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Question 1
Incorrect
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Left ventricular afterload is mostly calculated from systemic vascular resistance.
Which one of the following factors has most impact on systemic vascular resistance?Your Answer: Capillaries
Correct Answer: Small arterioles
Explanation:Systemic vascular resistance (SVR), also known as total peripheral resistance (TPR), is the amount of force exerted on circulating blood by the vasculature of the body. Three factors determine the force: the length of the blood vessels in the body, the diameter of the vessels, and the viscosity of the blood within them. The most important factor that determines the systemic vascular resistance (SVR) is the tone of the small arterioles.
These are otherwise known as resistance arterioles. Their diameter ranges between 100 and 450 µm. Smaller resistance vessels, less than 100 µm in diameter (pre-capillary arterioles), play a less significant role in determining SVR. They are subject to autoregulation.
Any change in the viscosity of blood and therefore flow (such as due to a change in haematocrit) might also have a small effect on the measured vascular resistance.
Changes of blood temperature can also affect blood rheology and therefore flow through resistance vessels.
Systemic vascular resistance (SVR) is measured in dynes·s·cm-5
It can be calculated from the following equation:
SVR = (mean arterial pressure − mean right atrial pressure) × 80 cardiac output
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This question is part of the following fields:
- Physiology
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Question 2
Correct
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Regarding the anatomical relations of the scalenus muscles, which of these is true?
Your Answer: The trunks of the brachial plexus emerge from the lateral border of scalenus anterior
Explanation:The ascending cervical artery lies media the phrenic nerve on scalenus anterior and can easily be mistaken for the phrenic nerve at operation.
The phrenic nerve passes across scalenus anterior and medius inferiorly.
The subclavian artery is separated from the vein by the scalenus anterior.
The brachiocephalic vein is formed at the medial border of scalenus anterior by the subclavian vein and the internal jugular vein.
Emerging from the lateral border of scalenus anterior are the trunks of the brachial plexus .
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This question is part of the following fields:
- Anatomy
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Question 3
Correct
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A 48-year-old woman has presented to the emergency with abdominal pain and distension complaints. She is a known case of diabetes mellitus type 2 and has a BMI of 28 kg/m². On investigations, the liver function tests (LFTs) show raised alanine transaminase (ALT).
Liver ultrasound is performed next to visualize the blood flow into and out of the liver.
Which blood vessel supplies approximately one-third of the blood supply to the liver?
Your Answer: Hepatic artery proper
Explanation:The liver receives blood supply from two sources.
1. Hepatic artery proper
It arises from the celiac trunk via the common hepatic artery and brings oxygenated blood to the liver.
It contributes to approximately 30% of the blood supply of the liver.
2. Hepatic portal vein – supplies the liver with partially deoxygenated blood, carrying nutrients absorbed from the small intestine. It gets tributaries from the inferior mesenteric vein, splenic vein, and superior mesenteric veinThe inferior mesenteric artery supplies the hindgut.
The superior mesenteric artery supplies the pancreas and intestine up to the proximal two-thirds of the transverse colon.
The inferior phrenic artery supplies the inferior surface of the diaphragm and oesophagus. -
This question is part of the following fields:
- Anatomy
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Question 4
Incorrect
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Regarding anti diuretic hormone (ADH), one of the following statements is correct:
Your Answer: Increases the volume of urine passed
Correct Answer: Increases the total amount of electrolyte free water in the body
Explanation:The major action of ADH is to increase reabsorption of osmotically unencumbered water from the glomerular filtrate and decreases the volume of urine passed. The osmolarity of urine is increased to a maximum of four times that of plasma (approx. 1200 mOsm/kg) by Increasing water reabsorption.
Chronic water loading, Lithium, potassium deficiency, cortisol and calcium excess, all blunt the action of ADH. This leads to nephrogenic diabetes insipidus.
ADH’s primary site of action is the distal tubule and collecting duct.
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This question is part of the following fields:
- Physiology
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Question 5
Correct
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The cardiac tissue type that that has the highest conduction velocity is:
Your Answer: Purkinje fibres
Explanation:Potassium maintains the resting potential of cardiac myocytes, with depolarization triggered by a rapid influx of sodium ions, and repolarization due to efflux of potassium. A slow influx of calcium is responsible for the longer duration of a cardiac action potential compared with skeletal muscle.
The cardiac action potential has several phases which have different mechanisms of action as seen below:
Phase 0: Rapid depolarisation – caused by a rapid sodium influx.
