-
Question 1
Correct
-
During her pregnancy, a 28-year-old lady was given an antibiotic. The neonate is born with bilateral deafness as a result of this.From the  following antibiotics, which one is most likely to cause this side effect?Â
Your Answer: Gentamicin
Explanation:Aminoglycosides cross the placenta and are linked to poisoning of the 8th cranial nerve in the foetus, as well as permanent bilateral deafness.
-
This question is part of the following fields:
- Infections
- Pharmacology
-
-
Question 2
Correct
-
A 50-year-old man presents with headaches, lethargy, hypertension, and electrolyte disturbance. A diagnosis of primary hyperaldosteronism is made.Which biochemical pictures would best support this diagnosis?
Your Answer: Hypokalaemic metabolic alkalosis
Explanation:When there are excessive levels of aldosterone outside of the renin-angiotensin axis, primary hyperaldosteronism occurs. High renin levels will lead to secondary hyperaldosteronism. The classical presentation of hyperaldosteronism when symptoms are present include:HypokalaemiaMetabolic alkalosisHypertensionNormal or slightly raised sodium levels
-
This question is part of the following fields:
- Endocrine Physiology
- Physiology
-
-
Question 3
Correct
-
A 67-year-old man complains of chest pain and goes to the emergency room. He takes several medications, including amiodarone.Which of the following is amiodarone mechanism of action?
Your Answer: Blocks Na + and K + channels and beta-adrenoreceptors in the heart
Explanation:Amiodarone is an anti-arrhythmic medication that can be used to treat both ventricular and atrial arrhythmias. It’s a class III anti-arrhythmic that works by blocking a variety of channels, including Na+ and K+ channels, as well as beta-adrenoreceptors. As a result, it slows conduction through the SA and AV nodes and prolongs phase 3 of the cardiac action potential (slowing repolarisation).
-
This question is part of the following fields:
- Cardiovascular Pharmacology
- Pharmacology
-
-
Question 4
Incorrect
-
Parathyroid hormone is released by which of the following:
Your Answer: Parafollicular cells of the thyroid gland
Correct Answer: Chief cells of the parathyroid gland
Explanation:Parathyroid hormone (PTH) is a peptide hormone synthesised by the chief cells of the parathyroid glands, located immediately behind the thyroid gland. PTH is primarily released in response to decreasing plasma [Ca2+] concentration. PTH acts to increase plasma calcium levels and decrease plasma phosphate levels.Parathyroid hormone (PTH) acts to increase calcium reabsorption in the distal tubule of the nephron (by activating Ca2+entry channels in the apical membrane and the Ca2+ATPase pump in the basolateral membrane) and increase phosphate excretion by inhibiting reabsorption in the proximal tubule of the nephron.
-
This question is part of the following fields:
- Endocrine
- Physiology
-
-
Question 5
Incorrect
-
Which of the following conditions require IV Lidocaine administration?
Your Answer: Terminating paroxysmal supraventricular tachycardia
Correct Answer: Refractory ventricular fibrillation in cardiac arrest
Explanation:IV Lidocaine is indicated in Ventricular Arrhythmias or Pulseless Ventricular Tachycardia (after defibrillation, attempted CPR, and vasopressor administration)
-
This question is part of the following fields:
- Cardiovascular
- Pharmacology
-
-
Question 6
Incorrect
-
At the start of the cardiac cycle, towards the end of diastole, all of the following statements are true EXCEPTÂ for:
Your Answer: The atrial pressure is greater than the ventricular pressure.
Correct Answer: The semilunar valves are open.
Explanation:At the start of the cardiac cycle, towards the end of diastole, the whole of the heart is relaxed. The atrioventricular (AV) valves are open because the atrial pressure is still slightly greater than the ventricular pressure. The semilunar valves are closed, as the pressure in the pulmonary artery and aorta is greater than the ventricular pressures. The cycle starts when the sinoatrial node (SAN) initiates atrial systole.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 7
Correct
-
A 55 year old man presents to ED complaining of leg weakness. Your colleague has examined the patient and suspects femoral nerve palsy. Which of the following clinical features would you LEAST expect to see on examination of this patient:
Your Answer: Weakness of hip extension
Explanation:Damage to the femoral nerve results in weakness of hip flexion and knee extension and loss of sensation over the anterior thigh and the anteromedial knee, medial leg and medial foot.
