-
Question 1
Incorrect
-
When conducting an exploratory laparotomy procedure of a patient diagnosed with a bleeding ulcer of the lesser curvature of the stomach, which artery in this patient are you most likely to ligate to control the bleeding?
Your Answer: Left gastroepiploic
Correct Answer: Left gastric
Explanation:The lesser curvature of the stomach is supplied by the left gastric artery along with the right gastric artery. These two arteries are the ones to most likely be ligated if bleeding was to be stopped at the lesser curvature of the stomach. The splenic artery branches from the celiac branch and supplies the spleen. The left gastro-omental, the right and left gastroepiploic arteries supply the greater curvature.
-
This question is part of the following fields:
- Anatomy
- Basic Sciences
-
-
Question 2
Incorrect
-
A 15 year old girl is diagnosed with familial adenomatous polyposis. Which of the following is the most appropriate recommended step in management?
Your Answer: Surveillance colonoscopy every 1–2 years from the age of 16 years onwards
Correct Answer: Surveillance annual flexible sigmoidoscopy from age 13 years until age 30 years
Explanation:Answer: Surveillance annual flexible sigmoidoscopy from age 13 years until age 30 years.
Familial adenomatous polyposis (FAP) is the most common adenomatous polyposis syndrome. It is an autosomal dominant inherited disorder characterized by the early onset of hundreds to thousands of adenomatous polyps throughout the colon. If left untreated, all patients with this syndrome will develop colon cancer by age 35-40 years. In addition, an increased risk exists for the development of other malignancies.
Most patients with FAP are asymptomatic until they develop cancer. As a result, diagnosing presymptomatic patients is essential.Of patients with FAP, 75%-80% have a family history of polyps and/or colorectal cancer at age 40 years or younger.
Nonspecific symptoms, such as unexplained rectal bleeding (haematochezia), diarrhoea, or abdominal pain, in young patients may be suggestive of FAP.
In a minority of FAP families a mutation cannot be identified and so annual flexible sigmoidoscopy should be offered to at risk family members from age 13–15 years until age 30, and at three to five year intervals thereafter until age 60 years.
-
This question is part of the following fields:
- Colorectal Surgery
- Generic Surgical Topics
-
-
Question 3
Incorrect
-
A 22-year-old male is diagnosed with an intersphincteric fistula-in-ano during an examination under anaesthetic. Which is the most appropriate treatment?
Your Answer: Biopsies of rectal mucosa only
Correct Answer: Insertion of a ‘loose’ seton
Explanation:An anal fistula is an abnormal tract between the anal canal and the skin around the anus.
Anal fistulas can be classified according to their relationship with the external sphincter. A fistula may be complex, with several openings onto the perianal skin. Intersphincteric fistulas are the most common type and cross only the internal anal sphincter. Trans-sphincteric fistulas pass through both the internal and external sphincters.The aim is to drain the infected material and encourage healing.
For simple intersphincteric and low trans-sphincteric anal fistulas, the most common treatment is a fistulotomy or laying open of the fistula tract.
For high and complex (deeper) fistulas that involve more muscle, with a high risk of faecal incontinence or recurrence, surgery aims to treat the fistula and preserve sphincter-muscle function. Techniques include a 1‑stage or 2‑stage seton (suture material or rubber sling) either alone or in combination with fistulotomy, ligation of an intersphincteric fistula tract, creating a mucosal advancement flap, injecting glue or paste, or inserting a fistula plug . -
This question is part of the following fields:
- Colorectal Surgery
- Generic Surgical Topics
-
-
Question 4
Incorrect
-
A 29-year-old woman presents to the doctor complaining of cough, shortness of breath, fever and weight loss. Chest X-ray revealed bilateral hilar and mediastinal lymph node enlargement and bilateral pulmonary opacities. Non-caseating granulomas were found on histological examination. The most likely diagnosis is:
Your Answer: Tuberculosis
Correct Answer: Sarcoidosis
Explanation:Sarcoidosis is an inflammatory disease of unknown aetiology that affects multiple organs but predominantly the lungs and intrathoracic lymph nodes. Systemic and pulmonary symptoms may both be present. Pulmonary involvement is confirmed by a chest X-ray and other imaging studies. The main histological finding is the presence of non-caseating granulomas.