These channels automatically deactivate after a few ms.Phase 1: caused by early repolarisation and an efflux of potassium.
Phase 2: Plateau – caused by a slow influx of calcium.
Phase 3 – Final repolarisation – caused by an efflux of potassium.
Phase 4 – Restoration of ionic concentrations – The resting potential is restored by Na+/K+ATPase.
There is slow entry of Na+into the cell which decreases the potential difference until the threshold potential is reached. This then triggers a new action potentialOf note, cardiac muscle remains contracted 10-15 times longer than skeletal muscle.
Different sites have different conduction velocities:
1. Atrial conduction – Spreads along ordinary atrial myocardial fibres at 1 m/sec2. AV node conduction – 0.05 m/sec
3. Ventricular conduction – Purkinje fibres are of large diameter and achieve velocities of 2-4 m/sec, the fastest conduction in the heart. This allows a rapid and coordinated contraction of the ventricles
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 6
Incorrect
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The typical fluid compartments in a normal 70kg male are:
Your Answer: intracellular>interstitial
Correct Answer: intracellular>extracellular
Explanation:Body fluid compartments in a 70kg male:
Total volume=42L (60% body weight)
Intracellular fluid compartment (ICF) =28L
Extracellular fluid compartment (ECF) = 14LECF comprises:
Intravascular fluid (plasma) = 3L
Extravascular fluid = 11LExtravascular fluids comprises:
Interstitial fluid = 10.5L
Transcellular fluid = 0.5L -
This question is part of the following fields:
- Physiology
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Question 7
Incorrect
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The Medical Admissions unit receives a 71-year-old woman. She has type 2 diabetes, which she manages with diet, but she has been feeling ill for the past 48 hours.
Her pulse rate is 110 beats per minute, her blood pressure is 90/50 mmHg, and she is clinically dehydrated. Her respiratory rate is 20 breaths per minute, and chest auscultation reveals no focal signs.
The following are the lab results:
Glucose 27.4 mmol/L (3.5-5.5)
Ketones 2.5 mmol/L (<0.1)
Urinary glucose is zero (dipstick) with ketones
A random blood glucose of 15.3 mmol/L was measured during a visit to the diabetic clinic one month prior to admission, according to her notes, and a urinary dipstick registered a high glucose and ketones++.
The discrepancy between plasma and urinary glucose measurements is best explained by which of the following physiological mechanisms?Your Answer: The glucose transport maximum (Tm) is abnormally high
Correct Answer: The glomerular filtration rate is abnormally low
Explanation:The glucose molecule enters the Bowman’s capsule freely and becomes part of the filtrate.
All glucose is reabsorbed in the proximal convoluted tubule when blood glucose concentrations are below a certain threshold (approximately 11 mmol/L) (PCT). Active transportation makes this possible. In the proximal tubular cells, sodium/glucose cotransporters (SGLT1 and SGLT2) are the proteins responsible.
Glucose does not normally appear in the urine below the renal threshold.
The renal glucose threshold is not set in stone and is affected by a variety of factors, including GFR, TmG, and the quantity of splay.
The different absorptive and filtering capacities of individual nephrons cause splay, which is the rounding of a glucose reabsorption curve.
The SGLT proteins have a high affinity for glucose, but not an infinite affinity. As a result, some glucose may escape reabsorption before the TmG. A decrease in renal threshold may be caused by an increase in splay.
Because the filtered glucose load is reduced and the PCT can reabsorb all of the filtered glucose despite hyperglycaemia, a low GFR causes an increase in TmG. In contrast, lowering the TmG lowers the threshold because the tubules’ ability to reabsorb glucose is reduced.
A reduction in GFR caused by severe dehydration and reduced perfusion pressure is the most obvious cause of the discrepancy between plasma and urinary glucose in this scenario.
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This question is part of the following fields:
- Pathophysiology
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Question 8
Correct
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Which of the following is true about number needed to harm?
Your Answer: The number of patients that must receive a particular treatment for one additional patient to experience an adverse outcome.
Explanation:Number needed to harm are a measure of the impact of a treatment or intervention that is often used to communicate results to patients, clinicians, the public and policymakers. It states how many patients need to be treated for one additional patient to experience an adverse outcome (e.g. a death). It is calculated as the inverse of the absolute risk reduction. It can equally well be applied to harmful outcomes as well as beneficial ones, where it becomes numbers needed to treat (NNT) instead.
