Instruction:
A 23 year old profession footballer suffered a twisting injury to his right ankle. On examination, there is a lot of swelling around the medial malleolus but x-ray doesn't show any fracture. The structure injured could be
A. Deltoid Ligament
B. Anterior talofibular ligament
C. Spring ligament
D. Tendo Achilles

Response:
(Refer: Manish Kumar Varshney's Essential Ohopedics Principles & Practice, 1st edition, pg no. 1265 - 1267)

So the answer to this question is A


Instruction:
Not a synthetic pyrethroid insectiside -
A. Propoxur
B. Permethrin
C. Cypermethrin
D. Etofenoprox

Response:
Ans. is 'a' i.e., Propoxuro Inectisides are divided into:-A) Fumigants : Hydrogen cyanide, SO2, methyl bromide, carbon disulphate.B) Stomach poisons: Paris green, Sodium fluoride.C) Contact (nerve) poisons: These are divided into -1) Natural: Pyrethrum, rotenone, Derris, nicotine, mineral oil.2) Synthetic: These are -i) Organophosphates: Malathion, parathion, fenthion, diazinon, fenitrothion, abate, chlorthion, dioxathion, chlorpyrifos, trichlorfon.ii) Carbamates : Carbaryl, propoxur, dimetilan, pyrolon.iii) Organochlorine: DDT, BHC (HCH), dieldrin, aldrin, chlordane, lindane, heptachlor.iv) Synthetic pyrethroid (pyrethrum derivatives) : Pothrin, resmethrin, permethrin, biopermethrin, cypermethrin, cyphenothrin, cycloprothrin, deltamethrin, cyfluthrin, etofenoprox.

So the answer to this question is A


Instruction:
Harlequin skin change is seen due to mutation of which gene?
A. ABC A 12
B. FAD
C. Keratin 1
D. ALOXE 3

Response:
Ans. is 'a' i.e., ABCA 12 Harlequin Ichthyosis* Harlequin ichthyosis (HI) is caused by mutations in the ABCA 12 gene.* Mutation in the gene leads to defective lipid transport and ABCA 12 activity is required for the generation of long- chain ceramides that are essential for the development of the normal skin barrier. It is inherited by autosomal recessive mode of inheritance.Clinical Manifestations* At birth, markedly thickened, ridged, and cracked skin forms horny plates over the entire body, disfiguring the facial features and constricting the digits.Treatment* Initial treatment includes high fluid intake to avoid dehydration from transepidermal water loss and use of a humidified heated incubator, emulsifying ointments, careful attention to hygiene, and oral retinoids (1 mg/kg/ day).* Prenatal diagnosis has been accomplished by fetoscopy, fetal skin biopsy, and microscopic examination of cells from amniotic fluid.Extra edge* N.B. The fetus in ichthyosis appears grostesque and the appearance is called harlequin fetus. But, it is different from Harlequin color change.Harlequin color change* This rare but dramatic vascular event occurs in immediate newborn period and is most common in low birth weight infants.* If reflects an imbalance in the autonomic vascular regulatory mechanism.* When the infant is placed on the side, the body is bisected longitudinally into pale upper half and a deep red dependent half.

So the answer to this question is A


Instruction:
The mandibular labial frenum is a fibrous band that can be influenced by
A. Incisivus
B. Orbicularis oris
C. Caninus and incisivus
D. Incisivus and orbicularis oris

Response:
None

So the answer to this question is D


Instruction:
Which antigen is not of prognostic significance in carcinoma breast
A. Her 2 neu
B. Epithelial membrane antigen
C. ER
D. PR

Response:
The presence of hormone receptors confers a slightly better prognosis.The highest rate of response (approximately 80%) to antiestrogen therapy (oophorectomy or tamoxifen) is seen in women whose tumor cells express both estrogen and progesterone receptors Overexpression of HER2/NEU:Overexpression of this membrane-bound protein is almost always caused by gene amplification ref : Robbins Basic Pathology, 9E,page-713

So the answer to this question is B


Instruction:
What is the concentration of lignocaine used for topical anaesthesia?
A. 2%
B. 2-10%
C. 7-12 %
D. 12-15%

Response:
Anesthesia of mucous membranes of the nose, mouth, throat, tracheobronchial tree, esophagus, and genitourinary tract can be produced by direct application of aqueous solutions of salts of many local anesthetics or by suspension of the poorly soluble local anesthetics. Tetracaine (2%), lidocaine (2-10%), and cocaine (1-4%) typically are used. Maximal safe total dosages for topical anesthesia in a healthy 70-kg adult are 300 mg for lidocaine, 150 mg for cocaine, and 50 mg for tetracaine. Ref: Catterall W.A., Mackie K. (2011). Chapter 20. Local Anesthetics. In L.L. Brunton, B.A. Chabner, B.C. Knollmann (Eds), Goodman & Gilman's The Pharmacological Basis of Therapeutics, 12e.

So the answer to this question is B


Instruction:
Most common histological type of the thyroid carcinoma is
A. Medullary type
B. Follicular type
C. Papillary type
D. Anaplastic type

Response:
Papillary thyroid cancer is the most common cancer. Occurs in 70-80% of patients diagnosed with thyroid cancer. Common in females and younger age group. According to Woolner's Classification there are 3 types. 1)Occult primary(less than 1.5cm) 2) Intrathyroidal 3) Extrathyroidal Reference: SRB's Manual of Surgery, 6th Edition, page no = 468.

So the answer to this question is C


Instruction:
During which gestational age is peak serum HCG levels attained ?
A. 7-9 weeks
B. 11-13 weeks
C. 20 weeks
D. 25 weeks

Response:
Human chorionic gonadotropin Glycoprotein with alpha and beta chains 3 subunit is specific to hCG Secreted by trophoblastic tissue, some amount by fetal tissues, adult ant pituitary Many isoforms of hCG with variable crossreactivity between various assays More than loo commercial assays Sandwich type immunoassay: Sens - 1 IU/ml Immuno radiometric assays: more Sensitive Ref : Dutta book of obstetrics 8th Ed.

So the answer to this question is A


Instruction:
To prevent the spread of yellow fever across borders, the area surrounding airpos and seapos should be free from breeding of aedes mosquito. What is the area to be kept free of aedes mosquito/insect vector?
A. At Least 200m around the sea/airpos
B. At Least 300m around the sea/airpos
C. At Least 400m around the sea/airpos
D. At Least 500m around the sea/airpos

Response:
As per the guidelines an area extending at least 400 m around an airpo/sea po must be kept free from the breeding of insect vector. Also know: Aedes aegypti index - is the % of houses and their premises in a well defined area, showing actual breeding of aedes aegypti larvae. It should not be more than 1% in towns and seapos in endemic areas to ensure freedom from yellow fever. Ref: Park, Edition 21, Page - 259.

So the answer to this question is C


Instruction:
Predisposing factor to carcinoma endometrium is
A. Multiparity
B. Early marriage
C. DM
D. OCP

Response:
i.e. (D M): (416-18-Shaw's 15th)PREDISPOSING FACTORS IN CARCINOMA ENDOMETRIUM* Un opposed and unsupervised administration of hormone replacement therapy (HRT) after menopause predispose to endometrial hyperplasia and cancer. Fortunately the malignancy is well-differentiated with good prognosis* Chronic non ovulatory cycle as seen in DUB* In some families, a strong familial predispotion is noticed. This may be due to genetic or dietic habits such as animal protein and fat. The oestrone is derived by peripherlal aromatization in the fat tissue from androstenedione and contributes to high level of oestrogen. Women with familial Lynch II syndrome suffering from anorectal and breast cancer are also likely to suffer from endometrial cancer* Tamoxifen given to women with breast cancer increases the risk of endometrial hyperplasia and cancer to 2- 3 folds* Combined oral hormonal pills have protective effect and reduce its risk by 40-50% and also adding progestogens for 12 days each cycle to oestrogen in HRT reduces its risks to 20%**** Obesity, hypertension and diabetes characterize this cancer in 30%* Infertile women and women with PCOS on account of non-ovulation have high oestrogen. They develop endometrial hyperplasia and endometrial cancer more than normal women. The uterine fibroid is associated with endometrial cancer in 3% after the age of 40 years* 15% women with endometrial cancer have feminizing ovarian tumour at the time of diagnosisOCP|| |Tumours associated Provides protection1. Cervical cancer 1. Ovarian tumours2. Hepatic adenoma 2. Uterine tumours3. Pituitary adenoma 3. Benign breast disease

So the answer to this question is C


Instruction:
Iris pearls are seen in uveiitis due to -
A. Tuberculosis
B. Syphilis
C. Toxoplasma
D. Leprosy

Response:
Ans. is 'd' i.e., Leprosy * Lepromatous uveitis may occur as acute iritis (non-granulomatous) or chronic iritis (granulomatous).1) Acute iritis : It is caused by antigen-antibody deposition and is characterised by severe exudative reaction.2) Chronic granulomatous iritis : It occurs due to direct organismal invasion and is characterised by presence of small glistening'iris pearls' near the pupillary margin in a necklace form; small pearls enlarge and coalesce to form large pearls. Rarely, a nodular lepromata may be seen.

So the answer to this question is D


Instruction:
Which of the following effects is unlikely to occur during treatment with imipramine ?
A. Elevation of seizure threshold
B. Mydriasis
C. Sedation
D. Urinary retention

Response:
None

So the answer to this question is A


Instruction:
Microscopy of corneal ulcer showed branching septate hyphae. The probable diagnosis is:
A. Candida
B. Mucormycosis
C. Aspergillus
D. Histoplasma

Response:
Ans. Aspergillus

So the answer to this question is C


Instruction:
A 72-year-old man undergoes a subtotal colectomy for a cecal perforation due to a sigmoid colon obstruction. He has had a prolonged recovery and has been on total parenteral nutrition (TPN) for 2 weeks postoperatively. After regaining bowel function, he experienced significant diarrhea. Examination of his abdominal wound demonstrates minimal granulation tissue. He complains that he has lost his taste for food. He also has increased hair loss and a new perioral pustular rash. Which of the following deficiencies does he most likely have?
A. Zinc
B. Selenium
C. Molybdenum
D. Chromium

Response:
The patient has a zinc deficiency. Alopecia, poor wound healing, night blindness or photophobia, anosmia, neuritis, and skin rashes are all characteristic of patients with zinc deficiency, which often results in the setting of excessive diarrhea. Selenium deficiency is characterized by the development of a cardiomyopathy. Molybdenum deficiency is manifested by encephalopathy due to toxic accumulation of sulfur-containing amino acids. Chromium deficiency can occur in patients on long-term TPN and is characterized by difficult-to-control hyperglycemia and peripheral neuropathy and encephalopathy. Thiamine deficiency results in beriberi, which includes symptoms of encephalopathy and peripheral neuropathy; patients with beriberi can also develop cardiovascular symptoms and cardiac failure.

So the answer to this question is A


Instruction:
Amsel criteria does not include:
A. Thick, curdy white discharge
B. Clue cells
C. Whiff test positive
D. Vaginal pH >4.5

Response:
Amsel criteria include thin, white homogenous discharge not thick discharge.
Option b,c,d are included in Amsel criteria.
3 out of 4 should be present to diagnose it as Bacterial vaginosis.

So the answer to this question is A


Instruction:
Which of the following types of glomerulonephritis is least likely to cause CRF:
A. Post streptococcal glomerulonephritis
B. Membranous GN
C. Membrano proliferative GN
D. Focal segmental glomerulosclerosis

Response:
Answer is A (Post streptococcal GN): Acute post streptococcal GN is a cause of acute Renal failure and not Chronic Renal Failure. The other types of GN listed in the question are causes of CRF. In Post Streptococcal Gn (Acute GN): 95% of the affected children recover. The remaining of who do not recover, develop a Rapidly Progressive form of Glomerulonephritis and land up in ACUTE RENAL FAILURE.