-
This question is part of the following fields:
- Basic Sciences
- Pathology
-
-
Question 5
Incorrect
-
A 55 year old lady underwent an uneventful appendicectomy. Two hours later, her arterial blood gas analysis on room revealed pH: 7.30, p(CO2): 53 mmHg and p(O2): 79 mmHg. What is the most likely cause of these findings?
Your Answer: Pulmonary embolus
Correct Answer: Alveolar hypoventilation
Explanation:In the given problem, there is respiratory acidosis due to hypercapnia from a low respiratory rate and/or volume (hypoventilation). Causes of hypoventilation include conditions impairing the central nervous system (CNS) respiratory drive, impaired neuromuscular transmission and other causes of muscular weakness (drugs and sedatives), along with obstructive, restrictive and parenchymal pulmonary disorders. Hypoventilation leads to hypoxia and hypercapnia reduces the arterial pH. Severe acidosis leads to pulmonary arteriolar vasoconstriction, systemic vascular dilatation, reduced myocardial contractility, hyperkalaemia, hypotension and cardiac irritability resulting in arrhythmias. Raised carbon dioxide concentration also causes cerebral vasodilatation and raised intracranial pressure. Over time, buffering and renal compensation occurs. However, this might not be seen in acute scenarios where the rise in p(CO2) occurs rapidly.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 6
Incorrect
-
A patient presented with continuous bleeding several hours after dental extraction. Which of the following findings is most often associated with clinical bleeding?
Your Answer: Factor XII deficiency
Correct Answer: Factor IX deficiency
Explanation:Factor IX deficiency, also called Haemophilia B or Christmas disease, is a disorder caused by missing or defective clotting factor IX. Deficiency of the factor IX causes irregular bleeding that can happen spontaneously or after mild trauma, surgery and dental extractions.
-
This question is part of the following fields:
- Basic Sciences
- Pathology
-
-
Question 7
Correct
-
A 32-year-old man presented with a metabolic acidosis and increased anion gap. What is the most likely cause of the changes of the anion gap in this patient?
Your Answer: Lactic acidosis
Explanation:High anion gap in metabolic acidosis is caused generally by the elevation of the levels of acids like ketones, lactate, sulphates in the body, which consume the bicarbonate ions. Other causes of a high anion gap include overdosing on salicylates, uraemia, rhabdomyolysis, hypocalcaemia, hypomagnesaemia, or ingestion of toxins such as ethylene glycol, methanol, propyl alcohol, cyanide and iron.
-
This question is part of the following fields:
- Basic Sciences
- Pathology
-
-
Question 8
Correct
-
Which of the following is the cause of flattened (notched) T waves on electrocardiogram (ECG)?
Your Answer: Hypokalaemia
Explanation:The T-wave is formed due to ventricular repolarisation. Normally, it is seen as a positive wave. It can be normally inverted (negative) in V1 (occasionally in V2-3 in African-Americans/Afro-Caribbeans). Hyperacute T-waves are the earliest ECG change of acute myocardial infarction. ECG findings of hyperkalaemia include high, tent-shaped T-waves, a small P-wave and a wide QRS complex. Hypokalaemia results in flattened (notched) T-waves, U-waves, ST-segment depression and prolonged QT interval.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 9
Incorrect
-
A brain tumour causing blockage of the hypophyseal portal system is likely to result in an increased secretion of which of the following hormones?