In this way, they are both calculated the same but NNT usually refers to a therapeutic treatment whereas NNH refers to a risk-factor for disease.
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This question is part of the following fields:
- Statistical Methods
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Question 9
Incorrect
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A 30-year-old woman with a BMI of 24 kg/m2 consumes four glasses of wine on an empty stomach. Her serum alcohol are levels measured over the following five hours. The serum alcohol level of 30-year-old man with the same BMI and alcohol consumption is also measured over the same duration.
The peak concentration of alcohol is found to be greater in the woman than in the man.
Which of these offers best explanation for this observation?Your Answer: Greater clearance of ethanol per unit lean body mass
Correct Answer: Lower volume of distribution
Explanation:The blood alcohol concentration depends on:
-The rate of alcohol absorption from the gastrointestinal tract
-The volume of distribution of alcohol in the body, and
-The rate of elimination of alcohol from the body.Total body water is approximately 50% in a female as compared to 60% in a typical male. This means that the volume of distribution of alcohol is lower in female compared with men. This is the principal reason for higher peak in alcohol levels.
About 4% of ingested alcohol is metabolised by the liver accounting for first pass metabolism and 0.4% is metabolised by gastric alcohol dehydrogenase (ADH). The absorbed alcohol is NOT distributed to fat cells but it is distributed throughout the water compartments (plasma, interstitial and intracellular) of the body. Women have very little gastric ADH, which further influences this exaggerated rise.
85-98% of the alcohol is oxidised by the liver to acetaldehyde and then to acetate. The metabolic pathway initially observes first order kinetics and then saturation or zero order kinetics leading to peaks in alcohol levels.
Clearance of ethanol per unit lean body mass is lower in male. The calculated alcohol elimination rate and liver volume per kilogram of lean body mass were 33% and 38% higher in women than in men, respectively.
Available evidence in the literature about the relationship of alcohol metabolism to the phases of the menstrual cycle is conflicting.
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This question is part of the following fields:
- Pathophysiology
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Question 10
Incorrect
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Regarding the treatment of bladder cancer, a study concerned with the usage of a combined or monotherapy was conducted. A forest plot was used for the visual representation of the data.
Which of the following is true regarding forest plots?Your Answer: Forest plots can only be used if the outcome is significant
Correct Answer: Forest plots can present data from multiple studies
Explanation:Being the part of a meta analysis, forest plots are more valued as evidence then randomised control trials.
The notion that forest plots can only be used if the results are substantial is not true. They are good indicators of the significance of the data. If the diamond intersects the central line, the data is rendered significant. It also aggregates means and confidence intervals from studies conducted in the past which makes the study much more reliable as errors associated with individual studies tend to have less of an impact in this way.
The suggestion that forest plots are primarily used for qualitative data is factually incorrect. Forest plots require numerical values to function.
All in all, forest plots help us in determining whether or not there is a significant trend in that particular field of study.
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This question is part of the following fields:
- Statistical Methods
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Question 11
Correct
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Calcium homeostasis is regulated by parathormone (PTH).
Which of the following PTH actions is most likely to cause calcium to be released from bone?Your Answer: Indirect stimulation of osteoclasts
Explanation:The hormone parathyroid hormone (PTH) and the receptor parathyroid hormone type 1 (PTH1-Rc) are important regulators of blood calcium homeostasis.
PTH can cause a rapid release of calcium from the matrix in bone, but it also affects long-term calcium metabolism by acting directly on bone-forming osteoblasts (by binding to PTH1-Rc) and indirectly on bone-resorbing osteoclasts.
PTH causes changes in the synthesis and/or activity of several proteins, including osteoclast-differentiating factor, also known as TRANCE or RANKL, when it acts on osteoblasts.
RANK receptors are found on the cell surfaces of osteoclast precursors. The osteoclasts are activated when RANKL binds to the RANK receptors. Osteoclasts lack PTH receptors, whereas osteoblasts do. Osteoclasts are activated indirectly when the RANK receptor binds to the RANKL secreted by osteoblasts, resulting in bone resorption. PTH1 receptors are found in osteoclasts, but they are few.
PTH activates G-protein coupled receptors in all target cells via adenylate cyclase.
The PTH2 receptor is most abundant in the nervous system and pancreas, but it is not a calcium metabolism regulator. It is abundant in the septum, midline thalamic nuclei, several hypothalamic nuclei, and the dorsal horn of the spinal cord, as well as the cerebral cortex and basal ganglia. Expression in pancreatic islet somatostatin cells is the most prominent on the periphery.
