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Question 1
Correct
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In a patient with an ongoing seizure, after what time period should treatment be commenced?
Your Answer: 5 minutes
Explanation:Immediate emergency care and treatment should be given to children, young people and adults who have prolonged or repeated convulsive seizures.
Prolonged seizures last 5 minutes or more.
Repeated seizures refer to 3 or more seizures in an hour. -
This question is part of the following fields:
- Central Nervous System
- Pharmacology
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Question 2
Correct
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Hepatitis A is transmitted by which of the following routes:
Your Answer: Faecal-oral route
Explanation:Hepatitis A transmission is by the faecal-oral route; the virus is excreted in bile and shed in the faeces of infected people.
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This question is part of the following fields:
- Microbiology
- Pathogens
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Question 3
Correct
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Question 4
Correct
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Regarding dermatophytes, which of the following statement is CORRECT:
Your Answer: Diagnosis is made from microscopy and culture of skin scrapings, hair samples or nail clippings.
Explanation:Diagnosis is made from microscopy and culture of skin scrapings, hair samples or nail clippings depending on the site of infection. The lesions of ringworm typically have a dark outer ring with a pale centre. Tinea capitis is ringworm affecting the head and scalp. Spread is via direct skin contact. Treatment is usually topical, oral antifungals are reserved for refractory infection.
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This question is part of the following fields:
- Microbiology
- Pathogens
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Question 5
Correct
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A 28-year-old female arrives after taking an unknown chemical in excess. She is tired and her speech is slurred. The following are her observations and results:HR 118,BP 92/58,SaO2 96%
HR 118, 11/15 The following are the results of his arterial blood gas (ABG):
pH: 7.24pO 2 : 9.4kPa PCO2 : 3.3kPa HCO 3 -: 22 mmol/lNa + : 143 mmol/lCl – : 99 mmol/lLactate: 5 IU/l
Which of the following statements about this patient is TRUE?Your Answer: Her anion gap is elevated
Explanation:The interpretation of arterial blood gas (ABG) aids in the measurement of a patient’s pulmonary gas exchange and acid-base balance.
The normal values on an ABG vary a little depending on the analyser, but they are roughly as follows:
Variable
Range
pH
7.35 – 7.45
PaO2
10 – 14 kPa
PaCO2
4.5 – 6 kPa
HCO3-
22 – 26 mmol/l
Base excess
-2 – 2 mmol/lThe patient’s history indicates that she has taken an overdose in this case. Because her GCS is 11/15 and she can communicate with slurred speech, she is clearly managing her own airway, there is no current justification for intubation.
The following are the relevant ABG findings:
Hypoxia (mild)
pH has been lowered (acidaemia)
PCO2 levels are low.
bicarbonate in its natural state
Lactate levels have increasedThe anion gap represents the concentration of all the unmeasured anions in the plasma. It is the difference between the primary measured cations and the primary measured anions in the serum. It can be calculated using the following formula:
Anion gap = [Na+] – [Cl-] – [HCO3-]The reference range varies depending on the technique of measurement, but it is usually between 8 and 16 mmol/L.
The following formula can be used to compute her anion gap:
Anion gap = [143] – [99] – [22]
Anion gap = 22As a result, it is clear that she has a metabolic acidosis with an increased anion gap.
The following are some of the causes of type A and type B lactic acidosis:
Type A lactic acidosis
Type B lactic acidosis
Shock (including septic shock)
Left ventricular failure
Severe anaemia
Asphyxia
Cardiac arrest
CO poisoning
Respiratory failure
Severe asthma and COPD
Regional hypoperfusion
Renal failure
Liver failure
Sepsis (non-hypoxic sepsis)
Thiamine deficiency
Alcoholic ketoacidosis
Diabetic ketoacidosis
Cyanide poisoning
Methanol poisoning
Biguanide poisoning -
This question is part of the following fields:
- Physiology
- Renal Physiology
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Question 6
Incorrect
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A 21-year-old student presents with fever, headache, malaise, fatigue, and muscle aches after returning from a trip to India. A diagnosis of malaria was suspected.
