to pituitary adenoma. Pheochromocytoma will not lead to hypokalaemia even though hypertension can be seen.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 26
Incorrect
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A 43-year-old diabetic man complains of headaches, palpitations, anxiety, abdominal pain and weakness. He is administered sodium bicarbonate used to treat:
Your Answer: Respiratory acidosis
Correct Answer: Metabolic acidosis
Explanation:Sodium bicarbonate is indicated in the management of metabolic acidosis, which may occur in severe renal disease, uncontrolled diabetes, circulatory insufficiency due to shock or severe dehydration, extracorporeal circulation of blood, cardiac arrest and severe primary lactic acidosis. Bicarbonate is given at 50-100 mmol at a time under scrupulous monitoring of the arterial blood gas readings. This intervention, however, has some serious complications including lactic acidosis, and in those cases, should be used with great care.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 27
Incorrect
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A chloride sweat test was performed on a 13-year-old boy. Results indicated a high likelihood of cystic fibrosis. This diagnosis is associated with a higher risk of developing which of the following?
Your Answer: Lymphangiectasis
Correct Answer: Bronchiectasis
Explanation:Cystic fibrosis is a life-threatening disorder that causes the build up of thick mucus in the lungs, digestive tract, and other areas of the body. It is a hereditary autosomal-recessive disease caused by mutations of the CFTR gene. Cystic fibrosis eventually results in bronchiectasis which is defined as a permanent dilatation and obstruction of bronchi or bronchioles.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 28
Incorrect
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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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 29
Correct
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In which of the following situations will fat necrosis occur?
Your Answer: Trauma to the breast
Explanation:Fat necrosis is necrosis of adipose tissue with subsequent deposition of calcium, giving it a white chalky appearance. It is seen characteristically in trauma to the breast and the pancreas with subsequent involvement of the peripancreatic fat. In the breast it may present as a palpable mass with is usually painless or as an incidental finding on mammogram. Fatty acids are released from the traumatic tissue which combine with calcium in a process known as saponification, this is an example of dystrophic calcification in which calcium binds to dead tissue. The central focus is surrounded by macrophages and neutrophils initially, followed by proliferation of fibroblasts, neovascularization and lymphocytic migration to the site of the insult.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 30
Incorrect
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What is the normal amount of oxygen that is carried in the blood?
Your Answer: 5 ml oxygen/100 ml blood
Correct Answer: 20 ml oxygen/100 ml blood
Explanation:Normally, 100 ml of blood contains 15g haemoglobin and a single gram of haemoglobin can bind to 1.34 ml oxygen when 100% saturated. Thus, 15 × 1.34 = 20 ml O2/100 ml blood. The haemoglobin in venous blood that is leaving the tissues is about 75% saturated with oxygen, and hence it carries about 15 ml O2/100 ml venous blood. This implies that for each 10 ml of blood, 5 ml oxygen is transported to the tissues. With a p(O2) > 100 mm Hg, only 3 ml of oxygen is dissolved in every one litre of plasma. By increasing the pA(O2) by breathing 100% oxygen, one can add an extra amount of oxygen in the plasma, but the amount of oxygen carried by haemoglobin will not increase significantly as it is already > 95% saturated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 31
Incorrect
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A cyclist fell and sustained a laceration to his elbow which was shortly sutured in the emergency department. Which of the following factors will aid in the wound healing process?