The distribution of the receptor is being used to test functional hypotheses. It may play a role in pain modulation and hypothalamic releasing-factor secretion control.
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This question is part of the following fields:
- Pathophysiology
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Question 12
Incorrect
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Which of the following statements is true regarding Entonox?
Your Answer: Is supplied in cylinders filled to a pressure is 140 bar
Correct Answer: Exists only in gaseous form in a cylinder (under normal working conditions)
Explanation:Entonox is a gas that consists of 50% oxygen and 50% Nitrous oxide. Nitrous oxide is sometimes used for anaesthetics but in this combination, it works as a short-acting painkiller.
Under normal working conditions, it exists only in gaseous form in a cylinder. The gauge pressure of a full Entonox cylinder is 137 bar.
Entonox cylinders should be stored horizontally at a temperature above 0 C. At temperatures below this the nitrous oxide component may separate.
Pseudocritical temperature and pseudocritical pressure can be defined as the molal average critical temperature and pressure of mixture components. In other words, the pseudo-critical temperature is the temperature at which the two gases separate. The pseudo-critical temperature of Entonox is approximately -5.50 C
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This question is part of the following fields:
- Pharmacology
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Question 13
Correct
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A 71-year-old man, presents with central crushing pain in his chest to the emergency department. On examination, he complains of nausea and is notably sweating. On ECG, elevation in the ST-segment is noted in multiple chest leads, as well as sinus bradycardia. A myocardial infarction can cause a sinus bradycardia.
The sinoatrial (SA) node and the atrioventricular (AV) node receive arterial supply from which vessel?Your Answer: Right coronary artery
Explanation:The left marginal artery comes off the left circumflex artery, and runs alongside the heart.
The left circumflex artery is one of the bifurcations of the left coronary artery, and eventually forms the left marginal artery.
An occlusion in the left circumflex artery often results in a lateral MI.
The right marginal artery originates from the right coronary artery.
The left anterior descending artery (LAD) is another bifurcation of the left coronary artery. An occlusion in the LAD would often result in an anteroseptal MI as is diagnosed on ECG by noting changes in leads V1-V4.
The right coronary artery originates from the right aortic sinus of the ascending aorta, and bifurcates to give rise to many branches, including the sinoatrial artery which supplies the sinoatrial (SA) node in 50-70% of cases, the artery of the atrioventricular (AV) node in 50-60% of cases, the right acute marginal artery which supplies the right ventricle. It also supplies the right atrium, interatrial septum and the posterior inferior third of the interventricular septum.
Arrhythmias and inferior MI often occurs as a result of an occlusion in the right coronary artery, and can be diagnosed by ECG changes in leads II, III and aVF.
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This question is part of the following fields:
- Anatomy
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Question 14
Incorrect
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At which of the following location is there no physiological oesophageal constriction?
Your Answer: Arch of the aorta
Correct Answer: Lower oesophageal sphincter
Explanation:The oesophagus is a muscular tube that connects the pharynx to the stomach. It begins at the lower border of the cricoid cartilage and C6 vertebra. It ends at T11.
The oesophagus has physiological constrictions at the following levels:
1. Cervical constriction: Pharyngo-oesophageal junction (15 cm from the incisor teeth) produced by the cricopharyngeal part of the inferior pharyngeal constrictor muscle
2. Thoracic constrictions:
i. where the oesophagus is first crossed by the arch of the aorta (22.5 cm from the incisor teeth)
ii. where the oesophagus is crossed by the left main bronchus (27.5 cm from the incisor teeth)
3. Diaphragmatic constriction: where the oesophagus passes through the oesophageal hiatus of the diaphragm (40 cm from the incisor teeth)Awareness of these constrictions is important for clinical purposes when it is required to pass instruments through the oesophagus into the stomach or when viewing radiographs of patients’ oesophagus.
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This question is part of the following fields:
- Anatomy
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Question 15
Correct
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A new proton pump inhibitor (PPI) is being evaluated in elderly patients who are taking aspiring. Study designed has 120 patients receiving the PPI, while a control group of 240 individuals is given the standard PPI. Over a span of 6 years, 24 of the group receiving the new PPI had an upper GI bleed compared to 60 individuals who received the standard PPI.