Which of the following statements is considered correct regarding malaria?Your Answer: Plasmodium falciparum has the longest incubation period
Correct Answer: Haemoglobinuria and renal failure following treatment is suggestive of Plasmodium falciparum
Explanation:Malaria results from infection with single-celled parasites belonging to the Plasmodium genus. Five species of Plasmodium are known to cause disease in humans: P. falciparum, P. vivax, P. ovale, P. malariae, and P. knowlesi.
Chloroquine remains the mainstay of treatment for uncomplicated vivax malaria.
The female Anopheles mosquito serves as the biologic vector and definitive host.
A complication of infection with P. falciparum is blackwater fever, a condition characterized by haemoglobinuria.
Plasmodium ovale has the longest incubation period, which can be up to 40 days. Plasmodium falciparum has a shorter incubation period of 7-14 days.
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This question is part of the following fields:
- Microbiology
- Specific Pathogen Groups
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Question 7
Correct
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You examine a 72-year-old man who has recently begun bumetanide treatment for worsening heart failure.
Which of these statements about bumetanide is correct?Your Answer: It has better intestinal absorption than furosemide
Explanation:Bumetanide is a loop diuretic that is used to treat congestive heart failure. It is frequently used in patients who have failed to respond to high doses of furosemide.
It has a potency of about 40 times that of furosemide, with a 1 mg dose being roughly equivalent to a 40 mg dose of furosemide.
Seizures are not known to be triggered by bumetanide. In fact, it lowers the concentration of neuronal chloride, making GABA’s action more depolarizing, and it’s currently being tested as an antiepileptic in the neonatal period.It takes effect after 1 hour of oral administration, and diuresis takes 6 hours to complete.
Bumetanide absorbs much better in the intestine than furosemide. Because it has a higher bioavailability than furosemide, it is commonly used in patients with gut oedema.
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This question is part of the following fields:
- Cardiovascular Pharmacology
- Pharmacology
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Question 8
Correct
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The pelvic bone is formed by which of the following:
Your Answer: Ilium, ischium and pubis
Explanation:Each pelvic bone is formed by three elements: the ilium (superiorly), the pubis (anteroinferiorly) and the ischium (posteroinferiorly).
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This question is part of the following fields:
- Abdomen
- Anatomy
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Question 9
Correct
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You suspect that your patient with polycystic kidney disease has developed a berry aneurysm as a complication of his disease. The patient complains of a sudden, severe headache. You are guessing subarachnoid haemorrhage secondary to a ruptured berry aneurysm as the cause of his severe headaches. What is the most likely location of his aneurysm?
Your Answer: Anterior communicating artery
Explanation:One of the complications that polycystic kidney disease may cause is the development of a brain aneurysm.
A berry aneurysm is the most common type of brain aneurysm.
The Circle of Willis, where the major blood vessels meet at the base of the brain, is where it usually appears. The most common junctions of the Circle of Willis where an aneurysm may occur include the anterior communicating artery (35%), internal carotid artery (30%), the posterior communicating artery and the middle cerebral artery (22%), and finally, the posterior circulation sites, most commonly the basilar artery tip.
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This question is part of the following fields:
- Anatomy
- Central Nervous System
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Question 10
Incorrect
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A 70-year-old patient presents with a chronic digoxin overdose. She has vomited several times, is extremely tired, and her vision 'appears yellow,' according to her.
Which of the following is a reason for this patient's use of Digifab?Your Answer: Digoxin level of 3.5 ng/ml
Correct Answer: Coexistent renal failure
Explanation:An antidote for digoxin overdose is digoxin-specific antibody (Digifab). It’s a lyophilized preparation of digoxin-immune ovine Fab immunoglobulin fragments that’s sterile, purified, and lyophilized. These fragments were extracted from the blood of healthy sheep that had been immunised with a digoxin derivative called digoxin-dicarboxymethoxylamine (DDMA), a digoxin analogue that contains the functionally important cyclopentaperhydrophenanthrene:lactone ring moiety coupled to keyhole limpet hemocyanin (KLH).