So the answer to this question is A


Instruction:
Methylene blue reduction test is done for estimating –
A. Fat content of milk
B. Sugar content of milk
C. Protein content of milk
D. Bacterial activity in milk

Response:
None

So the answer to this question is D


Instruction:
Drug of choice of paroxysmal supraventricular tachycardia is ?
A. Adenosine
B. Digitalis
C. Procainamide
D. Mexiletine

Response:
Ans. is 'a' i.e., Adenosine Treatment of paroxysmal supraventricular tachycardia Valsalva maneuver, carotid sinus massage - T vagal discharge - AV node conduction & T refractoriness. Adenosine CCBs b-blockers Adenosine is prefered because of very sho duration of action and lack of negative ionotropic effect. Prevention of PSVT - Drugs that slow conduction in the antegrade slow pathway, e.g. - Digitalis, P-blockers, CCBs.

So the answer to this question is A


Instruction:
Vitamin A prophylaxis to a child is
A. Health promotion
B. Specific protection
C. Primordial prevention
D. Secondary prevention

Response:
Ans. is 'b' i.e. Specific protection Vitamin A prophylaxis is a form of specific protection against vitamin A deficiency resulting in xerophthalmiaSpecific protection is an interventional strategy for primary prevention in a disease process.Disease process, prevention & interventions can be summarised as below:* Health education is a part of health promotion Levels of preventionPrimordial prevention - Prevention of the development or emergence of Risk Factors in population groups in which they have not yet appeared Primary prevention - Action taken prior to the onset of disease which removes the possibility that the disease will ever occur. It signifies intervention in the pathogenesis phase of the disease.Secondary prevention - Action which halts the progress of disease at its incipient stage and prevents complications i.e. early diagnosis & treatment.Tertiary treatment - Intervention in the late pathogenic phase all measures to reduce or limit impairments, Disabilities and handicap Specific protection:Specific protection is possible in a limited number of cases.The interventions available areImmunizationUse of specific nutrients (For e.g. Vit. A, Iron & FA)ChemoprophylaxisProtection against occupational hazardsProtection against accidentsProtection from carcinogensAvoidance of allergensControl of environmental hazards e.g. air pollution, noise control etc.

So the answer to this question is B


Instruction:
An infant presents with history of seizures and skin rashes. Investigations show metabolic acidosis increased blood ketone levels. This child is likely to be suffering from:
A. Propionic aciduraria
B. Urea cyclic disorder
C. Phenylketonuria
D. Multiple carboxylase deficiency

Response:
Each child with Homo Carboxylase Synthatase Deficiency (HCSD) is likely to experience slightly different effects. Many babies with this condition sta to have symptoms within hours of bih or during the first few days or weeks of life. Other babies have their first symptoms sometime in infancy, usually before two years of age. A small number of people with HCSD never show symptoms and are only found to be affected after a brother or sister is diagnosed. HCSD causes episodes of illness called metabolic crises. Some of the first symptoms of a metabolic crisis are: poor appetite vomiting extreme sleepiness or lack of energy irritability low muscle tone (floppy muscles and joints) severe peeling skin rash Common lab findings are: low blood sugar, called hypoglycemia high levels of acidic substances in the blood, called metabolic acidosis slightly high levels of ammonia in the blood low platelets ketones in the urine high levels of substances called organic acids in the urine If a metabolic crisis is not treated, a child with HCSD can develop: breathing problems seizures swelling of the brain coma, sometimes leading to death Ref: Kliegman, Behrman, Jenson, Stanton, (2007), Chapter 85, "Defects in Metabolism of Amino Acids", In the book, "Nelson's Textbook of Pediatrics", Volume 1, 18th Edition, New Delhi, Page 543

So the answer to this question is D


Instruction:
Hyperprolactinoma causes
A. Inter menstrual bleeding
B. Prolonged menstruation
C. Oligomenorrhea
D. Polymenorrhea

Response:
(Oligomenorhea): (2287-88- H18th edition; 788-Davidson 21st)HYPER PROLACTINEMIA* Hyper prolactinemia is the most common pituitary hormone hyper secretion syndrome in both men and women* PRL- secreting pituitary adenoma (Prolactinomas) are the most common cause of PRL levels > 200 mg/L* Pregnancy and lactation are the important physiologic causes of hyperprolactinemia* Amenorrhea, galactorrhea and infertility are the hall marks of hyperprolactinemia in women* More commonly hyperprolactinemia develops later in life and leads to oligo menorrhea** and ultimately to amenorrhea* Decreased libido, weight gain and mild hirsutism may also complainMen-* Diminished libido , infertility and visual loss, impotence and oligospermia* True galactorrhea is uncommon in men hypogonadism, osteopenia, reduced muscle mass and decreased beard growth

So the answer to this question is C


Instruction:
Best level of anesthesia for LSCS:
A. T8
B. T10
C. T6
D. T4

Response:
Ans. is d, i.e. T4Ref Dutta Obs. 7/e, p 519Spinal anesthesia:Spinal anesthesia is done by injection of local anesthetic agent into the subarachnoid space. It has less procedure time and high success rate. Spinal anesthesia can be employed to alleviate the pain of delivery and during the third stage of labor. For normal delivery or for outlet forceps with episiotomy, ventouse delivery, block should extend from T10 (umbilicus) to S1. For cesarean delivery, level of sensory block should be up to T4 dermatome. Hyperbaric bupivacaine (10-12 mg) or lignocaine (50-70 mg) is used.

So the answer to this question is D


Instruction:
A 12-year-old girl presents for a routine physical examination. The patient has numerous freckles over her upper trunk and face. Which of the following terms best describes the morphologic appearance of her freckles?
A. Comedo
B. Macule
C. Nodule
D. Papule

Response:
Freckles, or ephelides, are small, brown macules that occur on sun-exposed skin, especially in people with fair skin. Freckles usually appear at about age 5 years. The pigmentation of a freckle deepens with exposure to sunlight and fades when light exposure ceases. Freckles show hyperpigmentation of the basal keratinocytes without a concomitant increase in the number of melanocytes. The other choices represent raised lesions of the skin.Diagnosis: Freckles, ephelides

So the answer to this question is B


Instruction:
Examination of the fundus shows Cherry red spot in all of the following conditions, EXCEPT:
A. Gaucher's disease
B. Tay Sach's disease
C. Niemann Pick Disease
D. Central retinal vein occlusion

Response:
Cherry red spot is an ophthalmoscopic finding caused by the abnormal accumulation of products like gangliosides in the ganglion layer of the retina. The absence of ganglion cells in the area gives rise to a cherry-red spot surrounded by white cells filled with storage material. As the cells sta to die, optic atrophy becomes visibly apparent. Cherry red spot is seen in: Berlin's oedema Gaucher's disease Tay sach's disease Neimann pick's disease Central retinal aery occlusion Ref: Ophthalmology By A.K Khurana, 4th Edition, Page 257, 478; Harrison's Principles of Internal Medicine By Robe J. Hopkin, Gregory A. Grabowski, Chapter 355, Page 2452

So the answer to this question is D


Instruction:
Circumcision cannot be used in management of
A. Phimosis
B. Severe balanitis xerotica
C. Paraphimosis
D. Penile Cancer

Response:
Answer- D. Penile CancerPhimosisBalanitis PosthitisParaphimosisSevere UTIs

So the answer to this question is D


Instruction:
Hypersensitivity vasculities seen in -
A. Capillaries
B. Arterioles
C. Post capillary venules
D. Medium sized vessels

Response:
None

So the answer to this question is C


Instruction:
Which of the following is Venodilator
A. Hydralizine
B. Minoxidil
C. Nitroprusside
D. Nifedipine

Response:
Hydralizine, minoxidil and nifedipine are primarily aeriolar dilator whereas nitroprusside is a mixed aerial and venous dilator. Therefore best answer seems to be is a nitroprusside Refer kDT 6/e p540

So the answer to this question is C


Instruction:
Following admission to the hospital for intestinal obstruction, a 48-year-old woman states that she previously had undergone cholecystectomy and choledochoduodenostomy. The most likely indication for the performance of the choledochoduodenostomy was:
A. Hepatic metastasis were present.
B. Multiple stones were present in the gallbladder at the previous operation.
C. Multiple stones were present in the CBD at the previous operation.
D. The common hepatic duct had a stricture.

Response:
Multiple stones were present in the CBD at the previous operation. During exploration of the CBD, most stones can be removed by using Desjardin's forceps or under direct vision using a choledochoscope and Dormia basket. However, if there are multiple stones impacted in the lower part of the CBD, a drainage procedure may be indicated. The CBD must be dilated before considering performing a choledochoduodenostomy at the time of gallbladder surgery (Figure below). If a stone is present in a dilated CBD after previous cholecystectomy, a choledochoduodenostomy is performed, because the rate of recurrent jaundice is high (>20%). Alternatively ERCP and sphincterotomy could be considered.Treatment for invasive gallbladder cancer is cholecystectomy and a wedge resection of the liver along with a regional lymphadenectomy. The wedge resection of the liver is illustrated. Segments 4 and 5 together with the lymph node regions should be removed.

So the answer to this question is C


Instruction:
All muscles are useful in Supination and Pronation of Hand EXCEPT
A. Anconeus
B. Brachio-radialis
C. Supinator
D. Biceps brachii

Response:
(A) Anconeus # Supination & Pronation are rotatory movements of forearm (and hand) around a vertical axis.> In a semiflexed elbow, the palm is turned upwards in supination, and downwards in pronation Remember: Kings pronate, Beggars supinate> The movements are permitted at the superior and inferior radioulnar joints, there is no middle radioulnar joint, and so no role. The radius is pronated by the pronator teres and pronator quadratus muscles. Pronation is performed by pronator quadratus and pronator teres muscle. Brachioradialis puts the forearm into a midpronated/supinated position from either full pronation or supination. Hand is supinated by the brachioradialis, supinator (brevis), and biceps muscles. The hand is supine in the anatomical position (i.e., palms facing up during autopsy).This action is performed by the Biceps brachii and the Supinator muscle

So the answer to this question is A


Instruction:
Boundaries of quadrilateral space include all except?
A. Teres major
B. Long head of triceps
C. Neck of humerus
D. Deltoid

Response:
Deltoid

So the answer to this question is D


Instruction:
Most common site of endometriosis:
A. Fallopian tube
B. Cervix
C. Ovary
D. Vagina

Response:
Most common site of endometriosis is ovary other common sites are: fallopian tubes, pelvic peritoneum, uterosacral ligaments Others sites: Bowel Lung (periodic hemoptysis) Nose (periodic epistaxis) Eyes (periodic subconjuctival hemorrhage)

So the answer to this question is C


Instruction:
Slap cheek appearance is seen in -
A. Erythema infectiosum
B. Exanthem subitum
C. Roseola infantun
D. None of above

Response:
Ans. is 'a' i.e., Erythema infectiosum o Erythema infectiosum (also known as fifth disease) is usually a benign childhood condition characterized by aclassic slapped-cheek appearance (see the image below) and lacy exanthem. It results from infection with human parvovirus (PV) B19. an erythrovirus.o Human herpesvirus 6 (HHV-6) is a dominant cause of roseola infantum (exanthem subitum or sixth disease).

So the answer to this question is A


Instruction:
The buffering capacity of a buffer is maximum at pH equal to
A. 0.5 pKa
B. pKa
C. pKa+1
D. 2pKa

Response:
The pH of a buffer is determined by the Henderson-Hasselbalch equation So, at the max buffering capacity, = . This is because a buffer reacts in the following manner--it is able to buffer against equal amounts of acid (H+) or base (OH-) added. A- + H+ - HA HA + OH- - H2O + A- Finally, we can see from the above Henderson-Hasselbalch equation that when = , pH = pKa.