Your Answer: Poor tissue perfusion
Correct Answer: Presence of sutures
Explanation:Foreign bodies including sutures will delay wound healing, however due to the net affect being helpful they are used. Secondary wound infection will delay healing and is a potential post op complication. Corticosteroids depresses the wound healing ability of the body. Poor nutrition will also delay healing leading to decreased albumin, vit D and vit C. Diabetic patients with atherosclerosis with poor perfusion of tissues have notoriously delayed/poor healing.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 32
Incorrect
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A 56-year-old woman weighs 75 kg. In this patient, total body water, intracellular fluid and extracellular fluid are respectively:
Your Answer: 45 l, 35 l, 10 l
Correct Answer: 45 l, 30 l, 15 l
Explanation:The percentages of body water contained in various fluid compartments add up to total body water (TBW). This water makes up a significant fraction of the human body, both by weight and by volume. The total body water (TBW) content of humans is approximately 60% of body weight. Two-thirds is located in the intracellular and one-third in the extracellular compartment. So, in a 75-kg individual, TBW = 60 × 75/100 = 45 l. Intracellular content = 2/3 × 45 = 30 l and extracellular content = 1/3 × 45 = 15 l.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 33
Incorrect
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A 30-year old lady presented to her GP with complaints of tremors, excessive emotional outbursts, weight loss and increased sweating over 20 days. On examination, she had warm and moist skin, a fine tremor of the fingers and hyperreflexia. Her vital signs were normal. What is the likely diagnosis?
Your Answer: Alcohol withdrawal
Correct Answer: Hyperthyroidism
Explanation:Excess of circulating free thyroid hormones (thyroxine and/or triiodothyronine) leads to hyperthyroidism. Common causes include Graves’ disease, toxic thyroid adenoma and toxic multinodular goitre. Grave’s disease is the most common cause and is responsible for 70-80% cases of hyperthyroidism. Other causes include excess intake of thyroid hormone, amiodarone-related. It is important that hyperthyroidism is not confused with hyperthyroxinaemia (high levels of thyroid hormone in blood), which includes causes like thyroiditis. Both the conditions lead to thyrotoxicosis (symptoms due to hyperthyroxinemia). Symptoms include weight loss associated with increased appetite, anxiety, weakness, heat intolerance, depression, increased sweating, dyspnoea, loss of libido, diarrhoea, palpitations and occasionally arrhythmias. If there is an acute increase in metabolic rate, the condition is known as ‘thyroid storm’. Elderly sometimes present only with fatigue and weight loss and this is called apathetic hyperthyroidism. Neurological symptoms are also seen in hyperthyroidism and these are tremor, chorea, myopathy and periodic paralysis. One of the most serious complications of hyperthyroidism is stroke of cardioembolic origin due to coexisting atrial fibrillation.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 34
Incorrect
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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 reaction time of O2 with haemoglobin
Correct Answer: Is normal and due to shunted blood
Explanation:Blood that bypasses the ventilated parts of lung and enters the arterial circulation directly is known as shunted blood. It happens in normal people due to mixing of arterial blood with bronchial and some myocardial venous blood (which drains into the left heart). Diffusion limitation and reaction velocity with haemoglobin are immeasurably small. CO2 unloading will not affect the difference between alveolar and arterial p(O2). A large VSD will result in much lower arterial O2 as compared to alveolar O2.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 35
Incorrect
-
There are several mechanisms involved in the transport of sodium ions from blood to interstitial fluid of the muscle cells. Which of the following mechanisms best describes this phenomenon?
Your Answer: Active transport through endothelial cell membranes
Correct Answer: Diffusion through channels between endothelial cells
Explanation:Capillaries are the smallest of the body’s blood vessels, measuring 5–10 μm and they help to enable the exchange of water, oxygen, carbon dioxide, and many other nutrients and waste substances between the blood and the tissues surrounding them. The walls of capillaries are composed of only a single layer of cells, the endothelium. Ion channels are pore-forming proteins that help to establish and control the small voltage gradient that exists across the plasma membrane of all living cells by allowing the flow of ions down their electrochemical gradient. An ion channel is an integral membrane protein or more typically an assembly of several proteins. The archetypal channel pore is just one or two atoms wide at its narrowest point. It conducts a specific ion such as sodium or potassium and conveys them through the membrane in single file.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 36
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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 37
Incorrect
-
A cancer patient was found to have a radio resistant tumour. Which tumour does the patient most likely have?