How would you calculate the absolute risk reduction?Your Answer: 5%
Explanation:Absolute risk reduction = (Control event rate) – (Experimental event rate)
Experimental event rate = 24 / 120 = 0.2
Control event rate = 60 / 240 = 0.25
Absolute risk reduction = 0.25 – 0.2 = 0.05 = 5% reduction
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This question is part of the following fields:
- Statistical Methods
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Question 16
Correct
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A 21-year-old woman presents to ER following the deliberate ingestion of 2 g of amitriptyline. On clinical examination:
Glasgow coma score: 10
Pulse rate: 140 beats per minute
Blood pressure: 80/50 mmHg.
ECG showed a QRS duration of 233 Ms.
Which of the following statement describes the most important initial course of action?Your Answer: Give fluid boluses
Explanation:The first line of treatment in case of hypotension is fluid resuscitation.
Activated charcoal can be used within one hour of tricyclic antidepressant ingestion but an intact and secure airway must be checked before intervention. The risk of aspiration should be assessed.
Vasopressors are indicated for the treatment of hypotension following (Tricyclic Antidepressant) TCA overdose when patients fail to respond to fluids and bicarbonate.
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This question is part of the following fields:
- Pharmacology
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Question 17
Correct
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A trail has analysed that a new screening test may increase the survival time of ovarian cancer patients. But analyst say that the apparent increase in the patients survival time is just because of earlier detection instead of actual improvement.
What kind of bias is in this experiment?Your Answer: Lead time bias
Explanation:Observation bias occurs when the behaviour of an individual changes that results from their awareness of being observed.
Recall bias introduced when participants in a study are systematically more or less likely to recall and relate information on exposure depending on their outcome status.
Attrition bias is a systematic error caused by unequal loss of participants from a randomized controlled trial (RCT). In clinical trials, participants might dropout due to unsatisfactory treatment or efficacy, intolerable adverse events, or even death.
Selection bias introduced when the individuals are not chosen randomly to take a part in the study. It usually occurs when the research decides who is going to be studied, they are not the representative of the population.
Lead-time bias occurs when a disease is detected by a screening test at an earlier time point rather than it would have been diagnosed by its clinical appearance. In this bias, earlier detection improves the survival time in the intervention group.
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This question is part of the following fields:
- Statistical Methods
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Question 18
Correct
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The required sample size in a trial of a new therapeutic agent varies with?
Your Answer: Level of statistical significance required
Explanation:The level of statistical significance required influences the sample size used. This is because sample size is used in the calculation of SD/SE.
Sample size does not affect
The level of acceptance
The alternative hypothesis with a general level set at p<0.05
The test to be used.Experience of the investigator and the type of patient recruited should have no bearing on the required sample size.
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This question is part of the following fields:
- Statistical Methods
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Question 19
Correct
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The coronary sinus is a collection of veins joined together to form a large vessel that collects blood from the myocardium. Which structure forms the largest tributary of the coronary sinus?
Your Answer: Great cardiac vein
Explanation:The coronary sinus is large venous structure located on the posterior aspect of the left atrium, coursing within the left atrioventricular groove. The function of the coronary sinus is to drain the venous blood from the majority of the heart. It opens into the right atrium between the opening of inferior vena cava, the fossa ovalis and the right atrioventricular orifice. The coronary sinus is often guarded by a thin, semicircular endocardial fold, also known as the thebesian valve.
Tributaries include: Great cardiac vein, middle cardiac vein, small cardiac vein, posterior vein of left ventricle, oblique vein of left atrium. The great cardiac vein is the largest tributary of the coronary sinus. -
This question is part of the following fields:
- Anatomy
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Question 20
Correct
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Which of the following statements is true regarding prazosin?
Your Answer: Is a selective alpha 1 adrenergic receptor antagonist.
Explanation:Selective ?1 -Blockers like prazosin, terazosin, doxazosin, and alfuzosin cause a decrease in blood pressure with lesser tachycardia than nonselective blockers (due to lack of ?2 blocking action.
The major adverse effect of these drugs is postural hypotension. It is seen with the first few doses or on-dose escalation (First dose effect).
Its half-life is approximately three hours.
It is excreted primarily through bile and faeces (not through kidneys)
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This question is part of the following fields:
- Pharmacology
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Question 21
Correct
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Of the following, which of these oxygen carrying molecules causes the greatest shift of the oxygen-dissociation curve to the left?
Your Answer: Myoglobin (Mb)
Explanation:Myoglobin is a haemoglobin-like, iron-containing pigment that is found in muscle fibres. It has a high affinity for oxygen and it consists of a single alpha polypeptide chain. It binds only one oxygen molecule, unlike haemoglobin, which binds 4 oxygen molecules.
