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Transitional epithelium is present in all except:
Answer C. Membranous urethra
- Transitional epithelium is a type of stratified epithelium.
- This tissue consists of multiple layers of epithelial cells that can contract and expand in order to adapt to the degree of distension needed.
- Transitional epithelium lines the organs of the urinary system and is known here as urothelium.
- Transitional epithelium is a layer of cells that forms the mucosal lining of your ureters, a portion of your urethra, and your urinary bladder.
- These cells are called transitional because they can undergo a change in their shape and structure.
- The appearance of transitional epithelium depends on the layers in which it resides.
- Cells of the basal layer are cuboidal, or cube-shaped, and columnar, or column-shaped, while the cells of the superficial layer vary in appearance depending on the degree of distension.
- These cells appear to be cuboidal with a domed apex when the organ or the tube in which they reside is not stretched.
- When the organ or tube is stretched (e.g. when the bladder is filled with urine), the tissue compresses and the cells become stretched.
- When this happens, the cells flatten, and they appear to be squamous and irregular.
- The transitional epithelium cells stretch readily in order to accommodate fluctuation of the volume of the liquid in an organ (the distal part of the urethra becomes non-keratinized stratified squamous epithelium in females; the part that lines the bottom of the tissue is called the basement membrane).
- The transitional epithelium also functions as a barrier between the lumen, or inside the hollow space of the tract that it lines and the bloodstream.
Pretracheal layer of cervical fascia encloses all except :
Answer A. Sternocleidomastoid
- The pretracheal fascia (encloses the visceral region of the neck)
- The pretracheal fascia is a portion of the structure of the human neck.
- It extends medially in front of the carotid vessels and assists in forming the carotid sheath.
- It is continued behind the depressor muscles of the hyoid bone, and, after enveloping the thyroid gland, is prolonged in front of the trachea to meet the corresponding layer of the opposite side.
- The pretracheal layer of the deep cervical fascia passes in front of the carotid sheath (i.e., common carotid artery, internal jugular vein and vagus) and in front of the cervical viscera (larynx, esophagus, and pharynx).
- Above, it is fixed to the hyoid bone, while below it is carried downward in front of the trachea and large vessels at the root of the neck, and ultimately blends with the fibrous pericardium.
- This layer is fused on either side with the prevertebral fascia, and with it completes the compartment containing the larynx and trachea, the thyroid gland, and the pharynx and esophagus.
- It encloses the thyroid and is responsible for its movement during deglutition.
- The pretracheal fascia has two components which are continuous layers of fascia.
- A cervical layer that ensheaths cervical viscera including the larynx/trachea, pharynx/esophagus, thyroid and parathyroid glands, and then a muscular layer which ensheaths the infrahyoid muscles.
All of the following are the contents of the carotid sheath. Except?
|A.||Internal carotid artery
|B.||Internal jugular vein
|C.||Cranial nerve IX
Answer C. Cranial nerve IX
- It’s a tubular consolidation of deep cervical fascia around the (a) common carotid and internal carotid arteries and (b) internal jugular vein and vagus nerve (X).
- The vagus nerve is located posteriorly between the vein and arteries.
- The carotid sheath is wedged between the 3 layers of the deep cervical fascia, i.e., investing layer, pretracheal fascia, and prevertebral fascia and connected to all these layers by free areolar tissue.
- The sheath is thick around common and internal carotid arteries but thin over the internal jugular vein to be able to enable its growth during increased venous return.
- Carotid sheath goes from the base of the skull above to the arch of aorta below.
- At the base of the skull, it’s connected to the margins of the carotid canal and jugular fossa. At the arch of aorta, it combines with its tunica adventitia.
- The ansa cervicalis is embedded in the anterior wall of the carotid sheath.
- The cervical sympathetic chain is closely linked to the posterior wall of the sheath, plastered to the prevertebral fascia.
- The malignant and tuberculous deep cervical lymph nodes often become adherent to the internal jugular vein.
- For that reason, during block dissection of the neck, the sheath is exposed and a portion of the vessel is resected to ease removal of these lymph nodes.
Level of decussation of superior cerebellar peduncle:
Answer A. Midbrain
- The decussation of superior cerebellar peduncle is the crossing of fibers of the superior cerebellar peduncle across the midline, and is located at the level of the inferior colliculi.
- It comprises the cerebellothalamic tract, which arises from the dentate nucleus, as well as the cerebellorubral tract, which arises from the globose and emboliform nuclei and project to the contralateral red nucleus to eventually become the rubrospinal tract.
- It is important as an anatomical landmark, as lesions above it cause contralateral cerebellar signs, while lesions below it cause ipsilateral cerebellar signs.
Fornix Fibers is the example of :
|B.||Association and Projection
Ans D. Commissural
Key facts about Fornix Fibers
How will you differentiate exocrine pancreas and parotid?
|A.||Serous acini supported by basal lamina
|B.||Presence of acidophilic serous acini at the tip
|C.||Absence of striated duct
|D.||Apical acinar villi
Answer C. Absence of striated duct
The parotid gland has striated and intercalated intralobular ducts, whereas the pancreas has ONLY the intercalated type.
The excretory ducts of the parotid gland are STRATIFIED columnar, whereas the excretory duct of the pancreas is SIMPLE columnar.
The pancreas has ISLETS of LANGERHANS, whereas the parotid does not.
Ito cells in Liver present in:
|B.||Space of Disse
|C.||Space of mall
Answer B. Space of Disse
- Hepatic stellate cells (here HSC), also known as perisinusoidal cells or Ito cells (earlier lipocytes or fat-storing cells), are pericytes found in the perisinusoidal space of the liver, also known as the space of Disse (a small area between the sinusoids and hepatocytes).
- The stellate cell is the major cell type involved in liver fibrosis, which is the formation of scar tissue in response to liver damage.
- In normal liver, stellate cells are described as being in a quiescent state.
- Quiescent stellate cells represent 5-8% of the total number of liver cells.
- Each cell has several long protrusions that extend from the cell body and wrap around the sinusoids.
- The lipid droplets in the cell body store vitamin A as retinol ester.
Injury to which part of the internal capsule causes C/L Hemiplegia?
Answer D. Posterior Limb
Anatomy of the Internal Capsule
- The internal capsule is one of the subcortical structures of the brain.
- Subcortical structures: internal capsule, caudate, putamen, globus pallidus, thalamus, brainstem.
- The anterior limb of the internal capsule separates the caudate nucleus and the lenticular nucleus.
- The posterior limb separates the thalamus and lenticular nucleus
Types of fibers
- Anterior limb: frontopontine fibers (frontal cortex to pons), thalamocortical fibers (thalamus to frontal lobe)
- Genu (angle): corticobulbar fibers (cortex to the brainstem)
- Posterior limb: corticospinal fibers (cortex to spine), sensory fibers
- Anterior limb: mainly fed by the lenticulostriate branches of the middle cerebral artery(MCA), less often branches of the anterior cerebral artery (ACA)
- The lenticulostriate arteries are small penetrating blood vessels that supply blood flow to most of the subcortical structures.
- Genu: lenticulostriate branches of MCA
- Posterior limb: lenticulostriate branches of MCA & anterior choroidal artery (AChA) of internal carotid artery
Clinical Findings in Internal Capsular Stroke
Symptoms and Signs
The weakness of the face, arm, and/or leg (pure motor stroke)
- Known as one of the classic types of lacunar infarcts, a pure motor stroke is the result of an infarct in the internal capsule.
- Pure motor stroke caused by an infarct in the internal capsule is the most common lacunar syndrome.
Upper motor neuron signs
- hyperreflexia, Babinski sign, Hoffman present, clonus, spasticity
Mixed sensorimotor stroke
- Since both motor and sensory fibers are carried in the internal capsule, a stroke to the posterior limb of the internal capsule (where motor fibers for the arm, trunk and legs and sensory fibers are located) can lead to contralateral weakness and contralateral sensory loss.
Hemiplegia is a condition caused by brain damage or spinal cord injury that leads to paralysis on one side of the body. It causes weakness, problems with muscle control, and muscle stiffness.
Medial lemniscus is formed from:
Answer A. Fasciculus gracilis
The medial lemniscus, also known as Reil’s band or Reil’s ribbon, is a large ascending bundle of heavily myelinated axons that decussate in the brainstem, specifically in the medulla oblongata.
The medial lemniscus is formed by the crossings of the internal arcuate fibers. The internal arcuate fibers are composed of axons of nucleus gracilis and nucleus cuneatus.
- The axons of the nucleus gracilis and nucleus cuneatus in the medial lemniscus have cell bodies that lie contralaterally.
- The medial lemniscus is part of the dorsal column–medial lemniscus pathway, which ascends from the skin to the thalamus, which is important for somatosensation from the skin and joints, therefore, lesion of the medial lemnisci causes impairment of vibratory and touch-pressure sense.
Patient with hemiplegia, Horner syndrome, and diagnosed with Wallenberg Syndrome. Which Artery will be involved?
|A.||Anterior inferior cerebellar artery
|B.||Posterior inferior cerebellar artery
|D.||Posterior cerebral artery
Answer B. Posterior inferior cerebellar artery
- Wallenberg syndrome is also known as lateral medullary syndrome or the posterior inferior cerebellar artery syndrome.
- Common symptoms with lateral medullary syndrome may include: difficulty swallowing, or dysphagia.
- This can be caused by the involvement of the nucleus ambiguus, as it supplies the vagus and glossopharyngeal nerves.
- Slurred speech (dysarthria), and disordered vocal quality (dysphonia) are also common.
- The damage to the cerebellum or the inferior cerebellar peduncle can cause ataxia.
- Damage to the hypothalamospinal fibers disrupts sympathetic nervous system relay and gives symptoms that are similar to the symptoms caused by Horner syndrome – such as miosis, anhidrosis and partial ptosis.
What is the mechanism of Transcutaneous electrical nerve stimulation (TENS)?
|A.||Inhibitory neurotransmitter at the spinal cord.
|B.||Adrenergic receptor stimulation
|C.||Gating at spinal cord
Ans. C. Gating at spinal cord
- Modulation of the Gate-control mechanism of pain serves as a rationale behind the use of Transcutaneous electrical nerve stimulation (TENS) for pain relief.
- The method uses electrodes to activate Aα, Aβ fibers near the injury.
18. Gibbs-Donnan equilibrium is mainly due to?
Ans. D Proteins
- Donnan & Gibbs showed that in the presence of a non-diffusible ion (protein), diffusible ions distribute themselves differentially at equilibrium.
Which of the following is true regarding G6PD deficiency?
|A.||RBCs are resistant to hypotonic solutions.
|B.||RBCs undergo lysis in hypotonic solutions.
|D.||Presence of spherocytes.
Ans. B RBCs undergo lysis in hypotonic solutions.
- The most common medical problem associated with glucose-6-phosphate dehydrogenase deficiency is hemolytic anemia, which occurs when red blood cells are destroyed faster than the body can replace them.
- The more nearly spherical are erythrocytes, the greater is their susceptibility to hemolysis by swelling in hypotonic solutions.
- There has been no satisfactory explanation, however, for the increased destruction of red blood cells of such increased “fragility” in vitro (e. g., congenital hemolytic jaundice) under presumably isotonic conditions in vivo.
- It has now been demonstrated that sterile incubation in vitro and intravascular stasis in vivo produces progressive swelling of erythrocytes, spheroidicity, and increasing osmotic fragility, such that hemolysis eventually occurs in isotonic solutions.
Calculate ECF volume from the following. Mannitol injection given (100mg), urinary excretion 1% & plasma concentration of mannitol is 6.4mg/100ml.
Ans: A 14L
- This is based on formula
- V=(Q-e) / c =
- 10 – 1 g / (65 mg / 100 ml) =
- 900 mg / 65 mg/ml =
- 13.8 L
In negative feedback, feedback gain is infinity , which of the following?
Temperature control in hypothalamus
Blood volume control by kidney
|C.||Blood pressure control by baroreflex
No such infinite feedback gain is not possible
Ans: B. Blood volume control by kidney
- For a given intake of salt and a given functional state of the kidneys, there is only one single atrial pressure that will provide a balance between intake and output of salt.
- Because of this, the atrial pressure, in the long run, will always attempt to return exactly to this pressure level (Error = 0).
In skeletal muscle, signaling mechanisms between Ryanodine receptors & DHPR receptors are coupled by?
|C.||Electrically followed by mechanically
Ans. C Electrically followed by mechanically
- The electrical activation of T-tubules activates the DHPR channel on it, which interacts mechanically with RyR on the SR membrane for the release of calcium.
- Next, best option will be mechanically (If not provided in options).
Which of the following acts through intracellular receptor?
Ans. A Thyroxin
- Thyroxine has an intranuclear receptor. All other options have membrane receptors.
- Pg. 267 – Gobind Rai Garg 14 th edition.
Single neuron is receiving an impulse from 400 other sensory cells? Which of the following best explains the mechanism behind this?
Ans. C Convergence
This mechanism is an example of the convergence of input signals (Excitatory or inhibitory) from multiple sources
Which of the following is not required for the calculation of creatinine clearance:
|C.||Volume of Urine in 24hrs
|D.||Volume of serum in the kidney in 24hrs.
Ans- D- Volume of serum in the kidney in 24hrs
- Creatinine is an endogenous metabolite that is free. filtered by the glomerulus but is also actively secreted by the proximal tubular cells into the renal tubules in very small amounts.
- In the steady-state, such ‘Secretion’ of creatinine into the tubules is minimal, however as GFR falls creatinine secretion is increased.
- Creatinine clearance is calculated by Cockcroft-Gault equation :-
- Creatinine clearance = [(140 – Age in yrs) x weight in kg ] / 72 x serum creatinine in mg/dl.
- Creatinine clearance in a healthy young person is about 95 milliliters per minute for women/120 milliliters per minute for men. This means that each minute, that person’s kidneys clear 95-120 mL of blood free of creatinine
Type IV Hypersensitivity reaction is due to:
|A.||Innate immune response
|B.||Antibody and cell-mediated
Ans- C- Cell-mediated immunity
- Delayed-type hypersensitivity, cell-mediated immune memory response, antibody-independent-
- Hypersensitivity is the over-reaction of the immune system, often resulting in unwanted tissue destruction. This is
- responsible for caseation and lung cavity formation in the case of tuberculosis, granulomatous skin lesions in
- tuberculoid leprosy, skin lesions in herpes simplex and contact hypersensitivity to chemicals and plant
Choose the correct statement regarding why glucose is stored in glycogen form:
|A.||It is compact
|B.||Multiple Reducing Ends
|C.||Can Provide Glucose as and when needed for 1 Week
|D.||It can be stored at multiple sites
Ans- A- It is Compact
- Glucose is a monosaccharide while glycogen is a polysaccharide. It is therefore a more complex sugar than glucose. … If there is an excess of glucose in the system then it will be converted and then stored as glycogen in the liver. This stored form of glucose is made up of many connected glucose molecules and is called glycogen.
Enzyme not included in cholesterol synthesis.?
|A.||HMG CoA Synthase
|B.||HMG CoA Reductase
|C.||HMG CoA Lyase
Ans-c- HMG CoA Lyase
HMG CoA lyase is an enzyme for ketogenesis. HMG-CoA reductase is the rate-limiting enzyme of cholesterol synthesis. HMG CoA is reduced to mevalonate by NADPH in this reaction
Ammonia is detoxified in brain as:
Ans. B Glutamine
The brain’s uptake of glutamate is approximately balanced by its output of glutamine.
Glutamate entering the brain associates with ammonia and leaves as glutamine.
The glutamate-glutamine conversion in the brain—the opposite of the reaction in the kidney that produces some of the ammonia entering the tubules—serves as a detoxifying mechanism to keep the brain free of ammonia.
Ammonia is very toxic to nerve cells, and ammonia intoxication is believed to be a major cause of the bizarre neurologic symptoms in hepatic coma.
18. Intercellular Communication in Eukaryotes meditated by:
Ans- A. Exosomes
Cellular communication in mammals can be mediated by the exchange of genetic information mainly microRNA. This can occur when extracellular vesicles such as exosomes secreted by donor cells and are taken up by an acceptor cell. Specialized cell-cell contact, such as synapses have the potential to combine these modes of genetic transfer.