Your Answer: Lymphoma
Correct Answer: Liposarcoma
Explanation:Liposarcoma is a cancer that arises in fat cells in deep soft tissue. Commonly it occurs inside the thigh or retroperitoneum. It usually affects middle-aged and older adults, over 40 years. Liposarcoma is the most common soft-tissue sarcoma. It is very radio resistant. Five-year survival rates vary from 100% to 56% based on histological subtype.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 38
Incorrect
-
Diuretics that act on the ascending limb of the loop of Henle produce:
Your Answer: Reduced production of erythropoietin
Correct Answer: Reduced active transport of sodium
Explanation:The loop of Henlé leads from the proximal convoluted tubule to the distal convoluted tubule. Its primary function uses a counter current multiplier mechanism in the medulla to reabsorb water and ions from the urine. It can be divided into four parts:
1. Descending limb of loop of Henlé – low permeability to ions and urea, while being highly permeable to water
2. Thin ascending limb of loop of Henlé – not permeable to water, but it is permeable to ions
3. Medullary thick ascending limb of loop of Henlé – sodium (Na+), potassium (K+) and chloride (Cl–) ions are reabsorbed by active transport. K+ is passively transported along its concentration gradient through a K+ channel in the basolateral aspect of the cells, back into the lumen of the ascending limb.
4. The cortical thick ascending limb – the site of action where loop diuretics such as furosemide block the K+/Na+/2Cl− co-transporters = reduced active transport.
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This question is part of the following fields:
- Physiology
- Renal
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Question 39
Incorrect
-
Calculate the resistance of the artery if the pressure at one end is 60 mmHg, pressure at the other end is 20 mm Hg and the flow rate in the artery is 200 ml/min.
Your Answer: 0.3
Correct Answer: 0.2
Explanation:Flow in any vessel = Effective perfusion pressure divided by resistance, where effective perfusion pressure is the mean intraluminal pressure at the arterial end minus the mean pressure at the venous end. Thus, in the given problem, resistance = (60 − 20)/200 = 0.2 mmHg/ml per min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 40
Incorrect
-
A histology report of a cervical biopsy taken from a patient with tuberculosis revealed the presence of epithelioid cells. What are these cells formed from?
Your Answer: Epithelial cells
Correct Answer: Macrophages
Explanation:Granulomas formed in tuberculosis are called tubercles and are made up polynuclear phagocytes, Langhans cells and epithelioid cells. Macrophages when enlarged, consist of abundant cytoplasm and have a tendency of arranging themselves very closely to each other representing epithelial cells. These enlarged macrophages are therefore termed as epithelioid cells.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 41
Incorrect
-
Which of the following statements is true of Colles’ fracture?
Your Answer: Involves the ulnar styloid process
Correct Answer: Is a cause of carpal tunnel syndrome
Explanation:Colles’ fracture is a distal fracture of the radius that is a known cause of carpal tunnel syndrome (compression of the median nerve in the carpal tunnel). It rarely results in ulnar nerve compression. A Colles’ fracture is extra-articular and does not extend into the wrist joint, otherwise this would make it an intra-articular fracture (Barton’s fracture). The distal fragment in a Colles’ fracture is displaced dorsally, unlike in a Smith’s fracture where the distal fragment is displaced volarly (ventrally). Associated fracture of the ulnar styloid process may occur and is a common associated injury.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 42
Incorrect
-
The accumulation of eosinophils within tissues is mostly regulated by which of the following cytokines?
Your Answer: Interleukin-1
Correct Answer: Interleukin-5
Explanation:IL-5 is produced by TH2 helper cells and by mast cells. They stimulate increased secretion of immunoglobulins and stimulate B cell growth. They are the major regulators in eosinophil activation and control. They are also released from eosinophils and mast cells in asthmatic patients and are associate with a many other allergic conditions.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 43
Incorrect
-
Which of the following morphological characteristic is a salient feature of a pure apoptotic cell?