So the answer to this question is B


Instruction:
A 10 year old male patient broke his tooth while playing. He gives a history of Endocardial cushion defect. Extra-oral examination reveals an underdeveloped midface, brachycephalic skull, flat nasal bridge and oblique palpebral fissures. What would be the common cardiac finding in patients with this condition?
A. Atrioventricular septal defects
B. Ventricular septal defects
C. Patent ductus arteriosus
D. All of the above

Response:
Affected individuals are more prone to congenital heart defects (50%) such as atrioventricular septal defects, ventricular septal defects, isolated secundum atrial septal defects, patent ductus arteriosus, and tetralogy of Fallot.

So the answer to this question is D


Instruction:
A Syndrome is characterized by Pancreaticneoplasm (insulinoma), Pituitary, and Parathyroid involvement. The syndrome is:
A. Polyglandular Type I
B. Polyglangular Type 2
C. Wermer
D. Sipple

Response:
Wermer

So the answer to this question is C


Instruction:
The shortest diameter of the pelvic inlet is -
A. Obstetric conjugate
B. True conjugate
C. Diagonal conjugate
D. Transverse diameter

Response:
Ans-A

So the answer to this question is A


Instruction:
Ante's law provides an aid for
A. Selection of pontic
B. Selection of connector
C. Selection of retainers
D. Selection of abutments

Response:
None

So the answer to this question is D


Instruction:
A 12-year-old boy with a femur fracture after a motor vehicle collision undergoes operative repair. After induction of anesthesia, he develops a fever of 40degC (104degF), shaking rigors, and blood-tinged urine. Which of the following is the best treatment option?
A. Alkalinization of the urine, administration of mannitol, and continuation with the procedure
B. Administration of dantrolene sodium and continuation with the procedure
C. Administration of dantrolene sodium and termination of the procedure
D. Administration of intravenous steroids and an antihistamine agent with continuation of the procedure

Response:
The patient is manifesting symptoms of malignant hypehermia (fevers, rigors, and myoglobinuria), which is treated by administration of dantrolene, immediate discontinuation of offending medication (which can include succinylcholine or halothanebased inhalational anesthetics), and suppoive cooling measures. While urine alkalinization, loop diuretics, and mannitol are appropriate treatment measures for rhabdomyolysis, the underlying problem in this patient is malignant hypehermia, which, because of its associated moality of 30% in severe cases, should be treated first and foremost. Malignant hypehermia is not a manifestation of anaphylactic shock, and therefore steroids and antihistamines have no role in its treatment.

So the answer to this question is C


Instruction:
1 and 1/2 year old child with history of chronic diarrhea was brought to the hospital with complaints of perioral and perineal rash. What is your diagnosis?
A. Epidermolysis bullosa
B. Staphylococcal scalded skin syndrome
C. Acrodermatitis enteropathica
D. Toxic epidermal necrolysis

Response:
Ans. C. Acrodermatitis enteropathicaAcrodermatitis enteropathica:* Autosomal recessive disorder seen in infants due to defect in intestinal zinc specific transporter - Human ZIP4 protein* Genetic locus - Chromosome 8q24.3* Classically presents while weaning from breast milk to formula, which has lower zinc bioavailability than breast milk.* C/F - Perioral and perineal rashes, periungual dermatitis- Beau's lines on nails, chronic diarrhea, alopecia.* Labs - Low serum zinc, ALP* Paneth cell inclusion in intestinal mucosal biopsy* Treatment - Zinc - 50mg/day for infants, upto 150mg/day for children

So the answer to this question is C


Instruction:
Postmortem rigidity disappears first in:
A. Upper limbs
B. Neck
C. Lower limbs
D. Eyelids

Response:
Ans. (d) EyelidsRef: The Essentials of FSM by K.S. Narayan Reddy 31st ed. /140-43, 147; Reddy's 2Th/ 142Order of appearance and disappearance (in sequence)* Heart (left chamber in 1 hour)* Eyelids (3-4 hours)* Face muscles* Neck and trunk* Upper extremities* Legs* Small muscle of finger and toes (last to be affected, 11-12 hours)Also Know* Postmortem rigidity lasts for 18-36 hours in summer and 24-48 hours in winter.* Postmortem rigidity disappear in the same order as that of the appearance

So the answer to this question is D


Instruction:
DOC for Tourette syndrome
A. Haloperidol
B. Valproate
C. B complex
D. Clonidine

Response:
Ans. is 'd' i.e., Clonidine Earlier Haloperidol was considered as DOC for Tourette syndrome. Clonidine is considered as DOC for Tourette syndrome

So the answer to this question is D


Instruction:
Which of the following is known as Preisz Nocard bacillus
A. Corynebacterium diphtheria
B. Corynebacterium pseudotuberculosis
C. Influenza bacillus
D. Salmonella

Response:
Corynebacteria of veterinary impoance are Corynebacterium pseudotuberculosis. Also called as Preisz - Nocard bacilli. Causes pseudotuberculosis in sheeps and suppurative lymphadenitis in horses. Reference: Textbook of Microbiology; Anathanarayan and panicker&;s; 9th edition; Page no: 242

So the answer to this question is B


Instruction:
Which amongst these is a tuberculid?
A. Erythema nodosum
B. Scrofuloderma
C. Lupus vulgaris
D. Erythema induratum

Response:
It is a nodular type of hypersensitivity reaction typically presents as red tender nodules.

So the answer to this question is D


Instruction:
The length of distal convoluted tubule is
A. 5 mm
B. 2 mm
C. 12 mm
D. 8 mm

Response:
Although the DCT is the shoest segment of the nephron, spanning only about 5 mm in length in humans (1), it plays a critical role in a variety of homeostatic processes, including sodium chloride reabsorption, potassium secretion, and calcium and magnesium handling Ref: guyton and hall textbook of medical physiology 12 edition page number:460,461,462

So the answer to this question is A


Instruction:
Which is the prognostic scoring system for head injury in children?
A. CCS
B. AUDIT
C. Injury severity score
D. Pediatric Trauma Score

Response:
Ans. is 'a' i.e., CCS * Scoring scales such as the Abbreviated Injury Scale (AIS), Injury Severity Score (ISS), Pediatric Trauma Score and Revised Trauma Score are used to predict mortality as well as morbidity in patients sustaining trauma in general.* They use physiological parameters to predict patient outcome.* CCS (Children Coma Score) is the paediatric counterpart of Glasgow Coma Score for patient with head injury.Children Coma Score:Eye Opening (Total Possible Points 4)Spontaneous4 To voice3 To pain2 None1 Verbal Response (Total Possible Points 5)Older Children Anfants and Young Children Oriented5Appropriate words, smiles, fixes, and follows5Confused4Consolable crying4Inappropriate3Persistently imitable3Incomprehensible2Restiess, agitated2None1None1Motor Response (Total Possible Points 6)Obeys6 Localized pain5 Withdraws4 Flexion3 Extension2 None1

So the answer to this question is A


Instruction:
Single file pattern is seen in breast cancer type-
A. Intraductal
B. Infiltrating lobular
C. Infiltrating ductular
D. All of the above

Response:
None

So the answer to this question is B


Instruction:
Which of the following carriers recieved source of infection from other carrier is called-
A. Incubatory carrier
B. Convulescent carrier
C. Healthy carrier
D. Chronic carrier

Response:
.

So the answer to this question is C


Instruction:
A 35-year-old man is admitted after severing his arm on industrial machinery. His airway is patent and there is no identifiable hindrance to breathing. Hispulse is 110 beats/min, blood pressure is 130/105 mmHg, and respiratory rate is 25 breaths/min. Which stage of shock is this patient in?
A. Class I
B. Class II
C. Class III
D. Class IV

Response:
The following table illustrates the clinical signs of the different stages of hypovolaemic shock. Class I Class II Class III Class IV Percent blood loss Up to 15 15-30 30-40 >40 Pulse rate <100 >100 >120 >140 Blood pressure Normal Normal | | Pulse pressure Normal | | | Respiratory rate 14-20 20-30 30-40 >35 Urine output (mL/h) >30 20-30 5-15 <5 Mental status Slight anxiety Mild anxiety Anxious, confused Confused

So the answer to this question is B


Instruction:
Blood is stored at what temperature in blood bank?
A. -2 to -4 degrees Celsius
B. -2 to 0 degrees Celsius
C. 1 to 6 degrees Celsius
D. 6 to 12 degrees Celsius

Response:
Ans. is 'c' i.e., 1 to 6 degrees Celsius

So the answer to this question is C


Instruction:
The superficial external pudendal artery is a branch of -
A. Femoral artery
B. External iliac artery
C. Internal iliac artery
D. Aorta

Response:
Femoral artery branches
-        Superficial

Superficial external pudendal
Superficial epigastric
Superficial circumflex iliac

-        Deep

Profunda femoris
Deep external pudendal
Muscular branches
Descending genicular branch

So the answer to this question is A


Instruction:
The most useful incision in the operating room for patients with penetrating pericardium injury is
A. Left anterior thoracotomy
B. Right anterior thoracotomy
C. Subxyphoid
D. Median sternotomy

Response:
The subxiphoid incision is useful for determining if there is blood in the pericardium and if there is an intracardiac injury; however, exposure is extremely limited, and definitive repair can rarely be performed through the incision. Left (or right) anterior thoracotomy is easily performed, especially in the emergency room, and gives adequate exposure to ceain areas of the hea. However, each has significant limitations in exposure. Either may be extended across the thoracotomy into the other side of the chest, thus producing a bilateral anterior thoracotomy. Exposure is excellent through this incision, and most injuries can be satisfactorily repaired through this approach. Most cardiac operations today are performed through median sternotomy incisions. If the patient is in the operating room, this incision is easily performed and always provides excellent exposure for all areas of the hea.

So the answer to this question is D


Instruction:
Reverse 3 sign in barium swallow is seen in:
A. Coarctation of aorta
B. Aberrant right subclavian artery
C. Patent ductus arteriosus
D. Double aortic arch

Response:
Ans-A Ref: OP Ghai, Textbook of Pediatrics: 7th edition, page: 421 Nelson's Textbook of Pediatrics 18th edition: page no 1901Explanation:Radiological signs of coarctation of aorta:Rip notching or dock sign is the notching of the under surfaces of posterior ribs extending from third to ninth ribs. It may be unilateral or bilateral and is seen only after the age of 6 yearsPost stenotic dilatation is the dilation of the aorta distal to coarctation3-sign is a combination of dilated left subclav ian artery above.stenosed coarcted segment in the middle and dilated post-stenotic aorta belowE-sign is seen on the lateral view on a barium swallow. This results from the dilated left subclavian artery and dilated post stenotic aorta indenting the barium filled oesophagus.Coarctation of the aortais produced by a deformity of the aortic media and intima. which causes a prominent posterior infolding of the aortic lumen.This deformity characteristically occurs at or near the junction of the aortic arch and the descending thoracic aorta.Cystic medial necrosis is a common occurrence at the level of coarctation and may predispose the aorta to infective endarteritis, intimal dissection, and aneurysm.Coarctation of the aorta accounts for 5% -10% of congenital cardiac lesions.It occurs with increased frequency among patients with Turner syndrome. 20%-36% of whom are affected.Clinical manifestations vary from congestive heart failure in infancy to hypertension with differential pressures between the upper and lower extremities in adulthood.Two classic radiologic signs associated with aortic coarctation are the figure-of-three sign and the reverse figure-of-three sign.The aortic segment affected by coarctation has a shape that resembles the number 3 on frontal chest radiographs.The number 3 is formed by dilatation of the left subclavian artery and aorta proximal to the site of coarctation, indentation of the site, and dilatation of the aorta distal to the site.This sign is seen in 50%-66% of adults with aortic coarctation.The reverse figure-of-threesign. a mirror image of the number 3, is observedon the left anterior oblique view during bariumesophagography in patients with aortic coarctation.Common radiologic signs in Congenital Heart DiseaseTransposition of Great arteriesEgg on side or Apple on a string apoearanceTotal anomalous pulmonary venous return5nowman or Figure of esght appearancePartial anomalous pulmonary venous returnScimitar signTetralogy of FallotBoot shaped HeartEbstein's anomalyBox shaped heartEndocardial Cushion defectGoose neck sign