-
This question is part of the following fields:
- Anatomy
- Lower Limb
-
-
Question 8
Correct
-
A 18 year old male presents to the GP with painless asymmetrical cervical lymphadenopathy. Histological examination of a biopsied lymph node demonstrates Reed-Sternberg cells. What is the most likely diagnosis:
Your Answer: Hodgkin lymphoma
Explanation:Hodgkin’s lymphoma is a malignant tumour of the lymphatic system that is characterised histologically by the presence of Reed-Sternberg cells (multinucleated giant cells). The annual incidence of Hodgkin’s lymphoma in the UK is approximately 3 per 100,000 per year. The peak incidence is in young adults aged 20-35, and there is a slight male predominance.The following are recognised risk factors for Hodgkin’s lymphoma:Male genderAge 20-35Positive family historyEpstein-Barr virus infectionImmunosuppression including HIV infectionProlonged use of human growth hormoneMost patients present with an enlarged, but otherwise asymptomatic lymph node. The most commonly affected lymph nodes are in the supraclavicular and lower cervical areas. Other common clinical features include shortness of breath and chest discomfort secondary to mediastinal mass. Mediastinal masses are sometimes discovered as incidental findings on routine chest X-rays. Approximately 30% of patients with Hodgkin’s lymphoma develop splenomegaly.‘B’ symptoms occur in approximately 25% of patients. The ‘B’ symptoms of Hodgkin’s lymphoma are:Fever (>38ºC)Night sweatsWeight loss (>10% over 6 months)Pain after alcohol consumption is a pathognomonic sign of Hodgkin’s lymphoma, it is, however, not a ‘B’ symptom. It is rare though, only occurring in 2-3% of patients with Hodgkin’s lymphoma.The Ann Arbour clinical staging is as follows:Stage I: one involved lymph node groupStage II two involved lymph node groups on one side of the diaphragmStage III: lymph node groups involved on both sides of the diaphragmStage IV: Involvement of extra-nodal tissues, such as the liver or bone marrowDiagnosis is made by lymph node biopsy, which should be taken from a sufficiently large specimen or excisional biopsy, as opposed to a fine needle biopsy. The Reed-Sternberg cell is the most useful diagnostic feature. This is a giant cell with twin mirror-image nuclei and prominent ‘owl’s eye’ nucleoli.The Reed-Sternberg cell of Hodgkin’s LymphomaHistological typing depends upon the other cells within the diseased tissue. Nodular sclerosing is the most common type of Hodgkin’s lymphoma. Lymphocyte-depleted and lymphocyte-predominant are rare subtypes.The majority of cases can be successfully treated, and unlike many other malignancies even if the first-line treatment fails, a cure can often be achieved with second-line therapies. Stage 1 Hodgkin’s lymphoma is usually treated with radiotherapy alone, but more advanced stages require combination chemotherapy. In localised disease treated with irradiation, there is a 5-year survival rate of greater than 80%. In disseminated disease treated with chemotherapy, the 5-year survival falls to around 50%. Overall, a 5-year survival of >70% should be achieved.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 9
Incorrect
-
Where is angiotensin I primarily converted to angiotensin II:
Your Answer: Kidneys
Correct Answer: Lungs
Explanation:Angiotensin I is converted to angiotensin II by the removal of two C-terminal residues by the enzyme angiotensin-converting enzyme (ACE). This primarily occurs in the lungs, although it does also occur to a lesser degree in endothelial cells and renal epithelial cells.The main actions of angiotensin II are:Vasoconstriction of vascular smooth muscle (resulting in increased blood pressure)Vasoconstriction of the efferent arteriole of the glomerulus (resulting in an increased filtration fraction and preserved glomerular filtration rate)Stimulation of aldosterone release from the zona glomerulosa of the adrenal cortexStimulation of anti-diuretic hormone (vasopressin) release from the posterior pituitaryStimulation of thirst via the hypothalamusActs on the Na+/H+ exchanger in the proximal tubule of the kidney to stimulate Na+reabsorption and H+excretion
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 10
Incorrect
-
Cardiac myocytes contract by excitation-contraction coupling, very much like skeletal myocytes. On the other hand, calcium-induced calcium release (CICR) is a mechanism that is unique to Cardiac myocytes. The influx of calcium ions (Ca 2+) into the cell causes a 'calcium spark,' which causes more ions to be released into the cytoplasm.In CICR, which membrane protein in the sarcoplasmic reticulum is involved?Â
Your Answer: Ligand-gated calcium channel
Correct Answer: Ryanodine receptor
Explanation:Cardiac myocytes contract by excitation-contraction coupling, just like skeletal myocytes. Heart myocytes, on the other hand, utilise a calcium-induced calcium release mechanism that is unique to cardiac muscle (CICR). The influx of calcium ions (Ca2+) into the cell causes a ‘calcium spark,’ which causes more ions to be released into the cytoplasm.An influx of sodium ions induces an initial depolarisation, much as it does in skeletal muscle; however, in cardiac muscle, the inflow of Ca2+ sustains the depolarisation, allowing it to remain longer. Due to potassium ion (K+) inflow, CICR causes a plateau phase in which the cells remain depolarized for a short time before repolarizing. Skeletal muscle, on the other hand, repolarizes almost instantly.The release of Ca2+ from the sarcoplasmic reticulum is required for calcium-induced calcium release (CICR). This is mostly accomplished by ryanodine receptors (RyR) on the sarcoplasmic reticulum membrane; Ca2+ binds to RyR, causing additional Ca2+ to be released.