Your Answer: Nuclear stabilisation
Correct Answer: Chromatin condensation
Explanation:Apoptosis is the programmed death of cells which occurs as a normal and controlled part of an organism’s growth or development. The changes which occur in this process include blebbing, cell shrinkage, nuclear fragmentation, chromatin condensation, chromosomal DNA fragmentation, and global mRNA decay. The cell membrane however remains intact and the dead cells are phagocytosed prior to any content leakage and thus inflammatory response.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 44
Incorrect
-
Which of the following is responsible for the maximum increase in total peripheral resistance on sympathetic stimulation?
Your Answer: Capillaries
Correct Answer: Arterioles
Explanation:Arterioles are also known as the resistance vessels as they are responsible for approximately half the resistance of the entire systemic circulation. They are richly innervated by the autonomic nervous system and hence, will bring about the maximum increase in peripheral resistance on sympathetic stimulation.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 45
Incorrect
-
Which portion of the renal tubule absorbs amino acids and glucose?
Your Answer: Distal convoluted tubule
Correct Answer: Proximal convoluted tubule
Explanation:In relation to the morphology of the kidney as a whole, the convoluted segments of the proximal tubules are confined entirely to the renal cortex. Glucose, amino acids, inorganic phosphate and some other solutes are reabsorbed via secondary active transport in the proximal renal tubule through co-transport channels driven by the sodium gradient.
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This question is part of the following fields:
- Physiology
- Renal
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Question 46
Correct
-
What is the normal duration of PR interval on an electrocardiogram of a healthy individual?
Your Answer: 0.12–0.20 s
Explanation:PR interval extends from the beginning of the P-wave until the beginning of the QRS complex. The normal duration of the PR interval is 0.12-0.20 s. It can be prolonged in first degree heart block, and reduced in Wolff-Parkinson-White syndrome.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 47
Incorrect
-
Acute respiratory distress syndrome (ARDS) is a medical condition that occurs in critically ill patients, and can be triggered by events such as trauma and sepsis. Which of the following variables is most likely to be lower than normal in a patient with ARDS?
Your Answer: Work of breathing
Correct Answer: Lung compliance
Explanation:Acute (or Adult) respiratory distress syndrome (ARDS) is a medical condition occurring in critically ill patients characterized by widespread inflammation in the lungs. The development of acute respiratory distress syndrome (ARDS) starts with damage to the alveolar epithelium and vascular endothelium, resulting in increased permeability to plasma and inflammatory cells. These cells pass into the interstitium and alveolar space, resulting in pulmonary oedema. Damage to the surfactant-producing type II cells and the presence of protein-rich fluid in the alveolar space disrupt the production and function of pulmonary surfactant, leading to micro atelectasis and impaired gas exchange. The pathophysiological consequences of lung oedema in ARDS include a decrease in lung volumes, compliance and large intrapulmonary shunts. ARDS may be seen in the setting of pneumonia, sepsis, following trauma, multiple blood transfusions, severe burns, severe pancreatitis, near-drowning, drug reactions, or inhalation injuries.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 48
Incorrect
-
During a normal respiratory exhalation, what is the recoil alveolar pressure?
Your Answer: +15 cmH2O
Correct Answer: +10 cmH2O
Explanation:To determine compliance of the respiratory system, changes in transmural pressures (in and out) immediately across the lung or chest cage (or both) are measured simultaneously with changes in lung or thoracic cavity volume. Changes in lung or thoracic cage volume are determined using a spirometer with transmural pressures measured by pressure transducers. For the lung alone, transmural pressure is calculated as the difference between alveolar (pA; inside) and intrapleural (ppl; outside) pressure. To calculate chest cage compliance, transmural pressure is ppl (inside) minus atmospheric pressure (pB; outside). For the combined lung–chest cage, transmural pressure or transpulmonary pressure is computed as pA – pB. pA pressure is determined by having the subject deeply inhale a measured volume of air from a spirometer. Under physiological conditions the transpulmonary or recoil pressure is always positive; intrapleural pressure is always negative and relatively large, while alveolar pressure moves from slightly negative to slightly positive as a person breathes.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 49
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: 100 ml
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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 50
Incorrect
-
A cell is classified on the basis of its regenerative ability. Which of the following cells represent a permanent cell?