Which enzyme is common to both gluconeogenesis and glycolysis pathway:
Ans- d. Phosphoglycerate kinase
- Seven of the reactions of glycolysis are reversible and are used in the synthesis of glucose by gluconeogenesis. Thus, seven enzymes are common to both glycolysis and gluconeogenesis
- Phosphohexose isomerase; (ii) Aldolase; (iii) Phosphotriose isomerase; (iv) Glyceraldehyde 3-phosphate dehydrogenase; (v) Phosphoglycerate kinase; (vi) Phosphoglycerate mutase; (vii) Enolase
- Three of the reactions of glycolysis are irreversible and must be circumvented by four special reactions which are unique to gluconeogenesis and catalyzed by : (i) Pyruvate carboxylase, (ii) PEP carboxykinase, (iii) Fructose-1, 6- bisphosphatase, (iv) Glucose-6-phosphatase
Hayflick limit is:
|A.||Free radical oxidation levels in the cell
|B.||Total number of time a cell can divide in his lifetime
|C.||Induction of lactose to form allolactose
|D.||Coordinated regulation that keeps the synthesis of r-proteins in balance with the transcription of rRNA.
Ans B. Total number of time a cell can divide in his lifetime.
Cellular senescence, a phenomenon also referred to as the “Hayflick limit”, is defined as a process limiting the proliferation of normal human cells in culture. It was first described by Hayflick and Moorhead in 1961 in a seminal study on human fibroblasts. They identified one particular type of cellular senescence produced after extensive proliferation by the loss of telomeres following a significant drop in endogenous telomerase activity
Calculate total sodium deficit in 60kg male patient with serum Na+ 120meq/L, serum K+ 4meq/L and serum Cl– 90meq/L:
Ans- d 720
Sodium deficit- Estimates the total amount of sodium that needs to be replaced in hyponatremia patients.
Sodium deficit (meq) = Normal TBW * (140 – sNa)
(The easy way to calculate total body water is simply to multiply 0.6 times your weight in kilograms, since roughly 2/3 of your body weight is water)
= 0.6 х 60 * (140-120)
= 36 * 20
|D.||Both B and C.
Ans A- DNA Replication
DNA methylation alters the gene expression pattern in cells-for example it may cause X-chromosome inactivation, tumor suppressor gene silencing (in cancer cells)Q, suppression of expression of repetitive elements / deleterious elements / viral genes that have been incorporated into genome of the host over time E. coli has 2 prominent DNA methylation system, both using S-adenosyl methionine (SAM) as a methyl group donor.
Adenine or cytosine methylation in the restriction-modification system serves as a part of the defence mechanism against cleavage by a restriction endonuclease.
It helps the cell to distinguish its DNA from foreign DNA by marking its own DNA with a methyl group and destroying (foreign) DNA without the methyl group.
Mismatch repair mediated by Dam methylase improves the overall fidelity of DNA replication by fctor of 102 -103.
Dam (DNA adenine Methylation) methylase, methylates adenosine residue to N6– methyl adenosine within the palindromic (51) GATC (3) sequence.
This can serve to distinguish parent/template strand (which is methylated) from unmethylated (transient state) newly synthesized strand.
Which pathway describes the synthesis of metabolically active vitamin D
|A.||Liver -> Skin -> Kidney
|B.||Skin -> Liver -> Kidney
|C.||Skin -> Lungs -> Kidney
|D.||Kidney -> Lungs -> Skin
Ans B- Skin-Liver-Kidney
Vitamin D is derived either from 7-dehydrocholesterol or ergosterol by the action of ultraviolet radiations. 7-dehydrocholesterol, an intermediate of a minor pathway of cholesterol synthesis, is available in the Malpighian layer of epidermis.
In the skin, ultraviolet light (290-315 nm) breaks the bond between positions 9 and 10 of the steroid ring. So, the ring B is opened, to form the provitamin, secosterol.
The cis double bond between 5th and 6th carbon atoms is then isomerized to a trans double bond (rotation on the 6th carbon atom) to give rise to vitamin D3 or cholecalciferol. So, vitamin D is called the “sun-shine vitamin”
Which of the following will be incorrect in Rh incompatibility:
|A.||Raised unconjugated bilirubin
|B.||Raised stool urobilin
|C.||Raised urine urobilin
|D.||Increased urine bilirubin
Ans. D-Increase urine bilirubin
Higher-than-normal levels of bilirubin in your infant’s blood is a sign of Rh incompatibility. In a full-term baby who is less than 24 hours old, the levels of bilirubin should be less than 6.0 milligrams per deciliter. Signs of red blood cell destruction in your infant’s blood may indicate Rh incompatibility
Components of Apoptosome complex with cyt c and APAF 1
answer: b Procaspase 9
cytochrome c from mitochondria to cytosol, where it binds to Apaf-1 to form a procaspase-9-activating heptameric protein complex named apoptosome.
Which of the following is true about celiac disease
|A.||MC in developing countries
|B.||Improved with complex carbohydrates such as bread and pasta
|C.||Associated with increased risk of lymphomas
|D.||It is not associated with dermatitis herpetiformis
Answer: c Associated with increased risk of lymphomas
- It is an inflammatory disorder of the small bowel occurring in genetically susceptible individuals, which results from intolerance to wheat gluten and similar proteins found in rye, barley and, to a lesser extent, oats.
- Serum antibodies-IgA antigliadin, anti endomysial, and anti-tTG antibodies- are present.
- Celiac disease is associated with dermatitis herpetiformis (DH).
- The most important complication of celiac disease is the development of cancer.
Which of the following is not true regarding stem cells
|A.||Replication via an unregulated pathway
|B.||Regeneration of damaged organs
|C.||ability to differentiate
|D.||Ability to replicate
Answer: A Replication via an unregulated pathway
- capability of dividing and renewing themselves for long periods being unspecialized and basic cells
- they can give rise to any type of specialized cell because they are unspecialized
- Ability to differentiate, replicate, regeneration of damaged organs
Which of the following are cross-linking fixatives except
|D.||None of the above
Answer: A Methanol
Cross-Linking fixatives act by creating covalent chemical bonds between proteins in tissue. This anchors soluble proteins to the cytoskeleton and lends additional rigidity to the tissue.
The most commonly used fixative in histology is formaldehyde. Another popular aldehyde for fixation is glutaraldehyde.
Methanol is precipitating fixative
Genexpert tells resistance due to mutation in?
Answer : (D) rPOB (A) INH A
Gene expert tells resistance to rifampicin and isoniazid
Rpob – tells mutation to rifampicin
Which of the following is not alpha synucleinopathy?
|C.||Multisystem atrophy and frontotemporal dementia
|D.||Lewy body dementia
Answer: a -Alzheimer’s disease
Synucleinopathies (also called α-Synucleinopathies) are neurodegenerative diseases characterized by the abnormal accumulation of aggregates of alpha-synuclein protein in neurons, nerve fibers or glial cells.
There are three main types of synucleinopathy:
- Parkinson’s disease (PD),
- dementia with Lewy bodies (DLB), and
- multiple system atrophy (MSA).
- Other rare disorders, such as various neuroaxonal dystrophies, also have α-synuclein pathologies.
Appropriate FNAC specimen in case of thyroid
|A.||6-8 follicular cell contain 10-12 cell
|B.||10 follicular cell contain 6-8 cell
|C.||C)3 follicular cell groups with each 10-15 cells
|D.||6 follicular cell groups with each 10-15 cells
Answer- D 6 follicular cell groups with each 10-15 cells
For a thyroid FNA specimen to be satisfactory for evaluation (and benign), at least 6 groups of benign follicular cells are required, each group composed of at least 10 cells.
American society of for clinical pathology abstract, The Bethesda System for Reporting Thyroid Cytopathology
True regarding liquid-based cytology is?
|A.||Study of fluid in the tumor microenvironment
|B.||A urine specimen is used
|C.||Sanger sequencing is used
|D.||DNA collected in microvessels
Ans: B. A Urine specimen is used
Though urine specimen is not preferred but it is used.
Liquidbased cytology (LBC) is a thinlayer or monolayer slide preparation technology that has been introduced as a potential solution to overcome the shortcomings of conventional Papanicolaou (Pap) smears in cervical cancer screening.
Liquidbased cytology from gynecologic samples uses a collection fluid that fixes, homogenizes, and rinses the cells.
Preventive & Social Medicine
Due to the use of a spurious drug, the person died. Penalty to the owner of the company is
Ans: A: Life
The penalty for manufacture and sale of spurious drugs should be enhanced from five years to not less than seven years.
The maximum penalty for the manufacture and sale of spurious drugs causing grievous hurt or death should be enhanced from life imprisonment to death.
Which of the following are not included in sentinel surveillance of HIV?
|B.||STD clinic attendees
Answer is B. STD Clinic Attendees.
- Sentinel surveillance includes High-risk groups, Bridge population, and General Population
- High-risk groups include Intravenous drug abusers, Males having sex with males, Female sex workers and transgender
- Bridge population includes single male migrants and long-distance truckers
- The general population includes pregnant women attending antenatal clinic
Park 25 th edition, Pg. No. 467
Rashtriya Bal Suraksha prevention against all except
|C.||Retinopathy of prematurity
|D.||Vitamin A deficiency
Ans: A (Congenital glaucoma)
Child Health Screening and Early Intervention Services under RBSK envisages to cover 30 selected health conditions for screening, early detection, and free management in the age group birth to 16 years.
The eye conditions that are covered are:
- Vitamin A deficiency
- Congenital cataract
- Vision impairment
- Retinopathy of prematurity
Glasgow Coma scale is
Answer is B. Ordinal scale
Ordinal Scale. Glasgow Coma Scale has perfect ranking and order.
Score of ≤ 8 – Severe Brain Injury; 9 – 12 – Moderate brain injury; ≥13 – Mild brain Injury
The dose of IFA tablet for an adolescent girl under Anemia Mukt Bharat
|A.||60mg of elemental iron and 500 mcg of Folic acid daily once
|B.||60 mg of elemental iron and 500 mcg of folic acid weekly once
|C.||60 mg of elemental iron and 400 mcg of folic acid daily once
|D.||60 mg of elemental iron and 400 mcg of folic acid weekly once
Ans: B. 60 mg of elemental iron & 500 mcg of folic acid weekly once.
Blue colored tablet with a dose of 60 mg of elemental iron and 500 mcg of folic acid is given weekly once to adolescents (10 – 19 years) under the Anemia Mukt Bharat programme.
Rubber latex gloves discarded in
Ans: A. Red bag
Red bag (Gloves – recyclable waste; Recyclable waste – Red bag)
Pacemaker from the cardiac wing in hospitals is discarded in
Ans: B. Blue bag (Pacemaker – Metallic Implant; Metallic Implant & glassware – Blue bag)
Biosafety level for SARS 2 Stability Level?
Answer is D. BSL 4
- Activities and projects conducted in biological laboratories are categorized by biosafety level.
- The four biosafety levels are BSL-1, BSL-2, BSL-3, and BSL-4, with BSL-4 being the highest (maximum) level of containment.
> Safety level of lab Used for
>Biosafety Level 1 (BSL 1) Study infectious agents or toxins not known to consistently cause diseases in healthy adults
> Biosafety Level 2 (BSL 2) Study moderate-risk infectious agents or toxins that pose a risk if accidentally inhaled, swallowed, or exposed to the skin.
> Biosafety Level 3 (BSL 3) Study infectious agents or toxins that may be transmitted through the air and cause potentially lethal infection through inhalation exposure
>Biosafety Level 4 (BSL 4) Study infectious agents or toxins that pose a high risk of aerosol-transmitted laboratory infections and life-threatening disease for which no vaccine or therapy is available.
Formula for maternal mortality ratio is
|A.||Total no. of Maternal deaths/Total no. of live births ×1000
|B.||Total no. of Maternal deaths/ Total no. of pregnant women ×100000
|C.||Total no. of Maternal deaths/Total no. of live births ×100000
|D.||Total no. of Maternal deaths/ Total no. of women in reproductive age group ×100000
Answer is C. Total no. of Maternal deaths/Total no. of live births ×100000
Total no. of Maternal deaths/Total no. of live births×100000
Current MMR of India (SRS 2019) – 122/ 1lakh live births
Nikshay Poshan Yojana is
|B.||Mid-day meal everyday
|C.||Rs. 500/Month financial benefit
|D.||Rs. 500/week financial benefit
Ans: C. Rs. 500/Month financial benefit
Rs. 500/month. The financial benefit of Rs. 500/month for nutritional support for tuberculosis patients until completion of treatment is being provided.
In women to calculate RDA which parameters are used?
|A.||1.73 m height
|C.||10 hours of work
|D.||18 – 30 year age group
Ans: B. 55kgs weight.
Reference women – 1.61 meters height, 55 kgs weight, 18 – 29 years age group and 8 hours of work.
Which among the following is a non-parametric test
|C.||Pearson correlation test
Answer is D. Friedman test.
The Friedman test is the non-parametric alternative to the one-way ANOVA with repeated measures. It is used to test for differences between groups when the dependent variable being measured is ordinal. It can also be used for continuous data that has violated the assumptions necessary to run the one-way ANOVA with repeated measures (e.g., data that has marked deviations from normality)
Diagnostic test for Vertigo?
|C.||CT of the head
Answer : A. Dix Hallpike
The Dix–Hallpike test — or Nylen–Barany test — is a diagnostic maneuver used to identify benign paroxysmal positional vertigo (BPPV).
The maneuver, when properly employed, can identify a common, benign cause of vertigo, which can then be treated with bedside maneuvers, often providing instant relief to patients.
The Dix-Hallpike maneuver is the gold standard for diagnosing benign positional paroxysmal vertigo caused by free-floating debris (often in the form of calcium carbonate stones, termed otoliths) in the semicircular canals of the inner ear.
The patient is positioned recumbent with the head back and toward the affected ear, causing the otolith to progress superiorly along the natural course of the canal.
Typically, after a five to 20-second delay, this will cause vertigo and rotary or up-beating nystagmus, which will resolve within 60 seconds.
All of the following are features of corneal epithelium except?
|A.||Lined by stratified squamous epithelium
|B.||Bowman’s membrane regenerate
|C.||The external surface is having ciliated epithelium
|D.||Dividing cells are present at the limbus
Ans-B- Bowman’s membrane regenerate
Layers of the cornea:
Epithelium (Stratified squamous non-keratinized epithelium)
- A single layer of columnar basal cells attached by hemidesmosomes to an underlying basement membrane
- Two to three strata of “wings” cells
- Two layers of squamous surface cells
- Limbal stem cells are located at the corneoscleral limbus, possibly in the palisades of Vogt. Deficiency may result in chronic epithelial defects and & conjunctivalization (epithelial instability, vascularization, and the appearance of goblet cells)
- Complete turnover of corneal epithelial cells occurs in about 7–10 days.
- An acellular superficial layer of the stroma formed from collagen fibers.
- Does not regenerate
Stroma (Thickest layer):
- Makes up 90% of corneal thickness
- Contains chondroitin sulphate (dermatan sulphate) and keratan sulphate.
- The stroma can scar, but cannot regenerate the subsequent damage.
Dua’s layer (Newly discovered, toughest layer)
- Ends peripherally as Schwalbe’s ring
- Has regenerative potential
Endothelium (Most metabolically active layer)
- Does not regenerate
- Endothelial cells maintain corneal deturgescence
- The young adult cell density is about 3000 cells/mm 2
- The number of cells decreases at about 0.6% per year and neighboring cells enlarge to fill the space; the cells cannot regenerate.
- At a density of about 500 cells/mm 2 corneal edema develops and transparency is impaired. This is called a critical density.
If the numerical aperture of the objective lens is increased, what is the change in the image observed?
|D.||Increased working distance
Ans: C. Increased resolution
The bigger a cone of light that can be brought into the lens, the higher its numerical aperture is. Therefore the higher the numerical aperture of a lens, the better the resolution of a specimen will be which can be obtained with that lens. All high dry lenses work in the air which has a refractive index of 1.0.
Which anticonvulsant drug causes bilateral concentric visual field loss
Ans: A. Vigabatrin
The majority of epilepsy patients need long‐term antiepileptic drug (AED) therapy during some phase of their life.