Your Answer: Growth hormone
Correct Answer: Prolactin
Explanation:The hypophyseal portal system links the hypothalamus and the anterior pituitary. With the help of this system, the anterior pituitary receives releasing and inhibitory hormones from the hypothalamus and regulates the action of other endocrine glands. One of the inhibitory hormones carried by this system is the prolactin-inhibitory hormone. In the absence of this hormone which might occur in case of a blockage of the system, prolactin secretion increases to about three times normal levels.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 10
Incorrect
-
A 26 year old lady slips in her house and lands on her right arm. She has anatomical snuffbox tenderness but no x-rays either at the time or subsequently have shown evidence of a scaphoid fracture. She has been immobilised in a futura splint for two weeks and is now asymptomatic. What is the best course of action?
Your Answer:
Correct Answer: Discharge with reassurance
Explanation:The College of Emergency Medicine states that the patient should be discharged with no follow-up required if the patient presents with anatomical snuffbox tenderness and plain radiographs show no fracture when done initially and after when wrist splint has been used.
A scaphoid fracture is a break of the scaphoid bone in the wrist. Symptoms generally includes pain at the base of the thumb which is worse with use of the hand. The anatomic snuffbox is generally tender and swelling may occur. Complications may include non-union of the fracture, avascular necrosis, and arthritis.
Scaphoid fractures are most commonly caused by a fall on an outstretched hand. Diagnosis is generally based on examination and medical imaging. Some fractures may not be visible on plain X-rays. In such cases a person may be casted with repeat X-rays in two weeks or an MRI or bone scan may be done.
-
This question is part of the following fields:
- Generic Surgical Topics
- Orthopaedics
-
-
Question 11
Incorrect
-
A 35-year-old aid worker becomes unwell whilst helping at the scene of a recent earthquake. He develops vomiting and soon afterwards, profuse watery diarrhoea. What is the most likely infective organism?
Your Answer:
Correct Answer: Vibrio cholera
Explanation:The passage of extremely loose and watery stools is characteristic of Vibrio cholera infection. Most of the other gastroenteric infections do not produce such watery motions.
Vibrio cholerae is a Gram-negative, comma-shaped bacterium. It’s natural habitat is brackish or saltwater. Some strains of V. cholerae cause the disease cholera.
Transmission occurs through the ingestion of contaminated water or food. Sudden large outbreaks are usually caused by a contaminated water supply. Outbreaks and endemic and sporadic cases are often attributed to raw or undercooked seafood.
Symptoms of the disease include sudden onset of effortless vomiting and profuse watery diarrhoea. Correction of fluid and electrolyte losses are the mainstay of treatment. Most cases resolve shortly afterwards, and therefore, antibiotics are not generally indicated.
-
This question is part of the following fields:
- Clinical Microbiology
- Principles Of Surgery-in-General
-
-
Question 12
Incorrect
-
An experiment is carried out to observe engulfment and phagocytosis of microbes. Strep pneumoniae are added to a solution of leukocytes with a substance added to enhance the process of phagocytosis. What is this substance?
Your Answer:
Correct Answer: Complement C3b
Explanation:C3b is cleaved from C3 complement with the help of the enzyme C3- convertase. It binds to the cell surface of the offending substance and opsonizes it. This makes it easy for the phagocytes to detect and eliminate them.
IgM does not act as an opsonin but igG does.
Selectins aid leukocytes to bind to the endothelial surfaces.
C5a is a chemo-attractant and histamine a vasodilator.
NADPH oxidises offending substance after phagocytosis within the macrophage. -
This question is part of the following fields:
- Basic Sciences
- Pathology
-
-
Question 13
Incorrect
-
Chest X-ray of a 45-year old gentleman with a week history of pleurisy showed a small pneumothorax with moderate-sized pleural effusion. Arterial blood gas analysis showed p(CO2) = 23 mmHg, p(O2) = 234.5 mmHg, standard bicarbonate = 16 mmol/l. What are we most likely dealing with?