So the answer to this question is A


Instruction:
Which of the following is not a feature of Alcoholic liver disease?
A. Macrovesicular fat within hepatocytes
B. Lipogranuloma
C. Lymphocytic infiltration of portal tracts
D. Portal & sinusoidal collagen deposits

Response:
Ans. (c) Lymphocytic infiltration of portal tracts(Ref: Robbins 9th/pg 837; 8th/pg 850; Rosai and Ackerman's Surgical Pathology, Volume 1, chapter 13)Lipogranuloma (or fat granuloma) represents a focal response to rupture of lipid-laden hepatocytes. It contains macrophages, occasional lymphocytes, eosinophils, and sometimes giant cells; It may be seen in Alcoholic liver disease; Neutrophil rather than lymphocytic infiltration is seen in alcoholic liver disease

So the answer to this question is C


Instruction:
In ductus venous, the fetal blood from the placenta reaches to the liver through
A. Poal vein
B. Right umbilical vein
C. Inferior vena cava
D. Left umbilical vein

Response:
when there is an increase in placental circulation, to facilitate the rapid transpo of blood through the liver ,a large passage is formed between left umbilical vein and right hepato cardiac channel.This passage is called ductus venosus REFERENCE: TEXTBOOK OF CLINICAL EMBRYOLOGY... VISHRAM SINGH SECOND EDITION....PAGE NO:241

So the answer to this question is D


Instruction:
True regarding Sjogren's syndrome are all of the following except: September 2009
A. Autoimmune condition
B. Males are commonly affected
C. Progressive destruction of lacrimal and salivary gland
D. No single laboratory investigation is pathognomic

Response:
Ans. B: Males are commonly affected SjOgren's Syndrome (SS) is a systemic autoimmune disease characterised by lymphocytic infiltration, acinar cell destruction and proliferation of duct epithelium in all salivary and larimal gland tissue. Extra glandular involvement of muscles, blood vessels, lungs, kidneys may also occur. Females are affected more than the males It may be primary and secondary (when associated with other connective tissue diseases). There is a risk of progression to lymphoid malignancy. Dry mouth, dry eyes and ahritis/ahralgia are the 3 common presenting features. Of the extraaicular manifestations, Raynaud's phenomena is the most common skin manifestation seen in 35% of patients. Vasculitis has been repoed in 5% of patients with Sjogren's syndrome. This includes small vessel leucocytoclastic vasculitis and medium vessel necrotising vasculitis. Distal A may be silent or lead to renal stones, nephrocalcinosis and compromised renal function. Hypergammaglobulinemia may be due to polyclonal activation of B cells. The diagnosis is base on the history as no single laboratory investigation is pathognomic of either primary or secondary Sjogren's syndrome

So the answer to this question is B


Instruction:
Not a family of selectin
A. P selectin
B. L selectin
C. A selectin
D. E selectin

Response:
None

So the answer to this question is C


Instruction:
Penile ulcer is painful in:
A. Syphilis
B. LGV
C. Donovanosis
D. Chancroid

Response:
Ans. is. 'd' i. e., Chancroid

So the answer to this question is D


Instruction:
Slope of pressure volume curve of lung represents:
A. Compliance of lung
B. Conductance
C. Surface tension
D. Transpulmonary pressure

Response:
Ans. a (Compliance of lung). (Ref. Ganong Physiology 23rd/ pg. 654-657; Figure 35-10)Lung compliance is an index of lung distensibility or the ease with which the lungs are expanded; thus, compliance is the inverse of elastic recoil. Compliance is defined as the ratio of change of lung volume to the change in pressure required to inflate the lung (dV/dP). Compliance decreases in patients with pulmonary edema or surfactant deficiency and increases when there is a loss of elastic fibers in the lungs, such as occurs in patients with emphysema and with aging.MECHANICS OF BREATHING - Compliance of the lungs & chest wall# Curve of airway pressure plotted against volume, is relaxation pressure curve of the total respiratory system.# The pressure is zero at a lung volume that corresponds to the volume of gas in the lungs at the end of quiet expiration (relaxation volume, which equals the functional residual capacity).# It is positive at greater volumes and negative at smaller volumes. The change in lung volume per unit change in the airway pressure (DV/DP) is the stretchability (compliance) of the lungs and chest wall.Q# It is normally measured in the pressure range where the relaxation pressure curve is steepest, and the normal value is approximately 0.2 L/cm H2O.# The curve is shifted downward and to the right (compliance is decreased) by pulmonary congestion and interstitial pulmonary fibrosis; pulmonary fibrosis is a progressive disease of unknown cause in which there is stiffening and scarring of the lung.# The curve is shifted upward and to the left (compliance is increased) in emphysema.# It should be noted that compliance is a static measure of lung and chest recoil.# Two major forces contribute to lung compliance:o tissue elastic forces ando surface tension forces.Q

So the answer to this question is A


Instruction:
All are risk factors for ca endometrium except -
A. Tamoxifen
B. Multiparity
C. Obesity 8c hypertension
D. Late menopause

Response:
Ans. is 'b' i.e., Multiparity Rick Factors for Endometrial CA (Estrogen-dependent tumor):* Nulliparity* Early menarche, Late menopause* Obesity* Diabetes mellitus and hypertension* Unopposed estrogen therapy/Anovulation* Tamoxifen therapy* Atypical endometrial hyperplasia* Obesity, hypertension, and diabetes mellitus associated with CA endometrium = corpus CA syndrome* OC PILLS, POP, DMPA, MIRENA, Multiparity are all protective for CA endometrium

So the answer to this question is B


Instruction:
The largest organ of the body is
A. Liver
B. skin
C. Gluteus maximus
D. Femur

Response:
The liver is the largest gland in the body and has a wide variety of functions (265-Snell 7th )
The liver is the largest gland in the body. (283-BDC-2)
Femer is the longest and the strongest bone of the body (16-B DC-2)
Skin is the largest organ of the body (278-K- Sembulingam physiology)
Skin is considered as the largest sense organ in the body it has many nerve endings, which are specialized to form cutaneous receptors (281-K. Sembulingam physiology 3rd )

So the answer to this question is B


Instruction:
Patients with C5 through C9 complement deficiencies would be most likely to have which of the following infections?
A. AIDS
B. Meningococcal infection
C. Pneumococcal infection
D. Giardiasis

Response:
Patients with complement deficiencies such as C5 through C9, which form the membrane attack complex (MAC), are predisposed to disseminated meningococcal disease. These patients may also be susceptible to gonococcal infection. There appears to be no disposition to AIDS or to fungal, parasitic, or pneumococcal infections.

So the answer to this question is B


Instruction:
Horner's syndrome is seen in all except -
A. Carotid artery aneurysm
B. Medial medullary syndrome
C. Can occur following surgery for raynaud's syndrome
D. Multiple sclerosis

Response:
Ans. is 'b' i.e., Medial medullary syndrome Horner*s syndrome is a classic neurologic syndrome whose signs include : -Miosis,Ptosis, andAnhidrosisAlso called ocul sympathetic paresis, a Horner's syndrome can be produced by a lesion anywhere along the sympathetic pathway that supplies the head, eye, and neck.NeuroanatomyHorner's syndrome can result from a lesion anywhere along a three-neuron sympathetic (adrenergic) pathway that originates in the hypothalamusThe first-order neuronDescends caudally from the hypothalamus to the first synapse, which is located in the cervical spinal cord (levels C8-T2, also called ciliospinal center of Budge).The second-order neuronTravels from the sympathetic trunk, through the brachial plexus, over the lung apex. It then ascends to the superior cervical ganglion, located near the angle of the mandible and the bifurcation of the common carotid artery.The third-order neuronThen ascends within the adventitia of the internal carotid artery, through the cavernous sinus, where it is in close relation to the sixth cranial nerve.The ocul sympathetic pathway then joins the ophthalmic (VI) division of the fifth cranial nerve (trigeminal nerve).In the orbit and the eye, the ocul sympathetic fibers innervate the iris dilator muscle as well as Muller's muscle, a small smooth muscle in the eyelids responsible for a minor portion of the upper lid elevation and lower lid retraction.EtiologyThe etiology of Horner's syndrome in adults relates to the lesion location.First-order syndrome (central)* Lesions of the sympathetic tracts in the brainstem or cervicothoracic spinal cord can produce a first-order Horner's syndromeLateral medullary syndromeQ (M.C.)Brainstem disease (stroke, tumour, demyelinating disorder)SyringomyeliaCervical cord trauma (including surgery)Spinal cord tumourDiabetic autonomic neuropathyMultiple sclerosisSecond-order syndromeTrauma or surgery involving spinal cord, thoracic outlet lung apex.Pancoast tumourCarotid and Aortic aneurysm and dissectionLumbar epidural anaesthesia.Third-order syndrome (Postganglionic)Lesions of internal carotid artery such asDissection, thrombosisOtitis mediaCavernous sinus mass Surgery for Raynaud's syndrome (cervical sympathectomy) may cause trauma to cervical spinal cord.

So the answer to this question is B


Instruction:
Hypocretin system plays a critical role in which of the following disorders?
A. Insomnia
B. Depression
C. OCD
D. Narcolepsy

Response:
Ans. D. NarcolepsyIt has become apparent that the hypo cretin system plays a critical role in narcolepsy. Narcolepsy is the prototypical example of sleepiness produced by a basic central nervous system dysfunction of sleep mechanisms. The etiology stems from a genetically triggered hypo cretin dysfunction and deficit. Insomnia is defined as difficulty initiating sleep or maintaining sleep or having nonrestorative sleep for 1 month or more. The insomnia or resulting sleepiness must cause clinically significant impairment or distress. To qualify as primary insomnia, the etiology must not be rooted in psychiatric conditions, parasomnias, substance use or abuse, sleep-disordered breathing, or circadian rhythm disorders.

So the answer to this question is D


Instruction:
Parasympathetic postganglionic fibres are carried by the
A. Long ciliary nerves
B. Greater petrosal nerves
C. Auriculotemporal nerve
D. Deep petrosal nerve

Response:
Long ciliary nerves are mixed sensory and sympathetic. Greater petrosal nerves carries preganglionic parasympathetic fibers from the facial nerve to the pterygopalatine ganglion. The deep petrosal nerve carries mainly sympathetic fibres from the superior cervical ganglion and internal carotid plexus.

So the answer to this question is C


Instruction:
What is the use of Ballard's scoring system in newborn babies?
A. Assessment of gestational age
B. Assessment of bih weight
C. Prediction of neonatal death
D. Screening for sepsis

Response:
Ballard's scoring system in newborn babies is used for assessment of gestational age It is based on neuromuscular criteria and physical criteria for maturity It is accurate to +-2 weeks

So the answer to this question is A


Instruction:
A 35-year-old smoker is involved in a house fire and receives a 45% total surface area burn. One half of the burned surface appears to be third degree. On the third post-burn day, the patient is noted to have bloody drainage from a nasogastric tube and a decrease of 5% in his hematocrit. Appropriate management should include which of the following?
A. Urgent upper gastrointestinal contrast study to delineate site of bleeding
B. Immediate selective aeriography the left gastric aery to diagnose and treat presumed stress ulceration
C. Urgent esophagogastroduodenoscopy to diagnosis the cause of bleeding
D. Urgent intravenous infusion of vasopressin at 0.2-0.4 IU/min

Response:
Patients who have sustained a major thermal burn of 35% or more of their body surface area are at a predictably high risk for the development of gastric erosions and hemorrhage. Endoscopy has demonstrated that gastric erosions are present in 93% of these patients, whereas the occurrence of severe acute upper gastrointestinal hemorrhage in severely burned patients ranges between 25% to 50%. At least 60% of patients at risk develop stress erosions within 1 to 2 days after the precipitating event. Painless upper gastrointestinal bleeding may be the only clinical sign. The onset of hemorrhage is often delayed, usually occurring 3 to 10 days after the onset of the primary disease. Esophagogastroduodenoscopy is the diagnostic modality of choice to confirm the diagnosis and to differentiate stress erosion from other sources of upper gastrointestinal hemorrhage. Correct identification of the bleeding source is made in greater than 90% of instances. If endoscopy is not diagnostic, visceral angiography through selective catheterization of the left gastric or splenic vessels may provide information regarding the primary vessel supplying the bleeding site. In contrast, barium examinations are usually of little value, due to the superficial nature of stress erosions, and in fact may be detrimental by interfering with the interpretation of subsequent aeriography.