-
This question is part of the following fields:
- Basic Cellular Physiology
- Physiology
-
-
Question 11
Incorrect
-
A 40-year-old woman presents with a red, scaly, itchy rash around her navel that occurred after contact with a nickel belt buckle. A diagnosis of allergic contact dermatitis is made. Which type of hypersensitivity reaction is this?
Your Answer: Type I hypersensitivity reaction
Correct Answer: Type IV hypersensitivity reaction
Explanation:A type IV hypersensitivity reaction occurred in this patient. Allergic contact dermatitis is an inflammatory skin reaction occurring in response to an external stimulus, acting either as an allergen or an irritant, caused by a type IV or delayed hypersensitivity reaction. They usually take several days to develop.
-
This question is part of the following fields:
- General Pathology
- Pathology
-
-
Question 12
Correct
-
Regarding Escherichia coli, which of the following statements is INCORRECT:
Your Answer: It is a predominant member of the normal flora of the skin.
Explanation:Escherichia coli is a Gram-negative bacilli that is an important member of the intestinal flora. It is the most common cause of UTI in adults (about 70 – 95% of cases), followed by Staphylococcus saprophyticus (about 5 – 10% of cases), and an important cause of neonatal meningitis. E. coli O157 strain is implicated in the development of dysentery associated with haemolytic uraemic syndrome characterised by haemolytic anaemia, thrombocytopenia and acute renal failure.
-
This question is part of the following fields:
- Microbiology
- Pathogens
-
-
Question 13
Incorrect
-
A 22-year-old man arrives at the emergency department with a sore throat, low-grade fever, and malaise. His partner has infectious mononucleosis, which was recently diagnosed. In this situation, which of the following cells is the most proliferative:
Your Answer: Monocytes
Correct Answer: Lymphocytes
Explanation:Histologic findings in EBV infectious mononucleosis: Oropharyngeal epithelium demonstrates an intense lymphoproliferative response in the cells of the oropharynx. The lymph nodes and spleen show lymphocytic infiltration primarily in the periphery of a lymph node.Relative lymphocytosis (≥ 60%) plus atypical lymphocytosis (≥ 10%) are the characteristic findings of Epstein Barr virus (EBV) infectious mononucleosis.
-
This question is part of the following fields:
- Immune Responses
- Pathology
-
-
Question 14
Correct
-
Regarding relationships between two variables, what does a negative correlation coefficient indicate:
Your Answer: The two variables are inversely related
Explanation:A negative correlation coefficient means that the two variables are inversely related e.g. socio-economic class and mortality.
-
This question is part of the following fields:
- Evidence Based Medicine
- Statistics
-
-
Question 15
Incorrect
-
Which of the following statements regarding hookworm is FALSE:
Your Answer:
Correct Answer: Transmission of hookworm is via ingestion of contaminated food and water.
Explanation:The hookworm life cycle begins with the passage of eggs from an adult host into the stool. Hookworm eggs hatch in the soil to release larvae that mature into infective larvae. Infection is usually transmitted by larval penetration into human skin (duodenal infection may also be transmitted by the oral route). From the skin, larvae migrate into the blood vessels and are carried to the lungs, where they penetrate the pulmonary alveoli, ascend the bronchial tree to the pharynx, and are swallowed.
-
This question is part of the following fields:
- Microbiology
- Pathogens
-
-
Question 16
Incorrect
-
An analytical study is conducted to compare the risk of stroke between Ticagrelor therapy and Warfarin therapy among patients with atrial fibrillation. The following is obtained from the study:No. of patients who took Ticagrelor: 300No. of patients who took Ticagrelor and suffered a stroke: 30No. of patients who took Warfarin: 500No. of patients who took Warfarin and suffered a stroke: 20Compute for the absolute risk reduction of a stroke, with Warfarin as the standard of treatment.