Vigabatrin is a selective irreversible inhibitor of the metabolizing enzyme, GABA‐transaminase.
Vigabatrin, a gabaergic antiepileptic drug, causes concentric visual field defects.
Pupil do not change in dark:
|D.||Argyll Robertson’s pupil
Ans: b. Adie’s pupil
Adie’s tonic pupil:
- Light reflex absent or sluggish.
- Near reflex: Slow and tonic
- Pupil is larger
- Constricts with weak pilocarpine.
Patient walks with left head tilted and exaggerated on dextroversion & right head tilt and what is the diagnosis?
|A.||Right Superior oblique
|B.||Right inferior oblique
|C.||Right Superior rectus
|D.||Left lateral rectus
Ans: A. Right Superior oblique
In a patient with superior oblique paralysis, tilting the head to the affected side induces otolith
stimulation for ocular counter-rolling, resulting in the intorsion of the lower eye. In patients without
ocular disorders, the superior rectus and oblique muscles are both incyclotortors of the lower eye.
When there is paralysis of the superior oblique, the exaggerated tilt and turn test activates another
intortor of the eye (the superior rectus), resulting in hypertropia of the lower eye.
Simultaneously turning the face to the same side as the head tilt promotes the adduction of the lower eye and activation of the inferior oblique muscles the antagonist of the superior oblique; this results in hypertropia of the adducted eye.
In this new position, both the superior rectus and inferior oblique muscles are activated and result in hypertropia of the eye affected with superior oblique muscle paresis.
In the traditional three-step test, the second step stimulates only the inferior oblique of the affected eye, resulting in hypertropia. In the third step, only the superior rectus muscle of the affected eye is stimulated, also resulting in hypertropia.
However, in this new test, both the superior rectus and inferior oblique muscles of the affected eye are simultaneously stimulated and result in noticeably more pronounced hypertropia than the second or third step of the three-step test alone.
A patient underwent LASIK for Myopia and is now undergoing cataract surgery. Formula used for lens power calculation
Ans: A. SRL 1
For a patient undergoing cataract surgery who already underwent Lasik for myopia Haigis formula is used to calculate the power of the lens.
A patient was prescribed Eye ointment and eye drop for the right eye. What should nurse do
|A.||Apply eye ointment and then after 15-minute eye drop
|B.||Apply eye ointment and then eye drop immediately
|C.||Apply eye drops and then 15 minutes after eye ointment
|D.||Apply eye drop and then eye ointment immediately
Answer :- is C. Apply eye drops and then 15 minutes after eye ointment
If you’re using eye drops and an eye ointment, use the eye drops first. Then apply the ointment at least 15 minutes later.
Do understand it’s normal for some liquid to flow onto the skin around your eyes after using an eye drop.
Nerve which is least likely to be injured in thyroid surgery
Ans- A. Marginal mandibular
If an RLN is injured during surgery and the transected ends are identified, they should be re- anastomosed. In the event that a length of the nerve is excised (due to invasion by malignancy for example), anastomosis of the ansa cervicalis may be considered.
Injury to the external branch of the superior laryngeal nerve is more common because of its proximity to the superior thyroid artery. This leads to loss of tension in the vocal cord with diminished power and range in the voice.
Reference – Bailey and Love, Short practice of surgery, 27th edition published in 2018 Pg 815
A 26 year male patient with RTA complicated with pelvic fracture has been resuscitated with blood transfusion but after a few hours he develops fever, hypotension, normal CVP, oliguric and bleeding from IV line and Nasogastric tube. What will be the diagnosis?
|D.||Acute adrenal insufficiency
Answer- B. Fat Embolism
Fat embolism may follow major bony fractures.
Clinical signs are evident hours to days after an injury involving long bone or pelvic fractures which are characterized by respiratory distress, altered mental status, and skin petechiae.
Transfusion-related complications are primarily related to blood-induced proinflammatory responses.
The most common cause of shock in the surgical or trauma patient is loss of circulating volume from hemorrhage
Adrenal insufficiency- The most commonly encountered causes of primary adrenal insufficiency are autoimmune disease, infections, and metastatic deposits.
Acute adrenal insufficiency should be suspected in stressed patients.
Sabiston textbook of surgery, The biological basis of modern surgical practice, 19th edition, Pg 517
Schwartz’s Principles of Surgery, 10th edition, Pg 101, 119, 1509
Mesh for hernioplasty should be
|A.||Heavyweight large pore’
|B.||Lightweight small pore
|C.||Lightweight large pore
|D.||Heavyweight small pore
Ans- C. Lightweight large pore
The term ‘mesh’ refers to prosthetic material, either a net or a flat sheet, which is used to strengthen a hernia repair.
Meshes with thinner strands and larger spaces between them, ‘lightweight, large-pore meshes’, are preferred for hernioplasty because they have better tissue integration, less shrinkage, more flexibility and improved comfort.
Bailey and Love, Short practice of surgery, 27th edition published in 2018 Pg 1028
A 23 yr old male was burnt due to a burst stove. There is singeing of hairs in his face and neck area. And is covered on both sides and both arms are burnt. Calculate the percentage
Answer B. 27%
Wallace’s Rule of 9
Face and neck = 9%. Both arms = 18%
Bailey and Love, Short practice of surgery, 27th edition published in 2018 Pg 621
A patient after parotid surgery on awakening noticed lower lip paralysis. Which of the following is most likely to be injured?
|A.||Cervical branch of the facial nerve
|B.||Facial nerve main trunk
|D.||The temporal branch of facial nerve
Answer- A. Cervical branch of the facial nerve
This complication was most likely due to injury to which of the following:
Facial nerve – temporal branch
Facial nerve – cervical branch
Facial nerve – main trunk
The cervical branch of the facial nerve innervates the lower lip through the marginal mandibular branch of the nerve.
As no cross innervation exits to other branches of the facial nerve, marginal mandibular branch injuries always yield paralysis of the same side of the lower lip.
Injuries of the main trunk of the facial nerve or its temporal branch would usually produce upper facial paralysis as well.
ERCP is indicated in all except
|A.||Pancreatitis without features of Cholangitis or choledochal cyst
Answer- A. Pancreatitis without features of Cholangitis or choledochal cyst
ERCP indications include obstructive jaundice, biliary or pancreatic ductal system disease treatment or tissue sampling, suspicion for pancreatic cancer, pancreatitis of unknown cause, manometry for sphincter of Oddi, nasobiliary drainage, biliary stenting for strictures and leakage, drainage of pancreatic pseudocysts, and balloon dilation of the duodenal papilla and ductal strictures and in poor surgical candidates with ampullary carcinoma.
A person with recurrent abdominal pain USG has done is normal. On further evaluation by other investigations, he was found to have Biliary dyskinesia. Which of the following is not done for the diagnosis of biliary dyskinesia?
|D.||Post-surgery follow up
Answer- C. USG
The patient should have right upper quadrant pains similar to biliary colic but have a normal ultrasound examination of the gallbladder.
HIDA scan is considered before cholecystectomy.
Cholecystectomy is the only known effective treatment for the diagnosis of biliary dyskinesia. A period of observation can and should be offered however if the symptom complex has been of short duration or there remains concerned that other etiologies may be the primary contributor to the patient’s symptoms.
ERCP is good for the investigation of patients with biliary dyskinesia to characterize abnormalities of the sphincter (stenosis, dyskinesia).
General Surgery Outpatient Decisions By Gaunt Michael, Tjun Tang, Stewart Walsh
A Woman went for a thyroidectomy. After 6 hrs of operation, she starts getting shortness of breath, wound drainage on 2 sides is intact and nothing is coming out into the drainage. The wound area is swollen. What will be the appropriate next step management after giving Oxygen?
|B.||Suture out superficial and deep wounds
|C.||Wound drainage reposition
Answer- C. Wound drainage reposition
Post respiratory distress-
- Laryngeal edema
- Bilateral RLN injury
Tension hematoma –
- Treatment- Removal of sutures & evacuation of the hematoma
30 year old woman had fallen from height and was confirmed brain dead. Incase of organ donation, the ideal candidate will be-
|A.||30 yr old female with chronic renal failure
|B.||55yr old male with liver cirrhosis
|C.||70yr old man with Alzheimer’s and dialysis for CRF
|D.||14 yr old boy with MODS
Answer- A. 30 yr old female with chronic renal failure
Answer conclusion is based on the age and number of organs involved, the best answer will be a 30 yr old female with chronic renal failure
40yr female with 2*2 cm nodule in left lobe of thyroid which on biopsy is papillary. Most appropriate management?
|B.||Total thyroidectomy with central node dissection
Answer- C. Hemithyroidectomy
If <40- 45yrs
< 2cm size tumor
No lymphovascular invasion
No distant metastasis
Management is hemithyroidectomy
A male patient is diagnosed with carcinoma colon. Which of the following gene mutations likely to be present?
|A.||Kras, APC, DCC
|B.||APC, K-ras, p53
|C.||BRCA1, K-ras, APC
|D.||APC, K-ras, loss of 10 on myc
Answer- B. APC, K-ras, p53
This is a Fearon Vogelstein sequence based on order of mutation.
APC pathway of CRC formation often includes activation of oncogenes such as COX-2 and K-RAS and inactivation of additional TSGs such as DCC/DPC4 and p53
Prostate needle biopsy by single-core shows acinar adenocarcinoma predominant in a cribriform pattern followed by crowded but separate glands and a minor component 5% of single cell infiltration. What is the grade it belongs to?
Answer- B. Grade 3
Based Gleason score
|Grade group||Glean Score||Definition|
|1||3+3||only inndividual, discretc, well- formed glands|
|2||3+4=7||Predominatly well-formed glands with lesser component of poorly-fomed/fused/ cribriform glands|
|3||4+3=7||Predominantly poorly – formed /fused/ cribriform gland with lesser componentof well formed glandst|
|4||8||only poorlyformed /fused /cribriform glands or predominantly well formed glands and lesser component lacking glands++, or predominaatly lacking glands and lesser componentof well – formed gland++|
|5||9-10||Lack gland formation (or with necrosis), with oo without poorly formed/ fused//cribriform glands+|
100 g dextrose, 30g protein and 40 g fat in TPN. Calculate the calories of TPN
Kilocalories (dextrose) + kilocalories (protein) + kilocalories (lipids) weight (kg) = kcal/kg/day from TPN.
1g dextrose = 4calories
1g amino acids= 4 calories
1g fats = 9 calories
4✖100 + 30✖4 + 9✖40 = 880
In retroperitoneal hematoma following a trauma- True statement is-
|A.||In Zone 3 hematomas should always be explored
|B.||Mattox maneuver to explore aorta involves moving of splenic flexure of the colon and all structures to the right
|C.||Penetrating trauma in the pelvic region should be explored
|D.||In Zone 1 injured vessels embolization is done
Answer- B. Mattox maneuver to explore aorta involves moving of splenic flexure of the colon and all structures to the right
|Zone1-midline retroperitoneum||Extends from the aortic hiatus to the sacral promontory and is divided into supramesocolic and inframesocolic and inframesocolic zone||All injuries
|Zone ii- Lateralretroperitoneum||Extends on either side from the renal hila to the pericolic gutters Avoid exploration in blunt trauma||Explore all penetrating
|Zone iii-perivic retroperitoneum||Sacral promontory and
encompasses the pelvis.
Explore all penetrating trauma
|Explore for expending hematomain penetrating trauma|
Mattox Maneuver, also known as a left medial visceral rotation (LMVR), is a surgical step to explore and handle Zone 1 and 2 retroperitoneal injuries (aorta, left iliac and pelvic vessels). It starts with incising the parietal peritoneum at the white line of Toldt from the sigmoid colon to the splenic flexure
Patient with esophageal cancer underwent operation Ivory Lewis. On postoperative day 6, patients developed bilious drain output. Mild tachycardia was there but blood investigation is normal. CT suggests anastomosis site leak. What is the management?
|B.||Surgery with colon conduit
Answer- C. Esophageal stenting
It’s an Intrathoracic leak so esophageal stenting is the best management.
Stent implantation for intrathoracic esophageal anastomotic leaks is feasible and compares favourably with surgical re-exploration. It is an easily available, minimally invasive procedure that may reduce leak related mortality.
A diabetic nephropathy patient is a candidate for renal graft transplant. Which of the following statements is false?
|A.||1 yr survival of graft is 90%
|B.||The transplantation is cost-effective after the second year of transplantation
|C.||The treatment of chronic rejection has improved over the last 10 yrs
|D.||The life expectancy is doubled in a diabetic patient with renal transplantation
Answer-C. The treatment of chronic rejection has improved over the last 10 yrs
Patient survival after deceased donor renal transplantation is >95% at 1 year and >85% at 5 years.
Graft survival is around 90% at 1yr and 80% at 5 years.
Transplant outcome has improved progressively over the last two decades and continues to improve.
Chronic rejection is the most common cause of graft failure after all types of solid-organ transplant.
As many as one-third of cadaveric kidney transplant recipients suffer graft loss within five years of transplantation.
Changes in immunotherapy have proven to be mostly ineffective in changing the prognosis for patients with established chronic allograft nephropathy.
Transplantation is also more cost-effective than dialysis and improves patient survival.
DM is the most frequent causes of end-stage renal disease.
65 yrs old male with abdominal pain for 6 hrs, BP= 90/50 mm Hg. Respiratory rate = 24/min, saturation – 92% in 10 litres of Oxygen. Abdominal X-ray shows extraluminal air in the abdomen. Which of the following must be done in shifting to OT?
|A.||Intubate the patient
|B.||Initial IV access of choice is central venous catheter
|C.||Confirm the diagnosis with CT
|D.||2- 3 lts of crystalloid infused
Answer- D. 2- 3 lts of crystalloid infused
This is a case of perforation, usually we resuscitate before surgery
Hand to hand shift development in a 6 months old baby. What does it indicate?
|C.||Hand brain coordination
Answer- C. Hand brain coordination
Reaches for object – Development implication
Palmar Grasp gone- visual-motor coordination
Transfer object hand to hand – comparison of object
REFERENCE- Nelson Pg 132
Not a criteria for severe acute malnutrition (SAM)
|A.||Weight for age less than 3 SD
|B.||Weight for height less than 3SD
|C.||Mid arm circumstances <110m
Answer- A. Weight for age less than 3 SD
Diagnosis of SAM-
- Weight for height <70% of expected or < – 3Z score
- Mid upper arm circumference < 11.5 cm
2 months back history of fever with bloody diarrhea in a child followed by swollen, red and erythematous joint-
Answer- B. Shigella
Post-infectious complication in enteric infections-
- Reactive arthritis
- Guillain Barre syndrome
- Hemolytic uremic syndrome
- Shigella dysenteriae 1
Which vitamin deficiency causes neonatal seizures
Pyridoxine dependency and pyridoxal 5- phosphate dependency causes neonatal seizures
All are indicative of pediatric asthma except-
|A.||Increase in FEV1 > 12% after bronchodilator
|B.||AM :PM variation > 15%
|C.||FEV1 decrease more than 15% after exercise
Answer- B. AM :PM variation > 15%
Lung function abnormality in Asthma and Assessment of Airway Inflammation-
i) Low FEV1
ii) FEV1/ FVC <0.80
- Bronchodilator response (to inhaled beta-agonist) assesses the reversibility of airflow limitation.
- Reversibility is determined by an increase in either FEV1 > 12% or predicted FEV1) 10% after inhalation of a short-acting
- Beta agonist (SABA)
Exercise challenge –
i) Worsening in FEV1 > 15%
- Daily peak expiratory flow (PEF) or FEV1 monitoring- day to day and or AM to PM variation > 20%
Surgery is done in VSD to prevent
|D.||Failure to thrive
Answer- D. Failure to thrive
Indications for surgical closure of VSD-
- Patients at any age with large defects in whom clinical symptoms and failure to thrive cannot be controlled medically.
- Infants between 6- 12 months of age with moderate to large defect associated with pulmonary hypertension even if the symptoms are controlled by medication
- Patient older than 24 months with Qp: Qs ratio greater than 2:1
- Supracristal VSD
Ring sequestrum is seen in:
|A.||Skeletal fixation pin
Answer A. Skeletal fixation pin
Ring Sequestrum- Plain Film
- Late finding which is said to be pathognomonic of pin tract infection is “ring” sequestrum.