Digifab has a higher affinity for digoxin than digoxin does for its sodium pump receptor, which is thought to be the site of its therapeutic and toxic effects. When given to a patient who is intoxicated, Digifab binds to digoxin molecules, lowering free digoxin levels and shifting the equilibrium away from binding to receptors, reducing cardio-toxic effects. The kidney and reticuloendothelial system then clear the Fab-digoxin complexes.
The following summarises the indications for Digifab in acute and chronic digoxin toxicity:
Acute digoxin toxicity
Chronic digoxin toxicity
Cardiac arrest
Life-threatening arrhythmia
Potassium level > 5 mmol/l
>10 mg digoxin ingested (adult)
>4 mg digoxin ingested (child)
Digoxin level > 12 ng/ml
Cardiac arrest
Life-threatening arrhythmia
Significant gastrointestinal symptoms
Symptoms of digoxin toxicity -
This question is part of the following fields:
- Cardiovascular Pharmacology
- Pharmacology
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Question 11
Correct
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The proximal convoluted tubule (PCT) is the first part of the renal tubule and lies in the renal cortex. The bulk of reabsorption of solute occurs is the PCT and 100% of glucose is reabsorbed here.
Which of the following is the mechanism of glucose reabsorption in the PCT?Your Answer: Secondary active transport
Explanation:Glucose reabsorption occurs exclusively in the proximal convoluted tubule by secondary active transport through the Na.Glu co-transporters, driven by the electrochemical gradient for sodium.
The co-transporters transport two sodium ions and one glucose molecule across the apical membrane, and the glucose subsequently crosses the basolateral membrane by facilitated diffusion. -
This question is part of the following fields:
- Physiology
- Renal Physiology
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Question 12
Incorrect
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Which of the following is not true regarding the structure and function of glomerular filtration membrane?
Your Answer: Fenestrated capillaries give greater permeability
Correct Answer: The absence of a basement membrane reduces impedance to filtration
Explanation:The glomerular filtration membrane is composed of fenestrated capillary endothelium, basement membrane, and filtration slits. It is an organized, semipermeable membrane preventing the passage of most of the proteins into the urine.
The anatomical arrangement of the glomerular filtration membrane maximizes the surface area available for filtration. The arrangement of its arterioles results in high hydrostatic pressures and facilitates filtration.
Fenestrated capillary endothelium of the glomerular filtration membrane is with relatively large pores. It allows the free movement of plasma proteins and solutes but still restricts the movement of blood cells.
Filtration slits are the smallest filters and restrict the movement of plasma proteins but still allow free movement of ions and nutrients.
The glomerular basement membrane (GBM) is a critical component of the glomerular filtration membrane. Thus, it is not true that its absence will reduce the resistance of flow. The basement membrane is true to be more selective and contains negatively charged glycoproteins. However, it still allows free passage of water, nutrients, and ions. Severe structural abnormalities of the GBM can result in protein (albumin) leakage.
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This question is part of the following fields:
- Physiology
- Renal Physiology
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Question 13
Correct
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In patients who are not at risk of hypercapnic respiratory failure but are requiring oxygen, the oxygen saturations should be maintained at which of the following?
Your Answer: 94 - 98%
Explanation:Oxygen saturation should be 94 – 98% in most acutely ill patients with a normal or low arterial carbon dioxide (PaCO2). In some clinical situations, however, like cardiac arrest and carbon monoxide poisoning, it is more appropriate to aim for the highest possible oxygen saturation until the patient is stable.
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This question is part of the following fields:
- Pharmacology
- Respiratory
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Question 14
Incorrect
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A 68-year-old female has presented to the Emergency Department with chest pain, palpitations, and breathlessness complaints. On ECG, she is diagnosed with ventricular arrhythmia and is administered lidocaine.
