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Question 1
Incorrect
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Regarding restriction fragment length polymorphisms (RFLP) and Gene Knockout Mouse Models. All are true except:
Your Answer: Sequence changes involved in RFLP can be analysed more quickly by PCR.
Correct Answer: In knockout mouse models a gene is turned on through targeted mutation
Explanation:In RFLP, polymorphism occurs in 98% of the non coding genome, resulting in no phenotypical change in the organism. A gene is not turned on by a mutation, rather the mutation at the restriction site will alter the DNA and the DNA will now form fragments of different lengths. PCR is a better technique than RFLP.
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This question is part of the following fields:
- Genetics
- Physiology
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Question 2
Incorrect
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Which one of these equations best defines lung compliance?
Your Answer: Tidal volume / change in pressure
Correct Answer: Change in volume / change in pressure
Explanation:Lung compliance is defined as change in volume per unit change in distending pressure. Lung compliance is calculated using the equation: Lung compliance = ΔV / ΔP Where: ΔV is the change in volume ΔP is the change in pleural pressure. Static compliance is lung compliance in periods without gas flow, and is calculated using the equation: Static compliance = VT / Pplat − PEEP Where: VT = tidal volume Pplat = plateau pressure PEEP = positive end-expiratory pressure
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 3
Incorrect
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Choose the correct statement regarding trabecular bone:
Your Answer: Also known as spongy bone
Correct Answer: All of the options are correct
Explanation:Trabecular, spongy or cancellous bone. It is located inside the cortical bone and makes up around 20% of all bone in the body. It is made of spicules or plates with a high surface to volume ratio, where many cells sit on the surface of the end plates. It receives its nutrients from the extracellular fluid (ECF), exchanging about 10 mmol of calcium every 24 hours.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 4
Correct
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Which of the following cytokines is important for the maintenance of granulomatous inflammation:
Your Answer: TNF-alpha
Explanation:Granulomatous inflammation is a distinctive pattern of chronic inflammation that is encountered in a limited number of infectious and some non-infectious conditions. Briefly, a granuloma is a cellular attempt to contain an offending agent that is difficult to eradicate. In this attempt, there is often strong activation of T lymphocytes leading to macrophage activation, which can cause injury to normal tissues. IL-1 is important in initiating granuloma formation, IL-2 can cause them to enlarge and TNF-α maintains them.
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This question is part of the following fields:
- Inflammatory Responses
- Pathology
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Question 5
Correct
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Inactive cellular marrow is called
Your Answer: Yellow marrow
Explanation:The two types of bone marrow are red marrow, which consists mainly of hematopoietic tissue, and yellow marrow, which is mainly made up of fat cells.
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This question is part of the following fields:
- Haematology
- Physiology
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Question 6
Incorrect
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Question 7
Correct
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The residual cleft of the intermediate lobe of the pituitary is lined by follicles resembling which human gland?
Your Answer: Thyroid
Explanation:The cells located between the two main pituitary lobes form what is known as the intermediate pituitary. This area secretes melanocyte-stimulating hormone and it is only a few cell layers thick. It is rich in follicles filled with colloid, and lined by basophilic cells. This configuration resembles that of another important gland: the thyroid.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 8
Correct
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What is the function of secondary messengers?
Your Answer: Communication from extracellular to intracellular signalling pathways
Explanation:First messengers may not physically cross the phospholipid bilayer to initiate changes within the cell directly. This functional limitation necessitates the cell to devise signal transduction mechanisms to transduce first messenger into second messengers, so that the extracellular signal may be propagated intracellularly. Second messengers are intracellular signalling molecules released by the cell to trigger physiological changes such as proliferation, differentiation, migration, survival, and apoptosis. Secondary messengers are therefore one of the initiating components of intracellular signal transduction cascades.
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This question is part of the following fields:
- Cell Biology
- Physiology
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Question 9
Correct
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Which of the following statements is correct with regards to natural killer cells?
Your Answer: They are part of the innate immune system.
Explanation:Natural Killer (NK) Cells are lymphocytes in the same family as T and B cells, which come from a common progenitor. They are cytotoxic CD8 positive cells that do not have the T-cell receptor. They are very big cells with cytoplasmic granules and are designed to kill target cells with a low level of expression of HLA class I molecules. Examples include during a viral infection or on a malignant cell. NK cells display several receptors for HLA molecules on their surface, and when HLA is expressed on the target cell, these deliver an inhibitory signal into the NK cell. Absent HLA molecules on the target cell cause this inhibitory signal to get lost and as a result, the NK cell can then kill its target. Also, NK cells display antibody-dependent cell-mediated cytotoxicity, where antibody binds to antigen on the surface of the target cell. The NK cells then bind to the Fc portion of the bound antibody and kill the target cell.
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This question is part of the following fields:
- Immunology
- Pathology
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Question 10
Correct
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Question 11
Incorrect
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A 60-year-old man presents with marked breathlessness. He has with a history of ischaemic heart disease. On examination, there is coarse bibasal crackles, marked peripheral oedema and chest X-ray taken is consistent with severe pulmonary oedema. RR is 28 per minute. Which receptor is responsible for detecting pulmonary oedema and the subsequent increase in respiratory rate?
Your Answer: Atrial volume receptors
Correct Answer: Juxtacapillary receptors
Explanation:Pulmonary oedema causes stimulation of the Juxtacapillary receptors (J receptors) leading to a reflex increase in breathing rate. These receptors are also thought to be involved in the sensation of dyspnoea. The J receptors are sensory cells and are located within the alveolar walls in juxtaposition to the pulmonary capillaries. Aortic baroreceptor are involved in detecting blood pressure Central chemoreceptors detect changes in CO2 and hydrogen ion within the brain Atrial volume receptors regulate plasma volume
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 12
Incorrect
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The tegmentum as part of the midbrain, contains which cranial nerve nuclei?