Your Answer:
Correct Answer: Compensated respiratory alkalosis
Explanation:Normal pH with low p(CO2) and low standard bicarbonate could indicate either compensated respiratory alkalosis or a compensated metabolic acidosis. However, the history of hyperventilation for 5 days (pleurisy) favours compensated respiratory alkalosis. Compensated metabolic acidosis would have been likely in a diabetic patient with fever, vomiting and high glucose (diabetic ketoacidosis).
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 14
Incorrect
-
A 53 year old women with a history of atrial fibrillation developed an acute abdomen. On laparoscopic examination her bowels appeared to be dusky to red-purple in colour and her mesenteric veins appeared to be patent. Which of the following is most likely to occur in this situation?
Your Answer:
Correct Answer: Wet gangrene
Explanation:Infarction of the small bowel following a sudden and complete occlusion of the mesenteric artery can involve any portion of the bowel, whether small or a large. The splenic flexure is at most risk for infarction as it is the watershed area between the superior and inferior mesenteric vessels. Regardless of whether the arterial or the venous blood vessels are occluded, because of the blood reflow into the damaged portion, it will appear haemorrhagic. The bowel appearing congested at first and then becoming oedematous. If the artery is occluded then there will be a clear cut demarcation and in venous occlusion the dusky colour fades with the rest of the normal bowel. Wet gangrene is characteristic of ischaemic injury to the gut.
-
This question is part of the following fields:
- Basic Sciences
- Pathology
-
-
Question 15
Incorrect
-
When a patient that is less than 21 years of age develops a bone tumour. What is the most common benign bone tumour that would be considered in individuals below 21 years?
Your Answer:
Correct Answer: Osteochondroma
Explanation:Osteochondroma is a benign new bone growth that protrudes from the outer contour of bones and is capped by growing cartilage. Nearly 80% of these lesions are noted before the age of 21 years.
-
This question is part of the following fields:
- Basic Sciences
- Pathology
-
-
Question 16
Incorrect
-
A 5 year old boy accidentally spills boiling water over his legs and is diagnosed with superficial partial thickness burns. Which of the following is least likely to occur?
Your Answer:
Correct Answer: Damage to sweat glands
Explanation:Superficial burns are confined in the depth of the epidermis, and all dermal appendages and nerve endings are intact. Superficial burns generally heal in 3 to 5 days with minimal intervention and do not leave significant scarring as they typically heal by re epithelialization. Therefore the sweat glands will be intact.
-
This question is part of the following fields:
- Emergency Medicine And Management Of Trauma
- Principles Of Surgery-in-General
-
-
Question 17
Incorrect
-
A 64 year old man who has undergone a total hip replacement is given an infusion of packed red cells. Which of the following adverse effects is most likely to occur as a result of this treatment?
Your Answer:
Correct Answer: Pyrexia
Explanation:Acute transfusion reactions present as adverse signs or symptoms during or within 24 hours of a blood transfusion. The most frequent reactions are fever, chills, pruritus, or urticaria, which typically resolve promptly without specific treatment or complications.
-
This question is part of the following fields:
- Post-operative Management And Critical Care
- Principles Of Surgery-in-General
-
-
Question 18
Incorrect
-
What is a major source of fuel being oxidised by the skeletal muscles of a man who has undergone starvation for 7 days?
Your Answer:
Correct Answer: Serum fatty acids
Explanation:Starvation is the most extreme form of malnutrition. Prolonged starvation can lead to permanent organ damage and can be fatal. Starved individuals eventually lose significant fat and muscle mass as the body uses these for energy.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 19
Incorrect
-
A 56-year-old woman weighs 75 kg. In this patient, total body water, intracellular fluid and extracellular fluid are respectively:
Your Answer:
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.
-
This question is part of the following fields:
- Basic Sciences
- Physiology
-
-
Question 20
Incorrect
-
A 45 year old man who complains of chronic post prandial, burning epigastric pain undergoes a gastrointestinal endoscopy. There is no apparent mass or haemorrhage and a biopsy is taken from the lower oesophageal mucosa just above the gastro-oesophageal junction. The results reveal the presence of columnar cells interspersed with goblet cells. Which change best explains the above mentioned histology?