The myoglobin ODC is a rectangular hyperbola. There is a very low P50 0.37 kPa (2.75 mmHg). This means that it needs a lower P50 to facilitate oxygen offloading from haemoglobin. It is low enough to be able to offload oxygen onto myoglobin where it is stored. Myoglobin releases its oxygen at the very low PO2 values found inside the mitochondria.
P50 is defined as the affinity of haemoglobin for oxygen: It is the PO2 at which the haemoglobin becomes 50% saturated with oxygen. Normally, the P50 of adult haemoglobin is 3.47 kPa(26 mmHg).
Foetal haemoglobin has 2 ? and 2 ?chains. The ODC is left shifted – this means that P50 lies between 2.34-2.67 kPa [18-20 mmHg]) compared with the adult curve and it has a higher affinity for oxygen. Foetal haemoglobin has no ? chains so this means that there is less binding of 2.3 diphosphoglycerate (2,3 DPG).
Carbon monoxide binds to haemoglobin with an affinity more than 200-fold higher than that of oxygen. This therefore decreases the amount of haemoglobin that is available for oxygen transport. Carbon monoxide binding also increases the affinity of haemoglobin for oxygen, which shifts the oxygen-haemoglobin dissociation curve to the left and thus impedes oxygen unloading in the tissues.
In sickle cell disease, (HbSS) has a P50 of 4.53 kPa(34 mmHg).
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This question is part of the following fields:
- Physiology
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Question 22
Correct
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A graph is created to show the exponential relationship between bacterial growth (y-axis) and time (x-axis).
Which of the following statements is most true about this kind of exponential relationship?Your Answer: y = ex
Explanation:The relationship between bacterial growth and time is a tear-away exponential. The mathematical relationship between y and x in this case is:
y = ex
Where: the power is x, and the base is e.
Euler’s number (e) is a mathematical constant that is the base for all logarithms occurring naturally. Its value is 2.718.
The statement X increasing with an increase in Y is proportional to Y refers to the change in y in terms of x when considering any exponential relationship.
This is not a build-up exponential, and that is mathematically stated as y = 1-e-kt.
The negative x axis being a horizontal asymptote and the y intercept being 0, 1 are examples of tearaway exponentials , but do not describe an exponential process.
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This question is part of the following fields:
- Statistical Methods
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Question 23
Correct
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Health workers are at increased risks of anaesthetic exposure. Therefore, The Control of Substances Hazardous to Health (COSHH) regulations require measures implemented to assess and control the risks related to this exposure.
Among the following control measures, which one is most likely to limit potentially harmful exposure to anaesthetic agents within an operating theatre?Your Answer: Total intravenous anaesthesia (TIVA)
Explanation:Exposure to anaesthetic hazards is one among the occupational exposures in manipulating toxic agents or inhaling toxic gases during anaesthetic practices.
Toxic gases mainly nitrous oxide, is one of the most gaseous anaesthetic agents that constitutes an important source of pollution. One of the safe and effective technics used in anaesthesia and reducing the amount of pollution is the Total Intravenous Anaesthesia (TIVA) which consists of using opioids in analgesia and propofol for the induction and the maintenance of anaesthesia. It refers to the administration intravenously of an anaesthetic, sedative, and/or tranquilizer. A less polluting but not the best way to get rid of the toxic aesthetic agents is the scavenger system that collects and expels the gas outside the medical environment. Yet, this technique still represents a hazard for the environment and still increase the risk of exposure for the health workers and clinical staff.
Fume cupboards are also not recommended to use because of their high pollution potency, mainly of the air resulting in a great harm for medical workers.
Supraglottic airways as well as the Air Changes per Hour technics could be harmful for both patients and health workers, increasing the risks of transmitted diseases, namely nosocomial infections.
Therefore, the Total Intravenous Anaesthesia technique (TIVA) is most likely to be safe and recommended to use.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 24
Incorrect
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Which of these anaesthetics has the best chance of preventing HPV (hypoxic pulmonary vasoconstriction)?
Your Answer: Sevoflurane 1 MAC
Correct Answer: Desflurane 2 MAC
Explanation:Resistance pulmonary arteries constrict in response to alveolar and airway hypoxia, diverting blood to better-oxygenated alveoli.