Your Answer:
Correct Answer: -0.06
Explanation:Absolute risk reduction (ARR) is computed as the difference between the absolute risk in the control group (ARC) and the absolute risk in the treatment group (ART).Since Warfarin is the standard of treatment, Warfarin is considered as the control group.ARR = ARC-ARTARR = (20/500) – (30/300)ARR = -0.06This means that there is increased risk of stroke in the treatment group, which is the Ticagrelor group.
-
This question is part of the following fields:
- Evidence Based Medicine
-
-
Question 17
Incorrect
-
For which of the following infections is phenoxymethylpenicillin (penicillin V) primarily used?
Your Answer:
Correct Answer: Streptococcal tonsillitis
Explanation:Phenoxymethylpenicillin (penicillin V) is less active than benzylpenicillin but both have similar antibacterial spectrum. Because penicillin V is gastric-acid stable, it is suitable for oral administration, but should not be used for serious infections as absorption can be unpredictable and plasma concentrations can be variable. Its uses are:1. mainly for respiratory tract infections in children2. for streptococcal tonsillitis 3. for continuing treatment after one or more injections of benzylpenicillin when clinical response has begun. 4. for prophylaxis against streptococcal infections following rheumatic fever and against pneumococcal infections following splenectomy or in sickle-cell disease. It should not be used for meningococcal or gonococcal infections.
-
This question is part of the following fields:
- Infections
- Pharmacology
-
-
Question 18
Incorrect
-
The least likely feature expected to be seen in a lesion of the frontal lobe is which of the following?
Your Answer:
Correct Answer: Loss of two-point discrimination
Explanation:Lesions in different areas give rise to different symptoms. Lesions of the parietal lobe give rise to loss of two-point discrimination. Lesions to Broca’s area give rise to expressive dysphasia results from damage Lesions to the primary motor cortex give rise to contralateral weakness of the face and arm. Lesions to the prefrontal cortex give rise to personality change. Lesions to the frontal eye field give rise to conjugate eye deviation towards side of lesion.
-
This question is part of the following fields:
- Anatomy
- Central Nervous System
-
-
Question 19
Incorrect
-
Which of these structures is the smallest and deepest component of muscle connective tissue?
Your Answer:
Correct Answer: Endomysium
Explanation:There are three types of muscle:Skeletal muscleCardiac muscleSmooth muscleIndividual muscle is enveloped in a layer of dense irregular connective tissue called the epimysium. The epimysium protects the muscles from friction against bones and other muscles.Skeletal muscle is composed of muscle fibres, referred to as myofibers which is ensheathed by a wispy layer of areolar connective tissue called the endomysium. The endomysium is the smallest and deepest component of muscle connective tissue. Myofibers grouped together in bundles form fascicles, or fasciculi. These are surrounded by a type of connective tissue called the perimysium.Beneath the endomysium lies the sarcolemma, an elastic sheath with infoldings that invaginate the interior of the myofibers, particularly at the motor endplate of the neuromuscular junction.
-
This question is part of the following fields:
- Basic Cellular Physiology
- Physiology
-
-
Question 20
Incorrect
-
Through which of the following anatomical structures does an indirect inguinal hernia pass?
Your Answer:
Correct Answer: External oblique
Explanation:Inguinal hernias are subdivided into direct and indirect.An indirect hernia occurs when abdominal contents protrude through the internal inguinal ring and into the inguinal canal. This occurs lateral to the inferior epigastric vessels. The hernia contents may extend into the scrotum.A direct inguinal hernia is protrusion of abdominal contents through the transversalis fascia within Hesselbach’s triangle. The borders of Hesselbach’s triangle are the inferior epigastric vessels superolaterally, the rectus sheath medially, and inguinal ligament inferiorly.The deep (internal) inguinal ring is located above and halfway between the pubic tubercle and the anterior superior iliac spine. This serves as the entrance to the inguinal canal. The superficial (external) inguinal ring lies immediately above and medial to the pubic tubercle. This triangular opening is a defect in the external oblique aponeurosis, and forms the exit of the inguinal canal.
-
This question is part of the following fields:
- Abdomen And Pelvis
- Anatomy
-
SESSION STATS - PERFORMANCE PER SPECIALTY