Your Answer:
Correct Answer: Metaplasia
Explanation:Metaplasia is the transformation of one type of epithelium into another as a means to better cope with external stress on that epithelium. In this case metaplasia occurs due to the inflammation resulting from gastro-oesophageal reflux disease. Dysplasia is disordered cellular growth. Hyperplasia is an increase in cell number but not cell type i.e. transformation. Carcinoma is characterized by cellular atypia. Ischaemia would result in necrosis with ulceration. Carcinoma insitu involves dysplastic atypical cells with the basement membrane intact and atrophy would mean a decrease in number of cells.
-
This question is part of the following fields:
- Basic Sciences
- Pathology
-
-
Question 21
Incorrect
-
An enlarged lymph node excised from a patient and submitted for histopathological evaluation showed well-defined, prominent paracortical follicles with germinal centres? Which of the following patients are we likely talking about?
Your Answer:
Correct Answer: A 5-year-old boy with a sore throat and runny nose
Explanation:Lymphadenopathy is common in children and is usually reactive in nature. The description fits that of a benign, reactive lymph node.
-
This question is part of the following fields:
- Basic Sciences
- Pathology
-
-
Question 22
Incorrect
-
In order to arrest sudden bleeding during pericardiectomy that started after accidental injury to a major vasculature in the pericardium; the surgeon inserted his left index finger through the transverse pericardial sinus, pulled forward on the two large vessels lying ventral to his finger, and compressed these vessels with his thumb to control the bleeding. Which vessels were these?
Your Answer:
Correct Answer: Pulmonary trunk and aorta
Explanation:Transverse pericardial sinus: located behind two great vessels (aorta and pulmonary trunk) and in front of the superior vena cava and is accessed from above as in this case. The brachiocephalic trunk is located above the pericardium and the right pulmonary artery is above the pericardial reflections.
-
This question is part of the following fields:
- Anatomy
- Basic Sciences
-
-
Question 23
Incorrect
-
A 30 year old mother presents with a tender indurated mass in her left breast which developed 6 days after giving birth and breastfeeding the baby. Which organism is most likely to have caused this infection?
Your Answer:
Correct Answer: Staphylococcus aureus
Explanation:Answer: Staphylococcus aureus
Mastitis is defined as inflammation of the breast. Although it can occur spontaneously or during lactation, this discussion is limited to mastitis in breastfeeding women, with mastitis defined clinically as localized, painful inflammation of the breast occurring in conjunction with flu-like symptoms (e.g., fever, malaise). Mastitis is especially problematic because it may lead to the discontinuation of breast-feeding, which provides optimal infant nutrition. Mastitis occurs in approximately 10 percent of U.S. mothers who are breastfeeding, and it can lead to the cessation of breastfeeding. The risk of mastitis can be reduced by frequent, complete emptying of the breast and by optimizing breastfeeding technique. Sore nipples can precipitate mastitis. The differential diagnosis of sore nipples includes mechanical irritation from a poor latch or infant mouth anomalies, such as cleft palate or bacterial or yeast infection. The diagnosis of mastitis is usually clinical, with patients presenting with focal tenderness in one breast accompanied by fever and malaise. Treatment includes changing breastfeeding technique, often with the assistance of a lactation consultant. When antibiotics are needed, those effective against Staphylococcus aureus (e.g., dicloxacillin, cephalexin) are preferred. As methicillin-resistant S. aureus becomes more common, it is likely to be a more common cause of mastitis, and antibiotics that are effective against this organism may become preferred. Continued breastfeeding should be encouraged in the presence of mastitis and generally does not pose a risk to the infant. Breast abscess is the most common complication of mastitis. It can be prevented by early treatment of mastitis and continued breastfeeding. Once an abscess occurs, surgical drainage or needle aspiration is needed. Breastfeeding can usually continue in the presence of a treated abscess.