- The appearance of this finding is due to the particular geometry of a pin and pin tract.
- As a pin tract becomes infected, the bone immediately adjacent to the pin becomes infected first, and a certain amount of it dies.
- The viable bone adjacent to this infected dead bone then becomes hyperemic and becomes relatively osteopenic.
- The infected dead bone remains at its original density.
- Once the pin has been removed, if one looks directly down the pin tract with a radiograph, this cylinder of dead bone looks like a “ring”.
Osteomalacia with high urinary phosphate levels indicates:
|A.||Vitamin D deficiency
|B.||Vitamin D dependant deficiency
|D.||Familial Vitamin D resistant deficiency
Answer D. Familial Vitamin D resistant deficiency
|table 15.5 : biochemical abnormalities in various types of rickets|
|Nutritional Rickets||Vitamins D-dependent
Rickets I (hydraxylations Problem)
|Vitamin D- dependent
rickets ii(end organ insensitivity)
|Vitamin D- dependent
Rickets (end organ insensitivity)
|Vitamin D- dependent
Rickets(renal tubular rickets)
|25(OH) vitamins D||↓↓||↑↑||↑↑||↑↑||N|
|1,25 (OH), vitamins D||↓||↑↑||↑↑↑||↑↑↑||N|
Counterforce braces are used in?
Answer A. Tennis elbow
Counterforce braces are used in an attempt to reduce the tension forces on the wrist extensor tendons, and these orthotics may be superior to lateral epicondyle bandages in reducing resting pain.
The brace should be applied firmly approximately 10 cm distal to the elbow joint.
In case of meniscal tear repair, in which of the following zones healing is seen best?
|D.||White gray zone
Answer A. Red zone
Vascularity of Meniscus
The Meniscus is divided into three zones.
- The white-white zone (more than 5 mm from the periphery) is the central zone, which has no blood supply at all. Tears in this zone cannot be repaired.
- The Red-White zone (within 3 mm to 5 mm from the periphery) is the mid-zone of the Meniscus which has a minimal blood supply. Tears in this zone may be repaired depending on the type of tear and duration since the injury.
- Red- Red Zone (within 3 mm of the periphery) is the peripheral zone which is at the Meniscocapsular Junction and is rich in blood supply. Tears in this zone may be repaired with good results.
A 35 yr old male patient sustained a road traffic accident. Landed up in an emergency, on examination the hip is abducted and externally rotated. The likely damage caused is:
|A.||Anterior dislocation of the hip
|B.||Posterior dislocation of the hip
|C.||Fracture neck of the femur
|D.||Subtrochanteric fracture femur
Answer A. Anterior dislocation
Anterior hip dislocations are usually the result of forced abduction with external rotation of the thigh and often related to a motor vehicle accident or fall.
There are three types of anterior hip dislocations: obturator, an inferior dislocation due to simultaneous abduction; hip flexion; and external rotation.
Iliac and pubic dislocations are superior dislocations due to simultaneous abduction, hip extension, and external rotation.
Most specific Tests for Carpal Tunnel syndrome:
Answer B) Durkan’s test
- Durkan’s test is a medical procedure to diagnose a patient with carpal tunnel syndrome.
- It is a new variation of Tinel’s sign that was proposed by JA Durkan in 1991.
- Examiner presses thumbs over the carpal tunnel and holds pressure for 30 seconds.
- Onset of pain or paresthesia in the median nerve distribution within 30 seconds is a positive result of the test.
All are indications for surgery or biopsy in case of TB spine except:
|B.||Cold abscess without neurological involvement
|C.||Evolving Cauda equina syndrome
Answer B) Cold abscess without neurological involvement
Cold abscess without neurological involvement is not an indication for surgery.
Treatment (Middle path regimen)
- Strict bed rest
Indications for surgery in spinal TB should be limited to tissue sampling when the diagnosis is doubtful, drainage of an abscess in cervical spine (causing difficulty in swallowing and breathing), drainage of a large paravertebral abscess (that does not respond to 3 to 6 months antituberculosis treatment), persistent or deteriorating neurologic deficits in spite of antituberculosis treatment, recurrent neurologic complications, presence of instability in spinal column, and severe kyphotic deformity.
Which injury to the child progresses to show “Fish Tail sign” on x-ray?
|A.||Distal radius fracture
|B.||Fracture distal humerus
Answer B) Fracture distal humerus
- Fishtail deformity of the elbow is characterized by a contour abnormality of the distal humerus, which develops when the lateral trochlear ossification centers fail to develop or resorbs.
- It is an uncommon complication usually following a distal humeral fracture in childhood. Whilst initially presumed to be a benign condition, long-term follow-up suggests that patients with fishtail deformity are prone to:
- functional impairment
- ongoing pain
- development of early osteoarthrosis
How to screen for scoliosis in school children? Which test should be done?
|A.||Ask the patient to bend posteriorly
|B.||Ask the patient to bend forward
|C.||Stand on one leg
|D.||Ask patient to elevate one leg
Answer B) Ask the patient to bend forward
How to detect a Scoliosis and what to do if you identify one?
- Scoliosis is a lateral curvature of the spine greater than 10 degrees.
- When we look at a spine from behind the person it should appear that the spine runs in a straight line up and down.
- When a person has scoliosis this straight line will instead be curved.
- The curve can appear as either an “S” or a “C” shape.
- Parents need to be extra vigilant and aware of the warning signs.
- It is often overlooked as it may not be painful.
- If you suspect that you or someone you know may have a scoliosis
- The Adams Forward Bend Test can be used as an identification tool.
- Stand behind the person and have them flex their body forwards.
- If there is a notable hump on one side (if one side is raised) this is indicative of a scoliosis and the individual should seek advice from a medical professional.
|A.||High pitched bowel sounds with flatus present
|B.||High pitched bowel sounds with flatus present
|C.||Dilated bowel with prominent haustra
|D.||Constricted bowel with prominent haustra
Answer is C. Dilated bowel with prominent haustra
Large bowel obstructions are characterized by colonic distension proximal to the obstruction, with collapse distally. It should be noted that in some cases the point of obstruction and site of obstruction are not the same, with the point of obstruction located distal to the apparent cut-off point, e.g. an obstructing sigmoid tumor may present with an apparent cut-off at the splenic flexure.
Although there is no universally agreed upon cut-off for what constitutes dilatation of the large bowel, 6 cm is a reasonable value for the colon, with the cecum having an upper limit of 9 cm 1. This is known as the 3-6-9 rule.
- Colonic distension: gaseous secondary to gas-producing organisms in feces
- Collapsed distal colon
- Small bowel dilatation, which depends on
o duration of obstruction
In advanced cases one may see the stigmata of an ischemic colon, namely:
- Intramural gas (pneumatosis coli)
- Portal venous gas
- Free intra-abdominal gas (pneumoperitoneum)
|B.||GI more than Bone marrow
|C.||Bone marrow more than GI
|D.||Only 1 part receive large dose
Answer C. Bone marrow more than GI
Radiation sickness — known as acute radiation syndrome (ARS) — occurs after exposure to a large amount of radiation within a short period. Early symptoms can include skin irritation, nausea, vomiting, high fever, hair loss and burns to the skin. Other symptoms can include diarrhea, weakness, fatigue, loss of appetite, fainting, dehydration, inflammation of tissues, bleeding from the nose, mouth, gums or rectum, and anemia.
There are 4 stages of acute radiation syndrome
People exposed to radiation will get ARS only if:
- The radiation dose was high.
- The radiation was penetrating, reaching internal organs.
- The person’s entire body, or most of it, received the dose.
The radiation was received in a short time, usually within minutes.
FAST cannot diagnose
Answer -B. Pneumothorax
Focused assessment with sonography in trauma (commonly abbreviated as FAST) is a rapid bedside ultrasound examination performed by surgeons, emergency physicians, and certain paramedics as a screening test for blood around the heart (pericardial effusion) or abdominal organs (hemoperitoneum) after trauma.
The four classic areas that are examined for free fluid are the perihepatic space (including Morison’s pouch or the hepatorenal recess), perisplenic space, pericardium, and the pelvis. With this technique it is possible to identify the presence of intraperitoneal or pericardial free fluid. In the context of traumatic injury, this fluid will usually be due to bleeding.
USG least useful for
Answer:- B. Pneumothorax
- Ultrasound waves are disrupted by air or gas.
- Therefore, ultrasound is not an ideal imaging technique for the air-filled bowel or organs obscured by the bowel.
- Ultrasound is not as useful for imaging air-filled lungs, but it may be used to detect fluid around or within the lungs.
- Similarly, ultrasound cannot penetrate bone, but may be used for imaging bone fractures or for infection surrounding a bone.
- Large patients are more difficult to image by ultrasound because greater amounts of tissue attenuate (weaken) the sound waves as they pass deeper into the body and need to be returned to the transducer for analysis.
- Ultrasound has difficulty penetrating bone and, therefore, can only see the outer surface of bony structures and not what lies within (except in infants who have more cartilage in their skeletons than older children or adults).
- For visualizing internal structure of bones or certain joints, other imaging modalities such as MRI are typically used.
FAST full form
|A.||Focused assessment with sonography in trauma
|B.||Face Arm Speech Test
|C.||Fine Assessment of Specific tissue
|D.||Fine assessment of sonography in Trauma
Answer :- A. Focused assessment with sonography in trauma
Focused assessment with sonography in trauma (commonly abbreviated as FAST) is a rapid bedside ultrasound examination performed by surgeons, emergency physicians, and certain paramedics as a screening test for blood around the heart (pericardial effusion) or abdominal organs (hemoperitoneum) after trauma.
The four classic areas that are examined for free fluid are the perihepatic space (including Morison’s pouch or the hepatorenal recess), perisplenic space, pericardium, and the pelvis. With this technique it is possible to identify the presence of intraperitoneal or pericardial free fluid. In the context of traumatic injury, this fluid will usually be due to bleeding.
Which of the following is an adverse effect of a psychoactive drug that acts as a SSRI ?
|C.||Blurring of vision
Side effects of SSRIs include:
- GI side effects-Nausea(most common),diarrhea,constipation,anorexia
- Sexual dysfunction(most common)
- QTc prolongation
- CNS-Vivid dreams,sweating,anxiety,insomnia
- Weight gain
Psychiatry review book-Dr.Praveen Tripathi
A bipolar patient is on lithium therapy.As a doctor when are you going to ask a nurse to get blood lithium checked?
|A.||24 hours after last dose
|B.||8 hours after last dose
|C.||12 hours after last dose
Ans- C. 12 hrs. After last dose
The level should be a 12-hour trough i.e. taken approximately 12 hours after the last dose of lithium (In twice-daily dosing the morning dose should be withheld until the level has been taken.)
Psychiatry review book-Dr.Praveen Tripathi
In a physiology viva, teacher asks to the student, “who recorded first human EEG” student responded with I don’t know sir” teacher feels sorry for the student and gives a hint to student, “the word rhymes with what you get at McDonald’s”, then the student gives a right answer. This type of memory recall is used in – (physiology +psychiatry)
Priming is a phenomenon in which exposure to one stimulus influences how a person responds to a subsequent, related stimulus. These stimuli are often conceptually related words or images.
Priming effects appear in a person’s responses to stimuli, such as the speed with which the person is able to categorize a string of letters as a word or non-word
A 26 old woman presents to a physician with complaints of fatigue and insomnia for the past 8 months. She also reports difficulty in concentrating, irritability and a continued sense of worry. She performs well at her job as a lawyer where she was promoted but she states that she often lies awake at night worrying that she might be fired. She no longer goes out with friends or family because she is worried that she is not welcomed. She denies heart palpitations or changes in weight. Her medical history is unremarkable. Her vital signs are within normal limits. Physical examination shows no focal findings, she does not smoke but occasionally uses alcohol. She went through an uncomplicated divorce one year ago. Which of the following in this patient is most likely to occur:
|B.||Generalized anxiety disorder
|D.||Post-traumatic stress disorder
Ans-B-Generalized anxiety disorder
- Generalized Anxiety Disorder (GAD) is characterized by persistent and excessive worry about a number of different things.
- People with GAD may anticipate disaster and may be overly concerned about money, health, family, work, or other issues. Individuals with GAD find it difficult to control their worry. They may worry more than seems warranted about actual events or may expect the worst even when there is no apparent reason for concern.
- GAD is diagnosed when a person finds it difficult to control worry on more days than not for at least six months and has three or more Symptoms. This differentiates GAD from worry that may be specific to a set stressor or for a more limited period of time.
Signs and Symptoms of GAD
- Feeling nervous, irritable, or on edge
- Having a sense of impending danger, panic or doom
- Having an increased heart rate
- Breathing rapidly (hyperventilation), sweating, and/or trembling
- Feeling weak or tired
- Difficulty concentrating
- Having trouble sleeping
- Experiencing gastrointestinal (GI) problems
A known chronic alcoholic who used to drink 2 bottles of whiskey daily,now is not consuming for the last 2 days,presents to OPD with the symptoms of anxiety, hypertension, discomfort, he is seeing snakes on the floor, no seizures. Which of the following treatment should be given to control the symptoms immediately?
Since the patient has presented to the OPD with anxiety and discomfort he should be given lorazepam to control these symptoms.Lorazepam belongs to a class of drugs known as benzodiazepines which act on the brain and nerves (central nervous system) to produce a calming effect. This drug works by enhancing the effects of a certain natural chemical in the body (GABA).
Chronic alcoholism leads to thiamine deficiency, so injecting thiamine is essential for the normal function of heart,muscles & nervous system
A patient of severe depression was given ECT. Anaesthetist gave succinylcholine and thiopental to the patient.What did these agents do to the patient?
|A.||Mood elevation and soothing effect
|B.||General anesthesia and get muscle relaxation
|C.||To minimize memory loss
|D.||To prevent seizure
Ans-B. General anesthesia & get muscle relaxation
Succinylcholine is indicated as an adjunct to general anesthesia, to facilitate tracheal intubation, and to provide skeletal muscle relaxation during surgery or mechanical ventilation. Succinylcholine is a depolarizing skeletal muscle relaxant
Thiopental is a barbiturate (bar-BIT-chur-ate). Thiopental slows the activity of the brain and nervous system. Thiopental is used to help the patient relax before he receives general anesthesia with an inhaled medication.
The patient is on lithium therapy. For which of these clinical symptoms, should nurse warn the patient to be careful of :
|A.||Skin rash, bradycardia, elevated BP
|B.||Palpitations, occipital headache, chest pain
|C.||Fever, malaise, sore throat
|D.||Ataxia, diarrhea, tinnitus
Answer- D. Ataxia, diarrhea, tinnitus
Common side effects of lithium can include:
- Hand tremor (If tremors are particularly bothersome, dosages can sometimes be reduced, or an additional medication can help.)
- Increased thirst
- Increased urination
- Weight gain
- Impaired memory
- Poor concentration
- Muscle weakness
- Hair loss
- Decreased thyroid function (which can be treated with thyroid hormone)
Which of the following antifungal drugs has antipruritic action?
Ans. D. Sertaconazole
- Sertaconazole (Dermofix, Ertaczo, Ginedermofix, Monazol, Mykosert or Zalain), an imidazole antifungal agent, inhibits the synthesis of ergosterol, an essential cell wall component of fungi.
- It is indicated in the EU for the treatment of superficial skin mycoses such as dermatophytosis (including tinea corporis, tinea cruris, tinea manus, tinea barbae and tinea pedis), cutaneous candidiasis, pityriasis versicolor and seborrhoeic dermatitis of the scalp, and in the US for tinea pedis only.
- Sertaconazole has broad-spectrum antifungal activity against dermatophytes of the Trichophyton, Epidermophyton and Microsporum genera, and yeasts of the genera Candida and Cryptococcus; additionally, it is effective against opportunistic filamentous fungi and Gram-positive bacteria.
- Moreover, the antifungal activity of sertaconazole is maintained in clinical isolates of dermatophytes that show reduced susceptibility to other azoles.
- Mainly acts by inhibiting inflammatory cytokines.