Which of the following is the correct mechanism of action of lidocaine?Your Answer: Inhibits the Na/K ATPase in cardiac myocytes
Correct Answer: Blocks Na+ channels in the heart
Explanation:Lidocaine is a tertiary amide local anaesthetic and also a class IV antiarrhythmic.
Like other local anaesthetics, lidocaine works on the voltage-gated sodium ion channel on the nerve cell membranes. It works in the following steps:
1. diffuses through neural sheaths and the axonal membrane into the axoplasm
2. binds fast voltage-gated Na+ channels in the neuronal cell membrane and inactivates them
3. With sufficient blockage, the membrane of the postsynaptic neuron will not depolarise and will be unable to transmit an action potential, thereby preventing the transmission of pain signalsThe same principle applies to Lidocaine’s actions in the heart as it blocks the sodium channels in the conduction system and the myocardium. This raises the threshold for depolarizing, making it less likely for the heart to initiate or conduct any action potential that can cause arrhythmia.
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This question is part of the following fields:
- Cardiovascular Pharmacology
- Pharmacology
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Question 15
Correct
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A 55-year-old female is urgently rushed into the Emergency Department as she complains of chest pain that is worse on breathing, shortness of breath, palpitations, and haemoptysis.
She undergoes a CT pulmonary angiogram, which reveals a large pulmonary embolus. She is immediately started on heparin and shifted to the acute medical ward.
Which of the following does heparin activate?Your Answer: Antithrombin III
Explanation:Heparin works by binding to and activating the enzyme inhibitor antithrombin III.
Antithrombin III inactivates thrombin (factor IIa) by forming a 1:1 complex with thrombin. The heparin-antithrombin III complex also inhibits factor Xa and some other proteases involved with clotting. The heparin-ATIII complex can also inactivate IX, XI, XII, and plasmin.
Heparin is a polymer of glycosaminoglycan. It occurs naturally and is found in mast cells. Clinically, it is used in two forms:
1. Unfractionated: widely varying polymer chain lengths
2. Low molecular weight: Smaller polymers onlyHeparin is not thrombolytic or fibrinolytic. It prevents the progression of existing clots by inhibiting further clotting. The lysis of existing clots relies on endogenous thrombolytics.
Heparin is used for:
1. Prevention and treatment of venous thromboembolism
2. Treatment of disseminated intravascular coagulation
3. Treatment of fat embolism
4. Priming of haemodialysis and cardiopulmonary bypass machines -
This question is part of the following fields:
- Cardiovascular Pharmacology
- Pharmacology
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Question 16
Correct
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A 35-year-old man presents with haemoptysis, night sweats, and weight loss. Further examinations were done and a diagnosis of tuberculosis is suspected.
Which of the following statements is considered correct regarding Mycobacterium tuberculosis?Your Answer: It is impervious to decolourisation with acid
Explanation:Mycobacterium tuberculosis are part of the Mycobacteriaceae family. They are described to have the characteristics of a Gram-positive cell wall but they are not easily stained with Gram stain. This is because their cell wall contains a high lipid content, and this lipid allows the Mycobacteria to bind to alkaline stains with the application and help of heat. Once stained, they are able to resist decolorization even with the use of acid alcohol as the decolourizer, making them very difficult to decolorize, that is why they are known to be acid-fast.
The Ghon complex is a non-pathognomonic radiographic finding on a chest x-ray that is significant for pulmonary infection of tuberculosis. The location of the Ghon’s focus is usually subpleural and predominantly in the upper part of the lower lobe and lower part of the middle or upper lobe.
Skeletal tuberculosis of the spine is referred to as Pott disease.