-
This question is part of the following fields:
- Clinical Microbiology
- Principles Of Surgery-in-General
-
-
Question 24
Incorrect
-
A 57 year old woman, known case of idiopathic thrombocytopenic purpura is scheduled to undergo a splenectomy. Her platelet count has dropped to 50. Which of the following is the optimal timing for a platelet transfusion in this case?
Your Answer:
Correct Answer: After ligation of the splenic artery
Explanation:The timing of platelet transfusion in a patient of ITP undergoing splenectomy should be carefully monitored, it should neither be done too soon or too late. If done too soon it will be rendered ineffective, if too late it will result in unnecessary bleeding. The ideal time for a platelet transfusion would be when the splenic artery has been ligated.
-
This question is part of the following fields:
- Peri-operative Care
- Principles Of Surgery-in-General
-
-
Question 25
Incorrect
-
Question 26
Incorrect
-
A 45 years old women was diagnosed with breast cancer. After a radical mastectomy to remove all axillary lymph nodes from her right breast, it was noticed that she had a winged scapula. Which nerve injury would explain this?
Your Answer:
Correct Answer: Long thoracic nerve
Explanation:The long thoracic nerve supplies the serratus anterior muscle. When this nerve is injured the muscle undergoes paralysis, which is seen clinically as winging of the scapula most apparent when the arm is lifted forward. The long thoracic nerve is greatly susceptible to injury during breast surgery because of its long and superficial course along the thorax from its origin, the brachial plexus, to the lower border of the serratus anterior muscle.
-
This question is part of the following fields:
- Anatomy
- Basic Sciences
-
-
Question 27
Incorrect
-
A 40 year old woman, who is under anaesthesia for an elective procedure, received an antibiotic injection. She immediately developed a rash and her airway constricted raising the airway pressure. Which mechanism is responsible for this reaction?
Your Answer:
Correct Answer: Type I hypersensitivity
Explanation:Type I hypersensitivity reaction is an immediate reaction that occurs due to binding of the antigen with antibodies attached to mast cells in a previously sensitized person. It has an immediate phase, which is characterised by vasodilation, leakage of plasma, smooth muscle spasm, or glandular secretions. This manifests in about 5-30 min and usually resolves within 60 mins. The delayed phase follows after 24 hours and can persist up to several days. It is due to infiltration of eosinophils, neutrophils, basophils and CD4+ cells and leads to tissue destruction. The nature of the reaction varies according to the site. It can take the form of skin allergy, hives, allergic rhinitis, conjunctivitis, bronchial asthma or food allergy.
-
This question is part of the following fields:
- Basic Sciences
- Pathology
-
-
Question 28
Incorrect
-
A 28-year-old male injured his ankle playing football. On examination, he has tenderness over both medial and lateral malleoli. X-ray demonstrates a bimalleolar fracture with a displaced distal fibula fracture, at the level of the syndesmosis and fracture of the medial malleolus with talar shift. The ankle has been provisionally reduced and splinted in the emergency department. What is the most appropriate management?