In atelectasis, pneumonia, asthma, and adult respiratory distress syndrome, hypoxic pulmonary vasoconstriction optimises O2 uptake. Hypoxic pulmonary vasoconstriction helps maintain systemic oxygenation during single-lung anaesthesia.
A redox-based O2 sensor within pulmonary artery smooth muscle cells is involved in hypoxic pulmonary vasoconstriction. The production of reactive oxygen species by smooth muscle cells in the pulmonary artery varies in proportion to PaO2. Hypoxic removal of these redox second messengers inhibits voltage-gated potassium channels, depolarizing smooth muscle cells in the pulmonary artery.
L-type calcium channels are activated by depolarization, which raises cytosolic calcium and causes hypoxic pulmonary vasoconstriction. Some anaesthetics suppress this response, increasing the risk of further deterioration in ventilation perfusion mismatch.
Agents that inhibit HPV are ether, halothane, and desflurane (>1.6 MAC).
Agents with no effect on HPV include thiopentone, fentanyl, desflurane (1MAC), isoflurane (<1.5MAC), sevoflurane(1MAC), and propofol. -
This question is part of the following fields:
- Pharmacology
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Question 25
Incorrect
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A 28-year-old girl complained of severe abdominal pain and hematemesis and was rushed into the emergency department. She has an increased heart rate of 120 beats per minute and blood pressure of 90/65. She has a history of taking Naproxen for her Achilles tendinopathy. On urgent endoscopy, she is diagnosed with a bleeding peptic ulcer.
The immediate treatment is to permanently stop the bleeding by performing embolization of the left gastric artery via an angiogram.
What level of the vertebra will be used as a radiological marker for the origin of the artery that supplies the left gastric artery during the angiogram?Your Answer: T10
Correct Answer: T12
Explanation:The left gastric artery is the smallest branch that originates from the coeliac trunk—the coeliac trunk branches of the abdominal aorta at the vertebral level of T12.
The left gastric artery runs along the superior portion of the lesser curvature of the stomach. A peptic ulcer that is serious enough to erode through the stomach mucosa into a branch of the left gastric artery can cause massive blood loss in the stomach, leading to hematemesis. The patient also takes Naproxen, a non-steroidal anti-inflammatory drug that is a common cause for peptic ulcers in otherwise healthy patients.
The left gastric artery is responsible for 85% of upper GI bleeds. In cases refractory to initial treatment, angiography is sometimes needed to embolise the vessel at its origin and stop bleeding. During an angiogram, the radiologist will enter the aorta via the femoral artery, ascend to the level of the 12th vertebrae and then enter the left gastric artery via the coeliac trunk.
The important landmarks of vessels arising from the abdominal aorta at different levels of vertebrae are:
T12 – Coeliac trunk
L1 – Left renal artery
L2 – Testicular or ovarian arteries
L3 – Inferior mesenteric artery
L4 – Bifurcation of the abdominal aorta
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This question is part of the following fields:
- Anatomy
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Question 26
Correct
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A post-operative patient was brought to the recovery room after completion of dilation and curettage. Her medical history revealed that she was maintained on levodopa for Parkinson's disease. The nurses administered ondansetron 4 mg and dexamethasone 8 mg prior to transfer from the operating room to the recovery room. However, an additional antiemetic agent is warranted.
Which of the following agents should be prescribed to the patient?Your Answer: Cyclizine 50 mg IV
Explanation:The Beers criteria, a US set of criteria for good prescribing in the older patient, preclude the use of metoclopramide in Parkinson’s disease. The Adverse Reactions Register of the UK Committee on Safety of Medicines (CSM) for the years 1967 to 1982 contained 479 reports of extrapyramidal reactions in which metoclopramide was the suspected drug; 455 were for dystonic-dyskinetic reactions, 20 for parkinsonism and four for tardive dyskinesia. Effects can occur within days of initiation of treatment and may take months to wear off.
Other antiemetics are available, such as cyclizine (Valoid), domperidone and ondansetron, which would be more appropriate to use in those with Parkinson’s disease.
Cyclizine is a piperazine derivative with histamine H1 receptor antagonist and anticholinergic activity. It is used for the treatment of nausea, vomiting, (particularly opioid-induced vomiting), vertigo, motion sickness, and labyrinthine disorders.
Prochlorperazine is an antipsychotic known to cause tardive dyskinesia, tremor and parkinsonian symptoms and is therefore likely to exacerbate Parkinson’s disease. Prochlorperazine is not favoured for older patients because of the increased risk of stroke and transient ischaemic attack (TIA).