The risk of reactivation TB is about 3.3% during the first year after a positive PPD skin test and a total of 5% to 15% thereafter in the person’s lifetime. Progression from infection to active disease varies with age and the intensity and duration of exposure. Reactivation TB occurs when there is an alteration or suppression of the cellular immune system in the infected host that favours
replication of the bacilli and progression to disease. -
This question is part of the following fields:
- Microbiology
- Specific Pathogen Groups
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Question 17
Correct
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Which of the following is NOT a typical clinical feature of hypoglycaemia:
Your Answer: Polyuria
Explanation:Clinical features of hypoglycaemia:
Autonomic symptoms: Sweating, feeling hot, anxiety/agitation, palpitations, shaking, paraesthesia, dizziness
Neuroglycopaenic symptoms: Weakness, blurred vision, difficulty speaking, poor concentration, poor coordination, drowsiness, confusion, seizures, coma
Other symptoms: Nausea, fatigue, hunger -
This question is part of the following fields:
- Endocrine
- Physiology
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Question 18
Incorrect
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Among the following microorganisms, which is considered to be transmitted by invasion of intact skin?
Your Answer: HIV
Correct Answer: Leptospira spp.
Explanation:Rodents and domestic animals are the primary reservoirs for the Leptospira spp, although other animals, including cows, horses, mongooses, and frogs, can also harbour the leptospires. Humans may be directly infected from animal urine or indirectly by contact with soil or water that is contaminated with urine from infected animals. Infected humans can shed leptospires in urine for up to 11 months, infected cows for 3.5 months, infected dogs for 4 years, and infected rodents possibly for their entire lifetime.
The organisms enter the host through mucous membranes or abraded skin. The incubation period ranges from 5 to 14 days.
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This question is part of the following fields:
- Microbiology
- Principles Of Microbiology
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Question 19
Incorrect
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A 60-year-old female has a past medical history of diverticular disease. She now presents in the clinic with crampy abdominal pain. The nurse at the triage suggests prescribing hyoscine butyl bromide to help relieve the abdominal pain. However, after administering this treatment, the patient develops a side-effect to the medication.
What side-effect of using hyoscine butyl bromide is she MOST likely to develop out of the following?Your Answer: Diarrhoea
Correct Answer: Dry mouth
Explanation:Hyoscine butylbromide is an antispasmodic drug that blocks muscarinic receptors and reduces intestinal motility. It is used for gastrointestinal and genitourinary smooth muscle spasms and symptomatic relief of IBS.
It has the following side-effects:
1. Constipation
2. Dizziness
3. Drowsiness
4. Dry mouth
5. Dyspepsia
6. Flushing
7. Headache
8. Nausea and vomiting
9. Palpitations
10. Skin reactions
11. Tachycardia
12. Urinary disorders
13. Disorders of vision -
This question is part of the following fields:
- Gastrointestinal Pharmacology
- Pharmacology
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Question 20
Incorrect
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Gastrin is secreted by which of the following cell types in the stomach:
Your Answer: Parietal cells
Correct Answer: G-cells
Explanation:Gastrin is secreted by antral G-cells and acts on cholecystokinin B (CCKB) receptors.
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This question is part of the following fields:
- Gastrointestinal
- Physiology
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Question 21
Incorrect
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Which of the following statements regarding the infectivity periods of these corresponding diseases is correct?
Your Answer: Mumps is infectious until the onset of salivary gland swelling
Correct Answer: Rubella is infectious until 5 days after the rash appears
Explanation:Rubella can be contagious from 7 days before to 7 days after the rash appears.
Patients with measles are contagious from 1-2 days before the onset of symptoms.
A person with chickenpox is considered contagious beginning 1 to 2 days before rash onset until all the chickenpox lesions have crusted (scabbed).
The infectious period of mumps is considered from 2 days before to 5 days after parotitis onset.
Hepatitis A is highly transmissible and has an average incubation period of 28 to 30 days (range 15–50 days). The maximum infectivity is during the second half of the incubation period (i.e. while asymptomatic) and most cases are considered non-infectious after the first week of jaundice.