Your Answer: Hepatocyte
Correct Answer: Erythrocyte
Explanation:An erythrocyte is the last cell in the progeny of RBC cell division and is not capable of further division and regeneration. Hepatocytes, osteocytes and epithelium of kidney tubules are all stable cells. Colonic mucosa and pluripotent hematopoietic stem cells are all labile cells.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 51
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: 10.00 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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 52
Incorrect
-
A 40 year old man suffered severe trauma following a MVA. His BP is 72/30 mmhg, heart rate of 142 beats/mins and very feeble pulse. He was transfused 3 units of blood and his BP returned to 100/70 and his heart rate slowed to 90 beats/min. What decreased after transfusion?
Your Answer: Cardiac output
Correct Answer: Total peripheral resistance
Explanation:The patient is in hypovolemic shock, he is transfused with blood, this fluid resuscitation will result in a decreased sympathetic discharge and adequate ventricular filling which will result in the decreases TPR with an increased CO and cardiac filling pressures
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 53
Incorrect
-
A 76-year-old man with a urinary tract obstruction due to prostatic hyperplasia develops acute renal failure. Which of the following physiological abnormalities of acute renal failure will be most life threatening for this patient?
Your Answer: Raised serum creatinine level
Correct Answer: Acidosis
Explanation:Acute renal failure (ARF) is a rapid loss of renal function due to damage to the kidneys, resulting in retention of nitrogenous (urea and creatinine) and non-nitrogenous waste products that are normally excreted by the kidney. This accumulation may be accompanied by metabolic disturbances, such as metabolic acidosis and hyperkalaemia, changes in body fluid balance and effects on many other organ systems. Metabolic acidosis and hyperkalaemia are the two most serious biochemical manifestations of acute renal failure and may require medical treatment with sodium bicarbonate administration and antihyperkalaemic measures. If not appropriately treated these can be life-threatening. ARF is diagnosed on the basis of characteristic laboratory findings, such as elevated blood urea nitrogen and creatinine, or inability of the kidneys to produce sufficient amounts of urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 54
Incorrect
-
Which of the following is not seen as a complication of wound healing?
Your Answer: Proud flesh
Correct Answer: Malignancy
Explanation:Complications in wound healing can originate due to abnormalities in the repair process. These abnormalities are :
1) Deficient scar formation: insufficient granulation tissue can lead to wound dehiscence and ulceration. Dehiscence or wound rupture is seen most commonly in abdominal surgery due to increased intraabdominal pressure. Ulcerations are common in extremity wounds due to inadequate vascularization.
2) Excessive formation of repair components: collagen being laid down may begin normally however later lead to a raised scar also called a hypertrophic scar, which can extend beyond its boundaries to form a keloid and
3) Formation of contractures.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 55
Incorrect
-
Which tumour occurs in young adults, affecting the epiphyses of the bones and sometimes extending to the soft tissues?
Your Answer: Osteoid osteoma
Correct Answer: Benign giant-cell tumour
Explanation:Benign giant-cell tumours tend to affect adults in their twenties and thirties, occur in the epiphyses and can erode the bone and extend into the soft tissues. These tumours have a strong tendency to recur.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 56
Incorrect
-
During cardiac catheterisation in a 20-year old man, the following data is obtained: Pressure (mmHg), O2 saturation (%) Right atrium 7 (N = 5) 90 (N = 75), Right ventricle 35/7 (N = 25/5) 90 (N = 75), Pulmonary artery 35/8 (N = 25/15), 90 (N = 75), Left atrium 7 (N = 9) 95 (N = 95), Left ventricle 110/7 (N = 110/9) 95 (N = 95), Aorta 110/75 (N = 110/75) 95 (N = 95) where N = Normal value. What is the likely diagnosis?