So the answer to this question is C


Instruction:
Fixation reflex develops at the age of
A. 2-4 months
B. 4-6 months
C. 6-8 months
D. 5-7 years

Response:
Fixation reflex 2-4 months.
Macular differentiation 4-6 months.
Stereopsis 5-7 years.

So the answer to this question is A


Instruction:
All are true about Peyronie's disease EXCEPT:
A. Self limiting
B. Medial treatment is effective
C. Association with Dupvytren's contracture
D. Calcified plaques

Response:
(Medical treatment is effective) (1399-1400-Baily & 24th) (1373-B &L 25th)PEYRONIE'S DISEASE - Common cause of erect penis.* Hard plaques of fibrosis can usually be palpated in the tunica of one or both corpora covemosa. The plaque may be calcified.* May result of past trauma, there is association with dupuytren *s contracture.Treatment - continue to progress or remit after 3-5 years* When the deformity of the penis is causing distress, it may be possible to straighten it by placing non absorbable sutures in corpus cavernous opposite the plaque. This reduces the elasticity in the region to balance that caused by the plaque (NES BITT'S OPERATION)Peyronie's disease* Medical treatment are often ineffective* The disease is typically self-limiting* Surgery may be indicated to correct deformity with sexual function.

So the answer to this question is B


Instruction:
The apparent volume of distribution (Vd) of a drug exceeds total body fluid volume, if a drug is:
A. Sequestrated in body tissues
B. Slowly eliminated from body
C. Poorly soluble in plasma
D. Highly bound to plasma proteins

Response:
- The apparent volume of Distribution (Vd) = Dose administered by IV route/Plasma concentration.
- Consider the body as a single compartment in which the drug gets homogenously distributed. So it can be defined as, “The volume that would accommodate all the drug in the body if the concentration throughout was the same as in plasma.”
- In the case of lipid insoluble drugs, drugs do not enter the cells. So their V will be similar to ECF volume.
- Drugs sequestrated in other tissues may have V much more than total body water or even body mass.

So the answer to this question is A


Instruction:
Chronic active hepatitis is distinguished from chronic persistent hepatitis by -
A. Characteristic liver pathology
B. HBS Ag in serum
C. Extrahepatic manifestations
D. Presence of anti smooth muscle antibody

Response:
None

So the answer to this question is A


Instruction:
Fibrosis is due to-
A. TGF-b
B. TNF-a
C. IL - 7
D. IL -10

Response:
Robbins and Cotran pathologic basis of disease South Asia edition. Robbins 9/e p46 *activated macrophages secrete 1 growth factors ,TGF Beta:tissue repair and fibrosis. 2 IL 10 ,TGF Beta: anti inflammatory effect.

So the answer to this question is A


Instruction:
Retained mandibular deciduous central incisors will 10
result in
A. Lingual eruption of mandibular permanent incisors
B. Labial eruption of mandibular permanent incisors
C. Impaction of mandibular permanent incisors
D. Ankylosis of mandibular permanent incisors

Response:
None

So the answer to this question is A


Instruction:
ANA (antinuclear antibody) is seen in all except:
A. SLE
B. RA
C. Sjogren's syndrome
D. Systemic sclerosis

Response:
Ans. (b) RA(Ref: Harrison 18th/chapter 321, 323)Antibodies seen in RA are rheumatoid factors (RFs) and anti-cyclic citrullinated peptides (CCP) antibodies ANA is seen in:Almost all (>95%) cases of SLEIn virtually all patients with SSc (Systemic sclerosis)50% to 80% cases of Sjogren syndrome

So the answer to this question is B


Instruction:
Pharmacodynamics deals with -
A. Effect of drugs on body
B. Effect of body on drugs
C. Absorption of drugs
D. Metabolism of drugs

Response:
Ans. is 'a' i.e., Effect of drugs on body o Pharmacodynamics is the study dealing with effect of drugs on body. It includes action of drugs as well as their mechanism.o Pharmakokinetics is the quantitative study of drug movement, in through and out of the body. That means, it is the study of the effect of body on the drug. So, it is also known as ADME study as it deals with Absorption, Distribution, Metabolism and Excretion of a drug.RememberPharmacodynamics (D) - Drug does to bodyPharmacokinetics -Body does to drug

So the answer to this question is A


Instruction:
Transoesophageal Echocardiography is useful in -
A. Sinus venosus type of ASD
B. Dissection of arch of aoa
C. Prosthetic valve Thrombosis
D. Vascular ring

Response:
Transoesophageal echocardiography (TOE) involves passing an endoscope-like ultrasound probe into the oesophagus and upper stomach under light sedation and positioning it behind the LA. It is paicularly useful for imaging structures such as the left atrial appendage, pulmonary veins, thoracic aoa and interatrial septum, which may be poorly visualised by transthoracic echocardiography, especially if the patient is overweight or has obstructive airway disease. The high-resolution images that can be obtained makes TOE paicularly valuable for investigating patients with prosthetic (especially mitral) valve dysfunction, congenital abnormalities such as atrial septal defects, aoic dissection, infective endocarditis (vegetations that are too small to be detected by transthoracic echocardiography) and systemic embolism (intracardiac thrombus or masses) A vascular ring is a congenital defect in which there is an abnormal formation of the aoa and/or its surrounding blood vessels. The trachea and esophagus are completely encircled and sometimes compressed by a "ring" formed by these vessels, which can lead to breathing and digestive difficulties Ref Harrison 20edition pg 1441

So the answer to this question is D


Instruction:
All are hepatotoxic except
A. Halothane
B. Chlorform
C. Methoxyflurane
D. Isoflurane

Response:
Isoflurane is metabolized to trifluoroacetic acid. Although serum fluoride fluid levels may rise, nephrotoxicity is extremely unlikely, even in the presence of enzyme inducers. Prolonged sedation (>24 h at 0.1-0.6% isoflurane) of critically ill patients has resulted in elevated plasma fluoride levels (15-50 mmol/L) without evidence of kidney impairment. Similarly, up to 20 MAC-hours of isoflurane may lead to fluoride levels exceeding 50 mmol/L without detectable postoperative kidney dysfunction. Its limited oxidative metabolism also minimizes any possible risk of significant hepatic dysfunction. Total hepatic blood flow (hepatic aery and poal vein flow) may be reduced during isoflurane anesthesia. Hepatic oxygen supply is better-maintained with isoflurane than with halothane, however, because hepatic aery perfusion is preserved. Liver function tests are usually not affected. The major effect of acute inhalation of chloroform is central nervous system depression. At concentrations from 1,500-30,000 ppm, chloroform exposure can induce anesthesia; at concentrations exceeding 40,000 ppm, it can be fatal. Chronic inhalation of chloroform in humans results in hepatotoxicity and central nervous symptoms such as depression and irritability. Meanwhile, chronic oral exposure to chloroform in humans results in effects on the blood, livers, and kidneys. Ref: Miller's anesthesia 8th edition Ref: Morgan & Mikhail's clinical anesthesiology 6e

So the answer to this question is D


Instruction:
Which of the following drug is most effective in preventing emergence delirium with ketamine:
A. Atropine
B. Droperidol
C. Thiopentone
D. Midazolam

Response:
Benzodiazepine have proved the most effective in preventing emergence delirium with ketamine midazolam is drug of choice. Concomitant intravenous propofol and thiopentone decreases incidence of emergence delirium seen with propofol. Previous intravenous glycopyrolate increases the incidence of emergence delerium.

So the answer to this question is D


Instruction:
True about psoriatic ahritis are all expect ?
A. HLA-Cw6 association
B. Involvement of DIP joint
C. More common in males
D. DOC is methotrexate

Response:
Ans. is 'c' i.e., More common in males Psoriatic ahritis does not have sex predilection. The largest and most consistent repoed relative risk has been with HLA-Cw6. Psoriatic ahritis involves DIP joints and the DOC for psoriatic ahritis is methotrexate. Psoriatic ahritis Psoriatic ahritis refers to an inflammatory ahritis that characteristically occurs in individuals with psoriasis. Only 5% of patients with psoriasis develop ahritis. In 60-70% of cases, psoriasis precedes joint disease. In 15-20%, the two manifestations appear within 1 year of each other. In about 15-20% of cases, the ahritis precedes the onset of psoriasis and can present a diagnostic challange. Clinical features The usual age at onset is 30-50 years. The patient may present with one of several patterns of joint involvement. These include : Ahritis of distal interphalangeal (DIP) joints Assymetrical oligoahritis : Most common pattern Symmetrical polyahritis similar to RA Axial involvement (sacroiliac and spine) similar to ankylosing spondylitis Ahritis multilans So, Psoriatic ahritis can present with symptoms similar to any other inflammatory ahritis (RA, AS) or with any other different pattern. Shoening of digitis, called telescoping, because of underlying osteolysis is paicularly characteristic of Psoriatic ahritis and there is much greater tendency than in RA for both fibrous and bony ankylosis. Other findings are enthesitis, nail changes, and tenosynovitis. Treatment of Psoriatic ahritis Anti TNF - a agents eg etanercept and infliximab are newer drugs and are effective even in longstanding restistent PsA cases to previous therapy and extensive skin lesion. Methotrexate is drug of choice. Other effective agents are sulfasalazine, cyclosporine, retinoic acid & psoralen & UV-A (PUV A). Use of immunosuppressive therapy including anti TNF a agents, methotrexate and cycloserine are contraindi cated in HIV associated PsA.

So the answer to this question is C


Instruction:
All of the following statements about Neurofibromatosis are true, except?
A. Cataract
B. Scoliosis
C. Cutaneous neurofibromas
D. Autosomal Recessive Inheritance

Response:
Neurofibromatoses are a group of clinically and genetically different autosomal dominant hereditary diseases that predispose to benign and malignant tumors of the nervous system. It is of two types NF 1 and NF2. Cutaneous neurofibromas occur in NF1, and NF 2 is associated with posterior subcapsular cataract. Ref: Harrison's Internal Medicine, 18th Edition, Chapter 379; Color Atlas of Genetics By Eberhard Passarge, 3rd Edition, Page 338

So the answer to this question is D


Instruction:
Rhomboids major is inseed on ?
A. Medial border of scapula
B. Lateral border of scapula
C. Inferior angle of scapula
D. Inteubercular groove of humerus

Response:
Ans. is 'a' i.e., Medial border of scapula(Ref: Garg 4Uh/e p. 803, 810-812))Origin:Spinous processes of T2 to T5 veebraeInseionMedial border of the scapula, inferior to the inseion of rhomboid minor muscle.