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This question is part of the following fields:
- Microbiology
- Specific Pathogen Groups
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Question 22
Incorrect
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Regarding Gaussian sample distribution, which of the following statements is false?
Your Answer: The median is the middle point of the observations
Correct Answer: It can be skewed left or right
Explanation:Gaussian distribution, also known as normal distribution, is the most important probability distribution in statistics because it fits many natural phenomena. The normal distribution is a probability function that describes how the values of a variable are distributed. Below are its characteristics:
(1) Bell-shaped and symmetrical about the mean
(2) The mean, median, mode are all equal
(3) The total area under the curve and above the x-axis is equal to 1
(4) It has long tapering tails extending infinitely but never touching the x-axis
(5) It is determined by its parameters: its mean and standard deviation
(6) The standard deviation becomes a more meaningful quality than merely being a measure of dispersionThe mean is the most common measure of central tendency. It is the sum of all observed values divided by the number of observation, and is also known as the ‘average’.
The median is the value that falls in the middle position when the observations are ranked in order from the smallest to the largest. If the number of observations is odd, the median is the middle number. If it is even, the median is the average of the two middle numbers.
The mode is the value that occurs with the greatest frequency in a set of observations, and is commonly used in public health statistics, such as the top 10 causes of mortality.
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This question is part of the following fields:
- Evidence Based Medicine
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Question 23
Correct
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A 75-year-old man with rheumatoid arthritis had gained weight, developed resistant hypertension, muscle weakness, and ankle oedema. This patient is most likely suffering from what condition?
Your Answer: Cushing's syndrome
Explanation:Overuse of cortisol medication, as seen in the treatment of patients with chronic asthma or rheumatoid arthritis, can cause Cushing’s syndrome.
Weight gain, thin arms and legs, a round face, increased fat around the base of the neck, a fatty hump between the shoulders, easy bruising, wide purple stretch marks primarily on the abdomen, breasts, hips, and under the arms, weak muscles, hirsutism, hypertension, erectile dysfunction, osteoporosis, frontal alopecia, acne, depression, poor wound healing, and polycythaemia are all clinical features of Cushing’s syndrome.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 24
Incorrect
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Regarding hepatitis A, which of the following statements is CORRECT:
Your Answer: Infection does not confer subsequent immunity.
Correct Answer: Anti-HAV IgM antibodies are diagnostic.
Explanation:Anti-HAV IgM antibodies are diagnostic. Disease in children is more commonly asymptomatic, risk of symptomatic disease increases with age. Transmission is by the faecal-oral route. Faecal shedding has been demonstrated for 2 – 3 weeks before and about a week after, the onset of jaundice. Infection confers lifelong immunity.
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This question is part of the following fields:
- Microbiology
- Pathogens
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Question 25
Correct
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A new chemotherapy drug is being tested. The intervention reduces the risk of death from 10 in 1000 to 5 in 1000. What is the number needed to treat to prevent one death:
Your Answer: 200
Explanation:Absolute risk reduction (ARR) of treatment
= risk of death in control group – risk of death in treatment group
ARR = (10/1000) – (5/1000)
= 5/1000 = 0.005
Number needed to treat (NNT)
= 1/ARR
= 1/0.005
= 200
Therefore 200 people would need to be treated to prevent one extra death. -
This question is part of the following fields:
- Evidence Based Medicine
- Statistics
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Question 26
Correct
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Which of the following diseases is caused by a build-up of lymphoblasts in the bone marrow?
Your Answer: Acute lymphoblastic leukaemia
Explanation:Acute lymphoblastic leukaemia (ALL) is a clonal (malignant) bone marrow disorder in which early lymphoid precursors multiply and replace the marrow’s normal hematopoietic cells. ALL is most common between the ages of 3 and 7, with 75 percent of cases occurring before the age of 6.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 27
Correct
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Since the fluid that enters the loop of Henle is isotonic, what is its estimated osmolality?