Your Answer: CN 3 to 5
Correct Answer: CN 5 to 8
Explanation:The pontine tegmentum also known as dorsal pons is located within the brain stem. Several cranial nerve nuclei are located in the pontine tegmentum. The nuclei of CN V, CN VI, CN VII and CNVIII are located in the pontine tegmentum.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 13
Incorrect
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Which of the following is true with regard to the acini?
Your Answer: Zone 1 is the most susceptible to anoxic stimuli
Correct Answer: The central zone (Zone III) is the least oxygenated
Explanation:The acini are divided into zone I (periportal), zone II (transition zone), and zone III (pericentral). Cells located close to the portal triad or peripheral zone which consist of an arteriole from the hepatic artery are the most oxygenated cells. Cells in the central zone (III) are least oxygenated and mainly receive blood supply from the central vein which is a branch of hepatic vein.
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This question is part of the following fields:
- Gastrointestinal
- Physiology
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Question 14
Correct
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What is the principle site of action of adrenocorticotrophic hormone (ACTH)?
Your Answer: Adrenal gland
Explanation:Adrenocorticotropic hormone, also known as ACTH or corticotropin, is a polypeptide tropic hormone. It is synthesized by the corticotropic cells of the anterior pituitary. It works by regulating the secretion of glucocorticoid hormones from the cortex cells in the adrenal gland. It binds to the melanocortin (MC) 2 receptors on the surface of the adrenal zona glomerulosa cells, producing cortisol.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 15
Incorrect
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A 60-year-old man with trauma to his cervical spine suffers from damage to the ansa cervicalis, resulting to paresis of his infrahyoid muscles. All of the following are considered infrahyoid muscles, except:
Your Answer: Omohyoid
Correct Answer: Mylohyoid
Explanation:Infrahyoid muscles are also known as “strap muscles” which connect the hyoid, sternum, clavicle and scapula. They are located below the hyoid bone on the anterolateral surface of the thyroid gland and are involved in movements of the hyoid bone and thyroid cartilage during vocalization, swallowing and mastication. They are composed of four paired muscles, organized into two layers. Superficial layer consists of the sternohyoid and omohyoid Deep layer consists of the sternothyroid and thyrohyoid.
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This question is part of the following fields:
- Anatomy
- Head & Neck
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Question 16
Incorrect
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Carcinogens found in cigarette smoke can transform proto-oncogenes to oncogenes through:
Your Answer: Inhibiting tumour suppressor genes.
Correct Answer: Point mutations in genomic DNA
Explanation:80% of the pancreatic cancers are environmentally influenced by smoking which increases the risk by 50%. Mutation can occur in the KRAS, p16, SMAD4, and TP53 genes among other tumour suppression genes. Smoking can be implicated in transformation of all these genes. Apart from KRAS all the genes are inactivated in pancreatic cancer. As KRAS is the most commonly altered gene, mutation occurs as point mutation. As smoking is the most common aetiology in pancreatic cancer, and KRAS is the most important gene that is altered. Most commonly cigarette smoke causes point mutation.
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This question is part of the following fields:
- Genetics
- Physiology
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Question 17
Correct
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Which of the following is NOT one of the cardinal features of acute inflammation:
Your Answer: Discharge
Explanation:Classic signs: Rubor (redness) Calor (heat) Dolour (pain) Tumour (swelling) Functio laesa (loss of function) These classic signs are produced by a rapid vascular response and cellular events. The main function of these events is to bring elements of the immune system to the site of injury and prevent further tissue damage.
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This question is part of the following fields:
- Inflammatory Responses
- Pathology
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Question 18
Correct
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Southern Blotting and DNA probes:
Your Answer: DNA fragments are separated by gel electrophoresis and transferred onto membrane sheets in southern blotting
Explanation:A Southern blot is a method used in molecular biology for detection of a specific DNA sequence in DNA samples. Southern blotting combines transfer of electrophoresis-separated DNA fragments to a filter membrane and subsequent fragment detection by probe hybridization. The other forms of blotting involve the use of RNA and proteins.
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This question is part of the following fields:
- Genetics
- Physiology
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Question 19
Correct
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Which of the following normally has a slow depolarizing “prepotential”?
Your Answer: Sinoatrial node
Explanation:There are 2 main types of action potentials (AP) in the heart, the slow response and the fast response:The slow response is initiated by the slow calcium-sodium channels, found in the SA node (which is the natural pacemaker of the heart) and the conduction fibers of the AV node.The fast response occurs in the atrial and ventricles muscle cells and the purkinje fibers.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 20
Incorrect
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What is the normal amount of oxygen that is carried in the blood?
Your Answer: 40 ml oxygen/100 ml blood
Correct Answer: 20 ml oxygen/100 ml blood
Explanation:Normally, 100 ml of blood contains 15g haemoglobin and a single gram of haemoglobin can bind to 1.34 ml oxygen when 100% saturated. Thus, 15 × 1.34 = 20 ml O2/100 ml blood. The haemoglobin in venous blood that is leaving the tissues is about 75% saturated with oxygen, and hence it carries about 15 ml O2/100 ml venous blood. This implies that for each 10 ml of blood, 5 ml oxygen is transported to the tissues. With a p(O2) > 100 mm Hg, only 3 ml of oxygen is dissolved in every one litre of plasma. By increasing the pA(O2) by breathing 100% oxygen, one can add an extra amount of oxygen in the plasma, but the amount of oxygen carried by haemoglobin will not increase significantly as it is already > 95% saturated.
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This question is part of the following fields:
- Haematology
- Physiology
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SESSION STATS - PERFORMANCE PER SPECIALTY