Methyl iso xanthyl in cosmetic is
|D.||To increase SP factor
Ans. C – Preservative
- Methylisothiazolinone (MIT) and Methylchloroisothiazolinone (CMIT) are widely used preservatives found in liquid cosmetic and personal care products.
- Both chemicals inhibit bacterial growth in cosmetic products on their own, but they are most commonly used as a mixture of products.
- Methylisothiazolinone (MI) is listed in Annex V/57 of the Cosmetic Regulation (EC) 1223/2009, as amended, meaning it can be used as a preservative in cosmetics.
- It is contained in products such as makeup, moisturizers, baby wipes, sun creams, shampoos, and shower gels as well as household products and paints.
Pregnant female with hyperthyroidism due to graves disease is taking medication during pregnancy. The child born has aplasia cutis congenita. Which drug female was on:
Ans. A. Carbimazole
- Aplasia cutis congenita (ACC) following in utero exposure to antithyroid drugs such as methimazole/carbimazole (MTZ/CMZ).
- Aplasia cutis congenita (ACC)is a heterogeneous group of disorders characterized by the absence of a portion of skin in a localized or widespread area at birth.
- Although most commonly seen on the scalp, aplasia cutis congenita can affect any part of the body, including the trunk and limbs.
Patient was on metoprolol therapy for hypertension. He is also administered with verapamil. Which of the following is likely to happen with the above combination therapy.
|B.||Torsades de pointes
|C.||Bradycardia following av block
Ans. C Bradycardia following av block
- The combination of verapamil or diltiazem with beta‐blockers should be avoided because of potentially profound adverse effects on AV (atrioventricular) nodal conduction, heart rate, or cardiac contractility.
- This effect is unpredictable but may be enhanced due to CYP2D6 poor metabolizer status which could be a special vulnerability factor.
- Calcium channel blockers (CCB) are prescribed for the treatment of arrhythmia and hypertension.
○ Verapamil is a class IV antidysrhythmic drug, which acts by blocking voltage‐sensitive calcium channels.
- Beta‐adrenergic blockers (BB) are used in the treatment of hypertension and heart failure.
○ Metoprolol is a selective beta 1‐adrenergic blocking agent, and it is lipophilic and predominately metabolized in the liver via cytochrome CYP2D6.
○ The blockade of the myocardial b1 receptor reduces heart rate, myocardial contractility, and cardiac output.
○ Dizziness, bradycardia, and hypotension are observed as adverse reactions at therapeutic plasma levels.
- Combined use of CCBs and BBs may cause adverse cardiovascular effects.
Dicoumarol antagonizes the action of Vitamin K by?
Ans. A. Competitive inhibition.
- Dicoumarol is a hydroxycoumarin,
- Dicoumarol is a competitive inhibitor of vitamin K epoxide reductase; thus, it inhibits vitamin K recycling and causes depletion of active vitamin K in blood.
- When the active/catalytic site of an enzyme is occupied by a substance other than the substrate of that enzyme, its activity is inhibited.
- Inhibitor must be a structural analog of the substrate.
- In such inhibition, both ES & EI (Enzyme-Inhibitor) complexes are formed.
- With the increase in the concentration of inhibitor lowers the rate of enzymatic reaction.
- Thus, Km is high, but Vmax is same.
- Also, when concentration of the substrate is increased, the inhibitor effect reversed forcing it out from the EI complex.
Which of the following is not a Prodrug
Ans. D –Sulindac
- A prodrug can be defined as a drug substance that is inactive in the intended pharmacological actions and is must to be converted into the pharmacologically active agent by the metabolic or physico-chemical transformation.
- Prodrugs can exist naturally such as many phytochemicals/botanical constituents and endogenous substances, or they can result from synthetic or semisynthetic processes – produced intentionally as part of a rational drug design or unintentionally during drug development.
- Examples of prodrugs that exist naturally or were produced unintentionally during drug development include aspirin, psilocybin, parathion, irinotecan, codeine, heroin, L-dopa, and various antiviral nucleosides.
- Examples of products resulting from pharmaceutical processes as part of strategically targeted drug design include sulfasalazine, oseltamivir, various nonsteroidal anti-inflammatory drugs (ketoprofen, diclofenac), statins (lovastatin, simastatin), ACE inhibitors (captopril, lisinopril) and penicillin-related agents (bacampicillin, sarmoxicillin).,
A female with unilateral breast cancer is on Tamoxifen therapy. Which of the following adverse effects is associated with the use of Tamoxifen.
|C.||Carcinoma is opposite breast
Ans- B.Endometrial cancer
- Tamoxifen has been prescribed to millions of females for breast cancer prevention or treatment.
- However, tamoxifen is known to significantly enhance the risk of developing endometrial lesions, including hyperplasia, polyps, carcinomas, and sarcoma.
- Notably, tamoxifen-associated endometrial cancer often has a poor clinical outcome.
- Tamoxifen appears to mediate its effect on endometrial cells through estrogenic and non-genomic pathways, rather than introducing a genomic alteration as a carcinogen.
- Although tamoxifen functions as an agonist and promotes cell proliferation in endometrial cancer, it also displays antagonist activity towards some estrogen targets.
- Alterations in estrogen receptor-α and its isoforms, as well as the membrane-associated estrogen receptor G protein-coupled receptor 30, have been observed with tamoxifen-exposed endometrial cells, and likely mediate the effects of tamoxifen on endometrial cancer cell proliferation and invasion.
Which of the following is a second line anti-tubercular drug?
- First-line antituberculosis drugs- Isoniazid (INH), rifampicin (RIF), ethambutol (EMB), pyrazinamide (PZA) and streptomycin (SM).
- Second-line antituberculosis drugs- Subdivided into two
- Fluoroquinolones- Ofloxacin (OFX), levofloxacin (LEV), moxifloxacin (MOX) and ciprofloxacin (CIP).
- Injectable antituberculosis drugs- Kanamycin (KAN), amikacin (AMK) and capreomycin (CAP).
- Less-effective second-line antituberculosis drugs- Ethionamide (ETH)/Prothionamide (PTH), Cycloserine (CS)/Terizidone, P-aminosalicylic acid (PAS).
Which among the ATT is not bactericidal?
- The bactericidal activity of isoniazid was strongest.
- The bactericidal activity of isoniazid and streptomycin was most marked in the log phase.
- M. avium complex and M. kansasii resisted the bactericidal activity, but some strains of M. avium complex were killed by streptomycin and enviomycin, and the activities of these two drugs were most marked in the lag phase.
- Thioacteozone is a bacteriostatic drug.
Which of the following actions solely depends on muscarinic receptors?
|C.||Increased contraction of cardiac muscle
Ans. A Erection
- Autonomic response when muscarinic receptor-stimulated is erection – Due to muscarinic or parasympathetic actions.
- Other actions are all maThioacteozoneinly sympathetic effects.
All of the following are advantages of Enteric-coated tablets except?
|B.||Protect acid-labile drugs from gastric pH
|C.||Deliver drug distal to stomach
Prevents the irritant effect on the stomach
Ans. A.Increased half-life
- Half life mainly depends on volume of distribution (Vd) and clearance rate (CL).
- Enteric coating was carried out using different polymers like Eudragit L-30 D-55, hydroxypropyl methylcellulose phthalate, cellulose acetate phthalate, and Acryl-EZE® to achieve 5% weight gain.
The following are procedure/treatment of no proven clinical benefit in a case of organophosphate poisoning0 except:
|A.||Gastric lavage and activated charcoal
|C.||Oxygen supplementation- ANS
Ans : A Gastric lavage and activated charcoal
- The first step in the management of patients with organophosphate poisoning is putting on personal protective equipment.
- These patients may still have the compound on them, and you must protect yourself from exposure.
- Secondly, you must decontaminate the patient.
- This means removing and destroying all clothing because it may be contaminated even after washing. The patient’s skin needs to be flushed with water.
- Dry agents such as flour, sand, or bentonite also can be used to decontaminate the skin.
- In the case of ingestion, vomiting and diarrhea may limit the amount of substance absorbed but should never be induced.
- Activated charcoal can be given if the patient presents within 1 hour of ingestion, but studies have not shown a benefit.
- The definitive treatment for organophosphate poisoning is atropine, which competes with acetylcholine at the muscarinic receptors.
- The initial dose for adults is 2 to 5 mg IV or 0.05 mg/kg IV for children until reaching the adult dose.
- If the patient does not respond to the treatment, double the dose every 3 to 5 minutes until respiratory secretions have cleared and there is no bronchoconstriction.
- In patients with severe poisoning, it may take hundreds of milligrams of atropine given in bolus or continuous infusion over several days before the patient improves.
- Pralidoxime (2-PAM) also should be given to affect the nicotinic receptors since atropine only works on muscarinic receptors.
- Atropine must be given before 2-PAM to avoid worsening of muscarinic-mediated symptoms.
- A bolus of at least 30 mg/kg in adults or 20 to 50 mg/kg for children should be given over 30 minutes. Rapid administration can cause cardiac arrest.
- After the bolus, a continuous infusion of at least 8 mg/kg/hr for adults and 10 to 20 mg/kg/hr for children should be started and may be needed for several days.
Which of the following is an antidote for fibrinolytic agent?
|D.|| Epsilon amino Caproic acid.
Ans. D. Epsilon amino Caproic acid.
- Both Epsilon amino Caproic acid.& Tranexamic both used as an antidote for fibrinolytic agents.
A 70-year-old patient has diabetes mellitus and hypertension. He presents with late-stage chronic kidney disease. Which of the following anti-diabetic drugs require least dose modification in renal disease?
- Linagliptin is a recently approved oral antidiabetic drug that acts by inhibiting the enzyme dipeptidyl peptidase-4 (DPP-4).
- Unlike other DPP-4 inhibitors, linagliptin is excreted chiefly via the enterohepatic system, and can be used without dose adjustment in patients with renal or hepatic impairment.
- Linagliptin was associated with significant improvements in glycosylated hemoglobin, fasting plasma glucose and postprandial glucose, and more patients receiving linagliptin showed meaningful improvements and achieved targets for glycosylated hemoglobin.
- Linagliptin was well tolerated, with an adverse event profile similar to that of placebo, and low rates of hypoglycemic events.
Which of the following drug should be avoided in a patient on rosuvastatin therapy?
- CYP 3A4 inhibitors like erythromycin, clarithromycin, ketoconazole, etc. increase the risk of myopathy by inhibiting the metabolism of statins.
54. Wrong about Ebola virus?
Answer- A-Mosquito transmission
- The virus is transmitted through direct contact with blood, organs, body secretions, or other body fluids of infected animals like chimpanzees, gorillas, monkeys, fruit bats, etc.
- Human to human transmission is through blood or body fluids of an infected symptomatic person or through exposure to objects (such a needle) that have been contaminated with infected secretions.
- It is not transmitted through air, water, or food.
- The virus is transmitted through direct contact with blood, organs, body secretions, or other body fluids of infected animals like chimpanzees, gorillas, monkeys, fruit bats etc.
- Human to human transmission is through blood or body fluids of an infected symptomatic person or through exposure to objects (such as a needle) that have been contaminated with infected secretions
- It is not transmitted through air, water, or food
- The illness is characterized by sudden onset of fever, intense weakness, muscle pain, headache, sore throat, vomiting, diarrhea, rash, impaired kidney and liver function and in some both internal and external bleeding.
- The virus family Filoviridae includes three genera: Cuevavirus, Marburgvirus, and Ebolavirus.
- Within the genus Ebolavirus, five species have been identified: Zaire, Bundibugyo, Sudan, Reston and TalForest.
- The first three, Bundibugyo ebolavirus, Zaire ebolavirus, and Sudan ebolavirus have been associated with large outbreaks in Africa.
COVID precautions does not include
Ans-c Glutaraldehyde 1%
COVID-19 virus can survive for up to 72 hours on plastic and stainless steel, less than 4 hours on copper, and less than 24 hours on cardboard.
- If soap and water are not readily available, use a hand sanitizer that contains at least 70% alcohol. Cover all surfaces of your hands and rub them together until they feel dry.
- Avoid touching your eyes, nose, and mouth with unwashed hands.
- Medical masks are recommended primarily in health care settings, but can be considered in other circumstances (see below). Medical masks should be combined with other key infection prevention and control measures such as hand hygiene and physical distancing.
True about Corynebacterium diphtheriae are all except:
|A.||Iron is required for toxin production
|B.||Toxin production is responsible for local reaction
|C.||Non-Sporing, noncapsular and nonmotile
|D.||Toxin production is by lysogenic conversion
Answer-B Toxin production is responsible for local reaction
- “Mechanical problem in Diptheria is due to the membrane while the systemic effects are due to the toxin”
- Toxin acts mainly systemically though there are partial local effects.
- It has an affinity for myocardium, adrenals and nerve endings.
- Toxin acts by inactivating EF-2 thus inhibiting protein synthesis.
- Toxin production is influenced by iron concentration in the medium. Toxin production is optimal at 0.14 µg/ml and is suppressed at 0.5 µg/ml.
- Toxigenicity of diphtheria bacillus depends on symbiotic bacteriophages, so it shows lysogenic or phage conversion i.e nontoxigenic strain → toxigenic strain by infecting with beta phage.
- The strain almost universally used for toxin production is the ‘ParkWilliam 8’ Strain.
- Exotoxin is also produced by C. ulcerans, C. pseudotuberculosis.
Which of the following spreads as Droplet nuclei Infection
Answer- D Influenza
- ‘aerosols’ refer to particles in suspension in a gas, such as small droplets in air.
- Aerosols’ would also include ‘droplet nuclei’ which are small particles with an aerodynamic diameter of 10 μm or less, typically produced through the process of rapid desiccation of exhaled respiratory droplets.
- emission of influenza RNA from the exhaled breath of naturally influenza-infected human subjects and have detected influenza RNA in the environmental air.
- More recently, some of these studies have shown the absence of, or significantly reduced numbers of viable viruses in air-samples with high influenza RNA levels (as tested by PCR)
About Weil’s disease all are true except ?
|A.||it’s a spirochete
|B.||grows in EMJH
|D.||can be seen by light microscopy
Ans-d Can be seen by light microscopy
- Weil’s disease (leptospirosis) Weil’s disease is a form of a bacterial infection also known as Leptospirosis that is carried by animals, most commonly in rats and cattle.
- Weil’s disease, the acute, severe form of leptospirosis, causes the infected individual to become jaundiced (skin and eyes become yellow), develop kidney failure, and bleed.
- Bleeding from the lungs associated with leptospirosis is known as “severe pulmonary hemorrhage syndrome”
- The microscopic agglutination test (MAT) is the reference test for the diagnosis of leptospirosis.
- MAT is a test where human or animal serum is mixed with various types of Leptospira antigens serovars. The mixture is then examined under a microscope to look for agglutination. The MAT is then read by darkfield microscopy.
All the following are true about Mycoplasma pneumonia infection except
|A.||Cannot be cultured routinely
|B.||Serology is useful in diagnosis
|C.||Causes many extrapulmonary manifestations
|D.||Are sensitive to beta-lactam group of drugs
Answer- D-Are sensitive to beta-lactam group of drugs
Doxycycline is the drug of choice for mycoplasma
- For mycoplasma, tetracyclines are the drugs of choice (doxycycline 100 mg 2 times a day for 10 to 14 days).
- Erythromycin, Azithromycin, and Fluoroquinolones can also be used.
- Tetracyclines are contraindicated in pregnancy, therefore erythromycin or spiramycin can be used.
Antibody testing for cov-2 depends upon
Time of incubation
|C.||Past infection and asymptomatic
Currently infected, asymptomatic
Answer-C Past infection and asymptomatic
- Serologic tests for antibodies are becoming widely available and have the potential to make widespread testing much faster.
- The presence of antibodies can indicate either that a person is infected or has been infected, and so does not necessarily indicate the presence of a live virus.
- Following infection, IgM is the first antibody class produced, and so is more of an indicator of recent infection than is IgG, which is produced later. IgG tends to persist longer than IgM, and more sensitive to detect than past infections.