So the answer to this question is A


Instruction:
Which of the following is the most common location of hypeensive intracranial hemorrhage?
A. Pons
B. Thalamus
C. Putamen
D. Subcoical white matter

Response:
In order of frequency, the most common sites of hypeensive intracerebral bleed are the lentiform nucleus, especially the putamen, supplied by the lenticulostriate aeries, the thalamus, supplied by posterior perforating aeries off the posterior cerebrobasilar aery bifurcation, the white matter of the cerebral hemispheres (lobar hemorrhages), the pons, supplied by small perforating aeries from the basilar aery; and the cerebellum, supplied by branches of the cerebellar aeries. Ref: Clinical Neuroanatomy, 26th Edition By Stephen G. Waxman, Chapter 12

So the answer to this question is C


Instruction:
Which of the following type of cells are NOT found in stomach?
A. Chief cells
B. Parietal cells
C. Argentaffin cells
D. Goblet cells

Response:
Cardia and pyloric region of the stomach consist of glands which secretes mucus and body of the stomach including the fundus consist of parietal cells and chief cells. Goblet cells are found in the epithelium lining the small intestine.The parietal cell secretes hydrochloric acid and are located the middle and basilar poion of the gastric glands.Chief cells or zymogen cells are found near the base of the gastric glands. They synthesize and secrete pepsinogen.Enteroendocrine (argentaffin) cells secrete different neuropeptides and regulatory molecules. Gastrin producing G cells are found in the antrum, enterochromaffin cells secreting serotonin, substance P or motilin are found throughout the gastric mucosa, glucagon secreting A cells are found in the proximal third of the stomach and somatostatin secreting D cells are found in the upper third of stomach and antrum.

So the answer to this question is D


Instruction:
Acute onset areflexic quadriparesis with blurred vision and absent pupillary response in a 20 years old male, the most probable diagnosis is?
A. Porphyria
B. Botulism
C. Polio
D. Diphtheria

Response:
ANSWER: (B) BotulismREF: Harrison s 17th ed p. 901, CMDT - 2006/1410The clinical picture is most consistent with diagnosis of botulismBotulismh/o recent injection of canned or smoked foods or injection of drugSymmetric descending paralysis with prominent bulbar palsyDiplopia , dysarthria , dysphonia , dysphagiaPupils are fixed and dilatedPtosis and cranial nerve palsy with impairment of extra ocular muscles and fixed dilated pupils are characteristic signsSensory examination is normalNormal sensoriumAfebrile

So the answer to this question is B


Instruction:
The foetal circulation is first intact and functional, separated from maternal circulation at the age of:
A. 8 days
B. 12 days
C. 17 days
D. 21 days

Response:
None

So the answer to this question is D


Instruction:
Refeeding syndrome is due to increased release of the following hormone:
A. Growth hormone
B. Glucocoicoids
C. Insulin
D. Thyroxine

Response:
Refeeding syndrome occur with rapid and excessive feeding of patients with severe underlying malnutrition due to starvation, alcoholism, delayed nutritional suppo, anorexia nervosa, or massive weight loss in obese patients. With refeeding, a shift in metabolism from fat to carbohydrate substrate stimulates insulin release, which results in the cellular uptake of electrolytes, paicularly phosphate, magnesium, potassium, and calcium. However, severe hyperglycemia may result from blunted basal insulin secretion. The refeeding syndrome can be associated with enteral or parenteral refeeding, and symptoms from electrolyte abnormalities include cardiac arrhythmias, confusion, respiratory failure, and even death. Ref: Shires G. (2010). Chapter 3. Fluid and Electrolyte Management of the Surgical Patient. In F.C. Brunicardi, D.K. Andersen, T.R. Billiar, D.L. Dunn, J.G. Hunter, J.B. Matthews, R.E. Pollock (Eds), Schwaz's Principles of Surgery, 9e.

So the answer to this question is C


Instruction:
All of the following are major complications of massive transfusion, except -
A. Hypokalemia
B. Hypothermia
C. Hypomagnesemia
D. Hypocalcemia

Response:
Major complications in blood transfusion are fever,chills,tachycardia ,flushing ,rigor,uicaria,abdominal pain,shoeness of breath,anaphylactic shock,transmission of infection,air embolism ,thrombophlebitis,circulatory overload. Since very rapid hemolysis is a compliacation of transfusion ,hyperkalemia occurs.Reference:Davidson's principles & practice of medicine 22nd edition pg no 1016.

So the answer to this question is A


Instruction:
Pancytopenia with a cellular marrow is seen in all except -
A. PNH
B. Megaloblastic anemia
C. Myelodysplastic syndrome
D. Congenital dyserythropoietic anemia

Response:
None

So the answer to this question is D


Instruction:
Spermatogenesis occurs at:
A. Body temperature
B. Temperature lower than core body temperature
C. Temperature higher than core body temperature
D. Temperature does not play a role

Response:
Spermatogenesis requires a temperature considerably lower than that of the interior of the body. Ref: Ganong's Review of Medical Physiology, 22nd Edition, Page 427; Guyton and Hall - Textbook of Medical Physiology, 10th Edition, Page 920

So the answer to this question is B


Instruction:
Brill Zinsser disease is caused by
A. C. burnetti
B. R. typhi
C. R. slovaca
D. R. prowazekii

Response:
DiseaseEtiologyDescriptionGroup Affected/Epidemiologic FactorsClinical SyndromeEpidemic typhusRickettsia prowazekiiMaculopapular eruption appearing inaxillae, spreading to trunk and later toextremities; usually spares face, palms, soles;evolves from blanchable macules to confluent eruption with petechiae; rash evanescentin recrudescent typhus (Brill-Zinsserdisease)Exposure to body lice;occurrence of recrudescenttyphus as relapseafter 30-50 yearsHeadache, myalgias;moality rates 10-40% ifuntreated; milder clinicalpresentation in recrudescentformRef: Harrison; 19th edition; Page no: 128

So the answer to this question is D


Instruction:
Attitude of lower limb seen in posterior dislocation of hip is?
A. Flexion, Abduction, Internal rotation
B. Flexion, Adduction, Internal rotation
C. Flexion, Abduction, External rotation
D. Flexion, Adduction, External rotation

Response:
ANSWER: (B) Flexion, Adduction, Internal rotationREF: Apley's 8th ed p. 681Flexion , adduction , internal rotationPosterior dislocation of hipFlexion , adduction , lateral rotation with shorteningIntracapsular fracture of hipFlexion , abduction and lateral rotationAnterior dislocation of hip

So the answer to this question is B


Instruction:
80 years old female after fall developed inability to walk with external rotation deformity on examination SLR is not possible and broadening of trochenter is present. The possible diagnosis is:
A. Neck femur
B. Inter trochanteric femur
C. Subtrochanteric femur
D. Greater trochanter

Response:
B i.e. Inter trochanteric femur

So the answer to this question is B


Instruction:
Regarding HIV which of the following statement is not TRUE?
A. It is a DNA retrovirus
B. Contains Reverse Transcriptase
C. May infect host CD4 cells other than T-lymphocytes
D. Causes a reduction in host CD4 cells at late stage of disease

Response:
The genome of HIV is diploid, composed of 2 identical single stranded positive sense RNA copies. In association with viral RNA is the reverse transcriptase enzyme which is the characteristic feature of retroviruses. Pathogenesis: When virus infects a cell RNA, viral RNA is transcribed by the enzyme, first to single stranded DNA and then to double stranded DNA which is integrated into the host cell chromosome. The provirus initiate viral replication by directing synthesis of viral RNA and other components. When naked virus buds outs through host cell surface membrane, it acquires a lipoprotein envelope which consist of lipid derived from the host cell membrane and glycoprotein which are virus coded. The spikes which constitute the major surface component of the virus, binds to CD4 receptors on susceptible host cells. The transmembrane pedicle cause cell fusion. Ref: Ananthanarayan and Panicker's Textbook of Microbiology, 8th Edition, Page 395

So the answer to this question is A


Instruction:
A 40 year old female presents with swelling in the neck. O/E it is a 2x2 cm nodule on the right lobe of thyroid. No palpable cervical nodes. FNAC shows features of papillary carcinoma. Treatment of choice
A. Total thyroidectomy
B. Right hemithyroidectomy
C. Hartley-Dunhill operation
D. Subtotal thyroidectomy

Response:
Answer: a) Total thyroidectomy (SABISTON 19TH ED, P-908)PAPILLARY CARCINOMA (80%)Most common type of thyroid carcinomaMore common in womenMostly in the age group 25 - 50Mutations in RET or BRAF geneAssociated with previous exposure to ionizing radiationHistologyNuclei: optically clear or empty appearance - groundglass or Orphan Annie eye nucleiCrowded nuclei may demonstrate groovingPsammoma bodiesSeldom encapsulatedMultiple foci seenHistologic variants - follicular (mc), tall cell, diffuse sclerosing, insular, columnarSlowly progressive and less aggressive tumorThyroglobulin is used as tumor markerSpread to lymph nodes is common, blood borne metastasis is unusualLymph node status does not alter the prognosisSeen as cold masses on scintiscansClassic papillary carcinoma and follicular variant - Excellent prognosisAll other subtypes - aggressiveDiffuse sclerosing variantOccurs in younger individuals, including childrenExtensive, diffuse fibrosis throughout the thyroid glandOften associated with a prominent lymphocytic infiltrate, simulating Hashimoto thyroiditisLack BRAF mutations, but RET/PTC translocations are found in 50% casesMicrocarcinoma (Occult carcinoma)Papillary cancers less than 1 cmMost commonly come to attention as an incidental finding in patients undergoing surgeryMajority of such tumours never progress to become a clinically significant entityHave a uniformly excellent prognosisFactors associated with worse prognosisMale genderAge > 40 yarsSize > 3 cmTall cell variantAge at diagnosis - most important prognostic factor in well-differentiated thyroid cancer Treatment of differentiated thyroid cancerSolitary intrathyroidal papillary microcarcinoma: no clinically involved cervical lymph nodes, no history of head and neck radiation - unilateral thyroid lobectomy and isthmusectomyTumors > 1 cm with no palpable nodes: Total thyroidectomyTumors > 2 cm with nodal involvement and/or metastases: Total thyroidectomy + node dissection + radioiodine ablation + long term TSH suppressionThyroid cancer > 1 cm or papillary cancer < 1 cm: clinically positive nodal disease, multicentricity, or a history of head and neck radiation - total or near-total thyroidectomy, followed by radioablationTSH suppression: thyroxine 0.1- 0.2 mg daily for all patients after operation for differentiated thyroid carcinoma on the basis that most tumours are TSH dependent.

So the answer to this question is A


Instruction:
Laryngeal cartilage forming complete circle -
A. Arytenoid
B. Cricoid
C. Thyroid
D. Hyoid

Response:
Ans. is 'b' i.e., Cricoid [Ref: Dhingra 5th/ep. 299\Lvrangeal cartilagesThyroid (unpaired)o It is the largest of all laryngeal cartilages. It is 'V* shaped with right and left lamina. Both laminae (alae) meet anteriorly forming an angle of 90deg in males and 120deg in females. Vocal cords are attached to the middle of thyroid angle.Cricoid (unpaired)o It is the only cartilage forming a complete ring, therefore is shaped like a ring. It articulates with arytenoid cartilage to form cricoarytenoid joint, a type of synovial joint.Epiglottis (unpaired)o It is leaf-shaped elastic cartilage (in adults). It is omega shaped in children. It forms the anterior wall of laryngeal inlet.Arytenoid cartilage (paired)o Each Arytenoid cartilage is pyramidal in shape. Base articulates with cricoid cartilage, and apex supports the comiculate cartilage. A vocal process directed anteriorly and gives attachement to vocal cord. A muscular process directed laterally and gives attachment to intrinsic laryngeal muscles.Corniculate cartilage (of Santorini): Pairedo Articulates with apex of Arytenoid cartilageCuneiform cartilage (of Wrisberg): Pairedo Situated in aryepiglottic fold in front of comiculate cartilage.

So the answer to this question is B


Instruction:
False about Randomised Control Trials is
A. Randomisation is done while selecting subjects for the study
B. Results of attrition are included in the analysis
C. Double blinding is the most common form of blinding observed
D. Cross-over design helps removing ethical concerns

Response:
The essential purposes of randomization in a randomized controlled trial are:
- Participants have ‘Equal and Known Chance’ of falling into either ‘Experimental Group’ or ‘Reference Group’
- To eliminate Selection Bias (Selection Bias or ‘Susceptibility Bias’ is the bias due to differential susceptibility of two groups to outcome, even before intervention/experiment is performed; Thus two groups are not comparable)
- To ensure comparability among two groups
- To have ‘similar prognostic factors’ among two groups
- ‘Randomisation is done while dividing patients into the Experimental (Intervention) Group and the Reference Group’ AND not while selecting patients for RCT
- Randomisation is known as ‘Heart of a trial’
- Randomisation is superior to Matching.