Your Answer: Patent ductus arteriosus
Correct Answer: Atrial septal defect
Explanation:A congenital heart disease, ASD or atrial septal defect leads to a communication between the right and left atria due to a defect in the interatrial septum. This leads to mixing of arterial and venous blood from the right and left side of the heart. The hemodynamic significance of this defect depends on the presence of shunting of blood. Normally, the left side of the heart has higher pressure than the right as the left side has to pump blood throughout the body. A large ASD (> 9 mm) will result in a clinically significant left-to-right shunt, causing volume overload of the right atrium and ventricle, eventually leading to heart failure. Cardiac catheterization would reveal very high oxygen saturation in the right atrium, right ventricle and pulmonary artery. Eventually, the left-to-right shunt will lead to pulmonary hypertension and increased afterload in the right ventricle, along with the increased preload due to the shunted blood. This will either cause right ventricular failure, or raise the pressure in the right side of the heart to equal or more than that in the left. Elevation of right atrial pressure to that of left atrial pressure would thus lead to diminishing or complete cessation of the shunt. If left uncorrected, there will be reversal of the shunt, known as Eisenmenger syndrome, resulting in clinical signs of cyanosis as the oxygen-poor blood form right side of the heart will mix with the blood in left side and reach the peripheral vascular system.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 57
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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 58
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: Endocardial fibroelastosis
Correct Answer: Tetralogy of Fallot
Explanation:The most common congenital cyanotic heart disease and the most common cause of blue baby syndrome, Tetralogy of Fallot shows four cardiac malformations occurring together. These are ventricular septal defect (VSD), pulmonary stenosis (right ventricular outflow obstruction), overriding aorta (degree of which is variable), and right ventricular hypertrophy. The primary determinant of severity of disease is the degree of pulmonary stenosis. Tetralogy of Fallot is seen in 3-6 per 10,000 births and is responsible for 5-7% congenital heart defects, with slightly higher incidence in males. It has also been associated with chromosome 22 deletions and DiGeorge syndrome. It gives rise to right-to-left shunt leading to poor oxygenation of blood. Primary symptom is low oxygen saturation in the blood with or without cyanosis at birth of within first year of life. Affected children ay develop acute severe cyanosis or ‘tet spells’ (sudden, marked increase in cyanosis, with syncope, and may result in hypoxic brain injury and death). Other symptoms include heart murmur, failure to gain weight, poor development, clubbing, dyspnoea on exertion and polycythaemia. Chest X-ray reveals characteristic coeur-en-sabot (boot-shaped) appearance of the heart. Treatment consists of immediate care for cyanotic spells and Blalock–Taussig shunt (BT shunt) followed by corrective surgery.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 59
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: Work of breathing
Correct Answer: Compliance of the lung
Explanation:Acute or adult respiratory distress syndrome (ARDS) is a reaction to several forms of lung injuries and is commonly associated with sepsis and SIRS (systemic inflammatory response syndrome), severe traumatic injury, severe head injury, narcotics overdose, drowning, pulmonary contusion, and multiple blood transfusions. There is an increase in risk due to pre-existing liver disease or coagulation abnormalities. It results due to indirect toxic effects of neutrophil-derived inflammatory mediators in the lungs. ARDS is defined by the 1994 American–European Consensus Committee as the acute onset of bilateral infiltrates on chest X-ray, a partial pressure of arterial oxygen (pa(O2)) to fraction of inspired oxygen Fi(O2) ratio of less than 200 mmHg and a pulmonary artery occlusion pressure of less than 18 or the absence of clinical evidence of left arterial hypertension. ARDS is basically pulmonary oedema in the absence of volume overload or poor left ventricular function. This is different from acute lung injury, which shows a pa(O2)/Fi(O2) ratio of less than 300 mmHg. Pathogenesis of ARDS starts from damage to alveolar epithelium and vascular endothelium, causing increased permeability. Damage to surfactant-producing type II cells disrupts the production and function of pulmonary surfactant, causing micro atelectasis and poor gas exchange. There is a decrease in lung compliance and increase in work of breathing. Eventually, there is resorption of alveolar oedema, regeneration of epithelial cells, proliferation and differentiation of type II alveolar cells and alveolar remodelling. Some show resolution and some progress to fibrosing alveolitis, which involves the deposition of collagen in alveolar, vascular and interstitial spaces.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 60