- The two antibodies used in most available antibody tests are IgM and IgG. When a SARS-CoV-2 infection occurs, IgM appears to be the first responder, arriving on the scene within a week or so of infection. A second antibody, IgG, generates a more specific and longer-lasting response—but it shows up later, when the viral RNA signal may have already faded away.
- Viral load, also known as a viral burden, viral titre or viral titer, is a numerical expression of the quantity of virus in a given volume of fluid; sputum, and blood plasma being two bodily fluids.
- High viral loads and successful isolation from early throat swabs suggested potential virus replication in tissues of the upper respiratory tract.
Scrub typhus is transmitted by
Ans. B: Mite
Vector for Scrub typhus is trombiculid mite
Scrub typhus/Bush typhus
- It is a form of typhus caused by Orientia tsutsugamushi first isolated and identified in 1930 in Japan
- Scrub typhus is transmitted by some species of trombiculid mites (“chiggers”, particularly Leptotrombidium deliense), which are found in areas of heavy scrub vegetation.
- The bite of this mite leaves a characteristic black eschar that is useful for making the diagnosis.
- Scrub typhus is endemic to a part of the world known as the “tsutsugamushi triangle” (after the name “Orientia tsutsugamushi” (formerly “Rickettsia tsutsugamushi”), the obligate intracellular gram-negative bacterium causing same), which extends from northern Japan and far-eastern Russia in the north, to the territories around the Solomon Sea into northern Australia in the south, and to Pakistan and Afghanistan in the west.
- – Incubation period – 10–12days
– Weil Felix reaction – is strongly positive with Proteus strain OXK
– Treatment – Tetracycline
which is not a part of pap smear
Answer C. Methylene Blue
: Papanicolaou stain (also Papanicolaou’s stain and Pap stain) is a multichromatic (multicolored) cytological staining technique developed by George Papanicolaou in 1942.
- The classic form of the Papanicolaou stain involves five stains in three solutions.
- The first staining solution contains hematoxylin which stains cell nuclei
- Papanicolaou used Harris’s hematoxylin in all three formulations of the stain he published.
- The second staining solution (designated OG-6), contains Orange G in 95% ethyl alcohol with a small amount of phosphotungstic acid. In the OG-6, the OG signifies Orange G and the ‘6’ denotes the concentration of phosphotungstic acid added; other variants are OG-5 and OG-8).
- The third staining solution is composed of three dyes, Eosin Y, Light Green SF yellowish, and Bismarck brown Y in 95% ethyl alcohol with a small amount of phosphotungstic acid and lithium carbonate
- This solution, designated EA, followed by a number which denotes the proportion of the dyes, other formulations include EA-36, EA-50, and EA-65.
- The counterstains are dissolved in 95% ethyl alcohol which prevents cells from over staining which would obscure nuclear detail and cell outlines, especially in the case when cells are overlapping on the slide. Phosphotungstic acid is added to adjust the PH of counterstains and helps to optimize the color intensity. The EA counterstain contains Bismarck brown and phosphotungstic acid, which when in combination, cause both to precipitate out of solution, reducing the useful life of the mixture.
Not a feature of dehydration
|C.||Skin retracts slowly
answer: a Child thirsty
Signs of dehydration are:
- Peripheral pulses either rapid and
- weak or absent
- Decreased blood pressure
- No urine output
- Very sunken eyes and fontanel
- No tears
- Parched mucous membrane
- Delayed elasticity (poor skin turgor)
- Very delayed capillary refill (> 3 sec)
- Cold and mottled
- Depressed consciousness
Not included in MELD score
answer -c -Albumin
- The Model for End-stage Liver Disease (MELD) is a prospectively developed and validated chronic liver disease severity scoring system that uses a patient’s laboratory values for –
- Serum bilirubin
- Serum creatinine
- The international normalized ratio (INR) for prothrombin time to predict three-month survival.
- Patients with cirrhosis and increasing MELD score is associated with increasing severity of hepatic dysfunction and increased three-month mortality risk.
- Given its accuracy in predicting short-term survival among patients with cirrhosis, MELD was adopted by the united network for organ sharing (UNOS) in 2002 for prioritization or patients awaiting liver transplantation in the United States.
Complications of massive blood transfusion are all except
Answer- A. Hypercalcemia
- Hypothermia- Due to cold blood transfusion
- Hypothermia- Due to cold blood transfusion
- Hyperkalemia- K+ moves out of RBC during storage
- Hypokalemia- K+ taken back by depleted RBC
- important complications of massive blood transfusion
- Fluid overload
- Acidosis / Alkalosis
treatment of no use in Guillain barre syndrome
- Guillain-Barré syndrome is one of the most common life-threatening diseases of the peripheral nervous system.
- It is a rapidly progressive acute demyelinating disorder affecting motor axons that result in ascending weakness that may lead to death from failure of respiratory muscles over a period of only several days. It appears to be triggered by an infection or a vaccine that breaks down self-tolerance, thereby leading to an autoimmune response.
Treatments include :
- plasmapheresis (to remove offending antibodies),
- intravenous immunoglobulin infusions (which suppress immune responses through unclear mechanisms) and
- supportive care, such as ventilatory support. Patients who survive the initial acute phase of the disease usually recover over time.
Which of the following is not corrected by dialysis
Conditions corrected by dialysis are:
- Pericarditis or pleuritis (urgent indication).
- Progressive uremic encephalopathy or neuropathy
- Persistent metabolic disturbances
- Fluid overload refractory to diuretics.
- Hypertension poorly responsive to antihypertensive medications.
- Persistent nausea and vomiting.
Patient with HIV best P.E.P
|A.||Zidovudine + Lamivudine + Indinavir
|B.||Zidovudine + Lamivudine + Tenofovir
|C.||Zidovudine + Lamivudine
|D.||Zidovudine + Nevirapine
Answer – A Zidovudine + Lamivudine + Indinavir
A combination of drugs (two or three drugs) is started as soon as possible, ideally within one to two hours of exposure.
The drugs are zidovudine (300 mg BD), lamivudine (150 mg BD), and PI (nelfinavir 1250 mg BD or boosted lopinavir or indinavir).
The treatment is given for 4 weeks
Zidovudine 300 mg twice a day
Lamivudine 150 mg twice a day
Indinavir 800 mg thrice a day or Boosted indinavir (Indinavir 800 plus ritonavir 100 mg BD)
Boosted Lopinavir (400 mg plus ritonavir 100 mg BD)
Nelfinavir 1250 mg twice a day
Which of the following not used in NSTEMI
Answer- B streptokinase
Treatment for NSTEMI include:
2.Antiplatelet drugs – clopidogrel, ticagrelor, prasugrel
3.Anti ischemic drugs: nitroglycerin, beta-blockers
Thrombolysis is not indicated in NSTEMI.
The maximum dose for immunity from tetanus is-
Answer. B. 0.01 IU
- Standard treatment consists of a single IM dose of tetanus immune globulin
- (TIG; 3000-5000 IU) or equine antitoxin (10,000-20,000 IU). However, there is evidence that intrathecal TIG (50-1500 IU) inhibits disease progression and leads to a better outcome than IM-administered TIG.
|A.||BP more than 140/90 on 3 or more drugs
|B.||BP more than 160/90 0n 3 or more drugs
|C.||BP more than 140/90 on 2 more drugs
|D.||BP more than 160./90 0n 2 more drugs
answer : a BP more than 140/90 on 3 or more drugs
Resistant hypertension is generally defined as uncontrolled clinic blood pressure (>140/90 mm Hg) after treatment with three or more antihypertensives.
The National Institute for Health and Care Excellence (NICE) guidelines specify that these three antihypertensives should include optimal doses of an angiotensin-converting enzyme (ACE) inhibitor (or an angiotensin receptor blocker), a calcium channel blocker, and a diuretic.
Treatment of resistant hypertension is focused on the addition of fourth-line therapy where blood pressure is not controlled by treatment with three drugs, described by NICE as A+C+D: that is, an ACE inhibitor or an angiotensin II receptor blocker (A), a calcium channel antagonist (C), and a thiazide or a thiazide-like diuretic (D).
In a child with asthma what is true?
|A.||FEV1/FVC decrease to 80%
|B.||>15% decrease after exercise
|C.||Improvement of 15% after bronchodilator
|D.||More than 15% AM to PM variation
answer: c – Improvement of 15% after bronchodilator
- In bronchial asthma, FEV1, FEV1/VC ratio, and peak expiratory flow rate (PEFR) are reduced.
- An improvement of at least 15% in FEV1 or PEFR following administration of bronchodilator is diagnostic of bronchial asthma. PEFR is generally used for long-
- term home monitoring
- In asymptomatic individuals, exercise, histamine or methacholine can be used to provoke bronchospasm for the diagnosis of asthma.
Most reliable investigation of choice to be done in 50-year-old smoker with mild recurrent hemoptysis without fever or weight loss
|B.||Fibre optic bronchoscope
Answer- B Fibre optic Bronchoscope
- Bronchoscopy, performed with either a rigid or flexible endoscope, is helpful for identifying active bleeding and for checking the airways in patients with massive hemoptysis.
- Although rigid bronchoscopy may have a role in massive hemoptysis due to its ability to maintain airway patency, flexible fiberoptic bronchoscopy has the advantage of being carried out at the patient’s bedside without anesthesia and in the intensive care unit and is, therefore, more frequently used.
- The capability and success of bronchoscopy in localizing the bleeding site may vary according to the rate and severity of the hemorrhage. Hirshberg et al. found that bronchoscopy was more effective in finding the bleeding site in patients with moderate to severe hemoptysis (64% and 67%) than in those with mild hemoptysis (49%). Nevertheless, severe hemorrhage can hinder the visualization of airways, in particular at the level of distal bronchi. Another limitation of bronchoscopy is that lavage or the use of endoscope itself may cause bronchial mucosa irritation and recurrent bleeding
A 65-year-old person came for a routine checkup and his BP readings were 190/100 and no other symptoms. Decrease on which of the following is responsible for such high pulse pressure
|D.||Arterial wall compliance
answer : d Arterial wall compliance
Two major factors affect pulse pressure:
(1) the stroke volume output of the heart and
(2) the compliance (total distensibility) of the arterial tree.
A third, less important factor is the character of ejection from the heart during systole.
Which of the following is DOC for hairy cell leukemia
- The treatment of choice is Cladribine Q (2 chloro deoxy adenosine) 0.14 mg /kg daily for 7 days” – CMDT
- This is a relatively non-toxic drug that produces benefit in 95% of cases, and complete remission in more than 80%
A patient presented with status epilepticus which of the following parameters would suggest higher risk of him developing stroke in future according to abcd2 scoring
i. Age > 60 years
ii. Systolic BP> 140
iii. duration> 5 minutes
|A.|| i & ii are correct
|B.|| ii & iii are correct
|C.||i ii &iv are correct
|D.||All the above are correct
answer: c- i ii &iv are correct
Criteria in abcd2 score
- Age ≥ 60
- Elevated blood pressure
- Systolic ≥140 mm Hg
- Diastolic ≥90 mm Hg
- Unilateral weakness
- Speech impairment
- Symptom duration
- ≥ 60 minutes – 2
- 10–59 minutes – 1
- < 10 minutes – 0.
Obs / Gyne
A 50 years old female patient with Fibroids asymptomatic. Next step of Management will be –
|B.||No treatment required
Answer:- b-No Treatment Required
Fibroids need the hormone estrogen to grow. After menopause, a woman’s estrogen levels decrease dramatically, which usually reduces the risk of developing fibroids.
In many cases, fibroids actually shrink and cause fewer symptoms for women who have reached menopause.
Women who are taking Hormone Replacement Therapy (HRT) during perimenopause or after menopause may not see a decrease in their symptoms. This is because HRT usually contains a combination of estrogen and Progesterone, which are the same hormones that allow fibroids to grow in younger women.
A female came with a history of spotting. On USG no fluid in pouch of Douglas was found; there was no gestational sac seen in the uterus, ovaries normal. UPT positive. What will be the best management?
|A.||Follow with hCG
|B.||Diagnose complete abortion & call after 15 days
|C.||admit & observe
|D.||declare as missed abortion
Answer :- A- Follow with hCG
:- Since the question says UPT i.e. Urine Pregnancy Test is Positive, & No fluid in pouch of Douglas on USG. This rules out the Diagnosis of Ectopic Pregnancy ( Ruptured/Unruptured).
Also, Question says there is no Gestational Sac on Usg But to confirm the intrauterine Pregnancy hCG levels need to be done ( Critical Titre).
As far as Spotting is concerned during implantation spot bleeding is a normal feature/sign.
So in the present case, we first need to know the levels of Hcg to confirm Intrauterine Pregnancy.
Also, Remember in case of abortion Hcg levels decrease after 2 weeks.
Criteria for abdominal pregnancy
|D.||Ring of fire
In primary abdominal pregnancy, which is the rarest type of ectopic gestation, the conceptus implants on the peritoneal surface.
Studdiford’s criteria used to diagnose primary abdominal pregnancy are described as:
- The presence of normal bilateral tubes and ovaries with no evidence of recent or past pregnancy.
- No evidence of a uteroperitoneal fistula.
- The presence of pregnancy is related exclusively to the peritoneal surface, early enough to eliminate the possibility of secondary implantation after primary tubal nidation.
K. Sujata Primary abdominal pregnancy following intra-uterine insemination; J Hum Reprod Sci. 2011 May-Aug; 4(2): 95–99.
For PCOD all of the following are management done, except?
Answer b. Ulipristal
Ulipristal is a selective progesterone receptor modulator used for the purposes of emergency contraception (Ella) and for the treatment of uterine fibroids (Fibristal). It is a derivative of 19-norprogesterone and has both antagonistic and partial agonist activity at the progesterone receptor.
MANAGEMENT OF PCOS
|Life Style Modifications||
|Diet Restrictions||HORMONAL PILLS||ANTIANDROGENS||INSULIN SENSITIZER||Eflornithine Hydrochloride|
|Weight Reduction||Progesterone With Low Androgenic Activity||CPA||Waxing|
- Ovarian drilling: Tiny holes made in the ovaries can reduce the levels of androgens being produced.
- Oophorectomy: Surgery removes one or both ovaries.
- Hysterectomy: This involves the removal of all or part of the uterus.
- Cyst aspiration: Fluid is removed from the cyst
Which structure is not cut in episiotomy?
|D.||Superior deep perineal muscles
Ans- b-Obturator muscle
The incision can be performed on either side and is generally 3-4 cm in length. The anatomic structures involved in a mediolateral episiotomy include the vaginal epithelium, transverse perineal muscle, bulbocavernosus muscle, and perineal skin. A deep or large mediolateral episiotomy may expose the ischiorectal fossa.
In the oblique technique, the perineal body is avoided, cutting only the vagina epithelium, skin, and muscles (transversalis and Bulbospongiosus). This technique aids in avoiding trauma to the perineal body by either surgical or traumatic means.
Which of the following is not the function of OC pills as a contraceptive?
|C.||Cervical mucous thickening
Answer -d- Delayed Menstruation
:- The progesterone is primarily responsible for preventing pregnancy.
The main mechanism of action is the prevention of ovulation; they inhibit follicular development and prevent ovulation.
Progesterone negative feedback works at the hypothalamus to decrease the pulse frequency of gonadotropin-releasing hormone.
This, in turn, will decrease the secretion of follicle-stimulating hormone (FSH) and decreases the secretion of luteinizing hormone (LH).
If the follicle isn’t developing, then there is no increase in the estradiol levels (the follicle makes estradiol). The progestogen’s negative feedback and lack of estrogen positive feedback on LH secretion stop the mid-cycle LH surge.
With no follicle developed and no LH surge to release the follicle, there is the prevention of ovulation. Estrogen has some effect with inhibiting follicular development because of its negative feedback on the anterior pituitary with slow FSH secretion; it’s just not as prominent as the progesterone’s effect. Another primary mechanism of action is progesterone’s ability to inhibit sperm from penetrating through the cervix and upper genital tract by making the cervical mucous unfriendly.
Progesterone induced endometrial atrophy should deter implantation, but there is no proof that this occurs.
A 50-year-old lady presented with Atypical vaginal bleeding (AUB). What is the next step of management?
|A.||Ethinyl estradiol tablet
Answer:- B Hysterectomy
In postmenopausal women, atypical adenomatous endometrial hyperplasia is usually treated with hysterectomy.