Your Answer: 300 mOsm
Explanation:The loop of Henle connects the proximal tubule to the distal convoluted tubule and lies parallel to the collecting ducts. It is consists of three major segments, the thin descending limb, the thin ascending limb, and the thick ascending limb.
The segments are differentiated based on structure, anatomic location, and function. The main action of the loop of Henle is to recover water and sodium chloride from urine. The liquid entering the loop of Henle is a solution of salt, urea, and other substances traversed along by the proximal convoluted tubule, from which most of the dissolved components are needed by the body, particularly glucose, amino acids, and sodium bicarbonate that have been reabsorbed into the blood.
This fluid is isotonic. Isotonic fluids generally have an osmolality ranging from 270 to 310 mOsm/L. With the fluid that enters the loop of Henle, it is estimated to be 300 mOsm/L. However, after passing the loop, fluid entering the distal tubule is hypotonic to plasma since it has been diluted during its passage.
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This question is part of the following fields:
- Physiology
- Renal Physiology
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Question 28
Incorrect
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If the afferent arteriole's diameter is smaller than the efferent arteriole's diameter in the glomerulus:
Your Answer: The glomerular oncotic pressure will increase
Correct Answer: The net filtration pressure will decrease
Explanation:The relative resistance of the afferent and efferent arterioles substantially influences glomerular capillary hydrostatic pressure and consequently GFR. Filtration is forced through the filtration barrier due to high pressure in the glomerular capillaries. Afferent arteriolar constriction lowers this pressure while efferent arteriolar constriction raises it.
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This question is part of the following fields:
- Physiology
- Renal
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Question 29
Incorrect
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The most common type of adult leukaemia is chronic lymphocytic leukaemia (CLL). It develops as a result of lymphocyte clonal proliferation. Which of the following statements about CLL is CORRECT?
Your Answer: A predominance of immature lymphocytes are seen on the blood film
Correct Answer: It is most commonly discovered as an incidental finding
Explanation:CLL (chronic lymphocytic leukaemia) is the most common type of chronic lymphoid leukaemia, with a peak incidence between the ages of 60 and 80. It is the most common type of leukaemia in Europe and the United States, but it is less common elsewhere.
The CLL tumour cell is a mature B-cell with low immunoglobulin surface expression (IgM or IgD). The average age at diagnosis is 72 years, with only 15% of cases occurring before the age of 50.
The male-to-female ratio is about 2:1. Over 80% of cases are identified by the results of a routine blood test, which is usually performed for another reason.
Lymphocytic anaemia, thrombocytopenia, and normochromic normocytic anaemia are common laboratory findings. Aspiration of bone marrow reveals up to 95% lymphocytic replacement of normal marrow elements.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 30
Incorrect
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An very unwell patient is receiving treatment in your hospital's HDU and is found to have an Escherichia coli O157 infection.
Which one of these statements about Escherichia coli O157 is true?Your Answer: It is enteroaggregative
Correct Answer: Haemolytic uraemic syndrome develops in approximately 6% of patients
Explanation:Escherichia coli O157 is a serotype of Escherichia coli.
The Escherichia coliO157 strain is ‘enterohaemorrhagic’ and causes severe forms of acute haemorrhagic diarrhoea. It can also cause non-haemorrhagic diarrhoea.Incubation period of Escherichia coli O157 is usually 3-4 days and bloody diarrhoea usually begins on the 3rd or 4th day of the infection.
Infections with Escherichia coliO157 are more common during the warmer months than in winter.
Haemolytic uraemic syndrome develops in approximately 6% of patients. It is commonly seen in children and in the elderly.
Escherichia coli O157 can also cause:
Haemorrhagic colitis
Haemolytic uraemic syndrome
Thrombotic thrombocytopenic purpura but not immune thrombocytopenic purpura. -
This question is part of the following fields:
- Microbiology
- Specific Pathogen Groups
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SESSION STATS - PERFORMANCE PER SPECIALTY