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: Decreased residual volume
Correct Answer: Increased total lung capacity
Explanation:In patients suspected of having chronic obstructive pulmonary disease, pulmonary function testing (PFT) is useful to confirm airway obstruction, to quantify its severity and reversibility, for following disease progression and monitoring response to treatment. These tests include:
FEV1 – volume of air forcefully expired during the first second after a full breath
Forced vital capacity (FVC) – total volume of air expired with maximal force and flow-volume loops. The hallmark of airway obstruction is reduction of FEV1, FVC and the ratio of FEV1/FVC, with a concave pattern in expiratory tracing on the flow-volume loop. FEV1 and forced vital capacity (FVC) are easily measured with office spirometry and are useful to assess the severity of disease. Other parameters include increased total lung capacity, functional residual capacity and residual volume, which can help distinguish chronic obstructive pulmonary disease (COPD) from restrictive pulmonary disease where these values are lower than normal. Along with these, other tests are decreased vital capacity; and decreased diffusing capacity for carbon monoxide (DLCO). DLCO is non-specific and can be low in other disorders that affect the pulmonary vascular bed, such as interstitial lung disease. DLCO is however useful to distinguish COPD from asthma, in which DLCO is normal or elevated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 61
Incorrect
-
Diffusion is the movement of molecules from a region of high concentration to a region of low concentration. Which of these changes will decrease the rate of diffusion of a substance?
Your Answer: An increase in temperature
Correct Answer: An increase in the molecular weight of the substance
Explanation:Unless given IV, a drug must cross several semipermeable cell membranes before it reaches the systemic circulation. Drugs may cross cell membranes by diffusion, amongst other mechanisms. The rate of diffusion of a substance is proportional to the difference in the concentration of the diffusing substance between the two sides of the membrane, the temperature of the solution, the permeability of the membrane and, in the case of ions, the electrical potential difference between the two sides of the membrane.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 62
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: 50 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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 63
Incorrect
-
Which of the following diseases affects young adults, causing pain in any bone -particularly long bones- which worsens at night, and is typically relieved by common analgesics, such as aspirin?
Your Answer: Chondrosarcoma
Correct Answer: Osteoid osteoma
Explanation:Osteoid osteoma, which tends to affect young adults, can occur in any bone but is most common in long bones. It can cause pain (usually worse at night) that is typically relieved by mild analgesics, such as non-steroidal anti-inflammatory drugs. X-ray findings include a small radiolucent zone surrounded by a larger sclerotic zone.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 64
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: 4.5 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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 65
Incorrect
-
Which of the following morphological features is most characteristic of hyaline degeneration?
Your Answer: Accumulation of lipofuscin
Correct Answer: Homogeneous, ground-glass, pink-staining appearance of cells
Explanation:The characteristic morphological features of hyaline degeneration is ground-glass, pinking staining cytoplasm with an intact cell membrane. The accumulation of lipids, calcium salts, lipofuscin and an amorphous cytoplasm with an intact cell membrane are all characteristically found in different situations.
Pyknotic nucleus and orphan Annie eye nucleus are not seen in hyaline degeneration.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 66
Incorrect
-
The blood investigations of a 30-year old man with jaundice revealed the following : total bilirubin 6.5 mg/dl, direct bilirubin 1.1 mg/dl, indirect bilirubin 5.4 mg/dl and haemoglobin 7.3 mg/dl. What is the most likely diagnosis out of the following?