So the answer to this question is A


Instruction:
Daisy head colonies are seen with -
A. Staph. Aureus
B. Corynebacterium diphtheriae
C. Staph. Pyogenes
D. Anthrax

Response:
Ans. is 'b' i.e., Corynebacterium Diphtheriae

So the answer to this question is B


Instruction:
TMJ development starts at:
A. 2 weeks
B. 10 weeks
C. 29 weeks
D. 25 weeks

Response:
Three phases in the development of the TMJ were identified:

The first is the blastematic stage (weeks 7-8 of development), which corresponds with the onset of the organization of the condyle and the articular disc and capsule. During week 8, intramembranous ossification of the temporal squamous bone begins.
The second stage is the cavitation stage (weeks 9-11 of development), corresponding to the initial formation of the inferior joint cavity (week 9) and the start condylar chondrogenesis. Week 11 marks the initiation of organization of the superior joint cavity.
And the third stage is the maturation stage (after week 12 of development).

So the answer to this question is B


Instruction:
A group of 8 expes discussing and interacting about a topic in front of large audience is?
A. Workshop
B. Symposium
C. Semin ar
D. Panel discussion

Response:
- panel discussion is a group approach in health communication. - in a panel discussion, 4-8 people who are qualified to talk about a topic sit and discuss a given problem or topic in front of a large group or audience. Reference : Park's textbook of preventive and social medicine, 23rd edition, pg no:865 <\p>

So the answer to this question is D


Instruction:
Reverse cold chain is used for
A. Carrying stool samples of polio patients from a PHC to the lab
B. Transpoing outdated vaccines from PHC to district hospital
C. Transpoing vaccine to lab, to check its potency
D. Transpoing vaccine from camps to sub - centre

Response:
The cold chain is a system of storage and transpo of vaccines at low temperatures from the manufacturer to the actual vaccination site. The cold chain system is necessary because vaccine failure may occur due to failure to store and transpo under strict temperature control (refer pgno:109 park 23 rd edition)

So the answer to this question is A


Instruction:
Metformin is used as a antihyperglycemic agent in DM, but there are some other uses for this drug. All the following are such uses of the drug, except:
A. PCOD
B. Non-alcoholic fatty liver disease
C. Alcoholic fatty liver disease
D. Metabolic abnormalities associated with HIV disease

Response:
Other uses of metformin: PCOD Impaired glucose tolerance Obesity Metabolic abnormalities associated with HIV disease Non alcoholic fatty liver disease Ref: Harrison, 18th Edition, Page 2995

So the answer to this question is C


Instruction:
The pathways followed by corrosive acids in stomach is called -
A. Curling ulcer
B. Cushing ulcer
C. Magenstrasse
D. None

Response:
Ans. is 'c' i.e., Magenstrasse o Magenstrasse is the term applied to the pathway acidic agents follow in stomach.o The pathway of acids and alkalis in food filled stomach starts along the lesser curvature of the stomach and leads to the pylorus, which explains the location of greatest damage in food filled stomach,o Stomach without food have significant injury in the lower half of two thirds and may have sparing of fundus.Magenstrasse is the term applied to the pathway acidic agents followShaded region indicates the area of greatest injury caused by acid ingestion irt a food-frlled stomachShaded region indicates the area of greatest injury caused by acid ingestion in an empty stomach

So the answer to this question is C


Instruction:
Lung injury with bad prognosis is?
A. Open pneumothorax
B. Closed pneumothorax
C. Tension pneumothorax
D. All have same prognosis

Response:
ANSWER: (C) Tention pneumothoraxREF: Handbook of practical medicine: Volume 1 by Hermann Eichhorst page 29As a general thing, the prognosis of open and dosed pneumothorax is more favorable than that of tention pneumothorax, because the latter is apt to be attended with great tension of the gases, and marked displacement of the adjacent structures

So the answer to this question is C


Instruction:
A 15 year old boy presents was brought to OPD with a history of pain and swelling in the right thigh. Biopsy of the mass demonstrates Osteosarcoma. His mother was diagnosed with breast cancer 1 year ago and his maternal grandmother died of the breast cancer 10 years ago. The child has 3 younger siblings. Siblings are at an increased risk of developing which of the following cancer ?
A. Wilms
B. Neuroblastoma
C. Hepatoblastoma
D. Glioma

Response:
This is a likely the case of Li Fraumeni syndrome. Li Fraumeni syndrome Cancer predisposition syndrome before the age of 45 years Associated with high risks for a diverse spectrum of childhood- and adult-onset malignancies due to germline p53 gene mutation as per Chompret criteria. 5 cancer types are : Adrenocoical carcinomas, Breast cancer, Central nervous system tumors, Osteosarcomas, Soft-tissue sarcomas. Out of the given options central nervous system tumor is glioma

So the answer to this question is D


Instruction:
The water channel presented in proximal convoluted tubule is:
A. Aquaporin 1
B. Aquaporin 2
C. Aquaporin 5
D. Aquaporin 9

Response:
Rapid diffusion of water across cell membranes depends on the presence of water channels, integral membrane proteins called aquaporins. Aquaporin-1 is localized to both the basolateral and apical membrane of the proximal tubules and its presence allows water to move rapidly out of the tubule along the osmotic gradients set up by active transpo of solutes, and isotonicity is maintained. Ref: Barrett K.E., Barman S.M., Boitano S., Brooks H.L. (2012). Chapter 37. Renal Function & Micturition. In K.E. Barrett, S.M. Barman, S. Boitano, H.L. Brooks (Eds), Ganong's Review of Medical Physiology, 24e.

So the answer to this question is A


Instruction:
Patent foramen ovale is due to
A. Failure of closure of sinoatrial orifice
B. Failure of fusion of septum primum and endocardial cushion
C. Failure of fusion of septum primum and septum intermedium
D. Failure of fusion of septum primum and septum secondum

Response:
The primitive atrium is divided into left and right atria by an interatrial septum which is formed by fusion of septum primum and septum secundum. The sequence of events are as follows

Septum primum arises from the roof of the common atrium (primitive atrium) and grows caudally towards septum intermedium (fused AV cushions).
Foramen primum is formed between the free border of growing septum primum and fused AV cushions (septum intermedium).
Shortly afterwards the septum primum fuses with septum intermedium (fused AV cushions), obliterating foramen primum
Foramen secundum is formed by degeneration of cranial part of septum primum.
Septum secundum grows caudally to the right of septum primum and overlaps the foramen secundum. The passage between septum primum and septum secundum is called foramen ovale. After birth, foramen ovale closes by fusion of septum primum with septum secundum. Failure of fusion of these two septa results in patent foramen ovale.

So the answer to this question is D


Instruction:
All findings are seen in a patient with isolated aoic stenosis except?
A. Left ventricular impulse is displaced laterally
B. Pulsus bisferiens is hallmark finding
C. Systemic BP is normal in early stages of disease
D. Carotid thrill

Response:
Aoic stenosis(AS) Results in LVH -Double apical impulse and shift of apex beat laterally. Characteristic pulse -Pulsus parvus et tardus. Early stages -BP -Normal. Later stages -Low SBP , | Pulse pressure. Murmur - Ejection systolic murmur with carotid thrill. Pulsus bisferiens is seen in HOCM & severe AR.

So the answer to this question is B


Instruction:
After which of the following gestational age is a dead baby said to be a still born ?
A. 20 days
B. 28 days
C. 20 weeks
D. 28 weeks

Response:
The term stillborn refers to the death of a fetus after 28 weeks of gestation or death of a fetus weighing 1000gms.

So the answer to this question is D


Instruction:
Sternocleidomastoid is supplied by all of the following aeries except
A. Occipital
B. Posterior auricular
C. Thyrocervical trunk
D. Superior thyroid

Response:
Superior thyroid aery is a branch of ECA, inferior thyroid aery is a branch of thyrocervical trunk of subclan aery Ref: Gray's40e/262

So the answer to this question is C


Instruction:
The following tests are used to check the efficiency of pasteurization of milk except :
A. Phosphatase test
B. Standard plate count
C. Coliform count
D. Methylene blue reduction test

Response:
Methylene blue Reduction test is a indirect test for detection of microorganism in milk. It is carried out on milk prior to the process of pasteurization and is not used to check the efficiency of pasteurization. Ref: Alcamo's Laboratory Fundamentals of Microbiology By Jeffrey Pommerville, 2010, Page 297

So the answer to this question is D


Instruction:
Group B fibers are present in:
A. Intrafusal fibers of the muscle spindle
B. Golgi tendon apparatus
C. Autonomic preganglionic fibers
D. Spinothalamic tracts

Response:
None

So the answer to this question is C


Instruction:
This virus is the leading cause of bronchiolitis and community-acquired pneumonia in infants.
A. Measles virus
B. Influenza virus
C. Respiratory syncytial virus
D. Parainfluenza virus

Response:
Orthomyxoviruses and paramyxoviruses are RNA viruses that contain a single-stranded RNA genome. The influenza viruses belong to the orthomyxoviruses. They cause acute respiratory tract infections that usually occur in epidemics. Isolated strains of influenza virus are named after the virus type (influenza A, B, or C) as well as the host and location of initial isolation, the year of isolation, and the antigenic designation of the hemagglutinin and neuraminidase. Both the hemagglutinin and neuraminidase are glycoproteins under separate genetic control, and because of this they can and do vary independently. The changes in these antigens are responsible for the antigenic drift characteristic of these viruses. The paramyxoviruses include several important human pathogens (mumps virus, measles virus, respiratory syncytial virus, and parainfluenza virus). Both paramyxoviruses and orthomyxoviruses possess an RNA-dependent RNA polymerase that is a structural component of the virion and produces the initial RNA. Respiratory syncytial viruses (RSV) are not related to the paramyxoviruses. They are 150-nm single-stranded RNA viruses. There are 2 antigen groups, A and B, which play no role in diagnosis and treatment. While the overall mortality is 0.5%, at-risk groups may be 25 to 35% mortality if untreated. Some parainfluenza virus infections (type 3) may be indistinguishable from RSV, but most parainfluenza infections produce a laryngotracheobronchitis known as croup.

So the answer to this question is C


Instruction:
Prostatic calculi is made up of ?
A. Calcium carbonate
B. Calcium oxalate
C. Calcium phosphate
D. Triple phosphate

Response:
Prostatic calculi are of two varieties: endogenous, which are common, and exogenous, which are comparatively rare. An exogenous prostatic calculus is a urinary (commonly ureteric) calculus that becomes arrested in the prostatic urethra. Endogenous prostatic calculi are usually composed of calcium phosphate combined with about 20% organic material.Reference : page 1353 Bailey and Love's sho practice of surgery 25th edition

So the answer to this question is C


Instruction:
Insulin acts on glucose metabolism by:
A. T permeability of glucose across cell membrane
B. permeability across cell membrane against glucose gradient
C. T permeability of renal cells
D. T glucose transpo to brain

Response:
B i.e. T permeability across cell membrane against glucose gradient Insulin increases glucose uptake by increasing the number of glucose transpoer in cell membrane. Direct insulin stimulated glucose uptake is mediated by Glut 4 and is seen only in: Muscle: Skeletal muscleQ and cardiac muscleQ Adipose tissueQ Glucose Transpoers In Mammals Facilitated Bidirection Diffusion Function Major Sites of Expression Km GLUT 1 Basal glucose uptake - Placenta, blood - brain barrier, Red cells, kidneys 1-2 GLUT 2 B cell glucose sensor; transpo out of intestinal & renal epithelial cells. - B cells of islets, Liver, Small intestine, Kidney 12-20 GLUT 3 Basal glucose uptake - Brain, placenta, kidney <1 GLUT 4 Insulin stimulated glucose uptakeQ - Skeletal & cardiac muscleQ - Adipose tissueQ 5 GLUT 5 Fructose transpo - Jejunum, Sperm 1-2 GLUT 6 None - Pseudogene ? GLUT 7 Glucose 6-PO4 transpoer in endoplasmic reticulum - Liver ? Secondary Active Transpo Na+ glucose cotranspo (unidirectional -transpoer) SGLT-1 Active uptake of glucose against Small intestine, renal tubule 0.1 - 1.0 SGLT-2 concentration gradient Renal tubule 1.6

So the answer to this question is B


Instruction:
Fetal growth is maximally affected by –
A. Insulin
B. Growth hormone
C. Cortisol
D. Thyroxine

Response:
Insulin and insulin like growth factor stimulate fetal growth.
Thyroxine has no effect on linear growth but it has effect on skeletal maturation.
Fetal or maternal GH is not essential for fetal growth in utero.