Droperidol and phenothiazine are also potent antagonists on D2 receptors and must also be avoided.
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This question is part of the following fields:
- Pharmacology
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Question 27
Correct
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An inguinal hernia repair under general anaesthesia is scheduled for a fit 36-year-old man (75 kg). For perioperative and postoperative analgesia, you decide to perform an inguinal field block.
Which of the following local anaesthetic solutions is the most appropriate?Your Answer: 30 mL bupivacaine 0.5%
Explanation:Perioperative and postoperative analgesia can both be provided by an inguinal hernia field block. The Iliohypogastric and ilioinguinal nerves, as well as the skin, superficial fascia, and deeper structures, must be blocked for maximum effectiveness. The local anaesthetic should ideally have a long duration of action, be highly concentrated, and have a volume of at least 30 mL.
Plain bupivacaine has a maximum safe dose of 2 mg/kg body weight.
Because the patient weighs 75 kg, 150 mg bupivacaine can be safely administered. Both 30 mL 0.5 percent bupivacaine (150 mg) and 60 mL 0.25 percent bupivacaine (150 mg) are acceptable doses, but 30 mL 0.5 percent bupivacaine represents the optimal volume and strength, potentially providing a denser and longer block.
The maximum safe dose of plain lidocaine has been estimated to be between 3.5 and 5 mg/kg. The patient weighs 75 kg and can receive a maximum of 375 mg using the higher dosage regimen:
There are 200 mg of lidocaine in 10 mL of 2% lidocaine (and therefore 11 mL contains 220 mg)
200 mg of lidocaine is contained in 20 mL of 1% lidocaine.While alternatives are available, Although the doses of 11 mL lidocaine 2% and 20 mL lidocaine 1% are well within the dose limit, the volumes used are insufficient for effective field block for this surgery.
With 1 in 200,000 epinephrine, the maximum safe dose of lidocaine is 7 mg/kg. The patient can be given 525 mg in this case. Even with epinephrine, 60 mL of 1% lidocaine is 600 mg, which could be considered an overdose.
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This question is part of the following fields:
- Pharmacology
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Question 28
Incorrect
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If the speed of an ultrasound in soft tissue is 1540 meters per second, what is the estimated wavelength produced if the frequency of a generated ultrasound is 10 megahertz?
Your Answer: 0.15 metre
Correct Answer: 0.15 millimetre
Explanation:Wavelength can be computed as follows:
Wavelength = velocity/frequency
In the given problem, the values stated are:
Frequency = 10 x 10^6
Velocity = 1540 meters per secondWavelength = 1540/(10×10^6)
Wavelength = 1540/10,000,000 meters
Wavelength = 0.15 millimetres -
This question is part of the following fields:
- Clinical Measurement
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Question 29
Correct
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The thyroid gland:
Your Answer: Internalises iodine through active transport
Explanation:The thyroid gland is a gland shaped like a butterfly which lies at the base of the anterior neck. It controls metabolism using hormone secretion.
Iodine is extremely important for the synthesis of hormones within the thyroid. It is internalised into the thyroid follicular cells via the sodium/iodide symporter (NIS).
The parathyroid glands are found posterior to the thyroid gland, with the recurrent laryngeal nerves running posteromedially.
The expected weight of a normal thyroid gland is about 30 grams.
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This question is part of the following fields:
- Pathophysiology
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Question 30
Incorrect
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In the United Kingdom, a new breast cancer screening test is being conducted compared to the conventional use of mammography. This test predicts that if the breast cancer is diagnosed at an earlier stage, it could improve the survival rate but the overall results remains constant.
This is an example of what kind of bias?Your Answer: Procedure Bias
Correct Answer: Lead time bias
Explanation:Recall bias introduced when participants in a study are systematically more or less likely to recall and relate information on exposure depending on their outcome status.
In procedure bias, the researcher decides assignment of a treatment versus control and assigns particular patients to one group or the other non-randomly. This is unlikely to have occurred in this case, although it is not mentioned specifically.
Self Selection or volunteer bias occur when those subjects are selected to participate in the study who are not the representative of the entire target population. those subjects may be from high socio-economic status and practice those activities or lifestyle that improves their health.
Lead-time bias occurs when a disease is detected by a screening test at an earlier time point rather than it would have been diagnosed by its clinical appearance. In this bias, earlier detection improves the survival time in the intervention group.
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This question is part of the following fields:
- Statistical Methods
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