In premenopausal women, atypical adenomatous endometrial hyperplasia may be treated with medroxyprogesterone acetate 40 mg orally once a day for 3 to 6 months or a levonorgestrel-releasing IUD. After 3 to 6 months of treatment, endometrial sampling is repeated. If repeat endometrial sampling indicates the resolution of hyperplasia, women may be given cyclic medroxyprogesterone acetate (5 to 10 mg orally once a day for 10 to 14 days each month) or, if pregnancy is desired, clomiphene. If repeat endometrial sampling shows persistent or progressive atypical hyperplasia, hysterectomy may be necessary
10 weeks pregnant female has done USG which shows a snowstorm appearance. What will be the management?
|B.||a missed abortion do nothing
|D.||Follow up after 15 days by repeat USG
Snowstorm Appearance is a feature of Molar Pregnancy & management of molar pregnancy is evacuation & curettage irrespective of the gestational age.
Young female with post-coital bleeding examination showed no lesion. Pap smear shows carcinoma in situ. Next line of management
The following are generally considered to be indications for colposcopy
- Abnormal Pap test result (the primary indication for colposcopy)
- Abnormal-appearing tissue in the vagina, on the cervix or vulva, perineum, perianal area, or male genitalia
- Abnormal-appearing cervix, even if cervical cytology is normal
- Intrauterine exposure to diethylstilbestrol
- Child abuse and rape cases
- Patient history indicates a high risk for cervical cancer, such as a male partner who has had previous or current sex partners who developed cervical cancer
- Follow-up examinations after treatment for high-grade squamous intraepithelial lesion (HGSIL) or lower genital tract cancer
- Follow-up examination after a positive human papillomavirus (HPV) test result when the Pap test is normal
A 28 old G2P1 with 8 months of amenorrhea attains ANC OPD with history of pain in the abdomen, bleeding PV, and loss of fetal movements.She gives a history of hypertension and she is on medication. What is the probable diagnosis?
Answer:a– Abruptio Placenta
Placental abruption is most likely to occur in the last trimester of pregnancy, especially in the last few weeks before birth. Signs and symptoms of placental abruption include:
- Vaginal bleeding, although there might not be any
- Abdominal pain
- Back pain
- Uterine tenderness or rigidity
- Uterine contractions, often coming one right after another
Risk factors:-Factors that can increase the risk of placental abruption include:
- Placental abruption in a previous pregnancy that wasn’t caused by abdominal trauma
- Chronic high blood pressure (hypertension)
- Hypertension-related problems during pregnancy, including preeclampsia, HELLP syndrome or eclampsia
- A fall or other type of blow to the abdomen
- Cocaine use during pregnancy
- Early rupture of membranes, which causes leaking amniotic fluid before the end of pregnancy
- Infection inside of the uterus during pregnancy (chorioamnionitis)
- Being older, especially older than 40
Fetal biophysical profile test include all except
Answer:a– Contraction test
Several techniques for antepartum fetal surveillance currently in use are discussed in the ACOG bulletin. These include fetal movement assessment, nonstress test, contraction stress test, fetal biophysical profile, modified biophysical profile, and umbilical artery Doppler velocimetry.
The five components of the biophysical profile are as follows:
1. Nonstress test;
2. Fetal breathing movements (one or more episodes of rhythmic fetal breathing movements of 30 seconds or more within 30 minutes);
3. Fetal movement (three or more discrete body or limb movements within 30 minutes);
4. Fetal tone (one or more episodes of extension of a fetal extremity with the return to flexion, or opening or closing of a hand; and
5. Determination of the amniotic fluid volume (a single vertical pocket of amniotic fluid exceeding 2 cm is considered evidence of adequate amniotic fluid).
Each of the components is given a score of 2 (normal or present as defined previously) or 0 (abnormal, absent or insufficient). A composite score of 8 or 10 is normal, a score of 6 is equivocal and a score of 4 or less is abnormal. In the presence of oligohydramnios, further evaluation is warranted regardless of the composite score.
Folic acid and iron supplementation dose for reproductive age group
|A.||500mg Folic acid & 60mg elemental iron weekly
|B.||500mg Folic acid & 60mg elemental iron daily
|C.||400mg Folic acid & 100mg elemental iron weekly
|D.||400mg Folic acid & 100mg elemental iron daily
Answer:- B 500mg folic acid & 60mg elemental iron daily
- In all pregnant females: 60 mg elemental iron + 500 mcg of Folic acid is to be given daily.
- Starting from the 4th month of Pregnancy & Continued throughout pregnancy, minimum for 180 days.
- To be continued for 180 days, Postpartum (to replenish iron stores)
- Iron + folic acid is given because in India most common cause of anemia during pregnancy is iron & folic acid deficiency k/a dimorphic anemia
Note: if it is to prevent neural tube defect then folic acid supplementation is given before conception & first trimester. Iron salt present in this is ferrous sulphate.
- MALA-D & MALA-N:- ferrous fumarate
Esr increased in pregnancy due to?
Answer:- C Fibrinogen
- Blood volume Increases during pregnancy 40-45 % @ 10 weeks & max. @ mid trimester.
- In twin pregnancy 40-80%.
- Red blood cell volume – (20-30% starts at 8 weeks) since the increase in RBC volume is less in comparison to plasma volume there is hemodilution during pregnancy.
- No. of erythrocytes, reticulocytes & maternal erythropoietin volume also ↑
- In Physiological anemia, Hb is never <11gm/dl else it is pathological.
- Normally=A: G=1.7:1
- In pregnancy: A: G = 1:1
- Gestational age and hemoglobin concentration both significantly influenced the erythrocyte sedimentation rate. (P < 0.0001).
- For non-anemic women, the 95% reference range rose from 18-48 mm/h in the first half of pregnancy to 30-70 mm/h in the second half of pregnancy.
A 32-year female goes to an infertility clinic with a regular cycle of 28 days. What should be evaluated?
|A.||LH level at 21 days
|B.||Progesterone at 21 days
|C.||LH at 14 days
|D.||FSH at 21days
Answer: B- Progesterone at 21 days
Cycle Day 21 Progesterone Level & Ovulation Tests
- Day 21 testing checks a woman’s progesterone level to confirm that ovulation has occurred.
- A low Day 21 progesterone level suggests the cycle you are in currently was anovulatory (no egg was produced).
- When no egg is produced, pregnancy cannot be achieved that month.
In which age group should screen of Carcinoma Cervix is done?
Answer: A ; 22-64 years
The American Cancer Society recommends that women follow these guidelines to help find cervical cancer early. Following these guidelines can also find pre-cancers, which can be treated to keep cervical cancer from starting.
- All women should begin cervical cancer testing (screening) at age 21.
- Women aged 21 to 29, should have a Pap test every 3 years. HPV testing should not be used for screening in this age group unless it is needed after an abnormal Pap test result.
- Beginning at age 30, women should be screened with a Pap test combined with an HPV test every 5 years as long as the test results are normal. This is called co-testing and should continue until age 65.
Another reasonable option for women 30 to 65 is to get tested every 3 years with only the Pap test.
- Women over age 65 who have had the regular screening in the past 10 years with normal results should stop cervical cancer screening. Once stopped, it should not be started again.
- Women with a history of a serious pre-cancer, such as CIN2 or CIN3 should continue to have testing for at least 20 years after that condition was found, even if the testing goes past age 65.
- Women who have had a total hysterectomy (removal of the uterus and cervix) should stop screening (such as Pap tests and HPV tests) unless the hysterectomy was done as a treatment for cervical pre-cancer (or cancer). Women who have had a hysterectomy without removal of the cervix (called a supracervical hysterectomy) should continue cervical cancer screening according to the guidelines above.
- Women who are at high risk of cervical cancer because of a suppressed immune system (for example from HIV infection, organ transplant, or long-term steroid use) or because they were exposed to DES in utero may need to be screened more often. They should follow the recommendations of their health care team.
- Women of any age should NOT be screened every year by any screening method if their Pap tests have been normal and they do not have HIV infection or other cause for a weakened immune system.
Bishops Score includes all except:
|A.||Position of Cervix
|B.||Dilatation of Cervix
|C.||Consistency of Cervix
|D.||Maternal Pelvis Type
Answer :– D. Maternal Pelvis Type
Bishops scoring system
|Position of cervix||Effacement
(-3 to +3)
A 30yrs female with an enlarged uterus of 14 weeks size. USG showed a single fibroid, asymptomatic, best management is
|B.||as she is asymptomatic don’t do anything
Answer B. as she is asymptomatic don’t do anything
- Leiomyomata uteri (uterine fibroids) are benign tumours of the smooth muscle of the uterus. These tumours have prevalence ranging from 20 to 50% of women depending on the age, ethnicity, parity, and methods use to assess their presence.
- Uterine fibroids grow under the influence of the hormone estrogen and are most often seen after the menarche, and tend to shrink after the menopause.
- Fibroids are very common and women with small fibroids who are asymptomatic are best left untreated. Women with symptoms who have small fibroids but are close to the menopause or who are trying to conceive should be treated conservatively with analgesics and hematinics. Women who have severe symptoms or very large fibroids usually need surgical intervention. This may be conservative with myomectomy done by laparotomy (all fibroids), laparoscopy (subserous fibroids), or hysteroscopy (submucous fibroids).
54. A young girl came to AIIMS emergency OPD with her mother. Her DOB was 31/1/2007 and the history of peno- vaginal penetration by their neighbour was given by the patient. She also said that she gave consent for the same and refused for further medical examination. What could you do as a medical practitioner attending such patients?
1) Should call the police
2) Should counsel mother & daughter
3) Should do examination even after refusal
4) Document refusal
2, 3, 5 true
1, 2, 3 true
3, 4, 5, true
2, 3, 4 true
Answer- B– 1,2 and 3 are true
After 3 February 2013, the age to define rape has been changed.
- The age of consent in India has been increased to 18 years, which means any sexual activity irrespective of presence of consent with a woman below the age of 18 will constitute statutory rape.
- Here the age of the girl is 13 years, it is a case of statutory rape
- If she is less than 12 years of age or if she is mentally unsound, the written consent of parent/guardian should be taken (Sec. 90 IPC).
- Rape.—A man is said to commit “rape” who, except in the case hereinafter excepted, has sexual intercourse with a woman under circumstances falling under any of the six following descriptions:—
- (First) — Against her will.
- (Secondly) —Without her consent.
- (Thirdly) — With her consent, when her consent has been obtained by putting her or any person in whom she is interested in fear of death or of hurt.
- (Fourthly) —With her consent, when the man knows that he is not her husband, and that her consent is given because she believes that he is another man to whom she is or believes herself to be lawfully married.
- (Fifthly) — With her consent, when, at the time of giving such consent, by reason of unsoundness of mind or intoxication or the administration by him personally or through another of any stupefying or unwholesome substance, she is unable to understand the nature and consequences of that to which she gives consent.
- (Sixthly) — With or without her consent, when she is under sixteen years of age. Explanation.—Penetration is sufficient to constitute the sexual intercourse necessary to the offense of rape
54. A patient has been brought to your emergency department with the following signs and symptoms:
Dry mouth, irrelevant slurred speech, dry and hot skin, dilated pupils. What is the most probable diagnosis:
Answer-B- Datura Poisoning
- All parts of these plants are poisonous— fruit, flowers and seeds (highest concentrations of alkaloids are found in roots and seeds)
- Poisoning occurs only if seeds are masticated and swallowed
- Dryness of the mouth (dry as a bone), bitter taste,burning pain in stomach and vomiting.
- Dysphagia (difficulty in swallowing).
- Dysarthria (difficulty in talking) due to inhibition of salivation.
- Dilatation of cutaneous blood vessels (red as a beet).
- Face is flushed and conjunctiva congested
- Dry hot skin (hot as a hare) due to inhibition of sweat and stimulation of heat-regulating centre. Temperature is raised by 1-2°C
A patient died and relatives complain that it is due to negligence of the doctor. According to a recent Supreme court judgement, the doctor can be charged for Medical Negligence under section 304-4, only if
|A.||If he is from corporate hospital
|B.|| Negligence is from inadvertent error
|D.||Res Ipsa Loquitur
Answer- C Gross Negligence
- According to recent supreme court judgement, doctor can be charged for medical negligence under section 304- A IPC (i.e. criminal negligence), only if he shows gross lack of competence/skill, gross inattention, or inaction or recklessness or wanton (gross) indifference to the patient’s safety or gross negligence.
Image Based Question
Which artery passes deep to the marked area in the given image:
|A.||Deep temporal Artery
|B.||Middle meningeal Artery
|C.||Anterior cerebral Artery
|D.||Internal Carotid Artery
Answer B. Middle meningeal artery
- The pterion is the region where the frontal, parietal, temporal, and sphenoid bones join together.
- It is located on the side of the skull, just behind the temple.
- It is the junction between four bones:
- the parietal bone
- the squamous part of the temporal bone
- the greater wing of the sphenoid bone
- the frontal bone
- The pterion is known as the weakest part of the skull.
- The anterior division of the middle meningeal artery runs underneath the pterion.
- Consequently, a traumatic blow to the pterion may rupture the middle meningeal artery causing an epidural hematoma.
The image of a developing heart is given. Which structure developed from the given structure?
Answer C. Right ventricle
- The bulbus cordis develops into the right ventricle.
- The primitive ventricle forms the left ventricle.
- The primitive atrium becomes the anterior portions of both the right and left atria, and the two auricles.
Action of muscle at the hip attached to the marked ‘A’ area in the given image:
|A.||Abduction and External rotation
|B.||Adduction and Lateral Rotation
|C.||Extension and Lateral Rotation
|D.||Flexion and Medial Rotation
Answer C. Extension and Lateral Rotation
Marked area- Gluteal tuberosity
- The gluteal tuberosity gives attachment to part of the Gluteus maximus muscle.
- Function. Gluteus maximus main actions are to extend and laterally rotate the hip joint.
Neurotransmitter released by the marked structure:
Answer B. Dopamine
Structure marked in the image is – Substantia nigra
- The substantia nigra (SN) is a basal ganglia structure located in the midbrain that plays an important role in reward and movement.
- Substantia nigra is Latin for “black substance”, reflecting the fact that parts of the substantia nigra appear darker than neighboring areas due to high levels of neuromelanin in dopaminergic neurons.
- Parkinson’s disease is characterized by the loss of dopaminergic neurons in the substantia nigra pars compacta.
During deep inspiration, upto what level does the diaphragm descend?
|B.||Upper 1/3rd lower
Answer B. Upper 1/3rd lower
Lateral wall of the nose as shown in the image. The anterior most cell in the ethmoid air cell group is?
Answer A. Agger Nasi
Agger nasi air cells are the most anterior ethmoidal air cells lying anterolateral and inferior to the frontal recess and anterior and above the attachment of the middle turbinate.
They are located within the lacrimal bone and therefore have as lateral relations the orbit, the lacrimal sac and the nasolacrimal duct.
They are identified in 90% of patients, and should not be confused with Haller cells (located along the medial floor of the orbit) or supraorbital air cells (located in the orbital roof).
The following type of blood smear is due to?
|D.||Defect with glycoprotein
Ans. A. Spectrin deficiency
Pathogenesis of hereditary spherocytosis:
- Loss of membrane cytoskeleton proteins (ankyrin, spectrin, Band 3, 4.2) results in reduced membrane stability.
- Reduced membrane stability leads to spontaneous loss of membrane fragments during exposure to shear stresses in the circulation.
- The loss of membrane relative to cytoplasm forces the cells to assume the smallest possible diameter for a given volume —> Cells become microspherocytes.
- Because of their spheroidal shape and reduced membrane plasticity, red cells become less deformable and are trapped into spleen as they are unable to pass through the interendothelial fenestrations of the venous sinusoids.
- In the splenic sinusoids, red cells are phagocytosed by RE cells –> Extravascular hemolysis.
The graph below shows the normal rate of synthesis of various hemoglobin chains in utero, and after birth. Gamma chains of hemoglobin are indicated by which of the following?
Ans. B. (B)
The fetal hemoglobin (Hemoglobin F) is composed of α2γ2 (Above graph, Chain A:α; Chain B:γ).