So the answer to this question is A


Instruction:
Active tension in a muscle depends upon which of the following
A. Number of muscle fibres
B. Number of motor units recruited
C. Aerobic capacity of muscle
D. Action potential amplitude

Response:
Active Tension in a muscle depends upon the number of motor units recruited. The greater the number of motor units recruited, the greater will be the active tension. Factors Determining Muscle Tension (Active Tension) Tension developed by each fiber Action potential frequency Fibre length (Length-tension relationship) Fibre length Fatigue Number of active fibers Number of fibers per motor unit Number of active motor units (recruitment) Ref: Vander's Human physiology: The Mechanisms of Body function 11th edition Pgno: 276

So the answer to this question is B


Instruction:
Standard views for mammography are:
A. AP and Lateral
B. AP and Medio-lateral oblique
C. Cranio-caudal and Medio-lateral view
D. Cranio-caudal and Medio-lateral oblique

Response:
The two standard mammographic projections are:
A mediolateral oblique (MLO) view and a craniocaudal (CC) view.
The MLO is taken with the X-ray beam directed from superomedial to inferolateral, usually at an angle of 30°-60°.
The MLO projection is the only projection in which all the breast tissue can be demonstrated in a single image.
In CC view, the X-ray beam travels from superior to inferior. Positioning is achieved by pulling the breast up and forward away from the chest wall, with compression applied from above.

So the answer to this question is D


Instruction:
Homer wright rossette are seen in-a)  Neuroblastomab)  Astrocytomac)  Meningiomad)  Pinealoma
A. ad
B. c
C. ac
D. ab

Response:
Rosettes

Rosettes are pathological findings characterized by a halo or spoke wheel arrangement of cells surrounding a central core or hub.

Type of Rosette
1. Flexner - Wintersteiner Rosettes

A halo of cells surrounds a largely empty central hub but small cytoplasmic extension from cells project into the lumen.
It is characteristic of retinoblastoma.
May also be seen in → Medulloblastoma, Primitive neuroectodermal tumor, Pineoblastoma.

2. Homer Wright rosettes

A halo of cells surrounds a central hub that contains a meshwork of fibres (neutrophil rich centre).
Homer-Wright rosettes are characteristic of neuroblastomas and medulloblastomas.
May also be seen in → Primitive neuroectodermal tumors (PNET), Pineoblastomas, Retinoblastomas.

3. True ependymal Rosette

A halo of cells surrounds an empty lumen.
It is characteristic of ependymomas.

4. Perivascular pseudorosettes

A halo of cells surrounds a blood vessel.
The term pseudo is used because the central structure is not formed from the tumor itself, and represents a nonneoplastic element.
May be seen in
	Ependymonas
	Glioblastomas
	Medulloblastomas
	Retinoblastoma
	Primitive neuroectodermal tumors
	Monimorphous pilimyxoid astrocytoma
	Central neurocytoma

So the answer to this question is A


Instruction:
Midyear population calculated on?
A. 1st june
B. 30th june
C. 1st july
D. 31st june

Response:
ANSWER: (C) 1st JulyREF: Park 18th ed p.639Mid year population is estimated population on 1st of julyCensus estimation is done at end of first quarter in the first year after a decade (end of march)

So the answer to this question is C


Instruction:
Mulberry molar is associated with:
A. Initiation.
B. Histodifferentiation.
C. Morpho differentiation.
D. Apposition.

Response:
None

So the answer to this question is C


Instruction:
Sacrotuberous ligament is pierced by ?
A. SI nerve
B. L5 nerve
C. Coccygeal nerve
D. None

Response:
Ans. is 'c' i.e., Coccygeal nerve Sacrotuberous ligament is pierced by :-Perforating cutaneous nerveFifth sacral (S5) and cocygeal nervesBranches of coccygeal plexusCoccygeal branch of inferior gluteal aery

So the answer to this question is C


Instruction:
Following is not true regarding summons -
A. It is a legal document
B. No conduct money is paid to the doctor in criminal cases
C. Section 61-69 CrPC deal with summons
D. None of the above

Response:
It is a legal document.sections. 61 to 69 of Cr.P.C.del with the summons. Conduct is paid to doctor in criminal cases. Ref:The synopsis of Forensic Medicine (Dr.K.S.Narayan Reddy), pg no-5,29th edition.

So the answer to this question is B


Instruction:
A 25 years old female develops serous otitis media of left ear with cough and occasional hemoptysis and hematuria and epistaxis for one and half months her hemoglobin is 7 gm. B.P is > 170/100, proteinuria +++, RA - ve and ANCA - ve, the likely cause is?
A. Wegener's granulomatosis
B. Rheumiatoid ahritis
C. Rapidly proliferative glomerulonephritis
D. Goodpasteur's syndrome

Response:
Answer is A>D (Wegener's granulumatosis > Goodpasture's Syndrome); The patient in question is presenting with characteristic clinical triad of Wegener's granulomatosis, namely, Vasculitis of upper respiratory tract (serous otitis media, epistaxis), Vasculitis of lower respiratory tract (cough and hemoptysis) and vasculitis of the kidney (proteinuria)Although positive ANCA assays are often instrumental in suggesting the diagnosis of WG, Negative ANCA assays do not preclude the diagnosis, 10% of patients with disseminated WG and upto 30% of patients with limited WG may be negative for ANCAThe clinical picture is classical of Wegner's Granulomatosis. However, the absence of ANCA confuses the diagnosis. Neveheless the presence or absence of ANCA should be adjunctive and in the presence of a classical clinical picture of WG, tissue diagnosis (biopsy) should be performed to confirm the diagnosis.Definitive diagnosis of WG is established by tissue biopsy and not by presence or absence of ANCAPulmonary + Renal Syndrome Pulmonary (Hemoptysis) + Renal (Proteinuria) manifestationsWegener's granulomatosisClassical Clinical presentation is one of a young Male/Female (male: female=1:1) presenting with upper respiratory tract features (Otitis media, Epistaxis) along with hemoptysis and acute glomerulonephritis+ANCA is usually positive and suggests the diagnosisNegative ANCA does not preclude the diagnosis. In the presence of classical clinical picture a provisional diagnosis of WG should be made on clinical ground and confirmed/ excluded by tissue biopsyThe patient in question is presenting with classical clinical triad of Wegener's granulomatosis. Although ANCA is negative a negative ANCA does not preclude the diagnosis of WG. Based on the strong clinical suspicion, WG should the single best initial provisional diagnosis and tissue biopsy should be performed.Goodpasture's syndrome Classical clinical presentation is one of a young male smoker (male female=3-4:1) presenting with hemoptysis and signs of acute glomerulonephritis Although disease may be preceeded by an upper respiratory tract infection, upper respiratory tract involvement features like otitis media and epistaxis are not characterstic of good pasture's syndrome ANCA is usually negative Diagnosis is made by detecting serum levels of Anti GBM antibodies The patient in question is not presenting with classical clinical features of good pasture's syndrome and serum levels of Anti GBM antibodies have not been provided in the question. These make a diagnosis of Good pasture's syndrome less likely.

So the answer to this question is A


Instruction:
All are cluster C disorders of personality except
A. anxious
B. avoidant
C. scchizoid
D. dependent

Response:
personality disorders are of several types they can be named mainly as 3 clusters knowledgeof this is necessary for answering a lot of questions CLUSTER A=== paranoid, schizoid, schizotypal CLUSTERB===histironic, borederline, narcisstic, antisocial CLUSTERC====anxious, dependent, anakastic Ref.kaplon and sadock, synopsis of psychiatry,11 th edition, pg no.742

So the answer to this question is C


Instruction:
Tectal breaking is seen in:
A. Dandy-walker malformation
B. Amold-chiari malformation
C. Aqueductal stenosis
D. Third ventricular tumor

Response:
. Amold-chiari malformation

So the answer to this question is B


Instruction:
Which of the following can be seen in SCID?
A. Autoimmune ds
B. Granuloma formation
C. GVHD
D. Graft rejection

Response:
SCID (Severe Combined Immunodeficiency) Combined deficiency of B & T cells seen; so autoimmune ds, granuloma formation & graft rejection are absent GVHD which is due to immune-competent cells attacking an immunosuppressed host can still be seen

So the answer to this question is C


Instruction:
Which of the following is not transmitted by Arthropod-
A. Q fever
B. Rickettsial pox
C. Rocky mountain spotted fever
D. Relapsing feve

Response:
None

So the answer to this question is A


Instruction:
Drug of choice for anaphylactic shock is:
A. Adrenaline.
B. Dopamine.
C. Non-adrenaline.
D. All of the above.

Response:
Dopamine is the drug of choice for cardiogenic shock with oliguric renal failue.

So the answer to this question is A


Instruction:
Elimination after 3 half lives in first order kinetics is
A. 12.50%
B. 75%
C. 87.50%
D. 94%

Response:
Cp - > Cp/2 in 1 half-life i.e. 50.0 % lost 50.0 % Cp - > Cp/4 in 2 half-lives i.e. 25.0 % lost 75.0 % Cp - > Cp/8 in 3 half-lives i.e. 12.5 % lost 87.5 % Cp - > Cp/16 in 4 half-lives i.e. 6.25 % lost 93.75 % Cp - > Cp/32 in 5 half-lives i.e. 3.125 % lost 96.875 % Cp - > Cp/64 in 6 half-lives i.e. 1.563 % lost 98.438 % Cp - > Cp/128 in 7 half-lives i.e. 0.781 % lost 99.219 % Ref-KDT 7/e p30

So the answer to this question is C


Instruction:
Which among the following is not formed by kidney?
A. Erythropoietin
B. Renin
C. Vitamin D
D. Aldosterone

Response:
Ans: d (Aldosterone) Ref: Ganong 22nd ed/ p.454)Kidney produces 3 hormones1) 1,25(OH) cholecalciferol (vitamin D)2) Renin3) ErythropoietinAldosterone: is produced by Zona Glomerulosa of adrenal cortex.Erythropoietin: by the interstitial cells in the peritubular capillaries in kidneys (85%) and perivenous hepatocytes of liver (15%).Renin: by the juxta glomerular cells of the kidney.Renin Angiotensin SystemAngiotensinogen (from liver) - found in alpha 2 globulin fraction of plasma Renin (from JG cells)Angiotensin I ACE (Angiotensin converting enz. from Pulm. endothelium)Angiotensin IIMetabolism of vitamin D

So the answer to this question is D


Instruction:
Focal length of lens is 0.75 m, what will be its refreactive power -
A. 0.75D
B. 1.5D
C. 1.25 D
D. 1.33 D

Response:
The most convenient system for ophthalmic purpose is the one which takes a lens with focal distance of 1 meter as standard. Such a lens has a power of 1 dioptre (D). Thus , simple formula to calculate the refreactive power of a lens is - 1 Refractive power (D)= -------------------------------- Focal length (meter) 1 In this question refractive power =---------------------=1.33D 0.75 Reference : Parson&;s 22nd edition page 47

So the answer to this question is D