Your Answer: Schistosomiasis
Correct Answer: Haemolysis
Explanation:Hyperbilirubinemia can be caused due to increased bilirubin production, decreased liver uptake or conjugation, or decreased biliary excretion. Normal bilirubin level is less than 1.2 mg/dl (<20 μmol/l), with most of it unconjugated. Elevated unconjugated bilirubin (indirect bilirubin fraction >85%) can occur due to haemolysis (increased bilirubin production) or defective liver uptake/conjugation (Gilbert syndrome). Such increases are less than five-fold usually (<6 mg/dl or <100 μmol/l) unless there is coexistent liver disease.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 67
Incorrect
-
A 33-year old lady presented to the gynaecology clinic with amenorrhoea for 6 months and a recent-onset of milk discharge from her breasts. She was not pregnant or on any medication. On enquiry, she admitted to having frequent headaches the last 4 months. Which of the following findings would you expect to see in her condition?
Your Answer: Increased serum cortisol
Correct Answer: Hyperprolactinaemia
Explanation:Excessively high levels of prolactin in the blood is called hyperprolactinaemia. Normally, prolactin levels are less than 580 mIU/l in females and less than 450 mIU/l in men. The biologically inactive macroprolactin can lead to a false high reading. However, the patient remains asymptomatic. Dopamine down-regulates prolactin whereas oestrogen upregulates it. Hyperprolactinaemia can be caused due to lack of inhibition (compression of pituitary stalk or low dopamine levels), or increased production due to a pituitary adenoma (prolactinoma). Either of these causes can lead to a prolactin level of 1000-5000 mIU/l. However, levels more than 5000mIU/l are usually associated due to an adenoma and >100,000 mIU/l are seen in macroadenomas (tumours < 1cm in diameter). Increased prolactin causes increased dopamine release from the arcuate nucleus of hypothalamus. This increased dopamine in turn, inhibits the GnRH (Gonadotrophin Releasing Hormone) thus blocking gonadal steroidogenesis resulting in the symptoms of hyperprolactinaemia. In women, it includes hypoestrogenism, anovulatory infertility, decreased or irregular menstruation or complete amenorrhoea. It can even cause production of breast milk, loss of libido, vaginal dryness and osteoporosis. In men, the symptoms include impotence, decreased libido, erectile dysfunction and infertility. In men, treatment can be delayed due to late diagnosis as they have no reliable indicator such as menstruation that might indicate a problem. Most of the male patients seek help only when headaches and visual defects start to surface.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 68
Correct
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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, Decrease
Explanation:Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. Respiratory acidosis can be acute or chronic. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg). The given problem represents acute respiratory acidosis and thus, will show a increase in arterial p(CO2) and decrease in pH.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 69
Incorrect
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A 76-year-old woman is diagnosed with diabetes mellitus after a urine test revealed she has glucosuria. Glucosuria may occur due to inadequate glucose reabsorption at:
Your Answer:
Correct Answer: Proximal convoluted tubule
Explanation:Glucose is reabsorbed almost 100% via sodium–glucose transport proteins (apical) and GLUT (basolateral) in the proximal convoluted tubule. Glycosuria or glucosuria is a condition of osmotic diuresis typical in those suffering from diabetes mellitus. Due to a lack of insulin, plasma glucose levels are above normal. This leads to saturation of receptors in the kidneys and glycosuria usually at plasma glucose levels above 11 mmol/l. Rarely, glycosuria is due to an intrinsic problem with glucose reabsorption within the kidneys (such as Fanconi syndrome), producing a condition termed renal glycosuria.
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This question is part of the following fields:
- Physiology
- Renal
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Question 70
Incorrect
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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:
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.
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This question is part of the following fields:
- Physiology
- Renal
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