Your Answer:
Correct Answer: Surgical fixation
Explanation:The patient has Denis B fracture, unstable fracture requiring open reduction and internal fixation (ORIF)
open reduction internal fixation indications:
– any talar displacement
– displaced isolated medial malleolar fracture
– displaced isolated lateral malleolar fracture
– bimalleolar fracture and bimalleolar-equivalent fracture
– posterior malleolar fracture with > 25% or > 2mm step-off
– Bosworth fracture-dislocations
– open fractures
– malleolar non-unionDanis-Weber classification:
type A
below the level of the talar dome
usually transverse
tibiofibular syndesmosis intact
deltoid ligament intact
medial malleolus occasionally fractured
usually stable if medial malleolus intact
type B
the distal extent at the level of the talar dome; may extend some distance proximally
usually spiral
tibiofibular syndesmosis usually intact, but a widening of the distal tibiofibular joint (especially on stressed views) indicates syndesmotic injury
medial malleolus may be fractured
the deltoid ligament may be torn, indicated by a widening of the space between the medial malleolus and talar dome
variable stability, dependent on the status of medial structures (malleolus/deltoid ligament) and syndesmosis; may require ORIF
Weber B fractures could be further subclassified as 9
B1: isolated
B2: associated with a medial lesion (malleolus or ligament)
B3: associated with a medial lesion and fracture of the posterolateral tibia
type C
above the level of the ankle joint
tibiofibular syndesmosis disruption with a widening of the distal tibiofibular articulation
medial malleolus fracture or deltoid ligament injury often present
a fracture may arise as proximally as the level of the fibular neck and not visualized on ankle films, requiring knee or full-length tibia-fibula radiographs (Maisonneuve fracture)
unstable: usually requires ORIF
Weber C fractures can be further subclassified as 6
C1: diaphyseal fracture of the fibula, simple
C2: diaphyseal fracture of the fibula, complex
C3: proximal fracture of the fibula
a fracture above the syndesmosis results from external rotation or abduction forces that also disrupt the joint
usually associated with an injury to the medial side -
This question is part of the following fields:
- Generic Surgical Topics
- Orthopaedics
-
-
Question 29
Incorrect
-
A 44 year old actor presents with an attack of mild acute pancreatitis. Imaging identifies gallstones but a normal calibre bile duct, and a peripancreatic fluid collection. Which management option would be the most appropriate?
Your Answer:
Correct Answer: Cholecystectomy once the attack has settled
Explanation:Pancreatitis is inflammation of the pancreas with variable involvement of regional tissues or remote organ systems. Acute pancreatitis (AP) is characterized by severe pain in the upper abdomen and elevation of pancreatic enzymes in the blood. In the majority of patients,
Biliary pancreatitis should always be treated eventually with a cholecystectomy after the process has subsided.
Feeding should be introduced enterally as the patient’s anorexia and pain resolves.
The use of nasogastric aspiration offers no clear advantage in patients with mild AP, but is beneficial in patients with profound pain, severe disease, paralytic ileus, and intractable vomiting.
AP is a mild, self-limiting disease that resolves spontaneously without complications. Patients can be initiated on a low-fat diet initially and need not invariably start their dietary advancement using a clear liquid diet. Systematic reviews and meta-analyses have shown that administration of enteral nutrition may reduce mortality and infectious complications compared with parenteral nutrition. Although the ideal timing to initiate enteral feeding remains undetermined, administration within 48 hours appears to be safe and tolerated. -
This question is part of the following fields:
- Generic Surgical Topics
- Hepatobiliary And Pancreatic Surgery
-
-
Question 30
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:
Correct Answer: Increased total lung capacity
Explanation:In patients suspected of having chronic obstructive pulmonary disease, pulmonary function testing (PFT) is useful to confirm airway obstruction, to quantify its severity and reversibility, for following disease progression and monitoring response to treatment. These tests include:
FEV1 – volume of air forcefully expired during the first second after a full breath
Forced vital capacity (FVC) – total volume of air expired with maximal force and flow-volume loops. The hallmark of airway obstruction is reduction of FEV1, FVC and the ratio of FEV1/FVC, with a concave pattern in expiratory tracing on the flow-volume loop. FEV1 and forced vital capacity (FVC) are easily measured with office spirometry and are useful to assess the severity of disease. Other parameters include increased total lung capacity, functional residual capacity and residual volume, which can help distinguish chronic obstructive pulmonary disease (COPD) from restrictive pulmonary disease where these values are lower than normal. Along with these, other tests are decreased vital capacity; and decreased diffusing capacity for carbon monoxide (DLCO). DLCO is non-specific and can be low in other disorders that affect the pulmonary vascular bed, such as interstitial lung disease. DLCO is however useful to distinguish COPD from asthma, in which DLCO is normal or elevated. -
This question is part of the following fields:
- Basic Sciences
- Physiology
-
SESSION STATS - PERFORMANCE PER SPECIALTY