In young embryos, there are, in addition, δ and ε chains (Chains D in the above graph), forming Gower 1 hemoglobin (δ2ε2) and Gower 2 hemoglobin (α2ε2).
Identify the proteoglycan:
Ans D- Chondroitin sulfate
It is present in the ground substance of connective tissues widely distributed in cartilage, bone, tendons, cornea, and skin. It is composed of repeating units of glucuronic acid → beta-1,3-N-acetyl galactosamine sulphate → beta-1, 4 and so on
Correct about Given vacutainer is?
Answer: D. Clot activator
Red colour – serum tube – Clot activator
Red tube – plain tube – no additive
Which of the following drugs can cause the side effect given in the picture?
Ans. A. Bleomycin
- It is flagellate dermatitis.
- Usage of bleomycin causes the formation of rash after 2 weeks of drug use which is the linear rash on back.
- Flagellate dermatitis (Toxicoderma) is a rare, presumedly toxic epidermal damage following consumption of raw or half-cooked Shiitake mushrooms.
- Characteristically the patients develop a parallel, whiplash-like infiltrated erythemas with papulae or papulovesicules, arranged like scratches and associated with severe itching 1–2 days after eating Shiitake mushrooms.
- The localizations are preferentially the trunk, the extremities, and the nape of the neck.
- The clinical pattern gives rise to the name flagellate dermatitis, since the skin changes appear like skin damage caused by whiplashes.
- The Flagellants (Lat. flagellum = whip) were a lay Christian movement in the Middle Ages.
- The religious practices of the Flagellants included public self-flagellation as penance and purification from sin.
- Flagellate dermatitis, erythema or hyperpigmentation is a cutaneous side effect of bleomycin or its derivative peplomycin.
Identify the image
Ans: D. Nigella Sativa
- Nigella sativa (N. sativa) (Family Ranunculaceae) is a widely used medicinal plant throughout the world.
- The most important active compounds are thymoquinone (30%-48%), thymohydroquinone, dithymoquinone, p-cymene (7%-15%), carvacrol (6%-12%), 4-terpineol (2%-7%), t-anethol (1%-4%), sesquiterpene longifolene (1%-8%) α-pinene and thymol, etc.
- inhibition of the growth of Staphylococcus aureus was observed by a concentration of 300 mg/mL with distilled water as control, this inhibition was confirmed by using the positive control Azithromycin.
- Methanolic extracts of N. sativa have the strongest antifungal effect followed by the chloroform extracts against different strains of Candida albicans.
Identify the type of fracture shown in the image:
Comminuted fracture (spider-web/mosaic fracture):
Two or more intersecting lines of fracture divide the bone into three or more fragments.
- Skull bone gets broken into multiple pieces by fracture lines, which are haphazardly or concentrically arranged, or stellate if they radiate from the site of impact.
- It is caused by vehicular accidents, fall from a height on a hard surface or by blows with weapons having large striking surface, such as heavy iron bar, or from a bullet
Identify the type of wound ?
|C.||Incised looking lacerated wound
Answer-C. Incised looking lacerated wound
- Lacerations are caused by blunt-force trauma. The hallmark of lacerations is the presence of tissue bridging
- Incised-looking lacerated wounds: When the skin is closely applied to the bone and the subcutaneous tissue is scanty, blunt force may produce a wound which by the linear splitting of the tissues resembles an incised wound.
The Patient is a teenage girl with a history of seizure, an MRI Image was, Blood investigation reveals elevated Lactate levels. Causative organism is
Answer- A. Toxoplasmosis
- Toxoplasmosis results from infection with a common parasite found in cat faeces and contaminated food. It can cause serious complications for pregnant women and people with weakened immune systems. Symptoms include muscle pain, fever, and headache, all of which can last for weeks.
Toxoplasmosis is especially serious for people who have weakened immune systems. For these people, they’re at risk of developing:
- brain inflammation, causing headaches, seizures, confusion and coma.
- a lung infection, causing cough, fever, and shortness of breath
- an eye infection, causing blurry vision and eye pain
In Formal ether sedimentation Technique parasite eggs settle in
Sedimentation techniques use solutions of lower specific gravity than the parasitic organisms, thus concentrating the latter in the sediment.
It takes advantage of the high specific gravity of protozoan cysts and helminth eggs compared to water. Their natural tendency to settle out in aqueous solutions can be accelerated by light centrifugation.
Formalin fixes the eggs, larvae, oocysts, and spores, so that they are no longer infectious, as well as preserves their morphology. Fecal debris is extracted into the ethyl acetate phase of the solution. Parasitic elements are sedimented at the bottom.
A patient comes with a complaint of diarrhea for 7 days, weight loss, reduced appetite. On stool examination, there are acid-fast oocyst of size 28 micrometers, symptoms subsided with clindamycin, the organism is
Answer- B- Cystoisospora
- These parasites normally infect the enterocytes of the small intestine and are spread by the oral faecal route. The definitive hosts are cats but other species including various species of rodents may be infected.
- Clinical signs include watery diarrhea, vomiting, fever, and weight loss. The diagnosis is made by microscopic examination of the stool. Distinguishing between the species of Cystoisospora is most easily done with PCR. This method can also be used to make a diagnosis.
- Treatment is based on trimethoprim-sulfonamides with clindamycin or toltrazuril for resistant strains
Identify the structure (marked by the arrow) in the given image:
Answer B. Pyriform fossa
Larynx: Pyriform recess/sinus/fossa
- On either side of the laryngeal orifice in humans is a recess, termed the pyriform sinus (also piriform recess, piriform sinus, piriform fossa, or smuggler’s fossa), which is bounded medially by the aryepiglottic fold, laterally by the thyroid cartilage and thyrohyoid membrane. The fossae are involved in speech.
- The term “pyriform,” which means “pear-shaped,” is also sometimes spelled “piriform”.
The mode used for voice production is?
Ans: A.Tracheoesophageal Prosthesis
Tracheoesophageal puncture and tracheoesophageal prosthesis placement. Tracheoesophageal puncture is a surgical method of voice restoration after total laryngectomy that implies the creation of a fistulous tract in the tracheoesophageal wall—the wall that separates the trachea and esophagus, in the level of tracheostomy.
Image of elderly lady undergoing the procedure. Identify the procedure?
|A.||Anterior ethmoidal nerve block
|B.||Vidian nerve block
|D.||Diagnostic probing of Nasolacrimal gland
Ans: D. Diagnostic probing of Nasolacrimal gland
- Whenever a patient complains of excessive lacrimation it indicates the blockage in the lacrimal apparatus.
- So to determine the site of blockage and If pathology of the lacrimal drainage system is suspected, probing and irrigation of the system is a useful diagnostic tool.
- The above picture illustrates diagnostic probing.
- Probing: we insert a lacrimal probe through inferior puncta and it passes through the passages of the lacrimal apparatus.
- The blockage of nasolacrimal duct is the most common blockage in the Nasolacrimal duct.
A lady fell in the bathroom with an outstretched hand. She was managed conservatively. The clinical picture of the patient given below. The deformity is due to:
|A.||I/A fracture DER
|B.||E/A fracture DER
|D.||Dislocation of wrist
Answer B. Distal radius fracture (Extra Articular)
- A dinner fork deformity, also known as a bayonet deformity, occurs as the result of a malunited distal radial fracture, usually a Colles fracture.
- The distal fragment is dorsally angulated, displaced, and often also impacted.
- The term is descriptive, as the lateral view of the wrist is similar to the shape of a fork, seen from the side, tines down.
- Colles fractures are very common extra-articular fractures of the distal radius that occur as the result of a fall onto an outstretched hand.
- They consist of a fracture of the distal radial metaphyseal region with dorsal angulation and impaction, but without the involvement of the articular surface.
An old female Diabetic patient came to you and wished you Namaste. While doing namaste you as a doctor observes her hand. Identify the deformity as shown in the image:
|A.||Flexor tendon tenosynovitis
Answer C) Chieroarthropathy
- The “prayer sign” indicates the presence of diabetic cheiroarthropathy.
- It is characterized by patients’ inability to completely close gaps between opposed palms and fingers when pressing their hands together.
- Diabetic amyotrophy- Nerve disorder complication of diabetes mellitus. It affects the thighs, hips, buttocks, and legs, causing pain and muscle wasting.
Table 1.musculoskeletal and rheumatological complications of diabets
conditions affecting the hands
- diabetic cheiroarthropathy(stiff
- hand syndrome of syndrome of limited joint mobility)
- flexor tenosynovitis (trigger finger)
- dupuytren’s contracture
- carpal tunnel syndrome
conditions affecting the shoulders
- adhesiv capsulities (frozen shoulder )
- calcife periarthritis
- reflex sympathetic dystrophy
conditions affecting the feet
- diabetic asteoarthropathy (chacot
- or nuropathic arthropathy )
conditions affecting the muscles
- diabetic muscle infarction
conditions affecting the skeleton
- diffuse idiopathic skeletal hyper-
The given X-ray can be seen in:
Answer A) Ankylosing spondylosis
- Bamboo spine is a radiographic feature seen in ankylosing spondylitis that occurs as a result of vertebral body fusion by marginal syndesmophytes.
- It is often accompanied by fusion of the posterior vertebral elements as well.
- A bamboo spine typically involves the thoracolumbar and/or lumbosacral junctions and predisposes to unstable vertebral fractures and Andersson lesions.
A 60 years old female underwent hip replacement. After 1 year she landed up with loosening and the implant had to be removed. The slide (eosinophilic pigmentation and black pigment) of the implant is shown below. What is the likely cause?
Answer A) Metallosis
Metallosis is a type of metal poisoning that can occur as a side effect of joint replacement devices with metal components, such as metal-on-metal hip replacements or other metal implants.
These devices are made from a blend of several metals, including chromium, cobalt, nickel, titanium, and molybdenum.
When the metal parts rub against each other, they release microscopic metal particles into the blood and surrounding tissues.
Metallosis develops as these metal ions build up in the bone, muscle and other tissue around the implant. This can result in death of bone or other tissue.
Identify the instrument-
|A.||Perkins mastoid retractor
|B.||Mollison’s mastoid retractor
|D.||Jansen’s self-retaining mastoid retractor
Ans: A. Perkins self-retaining mastoid retractor
This self-retaining mastoid has three prongs of equal sizes in one blade and a single flat broad prong on the other blade.
The single flat broad prong is useful in retracting skin away from the surgical site.
It is very useful in retracting canal skin during ear surgeries.
Choose the correct option by looking at the given image?
|A.||Vidian nerve block
|B.||Frontal sinus trephination
|C.||Anterior ethmoidal block
Ans. C. Frontal sinus trephination
Anterior ethmoidal block
- The anterior ethmoidal nerve passes through the anterior ethmoidal foramen and canal and enters the cranial cavity superior to the cribriform plate of the ethmoid bone.
- It then descends through a slit at the side of the crista Galli to enter the nasal cavity where it divides into medial and lateral internal nasal branches (Standring, 2008).
- The lateral internal nasal branch supplies the anterior part of the lateral wall of the nasal cavity, while the medial internal nasal branch supplies the anterior and upper parts of the nasal septum.
- The anterior ethmoidal nerve continues anteriorly, emerging at the inferior margin of the nasal bone as the external nasal nerve, which supplies the skin of the nasal tip and external lateral nose.
- Effective anesthesia of all structures surrounding the nasal fracture, both internal and external, can be achieved reliably by a bilateral regional block of infratrochlear and ethmoid branches of the nasociliary nerve att he anterior ethmoid foramen.
Identify the given image with multiple calcifications, lipid, choline peaks
Answer . C. Neurocysticercosis
Neurocysticercosis (NCC) is the more frequent parasitic disease of the CNS in immunocompetent patients associated with infection due to its larval form of Taenia solium.
It spreads through intermediary hosts, pigs and eventually men to his definitive host: humans.
It causes lesions in the CNS which have different evolutive stages and can be asymptomatic or clinically evident.
Imaging studies play an important role in the diagnosis of the disease and in its appropriate control.
The most common symptoms are not specific, like headaches and convulsions due to perilesional and edema(3), and less frequently brain infarcts and vasculitis. If the cysts are located near eloquent areas there will be focal manifestations, like motor deficit, ataxia, etc.
NCC is classified according to imaging studies in five stages:- Non-Cystic, Vesicular, Colloidal Vesicular, Granular Nodular, Calcified Nodular
|D.||Klippel Feil syndrome
Diffuse idiopathic skeletal hyperostosis (DISH), also referred to as Forestier disease, is a common condition characterized by bony proliferation at sites of tendinous and ligamentous insertion of the spine affecting elderly individuals.
On imaging, it is typically characterized by the flowing ossification of the anterior longitudinal ligament involving the thoracic spine and enthesopathy (e.g. at the iliac crest, ischial tuberosities, and greater trochanters). There is no involvement of the sacroiliac synovial joints.
Radiographic features: Plain radiograph and CT
- flowing ossifications: florid, flowing ossification along the anterior or right 7 anterolateral aspects of at least four contiguous vertebrae
- Disc spaces are usually well preserved
- Ankylosis is more common in the thoracic than cervical or lumbar spine
- frequently incomplete
- can have interdigitating areas of protruding disc material in the flowing ossifications
- no sacroiliitis or facet joint ankylosis
- enthesopathy of the iliac crest, ischial tuberosities, and greater trochanters
- spur formation in the appendicular skeleton (olecranon, calcaneum, patellar ligament) frequently present
- ‘whiskering’ enthesophytes
Identify this condition of hypopigmented patches?
Ans. D-Pityriasis alba
- Pityriasis alba is a common, benign skin disorder occurring predominantly in children and adolescents.
- The name refers to its appearance: pityriasis refers to its fine scales and alba to its pale color (hypopigmentation).
- Most patients have a history of atopy, and pityriasis alba may be a minor manifestation of atopic dermatitis.
- It is characterized by ill-defined macules and patches (or thin plaques), round or oval, often with mild scaling, and sometimes with mild pruritus.
- The lesions initially may be mildly erythematous, and over time they become hypopigmented. They are most commonly located on the face (especially the cheeks), arms, and upper trunk, and they are more noticeable in people with darker skin types.
A pt.with joint pain, fever for one week, and to relieve pain he had NSAIDS,following which He had developed Rash and hyperpigmentation on nose.what is the most likely diagnosis?
|C.||Fixed drug eruption
Ans. B. Chikungunya rash
- The picture depicts a chick sign seen in Chikungunya.
- Generalized morbilliform maculopapular rash with normal islands of intervening skin and characteristically sparing the face has been the most commonly (30–55%) reported manifestation.
- The rash develops within 3–4 days of the onset of fever and tends to subside in about a week without any sequel.
- Skin hyperpigmentation is the next most commonly encountered finding.
- It occurs in the form of centrofacial pigmentation predominantly involving the nose (Chick sign).
- It is the most commonly described pigmentation developing during the acute phase of the disease and helps in the retrospective diagnosis of CHK infection.
Which of the following conditions does the picture depict?
Ans. A. Sporotrichosis
Sporotrichosis is caused by infection with Sporothrix schenkii, a dimorphic fungus, found in soil, wood, and plant surfaces.
Lymphocutaneous sporotrichosis, the most common form, presents as a small, nontender, erythematous papulonodule at the site of the primary injury.
This lesion may be smooth or verrucous, often ulcerates, and develops raised red borders.
Over days to weeks, proximal subcutaneous nodules form along the lymphatic drainage, and may ulcerate. Fungal cultures and tissue biopsies aid in the diagnosis.
Physical examination showed an ulcerated, raised, dry, crusted lesion on the lateral surface of the left thumb, with four proximal raised, erythematous, subcutaneous nodules, without epitrochlear or axillary lymphadenopathy.
Purulent material was aspirated from one of the nodules; gram, fungal and mycobacterial stains showed no organisms.
Which of the following conditions is presented in the picture below? History of death of sibling due to the same lesion reported.
Ans. A. Epidermolysis bullosa
This child presents with blisters & skin lesions on face.
Epidermolysis bullosa simplex (EBS) is characterized by fragility of the skin (and mucosal epithelia in some cases) that results in non-scarring blisters and erosions caused by minor mechanical trauma.